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Learning Goal: Doppler Effect - Both move A cop car moving at a speed vcop =44.9 m/s emits a siren whose buit-in frequency is fs=910.Hz. You are driving at vcar=39.6 m/s. The speed of sound in air is 343.m/sCase 3Here, both the source (cop car) and the observer are moving. You need to choose a proper sign in the numeritor, and from the denominator. Keep 2
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Question: Learning Goal: Doppler Effect - Both move A cop car moving at a speed vcop =44.9 m/s emits a siren whose buit-in frequency is fs=910.Hz. You are driving at vcar=39.6 m/s. The speed of sound in air is 343.m/sCase 3Here, both the source (cop car) and the observer are moving. You need to choose a proper sign in the numeritor, and from the denominator. Keep 2
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There are 4 steps to solve this one.Solution 100%(6 ratings) Share Share Share done loading Copy link Step 1 The problem is based on the
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Learning Goal: Doppler Effect - Both move A cop car moving at a speed v cop=44.9 m/s emits a siren whose buit-in frequency is f s=910.Hz. You are driving at v car=39.6 m/s. The speed of sound in air is 343.m/s Case 3 Here, both the source (cop car) and the observer are moving. You need to choose a proper sign in the numeritor, and from the denominator. Keep 2 digits after the decimal point, in Hz. The accepted range of answer is set to be narrow. Part B - Case2: Calculate the frequency of the siren heard by you. Here, both the source (cop car) and the observer are moving. You need to choose a proper sign in the numeritor, and from the denominator. Keep 2 digits after the decimal point, in Hz. The accepted range of answer is set to be narrow. Part C - Case 3: Calculate the frequency of the siren heard by you. Here, both the source (cop car) and the observer are moving. You need to choose a proper sign in the numeritor, and from the denominator. Keep 2 digits after the decimal point, in Hz. The accepted range of answer is set to be narrow. Part D - Case4: Calculate the frequency of the siren heard by you. Here, both the source (cop car) and the observer are moving. You need to choose a proper sign in the numeritor, and from the denominator. Keep 2 digits after the decimal point, in Hz. The accepted range of answer is set to be narrow.
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8801 | https://www.merriam-webster.com/thesaurus/conveniently | Synonyms of conveniently
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RIGHTS TALK
THE IMPOVERISHMENT OF POLITICAL DISCOURSE
by Mary Ann Glendon ‧RELEASE DATE: Sept. 1, 1991
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Here, Harvard Law School professor Glendon argues eloquently and persuasively that modern American political discourse, by emphasizing an ever-expanding catalogue of rights to the exclusion of duties and responsibilities, has lost the central role in civic life envisioned for it by the Founding Fathers. Glendon shows that, in American society, both sides in political debates frame issues in terms of individual rights—flag- burning, domestic relations, and human reproduction, for example- -and that this tendency impedes understanding and compromise. Such stark formulations, she says, ultimately lead to coerced, and often unsatisfying, social arrangements. Glendon makes a compelling case that the American political lexicon lacks a vocabulary for expressing normative and moral concepts that individual Americans understand and value highly, and that the legal culture, with its single-minded emphasis on obtaining civil rights (as opposed to cultivating moral norms), has actually contributed, albeit unwittingly, to the debasement of American political and legal discourse. Glendon calls for the inclusion of the missing language of responsibility'' and themissing language of sociality'' in American political dialogue, and for an increasing emphasis on individuals' responsibilities to their communities as a necessary concomitant to the rights they exercise. A forceful and valuable analysis of the banality of modern American public debate.
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Pub Date: Sept. 1, 1991
ISBN: 0-02-911825-5
Page Count: 256
Publisher: Free Press
Review Posted Online: May 19, 2010
Kirkus Reviews Issue: July 15, 1991
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A PEOPLE'S HISTORY OF THE UNITED STATES
by Howard Zinn ‧RELEASE DATE: Jan. 1, 1979
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For Howard Zinn, long-time civil rights and anti-war activist, history and ideology have a lot in common. Since he thinks that everything is in someone's interest, the historian—Zinn posits—has to figure out whose interests he or she is defining/defending/reconstructing (hence one of his previous books, The Politics of History). Zinn has no doubts about where he stands in this "people's history": "it is a history disrespectful of governments and respectful of people's movements of resistance." So what we get here, instead of the usual survey of wars, presidents, and institutions, is a survey of the usual rebellions, strikes, and protest movements. Zinn starts out by depicting the arrival of Columbus in North America from the standpoint of the Indians (which amounts to their standpoint as constructed from the observations of the Europeans); and, after easily establishing the cultural disharmony that ensued, he goes on to the importation of slaves into the colonies. Add the laborers and indentured servants that followed, plus women and later immigrants, and you have Zinn's amorphous constituency. To hear Zinn tell it, all anyone did in America at any time was to oppress or be oppressed; and so he obscures as much as his hated mainstream historical foes do—only in Zinn's case there is that absurd presumption that virtually everything that came to pass was the work of ruling-class planning: this amounts to one great indictment for conspiracy. Despite surface similarities, this is not a social history, since we get no sense of the fabric of life. Instead of negating the one-sided histories he detests, Zinn has merely reversed the image; the distortion remains.
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ISBN: 0061965588
Page Count: 772
Publisher: Harper & Row
Review Posted Online: May 26, 2012
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WHEN BREATH BECOMES AIR
by Paul Kalanithi ‧RELEASE DATE: Jan. 19, 2016
A moving meditation on mortality by a gifted writer whose dual perspectives of physician and patient provide a singular...
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A neurosurgeon with a passion for literature tragically finds his perfect subject after his diagnosis of terminal lung cancer.
Writing isn’t brain surgery, but it’s rare when someone adept at the latter is also so accomplished at the former. Searching for meaning and purpose in his life, Kalanithi pursued a doctorate in literature and had felt certain that he wouldn’t enter the field of medicine, in which his father and other members of his family excelled. “But I couldn’t let go of the question,” he writes, after realizing that his goals “didn’t quite fit in an English department.” “Where did biology, morality, literature and philosophy intersect?” So he decided to set aside his doctoral dissertation and belatedly prepare for medical school, which “would allow me a chance to find answers that are not in books, to find a different sort of sublime, to forge relationships with the suffering, and to keep following the question of what makes human life meaningful, even in the face of death and decay.” The author’s empathy undoubtedly made him an exceptional doctor, and the precision of his prose—as well as the moral purpose underscoring it—suggests that he could have written a good book on any subject he chose. Part of what makes this book so essential is the fact that it was written under a death sentence following the diagnosis that upended his life, just as he was preparing to end his residency and attract offers at the top of his profession. Kalanithi learned he might have 10 years to live or perhaps five. Should he return to neurosurgery (he could and did), or should he write (he also did)? Should he and his wife have a baby? They did, eight months before he died, which was less than two years after the original diagnosis. “The fact of death is unsettling,” he understates. “Yet there is no other way to live.”
A moving meditation on mortality by a gifted writer whose dual perspectives of physician and patient provide a singular clarity.
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Pub Date: Jan. 19, 2016
ISBN: 978-0-8129-8840-6
Page Count: 248
Publisher: Random House
Review Posted Online: Sept. 29, 2015
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8805 | https://www.ck12.org/flexi/math-grade-7/the-percent-equation/how-do-you-do-percent-increase-and-decrease-problems/ | How do you do percent increase and decrease problems?
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Explanation about product of two negative numbers being positive. [duplicate]
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$\begingroup$
I have been having a struggle finding an explanation why $-3 \cdot (-3) = 9$. Why does this question equal a positive number? Any explanations?
And btw, if $-3 \cdot (-3) = 9$. Why does $-3 + (-3) + (-3) = -9$ and not $9$?
arithmetic
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edited Sep 10, 2015 at 13:54
Alex M.
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asked Sep 10, 2015 at 13:40
Alexander SäfströmAlexander Säfström
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$\endgroup$
2
2
$\begingroup$ When I was young I remembered by thinking: If you think of a negative number as the opposite of a positive number, then multiplying two negatives gives you the opposite of the opposite of their product. Hence a positive number. $\endgroup$
graydad
– graydad
2015-09-10 13:46:02 +00:00
Commented Sep 10, 2015 at 13:46
$\begingroup$ Here is a set of equations which may help a bit $0=(-3)\times 0=(-3)\times (3-3)=-3\times (3+(-3))=(-3)\times 3+(-3)\times (-3)$ $\endgroup$
Mark Bennet
– Mark Bennet
2015-09-10 14:25:52 +00:00
Commented Sep 10, 2015 at 14:25
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4 Answers 4
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$\begingroup$
The best intuitive explanation I came across is to think of $x=vt$ where $x$ is the displacement, $v$ is the velocity and $t$ is the time. Now suppose you are moving at a velocity of 3 m/sec backwards (hence $v=-3$) and you want to calculate where you were 3 seconds ago ($t=-3$).
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answered Sep 10, 2015 at 13:45
Shahar Even-Dar MandelShahar Even-Dar Mandel
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$\endgroup$
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$\begingroup$
$3\times-5=-15$
$2\times-5=-10$
$1\times-5=-5$
$0\times-5=0$
$-1\times-5=5$
$-2\times-5=10$
See how they form a pattern? It's because subtracting a negative number gives you a positive, and having a negative amount of negative numbers means your total is positive.
Sure, this isn't rigorous, but hopefully you can get some intuition.
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answered Sep 10, 2015 at 14:22
DeusoviDeusovi
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$\begingroup$
To clear your second doubt,
$$(-3)+ (-3)+ (-3)=(-3)\times 3=-9$$ $$(-3)+ (-3)+ (-3)\neq(-3)\times(-3)$$
And your main doubt, as to why negative times negative is positive,
Let us assume $$(-3)\times (-3)=-9$$. We also know that $$(-3) \times 3=-9$$. Hence $$(-3) \times(-3)=(-3)\times 3$$, or $$-3=3$$, which is a problem.
There is no fundamental rule why negative times negative is positive. It is just that all our operators must be 'consistent'. Once we determine basic rules of addition of negative numbers, it becomes necessary to adopt specific rules for multiplication.
If we decided to adopt a system where negative times negative is negative, then the multiplication operator (and some other operators as well) will have a different meaning. It will no longer signify repeated addition. However, there is nothing wrong in doing so.
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edited Sep 10, 2015 at 14:31
answered Sep 10, 2015 at 14:09
ghosts_in_the_codeghosts_in_the_code
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$\endgroup$
2
$\begingroup$ I am pretty certain that you meant $(-3) + (-3) + (-3) = (-3)\times 3=-9$ $\endgroup$
Steven Alexis Gregory
– Steven Alexis Gregory
2015-09-10 14:28:22 +00:00
Commented Sep 10, 2015 at 14:28
$\begingroup$ @steven Yes, I did; careless of me.. $\endgroup$
ghosts_in_the_code
– ghosts_in_the_code
2015-09-10 14:31:24 +00:00
Commented Sep 10, 2015 at 14:31
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$\begingroup$
The definition of $3\times 2$ is to sum the number 3 exactly 2 times, so
$$ 3\times 2 = 0 + 3 + 3 = 6 $$ no sweat. Notice that I added a $0$ in the front, just to emphasize the sign of the first $3$. If you have $(-3)\times 2$, it's the same thing: sum the number $-3$ exactly two times. Now, the question is, what does it mean to sum a negative number? If you think about money, a negative number is an expense, while a positive is an income. To add an expense to your budget is the same as spending money, so it is the same as subtracting that amount (without the minus). Hence
$$ (-3)\times 2 = 0 + (-3) + (-3) = -3 -3 = -6 $$ So far so good. Now we get to the troubling part. What does it mean to compute $(-3)\times (-2)$? Well, if multiplying $-3$ by a positive number, say $2$, means to add the number $-3$ two times, it makes sense to think about multiplying $3$ by a negative number, sau $-2$, as subtracting the number $-3$ two times. Therefore,
$$ (-3)\times (-2) = 0 - (-3) - (-3). $$
But we just shifted the question: what does it mean to subtract a negative number? Well, thinking again in terms of your money budget, if you remove an expense (i.e., subtract a negative number), you're effectively increasing your budget. As they say: money saved is money earned. Hence
$$ (-3)\times (-2) = 0 - (-3) - (-3) = 0 + 3 + 3 = 6 $$
Hope this helps. When thinking about negative numbers, examples with money always helped me.
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answered Sep 10, 2015 at 14:49
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8807 | https://arxiv.org/abs/1503.05242 | [1503.05242] Colored partitions of a convex polygon by noncrossing diagonals
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arXiv:1503.05242 (math)
[Submitted on 17 Mar 2015]
Title:Colored partitions of a convex polygon by noncrossing diagonals
Authors:Daniel Birmajer, Juan B. Gil, Michael D. Weiner
View a PDF of the paper titled Colored partitions of a convex polygon by noncrossing diagonals, by Daniel Birmajer and 2 other authors
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Abstract:For any positive integers $a$ and $b$, we enumerate all colored partitions made by noncrossing diagonals of a convex polygon into polygons whose number of sides is congruent to $b$ modulo $a$. For the number of such partitions made by a fixed number of diagonals, we give both a recurrence relation and an explicit representation in terms of partial Bell polynomials. We use basic properties of these polynomials to efficiently incorporate restrictions on the type of polygons allowed in the partitions.
Comments:11 pages, 3 figures, Submitted for publication
Subjects:Combinatorics (math.CO)
MSC classes:05A17, 05A19
Cite as:arXiv:1503.05242 [math.CO]
(or arXiv:1503.05242v1 [math.CO] for this version)
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Journal reference:Discrete Math. 340 (2017), no. 4, 563-571
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From: Juan B. Gil [view email]
[v1] Tue, 17 Mar 2015 23:01:45 UTC (9 KB)
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8808 | https://library.achievingthedream.org/austinccphysics2/chapter/25-6-image-formation-by-lenses/ | Skip to content
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75 Image Formation by Lenses
Learning Objectives
By the end of this section, you will be able to:
List the rules for ray tracking for thin lenses.
Illustrate the formation of images using the technique of ray tracking.
Determine power of a lens given the focal length.
Lenses are found in a huge array of optical instruments, ranging from a simple magnifying glass to the eye to a camera’s zoom lens. In this section, we will use the law of refraction to explore the properties of lenses and how they form images.
The word lens derives from the Latin word for a lentil bean, the shape of which is similar to the convex lens in Figure 1. The convex lens shown has been shaped so that all light rays that enter it parallel to its axis cross one another at a single point on the opposite side of the lens. (The axis is defined to be a line normal to the lens at its center, as shown in Figure 1.) Such a lens is called a converging (or convex) lens for the converging effect it has on light rays. An expanded view of the path of one ray through the lens is shown, to illustrate how the ray changes direction both as it enters and as it leaves the lens. Since the index of refraction of the lens is greater than that of air, the ray moves towards the perpendicular as it enters and away from the perpendicular as it leaves. (This is in accordance with the law of refraction.) Due to the lens’s shape, light is thus bent toward the axis at both surfaces. The point at which the rays cross is defined to be the focal point F of the lens. The distance from the center of the lens to its focal point is defined to be the focal length f of the lens. Figure 2 shows how a converging lens, such as that in a magnifying glass, can converge the nearly parallel light rays from the sun to a small spot.
Converging or Convex Lens
The lens in which light rays that enter it parallel to its axis cross one another at a single point on the opposite side with a converging effect is called converging lens.
Focal Point F
The point at which the light rays cross is called the focal point F of the lens.
Focal Length f
The distance from the center of the lens to its focal point is called focal length f.
The greater effect a lens has on light rays, the more powerful it is said to be. For example, a powerful converging lens will focus parallel light rays closer to itself and will have a smaller focal length than a weak lens. The light will also focus into a smaller and more intense spot for a more powerful lens. The power P of a lens is defined to be the inverse of its focal length. In equation form, this is .
Power P
The power P of a lens is defined to be the inverse of its focal length. In equation form, this is , where f is the focal length of the lens, which must be given in meters (and not cm or mm). The power of a lens P has the unit diopters (D), provided that the focal length is given in meters. That is, . (Note that this power (optical power, actually) is not the same as power in watts defined in the chapter Work, Energy, and Energy Resources. It is a concept related to the effect of optical devices on light.) Optometrists prescribe common spectacles and contact lenses in units of diopters.
Example 1. What is the Power of a Common Magnifying Glass?
Suppose you take a magnifying glass out on a sunny day and you find that it concentrates sunlight to a small spot 8.00 cm away from the lens. What are the focal length and power of the lens?
Strategy
The situation here is the same as those shown in Figure 1 and Figure 2. The Sun is so far away that the Sun’s rays are nearly parallel when they reach Earth. The magnifying glass is a convex (or converging) lens, focusing the nearly parallel rays of sunlight. Thus the focal length of the lens is the distance from the lens to the spot, and its power is the inverse of this distance (in m).
Solution
The focal length of the lens is the distance from the center of the lens to the spot, given to be 8.00 cm. Thus,
f= 8.00 cm.
To find the power of the lens, we must first convert the focal length to meters; then, we substitute this value into the equation for power. This gives
.
Discussion
This is a relatively powerful lens. The power of a lens in diopters should not be confused with the familiar concept of power in watts. It is an unfortunate fact that the word “power” is used for two completely different concepts. If you examine a prescription for eyeglasses, you will note lens powers given in diopters. If you examine the label on a motor, you will note energy consumption rate given as a power in watts.
Figure 3 shows a concave lens and the effect it has on rays of light that enter it parallel to its axis (the path taken by ray 2 in the Figure is the axis of the lens). The concave lens is a diverging lens, because it causes the light rays to bend away (diverge) from its axis. In this case, the lens has been shaped so that all light rays entering it parallel to its axis appear to originate from the same point, F, defined to be the focal point of a diverging lens. The distance from the center of the lens to the focal point is again called the focal length f of the lens. Note that the focal length and power of a diverging lens are defined to be negative.
For example, if the distance to F in Figure 3 is 5.00 cm, then the focal length is f = –5.00 cm and the power of the lens is P = –20 D. An expanded view of the path of one ray through the lens is shown in the Figure to illustrate how the shape of the lens, together with the law of refraction, causes the ray to follow its particular path and be diverged.
Diverging Lens
A lens that causes the light rays to bend away from its axis is called a diverging lens.
As noted in the initial discussion of the law of refraction in The Law of Refraction, the paths of light rays are exactly reversible. This means that the direction of the arrows could be reversed for all of the rays in Figure 1 and Figure 3. For example, if a point light source is placed at the focal point of a convex lens, as shown in Figure 4, parallel light rays emerge from the other side.
Ray Tracing and Thin Lenses
Ray tracing is the technique of determining or following (tracing) the paths that light rays take. For rays passing through matter, the law of refraction is used to trace the paths. Here we use ray tracing to help us understand the action of lenses in situations ranging from forming images on film to magnifying small print to correcting nearsightedness. While ray tracing for complicated lenses, such as those found in sophisticated cameras, may require computer techniques, there is a set of simple rules for tracing rays through thin lenses.
A thin lens is defined to be one whose thickness allows rays to refract, as illustrated in Figure 1, but does not allow properties such as dispersion and aberrations. An ideal thin lens has two refracting surfaces but the lens is thin enough to assume that light rays bend only once. A thin symmetrical lens has two focal points, one on either side and both at the same distance from the lens. (See Figure 6.)
Another important characteristic of a thin lens is that light rays through its center are deflected by a negligible amount, as seen in Figure 5.
Thin Lens
A thin lens is defined to be one whose thickness allows rays to refract but does not allow properties such as dispersion and aberrations.
Take-Home Experiment: A Visit to the Optician
Look through your eyeglasses (or those of a friend) backward and forward and comment on whether they act like thin lenses.
Using paper, pencil, and a straight edge, ray tracing can accurately describe the operation of a lens. The rules for ray tracing for thin lenses are based on the illustrations already discussed:
A ray entering a converging lens parallel to its axis passes through the focal point F of the lens on the other side. (See rays 1 and 3 in Figure 1.)
A ray entering a diverging lens parallel to its axis seems to come from the focal point F. (See rays 1 and 3 in Figure 2.)
A ray passing through the center of either a converging or a diverging lens does not change direction. (See Figure 5, and see ray 2 in Figure 1 and Figure 2.)
A ray entering a converging lens through its focal point exits parallel to its axis. (The reverse of rays 1 and 3 in Figure 1.)
A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to the axis. (The reverse of rays 1 and 3 in Figure 2.)
Rules for Ray Tracing
A ray entering a converging lens parallel to its axis passes through the focal point F of the lens on the other side.
A ray entering a diverging lens parallel to its axis seems to come from the focal point F.
A ray passing through the center of either a converging or a diverging lens does not change direction.
A ray entering a converging lens through its focal point exits parallel to its axis.
A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to the axis.
Image Formation by Thin Lenses
In some circumstances, a lens forms an obvious image, such as when a movie projector casts an image onto a screen. In other cases, the image is less obvious. Where, for example, is the image formed by eyeglasses? We will use ray tracing for thin lenses to illustrate how they form images, and we will develop equations to describe the image formation quantitatively.
Consider an object some distance away from a converging lens, as shown in Figure 7. To find the location and size of the image formed, we trace the paths of selected light rays originating from one point on the object, in this case the top of the person’s head. The Figure shows three rays from the top of the object that can be traced using the ray tracing rules given above. (Rays leave this point going in many directions, but we concentrate on only a few with paths that are easy to trace.) The first ray is one that enters the lens parallel to its axis and passes through the focal point on the other side (rule 1). The second ray passes through the center of the lens without changing direction (rule 3). The third ray passes through the nearer focal point on its way into the lens and leaves the lens parallel to its axis (rule 4). The three rays cross at the same point on the other side of the lens. The image of the top of the person’s head is located at this point. All rays that come from the same point on the top of the person’s head are refracted in such a way as to cross at the point shown. Rays from another point on the object, such as her belt buckle, will also cross at another common point, forming a complete image, as shown. Although three rays are traced in Figure 7, only two are necessary to locate the image. It is best to trace rays for which there are simple ray tracing rules. Before applying ray tracing to other situations, let us consider the example shown in Figure 7 in more detail.
The image formed in Figure 7 is a real image, meaning that it can be projected. That is, light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye, for example. Figure 8 shows how such an image would be projected onto film by a camera lens. This Figure also shows how a real image is projected onto the retina by the lens of an eye. Note that the image is there whether it is projected onto a screen or not.
Real Image
The image in which light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye is called a real image.
Several important distances appear in Figure 7. We define do to be the object distance, the distance of an object from the center of a lens. Image distance di is defined to be the distance of the image from the center of a lens. The height of the object and height of the image are given the symbols ho and hi, respectively. Images that appear upright relative to the object have heights that are positive and those that are inverted have negative heights. Using the rules of ray tracing and making a scale drawing with paper and pencil, like that in Figure 7, we can accurately describe the location and size of an image. But the real benefit of ray tracing is in visualizing how images are formed in a variety of situations. To obtain numerical information, we use a pair of equations that can be derived from a geometric analysis of ray tracing for thin lenses. The thin lens equations are
and .
We define the ratio of image height to object height to be the magnification m. (The minus sign in the equation above will be discussed shortly.) The thin lens equations are broadly applicable to all situations involving thin lenses (and “thin” mirrors, as we will see later). We will explore many features of image formation in the following worked examples.
Image Distance
The distance of the image from the center of the lens is called image distance.
Thin Lens Equations and Magnification
Example 2. Finding the Image of a Light Bulb Filament by Ray Tracing and by the Thin Lens Equations
A clear glass light bulb is placed 0.750 m from a convex lens having a 0.500 m focal length, as shown in Figure 9. Use ray tracing to get an approximate location for the image. Then use the thin lens equations to calculate both the location of the image and its magnification. Verify that ray tracing and the thin lens equations produce consistent results.
Strategy and Concept
Since the object is placed farther away from a converging lens than the focal length of the lens, this situation is analogous to those illustrated in Figure 7 and Figure 8. Ray tracing to scale should produce similar results for di. Numerical solutions for di and m can be obtained using the thin lens equations, noting that do = 0.750 m and f = 0.500 m.
Solutions (Ray Tracing)
The ray tracing to scale in Figure 9 shows two rays from a point on the bulb’s filament crossing about 1.50 m on the far side of the lens. Thus the image distance di is about 1.50 m. Similarly, the image height based on ray tracing is greater than the object height by about a factor of 2, and the image is inverted. Thus m is about –2. The minus sign indicates that the image is inverted.
The thin lens equations can be used to find di from the given information:
.
Rearranging to isolate di gives
.
Entering known quantities gives a value for:
.
This must be inverted to find di:
.
Note that another way to find di is to rearrange the equation:
.
This yields the equation for the image distance as:
Note that there is no inverting here.
The thin lens equations can be used to find the magnification m, since both di and do are known. Entering their values gives
Discussion
Note that the minus sign causes the magnification to be negative when the image is inverted. Ray tracing and the use of the thin lens equations produce consistent results. The thin lens equations give the most precise results, being limited only by the accuracy of the given information. Ray tracing is limited by the accuracy with which you can draw, but it is highly useful both conceptually and visually.
Real images, such as the one considered in the previous example, are formed by converging lenses whenever an object is farther from the lens than its focal length. This is true for movie projectors, cameras, and the eye. We shall refer to these as case 1 images. A case 1 image is formed when do > f and f is positive, as in Figure 10a. (A summary of the three cases or types of image formation appears at the end of this section.)
A different type of image is formed when an object, such as a person’s face, is held close to a convex lens. The image is upright and larger than the object, as seen in Figure 10b, and so the lens is called a magnifier. If you slowly pull the magnifier away from the face, you will see that the magnification steadily increases until the image begins to blur. Pulling the magnifier even farther away produces an inverted image as seen in Figure 10a. The distance at which the image blurs, and beyond which it inverts, is the focal length of the lens. To use a convex lens as a magnifier, the object must be closer to the converging lens than its focal length. This is called a case 2 image. A case 2 image is formed when do < f and f is positive.
Figure 11 uses ray tracing to show how an image is formed when an object is held closer to a converging lens than its focal length. Rays coming from a common point on the object continue to diverge after passing through the lens, but all appear to originate from a point at the location of the image. The image is on the same side of the lens as the object and is farther away from the lens than the object. This image, like all case 2 images, cannot be projected and, hence, is called a virtual image.
Light rays only appear to originate at a virtual image; they do not actually pass through that location in space. A screen placed at the location of a virtual image will receive only diffuse light from the object, not focused rays from the lens. Additionally, a screen placed on the opposite side of the lens will receive rays that are still diverging, and so no image will be projected on it. We can see the magnified image with our eyes, because the lens of the eye converges the rays into a real image projected on our retina. Finally, we note that a virtual image is upright and larger than the object, meaning that the magnification is positive and greater than 1.
Virtual Image
An image that is on the same side of the lens as the object and cannot be projected on a screen is called a virtual image.
Example 3. Image Produced by a Magnifying Glass
Suppose the book page in Figure 11a is held 7.50 cm from a convex lens of focal length 10.0 cm, such as a typical magnifying glass might have. What magnification is produced?
Strategy and Concept
We are given that do = 7.50 cm and f= 10.0 cm, so we have a situation where the object is placed closer to the lens than its focal length. We therefore expect to get a case 2 virtual image with a positive magnification that is greater than 1. Ray tracing produces an image like that shown in Figure 11, but we will use the thin lens equations to get numerical solutions in this example.
Solution
To find the magnification m, we try to use magnification equation, . We do not have a value for di, so that we must first find the location of the image using lens equation. (The procedure is the same as followed in the preceding example, where do and f were known.) Rearranging the magnification equation to isolate di gives
.
Entering known values, we obtain a value for :
.
This must be inverted to find di:
.
Now the thin lens equation can be used to find the magnification m, since both di and do are known. Entering their values gives
Discussion
A number of results in this example are true of all case 2 images, as well as being consistent with Figure 11. Magnification is indeed positive (as predicted), meaning the image is upright. The magnification is also greater than 1, meaning that the image is larger than the object—in this case, by a factor of 3. Note that the image distance is negative. This means the image is on the same side of the lens as the object. Thus the image cannot be projected and is virtual. (Negative values of di occur for virtual images.) The image is farther from the lens than the object, since the image distance is greater in magnitude than the object distance. The location of the image is not obvious when you look through a magnifier. In fact, since the image is bigger than the object, you may think the image is closer than the object. But the image is farther away, a fact that is useful in correcting farsightedness, as we shall see in a later section.
A third type of image is formed by a diverging or concave lens. Try looking through eyeglasses meant to correct nearsightedness. (See Figure 12.) You will see an image that is upright but smaller than the object. This means that the magnification is positive but less than 1. The ray diagram in Figure 13 shows that the image is on the same side of the lens as the object and, hence, cannot be projected—it is a virtual image. Note that the image is closer to the lens than the object. This is a case 3 image, formed for any object by a negative focal length or diverging lens.
Example 4. Image Produced by a Concave Lens
Suppose an object such as a book page is held 7.50 cm from a concave lens of focal length –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced?
Strategy and Concept
This example is identical to the preceding one, except that the focal length is negative for a concave or diverging lens. The method of solution is thus the same, but the results are different in important ways.
Solution
To find the magnification m, we must first find the image distance di using thin lens equation , or its alternative rearrangement .
We are given that f= –10.0 cm and do= 7.50 cm. Entering these yields a value for :
This must be inverted to find di:
Or
Now the magnification equation can be used to find the magnification m, since both di and do are known. Entering their values gives
Discussion
A number of results in this example are true of all case 3 images, as well as being consistent with Figure 13. Magnification is positive (as predicted), meaning the image is upright. The magnification is also less than 1, meaning the image is smaller than the object—in this case, a little over half its size. The image distance is negative, meaning the image is on the same side of the lens as the object. (The image is virtual.) The image is closer to the lens than the object, since the image distance is smaller in magnitude than the object distance. The location of the image is not obvious when you look through a concave lens. In fact, since the image is smaller than the object, you may think it is farther away. But the image is closer than the object, a fact that is useful in correcting nearsightedness, as we shall see in a later section.
Table 1 summarizes the three types of images formed by single thin lenses. These are referred to as case 1, 2, and 3 images. Convex (converging) lenses can form either real or virtual images (cases 1 and 2, respectively), whereas concave (diverging) lenses can form only virtual images (always case 3). Real images are always inverted, but they can be either larger or smaller than the object. For example, a slide projector forms an image larger than the slide, whereas a camera makes an image smaller than the object being photographed. Virtual images are always upright and cannot be projected. Virtual images are larger than the object only in case 2, where a convex lens is used. The virtual image produced by a concave lens is always smaller than the object—a case 3 image. We can see and photograph virtual images only by using an additional lens to form a real image.
| Table 1. Three Types of Images Formed By Thin Lenses | | | | |
--- ---
| Type | Formed when | Image type | di | m |
| Case 1 | f positive, do>f | real | positive | negative |
| Case 2 | f positive, do<f | virtual | negative | positive m > 1 |
| Case 3 | f negative | virtual | negative | positive m < 1 |
In Image Formation by Mirrors, we shall see that mirrors can form exactly the same types of images as lenses.
Take-Home Experiment: Concentrating Sunlight
Find several lenses and determine whether they are converging or diverging. In general those that are thicker near the edges are diverging and those that are thicker near the center are converging. On a bright sunny day take the converging lenses outside and try focusing the sunlight onto a piece of paper. Determine the focal lengths of the lenses. Be careful because the paper may start to burn, depending on the type of lens you have selected.
Problem-Solving Strategies for Lenses
Step 1. Examine the situation to determine that image formation by a lens is involved.
Step 2. Determine whether ray tracing, the thin lens equations, or both are to be employed. A sketch is very useful even if ray tracing is not specifically required by the problem. Write symbols and values on the sketch.
Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns).
Step 4. Make alist of what is given or can be inferred from the problem as stated (identify the knowns). It is helpful to determine whether the situation involves a case 1, 2, or 3 image. While these are just names for types of images, they have certain characteristics (given in Table 1) that can be of great use in solving problems.
Step 5. If ray tracing is required, use the ray tracing rules listed near the beginning of this section.
Step 6. Most quantitative problems require the use of the thin lens equations. These are solved in the usual manner by substituting knowns and solving for unknowns. Several worked examples serve as guides.
Step 7. Check to see if the answer is reasonable: Does it make sense? If you have identified the type of image (case 1, 2, or 3), you should assess whether your answer is consistent with the type of image, magnification, and so on.
Misconception Alert
We do not realize that light rays are coming from every part of the object, passing through every part of the lens, and all can be used to form the final image.
We generally feel the entire lens, or mirror, is needed to form an image. Actually, half a lens will form the same, though a fainter, image.
Section Summary
Light rays entering a converging lens parallel to its axis cross one another at a single point on the opposite side.
For a converging lens, the focal point is the point at which converging light rays cross; for a diverging lens, the focal point is the point from which diverging light rays appear to originate.
The distance from the center of the lens to its focal point is called the focal length f.
Power P of a lens is defined to be the inverse of its focal length, .
A lens that causes the light rays to bend away from its axis is called a diverging lens.
Ray tracing is the technique of graphically determining the paths that light rays take.
The image in which light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye is called a real image.
Thin lens equations are
and (magnification).
The distance of the image from the center of the lens is called image distance.
An image that is on the same side of the lens as the object and cannot be projected on a screen is called a virtual image.
Conceptual Questions
It can be argued that a flat piece of glass, such as in a window, is like a lens with an infinite focal length. If so, where does it form an image? That is, how are di and do related?
You can often see a reflection when looking at a sheet of glass, particularly if it is darker on the other side. Explain why you can often see a double image in such circumstances.
When you focus a camera, you adjust the distance of the lens from the film. If the camera lens acts like a thin lens, why can it not be a fixed distance from the film for both near and distant objects?
A thin lens has two focal points, one on either side, at equal distances from its center, and should behave the same for light entering from either side. Look through your eyeglasses (or those of a friend) backward and forward and comment on whether they are thin lenses.
Will the focal length of a lens change when it is submerged in water? Explain.
Problems & Exercises
What is the power in diopters of a camera lens that has a 50.0 mm focal length?
Your camera’s zoom lens has an adjustable focal length ranging from 80.0 to 200 mm. What is its range of powers?
What is the focal length of 1.75 D reading glasses found on the rack in a pharmacy?
You note that your prescription for new eyeglasses is –4.50 D. What will their focal length be?
How far from the lens must the film in a camera be, if the lens has a 35.0 mm focal length and is being used to photograph a flower 75.0 cm away? Explicitly show how you follow the steps in the Problem-Solving Strategy for lenses.
A certain slide projector has a 100 mm focal length lens. (a) How far away is the screen, if a slide is placed 103 mm from the lens and produces a sharp image? (b) If the slide is 24.0 by 36.0 mm, what are the dimensions of the image? Explicitly show how you follow the steps in the Problem-Solving Strategy for Lenses(above).
A doctor examines a mole with a 15.0 cm focal length magnifying glass held 13.5 cm from the mole (a) Where is the image? (b) What is its magnification? (c) How big is the image of a 5.00 mm diameter mole?
How far from a piece of paper must you hold your father’s 2.25 D reading glasses to try to burn a hole in the paper with sunlight?
A camera with a 50.0 mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens must the film be? (b) If the film is 36.0 mm high, what fraction of a 1.75 m tall person will fit on it? (c) Discuss how reasonable this seems, based on your experience in taking or posing for photographs.
A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is 33.0 mm. (a) What is the closest object that can be photographed? (b) What is the magnification of this closest object?
Suppose your 50.0 mm focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus? (b) What is the height of the object if its image is 2.00 cm high?
(a) What is the focal length of a magnifying glass that produces a magnification of 3.00 when held 5.00 cm from an object, such as a rare coin? (b) Calculate the power of the magnifier in diopters. (c) Discuss how this power compares to those for store-bought reading glasses (typically 1.0 to 4.0 D). Is the magnifier’s power greater, and should it be?
What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 25.0 cm away?
In Example 3, the magnification of a book held 7.50 cm from a 10.0 cm focal length lens was found to be 3.00. (a) Find the magnification for the book when it is held 8.50 cm from the magnifier. (b) Do the same for when it is held 9.50 cm from the magnifier. (c) Comment on the trend in m as the object distance increases as in these two calculations.
Suppose a 200 mm focal length telephoto lens is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1000 m high cliff on one of the mountains?
A camera with a 100 mm focal length lens is used to photograph the sun and moon. What is the height of the image of the sun on the film, given the sun is 1.40 × 106 km in diameter and is 1.50 × 108 km away?
Combine thin lens equations to show that the magnification for a thin lens is determined by its focal length and the object distance and is given by .
Glossary
converging lens: a convex lens in which light rays that enter it parallel to its axis converge at a single point on the opposite side
diverging lens: a concave lens in which light rays that enter it parallel to its axis bend away (diverge) from its axis
focal point: for a converging lens or mirror, the point at which converging light rays cross; for a diverging lens or mirror, the point from which diverging light rays appear to originate
focal length: distance from the center of a lens or curved mirror to its focal point
magnification: ratio of image height to object height
power: inverse of focal length
real image: image that can be projected
virtual image: image that cannot be projected
Selected Solutions to Problems & Exercises
2. 5.00 to 12.5 D
4. –0.222 m
6. (a) 3.43 m; (b) 0.800 by 1.20 m
7. (a) −1.35 m (on the object side of the lens); (b) +10.0; (c) 5.00 cm
8. 44.4 cm
10. (a) 6.60 cm; (b) –0.333
12. (a) +7.50 cm; (b) 13.3 D; (c) Much greater
14. (a) +6.67; (b) +20.0; (c) The magnification increases without limit (to infinity) as the object distance increases to the limit of the focal distance.
16. −0.933 mm |
8809 | https://pubmed.ncbi.nlm.nih.gov/6332871/ | Interleukin 1 secretion by human monocytes and macrophages - PubMed
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. 1984 Oct;36(4):551-7.
doi: 10.1002/jlb.36.4.551.
Interleukin 1 secretion by human monocytes and macrophages
D G Mayernik,A Haq,J J Rinehart
PMID: 6332871
DOI: 10.1002/jlb.36.4.551
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Interleukin 1 secretion by human monocytes and macrophages
D G Mayernik et al. J Leukoc Biol.1984 Oct.
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. 1984 Oct;36(4):551-7.
doi: 10.1002/jlb.36.4.551.
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D G Mayernik,A Haq,J J Rinehart
PMID: 6332871
DOI: 10.1002/jlb.36.4.551
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Interleukin 1 (IL-1) is generally regarded as a major regulator of T lymphocyte proliferation. Macrophages from animals and cloned tumor cell lines have been shown to produce this monokine in response to a variety of stimuli. The ability of human monocytes and macrophages to generate IL-1 is much less well characterized. We previously demonstrated that human monocytes cultured for 1-6 days transformed to macrophages but retained their capacity to support concanavalin A-driven T cell proliferation. However, cultured macrophage capacity to support antigen-driven T cell proliferation began to decline after 3 days of culture and was markedly deficient by 6 days of culture. To determine if this loss of accessory cell function was due to the inability to secrete IL-1, we measured the monokine produced by normal fresh human monocytes and macrophages cultured in vitro from monocytes. IL-1 was assayed by the mouse thymocyte proliferation method. Fresh monocytes secreted IL-1 readily in response to lipopolysaccaride and latex particles. Macrophages cultured from fresh monocytes, however, lost this ability after greater than or equal to 2 days in culture. Mixing experiments failed to demonstrate an inhibitor present in the macrophage supernatants that would suppress thymocyte proliferation. Stimulated T cells incubated with monocytes and 3-day cultured macrophages failed to prolong or promote IL-1 secretion.
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Interleukin 1 secretion is not required for human macrophage support of T-cell proliferation.Haq AU, Mayernik DG, Orosz C, Rinehart JJ.Haq AU, et al.Cell Immunol. 1984 Sep;87(2):517-27. doi: 10.1016/0008-8749(84)90020-0.Cell Immunol. 1984.PMID: 6331896
Inability of human alveolar macrophages to stimulate resting T cells correlates with decreased antigen-specific T cell-macrophage binding.Lyons CR, Ball EJ, Toews GB, Weissler JC, Stastny P, Lipscomb MF.Lyons CR, et al.J Immunol. 1986 Aug 15;137(4):1173-80.J Immunol. 1986.PMID: 2426354
The accessory cell function of human alveolar macrophages in specific T cell proliferation.Toews GB, Vial WC, Dunn MM, Guzzetta P, Nunez G, Stastny P, Lipscomb MF.Toews GB, et al.J Immunol. 1984 Jan;132(1):181-6.J Immunol. 1984.PMID: 6228577
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8810 | http://boomeria.org/labsphys/physlabook/lab22.pdf | LAB 22, MECHANICAL EQUIVALENT OF HEAT Name _____ Period We shall drop a specific mass of lead shot inside a PVC pipe and measure the resultant increase in temperature. When the lead shot falls the length of the pipe and hits the other end, virtually all the kinetic energy of the shot will be converted into heat energy. Because of the insulating properties of the cork and the PVC walls of the pipe, this heat energy will be confined mostly to the shot and consequently raise its temperature.
OBJECTIVE: After completing this experiment, you should be able to make an approximate determination of the mechanical equivalent of heat.
PROCEDURE (SHOW ALL CALCULATIONS, 1,2,3,4): 1. Measure the mass of the calorimeter cup and record it in the data table: 2. pour the shot into the cup and measure the mass of the cup and shot. Record it.
Carefully 3. Calculate the mass of the shot by subtracting the mass of the cup from the total mass. Enter the result into the data table.
4. Carefully measure the temperature of the shot. ( .) Record in the data table.
Be careful not to break the thermometer 5. Pour the shot into the pipe. Close the open end of the tube with the stopper. Then invert the pipe 100 times. You must invert the pipe such as to make the shot fall through the entire length of the tube.
WARNING! Firmly hold the cork at all times to prevent the shot from shooooting all over the place!
6. After the last inversion of the tube, quickly pour the shot back into the cup and take the temperature. Record in the data table. The tube is 1 meter long, so the total distance dropped is 100 meters.
CALCULATIONS: 1. By using the equation, , calculate the potential energy of the shot.
Record in the Calculations Table.
PE = mgh, where PE is the potential energy in joules, m is the mass in KILOGRAMS, g = 9.8m/s , and h is the height in meters 2 (100m) NOTE: Kilograms = grams/1000g/kg. .
.
2. Determine the temperature change of the shot, , by taking the difference between the first and second temperatures. Record in Calculations Table.
Δt 3/12/05 9:41 AM Lab 22 Page 1 of 2 file://localhost/Documents/Internet/Adobe%20PageMill/slvhs.boom/labsphys/physlabook/lab22.html 3. From a Specific Heat Table we find that the specific heat, , of lead shot is . Calculate the heat gained by the shot.
Record in Calculations Table.
c 0.03 cal/gCo HINT: Q = mcΔt, Q is in calories, m is in GRAMS, and Δt is in C . o .
.
4. Determine the mechanical equivalent of heat in the experiment by dividing the joules of PE by the calories of heat. It's in joules/calorie. Record in Calculations Table.
.
5. Find the percentage error. Record in Calculations Table.
HINT: Percent error = your error/accepted value X 100%. The accepted value is (4.18j/cal). .
Data & Calculations Table Mass of empty cup _ g Distance Dropped _ m 100 Mass of cup + shot _ g PE at 100 meters _ J Mass of shot __ g Sp Ht of shot = .......... cal/gC 0.03 o Initial temp of shot _ C o Heat gained by shot _ cal Final temp of shot _ C o Mech equivalent ___ j/cal Temp change of shot _ Co Your error __j/cal .
Percent error _____ % QUESTIONS: 1. Could this experiment be completed without measuring the mass of the shot? Explain.
.
.
2. What are the main sources of error in this experiment?
.
.
CRITIQUE: 3/12/05 9:41 AM Lab 22 Page 2 of 2 file://localhost/Documents/Internet/Adobe%20PageMill/slvhs.boom/labsphys/physlabook/lab22.html |
8811 | https://iastate.pressbooks.pub/teachingmath/chapter/translanguaging-word-problems/ | Skip to content
5 Translanguaging & Word Problems
Ji-Yeong I and Ricardo Martinez
BUMBLEBEE Pre-Reading Questions
Have you observed EBs working on mathematical word problems? What were they doing to solve the problem?
Have you ever thought that certain mathematical tasks are not appropriate for EBs? If so, what type of task was it and why did you think it was not appropriate for EBs?
Have you seen your EBs talking in their home languages? What did you do or what would you do if EBs used their non-English language in class or during recess?
The Dilemma with Word Problems
Consider the following vignette:
Ms. A is a 6th-grade math teacher. When her personal learning community (PLC) looked ahead at the unit assessment set for her next math unit, she sighed because it was full of word problems. In Ms. Smith’s 6th-grade math classes, she has about 30% Emergent Bilinguals, and about half of them are recent immigrants whose English proficiency is at the lowest level. Ms. Smith found some EBs to be very capable of mathematics and had been working hard to receive a good grade in mathematics. However, she was worried about whether they could demonstrate their mathematics knowledge properly with these word problems, many of which included complex sentences and difficult non-mathematical words.
What do you think about this story? Have you had a similar experience? Do you think this is common?
We agree that Emergent Bilinguals (EBs) can do mathematics. But, many teachers are likely to believe that word problems are difficult for EBs. Why? It may be a pervasive belief among teachers that word problems are inappropriate for EBs, and EBs may not want to be assigned word problems because they may not understand the problem statement or have anxiety from reading heavy texts. The story of Ms. A was made up, but consider this real story below.
In one of my previous studies, Dr. I worked with preservice mathematics teachers. I recruited promising teacher candidates, and they volunteered to prepare a math lesson and work with one middle school EB in a one-on-one setting. One of the EBs was in an 8th-grade algebra class and the teacher who paired up with this student (let’s call her Ms. B) knew about this fact. The first task Ms. B provided was simple addition and subtraction worksheet, with tasks such as 9+11, which is aligned with 1st-grade mathematics standards according to the Common Core State Standards for Mathematics. The EB, who was in an early algebra class, did all computation tasks with high speed and accuracy and left her teacher bewildered. Later, I asked Ms. B why she prepared those easy tasks for this EB. Ms. B answered, “I knew she was in early algebra and she is capable, but I thought she may have difficulty doing mathematics due to language.”
What do you think about this story? Do you agree with Ms. B? The idea that “EBs cannot do well in mathematics due to language barriers” is a common belief and this belief makes teachers worry about not only word problems but also any mathematical tasks. If a mathematics teacher has this belief, they may think to themselves, “These tasks look too difficult for EBs, so I’ll assign them easier worksheets.” Research has in fact indicated that this belief is not uncommon. In fact, Reeves (2006) found that content teachers, including secondary math teachers, typically believe it is not appropriate to provide word problems to EBs due to their lack of English proficiency. What then can be done about these beliefs?
How to Make Word Problems Accessible to EBs
We would like to suggest rethinking these beliefs as the diagram below suggests. Teachers may observe EBs struggle in understanding word problems because they cannot fully understand the language. However, providing easy tasks or textless problems is not the only solution we can derive from this observation.
You identified that your EB students have difficulties reading word problems. What will you do to support them? Are you not going to give any word problems to EBs or give them only textless math tasks? Can you remove or reduce language demand without removing an opportunity to learn high-quality mathematics?
From here, we will talk about how we can make a word problem more accessible to EBs.
Generally, there are two directions.
First, we can reduce the language demand using following strategies:
Identify unnecessary parts and simplify the language
Provide more blank space on the worksheet, so students don’t feel overwhelmed by heavy dense text, and so they will feel free to jot down their ideas in the space provided.
Break down the word problem statement into multiple steps.
We need to be careful though. When you reduce the language demand, you must not reduce the mathematical demand. Also, be careful not to reduce the language demand too much because your EBs need to learn language through rich and varied English texts.
Second, we can provide linguistic support:
Provide mathematically meaningful visuals: You have seen what kind of visuals work better in Chapter 3. Just adding a picture related to the story or context does not help EBs reason mathematically. To learn more about mathematically meaningful visuals for EBs, you can check out the following MTE article and Podcast.
MTE article: I, J.-Y., & Stanford, J. (2018). Preservice teachers’ mathematical visual implementation for emergent bilinguals. Mathematics Teacher Educator, 7(1), 8–33. 10.5951/mathteaceduc.7.1.0008
Podcast at AMTE site
Have a set-up time before showing a word problem (I & de Araujo, 2019): During the set-up, a teacher can assess students’ understanding in relation to both mathematics and language to identify what needs to be taught beforehand. In addition, a teacher should assess students’ prior knowledge about the context. For example, when we taught EBs who were from refugee families, we changed a mathematics problem’s context from receiving a birthday gift to receiving an allowance because the EBs did not know their real birth dates and had never celebrated their birthday. If the result of this informal assessments indicates that you need to teach some words, cultural context, and/or prior math concepts, try to reteach or address this before you give the word problem so all students can be on the same page when they are given the word problem.
Utilize multimedia. The word problem does not always have to be in written form. For example, the story could be delivered through a video as you saw in the 3 Act Task and 5 Act Task introduced in Chapter 3. When you use multimedia that does not include any language, you can include other language activities such as discussion, presentation, or writing a journal to provide rich opportunities for EBs to develop their English skills.
Allow EBs to use their assets. Let EBs use their most comfortable communicative methods (e.g., written, verbal, drawing) to allow them to ask and respond to each other (if there are proficient bilinguals in the language of your EBs or if you have created a safe environment in your classroom). Most importantly, let EBs use their most comfortable language. Here, the concept of Translanguaging may help.
Translanguaging
[The content from here is adapted from Ofelia García’s presentation in 2015]
What do you think translanguaging means? To explain what it means, let’s start by reflecting on the following two short discourses:
Situation 1: Two 5-year-old Emergent Bilinguals who are speakers of Spanish in a bilingual classroom.
T: This tree is bigger. That tree is smaller.
Alicia: [Tries out under her breath]. This tree is grander.
Situation 2: Snack Time for a 5-year-old Emergent Bilingual.
Student A: [Looking out the window and talking to himself] Está lloviendo mucho. [It is raining a lot] Look [telling the others]. It’s washing. There washing afuera [outside]
Question: What did you notice from the EBs’ discourse?
In these dialogues, they used two languages in one sentence or in one situation.
What did you notice from the dialogue above? Did you notice the following?
EBs are not simply adding English, but using a whole autonomous language with Spanish.
They are using their own language features in an interrelationship with new ones to make meaning and communicate.
They are constructing their dynamic bilingual repertoire by adding features to those they already have.
The concept of code-switching, which describes how bilinguals perform discourse as they move from one language to another, is based on the belief that two separate languages exist in the brain of bilinguals. The term, code-switching has often been used negatively, indicating that bilinguals’ use of two languages within a sentence represents imperfect language acquisition.
However, we see the discourse pattern of bilinguals in the examples above as translanguaging, which is a natural way of how bilinguals/multilinguals use their multiple languages with their one language system. We can see that the speakers are actually combining two (or more) languages to communicate as meaningfully as possible. If you find EBs using two languages in one sentence, it does not mean a lack of language acquisition. Rather, they use two languages in an intertwined manner to communicate in the most meaningful way possible at their level. Hence, it is not necessary for teachers (or parents) to push EBs to use only one language in a sentence. If we force them to use only one language, it denies their whole language system and rejects part of who they are and what they can do.
So, which way would you guide your Emergent Bilinguals?
→ Utilize ONLY FEATURES THAT HAVE BEEN APPROVED BY SCHOOL for the task
→ Utilize ANY OF THE FEATURES in their language repertoire to show what they know and can do.
Learners ALWAYS leverage their entire language repertoire in the process of communication. With an understanding of translanguaging, we can change the question, “How do I support students to learn English as a second language?” to “How do I engage students in appropriating the language features associated with English into their own unique language repertoire?”
Translanguaging Pedagogy
Translanguaging can support EBs in more than the two ways discussed above. Translanguaging approaches do not reduce language demand but provide a richer and safer language environment so that EBs do not feel overwhelmed by the language demand. By valuing and enacting translanguaging, teachers can value not only students’ culture and language but also their identities as bilinguals (or multilinguals). This section includes how to employ a translanguaging pedagogy for teaching. Translanguaging Pedagogy means the deployment of a speaker’s full linguistic repertoire to learn and develop ways of using language and extend their repertoire and to equalize the positions of learners. For EBs, a Translanguaging Pedagogy removes the handicap of English-only instruction that has been shown to harmful for EBs (Garcia, 2015; Fu, Hadjioannou, & Zhou, 2019).
In the Foreword of the book, The Translanguaging Classroom: Leveraging Student Bilingualism for Learning (García, Johnson, & Seltzer, 2017), Guadalupe Valdés, a researcher in bilingualism at Stanford University, noted that this book is “by far the most compelling example proposed to date of a culturally sustaining pedagogy” (p. vii) with the following explanation.
Too often, proposed pedagogies for cultural responsiveness or relevance have not necessarily invited students to value what they bring or to proudly continue to use features of their full linguistic repertoires in both formal and informal oral and written production for a variety of purposes in and out of school. This book [translanguaging] is different. It explicitly takes the position that past scholarship on language has misunderstood the nature of bilingualism and bilingual practices. It insists that students be invited to foster, maintain, and develop their complex repertoires. It invites teachers to reject static views of Language A versus Language B kept separate and pristine. It urges them to engage thoughtfully and joyfully with the richness of multicompetence in children’s lives.
– Guadalupe Valdés within García, Johnson, & Seltzer (2017, p. vii-viii)
Emphasizing that translanguaging is a pedagogical model grounded on pluralist theory, Fu, Hadjioannou, and Zhou (2019) propose three key tenets for translanguaging practice that can be used in any setting with both emergent and experienced/proficient bilingual students.
Individuals have a single, unified linguistic repertoire.
Teachers are co-learners in their classrooms, willing to learn from students, their languages, and their cultures, rather than functioning as the sole possessors of knowledge, “the expert” or the only language instructor in the classroom.
Translanguaging practice is purposefully and systematically incorporated in both instructional planning and practices.
To help gain a better understanding of how to apply translanguaging pedagogy to a classroom, we will introduce a case of a 5th grade class, which was included in the book, Translanguaging for Emergent Bilinguals (Fu, Hadjioannou, & Zhou, 2019). Mr. Miller is a native English speaker who learned Spanish as a foreign language and the students in the class speak English, Spanish, Polish, Arabic, and Russian. In the classroom, the learning objectives, class schedule, and homework assignments are written in five languages. There is also a word wall of the five languages that Mr. Miller asked students to create, so the students completed the word wall with help from parents and family members.
To prepare for the lesson, Mr. Miller asks the students to do research about school bullying using all of the assets they have including their family members, books and internet resources written in any language and to take notes in the language of student’s choice.
During the lesson, students share their notes in groups in various forms, such as stories, blogs, or real experiences. The students switch back and forth between their home languages and English depending on their audience to make sure everyone understands the conversation.
Next, Mr. Miller reads a storybook on school bullying aloud in English and shows the pictures to the class without stopping the first time. During a second reading, he reads a section and then pauses, letting students talk to their neighbor and discuss what they have learned using any language they choose.
After this, Mr. Miller writes key words related to bullying in English on the board, and invites students to add different home language words under the English words. If they don’t know the word in their home language, they can check the words online or in dictionaries. Phonemic spelling of the words using the Latin alphabet is acceptable.
After checking the key vocabulary words, Mr. Miller asks the students to write individually in any language on their thoughts about the book and any connections they might have between the story and their prior knowledge. After this quick-write, students have a group discussion using translanguaging. When the teacher comes to a group with a language he doesn’t know, the students immediately shift their language to English so the teacher can understand them.
After the group discussion, each group creates a list of questions about bullying they want to investigate and presents them in class. Most questions are written in English, though some questions include words in other languages. Mr. Miller writes the questions on the board including significant non-English words in parentheses. The teacher and students go over the questions together and choose four questions they want to explore, and a new set of student groups discuss the questions.
Before the end of the class, the groups plan how they will find the information to respond to the chosen questions. Mr. Miller closes the class with this statement, “Your resources can be in any language you want, but please share your information tomorrow in English so we can all understand you.”
This is an example of a social studies translanguaging classroom, but you may find a way to apply similar strategies for math classes. Here are some examples of translanguaging strategies for mathematics teachers.
Give a word problem in English and give students time to discuss and write their solutions in their own languages (or vice versa).
Let students to see or listen for synonyms and homophones for certain words in English, their home language or both.
Give EBs an opportunity to demonstrate their knowledge and skills in mathematics in a variety of ways (multiple languages and multiple modes).
Encourage EBs to use various resources in their home languages including their family members.
You may want to see what a math lesson looks like when it employs a translanguaging pedagogy. You can find one example at the website of CUNY-NYS Initiative on Emergent Bilinguals (CUNY-NYSIEB, ).
Word Problem Analysis Framework
We say teachers need to make mathematical problems accessible to EBs, but what exactly does that mean? According to Luciana de Oliveira (2012), accessible in this statement means providing EBs with “access to the ways in which knowledge is constructed in the content areas—not by simplifying the texts, but by developing teachers’ understanding about how mathematical disciplinary discourse is constructed (p. 195). This section introduces one accessible way to use word problems for EBs by applying the Translanguaging Pedagogy and Word Problem Analysis Framework (Oliveira, 2012).
Here is one example of a word problem:
Moises is saving money to buy a book, which costs $20. He can save $3 per week. How many weeks will it take him to save enough money to buy the book?
This word problem can be divided into three clauses:
Clause 1: Moises is saving money to buy a book, which costs $20.
Clause 2: He can save $3 per week.
Clause 3: How many weeks will it take him to save enough money to buy the book?
de Oliveira (2012) recommends analyzing each “clause” of a mathematical word problem to identify the language demand needed to make the problem accessible for EBs. She proposed five questions based on Huang and Normandia (2008).
What task is the student asked to perform?
What relevant information is presented in the word problem?
What mathematical concepts are presented in the information?
What mathematical representations and procedures can students use to solve the problem, based on the information presented and the mathematical concepts identified?
What additional language demands exist in this problem?
The table below is a framework that can be used by teachers or students. Teachers are encouraged to fill out the form in advance, before asking students to analyze the given word problem. We adapted the Framework for Analyzing Word Problems (de Oliveira, 2012) by adding translanguaging components.
Table 5.1: Translanguaging Framework of Analyzing Word Problems
| Information Provided in English | Information in a language of choice | Mathematical concept in a language of choice | Mathematical Representation and Procedures |
| Clause 1: Moises is saving money to buy a book, which costs $20. | 모세는 $20짜리 책을 사기 위해 저금하고 있다. | The cost of the book: $20 | |
| Clause 2: He can save $3 per week. | 모세는 매주 3달러 모을 수 있다. | Money that Moises can save for 1 week: $3 Per week: 1주 Therefore, $3 can be saved each week. | |
| Clause 3: How many weeks will it take him to save money to buy the book? | 모세가 책을 살 수 있을 때까지 몇 주나 걸릴까? | Money to buy the book ≥ $20 모인 돈이 $20보다 많아야 한다. | |
The goal of using this framework is to enable teachers to be more proactive in helping EBs learn the ways in which language is used to construct mathematical knowledge. Although this framework shows reading, writing and representation modes, teachers can utilize other language modes, such as speaking or gesturing in classrooms. The teachers should clearly provide directions for using this form so that students can use any language they choose, as this tool is for them to understand the language and mathematical information embedded in a word problem. Sometimes, teachers may give a word problem written in their EBs’ first language and change the second column to translating each clause into English. In this way, EBs can develop not only English but also their first language, and by comparing the two languages, EBs can deepen their understanding of English. After students fill out this form, the teacher can ask students to share their analysis with their group. Students can start solving the problem after the teacher makes sure everyone understands the problem’s language through this framework.
We would like to close this chapter with the following quote from de Oliveira (2012).
Content is never separate from the language through which that content manifests itself. Learning mathematics means learning the language that expresses mathematics. The language demands that this chapter addresses highlight the kind of discipline-specific academic support in language and literacy development that would enable English language learners to be more successful in their mathematics learning (p. 204).
Reference
de Oliveira, L. C. (2012). The language demands of word problems for English language learners. In S. Celedon-Pattichis & N. G. Ramirez (Eds.), Beyond good teaching: Advancing mathematics education for ELLs (pp. 195–205). National Council of Teachers of Mathematics.
Fu, D., Hadjioannou, X., & Zhou, X. (2019). Translanguaging for emergent bilinguals: Inclusive teaching in the linguistically diverse classroom (First edition). Teachers College Press.
García, O. (2015), The translanguaging current in language education, Presented at åhörarkopior från Symposium 2015. Retrieved from
García, O., Johnson, S. I., & Seltzer, K. (2017). The translanguaging classroom: Leveraging student bilingualism for learning. Caslon.
Huang, J., & Normandia, B. (2008). Comprehending and solving word problems in mathematics: Beyond key words. In Z. Fang & M. Schleppegrell (Eds.), Reading in secondary content areas: A language-based pedagogy. University of Michigan Press.
I, J. Y., & de Araujo, Z. (2019). An examination of monolingual preservice teachers’ set-up of cognitively demanding mathematics tasks with emergent multilingual students. Research in Mathematics Education, 1–21.
I, J. Y., & Stanford, J. (2018). Preservice teachers’ mathematical visual implementation for emergent bilinguals. Mathematics Teacher Educator, 7(1), 8–33. 10.5951/mathteaceduc.7.1.0008
Reeves, J. (2006). Secondary teacher attitudes toward including English-language learners in mainstream classrooms. The Journal of Educational Research, 99(3), 131–142. |
8812 | https://www.analyzemath.com/Geometry/trapezoid_solve.html | Solve a Trapezoid Given its Bases and Legs
A trapezoid with bases b and d (d > b), legs a and c, and AD and BC are parallel is shown below . Calculate all its angles and its height h.
Angles of a Trapezoid
Using the trapezoid above, we draw BB' parallel to CD.
Using the triangle ABB', we use the cosine rule to write
c2 = a2 + (d-b)2 - 2 a (d - b) cos(∠BAD)
cos(∠BAD) = (a2 + (d-b)2 - c2) / (2 a (d - b) )
∠BAD = arccos ( (a2 + (d-b)2 - c2) / (2 a (d - b) ) )
In the same figure, ∠BB'A and ∠CDA have the same size.
Using the same triangle, we use the cosine rule again to write
a2 = c2 + (d-b)2 - 2 c (d - b) cos(∠BB'A)
cos(∠BB'A) = (c2 + (d-b)2 - a2) / (2 c (d - b) )
∠CDA = ∠BB'A = arccos ( (c2 + (d-b)2 - a2) / (2 c (d - b) ) )
In the given trapezoid, AD and BC are parallel. Hence the pairs of angles BAD and ABC and CDA and DCB are supplementary. Hence
ABC = 180° - BAD and DCB = 180° - CDA
Angles of a Trapezoid
Height and Area of a Trapezoid
h = a cos (∠BAD)
area = (1/2)(b + d) h
Height and Area of a Trapezoid
Diagonals of a Trapezoid
Use cosine rule in triangles DAB and BCD to write:
BD2 = a2 + d2 - 2 a d cos (∠ BAD)
BD = √ (a2 + d2 - 2 a d cos (∠ BAD))
CA2 = c2 + d2 - 2 c d cos (∠ CDA)
CA = √ (c2 + d2 - 2 c d cos (∠ CDA))
Diagonals of a Trapezoid
More References and Links to Geometry
Geometry Tutorials, Problems and Interactive Applets.
Trapezoid Area Calculator. Calculator to calculate the area of a trapezoid given the bases and the height.
Trapezoid Calculator and Solver. An easy to use online calculator to solve trapezoid problems. The area, the angles and the diagonals of a Trapezoid are calculated given its 4 sides.
More References and Links to Geometry |
8813 | https://docenti.ing.unipi.it/artoni-a/Di_Puccio-Cinematica.pdf | Capitolo 3 Cinematica La cinematica studia il moto di punti e corpi a prescindere dalle cause che lo determinano.
La relazione tra moto e azioni sara oggetto della dinamica. In questo capitolo si descrivono velocit a ed accelerazioni prima di un punto materiale e poi di un corpo rigido e di sistemi di corpi rigidi.
3.1 Cinematica del punto materiale Per determinare la velocita e l’accelerazione di un punto P, si deve innanzitutto esprimere il vettore posizione del punto − − − → OP(t) rispett o ad un altro punto O fisso; per definizione la veloci a di P e il vettore vP vP = d− − − → OP(t) dt = ˙ − − − → OP(t) (3.1) mentre l’accelerazione e aP = dvP dt = d2− − − → OP(t) dt2 = ˙ vP = ¨ − − − → OP(t) (3.2) Nelle precedenti relazioni si e fatto uso dell’analisi di funzioni vettoriali, per effettuare la derivata del vettore − → OP(t). Come per le funzioni scalari la derivata e il limite di un rapporto incrementale e molte proprieta delle funzioni scalari possono essere applicate al caso di finzioni vettoriali (derivata della somma, differenza, prodotto, funzioni composte etc.).
Come gi a fatto nell’algebra vettoriale, per effettuare i calcoli si esprimono le relazioni vettoriali attraverso equazioni scalari, sfruttando le componenti. In questo caso e conve-niente scrivere il vettore posizione del punto in un sistema di riferimento cartesiano fisso, secondo la relazione − − − → OP(t) = x(t) y(t) z(t) (3.3) cosi che le 3.1 e 3.2 diventano rispettivamente vP = ˙ x(t) ˙ y(t) ˙ z(t) = ˙ x i + ˙ y j + ˙ z k (3.4) e aP = ¨ x(t) ¨ y(t) ¨ z(t) = ¨ x i + ¨ y j + ¨ z k (3.5) 3.1.1 Moto lungo una traiettoria Nel caso in cui sia nota la traiettoria del punto, possono risultare utili altre espressioni per la determinazione di velocit a ed accelerazioni del punto.
x y z s O P γ P s n τ n τ Figura 3.1: Moto di un punto lungo una traiettoria.
Sia γ una curva regolare nello spazio, corrispondente alla traiettoria del punto P. Su tale curva si fissa un verso di percorrenza ed un’ascissa curvilinea s che rappresenta la posizione del punto rispetto alla curva. Il vettore posizione del punto risulta pertanto definito in funzione dell’ascissa curvilinea come − − − − → OP(s) = x(s) y(s) z(s) (3.6) Sono utili anche per la cinematica alcune proprieta differenziali delle curve, legate alle derivate di − − − − → OP(s). Si pu o dimostrare che d− → OP(s) ds = τ (3.7) ossia la derivata prima del vettore posizione rispetto all’ascissa curvilinea fornisce il versore tangente τ alla curva con verso concorde ad s. Ricorda che |τ| = 1. La derivata seconda di − − − − → OP(s) coinvolge un’altra caratteristica importante della curva γ, la curvatura c o al raggio di curvatura ρ d2− → OP(s) ds2 = dτ ds = c n = n ρ (3.8) Il raggio di curvatura puo essere interpretato come il raggio della circonferenza che ap-prossima localmente, nell’intorno di P la curva γ; con n si e invece indicato il versore normale alla curva diretto verso il centro di curvatura, ossia il centro della circonferenza approssimante.
Il moto del punto lungo la curva, come potrebbe accadere per quello di un’automobile lungo un tragitto, si esprime attraverso una legge del tipo s(t), ossia con la funzione che lega l’ascissa curvilinea al tempo. Per l’equazione 3.1 e per le regole di derivazione delle funzioni composte si ha vP = d− → OP(s(t)) dt = d− → OP(s) ds ˙ s = ˙ s τ (3.9) che indica che la velocita e sempre tangente alla traiettoria. Per le accelerazioni, in base alla precedente e alla 3.2, considerando che anche τ cambia nel tempo e applicando le regole di derivazione del prodotto di funzioni, si ottiene aP = dvP dt = d (˙ s τ) dt = ¨ sτ + ˙ sdτ dt = ¨ sτ + ˙ s µdτ ds ˙ s ¶ = ¨ sτ + ˙ s2 ρ n (3.10) La relazione 3.10 evidenzia che l’accelerazione di un punto e data da una componente diretta come n, pertanto denominata centripeta e da una parallela a τ, che rappresenta la componente tangenziale. Si osserva inoltre che l’accelerazione del punto pu o essere diversa da zero anche se questo si muove lungo la curva con velocita ˙ s costante; infatti la curvatura fa si che il vettore velocit a cambi in direzione, grazie alla componente normale dell’accelerazione.
Moto lungo una traiettoria circolare Un caso di particolare interesse come applicazione del paragrafo precedente e il moto di un punto lungo una traiettoria circolare. In questo caso e conveniente esprimere l’ascissa curvilinea s, rappresentante la lunghezza dell’arco, in funzione dell’angolo al centro θ, ossia s = r θ, come mostrato in Fig. 3.2.
Ne segue facilmente dalle 3.9 e 3.10 che velocita ed accelerazione sono date da vP = r ˙ θ τ (3.11) aP = r¨ θ τ + r ˙ θ2n (3.12) 3.2 Cinematica del corpo rigido Le relazioni cinematiche per un corpo rigido possono essere ottenute da quelle del punto materiale, tenendo conto che i punti che costituiscono il corpo si muovono compatibil-mente con il vincolo di rigid a. Ossia, ad ogni istante durante il movimento P O θ s r Figura 3.2: Moto di un punto lungo una traiettoria circolare.
a) deve rimanere costante la distanza tra due punti qualsiasi del corpo rigido b) deve rimanere costante l’angolo formato tra due segmenti qualsiasi solidali al corpo rigido.
3.2.1 Moto rotatorio attorno ad un asse fisso Si consideri adesso un corpo rigido vincolato al telaio con una coppia rotoidale in un punto A. Il corpo si muove di moto rotatorio ed ogni suo punto, dovendo per l’ipotesi di rigidita avere distanza costante da A, descrive una traiettoria circolare di centro A e raggio pari alla distanza dal punto A stesso, come mostrato in Fig.3.3. Possono quindi essere usate per ogni punto le formule 3.11 vP = rP ˙ θ τ P vQ = rQ ˙ φ τ Q aP = rP ¨ θ τ P + rP ˙ θ2nP aQ = rQ ¨ φ τ Q + rQ ˙ φ2nQ Poich´ e, sempre per l’ipotesi di rigidit a, l’angolo tra AP e AQ –pari a γ = θ −φ– rimane P θ A Q φ traiettoria di P traiettoria di Q Figura 3.3: Moto rotatorio attorno ad un asse fisso per A.
costante durante il moto, si ha che ˙ θ = ˙ φ ¨ θ = ¨ φ Questo permette di definire due vettori, denominati rispettivamente velocita angolare e accelerazione angolare ω = ˙ θ k (3.13) ˙ ω = ¨ θ k (3.14) che in questo caso sono entrambi paralleli all’asse di rotazione k; in un moto qualsiasi i due vettori non sono tra loro paralleli ma rimane comunque vero che essi sono grandezze rappresentative della rotazione del corpo rigido. E importante quindi sottolineare che si parla di velocita ed accelerazione angolare di un corpo rigido e non di un punto del corpo rigido.
Avendo introdotto questi vettori, si possono scrivere velocit a ed accelerazione dei punti di un corpo rigido in rotazione attorno ad un asse fisso di direzione k, come vP = ω ∧− → AP (3.15) aP = ˙ ω ∧− → AP −ω2− → AP (3.16) Moto rotatorio attorno ad un punto fisso Si analizza il caso di rotazione attorno ad un punto, ossia si sostituisce la coppia rotoidale in A con una coppia sferica.
Si osserva innanzitutto che, mentre nel caso di coppia rotoidale la traiettoria di un punto qualsiasi giace su un piano ortogonale all’asse di rotazione, nel caso di coppia sferica la traiettoria e una curva nello spazio appartenente ad una sfera di centro A. Sottolineando che in questo caso le direzioni di ω e ˙ ω possono essere qualsiasi, le relazioni precedenti divengono vP = ω ∧− → AP (3.17) aP = ˙ ω ∧− → AP + ω ∧(ω ∧− → AP) = ˙ ω ∧− → AP −ω2− − → AP ∗ (3.18) dove P ∗ e il punto proiezione di P sulla retta per A parallela ad ω.
3.2.2 Moto traslatorio Un corpo rigido si muove di moto traslatorio quando la sua velocita angolare rimane nulla nel tempo e quindi e nulla anche la sua accelerazione angolare. Puo essere utile osservare che il moto traslatorio pu o essere sia rettilineo che curvilineo, come mostrato in Fig.3.4.
In entrambi i casi tutti i punti descrivono traiettorie omologhe per cui il vettore sposta-mento di un punto tra la posizione ad un istante t e quella ad un istante t′ e lo stesso per tutti i punti − − → AA′ = − − → PP ′ = d Ne consegue inoltre che tutti i punti di un corpo rigido che si muove di moto traslatorio hanno ugual velocit a e accelerazione vA = vP aA = aP (3.19) P A Q P' A' Q' t t' P A Q P' A' Q' t t' moto traslatorio rettilineo moto traslatorio curvilineo Figura 3.4: Moto traslatorio rettilineo o curvilineo.
3.2.3 Moto rototraslatorio Il moto piu generale di corpo rigido e dato dalla combinazione di traslazione e rotazione e si definisce pertanto rototraslatorio. Come sara chiarito nello studio dei moti composti, dato un punto fisso O e due punti A e B si ha che − − → OB = − → OA + − → AB che per derivazione fornisce vB = vA + vBA aB = aA + aBA dove vBA e aBA rappresentano la velocit a e l’accelerazione di B rispetto ad A. Quando A e B appartengono allo stesso corpo rigido, possono essere applicati per vBA e aBA le relazioni 3.15 per moto piano o 3.17 per moto nello spazio. Si ottengono cosı le relazioni che legano velocit a ed accelerazioni di punti di un corpo rigido ossia la formula fondamentale della cinematica rigida valida sia nel piano che nello spazio vB = vA + ω ∧− → AB (3.20) il teorema di Rivals per accelerazioni nel caso di moto piano aB = aA + ˙ ω ∧− → AB −ω2− → AB (3.21) che per moti nello spazio diviene aB = aA + ˙ ω ∧− → AB −ω2− − → AB∗ (3.22) .
3.2.4 Centro delle velocita Dal confronto tra la 1.22 e la formula fondamentale, ossia vB = ω ∧− → AB M B = M A + R ∧− → AB il campo di velocit a di un corpo rigido puo essere studiato in modo analogo a quanto fatto per i sistemi equivalenti.
In particolare, per un modo piano non traslatorio (l’ipotesi R ̸= 0 diviene ω ̸= 0), si pu o individuare un punto CV avente velocita nulla. Questo punto prende il nome di centro delle velocit a del corpo rigido, rispetto al quale la formula fondamentale diviene vB = ω ∧− − → CV B (3.23) che esprime il fatto che il moto del corpo e equivalente ad una rotazione attorno al centro delle velocit a. L’individuazione del punto CV , che cambia nel tempo, si ottiene conoscendo la direzione delle velocita di due suoi punti; infatti, in base al teorema di Chasles, il centro delle velocit a giace sull’ortogonale alle velocita nei punti.
Nel caso di moto della spazio, come per i sistemi di vettori, esiste un punto Ωavente velocit a parallela alla velocita angolare ω, ossia, analogamente alla (1.26), vΩ∧ω = 0 (3.24) L’asse per Ωparallelo a ω prende il nome di asse elicoidale del moto o asse di Mozzi, proprio perch´ e i punti del corpo hanno velocit a come una vite con quest’asse. La de-terminazione dell’asse elicoidale del moto segue la stessa procedura descritta per l’asse centrale.
3.3 Moti relativi Le espressioni precedenti permettono di determinare velocita ed accelerazione di un punto o di un corpo, che vengono definite assolute in quanto valutate rispetto ad un punto O fisso o ad un osservatore fisso. Talvolta per o risulta piu semplice descrivere la cinematica attraverso un punto o un osservatore mobile; ad esempio, se si considera una pallina che pu o scorrere in una guida ricavata in un disco -Fig. 3.5 , un osservatore solidale al disco vede la pallina muoversi nella guida secondo una traiettoria semplice, mentre per un osservatore fisso la traiettoria della pallina e di difficile determinazione.
Figura 3.5: Moto rotatorio attorno ad un asse fisso per A.
Occorre pertanto acquisire strumenti per poter trattare opportunamente anche questi problemi, ossia casi in cui ci sia la composizione di un movimento (come quello della pallina e quello del disco nell’esempio precedente).
Si fa innanzitutto riferimento a due situazioni diverse: -moto rispetto ad un punto, a cui si e gia accennato e nel quale si considera il moto relativo fra due punti; -moto rispetto ad un osservatore ausiliario o moto rispetto ad un corpo rigido, come quel-lo della pallina rispetto al disco.
3.3.1 Moto rispetto ad un punto Come anticipato, si considerano due punti A e B indipendenti o appartenenti allo stesso corpo rigido; si hanno le seguenti relazioni rispettivamente per la velocit a e per l’accele-razione dei punti − − → OB = − → OA + − → AB (3.25) vB = vA + vBA (3.26) aB = aA + aBA (3.27) (3.28) 3.3.2 Composizione delle velocita Si consideri adesso un osservatore ausiliario, mobile, per il quale sono facilmente deter-minabili velocit a v(rel) P ed accelerazione a(rel) P relative di un punto o di un corpo. Noto il moto dell’osservatore e possibile risalire alla velocit a assoluta del punto attraverso il teorema di composizione delle velocita che afferma che vP = v(tr) P + v(rel) P (3.29) ossia la velocit a assoluta di un punto e data dalla somma vettoriale della velocit a relativa e della velocita di trascinamento v(tr) P , ossia la velocit a che avrebbe P se fosse solidale all’osservatore ausiliario e quindi fermo rispetto ad esso.
3.3.3 Composizione delle accelerazioni Analogamente al precedente si ha anche il teorema di composizione delle accelera-zioni in base al quale aP = a(tr) P + a(rel) P + a(Cor) P (3.30) dove con a(tr) P si intende l’accelerazione di trascinamento del punto e con a(Cor) P l’accele-razione di Coriolis definita come a(Cor) P = 2ω(tr) ∧v(rel) P (3.31) Dalla equazione precedente si ha che l’accelerazione di Coriolis e nulla se l’osservatore ausiliario non ruota ossia se si muove di moto traslatorio.
3.3.4 Composizione delle velocit a angolari Una ulteriore importante relazione si ottiene anche per le velocita angolari nei moti relativi ω = ω(tr) + ω(rel) (3.32) e quindi la velocit a angolare assoluta di un corpo e data dalla somma vettoriale della velocit a angolare relativa e della velocita angolare di trascinamento ω(tr), ossia la velocit a angolare dell’osservatore ausiliario.
Si sottolinea che tutte queste relazioni sono applicabili sia a moti nel piano che a moti nello spazio. Possono essere ottenute prendendo un sistema di riferimento fisso S(O, x, y, z) ed uno mobile S′(O′, x′, y′, z′) ed esprimendo il vettore posizione del punto P come − → OP(t) = − − → OO′(t) + − − → O′P(t) e quindi in componenti − → OP(t) = x(t)i + y(t)j + z(t)k − − → OO′(t) = xO′(t)i + yO′(t)j + zO′(t)k − − → O′P(t) = x′(t)i′(t) + y′(t)j′(t) + z′(t)k′(t) quindi xi + yj + zk = xO′i + yO′j + zO′k + x′i′ + y′j′ + z′k′ Per derivazione si ottengono le velocita vP = ˙ xi + ˙ yj + ˙ zk vP = ˙ xO′i + ˙ yO′j + ˙ zO′k + ω ∧(x′i′ + y′j′ + z′k′) + ˙ x′i′ + ˙ y′j′ + ˙ z′k′ in cui si riconoscono v(tr) P = ˙ xO′i + ˙ yO′j + ˙ zO′k + ω ∧(x′i′ + y′j′ + z′k′) v(rel) P = ˙ x′i′ + ˙ y′j′ + ˙ z′k′ Nelle precedenti relazioni si e fatto uso delle formule di Poisson per ricavare le derivate dei versori del riferimento mobile. In generale si ha che il vettore derivata di un versore -o di un vettore con modulo costante- e ortogonale al versore stesso; in particolare, se ω e la velocita angolare dell’osservatore ausiliario e quindi anche del sistema di riferimento S′ ad esso solidale, le formule di Poisson mostrano che di′ dt = ω ∧i′ dj′ dt = ω ∧j′ dk′ dt = ω ∧k′ (3.33) 3.4 Gradi di libert a Ci si pone adesso il problema di determinare la posizione di un punt o di un corpo nel piano e nello spazio. Chiaramente per definire la posizione di un punto basta conoscere le sue coordinate, due o tre rispettivamente nel piano o nello spazio. Il numero minimo di parametri necessari e sufficienti a definire la posizione di un punto/corpo prende il nome di gradi di liberta. Quanti sono i gradi di libert a di un corpo rigido nel piano?
Per rispondere basta considerare che la posizione del corpo e nota quando lo e quella di due suoi punti A e B. Poich´ e la distanza tra A e B e costante e nota, sar a sufficiente conoscere la posizione di A e l’inclinazione del segmento AB rispetto ad una direzione di riferimento, quindi i gradi di liberta di un corpo rigido nel piano sono 3. Analogamente nello spazio con tre coordinate siamo in grado di individuare il punto A e con altri tre parametri l’orientazione nello spazio del segmento AB, quindi i gradi di libert a di un corpo rigido nello spazio sono sei.
Queste considerazioni ipotizzano che il corpo sia libero, sia cioe senza vincoli che possono limitare i movimenti riducendone i gradi di libert a.
3.5 Analisi dei vincoli Cosı come in statica i vincoli giocano in cinematica un ruolo fondamentale nella deter-minazione del moto di un corpo o di un punto ed in questo paragrafo si analizzano in dettaglio i vari tipi di vincolo limitandoci al caso piano.
3.5.1 Appoggio semplice o carrello Appoggio semplice o carrello su telaio Si consideri innanzitutto un corpo rigido vincolato al telaio con un carrello nel punto A.
In questo caso l’ordinata del punto A non pu o variare per cui sara sufficiente conoscerne l’ascissa per individuare la sua posizione, che con l’angolo θ costituiscono i due gradi di libert a del corpo. Si ha quindi che un vincolo carrello toglie un grado di liberta al corpo o alternativamente, fornisce una condizione di vincolo yA = cost che implica inoltre: - per le velocit a vA = v i - per le accelerazioni aA = a i A A B θ x y B θ x y Figura 3.6: Corpo rigido vincolato con carrello al telaio in A.
3.5.2 Coppia rotoidale Coppia rotoidale con telaio Nel caso in cui il corpo rigido sia vincolato al telaio con una coppia rotoidale in A, che ne mantiene bloccate ascissa ed ordinata xA = cost yA = cost . Il corpo ha percio un solo grado di libert a corrispondente all’angolo θ. Quindi una coppia rotoidale toglie due gradi di liberta al corpo con due condizioni di vincolo, dalle quali si ha - per le velocit a vA = 0 - per le accelerazioni aA = 0 Coppia rotoidale mobile Come vincolo mobile la coppia rotoidale fa si che due punti di due corpi rimangano uniti durante il movimento lasciando pero liberi due corpi di ruotare. Le condizioni che fornisce questo vincolo sono dunque due, cio e x(1) B = x(2) B y(1) B = y(2) B dalle quali si ottengono - per le velocita v(1) A = v(2) A - per le accelerazioni a(1) A = a(2) A Considerando il moto relativo di A(2) rispetto al corpo (1) - velocit a v(rel) A = 0 - accelerazione a(rel) A = 0 che significa che il moto relativo del corpo (2) rispetto al corpo (1) e rotatorio attorno al punto A.
3.5.3 Coppia prismatica Coppia prismatica con telaio Un corpo vincolato con coppia prismatica al telaio, pu o semplicemente traslare nella direzione della coppia prismatica lasciando al corpo un solo grado di liberta. Le condizioni fornite dal vincolo sono due, cio e ya = cost θ = cost da cui - il moto e traslatorio, per cui ω = ˙ ω = 0 - per le velocit a v = v i - per le accelerazioni a = a i Coppia prismatica mobile Quando la coppia prismatica collega due corpi come vincolo interno le relazioni precedenti continuano ad essere valide in un moto relativo; le condizioni del vincolo sono ancora due y′ a = cost θ′ = cost da cui - il moto relativo e traslatorio, per cui ω(rel) = ˙ ω(rel) = 0 - per le velocit a v(rel) = v i′ - per le accelerazioni a(rel) = a i′ |
8814 | https://www.youtube.com/watch?v=8xkP6rX4TqM | Finding a Cusp. Sketch the Graph. Concavity y = 2- x^(2/3)
Ms Shaws Math Class
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Intro hi everyone we're going to sketch the graph of y equals 2 minus X to the 2/3 anytime it looks like this a little bit it's probably going to be a cusp type graph and we're first going to test the function that means finding for x and y-intercepts take the first derivative second derivative and test in behavior so basically here we go and let's first find the x and Find x yintercepts y-intercepts so we set to find the x-intercept set Y to 0 so you get 2 minus X to the 2/3 which is X to the 2/3 equals 2 and then just do this three halves three halves so basically this is going to x equals the square root of 2 cubed which is square root of 8 which equals to square root of 2 and don't forget this is going to be plus and minus all right so when you do the square root plus and minus so basically our x-intercepts are at plus or minus or just put plus 2 square root of 2 comma 0 and negative 2 square root of 2 comma 0 all right when we're square root of 2 equals 1.41 approximately approximately all right now we have to find the y-intercepts so that's when you let X be 0 so we have y equals 2 minus 0 to the 2/3 so that just means y equals 2 so when x is 0 Y is 2 all right so that's going to help us with the graph next let's test the first derivative so when First derivative you do that you're going to get this to goes away so y prime equals let's see just to - so this is to the 2/3 I thought I did something wrong ok so that's positive 2/3 so you're going to get negative 2/3 X minus 1/3 to the negative 1/3 power now set this to 0 now there are no values for X which this equation is zero so we have new stuff we have to deal with at x equals 0 then we're going to have to look at the limits as X approaches 0 from the right of negative 2/3 X to the negative 1/3 that's going to lead to negative infinity all right and as the limit as X approaches 0 from the left of negative 2/3 X to the negative 1/3 that leads to infinity therefore when you have something like this that leads to a cusp and the cusp is going to be at 0 comma 2 because we did it at 0 here all right now we have to take the second derivative and so the Y double prime is Second derivative going to equal to 9 X to the negative 4/3 again you set this to zero there is no way for this to equal zero so the second derivative is positive for all values of X except at 0 because that's going to be stationary there so therefore the graph is concave up everywhere except zero it's positive at all values except zeros so it's always concave up so if we graph this we have this we have Graphing our y-intercept these are X intercepts here and it's concave up everywhere so it's concave up here graphing from the left taking the limit and then from the right okay it's always going positive here positive here and positive there and there's your cusp thank you guys have a nice day bye bye [Music] |
8815 | https://brainly.com/question/39715522 | [FREE] Find the exact value of \cos\left(\frac{2\pi}{3}\right) in its simplest form with a rational denominator. - brainly.com
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Find the exact value of cos(3 2 π) in its simplest form with a rational denominator.
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The exact value of cos (2 π/3) is -1/2. This is solved either by referencing the unit circle or invoking the symmetries of the cosine function, both of which confirm a cosine of -1/2 for an angle of 2 π/3 radians.
Explanation
The question is asking for the exact value of the cosine of an angle represented by 2 π/3. To solve, we can use the unit circle or the symmetries of the cosine function.
In the unit circle with a radius of 1 unit, the x-coordinate is the cosine of the angle. For an angle of π/3 radians (or 60 degrees), the x-coordinate is 1/2. But the angle 2 π/3is in the second quadrant, where cosine (the x-coordinate) is negative. So, cos (2 π/3) = -1/2.
Alternatively, we can use the fact that cosine is an even function, meaning cos(θ) = cos(-θ). But, 2 π/3 = π - π/3. Because cosine of an angle in the second quadrant is negative and cos(π - θ) = - cos(θ), we can confirm that cos(2 π/3) = -1/2. This answer is in simplest form, with a rational denominator.
Learn more about Cosine Value of an Angle here:
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The exact value of cos(3 2 pi) is −2 1, derived from the unit circle. This result indicates that the cosine of an angle in the second quadrant is negative. The calculation utilized the reference angle of 3 pi, where cos(3 pi)=2 1.
Explanation
To find the exact value of cos(3 2 π), we can utilize the unit circle and the properties of trigonometric functions. 3 2 π radians corresponds to an angle of 120 degrees, which lies in the second quadrant of the unit circle.
In the unit circle, each angle corresponds to a point with coordinates (x,y), where x represents the cosine of the angle and y represents the sine.
For the reference angle corresponding to 3 2 π, we can subtract π (or 180 degrees). The reference angle for 3 2 p i is thus 3 pi (or 60 degrees), for which cos(3 pi)=2 1.
However, in the second quadrant, the cosine value is negative. Therefore, cos(3 2 pi)=−2 1.
In conclusion, the exact value of cos(3 2 π) is −2 1, expressed in its simplest form with a rational denominator.
Examples & Evidence
You can use the unit circle to visualize how cosine values change in different quadrants. For instance, the values of cosine are positive in the first quadrant and negative in the second quadrant, as illustrated by the angles 60 degrees (or 3 pi) and 120 degrees (or 3 2 pi).
The trigonometric values derived from the unit circle are well-documented, confirming that the cosine value for angles in the second quadrant is indeed negative. For instance, standard trigonometric tables and the unit circle depict cos(6 0 o)=2 1 and cos(12 0 o)=−2 1.
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The exact value of cos(2π/3) is -1/2.
To find the exact value of cos(2π/3) in its simplest form with a rational denominator, we follow these steps:
Recall that cosine is an even function, which means cos(-θ) = cos(θ).
Recognize that 2π/3 is in the second quadrant (between π/2 and π radians).
In the second quadrant, cosine is negative, so
cos(2π/3) = -cos(π - 2π/3).
Simplify the angle: π - 2π/3 = π/3.
cos(π/3) = 1/2.
Therefore, cos(2π/3) = -1/2.
The exact value of cos(2π/3) is -1/2.
Answered by WilliamClarkGable •7.3K answers•1.4M people helped
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8816 | https://www.fullpotentialtutor.com/what-is-a-linear-cost-function/ | What Is A Linear Cost Function – Full Potential Learning Academy
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PreviousNext
What Is A Linear Cost Function
What are Linear Cost Functions?
By Robert O
Explain linear cost functions with examples:
There is nothing unique about linear cost functions if you are already familiar with general linear functions, which are of the form y = m x+c The equation is not new to you, right?
When it comes to linear cost functions, we will be dealing with the cost of a commodity or service that has two parts. To get what it will cost to purchase the product or receive the service, you need to consider both the variable cost and a fixed cost. The equation is of the form:
Where:
Y is the total cost
A is the cost per unit, which depends on the number of units x.
B is the fixed cost or fixed charge.
Think of your utility bills as an example. There is usually a fixed charge that does not depend on the units that you use. So, whether you used energy in a month or not, you have to pay the fixed charges.
How do we solve linear cost problems?
The key to solving problems involving linear cost is to understand the question statement. You calculate the value of B and A using the information from it. With these constants known, you can calculate the total cost y for each input variable x. Let’s understand it by using examples.
Example 1
The total cost of producing two dresses is 130 dollars, and the production cost of 5 similar dresses is 190 dollars. By assuming a linear cost function, what is the cost of producing 8 such dresses?
Solution
We use the linear cost function to form two linear equations and then solve for the unknown.
From the general equation:
It is now clear that we have a system of two linear equations. We can solve it by Gaussian elimination, algebraic substitution, or graphical method. In this case, we will use algebraic substitution.
Using the first equation:
Now, substitute the value of A in any equation to find B.
Form a linear cost function:
The total cost of producing 8 dresses is:
The total cost of producing 8 dresses is 250 dollars.
What if we graph our linear cost function
Graph generated from
The line crosses the y-axis (y-intercept) at the fixed cost point. The value is 90. The x-axis represents the units produced, and the y-axis represents the total cost.
Example 2
A fixed cost of running machine A is 75 dollars and a variable cost of 3 dollars for every unit of product from this machine. Another machine, B, has a fixed cost of 60 dollars and a variable cost for producing an item of 4.5 dollars. What number of items will make the running cost of both machines the same?
Solution
Let us use the information from the problem statement to create two linear equations.
We solve for x, which is the value we need, by equating the right side of the equations.
The cost of running the two machines is the same when we produce the 10 th item.
Remarks
The step to solving linear cost functions is to understand the statement to form a system of two linear equations. We then solve to find the unknown values, after which we can compute the cost of producing any unit. You only need to be keen on interpreting the problem statement.
About the Author
This lesson was prepared by Robert O. He holds a Bachelor of Engineering (B.Eng.) degree in Electrical and electronics engineering. He is a career teacher and headed the department of languages and assumed various leadership roles. He writes for Full Potential Learning Academy.
Subbarayan Pochi2021-06-11T00:55:46-04:00 May 7th, 2021|Brain teaser, FPLA Winning Strategies, Home School, Math-e-Magics, SAT, ACT, PERT, HSPT|
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8817 | https://math-gpt.org/problems/question/using-bezout-s-identity-prove-that-gcd-2-n-1-2-m-1-gcd-n-m | To prove that (2^n - 1, 2^m - 1) = (n, m) using Bézout's identity, we start by using the property of exponents and the d...
Question:
To prove that (2^n - 1, 2^m - 1) = (n, m) using Bézout's identity, we start by using the property of exponents and the definition of greatest common divisor (gcd).Let d = (n, m). By the properties of gcd, we can express n and m in terms of d:n = kd and m = ldfor some integers k and l.Now we will show that 2^d - 1 divides both 2^n - 1 and 2^m - 1.By the property of exponents:2^n - 1 = 2^kd - 1 = (2^d)^k - 1Using the difference of squares (or the general factorization), we have:(2^d - 1) ( (2^d)^k-1 + (2^d)^k-2 + + 1 )Thus, it follows that 2^d - 1 divides 2^n - 1. Similarly,2^m - 1 = 2^ld - 1 = (2^d)^l - 1Also implies that 2^d - 1 divides 2^m - 1.Now, we need to prove that 2^n - 1 and 2^m - 1 are divisible by 2^(n, m) - 1. Let g = (2^n - 1, 2^m - 1). By the property of the gcd of the two exponents:(n, m) = d,We employ the result:g divides 2^kd - 1 and 2^ld - 1,which implies:g divides 2^d - 1.Thus, we conclude that g divides 2^(n, m) - 1.Since 2^d - 1 divides both 2^n - 1 and 2^m - 1, we can say:(2^n - 1, 2^m - 1) = 2^(n, m) - 1.This leads us to conclude,(2^n - 1, 2^m - 1) = 2^d - 1 = 2^(n, m) - 1.Hence, we have shown that:(2^n - 1, 2^m - 1) = (n, m).This completes the proof.
Answer: |
8818 | https://www.youtube.com/watch?v=SuWD39jh1Xc | Establish sum and difference identity sin(pi/2 + x) = cos x.
Ms Shaws Math Class
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Posted: 6 Feb 2021
Transcript:
hi everyone we're going to establish this identity using our sum and difference formulas so the first one we have is this so we have sine of alpha plus beta in our case our alpha equals pi divided by 2 and our beta equals theta so what we're looking for is this format now let's use substitution and write this right hand side so we have sine of alpha or alpha is pi divided by 2 times cosine of beta so our beta is theta and then we write the plus cosine of alpha pi divided by 2 times sine of beta which is theta now we want to make sure it equals cosine of theta so we have sine of pi divided by two that equals one so one times cosine of theta plus cosine of pi divided by 2 is 0. so you have 0 times sine of theta this 0 times sine of theta is cancelled out and 1 times cosine of theta equals just cosine of theta which is our desired result and that's it thank you have a nice day bye bye [Applause] [Music] |
8819 | https://www3.nd.edu/~dgalvin1/10120/10120_S16/Topic20_8p7_Galvin.pdf | Normal Distributions So far we have dealt with random variables with a finite number of possible values. For example; if X is the number of heads that will appear, when you flip a coin 5 times, X can only take the values 0, 1, 2, 3, 4, or 5.
Some variables can take a continuous range of values, for example a variable such as the height of 2 year old children in the U.S. population or the lifetime of an electronic component. For a continuous random variable X, the analogue of a histogram is a continuous curve (the probability density function) and it is our primary tool in finding probabilities related to the variable. As with the histogram for a random variable with a finite number of values, the total area under the curve equals 1.
Normal Distributions Probabilities correspond to areas under the curve and are calculated over intervals rather than for specific values of the random variable.
Although many types of probability density functions commonly occur, we will restrict our attention to random variables with Normal Distributions and the probabilities will correspond to areas under a Normal Curve (or normal density function).
This is the most important example of a continuous random variable, because of something called the Central Limit Theorem: given any random variable with any distribution, the average (over many observations) of that variable will (essentially) have a normal distribution. This makes it possible, for example, to draw reliable information from opinion polls.
Normal Distributions The shape of a Normal curve depends on two parameters, µ and σ, which correspond, respectively, to the mean and standard deviation of the population for the associated random variable. The graph below shows a selection of Normal curves, for various values of µ and σ. The curve is always bell shaped, and always centered at the mean µ.
Larger values of σ give a curve that is more spread out.
The area beneath the curve is always 1.
Properties of a Normal Curve 1. All Normal Curves have the same general bell shape.
2. The curve is symmetric with respect to a vertical line that passes through the peak of the curve.
3. The curve is centered at the mean µ which coincides with the median and the mode and is located at the point beneath the peak of the curve.
4. The area under the curve is always 1.
5. The curve is completely determined by the mean µ and the standard deviation σ. For the same mean, µ, a smaller value of σ gives a taller and narrower curve, whereas a larger value of σ gives a flatter curve.
6. The area under the curve to the right of the mean is 0.5 and the area under the curve to the left of the mean is 0.5.
Properties of a Normal Curve 7. The empirical rule (68%, 95%, 99.7%) for mound shaped data applies to variables with normal distributions.
For example, approximately 95% of the measurements will fall within 2 standard deviations of the mean, i.e.
within the interval (µ −2σ, µ + 2σ).
8. If a random variable X associated to an experiment has a normal probability distribution, the probability that the value of X derived from a single trial of the experiment is between two given values x1 and x2 (P(x1 ⩽X ⩽x2)) is the area under the associated normal curve between x1 and x2. For any given value x1, P(X = x1) = 0, so P(x1 ⩽X ⩽x2) = P(x1 < X < x2).
Properties of a Normal Curve Here are a couple of pictures to illustrate items 7 and 8.
xx x 1 2 Area approx. 0.95 μ μ − 2σ μ + 2σ The standard Normal curve The standard Normal curve is the normal curve with mean µ = 0 and standard deviation σ = 1.
We will see later how probabilities for any normal curve can be recast as probabilities for the standard normal curve.
For the standard normal, probabilities are computed either by means of a computer/calculator of via a table.
Areas under the Standard Normal Curve z Area = A(z) = P(Z ⩽z) Probabilities for the standard Normal The table consists of two columns. One (on the left) gives a value for the variable z, and one (on the right) gives a value A(z), which can be interpreted in either of two ways: z A(z) 1 0.8413 A(z) = the area under the standard normal curve (µ = 0 and σ = 1) to the left of this value of z, shown as the shaded region in the diagram on the next page.
A(z) = the probability that the value of the random variable Z observed for an individual chosen at random from the population is less than or equal to z.
A(z) = P(Z ⩽z).
Probabilities for the standard Normal The shaded area is A(1) = 0.8413, correct to 4 decimal places.
The section of the table shown above tells us that the area under the standard normal curve to the left of the value z = 1 is 0.8413. It also tells us that if Z is normally distributed with mean µ = 0 and standard deviation σ = 1, then P(Z ⩽1) = .8413.
Examples If Z is a standard normal random variable, what is P(Z ⩽2)? Sketch the region under the standard normal curve whose area is equal to P(Z ⩽2). Use the table to find P(Z ⩽2).
P(Z ⩽2) = 0.9772.
Examples If Z is a standard normal random variable, what is P(Z ⩽−1)? Sketch the region under the standard normal curve whose area is equal to P(Z ⩽−1).
P(Z ⩽−1) = 0.1587.
Area to the right of a value Recall now that the total area under the standard normal curve is equal to 1. Therefore the area under the curve to the right of a given value z is 1 −A(z). By the complement rule, this is also equal to P(Z > z).
z Area = 1 −A(z) = P(Z ⩾z) Examples If Z is a standard normal random variable, use the above principle to find P(Z ⩾2). Sketch the region under the standard normal curve whose area is equal to P(Z ⩾2).
P(Z ⩽2) = 0.9772 so P(Z ⩾2) = 1 −0.9772 = 0.0228.
Examples If Z is a standard normal random variable, find P(Z ⩾−1). Sketch the region under the standard normal curve whose area is equal to P(Z ⩾−1).
P(Z ⩽ −1) = 0.1587 so P(Z ⩾−1) = 1 −0.1587 = 0.8413.
The area between two values We can also use the table to compute P(z1 < Z < z2) = P(z1 ⩽Z < z2) = P(z1 < Z ⩽z2) = P(z1 ⩽Z ⩽z2) = A(z2) −A(z1).
z1 z2 Area = A(z2) −A(z1) = P(z1 < Z < z2) Our previous examples can be thought of like this: P(Z ⩽z) = P(−∞< Z ⩽z) = A(z) −A(−∞) = A(z) P(z < Z) = P(z < Z < ∞) = A(∞) −A(z) = 1 −A(z) Example If Z is a standard normal random variable, find P(−3 ⩽Z ⩽3). Sketch the region under the standard normal curve whose area is equal to P(−3 ⩽Z ⩽3).
P(−3 ⩽Z ⩽3) = P(Z ⩽3) −P(Z ⩽−3) = 0.9987 −0.0013 = 0.9973.
Empirical Rule for the standard normal If data has a normal distribution with µ = 0, σ = 1, we have the following empirical rule: ▶Approximately 68% of the measurements will fall within 1 standard deviation of the mean or equivalently in the interval (−1, 1).
▶Approximately 95% of the measurements will fall within 2 standard deviations of the mean or equivalently in the interval (−2, 2).
▶Approximately 99.7% of the measurements (essentially all) will fall within 3 standard deviations of the mean, or equivalently in the interval (−3, 3).
Verifying the empirical rule P(−1 ⩽Z ⩽1) = P(Z ⩽ 1)−P(Z ⩽−1) = 0.8413− 0.1587 = 0.6827.
P(−2 ⩽Z ⩽2) = P(Z ⩽ 2)−P(Z ⩽−2) = 0.9772− 0.0228 = 0.9545.
Examples (a) Sketch the area beneath the density function of the standard normal random variable, corresponding to P(−1.53 ⩽Z ⩽2.16), and find the area.
P(−1.53 ⩽Z ⩽2.16) = P(Z ⩽2.16) −P(Z ⩽−1.53) = 0.9846 −0.0630 = 0.9216.
(b) Sketch the area beneath the density function of the standard normal random variable, corresponding to P(−∞⩽Z ⩽1.23) and find the area.
P(−∞⩽Z ⩽1.23) = 0.8907.
(c) Sketch the area beneath the density function of the standard normal random variable, corresponding to P(1.12 ⩽Z ⩽∞) and find the area.
P(1.12 ⩽Z ⩽∞) = 1 − P(Z ⩽1.12) = 1 −0.8686 = 0.1314.
General Normal Random Variables Recall how we used the empirical rule to solve the following problem: The scores on the LSAT exam, for a particular year, are normally distributed with mean µ = 150 points and standard deviation σ = 10 points. What percentage of students got a score between 130 and 170 points in that year (or what percentage of students got a Z-score between -2 and 2 on the exam)?
LSAT Scores distribution and US Law Schools General Normal Random Variables LSAT Scores distribution and US Law Schools We will now use normal distribution tables to solve this kind of problem. We do not have a table for every normal random variable (there are infinitely many of them!). So we will convert problems about general normal random to problems about the standard normal random variable, by standardizing — converting all relevant values of the general normal random variable to z-scores, and then calculating probabilities of these z-scores from a standard normal table (or using a calculator).
Standardizing If X is a normal random variable with mean µ and standard deviation σ, then the random variable Z defined by Z = X −µ σ “z-score of Z” has a standard normal distribution. The value of Z gives the number of standard deviations between X and the mean µ (negative values are values below the mean, positive values are values above the mean).
Standardizing To calculate P(a ⩽X ⩽b), where X is a normal random variable with mean µ and standard deviation σ: ▶Calculate the z-scores for a and b, namely (a −µ)/σ and (b −µ)/σ ▶ P(a ⩽X ⩽b) =P a −µ σ ⩽X −µ σ ⩽b −µ σ = P a −µ σ ⩽Z ⩽b −µ σ where Z is a standard normal random variable.
▶If a = −∞, then a−µ σ = −∞and similarly if b = ∞, then b−µ σ = ∞.
▶Use a table or a calculator for standard normal probability distribution to calculate the probability.
Examples If the length of newborn alligators, X, is normally distributed with mean µ = 6 inches and standard deviation σ = 1.5 inches, what is the probability that an alligator egg about to hatch, will deliver a baby alligator between 4.5 inches and 7.5 inches?
P(4.5 ⩽X ⩽7.5) = P 4.5 −6 1.5 ⩽Z ⩽7.5 −6 1.5 = P(−1 ⩽z ⩽1) = 0.6827 or about 68%.
Examples Time to failure of a particular brand of light bulb is normally distributed with mean µ = 400 hours and standard deviation σ = 20 hours.
(a) What percentage of the bulbs will last longer than 438 hours?
P(438 ⩽X < ∞) = P 438 −400 20 ⩽Z ⩽∞ = P(1.9 ⩽ z) = 1 −P(Z ⩽1.9) = 1 −0.9713 = 0.0287 or about 2.9%.
(b)What percentage of the bulbs will fail before 360 hours?
P(−∞< X ⩽438) = P −∞⩽Z ⩽360 −400 20 = P(Z ⩽−2) = 0.0228 or about 2.9%.
Examples Let X be a normal random variable with mean µ = 100 and standard deviation σ = 15. What is the probability that the value of X falls between 80 and 105; P(80 ⩽X ⩽105)?
P(80 ⩽X ⩽105) = P 80 −100 15 ⩽Z ⩽105 −100 15 = P(−1.3333 ⩽Z ⩽0.3333) = 0.6305 −0.0912 = 0.5393.
Example Dental Anxiety Assume that scores on a Dental anxiety scale (ranging from 0 to 20) are normal for the general population, with mean µ = 11 and standard deviation σ = 3.5.
(a) What is the probability that a person chosen at random will score between 10 and 15 on this scale?
P(80 ⩽X ⩽105) = P 10 −11 3.5 ⩽Z ⩽15 −11 3.5 = P(−0.2857 ⩽Z ⩽1.1429) = 0.8735 −0.3875 = 0.4859.
Examples (b) What is the probability that a person chosen at random will have a score larger then 10 on this scale?
P(10 ⩽X < ∞) = P 10 −11 3.5 ⩽Z < ∞ = P(−0.2857 ⩽ Z < ∞) = 1 −(0.3875) = 0.6125.
(c) What is the probability that a person chosen at random will have a score less than 5 on this scale?
P(−∞< X ⩽5) = P ∞< Z ⩽5 −11 3.5 = P(Z ⩽ −1.7143) = 0.0432.
Examples Let X denote scores on the LSAT for a particular year.
The mean of X µ = 150 and the standard deviation is σ = 10. The histogram for the scores looks like: AT Scores distribution and US Law Schools Although, technically, the variable X is not continuous, the histogram is very closely approximated by a normal curve and the probabilities can be calculated from it.
Examples What percentage of students had a score of 165 or higher on this LSAT exam?
P(165 ⩽X < ∞) = P 165 −150 10 ⩽Z < ∞ = P(1.5 ⩽ Z < ∞) = 1 −P(Z ⩽1.5) = 1 −(0.9332) = 0.0668.
Examples Let X denote the weight of newborn babies at Memorial Hospital. The weights are normally distributed with mean µ = 8 lbs and standard deviation σ = 2 lbs.
(a) What is the probability that the weight of a newborn, chosen at random from the records at Memorial Hospital, is less than or equal to 9 lbs?
P(X ⩽9) = P Z ⩽9 −8 2 = P(Z ⩽0.5) = 0.6915.
(b) What is the probability that the weight of a newborn baby, selected at random from the records of Memorial Hospital, will be between 6 lbs and 8 lbs?
P(6 ⩽X ⩽8) = P 6 −8 2 ⩽Z < 8 −8 2 = P(1 ⩽Z < 0) = 0.5 −0.1587 = 0.3413.
Examples Example Let X denote Miriam’s monthly living expenses.
X is normally distributed with mean µ = $1, 000 and standard deviation σ = $150. On Jan. 1, Miriam finds out that her money supply for January is $1,150. What is the probability that Miriam’s money supply will run out before the end of January?
If Miriam’s monthly expenses exceed $1, 150 she will run out of money before the end of the month. Hence we want P(1, 150 ⩽X): P 1150 −1000 150 ⩽X = P(1 ⩽Z) = 1 −P(Z ⩽1) = 1 −(0.8413) = 0.1587.
Calculating Percentiles/Using the table in reverse Recall that xp is the pth percentile for the random variable X if p% of the population have values of X which are at or lower than xp and (100 −p)% have values of X at or greater than xp. To find the pth percentile of a normal distribution with mean µ and standard deviation σ, we can use the tables in reverse (or use a function on a calculator).
Calculating Percentiles/Using the table in reverse Example Calculate the 95th, 97.5th and 60th percentile of a normal random variable X, with mean µ = 400 and standard deviation σ = 35.
▶95th-percentile: From the table we see that 95% of the area under a standard normal curve is to the left of 1.65. Which reading x of X has z-score 1.65? Want 1.65 = (x −140)/35, so x = 35 · 1.65 + 400 = 457.75.
This is the 95th-percentile of X; 95% of all readings of X give a value at or below 457.75.
▶97.5th-percentile: 35 · 1.95 + 400 = 468.25.
▶60th-percentile: 35 · 0.27 + 400 = 409.45.
Calculating Percentiles/Using the table in reverse The scores on the LSAT for a particular year have a normal distribution with mean µ = 150 and standard deviation σ = 10. The distribution is shown below.
res distribution and US Law Schools (a) Find the 90th percentile of the distribution of scores.
90th-percentile a = 162.8155.
The table in the back of the book In the back of the book there is a table like the one we have used. The z values run from 0 to 3.19 and look different to our values. The difference is that the function in the book is defined for positive z, and measures the area under the standard normal curve from 0 to z.
Let’s see how the two tables are related. Let’s use B(z) to denote the values of the table in the book.
▶If 0 ⩽z < ∞, A(z) = P(Z ≤z) = P(−∞< Z < 0) + P(0 ≤Z ≤z) = 0.5 + B(z) ▶So for 0 ⩽z < ∞, A(z) = 0.5 + B(z) ▶If −∞< z < 0, A(z) = P(Z ≤z) = P(Z ≥−z) = P(0 < Z < ∞) −P(0 ≤Z ≤−z) = 0.5 −B(−z) ▶So for −∞< z < 0, A(z) = 0.5 −B(−z) Old exam questions The lifetime of Didjeridoos is normally distributed with mean µ = 150 years and standard deviation σ = 50 years.
What proportion of Didjeridoos have a lifetime longer than 225 years?
(a) 0.0668 (b) 0.5668 (c) 0.9332 (d) 0.5 (e) 0.4332 P(225 ⩽X) = P 225 −150 50 ⩽Z = P(1.5 ⩽Z) = 1 −P(Z ⩽1.5) = 1 −0.9332 = 0.0668.
Old exam questions Test scores on the OWLs at Hogwarts are normally distributed with mean µ = 250 and standard deviation σ = 30 . Only the top 5% of students will qualify to become an Auror. What is the minimum score that Harry Potter must get in order to qualify?
(a) 200.65 (b) 299.35 (c) 280 (d) 310 (e) 275.5 We need to find a so that P(a ⩽X) = 0.05. Let α = a −µ σ . Then P(a ⩽X) = P(α ⩽Z) = 0.05 so P(α ⩽Z) = 1 −P(Z ⩽α) so P(Z ⩽α) ⩽1 −0.05 = 0.95.
From the table P(α ⩽Z) = 0.95 so α ≈1.65. Hence a = 250 + 30 · 1.65 = 299.3456 to four decimal places so (b) is the correct answer.
Old exam questions Find the area under the standard normal curve between z = −2 and z = 3.
(a) 0.9759 (b) 0.9987 (c) 0.0241 (d) 0.9785 (e) 0.9772 P(−2 ⩽Z ⩽3) = P(Z ⩽3) −P(Z ⩽−2) = 0.9987 −0.0228 = 0.9759.
Old exam questions The number of pints of Guinness sold at “The Fiddler’s Hearth” on a Saturday night chosen at random is Normally distributed with mean µ = 50 and standard deviation σ = 10. What is the probability that the number of pints of Guinness sold on a Saturday night chosen at random is greater than 55?
(a) .6915 (b) .3085 (c) .8413 (d) .1587 (e) .5 P(55 ⩽X) = P 55 −50 10 ⩽Z < ∞ = P(0.5 ⩽Z) = 1 −P(Z ⩽0.5) = 1 −(0.6915) = 0.3085.
Approximating Binomial with Normal Recall that a binomial random variable, X, counts the number of successes in n independent trials of an experiment with two outcomes, success and failure.
Below are histograms for a binomial random variable, with p = 0.6, q = 0.4, as the value of n (= the number of trials ) varies from n = 10 to n = 30 to n = 100 to n = 200.
Superimposed on each histogram is the density function for a normal random variable with mean µ = E(X) = np and standard deviation σ = σ(X) = √npq. Even at n = 10, areas from the histogram are well approximated by areas under the corresponding normal curve. As n increases, the approximation gets better and better and the Normal distribution with the appropriate mean and standard deviation gives a very good approximation to the probabilities for the binomial distribution.
Approximating Binomial with Normal n = 10: The histogram below shows the n = 10, p = 0.6 Binomial distribution histogram, P(X = k) = 10 k (0.6)k(0.4)10−k for k = 0, 1, . . . , 10, along with a normal density curve with µ = 6 = np = E(X) and σ = 1.55 = √npq = σ(X).
2 4 6 8 10 0.05 0.10 0.15 0.20 0.25 Approximating Binomial with Normal n = 30: Here’s the histogram of the n = 30, p = 0.6 Binomial distribution for k = 0, 1, . . . , 30, along with a normal density curve with µ = 18 = E(X) and σ = 2.68 = √npq = σ(X).
5 10 15 20 25 30 -0.10 -0.05 0.05 0.10 0.15 0.20 Approximating Binomial with Normal n = 100: Here’s the histogram of the n = 100, p = 0.6 Binomial distribution for k = 0, 1, . . . , 100, along with a normal density curve with µ = 60 = E(X) and σ = 4.9 = √npq = σ(X).
40 50 60 70 80 Approximating Binomial with Normal n = 200: Finally, here’s the histogram of the n = 200, p = 0.6 Binomial distribution for k = 0, 1, . . . , 200, along with a normal density curve with µ = 120 = E(X) and σ = 6.93 = √npq = σ(X).
90 100 110 120 130 140 150 Using the approximation — continuity correction Given a binomial distribution X with n trials, success probability p, we can approximate it using a Normal random variable N with mean np, variance np(1 −p).
E.g., suppose n = 10, p = 0.5, and we want to know P(X ≥3).
It is tempting to estimate this by calculating P(N ≥3) where N is Normal, mean 5 and variance 2.5. But as the picture below shows, that will give us an answer that is too small.
To best match up the Binomial histogram area and the Normal curve area, we should calculate P(N ≥2.5). This is called the continuity correction.
P(X ≥3) ≈.945, P(N ≥3) ≈.897, P(N ≥2.5) ≈.943.
Continuity correction Given a binomial distribution X with n trials, success probability p, we can approximate it using a Normal random variable N with mean np, variance np(1 −p).
The continuity correction tells us that when we move from X to N, we should make the following changes to the probabilities we are calculating: ▶X ≥a changes to N ≥a −0.5 ▶X > a changes to N ≥a + 0.5 ▶X ≤a changes to N ≤a + 0.5 ▶X < a changes to N ≤a −0.5 Example An aeroplane has 200 seats. Knowing that passengers show up to flights with probability only 0.96, the airlines sells 205 seats for each flight. What is the probability that a given flight will be oversold (i.e., that more than 200 passengers will show up)?
We model the number of passengers who show up as a Binomial random variable X with n = 205, p = 0.96. We want to know that probability that X > 200.
We estimate X using a Normal random variable N with mean 205 × 0.96 = 196.8, variance 205 × 0.96 × 0.04 = 7.872, standard deviation ≈2.8.
The continuity correction says that we should estimate P(X > 200) by P(N ≥200.5). The z-score of 200.5 is ≈1.32.
So P(X > 200) ≈P(Z ≥1.32) ≈0.09.
From a Binomial calculator, the exact probability is ≈0.084.
Polling example I Melinda McNulty is running for the city council this May, with one opponent, Mark Reckless. She needs to get more than 50% of the votes to win.
I take a random sample of 100 people and ask them if they will vote for Melinda or not. Now assuming the population is large, the variable X = number of people who say “yes” has a distribution which is basically a binomial distribution with n = 100.
We do not know what p is. Suppose that in our poll, we found that 40% of the sample say that they will vote for Melinda. This is not good news, as it suggests p ≈.4, but this may be just due to variation in sample statistics.
Polling example I We can use our normal approximation to the binomial to see how hopeless the situation is, by asking the question: suppose in reality 50% of the population will vote for Melinda. How likely is it that in a sample of 100 people, we find 40 or fewer people who support Melinda?
Assuming p = 0.5, the distribution of X, the number of Melinda supporters we find in a sample of 100 is approximately normal with mean µ = np = 50 and standard deviation σ = √npq = √ 25 = 5.
P(X ⩽40) = P Z ⩽40 −50 5 = P(Z ⩽−2) ≈0.0228 (so things don’t look so good for Melinda...) Polling example II In a large population, some unknown proportion p of the people hold opinion o. A pollster, wanting to estimate p, polls 1000 people chosen at random, and asks each if they hold opinion o. She lets X be the number that say “yes”.
X is a Binomial random variable with n = 1000, some unknown mean 1000p and unknown variance 1000p(1 −p).
So it is very closely approximated by a normal random variable with mean 1000p, variance 1000p(1 −p).
Question: If the pollster uses the proportion X/1000 as an estimate for p, how likely is it that she gets an answer that within ±3.1% of the truth?
I.e., what is P −0.031 ≤ X 1000 −p ≤0.031 ?
Polling example II P(−0.031 ≤ X 1000 −p ≤0.031) = P(1000p−31 ≤X ≤1000p+31) z-score of 1000p −31 is −31 √ 1000p(1−p) ≈ −0.98 √ p(1−p).
z-score of 1000p + 31 is 31 √ 1000p(1−p) ≈ 0.98 √ p(1−p).
So P(−0.031 ≤ X 1000 −p ≤0.031) ≈P( −0.98 √ p(1−p) ≤Z ≤ 0.98 √ p(1−p)) P( −0.98 √ p(1−p) ≤Z ≤ 0.98 √ p(1−p)) is smallest when p(1 −p) is biggest, which is when p = 0.5 and 0.98/ p p(1 −p) = 1.96 Polling example II When it is at its smallest, P(−0.031 ≤ X 1000−p ≤0.031) ≈P(−1.96 ≤Z ≤1.96) ≈.95 Conclusion: When using the results of a 1000-person opinion poll to estimate some unknown population proportion, we can be at least 95% confident that our estimate will be within ±3.1% of the true proportion, meaning that at least 95 out of every 100 (or 19 out of every 20) opinion polls conducted will result in an observed proportion that is within ±3.1% of the true proportion.
▶But 1 out of every 20 polls will be wrong!
▶±3.1% is called the “margin or error” ▶All this assumes that the polling was done randomly ▶Works regardless of the size of the population being polled |
8820 | https://askfilo.com/user-question-answers-smart-solutions/find-the-monotonocity-of-f-x-x-lnx-3132323732303939 | Question asked by Filo student
find the monotonocity of f(x)= x/lnx
Views: 5,733 students
Updated on: Aug 29, 2024
Text SolutionText solutionverified iconVerified
Concepts:
Monotonicity, Derivative, Critical points
Explanation:
To determine the monotonicity of the function f(x)=lnxx, we need to find its first derivative and analyze the sign of the derivative to determine where the function is increasing or decreasing.
Step by Step Solution:
Step 1
Let f(x)=lnxx. To find the monotonicity, we first need to find the first derivative f′(x).
Step 2
Using the quotient rule, (vu)′=v2u′v−uv′, where u=x and v=lnx, we get: f′(x)=(lnx)2(1)(lnx)−(x)(x1)=(lnx)2lnx−1.
Step 3
To determine where f(x) is increasing or decreasing, we need to analyze the sign of f′(x). Set f′(x)>0: (lnx)2lnx−1>0.
Step 4
The numerator lnx−1 is positive when lnx>1, i.e., x>e. The denominator (lnx)2 is always positive for x>1. Therefore, f′(x)>0 when x>e, meaning f(x) is increasing for x>e.
Step 5
Similarly, f′(x)<0 when lnx−1<0, i.e., x<e. Therefore, f(x) is decreasing for 1<x<e.
Final Answer:
The function f(x)=lnxx is decreasing on the interval (1,e) and increasing on the interval (e,∞).
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| Question Text | find the monotonocity of f(x)= x/lnx |
| Updated On | Aug 29, 2024 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 12 |
| Answer Type | Text solution:1 |
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8821 | https://www.alistairstutorials.co.uk/mechanical_vibrations_content.html | Mechanical vibrations - Contents
Mechanical vibrations - Contents
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Strength of materialsMechanics of machinesRoboticsMechanical vibrationsAll tutorials
This series of tutorials covers the basic theory of mechanical vibrations, an important subject in mechanical engineering. Establishing the fundamentals of the subject is largely an exercise in applied mathematics, in particular finding and interpreting solutions to homogeneous and non-homogeneous, second order differential equations. For this reason I've included a maths tutorial covering this aspect.
Tutorials in the Mechanical vibrations series are as follows.
Mechanical vibrations - introduction and overview
Establishing free body diagrams and equations of motion for spring and mass vibrating system with one degree of freedom for: (a) free vibrations, (b) free vibrations with damping, (c) forced free vibrations, (d) forced free vibrations with damping.
Mechanical vibrations - maths tutorial
Methods for solving ordinary, second order, linear, homogeneous and non-homogeneous equations, specifically: general solutions for homogeneous equations of form y = e rx obtained from the roots of characteristic equation ar 2 + br + c = 0; particular solution derived from solving for constant terms of the general solution using two initial conditions; general solution for non-homogeneous equations obtained from sum of the general solution of the complementary homogeneous equation and the particular solution of the non-homogeneous equation obtained by the method of undetermined coefficients.
Free vibrations
Deriving the expression for displacement x(t) = R.cos(ωt - φ) from solutions to the homogeneous differential equation of motion of a free vibrating spring and mass system where R is the amplitude and φ the phase angle; establishing the angular frequency of a free vibrating system ω is the natural frequency ω n for all initial conditions of the system.
Free vibrations with damping
Deriving an expression for displacement from solutions to the homogeneous differential equation of motion of a free vibrating spring and mass system with damping; definition of dimensionless damping ratio ζ and derivation of solutions for ζ > 1, ζ < 1 and ζ = 1 ; worked examples and plots illustrating heavily damped (ζ > 1), critically damped (ζ = 1) and lightly damped (ζ < 1) behaviour; method of estimating value of ζ from experimentally derived plots.
Forced vibrations without damping
Deriving an expression for displacement from solutions to the non-homogeneous differential equation of motion of a free vibrating spring and mass system without damping subjected to a harmonic forcing function F 0 cos(ωt) ; interpretation of the solution as separable transient and steady state responses; plots illustrating amplification factor and phase relationship for steady state response where ω < ω n and ω > ω n ; illustration of resonance condition where ω ≅ ω n ; development of solution for combined transient and steady state response illustrating the harmonic phenomenon of beating;
Forced vibrations with damping
Deriving expressions for steady state displacement and phase angle using the non-homogeneous differential equation of motion of a free vibrating spring and mass system with damping subjected to a harmonic forcing function F 0 cos(ωt) ; plots illustrating amplification factor and phase relationship for steady state response where ω < ω n and ω > ω n ; plot illustrating variation of amplification factor with damping ratio ζ .
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8822 | https://pollack.uga.edu/radicalrefinement2.pdf | Multiplicative partitions of numbers with a large squarefree divisor Paul Pollack Abstract. For each positive integer n, let f(n) denote the number of multiplicative partitions of n, meaning the number of ways of writing n as a product of integers larger than 1, where the order of the factors is not taken into account. It was shown by Oppenheim in 1926 that, as x →∞, max n≤x n squarefree f(n) = x/L(x)2+o(1), where L(x) = exp(log x · log log log x log log x ). Without the restriction to squarefree n, the maximum is the significantly larger quantity x/L(x)1+o(1); this was proved by Canfield, Erd˝ os, and Pomerance in 1983. We prove the following theorem that interpolates between these two results: For each fixed α ∈[0, 1], max n≤x rad(n)≥nα f(n) = x/L(x)1+α+o(1).
We deduce, on the abc-conjecture, a nontrivial upper bound on how often values of certain polynomials appear in the range of Euler’s ϕ-function.
1. Introduction.
By a multiplicative partition (or unordered factorization) of n, we mean a way of decomposing n as a product of integers larger than 1, where two decompositions are considered the same if they differ only in the order of the factors. Let f(n) denote the number of multiplicative partitions of n. For example, f(12) = 4, corresponding to the factorizations 2 · 6, 2 · 2 · 3, 3 · 4, and 12.
The function f(n) was introduced by MacMahon in 1923 and was shortly afterwards the subject of two papers by Oppenheim [10, 11]. The main result of Oppenheim’s first paper concerns the maximum size of f(n). Let logk x denoting the kth iterate of the natural logarithm, and put L(x) = exp log x · log3 x log2 x .
In , Oppenheim claims to prove that f(n) ≤n/L(n)2+o(1), as n →∞, and that this is optimal: there is an infinite, increasing sequence of positive integers n along which f(n) = n/L(n)2+o(1). However, in 1983, Canfield, Erd˝ os, and Pomerance disproved 2010 Mathematics Subject Classification. Primary 11A51; Secondary 11N32, 11N56, 11N64.
Key words and phrases. multiplicative partition, factorization, divisor functions, factorisatio numerorum, abc conjecture.
1 2 PAUL POLLACK Oppenhein’s “theorem”, showing that the true maximal order is n/L(n)1+o(1); more precisely, (1) max n≤x f(n) = x/L(x)1+o(1), as x →∞.
Oppenheim’s “proof” that f(n) ≤n/L(n)2+o(1) rests on a mistaken assertion concerning the maximal order of the k-fold divisor function dk(n).
Specifically, Oppenheim claims that (2) dk(n) < klog n/ log2 n+log n/(log2 n)2+O(log n/(log2 n)3) for all large n and all k. Now (2) is true when k = 2 (a result of Ramanujan ), and in fact true for each fixed k (see for sharper results), but it is not true uniformly in k, and this invalidates his argument. Upper bounds for dk(n) which are uniform in k were eventually supplied by Usol′tsev and Norton , and using Norton’s work one can prove that f(n) ≤n/L(n)1+o(1) along the lines envisioned by Oppenheim.
(The proof in is different, not relying on bounds for dk(n).) It is worth observing that (2) does hold uniformly in k under the restriction that n is squarefree. In that case, dk(n) = kω(n), and it is known that (3) ω(n) ≤log n log2 n + log n (log2 n)2 + O log n (log2 n)3 .
(The estimate (3) follows from the prime number theorem with error term. See [17, Th´ eoreme 16] for an explicit determination of the O-constant.) Following Oppenheim’s arguments leads one to a correct proof that (4) max n≤x n squarefree f(n) = x/L(x)2+o(1), as x →∞. This asymptotic formula can also be obtained in other ways. For instance, one can note that when n is squarefree with k prime factors, f(n) is the number of set partitions of a k-element set, i.e., the kth Bell number. A sharp form of the prime number theorem, together with known bounds on Bell numbers (as in Lemma 6 below), easily yields (4).
The main result of this note is the following “convex combination” of the estimates (1) and (4). As usual, rad(n) denotes the radical of n, i.e., its largest squarefree divisor.
Theorem 1. Fix α ∈[0, 1]. As x →∞, max n≤x rad(n)≥nα f(n) = x/L(x)1+α+o(1).
Theorem 1 emerged during the author’s investigations into the value-distribution of Euler’s ϕ-function. Let F(n) = #ϕ−1(n) denote the number of ϕ-preimages of n.
It was shown by Pomerance that max n≤x F(n) ≤x/L(x)1+o(1), and that equality holds if one assumes plausible conjectures on the distribution of shifted primes without large prime factors (see also ). Under the same conjectures, MULTIPLICATIVE PARTITIONS 3 arguments of Banks, Friedlander, Pomerance, and Shparlinski establish that for each fixed positive integer k, max nk≤x F(nk) = x/L(x)1+o(1).
(A proof is carried out explicitly in ; be careful to note the Remark at the top of that article’s page 4.) Thus, for the polynomial P(T) = T k, there are values of P(T) that are essentially “as popular as possible” in terms of the multiplicity with which they appear in the range of ϕ. Under the abc conjecture, we deduce from Theorem 1 a contrasting result when P(T) has at least two distinct roots.
Theorem 2 (conditional on the abc conjecture). Let P(T) ∈Z[T] be a nonzero polynomial with at least two distinct complex roots. There is a constant cP > 0 such that, as x →∞, max n: 0<|P(n)|≤x F(|P(n)|) ≤x/L(x)1+cP +o(1).
2. ‘Radically’ refining (1) and (4): Proof of Theorem 1 Since (1) and (4) cover the cases α = 0 and α = 1 of Theorem 1, we will assume that 0 < α < 1. We treat the upper bound half of Theorem 1 first. The following estimate of Oppenheim plays a central role.
Lemma 3 (see eq. (1.52) in ). There are constants C1, C2 > 0 such that, for all positive integers n ≥16 (> ee), (5) f(n) ≤C1 log n · max 1≤k≤log n log 2 dk(n) k!
(C2 log3 n)k.
The next two lemmas are due to Norton.
Lemma 4 (see eq. (1.32) in ). For all integers n ≥16 and all integers k with 2 ≤k ≤2 log n log2 n , we have log dk(n) ≤log k log n log2 n 1 + log3 n log2 n + O 1 log2 n + k log3 n log n .
Lemma 5 (see eq. (1.34) in ). For all k ≥2 and all positive integers n, (6) dk(n) < n2ek.
Remark. While Norton only claims (6) when k ≥log n, it is clear from eq. (5.6) in that this result holds for all k ≥2.
Proof of the upper bound in Theorem 1. It suffices to show that for each ϵ ∈(0, 1/2), the maximum appearing in Theorem 1 is Oϵ(x/L(x)1+α−2ϵ) for all x > x0(ϵ). Of course, f(n) ≤x/L(x)1+α when n is bounded and x →∞, so we can and will assume when convenient that n is sufficiently large.
If the maximum in (5) occurs at k where k ≥2 log n/ log2 n, then (keeping in mind (6)) f(n) ≪log n · dk(n) k!
(C2 log3 n)k ≤log n · n2 · ek k!
(C2 log3 n)k ≤n2 log n · (C2e2 log3 n/k)k, 4 PAUL POLLACK where we used in the last step the elementary inequality k! ≥(k/e)k. Our lower bound on k implies that the last displayed quantity is of size no(1), as n →∞, which is smaller than x/L(x)2 for large x. So the upper bound of the theorem holds in this case, with much room to spare.
Now suppose the maximum occurs at a value k with 1.1 log3 n log n (log2 n)2 ≤k < 2 log n/ log2 n.
Write k = Z log n/(log2 n)2, so that 1.1 log3 n ≤Z < 2 log2 n.
Then log k log n log2 n = log n −2 log L(n) + O log Z log n log2 n , log k log n log2 n · log3 n log2 n = (1 + o(1)) log L(n), and log k log n log2 n · 1 log2 n + k log3 n log n = o(log L(n)) + O Z log2 n log L(n) .
(Here and elsewhere in this paragraph, the limit implicit in the o(·) terms is as n →∞.) Collecting these estimates and appealing to Lemma 4 reveals that log dk(n) ≤log n −(1 + o(1)) log L(n) + O Z log2 n log L(n) + O log Z log n log2 n .
Also, log k! = k log k + o(log L(n)) ≥k(log2 n −2 log3 n) + o(log L(n)) = Z log n log2 n(1 −O(log3 n/ log2 n)) + o(log L(n)).
Moreover, log((C2 log3 n)k) = o(log L(n)). So from (5), as n →∞, log f(n) ≤O(1) + log2 n + log dk(n) −log k! + log((C2 log3 n)k) ≤log n −(1 + o(1)) log L(n) −Z log n log2 n(1 + o(1)).
Inserting our lower bound on Z and exponentiating, we find that for large n, f(n) ≤n/L(n)2 ≤x/L(x)2.
Thus, the upper bound in the theorem holds in this case as well.
We may therefore suppose the maximum occurs at k < 1.1 log n log3 n (log2 n)2 .
Write n = AB, where A = rad(n). Recall that dk is a submultiplicative function, meaning that dk(ab) ≤dk(a)dk(b) for every pair of positive integers a, b (see for instance ).
Thus, f(n) = f(AB) ≪log n · dk(A)dk(B) k!
· (C2 log3 n)k.
MULTIPLICATIVE PARTITIONS 5 Applying (3) with n replaced by A shows that (7) log dk(A) = ω(A) log k ≤ log A log2 A + O(log A/(log2 A)2) log k.
Recall that A ≥nα. Thus, assuming (as we may) that n is large, we have that log k ≤log2 n −2 log3 n + O(log4 n) ≤log2 A −2 log3 A + O(log4 A), so that (7) yields log dk(A) ≤log A −(2 + o(1)) log L(A), as n →∞. So for large n, dk(A) < A/L(A)2−ϵ.
Also, log n · (C2 log3 n)k ≪L(x)ϵ, and a moment’s thought reveals that dk(B)/k! ≤ f(B). Hence, if n is large, then f(n) ≪ A L(A)2−ϵf(B)L(x)ϵ.
If B < 16, then this upper bound is O(x/L(x)2−2ϵ), completing the proof of the theorem. If B ≥16, we have from (1) that f(B) ≪B/L(B)1−ϵ, making f(n) ≪ n L(A)2−ϵL(B)1−ϵL(x)ϵ.
To finish things off, notice that log(L(A)2−ϵL(B)1−ϵ) = (2 −ϵ)log A log3 A log2 A + (1 −ϵ)log B log3 B log2 B ≥(2 −ϵ)log A log3 n log2 n + (1 −ϵ)log B log3 n log2 n = log A log3 n log2 n + (1 −ϵ) log L(n) ≥(1 + α −ϵ) log L(n).
Hence, f(n) ≪ n L(n)1+α−ϵL(x)ϵ ≪ x L(x)1+α−2ϵ, as desired.
□ The lower bound half of Theorem 1 is easier. We use the following asymptotic estimate for Bell numbers, which is a weak form of a result proved in [3, Chapter 6].
Lemma 6. The kth Bell number Bk satisfies, for k ≥3, log Bk = k log k −k log log k + O(k).
Proof of the lower bound in Theorem 1. It is enough to show that for all ϵ ∈(0, 1 2(1 −α)), the maximum indicated in Theorem 1 is at least x/L(x)1+α+4ϵ once x > x0(ϵ). Let B be a positive integer in [1, x1−α−ϵ] for which f(B) is as as large as possible; from (1), we know that for large x, (8) f(B) > x1−α−ϵ/L(x1−α−ϵ)1+ 1 2 ϵ ≥B/L(B)1+ 1 2 ϵ.
Since the value of f(B) depends only on the array of exponents in the prime factoriza-tion of B, and not on the primes themselves, we may assume that the primes dividing B are precisely the primes not exceeding its largest prime factor q. Since B ≤x1−α−ϵ, we have by the prime number theorem that q < log x (for large x). Let A be the product of the consecutive primes exceeding log x, with the product extending as far 6 PAUL POLLACK as possible with A ≤x/B. Then A and B are relatively prime, and so (concatenating factorizations) we see that (9) f(AB) ≥f(A)f(B).
The proof will be completed by showing that rad(AB) ≥xα, and that f(A)f(B) ≥ x/L(x)1+α+4ϵ.
To get started, we observe that (log x)ω(A) ≤A ≤x/B, so that ω(A) ≤log (x/B)/ log2(x).
For large x, this implies that the first ω(A) + 1 primes exceeding log x all belong to the interval (log x, 3 log x]. Now the choice of A implies that rad(AB) ≥rad(A) = A ≥ x 3B log x ≥xα.
Moreover, since (3 log x)ω(A) ≥A ≥x/(3B log x), we have ω(A) ≥log(x/(3B log x))/ log(3 log x) = log(x/B) log2 x (1 + O(1/ log2 x)) .
Combining this with our earlier upper bound for ω(A), we see that ω(A) = log(x/B) log2 x (1 + O(1/ log2 x)) .
Hence log ω(A) = log2 x −log3 x + O(1), log2 ω(A) = log3 x + O(log3 x/ log2 x).
Now a straightforward calculation using Lemma 6 reveals that log f(A) = log Bω(A) = log(x/B) −2 log(x/B)log3 x log2 x + o(log L(x)), as x →∞. Recalling (8) and (9), and observing that (8) implies that B > x1−α−2ϵ, we find that log f(AB) ≥log f(A) + log f(B) ≥log x −2 log(x/B)log3 x log2 x − 1 + 1 2ϵ log B log3 B log2 B + o(log L(x)) ≥log x −2 log(x/B)log3 x log2 x −(1 + ϵ) log B log3 x log2 x + o(log L(x)) = log x −(1 + o(1)) log L(x) −log(x/B)log3 x log2 x −ϵ log B log3 x log2 x.
Since x B ≤xα+2ϵ and B ≤x, we deduce that log f(AB) ≥log x −(1 + α + 3ϵ + o(1)) log L(x), and so for large x, we have f(AB) ≥x/L(x)1+α+4ϵ, as desired.
□ MULTIPLICATIVE PARTITIONS 7 3. (Un)popular polynomial values: Proof of Theorem 2 We use the following consequence of the abc conjecture, due to Langevin (see also Granville ).
Proposition 7 (conditional on abc). Fix a polynomial P(T) ∈Z[T] with degree d ≥2 and no repeated roots. Then rad(P(n)) ≥|n|d−1−o(1), for integers n with |n| →∞.
We also need the following lemma comparing F(n) and f(n). A similar result, for the Dedekind ψ-function in place of ϕ, was given by Pomerance .
Lemma 8. For every positive integer n, we have F(n) ≤4f(n).
Proof. By an extended factorization of n, we mean a multiplicative partition of n, but with 1 now allowed to appear as a factor at most once. Clearly, the number of extended factorizations of n is exactly 2f(n).
Suppose that ϕ(m) = n, and write m = pe1 1 · · · pek k , where the pi are distinct primes and the ei are positive integers; then n = k Y i=1 ei−1 times z }| { pi · · · pi · · · pi ·(pi −1).
We view the right-hand side as describing an extended factorization of n into Pk i=1((ei −1) + 1) = Ω(m) parts.
It suffices to show that each such extended factorization of ϕ(m) corresponds to at most two different preimages m.
Starting from the (extended) factorization of ϕ(m), one might attempt to recover m as follows: Such a factorization contains a unique smallest term p1 −1, where p1 is the smallest prime factor of m. We read offthe multiplicity of p1 in m as 1 + the multiplicity of p1 in our factorization. We then remove p1 −1 and the copies of p1 from our factorization and start over to determine the next largest prime factor of m and the multiplicity with which it appears. We continue in this way until the entire factorization of ϕ(m) is exhausted.
However, this procedure can fail if p1 = 2, since the number of 2’s appearing in the factorization of ϕ(m) depends not only on the power of 2 in m but also on whether or not 3 | m. We work around this as follows: • If p1 = 2, and the factorization of ϕ(m) contains 3 as a term, we know 3 | m, which is enough to resolve all ambiguity: If 2 appears in the given factorization of ϕ(m) k times, then 2k ∥m. Having figured out the power of 2 dividing m, we remove the factor 1 and k −1 factors of 2 from the factorization of ϕ(m) and proceed to determine the remaining components of m by the procedure of the last paragraph.
• Suppose p1 = 2 and 3 does not appear in the factorization of ϕ(m). If 2 appears in the factorization k times, then either 2k+1 ∥m or 2k · 3 ∥m. In either case, we may remove 1 and all factors of 2 from the factorization of ϕ(m) and continue with the algorithm above to determine the remaining components of m.
In either case, the factorization of ϕ(m) determines m in at most two ways, finishing the proof.
□ 8 PAUL POLLACK Proof of Theorem 2. Write P(T) = ± Q i Pi(T)ei, where the Pi(T) are dis-tinct nonconstant irreducibles in Z[T], each with positive leading coefficient. Let Q(T) = Q i Pi(T). Then Q(T) has distinct roots, and d := deg Q(T) ≥2. By Proposition 7, as |n| →∞, rad(P(n)) ≥rad(Q(n)) ≥|n|d−1−o(1) ≥|P(n)|1−1 d −o(1).
It follows from Lemma 8 and Theorem 1 that F(P(n)) ≤4f(P(n)) ≤|P(n)|/L(|P(n)|)2−1 d +o(1).
Theorem 2 follows with cP = 1 −1 d.
□ Probably the conclusion of Theorem 2 is still quite far from the truth. For each fixed α ∈[0, 1], (10) #{n ≤x : F(n) ≥xα} ≤x1−α+o(1), as x →∞. This follows from the easy estimate #ϕ−1([1, x]) ≪x log2 x. (In fact, x log2 x can be improved to x .) Now a naive probabilistic argument suggests that if P(T) is a polynomial of degree d, then (11) max n: 0<|P(n)|≤x F(|P(n)|) ≤x 1 d +o(1), as x →∞. This conclusion should be taken with a grain of salt; for the polynomials P(T) = T k, this maximum can be shown rigorously to be at least x0.7038 (and as pointed out in the introduction, we expect it to be x1−o(1)). (See , which develops arguments of .) But it may be that (11) holds generically, perhaps whenever P(T) has distinct roots.
Remark. It would also be sensible to study f(|P(n)|) rather than F(|P(n)|).
Clearly, Theorem 2 remains valid in this context.
Moreover, one can show rig-orously that maxnk≤x f(nk) = x/L(x)1+o(1), as x →∞.1 Since P n≤x f(n) ≤ x exp(O(√log x)) (see or for an asymptotic formula), one has the ana-logue of (10), and our probabilistic heuristic suggests that the analogue of (11) holds for a generic choice of P(T).
Acknowledgements The author is supported by NSF award DMS-1402268. He thanks Carl Pomerance for helpful comments.
References 1. W. D. Banks, J. B. Friedlander, C. Pomerance, and I. E. Shparlinski, Multiplicative structure of values of the Euler function, High primes and misdemeanours: lectures in honour of the 60th birthday of Hugh Cowie Williams, Fields Inst. Commun., vol. 41, Amer. Math. Soc., Providence, RI, 2004, pp. 29–47.
2. E. R. Canfield, P. Erd˝ os, and C. Pomerance, On a problem of Oppenheim concerning “factorisatio numerorum”, J. Number Theory 17 (1983), 1–28.
1In view of (1), only the implicit lower bound needs discussion. One modifies the proof of Theorem 1 in as follows: Rather than choose t distinct primes p ≤X with p−1 y-smooth, choose t y-smooth integers m1 < · · · < mt ≤X. Each choice of m1, . . . , mt corresponds to a factorization of N := m1 · · · mt, where N ≤x. The argument for Theorem 1 in shows that there are at least x/L(x)1+o(1) choices of the mi for which N is a kth power. But N is y-smooth, and there are only L(x)o(1) y-smooth integers in [1, x]. Thus, f(N) ≥x/L(x)1+o(1) for some N.
MULTIPLICATIVE PARTITIONS 9 3. N. G. de Bruijn, Asymptotic methods in analysis, third ed., Dover Publications, Inc., New York, 1981.
4. J.-L. Duras, J.-L. Nicolas, and G. Robin, Grandes valeurs de la fonction dk, Number theory in progress, Vol. 2 (Zakopane-Ko´ scielisko, 1997), de Gruyter, Berlin, 1999, pp. 743–770.
5. P. Erd˝ os, Some remarks on Euler’s ϕ-function and some related problems, Bull. Amer. Math.
Soc. 51 (1945), 540–544.
6. T. Freiberg, Products of shifted primes simultaneously taking perfect power values, J. Aust. Math.
Soc. 92 (2012), 145–154.
7. A. Granville, ABC allows us to count squarefrees, Internat. Math. Res. Notices (1998), no. 19, 991–1009.
8. M. Langevin, Partie sans facteur carr´ e d’un produit d’entiers voisins, Approximations diophanti-ennes et nombres transcendants (Luminy, 1990), de Gruyter, Berlin, 1992, pp. 203–214.
9. K. K. Norton, Upper bounds for sums of powers of divisor functions, J. Number Theory 40 (1992), 60–85.
10. A. Oppenheim, On an Arithmetic Function, J. London Math. Soc. 1 (1926), 205–211.
11.
, On an Arithmetic Function (II), J. London Math. Soc. 2 (1927), 123–130.
12. P. Pollack, How often is Euler’s totient a perfect power?, J. Number Theory 197 (2019), 1–12.
13. C. Pomerance, Popular values of Euler’s function, Mathematika 27 (1980), 84–89.
14.
, On the distribution of amicable numbers. II, J. Reine Angew. Math. 325 (1981), 183–188.
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, Two methods in elementary analytic number theory, Number theory and applications (Banff, AB, 1988), NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., vol. 265, Kluwer Acad. Publ., Dordrecht, 1989, pp. 135–161.
16. S. Ramanujan, Highly composite numbers, Proc. London Math. Soc. 14 (1915), 347–409.
17. G. Robin, Estimation de la fonction de Tchebychef θ sur le k-i eme nombre premier et grandes valeurs de la fonction ω(n) nombre de diviseurs premiers de n, Acta Arith. 42 (1983), 367–389.
18. J. S´ andor, On the arithmetical functions dk(n) and d∗ k(n), Portugal. Math. 53 (1996), 107–115.
19. G. Szekeres and P. Tur´ an, ¨ Uber das zweite Hauptproblem der “Factorisatio Numerorum”, Acta Litt. Sci. Szeged 6 (1933), 143–154.
20. L. P. Usol′tsev, On an estimate for a multiplicative function, Additive problems in number theory (Russian), Ku˘ ıbyshev. Gos. Ped. Inst., Kuybyshev, 1985, pp. 34–37.
Department of Mathematics, University of Georgia, Athens, GA 30602 E-mail address: pollack@uga.edu |
8823 | https://www.khanacademy.org/math/in-class-9-math-foundation/x6e1f683b39f990be:circles/x6e1f683b39f990be:circumference-of-a-circle/e/radius_diameter_and_circumference | Radius and diameter (practice) | Circles | Khan Academy
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NCERT Math Class 9 (Bridge)
Course: NCERT Math Class 9 (Bridge)>Unit 11
Lesson 1: Circumference of a circle
Radius, diameter, circumference & π
Radius and diameter
Radius & diameter from circumference
Circumference of a circle
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Circumference of a circle
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Radius and diameter
FL.BEST.Math: MA.7.GR.1.3
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What is the radius and diameter of the following circle?
11 cm
Radius =cm
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Video 11 minutes 5 seconds 11:05 Radius, diameter, circumference & π
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8824 | https://www.academia.edu/23202815/Effects_of_flooding_and_drought_on_stomatal_activity_transpiration_photosynthesis_water_potential_and_water_channel_activity_in_strawberry_stolons_and_leaves | (PDF) Effects of flooding and drought on stomatal activity, transpiration, photosynthesis, water potential and water channel activity in strawberry stolons and leaves
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Effects of flooding and drought on stomatal activity, transpiration, photosynthesis, water potential and water channel activity in strawberry stolons and leaves
Michael Blanke
2000, Plant Growth Regulation
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Abstract
Transpiration, xylem water potential and water channel activity were studied in developing stolons and leaves of strawberry (Fragaria  ananassa Duch.) subjected to drought or flooding, together with morphological studies of their stomata and other surface structures. Stolons had 0.12 stomata mm À2 and a transpiration rate of 0.6 mmol H 2 O m À2 s À1 , while the leaves had 300 stomata mm À2 and a transpiration rate of 5.6 mmol H 2 O m À2 s À1 . Midday water potentials of stolons were always less negative than in leaves enabling nutrient ion and water transport via or to the strawberry stolons. Drought stress, but not flooding, decreased stolon and leaf water potential from À0.7 to À1 MPa and from À1 to À2 MPa, respectively, with a concomitant reduction in stomatal conductance from 75 to 30 mmol H 2 O m À2 s À1 . However, leaf water potentials remained unchanged after flooding. Similarly, membrane vesicles derived from stolons of flooded strawberry plants showed no change in water channel activity. In these stolons, turgor may be preserved by maintaining root pressure, an electrochemical and ion gradient and xylem differentiation, assuming water channels remain open. By contrast, water channel activity was reduced in stolons of drought stressed strawberry plants. In every case, the effect of flooding on water relations of strawberry stolons and leaves was less pronounced than that of drought which cannot be explained by increased ABA. Stomatal closure under drought could be attributed to increased delivery of ABA from roots to the leaves. However, stomata closed more rapidly in leaves of flooded strawberry despite ABA delivery from the roots in the xylem to the leaves being strongly depressed. This stomatal closure under flooding may be due to release of stress ethylene. In the relative absence of stomata from the stolons, cellular (apoplastic) water transport in strawberry stolons was primarily driven by water channel activity with a gradient from the tip of the stolon to the base, concomitant with xylem differentiation and decreased water transport potential from the stolon tip to its base. Reduced water potential in the stolons under drought are discussed with respect to reduced putative water channel activity.
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Figures (8)
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Table 1. Properties of stolons and trichomes, compared with leaves of strawberry. n/a, not applicable. Stansted, Essex, UK) and colloidal graphite, and frozen by inserting into the pre-chamber stage of the cryo unit at —150 °C at Long Ashton Research Station, University of Bristol, UK. Cryopreserved samples were transferred to the scanning electron microscopy (SEM) cold stage and examined for superficial ice contamination. If present, this was removed by warming the specimen to —70 °C. When the surface was clean, the sample was returned o the pre-chamber and sputter-coated with ~20 nm of gold. Samples for sectioning were frozen by plun- ging the holder with the specimen attached into iquid nitrogen slush at —210 °C, transferred to the cooled pre-chamber and gold-coated. Coated speci- mens were studied at temperatures of —130 °C o —150 °C in a Philips SEM 505 microscope equipped with a Hexland cryo stage (Blanke et al. 994). Micrographs were taken with a Leitz Leica 2 camera using 21 DIN Kodak TMX 100 film.
Figures 1-4. | and 2 (top row): SEM micrograph of stomata on strawberry stolons. Magnification x963 and x 1425, respectively. 3 and 4 (bottom row): SEM micrograph of the dense trichome cover at the stolon tip. Magnification x41 and x240, respectively.
Measuring conditions: PFD, 600 jzmol photons m~? s~!, temperature of 20 °C and VPD 1.3 kPa. Table 2. Transpiration and stomatal conductance of non-stressed strawberry stolons and leaves.
Figures 5 and 6. Diurnal course of stomatal conductance (as a measure of transpiration) and environmental conditions during the water relation measurements.
Figures 7 and 8. Photosynthesis (mol CO, m-? s!) of strawberry leaves as affected by 4 days flooding or drought. Intercellular CO, concentration (yl I~) of strawberry leaves as affected by 4 days flooding or drought.
n/a, not applicable. "Jackson M.B. (2003, pers. comm.). Bangerth F. (2003, pers. comm.). Table 5. Proposed regulatory mechanisms of strawberry responses to water stress.
Table 3. Midday xylem water potentials (MPa) of strawberry cv. ‘Florika’ stolons and leaves subjected to 4 days of water stress as drought or flooding. Means of three measurements (+SE, n = 15), 6 cm from stolon tip where the xylem is developed. Letters denote significant differences in an unpaired t-test at the 0.05% confidence level or better.
Table 4. Effects of changes in osmoticum and of HgCl or absorbance (shrinking) of plasma membrane vesicles from strawberry stolons subjected to drought or flooding (n = 4).
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Mark Rieger
Tree Physiology, 1995
Root hydraulic conductivity (L p) and leaf osmotic potential at full turgor (Ψ π,o) were measured in young, droughtstressed and nonstressed peach (Prunus persica (L.) Batsch), olive (Olea europaea L.), citrumelo (Poncirus trifoliata Raf. × Citrus paradisi Macf.) and pistachio (Pistachia integerrima L.). Drought stress caused a 2.5-to 4.2-fold reduction in L p , depending on species, but Ψ π,o was reduced only in citrumelo and olive leaves by 0.34 and 1.4 MPa, respectively. No differences existed in L p among species for nonstressed plants. A simple model linking L p to osmotic adjustment through leaf water potential (Ψ) quantified the offsetting effects of reduced L p and osmotic adjustment on the hypothetical turgor pressure difference between drought-stressed and nonstressed plants (∆Ψ p). For olive, the 2.5-fold reduction in L p caused a linear decrease in ∆Ψ p such that the effect of osmotic adjustment was totally negated at Ψ = −3.2 MPa. Thus, no stomatal closure would be required to maintain higher turgor in drought-stressed olive plants than in nonstressed plants over their typical diurnal range of Ψ (−0.6 to −2.0 MPa). For citrumelo, osmotic adjustment was offset by reduced L p at Ψ ≈ −0.9 MPa. Unlike olive, stomatal closure would be necessary to maintain higher turgor in drought-stressed citrumelo plants than in nonstressed plants over their typical diurnal range of Ψ (0 to −1.5 MPa). Regardless of species or the magnitude of osmotic adjustment, my analysis suggests that a drought-induced reduction in L p reduces or eliminates turgor maintenance through osmotic adjustment.
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The response of transgenic strawberry plants overexpressing a drought induced gene to water stress
Mahmoud Raeini
Transgenic strawberry plants expressing a chitinase gene were evaluated for their performance during water stress. Transgenic and non-transgenic plants were assigned to three different soil water contents (SWC). They were kept under well-watered, moderately watered or stressed water conditions. At fi nal stage of experiment, dry matter components, leaf area, photosynthesis rate, water-use effi ciency (WUE) and water use per leaf area (WULA) were measured. Transgenic lines showed vigorous growth as compared with non-transgenic plants. Leaf area (LA), leaf dry matter (LDM), root dry matter (RDM) and total dry matter (TDM) of well-watered and water-stressed plants of transgenic lines were signifi cantly higher than those of non-transgenic plants. The WUE increased signifi cantly in transgenic lines, while water use (WU) per leaf area reduced in transgenic plants relative to control plants. Photosynthetic rates were not different between transgenic and non-transgenic plants. Soil water contents signifi cantly affected dry matter production, and photosynthetic rates. Transgenic plants also showed vigorous growth in comparison to non-transgenic plants when grown in vitro. Shoot, root and total fresh and dry weight of in vitro transgenic lines were signifi cantly higher than those of nontransgenic plants.
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Physiological and Biochemical Responses of Commercial Strawberry Cultivars under Optimal and Drought Stress Conditions
Saeid Kadkhodaei
Plants, 2023
This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY
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The Effect of Antagonist Abiotic Stress on Bioactive Compounds from Basil (Ocimum basilicum)
Dana Copolovici
Applied Sciences
Drought and flooding are some of the most common stressful conditions for plants. Due to the recent climate changes, they can occur one after another. This study is focused on the effect of antagonistic abiotic stress such as drought and flooding on the different metabolites from Ocimum basilicum leaves. Six-week-old plants of Ocimum basilicum were exposed to drought or flooding stress for 15 days, followed by antagonist stress for 14 days. The assimilation rates decrease drastically for plants under consecutive stresses from 18.9 to 0.25 µmol m−2 s−1 starting at day 3 of treatment. The stomatal conductance to water vapor gs was also reduced from 86 to 29 mmol m−2 s−1. The emission of green leaf volatiles compounds increases from 0.14 to 2.48 nmol m−2 s−1, and the emission of monoterpenes increased from 2.00 to 7.37 nmol m−2 s−1. The photosynthetic pigment concentration (chlorophyll a and b, and β-carotene), the flavonoid content, and total phenolic content decrease for all stressed...
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Effects of silicon on growth processes and adaptive potential of barley plants under optimal soil watering and flooding
Tamara Balakhnina
Plant Growth Regulation, 2012
Barley (Hordeum vulgare L.) was grown in pots with brown loess soil and highly soluble amorphous silicon dioxide as the source of monosilicic acid to examine its influence on plant growth and adaptive potential under optimal soil watering and flooding. The adaptive potential of plants was estimated by the concentration of the thiobarbituric acid reactive substances (TBARs) as well as superoxide dismutase (SOD), guaiacol peroxidase (GPX) and ascorbate peroxidase (AsP) activities. Application of amorphous silica to the soil increased the Si content in barley shoots and roots and stimulated their growth and biomass production under optimal soil watering. Soil flooding suppressed the growth both of the (-Si)-and (?Si)-plants. The intensity of oxidative destruction estimated by the concentration of TBARs was lower in the roots and leaves of the (?Si)-plants. Soil flooding induced SOD activity in the roots and in the leaves of the (-Si;?flooding) and (?Si;?flooding)-plants, but no significant differences were observed due to the Si treatment. GPX activity in the roots of (?Si)-plants was higher than in the (-Si)-ones under optimal soil watering, but under soil flooding no differences between (?Si)-and (-Si)-treatments were observed. AsP activity was not influenced by Si treatment neither under optimal soil watering nor under flooding. Thus, application of Si stimulates growth processes of barley shoots and roots under optimal soil watering and decreases intensity of oxidative destruction under soil flooding without significant changes in the activities of antioxidant enzymes.
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Produção de estolhos de cultivares de morangueiro em função da condutividade elétrica da solução nutritiva
Odair Schmitt
Horticultura Brasileira, 2016
Hortic. bras., v. 34, n. 2, abr.-jun. 2016 horticultura horticultura brasileira brasileira S istemas de cultivo sem solo são recomendados e amplamente utilizados na produção de mudas de morangueiro com torrão na Europa e Estados Unidos (Bish et al., 2001; Durner et al., 2002; Armeflhor, 2006). Para a produção de mudas com torrão em sistemas de cultivo sem solo é necessário um grande número de estruturas vegetativas jovens (estolhos) viáveis (Schmitt et al., 2012), sendo cada estolho formador de uma única muda. Isso é diferente dos sistemas de produção de mudas de raízes nuas, em que um estolho emitido e enraizado origina várias mudas. Na região Sul do Brasil, produção de mudas de morangueiro é alternativa à importação de mudas do Chile e Argentina, as quais apresentam custo elevado, sofrem danos no transporte, com qualidade fisiológica e sanitária
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Phytohormone-Mediated Stomatal Response, Escape and Quiescence Strategies in Plants under Flooding Stress
Md. Abu Sadat
Agronomy
Generally, flooding causes waterlogging or submergence stress which is considered as one of the most important abiotic factors that severely hinders plant growth and development. Plants might not complete their life cycle even in short duration of flooding. As biologically intelligent organisms, plants always try to resist or survive under such adverse circumstances by adapting a wide array of mechanisms including hormonal homeostasis. Under this mechanism, plants try to adapt through diverse morphological, physiological and molecular changes, including the closing of stomata, elongating of petioles, hollow stems or internodes, or maintaining minimum physiological activity to store energy to combat post-flooding stress and to continue normal growth and development. Mainly, ethylene, gibberellins (GA) and abscisic acid (ABA) are directly and/or indirectly involved in hormonal homeostasis mechanisms. Responses of specific genes or transcription factors or reactive oxygen species (ROS)...
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Physiological response and susceptibility of strawberry cultivars to the charcoal rot caused by Macrophomina phaseolina under drought stress conditions
Marina Gambardella
Journal of Berry Research, 2019
BACKGROUND: Charcoal rot of strawberry (Macrophomina phaseolina) is an emerging disease difficult to manage, a desirable alternative is the use of resistant cultivars. However, little is known regarding the reaction of cultivars to the pathogen under water stress conditions. OBJECTIVE: The aims of this work were to study the effect of water stress on the physiology of four strawberry cultivars during the infection, and to determine the relationship between water stress and cultivar susceptibility. METHODS: Healthy and inoculated plants of 'Monterey', 'Albion', 'Camarosa' and 'Sabrina' were maintained under no irrigation and full irrigation regimes, in greenhouse conditions. Stem water potential (SWP) and stomatal conductance (gs) were evaluated. The disease severity was recorded weekly for seven weeks. RESULTS: The disease detrimentally affected the water relations in 'Sabrina', 'Albion' and 'Monterey'. A significant correlation was detected between the evaluated parameters and the disease severity. The disease severity increases in plants with no irrigation, regardless of cultivar. CONCLUSIONS: Our results show that the infection caused by M. phaseolina increases the negative effects of water stress, depending on the genotype, and that the cultivars that were able to maintain more stable water relations respond better to the disease.
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Waterlogging tolerance evaluation of fifteen poplar clones cultivated in the Jianghan Plain of China
TianHong Ni
Notulae Botanicae Horti Agrobotanici Cluj-Napoca, 2021
To provide references for poplar cultivation in waterlogged prone area of Jianghan Plain of China, the waterlogging tolerance of 15 poplar clones widely cultivated in these areas were evaluated based on their responses to 45-day waterlogging stress followed by 15-day drainage recovery in morphology, growth, biomass accumulation, leaf gas exchange and chlorophyll fluorescence parameters. The results showed that the normal watered seedlings (CK) of the 15 clones grew vigorously during the experiment, and no defoliation and death occurred. For the seedlings under waterlogging treatment (water 10 cm above the soil surface), its morphology changed markedly, including slowing growth, chlorosis and abscission of leaves, development of hypertrophied lenticels and adventitious roots etc. Waterlogging stress significantly inhibited the seedling growth of height and ground diameter, biomass accumulation, as well as leaf gas exchange and chlorophyll fluorescence parameters of the 15 clones with...
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Shading Reduces Water Deficit in Strawberry (Fragaria ×ananassa Duch.) Plants during Vegetative Growth
Liz Moreno
ABSTRACTStrawberry (Fragaria ×ananassa Duch.) is a commercially important crop with high water requirements, for which it is necessary to find strategies that mitigate the influence of water deficit on plant growth. This study was aimed to evaluate the effects of shading on the vegetative growth of strawberry cv. Sweet Ann under water deficit. The treatments consisted of the combination of two levels of shading (light intensity reduced on 47% vs. non-shaded plants) and two levels of water availability (water deficit vs. well-watered plants). The water deficit reduced the leaf water potential from −1.52 to −2.21 MPa, and diminished stomatal conductance, net photosynthetic rate (from 9.13 to 2.5 μmol m−2 s−1), photosystem II photochemical efficiency (from 0.79 to 0.67), and biomass accumulation, while increased the electrolyte leakage. The shading allowed the water-deficient plants to maintain water potential (−1.58 MPa) and photosystem II efficiency (0.79) and to increase water use e...
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Yield, quality and biochemical properties of various strawberry cultivars under water stress
Hamide Gubbuk
Journal of the science of food and agriculture, 2017
Although strawberry (Fragaria x ananassa Duch.) species are sensitive to abiotic stress conditions, some cultivars are known to be tolerant to different environmental conditions. We examined the response of different strawberry cultivars to water stress conditions in terms of yield, quality and biochemical features. The trial was conducted under two different irrigation regimes: in grow bags containing cocopeat (control: 30%; water stress: 15% drainage) with four different cultivars (Camarosa, Albion, Amiga, and Rubygem). Fruit weight was declined by 59.72% and the yield per unit area by 63.62% under water stress conditions as compared to control. Albion and Rubygem were found to be more tolerant and Amiga the most sensitive in terms of yield under stress conditions. Water stress increased all biochemical features in fruits such as total phenol, total anthocyanin, antioxidant activity and sugar contents. Among the cultivars, glucose and fructose was higher in Albion. Considering the...
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Yield and Fruit Quality of Strawberry Cultivars under Different Irrigation Regimes
Elsa Martinez-Ferri
Agronomy, 2021
This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY
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8825 | https://artofproblemsolving.com/wiki/index.php/Rational_root_theorem?srsltid=AfmBOorPpUidSMcMKfTtfeiNOFZu7fCCq8wToGqRXHrImEZ6cvoNk3q_ | Art of Problem Solving
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Rational root theorem
In algebra, the rational root theorem states that given an integer polynomial with leading coefficient and constant term , if has a rational root in lowest terms, then and .
This theorem is most often used to guess the roots of polynomials. It sees widespread usage in introductory and intermediate mathematics competitions.
Contents
[hide]
1 Proof
2 Examples
2.1 Example 1
2.2 Example 2
2.3 Example 3
3 See also
Proof
Let be a rational root of , where every is an integer; we wish to show that and . Since is a root of , Multiplying by yields Using modular arithmetic modulo , we have , which implies that . Because we've defined and to be relatively prime, , which implies by Euclid's lemma. Via similar logic in modulo , , as required.
Intro to Rational Roots theorem:
Examples
Here are some problems with solutions that utilize the rational root theorem.
Example 1
Find all rational roots of the polynomial .
Solution: The polynomial has leading coefficient and constant term , so the rational root theorem guarantees that the only possible rational roots are , , , , , , , and . After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots.
Example 2
Factor the polynomial .
Solution: After testing the divisors of 8, we find that it has roots , , and . Then because it has leading coefficient , the factor theorem tells us that it has the factorization .
Example 3
Using the rational root theorem, prove that is irrational.
Solution: The polynomial has roots . The rational root theorem guarantees that the only possible rational roots of this polynomial are , and . Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because is a root of the polynomial, it cannot be a rational number.
See also
Root
Polynomial
Retrieved from "
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Polynomials
Theorems
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8826 | https://www.echemi.com/community/test-to-distinguish-between-nahco3-and-na2co3_mjart2205241009_217.html | Test to distinguish between NaHCO3 and Na2CO3 - ECHEMI
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Home>Community>Test to distinguish between NaHCO3 and Na2CO3
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Chemistry
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Aarzoo Dhingra
Test to distinguish between NaHCO3 and Na2CO3
I just baked Sodium Bicarbonate to make Sodium Carbonate, but I need to be sure that I indeed have. The tolerance is between 400℉ - 450℉ (200℃ - 230℃) and I'm not sure I had the oven on high enough. I baked it for one hour, staring the powder half way.
What test can I perform to distinguish the two please?
There are many methods, some of them may be performed in home:
Solubility of both compounds are quite different,
NaHCO 3 does not form hydrates,
Na 2CO 3 change color of phenolphtalein, hydrogen carbonate usually change color to pale pink since small content od carbonate,
Titration.
And many others.
David BrowerFollowFollowing
There are many methods, some of them may be performed in home:
Solubility of both compounds are quite different,
NaHCO 3 does not form hydrates,
Na 2CO 3 change color of phenolphtalein, hydrogen carbonate usually change color to pale pink since small content od carbonate,
Titration.
And many others.
More
Upvote
VOTE
Downvote
Sodium hydrogen carbonate start decomposing at ~60 C - solid, or ~40 C - in water solution. This is an equilibrium reaction and carbon dioxide is removed to atmosphere. After many hours of warming at these temperatures you can obtain almost pure sodium carbonate. But air contains some amount of carbon dioxide, and in this way you will never get absolutely pure sodium carbonate, even at higher temperatures. The only simple and quantitative way to determine contents of you sample is titration.
May first three suggestions can be used qualitatively. Just one of these is sufficient, and it depends on you home lab possibility, you invention, and your knowledge.
In the same way past great chemists worked. But the did not wait for precise procedure from others that solve their problems.
Andrew KaronisFollowFollowing
Sodium hydrogen carbonate start decomposing at ~60 C - solid, or ~40 C - in water solution. This is an equilibrium reaction and carbon dioxide is removed to atmosphere. After many hours of warming at these temperatures you can obtain almost pure sodium carbonate. But air contains some amount of carbon dioxide, and in this way you will never get absolutely pure sodium carbonate, even at higher temperatures. The only simple and quantitative way to determine contents of you sample is titration.
May first three suggestions can be used qualitatively. Just one of these is sufficient, and it depends on you home lab possibility, you invention, and your knowledge.
In the same way past great chemists worked. But the did not wait for precise procedure from others that solve their problems.
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Thank you for your replies.
I did notice some colour change, but didn't consider that t be enough of a uniform indicator of the change.
Not being a chemist I'm not sure how to try your fourth suggestion "Titration" and would like to know, with home appliances, what would be the simplest method to uniformly confirm the change please?
Thanks again, in advance.
Chuck CobbFollowFollowing
Thank you for your replies.
I did notice some colour change, but didn't consider that t be enough of a uniform indicator of the change.
Not being a chemist I'm not sure how to try your fourth suggestion "Titration" and would like to know, with home appliances, what would be the simplest method to uniformly confirm the change please?
Thanks again, in advance.
More
Upvote
VOTE
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Quote from:
There are many methods, some of them may be performed in home:
Solubility of both compounds are quite different,
NaHCO3does not form hydrates,
Na2CO3change color of phenolphtalein, hydrogen carbonate usually change color to pale pink since small content od carbonate,
Titration.
And many others.
Agreed.
I think elemental analysis also works very well
BPS Glass Bottle Packaging SolutionsFollowFollowing
Quote from:
There are many methods, some of them may be performed in home:
Solubility of both compounds are quite different,
NaHCO3does not form hydrates,
Na2CO3change color of phenolphtalein, hydrogen carbonate usually change color to pale pink since small content od carbonate,
Titration.
And many others.
Agreed.
I think elemental analysis also works very well
More
Upvote
VOTE
Downvote
I ain't no chemist mate. I'm just trying to make laundry powder.
Chen TangFollowFollowing
I ain't no chemist mate. I'm just trying to make laundry powder.
More
Upvote
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Downvote
From GOOGLE
Arm & Hammer Super Washing Soda Detergent Booster & Household Cleaner, 55 oz Misplaced &Misplaced &0.75
0.75 / 16 = 0.046875
By the way
can we assume you have read
and
From WIKI
Quote
Thermal decomposition
Above 50 °C, sodium bicarbonate gradually decomposes into sodium carbonate, water and carbon dioxide. The conversion is fast at 200 °C:
2 NaHCO3Na2CO3+ H2O + CO2
Most bicarbonates undergo this dehydration reaction. Further heating converts the carbonate into the oxide (at over 850 °C):
Na2CO3Na2O + CO2
Bob McDonaldFollowFollowing
From GOOGLE
Arm & Hammer Super Washing Soda Detergent Booster & Household Cleaner, 55 oz Misplaced &Misplaced &0.75
0.75 / 16 = 0.046875
By the way
can we assume you have read
and
From WIKI
Quote
Thermal decomposition
Above 50 °C, sodium bicarbonate gradually decomposes into sodium carbonate, water and carbon dioxide. The conversion is fast at 200 °C:
2 NaHCO3Na2CO3+ H2O + CO2
Most bicarbonates undergo this dehydration reaction. Further heating converts the carbonate into the oxide (at over 850 °C):
Na2CO3Na2O + CO2
More
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Titration seems to be the easiest approach here, seeing how much of your solid is required to neutralize given amount of a strong acid.
See www.titrations.info for a general description of the titration itself, you need a variant of what is described in the acid-base section.
David LevyFollowFollowing
Titration seems to be the easiest approach here, seeing how much of your solid is required to neutralize given amount of a strong acid.
See www.titrations.info for a general description of the titration itself, you need a variant of what is described in the acid-base section.
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8827 | https://education.nationalgeographic.org/resource/resource-library-abiotic-factor/ | COLLECTION
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Abiotic Factors
Abiotic Factors
An abiotic factor is a non-living part of an ecosystem that shapes its environment. In a terrestrial ecosystem, examples might include temperature, light, and water. In a marine ecosystem, abiotic factors would include salinity and ocean currents. Abiotic and biotic factors work together to create a unique ecosystem.
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8828 | https://math.stackexchange.com/questions/978927/vector-proof-that-d-12-d-22-2a2-2b2-in-a-parallelogram | Vector proof that $d_1^2 + d_2^2 = 2a^2 + 2b^2$ in a parallelogram - Mathematics Stack Exchange
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Vector proof that d 2 1+d 2 2=2 a 2+2 b 2 d 1 2+d 2 2=2 a 2+2 b 2 in a parallelogram
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How would one prove the equality of the sum of squares of diagonals and twice the sum of squares of the two sides:
|p+q|2+|p−q|2=2|p|2+2|q|2|p+q|2+|p−q|2=2|p|2+2|q|2
where p p and q q are vectors, representing two intersecting sides of a parallelogram.
Am I supposed to take the LHS and prove it equals the RHS like in a normal proof?
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edited Oct 18, 2014 at 1:20
Fengyang Wang
1,784 1 1 gold badge 14 14 silver badges 23 23 bronze badges
asked Oct 18, 2014 at 0:18
KieranKieran
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Use ∥x∥2=⟨x,x⟩‖x‖2=⟨x,x⟩.copper.hat –copper.hat 2014-10-18 00:25:51 +00:00 Commented Oct 18, 2014 at 0:25
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The key is to use the dot product to simplify the proof. Remember that |v⃗|2=v⃗⋅v⃗|v→|2=v→⋅v→.
|p⃗+q⃗|2+|p⃗−q⃗|2=(p⃗+q⃗)⋅(p⃗+q⃗)+(p⃗−q⃗)⋅(p⃗−q⃗)=(p⃗⋅p⃗+2 p⃗⋅q⃗+q⃗⋅q⃗)+(p⃗⋅p⃗−2 p⃗⋅q⃗+q⃗⋅q⃗)=2 p⃗⋅p⃗+2 q⃗⋅q⃗=2|p⃗|2+|p⃗|2|p→+q→|2+|p→−q→|2=(p→+q→)⋅(p→+q→)+(p→−q→)⋅(p→−q→)=(p→⋅p→+2 p→⋅q→+q→⋅q→)+(p→⋅p→−2 p→⋅q→+q→⋅q→)=2 p→⋅p→+2 q→⋅q→=2|p→|2+|p→|2
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answered Oct 18, 2014 at 0:29
Fengyang WangFengyang Wang
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|p+q|2=⟨p+q,p+q⟩=⟨p,p+q⟩+⟨q,p+q⟩=⟨p,p⟩+⟨p,q⟩+⟨q,p⟩+⟨q,q⟩|p+q|2=⟨p+q,p+q⟩=⟨p,p+q⟩+⟨q,p+q⟩=⟨p,p⟩+⟨p,q⟩+⟨q,p⟩+⟨q,q⟩
Repeat for |p−q|2|p−q|2 and add.
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answered Oct 18, 2014 at 0:30
JessicaKJessicaK
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2008 AIME II Problems/Problem 7
Contents
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1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6
8 Solution 7
9 Solution 8
10 Solution 9
11 Solution 10
12 Video Solution by Punxsutawney Phil
13 See also
Problem
Let , , and be the three roots of the equation Find .
Solution 1
By Vieta's formulas, we have so Substituting this into our problem statement, our desired quantity is Also by Vieta's formulas we have so negating both sides and multiplying through by 3 gives our answer of
Solution 2
By Vieta's formulas, we have , and so the desired answer is . Additionally, using the factorization we have that . By Vieta's again,
Solution 3
Vieta's formulas gives . Since is a root of the polynomial, , and the same can be done with . Therefore, we have yielding the answer .
Also, Newton's Sums yields an answer through the application.
Solution 4
Expanding, you get: This looks similar to Substituting: Since , Substituting, we get or, We are trying to find . Substituting:
Solution 5
Write and let . Then Solving for and negating the result yields the answer
Solution 6
Here by Vieta's formulas: --(1)
--(2)
By the factorisation formula: Let , , , (By (1))
So
Solution 7
Let's construct a polynomial with the roots and .
sum of the roots:
pairwise product of the roots:
product of the roots:
thus, the polynomial we get is
as and are roots of this polynomial, we know that (using power reduction)
adding all of the equations up, we see that
Solution 8
We want to find what is which reminds us of Newton sum. So we can see that Notice that so it is just , the desired answer is
~bluesoul
Solution 9
This solution uses Vietas, as with everyone else's solution. Expanding the expression we get
Seeing the cubes, we try to find a and upon doing so, we get
Recall that . Thus, we get
Plugging in we get
~firebolt360
Solution 10
We want to find Let's call this result n.
From vieta's formulas, we find that , , and
Expanding and rearranging gives us
Let
Solving gives us
Therefore, the answer is
Also note that this is the only solution that still would have worked effectively if was nonzero.
Video Solution by Punxsutawney Phil
(unavailable)
See also
2008 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6Followed by
Problem 8
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
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12: Heat and Heat Transfer
Book: Physics (Boundless)
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Mon, 17 Jun 2019 22:10:52 GMT
12.2: Specific Heat
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[ "article:topic", "calorimeter", "Heat capacity", "authorname:boundless", "enthalpy", "specific heat", "fundamental thermodynamic relation", "adiabatic index", "constant-pressure calorimeter", "constant-volume calorimeter", "heat of reaction", "combustion", "showtoc:no", "transcluded:yes", "source-phys-14516" ]
[ "article:topic", "calorimeter", "Heat capacity", "authorname:boundless", "enthalpy", "specific heat", "fundamental thermodynamic relation", "adiabatic index", "constant-pressure calorimeter", "constant-volume calorimeter", "heat of reaction", "combustion", "showtoc:no", "transcluded:yes", "source-phys-14516" ]
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Contents
1. Home
2. Campus Bookshelves
3. Joliet Junior College
4. Physics 201 - Fall 2019
5. Book: Physics (Boundless)
6. 12: Heat and Heat Transfer
7. 12.2: Specific Heat
Expand/collapse global location
Book: Physics (Boundless)
00: Front Matter
1: The Basics of Physics
2: Kinematics
3: Vectors
4: Two-Dimensional Kinematics
5: The Laws of Motion
7: Work and Energy
8: Linear Momentum and Collisions
9: Static Equilibrium, Elasticity, and Torque
10: Rotational Kinematics, Angular Momentum, and Energy
11: Temperature and Kinetic Theory
12: Heat and Heat Transfer
13: Thermodynamics
6: Applications of Newton
Front Matter
12.2: Specific Heat
Last updated Jun 17, 2019
Save as PDF
12.1: Introduction
12.3: Phase Change and Latent Heat
Page ID 18052
Boundless
Boundless
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. The Measurement of Heat Capacity
2. Thermodynamic Relations and Definition of Heat Capacity
3. Specific Heat
1. Specific Heat
Calorimetry
Calorimetry
Overview
Basic Calorimetry at Constant Value
Measuring Enthalpy Change
Constant-Pressure Calorimetry
Specific Heat for an Ideal Gas at Constant Pressure and Volume
Specific Heat for an Ideal Gas at Constant Pressure and Volume
Solving Problems with Calorimetry
Coffee-Cup Calorimeters
Structure of the Constant Volume (or “Bomb”) Calorimeter
Solution
Key Points
Key Terms
learning objectives
Explain the enthalpy in a system with constant volume and pressure
Heat capacity (usually denoted by a capital C, often with subscripts), or thermal capacity, is the measurable physical quantity that characterizes the amount of heat required to change a substance’s temperature by a given amount. In SI units, heat capacity is expressed in units of joules per kelvin (J/K).
An object’s heat capacity (symbol C) is defined as the ratio of the amount of heat energy transferred to an object to the resulting increase in temperature of the object.
(12.2.1)C=Q ΔT.
Heat capacity is an extensive property, so it scales with the size of the system. A sample containing twice the amount of substance as another sample requires the transfer of twice as much heat (Q) to achieve the same change in temperature (ΔT). For example, if it takes 1,000 J to heat a block of iron, it would take 2,000 J to heat a second block of iron with twice the mass as the first.
The Measurement of Heat Capacity
The heat capacity of most systems is not a constant. Rather, it depends on the state variables of the thermodynamic system under study. In particular, it is dependent on temperature itself, as well as on the pressure and the volume of the system, and the ways in which pressures and volumes have been allowed to change while the system has passed from one temperature to another. The reason for this is that pressure-volume work done to the system raises its temperature by a mechanism other than heating, while pressure-volume work done by the system absorbs heat without raising the system’s temperature. (The temperature dependence is why the definition a calorie is formally the energy needed to heat 1 g of water from 14.5 to 15.5 °C instead of generally by 1 °C. )
Different measurements of heat capacity can therefore be performed, most commonly at constant pressure and constant volume. The values thus measured are usually subscripted (by p and V, respectively) to indicate the definition. Gases and liquids are typically also measured at constant volume. Measurements under constant pressure produce larger values than those at constant volume because the constant pressure values also include heat energy that is used to do work to expand the substance against the constant pressure as its temperature increases. This difference is particularly notable in gases where values under constant pressure are typically 30% to 66.7% greater than those at constant volume.
Thermodynamic Relations and Definition of Heat Capacity
The internal energy of a closed system changes either by adding heat to the system or by the system performing work. Recalling the first law of thermodynamics,
(12.2.2)dU=δQ−δW.
For work as a result of an increase of the system volume we may write,
(12.2.3)dU=δQ−PdV.
If the heat is added at constant volume, then the second term of this relation vanishes and one readily obtains
(12.2.4)(∂U∂T)V=(∂Q∂T)V=C V.
This defines the heat capacity at constant volume, C V. Another useful quantity is the heat capacity at constant pressure, C P. With the enthalpy of the system given by
(12.2.5)H=U+PV,
our equation for d U changes to
(12.2.6)dH=δQ+VdP,
and therefore, at constant pressure, we have
(12.2.7)(∂H∂T)P=(∂Q∂T)P=C P.
Specific Heat
The specific heat is an intensive property that describes how much heat must be added to a particular substance to raise its temperature.
learning objectives
Summarize the quantitative relationship between heat transfer and temperature change
Specific Heat
The heat capacity is an extensive property that describes how much heat energy it takes to raise the temperature of a given system. However, it would be pretty inconvenient to measure the heat capacity of every unit of matter. What we want is an intensive property that depends only on the type and phase of a substance and can be applied to systems of arbitrary size. This quantity is known as the specific heat capacity (or simply, the specific heat), which is the heat capacity per unit mass of a material. Experiments show that the transferred heat depends on three factors: (1) The change in temperature, (2) the mass of the system, and (3) the substance and phase of the substance. The last two factors are encapsulated in the value of the specific heat.
Heat Transfer and Specific Heat Capacity: The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved. (a) The amount of heat transferred is directly proportional to the temperature change. To double the temperature change of a mass m, you need to add twice the heat. (b) The amount of heat transferred is also directly proportional to the mass. To cause an equivalent temperature change in a doubled mass, you need to add twice the heat. (c) The amount of heat transferred depends on the substance and its phase. If it takes an amount Q of heat to cause a temperature change ΔT in a given mass of copper, it will take 10.8 times that amount of heat to cause the equivalent temperature change in the same mass of water assuming no phase change in either substance.
Specific Heat Capacity: This lesson relates heat to a change in temperature. We discuss how the amount of heat needed for a temperature change is dependent on mass and the substance involved, and that relationship is represented by the specific heat capactiy of the substance, C.
The dependence on temperature change and mass are easily understood. Because the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules. Since the transferred heat is equal to the change in the internal energy, the heat is proportional to the mass of the substance and the temperature change. The transferred heat also depends on the substance so that, for example, the heat necessary to raise the temperature is less for alcohol than for water. For the same substance, the transferred heat also depends on the phase (gas, liquid, or solid).
The quantitative relationship between heat transfer and temperature change contains all three factors:
(12.2.8)Q=mcΔT,
where Q is the symbol for heat transfer, m is the mass of the substance, and ΔT is the change in temperature. The symbol c stands for specific heat and depends on the material and phase.
The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00ºC. The specific heat c is a property of the substance; its SI unit is J/(kg⋅K) or J/(kg⋅C). Recall that the temperature change (ΔT) is the same in units of kelvin and degrees Celsius. Note that the total heat capacity C is simply the product of the specific heat capacity c and the mass of the substance m, i.e.,
(12.2.9)C=mc or c=C m=C ρV,
where ϱ is the density of the substance and V is its volume.
Values of specific heat must generally be looked up in tables, because there is no simple way to calculate them. Instead, they are measured empirically. In general, the specific heat also depends on the temperature. The table below lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak. The specific heat of water is five times that of glass and ten times that of iron, which means that it takes five times as much heat to raise the temperature of water the same amount as for glass and ten times as much heat to raise the temperature of water as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.
Specific Heats: Listed are the specific heats of various substances. These values are identical in units of cal/(g⋅C).3. cv at constant volume and at 20.0ºC, except as noted, and at 1.00 atm average pressure. Values in parentheses are cp at a constant pressure of 1.00 atm.
Calorimetry
Calorimetry is the measurement of the heat of chemical reactions or physical changes.
learning objectives
Analyze the relationship between the gas constant for an ideal gas yield and volume
Calorimetry
Overview
Calorimetry is the science of measuring the heat of chemical reactions or physical changes. Calorimetry is performed with a calorimeter. A simple calorimeter just consists of a thermometer attached to a metal container full of water suspended above a combustion chamber. The word calorimetry is derived from the Latin word calor, meaning heat. Scottish physician and scientist Joseph Black, who was the first to recognize the distinction between heat and temperature, is said to be the founder of calorimetry.
Calorimetry requires that the material being heated have known thermal properties, i.e. specific heat capacities. The classical rule, recognized by Clausius and by Kelvin, is that the pressure exerted by the calorimetric material is fully and rapidly determined solely by its temperature and volume; this rule is for changes that do not involve phase change, such as melting of ice. There are many materials that do not comply with this rule, and for them, more complex equations are required than those below.
Ice Calorimeter: The world’s first ice-calorimeter, used in the winter of 1782-83, by Antoine Lavoisier and Pierre-Simon Laplace, to determine the heat evolved in variouschemical changes; calculations which were based on Joseph Black’s prior discovery of latent heat. These experiments mark the foundation of thermochemistry.
Basic Calorimetry at Constant Value
Constant-volume calorimetry is calorimetry performed at a constant volume. This involves the use of a constant-volume calorimeter (one type is called a Bomb calorimeter). For constant-volume calorimetry:
(12.2.10)δQ=CVΔT=mcVΔT
where δQ is the increment of heat gained by the sample, C V is the heat capacity at constant volume, c v is the specific heat at constant volume, and ΔT is the change in temperature.
Measuring Enthalpy Change
To find the enthalpy change per mass (or per mole) of a substance A in a reaction between two substances A and B, the substances are added to a calorimeter and the initial and final temperatures (before the reaction started and after it has finished) are noted. Multiplying the temperature change by the mass and specific heat capacities of the substances gives a value for the energy given off or absorbed during the reaction:
\mathrm{δQ=ΔT(m_Ac_A+m_Bc_B)}
Dividing the energy change by how many grams (or moles) of A were present gives its enthalpy change of reaction. This method is used primarily in academic teaching as it describes the theory of calorimetry. It does not account for the heat loss through the container or the heat capacity of the thermometer and container itself. In addition, the object placed inside the calorimeter shows that the objects transferred their heat to the calorimeter and into the liquid, and the heat absorbed by the calorimeter and the liquid is equal to the heat given off by the metals.
Constant-Pressure Calorimetry
A constant-pressure calorimeter measures the change in enthalpy of a reaction occurring in solution during which the atmospheric pressure remains constant. An example is a coffee-cup calorimeter, which is constructed from two nested Styrofoam cups and a lid with two holes, allowing insertion of a thermometer and a stirring rod. The inner cup holds a known amount of a solute, usually water, that absorbs the heat from the reaction. When the reaction occurs, the outer cup provides insulation. Then
(12.2.11)C P=WΔH MΔT
where C p is the specific heat at constant pressure, ΔH is the enthalpy of the solution, ΔT is the change in temperature, W is the mass of the solute, and M is the molecular mass of the solute. The measurement of heat using a simple calorimeter, like the coffee cup calorimeter, is an example of constant-pressure calorimetry, since the pressure (atmospheric pressure) remains constant during the process. Constant-pressure calorimetry is used in determining the changes in enthalpy occurring in solution. Under these conditions the change in enthalpy equals the heat (Q=ΔH).
Specific Heat for an Ideal Gas at Constant Pressure and Volume
An ideal gas has different specific heat capacities under constant volume or constant pressure conditions.
learning objectives
Explain how to derive the adiabatic index
Specific Heat for an Ideal Gas at Constant Pressure and Volume
The heat capacity at constant volume of nR = 1 J·K−1 of any gas, including an ideal gas is:
(12.2.12)(∂U∂T)V=c v
This represents the dimensionless heat capacity at constant volume; it is generally a function of temperature due to intermolecular forces. For moderate temperatures, the constant for a monoatomic gas is c v=3/2 while for a diatomic gas it is c v=5/2 (see ). Macroscopic measurements on heat capacity provide information on the microscopic structure of the molecules.
Molecular internal vibrations: When a gas is heated, translational kientic energy of molecules in the gas will increase. In addition, molecules in the gas may pick up many characteristic internal vibrations. Potential energy stored in these internal degrees of freedom contributes to specific heat of the gas.
The heat capacity at constant pressure of 1 J·K−1 ideal gas is:
(12.2.13)(∂H∂T)V=c p=c v+R
where H=U+pV is the enthalpy of the gas.
Measuring the heat capacity at constant volume can be prohibitively difficult for liquids and solids. That is, small temperature changes typically require large pressures to maintain a liquid or solid at constant volume (this implies the containing vessel must be nearly rigid or at least very strong). It is easier to measure the heat capacity at constant pressure (allowing the material to expand or contract freely) and solve for the heat capacity at constant volume using mathematical relationships derived from the basic thermodynamic laws.
Utilizing the Fundamental Thermodynamic Relation we can show:
(12.2.14)C p−C V=T(∂P∂T)V,N(∂V∂T)p,N
where the partial derivatives are taken at: constant volume and constant number of particles, and at constant pressure and constant number of particles, respectively.
The heat capacity ratio or adiabatic index is the ratio of the heat capacity at constant pressure to heat capacity at constant volume. It is sometimes also known as the isentropic expansion factor:
(12.2.15)γ=C P C V=c p c v
For an ideal gas, evaluating the partial derivatives above according to the equation of state, where R is the gas constant for an ideal gas yields:
(12.2.16)pV=RT
(12.2.17)C p−C V=T(∂P∂T)V(∂V∂T)p
(12.2.18)C p−C V=−T(∂P∂V)V(∂V∂T)p 2
(12.2.19)P=RT Vn→(∂P∂V)T=−RT V 2=−P V
(12.2.20)V=RT Pn→(∂V∂T)p 2=R 2 P 2
substituting:
(12.2.21)−T(∂P∂V)V(∂V∂T)p 2=−T−P VR 2 P 2=R
This equation reduces simply to what is known as Mayer’s relation:
Julius Robert Mayer: Julius Robert von Mayer (November 25, 1814 – March 20, 1878), a German physician and physicist, was one of the founders of thermodynamics. He is best known for his 1841 enunciation of one of the original statements of the conservation of energy (or what is now known as one of the first versions of the first law of thermodynamics): “Energy can be neither created nor destroyed. ” In 1842, Mayer described the vital chemical process now referred to as oxidation as the primary source of energy for any living creature. His achievements were overlooked and credit for the discovery of the mechanical equivalent of heat was attributed to James Joule in the following year. von Mayer also proposed that plants convert light into chemical energy.
(12.2.22)C P−C V=R.
It is a simple equation relating the heat capacities under constant temperature and under constant pressure.
Solving Problems with Calorimetry
Calorimetry is used to measure the amount of heat produced or consumed in a chemical reaction.
learning objectives
Explain a bomb calorimeter is used to measure heat evolved in a combustion reaction
Calorimeters are designed to minimize energy exchange between the system being studied and its surroundings. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food.
Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity ). The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system). Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section.
A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature. The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.
Coffee-Cup Calorimeters
General chemistry students often use simple calorimeters constructed from polystyrene cups. These easy-to-use “coffee cup” calorimeters allow more heat exchange with their surroundings, and therefore produce less accurate energy values.
Structure of the Constant Volume (or “Bomb”) Calorimeter
Bomb Calorimeter: This is the picture of a typical setup of bomb calorimeter.
A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter, is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term “bomb” comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.) This type of calorimeter consists of a robust steel container (the “bomb”) that contains the reactants and is itself submerged in water. The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is trapped in the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction. Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known q, such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data.
Example 12.2.1: Identifying a Metal by Measuring Specific Heat
A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.
Solution
Assuming perfect heat transfer, the heat given off by metal is the negative of the heat taken in by water, or:
(12.2.23)q metal=−q water
In expanded form, this is:
(12.2.24)c metal×m metal×(T f,metal−T i,metal)=c water×m water×(T f,water−T i,water)
Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:
(12.2.25)(c metal)(59.7g)(28.5 oC−100.0 oC)=(4.18J/g oC)(60.0g)(28.5oC−22.0oC)
Solving this:
(12.2.26)c metal=−(4.184J/g oC)(60.0g)(6.5 oC)(59.7g)(−71.5 oC)=0.38J/g oC
Our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper.
Key Points
Heat capacity is the measurable physical quantity that characterizes the amount of heat required to change a substance’s temperature by a given amount. It is measured in joules per Kelvin and given by.
The heat capacity is an extensive property, scaling with the size of the system.
The heat capacity of most systems is not constant (though it can often be treated as such). It depends on the temperature, pressure, and volume of the system under consideration.
Unlike the total heat capacity, the specific heat capacity is independent of mass or volume. It describes how much heat must be added to a unit of mass of a given substance to raise its temperature by one degree Celsius. The units of specific heat capacity are J/(kg °C) or equivalently J/(kg K).
The heat capacity and the specific heat are related by C=cm or c=C m.
The mass m, specific heat c, change in temperature ΔT, and heat added (or subtracted) Q are related by the equation: Q=mcΔT.
Values of specific heat are dependent on the properties and phase of a given substance. Since they cannot be calculated easily, they are empirically measured and available for reference in tables.
A calorimeter is used to measure the heat generated (or absorbed) by a physical change or chemical reaction. The science of measuring these changes is known as calorimetry.
In order to do calorimetry, it is crucial to know the specific heats of the substances being measured.
Calorimetry can be performed under constant volume or constant pressure. The type of calculation done depends on the conditions of the experiment.
The specific heat at constant volume for a gas is given as (∂U∂T)V=c v.
The specific heat at constant pressure for an ideal gas is given as (∂H∂T)V=c p=c v+R.
The heat capacity ratio (or adiabatic index ) is the ratio of the heat capacity at constant pressure to heat capacity at constant volume.
Calorimetry is used to measure amounts of heat transferred to or from a substance.
A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process.
This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution.
Key Terms
heat capacity: The amount of heat energy needed to raise the temperature of an object or unit of matter by one degree Celsius; in units of joules per kelvin (J/K).
enthalpy: the total amount of energy in a system, including both the internal energy and the energy needed to displace its environment
specific heat capacity: The amount of heat that must be added (or removed) from a unit mass of a substance to change its temperature by one degree Celsius. It is an intensive property.
constant-pressure calorimeter: An instrument used to measure the heat generated during changes that do not involve changes in pressure.
calorimeter: An apparatus for measuring the heat generated or absorbed by either a chemical reaction, change of phase or some other physical change.
constant-volume calorimeter: An instrument used to measure the heat generated during changes that do not involve changes in volume.
Fundamental Thermodynamic Relation: In thermodynamics, the fundamental thermodynamic relation expresses an infinitesimal change in internal energy in terms of infinitesimal changes in entropy, and volume for a closed system in thermal equilibrium in the following way: dU=TdS-PdV. Here, U is internal energy, T is absolute temperature, S is entropy, P is pressure and V is volume.
adiabatic index: The ratio of the heat capacity at constant pressure to heat capacity at constant volume.
specific heat: The ratio of the amount of heat needed to raise the temperature of a unit mass of substance by a unit degree to the amount of heat needed to raise that of the same mass of water by the same amount.
heat of reaction: The enthalpy change in a chemical reaction; the amount of heat that a systems gives up to its surroundings so it can return to its initial temperature.
combustion: A process where two chemicals are combined to produce heat.
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12.1: Introduction
12.3: Phase Change and Latent Heat |
8831 | https://math.arizona.edu/~jwatkins/H3_composite.pdf | Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Chapter 8 Hypothesis Testing Multiple Testing 1 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Outline Partitioning the Parameter Space The Power Function One-Sided Test Mark and Recapture Two-Sided Test Sample Proportion The p-value 2 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Partitioning the Parameter Space Simple hypotheses limit us to a decision between one of two possible states of nature.
This limitation does not allow us, under the procedures of hypothesis testing to address the basic question: Does the parameter value θ0 increase, decrease or change at all under under a different experimental condition?
This leads us to consider composite hypotheses. In this case, the parameter space Θ is divided into two disjoint regions, Θ0 and Θ1. The hypothesis test is now written H0 : θ ∈Θ0 versus H1 : θ ∈Θ1.
Again, H0 is called the null hypothesis and H1 the alternative hypothesis.
3 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Partitioning the Parameter Space For the three alternatives to the question posed above, we have • increase would lead to the choices Θ0 = {θ; θ ≤θ0} and Θ1 = {θ; θ > θ0}, • decrease would lead to the choices Θ0 = {θ; θ ≥θ0} and Θ1 = {θ; θ < θ0}, and • change would lead to the choices Θ0 = {θ0} and Θ1 = {θ; θ ̸= θ0} for some choice of parameter value θ0. The effect that we are meant to show, here the nature of the change, is contained in Θ1. The first two options given above are called one-sided tests. The third is called a two-sided test.
Rejecting the null hypothesis, critical regions, and type I and type II errors have the same meaning for a composite hypotheses. Significance level and power will necessitate an extension of the ideas for simple hypotheses.
4 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The Power Function Power is now a function of the parameter value θ. If our test is to reject H0 whenever the data fall in a critical region C, then the power function is defined as π(θ) = Pθ{X ∈C}, the probability of rejecting the null hypothesis for a given parameter value.
• For θ ∈Θ0, π(θ) is the probability of making a type I error, i.e., rejecting the null hypothesis when it is indeed true.
• For θ ∈Θ1, 1 −π(θ) is the probability of making a type II error, i.e., failing to reject the null hypothesis when it is false.
The ideal power function has π(θ) ≈0 for all θ ∈Θ0 and π(θ) ≈1 for all θ ∈Θ1.
5 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The Power Function • The goal is to make the chance for error small.
• One strategy is to consider a method analogous to that employed in the Neyman-Pearson lemma. Thus, we must simultaneously, • fix a (significance) level α, now defined to be the largest value of π(θ) in the region Θ0 defined by the null hypothesis, By focusing on the value of the parameter in Θ0 that is most likely to result in an error, we insure that the probability of a type I error is no more that α irrespective of the value for θ ∈Θ0.
• and look for a critical region that makes the power function as large as possible for values of the parameter θ ∈Θ1.
6 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The Power Function Example. Let X1, X2, . . . , Xn be independent N(µ, σ0) random variables with σ0 known and µ unknown. For the composite hypothesis for the one-sided test H0 : µ ≤µ0 versus H1 : µ > µ0, we use the test statistic from the likelihood ratio test and reject H0 if the statistic ¯ x is too large. Thus, the critical region C = {x; ¯ x ≥k(µ0)}.
If µ is the true mean, then the power function π(µ) = Pµ{X ∈C} = Pµ{ ¯ X ≥k(µ0)}.
The value of k(µ0) depends on the level α of the test.
7 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The Power Function • As the actual mean µ increases, then the probability that the sample mean ¯ X exceeds a particular value k(µ0) also increases.
• In other words, π is an increasing function.
• Thus, the maximum value of π on Θ0 = {µ; µ ≤µ0} takes place for µ = µ0.
• Consequently, to obtain level α for the hypothesis test, set α = π(µ0) = Pµ0{ ¯ X ≥k(µ0)}.
We now use this to find the value k(µ0). When µ0 is the value of the mean, we standardize to give a standard normal random variable Z = ¯ X −µ0 σ0/√n .
Choose zα so that P{Z ≥zα} = α. Thus, Pµ0{Z ≥zα} = Pµ0{ ¯ X ≥µ0 + σ0 √nzα} and k(µ0) = µ0 + (σ0/√n)zα.
8 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The Power Function Exercise. If µ is the true state of nature, then Z = ¯ X −µ σ0/√n is a standard normal random vari-able.
Use this to show that the power function π(µ) = 1 −Φ zα −µ −µ0 σ0/√n where Φ is the distribution function for the standard normal.
-0.5 0.0 0.5 1.0 0.0 0.2 0.4 0.6 0.8 1.0 mu pi -0.5 0.0 0.5 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Power function for the one-sided test with alternative greater. The size of the test α is given by the height of the red segment. Notice that π(µ) < α for all µ < µ0 and π(µ) > α for all µ > µ0.
9 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The Power Function We have seen the expression π(µ) = 1 −Φ zα −µ −µ0 σ0/√n in several contexts.
• If we fix n, the number of observations and the alternative value µ = µ1 > µ0 and determine the power 1 −β as a function of the significance level α, then we have the receiving operator characteristic.
• If we fix µ1 the alternative value and the significance level α, then we can determine the power as a function of n the number of observations.
• If we fix n and the significance level α, then we can determine the power function, π(µ), as a function of the alternative value µ.
Exercise. Give the appropriate expression for π for a less than alternative and use this to plot the power function for the example with a model species and its mimic. Take α = 0.05, µ0 = 10, σ0 = 3, and n = 16 observations, 10 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The Power Function To compute sample size for chosen type I and type II errors, let µ1 > µ0.
β = Φ zα −|µ1 −µ0| σ0/√n Choose n so that zα + |µ1−µ0| σ0/√n has probability β. However, β = Φ(−zβ).
−zβ = zα −|µ1 −µ0| σ0/√n √n|µ1 −µ0| σ0 = zα + zβ √n = σ0 |µ1 −µ0|(zα + zβ) n = σ2 0 (µ1 −µ0)2 (zα + zβ)2 Choose n∗, any integer al least as large as n.
11 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Mark and Recapture Mark and recapture can be used as experimental procedure to test whether or not a population has reached a dangerously low level. The variables are • t be the number captured and tagged, • k be the number in the second capture, • r be the number in the second capture that are tagged, and • N be the total population.
If N0 is the level that a wildlife biologist say is dangerously low, then the natural hypothesis is one-sided.
H0 : N ≥N0 versus H1 : N < N0.
12 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Mark and Recaputure The data are used to compute r, the number in the second capture that are tagged.
The likelihood function for N is the hypergeometric distribution, L(N|r) = t r N−t k−r N k The maximum likelihood estimate is ˆ N = [tk/r]. Thus, higher values for r lead us to lower estimates for N. Let R be the (random) number in the second capture that are tagged, then, for an α level test, we look for the minimum value rα so that π(N) = PN{R ≥rα} ≤α for all N ≥N0.
As N increases, then recaptures become less likely and the probability above decreases.
Thus, we set the value of rα according to the parameter value N0, the minimum value under the null hypothesis.
13 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Mark and Recaputure To determine rα for α = 0.05, 0.02, 0.01, > N0<-4500;t<-400;k<-500 > alpha<-c(0.05,0.02,0.01) > ralpha<-qhyper(1-alpha,t,N0-t,k) > data.frame(alpha,ralpha) alpha ralpha 1 0.05 54 2 0.02 57 3 0.01 59 The power curve π(N) = PN{R ≥r0.05} is given using the R commands > N<-c(2500:5500) > power<-1-phyper(54,t,N-t,k) > plot(N,power,type="l",ylim=c(0,1)) 2500 3000 3500 4000 4500 5000 5500 0.0 0.2 0.4 0.6 0.8 1.0 N power 2500 3000 3500 4000 4500 5000 5500 0.0 0.2 0.4 0.6 0.8 1.0 Power curve. Vertical red segment at N = N0 = 4000 has height α = 0.05.
14 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Mark and Recaputure Note that we must capture al least rα = 54 that were tagged in order to reject H0 at the α = 0.05 level. In this case the estimate for N is ˆ N = kt rα = 3703 is well below N0 = 4500.
Exercise. Determine the type II error rate for N = 4500 with • k = 800 and α = 0.05, 0.02, 0.01, and • α = 0.05 and k = 600, 800, and 1000.
15 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Sample Proportion Honey bees store honey for the winter. This honey serves both as nourishment and insulation from the cold. Typically for a given region, the probability of survival of a feral bee hive over the winter is p0 = 0.7. To check whether this winter has a different survival probability, we consider the hypotheses H0 : p = p0 versus H1 : p ̸= p0.
If we use the central limit theorem, then, under the null hypothesis, z = ˆ p −p0 p p0(1 −p0)/n has a distribution approximately that of a standard normal random variable. We reject if |z| is too big.
16 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Sample Proportion For an α level test, the critical value is zα/2. The critical region C = ( ˆ p; ˆ p −p0 p p0(1 −p0)/n > zα/2 ) .
For this study, we examine 336 colonies and find that 250 survive.
Exercise. For α = 0.05, determine whether or not we reject H0.
Exercise. Show that −zα/2 < ˆ p −p0 p p0(1 −p0)/n < zα/2 if and only if p0 −p p p(1 −p)/n −zα/2 s p0(1 −p0) p(1 −p) < ˆ p −p p p(1 −p)/n < p0 −p p p(1 −p)/n + zα/2 s p0(1 −p0) p(1 −p) .
17 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value Sample Proportion 0.55 0.65 0.75 0.85 0.0 0.2 0.4 0.6 0.8 1.0 p power 0.55 0.65 0.75 0.85 0.0 0.2 0.4 0.6 0.8 1.0 p power 0.55 0.65 0.75 0.85 0.0 0.2 0.4 0.6 0.8 1.0 p power 0.55 0.65 0.75 0.85 0.0 0.2 0.4 0.6 0.8 1.0 p power 0.55 0.65 0.75 0.85 0.0 0.2 0.4 0.6 0.8 1.0 p power 0.55 0.65 0.75 0.85 0.0 0.2 0.4 0.6 0.8 1.0 p power Power curve. (left) n = 336, α = 0.05, 0.02, 0.01. (right) α = 0.05, n = 336, 672, 1008.
18 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The p-value • The report of reject the null hypothesis does not describe the strength of the evidence because it fails to give us the sense of whether or not a small change in the values in the data could have resulted in a different decision.
• Consequently, one common method is to report the value of the test statistic and to give all the values for α that would lead to the rejection of H0.
• The p-value is the probability of obtaining a result at least as extreme as the one that was actually observed, assuming that the null hypothesis is true. In this way, we provide an assessment of the strength of evidence against H0.
• Consequently, a very low p-value indicates strong evidence against the null hypothesis.
19 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The p-value We can see how this works with the example on winter survival of beehives using the R command prop.test.
prop.test(250,336,p=0.7) 1-sample proportions test with continuity correction data: 250 out of 336, null probability 0.7 X-squared = 2.8981, df = 1, p-value = 0.08868 alternative hypothesis: true p is not equal to 0.7 95 percent confidence interval: 0.6932518 0.7891515 sample estimates: p 0.7440476 The p-value states that we could reject H0 for any significance level above this value.
20 / 21 Partitioning the Parameter Space The Power Function One-Sided Test Two-Sided Test The p-value The p-value • Under the null hypothesis, ˆ p has approximately normal, mean p0 = 0.7.
• The p-value, 0.089, is the area under the density curve outside the test statistic values |z| = ˆ p −p0 p0(1 −p0)/n = 1.702 (indicated in red), • The critical value, 1.96, for an α = 0.05 level test. (indicated in blue).
• Because the p-value is greater than the significance level, we cannot reject H0 at the 5% level.
-3 -2 -1 0 1 2 3 0.0 0.1 0.2 0.3 0.4 z dnorm(z) -3 -2 -1 0 1 2 3 0.0 0.1 0.2 0.3 0.4 -3 -2 -1 0 1 2 3 0.0 0.1 0.2 0.3 0.4 -3 -2 -1 0 1 2 3 0.0 0.1 0.2 0.3 0.4 -3 -2 -1 0 1 2 3 0.0 0.1 0.2 0.3 0.4 21 / 21 |
8832 | https://www.mathway.com/popular-problems/Basic%20Math/17727 | Factor a^3-2a^2+2a-4 | Mathway
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Basic Math Examples
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Factor a^3-2a^2+2a-4
a 3−2 a 2+2 a−4 a 3-2 a 2+2 a-4
Step 1
Factor out the greatest common factor from each group.
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Step 1.1
Group the first two terms and the last two terms.
(a 3−2 a 2)+2 a−4(a 3-2 a 2)+2 a-4
Step 1.2
Factor out the greatest common factor (GCF) from each group.
a 2(a−2)+2(a−2)a 2(a-2)+2(a-2)
a 2(a−2)+2(a−2)a 2(a-2)+2(a-2)
Step 2
Factor the polynomial by factoring out the greatest common factor, a−2 a-2.
(a−2)(a 2+2)(a-2)(a 2+2)
a 3−2 a 2+2 a−4 a 3-2 a 2+2 a-4
from=0=0−5 a 2-5 a 2
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8833 | https://www.varsitytutors.com/hotmath/hotmath_help/topics/surface-area-of-a-cylinder | Skip to main content
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8834 | https://www.savemyexams.com/gcse/statistics/edexcel/17/higher/revision-notes/processing-representing-and-analysing-data/measures-of-central-tendency/linear-interpolation/ | Linear Interpolation | Edexcel GCSE Statistics Revision Notes 2017
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Revision Notes
The Collection of Data
2 Topics · 4 Revision Notes
1.
Planning & Types of Data
1. ##### Planning an Enquiry
2. ##### Types of Data
2.
Population, Sampling & Collecting Data
1. ##### Population & Sampling
2. ##### Collecting Data
Processing, Representing & Analysing Data
8 Topics · 33 Revision Notes
1.
Tabulation, Diagrams & Representation
1. ##### Bar Charts, Line Graphs & Pictograms
2. ##### Pie Charts
3. ##### Stem & Leaf Diagrams
4. ##### Two-way Tables & Venn Diagrams
5. ##### Population Pyramids
6. ##### Choropleth Maps
7. ##### Cumulative Frequency Charts
8. ##### Box Plots
9. ##### Histograms & Frequency Polygons
10. ##### Selecting & Interpreting Data Representations
2.
Measures of Central Tendency
1. ##### Mode, Median & Arithmetic Mean
2. ##### Mode & Mean from Grouped Data
3. ##### Linear Interpolation
4. ##### Transforming Data
5. ##### Other Types of Mean
3.
Measures of Dispersion
1. ##### Quartiles & Percentiles
2. ##### Types of Range
3. ##### Standard Deviation
4. ##### Outliers
4.
Using Summary Statistics
1. ##### Using Measures of Central Tendency
2. ##### Using Measures of Dispersion
3. ##### Skewness
5.
Index Numbers & Rates of Change
1. ##### Index Numbers
2. ##### Rates of Change
6.
Scatter Diagrams & Correlation
1. ##### Scatter Diagrams & Correlation
2. ##### Lines of Best Fit & Regression Lines
3. ##### Spearman's Rank Correlation Coefficient
4. ##### Pearson's Product Moment Correlation Coefficient (PMCC)
7.
Time Series
1. ##### Time Series Graphs
2. ##### Moving Averages
3. ##### Identifying & Interpreting Trends in Data
8.
Quality Assurance & Estimation
1. ##### Quality Assurance
2. ##### Estimation from Statistical Data
Probability
2 Topics · 6 Revision Notes
1.
Probability Basics
1. ##### Probabilities & Data
2. ##### Risk & Bias
3. ##### Probability Formulae
4. ##### Probability Diagrams
2.
Probability Distributions
1. ##### The Binomial Distribution
2. ##### The Normal Distribution
GCSEStatisticsEdexcelHigherRevision NotesProcessing, Representing & Analysing Data Measures of Central Tendency Linear Interpolation
Linear Interpolation(Edexcel GCSE Statistics):Revision Note
Exam code:1ST0
Download PDF
Written by:Roger B
Reviewed by:Dan Finlay
Updated on 7 May 2024
Exam board:
Edexcel
Edexcel Statistics: HigherEdexcel Statistics: Foundation
Median from Grouped Data
How do I find the median for grouped data?
Grouped datadoesn’t contain the individual data values
So we can’t find theexact median in the usual way
We can estimate the median for grouped data using linear interpolation
This assumes the data is evenly spread across the class containing the median
STEP 1
Identifythe class interval containing the median
This is the class containing the ‘Image 6: n over 2th’ data value
Divide the total number of values, _n_, by 2
e.g. if there are 80 data values, Image 7: n over 2 equals 80 over 2 equals 40
Consider cumulative frequencies until you find the class interval containing the 40 th value
STEP 2
Find ‘how far into’ that class interval the th data value is
e.g. if the interval with the median contains the 36 th through 43 rd data values
then the 40 th value is ‘5 values in’ (36, 37, 38, 39, 40)
and there are 8 values in the interval
so the 40 th value isImage 9: 5 over 8of the way into the interval
STEP 3
Multiplythe class width of the class interval containing the median by the fraction found in Step 2
e.g. if the interval with the median is Image 10: 50 less or equal than x less or equal than 70
the class width is Image 11: 70 minus 50 equals 20
Image 12: 20 cross times 5 over 8 equals 12.5
STEP 4
Addthe result from Step 3 to the lower boundary of the class interval containing the median
The result is the estimated median
e.g. lower bound of Image 13: 50 less or equal than x less or equal than 70 is 50
The estimated median is Image 14: 50 plus 12.5 equals 62.5
The estimated median can also be found using the following formula:
L is the lower boundary of the class interval containing the median
n is the total number of data values
C is the cumulative frequency of all the class intervals before the one containing the median
f is the frequency of (i.e. the number of values in) the class interval containing the median
w is the width of the class interval containing the median (upper boundary minus lower boundary)
This formula combines all four steps of the process
but it is not on the exam formula sheet
So if you want to use it you’ll need to remember it
Examiner Tips and Tricks
The formula can be tricky to remember correctly
It’s better to understand how the method works
Then you don’t need to remember the formula
Remember that the median found this way is an estimate
You can’t find the exact median without knowing all the data values
Worked Example
A student collected data about the length of time (x hours) students in his school spent listening to music in a given week. He collected data from 50 students in total. The following table summarises the data:
Time spent, x (hours)Number of students
0 ≤ x ≤ 10 3
10 <x ≤ 20 19
20 <x ≤ 30 12
30 <x ≤ 40 10
40 <x ≤ 50 5
50 <x ≤ 60 1
Work out an estimate for the median amount of time spent listening to music by the students.
STEP 1: Identify the class interval containing the median
Divide the total number of values, n, by 2
Here n = 50
So we’re looking for the interval with the 25 th value
Note that this is different from finding the median from a set of data values
In that case we would be looking for the value halfway between the 25 th and 26 th values
With linear interpolation we don’t have to worry about that!
Add a cumulative frequency column to the table and work out the cumulative frequencies
Time spent, x (hours)Number of students Cumulative Frequency
0 ≤ x ≤ 10 3 3
10 <x ≤ 20 19 22
20 <x ≤ 30 12 34
30 <x ≤ 40 10 44
40 <x ≤ 50 5 49
50 <x ≤ 60 1 50
The second class interval goes up to the 22 nd data value and the third class interval goes up to the 34 th data value
So the median is in the third class interval
The median is in the class interval
STEP 2: Find how far into the class interval the th data value is
The interval with the median contains the 23 rd through 34 th data values
The 25 th data value is ‘3 values in’ to the interval (23, 24, 25)
And there are 12 data values in the interval
So the median is 1/4 of the way into the interval
STEP 3: Multiply the class width by the fraction found in Step 2
Subtract the lower boundary from the upper boundary to find the class width of the interval
30-20=10
Multiply by the fraction in Step 2
STEP 4: Add the result from Step 3 to the lower boundary of the class interval
20 + 2.5 = 22.5
Don’t forget the units in your answer!
Estimated median = 22.5 hours
How do I find the median for data on a histogram?
A histogram is a way of representing grouped data as a diagram
See the 'Histograms & Frequency Polygons' revision note for full details
The connection between the frequency density shown on the histogram and the frequency that would be shown in a grouped data table is given by the formula
If you are asked to estimate a median for data in a histogram there are two options:
You can recreate the grouped data table using the frequency density formula, and then follow the method given above
Or you can work out the estimated median directly from the histogram
See the following Worked Example for how to do this
Worked Example
The histogram shows the weight, in kg, of 60 newborn bottlenose dolphins.
Find an estimate for the median weight of the dolphins in the sample, giving your answer correct to two decimal places.
, so to estimate the median we need to find the weight of the 30th dolphin
To find the frequencies represented by the different bars, rearrangeas
The first two classes have a cumulative frequency of
So the median is going to be '10 dolphins into' the 10-12 kg class
The height (frequency density) of the 10-12 kg bar is 9.5
We need to find what width would give a frequency of 10
Use and solve for width
Image 30: 9.5 equals 10 over width
9.5 cross times width equals 10
width equals fraction numerator 10 over denominator 9.5 end fraction equals 20 over 19 equals 1.0526... equals 1.05 space open parentheses 2 space straight d. straight p. close parentheses
That means that the median lies 1.05 kg into the 10-12 kg class interval
Estimated median = 11.05 kg (2 d.p.)
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Previous:Mode & Mean from Grouped DataNext:Transforming Data
More Exam Questions you might like
Measures of Central Tendency
Measures of Dispersion
Using Summary Statistics
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Scatter Diagrams & Correlation
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Probability Distributions
Planning & Types of Data
Author
Reviewer
Author:Roger B
Expertise:Maths Content Creator
Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.
Reviewer:Dan Finlay
Expertise:Maths Subject Lead
Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.
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8835 | https://uniformlyatrandom.wordpress.com/2009/01/25/the-erdos-ginzburg-ziv-theorem/ | Uniformly at Random
The Erdos-Ginzburg-Ziv Theorem
Here we present a proof of the Erdos-Ginzburg-Ziv Theorem. We shall deduce this result using the Cauchy-Davenport Theorem, which in turn can be derived from Noga Alon’s Combinatorial Nullstellensatz. We begin by stating without proof the Combinatorial Nullstellensatz.
Combinatorial Nullstellensatz. Let be a polynomial over a field in the variables . Suppose that the total degree of is and that the coefficient in of is non-zero. For , let be a set of elements of . Then there exists such that .
We wish to deduce the following classical theorem, following the proof of Noga Alon in his paper “Combinatorial Nullstellensatz”.
Theorem (Cauchy-Davenport). Let be a prime and let be subsets of . Then either or .
Proof. We first observe that if , then for every we have , so that is non-empty. This implies that can be written as the sum of elements and . Since this holds for all we have , as required.
Suppose now that and that , contrary to the claimed result. Let be any subset of of size containing . Define a polynomial . Clearly, for all . Moreover, the coefficient of in equals , and is therefore non-zero in , since . Applying the Combinatorial Nullstellensatz to , we conclude that there exists such that , contradicting our earlier observation. This contradiction establishes the desired result.
This result can also be proved directly by induction on the size of . Using the Cauchy-Davenport Theorem we now show the following result, following the presentation of Alon and Dubiner.
Theorem (Erdos-Ginzburg-Ziv). Let be an integer. For any sequence of (not necessarily distinct) elements of , there exists a set of size such that (in ).
Proof. The proof is by induction on the number of primes in the prime factorization of . Suppose that the result holds for prime and now consider for some prime and some integer . Consider any element subsequence of ; by the claimed result for prime values, this element subsequence contains elements whose sum is a multiple of . We can therefore remove this element subsequence and repeat the argument to obtain disjoint sets such that for , and is a multiple of . For , define . We now apply the induction hypothesis to the sequence to conclude that there exists of size such that is a multiple of . This implies that if , then and is a multiple of , as required.
It remains now to establish the base case of the induction, namely the case when equals a prime . Suppose that . If for some we have , then so that is a multiple of and we are done. Suppose then that for , , and define sets . By repeated application of the Cauchy-Davenport Theorem, we see that . Thus every element of can be written as a sum of exactly elements of our sequence. In particular, can be so written, so that , taken together with the elements of our sequence that sum to , gives a element subsequence whose sum is 0 in , as required. This completes the proof.
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[…] As a concluding remark, the result can be generalized that in a group of 2n – 1 numbers, there is always a subset of n numbers whose sum is divisible by n. This is known as the Erdos-Ginzburg-Ziv Theorem and is a bit more complicated to prove. You can read a proof of it here. […]
Monday puzzle: divisible by 3 - Mind Your Decisions
August 26, 2013 at 3:59 am
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8836 | https://brainly.com/question/34702204 | [FREE] The primary transcript in the figure is 15 kilobases (kb) long, but the mature mRNA is 7 kb in length. - brainly.com
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The primary transcript in the figure is 15 kilobases (kb) long, but the mature mRNA is 7 kb in length. Describe the modification that most likely resulted in the 8 kb difference in length of the mature mRNA molecule. Identify in your response the location in the cell where the change occurs.
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The 8 kb difference in length of the mature mRNA molecule is likely due to RNA splicing, which occurs in the nucleus of the cell.
The 8 kb difference in length of the mature mRNA molecule is most likely due to post-transcriptional modifications. One common modification is RNA splicing, which occurs in the nucleus of the cell. RNA splicing involves the removal of introns (non-coding regions) from the primary transcript, leaving behind only the exons (coding regions). This process results in a shorter mature mRNA molecule.
During splicing, a complex called the spliceosome recognizes specific nucleotide sequences at the boundaries of introns and exons. The introns are then cut out and the exons are joined together, forming a continuous coding sequence. This process is crucial for generating a functional mRNA that can be translated into protein.
It's important to note that the length difference between the primary transcript and mature mRNA can vary depending on the number and size of introns present in the gene. Additionally, other post-transcriptional modifications such as capping and polyadenylation may also contribute to the length difference.
In summary, the 8 kb difference in length of the mature mRNA molecule is likely due to RNA splicing, which occurs in the nucleus of the cell.
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The difference in length between the primary transcript and mature mRNA is due to RNA splicing, which removes non-coding introns and connects coding exons. This modification takes place in the nucleus of the cell before the mRNA is translated. The resulting mature mRNA is significantly shorter than the original primary transcript.
Explanation
The 8 kb difference in length between the primary transcript and the mature mRNA molecule is primarily due to a process called RNA splicing, which occurs in the nucleus of the cell.
Here's how this process works:
Primary Transcript: The primary transcript, also known as pre-mRNA, includes both exons (coding regions) and introns (non-coding regions).
Splicing Process: Once transcription is complete, the pre-mRNA undergoes splicing where a complex known as the spliceosome removes the introns. The spliceosome identifies specific sequences at the edges of introns and exons, cuts them out, and joins the exons together.
Resulting Mature mRNA: After splicing, only the exons remain, leading to a mature mRNA molecule that is shorter than the original primary transcript because all the intronic sequences have been excised. The mRNA is then modified further by adding a 5' cap and a poly-A tail, enhancing its stability and readiness for translation.
Overall, the difference of 8 kb indicates that a significant amount of non-coding sequence has been removed during the splicing process in the nucleus before the mRNA is transported to the cytoplasm for protein synthesis.
Examples & Evidence
For instance, if a gene has one coding exon and two introns, the original transcript might include both introns and the exon. After splicing, only the coding exon remains, resulting in a shorter mRNA molecule suitable for translation into a protein.
Research on eukaryotic gene expression has shown that splicing is a crucial step in mRNA processing, as it directly affects the coding potential of mRNA and the diversity of proteins that can be produced.
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8837 | https://www.youtube.com/watch?v=HIeU404nMgo | Argand Diagrams 15 • Loci - intersections • CP1 Ex2E • 🏅
Bicen Maths
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The best way to find videos for other topics is to go to my channel's homepage, then scroll down to the relevant section.
There are playlists per chapter, with videos linked to individual exercises. Thanks for watching!
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Transcript:
okay so we're still continuing with exercise 2e it's a pretty big one and but this time we're going to be looking at the intersections of complex loci so far we've just been doing one kind of loci drawing at a time this is the first time we're going to have two of them at the same time and we're going to try and find the solution to where these two complex loci are going to cross each other so it says find the complex number z that satisfies both this thing that we've got here which we know is going to be a circle and this thing here which we know is going to be a half line i'll talk about what we're actually doing once we've got the sketch done now you'll notice this one that we've got here is the same question that we've just done in the previous video so i'm actually going to kind of draw that from the previous one so it was at minus 3 minus 2 so minus 3 minus 2 was where the half line was being drawn and it was coming upwards like this okay this one also is a circle it's got a center of minus three minus two and it has a radius of ten so in fact i think i probably should have drawn this a little bit bigger but hey radius of 10 means it's definitely going to be crossing see if i can grab all of this and pull it across good it means it's definitely going to be crossing the axis here and here because it's going to be a lot bigger than this distance which is three that's minus three here and that's minus two okay now if you think about what's actually happening this one which is the circle is any of these points around here that could be any of those points but what we're actually trying to say is we want it to be any of those points on the circle plus we want it to be any of those points along there because we want it to satisfy both of them at the same time that means it has to be the one place where it is actually matching both in other words where do the lines cross over so there's two different ways you can do this method one is coordinate geometry drawing the diagram and thinking about the patterns that you know to do with shapes when i say coordinate geometry i really mean stuff to do with shapes the second method is by using cartesian cartesian equations and solving them simultaneously now what i've noticed in the past is that people usually prefer to do method 2 because they're very comfortable with algebra but i want to try and draw you away from that i try and want you to think about this one more because although there's nearly always two methods if you're good with geometry this method is far superior and is often a lot quicker at coming to the answer and it's a lot easier to make fewer mistakes so let's have a look at this diagram that we've got here first of all we know something about this line that we've got we know that the length of this line is the radius of the circle and the radius of the circle is 10. we also know that this angle is three pi over four which means that if i was to add in a little line here this angle is pi over four now pi over four you should know this already but pi over four is 45 degrees and as soon as i see a triangle that's got 45 degrees with it i start thinking of a very special triangle which is the isosceles triangle okay and we know that this length over here is 10 and this is going to be a right angle so i'm just going to get some of that information i'm going to kind of add this onto the diagram i know that this must be 90. in fact the angle at the top must also be pi over 4 but i don't think that's going to be that useful and i know that these two lengths here must be equal just to remind you what we're doing we are trying to find out what is this complex number here in other words what is the coordinate of that complex number so let's see i think to find out the x coordinate of this well i know it's minus 3 for this bit it's minus 3 across plus this extra journey here and the y coordinate looks like it's well minus 2 to begin with and then it's going to come up a certain amount so all we need to do is find out how long is this journey and this journey that we've got here but we can do that because of this property of the triangle that we have i'm going to simplify this a bit more okay i'm just going to draw a triangle it looks like this this is 10 and we know that these two sides here and here they must be the same length as each other because it's 45 degrees here and it's 45 degrees like this you can now see very clearly it's an isosceles triangle so these two sides must be equal in length so if i just call this one i don't know a or b or c they must be the same as each other i can just do pythagoras so a squared plus a squared equals 10 squared in other words two a squared is a hundred so a squared is 50 and so a is the square root of 50 which is the square root of 25 times the square root of 2 which is just 5 root 2. so this distance is 5 root 2 and this distance is 5 root 2 and we're trying to describe what is this coordinate that we've got here well that coordinate let's start off with the x coordinate to go from the origin you're coming three across and five root two across so it's going to be minus three because you're going to the left minus five root two that is the x coordinate the y coordinate to get here well we started off coming down minus two to get to this central point and then we're adding five root two on so it's going to be minus 2 plus 5 root 2. so that's what it is as a coordinate that doesn't quite answer the question we need to say what is the complex number said and if you're not so sure about how i got to this coordinate try and think about where you're at the origin and what you're doing in the left and right direction so it was minus three and then minus five root two this one was harder to see that to get to it you came down two and then up five root two down two and then up five root two so let's just actually write it in its complex number form it is going to be x which is this plus the y value multiplied by i so that complex number satisfies both of these things that we've got here and i guarantee if you took this complex number you added 3 and 2i to it and you found its modulus you would get 10. and also if you took this complex number you added 3 and 2i to it and you found its argument i guarantee that you would get 3 pi over 4. so we're going to now try and do this using cartesian equations which i think is going to be a bit slower because you may have to work out this cartesian equation and we're going to have to work out this cartesian equation first we have this one nice and easy it's ready to go but we're going to have to um i've made it a bit easier if you didn't have it you would have to work it out so we have y equals minus x minus five that's one of the cartesian equations and then for the circle if you remember from earlier on this would be x plus three squared plus y plus two squared equals the radius squared so i'm going to take this y value here and i'm going to sub it in here so begin by expanding these brackets i have x squared plus 6x plus 9 plus minus x minus five plus two squared equals a hundred i've just subbed in this in place of y so it's x squared plus six x plus nine plus x squared i should probably quickly simplify this over here so that's going to be minus x minus 3 squared that's x squared plus 6 x plus 9 as well yeah that's correct equals 100 so i've now got 2x squared plus 12x plus 18 equals 100. i'm going to half everything so that's x squared plus six x plus nine equals fifty and i'm going to subtract fifty so that's x squared plus six x minus forty one equals zero now i'm gonna go to my quadratic equations over just off screen here so i'm going to put in my 1 6 and -41 and we get that x is either equal to minus 3 plus 5 root 2 or x is equal to three minus five root two now that's odd because we only had um this one we only had the second one before we didn't have this first one so i'm kind of thinking i wonder why i've got this one well let's just continue let's find out what the value of y would be so we know that y is minus x minus five so y would be this one here negated and then subtract five so it'd be three minus five root two minus five so that's minus two minus five root two and here we have y equals this one negated so that's three plus five root two minus five which is minus two plus five root two now we can tell these are the correct ones look i've got this one and this one here which matches these ones that i've got here and here so what is going on with these ones that i'm going to highlight here in purple well we only have a half line here and when you're solving these equations simultaneously it doesn't know the math doesn't know that it's a half line they think when we're solving it in um simultaneously they're presuming that it's a long line they don't know that and so they're also giving us this solution that we've got here that solution that we've got here you need to try and work out that it's not going to be valid the bigger x value which is this one is not going to be valid so you have to do some problem solving at the end to decide whether you actually want to include all of the solutions now i know this one has more of like a jump of being able to recognize these kinds of triangles and different patterns but i do think this is the more superior one that's going to help you become a better mathematician i'm going to recommend some questions for you to do so i want you to leave out question 12 for a while but i'm going to help you with question 12 in a future video i want you to try question 11 and then 13 to 19. i would like to try maybe both methods and in some upcoming videos i'm only going to do the geometric approaches for questions 11 14 and 16 and 17 because i would like you to try working towards that kind of approach for this you can do them with algebra but you'll see here it's quite complicated and then you've also got to do some some reasoning at the end to decide which solutions you won't use if you can do this it's gonna be much better if you'd like me to help you through some of those questions stay tuned and i will be doing that in the next videos |
8838 | https://www.merriam-webster.com/thesaurus/unclear | UNCLEAR Synonyms: 96 Similar and Opposite Words | Merriam-Webster Thesaurus
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Est. 1828
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Synonyms of unclear
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unclear
adjective
ˌən-ˈklir
Definition of unclear
1
as in vague
not expressed in precise terms their suggestion for correcting the problem is a bit unclear
Synonyms & Similar Words
Relevance
vague
ambiguous
fuzzy
cryptic
confusing
indefinite
obscure
enigmatic
inexplicit
uncertain
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murky
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dark
misty
bleary
inscrutable
indistinguishable
confounding
dim
woozy
unfathomable
undefinable
mystifying
bewildering
gauzy
obfuscatory
Antonyms & Near Antonyms
specific
clear
explicit
definite
obvious
direct
understandable
open
straightforward
honest
frank
outspoken
unambiguous
forthright
unequivocal
straight
plain
candid
unguarded
distinct
comprehensible
intelligible
defined
plainspoken
unmistakable
openhearted
blatant
foursquare
fathomable
patent
well-defined
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2
as in faint
not seen or understood clearly obtained at best an unclear glimpse of the rarely seen bird from across the road
Synonyms & Similar Words
vague
faint
hazy
undetermined
indistinct
undefined
nebulous
murky
fuzzy
obscure
blurry
pale
foggy
shadowy
indefinite
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dim
bleary
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invisible
gauzy
intangible
gloomy
inexplicable
dusky
incomprehensible
indiscernible
inappreciable
indecipherable
impalpable
hieroglyphic
obfuscatory
hieroglyphical
Antonyms & Near Antonyms
clear
definite
obvious
evident
certain
distinct
sure
plain
bright
pellucid
strong
firm
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Example Sentences
Examples are automatically compiled from online sources to show current usage. Read More Opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback.
Recent Examples of unclear Though the context of Tuesday's picture was unclear, Disney+ announced a four-episode Malcolm in the Middle revival in December 2024, which Muniz followed with a video featuring himself, Cranston and Kaczmarek celebrating the news. —Meredith Wilshere, People.com, 23 Apr. 2025 The reasons for these moves are unclear; ICE did not respond to multiple requests for comment. —Jack Herrera, New Yorker, 23 Apr. 2025 The fate of the federal suit against RealPage remains unclear. —Joseph Wilkinson, New York Daily News, 23 Apr. 2025 How long those plates can keep spinning is unclear, especially as tariffs start nudging up costs for American consumers. —Anne Marie Drummond Lee, CBS News, 11 Apr. 2025 See All Example Sentences for unclear
Recent Examples of Synonyms for unclear
vague
faint
ambiguous
hazy
Adjective
In fact, those professions grew because the accountant must also talk to clients, ask them questions, make judgement calls about sometimes vague tax code interpretations and execute numerous other judgements on a daily basis.
— Ismail Amla, Forbes.com, 14 Apr. 2025
Instead of vague goals, such as align with human values, researchers and developers can talk about specific contexts and roles for AI more clearly.
— Aidan Kierans, The Conversation, 14 Apr. 2025
Definition of vague
Adjective
The Oscar nominee's glam included a soft-bronzed makeup look, with a faint pink blush and a light mauve lipstick.
— Catherine Santino, People.com, 25 Apr. 2025
The Accountant 2, arriving this week some eight long years after the original, does not earn even that faintest of praise.
— A.A. Dowd, Rolling Stone, 25 Apr. 2025
Definition of faint
Adjective
But the scene itself could be read as a little ambiguous, too: is Sammie trying to pierce the veil again, playing a song that could both foster community but also bring back the vampires?
— Fran Hoepfner, Vulture, 18 Apr. 2025
As the days turned into weeks, the reality of her situation remained ambiguous.
— Ashley Vega, People.com, 17 Apr. 2025
Definition of ambiguous
Adjective
The reasons for her travels—whether to marry a man at her father’s behest or to undertake a pilgrimage to Rome—are hazy.
— Eli Wizevich, Smithsonian Magazine, 24 Apr. 2025
The details of the bet have become hazy in the intervening 35 years.
— Leila Sloman, Quanta Magazine, 18 Apr. 2025
Definition of hazy
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8839 | https://cdn.ymaws.com/www.kcnpnm.org/resource/resmgr/2024_conference/handouts/116_acute_outpatient_emergen.pdf | 4/9/2024 1 Acute Outpatient Emergencies Brett Foster, MD, FAWM KANPNM Conference April 23rd, 2024 Disclosures Patient Assessment The Adult Blunt Trauma Patient Risk Stratification Head Injury Hemorrhage Acute Coronary Syndrome Disclaimers STEMI vs NSTEACS STEMI Features STEMI Management NSTEACS Features NSTEACS Management Clinical Decision Tools Outline Stroke Disclaimer Assessment Tools Pre-Imaging Management Post-Imaging Management Patient Assessment Our goal is to be thorough, systematic, and triage our interventions & diagnostics appropriately.
“Scene Survey” or “Scene Size-Up” What’s going on?
1-2-3-4-5 Mnemonic Primary Survey Identify and Intervene on Immediate Life Threats A-B-C-D-E or M-A-R-C-H Algorithms Deep Breath Secondary Survey Just like any other patient encounter.
HPI, PMH, Physical Exam (C-A-R-T-S) O-P-Q-R-S-T , C-O-L-D-E-R-R, S-A-M-P-L-E Mnemonics Vital Signs Provide objective data and establish a baseline for trending purposes 4/9/2024 2 Blunt Trauma - Risk Stratification We can use physiologic, anatomic, mechanism, and demographic criteria to risk stratify patients in conjunction with our assessment and gestalt.
Source: 2011 Guidelines for Field Triage of Injured Patients (CDC) Physiological Criteria - GCS ≤ 13 - SBP < 90mmHg - RR < 10 or > 29 bpm These patients should be transported preferentially to the highest level of trauma care within the system.
Anatomical Criteria - chest wall instability - 2 or more proximal long-bone fractures - pelvic fractures - depressed skull fractures - crushed, pulseless, mangled extremities - paralysis These patients should be transported preferentially to the highest level of trauma care within the system.
Blunt Trauma - Risk Stratification Mechanism Criteria - falls > 20ft - MVC’s with > 12in intrusion - MVC’s with ejection - MVC’s with death of passenger - auto vs pedestrian/cyclist run over or > 20mph - motorcycle crash > 20mph These patients should be transported to a trauma center, but not necessarily the highest level.
Demographic Criteria - adults > 65yoa with BP < 110mmHg or a ground-level fall - anti-coagulants, anti-platelet agents, or coagulopathy - pregnancy > 20 weeks These patients should be transported to a trauma center, but not necessarily the highest level.
Blunt Trauma – Head Injury TBI’s come in a spectrum of severity. Disposition for moderate and severe is clear cut, we’ll talk about interventions/management. For mild TBI’s, we’ll talk about decision tools to help assess and dispo.
Mild: GCS 14-15 Moderate: GCS 9-13 Severe: GCS 3-8 Moderate and Severe Head Injury The SIBCC Tier Zero Recommendations can help guide management of severe TBI’s. We can extrapolate them into an intuitive A-B-C-D-E format.
Secure optimize the airway.
Avoid hypoxia (goal SpO2 > 94%) Avoid hyperventilation (goal ETCO2 35-40mmHg) unless herniating.
Avoid hypotension (goal SBP >90) Control pain.
Avoid hyponatremia.
Consider seizure ppx.
Optimize positioning.
Avoid fever.
4/9/2024 3 Moderate and Severe Head Injury -Pharmacology Secure optimize the airway.
Avoid hypoxia (goal SpO2 > 94%) Avoid hyperventilation (goal ETCO2 35-40mmHg) Avoid hypotension (goal SBP >90) Control pain.
Avoid hyponatremia.
Consider seizure ppx.
Optimize positioning.
Avoid fever.
Ketamine is an acceptable induction agent. Be careful with propofol. In the event of herniation, hypertonic saline preferred over mannitol. TXA is not a silver bullet. Levophed acceptable 1st line pressor.
Don’t diurese.
Fentanyl preferred 1st line for pain control (more predictable hemodynamics). Empiric steroids not recommended. Seizure ppx - depends on who you ask.
Empiric tylenol prophylaxis.
Mild Head Injury Other decision tools can help us determine need for cross-sectional imaging and our ultimate disposition, or alternatively RTC instructions.
Blunt Trauma - Hemorrhage We’ll discuss areas susceptible to significant internal hemorrhage, clinical and diagnostic features, and some fluid/pharmacology management.
C-A-R-T-S Chest Abdomen/Pelvis Retroperitoneal Space Thigh Skin/Street entry/10.1007/978-3-642-36801-1_177-1 Why do we care?
Blunt Trauma - Hemorrhage 𝑺𝒉𝒐𝒄𝒌 𝑰𝒏𝒅𝒆𝒙= 𝑯𝒆𝒂𝒓𝒕 𝑹𝒂𝒕𝒆 𝑺𝑩𝑷 4/9/2024 4 Blunt Trauma – Hemorrhage Pharmacology The jury is still out on reversal agents, other blood products, and calcium in the pre-hospital setting.
Blunt Trauma – Hemorrhage Adjunctive Management Survival worsens with larger volume.
ONLY if signs of shock.
ONLY if blood not available. STOP once palpable pulse, improved mental status, or BP .
Exception: severe TBI.
Acute Coronary Syndrome A spectrum of conditions that result in myocardial ischemia or infarction due to an interruption in coronary blood flow.
Semantics Stable angina Unstable angina Non-ST elevation myocardial infarction (NSTEMI) ST-elevation myocardial infarction (STEMI) Current ACC/AHA Terminology Non-ST Elevation Acute Coronary Syndromes (NSTEACS) ST-Elevation Myocardial Infarction (STEMI) vs.
On the horizon: OMI and NOMI… Disclaimer #1 Not all chest pain is acute coronary syndrome.
“Can’t Miss” Causes of Chest Pain: - acute coronary syndrome - pulmonary embolism - aortic dissection - acute heart failure - pneumonia - tension physiology (PTX) - tamponade physiology - mediastinitis 4/9/2024 5 Disclaimer #2 Not all acute coronary syndromes are created equal.
Types of Myocardial Infarction Type 1 - spontaneous thrombotic event Type 2 - ischemic imbalance Type 3 - sudden cardiac death Type 4 - revascularization procedure Type 5 - CABG-associated ACS - Clinical Features Reassuring against ACS - sharp or stabbing pain - pleuritic pain - reproducible pain - positional pain Scrubs –My Tuscaloosa Heart Concerning for ACS Courtesy of EmojiTerra ACS - EKG Features Ideally it looks like this… Remember, EKG’s rule in myocardial infarction, they cannot rule out.
If it looks like this, at least our decisions become easy.
ACS - EKG Features (STEMI) What makes a STEMI a STEMI?
4/9/2024 6 ACS - EKG Features (STEMI) ST_elevation#/media/File:ST_ elevation_illustration.jpg Other EKG features typically seen with a STEMI: Pardee’s Sign - hyperacute T-waves - pathologic Q- Waves - Wellen’s morphology - de Winter’s segment - Aslanger criteria - aVR abnormalities Reciprocal Changes and of course, the left bundle branch block (LBBB) What about for an NSTEACS?
ACS Management - STEMI Non-pharmacologic Management - IV access - transport to PCI-capable CCU - monitor and pads on - supplemental O2 for SpO2 < 90% - decision on a reperfusion strategy PCI preferred, but time > 120min may necessitate thrombolytics prior ACS Management - STEMI Thrombolytics (aka fibrinolytics) fibrin non-selective vs fibrin-selective To date: no difference between the fibrin selective agents. Check your local listings.
Big Picture: STEMI + thrombolytic studies to date have a sample size of 60,000 patients, and every study has been positive, regardless of agent or timing (w/in 12 hours).
If we were to get REALLY granular….
Alteplase may allow for the fastest revascularization, but may cause more bleeding. Reteplase is not weight-based and may have better revascularization long-term. And Tenecteplase may have less bleeding risk.
ACS Management - STEMI Antiplatelet Therapies Aspirin P2Y12 Inhibitors: clopidogrel (Plavix) vs ticagrelor (Brilinta) Patients with STEMI should immediately chew 162 to 325 mg of aspirin on recognition of symptoms, unless there is an absolute contraindication.
…treatment with ticagrelor as compared with clopidogrel significantly reduced the rate of death from vascular causes, myocardial infarction, or stroke without an increase in the rate of overall major bleeding… 4/9/2024 7 ACS Management - STEMI Anticoagulation Heparin vs enoxaparin (Occasionally fondaparinux or bivalirudin) The data on A/C is murky, and seems to be trending towards no A/C, especially if prompt PCI plus DAPT .
Pain control No NSAIDs.
Opiates encouraged.
morphine vs fentanyl Follow-on medications - statin therapy - ACE/ARB - Beta blocker ACS Management - NSTEACS Similarities w/ STEMI: - clinical presentation of NSTEACS ≈ STEMI - importance of an initial EKG - importance of early chewable aspirin - similar trend away from anti-coagulation Differences w/ STEMI: - potential to work up alternative causes (labs, imaging) - troponin measurement becomes a possibility - serial EKG’s become (more) important - risk stratification ACS - EKG Features (NSTEACS) Reciprocal Changes ST Depression Hyperacute T-waves Pathologic Q Waves “Dynamic EKG Changes” NEW T-wave inversions ACS Management - NSTEACS Pharmacologic Management - 324mg ASA +/- P2Y12 inhibitor - no thrombolytics - opiate pain control - consider nitrates - anticoagulation unlikely… Non-pharmacologic Management - IV access - POC troponin?
- alternative etiologies?
- supplemental O2 for SpO2 < 90% - serial EKGs - disposition decision 4/9/2024 8 ACS Disposition - NSTEACS Harder - TIMI Score?
- GRACE Score?
Easy - active or ongoing pain - dynamic EKG changes - elevated/rising troponin - arrhythmias - vital sign instability Remember, your role is to be sensitive, not specific.
Stroke Cell death due to interruption in the blood supply to the brain. What about a TIA or “mini-stroke?” a/File:MCA_Territory_Infarct.svg - thrombotic - embolic - cryptogenic - hypoperfusion - intraparenchymal - subarachnoid Disclaimer Not all stroke-like symptoms are due to a stroke.
But given the costs to society, frequency, narrow window for intervention, it’s a stroke until proven otherwise (after checking a blood sugar).
Stroke - Assessment Is there a reliable way to differentiate between an ischemic stroke and a hemorrhagic stroke?
What about Mobile Stroke Units -can they increase fidelity?
abrupt onset of deficit (96%) extremity deficit (60-70%) speech deficit > 50% headache? only 14 % headache preceding deficit nausea and vomiting dramatic hypertension obtunded, combative, comatose meningeal signs Ischemic Hemorrhagic 4/9/2024 9 Stroke - Assessment Okay, so is there a reliable way to assess if someone is having a stroke, regardless of type?
Stroke - Assessment What about the NIH Stroke Scale?
It is a prognostic tool, not a screening tool. It helps gradate severity, predict outcomes, and inform decisions on interventions.
It has limitations, especially in the setting of posterior infarcts.
Stroke - Management We need to get patients moving towards neuroimaging.
“CODE STROKE” or “CODE EMA” vs EMS sweater Please obtain an accurate history but not at the expense of activating transport.
- At a minimum, establish a “last known normal” or “last known well” - recent trauma?
- recent surgery?
- AC or anti-platelet tx?
- previous stroke?
A - B - C - Don’t Forget the Glucose Stroke - Management Destination?
Acute Stroke Ready Hospital?
- can push thrombolytics Primary Stroke Center?
- can admit and has NSG Thrombectomy Capable Stroke Center Comprehensive Stroke Center Other Dos and Don’ts - IV access – yes, 20ga or bigger - Oxygen only for hypoxia - cardiac monitoring - please defer blood pressure management - no validated recommendations on positioning - hold AC, anti-platelets, reversal, blood products 4/9/2024 10 Stroke - Management What happens in the ED?
Why so serious?
Because time is brain, and our windows for interventions are narrow: Hemorrhagic Stroke Mangement & Pharmacology Environmental “Deficits” Coagulopathy Blood Pressure Airway Normothermia/f ever ppx HOB elevated Midline No collar - EEG monitoring - Empiric seizure prophylaxis - empiric steroids - anticoagulation reversal - consider desmopressin platelet transfusion 160/90 seems to be the golidlocks BP - titratable med - nicardipine vs labetalol Neuro-protective Intubation - lidocaine - fentanyl - induction based on BP Ischemic Stroke Management & Pharmacology Crux: thrombolytics tPA (aka alteplase or Activase) vs TNKase (Tenecteplase) ASA Antiplatelets Anticoagulation Questions?
Thank you for your time! |
8840 | https://www.gorillasun.de/blog/an-algorithm-for-polygons-with-rounded-corners/ | Gorilla Sun
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Tutorial
An Algorithm for Polygons with Rounded Corners
An approach for creating all sorts of different smooth shapes in p5, using the canvas rendering context.
By Ahmad Moussa
23 min read
Quick Index
Introduction: An algorithm for rounded corners
Credit where credit's due
An intuitive explanation of the algorithm
Breakdown of the algorithm
Two Vectors from three Points
Calculating the angle using the cross product
Ambiguity of the cross product
Positioning the circle
The rendering context
Drawing the final shape
Further Improvements
Blindman67's method using p5 vectors
Dave's acceleration method
Dave's Bezier curve method
Introduction: An algorithm for rounded corners
If you spent some time doing creative coding, you'll very quickly come to the realization that anything, which has a shape that is little more complicated than your average rectangle or circle, quickly starts requiring a fair amount of code to be summoned onto your canvas.
This post/tutorial was initially sparked by a snippet of code that I've found on stackoverflow. In the past months, this algorithm has unlocked a number shapes and sketches for me, that I otherwise wouldn't have been able to make. In essence, the algorithm allows you to create arbitrarily shaped polygons, with any number of vertices while at the same time being able to control the roundness (or curvature) of each vertex.
I came across this code when I attempted a sketch in which I wanted to round off the corners of some triangles:
In p5js there is no out-of-the-box method for doing so, except when it comes to rectangular shapes, where the 5th to 8th parameters can be used for that purpose. You could technically do it with a series of curveVertex() calls, however that strategy doesn't offer much control. Here's another sketch that I made during #GENUARY2022, which also makes use of the same algorithm, even if it might not be as obvious:
You can notice that you can even mix between pointy corners and round corners, which is another thing that isn't possible with pure p5js (without writing a lot of additional code towards that end). This definitely came in clutch this Genuary. The remainder of this post will walk you through this really cool and useful algorithm, step by step.
Credit where credit's due
The code discussed in this post stems from this stackoverflow answer by stackoverflow user Blindman67. (Go give him some upvotes on his answer if you find this useful!) Blindman67 explains his strategy succinctly, and is more than sufficient for you to start using his code.
But for the sake of REALLY understanding the algorithm, we'll recreate it from scratch based off of his explanation and analyzing his code! There are a number of noteworthy things going on, that are worth inspecting in more detail, and can be generalized to a number of other scenarios.
An intuitive explanation of the algorithm
We'll begin with an intuitive depiction of what the algorithm is accomplishing. The idea behind what we're trying to do is generally simple: We're going to take an imaginary circle of a certain radius, and push it as far as possible into a corner (any kind of corner, not necessarily 90 degrees) that we're trying to round. Next, we erase the lines that form the pointy corner, starting from the place where they are tangential to the imaginary circle, and close the shape again by tracing the outward facing portion of our circle.
And voila, we have obtained a rounded corner with a specific radius. Seems straightforward, right? However, to accomplish this there are a number of steps that we need to follow.
Breakdown of the algorithm
Given a specific radius, the difficulty lies within finding where to exactly position the circle that rounds the corner to that radius. If we can exactly pinpoint where this circle is positioned, we can also determine where it is tangential to the lines that form the pointy corner. Those two points will determine the start and end of the arc that will form the rounded corner. Overall, given three points A, B and C that form a corner, as well as a given radius, the steps are:
Converting 3 points into 2 vectors
Calculating the angle between those two vectors
Calculating and locating the half-angle between the two vectors
Finding the distance from the circle center to the corner
Drawing the arc
Two Vectors from three Points
First we'll need to find the value of the angle that is formed by three points. To do so we first need to turn these 3 points into two vectors. For three points A, B and C we can find the vectors BA and BC by using the following function:
// p1 -> first point // p2 -> second point // v -> container that we will fill // we could also not pass v and simply return it function toVec(p1, p2, v) { // center the vector based on the coordinates of p2 (the point B in this case) v.x = p2.x - p1.x; v.y = p2.y - p1.y; // compute magnitude (length) of the vector v.len = Math.sqrt(v.x v.x + v.y v.y); // alternatively and more concisely v.len = Math.hypot(v.x, v.y); // normalized coordinates v.nx = v.x / v.len; v.ny = v.y / v.len; // calculate the angle v.ang = Math.atan2(v.ny, v.nx); }
The comments should be sufficient to understand what is happening, we also use the atan2() function to calculate the angle of the vectors with respect to the zero angle. For simplicity's sake we'll make the origin at the center of the canvas. To fill our vector containers, we can do as such:
function setup() { w = min(windowWidth, windowHeight) createCanvas(w, w); } function draw() { background(220); // make origin at center of canvas translate(w/2,w/2) p1 = {x: 100, y: 0} p2 = {x: 0, y: 0} p3 = {x: 0, y: -100} strokeWeight(5) point(p1.x,p1.y) point(p2.x,p2.y) point(p3.x,p3.y) v1 = {} v2 = {} toVec(p1,p2,v1) toVec(p3,p2,v2) noLoop() } // p1 -> first point // p2 -> second point // v -> container that we will fill function toVec(p1, p2, v) { v.x = p2.x - p1.x; v.y = p2.y - p1.y; v.len = Math.hypot(v.x, v.y); v.nx = v.x / v.len; v.ny = v.y / v.len; v.ang = Math.atan2(v.ny, v.nx); }
You might ask, why we're not doing this with p5js's inbuilt vector class? And fret not, at the end of this post well go over an alternative version where we use the class with all it's functionalities. Now that we have converted our 3 points into vector form, we can tackle computing and locating the intermediary angle between the two vectors!
Calculating the angle using the cross product
Explanation
Getting the halfway angle is probably the trickiest part of the procedure, and requires a number of steps. There probably are multiple ways to do so (more on that later), but for starters we'll begin with the method deduced from Blindman67's code. It will require us to freshen up on our linear algebra a little bit, specifically on the cross product of two vectors. We'll be heavily exploiting the formula of the cross product to compute the value of the formed angle, as well as where it is located with respect to the two vectors.
What the cross product actually represents algebraically, is a bit outside of the scope of this post. A perfect resource for understanding it can be found in the form of this video by 3Blue1Brown (I don't think there's a need for me to introduce his channel).
The important part here is how the cross product can help us find the angle between two vectors, and what problems arise from this method. Given two vectors, we observe two angles, one that is less than 180 degrees, and another 'reflex' angle larger than 180 degrees. Our task is thus finding the angle that allows us to draw an arc, such that we can trace it and round off the pointy corner. We can only round off a corner if we select the angle that has a value less than 180 degrees.
In this manner, we also need to determine the correct drawing order. We're going around the polygonal shape in a clockwise manner, but depending if the angle is concave or convex (curved inwards vs being curved outwards), we'll have to sometimes draw our arcs in a counterclockwise manner. To this end, calculating the value of the angle is not sufficient, we need an additional indicator that tells us how the second vector BC is situated with respect to the first one BA. For this we can use the cross product of BC and the perpendicular line to BA. More on that in a bit.
The Math
Generally the formula for finding the cross product, is the product of the magnitudes of our two vectors, with the sine of the angle that they form. More concretely:
( \Vert BA \Vert \Vert BC \Vert sin( \Theta ) )
This means that, finding the cross product requires us to already have the angle... who's value we're trying to find. Now, if we were to somehow already have the numerical value of the cross product, we could solve for ( sin( \Theta ) ), since we also already have the magnitudes of the two vectors concerned ( \Vert BA \Vert ) and ( \Vert BC \Vert ).
Luckily, there is another way to find the numerical value of the cross product! It is actually equal to the determinant of the 2x2 matrix formed by these vectors. This means computing this determinant will allow us to find the angle by solving the previous formula! Computing the determinant of a matrix is done as follows:
( v_{1}x v_{2}y - v_{2}x v_{1}y )
Then the value of the angle can be calculated as follows:
( \Theta = sin^{-1}(\frac{v_{1}x v_{2}y - v_{2}x v_{1}y}{\Vert BA \Vert \Vert BC \Vert}) )
If you're not familiar with the inverse sine function, it simply does the reverse operation that a regular sine function does. So for example, a sine function will accept a value between -PI/2 and PI/2, and return a value that ranges between -1 and 1. The inverse function will accept values between -1 and 1 and return an angle that ranges between -PI/2 and PI/2.
This formula can be further simplified! Since our vectors are already normalized ( \Vert BA \Vert \Vert BC \Vert ), they will simply evaluate to 1, leaving us with:
( \Theta = sin^{-1}(det) )
Which is thus simply the inverse sine of the determinant. Now let's implement this.
The Code
The code for all of what we have discussed in the previous section is relatively... tame, and can essentially be summarised in a single line! Let's have a look at Blindman67's code:
// compute and store our 2 vectors toVec(p2, p1, v1); toVec(p2, p3, v2); // Cross product of the two vectors - what we discussed in the previous section sinA = v1.nx v2.ny - v1.ny v2.nx; // Cross product of v2 and the line at 90 degrees to v1 // This is necessary and will help determine where to exactly position the center of the circle sinA90 = v1.nx v2.nx - v1.ny -v2.ny; // Clamping the value of sinA to avoid NaNs angle = Math.asin(sinA < -1 ? -1 : sinA > 1 ? 1 : sinA);
We already discussed at length the first calculation that computes the variable sinA. The second line computes the cross product of the perpendicular line to BA and the vector BC. In actuality, we don't need to create a separate vector for this perpendicular, we can simply exploit the fact that the dot product of two vectors is zero when they're perpendicular. A vector perpendicular to BA would be simply one that has it's x and y values flipped, in addition to having one of their signs flipped. For example:
zeroDotProduct = v1.nx(-v1.ny) + v1.nyv1.nx
Of course it also matters which coordinate's sign we flip, because it determines which perpendicular we're selecting (the one at 90 degrees or the other one at -90 degrees). If we flip the other sign, we'll would have to proceed differently in the calculations that follow.
Another tricky part here, I found to be the nested ternary check inside the inverse sine function. By simply expanding it, we visually get some clarity:
if(sinA1){ angle = Math.asin(1) }else{ angle = Math.asin(sinA) }
Go ahead a try putting values lesser than -1 or greater than 1 into the inverse sine function. It'll yield a NaN value. To avoid such thing, we simply clamp the value and can be done compactly as a one liner. And if you noticed, in actuality, we have already obtained the value of the angle!
Ambiguity of the cross product
There's a couple more steps that we have to do to go through to get the correct half angle between our two vectors. We'll start with this if/else flurry to determine the orientation of our angle:
radDirection = 1; drawDirection = false; if (sinA90 < 0) { if (angle < 0) { angle = Math.PI + angle; } else { angle = Math.PI - angle; radDirection = -1; drawDirection = true; } } else { if (angle > 0) { radDirection = -1; drawDirection = true; } }
In essence, the information that we're trying to find here, is where exactly the center of the circle is located (the angle or it's reflex) as well as the correct drawing order of the arc (clockwise or counter clockwise). This is what the above if/else block takes care of. Let's for a second assume that the angle that we have calculated is sufficient, and simply halve it to find the bisector:
Run the snippet a couple of times and take note of where the red line falls, it only gets it right when the angle is acute. We do not get the correct bisector when the angle is obtuse.
Hence, we need another indicator to be able to determine where to position the center of the circle. For this purpose we can use the cross product of the angle formed by the perpendicular of BA and the vector BC that we previously calculated. Simultaneously observing where that perpendicular falls with respect to BC, by inspecting the sign of these two cross products we can pin point where the bisector is located. Let's examine the different scenarios that arise:
When sinA90 < 0 and sinA > 0
When sinA90 is larger than 0, and sinA is less than 0, then we can conclude that the perpendicular lies in between BA and BC, and the vectors BA and BC form an obtuse fan. In this case we need to add PI to the angle before halving it to obtain the correct half angle.
When sinA90 < 0 and sinA > 0
In this case the angle formed is also an obtuse fan, but the perpendicular does not lie in between BA and BC. In this case we need to subtract the angle from PI and also invert the drawing order (this is what radDirection and drawDirection are used for).
When sinA90 > 0
In this case the angle formed is an acute wedge, and we need to distinguish the two cases where the vector BC lies in between BA and it's perpendicular aka sinA < 0 and when it's not aka sinA > 0, and in the latter case we need to invert the drawing order.
If that explanation wasn't enough mental gymnastics, then feel welcome to sketch out a couple of angles and work them out by yourself. Important to keep in mind that if we were to select the other perpendicular, we then would have to flip some of the checks.
Example
Here's an example of this in action. Observe the position of BC, BA and it's perpendicular as well as the values of sinA and sinA90. The assumed drawing order is clockwise starting from BA, going to BC:
If you've followed until here, then congrats, the hardest part is past us!
Simpler way to calculating the angle?
Now these are a lot of hoops to go through for simply calculating. If you want to have a simpler way of finding that angle you can use:
// vectors should be normalized angle = atan2(v2.y, v2.x) - atan2(v1.y, v1.x) if (angle < 0) { angle += 2 PI;}
However, this isn't sufficient to figuring out the correct drawing order of our arc. We'll go over a simpler way to our strategy later in this post.
A visual example
To conclude this section, here's a visual example of why this is necessary. If we were to ignore the correct orientation and drawing order of the arcs we would end up with weird behaviour:
Positioning the circle
The next step is finding the distance from the circlecenter to the coordinate of the corner:
lenOut = abs(cos(halfAngle) radius / sin(halfAngle));
Here, instead of finding the distance of the circle center F to corner point B, we're directly finding the distance from the corner point B to the tangential points E and F. For this we're making use of the formulas that can be applied for finding the length of the hypotenuse, which is designated by BF. We then have ( BF cos(\Theta) = BE ) and ( BF sin( \Theta) = FE ).
Using these two formulas we find the formula that Blindman67 used:
( \frac{BE}{cos(\Theta)} = \frac{r}{sin(\theta)} )
And here we also have the value of the radius (that we specified):
( BE = abs( \frac{cos(\Theta) r}{sin(\theta)}) )
One problem here is that the radius that we specified as input, might not fit into the corner. Since it could techincally be longer than the edge itself. This can be remedied with the following check:
if (lenOut > min(v1.len / 2, v2.len / 2)) { lenOut = min(v1.len / 2, v2.len / 2); cRadius = abs(lenOut sin(halfAngle) / cos(halfAngle)); } else { cRadius = radius; }
Once this is dealt with, we can calculate the coordinates of one of the tangential points to the circle:
x = p2.x + v2.nx lenOut; y = p2.y + v2.ny lenOut;
Then move along the perpendicular to find the circle center:
x += -v2.ny cRadius radDirection; y += v2.nx cRadius radDirection;
Note that this also depends on the direction that we determined earlier.
The rendering context
For drawing purposes we will make use of the javascript canvas rendering context interface, which allows us to draw all sorts of shapes. You can find an extensive documentation here.
To do so we need to invoke the context, usually I do this in the setup function, as such:
function setup(){ ctx = canvas.getContext('2d'); }
And then we can make use of it to draw stuff:
For our purposes we'll only need to use one function, namely the ctx.arc() call:
ctx.arc(x, y, cRadius, v1.ang + Math.PI / 2 radDirection, v2.ang - Math.PI / 2 radDirection, drawDirection);
Where the first two parameters are the coordinates of the arc center, followed by the radius in third place, and the starting and end angle in fourth and fifth position. the last input specifies if this arc is drawn in clockwise or counter clockwise manner.
Drawing the final shape
The final shape will be drawn by looping through the set of points that make up the vertices of our polygon. Here's a condensed version of this loop with the calculations omitted:
ctx.beginPath(); len = points.length; // initial point is the last point of the shape p1 = points[len - 1]; for (i = 0; i < len; i++) { p2 = points[(i) % len]; p3 = points[(i + 1) % len]; // Calculations go here // get the next set of points p1 = p2; p2 = p3; } ctx.closePath(); ctx.stroke() // add an outline to the shape ctx.fill() // add color to the shape
In this manner, the first point is actually the last vertex of the polygon. The final shape would then be drawn by first positioning all of your vertices, and storing them in an array, in the following manner:
vertices = [] vertices.push({x: coordX, y:coordY}) // add more vertices etc...
And then passing the array to the drawing function:
drawRoundPoly(ctx, vertices, radius)
The final snippet of code that you would then use, is the one provided by Blindman67. Here's a minimal example:
Also notice that you can additionally pass a 'radius' to each individual vertex to override the main radius passed to the function call.
Further Improvements
First off, huge thanks to Dave and Clay from the birbsnest for the feedback! Thanks to Dave (who's a freaking coding wizard), you're getting three additional sections here, firstly we'll update the method that we've discussed so far using the inbuilt p5 vector class and it's functionalities, and secondly we'll have a look at the solution Dave came up with for rounding off corners. He also shared a version using Bezier Curves!
Blindman67's method using p5 vectors
Rather than defining our own asVec() function, we can make use of the p5 vector class and the functions that come along with it. In that manner we declare our vector BA and BC like this:
p2 = points[i % len]; p3 = points[(i + 1) % len]; A = createVector(p1.x, p1.y); B = createVector(p2.x, p2.y); C = createVector(p3.x, p3.y); BA = A.sub(B); BC = C.sub(B);
Next we can compute the angle and the angle of the perpendicular in the following manner:
// need to call copy() because most vector functions are in-place BAnorm = BA.copy().normalize(); BCnorm = BC.copy().normalize(); sinA = -BAnorm.dot(BCnorm.copy().rotate(PI / 2)); sinA90 = BAnorm.dot(BCnorm); angle = asin(sinA);
Note that we have to call the copy() function prior to functions such as normalize() and rotate(), as they will mutate and alter the vector itself. Hence we create two normalized copies of BA and BC, since we'll need to use their magnitudes later on again.
Also note how we're now calculating the dot product rather than the cross product! It's still the same calculation, but for convenience it's easier to do it in that manner. This is derived from the fact that the dot product of two vectors is equal to the cross product of one of these vectors and the perpendicular to the other vector. So we simply rotate one of them by 90 degress and compute the dot product. Same for sinA90.
The rest of the code is identical to what we had before, with the exception that we're solely using the p5 vector properties and functions:
This makes it a little more compact, but also less readable.
Dave's acceleration method
After asking for feedback on Raph's discord and discussing the previous method a little, Dave pointed out that an alternative way to knowing the drawing order would be using the curvature vector's direction. Rather than using the positions of BA and it's perpepndicular with respect to BC, and doing this check:
(radDirection = 1), (drawDirection = false); if (sinA90 < 0) { angle < 0 ? (angle += PI) : ((radDirection = -1), (drawDirection = true)); } else { angle > 0 ? ((radDirection = -1), (drawDirection = true)) : 0; }
We can do this:
accelDir = BAnorm.copy().add(BCnorm) radDirection = Math.sign(accelDir.dot(BCnorm.rotate(PI / 2))) drawDirection = radDirection === -1
We obtain the curvature vector by adding BA and BC, and this vector will generally always point into the direction of the circlecenter. Now we only need to determine if we need to draw the arc in a clockwise or counterclockwise direction. For this we recycle our previous check and observe where the curvature vector falls with respect to BC. The drawing order can then be deduced from the arc direction!
Here's a sketch where we visualize these curvature vectors:
Using Bezier vertices
And for the final alternative method, Dave wrote this method that relies on the bezierVertex() function in p5:
And that's a wrap! If there are any unclear notions, mistakes or questions, feel free to leave me a comment or reach out to me on social media! If you find this post useful, consider supporting me by sharing it or following me on my social medias. Otherwise, cheers and happy sketching! |
8841 | https://www.youtube.com/watch?v=dtMnHcaa-H8 | Fraction Decimal Percent Conversion with Desmos
David Petro
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Posted: 25 Jan 2021
In this activity, lead by the teacher, students can practice converting to and from fractions, decimals and percents using both number lines and numerical methods. For the actual activity, follow this link
Transcript:
so in this activity students are going to be able to demonstrate converting between percents decimals and fractions using number lines and using numerical values this doesn't work in the same way a normal desmos activity works in that it really depends on what you tell the students on exactly what they're going to do so let's take a look at what that looks like so i've got a fake classroom here set up and a few fake students in here and we're going to start on the first page uh the first page is converting fractions or decimals to percents so the way you would do this is is ideally because every one of these sort of sections is a different type of conversion so you're definitely going to be using teacher pacing and so i'm going to because i want to just worry about this single conversion right now i'm going to pace for that single slide and so students on their screen they are stuck so they they are stuck to that that only that slide and this is what they see and so they are basically waiting for your instructions they want to know what you want to convert to a percent and so in this case you can convert a fraction or a decimal to a percent and so the first thing you do is you tell them what you want to convert to a percent so i might say convert one-fifth for example to a and so we have them type the one-fifth in here and that's just a placeholder it's just sort of to keep track what uh you as the teacher told them to convert to a percent and then we asked them to slide their slider over to pick the individual value where they think uh one-fifth is as a percent and then and in the teacher dashboard you're going to see everybody else's and we want to wait until everybody in the class has a chance to put their point in their proper position so we can get some data and you may want to take a snapshot of anybody that is maybe way off from where you think it is like this student so i can take a snapshot of that and then take a look at that later now we don't want students to change the guess and so i'm going to use the pacing feature here to move everybody over to the next slide and when we go to the next slide what students see is they see themselves in red and they can see everybody else in blue as to what they picked um they are reminded of what they were supposed to convert to a percent and then they can actually see where that value actually is then we ask them to actually explain what they did so as students put things in what you might want to do is if you see something that you like then you can actually take a snapshot of that or something that you want to talk about maybe that is interesting or as an interesting mistake you might want to take a snapshot of that to look at later even to talk about at that time so the thing about this now is we want to con we probably want to continue doing this and so at this point you're going to move everybody back to the first screen and you're going to say okay now i want you to convert the fraction three quarters for example into a percent and so when they do that they again have to use their slider and they choose where they think three quarters is and and again in the teacher dashboard you're going to look for anything interesting to see if there's anybody that's way off so for example i can see this one's way off and so i'm going to maybe look at that either and put it in anonymize mode so i can talk about it at this point and i can look bring this up and show students or i'll look at it later because i put them in the snapshots for later and then uh just as we did before we're gonna once everybody's got their their info in we're going to move everybody's to the next slide and then student c again where everybody put their estimate they see themselves they can show where the actual value is and then they can type in what they've done now notice this is what was typed in before because we're reusing the same page over and over again and so that's why taking the snapshots on the fly is really important because students are going to be rewriting what they've they've already written in so this right now is what the student had for the previous number that we had in there and the way this is broken up is is broken up into sets of two slides so in this these first ones we're converting either fractions or decimals to percents in the next two we're converting fractions or decimals to percents and this is now numerical so in this case uh let's move our guys over there so what we it's the same idea we're asking them to convert and let's say a decimal at this time so let's convert the decimal 0.45 to a percent so we want them to type that value in here what we want them to convert and then instead of moving a slider we want them to type in the actual percent so in this case and they type it in without the percent sign and when they submit that it shows up on the screen and so as student put those in we'll see those in the teacher dashboard as well so let's look at the teacher dashboard we see what they've typed in in the first response and in the second response below and we see their screens up here so in the same way we can take some snapshots if we want to look at those further so the second one is converting using a numerical value as opposed to a slider so perhaps we're expecting something more exact and so in like i said they're in these pairs and the first two pairs are about converting uh fractions or decimals to percents the first pair is going to be estimation and the second pair is going to be numerical and then we're converting percents to decimals and we start with estimation for that um that's in the first pair of converting percents of decimals and then we can do it numerically where they have to type the value in and we can go through the various scenarios converting percents to decimals or converting percents to fractions and in each case whether we're talking about converting percents to fractions or converting decimals to fractions and all the way down to converting fractions to decimals there is an estimation page where they use the slider and there is a numerical page where they type the value in and in each case you can reuse these pages as part of the lesson to keep trying different conversions and students just keep reusing that and as a teacher you keep moving them to the individual page that you want them to be on by using the pacing feature so you're they're only ever going to be on one page and you're going to probably be moving back and forth from the the question page to the reveal page and then changing perhaps to a different conversion if you want to go there |
8842 | https://fsw01.bcc.cuny.edu/jorge.pineiro/Lecture%2020%20Permutations%20and%20Combinations.pdf | Based on K. H. Rosen: Discrete Mathematics and its Applications.
Lecture 20: Permutations and Combinations. Section 6.3 1 Permutations and Combinations 1.1 Permutations A permutation of a set of distinct objects is an ordered arrangement of these objects.
We also are interested in ordered arrangements of some of the elements of a set. An ordered arrangement of r elements of a set is called an r-permutation.
Example 1. In how many ways can we select three students from a group of five students to stand in line for a picture? In how many ways can we arrange all five of these students in a line for a picture?
Ans: First, note that the order in which we select the students matters. There are five ways to select the first student to stand at the start of the line. Once this student has been selected, there are four ways to select the second student in the line. After the first and second students have been selected, there are three ways to select the third student in the line. By the product rule, there are 5 · 4 · 3 = 60 ways to select three students from a group of five students to stand in line for a picture.
To arrange all five students in a line for a picture, we select the first student in five ways, the second in four ways, the third in three ways, the fourth in two ways, and the fifth in one way. Consequently, there are 5 · 4 · 3 · 2 · 1 = 120 ways to arrange all five students in a line for a picture.
Theorem 2. If n is a positive integer and r is an integer in the range 0 ≤r ≤n, then there are P(n, r) = n(n −1)(n −2) · · · (n −r + 1) r-permutations of a set with n distinct elements.
Corollary 3. If n and r are integers with 0 ≤r ≤n, then P(n, r) = n!
(n −r)!.
1.2 Combinations We now turn our attention to counting unordered selections of objects.
Example 4. How many different committees of three students can be formed from a group of four students.
Ans: To answer this question, we need only find the number of subsets with three elements from the set containing the four students. We see that there are four such 1 subsets, one for each of the four students, because choosing three students is the same as choosing one of the four students to leave out of the group. This means that there are four ways to choose the three students for the committee, where the order in which these students are chosen does not matter.
Definition 5. An r-combination of elements of a set is an unordered selection of r elements from the set. Thus, an r-combination is simply a subset of the set with r elements.
Theorem 6. The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer in the range 0 ≤r ≤n, equals C(n, r) = n!
r!(n −r)!.
2 |
8843 | https://radiopaedia.org/articles/parathyroid-adenoma?lang=us | Parathyroid adenoma
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Citation:
Weerakkody Y, Atkinson H, Knipe H, et al. Parathyroid adenoma. Reference article, Radiopaedia.org (Accessed on 16 Sep 2025)
DOI:
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rID:
23949
Article created:
19 Jul 2013,
Yuranga Weerakkody
At the time the article was created Yuranga Weerakkody had no recorded disclosures.
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Last revised:
16 Jul 2025,
Henry Knipe
Disclosures: At the time the article was last revised Henry Knipe had the following disclosures:
Micro-X Ltd, Shareholder (past)
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not be relevant to the changes that were made.
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Systems:
Head & Neck
Tags:
nuclear medicine, endocrine, ultrasound, rg_40_5_edit
Synonyms:
Parathyroid adenomas
Parathyroid adenomata
Adenoma of parathyroid gland
Adenomas of parathyroid gland
Adenoma of parathyroid glands
Parathyroid adenomas are benign tumors of the parathyroid glands and are the most common cause of primary hyperparathyroidism. Imaging is essential for preoperative localization, helping identify ectopic or multiglandular disease and facilitating minimally invasive surgery.
On this page:
Article:
Terminology
Epidemiology
Clinical presentation
Pathology
Radiographic features
Treatment and prognosis
Differential diagnosis
Practical points
Related articles
References
Images:
Cases and figures
Terminology
Primary hyperparathyroidism was traditionally classified as being due to either a parathyroid adenoma (typically solitary) or parathyroid hyperplasia (involving multiple glands). However, this distinction, based primarily on the number of enlarged glands, has proven unreliable on histology, especially when multiple glands are involved.
The WHO classification of endocrine and neuroendocrine tumors discourages the use of hyperplasia for primary hyperparathyroidism and instead recommends the term multiglandular parathyroid disease 36,37. The term hyperplasia is now reserved for secondary and tertiary hyperparathyroidism 37.
Atypical parathyroid adenoma is no longer recommended in the WHO Classification, and has been renamed as a separate clinical entity, atypical parathyroid tumor.
Epidemiology
Parathyroid adenomas are most common in women over 50 years and are usually sporadic 34.
Associations
Prior neck irradiation is a possible but weak risk factor. A minority are associated with hereditary syndromes, including MEN1 (more often multiglandular parathyroid disease than adenoma) and MEN4.
Clinical presentation
Many patients are diagnosed asymptomatically through biochemical screening. Symptomatic cases reflect primary hyperparathyroidism and the effects of elevated calcium and parathyroid hormone, including bone pain, fractures, renal calculi, constipation, pancreatitis, and neurocognitive symptoms (e.g. fatigue or poor concentration) 34,35.
Symptom severity does not always correlate with calcium levels, and normocalcaemic variants may occur 34. Rarely, parathyroid adenomas present with spontaneous hemorrhage, which may be large and cause airway compromise 21.
Pathology
Parathyroid adenomas are usually composed of chief cells, though oxyphil and mixed types occur. Most are solitary (up to 87% 2), well-circumscribed, and oval or bean-shaped; larger lesions may be multilobulated.
Location
Most parathyroid adenomas are found immediately posterior or inferior to the thyroid. Superior gland adenomas lie in the tracheo-esophageal or paraesophageal space, while inferior gland adenomas have more variable positions due to longer embryologic descent 12,35.
Supernumerary glands (more than four) occur in up to 13% of individuals and may contribute to persistent or recurrent hyperparathyroidism if unrecognised 34.
Ectopic adenomas account for up to 5% of cases, with locations including 1,12:
mediastinum (usually superior mediastinum)
carotid sheath
retropharyngeal space
intrathyroidal
submandibular region (less common 34)
Variants
parathyroid lipoadenoma 16
Markers
Parathyroid hormone (PTH) is typically elevated (normal range 10–55 pg/mL), often with raised serum calcium. However, even a “normal” PTH may be inappropriately elevated if calcium is high - values above ~25 pg/mL are generally considered non-suppressed in this context 45.
Adenoma size may correlate with PTH level, but biochemistry alone cannot distinguish adenomas from multiglandular disease or parathyroid carcinoma. PTH levels in carcinoma are, however, often elevated 5-10 times above the upper limit of normal (e.g. >300 pg/mL) 46.
Normocalcaemic primary hyperparathyroidism can also occur, and urinary calcium excretion may be assessed to support the diagnosis.
Radiographic features
Ultrasound
Ultrasound is a common first-line modality for parathyroid adenoma localization. Eutopic adenomas are often well seen if greater than 1 cm and typically lie adjacent to the posterior or inferior thyroid margin. Sensitivity is lower with smaller adenomas, ectopic position, adjacent thyroid nodules, or prior neck surgery.
Greyscale
Typical adenoma features include:
oval, well-defined nodule, 0.5-3 cm (most 1-1.5 cm)
homogeneously hypoechoic relative to thyroid
thin echogenic interface separating it from the thyroid
larger lesions may appear lobulated or partly cystic 34
Atypical features include 27:
cystic degeneration (ranging from minor to >50% cyst)
coarse calcification (~1.5%)
Doppler
internal or central vascularity (~85–97% in primary disease) 29-31
polar feeding vessel at one pole (usually from inferior thyroid artery) 1,6
low resistive index, correlating with elevated PTH 6,32
arc or rim vascularity pattern in some cases 29,30
Smaller adenomas and secondary hyperplasia are less likely to show these features 31. The polar vessel can help distinguish adenomas from lymph nodes (hilum vessel) and thyroid nodules (no polar vessel) 32,33.
Additional techniques
shear-wave elastography may show lower stiffness than thyroid 38
microvascular Doppler techniques (e.g. SMI or MVFI) can improve detection of slow flow and small feeders 39
contrast-enhanced ultrasound may show early peripheral enhancement followed by central washout 40
CT
4D parathyroid CT plays a key role in preoperative localization of parathyroid adenomas. It offers higher sensitivity than ultrasound or sestamibi, particularly in reoperative cases or when adenomas are ectopic or small 41,42.
See the separate article on 4D parathyroid CT for protocol details and interpretative approach.
Multiple phase-contrast categories comparing the attenuation of the candidate lesion to the thyroid have been described on 4D CT 12:
all types: lower attenuation than the thyroid on non-contrast due to low iodine content
type A: higher attenuation than thyroid on arterial phase regardless regardless of attenuation on delayed phase
type B (most common): lower attenuation than thyroid on arterial phase and lower attenuation than thyroid on delayed phase
type C: lower attenuation than thyroid on arterial phase and higher attenuation than thyroid on delayed phase
Other supportive morphologic features include 14:
oval or lobulated nodule adjacent to thyroid
polar feeding vessel (usually from inferior thyroid artery)
larger size (>13 mm) favoring single adenoma
Pitfalls of 4D CT include:
cystic adenomas or parathyroid cysts with atypical enhancement
intrathyroidal parathyroid glands mimicking thyroid nodules
lymph nodes mistaken for adenomas
progressive enhancement rather than washout
enhancing veins or small vessels seen in cross-section
small lesions (<7 mm) or multiglandular disease
Newer CT approaches include low-dose protocols, dual-energy CT, radiomics models, and hybrid imaging.
MRI
MRI is not commonly used as a first-line modality due to lower spatial resolution and vulnerability to motion artefact, particularly at the thoracic inlet. It can be useful in selected cases when other imaging is inconclusive or contraindicated, such as localization of ectopic glands or in reoperative settings.
Parathyroid adenoma signal characteristics include 6:
T1
typically intermediate to low signal
high signal may occur with subacute hemorrhage
low signal may reflect fibrosis or chronic hemorrhage
T2
typically hyperintense
low T2 signal may be seen with fibrosis or old hemorrhage
marked T2 hyperintensity may correlate with elevated PTH levels 35
T1 C+ (Gd)
usually demonstrate avid enhancement
DWI/ADC
adenomas usually show higher ADC values (~1.5–2.0 × 10⁻³ mm²/s) than thyroid or lymph nodes
It is reported that, since most adenomas demonstrate high T2 signal intensity, the addition of contrast for MRI does not significantly increase detection rates.
Dynamic contrast-enhanced MRI (4D MRI) may show enhancement patterns similar to 4D CT, with reported sensitivity up to 80-85%, although spatial resolution remains lower 34.
Nuclear medicine
Scintigraphy
Tc-99m sestamibi is a lipophilic radiotracer that accumulates in tissues with high mitochondrial activity, including parathyroid adenomas. It is commonly used for localization, particularly in ectopic glands or locations less accessible to ultrasound. Tc-99m tetrofosmin is used as an alternative to Tc-99m sestamibi in some practices.
The specific nuclear medicine protocols used are highly variable, with options including single-tracer dual-phase (washout), dual-isotope scanning and combinations of these approaches. Dual-isotope (tracer) techniques combining thyroid scintigraphy (e.g. using Tc-99m pertechnetate or I-123) and Tc-99m sestamibi are employed to increase specificity in patients with thyroid nodules 34 and distinguish small adenomas from adjacent physiological uptake in the thyroid. Whether Tc-99m pertechnetate or I-123 is used is determined by local availability and practice preference. Visual comparison or quantitative subtraction can be used for interpretation.
SPECT-CT imaging is now standard in most parathyroid scintigraphy protocols, imaging the neck and upper thorax 48. It offers superior anatomical localization and diagnostic imaging compared to 2D planar imaging alone 47.
A commonly used protocol is:
Intravenous administration of Tc-99m sestamibi
10-15 minute planar imaging
2 hour delayed/washout planar imaging with SPECT-CT of the neck and upper thorax
Intravenous administration of Tc-99m pertechnetate
20-30 minute planar imaging
Planar images should include views of the upper chest as well as the lower neck to evaluate for ectopic parathyroid adenomas. Protocol variations include 48:
performing thyroid scintigraphy before administering Tc-99m sestamibi
performing SPECT-CT at an earlier time point e.g. after the 10-15 minute early acquisition
performing SPECT-CT twice for comparison of early and delayed time points
Typical scintigraphic findings on Tc-99m sestamibi imaging include:
early uptake with retention on delayed (2-hour) images
focal tracer activity near the thyroid poles or in ectopic locations e.g. superior mediastinum
Limitations include:
false positives from thyroid nodules or lymph nodes
false negatives in small lesions, rapid washout, or low mitochondrial density (e.g. chief cell–predominant adenomas)
false negatives in mutli-gland disease (even with SPECT-CT)48
PET imaging
F-18 fluorocholine PET-CT is increasingly used, particularly when prior imaging is inconclusive or in reoperative cases 35. Fluorocholine is taken up by cells with elevated phospholipid metabolism, including hyperfunctioning parathyroid tissue. The technique has shown particular value in localizing small or recurrent adenomas, and in cases of negative or discordant conventional imaging, multi-gland disease and/or concurrent thyroid nodules 49.
Features include:
higher sensitivity and spatial resolution than sestamibi 43
early uptake with clear localization
Other PET tracers may be used in select settings:
C-11 methionine: limited availability; niche role 34
F-18 FDG: may show uptake but lacks specificity
Ga-68 DOTATATE: somatostatin receptor targeting; occasionally useful in MEN1 or atypical glands 44
Hybrid 4D PET-CT or PET-MRI may assist in complex or ectopic cases 34.
Treatment and prognosis
Surgical excision is the treatment of choice for parathyroid adenomas and is curative in over 95% of cases 17. Minimally invasive parathyroidectomy is preferred when preoperative localization is successful, with intraoperative PTH monitoring commonly used to confirm adequate resection (typically a >50% drop from baseline) 35.
Current guidelines support surgery not only for symptomatic patients but also selected asymptomatic individuals, including younger patients and those with elevated calcium or reduced bone mineral density 34.
Non-surgical treatments such as ethanol ablation are rarely used and are generally reserved for patients who are not surgical candidates 28,34.
Differential diagnosis
For non-ectopic adenoma consider:
atypical parathyroid tumor
parathyroid carcinoma (rare)
eccentric thyroid nodule: more variable echogenicity, often less vascular, and lacks a polar feeding vessel
lymph node: typically shows a fatty hilum, central (not polar) vascularity, and less enhancement
blood vessel: on 4D CT, small veins can mimic nodules in cross-section
esophagus: may mimic a lesion on ultrasound; look for movement, internal gas or mucosal stripe
In ectopic location also consider:
sequestered thyroid tissue: may appear similar; typically shows higher attenuation on non-contrast CT due to iodine
thymic tissue: located in anterior mediastinum, often triangular or bilobed, with less rapid enhancement
For multiple enlarged parathyroid glands consider:
multiglandular parathyroid disease
secondary parathyroid hyperplasia
Practical points
imaging approach varies by institutional preference
ultrasound is typically first-line; 4D CT or sestamibi added if ultrasound is non-localizing
small adenomas (<1 cm) may be occult on ultrasound
combining two modalities often improves localization and helps detect multiglandular disease
ectopic parathyroid glands are usually only identified on 4D CT or functional imaging
in patients under 50 years, consider a genetic cause (e.g. MEN), which increases the likelihood of multiglandular disease
Quiz questions
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References
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Eurorad teaching files : Case 3589
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Naik M, Khan S, Owusu D et al. Contemporary Multimodality Imaging of Primary Hyperparathyroidism. Radiographics. 2022;42(3):841-60. doi:10.1148/rg.210170 - Pubmed
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Gill AJ, Erickson LA, Perren A, Mete O, Williams MD, Johnson SJ, Juhlin CC. Parathyroid hyperplasia. In: WHO Classification of Tumours Editorial Board. Endocrine and Neuroendocrine Tumours (5th ed.). Lyon (France): International Agency for Research on Cancer. WHO Classification of Tumours online
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Żyłka A, Dobruch-Sobczak K, Piotrzkowska-Wróblewska H et al. 8402 The Usefulness Of The Contrast-Enhanced Ultrasound (CEUS) In Preoperative Localization Of Parathyroid Gland Adenomas. Journal of the Endocrine Society. 2024;8(Supplement_1). doi:10.1210/jendso/bvae163.2130
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Kattar N, Migneron M, Debakey M, Haidari M, Pou A, McCoul E. Advanced Computed Tomographic Localization Techniques for Primary Hyperparathyroidism. JAMA Otolaryngol Head Neck Surg. 2022;148(5):448. doi:10.1001/jamaoto.2022.0271 - Pubmed
Whitman J, Allen I, Bergsland E, Suh I, Hope T. Assessment and Comparison of 18F-Fluorocholine PET and 99mTc-Sestamibi Scans in Identifying Parathyroid Adenomas: A Metaanalysis. J Nucl Med. 2021;62(9):1285-91. doi:10.2967/jnumed.120.257303 - Pubmed
Lastoria S, Marciello F, Faggiano A et al. Role of 68Ga-DOTATATE PET/CT in Patients with Multiple Endocrine Neoplasia Type 1 (MEN1). Endocrine. 2015;52(3):488-94. doi:10.1007/s12020-015-0702-y - Pubmed
Wang R, Abraham P, Fazendin J, Lindeman B, Chen H. Hypercalcemia with a Parathyroid Hormone Level of ≤50 Pg/ML: Is This Primary Hyperparathyroidism? Surgery. 2023;173(1):154-9. doi:10.1016/j.surg.2022.05.043 - Pubmed
McInerney N, Moran T, O'Duffy F. Parathyroid Carcinoma: Current Management and Outcomes – A Systematic Review. Am J Otolaryngol. 2023;44(4):103843. doi:10.1016/j.amjoto.2023.103843 - Pubmed
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8844 | https://www.geeksforgeeks.org/maths/matrix-operations/ | Matrix Operations
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Matrix Operations are basic calculations performed on matrices to solve problems or manipulate their structure. Common operations include:
Addition: Add two matrices of the same size.
Subtraction: Subtract two matrices of the same size.
Scalar Multiplication: Multiply each element of a matrix by a constant.
Matrix Multiplication: Multiply two matrices to create a new matrix.
Transpose: Flip the rows and columns of a matrix.
Inverse: Find the inverse of a Matrix.
In this section, we will explore matrix operations in detail, covering their examples, properties, and visual representations.
Addition of Matrices
Adding matrices is as simple as adding numbers, but there’s one important rule: the matrices must have the same order (i.e., the same number of rows and columns). Once this condition is met, the addition is performed by adding corresponding elements of both matrices to form a new matrix.
Example: Take three matrices A, B, and C of order (2×2), (2×2), and (3×3), respectively. Find the sum of (A + B) and (A + C).
The addition of matrix A and matrix B is found as,
A + B = [2+15+29+76+3]
A + B = [37169]
Matrix A and matrix C can not be added as their order is not the same.
Properties of Matrix Addition
There are various properties associated with matrix addition that are, for matrices A, B, and C of the same order, then,
Commutative Law: A + B = B + A
Associative Law: (A + B) + C = A + (B + C)
Identity of Matrix: A + O = O + A = A, where O is a zero matrix, which is the Additive Identity of the Matrix
Additive Inverse: A + (-A) = O = (-A) + A, where (-A) is obtained by changing the sign of every element of A, which is the additive inverse of the matrix.
Subtraction of Matrices
To subtract two matrices, the matrices must have the same order (i.e., the same number of rows and columns). Subtraction is performed by subtracting the corresponding elements of the second matrix from the first matrix to form a new matrix. This completes the subtraction operation.
Example: For matrices A and B, subtract matrix B from matrix A
A=[2596]B=[1273]
Solution:
Matrix A and Matrix B can be easily subtracted as their order is the same. The subtraction of matrix A and matrix B is found as,
A - B = [2−15−29−76−3]
A - B = [1323]
Scalar Multiplication of Matrices
For any matrix A = [aij]m×n, if we multiply the matrix A by any scalar (say k), then the scalar is multiplied by each element of the matrix, and this is called the scalar multiplication of matrices.
Example: Multiply the matrix A by the scalar value k = 3.
A=[1324]
The scalar multiplication kA is computed by multiplying each element of A by 3:
kA=3×[1324]=[3×13×33×23×4]
So,
kA=[39612]
Properties of Scalar Multiplication
For any matrices A and B of the same order, and λ and μ are any two scalars, then,
λ(A + B) = λA + λB
(λ + μ)A = λA + μA
λ(μA) = (λμA) = μ(λA)
(-λA) = -(λA) = λ(-A)
Multiplication of Matrices
Matrix multiplication is the operation that helps us multiply two matrices. This is different from algebraic multiplication, and not all matrices can be multiplied. Only those matrices can be multiplied where the number of columns in the first is equal to the number of rows in the second, i.e, for matrix Am×n and matrix Bn×p, the multiplication is possible for any other matrices where the column of the first matrix is not equal to the row in the second matrix the multiplication is not possible.
Also, the multiplication of the matrices is not commutative; if matrix A and matrix B are taken, then A×B ≠ B×A.
Properties of Matrix Multiplication
Matrix Multiplication is not commutative in general i.e AB ≠ BA.
Matrix Multiplication is associative i.e (AB)C = A(BC).
Matrix Multiplication is distributive over matrix addition i.e A(B + C) = A.B + A.C.
The product of two matrices can be null matrix while neither of them is null i.e If AB = 0 it is not necessary that A = 0 or B =0.
If AB = AC then B ≠ C ( Cancellation law is not applicable).
There exists a multiplicative identity for every square matrix, such as AI = IA = A.
Transpose Operation of a Matrix
The transpose operation of a matrix rearranges its rows into columns and its columns into rows. For a matrix A of order m×n, denoted as A = [aij]m×n, its transpose is represented by AT and is defined as:
(A)T = [aji]n×m
Example: Find the Transpose of the Matrix A:
A=⎣⎡147258369⎦⎤
Solution:
AT=⎣⎡123456789⎦⎤
Inverse Operation of a Matrix
The inverse of a matrix A exists only if A is a square matrix (i.e., has the same number of rows and columns) and its determinant is non-zero.
A = [ij]n×nand |A| ≠ 0
Specifically, if ∣A∣ = 1, the matrix A is invertible. The inverse of A, denoted as A−1, is a matrix that satisfies:
A × A−1 = I
Where I is the identity matrix of the same order as A. The inverse operation finds this A−1.
Example: Find the inverse of the given matrix.
A=⎣⎡147258369⎦⎤
Solution:
Step 1: Caclulate the determinant of the matrix
The determinant will be -6 ( To learn about determinant, check how to calulate determinant )
Step 2: Calculate Adjoint of the matrix
Adjoint of the above matrix is: A=⎣⎡5−1−36−126−57−3⎦⎤ ( To learn about adjoint, check how to calculate adjoint )
Step 3: The Inverse is calculated by dividing the adjoint by the determinant
so, the inverse matrix is A−1=⎣⎡−65 61 21 −1 2 −1 65 −6721⎦⎤
Also Check
The inverse can also be found using the Gaussian Elemination Method.
Check Other Elementary Operations on Matrices
Solved Examples of Matrix Operations
Example 1: Find the sum of matrix A and B when,
A=[1537]B=[2648]
Solution:
Matrix A and Matrix B can be easily added as their order is the same. The addition of matrix A and matrix B is found as,
A + B = P = [1+25+63+47+8]
P = [311715]
Example 2: Find (A - B) when,
A=[2648]B=[1537]
Solution:
Matrix A and Matrix B can be easily subtracted as their order is the same. The value of (A-B) is found as,
(A - B) = P = [2−16−54−38−7]
P = [1111]
Example 2: Find the transpose of matrix A
A=⎣⎡26104812⎦⎤
Solution:
Transpose of matrix A is the matrix in which the rows of matrix A are its column and the column are its rows.
Tranpose of matrix A is represented as, (A)T
(A)T= [24681012]
Unsolved Practice Questions on Matrix Operations
Question 1: Given the matrices:A=[2−134]andB=[150−2].
Find A + B.
Find A - B.
Question 2: Given the matrix: A=[1340−25]. Find 3A (scalar multiplication by 3).
Question 3: Given the matrices: A=[1023]andB=[4105]. Find A ⨉ B.
Question 4: Given the Matrix: A=[123546]. Find AT.
Question 5: Given the Matrix: A=[4276]. Find the inverse of the matrix A if it exists.
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DŝĐŚĂĞůůůǁŽŽĚ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ dŝĂŚůƉŚŽŶƐŽ͕WƌŽŐƌĂŵDĂŶĂŐĞƌͶƵƌƌŝĐƵůƵŵWƌŽĚƵĐƚŝŽŶ ĂƚƌŝŽŶĂŶĚĞƌƐŽŶ͕WƌŽŐƌĂŵDĂŶĂŐĞƌͶ/ŵƉůĞŵĞŶƚĂƚŝŽŶ^ƵƉƉŽƌƚ ĞĂƵĂŝůĞLJ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ ^ĐŽƚƚĂůĚƌŝĚŐĞ͕>ĞĂĚDĂƚŚĞŵĂƚŝĐŝĂŶĂŶĚ>ĞĂĚƵƌƌŝĐƵůƵŵtƌŝƚĞƌ ŽŶŶŝĞĞƌŐƐƚƌĞƐƐĞƌ͕DĂƚŚƵĚŝƚŽƌ 'ĂŝůƵƌƌŝůů͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ ĞƚŚŚĂŶĐĞ͕^ƚĂƚŝƐƚŝĐŝĂŶ :ŽĂŶŶĞŚŽŝ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ :ŝůůŝŶŝnj͕WƌŽŐƌĂŵŝƌĞĐƚŽƌ >Žƌŝ&ĂŶŶŝŶŐ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ ůůĞŶ&Žƌƚ͕DĂƚŚƵĚŝƚŽƌ <ĂƚŚLJ&ƌŝƚnj͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ 'ůĞŶŶ'ĞďŚĂƌĚ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ <ƌLJƐƚĂ'ŝďďƐ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ tŝŶŶŝĞ'ŝůďĞƌƚ͕>ĞĂĚtƌŝƚĞƌͬ ĚŝƚŽƌ͕'ƌĂĚĞϴ WĂŵ'ŽŽĚŶĞƌ͕DĂƚŚƵĚŝƚŽƌ ĞďďLJ'ƌĂǁŶ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ ŽŶŶŝĞ,Ăƌƚ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ ^ƚĞĨĂŶŝĞ,ĂƐƐĂŶ͕>ĞĂĚtƌŝƚĞƌͬ ĚŝƚŽƌ͕'ƌĂĚĞϴ ^ŚĞƌƌŝ,ĞƌŶĂŶĚĞnj͕DĂƚŚƵĚŝƚŽƌ Žď,ŽůůŝƐƚĞƌ͕DĂƚŚƵĚŝƚŽƌ WĂƚƌŝĐŬ,ŽƉĨĞŶƐƉĞƌŐĞƌ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ ^ƵŶŝů<ŽƐǁĂƚƚĂ͕DĂƚŚĞŵĂƚŝĐŝĂŶ͕'ƌĂĚĞϴ ƌŝĂŶ<Žƚnj͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ ,ĞŶƌLJ<ƌĂŶĞŶĚŽŶŬ͕>ĞĂĚtƌŝƚĞƌͬ ĚŝƚŽƌ͕^ƚĂƚŝƐƚŝĐƐ ŽŶŶŝĞ>ĂƵŐŚůŝŶ͕DĂƚŚƵĚŝƚŽƌ :ĞŶŶŝĨĞƌ>ŽĨƚŝŶ͕WƌŽŐƌĂŵDĂŶĂŐĞƌͶWƌŽĨĞƐƐŝŽŶĂůĞǀĞůŽƉŵĞŶƚ EĞůůDĐŶĞůůLJ͕WƌŽũĞĐƚŝƌĞĐƚŽƌ ĞŶDĐĂƌƚLJ͕DĂƚŚĞŵĂƚŝĐŝĂŶ ^ƚĂĐŝĞDĐůŝŶƚŽĐŬ͕ŽĐƵŵĞŶƚWƌŽĚƵĐƚŝŽŶDĂŶĂŐĞƌ ^ĂŬŝDŝůƚŽŶ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ WŝĂDŽŚƐĞŶ͕ƵƌƌŝĐƵůƵŵtƌŝƚĞƌ :ĞƌƌLJDŽƌĞŶŽ͕^ƚĂƚŝƐƚŝĐŝĂŶ ŶŶEĞƚƚĞƌ͕>ĞĂĚtƌŝƚĞƌͬ ĚŝƚŽƌ͕'ƌĂĚĞƐϲʹϳ ^ĂƌĂŚKLJůĞƌ͕ŽĐƵŵĞŶƚŽŽƌĚŝŶĂƚŽƌ ZŽdžLJWĞĐŬ͕^ƚĂƚŝƐƚŝĐŝĂŶ͕>ĞĂĚtƌŝƚĞƌͬ ĚŝƚŽƌ͕^ƚĂƚŝƐƚŝĐƐ dĞƌƌŝĞWŽĞŚů͕DĂƚŚƵĚŝƚŽƌ <ƌŝƐƚĞŶZŝĞĚĞů͕DĂƚŚƵĚŝƚdĞĂŵ>ĞĂĚ ^ƉĞŶĐĞƌZŽďLJ͕DĂƚŚƵĚŝƚŽƌ <ĂƚŚůĞĞŶ^ĐŚŽůĂŶĚ͕DĂƚŚƵĚŝƚŽƌ ƌŝŬĂ^ŝůǀĂ͕>ĞĂĚtƌŝƚĞƌͬ ĚŝƚŽƌ͕'ƌĂĚĞϲʹϳ ZŽďLJŶ^ŽƌĞŶƐŽŶ͕DĂƚŚƵĚŝƚŽƌ ,ĞƐƚĞƌ^ƵƚƚŽŶ͕ĚǀŝƐŽƌͬZĞǀŝĞǁĞƌ'ƌĂĚĞƐϲʹϳ ^ŚĂŶŶŽŶsŝŶƐŽŶ͕>ĞĂĚtƌŝƚĞƌͬ ĚŝƚŽƌ͕^ƚĂƚŝƐƚŝĐƐ ůůŝƐŽŶtŝƚĐƌĂĨƚ͕DĂƚŚƵĚŝƚŽƌ :ƵůŝĞtŽƌƚŵĂŶŶ͕>ĞĂĚtƌŝƚĞƌͬ ĚŝƚŽƌ͕'ƌĂĚĞϳ ĂǀŝĚtƌŝŐŚƚ͕DĂƚŚĞŵĂƚŝĐŝĂŶ͕>ĞĂĚtƌŝƚĞƌͬ ĚŝƚŽƌ͕'ƌĂĚĞƐϲʹϳ Board of Trustees
LJŶŶĞDƵŶƐŽŶ͕WƌĞƐŝĚĞŶƚĂŶĚ džĞĐƵƚŝǀĞŝƌĞĐƚŽƌŽĨ'ƌĞĂƚDŝŶĚƐ EĞůůDĐŶĞůůLJ͕ŚĂŝƌŵĂŶ͕ŽͲŝƌĞĐƚŽƌ ŵĞƌŝƚƵƐŽĨƚŚĞ'ŽƌĚŽŶ͘ĂŝŶĞŶƚĞƌĨŽƌ^d D>ŝƚĞƌĂĐLJĂƚ>ŽƵŝƐŝĂŶĂ ^ƚĂƚĞhŶŝǀĞƌƐŝƚLJ tŝůůŝĂŵ<ĞůůLJ͕dƌĞĂƐƵƌĞƌ͕ŽͲ&ŽƵŶĚĞƌĂŶĚ KĂƚZĞĞůdž :ĂƐŽŶ'ƌŝĨĨŝƚŚƐ͕^ĞĐƌĞƚĂƌLJ͕ŝƌĞĐƚŽƌŽĨWƌŽŐƌĂŵƐĂƚƚŚĞEĂƚŝŽŶĂůĐĂĚĞŵLJŽĨĚǀĂŶĐĞĚdĞĂĐŚĞƌ ĚƵĐĂƚŝŽŶ WĂƐĐĂů&ŽƌŐŝŽŶĞ͕&ŽƌŵĞƌ džĞĐƵƚŝǀĞŝƌĞĐƚŽƌŽĨƚŚĞĞŶƚĞƌŽŶ<ͲϭϮƐƐĞƐƐŵĞŶƚĂŶĚWĞƌĨŽƌŵĂŶĐĞDĂŶĂŐĞŵĞŶƚ Ăƚ d^ >ŽƌƌĂŝŶĞ'ƌŝĨĨŝƚŚ͕dŝƚůĞ/ZĞĂĚŝŶŐ^ƉĞĐŝĂůŝƐƚĂƚtĞƐƚƵŶĐŽŵďĞ ůĞŵĞŶƚĂƌLJ^ĐŚŽŽůŝŶƐŚĞǀŝůůĞ͕EŽƌƚŚĂƌŽůŝŶĂ ŝůů,ŽŶŝŐ͕WƌĞƐŝĚĞŶƚŽĨƚŚĞŽŶƐŽƌƚŝƵŵŽŶZĞĂĚŝŶŐ džĐĞůůĞŶĐĞ;KZ Ϳ ZŝĐŚĂƌĚ<ĞƐƐůĞƌ͕ džĞĐƵƚŝǀĞĞĂŶŽĨDĂŶŶĞƐŽůůĞŐĞƚŚĞEĞǁ^ĐŚŽŽůĨŽƌDƵƐŝĐ Śŝ<ŝŵ͕&ŽƌŵĞƌ^ƵƉĞƌŝŶƚĞŶĚĞŶƚ͕ZŽƐƐ^ĐŚŽŽůŝƐƚƌŝĐƚ <ĂƌĞŶ>Ğ&ĞǀĞƌ͕ džĞĐƵƚŝǀĞsŝĐĞWƌĞƐŝĚĞŶƚĂŶĚŚŝĞĨĞǀĞůŽƉŵĞŶƚKĨĨŝĐĞƌĂƚŚĂŶĐĞ>ŝŐŚƚĞŚĂǀŝŽƌĂů,ĞĂůƚŚĂŶĚ ĚƵĐĂƚŝŽŶ DĂƌŝĂEĞŝƌĂ͕&ŽƌŵĞƌsŝĐĞWƌĞƐŝĚĞŶƚ͕EĞǁzŽƌŬ^ƚĂƚĞhŶŝƚĞĚdĞĂĐŚĞƌƐ This page intentionally left blank
GRADE ϲͻDKh> ϯ
6G R A D E DĂƚŚĞŵĂƚŝĐƐƵƌƌŝĐƵůƵŵ
DŽĚƵůĞϯ :ZĂƚŝŽŶĂůEƵŵďĞƌƐ
dĂďůĞŽĨŽŶƚĞŶƚƐ ϭ
ZĂƚŝŽŶĂůEƵŵďĞƌƐ
DŽĚƵůĞKǀĞƌǀŝĞǁ ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ Ϯ
A STORY OF RATIOS
1
dŽƉŝĐ͗hŶĚĞƌƐƚĂŶĚŝŶŐWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞ ƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ &͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϬ >ĞƐƐŽŶϭ͗WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶĂŶĚsĂůƵĞ͘͘͘͘͘͘͘͘͘ϭϮ >ĞƐƐŽŶƐϮʹϯ͗ ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϮϬ >ĞƐƐŽŶϰ͗ dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϯϳ >ĞƐƐŽŶϱ͗dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ͛ƐKƉƉŽƐŝƚĞ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϰϱ >ĞƐƐŽŶϲ͗ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚ ĞEƵŵďĞƌ>ŝŶĞ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϱϭ dŽƉŝĐ͗ KƌĚĞƌĂŶĚďƐŽůƵƚĞsĂůƵĞ &͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϲϭ >ĞƐƐŽŶƐϳʹϴ͗KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌ ƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϲϯ >ĞƐƐŽŶϵ͗ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϳϵ >ĞƐƐŽŶϭϬ͗tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶ Ő/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂůEƵŵďĞƌƐ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϴϵ >ĞƐƐŽŶϭϭ͗ďƐŽůƵƚĞsĂůƵĞ ͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϬϭ >ĞƐƐŽŶϭϮ͗dŚĞZĞůĂƚŝŽŶƐ ŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϬϵ >ĞƐƐŽŶϭϯ͗^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϭϴ
DŝĚ ,DŽĚƵůĞƐƐĞƐƐŵĞŶƚĂŶĚZƵďƌŝĐ ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϮϲ
!"#$%&'(')+",-'.'/0&&1&&213)'4'5067'+1),+3'4'5067'+1215$0)$"3'"+'8,+)1+'0##9$%0)$"3&'4'506: dŽƉŝĐ͗ZĂƚŝŽŶĂůEƵŵďĞƌƐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϰϭ >ĞƐƐŽŶϭϰ͗KƌĚĞƌĞĚWĂŝƌƐ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϰϯ >ĞƐƐŽŶϭϱ͗>ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐ ŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ϭϱϬ >ĞƐƐŽŶϭϲ͗^LJŵŵĞƚƌLJ ŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϱϴ >ĞƐƐŽŶϭϳ͗ƌĂǁŝŶŐƚŚĞ ŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ϭϲϲ >ĞƐƐŽŶϭϴ͗ ŝƐƚĂŶĐĞŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϳϱ >ĞƐƐŽŶϭϵ͗Wƌ ŽďůĞŵ^ŽůǀŝŶŐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ ϭϴϬ
ŶĚ ,./, DŽĚƵůĞƐƐĞƐƐŵĞŶƚ &ĂŶĚZƵďƌŝĐ ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘ϭϴϲ
!"#$%&'(')+",-';'/0&&1&&213)'4'5067'+1),+3'4'5067'+1215$0)$"3'"+'8,+)1+'0##9$%0)$"3&'4'506:
ϭ ĂĐŚůĞƐƐŽŶŝƐKE ĚĂLJ͕ĂŶĚ KE ĚĂLJŝƐĐŽŶƐŝĚĞƌĞĚĂϰϱͲŵŝŶƵƚĞƉĞƌŝŽĚ͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘͘
©201 8Great Minds ®. eureka-math.org
6ͻϯ DŽĚƵůĞKǀĞƌǀŝĞǁ
DŽĚƵůĞϯ :ZĂƚŝŽŶĂůEƵŵďĞƌƐ
'ƌĂĚĞϲ ͻ DŽĚƵůĞϯ
ZĂƚŝŽŶĂůEƵŵďĞƌƐ
Ks Zs/ t
A STORY OF RATIOS
2
^ƚƵĚĞŶƚƐĂƌĞĨĂŵŝůŝĂƌǁŝƚŚƚŚĞŶƵŵďĞƌůŝŶĞĂŶĚĚĞƚĞƌŵŝŶŝŶŐƚŚĞůŽĐĂƚŝŽŶŽĨƉŽƐŝƚŝǀĞĨƌĂĐƚŝŽŶƐ͕ĚĞĐŝŵĂůƐ͕ĂŶĚ ǁŚŽůĞŶƵŵďĞƌƐĨƌŽŵƉƌĞǀŝŽƵƐŐƌĂĚĞƐ͘^ƚƵĚĞŶƚƐĞdžƚĞŶĚƚŚĞŶƵŵďĞƌůŝŶĞ;ďŽƚŚŚŽƌŝnjŽŶƚĂůůLJĂŶĚǀĞƌƚŝĐĂůůLJͿŝŶ DŽĚƵůĞϯƚŽŝŶĐůƵĚĞƚŚĞŽƉƉŽƐŝƚĞƐŽĨǁŚŽůĞŶƵŵďĞƌƐ͘dŚĞŶƵŵďĞƌůŝŶĞƐĞƌǀĞƐĂƐĂŵŽĚĞůƚŽƌĞůĂƚĞŝŶƚĞŐĞƌƐ ĂŶĚŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŽƐƚĂƚĞŵĞŶƚƐŽĨŽƌĚĞƌŝŶƌĞĂůͲǁŽƌůĚĐŽŶƚĞdžƚƐ͘/ŶƚŚŝƐŵŽĚ ƵůĞ͛ƐĨŝŶ ĂůƚŽƉŝĐ͕ƚŚĞ ŶƵŵďĞƌůŝŶĞŵŽĚĞůŝƐĞdžƚĞŶĚĞĚƚŽƚǁŽͲĚŝŵĞŶƐŝŽŶƐĂƐƐƚƵĚĞŶƚƐƵƐĞƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽŵŽĚĞůĂŶĚƐŽůǀĞ ƌĞĂůͲǁŽƌůĚƉƌŽďůĞŵƐŝŶǀŽůǀŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘ /ŶdŽƉŝĐ͕ƐƚƵĚĞŶƚƐĞdžƚĞŶĚƚŚĞ ŝƌƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨƚŚĞŽƌĚĞƌŝŶŐŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶŽŶĞĚŝŵĞŶƐŝŽŶ;ŽŶĂ ŶƵŵďĞƌůŝŶĞͿƚŽƚŚĞƚǁŽͲĚŝŵĞŶƐŝŽŶĂůƐƉĂĐĞŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘dŚĞLJĐŽŶƐƚƌƵĐƚƚŚĞƉůĂŶĞ͛ƐǀĞƌƚŝĐĂůĂŶĚ ŚŽƌŝnjŽŶƚĂůĂdžĞƐ͕ĚŝƐĐŽǀĞƌŝŶŐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞĨŽƵƌƋƵĂĚƌĂŶƚƐĂŶĚƚŚĞƐŝŐŶƐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨ ƉŽŝŶƚƐƚŚĂƚůŝĞŝŶĞĂĐŚƋƵĂĚƌĂŶƚ͘^ƚƵĚĞŶƚƐďƵŝůĚƵƉŽŶƚŚĞŝƌĨŽƵŶĚĂƚŝŽŶĂůƵŶĚĞƌƐƚĂŶĚŝŶŐ
Top ic A f oc uses on the dev el op ment of the numb er l ine in the op p osite direc tion ( to the l ef t or b el ow z ero) . Students use p ositiv e integers to l oc ate negativ e integers, understanding that a numb er and its op p osite are on op p osite sides of z ero and that b oth l ie the same distanc e f rom z ero. S tudents rep resent the op p osite of a positiv e numb er as a negativ e numb er and v ic e v ersa. S tudents real iz e that z ero is its ow n op p osite and that the op p osite of the op p osite of a numb er is ac tual l y the numb er itsel f . T hey use p ositiv e and negativ e numb ers to rep resent real -w orl d q uantities, suc h as −50 to rep resent a $50 deb t or 50 to rep resent a $50
dep osit into a sav ings ac c ount. T op ic A c onc l udes w ith students f urthering their understanding of signed numb ers to inc l ude the rational numb ers. S tudents rec ogniz e that f inding the op p osite of any rational numb er is the same as f inding an integer’ s op p osite and that tw o rational numb ers that l ie on the same side of z ero hav e the same sign, w hil e those that l ie on op p osites sides of z ero hav e op p osite signs. In T op ic B , students ap p l y their understanding of a rational numb er’ s p osition on the numb er l ine to order rational numb ers. S tudents understand that w hen using a c onv entional horiz ontal numb er l ine, the numb ers inc rease as they mov e al ong the l ine to the right and dec rease as they mov e to the l ef t. T hey rec ogniz e that if
𝑎𝑎 and 𝑏𝑏 are rational numb ers and 𝑎𝑎 < 𝑏𝑏 , then it must b e true that −𝑎𝑎 > −𝑏𝑏 . S tudents c omp are rational numb ers using ineq ual ity sy mb ol s and w ords to state the rel ationship b etw een tw o or more rational numb ers. T hey desc rib e the rel ationship b etw een rational numb ers in real -w orl d situations and w ith resp ec t to numb ers’ p ositions on the numb er l ine. F or instanc e, students ex p l ain that −10°F is w armer than −11°F
bec ause −10 is to the right ( or ab ov e) −11 on a numb er l ine and w rite −10°F > −11°F . S tudents use the conc ep t of ab sol ute v al ue and its notation to show a numb er’ s distanc e f rom z ero on the numb er l ine and rec ogniz e that op p osite numb ers hav e the same ab sol ute v al ue. I n a real -w orl d sc enario, students interp ret ab sol ute v al ue as magnitude f or a p ositiv e or negativ e q uantity . T hey ap p l y their understanding of order and ab sol ute v al ue to determine that, f or instanc e, a c hec king ac c ount b al anc e that is l ess than −25 dol l ars rep resents a deb t of more than $25 .
ĨƌŽŵ 'ƌĂĚĞ ϱ
©201 8Great Minds ®. eureka-math.org
6ͻϯ DŽĚƵůĞKǀĞƌǀŝĞǁ
DŽĚƵůĞϯ :ZĂƚŝŽŶĂůEƵŵďĞƌƐ
ĨƌŽŵ'ƌĂĚĞϱŽĨƉůŽƚƚŝŶŐƉŽŝŶƚƐŝŶƚŚĞĨŝƌƐƚƋƵĂĚƌĂŶƚĂŶĚƚƌĂŶƐŝƚŝŽŶƚŽůŽĐĂƚŝŶŐƉŽŝŶƚƐŝŶĂůůĨŽƵƌƋƵĂĚƌĂŶƚƐ͘ ^ƚ ƵĚĞŶƚƐĂƉƉůLJƚŚĞĐŽŶĐĞƉƚŽĨĂďƐŽůƵƚĞǀĂůƵĞƚŽĨŝŶĚƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƉŽŝŶƚƐůŽĐĂƚĞĚŽŶ ǀĞƌƚŝĐĂů
ŽƌŚŽƌŝnjŽŶƚĂůůŝŶĞƐĂŶĚƐŽůǀĞƌĞĂůͲǁŽƌůĚƉƌŽďůĞŵƐƌĞůĂƚĞĚƚŽĚŝƐƚĂŶĐĞ͕ƐĞŐŵĞŶƚƐ͕ĂŶĚƐŚĂƉĞƐ ͘ dŚĞϮϱͲĚĂLJŵŽĚƵůĞĐŽŶƐŝƐƚƐŽĨϭϵůĞƐƐŽŶƐ͖ϲĚĂLJƐĂƌĞƌĞƐĞƌǀĞĚĨŽƌĂĚŵŝŶŝƐƚĞƌŝŶŐƚŚĞDŝĚͲĂŶĚ ŶĚͲŽĨͲDŽĚƵůĞ ƐƐĞƐƐŵĞŶƚƐ͕ƌĞƚƵƌŶŝŶŐĂƐƐĞƐƐŵĞŶƚƐ͕ĂŶĚƌĞŵĞĚŝĂƚŝŶŐŽƌƉƌŽǀŝĚŝŶŐĨƵƌƚŚĞƌĂƉƉůŝĐĂƚŝŽŶƐŽĨƚŚĞĐŽŶĐĞƉƚƐ͘dŚĞ DŝĚͲDŽĚƵůĞƐƐĞƐƐŵĞŶƚĨŽůůŽǁƐdŽƉŝĐ͕ĂŶĚƚŚĞ ŶĚͲŽĨͲDŽĚƵůĞƐƐĞƐƐŵĞŶƚĨŽůůŽǁƐdŽƉŝĐ͘
&ŽĐƵƐ^ƚĂŶĚĂƌĚƐ
ƉƉůLJĂŶĚĞdžƚĞŶĚƉƌĞǀŝŽƵƐƵŶĚĞƌƐƚĂŶĚŝŶŐƐ ŽĨŶƵŵďĞƌƐƚŽƚŚĞƐLJƐƚĞŵŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
hŶĚĞƌƐƚĂŶĚƚŚĂƚƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐĂƌĞƵƐĞĚƚŽŐĞƚŚĞƌƚŽĚĞƐĐƌŝďĞƋƵĂŶƚŝƚŝĞƐ ŚĂǀŝŶŐŽƉƉŽƐŝƚĞĚŝƌĞĐƚŝŽŶƐŽƌǀĂůƵĞƐ;Ğ͘Ő͕͘ƚĞŵƉĞƌĂƚƵƌĞĂďŽǀĞͬďĞůŽǁnjĞƌŽ͕ĞůĞǀĂƚŝŽŶ ĂďŽǀĞͬďĞůŽǁƐĞĂůĞǀĞů͕ĐƌĞĚŝƚƐͬĚĞďŝƚƐ͕ƉŽƐŝƚŝǀĞͬŶĞŐĂƚŝǀĞĞůĞĐƚƌŝĐĐŚĂƌŐĞͿ͖ƵƐĞƉŽƐŝƚŝǀĞĂŶĚ ŶĞŐĂƚŝǀĞŶƵŵďĞƌƐƚŽƌĞƉƌĞƐĞŶƚƋƵĂŶƚŝƚŝĞƐŝŶƌĞĂůͲǁŽƌůĚĐŽŶƚĞdžƚƐ͕ĞdžƉůĂŝŶŝŶŐƚŚĞŵĞĂŶŝŶŐŽĨ
ͲŝŶĞĂĐŚƐŝƚƵĂƚŝŽŶ͘
hŶĚĞƌƐƚĂŶĚĂƌĂƚŝŽŶĂůŶƵŵďĞƌĂƐĂƉŽŝŶƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ džƚĞŶĚŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵƐ ĂŶĚĐŽŽƌĚŝŶĂƚĞĂdžĞƐĨĂŵŝůŝĂƌĨƌŽŵƉƌĞǀŝŽƵƐŐƌĂĚĞƐƚŽƌĞƉƌĞƐĞŶƚƉŽŝŶƚƐŽŶƚŚĞůŝŶĞĂŶĚŝŶƚŚĞ ƉůĂŶĞǁŝƚŚŶĞŐĂƚŝǀĞŶƵŵďĞƌĐŽŽƌĚŝŶĂƚĞƐ͘
ZĞĐŽŐŶŝnjĞŽƉƉŽƐŝƚĞƐŝŐŶƐŽĨŶƵŵďĞƌƐĂƐŝŶĚŝĐĂƚŝŶŐůŽĐĂƚŝŽŶƐŽŶŽƉƉŽƐŝƚĞƐŝĚĞƐŽĨ ͲŽŶ ƚŚĞŶƵŵďĞƌůŝŶĞ͖ƌĞĐŽŐŶŝnjĞƚŚĂƚƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶƵŵďĞƌŝƐƚŚĞŶƵŵďĞƌ ŝƚƐĞůĨ͕Ğ͘Ő͕͘ െሺെ͵ሻ ൌ ͵ ͕ĂŶĚƚŚĂƚ ͲŝƐŝƚƐŽǁŶŽƉƉŽƐŝƚĞ͘
hŶĚĞƌƐƚĂŶĚƐŝŐŶƐŽĨŶƵŵďĞƌƐŝŶŽƌĚĞƌĞĚƉĂŝƌƐĂƐŝŶĚŝĐĂƚŝŶŐůŽĐĂƚŝŽŶƐŝŶƋƵĂĚƌĂŶƚƐŽĨ ƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͖ƌĞĐŽŐŶŝnjĞƚŚĂƚǁŚĞŶƚǁŽŽƌĚĞƌĞĚƉĂŝƌƐĚŝĨĨĞƌŽŶůLJďLJƐŝŐŶƐ͕ƚŚĞ ůŽĐĂƚŝŽŶƐŽĨƚŚĞƉŽŝŶƚƐĂƌĞƌĞůĂƚĞĚďLJƌĞĨůĞĐƚŝŽŶƐĂĐƌŽƐƐŽŶĞŽƌďŽƚŚĂdžĞƐ͘
&ŝŶĚĂŶĚƉŽƐŝƚŝŽŶŝŶƚĞŐĞƌƐĂŶĚŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶĂŚŽƌŝnjŽŶƚĂůŽƌǀĞƌƚŝĐĂů ŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵ͖ĨŝŶĚĂŶĚƉŽƐŝƚŝŽŶƉĂŝƌƐŽĨŝŶƚĞŐĞƌƐĂŶĚŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶ ĂĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘
hŶĚĞƌƐƚĂŶĚŽƌĚĞƌŝŶŐĂŶĚĂďƐŽůƵƚĞǀĂůƵĞŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
/ŶƚĞƌƉƌĞƚƐƚĂƚĞŵĞŶƚƐŽĨŝŶĞƋƵĂůŝƚLJĂƐƐƚĂƚĞŵĞŶƚƐĂďŽƵƚƚŚĞƌĞůĂƚŝǀĞƉŽƐŝƚŝŽŶŽĨƚǁŽ ŶƵŵďĞƌƐŽŶĂŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵ͘ For example, interpret െ͵ െ as a statement that െ͵ is located to the right of െ on a number line oriented from left to right.
tƌŝƚĞ͕ŝŶƚĞƌƉƌĞƚ͕ĂŶĚĞdžƉůĂŝŶƐƚĂƚĞŵĞŶƚƐŽĨŽƌĚĞƌĨŽƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶƌĞĂůͲǁŽƌůĚ ĐŽŶƚĞdžƚƐ͘ For example, write െ͵ι െι to express the fact that െ͵ι is warmer than െι .
A STORY OF RATIOS
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©201 8Great Minds ®. eureka-math.org
6ͻϯ DŽĚƵůĞKǀĞƌǀŝĞǁ
DŽĚƵůĞϯ :ZĂƚŝŽŶĂůEƵŵďĞƌƐ
hŶĚĞƌƐƚĂŶĚƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨĂƌĂƚŝŽŶĂůŶƵŵďĞƌĂƐŝƚƐĚŝƐƚĂŶĐĞĨƌŽŵ ͲŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞ͖ŝŶƚĞƌƉƌĞƚĂďƐŽůƵƚĞǀĂůƵĞĂƐŵĂŐŶŝƚƵĚĞĨŽƌĂƉŽƐŝƚŝǀĞŽƌŶĞŐĂƚŝǀĞƋƵĂŶƚŝƚLJ ŝŶĂƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶ͘ For example, for an account balance of െ͵Ͳ dollars, write
ȁെ͵Ͳȁ ൌ ͵Ͳ to describe the size of the debt in dollars.
ŝƐƚŝŶŐƵŝƐŚĐŽŵƉĂƌŝƐŽŶƐŽĨĂďƐŽůƵƚĞǀĂůƵĞĨƌŽŵƐƚĂƚĞŵĞŶƚƐĂďŽƵƚŽƌĚĞƌ͘ For example, recognize that an account balance less than െ͵Ͳ dollars represents a debt greater than
͵Ͳ dollars. ^ŽůǀĞƌĞĂůͲǁŽƌůĚĂŶĚŵĂƚŚĞŵĂƚŝĐĂůƉƌŽďůĞŵƐďLJŐƌĂƉŚŝŶŐƉŽŝŶƚƐŝŶĂůůĨŽƵƌƋƵĂĚƌĂŶƚƐŽĨƚŚĞ ĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘/ŶĐůƵĚĞƵƐĞŽĨĐŽŽƌĚŝŶĂƚĞƐĂŶĚĂďƐŽůƵƚĞǀĂůƵĞƚŽĨŝŶĚĚŝƐƚĂŶĐĞƐďĞƚǁĞĞŶ ƉŽŝŶƚƐǁŝƚŚƚŚĞƐĂŵĞĨŝƌƐƚĐŽŽƌĚŝŶĂƚĞŽƌƚŚĞƐĂŵĞƐĞĐŽŶĚĐŽŽƌĚŝŶĂƚĞ͘
&ŽƵŶĚĂƚŝŽŶĂů^ƚĂŶĚĂƌĚƐ
ĞǀĞůŽƉƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨĨƌĂĐƚŝŽŶƐĂƐŶƵŵďĞƌƐ͘
hŶĚĞƌƐƚĂŶĚĂĨƌĂĐƚŝŽŶĂƐĂŶƵŵďĞƌŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͖ƌĞƉƌĞƐĞŶƚĨƌĂĐƚŝŽŶƐŽŶĂŶƵŵďĞƌůŝŶĞ ĚŝĂŐƌĂŵ͘
ZĞƉƌĞƐĞŶƚĂĨƌĂĐƚŝŽŶ ܾͳȀ ŽŶĂŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵďLJĚĞĨŝŶŝŶŐƚŚĞŝŶƚĞƌǀĂůĨƌŽŵ ͲƚŽ ͳĂƐƚŚĞǁŚŽůĞĂŶĚƉĂƌƚŝƚŝŽŶŝŶŐŝƚŝŶƚŽ ܾ ĞƋƵĂůƉĂƌƚƐ͘ZĞĐŽŐŶŝnjĞƚŚĂƚĞĂĐŚƉĂƌƚŚĂƐƐŝnjĞ
ܾͳȀ ĂŶĚƚŚĂƚƚŚĞĞŶĚƉŽŝŶƚŽĨƚŚĞƉĂƌƚďĂƐĞĚĂƚ ͲůŽĐĂƚĞƐƚŚĞŶƵŵďĞƌ ܾͳȀ ŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞ͘
ZĞƉƌĞƐĞŶƚĂĨƌĂĐƚŝŽŶ ܽ ܾȀ ŽŶĂŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵďLJŵĂƌŬŝŶŐŽĨĨ ܽ ůĞŶŐƚŚƐ ܾͳȀ ĨƌŽŵ
Ͳ͘ ZĞĐŽŐŶŝnjĞƚŚĂƚƚŚĞƌĞƐƵůƚŝŶŐŝŶƚĞƌǀĂůŚĂƐƐŝnjĞ ܽ ܾȀ ĂŶĚƚŚĂƚŝƚƐĞŶĚƉŽŝŶƚůŽĐĂƚĞƐƚŚĞ ŶƵŵďĞƌ ܽ ܾȀ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘
ƌĂǁĂŶĚŝĚĞŶƚŝĨLJůŝŶĞƐĂŶĚĂŶŐůĞƐ͕ĂŶĚĐůĂƐƐŝĨLJƐŚĂƉĞƐďLJƉƌŽƉĞƌƚŝĞƐŽĨƚŚĞŝƌůŝŶĞƐĂŶĚ ĂŶŐůĞƐ͘
ZĞĐŽŐŶŝnjĞĂůŝŶĞŽĨƐLJŵŵĞƚƌLJĨŽƌĂƚǁŽͲĚŝŵĞŶƐŝŽŶĂůĨŝŐƵƌĞĂƐĂůŝŶĞĂĐƌŽƐƐƚŚĞĨŝŐƵƌĞƐƵĐŚ ƚŚĂƚƚŚĞĨŝŐƵƌĞĐĂŶďĞĨŽůĚĞĚĂůŽŶŐƚŚĞůŝŶĞŝŶƚŽŵĂƚĐŚŝŶŐƉĂƌƚƐ͘/ĚĞŶƚŝĨLJůŝŶĞͲƐLJŵŵĞƚƌŝĐ ĨŝŐƵƌĞƐĂŶĚĚƌĂǁůŝŶĞƐŽĨƐLJŵŵĞƚƌLJ͘
'ƌĂƉŚƉŽŝŶƚƐŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽƐŽůǀĞƌĞĂů -ǁŽƌůĚĂŶĚŵĂƚŚĞŵĂƚŝĐĂůƉƌŽďůĞŵƐ͘
hƐĞĂƉĂŝƌŽĨƉĞƌƉĞŶĚŝĐƵůĂƌŶƵŵďĞƌůŝŶĞƐ͕ĐĂůůĞĚĂdžĞƐ͕ƚŽĚĞĨŝŶĞĂĐŽŽƌĚŝŶĂƚĞƐLJƐƚĞŵ͕ǁŝƚŚ ƚŚĞŝŶƚĞƌƐĞĐƚŝŽŶŽĨƚŚĞůŝŶĞƐ;ƚŚĞŽƌŝŐŝŶͿĂƌƌĂŶŐĞĚƚŽĐŽŝŶĐŝĚĞǁŝƚŚƚŚĞ ͲŽŶĞĂĐŚůŝŶĞĂŶĚĂ ŐŝǀĞŶƉŽŝŶƚŝŶƚŚĞƉůĂŶĞůŽĐĂƚĞĚďLJƵƐŝŶŐĂŶŽƌĚĞƌĞĚƉĂŝƌŽĨŶƵŵďĞƌƐ͕ĐĂůůĞĚŝƚƐĐŽŽƌĚŝŶĂƚĞƐ͘ hŶĚĞƌƐƚĂŶĚƚŚĂƚƚŚĞĨŝƌƐƚŶƵŵďĞƌŝŶĚŝĐĂƚĞƐŚŽǁĨĂƌƚŽƚƌĂǀĞůĨƌŽŵƚŚĞŽƌŝŐŝŶŝŶƚŚĞĚŝƌĞĐƚŝŽŶ ŽĨŽŶĞĂdžŝƐ͕ĂŶĚƚŚĞƐĞĐŽŶĚŶƵŵďĞƌŝŶĚŝĐĂƚĞƐŚŽǁĨĂƌƚŽƚƌĂǀĞůŝŶƚŚĞĚŝƌĞĐƚŝŽŶŽĨƚŚĞƐĞĐŽŶĚ ĂdžŝƐ͕ǁŝƚŚƚŚĞĐŽŶǀĞŶƚŝŽŶƚŚĂƚƚŚĞŶĂŵĞƐŽĨƚŚĞƚǁŽĂdžĞƐĂŶĚƚŚĞĐŽŽƌĚŝŶĂƚĞƐĐŽƌƌĞƐƉŽŶĚ ;Ğ͘Ő͕͘ ݔͲĂdžŝƐĂŶĚ ݔͲĐŽŽƌĚŝŶĂƚĞ͕ ݕͲĂdžŝƐĂŶĚ ݕͲĐŽŽƌĚŝŶĂƚĞͿ͘
A STORY OF RATIOS
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©201 8Great Minds ®. eureka-math.org
6ͻϯ DŽĚƵůĞ KǀĞƌǀŝĞǁ
DŽĚƵůĞϯ :ZĂƚŝŽŶĂůEƵŵďĞƌƐ
ZĞƉƌĞƐĞŶƚƌĞĂůͲǁŽƌůĚĂŶĚŵĂƚŚĞŵĂƚŝĐĂůƉƌŽďůĞŵƐďLJŐƌĂƉŚŝŶŐƉŽŝŶƚƐŝŶƚŚĞĨŝƌƐƚƋƵĂĚƌĂŶƚŽĨ ƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͕ĂŶĚŝŶƚĞƌƉƌĞƚĐŽŽƌĚŝŶĂƚĞǀĂůƵĞƐŽĨƉŽŝŶƚƐŝŶƚŚĞĐŽŶƚĞdžƚŽĨƚŚĞ ƐŝƚƵĂƚŝŽŶ͘
&ŽĐƵƐ^ƚĂŶĚĂƌĚƐ ĨŽƌDĂƚŚĞŵĂƚŝĐĂůWƌĂĐƚŝĐĞ
ZĞĂƐŽŶ aďƐƚƌĂĐƚůLJĂŶĚ qƵĂŶƚŝƚĂƚŝǀĞůLJ͘ ^ƚƵĚĞŶƚƐƌĞĂĚĂǁŽƌĚƉƌŽďůĞŵŝŶǀŽůǀŝŶŐŝŶƚĞŐĞƌƐ͕ ĚƌĂǁĂŶƵŵďĞƌůŝŶĞŽƌĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŵŽĚĞů͕ĂŶĚǁƌŝƚĞĂďŽƵƚƚŚĞŝƌĐŽŶĐůƵƐŝŽŶƐ͘dŚĞLJ ƵŶĚĞƌƐƚĂŶĚƚŚĞŵĞĂŶŝŶŐŽĨƋƵĂŶƚŝƚŝĞƐĂƐƚŚĞLJƌĞůĂƚĞƚŽƚŚĞƌĞĂůǁŽƌůĚ͘&ŽƌŝŶƐƚĂŶĐĞ͕ĂůŽƐƐŽĨ
ͳͶ LJĂƌĚƐŝŶĂĨŽŽƚďĂůůŐĂŵĞĐĂŶďĞƌĞƉƌĞƐĞŶƚĞĚďLJ െͳͶ ͕ĂŶĚĂĚŝƐƚĂŶĐĞŽĨ ʹͷ ĨĞĞƚďĞůŽǁƐĞĂ ůĞǀĞůŝƐŐƌĞĂƚĞƌƚŚĂŶĂĚŝƐƚĂŶĐĞŽĨ ͷĨĞĞƚĂďŽǀĞƐĞĂůĞǀĞůďĞĐĂƵƐĞ ȁ െ ʹͷȁ ȁͷȁ ͘^ƚƵĚĞŶƚƐ ĚĞĐŽŶƚĞdžƚƵĂůŝnjĞǁŽƌĚƉƌŽďůĞŵƐƌĞůĂƚĞĚƚŽĚŝƐƚĂŶĐĞďLJĐƌĞĂƚŝŶŐŶƵŵďĞƌůŝŶĞƐĂŶĚĐŽŽƌĚŝŶĂƚĞ ƉůĂŶĞŵŽĚĞůƐ͘/ŶĚŽŝŶŐƐŽ͕ƚŚĞLJĐŽƵŶƚƚŚĞŶƵŵďĞƌŽĨƵŶŝƚƐďĞƚǁĞĞŶĞŶĚƉŽŝŶƚƐĂŶĚƵƐĞƚŚĞ ĐŽŶĐĞƉƚŽĨĂďƐŽůƵƚĞǀĂůƵĞƚŽũƵƐƚŝĨLJƚŚĞŝƌĂŶƐǁĞƌƐ͘&ŽƌŝŶƐƚĂŶĐĞ͕ǁŚĞŶŐŝǀĞŶƚŚĞĐŽŽƌĚŝŶĂƚĞ
ሺʹǡ ሻ ͕ƐƚƵĚĞŶƚƐĚĞƚĞƌŵŝŶĞƚŚĂƚƚŚĞƉŽŝŶƚ ሺʹǡ െሻ ǁŽƵůĚďĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵƚŚĞ
ݔͲĂdžŝƐďƵƚŝŶƚŚĞŽƉƉŽƐŝƚĞĚŝƌĞĐƚŝŽŶďĞĐĂƵƐĞďŽƚŚƉŽŝŶƚƐŚĂǀĞƚŚĞƐĂŵĞ ݔͲĐŽŽƌĚŝŶĂƚĞĂŶĚ ƚŚĞŝƌ ݕͲĐŽŽƌĚŝŶĂƚĞƐ; ĂŶĚ െ ͿŚĂǀĞƚŚĞƐĂŵĞĂďƐŽůƵƚĞǀĂůƵĞ͘ DŽĚĞůǁŝƚŚ mĂƚŚĞŵĂƚŝĐƐ͘ ^ƚƵĚĞŶƚƐƵƐĞǀĞƌƚŝĐĂůĂŶĚŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞƐƚŽǀŝƐƵĂůŝnjĞ ŝŶƚĞŐĞƌƐĂŶĚďĞƚƚĞƌƵŶĚĞƌƐƚĂŶĚƚŚĞŝƌĐŽŶŶĞĐƚŝŽŶƚŽǁŚŽůĞŶƵŵďĞƌƐ͘dŚĞLJĚŝǀŝĚĞŶƵŵďĞƌ ůŝŶĞŝŶƚĞƌǀĂůƐŝŶƚŽƐƵďͲŝŶƚĞƌǀĂůƐŽĨƚĞŶƚŚƐƚŽĚĞƚĞƌŵŝŶĞƚŚĞĐŽƌƌĞĐƚƉůĂĐĞŵĞŶƚŽĨƌĂƚŝŽŶĂů ŶƵŵďĞƌƐ͘^ƚƵĚĞŶƚƐŵĂLJƌĞƉƌĞƐĞŶƚĂĚĞĐŝŵĂůĂƐĂĨƌĂĐƚŝŽŶŽƌĂĨƌĂĐƚŝŽŶĂƐĂĚĞĐŝŵĂůƚŽďĞƚƚĞƌ ƵŶĚĞƌƐƚĂŶĚŝƚƐƌĞůĂƚŝŽŶƐŚŝƉƚŽŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŽǁŚŝĐŚŝƚŝƐďĞŝŶŐĐŽŵƉĂƌĞĚ͘dŽ ĞdžƉůĂŝŶƚŚĞŵĞĂŶŝŶŐŽĨĂƋƵĂŶƚŝƚLJŝŶĂƌĞĂůͲůŝĨĞƐŝƚƵĂƚŝŽŶ;ŝŶǀŽůǀŝŶŐĞůĞǀĂƚŝŽŶ͕ƚĞŵƉĞƌĂƚƵƌĞ͕ ŽƌĚŝƌĞĐƚŝŽŶͿ͕ƐƚƵĚĞŶƚƐŵĂLJĚƌĂǁĂĚŝĂŐƌĂŵĂŶĚͬŽƌŶƵŵďĞƌůŝŶĞƚŽŝůůƵƐƚƌĂƚĞƚŚĞůŽĐĂƚŝŽŶŽĨ ƚŚĞƋƵĂŶƚŝƚLJŝŶƌĞůĂƚŝŽŶƚŽnjĞƌŽŽƌĂŶĞƐƚĂďůŝƐŚĞĚůĞǀĞůƚŚĂƚƌĞƉƌĞƐĞŶƚƐnjĞƌŽŝŶƚŚĂƚƐŝƚƵĂƚŝŽŶ͘ Atten ĚƚŽƉƌĞĐŝƐŝŽŶ͘ /ŶƌĞƉƌĞƐĞŶƚŝŶŐƐŝŐŶĞĚŶƵŵďĞƌƐŽŶĂŶƵŵďĞƌůŝŶĞŽƌĂƐĂƋƵĂŶƚŝƚLJ͕ ƐƚƵĚĞŶƚƐƉĂLJĐůŽƐĞĂƚƚĞŶƚŝŽŶƚŽƚŚĞĚŝƌĞĐƚŝŽŶĂŶĚƐŝŐŶŽĨĂŶƵŵďĞƌ͘dŚĞLJƌĞĂůŝnjĞƚŚĂƚĂ ŶĞŐĂƚŝǀĞŶƵŵďĞƌŵƵƐƚůŝĞƚŽƚŚĞůĞĨƚŽĨnjĞƌŽŽŶĂŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞŽƌďĞůŽǁnjĞƌŽŽŶĂ ǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘dŚĞLJƌĞĐŽŐŶŝnjĞƚŚĂƚƚŚĞǁĂLJƚŚĞLJƌĞƉƌĞƐĞŶƚƚŚĞŝƌĂŶƐǁĞƌĚĞƉĞŶĚƐŽŶ ƚŚĞƉŚƌĂƐŝŶŐŽĨĂƋƵĞƐƚŝŽŶĂŶĚĐŽŶƚĞdžƚŽĨĂǁŽƌĚƉƌŽďůĞŵ͘&ŽƌŝŶƐƚĂŶĐĞ͕ĂƋƵĞƐƚŝŽŶƚŚĂƚ ĂƐŬƐĂƐƚƵĚĞŶƚ͞,ŽǁŵĂŶLJĨĞĞƚďĞůŽǁƐĞĂůĞǀĞůŝƐƚŚĞĚŝǀĞƌ͍͟ǁŽƵůĚƌĞƋƵŝƌĞƚŚĞĂŶƐǁĞƌƚŽ ďĞƐƚĂƚĞĚĂƐĂƉŽƐŝƚŝǀĞŶƵŵďĞƌ͕ǁŚĞƌĞĂƐ͕ĂƋƵĞƐƚŝŽŶƚŚĂƚŝƐƉŚƌĂƐĞĚ͞tŚŝĐŚŝŶƚĞŐĞƌǁŽƵůĚ ƌĞƉƌĞƐĞŶƚ ͶͲ ĨĞĞƚďĞůŽǁƐĞĂůĞǀĞů͍͟ǁŽƵůĚƌĞƋƵŝƌĞƚŚĞĂŶƐǁĞƌƚŽďĞǁƌŝƚƚĞŶĂƐ െͶͲ ͘ >ŽŽŬĨŽƌĂŶĚŵĂŬĞƵƐĞŽĨƐƚƌƵĐƚƵƌĞ͘ ^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĞƉůĂĐĞŵĞŶƚŽĨŶĞŐĂƚŝǀĞ ŶƵŵďĞƌƐŽŶĂŶƵŵďĞƌůŝŶĞďLJŽďƐĞƌǀŝŶŐƚŚĞƉĂƚƚĞƌŶƐƚŚĂƚĞdžŝƐƚďĞƚǁĞĞŶŶĞŐĂƚŝǀĞĂŶĚ ƉŽƐŝƚŝǀĞŶƵŵďĞƌƐǁŝƚŚƌĞƐƉĞĐƚƚŽnjĞƌŽ͘dŚĞLJƌĞĐŽŐŶŝnjĞƚŚĂƚƚǁŽŶƵŵďĞƌƐĂƌĞŽƉƉŽƐŝƚĞƐŝĨ ƚŚĞLJĂƌĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽĂŶĚƚŚĂƚnjĞƌŽŝƐŝƚƐŽǁŶŽƉƉŽƐŝƚĞ͘^ƚƵĚĞŶƚƐĞdžƚĞŶĚ ƚŚĞŝƌƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨƚŚĞŶƵŵďĞƌůŝŶĞ͛ƐƐƚƌƵĐƚƵƌĞƚŽƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽĚĞƚĞƌŵŝŶĞĂ ƉŽŝŶƚ͛ƐůŽĐĂƚŝŽŶ͘dŚĞLJƌĞĐŽŐŶŝnjĞƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞƐŝŐŶƐŽĨĂƉŽŝŶƚ͛ƐĐŽŽƌĚŝŶĂƚĞƐ ĂŶĚƚŚĞƋƵĂĚƌĂŶƚŝŶǁŚŝĐŚƚŚĞƉŽŝŶƚůŝĞƐ͘
A STORY OF RATIOS
5
©201 8Great Minds ®. eureka-math.org
6ͻϯ DŽĚƵůĞ KǀĞƌǀŝĞǁ
DŽĚƵůĞϯ :ZĂƚŝŽŶĂůEƵŵďĞƌƐ
dĞƌŵŝŶŽůŽŐLJ
EĞǁŽƌZĞĐĞŶƚůLJ/ŶƚƌŽĚƵĐĞĚdĞƌŵƐ
ďƐŽůƵƚĞsĂůƵĞ ;dŚĞ absolute value ŽĨĂŶƵŵďĞƌŝƐƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƚŚĞŶƵŵďĞƌĂŶĚnjĞƌŽŽŶ ƚŚĞŶƵŵďĞƌůŝŶĞ͘&ŽƌĞdžĂŵƉůĞ͕ ȁ͵ȁ ൌ ͵ ͕ ȁെͶȁ ൌ Ͷ ͕ĞƚĐ͘Ϳ
/ŶƚĞŐĞƌ ;Ŷ integer ŝƐĂŶƵŵďĞƌƚŚĂƚĐĂŶĞŝƚŚĞƌďĞƌĞƉƌĞƐĞŶƚĞĚĂƐĂǁŚŽůĞŶƵŵďĞƌŽƌĂƐƚŚĞŽƉƉŽƐŝƚĞ ŽĨĂǁŚŽůĞŶƵŵďĞƌ͘dŚĞ set of integers ŝƐƚŚĞŝŶĨŝŶŝƚĞůŝƐƚŽĨŶƵŵďĞƌƐ͕͗͘͘͘ െ͵ ͕ െʹ ͕ െͳ ͕ Ͳ͕ ͳ͕ ʹ͕ ͵͕ ͙͘Ϳ
DĂŐŶŝƚƵĚĞ ;dŚĞ magnitude of a measurement ŝƐƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨƚŚĞŵĞĂƐƵƌĞŽĨƚŚĞ ŵĞĂƐƵƌĞŵĞŶƚ͘&ŽƌĞdžĂŵƉůĞ͕ƚŚĞŵĂŐŶŝƚƵĚĞŽĨƚŚĞŵĞĂƐƵƌĞŵĞŶƚ െʹͷι ŝƐ ʹͷ ͘ Ϳ
EĞŐĂƚŝǀĞEƵŵďĞƌ ; negative number ŝƐĂŶƵŵďĞƌůĞƐƐƚŚĂŶnjĞƌŽ͘Ϳ
KƉƉŽƐŝƚĞ ;'ŝǀĞŶĂŶŽŶnjĞƌŽŶƵŵďĞƌ ܽ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͕ƚŚĞ opposite of ܽ ͕ ĚĞŶŽƚĞĚ ܽെ ͕ ŝƐƚŚĞ ŶƵŵďĞƌŽŶƚŚĞŶƵŵďĞƌůŝŶĞƐƵĐŚƚŚĂƚ;ϭͿ ͲŝƐďĞƚǁĞĞŶ ܽ ĂŶĚ ܽെ ͕ ĂŶĚ;ϮͿƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶ ͲĂŶĚ ܽ
ŝƐĞƋƵĂůƚŽƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶ ͲĂŶĚ ܽെ ͘ dŚĞŽƉƉŽƐŝƚĞŽĨ ͲŝƐ Ͳ͘Ϳ
WŽƐŝƚŝǀĞEƵŵďĞƌ ; positive number ŝƐĂŶƵŵďĞƌŐƌĞĂƚĞƌƚŚĂŶnjĞƌŽ͘Ϳ
YƵĂĚƌĂŶƚ ;ĚĞƐĐƌŝƉƚŝŽŶͿ ;/ŶƚŚĞĂƌƚĞƐŝĂŶƉůĂŶĞ͕ƚŚĞƚǁŽĂdžĞƐƐĞƉĂƌĂƚĞƚŚĞƉůĂŶĞŝŶƚŽĨŽƵƌƌĞŐŝŽŶƐ ĐĂůůĞĚ quadrants ͘dŚĞĨŝƌƐƚƋƵĂĚƌĂŶƚĐŽŶƐŝƐƚƐŽĨĂůůƚŚĞƉŽŝŶƚƐǁŚŽƐĞ ݔͲĂŶĚ ݕͲĐŽŽƌĚŝŶĂƚĞƐĂƌĞďŽƚŚ ƉŽƐŝƚŝǀĞ͘dŚĞĨŝƌƐƚ͕ƐĞĐŽŶĚ͕ƚŚŝƌĚ͕ĂŶĚĨŽƵƌƚŚƋƵĂĚƌĂŶƚƐĂƌĞŝĚĞŶƚŝĨŝĞĚĐŽƵŶƚĞƌĐůŽĐŬǁŝƐĞĂƌŽƵŶĚƚŚĞ ŽƌŝŐŝŶŝŶŽƌĚĞƌƐƚĂƌƚŝŶŐǁŝƚŚƚŚĞĨŝƌƐƚƋƵĂĚƌĂŶƚ͘Ϳ
ZĂƚŝŽŶĂůEƵŵďĞƌ ;ĚĞƐĐƌŝƉƚŝŽŶͿ ; rational number ŝƐĂŶƵŵďĞƌƚŚĂƚĐĂŶďĞƌĞƉƌĞƐĞŶƚĞĚĂƐĂĨƌĂĐƚŝŽŶ ŽƌƚŚĞŽƉƉŽƐŝƚĞŽĨĂĨƌĂĐƚŝŽŶ͘Ϳ
&ĂŵŝůŝĂƌ dĞƌŵƐĂŶĚ^LJŵďŽůƐ 2
ŽŽƌĚŝŶĂƚĞWĂŝƌ
ŽŽƌĚŝŶĂƚĞWůĂŶĞ
&ƌĂĐƚŝŽŶ
>ŝŶĞŽĨ^LJŵŵĞƚƌLJ
KƌĚĞƌĞĚWĂŝƌ
KƌŝŐŝŶ
YƵĂĚƌĂŶƚ
^LJŵŵĞƚƌLJ
tŚŽůĞEƵŵďĞƌƐ
ݔͲdžŝƐ
ݔͲŽŽƌĚŝŶĂƚĞ
ݕͲdžŝƐ
ݕͲŽŽƌĚŝŶĂƚĞ ϮdŚĞƐĞĂƌĞƚĞƌŵƐĂŶĚƐLJŵďŽůƐƐƚƵĚĞŶƚƐŚĂǀĞƐĞĞŶƉƌĞǀŝŽƵƐůLJ͘
A STORY OF RATIOS
6
©201 8Great Minds ®. eureka-math.org
6ͻϯ DŽĚƵůĞ KǀĞƌǀŝĞǁ
DŽĚƵůĞϯ :ZĂƚŝŽŶĂůEƵŵďĞƌƐ
^ƵŐŐĞƐƚĞĚdŽŽůƐĂŶĚZĞƉƌĞƐĞŶƚĂƚŝŽŶƐ
,ŽƌŝnjŽŶƚĂůĂŶĚsĞƌƚŝĐĂůEƵŵďĞƌ>ŝŶĞƐ
ŽŽƌĚŝŶĂƚĞWůĂŶĞ
^ƉƌŝŶƚƐ
^ƉƌŝŶƚƐĂƌĞĚĞƐŝŐŶĞĚƚŽĚĞǀĞůŽƉĨůƵĞŶĐLJ͘dŚĞLJƐŚŽƵůĚďĞĨƵŶ͕ĂĚƌĞŶĂůŝŶĞͲƌŝĐŚĂĐƚŝǀŝƚŝĞƐƚŚĂƚŝŶƚĞŶƚŝŽŶĂůůLJďƵŝůĚ ĞŶĞƌŐLJĂŶĚĞdžĐŝƚĞŵĞŶƚ͘ĨĂƐƚƉĂĐĞŝƐĞƐƐĞŶƚŝĂů͘ƵƌŝŶŐ^ƉƌŝŶƚĂĚŵŝŶŝƐƚƌĂƚŝŽŶ͕ƚĞĂĐŚĞƌƐĂƐƐƵŵĞƚŚĞƌŽůĞŽĨ ĂƚŚůĞƚŝĐĐŽĂĐŚĞƐ͘ƌŽƵƐŝŶŐƌŽƵƚŝŶĞĨƵĞůƐƐƚƵĚĞŶƚƐ͛ŵŽƚŝǀĂƚŝŽŶƚŽĚŽƚŚĞŝƌƉĞƌƐŽŶĂůďĞƐƚ͘^ƚƵĚĞŶƚƌĞĐŽŐŶŝƚŝŽŶ ŽĨŝŶĐƌĞĂƐŝŶŐƐƵĐĐĞƐƐŝƐĐƌŝƚŝĐĂů͕ĂŶĚƐŽĞǀĞƌLJŝŵƉƌŽǀĞŵĞŶƚŝƐĂĐŬŶŽǁůĞĚŐĞĚ͘;^ĞĞƚŚĞ^ƉƌŝŶƚĞůŝǀĞƌLJ^ĐƌŝƉƚ ĨŽƌƚŚĞƐƵŐŐĞƐƚĞĚŵĞĂŶƐŽĨĂĐŬŶŽǁůĞĚŐŝŶŐĂŶĚĐĞůĞďƌĂƚŝŶŐƐƚƵĚĞŶƚƐƵĐĐĞƐƐ͘Ϳ KŶĞ^ƉƌŝŶƚŚĂƐƚǁŽƉĂƌƚƐǁŝƚŚĐůŽƐĞůLJƌĞůĂƚĞĚƉƌŽďůĞŵƐŽŶĞĂĐŚ͘^ƚƵĚĞŶƚƐĐŽŵƉůĞƚĞƚŚĞƚǁŽƉĂƌƚƐŽĨƚŚĞ ^ƉƌŝŶƚŝŶƋƵŝĐŬƐƵĐĐĞƐƐŝŽŶǁŝƚŚƚŚĞŐŽĂůŽĨŝŵƉƌŽǀŝŶŐŽŶƚŚĞƐĞĐŽŶĚƉĂƌƚ͕ĞǀĞŶŝĨŽŶůLJďLJŽŶĞŵŽƌĞ͘ ^ƉƌŝŶƚƐĂƌĞŶŽƚƚŽďĞƵƐĞĚĨŽƌĂŐƌĂĚĞ͘dŚƵƐ͕ƚŚĞƌĞŝƐŶŽŶĞĞĚĨŽƌƐƚƵĚĞŶƚƐƚŽǁƌŝƚĞƚŚĞŝƌŶĂŵĞƐŽŶƚŚĞ^ƉƌŝŶƚƐ͘ dŚĞůŽǁͲƐƚĂŬĞƐŶĂƚƵƌĞŽĨƚŚĞĞdžĞƌĐŝƐĞŵĞĂŶƐƚŚĂƚĞǀĞŶƐƚƵĚĞŶƚƐǁŝƚŚĂůůŽǁĂŶĐĞƐĨŽƌĞdžƚĞŶĚĞĚƚŝŵĞĐĂŶ ƉĂƌƚŝĐŝƉĂƚĞ͘tŚĞŶĂƉĂƌƚŝĐƵůĂƌƐƚƵĚĞŶƚĨŝŶĚƐƚŚĞĞdžƉĞƌŝĞŶĐĞƵŶĚĞƐŝƌĂďůĞ͕ŝƚŝƐƌĞĐŽŵŵĞŶĚĞĚƚŚĂƚƚŚĞƐƚƵĚĞŶƚ ďĞĂůůŽǁĞĚƚŽŽƉƚŽƵƚĂŶĚƚĂŬĞƚŚĞ^ƉƌŝŶƚŚŽŵĞ͘/ŶƚŚŝƐĐĂƐĞ͕ŝƚŝƐŝĚĞĂůŝĨƚŚĞƐƚƵĚĞŶƚŚĂƐĂƌĞŐƵůĂƌ ŽƉƉŽƌƚƵŶŝƚLJƚŽĞdžƉƌĞƐƐƚŚĞĚĞƐŝƌĞƚŽŽƉƚŝŶ͘ tŝƚŚƉƌĂĐƚŝĐĞ͕ƚŚĞ^ƉƌŝŶƚƌŽƵƚŝŶĞƚĂŬĞƐĂďŽƵƚϴŵŝŶƵƚĞƐ͘
^ƉƌŝŶƚĞůŝǀĞƌLJ^ĐƌŝƉ t
'ĂƚŚĞƌƚŚĞĨŽůůŽǁŝŶŐ͗ƐƚŽƉǁĂƚĐŚ͕ĂĐŽƉLJŽĨ^ƉƌŝŶƚĨŽƌĞĂĐŚƐƚƵĚĞŶƚ͕ĂĐŽƉLJŽĨ^ƉƌŝŶƚĨŽƌĞĂĐŚƐƚƵĚĞŶƚ͕ ĂŶƐǁĞƌƐĨŽƌ^ƉƌŝŶƚĂŶĚ^ƉƌŝŶƚ͘dŚĞĨŽůůŽǁŝŶŐĚĞůŝŶĞĂƚĞƐĂƐĐƌŝƉƚĨŽƌĚĞůŝǀĞƌLJŽĨĂƉĂŝƌŽĨ^ƉƌŝŶƚƐ͘
dŚŝƐƐƉƌŝŶƚĐŽǀĞƌƐ͗ topic.
ŽŶŽƚůŽŽŬĂƚƚŚĞ SƉƌŝŶƚ ; ŬĞĞƉŝƚƚƵƌŶĞĚĨĂĐĞĚŽǁŶŽŶLJŽƵƌĚĞƐŬ͘
dŚĞƌĞĂƌĞdždž ƉƌŽďůĞŵƐŽŶƚŚĞ SƉƌŝŶƚ͘zŽƵǁŝůůŚĂǀĞ 60 ƐĞĐŽŶĚƐ͘ŽĂƐŵĂŶLJĂƐLJŽƵĐĂŶ͘/ĚŽŶŽƚĞdžƉĞĐƚ ĂŶLJŽĨLJŽƵƚŽĨŝŶŝƐŚ͘
KŶLJŽƵƌŵĂƌŬ͕ŐĞƚƐĞƚ͕'K͘
ϲϬ seconds of silence.
^dKW͘ ŝƌĐůĞƚŚĞůĂƐƚƉƌŽďůĞŵLJŽƵĐŽŵƉůĞƚĞĚ͘
/ǁŝůůƌĞĂĚƚŚĞĂŶƐǁĞƌƐ͘zŽƵƐĂLJ “YES” ŝĨLJŽƵ r ĂŶƐǁĞƌŵĂƚĐŚĞƐ͘DĂƌŬƚŚĞŽŶĞƐLJŽƵŚĂǀĞǁƌŽŶŐ͘ŽŶ͛ƚƚƌLJ ƚŽĐŽƌƌĞĐƚƚŚĞŵ͘
Energetically, rapid-fire call the answers ONLY. Stop reading answers after there are no more students answering, “Yes.”
&ĂŶƚĂƐƚŝĐ͊ŽƵŶƚƚŚĞŶƵŵďĞƌLJŽƵŚĂǀĞĐŽƌƌĞĐƚ͕ ĂŶĚǁƌŝƚĞŝƚŽŶƚŚĞƚŽƉŽĨƚŚĞƉĂŐĞ͘ dŚŝƐŝƐLJŽƵƌƉĞƌƐŽŶĂů ŐŽĂůĨŽƌ^ƉƌŝŶƚ͘
A STORY OF RATIOS
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©201 8Great Minds ®. eureka-math.org
6ͻϯ DŽĚƵůĞ KǀĞƌǀŝĞǁ
DŽĚƵůĞϯ :ZĂƚŝŽŶĂůEƵŵďĞƌƐ
ZĂŝƐĞLJŽƵƌŚĂŶĚŝĨLJŽƵŚĂǀĞ one or more ĐŽƌƌĞĐƚ͘dǁŽŽƌŵŽƌĞ͕ƚŚƌĞĞ or more ͕ ͘͘͘
ĞƚƵƐĂůůĂƉƉůĂƵĚŽƵƌƌƵŶŶĞƌ -ƵƉ͕ŝŶƐĞƌƚŶĂŵĞ͕ ǁŝƚŚdž ĐŽƌƌĞĐƚ͘ŶĚůĞƚƵƐĂƉƉůĂƵĚŽƵƌǁŝŶŶĞƌ͕ŝŶƐĞƌƚ ŶĂŵĞ͕ǁŝƚŚdž ĐŽƌƌĞĐƚ͘
zŽƵŚĂǀĞĂĨĞǁŵŝŶƵƚĞƐƚŽĨŝŶŝƐŚƵƉƚŚĞƉĂŐĞĂŶĚŐĞƚƌĞĂĚLJĨŽƌƚŚĞŶĞdžƚ SƉƌŝŶƚ͘
Students are allowed to talk and ask for help; let this part last as long as most are working seriously.
^ƚŽƉǁŽƌŬŝŶŐ͘/ǁŝůůƌĞĂĚƚŚĞĂŶƐǁĞƌƐĂŐĂŝŶƐŽLJŽƵĐĂŶĐŚĞĐŬLJŽƵƌǁŽƌŬ͘zŽƵƐĂLJ “YES” ŝĨLJŽƵƌĂŶƐǁĞƌ ŵĂƚĐŚĞƐ͘
Energetically, rapid-fire call the answers ONLY. Optionally, ask students to stand, and lead them in an energy-expanding exercise that also keeps the brain going. Examples are jumping jacks or arm circles, etc., while counting by ͳͷ ’s starting at ͳͷ , going up to ͳͷͲ
and back down to Ͳ. You can follow this first exercise with a cool-down exercise of a similar nature, such as calf raises with counting by one-sixths ቀ ଵ ǡ ଵଷ ǡ ଵଶ ǡ ଶଷ ǡ ହ ǡ ͳǡ ǥ ቁ.Hand out the second Sprint, and continue reading the script.
<ĞĞƉƚŚĞ SƉƌŝŶƚĨĂĐĞĚŽǁŶŽŶLJŽƵƌĚĞƐŬ͘
dŚĞƌĞĂƌĞdždžƉƌŽďůĞŵƐŽŶƚŚĞ SƉƌŝŶƚ͘zŽƵǁŝůůŚĂǀĞ 60 ƐĞĐŽŶĚƐ͘ŽĂƐŵĂŶLJĂƐLJŽƵĐĂŶ͘/ĚŽŶŽƚĞdžƉĞĐƚ ĂŶLJŽĨLJŽƵƚŽĨŝŶŝƐŚ͘
KŶLJŽƵƌŵĂƌŬ͕ŐĞƚƐĞƚ͕'K͘
ϲϬ seconds of silence.
^dKW͘ŝƌĐůĞƚŚĞůĂƐƚƉƌŽďůĞŵLJŽƵĐŽŵƉůĞƚĞĚ͘
/ǁŝůůƌĞĂĚƚŚĞĂŶƐǁĞƌƐ͘zŽƵƐĂLJ “YES” ŝĨLJŽƵ r ĂŶƐǁĞƌŵĂƚĐŚĞƐ͘DĂƌŬƚŚĞŽŶĞƐLJŽƵŚĂǀĞǁƌŽŶŐ͘ŽŶ͛ƚƚƌLJ ƚŽĐŽƌƌĞĐƚƚŚĞŵ͘
Quickly read the answers ONLY.
ŽƵŶƚƚŚĞŶƵŵďĞƌLJŽƵŚĂǀĞĐŽƌƌĞĐƚ͕ ĂŶĚǁƌŝƚĞŝƚŽŶƚŚĞƚŽƉŽĨƚŚĞƉĂŐĞ͘
ZĂŝƐĞLJŽƵƌŚĂŶĚŝĨLJŽƵŚĂǀĞ one or more ĐŽƌƌĞĐƚ͘dǁŽŽƌŵŽƌĞ͕ƚŚƌĞĞŽƌŵŽƌĞ͕ ͘͘͘
ĞƚƵƐĂůůĂƉƉůĂƵĚŽƵƌƌƵŶŶĞƌ -ƵƉ͕ŝŶƐĞƌƚ ŶĂŵĞ͕ ǁŝƚŚdžĐŽƌƌĞĐƚ͘ŶĚůĞƚƵƐĂƉƉůĂƵĚŽƵƌǁŝŶŶĞƌ͕ [ŝŶƐĞƌƚ ŶĂŵĞ͕ǁŝƚŚdžĐŽƌƌĞĐƚ͘
tƌŝƚĞƚŚĞĂŵŽƵŶƚďLJǁŚŝĐŚLJŽƵƌƐĐŽƌĞŝŵƉƌŽǀĞĚĂƚƚŚĞƚŽƉŽĨƚŚĞƉĂŐĞ͘
ZĂŝƐĞLJŽƵƌŚĂŶĚŝĨLJŽƵŚĂǀĞ one or more ĐŽƌƌĞĐƚ͘dǁŽŽƌŵŽƌĞ͕ƚŚƌĞĞŽƌŵŽƌĞ͕ ͘͘͘
ĞƚƵƐĂůůĂƉƉůĂƵĚŽƵƌƌƵŶŶĞƌ -ƵƉĨŽƌŵŽƐƚŝŵƉƌŽǀĞĚ͕ ŝŶƐĞƌƚŶĂŵĞ͘ŶĚůĞƚƵƐĂƉƉůĂƵĚŽƵƌǁŝŶŶĞƌĨŽƌŵŽƐƚ ŝŵƉƌŽǀĞĚ͕ŝŶƐĞƌƚŶĂŵĞ͘
zŽƵĐĂŶƚĂŬĞƚŚĞ^ƉƌŝŶƚŚŽŵĞĂŶĚĨŝŶŝƐŚŝƚŝĨLJŽƵǁĂŶƚ͘
A STORY OF RATIOS
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©201 8Great Minds ®. eureka-math.org
6ͻϯ DŽĚƵůĞ KǀĞƌǀŝĞǁ
DŽĚƵůĞϯ :ZĂƚŝŽŶĂůEƵŵďĞƌƐ
ƐƐĞƐƐŵĞŶƚ^ƵŵŵĂƌLJ
ƐƐĞƐƐŵĞŶƚdLJƉĞ ĚŵŝŶŝƐƚĞƌĞĚ Format DŝĚͲDŽĚƵůĞ ƐƐĞƐƐŵĞŶƚdĂƐŬ ĨƚĞƌdŽƉŝĐ ŽŶƐƚƌƵĐƚĞĚƌĞƐƉŽŶƐĞǁŝƚŚƌƵďƌŝĐ ŶĚͲŽĨͲDŽĚƵůĞ ƐƐĞƐƐŵĞŶƚdĂƐŬ ĨƚĞƌdŽƉŝĐ ŽŶƐƚƌƵĐƚĞĚƌĞƐƉŽŶƐĞǁŝƚŚƌƵďƌŝĐ
A STORY OF RATIOS
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©201 8Great Minds ®. eureka-math.org
6G R A D E Mathematics Curriculum
GRADE 6 ͻDKh> 3Topic A: hŶĚĞƌƐƚĂŶĚŝŶŐWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
dŽƉŝĐ
hŶĚĞƌƐƚĂŶĚŝŶŐWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞ EƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
Focus Standards: hŶĚĞƌƐƚĂŶĚƚŚĂƚƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐĂƌĞƵƐĞĚƚŽŐĞƚŚĞƌƚŽĚĞƐĐƌŝďĞ ƋƵĂŶƚŝƚŝĞƐŚĂǀŝŶŐŽƉƉŽƐŝƚĞĚŝƌĞĐƚŝŽŶƐŽƌǀĂůƵĞƐ;Ğ͘Ő͕͘ƚĞŵƉĞƌĂƚƵƌĞĂďŽǀĞͬďĞůŽǁ njĞƌŽ͕ĞůĞǀĂƚŝŽŶĂďŽǀĞͬďĞůŽǁƐĞĂůĞǀĞů͕ĐƌĞĚŝƚƐͬĚĞďŝƚƐ͕ƉŽƐŝƚŝǀĞͬŶĞŐĂƚŝǀĞĞůĞĐƚƌŝĐ ĐŚĂƌŐĞͿ͖ƵƐĞƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐƚŽƌĞƉƌĞƐĞŶƚƋƵĂŶƚŝƚŝĞƐŝŶƌĞĂůͲ ǁŽƌůĚĐŽŶƚĞdžƚƐ͕ĞdžƉůĂŝŶŝŶŐƚŚĞŵĞĂŶŝŶŐŽĨ ͲŝŶĞĂĐŚƐŝƚƵĂƚŝŽŶ͘
hŶĚĞƌƐƚĂŶĚĂƌĂƚŝŽŶĂůŶƵŵďĞƌĂƐĂƉŽŝŶƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ džƚĞŶĚŶƵŵďĞƌ ůŝŶĞĚŝĂŐƌĂŵƐĂŶĚĐŽŽƌĚŝŶĂƚĞĂdžĞƐĨĂŵŝůŝĂƌĨƌŽŵƉƌĞǀŝŽƵƐŐƌĂĚĞƐƚŽƌĞƉƌĞƐĞŶƚ ƉŽŝŶƚƐŽŶƚŚĞůŝŶĞĂŶĚŝŶƚŚĞƉůĂŶĞǁŝƚŚŶĞŐĂƚŝǀĞŶƵŵďĞƌĐŽŽƌĚŝŶĂƚĞƐ͘
ZĞĐŽŐŶŝnjĞŽƉƉŽƐŝƚĞƐŝŐŶƐŽĨŶƵŵďĞƌƐĂƐŝŶĚŝĐĂƚŝŶŐůŽĐĂƚŝŽŶƐŽŶŽƉƉŽƐŝƚĞ ƐŝĚĞƐŽĨ ͲŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͖ƌĞĐŽŐŶŝnjĞƚŚĂƚƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞ ŽƉƉŽƐŝƚĞŽĨĂŶƵŵďĞƌŝƐƚŚĞŶƵŵďĞƌŝƚƐĞůĨ͕Ğ͘Ő͕͘ െሺെ͵ሻ ൌ ͵ ͕ ĂŶĚƚŚĂƚ ͲŝƐ ŝƚƐŽǁŶŽƉƉŽƐŝƚĞ͘
&ŝŶĚĂŶĚƉŽƐŝƚŝŽŶŝŶƚĞŐĞƌƐĂŶĚŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶĂŚŽƌŝnjŽŶƚĂůŽƌ ǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵ͖ĨŝŶĚĂŶĚƉŽƐŝƚŝŽŶƉĂŝƌƐŽĨŝŶƚĞŐĞƌƐĂŶĚ ŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶĂĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘
Instructional Days: ϲ
ĞƐƐŽŶϭ͗ WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶĂŶĚsĂůƵĞ; Ϳ ϭ
ĞƐƐŽŶ s 2–3: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ;W͕ Ϳ
ĞƐƐŽŶϰ : dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ;WͿ
ĞƐƐŽŶϱ : dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ͛ƐKƉƉŽƐŝƚĞ;WͿ
ĞƐƐŽŶϲ : ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ;^Ϳ ϭ>ĞƐƐŽŶ^ƚƌƵĐƚƵƌĞ<ĞLJ͗ WͲWƌŽďůĞŵ^Ğƚ>ĞƐƐŽŶ͕ MͲDŽĚĞůŝŶŐLJĐůĞ>ĞƐƐŽŶ͕ EͲ džƉůŽƌĂƚŝŽŶ>ĞƐƐŽŶ͕ SͲ^ŽĐƌĂƚŝĐ>ĞƐƐŽŶ
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6ͻϯ Topic A Topic A: hŶĚĞƌƐƚĂŶĚŝŶŐWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ /ŶdŽƉŝĐ͕ƐƚƵĚĞŶƚƐĂƉƉůLJƚŚĞŝƌƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨƚŚĞŽƌĚĞƌŝŶŐŽĨǁŚŽůĞŶƵŵďĞƌƐ͕ƉŽƐŝƚŝǀĞĨƌĂĐƚŝŽŶƐ͕ĂŶĚ
ĚĞĐŝŵĂůƐƚŽĞdžƚĞŶĚƚŚĞŶƵŵďĞƌůŝŶĞŝŶƚŚĞŽƉƉŽƐŝƚĞĚŝƌĞĐƚŝŽŶ͘/Ŷ>ĞƐƐŽŶƐϭʹϯ͕ƐƚƵĚĞŶƚƐƵƐĞƉŽƐŝƚŝǀĞŝŶƚĞŐĞƌƐ
ƚŽůŽĐĂƚĞŶĞŐĂƚŝǀĞŝŶƚĞŐĞƌƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͕ŵŽǀŝŶŐŝŶƚŚĞŽƉƉŽƐŝƚĞĚŝƌĞĐƚŝŽŶĨƌŽŵnjĞƌŽ͕ƌĞĂůŝnjŝŶŐƚŚĂƚnjĞƌŽ
ŝƐŝƚƐŽǁŶŽƉƉŽƐŝƚĞ͘dŚĞLJƌĞƉƌĞƐĞŶƚƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƐǁŝƚŚŝŶƚĞŐĞƌƐĂŶĚƵŶĚĞƌƐƚĂŶĚƚŚĞǀŽĐĂďƵůĂƌLJĂŶĚ
ĐŽŶƚĞdžƚƌĞůĂƚĞĚƚŽŽƉƉŽƐŝƚĞƋƵĂŶƚŝƚŝĞƐ;Ğ͘Ő͕͘ĚĞƉŽƐŝƚͬǁŝƚŚĚƌĂǁ͕ĞůĞǀĂƚŝŽŶĂďŽǀĞͬďĞůŽǁƐĞĂůĞǀĞů͕ĚĞďŝƚͬĐƌĞĚŝƚͿ͘
^ƚƵĚĞŶƚƐƵƐĞƉƌĞĐŝƐĞǀŽĐĂďƵůĂƌLJƚŽƐƚĂƚĞ͕ĨŽƌŝŶƐƚĂŶĐĞ͕ƚŚĂƚ െͳͲ ǁŽƵůĚĚĞƐĐƌŝďĞĂŶĞůĞǀĂƚŝŽŶƚŚĂƚŝƐ ͳͲ ĨĞĞƚ
ďĞůŽǁƐĞĂůĞǀĞů͘/Ŷ>ĞƐƐŽŶƐϰĂŶĚϱ͕ƐƚƵĚĞŶƚƐĨŽĐƵƐŽŶůŽĐĂƚŝŶŐƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶƵŵďĞƌĂŶĚƚŚĞŽƉƉŽƐŝƚĞŽĨ
ĂŶŽƉƉŽƐŝƚĞ͕ƵƐŝŶŐnjĞƌŽĂŶĚƚŚĞƐLJŵŵĞƚƌLJŽĨƚŚĞŶƵŵďĞƌůŝŶĞƚŽďƵŝůĚĂĐŽŶĐĞƉƚƵĂůƵŶĚĞƌƐƚĂŶĚŝŶŐ͘/Ŷ>ĞƐƐŽŶϲ͕ ƐƚƵĚĞŶƚƐĞdžƚĞŶĚƚŚĞŝƌƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨŝŶƚĞŐĞƌƐƚŽůŽĐĂƚĞƐŝŐŶĞĚŶŽŶͲŝŶƚĞŐĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶƚŚĞŶƵŵďĞƌ
ůŝŶĞ͕ ƌĞĂůŝnjŝŶŐƚŚĂƚĨŝŶĚŝŶŐƚŚĞŽƉƉŽƐŝƚĞ ŽĨĂŶLJƌĂƚŝŽŶĂůŶƵŵďĞƌŝƐƚŚĞƐĂŵĞĂƐĨŝŶĚŝŶŐĂŶŝŶƚĞŐĞƌ͛ƐŽƉƉŽƐŝƚĞ͘
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Lesson 1: WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶ ĂŶĚsĂůƵĞ
6ͻϯ Lesson 1 Lesson 1: Positive and Negative Numbers on the Number Line—Opposite Direction and Value Student Outcomes
^ƚƵĚĞŶƚƐĞdžƚĞŶĚƚŚĞŝƌƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨƚŚĞŶƵŵďĞƌůŝŶĞ͕ǁŚŝĐŚŝŶĐůƵĚĞƐnjĞƌŽĂŶĚŶƵŵďĞƌƐƚŽƚŚĞƌŝŐŚƚŽƌ ĂďŽǀĞnjĞƌŽƚŚĂƚĂƌĞŐƌĞĂƚĞƌƚŚĂŶnjĞƌŽĂŶĚŶƵŵďĞƌƐƚŽƚŚĞůĞĨƚŽƌďĞůŽǁnjĞƌŽƚŚĂƚĂƌĞůĞƐƐƚŚĂŶnjĞƌŽ͘
^ƚƵĚĞŶƚƐƵƐĞƉŽƐŝƚŝǀĞŝŶƚĞŐĞƌƐƚŽůŽĐĂƚĞŶĞŐĂƚŝǀĞŝŶƚĞŐĞƌƐďLJŵŽǀŝŶŐŝŶƚŚĞŽƉƉŽƐŝƚĞĚŝƌĞĐƚŝŽŶĨƌŽŵnjĞƌŽ͘
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĂƚƚŚĞƐĞƚŽĨŝŶƚĞŐĞƌƐŝŶĐůƵĚĞƐƚŚĞƐĞƚŽĨƉŽƐŝƚŝǀĞǁŚŽůĞŶƵŵďĞƌƐĂŶĚƚŚĞŝƌŽƉƉŽƐŝƚĞƐ͕ĂƐ ǁĞůůĂƐnjĞƌŽ͘dŚĞLJĂůƐŽƵŶĚĞƌƐƚĂŶĚƚŚĂƚnjĞƌŽŝƐŝƚƐŽǁŶŽƉƉŽƐŝƚĞ͘
Lesson Notes
ĂĐŚƐƚƵĚĞŶƚŶĞĞĚƐĂĐŽŵƉĂƐƐƚŽĐŽŵƉůĞƚĞƚŚĞ džƉůŽƌĂƚŽƌLJŚĂůůĞŶŐĞŝŶƚŚŝƐůĞƐƐŽŶ͘
Classwork Opening Exercise ;ϯ minutes): Number Line Review
ŝƐƉůĂLJƚǁŽŶƵŵďĞƌůŝŶĞƐ;ŚŽƌŝnjŽŶƚĂůĂŶĚǀĞƌƚŝĐĂůͿ͕ĞĂĐŚŶƵŵďĞƌĞĚ ͲʹͳͲ ͘ ůůŽǁƐƚƵĚĞŶƚƐ ƚŽĚŝƐĐƵƐƐƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐŝŶĐŽŽƉĞƌĂƚŝǀĞůĞĂƌŶŝŶŐŐƌŽƵƉƐŽĨƚŚƌĞĞŽƌĨŽƵƌƐƚƵĚĞŶƚƐ ĞĂĐŚ͘ ^ƚƵĚĞŶƚƐƐŚŽƵůĚƌĞŵĂŝŶŝŶƚŚĞŐƌŽƵƉƐĨŽƌƚŚĞĞŶƚŝƌĞůĞƐƐŽŶ͘ ŝƐĐƵƐƐƚŚĞĨŽůůŽǁŝŶŐ͗
tŚĂƚŝƐƚŚĞƐƚĂƌƚŝŶŐƉŽƐŝƚŝŽŶŽŶďŽƚŚŶƵŵďĞƌůŝŶĞƐ͍
à Ͳ (zero)
tŚĂƚŝƐƚŚĞůĂƐƚǁŚŽůĞŶƵŵďĞƌĚĞƉŝĐƚĞĚŽŶďŽƚŚŶƵŵďĞƌůŝŶĞƐ͍
à ͳͲ (ten)
KŶĂŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͕ĚŽƚŚĞŶƵŵďĞƌƐŝŶĐƌĞĂƐĞŽƌĚĞĐƌĞĂƐĞĂƐLJŽƵŵŽǀĞ ĨĂƌƚŚĞƌƚŽƚŚĞƌŝŐŚƚŽĨnjĞƌŽ͍
à Increase
KŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͕ĚŽƚŚĞŶƵŵďĞƌƐŝŶĐƌĞĂƐĞŽƌĚĞĐƌĞĂƐĞĂƐLJŽƵŵŽǀĞ ĨĂƌƚŚĞƌĂďŽǀĞnjĞƌŽ͍
à Increase
Scaffolding:
ƌĞĂƚĞƚǁŽĨůŽŽƌŵŽĚĞůƐ ĨŽƌǀĞƌƚŝĐĂůĂŶĚŚŽƌŝnjŽŶƚĂů ŶƵŵďĞƌůŝŶĞƐƵƐŝŶŐ ƉĂŝŶƚĞƌ͛ƐƚĂƉĞĨŽƌǀŝƐƵĂů ůĞĂƌŶĞƌƐ͘
,ĂǀĞĂƐƚƵĚĞŶƚŵŽĚĞů ŵŽǀĞŵĞŶƚĂůŽŶŐĞĂĐŚ ŶƵŵďĞƌůŝŶĞĨŽƌŬŝŶĞƐƚŚĞƚŝĐ ůĞĂƌŶĞƌƐ͘
hƐĞƉŽůůŝŶŐƐŽĨƚǁĂƌĞĨŽƌ ƋƵĞƐƚŝŽŶƐƚŽŐĂŝŶ ŝŵŵĞĚŝĂƚĞĨĞĞĚďĂĐŬǁŚŝůĞ ĂĐĐĞƐƐŝŶŐƉƌŝŽƌŬŶŽǁůĞĚŐĞ͘
A STORY OF RATIOS
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Lesson 1: WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶ ĂŶĚsĂůƵĞ
6ͻϯ Lesson 1 Exploratory Challenge (10 minutes): Constructing the Number Line
dŚĞƉƵƌƉŽƐĞŽĨƚŚŝƐĞdžĞƌĐŝƐĞŝƐƚŽůĞƚƐƚƵĚĞŶƚƐĐŽŶƐƚƌƵĐƚƚŚĞŶƵŵďĞƌůŝŶĞ;ƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐĂŶĚnjĞƌŽͿƵƐŝŶŐ ĂĐŽŵƉĂƐƐ͘ ,ĂǀĞƐƚƵĚĞŶƚƐĚƌĂǁĂůŝŶĞ͕ƉůĂĐĞĂƉŽŝŶƚŽŶƚŚĞůŝŶĞ͕ĂŶĚůĂďĞůŝƚ Ͳ͘ ,ĂǀĞƐƚƵĚĞŶƚƐƵƐĞƚŚĞĐŽŵƉĂƐƐƚŽůŽĐĂƚĞĂŶĚůĂďĞůƚŚĞŶĞdžƚƉŽŝŶƚ ͳ͕ ƚŚƵƐĐƌĞĂƚŝŶŐƚŚĞƐĐĂůĞ͘^ƚƵĚĞŶƚƐĐŽŶƚŝŶƵĞƚŽůŽĐĂƚĞ ŽƚŚĞƌǁŚŽůĞŶƵŵďĞƌƐƚŽƚŚĞƌŝŐŚƚŽĨnjĞƌŽƵƐŝŶŐƚŚĞƐĂŵĞƵŶŝƚŵĞĂƐƵƌĞ͘ hƐŝŶŐƚŚĞƐĂŵĞƉƌŽĐĞƐƐ͕ŚĂǀĞƐƚƵĚĞŶƚƐůŽĐĂƚĞƚŚĞŽƉƉŽƐŝƚĞƐŽĨƚŚĞǁŚŽůĞŶƵŵďĞƌƐ͘,ĂǀĞƐƚƵĚĞŶƚƐůĂďĞůƚŚĞĨŝƌƐƚƉŽŝŶƚƚŽ ƚŚĞůĞĨƚŽĨnjĞƌŽ െͳ ͘ /ŶƚƌŽĚƵĐĞƚŽƚŚĞĐůĂƐƐƚŚĞĚĞĨŝŶŝƚŝŽŶŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶƵŵďĞƌ͘ ^ĂŵƉůĞƐƚƵĚĞŶƚǁŽƌŬŝƐƐŚŽǁŶďĞůŽǁ͘
dŚĞƐĞƚŽĨǁŚŽůĞŶƵŵďĞƌƐĂŶĚƚŚĞŝƌŽƉƉŽƐŝƚĞƐ͕ŝŶĐůƵĚŝŶŐnjĞƌŽ͕ĂƌĞĐĂůůĞĚ integers .ĞƌŽŝƐŝƚƐŽǁŶŽƉƉŽƐŝƚĞ͘ dŚĞŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵƐŚŽǁƐŝŶƚĞŐĞƌƐůŝƐƚĞĚŝŶŽƌĚĞƌĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚƵƐŝŶŐĞƋƵĂůƐƉĂĐĞƐ͘ DŽŶŝƚŽƌƐƚƵĚĞŶƚĐŽŶƐƚƌƵĐƚŝŽŶƐ͕ŵĂŬŝŶŐƐƵƌĞƐƚƵĚĞŶƚƐĂƌĞƉĂLJŝŶŐĐůŽƐĞĂƚƚĞŶƚŝŽŶƚŽƚŚĞĚŝƌĞĐƚŝŽŶĂŶĚƐŝŐŶŽĨĂŶƵŵďĞƌ͘
Example 1 (5 minutes): Negative Numbers on the Number Line
^ƚƵĚĞŶƚƐƵƐĞƚŚĞŝƌĐŽŶƐƚƌƵĐƚŝŽŶƐƚŽŵŽĚĞůƚŚĞůŽĐĂƚŝŽŶŽĨĂŶƵŵďĞƌƌĞůĂƚŝǀĞƚŽnjĞƌŽďLJƵƐŝŶŐĂĐƵƌǀĞĚĂƌƌŽǁƐƚĂƌƚŝŶŐĂƚ njĞƌŽĂŶĚƉŽŝŶƚŝŶŐĂǁĂLJĨƌŽŵnjĞƌŽƚŽǁĂƌĚƚŚĞŶƵŵďĞƌ͘WŽƐĞƋƵĞƐƚŝŽŶƐƚŽƐƚƵĚĞŶƚƐĂƐĂǁŚŽůĞŐƌŽƵƉ͕ŽŶĞƋƵĞƐƚŝŽŶĂƚĂ ƚŝŵĞ͘
'ŝǀĞŶĂŶŽŶnjĞƌŽŶƵŵďĞƌ͕ ͕ܽ ŽŶĂŶƵŵďĞƌůŝŶĞ͕ƚŚĞŽƉƉŽƐŝƚĞŽĨ ͕ܽ ůĂďĞůĞĚ ܽെ ͕ ŝƐƚŚĞŶƵŵďĞƌƐƵĐŚ ƚŚĂƚ
ͲŝƐďĞƚǁĞĞŶ ܽ ĂŶĚ ܽെ .
dŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶ ͲĂŶĚ ܽ ŝƐĞƋƵĂůƚŽƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶ ͲĂŶĚ ܽെ ͘ dŚĞŽƉƉŽƐŝƚĞŽĨ ͲŝƐ Ͳ͘
ͻͺͷͶ͵ʹͳͲ ͳʹ ͳͳ ͳͲ
A STORY OF RATIOS
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©201 8Great Minds ®. eureka-math.org
Lesson 1: WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶ ĂŶĚsĂůƵĞ
6ͻϯ Lesson 1
^ƚĂƌƚŝŶŐĂƚ Ͳ͕ ĂƐ/ŵŽǀĞƚŽƚŚĞƌŝŐŚƚŽŶĂŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͕ƚŚĞǀĂůƵĞƐŐĞƚůĂƌŐĞƌ͘dŚĞƐĞŶƵŵďĞƌƐĂƌĞ ĐĂůůĞĚ positive numbers ďĞĐĂƵƐĞƚŚĞLJĂƌĞŐƌĞĂƚĞƌƚŚĂŶnjĞƌŽ͘EŽƚŝĐĞƚŚĞĐƵƌǀĞĚĂƌƌŽǁŝƐƉŽŝŶƚŝŶŐƚŽƚŚĞƌŝŐŚƚƚŽ ƐŚŽǁĂƉŽƐŝƚŝǀĞĚŝƌĞĐƚŝŽŶ͘
,ŽǁĨĂƌŝƐƚŚĞŶƵŵďĞƌĨƌŽŵnjĞƌŽŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à ͵ units
/Ĩ Ͳ ǁĂƐĂŵŝƌƌŽƌĨĂĐŝŶŐƚŽǁĂƌĚƚŚĞĂƌƌŽǁ͕ǁŚĂƚǁŽƵůĚďĞƚŚĞĚŝƌĞĐƚŝŽŶŽĨƚŚĞĂƌƌŽǁŝŶƚŚĞŵŝƌƌŽƌ͍
à To the left
tŽƵůĚƚŚĞŶƵŵďĞƌƐŐĞƚůĂƌŐĞƌŽƌƐŵĂůůĞƌĂƐǁĞŵŽǀĞƚŽƚŚĞůĞĨƚŽĨnjĞƌŽ͍
à Smaller
^ƚĂƌƚŝŶŐĂƚ Ͳ ͕ĂƐ/ŵŽǀĞĨĂƌƚŚĞƌƚŽƚŚĞůĞĨƚŽĨnjĞƌŽŽŶĂŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͕ƚŚĞǀĂůƵĞƐŐĞƚƐŵĂůůĞƌ͘dŚĞƐĞ ŶƵŵďĞƌƐĂƌĞĐĂůůĞĚ negative numbers ďĞĐĂƵƐĞƚŚĞLJĂƌĞůĞƐƐƚŚĂŶnjĞƌŽ͘EŽƚŝĐĞƚŚĞĐƵƌǀĞĚĂƌƌŽǁŝƐƉŽŝŶƚŝŶŐƚŽ ƚŚĞůĞĨƚƚŽƐŚŽǁĂŶĞŐĂƚŝǀĞĚŝƌĞĐƚŝŽŶ͘dŚĞƉŽƐŝƚŝŽŶŽĨƚŚĞƉŽŝŶƚŝƐŶŽǁĂƚŶĞŐĂƚŝǀĞ ͵͕ ǁƌŝƚƚĞŶĂƐ െ͵ ͘
EĞŐĂƚŝǀĞŶƵŵďĞƌƐĂƌĞůĞƐƐƚŚĂŶnjĞƌŽ͘ƐLJŽƵŵŽǀĞƚŽƚŚĞůĞĨƚŽŶĂŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͕ƚŚĞǀĂůƵĞƐŽĨƚŚĞ ŶƵŵďĞƌƐĚĞĐƌĞĂƐĞ͘
tŚĂƚŝƐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶ ͵ĂŶĚ െ͵ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à ͵ and െ͵ are located on opposite sides of zero. They are both the same distance from zero. ͵ and െ͵ are called opposites.
ƐǁĞůŽŽŬĨĂƌƚŚĞƌƌŝŐŚƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͕ƚŚĞǀĂůƵĞƐŽĨƚŚĞŶƵŵďĞƌƐŝŶĐƌĞĂƐĞ͘ &ŽƌĞdžĂŵƉůĞ͕ െͳ ൏ Ͳ ൏ ͳ ൏ ʹ ൏ ͵ ͘
dŚŝƐŝƐĂůƐŽƚƌƵĞĨŽƌĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘KŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͕ƉŽƐŝƚŝǀĞŶƵŵďĞƌƐ ĂƌĞůŽĐĂƚĞĚĂďŽǀĞnjĞƌŽ͘ƐǁĞůŽŽŬƵƉǁĂƌĚŽŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͕ƚŚĞǀĂůƵĞƐŽĨƚŚĞ ŶƵŵďĞƌƐŝŶĐƌĞĂƐĞ͘KŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͕ŶĞŐĂƚŝǀĞŶƵŵďĞƌƐĂƌĞůŽĐĂƚĞĚďĞůŽǁnjĞƌŽ͘ ƐǁĞůŽŽŬĨĂƌƚŚĞƌĚŽǁŶŽŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͕ƚŚĞǀĂůƵĞƐŽĨƚŚĞŶƵŵďĞƌƐĚĞĐƌĞĂƐĞ͘
dŚĞƐĞƚŽĨǁŚŽůĞŶƵŵďĞƌƐĂŶĚƚŚĞŝƌŽƉƉŽƐŝƚĞƐ͕ŝŶĐůƵĚŝŶŐnjĞƌŽ͕ĂƌĞĐĂůůĞĚ integers ͘ĞƌŽŝƐ ŝƚƐŽǁŶŽƉƉŽƐŝƚĞ͘ŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵƐŚŽǁƐŝŶƚĞŐĞƌƐůŝƐƚĞĚŝŶŝŶĐƌĞĂƐŝŶŐŽƌĚĞƌĨƌŽŵ ůĞĨƚƚŽƌŝŐŚƚ͕ŽƌĨƌŽŵďŽƚƚŽŵƚŽƚŽƉ͕ƵƐŝŶŐĞƋƵĂůƐƉĂĐĞƐ͘ &ŽƌĞdžĂŵƉůĞ͗ െͶ ͕ െ͵ ͕ െʹ ͕ െͳ ͕ Ͳ ͕ ͳ ͕ ʹ ͕ ͵ ͕ Ͷ ͘
ůůŽǁƐƚƵĚĞŶƚƐƚŽĚŝƐĐƵƐƐƚŚĞĞdžĂŵƉůĞŝŶƚŚĞŝƌŐƌŽƵƉƐƚŽƐŽůŝĚŝĨLJƚŚĞŝƌƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨƉŽƐŝƚŝǀĞĂŶĚ ŶĞŐĂƚŝǀĞŶƵŵďĞƌƐ͘
WŽƐƐŝďůĞĚŝƐĐƵƐƐŝŽŶƋƵĞƐƚŝŽŶƐ͗
tŚĞƌĞĂƌĞŶĞŐĂƚŝǀĞŶƵŵďĞƌƐůŽĐĂƚĞĚŽŶĂŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͍
à Negative numbers are located to the left of Ͳ on a horizontal number line.
െ͵ െͶ െͷ െ െ െͺ െͻ െͳͲ െͳͳ െͳʹ Ͳെͳ െʹ
െ͵ െͶ െͷ െ െ െͺ െͻ െͳͲ െͳͳ െͳʹ Ͳെͳ െʹ ͻͺͷͶ͵ʹͳͳʹ ͳͳ ͳͲ
െ͵
Ͳ
െͳ
െʹ
െͶ
Ͷ
͵
ʹ
ͳ
A STORY OF RATIOS
14
©201 8Great Minds ®. eureka-math.org
Lesson 1: WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶ ĂŶĚsĂůƵĞ
6ͻϯ Lesson 1
tŚĞƌĞĂƌĞŶĞŐĂƚŝǀĞŶƵŵďĞƌƐůŽĐĂƚĞĚŽŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͍
à Negative numbers are located below Ͳ on a vertical number line.
tŚĂƚŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨ ʹ͍
à െʹ
tŚĂƚŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨ Ͳ͍
à Ͳ
ĞƐĐƌŝďĞƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶ ͳͲ ĂŶĚ െͳͲ ͘
à ͳͲ and െͳͲ are opposites because they are on opposite sides of Ͳ and are both ͳͲ units from Ͳ.
Example 2 (5 minutes): Using Positive Integers to Locate Negative Integers on the Number Line
,ĂǀĞƐƚƵĚĞŶƚƐĞƐƚĂďůŝƐŚĞůďŽǁƉĂƌƚŶĞƌƐ͕ĂŶĚƚĞůůƚŚĞŵƚŽŵŽǀĞƚŚĞŝƌĨŝŶŐĞƌƐĂůŽŶŐƚŚĞŝƌŶƵŵďĞƌůŝŶĞƐƚŽĂŶƐǁĞƌƚŚĞ ĨŽůůŽǁŝŶŐƐĞƚŽĨƋƵĞƐƚŝŽŶƐ͘^ƚƵĚĞŶƚƐĐĂŶĚŝƐĐƵƐƐĂŶƐǁĞƌƐǁŝƚŚƚŚĞŝƌĞůďŽǁƉĂƌƚŶĞƌƐ͘ŝƌĐƵůĂƚĞĂƌŽƵŶĚƚŚĞƌŽŽŵ͕ĂŶĚ ůŝƐƚĞŶƚŽƚŚĞƐƚƵĚĞŶƚʹƉĂƌƚŶĞƌĚŝƐĐƵƐƐŝŽŶƐ͘
ĞƐĐƌŝďĞƚŽLJŽƵƌĞůďŽǁƉĂƌƚŶĞƌŚŽǁƚŽĨŝŶĚ ͶŽŶĂŶƵŵďĞƌůŝŶĞ͘ĞƐĐƌŝďĞŚŽǁƚŽ ĨŝŶĚ െͶ ͘
à To find Ͷ, start at zero, and move right to Ͷ. To find െͶ , start at zero, and move left to െͶ .
DŽĚĞůŚŽǁƚŚĞůŽĐĂƚŝŽŶŽĨĂƉŽƐŝƚŝǀĞŝŶƚĞŐĞƌĐĂŶďĞƵƐĞĚƚŽůŽĐĂƚĞĂŶĞŐĂƚŝǀĞŝŶƚĞŐĞƌďLJ ŵŽǀŝŶŐŝŶƚŚĞŽƉƉŽƐŝƚĞĚŝƌĞĐƚŝŽŶ͘
džƉůĂŝŶĂŶĚƐŚŽǁŚŽǁƚŽĨŝŶĚ ͶĂŶĚƚŚĞŽƉƉŽƐŝƚĞŽĨ ͶŽŶĂŶƵŵďĞƌůŝŶĞ͘
à Start at zero, and move Ͷ units to the right to locate Ͷ on the number line. To locate െͶ , start at zero, and move Ͷ units to the left on the number line.
tŚĞƌĞĚŽLJŽƵƐƚĂƌƚǁŚĞŶůŽĐĂƚŝŶŐĂŶŝŶƚĞŐĞƌŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à Always start at zero.
tŚĂƚĚŽLJŽƵŶŽƚŝĐĞĂďŽƵƚƚŚĞĐƵƌǀĞĚĂƌƌŽǁƐƚŚĂƚƌĞƉƌĞƐĞŶƚƚŚĞůŽĐĂƚŝŽŶŽĨ ͶĂŶĚ െͶ ͍
à They are the same distance but pointing in opposite directions.
Exercises 1–5 (1 ϯ minutes)
ƌĞĂƚĞĂŶĚĚŝƐƉůĂLJƚǁŽůĂƌŐĞŶƵŵďĞƌůŝŶĞƐŽŶƚŚĞďŽĂƌĚ͕ŽŶĞŚŽƌŝnjŽŶƚĂůĂŶĚŽŶĞǀĞƌƚŝĐĂů͕ĞĂĐŚŶƵŵďĞƌĞĚ െͳʹ ƚŽ ͳʹ ͕ ĨŽƌ ƚŚĞĐůĂƐƐ͘ŝƐƚƌŝďƵƚĞĂŶŝŶĚĞdžĐĂƌĚǁŝƚŚĂŶŝŶƚĞŐĞƌĨƌŽŵ െͳʹ ƚŽ ͳʹ ŽŶŝƚƚŽĞĂĐŚŐƌŽƵƉ͘ ϭ^ƚƵĚĞŶƚƐǁŽƌŬŝŶŐƌŽƵƉƐĨŝƌƐƚ͕ ĐŽŵƉůĞƚŝŶŐƚŚĞĞdžĞƌĐŝƐĞƐŝŶƚŚĞŝƌƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐ͘ŽŶĐůƵĚĞƚŚĞĞdžĞƌĐŝƐĞďLJŚĂǀŝŶŐƐƚƵĚĞŶƚƐůŽĐĂƚĞĂŶĚůĂďĞůƚŚĞŝƌ ŝŶƚĞŐĞƌƐŽŶƚŚĞƚĞĂĐŚĞƌ͛ƐŶƵŵďĞƌůŝŶĞƐ͘ ϭĞƉĞŶĚŝŶŐŽŶƚŚĞĐůĂƐƐƐŝnjĞ͕ůĂďĞůĞŶŽƵŐŚŝŶĚĞdžĐĂƌĚƐĨŽƌƚĞŶŐƌŽƵƉƐ;ŽŶĞĐĂƌĚƉĞƌŐƌŽƵƉͿ͘sĂƌLJƚŚĞŶƵŵďĞƌƐƵƐŝŶŐƉŽƐŝƚŝǀĞƐĂŶĚ ŶĞŐĂƚŝǀĞƐ͕ƐƵĐŚĂƐ െͷ ƚŚƌŽƵŐŚ ͷ͕ ŝŶĐůƵĚŝŶŐnjĞƌŽ͘/ĨĞĂĐŚŐƌŽƵƉĨŝŶĚƐĂŶĚůŽĐĂƚĞƐƚŚĞŝŶƚĞŐĞƌĐŽƌƌĞĐƚůLJ͕ĞĂĐŚŐƌŽƵƉǁŝůůŚĂǀĞĂĐĂƌĚƚŚĂƚ ŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶŽƚŚĞƌŐƌŽƵƉ͛ƐĐĂƌĚ͘
െ͵ െͶ െͷ Ͳെͳ െʹ ͷͶ͵ʹͳ
Scaffolding:
ƐĂŶĞdžƚĞŶƐŝŽŶĂĐƚŝǀŝƚLJ͕ŚĂǀĞ ƐƚƵĚĞŶƚƐŝĚĞŶƚŝĨLJƚŚĞ unit ĚŝĨĨĞƌĞŶƚůLJŽŶĚŝĨĨĞƌĞŶƚŶƵŵďĞƌ ůŝŶĞƐ͕ĂŶĚĂƐŬƐƚƵĚĞŶƚƐƚŽ ůŽĐĂƚĞƚǁŽǁŚŽůĞŶƵŵďĞƌƐ ŽƚŚĞƌƚŚĂŶ ͳĂŶĚƚŚĞŝƌ ŽƉƉŽƐŝƚĞƐ͘
A STORY OF RATIOS
15
©201 8Great Minds ®. eureka-math.org
Lesson 1: WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶ ĂŶĚsĂůƵĞ
6ͻϯ Lesson 1 Exercises Complete the diagrams. Count by ones to label the number lines. 1. Plot your point on both number lines.
Answers may vary.
Show and explain how to find the opposite of your number on both number lines.
In this example, the number chosen was െ . So െ is the first number plotted, and the opposite is .Horizontal Number Line: I found my point by starting at zero and counting four units to the left to end on െ . Then, to find the opposite of my number, I started on zero and counted to the right four units to end on .Vertical Number Line: I found my point by starting at zero and counting four units down to end on
െ .I found the opposite of my number by starting at zero and counting four units up to end on .
ϯ͘ Mark the opposite on both number lines.
Answers may vary.
Choose a group representative to place the opposite number on the class number lines. 5. Which group had the opposite of the number on your index card?
Answers may vary. Jackie’s group had the opposite of the number on my index card. They had .
Closing (2 minutes)
'ŝǀĞĂŶĞdžĂŵƉůĞŽĨƚǁŽŽƉƉŽƐŝƚĞŶƵŵďĞƌƐ͕ĂŶĚĚĞƐĐƌŝďĞƚŚĞŝƌůŽĐĂƚŝŽŶƐĨŝƌƐƚŽŶĂŚŽƌŝnjŽŶƚĂůĂŶĚƚŚĞŶŽŶĂ ǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘
à For example, and െ are the same distance from zero but on opposite sides. Positive is located
units to the right of zero on a horizontal number line and units above zero on a vertical number line. Negative is located units to the left of zero on a horizontal number line and units below zero on a vertical number line.
Exit Ticket (7 minutes)
െ െ െ െ െ
െ
െ
െ
െ
A STORY OF RATIOS
16
©201 8Great Minds ®. eureka-math.org
Lesson 1: WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶ ĂŶĚsĂůƵĞ
6ͻϯ Lesson 1
EĂŵĞ ĂƚĞ
Lesson 1: Positive and Negative Numbers on the Number Line— Opposite Direction and Value Exit Ticket
ϭ͘ /ĨnjĞƌŽůŝĞƐďĞƚǁĞĞŶ ܽ ĂŶĚ ͕݀ ŐŝǀĞŽŶĞƐĞƚŽĨƉŽƐƐŝďůĞǀĂůƵĞƐĨŽƌ ͕ܽ ܾ͕ ͕ܿ ĂŶĚ ݀͘ Ϯ͘ ĞůŽǁŝƐĂůŝƐƚŽĨŶƵŵďĞƌƐŝŶŽƌĚĞƌĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͘hƐĞǁŚĂƚLJŽƵŬŶŽǁĂďŽƵƚƚŚĞŶƵŵďĞƌůŝŶĞƚŽĐŽŵƉůĞƚĞ ƚŚĞůŝƐƚŽĨŶƵŵďĞƌƐďLJĨŝůůŝŶŐŝŶƚŚĞďůĂŶŬƐǁŝƚŚƚŚĞŵŝƐƐŝŶŐŝŶƚĞŐĞƌƐ͘
െ ͕ െͷ ͕ ͕ െ͵ ͕ െʹ ͕ െͳ ͕ ͕ ͳ ͕ ʹ ͕͕ Ͷ͕ ͕ ϯ͘ ŽŵƉůĞƚĞƚŚĞŶƵŵďĞƌůŝŶĞƐĐĂůĞ͘ džƉůĂŝŶĂŶĚƐŚŽǁŚŽǁƚŽĨŝŶĚ ʹĂŶĚƚŚĞŽƉƉŽƐŝƚĞŽĨ ʹŽŶĂŶƵŵďĞƌůŝŶĞ͘ ܽ
ܾ ܿ ݀
Ͳ ʹ
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Lesson 1: WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶ ĂŶĚsĂůƵĞ
6ͻϯ Lesson 1 Exit Ticket Sample Solutions 1. If zero lies between ࢇ and ࢊ, give one set of possible values for ࢇ,࢈ ,ࢉ , and ࢊ.
Answers will vary. One possible answer is ࢇ: െ ; ࢈: െ ; ࢉ: ; ࢊ: Ǥ
Below is a list of numbers in order from least to greatest. Use what you know about the number line to complete the list of numbers by filling in the blanks with the missing integers.
െ , െ , െ , െ , െ , െ , , , , , , ,
ϯ͘ Complete the number line scale. Explain and show how to find and the opposite of on a number line.
I would start at zero and move units to the left to locate the number െ on the number line. So, to locate , I would start at zero and move units to the right (the opposite direction).
Problem Set Sample Solutions 1. Draw a number line, and create a scale for the number line in order to plot the points െ , , and .a. Graph each point and its opposite on the number line. b. Explain how you found the opposite of each point.
To graph each point, I started at zero and moved right or left based on the sign and number (to the right for a positive number and to the left for a negative number). To graph the opposites, I started at zero, but this time I moved in the opposite direction the same number of times.
Carlos uses a vertical number line to graph the points െ , െ , , and . He notices that െ is closer to zero than െ . He is not sure about his diagram. Use what you know about a vertical number line to determine if Carlos made a mistake or not. Support your explanation with a number line diagram.
Carlos made a mistake because െ is less than െ , so it should be farther down the number line. Starting at zero, negative numbers decrease as we look farther below zero. So, െ lies before െ on a number line since െ is units below zero and െ is units below zero.
ࢇ ࢈ ࢉ ࢊ
െ െ െ െ െ
െ െ െ െ െૠ െૡ െૢ െ െ െ െ െ ૢ ૡૠ
െ
െ
െ
െ
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Lesson 1: WŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞͶKƉƉŽƐŝƚĞŝƌĞĐƚŝŽŶ ĂŶĚsĂůƵĞ
6ͻϯ Lesson 1
ϯ͘ Create a scale in order to graph the numbers െ through on a number line. What does each tick mark represent?
Each tick mark represents unit.
Choose an integer between െ and െ . Label it ࡾ on the number line created in P ƌŽďůĞŵϯ , and complete the following tasks.
Answers may vary. Refer to the number line above for sample student work. െ , െૠ , െૡ , or െૢ
a. What is the opposite of ࡾ? Label it ࡽ.
Answers will vary.
b. State a positive integer greater than ࡽ. Label it ࢀ.
Answers will vary.
c. State a negative integer greater than ࡾ. Label it ࡿ.
Answers will vary. െ
d. State a negative integer less than ࡾ. Label it ࢁ.
Answers will vary. െૢ
e. State an integer between ࡾ and Ǥࡽ Label it ࢂ.
Answers will vary.
Will the opposite of a positive number always, sometimes, or never be a positive number? Explain your reasoning.
The opposite of a positive number will never be a positive number. For two nonzero numbers to be opposites, zero has to be in between both numbers, and the distance from zero to one number has to equal the distance between zero and the other number.
Will the opposite of zero always, sometimes, or never be zero? Explain your reasoning.
The opposite of zero will always be zero because zero is its own opposite.
Will the opposite of a number always, sometimes, or never be greater than the number itself? Explain your reasoning. Provide an example to support your reasoning.
The opposite of a number will sometimes be greater than the number itself because it depends on the given number. For example, if the number given is െ , then the opposite is , which is greater than െ . If the number given is ,then the opposite is െ , which is not greater than . If the number given is , then the opposite is , which is never greater than itself.
െ െ െ െ െૠ െૡ െૢ െ െ െ െ െ ૢ ૡૠ
ࢁࡾࢂࡿࡽࢀ
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6ͻϯ Lesson 2 Lesson 2: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
Lesson 2: Real-World Positive and Negative Numbers and Zero Student Outcomes
^ƚƵĚĞŶƚƐƵƐĞƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐƚŽŝŶĚŝĐĂƚĞĂĐŚĂŶŐĞ;ŐĂŝŶŽƌůŽƐƐͿŝŶĞůĞǀĂƚŝŽŶǁŝƚŚĂĨŝdžĞĚ ƌĞĨĞƌĞŶĐĞƉŽŝŶƚ͕ƚĞŵƉĞƌĂƚƵƌĞ͕ĂŶĚƚŚĞďĂůĂŶĐĞŝŶĂďĂŶŬĂĐĐŽƵŶƚ͘
^ƚƵĚĞŶƚƐƵƐĞǀŽĐĂďƵůĂƌLJƉƌĞĐŝƐĞůLJǁŚĞŶĚĞƐĐƌŝďŝŶŐĂŶĚƌĞƉƌĞƐĞŶƚŝŶŐƐŝƚƵĂƚŝŽŶƐŝŶǀŽůǀŝŶŐŝŶƚĞŐĞƌƐ͖ĨŽƌĞdžĂŵƉůĞ͕ ĂŶĞůĞǀĂƚŝŽŶŽĨ െͳͲ ĨĞĞƚŝƐƚŚĞƐĂŵĞĂƐ ͳͲ ĨĞĞƚďĞůŽǁƚŚĞĨŝdžĞĚƌĞĨĞƌĞŶĐĞƉŽŝŶƚ͘
^ƚƵĚĞŶƚƐĐŚŽŽƐĞĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞĨŽƌƚŚĞŶƵŵďĞƌůŝŶĞǁŚĞŶŐŝǀĞŶĂƐĞƚŽĨƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐ ƚŽŐƌĂƉŚ͘
Classwork Opening Exercise (5 minutes)
ŝƐƉůĂLJĂŶƵŵďĞƌůŝŶĞǁŝƚŚŽƵƚĂƐĐĂůĞůĂďĞůĞĚ͘WŽƐĞƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐƚŽƚŚĞǁŚŽůĞ ŐƌŽƵƉ͕ĂŶĚĂůůŽǁƐƚƵĚĞŶƚƐƚŚƌĞĞŵŝŶƵƚĞƐƚŽĚŝƐĐƵƐƐƚŚĞŝƌƌĞƐƉŽŶƐĞƐŝŶƉĂŝƌƐ͘ZĞĐŽƌĚ ĨĞĞĚďĂĐŬďLJůĂďĞůŝŶŐĂŶĚƌĞůĂďĞůŝŶŐƚŚĞŶƵŵďĞƌůŝŶĞďĂƐĞĚŽŶĚŝĨĨĞƌĞŶƚƌĞƐƉŽŶƐĞƐ͘ ŝƐĐƵƐƐƚŚĞĨŽůůŽǁŝŶŐ͗
džƉůĂŝŶŚŽǁLJŽƵǁŽƵůĚƐŚŽǁ ͳͷͲ ŽŶĂŶƵŵďĞƌůŝŶĞ͘
à I would start at zero and move to the right ͳͷͲ units.
tŚĂƚƐƚƌĂƚĞŐLJǁŽƵůĚLJŽƵƵƐĞƚŽŶƵŵďĞƌƚŚĞŶƵŵďĞƌůŝŶĞŝŶŽƌĚĞƌƚŽƐŚŽǁ ͳͷͲ ͍
à I would locate (place) zero as far to the left as possible and use a scale of ͳͲ . I could also label the first tick mark ͳͶͲ and count by ones.
/ĨLJŽƵǁĂŶƚƚŽŚĂǀĞnjĞƌŽĂŶĚ ͳͷͲ ŽŶƚŚĞŐŝǀĞŶŶƵŵďĞƌůŝŶĞ͕ǁŚĂƚƐĐĂůĞƐǁŽƵůĚǁŽƌŬǁĞůů;ǁŚĂƚƐŚŽƵůĚLJŽƵ ĐŽƵŶƚďLJͿ͍
à I could count by fives, tens, or twenty-fives.
Common Misconceptions
džƉůĂŝŶƚŽƐƚƵĚĞŶƚƐŚŽǁƚŽĐŚŽŽƐĞĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞ͘WĂLJĐĂƌĞĨƵůĂƚƚĞŶƚŝŽŶƚŽƐƚƵĚĞŶƚŐƌĂƉŚƐ͕ĂŶĚĂĚĚƌĞƐƐĐŽŵŵŽŶ ŵŝƐĐŽŶĐĞƉƚŝŽŶƐ͕ƐƵĐŚĂƐ͗
x hŶĞƋƵĂůŝŶƚĞƌǀĂůƐͶ/ŶƚĞƌǀĂůƐƐŚŽƵůĚďĞĞƋƵĂůĨƌŽŵŽŶĞŵĂƌŬƚŽƚŚĞŶĞdžƚ͘dŚŝƐƵƐƵĂůůLJŚĂƉƉĞŶƐǁŚĞŶƐƚƵĚĞŶƚƐ ƐƚŽƉƐŬŝƉͲĐŽƵŶƚŝŶŐŝŶŽƌĚĞƌƚŽŵĂŬĞƚŚĞŶƵŵďĞƌƐĨŝƚŽŶƚŚĞĚŝĂŐƌĂŵ;Ğ͘Ő Ǥ͕ ͷ ͕ ͳͲ ͕ ͳͷ ͕ ʹͲ ͕ ͷͲ ͕ ͳͲͲ ͕ ͳͷͲ Ϳ͘
Scaffolding:
&ŽƌŬŝŶĞƐƚŚĞƚŝĐůĞĂƌŶĞƌƐ͕ ƉƌŽǀŝĚĞƐƚƵĚĞŶƚƐǁŝƚŚǁŚŝƚĞ ďŽĂƌĚƐĂŶĚŵĂƌŬĞƌƐƚŽĐƌĞĂƚĞ ƚŚĞŝƌŶƵŵďĞƌůŝŶĞƐ͘ƐŬƚŚĞŵ ƚŽŚŽůĚƵƉƚŚĞŝƌďŽĂƌĚƐ͕ĂŶĚ ƐĞůĞĐƚĂĨĞǁƐƚƵĚĞŶƚƐƚŽĞdžƉůĂŝŶ ƚŚĞŝƌĚŝĂŐƌĂŵƐƚŽƚŚĞĐůĂƐƐ .
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6ͻϯ Lesson 2 Lesson 2: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
x DŝƐĐŽƵŶƚŝŶŐͶdŚŝƐŝƐƵƐƵĂůůLJƚŚĞƌĞƐƵůƚŽĨƐƚƵĚĞŶƚƐƌƵƐŚŝŶŐĂŶĚŶŽƚƉĂLJŝŶŐĂƚƚĞŶƚŝŽŶƚŽĚĞƚĂŝůƐ͘^ƚƵĚĞŶƚƐ ƐŚŽƵůĚĂůǁĂLJƐĐŚĞĐŬƚŚĞŝƌƐĐĂůĞƐĨŽƌĂĐĐƵƌĂĐLJďĞĨŽƌĞƉůŽƚƚŝŶŐƉŽŝŶƚƐ͘
x ůǁĂLJƐƐƚĂƌƚŝŶŐĂƚnjĞƌŽͶdŚĞƉƌŽďůĞŵƐŚŽƵůĚĚĞƚĞƌŵŝŶĞƚŚĞĂƉƉƌŽƉƌŝĂƚĞƐƚĂƌƚĂŶĚĞŶĚƉŽŝŶƚĨŽƌĂŶƵŵďĞƌůŝŶĞ͘ ,ĞůƉƐƚƌƵŐŐůŝŶŐƐƚƵĚĞŶƚƐďLJĐŽƵŶƚŝŶŐƚŚĞŶƵŵďĞƌŽĨƚŝĐŬŵĂƌŬƐ;ůŝŶĞƐͿĨŝƌƐƚŝŶŽƌĚĞƌƚŽĚĞƚĞƌŵŝŶĞĂƐƚĂƌƚŝŶŐ ƉŽŝŶƚ͘
x EŽƚƵƐŝŶŐƚŚĞĞŶƚŝƌĞŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵͶ^ƉĂĐŝŶŐƐŚŽƵůĚďĞĞǀĞŶůLJĚŝƐƚƌŝďƵƚĞĚƚŚƌŽƵŐŚŽƵƚĂŶƵŵďĞƌůŝŶĞ͘ dŚŝƐƵƐƵĂůůLJŚĂƉƉĞŶƐǁŚĞŶƐƚƵĚĞŶƚƐĂƌĞĐŽƵŶƚŝŶŐďLJĂǀĂůƵĞƚŚĂƚŝƐƚŽŽůĂƌŐĞ;Ğ͘Ő͕͘ĐŽƵŶƚŝŶŐďLJƚĞŶƐŝŶƐƚĞĂĚŽĨ ƚǁŽƐͿ͘
Example 1 (10 minutes): Take It to the Bank
dŚĞƉƵƌƉŽƐĞŽĨƚŚŝƐĞdžĂŵƉůĞŝƐĨŽƌƐƚƵĚĞŶƚƐƚŽƵŶĚĞƌƐƚĂŶĚŚŽǁŶĞŐĂƚŝǀĞĂŶĚƉŽƐŝƚŝǀĞŶƵŵďĞƌƐĐĂŶďĞƵƐĞĚƚŽƌĞƉƌĞƐĞŶƚ ƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƐŝŶǀŽůǀŝŶŐŵŽŶĞLJ͘^ƚƵĚĞŶƚƐĂƌĞŝŶƚƌŽĚƵĐĞĚƚŽďĂƐŝĐĨŝŶĂŶĐŝĂůǀŽĐĂďƵůĂƌLJͶĚĞƉŽƐŝƚ͕ĐƌĞĚŝƚ;ĐƌĞĚŝƚĞĚͿ͕ ĚĞďŝƚ;ĚĞďŝƚĞĚͿ͕ǁŝƚŚĚƌĂǁĂů͕ĂŶĚĐŚĂŶŐĞ;ŐĂŝŶŽƌůŽƐƐͿƚŚƌŽƵŐŚŽƵƚƚŚĞĞdžĂŵƉůĞ͘dŚĞƚĞĂĐŚĞƌƐŚŽƵůĚĂĐĐĞƐƐƉƌŝŽƌ ŬŶŽǁůĞĚŐĞďLJŚĂǀŝŶŐƐƚƵĚĞŶƚƐŝŶĚĞƉĞŶĚĞŶƚůLJĐŽŵƉůĞƚĞƚŚĞĨŝƌƐƚƚǁŽĐŽůƵŵŶƐŽĨƚŚĞŐƌĂƉŚŝĐŽƌŐĂŶŝnjĞƌŝŶƚŚĞŝƌ ƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐ͘DŽŶŝƚŽƌƐƚƵĚĞŶƚƌĞƐƉŽŶƐĞƐ͕ĂŶĚƐĞůĞĐƚĂĨĞǁƐƚƵĚĞŶƚƐƚŽƐŚĂƌĞŽƵƚůŽƵĚ͘
Example 1: Take It to the Bank Read Example 1 silently. In the first column, write down any words and definitions you know. In the second column, write down any words you do not know. For Tim’s ϭϯ th birthday, he received ̈́ in cash from his mom. His dad took him to the bank to open a savings account. Tim gave the cash to the banker to deposit into the account. The banker credited Tim’s new account ̈́ and gave Tim a receipt. One week later, Tim deposited another ̈́ that he had earned as allowance. The next month, Tim’s dad gave him permission to withdraw ̈́ to buy a new video game. Tim’s dad explained that the bank would charge a ̈́ fee for each withdrawal from the savings account and that each withdrawal and charge results in a debit to the account. Words I Already Know:
Bank account—place where you put your money Receipt—ticket they give you to show how much you spent Allowance—money for chores Charge—something you pay
Words I Want to Know:
Credited Debit Fee Deposit Withdraw
Words I Learned: In the third column, write down any new words and definitions that you learn during the discussion.
Exercises 1–2 (7 minutes)
dŚĞƐĞĞdžĞƌĐŝƐĞƐĂƐŬƐƚƵĚĞŶƚƐƚŽŶƵŵďĞƌƚŚĞĞǀĞŶƚƐŽĨƚŚĞƐƚŽƌLJƉƌŽďůĞŵŝŶŽƌĚĞƌƚŽƐŚŽǁŚŽǁĞĂĐŚĂĐƚŝŽŶĐĂŶďĞ ƌĞƉƌĞƐĞŶƚĞĚďLJĂŶŝŶƚĞŐĞƌĂŶĚŵŽĚĞůĞĚŽŶĂŶƵŵďĞƌůŝŶĞ͘ZĞĐŽƌĚƚŚĞĞǀĞŶƚƐŝŶƚŚĞĚŝĂŐƌĂŵďĞůŽǁ͘
ŽŵƉůĞƚĞƚŚĞĨŝƌƐƚĞdžĞƌĐŝƐĞ͕ĂŶĚƚŚĞŶǁĂŝƚĨŽƌĨƵƌƚŚĞƌŝŶƐƚƌƵĐƚŝŽŶ͘
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6ͻϯ Lesson 2 Lesson 2: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
Exercises 1–2 1. Read Example 1 again. With your partner, number the events in the story problem. Write the number above each sentence to show the order of the events. For Tim’s ϭϯ th birthday, he received ̈́ in cash from his mom. His dad took him to the bank to open a savings account. Tim gave the cash to the banker to deposit into the account. The banker credited Tim’s new account ̈́ and gave Tim a receipt. One week later, Tim deposited another ̈́ that he had earned as allowance. The next month, Tim’s dad gave him permission to withdraw ̈́ to buy a new video game. Tim’s dad explained that the bank would charge a ̈́ fee for each withdrawal from the savings account and that each withdrawal and charge results in a debit to the account.
ĨƚĞƌƐƚƵĚĞŶƚƐĐŽŵƉůĞƚĞ džĞƌĐŝƐĞϭ͕ƉƌĞĐŝƐĞůLJĚĞĨŝŶĞǀŽĐĂďƵůĂƌLJƚŽĚĞƐĐƌŝďĞĞĂĐŚƐŝƚƵĂƚŝŽŶĂƐĂŶŝŶƚĞŐĞƌ͕ĂŶĚŵŽĚĞůƚŚĞ ŝŶƚĞŐĞƌŽŶĂŶƵŵďĞƌůŝŶĞ͘WŽƐĞƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐƚŚƌŽƵŐŚŽƵƚƚŚĞĞdžĞƌĐŝƐĞ͘WŽŝŶƚŽƵƚƚŚĂƚnjĞƌŽƌĞƉƌĞƐĞŶƚƐƚŚĞ ďĂůĂŶĐĞďĞĨŽƌĞĞĂĐŚƚƌĂŶƐĂĐƚŝŽŶŝŶƚŚĞƐƚŽƌLJƉƌŽďůĞŵ͘ ŝƐĐƵƐƐƚŚĞĨŽůůŽǁŝŶŐ͗
dŝŵƌĞĐĞŝǀĞƐ ̈́ ͳͷͲ ĨŽƌŚŝƐďŝƌƚŚĚĂLJ͘ŽLJŽƵƚŚŝŶŬƚŚŝƐǁŝůůďĞĂƉŽƐŝƚŝǀĞŽƌŶĞŐĂƚŝǀĞŶƵŵďĞƌĨŽƌdŝŵ͛ƐŵŽŶĞLJ͍ džƉůĂŝŶ͘
à Positive; ̈́ ͳͷͲ is a gain for Tim’s money. Positive numbers are greater than ͲǤ
,ŽǁŵƵĐŚŵŽŶĞLJŝƐŝŶƚŚĞĂĐĐŽƵŶƚǁŚĞŶdŝŵŽƉĞŶĞĚŝƚ͍tŚĂƚĚŽĞƐƚŚŝƐŶƵŵďĞƌƌĞƉƌĞƐĞŶƚŝŶƚŚŝƐƐŝƚƵĂƚŝŽŶ͍
à The account has ̈́Ͳ in it because Tim had not put in or taken out any money. Zero represents the starting account balance.
dŚĞ ̈́ ͳͷͲ ƚŚĂƚdŝŵŐŝǀĞƐƚŚĞďĂŶŬĞƌŝƐĐĂůůĞĚĂ deposit ͘ĚĞƉŽƐŝƚŝƐƚŚĞĂĐƚŽĨƉƵƚƚŝŶŐŵŽŶĞLJŝŶƚŽĂďĂŶŬ ĂĐĐŽƵŶƚ͘dŽƐŚŽǁƚŚĞĂŵŽƵŶƚŽĨŵŽŶĞLJŝŶdŝŵ͛ƐƐĂǀŝŶŐƐĂĐĐŽƵŶƚ͕ǁŽƵůĚƚŚŝƐĚĞƉŽƐŝƚďĞůŽĐĂƚĞĚƚŽƚŚĞůĞĨƚŽƌ ƌŝŐŚƚŽĨnjĞƌŽŽŶƚŚĞŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͍
à This deposit is located to the right of zero because it increases the amount of money in the savings account.
dŚĞďĂŶŬĐƌĞĚŝƚĞĚƚŚĞĂĐĐŽƵŶƚ ̈́ ͳͷͲ ͘ credit ŝƐǁŚĞŶŵŽŶĞLJŝƐĚĞƉŽƐŝƚĞĚŝŶƚŽĂŶĂĐĐŽƵŶƚ͘dŚĞĂĐĐŽƵŶƚ ŝŶĐƌĞĂƐĞƐŝŶǀĂůƵĞ͘,ŽǁǁŽƵůĚLJŽƵƌĞƉƌĞƐĞŶƚĂĐƌĞĚŝƚŽĨ ̈́ ͳͷͲ ĂƐĂŶŝŶƚĞŐĞƌ͍ džƉůĂŝŶ͘
à Since a credit is a deposit and deposits are written as positive numbers, then positive ͳͷͲ represents a credit of ̈́ ͳͷͲ .
dŝŵŵĂŬĞƐĂŶŽƚŚĞƌĚĞƉŽƐŝƚŽĨ ̈́ ʹͷ ͘ tŽƵůĚƚŚŝƐďĞĂƉŽƐŝƚŝǀĞŽƌŶĞŐĂƚŝǀĞŶƵŵďĞƌĨŽƌdŝŵ͛ƐƐĂǀŝŶŐƐĂĐĐŽƵŶƚ͕ĂŶĚ ŚŽǁǁŽƵůĚLJŽƵƐŚŽǁŝƚŽŶĂŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͍
à A deposit increases the amount of money in the savings account, so ʹͷ is positive. I would place the point ʹͷ units to the right of zero.
dŚĞďĂŶŬĐƌĞĂƚĞƐĂĚĞďŝƚŽĨ ̈́ͷ ĨŽƌĂŶLJǁŝƚŚĚƌĂǁĂů͘tŚĂƚĚŽLJŽƵƚŚŝŶŬƚŚĞǁŽƌĚ debit ŵĞĂŶƐŝŶƚŚŝƐƐŝƚƵĂƚŝŽŶ͍
à A debit sounds like the opposite of a credit. It might be something taken away. Taking money out of the savings account is the opposite of putting money in.
ૠ
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6ͻϯ Lesson 2 Lesson 2: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
ĚĞďŝƚŵĞĂŶƐŵŽŶĞLJƉĂŝĚŽƵƚŽĨĂŶĂĐĐŽƵŶƚ͘/ƚŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨĂĐƌĞĚŝƚ͘ƌĞĚĞďŝƚƐƌĞƉƌĞƐĞŶƚĞĚĂƐƉŽƐŝƚŝǀĞ ŽƌŶĞŐĂƚŝǀĞŶƵŵďĞƌƐŽŶƚŚĞŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞĨŽƌƚŚĞĂŵŽƵŶƚŽĨŵŽŶĞLJŝŶĂƐĂǀŝŶŐƐĂĐĐŽƵŶƚ͍
à A debit is represented as a negative number to the left of zero on a number line because debits are the opposite of credits, which are positive numbers.
dŚĞďĂŶŬĐŚĂƌŐĞƐĂ ̈́ ͷ ƐĞƌǀŝĐĞĨĞĞĨŽƌĂŶLJǁŝƚŚĚƌĂǁĂůĨƌŽŵĂƐĂǀŝŶŐƐĂĐĐŽƵŶƚ͘ charge, ĂůƐŽĐĂůůĞĚĂ fee, ŝƐƚŚĞ ĂŵŽƵŶƚŽĨŵŽŶĞLJĂƉĞƌƐŽŶŚĂƐƚŽƉĂLJĨŽƌƐŽŵĞƚŚŝŶŐ͘ĂŶLJŽƵŶĂŵĞĂƐŝƚƵĂƚŝŽŶǁŚĞƌĞLJŽƵǁŽƵůĚŚĂǀĞƚŽƉĂLJ ĂĐŚĂƌŐĞ͍
à I would have to pay a charge at an amusement park, a concert, a basketball game, or a doctor’s office.
,ŽǁǁŽƵůĚLJŽƵƌĞƉƌĞƐĞŶƚĂĐŚĂƌŐĞŽĨ ̈́ ͷ ĨŽƌdŝŵ͛ƐƐĂǀŝŶŐƐĂĐĐŽƵŶƚŽŶƚŚĞŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͍
à A charge of ̈́ ͷ would be െͷ because money is being taken out of the account. I would find positive five on the number line by starting at Ͳ and moving ͷ units to the right. Then, I would count ͷ units going left of zero to end at െͷ .
dŝŵǁŝƚŚĚƌĞǁ ̈́ ͵ͷ ĨƌŽŵŚŝƐĂĐĐŽƵŶƚ͘ĂƐĞĚŽŶƚŚĞƐƚŽƌLJƉƌŽďůĞŵ͕ǁŚĂƚŝƐƚŚĞŵĞĂŶŝŶŐŽĨƚŚĞƚĞƌŵ withdraw ͍
à Since Tim wanted to buy something, he took money out of the account. I think withdraw means to take money out of an account.
dŽǁŝƚŚĚƌĂǁŵŽŶĞLJŝƐƚŽƚĂŬĞŵŽŶĞLJŽƵƚŽĨĂŶĂĐĐŽƵŶƚ͘,ŽǁǁŽƵůĚLJŽƵƌĞƉƌĞƐĞŶƚƚŚĞ ̈́ ͵ͷ ĨŽƌƚŚĞǀŝĚĞŽŐĂŵĞ ĂƐĂŶŝŶƚĞŐĞƌĨŽƌdŝŵ͛ƐƐĂǀŝŶŐƐĂĐĐŽƵŶƚ͍
à The money was taken out of Tim’s account; it would be represented as െ͵ͷ .
Write each individual description below as an integer. Model the integer on the number line using an appropriate scale. EVENT INTEGER NUMBER LINE MODEL Open a bank account with ̈́ .
Make a ̈́
deposit.
Credit an account for ̈́ .
Make a deposit of ̈́ .
A bank makes a charge of ̈́ .െ
Tim withdraws ̈́ .െ
ૢૡ ૠ ૢ ૡ ૠ ૢ ૡ ૠ
െ െ െૠ ૠ
െ െ െ
െ െ െ െ െ െ െ
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6ͻϯ Lesson 2 Lesson 2: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
Freezing point of water in
ι۱
Example 2 (7 minutes): How Hot, How Cold?
dŚŝƐĞdžĂŵƉůĞŐŝǀĞƐƐƚƵĚĞŶƚƐƉƌĂĐƚŝĐĞƌĞĂĚŝŶŐƚŚĞƌŵŽŵĞƚĞƌƐŝŶďŽƚŚ&ĂŚƌĞŶŚĞŝƚĂŶĚĞůƐŝƵƐƐĐĂůĞƐ͘^ƚƵĚĞŶƚƐǁƌŝƚĞ ƚĞŵƉĞƌĂƚƵƌĞƐĂƐŝŶƚĞŐĞƌƐĂŶĚĚĞƐĐƌŝďĞŚŽǁƚĞŵƉĞƌĂƚƵƌĞĐŽƵůĚďĞŵŽĚĞůĞĚŽŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘
Example 2: How Hot, How Cold? Temperature is commonly measured using one of two scales, Celsius or Fahrenheit. In the United States, the Fahrenheit system continues to be the accepted standard for nonscientific use. All other countries have adopted Celsius as the primary scale in use. The thermometer shows how both scales are related. a. The boiling point of water is ι۱ . Where is degrees Celsius located on the thermometer to the right?
It is not shown because the greatest temperature shown in Celsius is ι۱ .
b. On a vertical number line, describe the position of the integer that represents ι۱ .
The integer is , and it would be located units above zero on the Celsius side of the scale.
c. Write each temperature as an integer. i. The temperature shown on the thermometer in degrees Fahrenheit:
ii. The temperature shown on the thermometer in degrees Celsius:
ૡ
iii. The freezing point of water in degrees Celsius:
d. If someone tells you your body temperature is ૢ ૡ Ǥ ι , what scale is being used? How do you know?
Since water boils at ι۱ , they must be using the Fahrenheit scale.
e. Does the temperature degrees mean the same thing on both scales?
No. ι۱ corresponds to ι۴ , and ι۴ corresponds to approximately െૡι۱ .
ĚĚƌĞƐƐƚŚĞĐŽŵŵŽŶŵŝƐĐŽŶĐĞƉƚŝŽŶŽŶŚŽǁƚŽĚĞƐĐƌŝďĞŶĞŐĂƚŝǀĞƚĞŵƉĞƌĂƚƵƌĞƐ͘ െͳͲԨ ĐĂŶďĞƌĞĂĚĂƐ͞ŶĞŐĂƚŝǀĞƚĞŶĚĞŐƌĞĞƐĞůƐŝƵƐ͘͟/ƚĐĂŶĂůƐŽďĞƌĞĂĚĂƐ͞ƚĞŶĚĞŐƌĞĞƐďĞůŽǁ njĞƌŽ͘͟,ŽǁĞǀĞƌ͕ŝƚƐŚŽƵůĚŶŽƚďĞƌĞĂĚĂƐ͞ŶĞŐĂƚŝǀĞƚĞŶĚĞŐƌĞĞƐďĞůŽǁnjĞƌŽ͘͟
dĞŵƉĞƌĂƚƵƌĞƐƚŚĂƚĂƌĞĂďŽǀĞnjĞƌŽĐĂŶďĞƐƚĂƚĞĚĂƐƚŚĞŝƌŶƵŵĞƌŝĐĂůǀĂůƵĞ͘&ŽƌĞdžĂŵƉůĞ͕ ĚĞƐĐƌŝďŝŶŐĂĨĞǀĞƌŽĨ ͳͲʹԬ ĐĂŶďĞƐŝŵƉůLJƐƚĂƚĞĚĂƐ͞ŽŶĞŚƵŶĚƌĞĚƚǁŽĚĞŐƌĞĞƐ͘͟
Scaffolding:
WƌŽǀŝĚĞŬŝŶĞƐƚŚĞƚŝĐĂŶĚǀŝƐƵĂů ůĞĂƌŶĞƌƐǁŝƚŚĂƚŚĞƌŵŽŵĞƚĞƌƚŽ ƌĞŝŶĨŽƌĐĞƐĐĂůĞƐ͘
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6ͻϯ Lesson 2 Lesson 2: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
Exercises ϯ–5 (7 minutes)
dŚĞĨŽůůŽǁŝŶŐƉƌŽďůĞŵƐƉƌŽǀŝĚĞƐƚƵĚĞŶƚƐĂĚĚŝƚŝŽŶĂůƉƌĂĐƚŝĐĞǁŝƚŚƌĞĂůͲǁŽƌůĚƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐĂŶĚnjĞƌŽ͘ 'ŝǀĞƐƚƵĚĞŶƚƐƚŝŵĞƚŽƐŚĂƌĞƌĞƐƉŽŶƐĞƐƚŽƚŚĞǁŚŽůĞŐƌŽƵƉ͘
džĞƌĐŝƐĞƐϯ –5
ϯ͘ Write each word under the appropriate column, “Positive Number” or “Negative Number.” Gain Loss Deposit Credit Debit Charge Below Zero Withdraw Owe Receive Positive Number Negative Number
Gain Deposit Credit Receive Loss Debit Charge Below zero Withdraw Owe
Write an integer to represent each of the following situations: a. A company loses ̈́ ǡ in 2011. െǡ
b. You earned ̈́ for dog sitting.
c. Jacob owes his dad ̈́ . െ
d. The temperature at the sun’s surface is about ǡ ι۱
e. The temperature outside is degrees below zero. െ
f. A football player lost yards when he was tackled. െ
Describe a situation that can be modeled by the integer െ . Explain what zero represents in the situation.
Answers will vary. I owe my best friend ̈́ . In this situation, represents my owing nothing to my best friend.
Closing (2 minutes)
,ŽǁĚŝĚǁĞƌĞƉƌĞƐĞŶƚĚĞďŝƚĂŶĚĐƌĞĚŝƚŽŶĂŶƵŵďĞƌůŝŶĞ͍
à A debit is represented as a negative number that is located to the left of (or below) zero. A credit is represented as a positive number that is located to the right of (or above) zero.
ĂŶĂƚĞŵƉĞƌĂƚƵƌĞŽĨ െͻ ĚĞŐƌĞĞƐďĞĚĞƐĐƌŝďĞĚĂƐ͞EĞŐĂƚŝǀĞŶŝŶĞĚĞŐƌĞĞƐďĞůŽǁnjĞƌŽ͍͟tŚLJŽƌǁŚLJŶŽƚ͍
à No, because “below zero” already means that the temperature is negative.
Exit Ticket (7 minutes)
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6ͻϯ Lesson 2 Lesson 2: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
EĂŵĞ ĂƚĞ
Lesson 2: Real-World Positive and Negative Numbers and Zero Exit Ticket
ϭ͘ tƌŝƚĞĂƐƚŽƌLJƉƌŽďůĞŵƚŚĂƚŝŶĐůƵĚĞƐďŽƚŚŝŶƚĞŐĞƌƐ െͺ ĂŶĚ ͳʹ ͘ Ϯ͘ tŚĂƚĚŽĞƐnjĞƌŽƌĞƉƌĞƐĞŶƚŝŶLJŽƵƌƐƚŽƌLJƉƌŽďůĞŵ͍ ϯ͘ ŚŽŽƐĞĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞƚŽŐƌĂƉŚďŽƚŚŝŶƚĞŐĞƌƐŽŶƚŚĞǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘>ĂďĞůƚŚĞƐĐĂůĞ͘ ϰ͘ 'ƌĂƉŚďŽƚŚƉŽŝŶƚƐŽŶƚŚĞǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘
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6ͻϯ Lesson 2 Lesson 2: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
Exit Ticket Sample Solutions 1. Write a story problem that includes both integers െૡ and .
Answers may vary. One boxer gains pounds of muscle to train for a fight. Another boxer loses ૡ
pounds of fat.
What does zero represent in your story problem?
Zero represents no change in the boxer’s weight.
ϯ͘ Choose an appropriate scale to graph both integers on the vertical number line. Label the scale.
I chose a scale of .
Graph both points on the vertical number line. Problem Set Sample Solutions 1. Express each situation as an integer in the space provided. a. A gain of points in a game
b. A fee charged of ̈́ െ
c. A temperature of degrees below zero െ
d. A -yard loss in a football game െ
e. The freezing point of water in degrees Celsius
f. A ̈́ ǡ deposit ǡ
െૡ
െ
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6ͻϯ Lesson 2 Lesson 2: ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
For Problems 2–5, use the thermometer to the right. 2. Each sentence is stated incorrectly . Rewrite the sentence to correctly describe each situation. a. The temperature is െ degrees Fahrenheit below zero.
Correct: The temperature is െι۴ .OR The temperature is degrees below zero Fahrenheit.
b. The temperature is െ degrees Celsius below zero.
Correct: The temperature is െι۱ .OR The temperature is degrees below zero Celsius.
ϯ͘ Mark the integer on the thermometer that corresponds to the temperature given. a. ૠι۴
b. ι۱
c. ι۴
d. െι۱
The boiling point of water is ι۴ . Can this thermometer be used to record the temperature of a boiling pot of water? Explain.
No, it cannot because the highest temperature in Fahrenheit on this thermometer is ι .
Kaylon shaded the thermometer to represent a temperature of degrees below zero Celsius as shown in the diagram. Is she correct? Why or why not? If necessary, describe how you would fix Kaylon’s shading.
She is incorrect because she shaded a temperature of െι۴ . I would fix this by marking a line segment at െι۱
and shade up to that line.
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6ͻϯ >ĞƐƐŽŶϯ
ĞƐƐŽŶϯ : ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
>ĞƐƐŽŶϯ : Real-World Positive and Negative Numbers and Zero Student Outcomes
^ƚƵĚĞŶƚƐƵƐĞƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐƚŽŝŶĚŝĐĂƚĞĂĐŚĂŶŐĞ;ŐĂŝŶŽƌůŽƐƐͿŝŶĞůĞǀĂƚŝŽŶǁŝƚŚĂĨŝdžĞĚ ƌĞĨĞƌĞŶĐĞƉŽŝŶƚ͕ƚĞŵƉĞƌĂƚƵƌĞ͕ĂŶĚƚŚĞďĂůĂŶĐĞŝŶĂďĂŶŬĂĐĐŽƵŶƚ͘
^ƚƵĚĞŶƚƐƵƐĞǀŽĐĂďƵůĂƌLJƉƌĞĐŝƐĞůLJǁŚĞŶĚĞƐĐƌŝďŝŶŐĂŶĚƌĞƉƌĞƐĞŶƚŝŶŐƐŝƚƵĂƚŝŽŶƐŝŶǀŽůǀŝŶŐŝŶƚĞŐĞƌƐ͖ĨŽƌŝŶƐƚĂŶĐĞ͕ ĂŶĞůĞǀĂƚŝŽŶŽĨ െͳͲ ĨĞĞƚŝƐƚŚĞƐĂŵĞĂƐ ͳͲ ĨĞĞƚďĞůŽǁƚŚĞĨŝdžĞĚƌĞĨĞƌĞŶĐĞƉŽŝŶƚ͘
^ƚƵĚĞŶƚƐĐŚŽŽƐĞĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞĨŽƌƚŚĞŶƵŵďĞƌůŝŶĞǁŚĞŶŐŝǀĞŶĂƐĞƚŽĨƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐ ƚŽŐƌĂƉŚ͘
Classwork Example 1 (10 minutes): A Look at Sea Level
dŚĞƉƵƌƉŽƐĞŽĨƚŚŝƐĞdžĂŵƉůĞŝƐĨŽƌƐƚƵĚĞŶƚƐƚŽƵŶĚĞƌƐƚĂŶĚŚŽǁŶĞŐĂƚŝǀĞĂŶĚƉŽƐŝƚŝǀĞŶƵŵďĞƌƐĐĂŶďĞƵƐĞĚƚŽƌĞƉƌĞƐĞŶƚ ƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƐŝŶǀŽůǀŝŶŐĞůĞǀĂƚŝŽŶ͘ZĞĂĚƚŚĞĞdžĂŵƉůĞĂůŽƵĚ͘
Example 1: A Look at Sea Level The picture below shows three different people participating in activities at three different elevations. With a partner, discuss what you see. What do you think the word elevation means in this situation?
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6ͻϯ >ĞƐƐŽŶϯ
ĞƐƐŽŶϯ :ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
WŽƐĞƋƵĞƐƚŝŽŶƐƚŽƚŚĞĐůĂƐƐ͕ĂŶĚĚĞĨŝŶĞĞůĞǀĂƚŝŽŶ͘^ƚƵĚĞŶƚƐŐĂŝŶĂĚĚŝƚŝŽŶĂůƉƌĂĐƚŝĐĞǁŝƚŚĞůĞǀĂƚŝŽŶďLJĐŽŵƉůĞƚŝŶŐ džĞƌĐŝƐĞϭŝŶĚĞƉĞŶĚĞŶƚůLJ͘ WŽƐƐŝďůĞĚŝƐĐƵƐƐŝŽŶƋƵĞƐƚŝŽŶƐ͗
>ŽŽŬŝŶŐĂƚƚŚĞƉŝĐƚƵƌĞ͕ŝĨLJŽƵǁĞƌĞƚŽĚƌĂǁĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞƚŽŵŽĚĞůĞůĞǀĂƚŝŽŶ͕ǁŚŝĐŚƉĞƌƐŽŶ͛ƐĞůĞǀĂƚŝŽŶ ĚŽLJŽƵƚŚŝŶŬǁŽƵůĚďĞĂƚnjĞƌŽ͍ džƉůĂŝŶ͘
à Sea level should represent an elevation of zero. So, the person sailing would be at zero because he is sailing on the surface of the water, which is neither above nor below the surface. On a number line, zero is the point or number separating positive and negative numbers.
KŶƚŚĞƐĂŵĞǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͕ǁŚŝĐŚƉĞƌƐŽŶ͛ƐĞůĞǀĂƚŝŽŶǁŽƵůĚďĞƌĞƉƌĞƐĞŶƚĞĚĂďŽǀĞnjĞƌŽ͍
à The elevation of the person hiking would be above zero because she is moving higher above the water. On a vertical number line, this is represented by a positive value above zero because she is above the surface.
KŶƚŚĞƐĂŵĞǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͕ǁŚŝĐŚƉĞƌƐŽŶ͛ƐĞůĞǀĂƚŝŽŶĚŽLJŽƵƚŚŝŶŬǁŽƵůĚďĞďĞůŽǁnjĞƌŽ͍
à The elevation of the person scuba diving would be below zero because he is swimming below the surface of the water. On a vertical number line, this is represented by a negative value below zero because he is below the surface.
tŚĂƚĚŽĞƐnjĞƌŽƌĞƉƌĞƐĞŶƚŝŶƚŚŝƐƐŝƚƵĂƚŝŽŶ͍
à Zero represents the top of the water (the water’s surface).
/ŶƚŚŝƐĞdžĂŵƉůĞ͕ǁŚŝĐŚŶƵŵďĞƌƐĐŽƌƌĞƐƉŽŶĚƚŽĞůĞǀĂƚŝŽŶƐĂďŽǀĞƐĞĂůĞǀĞů͍
à Above sea level means to be above zero, which are positive numbers.
/ŶƚŚŝƐĞdžĂŵƉůĞ͕ǁŚŝĐŚŶƵŵďĞƌƐĐŽƌƌĞƐƉŽŶĚƚŽĞůĞǀĂƚŝŽŶƐďĞůŽǁƐĞĂůĞǀĞů͍
à Below sea level means to be below zero, which are negative numbers.
KŶĂŶƵŵďĞƌůŝŶĞ͕ǁŚĂƚĚŽĞƐŝƚŵĞĂŶƚŽďĞĂƚƐĞĂůĞǀĞů͍
à To be at zero means to be at sea level.
ůĞǀĂƚŝŽŶŝƐƚŚĞŚĞŝŐŚƚŽĨĂƉĞƌƐŽŶ͕ƉůĂĐĞ͕ŽƌƚŚŝŶŐĂďŽǀĞŽƌďĞůŽǁĂĐĞƌƚĂŝŶƌĞĨĞƌĞŶĐĞƉŽŝŶƚ͘/ŶƚŚŝƐĐĂƐĞ͕ǁŚĂƚ ŝƐƚŚĞƌĞĨĞƌĞŶĐĞƉŽŝŶƚ͍
à The reference point is sea level.
Exercises 1–2 (5 minutes) Exercises 1– ϯ
Refer back to Example 1. Use the following information to answer the questions.
The scuba diver is feet below sea level.
The sailor is at sea level.
The hiker is miles ( ǡ feet) above sea level. 1. Write an integer to represent each situation.
Scuba Diver: െ
Sailor:
Hiker: (to represent the elevation in miles) or ǡ (to represent the elevation in feet)
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6ͻϯ >ĞƐƐŽŶϯ
ĞƐƐŽŶϯ : ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
Use an appropriate scale to graph each of the following situations on the number line to the right. Also, write an integer to represent both situations. a. A hiker is feet above sea level.
b. A diver is feet below sea level.
െ
^ƚƵĚĞŶƚƐƐŚŽƵůĚŝĚĞŶƚŝĨLJĐŽŵŵŽŶŵŝƐĐŽŶĐĞƉƚŝŽŶƐŽĨŚŽǁƚŽƌĞƉƌĞƐĞŶƚĂŶĂŶƐǁĞƌďĂƐĞĚŽŶƚŚĞƉŚƌĂƐŝŶŐŽĨĂƋƵĞƐƚŝŽŶ͘ ^ƚƵĚĞŶƚƐƉƌĂĐƚŝĐĞƚŚŝƐƐŬŝůůŝŶ džĞƌĐŝƐĞϯ͘
,ŽǁŵĂŶLJĨĞĞƚďĞůŽǁƐĞĂůĞǀĞůŝƐƚŚĞĚŝǀĞƌ͍
à Students should answer using a positive number, such as Ͳ feet, because “below” already indicates that the number is negative.
tŚŝĐŚŝŶƚĞŐĞƌǁŽƵůĚƌĞƉƌĞƐĞŶƚ ͷͲ ĨĞĞƚďĞůŽǁƐĞĂůĞǀĞů͍
à Students should answer by saying “ െͷͲ ” and not “ െͷͲ below sea level.”
ϯ͘ For each statement, there are two related statements: (i) and (ii). Determine which related statement ((i) or (ii)) is expressed correctly, and circle it. Then, correct the other related statement so that both parts, (i) and (ii) , are stated correctly. a. A submarine is submerged ૡ feet below sea level. i. The depth of the submarine is െૡ feet below sea level.
The depth of the submarine is ૡ feet below sea level.
ii. ૡ feet below sea level can be represented by the integer െૡ .b. The elevation of a coral reef with respect to sea level is given as െ feet. i. The coral reef is feet below sea level. ii. The depth of the coral reef is െ feet below sea level.
The depth of the coral reef is feet below sea level.
െ
െ
െ
െ
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6ͻϯ >ĞƐƐŽŶϯ
ĞƐƐŽŶϯ :ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
Exploratory Challenge (20 minutes)
DĂƚĞƌŝĂůƐ͗
x ŽƉŝĞƐ;ŽŶĞƉĞƌƐƚƵĚĞŶƚͿŽĨƚŚĞ džƉůŽƌĂƚŽƌLJŚĂůůĞŶŐĞ^ƚĂƚŝŽŶZĞĐŽƌĚ^ŚĞĞƚ;ƐĞĞĂƚƚĂĐŚĞĚƌĞĐŽƌĚƐŚĞĞƚ͘Ϳ
x ^ŚĞĞƚƐŽĨůŽŽƐĞͲůĞĂĨƉĂƉĞƌ;ŽŶĞƉĞƌŐƌŽƵƉͿĨŽƌƚŚĞĂŶƐǁĞƌŬĞLJĨŽƌƚŚĞŝƌƉŽƐƚĞƌƐ
x ZƵůĞƌƐŽƌŵĞƚĞƌƐƚŝĐŬŽƌLJĂƌĚƐƚŝĐŬ;ŽŶĞƉĞƌŐƌŽƵƉͿ
x ŽŶƐƚƌƵĐƚŝŽŶƉĂƉĞƌŽƌǁĂůůͲƐŝnjĞĚŐƌŝĚƉĂƉĞƌ;ŽŶĞƐŚĞĞƚĨŽƌĞĂĐŚŐƌŽƵƉͿ
x DĂƌŬĞƌƐ;ŽŶĞƐĞƚŽƌĂĨĞǁĨŽƌĞĂĐŚŐƌŽƵƉͿ ^ƚƵĚĞŶƚƐǁŽƌŬŝŶŐƌŽƵƉƐŽĨƚŚƌĞĞƚŽĨŽƵƌƚŽĐƌĞĂƚĞƚŚĞŝƌŽǁŶƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƐŝŶǀŽůǀŝŶŐ ŵŽŶĞLJ͕ƚĞŵƉĞƌĂƚƵƌĞ͕ĞůĞǀĂƚŝŽŶ͕ĂŶĚŽƚŚĞƌƌĞĂůͲǁŽƌůĚƐĐĞŶĂƌŝŽƐ͘'ŝǀĞĞĂĐŚŐƌŽƵƉĂƐŚĞĞƚ ŽĨǁĂůůͲƐŝnjĞĚŐƌŝĚƉĂƉĞƌ;ŽƌĐŽŶƐƚƌƵĐƚŝŽŶƉĂƉĞƌͿŶƵŵďĞƌĞĚŽŶĞƚŽĨŝǀĞ͕ŵĂƌŬĞƌƐ͕ĂŶĚĂ ƌƵůĞƌ͘hƐŝŶŐƚŚĞƐĞŵĂƚĞƌŝĂůƐ͕ĞĂĐŚŐƌŽƵƉƉƌĞƐĞŶƚƐŝƚƐƐŝƚƵĂƚŝŽŶŽŶƚŚĞƉĂƉĞƌďLJŝŶĐůƵĚŝŶŐ ƚŚĞĐŽŵƉŽŶĞŶƚƐŝŶƚŚĞďƵůůĞƚĞĚůŝƐƚďĞůŽǁ͘ůůŽǁƐƚƵĚĞŶƚƐϭϬŵŝŶƵƚĞƐƚŽĐƌĞĂƚĞƚŚĞŝƌ ƉŽƐƚĞƌƐ͕ĂŶĚŚĂŶŐƚŚĞŵŽŶĂǁĂůůŝŶƚŚĞƌŽŽŵ͘
x dŝƚůĞ;Ğ͘Ő͕͘^ĞĂ>ĞǀĞů͕dĞŵƉĞƌĂƚƵƌĞͿ
x ǁƌŝƚƚĞŶƐŝƚƵĂƚŝŽŶďĂƐĞĚŽŶƚŚĞƚŝƚůĞ;ƵƐŝŶŐĂƚůĞĂƐƚƚǁŽƉŽŝŶƚƐͿ
x ďůĂŶŬǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ
x WŝĐƚƵƌĞ;ŽƉƚŝŽŶĂůŝĨƚŝŵĞƉĞƌŵŝƚƐͿ
x ŶƐǁĞƌŬĞLJ;ŽŶĂƐĞƉĂƌĂƚĞƐŚĞĞƚŽĨƉĂƉĞƌƐƚĂƉůĞĚƚŽƚŚĞƚŽƉďĂĐŬƌŝŐŚƚĐŽƌŶĞƌͿ 'ƌŽƵƉƐƌŽƚĂƚĞĞǀĞƌLJĨĞǁŵŝŶƵƚĞƐƚŽĐŽŵƉůĞƚĞƚŚĞƚŚƌĞĞƚĂƐŬƐŽŶƚŚĞ džƉůŽƌĂƚŽƌLJŚĂůůĞŶŐĞ^ƚĂƚŝŽŶZĞĐŽƌĚ^ŚĞĞƚǁŚŝůĞ ǀŝĞǁŝŶŐĞĂĐŚƉŽƐƚĞƌ͘
x tƌŝƚĞƚŚĞŝŶƚĞŐĞƌƐĨŽƌĞĂĐŚƐŝƚƵĂƚŝŽŶ͘
x ĞƚĞƌŵŝŶĞƚŚĞĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞƚŽŐƌĂƉŚƚŚĞƉŽŝŶƚƐ͘
x 'ƌĂƉŚƚŚĞƉŽŝŶƚƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘
Closing ;ϯ minutes)
,ŽǁĚŝĚǁĞƌĞĐŽƌĚŵĞĂƐƵƌĞƐŽĨĞůĞǀĂƚŝŽŶŽŶĂŶƵŵďĞƌůŝŶĞ͍
à Elevations above sea level are positive numbers, and they are above zero. Elevations below sea level are negative numbers, and they are below zero.
/Ɛ͞ െͻͲ ĨĞĞƚďĞůŽǁƐĞĂůĞǀĞů͟ĂŶĂƉƉƌŽƉƌŝĂƚĞĂŶƐǁĞƌƚŽĂƋƵĞƐƚŝŽŶ͍tŚLJŽƌǁŚLJŶŽƚ͍
à No. You do not need the negative sign to write ͻͲ feet below zero because the word “below” in this case means a negative number.
Exit Ticket (7 minutes)
Scaffolding:
ůůŽǁŐƌŽƵƉƐƚŽƉƌĞƐĞŶƚƚŚĞŝƌ ƉŽƐƚĞƌƐƚŽƚŚĞĐůĂƐƐ͕ĂŶĚƚŚĞ ĐůĂƐƐĂŶƐǁĞƌƐƚŚĞƋƵĞƐƚŝŽŶƐ ĚƵƌŝŶŐƚŚĞƉƌĞƐĞŶƚĂƚŝŽŶ͘
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6ͻϯ >ĞƐƐŽŶϯ
ĞƐƐŽŶϯ : ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
EĂŵĞ ĂƚĞ
>ĞƐƐŽŶϯ : Real-World Positive and Negative Numbers and Zero Exit Ticket
ϭ͘ tƌŝƚĞĂƐƚŽƌLJƉƌŽďůĞŵƵƐŝŶŐƐĞĂůĞǀĞůƚŚĂƚŝŶĐůƵĚĞƐďŽƚŚŝŶƚĞŐĞƌƐ െͳͳͲ ĂŶĚ ͳʹͲ ͘ Ϯ͘ tŚĂƚĚŽĞƐnjĞƌŽƌĞƉƌĞƐĞŶƚŝŶLJŽƵƌƐƚŽƌLJƉƌŽďůĞŵ͍ ϯ͘ ŚŽŽƐĞĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞƚŽŐƌĂƉŚďŽƚŚŝŶƚĞŐĞƌƐŽŶƚŚĞǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘ ϰ͘ 'ƌĂƉŚĂŶĚůĂďĞůďŽƚŚƉŽŝŶƚƐŽŶƚŚĞǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘
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6ͻϯ >ĞƐƐŽŶϯ
ĞƐƐŽŶϯ : ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
Exit Ticket Sample Solutions 1. Write a story problem using sea level that includes both integers െ and .
Answers may vary. On the beach, a man’s kite flies at feet above sea level, which is indicated by the water’s surface. In the ocean, a white shark swims at feet below the water’s surface.
What does zero represent in your story problem?
Zero represents the water’s surface level, or sea level.
ϯ͘ Choose and label an appropriate scale to graph both integers on the vertical number line.
I chose a scale of .
ϰ͘ Graph and label both points on the vertical number line. Problem Set Sample Solutions 1. Write an integer to match the following descriptions. a. A debit of ̈́ െ
b. A deposit of ̈́
c. ǡ feet above sea level ǡ
d. A temperature increase of ι۴
e. A withdrawal of ̈́ െ
f. ǡ feet below sea level െǡ
For Problems 2– ϰ͕ read each statement about a real-world situation and the two related statements in parts (a) and (b) carefully. Circle the correct way to describe each real-world situation ; possible answers include either (a), (b), or both (a) and (b) .2. A whale is feet below the surface of the ocean. a. The depth of the whale is feet from the ocean’s surface. b. The whale is െ feet below the surface of the ocean.
െ
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6ͻϯ >ĞƐƐŽŶϯ
ĞƐƐŽŶϯ : ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
ϯ͘ The elevation of the bottom of an iceberg with respect to sea level is given as െ feet. a. The iceberg is feet above sea level. b. The iceberg is feet below sea level.
ϰ͘ Alex’s body temperature decreased by ι۴ .a. Alex’s body temperature dropped ι۴ .b. The integer െ represents the change in Alex’s body temperature in degrees Fahrenheit. 5. A credit of ̈́ and a debit of ̈́ are applied to your bank account. a. What is an appropriate scale to graph a credit of ̈́ and a debit of ̈́ ? Explain your reasoning.
Answers will vary. I would count by ’s because both numbers are multiples of .
b. What integer represents “a credit of ̈́ ” if zero represents the original balance? Explain.
; a credit is greater than zero, and numbers greater than zero are positive numbers.
c. What integer describes “a debit of ̈́ ” if zero represents the original balance? Explain.
െ ; a debit is less than zero, and numbers less than zero are negative numbers.
d. Based on your scale, describe the location of both integers on the number line.
If the scale is multiples of , then would be ૠ units to the right of (or above) zero, and െ would be ૡ
units to the left of (or below) zero.
e. What does zero represent in this situation?
Zero represents no change being made to the account balance. In other words, no amount is either subtracted or added to the account.
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6ͻϯ >ĞƐƐŽŶϯ
ĞƐƐŽŶϯ :ZĞĂůͲtŽƌůĚWŽƐŝƚŝǀĞĂŶĚEĞŐĂƚŝǀĞEƵŵďĞƌƐĂŶĚĞƌŽ
EĂŵĞ ĂƚĞ
Exploratory Challenge Station Record Sheet
WŽƐƚĞƌηͺͺͺͺͺͺͺ /ŶƚĞŐĞƌƐ͗ͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺ EƵŵďĞƌ>ŝŶĞ^ĐĂůĞ͗ͺͺͺͺͺͺͺͺ WŽƐƚĞƌηͺͺͺͺͺͺͺ /ŶƚĞŐĞƌƐ͗ͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺ EƵŵďĞƌ>ŝŶĞ^ĐĂůĞ͗ͺͺͺͺͺͺͺͺ WŽƐƚĞƌηͺͺͺͺͺͺͺ /ŶƚĞŐĞƌƐ͗ͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺ EƵŵďĞƌ>ŝŶĞ^ĐĂůĞ͗ͺͺͺͺͺͺͺͺ WŽƐƚĞƌηͺͺͺͺͺͺͺ /ŶƚĞŐĞƌƐ͗ͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺ EƵŵďĞƌ>ŝŶĞ^ĐĂůĞ͗ͺͺͺͺͺͺͺͺ
WŽƐƚĞƌηͺͺͺͺͺͺͺ /ŶƚĞŐĞƌƐ͗ͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺͺ EƵŵďĞƌ>ŝŶĞ^ĐĂůĞ͗ͺͺͺͺͺͺͺͺ
# # #
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6ͻϯ Lesson 4 Lesson 4: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ
Lesson 4: The Opposite of a Number Student Outcomes
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĂƚĞĂĐŚŶŽŶnjĞƌŽŝŶƚĞŐĞƌ͕ ͕ܽ ŚĂƐĂŶŽƉƉŽƐŝƚĞ͕ĚĞŶŽƚĞĚ ܽെ ͕ ĂŶĚƚŚĂƚ ܽെ ĂŶĚ ܽ ĂƌĞ ŽƉƉŽƐŝƚĞƐŝĨƚŚĞLJĂƌĞŽŶŽƉƉŽƐŝƚĞƐŝĚĞƐŽĨnjĞƌŽĂŶĚĂƌĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘
^ƚƵĚĞŶƚƐƌĞĐŽŐŶŝnjĞƚŚĞŶƵŵďĞƌnjĞƌŽŝƐŝƚƐŽǁŶŽƉƉŽƐŝƚĞ͘
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĂƚƐŝŶĐĞĂůůĐŽƵŶƚŝŶŐŶƵŵďĞƌƐĂƌĞƉŽƐŝƚŝǀĞ͕ŝƚŝƐŶŽƚŶĞĐĞƐƐĂƌLJƚŽŝŶĚŝĐĂƚĞƐƵĐŚǁŝƚŚĂ ƉůƵƐƐŝŐŶ͘
Lesson Notes
/ŶƚŚŝƐůĞƐƐŽŶ͕ƐƚƵĚĞŶƚƐƉƌĂĐƚŝĐĞŐƌĂƉŚŝŶŐƉŽŝŶƚƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘/ŶƉĂƌƚŝĐƵůĂƌ͕ƐƚƵĚĞŶƚƐĚĞƚĞƌŵŝŶĞƚŚĞĂƉƉƌŽƉƌŝĂƚĞ ƐĐĂůĞŐŝǀĞŶĂƐĞƚŽĨŽƉƉŽƐŝƚĞƐŝŶƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƐ͘^ƚƵĚĞŶƚƐƉĂLJĐĂƌĞĨƵůĂƚƚĞŶƚŝŽŶƚŽƚŚĞŵĞĂŶŝŶŐŽĨnjĞƌŽŝŶƉƌŽďůĞŵ ƐŝƚƵĂƚŝŽŶƐĂŶĚŚŽǁŽƉƉŽƐŝƚĞƐĂƌĞƌĞůĂƚĞĚŝŶƚŚĞĐŽŶƚĞdžƚŽĨĂŐŝǀĞŶƐŝƚƵĂƚŝŽŶ͘ƌĞĂƚĞĂĨůŽŽƌŵŽĚĞůŽĨĂŶƵŵďĞƌůŝŶĞƉƌŝŽƌ ƚŽƚŚĞůĞƐƐŽŶ͘
Classwork Opening (5 minutes): What Is the Relationship?
^ƚƵĚĞŶƚƐǁŽƌŬŝŶƉĂŝƌƐƚŽĚĞƚĞƌŵŝŶĞƚŚĞƌĞůĂƚŝŽŶƐŚŝƉƐďĞƚǁĞĞŶƐĞƚƐŽĨǁŽƌĚƐǁŝƚŚŽƌ ǁŝƚŚŽƵƚƉŝĐƚƵƌĞƐ͘ŝƐƉůĂLJƚŚĞƚĂƐŬƚŽƚŚĞǁŚŽůĞŐƌŽƵƉ͘
&ŝŶĚƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞƐĞƚƐŽĨǁŽƌĚƐ͘ &ĂƐƚ ՜^ůŽǁ ZŽƵŐŚ ՜^ŵŽŽƚŚ KƉĞŶ ՜ůŽƐĞ &ŝĐƚŝŽŶ ՜EŽŶĨŝĐƚŝŽŶ >ŝŐŚƚ ՜ĂƌŬ ŵƉƚLJ ՜&Ƶůů ĐĐĞƉƚ ՜ZĞĨƵƐĞ ^ŚĂůůŽǁ ՜ĞĞƉ ŝƌƚLJ ՜ůĞĂŶ ƉĂƌƚ ՜dŽŐĞƚŚĞƌ YƵĞƐƚŝŽŶ ՜ŶƐǁĞƌ ŶĐŝĞŶƚ ՜DŽĚĞƌŶ ůŝŬĞ ՜ŝĨĨĞƌĞŶƚ ůů ՜EŽŶĞ ĂŶŐĞƌŽƵƐ ՜^ĂĨĞ ŽƌƌĞĐƚ ՜/ŶĐŽƌƌĞĐƚ ĞĨĞĂƚ ՜sŝĐƚŽƌLJ ĂƐLJ ՜,ĂƌĚ &ƵƚƵƌĞ ՜WĂƐƚ ƌĞĂŬ ՜&ŝdž /ŶƐŝĚĞ ՜KƵƚƐŝĚĞ hƉ ՜ŽǁŶ tĞƚ ՜ƌLJ ŶƚƌĂŶĐĞ ՜ džŝƚ
à The words are opposites of each other.
KŶĐĞLJŽƵŚĂǀĞĚĞƚĞƌŵŝŶĞĚƚŚĞƌĞůĂƚŝŽŶƐŚŝƉ͕ĐƌĞĂƚĞLJŽƵƌŽǁŶĞdžĂŵƉůĞƐ͕ ŝŶĐůƵĚŝŶŐĂŵĂƚŚĞdžĂŵƉůĞ͘
à Left ՜ Right
à Cold ՜ Hot
à ͷ ՜ െͷ
Scaffolding:
ŝĨĨĞƌĞŶƚŝĂƚĞůĞǀĞůƐďLJ ƉƌŽǀŝĚŝŶŐŐƌŽƵƉƐǁŝƚŚĂ ƐĞƚŽĨ ͺʹͳͲ ƉƌĞƐĞůĞĐƚĞĚ ǁŽƌĚƐĐƵƚŽƵƚŝŶĚŝǀŝĚƵĂůůLJ ŽŶĐĂƌĚƐƚŽĐŬ͘hƐĞŵŽƌĞ ĐŚĂůůĞŶŐŝŶŐǀŽĐĂďƵůĂƌLJ ǁŽƌĚƐĨŽƌĂĚǀĂŶĐĞĚ ůĞĂƌŶĞƌƐ͕ĂŶĚƉƌŽǀŝĚĞ ƉŝĐƚƵƌĞƐǁŝƚŚǁŽƌĚƐĨŽƌ ŶŐůŝƐŚůĂŶŐƵĂŐĞůĞĂƌŶĞƌƐ ŽƌŝŶĐůƵƐŝŽŶƐƚƵĚĞŶƚƐ͘
ƐŬůĂŶŐƵĂŐĞĂƌƚƐĂŶĚ ƐĐŝĞŶĐĞƚĞĂĐŚĞƌƐĨŽƌŝŶƉƵƚ ƚŽƉƌŽǀŝĚĞŵŽƌĞǀĂƌŝĂƚŝŽŶ ŝŶǀŽĐĂďƵůĂƌLJ͘
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6ͻϯ Lesson 4 Lesson 4: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ
Exercise 1 (10 minutes): Walk the Number Line
ŝƐƚƌŝďƵƚĞĂŶŝŶĚĞdžĐĂƌĚƚŽĞĂĐŚƐƚƵĚĞŶƚƚŚĂƚŝƐůĂďĞůĞĚǁŝƚŚĂŶŝŶƚĞŐĞƌƌĂŶŐŝŶŐĨƌŽŵ െͳͲ ƚŽ ͳͲ ͘ ƌĞĂƚĞĞŶŽƵŐŚĐĂƌĚƐďĂƐĞĚŽŶƚŚĞĐůĂƐƐƐŝnjĞ͘,ĂǀĞƐƚƵĚĞŶƚƐƐƚĂŶĚŽƌƉůĂĐĞƚŚĞŝƌ ŝŶĚĞdžĐĂƌĚƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞŽŶĞĂƚĂƚŝŵĞ͘tŚĞŶƉůĂĐŝŶŐƚŚĞŝƌŶƵŵďĞƌƐ͕ƐƚƵĚĞŶƚƐ ƐŚŽƵůĚƐƚĂƌƚĂƚnjĞƌŽĂŶĚŵŽǀĞŝŶƚŚĞĚŝƌĞĐƚŝŽŶŽĨƚŚĞŝƌŶƵŵďĞƌƐ͕ĐŽƵŶƚŝŶŐŽƵƚůŽƵĚ͘WŽƐĞ ĚŝƐĐƵƐƐŝŽŶƋƵĞƐƚŝŽŶƐĂĨƚĞƌƚŚĞĞdžĞƌĐŝƐĞ͘ ŝƐĐƵƐƐƚŚĞĨŽůůŽǁŝŶŐ͗
tŚĂƚƉĂƚƚĞƌŶƐĚŽLJŽƵƐĞĞǁŝƚŚƚŚĞŶƵŵďĞƌƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à For each number to the right of zero, there is a corresponding number the same distance from zero to the left.
tŚĂƚĚŽĞƐnjĞƌŽƌĞƉƌĞƐĞŶƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à Zero represents the reference point when locating a point on the number line. It also represents the separation of positive numbers from negative numbers.
tŚĂƚŝƐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶĂŶLJƚǁŽŽƉƉŽƐŝƚĞŶƵŵďĞƌƐĂŶĚnjĞƌŽŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à Opposite numbers are the same distance from zero, but they are on opposite sides of zero.
ZĞĂĚĂŶĚĚŝƐƉůĂLJƚŚĞƐƚĂƚĞŵĞŶƚďĞůŽǁƚŽƚŚĞĐůĂƐƐ͕ĂŶĚƚŚĞŶŵŽĚĞů džĂŵƉůĞϭ͘
Exercise 1: Walk the Number Line 1. Each nonzero integer has an opposite, denoted ࢇെ ;ࢇെ and ࢇare opposites if they are on opposite sides of zero and the same distance from zero on the number line.
Example 1 (5 minutes): Every Number Has an Opposite
ZĞĂĚƚŚĞĞdžĂŵƉůĞŽƵƚůŽƵĚ͘DŽĚĞůŚŽǁƚŽŐƌĂƉŚďŽƚŚƉŽŝŶƚƐ͕ĂŶĚĞdžƉůĂŝŶŚŽǁƚŚĞLJĂƌĞƌĞůĂƚĞĚ͘
Example 1: Every Number Has an Opposite Locate the number ૡand its opposite on the number line. Explain how they are related to zero.
&ŝƌƐƚ͕ƐƚĂƌƚĂƚnjĞƌŽ͕ĂŶĚŵŽǀĞ ͺƵŶŝƚƐƚŽƚŚĞƌŝŐŚƚƚŽůŽĐĂƚĞƉŽƐŝƚŝǀĞ ͺ͘ ^Ž͕ƚŚĞŽƉƉŽƐŝƚĞŽĨ ͺŵƵƐƚďĞ ͺƵŶŝƚƐƚŽ ƚŚĞůĞĨƚŽĨnjĞƌŽ͘tŚĂƚŶƵŵďĞƌŝƐ ͺƵŶŝƚƐƚŽƚŚĞůĞĨƚŽĨnjĞƌŽ͍
à െͺ
ͺĂŶĚ െͺ ĂƌĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽ͘^ŝŶĐĞďŽƚŚŶƵŵďĞƌƐĂƌĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽďƵƚŽŶ ŽƉƉŽƐŝƚĞƐŝĚĞƐŽĨnjĞƌŽŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͕ƚŚĞLJĂƌĞŽƉƉŽƐŝƚĞƐ͘
െ െ െ െ െ െ െૠ െૡ ૡૠ
Scaffolding:
ZĞŵŝŶĚƐƚƵĚĞŶƚƐŚŽǁƚŽ ůŽĐĂƚĞĂŶĞŐĂƚŝǀĞŶƵŵďĞƌ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞĨƌŽŵ >ĞƐƐŽŶϭ͘
&ŽƌĂĚǀĂŶĐĞĚůĞĂƌŶĞƌƐ͕ ƉƌŽǀŝĚĞƉŽƐŝƚŝǀĞĂŶĚ ŶĞŐĂƚŝǀĞĨƌĂĐƚŝŽŶƐ͘WŽƐĞ ƚŚĞůĂƐƚƚŚƌĞĞƋƵĞƐƚŝŽŶƐ ĨŽƌŝŶƋƵŝƌLJŽŶůLJ͘
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6ͻϯ Lesson 4 Lesson 4: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ
Exercises 2– ϯ (5 minutes)
^ƚƵĚĞŶƚƐǁŽƌŬŝŶĚĞƉĞŶĚĞŶƚůLJƚŽĂŶƐǁĞƌƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐ͘ůůŽǁϮʹϯŵŝŶƵƚĞƐĨŽƌƌĞǀŝĞǁĂƐĂǁŚŽůĞŐƌŽƵƉ͘
Exercises 2– ϯ
Locate and label the opposites of the numbers on the number line. a. ૢ
b. െ
c.
d. െૠ
ϯ͘ Write the integer that represents the opposite of each situation. Explain what zero means in each situation. a. feet above sea level
െ ; zero represents sea level.
b. ι۱ below zero
; zero represents degrees Celsius.
c. A withdrawal of ̈́
; zero represents no change, where no withdrawal or deposit is made.
ͳͲെͳ െʹ െ͵ െͶ െͷ െ െ െͺ ʹ ͵ Ͷ ͺͷ
dŚĞŽƉƉŽƐŝƚĞŽĨ ͺŝƐ െͺ ͘
dŚĞŽƉƉŽƐŝƚĞŽĨ െͺ ŝƐ ͺ͘
െ െ െ െ െ െ െૠ െૡ ૡૠെૢ െ ૢ
ࢇ ࢉ ࢈ ࢊ
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6ͻϯ Lesson 4 Lesson 4: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ
Example 2 ;ϴ minutes): A Real-World Example dŚĞƉƵƌƉŽƐĞŽĨƚŚŝƐĞdžĂŵƉůĞŝƐƚŽƐŚŽǁƐƚƵĚĞŶƚƐŚŽǁƚŽŐƌĂƉŚŽƉƉŽƐŝƚĞŝŶƚĞŐĞƌƐŐŝǀĞŶĂƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶ͘/ŶƉĂŝƌƐ͕ ŚĂǀĞƐƚƵĚĞŶƚƐƌĞĂĚƚŚĞƉƌŽďůĞŵĂůŽƵĚƚŽĞĂĐŚŽƚŚĞƌ͘/ŶƐƚƌƵĐƚƐƚƵĚĞŶƚƐƚŽĐŝƌĐůĞĂŶLJǁŽƌĚƐƚŚĂƚŵŝŐŚƚďĞŝŵƉŽƌƚĂŶƚƚŽ ƐŽůǀĞƚŚĞƉƌŽďůĞŵ͘WŽƐĞƋƵĞƐƚŝŽŶƐƚŽƚŚĞĐůĂƐƐǁŚŝůĞŐƵŝĚŝŶŐƐƚƵĚĞŶƚƐĂƐĂǁŚŽůĞŐƌŽƵƉƚŚƌŽƵŐŚƚŚĞĞdžĂŵƉůĞ͘
Example 2: A Real-World Example Maria decides to take a walk along Central Avenue to purchase a book at the bookstore. On her way, she passes the Furry Friends Pet Shop and goes in to look for a new leash for her dog. Furry Friends Pet Shop is seven blocks west of the bookstore. She leaves Furry Friends Pet Shop and walks toward the bookstore to look at some books. After she leaves the bookstore, she heads east for seven blocks and stops at Ray’s Pet Shop to see if she can find a new leash at a better price. Which location, if any, is the farthest from Maria while she is at the bookstore? Determine an appropriate scale, and model the situation on the number line below.
Answers will vary.
Explain your answer. What does zero represent in the situation?
The pet stores are the same distance from Maria, who is at the bookstore. They are each ૠ blocks away but in opposite directions. In this example, zero represents the bookstore.
ŝƐĐƵƐƐƚŚĞĨŽůůŽǁŝŶŐ͗
,ŽǁĚŝĚLJŽƵĚĞƚĞƌŵŝŶĞĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞĨŽƌƚŚĞƐŝƚƵĂƚŝŽŶŝĨĂůůďůŽĐŬƐŝŶƚŚĞĐŝƚLJĂƌĞƚŚĞƐĂŵĞůĞŶŐƚŚ͍
à Because both stores are seven blocks in opposite directions, I knew that I could count by ones since the numbers are not that large.
tŚĞƌĞǁŽƵůĚƚŚĞŬƐƚŽƌĞďĞůŽĐĂƚĞĚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à The bookstore would be located at zero.
tŚĞƌĞǁŽƵůĚZĂLJ͛ƐWĞƚ^ŚŽƉďĞůŽĐĂƚĞĚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍ džƉůĂŝŶ͘
à It would be seven units to the right of zero because it is seven blocks east of the bookstore.
tŚĂƚŝŶƚĞŐĞƌƌĞƉƌĞƐĞŶƚƐƚŚŝƐƐŝƚƵĂƚŝŽŶ͍
à
tŚĞƌĞǁŽƵůĚ&ƵƌƌLJ&ƌŝĞŶĚƐďĞůŽĐĂƚĞĚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍ džƉůĂŝŶ͘
à It would be seven units to the left of zero because it is seven blocks west of the bookstore.
tŚĂƚŝŶƚĞŐĞƌƌĞƉƌĞƐĞŶƚƐƚŚŝƐƐŝƚƵĂƚŝŽŶ͍
à െ
tŚĂƚĚŽLJŽƵŶŽƚŝĐĞĂďŽƵƚƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶďŽƚŚƐƚŽƌĞƐĨƌŽŵƚŚĞŬƐƚŽƌĞ͍
à Both stores are the same distance from the bookstore but in opposite directions.
െ െ െ െ െ െ െૠ െૡ ૡૠ
Furry Friends Bookstore Ray’s Pet Shop
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6ͻϯ Lesson 4 Lesson 4: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ
^ƚƵĚĞŶƚƐƐŚŽƵůĚƉƌĂĐƚŝĐĞĐůĂƌŝĨLJŝŶŐĂŶLJŵŝƐĐŽŶĐĞƉƚŝŽŶƐĂďŽƵƚŚŽǁƚŽƌĞƉƌĞƐĞŶƚƚŚĞƐĞƐŝƚƵĂƚŝŽŶƐĂƐŝŶƚĞŐĞƌƐ͘
͞ ^ĞǀĞŶďůŽĐŬƐƚŽƚŚĞůĞĨƚ͟ǁŽƵůĚŶŽƚďĞǁƌŝƚƚĞŶĂƐ͞ െ ďůŽĐŬƐĨƌŽŵƚŚĞŬƐƚŽƌĞ͟Žƌ͞ െ ƵŶŝƚƐĨƌŽŵ Ͳ͘͟
WŽƐŝƚŝǀĞŶƵŵďĞƌƐĂƌĞĐŽƵŶƚŝŶŐŶƵŵďĞƌƐĂŶĚĚŽŶŽƚŚĂǀĞĂƐŝŐŶ͘
Exercises 4–6 (5 minutes)
^ƚƵĚĞŶƚƐǁŽƌŬŝŶĚĞƉĞŶĚĞŶƚůLJƚŽĂŶƐǁĞƌƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐ͘ůůŽǁϮʹϯŵŝŶƵƚĞƐĨŽƌƌĞǀŝĞǁĂƐĂǁŚŽůĞŐƌŽƵƉ͘
Exercises 4–6 Read each situation carefully, and answer the questions. 4. On a number line, locate and label a credit of ̈́ and a debit for the same amount from a bank account. What does zero represent in this situation?
Zero represents no change in the balance.
On a number line, locate and label ι۱ below zero and ι۱ above zero. What does zero represent in this situation?
Zero represents ι۱ .
A proton represents a positive charge. Write an integer to represent protons. An electron represents a negative charge. Write an integer to represent electrons.
protons:
electrons: െ
Closing (2 minutes)
tŚĂƚŝƐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶĂŶLJŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞǁŚĞŶƉůŽƚƚĞĚŽŶĂŶƵŵďĞƌůŝŶĞ͍
à A nonzero number and its opposite are both the same distance away from zero on a number line, but they are on opposite sides of zero.
,ŽǁǁŽƵůĚLJŽƵƵƐĞƚŚŝƐƌĞůĂƚŝŽŶƐŚŝƉƚŽůŽĐĂƚĞƚŚĞŽƉƉŽƐŝƚĞŽĨĂŐŝǀĞŶŶƵŵďĞƌŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à I would use the given number to find the distance from zero on the opposite side.
tŝůůƚŚŝƐƉƌŽĐĞƐƐǁŽƌŬǁŚĞŶĨŝŶĚŝŶŐƚŚĞŽƉƉŽƐŝƚĞŽĨnjĞƌŽ͍
à No, because zero is its own opposite.
Exit Ticket (5 minutes)
െ െ െ ̈́ credit ̈́ debit
െ െ
degrees above degrees below
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6ͻϯ Lesson 4 Lesson 4: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ
EĂŵĞ ĂƚĞ
Lesson 4: The Opposite of a Number Exit Ticket
/ŶĂƌĞĐĞŶƚƐƵƌǀĞLJ͕ĂŵĂŐĂnjŝŶĞƌĞƉŽƌƚĞĚƚŚĂƚƚŚĞƉƌĞĨĞƌƌĞĚƌŽŽŵƚĞŵƉĞƌĂƚƵƌĞŝŶƚŚĞƐƵŵŵĞƌŝƐ ͺι ͘ ǁĂůůƚŚĞƌŵŽƐƚĂƚ͕ůŝŬĞƚŚĞŽŶĞƐƐŚŽǁŶďĞůŽǁ͕ƚĞůůƐĂƌŽŽŵ͛ƐƚĞŵƉĞƌĂƚƵƌĞŝŶĚĞŐƌĞĞƐ&ĂŚƌĞŶŚĞŝƚ͘ ^ĂƌĂŚ͛ƐhƉƐƚĂŝƌƐĞĚƌŽŽŵ ŽǁŶƐƚĂŝƌƐĞĚƌŽŽŵ Ă͘ tŚŝĐŚďĞĚƌŽŽŵŝƐǁĂƌŵĞƌƚŚĂŶƚŚĞƌĞĐŽŵŵĞŶĚĞĚƌŽŽŵƚĞŵƉĞƌĂƚƵƌĞ͍ ď͘ tŚŝĐŚďĞĚƌŽŽŵŝƐĐŽŽůĞƌƚŚĂŶƚŚĞƌĞĐŽŵŵĞŶĚĞĚƌŽŽŵƚĞŵƉĞƌĂƚƵƌĞ͍ Đ͘ ^ĂƌĂŚŶŽƚŝĐĞƐƚŚĂƚŚĞƌƌŽŽŵ͛ƐƚĞŵƉĞƌĂƚƵƌĞŝƐ Ͷι ĂďŽǀĞƚŚĞƌĞĐŽŵŵĞŶĚĞĚƚĞŵƉĞƌĂƚƵƌĞ͕ĂŶĚ ƚŚĞĚŽǁŶƐƚĂŝƌƐďĞĚƌŽŽŵ͛ƐƚĞŵƉĞƌĂƚƵƌĞŝƐ Ͷι ďĞůŽǁƚŚĞƌĞĐŽŵŵĞŶĚĞĚƚĞŵƉĞƌĂƚƵƌĞ͘^ŚĞ ŐƌĂƉŚƐ ʹ ĂŶĚ Ͷ ŽŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞĂŶĚĚĞƚĞƌŵŝŶĞƐƚŚĞLJĂƌĞŽƉƉŽƐŝƚĞƐ͘/Ɛ^ĂƌĂŚ ĐŽƌƌĞĐƚ͍ džƉůĂŝŶ͘ Ě͘ ĨƚĞƌĚĞƚĞƌŵŝŶŝŶŐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞƚĞŵƉĞƌĂƚƵƌĞƐ͕^ĂƌĂŚŶŽǁĚĞĐŝĚĞƐƚŽƌĞƉƌĞƐĞŶƚ
ʹι ĂƐ ͶĂŶĚ Ͷι ĂƐ െͶ ĂŶĚŐƌĂƉŚƐƚŚĞŵŽŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘'ƌĂƉŚ ͶĂŶĚ െͶ ŽŶƚŚĞ ǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞŽŶƚŚĞƌŝŐŚƚ͘ džƉůĂŝŶǁŚĂƚnjĞƌŽƌĞƉƌĞƐĞŶƚƐŝŶƚŚŝƐƐŝƚƵĂƚŝŽŶ͘
ʹԬ ͶԬ
Ͳ
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6ͻϯ Lesson 4 Lesson 4: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ
Exit Ticket Sample Solutions In a recent survey, a magazine reported that the preferred room temperature in the summer is ૡι۴ . A wall thermostat, like the ones shown below, tells a room’s temperature in degrees Fahrenheit. Sarah’s Upstairs Bedroom Downstairs Bedroom a. Which bedroom is warmer than the recommended room temperature?
The upstairs bedroom is warmer than the recommended room temperature.
b. Which bedroom is cooler than the recommended room temperature?
The downstairs bedroom is cooler than the recommended room temperature.
c. Sarah notices that her room’s temperature is ι۴ above the recommended temperature, and the downstairs bedroom’s temperature is ι۴ below the recommended temperature. She graphs ૠ
and on a vertical number line and determines they are opposites. Is Sarah correct? Explain.
No. Both temperatures are positive numbers and not the same distance from , so they cannot be opposites. Both numbers have to be the same distance from zero, but one has to be above zero, and the other has to be below zero in order to be opposites.
d. After determining the relationship between the temperatures, Sarah now decides to represent ૠι۴ as and
ι۴ as െ and graphs them on a vertical number line. Graph and െ on the vertical number line on the right. Explain what zero represents in this situation.
Zero represents the recommended room temperature of ࡲૡι . Zero could also represent not being above or below the recommended temperature.
Problem Set Sample Solutions 1. Find the opposite of each number, and describe its location on the number line. a. െ
The opposite of െ is , which is units to the right of (or above) .
b.
The opposite of is െ , which is units to the left of (or below) .
c. െ
The opposite of െ is , which is units to the right of (or above) .
ʹԬ ͶԬ
െ
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6ͻϯ Lesson 4 Lesson 4: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ
d.
The opposite of is െ , which is units to the left of (or below) .
Write the opposite of each number, and label the points on the number line. a. Point : the opposite of ૢ
െૢ
b. Point : the opposite of െ
c. Point : the opposite of െૠ
ૠ
d. Point ࡰ: the opposite of
e. Point ࡱ: the opposite of
െ
ϯ͘ Study the first example. Write the integer that represents the opposite of each real-world situation. In words, write the meaning of the opposite. a. An atom’s positive charge of ૠ െૠ , an atom’s negative charge of ૠ
b. A deposit of ̈́ െ , a withdrawal of ̈́
c. ǡ feet below sea level ǡ , ǡ feet above sea level
d. A rise of ι۱ െ , a decrease of ι۱
e. A loss of pounds , a gain of pounds
On a number line, locate and label a credit of ̈́ ૡ and a debit for the same amount from a bank account. What does zero represent in this situation?
Zero represents no change in the balance.
On a number line, locate and label ι۱ below zero and ι۱ above zero. What does zero represent in this situation?
Zero represents ι۱ .
െ െ െ െ െ െ െૠ െૡ ૡૠെૢ െ ૢ
ࡱ ࡰ
ૡ െૡ ̈́ ૡ credit ̈́ ૡ debit
െ െ
degrees above degrees below
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6ͻϯ Lesson 5 Lesson 5: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ͛ƐKƉƉŽƐŝƚĞ
Lesson 5: The Opposite of a Number’s Opposite Student Outcomes
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĂƚ͕ĨŽƌŝŶƐƚĂŶĐĞ͕ƚŚĞŽƉƉŽƐŝƚĞŽĨ െͷ ŝƐĚĞŶŽƚĞĚ െሺെͷሻ ĂŶĚŝƐĞƋƵĂůƚŽ ͷ͘ /ŶŐĞŶĞƌĂů͕ ƚŚĞLJŬŶŽǁƚŚĂƚƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŝƐƚŚĞŽƌŝŐŝŶĂůŶƵŵďĞƌ͖ĨŽƌĞdžĂŵƉůĞ͕ െሺെ ܽሻ ൌܽ .
^ƚƵĚĞŶƚƐůŽĐĂƚĞĂŶĚƉŽƐŝƚŝŽŶŽƉƉŽƐŝƚĞŶƵŵďĞƌƐŽŶĂŶƵŵďĞƌůŝŶĞ͘
Classwork Opening Exercise (7 minutes)
^ƚƵĚĞŶƚƐǁŽƌŬŝŶĚĞƉĞŶĚĞŶƚůLJĨŽƌϱŵŝŶƵƚĞƐƚŽĐŽŵƉůĞƚĞƚŚĞĨŽůůŽǁŝŶŐƌĞǀŝĞǁƉƌŽďůĞŵƐĂŶĚƚŚĞŶƌĞǀŝĞǁĂƐĂĐůĂƐƐ͘
Opening Exercise a. Locate the number െ and its opposite on the number line below. b. Write an integer that represents each of the following. i. ૢ feet below sea level
െૢ
ii. ̈́ of debt
െ
iii. ι۱ above zero
c. Joe is at the ice cream shop, and his house is blocks north of the shop. The park is blocks south of the ice cream shop. When he is at the ice cream shop, is Joe closer to the park or his house? How could the number zero be used in this situation? Explain.
He is the same distance from his house and the park because both are located blocks away from the ice cream shop but in opposite directions. In this situation, zero represents the location of the ice cream shop.
െ െ െ െ െ െ െૠ െૡ ૡૠ
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6ͻϯ Lesson 5 Lesson 5: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ͛ƐKƉƉŽƐŝƚĞ
Example 1 (8 minutes): The Opposite of an Opposite of a Number
tŚĂƚŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨ ͺ͍,ŽǁĐĂŶǁĞŝůůƵƐƚƌĂƚĞƚŚŝƐŶƵŵďĞƌŽŶĂŶƵŵďĞƌůŝŶĞ͍ (Example student responses are listed below.) ĞĨŽƌĞƐƚĂƌƚŝŶŐƚŚĞĞdžĂŵƉůĞ͕ĂůůŽǁƐƚƵĚĞŶƚƐƚŽĚŝƐĐƵƐƐƚŚĞŝƌƉƌĞĚŝĐƚŝŽŶƐŝŶŐƌŽƵƉƐĨŽƌŽŶĞŵŝŶƵƚĞ͘ŚŽŽƐĞĂĨĞǁƐƚƵĚĞŶƚƐ ƚŽƐŚĂƌĞƚŚĞŝƌƌĞƐƉŽŶƐĞƐǁŝƚŚƚŚĞƌĞƐƚŽĨƚŚĞĐůĂƐƐ͘
Example 1: The Opposite of an Opposite of a Number What is the opposite of the opposite of ૡ? How can we illustrate this number on a number line? a. What number is ૡunits to the right of ?ૡ
b. How can you illustrate locating the opposite of ૡon this number line?
We can illustrate the opposite of ૡon the number line by counting ૡunits to the left of zero rather than to the right of zero.
c. What is the opposite of ૡ?െૡ
d. Use the same process to locate the opposite of െૡ . What is the opposite of െૡ ?ૡ
e. The opposite of an opposite of a number is the original number .
Example 2 (8 minutes): Writing the Opposite of an Opposite of a Number
/ŶƚŚŝƐĞdžĂŵƉůĞ͕ƐƚƵĚĞŶƚƐƵƐĞƚŚĞ͞ െ͟ ƐLJŵďŽůƚŽŝŶĚŝĐĂƚĞ͞ƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶƵŵďĞƌ͘͟ŝƐƉůĂLJƚŚĞƚĂƐŬ͕͞ džƉůĂŝŶǁŚLJ
െሺെͷሻ ൌ ͷ ͟ ŽŶƚŚĞďŽĂƌĚ͘͞ െ͟ ƐLJŵďŽůŵĞĂŶƐ͞ƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶƵŵďĞƌ͘͟
െሺെͷሻ ൌ ͷ ^ŝŶĐĞƚŚĞŽƉƉŽƐŝƚĞŽĨ ͷŝƐŶĞŐĂƚŝǀĞ ͷĂŶĚƚŚĞŽƉƉŽƐŝƚĞŽĨŶĞŐĂƚŝǀĞ ͷŝƐƉŽƐŝƚŝǀĞ ͷ͕ ƚŚĞŶ െሺെͷሻ ൌ ͷ ͘ WŽƐĞƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐƚŽƚŚĞĐůĂƐƐƚŽĐŚĞĐŬĨŽƌƵŶĚĞƌƐƚĂŶĚŝŶŐ͘
tŚĂƚŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨ െ ͍
à
tŚĂƚŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨ ͳͲ ͍
à ͳͲ
,ŽǁǁŽƵůĚLJŽƵǁƌŝƚĞƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨ െͳʹ ͍
à The opposite of െͳʹ ͗ െሺെͳʹሻ ൌ ͳʹ . The opposite of ͳʹ ͗ െሺͳʹሻ ൌ െͳʹ . So, െ൫െሺെͳʹሻ൯ ൌ െͳʹ .
tŚĂƚĚŽĞƐĂ͞ െ͟ ƐLJŵďŽůŵĞĂŶ͍
à It can mean the opposite of a number or indicate a negative number.
െ െ െ െ െ െ െૠ െૡ ૡૠ
Scaffolding:
&ŽƌǀŝƐƵĂůůĞĂƌŶĞƌƐ͕ƵƐĞƚŚĞ ĐƵƌǀĞĚĂƌƌŽǁƐŝŶŝƚŝĂůůLJƚŽ ĚĞǀĞůŽƉĐŽŶĐĞƉƚƵĂů ƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨŚŽǁƚŽĨŝŶĚ ƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶŽƉƉŽƐŝƚĞŽĨ ĂŶƵŵďĞƌ͘
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6ͻϯ Lesson 5 Lesson 5: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ͛ƐKƉƉŽƐŝƚĞ
tŚĂƚŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨĂĚĞďŝƚŽĨ ̈́ ͳʹ ͍
à A debit of ̈́ ͳʹ is െͳʹ . The opposite of െͳʹ is ͳʹ , or a credit of ̈́ ͳʹ . The opposite of a credit of ̈́ ͳʹ is a debit of ̈́ ͳʹ .
/ŶŐĞŶĞƌĂů͕ƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶƵŵďĞƌŝƐƚŚĞŽƌŝŐŝŶĂůŶƵŵďĞƌ͖ĨŽƌĞdžĂŵƉůĞ͕ െሺെ ܽሻ ൌܽ ͘
Exercises 1– ϯ (12 minutes)
^ƚƵĚĞŶƚƐǁŽƌŬŝŶŐƌŽƵƉƐŽĨϯʹϰƚŽĨŝŶĚƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶŽƉƉŽƐŝƚĞŽĨĂƐĞƚŽĨŐŝǀĞŶŶƵŵďĞƌƐ͘ůůŽǁϰʹϱŵŝŶƵƚĞƐĨŽƌ ƌĞǀŝĞǁĂƐĂǁŚŽůĞĐůĂƐƐ͘ŝƐƚƌŝďƵƚĞĂĐĂƌĚǁŝƚŚĂŶŝŶƚĞŐĞƌŽŶŝƚƚŽĞĂĐŚŐƌŽƵƉ͘;hƐĞƐƚŝĐŬLJŶŽƚĞƐŽƌŝŶĚĞdžĐĂƌĚƐ͘Ϳ ^ƚƵĚĞŶƚƐĐŽŵƉůĞƚĞƚŚĞƚĂďůĞƵƐŝŶŐĂůůƚŚĞĐĂƌĚƐŝŶƚŚĞŝƌŐƌŽƵƉ͘
Exercises Complete the table using the cards in your group. Person Card ( ࢇ)Opposite of Card ( ࢇെ )Opposite of Opposite of Card െሺെࢇሻ
Jackson െሺሻ ൌ െ െሺെሻ ൌ
DeVonte െሺሻ ൌ െ െሺെሻ ൌ
Cheryl െ െሺെሻ ൌ െ൫െሺെሻ൯ ൌ െ
Toby െૢ െሺെૢ ሻ ൌૢ െ൫െሺെૢ ሻ൯ ൌ െૢ
1. Write the opposite of the opposite of െ as an equation.
The opposite of െ :െሺെሻ ൌ ; the opposite of :െሺሻ ൌ െ . Therefore, ቀെ൫െሺെሻ൯ቁ ൌ െ .
2. In general, the opposite of the opposite of a number is the original number .
ϯ͘ Provide a real-world example of this rule. Show your work.
Answers will vary. The opposite of the opposite of feet below sea level is feet below sea level.
െ is feet below sea level.
െሺെሻ ൌ , the opposite of െ
െሺሻ ൌ െ , the opposite of
Closing ;ϯŵŝŶƵƚĞƐͿ
tŚĂƚŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶŽƉƉŽƐŝƚĞŽĨĂŶƵŵďĞƌ͍^ƵƉƉŽƌƚLJŽƵƌĂŶƐǁĞƌǁŝƚŚĂŶĞdžĂŵƉůĞ͘
à The opposite of an opposite of a number is the original number. The opposite of the opposite of െ is
െ because the opposite of െ is . The opposite of is െ .
tŚĂƚŝƐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞůŽĐĂƚŝŽŶŽĨĂŶŽŶnjĞƌŽŶƵŵďĞƌŽŶƚŚĞŶƵŵďĞƌůŝŶĞĂŶĚƚŚĞůŽĐĂƚŝŽŶŽĨŝƚƐ ŽƉƉŽƐŝƚĞŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à A number and its opposite are located the same distance from Ͳ on a number line but on opposite sides of Ͳ.
Exit Ticket (7 minutes)
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6ͻϯ Lesson 5 Lesson 5: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ͛ƐKƉƉŽƐŝƚĞ
EĂŵĞ ĂƚĞ
Lesson 5: The Opposite of a Number’s Opposite Exit Ticket
ϭ͘ :ĂŶĞĐŽŵƉůĞƚĞƐƐĞǀĞƌĂůĞdžĂŵƉůĞƉƌŽďůĞŵƐƚŚĂƚĂƐŬŚĞƌƚŽƚŚĞĨŝŶĚƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨĂŶƵŵďĞƌ͕ĂŶĚĨŽƌ ĞĂĐŚĞdžĂŵƉůĞ͕ƚŚĞƌĞƐƵůƚŝƐĂƉŽƐŝƚŝǀĞŶƵŵďĞƌ͘:ĂŶĞĐŽŶĐůƵĚĞƐƚŚĂƚǁŚĞŶƐŚĞƚĂŬĞƐƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨ ĂŶLJŶƵŵďĞƌ͕ƚŚĞƌĞƐƵůƚǁŝůůĂůǁĂLJƐďĞƉŽƐŝƚŝǀĞ͘/Ɛ:ĂŶĞĐŽƌƌĞĐƚ͍tŚLJŽƌǁŚLJŶŽƚ͍ Ϯ͘ dŽƐƵƉƉŽƌƚLJŽƵƌĂŶƐǁĞƌĨƌŽŵƚŚĞƉƌĞǀŝŽƵƐƋƵĞƐƚŝŽŶ͕ĐƌĞĂƚĞĂŶĞdžĂŵƉůĞ͕ǁƌŝƚƚĞŶĂƐĂŶĞƋƵĂƚŝŽŶ͘/ůůƵƐƚƌĂƚĞLJŽƵƌ ĞdžĂŵƉůĞŽŶƚŚĞŶƵŵďĞƌůŝŶĞďĞůŽǁ͘
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6ͻϯ Lesson 5 Lesson 5: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ͛ƐKƉƉŽƐŝƚĞ
Exit Ticket Sample Solutions 1. Jane completes several example problems that ask her to the find the opposite of the opposite of a number, and for each example, the result is a positive number. Jane concludes that when she takes the opposite of the opposite of any number, the result will always be positive. Is Jane correct? Why or why not?
She is not correct. The opposite of the opposite of a number is the original number. So, if Jane starts with a negative number, she will end with a negative number.
To support your answer from the previous question, create an example, written as an equation. Illustrate your example on the number line below.
If Jane starts with െૠ , the opposite of the opposite of െૠ is written as െ൫െሺെૠሻ൯ ൌ െૠ or the opposite of െૠ :
െሺെૠሻ ൌ ૠ ; the opposite of ૠ: െሺૠሻ ൌ െૠ .
Problem Set Sample Solutions 1. Read each description carefully, and write an equation that represents the description. a. The opposite of negative seven
െሺെૠሻ ൌ ૠ
b. The opposite of the opposite of twenty-five
െ൫െሺሻ൯ ൌ
c. The opposite of fifteen
െሺሻ ൌ െ
d. The opposite of negative thirty-six
െሺെሻ ൌ
Jose graphed the opposite of the opposite of on the number line. First, he graphed point ࡼ on the number line
units to the right of zero. Next, he graphed the opposite of ࡼ on the number line units to the left of zero and labeled it ࡷ. Finally, he graphed the opposite of ࡷ and labeled it ࡽ.a. Is his diagram correct? Explain. If the diagram is not correct, explain his error, and correctly locate and label point ࡽ.
Yes, his diagram is correct. It shows that point ࡼ is because it is units to the right of zero. The opposite of
is െ , which is point ࡷ ( units to the left of zero). The opposite of െ is , so point ࡽ is units to the right of zero.
െૠ ૠ
ࡷ ࡼ
ࡽ
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6ͻϯ Lesson 5 Lesson 5: dŚĞKƉƉŽƐŝƚĞŽĨĂEƵŵďĞƌ͛ƐKƉƉŽƐŝƚĞ
b. Write the relationship between the points:
ࡼ and ࡷ
They are opposites.
ࡷ and ࡽ
They are opposites.
ࡼ and ࡽ
They are the same.
ϯ͘ Read each real-world description. Write the integer that represents the opposite of the opposite. Show your work to support your answer. a. A temperature rise of degrees Fahrenheit
െ is the opposite of (fall in temperature).
is the opposite of െ (rise in temperature).
െ൫െሺሻ൯ ൌ
b. A gain of yards
െ is the opposite of (loss of yards).
is the opposite of െ (gain of yards).
െ൫െሺሻ൯ ൌ
c. A loss of pounds
is the opposite of െ (gain of pounds).
െ is the opposite of (loss of pounds).
െ൫െሺെሻ൯ ൌ െ
d. A withdrawal of ̈́ ǡ
ǡ is the opposite of െǡ (deposit).
െǡ is the opposite of ǡ (withdrawal).
െ൫െሺെǡ ሻ൯ ൌ െǡ
Write the integer that represents the statement. Locate and label each point on the number line below. a. The opposite of a gain of െ
b. The opposite of a deposit of ̈́ െ
c. The opposite of the opposite of
d. The opposite of the opposite of
e. The opposite of the opposite of a loss of െ
െ െ െ
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6 Lesson 6: Rational Numbers on the Number Line Student Outcomes
^ƚƵĚĞŶƚƐƵƐĞŶƵŵďĞƌůŝŶĞƐƚŚĂƚĞdžƚĞŶĚŝŶďŽƚŚĚŝƌĞĐƚŝŽŶƐĂŶĚƵƐĞ ͲĂŶĚ ͳƚŽůŽĐĂƚĞŝŶƚĞŐĞƌƐĂŶĚƌĂƚŝŽŶĂů ŶƵŵďĞƌƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘^ƚƵĚĞŶƚƐŬŶŽǁƚŚĂƚƚŚĞƐŝŐŶŽĨĂŶŽŶnjĞƌŽƌĂƚŝŽŶĂůŶƵŵďĞƌŝƐƉŽƐŝƚŝǀĞŽƌ ŶĞŐĂƚŝǀĞ͕ĚĞƉĞŶĚŝŶŐŽŶǁŚĞƚŚĞƌƚŚĞŶƵŵďĞƌŝƐŐƌĞĂƚĞƌƚŚĂŶnjĞƌŽ;ƉŽƐŝƚŝǀĞͿŽƌůĞƐƐƚŚĂŶnjĞƌŽ;ŶĞŐĂƚŝǀĞͿ͕ĂŶĚ ƵƐĞĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞǁŚĞŶŐƌĂƉŚŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘
^ƚƵĚĞŶƚƐŬŶŽǁƚŚĂƚƚŚĞŽƉƉŽƐŝƚĞƐŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐĂƌĞƐŝŵŝůĂƌƚŽƚŚĞŽƉƉŽƐŝƚĞƐŽĨŝŶƚĞŐĞƌƐ͘^ƚƵĚĞŶƚƐŬŶŽǁ ƚŚĂƚƚǁŽƌĂƚŝŽŶĂůŶƵŵďĞƌƐŚĂǀĞŽƉƉŽƐŝƚĞƐŝŐŶƐŝĨƚŚĞLJĂƌĞŽŶĚŝĨĨĞƌĞŶƚƐŝĚĞƐŽĨnjĞƌŽĂŶĚƚŚĂƚƚŚĞLJŚĂǀĞƚŚĞ ƐĂŵĞƐŝŐŶŝĨƚŚĞLJĂƌĞŽŶƚŚĞƐĂŵĞƐŝĚĞŽĨnjĞƌŽŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘
Classwork Opening Exercise (5 minutes)
^ƚƵĚĞŶƚƐǁŽƌŬŝŶĚĞƉĞŶĚĞŶƚůLJĨŽƌϱŵŝŶƵƚĞƐƚŽƌĞǀŝĞǁĨƌĂĐƚŝŽŶƐĂŶĚĚĞĐŝŵĂůƐ͘
Opening Exercise a. Write the decimal equivalent of each fraction. i.
Ǥ
ii.
Ǥ ૡ
iii. ૠ
Ǥ ૠ
b. Write the fraction equivalent of each decimal. i. Ǥ
ൌ
ii. Ǥ ૠ
ૠ ൌ
iii. Ǥૢ
ૢ ൌ ૢ
Scaffolding:
hƐĞƉŽůůŝŶŐƐŽĨƚǁĂƌĞƚŽ ĞůŝĐŝƚŝŵŵĞĚŝĂƚĞĨĞĞĚďĂĐŬ ĨƌŽŵƚŚĞĐůĂƐƐƚŽĞŶŐĂŐĞ ĂůůůĞĂƌŶĞƌƐ͘
ŝƐƉůĂLJĞĂĐŚƉƌŽďůĞŵŽŶĞ ĂƚĂƚŝŵĞ͕ĂŶĚƵƐĞƉĞƌƐŽŶĂů ǁŚŝƚĞďŽĂƌĚƐĨŽƌ ŬŝŶĞƐƚŚĞƚŝĐůĞĂƌŶĞƌƐ͘
Scaffolding:
hƐĞĞĚŐĞƐŽĨƐƋƵĂƌĞƚŝůĞƐ ŽŶƚŚĞĨůŽŽƌĂƐĂŶƵŵďĞƌ ůŝŶĞƚŽŝůůƵƐƚƌĂƚĞŚŽǁƚŽ ĐŽŶŶĞĐƚƐĞŐŵĞŶƚƐŽĨĞƋƵĂů ůĞŶŐƚŚĨŽƌǀŝƐƵĂůĂŶĚ ŬŝŶĞƐƚŚĞƚŝĐůĞĂƌŶĞƌƐ͘
WƌŽǀŝĚĞŐƌĞĞŶĂŶĚƌĞĚ ƉĞŶĐŝůƐƚŽŚĞůƉǁŝƚŚ ŵŽĚĞůŝŶŐƚŚĞĞdžĂŵƉůĞĨŽƌ ǀŝƐƵĂůůĞĂƌŶĞƌƐ͘
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6 Example 1 (10 minutes): Graphing Rational Numbers
dŚĞƉƵƌƉŽƐĞŽĨƚŚŝƐĞdžĂŵƉůĞŝƐƚŽƐŚŽǁƐƚƵĚĞŶƚƐŚŽǁƚŽŐƌĂƉŚŶŽŶͲŝŶƚĞŐĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶĂƌĞĂůŶƵŵďĞƌůŝŶĞ͘ ^ƚƵĚĞŶƚƐĐŽŵƉůĞƚĞƚŚĞĞdžĂŵƉůĞďLJĨŽůůŽǁŝŶŐĂůŽŶŐǁŝƚŚƚŚĞƚĞĂĐŚĞƌ͘
>ŽĐĂƚĞĂŶĚŐƌĂƉŚƚŚĞŶƵŵďĞƌ ͵ͳͲ ĂŶĚŝƚƐŽƉƉŽƐŝƚĞŽŶĂŶƵŵďĞƌůŝŶĞ͘ ĞĨŽƌĞŵŽĚĞůŝŶŐƚŚĞĞdžĂŵƉůĞ͕ƌĞǀŝĞǁŐƌĂƉŚŝŶŐĂĨƌĂĐƚŝŽŶŽŶƚŚĞŶƵŵďĞƌůŝŶĞƚŽƚŚĞǁŚŽůĞĐůĂƐƐďLJĨŝƌƐƚƌĞǀŝĞǁŝŶŐ ĨƌĂĐƚŝŽŶĚĞĨŝŶŝƚŝŽŶƐǁŝƚŚƌĞƐƉĞĐƚƚŽƚŚĞŶƵŵďĞƌůŝŶĞ͘
Example 1: Graphing Rational Numbers If ࢈is a nonzero whole number, then the unit fraction ࢈is located on the number line by dividing the segment between
and into ࢈segments of equal length. One of the ࢈segments has as its left end point; the right end point of this segment corresponds to the unit fraction ࢈.
^ŝŶĐĞƚŚĞŶƵŵďĞƌŝƐĂ rational number ͕ ĂŶƵŵďĞƌƚŚĂƚĐĂŶďĞƌĞƉƌĞƐĞŶƚĞĚĂƐĂĨƌĂĐƚŝŽŶ͕ĚĞƚĞƌŵŝŶĞŚŽǁƚŚĞ ŶƵŵďĞƌůŝŶĞƐŚŽƵůĚďĞƐĐĂůĞĚ͘ ϭ
&ŝƌƐƚ͕ĚŝǀŝĚĞƚŚĞŶƵŵďĞƌůŝŶĞŝŶƚŽƚǁŽŚĂůǀĞƐƚŽƌĞƉƌĞƐĞŶƚƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐ͘
,ĂǀĞƐƚƵĚĞŶƚƐĐŽŵƉůĞƚĞƚŚŝƐƚĂƐŬŽŶƚŚĞŝƌƐƚƵĚĞŶƚƉĂŐĞƐ͘
EĞdžƚ͕ĚŝǀŝĚĞƚŚĞƌŝŐŚƚŚĂůĨŽĨƚŚĞŶƵŵďĞƌůŝŶĞƐĞŐŵĞŶƚďĞƚǁĞĞŶ ͲĂŶĚ ͳŝŶƚŽƚĞŶƐĞŐŵĞŶƚƐŽĨĞƋƵĂůůĞŶŐƚŚ͖ ĞĂĐŚƐĞŐŵĞŶƚŚĂƐĂůĞŶŐƚŚŽĨ ͳͳͲ ͘
^ƚƵĚĞŶƚƐĚŝǀŝĚĞƚŚĞŝƌŶƵŵďĞƌůŝŶĞƐŝŶƚŽƚĞŶĞƋƵĂůƐĞŐŵĞŶƚƐĂƐƐŚŽǁŶ͘ŚĞĐŬĨŽƌĂĐĐƵƌĂĐLJ͘
dŚĞƌĞĂƌĞ ͳͲ ĞƋƵĂůƐĞŐŵĞŶƚƐ͘ ĂĐŚƐĞŐŵĞŶƚŚĂƐĂůĞŶŐƚŚŽĨ ଵଵ ͘ dŚĞĨŝƌƐƚƐĞŐŵĞŶƚŚĂƐ ͲĂƐŝƚƐůĞĨƚĞŶĚƉŽŝŶƚ͕ ĂŶĚƚŚĞƌŝŐŚƚĞŶĚƉŽŝŶƚĐŽƌƌĞƐƉŽŶĚƐƚŽ ͳͳͲ ͘ ϭ^ƵƉƉůĞŵĞŶƚĂů džĞƌĐŝƐĞ͗
,ĂǀĞĨŽƵƌƐƚƵĚĞŶƚƐĞĂĐŚƐƚĂŶĚŝŶĂƐƋƵĂƌĞĨůŽŽƌƚŝůĞĨŽƌŵŝŶŐĂƐƚƌĂŝŐŚƚůŝŶĞĨĂĐŝŶŐƚŚĞĐůĂƐƐ͘'ŝǀĞĞĂĐŚƐƚƵĚĞŶƚĂŶƵŵďĞƌƚŽƚŝĞ ĂƌŽƵŶĚŚŝƐŶĞĐŬ͗ Ͳ͕ ଵଵ ͕ ଶଵ ͕Žƌ ଷଵ ͘;hƐĞŝŶĚĞdžĐĂƌĚƐŽƌĐŽŶƐƚƌƵĐƚŝŽŶƉĂƉĞƌ͘Ϳ
ƐŬĂĨŝĨƚŚƐƚƵĚĞŶƚƚŽĂƐƐŝƐƚďLJŐŝǀŝŶŐŽŶĞĞŶĚŽĨĂďĂůůŽĨƐƚƌŝŶŐƚŽƚŚĞƉĞƌƐŽŶĂƚ Ͳ͘ dŚŝƐƉĞƌƐŽŶŚŽůĚƐŽŶĞĞŶĚŽĨƚŚĞƐƚƌŝŶŐĂŶĚ ƉĂƐƐĞƐƚŚĞƌĞƐƚƚŽƚŚĞƉĞƌƐŽŶƚŽƚŚĞůĞĨƚ͘;^ŽƚŚĞĐůĂƐƐƐĞĞƐŝƚŵŽǀŝŶŐƚŽƚŚĞƌŝŐŚƚ͘Ϳ
ƐƚŚĞƐƚƌŝŶŐŐĞƚƐƉĂƐƐĞĚĚŽǁŶƚŚĞůŝŶĞ͕ĞĂĐŚƉĞƌƐŽŶĂŶŶŽƵŶĐĞƐŚĞƌŶƵŵďĞƌ͕͞ ଵଵ ͕ ଶଵ ͕ ଷଵ ǡ͟ ƐƚŽƉƉŝŶŐĂƚ ଷଵ ͘
dŚĞĂƐƐŝƐƚĂŶƚĐƵƚƐƚŚĞƐƚƌŝŶŐĂƚ ଷଵ ĂŶĚŐŝǀĞƐƚŚĂƚĞŶĚŽĨƚŚĞƐƚƌŝŶŐƚŽƚŚĞƉĞƌƐŽŶŚŽůĚŝŶŐ ଷଵ ͕ŵĂŬŝŶŐŽŶĞƐĞŐŵĞŶƚŽĨůĞŶŐƚŚ ଷଵ ͘
,ĂǀĞƐƚƵĚĞŶƚƐƚƵƌŶŽǀĞƌƚŚĞŝƌŶƵŵďĞƌƐƚŽƌĞǀĞĂůƚŚĞŝƌŽƉƉŽƐŝƚĞƐĂŶĚƌĞĂƌƌĂŶŐĞƚŚĞŵƐĞůǀĞƐƚŽƌĞƉƌĞƐĞŶƚƚŚĞŽƉƉŽƐŝƚĞŽĨ ଷଵ ƵƐŝŶŐƚŚĞƐĂŵĞƉƌŽĐĞƐƐ͘dŚŝƐƚŝŵĞ͕ƐƚƵĚĞŶƚƐƉĂƐƐƚŚĞƐƚƌŝŶŐƚŽƚŚĞƌŝŐŚƚ͘;^ŽƚŚĞĐůĂƐƐƐĞĞƐŝƚŵŽǀŝŶŐƚŽƚŚĞůĞĨƚ͘Ϳ
ͻͳͲ
ͺͳͲ
ͳͲ
ͳͲ
ͷͳͲ
ͶͳͲ
͵ͳͲ
ʹͳͲ
ͳͳͲ Ͳ ͳ
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6 The fraction ࢇ࢈ is located on the number line by joining ࢇ segments of length ࢈ so that (1) the left end point of the first segment is , and (2) the right end point of each segment is the left end point of the next segment. The right end point of the last segment corresponds to the fraction ࢇ࢈.
dŽůŽĐĂƚĞƚŚĞŶƵŵďĞƌ ŽŶĂŶƵŵďĞƌůŝŶĞ͕ƐƚƵĚĞŶƚƐƐŚŽƵůĚĚŝǀŝĚĞƚŚĞŝŶƚĞƌǀĂůďĞƚǁĞĞŶnjĞƌŽĂŶĚ ͳŝŶƚŽ ܾ ĞƋƵĂů ƉĂƌƚƐ͘^ƚĂƌƚŝŶŐĂƚ Ͳ͕ ŵŽǀĞĂůŽŶŐƚŚĞŶƵŵďĞƌůŝŶĞ ܽ ŶƵŵďĞƌŽĨƚŝŵĞƐ͘
dŚĞƌĞĂƌĞƚĞŶĞƋƵĂůƐĞŐŵĞŶƚƐ͘ ĂĐŚƐĞŐŵĞŶƚŚĂƐĂůĞŶŐƚŚŽĨ ͳͳͲ ͘dŚĞĨŝƌƐƚƐĞŐŵĞŶƚŚĂƐĂ ͲĂƐŝƚƐůĞĨƚĞŶĚƉŽŝŶƚ͕ ĂŶĚƚŚĞƌŝŐŚƚĞŶĚƉŽŝŶƚŽĨƚŚĞƚŚŝƌĚƐĞŐŵĞŶƚĐŽƌƌĞƐƉŽŶĚƐƚŽ ͵ͳͲ ͘dŚĞƉŽŝŶƚŝƐůŽĐĂƚĞĚĂƚ ͵ͳͲ ͘
dŚĞŽƉƉŽƐŝƚĞŽĨ ͵ͳͲ ŝƐůŽĐĂƚĞĚƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽĂƐ ͵ͳͲ ďƵƚŝŶƚŚĞŽƉƉŽƐŝƚĞĚŝƌĞĐƚŝŽŶŽƌƚŽƚŚĞůĞĨƚ͘ hƐŝŶŐLJŽƵƌŬŶŽǁůĞĚŐĞŽĨŽƉƉŽƐŝƚĞƐ͕ǁŚĂƚƌĂƚŝŽŶĂůŶƵŵďĞƌƌĞƉƌĞƐĞŶƚƐƚŚĞŽƉƉŽƐŝƚĞŽĨ ͵ͳͲ ͍
à െ ͵ͳͲ
dŽůŽĐĂƚĞƚŚĞŽƉƉŽƐŝƚĞŽĨ ͵ͳͲ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͕ĚŝǀŝĚĞƚŚĞŝŶƚĞƌǀĂůďĞƚǁĞĞŶnjĞƌŽĂŶĚ െͳ ŝŶƚŽƚĞŶĞƋƵĂů ƐĞŐŵĞŶƚƐ͘^ƚĂƌƚŝŶŐĂƚnjĞƌŽ͕ŚŽǁĨĂƌǁŽƵůĚǁĞŵŽǀĞƚŽůŽĐĂƚĞƚŚĞŽƉƉŽƐŝƚĞŽĨ ͵ͳͲ ǡĂŶĚŝŶǁŚĂƚĚŝƌĞĐƚŝŽŶ͍
à We would move ͵ units to the left of zero because that is the same distance but opposite direction we moved to plot the point ͵ͳͲ .
dŚĞƌĞĂƌĞƚĞŶĞƋƵĂůƐĞŐŵĞŶƚƐ͘ ĂĐŚƐĞŐŵĞŶƚŚĂƐĂůĞŶŐƚŚŽĨ ͳͳͲ ͘dŚƌĞĞĐŽŶƐĞĐƵƚŝǀĞƐĞŐŵĞŶƚƐ͕ƐƚĂƌƚŝŶŐĂƚ ͲĂŶĚŵŽǀŝŶŐƚŽƚŚĞůĞĨƚ͕ǁŽƵůĚŚĂǀĞĂƚŽƚĂůůĞŶŐƚŚŽĨ ͵ͳͲ ͘dŚĞƉŽŝŶƚŝƐůŽĐĂƚĞĚĂƚ െ ͵ͳͲ ͘
ŽƵŶƚŝŶŐƚŚƌĞĞĐŽŶƐĞĐƵƚŝǀĞƐĞŐŵĞŶƚƐŽĨůĞŶŐƚŚŽĨ ͳͳͲ ĨƌŽŵ ͲŵŽǀŝŶŐƚŽƚŚĞůĞĨƚĂŶĚƚĂŬŝŶŐƚŚĞĞŶĚƉŽŝŶƚŽĨƚŚĞ ůĂƐƚƐĞŐŵĞŶƚĐŽƌƌĞƐƉŽŶĚƐƚŽƚŚĞŶƵŵďĞƌ െ ͵ͳͲ ͘dŚĞƌĞĨŽƌĞ͕ƚŚĞŽƉƉŽƐŝƚĞŽĨ ͵ͳͲ ŝƐ െ ͵ͳͲ ͘
Locate and graph the number and its opposite on a number line.
ͻͳͲ
ͺͳͲ
ͳͲ
ͳͲ
ͷͳͲ
ͶͳͲ
͵ͳͲ
ʹͳͲ
ͳͳͲ Ͳ ͳ
െ ͻͳͲ െ ͺͳͲ െ ͳͲ െ ͳͲ െ ͷͳͲ െ ͶͳͲ െ ͵ͳͲ െ ʹͳͲ െ ͳͳͲ Ͳെͳ ૢ
ૡ
ૠ
െૢ െ ૡ െ ૠ െ െ െ െ െ െ െ
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6
Exercise 1 (5 minutes)
^ƚƵĚĞŶƚƐǁŽƌŬŝŶĚĞƉĞŶĚĞŶƚůLJƚŽƉƌĂĐƚŝĐĞŐƌĂƉŚŝŶŐĂŶŽŶͲŝŶƚĞŐĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ ůůŽǁϮʹϯŵŝŶƵƚĞƐĨŽƌƌĞǀŝĞǁĂƐĂǁŚŽůĞŐƌŽƵƉ͘
Exercise 1 Use what you know about the point െ ૠ and its opposite to graph both points on the number line below. The fraction
െ ૠ is located between which two consecutive integers? Explain your reasoning. On the number line, each segment will have an equal length of
. The fraction is located between െ and െ .Explanation:
ૠ is the opposite of െ ૠ. It is the same distance from zero but on the opposite side of zero. Since െ ૠ is to the left of zero,
ૠ is to the right of zero. The original fraction is located between െ (or െ ૡ) and െ (or െ ).
Example 2 (7 minutes): Rational Numbers and the Real World ŝƐƉůĂLJƚŚĞĨŽůůŽǁŝŶŐǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞŵŽĚĞůŽŶƚŚĞďŽĂƌĚ͘^ƚƵĚĞŶƚƐĂƌĞƚŽĨŽůůŽǁĂůŽŶŐŝŶƚŚĞŝƌƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐƚŽ ĂŶƐǁĞƌƚŚĞƋƵĞƐƚŝŽŶƐ͘WŽƐĞĂĚĚŝƚŝŽŶĂůƋƵĞƐƚŝŽŶƐƚŽƚŚĞĐůĂƐƐƚŚƌŽƵŐŚŽƵƚƚŚĞĞdžĂŵƉůĞ͘
Example 2: Rational Numbers and the Real World The water level of a lake rose Ǥ feet after it rained. Answer the following questions using the number line below. a. Write a rational number to represent the situation.
Ǥ or
b. What two integers is Ǥ between on a number line?
and
c. Write the length of each segment on the number line as a decimal and a fraction.
Ǥ and
d. What will be the water level after it rained? Graph the point on the number line.
Ǥ feet above the original lake level
ૠ
ૡ
െ ૠ െ െ െ െ െ െ െ ૡ
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6 e. After two weeks have passed, the water level of the lake is now the opposite of the water level when it rained. What will be the new water level? Graph the point on the number line. Explain how you determined your answer.
The water level would be Ǥ feet below the original lake level. If the water level was Ǥ , the opposite of
Ǥ is െǤ .
f. State a rational number that is not an integer whose value is less than Ǥ , and describe its location between two consecutive integers on the number line.
Answers will vary. A rational number whose value is less than Ǥ is Ǥ ૠ . It would be located between
and on a number line.
WŽƐƐŝďůĞĚŝƐĐƵƐƐŝŽŶƋƵĞƐƚŝŽŶƐ͗
tŚĂƚƵŶŝƚƐĂƌĞǁĞƵƐŝŶŐƚŽŵĞĂƐƵƌĞƚŚĞǁĂƚĞƌůĞǀĞů͍
à Feet
tŚĂƚǁĂƐƚŚĞǁĂƚĞƌůĞǀĞůĂĨƚĞƌƚŚĞƌĂŝŶ͍,ŽǁĚŽLJŽƵŬŶŽǁ͍
à If zero represents the original water level on the number line, the water level after rain is ͳǤʹͷ feet. From Ͳ to ͳ, there are four equal segments. This tells me that the scale is ଵସ. The top of the water is represented on the number line at one mark above ͳ, which represents ହସ feet or ͳǤʹͷ feet.
tŚĂƚƐƚƌĂƚĞŐLJĐŽƵůĚǁĞƵƐĞƚŽĚĞƚĞƌŵŝŶĞƚŚĞůŽĐĂƚŝŽŶŽĨƚŚĞǁĂƚĞƌůĞǀĞůŽŶƚŚĞŶƵŵďĞƌůŝŶĞĂĨƚĞƌŝƚƌĂŝŶĞĚ͍
à I started at Ͳ and counted by ଵସ for each move. I counted ଵସ five times to get ହସ, which is equivalent to ͳ ͳͶ
and ͳǤʹͷ . I know the number is positive because I moved up. Since the measurements are in feet, the answer is ͳǤʹͷ feet.
&ŽƌƚŚĞĨƌĂĐƚŝŽŶ ହସ͕ǁŚĂƚŝƐƚŚĞǀĂůƵĞŽĨƚŚĞ ŶƵŵĞƌĂƚŽƌĂŶĚĚĞŶŽŵŝŶĂƚŽƌ͍
à The numerator is ͷǡ and the denominator is Ͷ.
tŚĂƚĚŽƚŚĞŶĞŐĂƚŝǀĞŶƵŵďĞƌƐƌĞƉƌĞƐĞŶƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍
à They represent the number of feet below the original lake level.
Exercise 2 (10 minutes)
^ƚƵĚĞŶƚƐĂƌĞƐĞĂƚĞĚŝŶŐƌŽƵƉƐŽĨƚŚƌĞĞŽƌĨŽƵƌ͘ŝƐƚƌŝďƵƚĞŽŶĞƐŚĞĞƚŽĨŐƌŝĚƉĂƉĞƌĂŶĚ ĂƌƵůĞƌƚŽĞĂĐŚƐƚƵĚĞŶƚ͘ ĂĐŚŐƌŽƵƉĐŽŵƉůĞƚĞƐƚŚĞĨŽůůŽǁŝŶŐƚĂƐŬƐ͗
ϭ͘ tƌŝƚĞĂƌĞĂůͲǁŽƌůĚƐƚŽƌLJƉƌŽďůĞŵƵƐŝŶŐĂƌĂƚŝŽŶĂůŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞ͘ Ϯ͘ ƌĞĂƚĞĂŚŽƌŝnjŽŶƚĂůŽƌǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵƚŽƌĞƉƌĞƐĞŶƚLJŽƵƌƐŝƚƵĂƚŝŽŶ͘ Ă͘ ĞƚĞƌŵŝŶĞĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞ͕ĂŶĚůĂďĞůƚŚĞŶƵŵďĞƌůŝŶĞ͘ ď͘ tƌŝƚĞƚŚĞƵŶŝƚƐŽĨŵĞĂƐƵƌĞŵĞŶƚ;ŝĨŶĞĞĚĞĚͿ͘ Đ͘ 'ƌĂƉŚƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞƚŚĂƚƌĞƉƌĞƐĞŶƚƚŚĞƐŝƚƵĂƚŝŽŶ͘ ϯ͘ ĞƐĐƌŝďĞǁŚĂƚƉŽŝŶƚƐ ͲĂŶĚƚŚĞŽƉƉŽƐŝƚĞŶƵŵďĞƌƌĞƉƌĞƐĞŶƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ ϰ͘ EĂŵĞĂƌĂƚŝŽŶĂůŶƵŵďĞƌƚŽƚŚĞůĞĨƚĂŶĚƌŝŐŚƚŽĨƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌLJŽƵŝŶŝƚŝĂůůLJ ĐŚŽƐĞ͘
Scaffolding:
WƌŽũĞĐƚƚŚĞĚŝƌĞĐƚŝŽŶƐĨŽƌƚŚĞ ĞdžĂŵƉůĞĂƐĂǁĂLJĨŽƌŐƌŽƵƉƐƚŽ ŵĂŬĞƐƵƌĞƚŚĞLJĂƌĞĐŽŵƉůĞƚŝŶŐ ĂůůƚĂƐŬƌĞƋƵŝƌĞŵĞŶƚƐ͘
,ĂǀĞƐƚƵĚĞŶƚƐǁƌŝƚĞƚŚĞŝƌƐƚŽƌLJ ƉƌŽďůĞŵƐĂŶĚĚƌĂǁƚŚĞŝƌ ŶƵŵďĞƌůŝŶĞƐŽŶůĂƌŐĞǁĂůůŐƌŝĚ ƉĂƉĞƌ͘
,ĂŶŐƉŽƐƚĞƌƐĂƌŽƵŶĚƚŚĞƌŽŽŵ ƚŽƵƐĞĂƐĂŐĂůůĞƌLJǁĂůŬĨŽƌ ƐƚƵĚĞŶƚƐǁŚŽĨŝŶŝƐŚƚŚĞŝƌ džŝƚ dŝĐŬĞƚƐĞĂƌůLJ͕ŽƌƵƐĞƚŚĞŵĂƐ ƌĞǀŝĞǁĨŽƌƚŚĞDŝĚͲDŽĚƵůĞ ƐƐĞƐƐŵĞŶƚůĂƚĞƌŝŶƚŚĞŵŽĚƵůĞ͘
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6 Exercise 2 Our Story Problem
Answers will vary. Melissa and Samantha weigh the same amount. Melissa gained Ǥ pounds last month, while Samantha lost the same amount Melissa gained.
Our Scale:
Our Units: Pounds
Description: On the number line, zero represents Melissa and Samantha’s original weight. The point െǤ
represents the change in Samantha’s weight. The amount lost is Ǥ pounds.
Other Information: A rational number to the left of Ǥ is Ǥ . A rational number to the right of Ǥ is Ǥ ૠ .
Closing (2 minutes)
,ŽǁŝƐŐƌĂƉŚŝŶŐƚŚĞŶƵŵďĞƌ Ͷ͵ ŽŶĂŶƵŵďĞƌůŝŶĞƐŝŵŝůĂƌƚŽŐƌĂƉŚŝŶŐƚŚĞŶƵŵďĞƌ ͶŽŶĂŶƵŵďĞƌůŝŶĞ͍
à When graphing each number, you start at zero and move to the right (or up) Ͷ units.
,ŽǁŝƐŐƌĂƉŚŝŶŐƚŚĞŶƵŵďĞƌ Ͷ͵ŽŶĂŶƵŵďĞƌůŝŶĞĚŝĨĨĞƌĞŶƚĨƌŽŵŐƌĂƉŚŝŶŐƚŚĞŶƵŵďĞƌ ͶŽŶĂŶƵŵďĞƌůŝŶĞ͍
à When we graph Ͷ, the unit length is one, and when we graph Ͷ͵, the unit length is ͳ͵.
KŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͕ĚĞƐĐƌŝďĞƚŚĞůŽĐĂƚŝŽŶŽĨƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌƚŚĂƚƌĞƉƌĞƐĞŶƚƐ െͷǤͳ ĂŶĚŝƚƐŽƉƉŽƐŝƚĞ͘
à The number െͷǤͳ would be ͷǤͳ units below zero because it is negative. Its opposite, ͷǤͳ , would be ͷǤͳ
units above zero because it is positive.
Exit Ticket (6 minutes)
െ െ െ െ െ െ (pounds)
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6
EĂŵĞ ĂƚĞ
Lesson 6: Rational Numbers on the Number Line Exit Ticket
hƐĞƚŚĞŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵďĞůŽǁƚŽĂŶƐǁĞƌƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐ͘ ϭ͘ tŚĂƚŝƐƚŚĞůĞŶŐƚŚŽĨĞĂĐŚƐĞŐŵĞŶƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍ Ϯ͘ tŚĂƚŶƵŵďĞƌĚŽĞƐƉŽŝŶƚ ܭƌĞƉƌĞƐĞŶƚ͍ ϯ͘ tŚĂƚŝƐƚŚĞŽƉƉŽƐŝƚĞŽĨƉŽŝŶƚ ܭ͍ ϰ͘ >ŽĐĂƚĞƚŚĞŽƉƉŽƐŝƚĞŽĨƉŽŝŶƚ ܭŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͕ĂŶĚůĂďĞůŝƚƉŽŝŶƚ ܮ͘ ϱ͘ /ŶƚŚĞĚŝĂŐƌĂŵĂďŽǀĞ͕njĞƌŽƌĞƉƌĞƐĞŶƚƐƚŚĞůŽĐĂƚŝŽŶŽĨDĂƌƚŝŶ>ƵƚŚĞƌ<ŝŶŐDŝĚĚůĞ^ĐŚŽŽů͘WŽŝŶƚ ܭƌĞƉƌĞƐĞŶƚƐƚŚĞ ůŝďƌĂƌLJ͕ǁŚŝĐŚŝƐůŽĐĂƚĞĚƚŽƚŚĞĞĂƐƚŽĨƚŚĞŵŝĚĚůĞƐĐŚŽŽů͘/ŶǁŽƌĚƐ͕ĐƌĞĂƚĞĂƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƚŚĂƚĐŽƵůĚ ƌĞƉƌĞƐĞŶƚƉŽŝŶƚ ܮ͕ ĂŶĚĚĞƐĐƌŝďĞŝƚƐůŽĐĂƚŝŽŶŝŶƌĞůĂƚŝŽŶƚŽ ͲĂŶĚƉŽŝŶƚ ܭ͘
െͳ Ͳ ͳ
ܭ
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6 Exit Ticket Sample Solutions Use the number line diagram below to answer the following questions. 1. What is the length of each segment on the number line?
What number does point ࡷ represent?
ૡ , or
ϯ͘ What is the opposite of point ࡷ?
െ ૡ , or െ
Locate the opposite of point ࡷ on the number line, and label it point ࡸ.5. In the diagram above, zero represents the location of Martin Luther King Middle School. Point ࡷ represents the library, which is located to the east of the middle school. In words, create a real-world situation that could represent point ࡸ, and describe its location in relation to and point ࡷ.
Answers may vary. Point ࡸ is ૡ units to the left of , so it is a negative number. Point ࡸ represents the recreation center, which is located ૡ mile west of Martin Luther King Middle School. This means that the recreation center and library are the same distance from the middle school but in opposite directions because the opposite of ૡ is
െ ૡ .
Problem Set Sample Solutions
^ƚƵĚĞŶƚƐŐĂŝŶĂĚĚŝƚŝŽŶĂůƉƌĂĐƚŝĐĞǁŝƚŚŐƌĂƉŚŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘
In the space provided, write the opposite of each number. a. ૠ
െ ૠ
b. െ
c. Ǥ ૡ െǤ ૡ
d. െ
െ
ࡷࡸ
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6 2. Choose a non-integer between and . Label it point and its opposite point on the number line. Write values below the points.
(Answers may vary.)
a. To draw a scale that would include both points, what could be the length of each segment?
Answers may vary.
b. In words, create a real-world situation that could represent the number line diagram.
Answers may vary. Starting at home, I ran mile. My brother ran mile from home in the opposite direction.
ϯ͘ Choose a value for point ࡼ that is between െ and െૠ .
Answers may vary. െ , െǤ , െǤ ૡ
a. What is the opposite of point ࡼ?
Answers may vary. , Ǥ , Ǥ ૡ
b. Use the value from part (a), and describe its location on the number line in relation to zero.
is the same distance as െ from zero but to the right. is units to the right of (or above) zero.
c. Find the opposite of the opposite of point ࡼ. Show your work, and explain your reasoning.
The opposite of an opposite of the number is the original number. If point ࡼ is located at െ , then the opposite of the opposite of point ࡼ is located at െ . The opposite of െ is . The opposite of is െ .
െ ቀെ ቁ ൌ and െ ቆെ ቀെ ቁቇ ൌ െ
െ
െ
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Lesson 6: ZĂƚŝŽŶĂůEƵŵďĞƌƐŽŶƚŚĞEƵŵďĞƌ>ŝŶĞ
6ͻϯ Lesson 6 4. Locate and label each point on the number line. Use the diagram to answer the questions.
Jill lives one block north of the pizza shop. Janette’s house is block past Jill’s house. Jeffrey and Olivia are in the park blocks south of the pizza shop. Jenny’s Jazzy Jewelry Shop is located halfway between the pizza shop and the park.
a. Describe an appropriate scale to show all the points in this situation.
An appropriate scale would be because the numbers given in the example all have denominators of . I would divide the number line into equal segments of
.
b. What number represents the location of Jenny’s Jazzy Jewelry Shop? Explain your reasoning.
The number is െ . I got my answer by finding the park first. It is units below
. Since the jewelry shop is halfway between the pizza shop and the park, half of is . Then, I moved units down on the number line since the shop is south of the pizza shop before the park.
Janette’s house
Jill’s house
Pizza shop
Jewelry shop
Park
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Topic B: KƌĚĞƌĂŶĚďƐŽůƵƚĞsĂůƵĞ
GRADE 6 ͻDKh> 36G R A D E DĂƚŚĞŵĂƚŝĐƐƵƌƌŝĐƵůƵŵ
dŽƉŝĐ
KƌĚĞƌĂŶĚďƐŽůƵƚĞsĂůƵĞ
&ŽĐƵƐ^ƚĂŶĚĂƌĚ s: hŶĚĞƌƐƚĂŶĚĂƌĂƚŝŽŶĂůŶƵŵďĞƌĂƐĂƉŽŝŶƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ džƚĞŶĚŶƵŵďĞƌ ůŝŶĞĚŝĂŐƌĂŵƐĂŶĚĐŽŽƌĚŝŶĂƚĞĂdžĞƐĨĂŵŝůŝĂƌĨƌŽŵƉƌĞǀŝŽƵƐŐƌĂĚĞƐƚŽƌĞƉƌĞƐĞŶƚ ƉŽŝŶƚƐŽŶƚŚĞůŝŶĞĂŶĚŝŶƚŚĞƉůĂŶĞǁŝƚŚŶĞŐĂƚŝǀĞŶƵŵďĞƌĐŽŽƌĚŝŶĂƚĞƐ͘
&ŝŶĚĂŶĚƉŽƐŝƚŝŽŶŝŶƚĞŐĞƌƐĂŶĚŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶĂŚŽƌŝnjŽŶƚĂůŽƌ ǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵ͖ĨŝŶĚĂŶĚƉŽƐŝƚŝŽŶƉĂŝƌƐŽĨŝŶƚĞŐĞƌƐĂŶĚ ŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶĂĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘
hŶĚĞƌƐƚĂŶĚŽƌĚĞƌŝŶŐĂŶĚĂďƐŽůƵƚĞǀĂůƵĞŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
/ŶƚĞƌƉƌĞƚƐƚĂƚĞŵĞŶƚƐŽĨŝŶĞƋƵĂůŝƚLJĂƐƐƚĂƚĞŵĞŶƚƐĂďŽƵƚƚŚĞƌĞůĂƚŝǀĞ ƉŽƐŝƚŝŽŶŽĨƚǁŽŶƵŵďĞƌƐŽŶĂŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵ͘ For example, interpret െ͵ െ as a statement that െ͵ is located to the right of െ
on a number line oriented from left to right.
tƌŝƚĞ͕ŝŶƚĞƌƉƌĞƚ͕ĂŶĚĞdžƉůĂŝŶƐƚĂƚĞŵĞŶƚƐŽĨŽƌĚĞƌĨŽƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶ ƌĞĂůͲǁŽƌůĚĐŽŶƚĞdžƚƐ͘ For example, write െ͵ι െι to express the fact that െ͵ι is warmer than െιǤ
hŶĚĞƌƐƚĂŶĚƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨĂƌĂƚŝŽŶĂůŶƵŵďĞƌĂƐŝƚƐĚŝƐƚĂŶĐĞĨƌŽŵ
ͲŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͖ŝŶƚĞƌƉƌĞƚĂďƐŽůƵƚĞǀĂůƵĞĂƐŵĂŐŶŝƚƵĚĞĨŽƌĂ ƉŽƐŝƚŝǀĞŽƌŶĞŐĂƚŝǀĞƋƵĂŶƚŝƚLJŝŶĂƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶ͘ For example, for an account balance of െ͵Ͳ dollars, write ȁെ͵Ͳȁ ൌ ͵Ͳ to describe the size of the debt in dollars.
ŝƐƚŝŶŐƵŝƐŚĐŽŵƉĂƌŝƐŽŶƐŽĨĂďƐŽůƵƚĞǀĂůƵĞĨƌŽŵƐƚĂƚĞŵĞŶƚƐĂďŽƵƚŽƌĚĞƌ͘
For example, recognize that an account balance less than Ȃ ͵Ͳ dollars represents a debt greater than ͵Ͳ dollars.
/ŶƐƚƌƵĐƚŝŽŶĂůĂLJƐ͗ ϳ
ĞƐƐŽŶƐ 7–8: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ;W͕WͿ ϭ
ĞƐƐŽŶϵ͗ ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ;WͿ
ĞƐƐŽŶϭϬ͗ tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂůEƵŵďĞƌƐ;WͿ
ĞƐƐŽŶϭϭ͗ ďƐŽůƵƚĞsĂůƵĞͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ;WͿ ϭ>ĞƐƐŽŶ^ƚƌƵĐƚƵƌĞ<ĞLJ͗ PͲWƌŽďůĞŵ^Ğƚ>ĞƐƐŽŶ͕ DͲDŽĚĞůŝŶŐLJĐůĞ>ĞƐƐŽŶ͕ EͲ džƉůŽƌĂƚŝŽŶ>ĞƐƐŽŶ͕ SͲ^ŽĐƌĂƚŝĐ>ĞƐƐŽŶ
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Topic B: KƌĚĞƌĂŶĚďƐŽůƵƚĞsĂůƵĞ
6.3 Topic B
ĞƐƐŽŶϭϮ͗ dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ;WͿ
ĞƐƐŽŶϭϯ͗ ^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ;WͿ
/ŶdŽƉŝĐ͕ƐƚƵĚĞŶƚƐĨŽĐƵƐŽŶƚŚĞŽƌĚĞƌŝŶŐŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘dŚĞLJƵŶĚĞƌƐƚĂŶĚĂďƐŽůƵƚĞǀĂůƵĞĂƐĂŶƵŵďĞƌ͛Ɛ
ĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽŽŶƚŚĞŶƵŵďĞƌůŝŶĞĂŶĚĐŽŶƚŝŶƵĞƚŽĨŝŶĚĂŶĚƉŽƐŝƚŝŽŶƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶŚŽƌŝnjŽŶƚĂůĂŶĚ
ǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞƐ͘/Ŷ>ĞƐƐŽŶƐϳʹϭϬ͕ƐƚƵĚĞŶƚƐĐŽŵƉĂƌĞĂŶĚŽƌĚĞƌŝŶƚĞŐĞƌƐĂŶĚŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
dŚĞLJǁƌŝƚĞ͕ŝŶƚĞƌƉƌĞƚ͕ĂŶĚĞdžƉůĂŝŶƐƚĂƚĞŵĞŶƚƐŽĨŽƌĚĞƌĨŽƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶƌĞĂůͲǁŽƌůĚĐŽŶƚĞdžƚƐ͘&ŽƌŝŶƐƚĂŶĐĞ͕ ŝĨ ƚŚĞƚĞŵƉĞƌĂƚƵƌĞƌĞĂĐŚĞĚĂůŽǁŽĨ െͺԬ ŽŶ^ĂƚƵƌĚĂLJĞǀĞŶŝŶŐĂŶĚŝƚŝƐĞdžƉĞĐƚĞĚƚŽďĞĐŽůĚĞƌůĂƚĞƌ^ĂƚƵƌĚĂLJ
ŶŝŐŚƚ͕ƐƚƵĚĞŶƚƐĂƌĞĂďůĞƚŽĂƌƌŝǀĞĂƚĂƉŽƐƐŝďůĞǀĂůƵĞĨŽƌƚŚĞƚĞŵƉĞƌĂƚƵƌĞůĂƚĞƌƚŚĂƚŶŝŐŚƚ͕ƌĞĂůŝnjŝŶŐƚŚĂƚŝƚǁŽƵůĚ
ŚĂǀĞƚŽůŝĞďĞůŽǁŽƌƚŽƚŚĞůĞĨƚŽĨ െͺ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘^ƚƵĚĞŶƚƐŝŶƚĞƌƉƌĞƚŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƐĂďŽƵƚƚŚĞ
ƉŽƐŝƚŝŽŶŝŶŐŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐǁŝƚŚƌĞƐƉĞĐƚ ƚŽŽŶĞĂŶŽƚŚĞƌĂŶĚƌĞĐŽŐŶŝnjĞƚŚĂƚŝĨ ܽ ൏ܾ ͕ ƚŚĞŶ ܽെ ܾെ
ďĞĐĂƵƐĞĂŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞĂƌĞ ĞƋƵĂůĚŝƐƚĂŶĐĞƐĨƌŽŵnjĞƌŽ͘ /Ŷ>ĞƐƐŽŶƐϭϭʹϭϯ͕ƐƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚĂďƐŽůƵƚĞǀĂůƵĞƚŽďĞĂŶƵŵďĞƌ͛ƐĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽ͘dŚĞLJŬŶŽǁ
ŽƉƉŽƐŝƚĞŶƵŵďĞƌƐĂƌĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽĂŶĚ͕ƚŚĞƌĞĨŽƌĞ͕ŚĂǀĞƚŚĞƐĂŵĞĂďƐŽůƵƚĞǀĂůƵĞ͘^ƚƵĚĞŶƚƐ
ŝŶƚĞƌƉƌĞƚĂďƐŽůƵƚĞǀĂůƵĞĂƐŵĂŐŶŝƚƵĚĞĂŶĚĞdžƉƌĞƐƐƚŚĞŝƌĂŶƐǁĞƌƐƚŽƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƐďĂƐĞĚŽŶƚŚĞĐŽŶƚĞdžƚ͘
&ŽƌĞdžĂŵƉůĞ͕ŝĨĂƐĂǀŝŶŐƐĂĐĐŽƵŶƚƐƚĂƚĞŵĞŶƚƐŚŽǁƐĂƚƌĂŶƐĂĐƚŝŽŶĂŵŽƵŶƚŽĨ െͶͲͲǡ ƐƚƵĚĞŶƚƐŬŶŽǁƚŚĂƚƚŚĞƌĞ
ŵƵƐƚŚĂǀĞďĞĞŶĂ ͶͲͲ ĚŽůůĂƌǁŝƚŚĚƌĂǁĂůĂŶĚƚŚĂƚĂǁŝƚŚĚƌĂǁĂůŽĨŵŽƌĞƚŚĂŶ ͶͲͲ ĚŽůůĂƌƐŝƐƌĞƉƌĞƐĞŶƚĞĚďLJĂ
ŶƵŵďĞƌůĞƐƐƚŚĂŶ െͶͲͲ ͘KƌĚĞƌŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐĂŶĚƚŚĞƵƐĞŽĨĂďƐŽůƵƚĞǀĂůƵĞĂŶĚƚŚĞĐŽŶƚĞdžƚŽĨĂƐŝƚƵĂƚŝŽŶ ĐƵůŵŝŶĂƚĞƐŝŶ>ĞƐƐŽŶϭϯ͘^ƚƵĚĞŶƚƐĞdžĂŵŝŶĞƌĞĂůͲǁŽƌůĚƐĐĞŶĂƌŝŽƐĂŶĚĚĞƐĐƌŝďĞƚŚĞƌĞůĂƚŝŽŶƐŚŝƉƚŚĂƚĞdžŝƐƚƐĂŵŽŶŐ ƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶǀŽůǀĞĚ͘dŚĞLJĐŽŵƉĂƌĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐĂŶĚǁƌŝƚĞƐƚĂƚĞŵĞŶƚƐŽĨŝŶĞƋƵĂůŝƚLJďĂƐĞĚŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞŵŽĚĞůƵƐŝŶŐĂďƐŽůƵƚĞǀĂůƵĞƚŽĚĞƚĞƌŵŝŶĞ͕ĨŽƌŝŶƐƚĂŶĐĞ͕ƚŚĂƚ ĚĞƉƚŚŽĨ ͶͲ ĨĞĞƚďĞůŽǁƐĞĂůĞǀĞůŝƐ ͳͷ
ĨĞĞƚĨĂƌƚŚĞƌĨƌŽŵƐĞĂůĞǀĞůƚŚĂŶĂŚĞŝŐŚƚŽĨ ʹͷ ĨĞĞƚ ĂďŽǀĞƐĞĂůĞǀĞů͘
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6ͻϯ Lesson 7 Lesson 7: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Lesson 7: Ordering Integers and Other Rational Numbers Student Outcomes
^ƚƵĚĞŶƚƐǁƌŝƚĞ͕ŝŶƚĞƌƉƌĞƚ͕ĂŶĚĞdžƉůĂŝŶƐƚĂƚĞŵĞŶƚƐŽĨŽƌĚĞƌĨŽƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶƚŚĞƌĞĂůǁŽƌůĚ͘
^ƚƵĚĞŶƚƐƌĞĐŽŐŶŝnjĞƚŚĂƚŝĨ ܽ ܾ͕ ƚŚĞŶ ܽെ ܾെ ďĞĐĂƵƐĞĂŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞĂƌĞĞƋƵĂůĚŝƐƚĂŶĐĞƐĨƌŽŵ njĞƌŽ͘^ƚƵĚĞŶƚƐĂůƐŽƌĞĐŽŐŶŝnjĞƚŚĂƚŵŽǀŝŶŐĂůŽŶŐƚŚĞŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞƚŽƚŚĞƌŝŐŚƚŵĞĂŶƐƚŚĞŶƵŵďĞƌƐ ĂƌĞŝŶĐƌĞĂƐŝŶŐ͘
Classwork Opening Exercise (6 minutes): Guess My Integer and Guess My Rational Number
&ŽƌƚŚŝƐǀĞƌďĂůǁĂƌŵͲƵƉĞdžĞƌĐŝƐĞ͕ƐƚƵĚĞŶƚƐĂƌƌŝǀĞĂƚĂƌĂƚŝŽŶĂůŶƵŵďĞƌďĂƐĞĚŽŶƉĂƌĂŵĞƚĞƌƐƉƌŽǀŝĚĞĚďLJƚŚĞƚĞĂĐŚĞƌĂŶĚ ƚŚĞŝƌŽǁŶƋƵĞƐƚŝŽŶŝŶŐ͘^ƚƵĚĞŶƚƐĂƌĞĂůůŽǁĞĚƚŽĂƐŬ no more than three questions ƚŽŚĞůƉƚŚĞŵĨŝŐƵƌĞŽƵƚƚŚĞŶƵŵďĞƌ ƚŚĞƚĞĂĐŚĞƌŝƐƚŚŝŶŬŝŶŐŽĨ͘^ƚƵĚĞŶƚƐŵƵƐƚƌĂŝƐĞƚŚĞŝƌŚĂŶĚƐĂŶĚǁĂŝƚƚŽďĞĐĂůůĞĚƵƉŽŶƚŽĂƐŬĂƋƵĞƐƚŝŽŶ͘dŚŝƐŝƐĂǁŚŽůĞͲ ŐƌŽƵƉĂĐƚŝǀŝƚLJ͕ĂŶĚŽŶĐĞƐƚƵĚĞŶƚƐŬŶŽǁǁŚĂƚƚŚĞŶƵŵďĞƌŝƐ͕ƚŚĞLJƐŚŽƵůĚƌĂŝƐĞƚŚĞŝƌŚĂŶĚƐ͘ĨƚĞƌƚŚĞĂĐƚŝǀŝƚLJŝƐŵŽĚĞůĞĚ ƐĞǀĞƌĂůƚŝŵĞƐ͕ƐƚƵĚĞŶƚƐŝŶŐƌŽƵƉƐŽĨƚǁŽŽƌƚŚƌĞĞƐŚŽƵůĚƚƌLJƚŚĞĂĐƚŝǀŝƚLJĂŵŽŶŐƚŚĞŵƐĞůǀĞƐ͕ǁŝƚŚŽŶĞƉĞƌƐŽŶƚŚŝŶŬŝŶŐŽĨĂ ŶƵŵďĞƌĂŶĚƚŚĞŽƚŚĞƌƐĂƐŬŝŶŐƋƵĞƐƚŝŽŶƐĂŶĚŐƵĞƐƐŝŶŐƚŚĞŶƵŵďĞƌ͘ dŚĞŐĂŵĞƐŚŽƵůĚŝŶĐƌĞĂƐĞŝŶĚŝĨĨŝĐƵůƚLJƐŽƚŚĂƚƚŚĞĨŝƌƐƚƌŽƵŶĚŝŶǀŽůǀĞƐĂŶŝŶƚĞŐĞƌŶƵŵďĞƌ͕ƚŚĞƐĞĐŽŶĚƌŽƵŶĚŝŶǀŽůǀĞƐĂ ŶŽŶͲŝŶƚĞŐĞƌƉŽƐŝƚŝǀĞƌĂƚŝŽŶĂůŶƵŵďĞƌ͕ĂŶĚƚŚĞƚŚŝƌĚƌŽƵŶĚŝŶǀŽůǀĞƐĂŶŽŶͲŝŶƚĞŐĞƌŶĞŐĂƚŝǀĞƌĂƚŝŽŶĂůŶƵŵďĞƌ͘ ƐĂŵƉůĞĚŝĂůŽŐƵĞĨŽƌƚǁŽƌŽƵŶĚƐŽĨƚŚĞŐĂŵĞŝƐƐƚĂƚĞĚďĞůŽǁ͘ ZŽƵŶĚϭ͗'ƵĞƐƐDLJ/ŶƚĞŐĞƌ
/ĂŵƚŚŝŶŬŝŶŐŽĨĂŶŝŶƚĞŐĞƌďĞƚǁĞĞŶ െ7 ĂŶĚ 0͘ ;dŚĞƚĞĂĐŚĞƌŝƐƚŚŝŶŬŝŶŐŽĨƚŚĞŶƵŵďĞƌ െ1 ͘ Ϳ
à Student 1: Is the number greater than െ6 ?
zĞƐ
à Student 2: Is the number less than െʹ?
EŽ
à Student 3: Is the number െ2 ?
EŽ
zŽƵŚĂǀĞĂƐŬĞĚŵĞƚŚƌĞĞƋƵĞƐƚŝŽŶƐ͘ZĂŝƐĞLJŽƵƌŚĂŶĚŝĨLJŽƵŬŶŽǁƚŚĞŝŶƚĞŐĞƌƚŚĂƚ/ĂŵƚŚŝŶŬŝŶŐŽĨ͘
à െ1
zĞƐ͊
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6ͻϯ Lesson 7 Lesson 7: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
ZŽƵŶĚϮ͗'ƵĞƐƐDLJWŽƐŝƚŝǀĞEŽŶͲ/ŶƚĞŐĞƌZĂƚŝŽŶĂůEƵŵďĞƌ
/ĂŵƚŚŝŶŬŝŶŐŽĨĂƉŽƐŝƚŝǀĞƌĂƚŝŽŶĂůŶƵŵďĞƌďĞƚǁĞĞŶ 1 ĂŶĚ 2͘ ;dŚĞƚĞĂĐŚĞƌŝƐƚŚŝŶŬŝŶŐŽĨƚŚĞŶƵŵďĞƌ ହସ͘Ϳ
à Student 1: Is the number greater than 1 12?
EŽ
à Student 2: Is the number less than 1.4 ?
zĞƐ
à Student 3: Is the number less than 1.2 ?
EŽ
zŽƵŚĂǀĞĂƐŬĞĚŵĞƚŚƌĞĞƋƵĞƐƚŝŽŶƐ͘ZĂŝƐĞLJŽƵƌŚĂŶĚŝĨLJŽƵŬŶŽǁƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌƚŚĂƚ/ĂŵƚŚŝŶŬŝŶŐŽĨ͘
à 1.3
EŽ
à 1.35
EŽ
à 1.25
zĞƐ͊
Discussion ;ϯ minutes)
/ĨǁĞŚĂǀĞƚǁŽƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŚĂƚǁĞǁĂŶƚƚŽŐƌĂƉŚŽŶƚŚĞƐĂŵĞŶƵŵďĞƌůŝŶĞ͕ŚŽǁǁŽƵůĚǁĞĚĞƐĐƌŝďĞƚŽ ƐŽŵĞŽŶĞǁŚĞƌĞĞĂĐŚŶƵŵďĞƌůŝĞƐŝŶĐŽŵƉĂƌŝƐŽŶƚŽƚŚĞŽƚŚĞƌŶƵŵďĞƌ͍
à The lesser number is to the left of (or below) the greater number. (Or the greater number is to the right of [or above] the lesser number.) Negative numbers are to the left of (or below) zero, and positive numbers are to the right of (or above) zero.
ĞƐĐƌŝďĞƚŚĞůŽĐĂƚŝŽŶŽŶƚŚĞŶƵŵďĞƌůŝŶĞŽĨĂŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞ͘
à They are both the same distance from zero but on opposite sides of zero.
/ĨǁĞǁĞƌĞƚŽůŝƐƚƚŚĞŶƵŵďĞƌƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͕ǁŚŝĐŚǁŽƵůĚďĞĨĂƌƚŚĞƐƚƚŽƚŚĞůĞĨƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍ tŚLJ͍tŚŝĐŚǁŽƵůĚďĞĨĂƌƚŚĞƐƚƚŽƚŚĞƌŝŐŚƚ͍tŚLJ͍
à The numbers on the number line decrease as you move to the left (or down) and increase as you move to the right (or up). So, the number that is the least would be farthest left (or down), and the number that is the greatest would be farthest right (or up).
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6ͻϯ Lesson 7 Lesson 7: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Exercise 1 (5 minutes)
ůůŽǁƚŝŵĞĨŽƌĂǁŚŽůĞͲŐƌŽƵƉĚŝƐĐƵƐƐŝŽŶƚŽĨŽůůŽǁĐŽŵƉůĞƚŝŽŶŽĨƚŚĞĞdžĞƌĐŝƐĞ͘
Exercise 1 a. Graph the number ૠ and its opposite on the number line. Graph the number and its opposite on the number line. b. Where does ૠ lie in relation to on the number line?
On the number line, ૠ is units to the right of .
c. Where does the opposite of ૠ lie on the number line in relation to the opposite of ?
On the number line, െૠ is units to the left of െ .
d. I am thinking of two numbers. The first number lies to the right of the second number on a number line. What can you say about the location of their opposites? (If needed, refer to your number line diagram.)
On the number line, the opposite of the second number must lie to the right of the opposite of the first number. If we call the first number ࢌ and the second number ࢙ , then ࢌെ and ࢙െ will have the opposite order of ࢌ and ࢙ because ࢌെ and ࢙െ are opposites of ࢌ and ࢙ , so they lie on the opposite side of zero.
Example 1 (4 minutes)
Example 1 The record low temperatures for a town in Maine are െ°۴ for January and െૢ °۴ for February. Order the numbers from least to greatest. Explain how you arrived at the order.
x Read: January: െ and February: െૢ
x Draw: Draw a number line model.
x Write: Since െ is farthest below zero and െૢ is above െ on the vertical number line, െ is less than െૢ .
x Answer: െ , െૢ
െ െ െ െ െ െ െૠ െૢ ૡૠ െ ૢ െૡ
Scaffolding:
KŶĞƉŽƚĞŶƚŝĂůƐƚƌĂƚĞŐLJƚŽƵƐĞ ǁŚĞŶƚĂĐŬůŝŶŐǁŽƌĚƉƌŽďůĞŵƐ ŝƐƚŚĞZtĂƉƉƌŽĂĐŚ͕ǁŚŝĐŚ ƐƚĂŶĚƐĨŽƌ RĞĂĚ͕ DƌĂǁ͕ĂŶĚ
WƌŝƚĞ͘
െૢ
െ
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6ͻϯ Lesson 7 Lesson 7: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Exercises 2–4 (8 minutes)
ůůŽǁƚŝŵĞƚŽĚŝƐĐƵƐƐƚŚĞƐĞƉƌŽďůĞŵƐĂƐĂǁŚŽůĞĐůĂƐƐ͘
Exercises 2–4 For each problem, order the rational numbers from least to greatest by first reading the problem, then drawing a number line diagram, and finally, explaining your answer. 2. Jon’s time for running the mile in gym class is ૢ . minutes. Jacky’s time is ૢ . ૡ minutes. Who ran the mile in less time? ૢ . ૡ ,ૢ .
I drew a number line and graphed ૢ . and ૢ . ૡ ; ૢ . is to the right of ૢ . ૡ . So, ૢ . ૡ is less than ૢ . , which means Jacky ran the mile in less time than Jon.
ϯ͘ Mrs. Rodriguez is a teacher at Westbury Middle School. She gives bonus points on tests for outstanding written answers and deducts points for answers that are not written correctly. She uses rational numbers to represent the points. She wrote the following on students’ papers: Student A: െ points, Student B: െ. points. Did Student A or Student B perform worse on the test?
െ. , െ
I drew a number line, and െ and െ. are both to the left of zero, but െ. is to the left of െ . So, െ. is less than െ . That means Student B did worse than Student A.
A carp is swimming approximately ૡ feet beneath the water’s surface, and a sunfish is swimming approximately
feet beneath the water’s surface. Which fish is swimming farther beneath the water’s surface?
െૡ , െ
I drew a vertical number line, and െૡ is farther below zero than െ . So, െૡ is less than െ , which means the carp is swimming farther beneath the water’s surface.
Example 2 ;ϯ minutes)
Example 2 Henry, Janon, and Clark are playing a card game. The object of the game is to finish with the most points. The scores at the end of the game are Henry: െૠ , Janon: , and Clark: െ . Who won the game? Who came in last place? Use a number line model, and explain how you arrived at your answer.
x Read: Henry: െૠ , Janon: , and Clark: െ
x Draw:
x Explain: െૠ , െ ,
x Janon won the game, and Henry came in last place. I ordered the numbers on a number line, and െૠ is farthest to the left. That means െૠ is the smallest of the three numbers, so Henry came in last place. Next on the number line is െ , which is to the right of െૠ but to the left of . Farthest to the right is ; therefore, is the greatest of the three numbers. This means Janon won the game.
െ െ െ െ െ െ െૠ െૢ ૡૠ െ ૢ െૡ
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6ͻϯ Lesson 7 Lesson 7: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Exercises 5–6 (6 minutes) Exercises 5–6 For each problem, order the rational numbers from least to greatest by first reading the problem, then drawing a number line diagram, and finally, explaining your answer. 5. Henry, Janon, and Clark are playing another round of the card game. Their scores this time are as follows: Clark: െ , Janon: െ , and Henry: െ . Who won? Who came in last place?
െ , െ , െ
Clark won the game, and Henry came in last place. I ordered the numbers on a number line, and െ is farthest to the left. That means െ is the smallest of the three numbers, so Henry lost. Next on the number line is െ , which is to the right of െ and to the left of െ . Farthest to the right is െ , which is the greatest of the three negative numbers, so Clark won the game.
Represent each of the following elevations using a rational number. Then, order the numbers from least to greatest. Cayuga Lake meters above sea level Mount Marcy , ૢ meters above sea level New York Stock Exchange Vault . meters below sea level
െ. ; ; , ૢ
I drew a number line, and െ. is the only number to the left of zero, so it is the least (because as you move to the right, the numbers increase). Next on the number line is , which is to the right of zero. Last on the number line is , ૢ , which is to the right of , so , ૢ meters is the greatest elevation.
Closing (5 minutes): What Is the Value of Each Number, and Which Is Larger?
dŚŝƐůŽƐŝŶŐŝƐĂǀĞƌďĂůŐĂŵĞƐŝŵŝůĂƌƚŽƚŚĞKƉĞŶŝŶŐ džĞƌĐŝƐĞ͘,ŽǁĞǀĞƌ͕ŝŶƐƚĞĂĚŽĨĂƐŬŝŶŐ ƋƵĞƐƚŝŽŶƐ͕ƐƚƵĚĞŶƚƐƵƐĞƚŚĞƚǁŽĐůƵĞƐƚŚĞƚĞĂĐŚĞƌƉƌŽǀŝĚĞƐĂŶĚ͕ĂƐĂǀŝƐƵĂůŵŽĚĞů͕ƚŚĞ ŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵŝŶƚŚĞŝƌŵĂƚĞƌŝĂůƐ͘hƐŝŶŐƚŚĞĐůƵĞƐĂŶĚƚŚĞŶƵŵďĞƌůŝŶĞ͕ƐƚƵĚĞŶƚƐ ĚĞƚĞƌŵŝŶĞƚŚĞǀĂůƵĞŽĨĞĂĐŚŶƵŵďĞƌĂŶĚǁŚŝĐŚŶƵŵďĞƌŝƐŐƌĞĂƚĞƌ͘tŚĞŶĂƐƚƵĚĞŶƚƌĂŝƐĞƐ ŚŝƐŚĂŶĚĂŶĚŝƐĐĂůůĞĚƵƉŽŶ͕ŚĞƐƚĂƚĞƐƚŚĞǀĂůƵĞƐĂŶĚǁŚŝĐŚŝƐŐƌĞĂƚĞƌ͘ >ŝƐƚĞĚďĞůŽǁĂƌĞĨŝǀĞƌŽƵŶĚƐŽĨƉŽƚĞŶƚŝĂůƋƵĞƐƚŝŽŶƐ͘
Round 1: dŚĞĨŝƌƐƚŶƵŵďĞƌŝƐ 3ƵŶŝƚƐƚŽƚŚĞƌŝŐŚƚŽĨ െ1 ͘ dŚĞƐĞĐŽŶĚŶƵŵďĞƌŝƐ 5ƵŶŝƚƐƚŽƚŚĞůĞĨƚŽĨ 4͘ tŚĂƚŝƐ ƚŚĞǀĂůƵĞŽĨĞĂĐŚŶƵŵďĞƌ͕ĂŶĚǁŚŝĐŚŝƐŐƌĞĂƚĞƌ͍
à First Number: 2; Second Number: െ1 ; 2 is greater than െ1 .
Round 2: dŚĞĨŝƌƐƚŶƵŵďĞƌŝƐŶĞŝƚŚĞƌƉŽƐŝƚŝǀĞŶŽƌŶĞŐĂƚŝǀĞ͘dŚĞƐĞĐŽŶĚŶƵŵďĞƌ ŝƐ 1ƵŶŝƚƚŽƚŚĞůĞĨƚŽĨ െ1 ͘tŚĂƚŝƐƚŚĞǀĂůƵĞŽĨĞĂĐŚŶƵŵďĞƌ͕ĂŶĚǁŚŝĐŚŝƐ ŐƌĞĂƚĞƌ͍
à First Number: 0; Second Number: െ2 ; 0 is greater than െ2 .
ZŽƵŶĚϯ͗ dŚĞĨŝƌƐƚŶƵŵďĞƌŝƐ 8 12ƵŶŝƚƐƚŽƚŚĞƌŝŐŚƚŽĨ െ5 ͘ dŚĞƐĞĐŽŶĚŶƵŵďĞƌŝƐ 3ƵŶŝƚƐƚŽƚŚĞƌŝŐŚƚŽĨ 0͘ tŚĂƚ ŝƐƚŚĞǀĂůƵĞŽĨĞĂĐŚŶƵŵďĞƌ͕ĂŶĚǁŚŝĐŚŝƐŐƌĞĂƚĞƌ͍
à First Number: 3.5 ; Second Number: 3; 3.5 is greater than 3.
Scaffolding:
^ŽŵĞƐƚƵĚĞŶƚƐŶĞĞĚƚŽ ƌĞůLJŽŶƚŚĞŶƵŵďĞƌůŝŶĞĨŽƌ ƚŚŝƐĂĐƚŝǀŝƚLJ͘KƚŚĞƌƐĚŽ ŶŽƚ͕ĂƐƚŚĞLJŚĂǀĞŵŽǀĞĚ ĨƌŽŵƚŚĞƉŝĐƚŽƌŝĂůƚŽƚŚĞ ĂďƐƚƌĂĐƚƐƚĂŐĞ͘
ĚũƵƐƚƚŚĞǁĂŝƚƚŝŵĞĂŶĚ ĚŝĨĨŝĐƵůƚLJůĞǀĞůŽĨƚŚĞ ƋƵĞƐƚŝŽŶƐƚŽŵĞĞƚƚŚĞ ŶĞĞĚƐŽĨůĞĂƌŶĞƌƐ͘
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6ͻϯ Lesson 7 Lesson 7: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Round 4: dŚĞĨŝƌƐƚŶƵŵďĞƌŝƐ ଵସƵŶŝƚƚŽƚŚĞůĞĨƚŽĨ െ7 ͘dŚĞƐĞĐŽŶĚŶƵŵďĞƌŝƐ 8ƵŶŝƚƐƚŽƚŚĞůĞĨƚŽĨ 1͘ tŚĂƚŝƐ ƚŚĞǀĂůƵĞŽĨĞĂĐŚŶƵŵďĞƌ͕ĂŶĚǁŚŝĐŚŝƐŐƌĞĂƚĞƌ͍
à First Number: െ7.25 ; Second Number: െ7 ; െ7 is greater than െ7.25 .
Round 5: dŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞĨŝƌƐƚŶƵŵďĞƌŝƐ 2ƵŶŝƚƐƚŽƚŚĞƌŝŐŚƚŽĨ 3͘ dŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞƐĞĐŽŶĚŶƵŵďĞƌŝƐ 2ƵŶŝƚƐƚŽƚŚĞůĞĨƚŽĨ െ3 ͘ tŚĂƚŝƐƚŚĞǀĂůƵĞŽĨĞĂĐŚŶƵŵďĞƌ͕ĂŶĚǁŚŝĐŚŝƐŐƌĞĂƚĞƌ͍
à First Number: െ5 ; Second Number: 5; 5 is greater than െ5.
Closing: What Is the Value of Each Number, and Which Is Larger? Use your teacher’s verbal clues and this number line to determine which number is larger. Exit Ticket (5 minutes)
െ െ െ െ െ െ െૠ െૡ ૡૠ
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6ͻϯ Lesson 7 Lesson 7: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
EĂŵĞ ĂƚĞ
Lesson 7: Ordering Integers and Other Rational Numbers Exit Ticket
/ŶŵĂƚŚĐůĂƐƐ͕ŚƌŝƐƚŝŶĂĂŶĚƌĞƚƚĂƌĞĚĞďĂƚŝŶŐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚǁŽƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘ZĞĂĚƚŚĞŝƌĐůĂŝŵƐďĞůŽǁ͕ ĂŶĚƚŚĞŶǁƌŝƚĞĂŶĞdžƉůĂŶĂƚŝŽŶŽĨǁŚŽŝƐĐŽƌƌĞĐƚ͘hƐĞĂŶƵŵďĞƌůŝŶĞŵŽĚĞůƚŽƐƵƉƉŽƌƚLJŽƵƌĂŶƐǁĞƌ͘ ŚƌŝƐƚŝŶĂ͛ƐůĂŝŵ͗͞/ŬŶŽǁƚŚĂƚ 3ŝƐŐƌĞĂƚĞƌƚŚĂŶ 2 12͘ ^Ž͕ െ3 ŵƵƐƚďĞŐƌĞĂƚĞƌƚŚĂŶ െ2 12͘͟ ƌĞƚƚ͛ƐůĂŝŵ͗͞zĞƐ͕ 3ŝƐŐƌĞĂƚĞƌƚŚĂŶ 2 12͕ ďƵƚǁŚĞŶLJŽƵůŽŽŬĂƚƚŚĞŝƌŽƉƉŽƐŝƚĞƐ͕ƚŚĞŝƌŽƌĚĞƌǁŝůůďĞŽƉƉŽƐŝƚĞ͘^Ž͕ƚŚĂƚ ŵĞĂŶƐ െ2 12ŝƐŐƌĞĂƚĞƌƚŚĂŶ െ3 ͘͟
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6ͻϯ Lesson 7 Lesson 7: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Exit Ticket Sample Solutions In math class, Christina and Brett are debating the relationship between two rational numbers. Read their claims below, and then write an explanation of who is correct. Use a number line model to support your answer. Christina’s Claim: “I know that is greater than . So, െ must be greater than െ .” Brett’s Claim: “Yes, is greater than , but when you look at their opposites, their order will be opposite. So, that means െ is greater than െ .”
Brett is correct. I graphed the numbers on the number line, and െ is to the left of െ . The numbers increase as you move to the right, so െ is greater than െ .
Problem Set Sample Solutions 1. In the table below, list each set of rational numbers in order from least to greatest. Then, list their opposites. Finally, list the opposites in order from least to greatest. The first example has been completed for you. Rational Numbers Ordered from Least to Greatest Opposites Opposites Ordered from Least to Greatest
െૠ. , െૠ. െૠ. , െૠ. ૠ. , ૠ. ૠ. , ૠ.
, െ െ ,
, െ െ ,
, െ െ , , െ െ ,
, , , െ െ ,
െ , െ. െ. , െ . , , .
, , െ , െ െ , െ
െૢૢ .ૢ , െ െ , െૢૢ .ૢ ,ૢૢ .ૢ ૢૢ .ૢ ,
െ. , െ. െ. , െ. . , . . , .
െ. ૠ , െ. ૠ , . ૠ , , . ૠ
. , . . , . െ. , െ. െ. , െ.
For each row, what pattern do you notice between the numbers in the second and fourth columns? Why is this so?
For each row, the numbers in the second and fourth columns are opposites, and their order is opposite. This is because on the number line, as you move to the right, numbers increase. But as you move to the left, the numbers decrease. So, when comparing and , is to the right of ; therefore, is greater than . However, െ is to the left of െ ; therefore, െ is less than െ .
െ െ െ െ
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6ͻϯ Lesson 8 Lesson 8: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Lesson 8: Ordering Integers and Other Rational Numbers Student Outcomes
^ƚƵĚĞŶƚƐǁƌŝƚĞ͕ŝŶƚĞƌƉƌĞƚ͕ĂŶĚĞdžƉůĂŝŶƐƚĂƚĞŵĞŶƚƐŽĨŽƌĚĞƌĨŽƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶƚŚĞƌĞĂůǁŽƌůĚ͘
^ƚƵĚĞŶƚƐƌĞĐŽŐŶŝnjĞƚŚĂƚŝĨ ܽ ܾ൏ ͕ ƚŚĞŶ െ ܽ െܾ ďĞĐĂƵƐĞĂŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞĂƌĞĞƋƵĂůĚŝƐƚĂŶĐĞƐĨƌŽŵ njĞƌŽ͕ĂŶĚŵŽǀŝŶŐĂůŽŶŐƚŚĞŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞƚŽƚŚĞƌŝŐŚƚŵĞĂŶƐƚŚĞŶƵŵďĞƌƐĂƌĞŝŶĐƌĞĂƐŝŶŐ͘
Lesson Notes
ƐĂĐŽŶƚŝŶƵĂƚŝŽŶŽĨ>ĞƐƐŽŶϳ͕ƐƚƵĚĞŶƚƐŽƌĚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚĂŶĚĨƌŽŵŐƌĞĂƚĞƐƚƚŽůĞĂƐƚ͘dŚĞLJ ƌĞůĂƚĞƚŚĞŽƌĚĞƌŝŶŐƐƚŽŶƵŵďĞƌƐ͛ůŽĐĂƚŝŽŶƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘
Classwork Opening Exercise (6 minutes)
&ŽƌƚŚŝƐǁĂƌŵͲƵƉĞdžĞƌĐŝƐĞ͕ƐƚƵĚĞŶƚƐǁŽƌŬŝŶŐƌŽƵƉƐŽĨƚŚƌĞĞŽƌĨŽƵƌƚŽŽƌĚĞƌƚŚĞĨŽůůŽǁŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐĨƌŽŵůĞĂƐƚƚŽ ŐƌĞĂƚĞƐƚ͘ ĂĐŚŐƌŽƵƉŽĨƐƚƵĚĞŶƚƐŵĂLJďĞƉƌŽǀŝĚĞĚǁŝƚŚĐĂƌĚƐƚŽƉƵƚŝŶŽƌĚĞƌ͕ŽƌƚŚĞŶƵŵďĞƌƐŵĂLJďĞĚŝƐƉůĂLJĞĚŽŶƚŚĞ ďŽĂƌĚǁŚĞƌĞƐƚƵĚĞŶƚƐǁŽƌŬĂƚƚŚĞŝƌƐĞĂƚƐ͕ƌĞĐŽƌĚŝŶŐƚŚĞŵŝŶƚŚĞĐŽƌƌĞĐƚŽƌĚĞƌ͘ƐĂŶĂůƚĞƌŶĂƚŝǀĞ͕ƚŚĞŶƵŵďĞƌƐŵĂLJďĞ ĚŝƐƉůĂLJĞĚŽŶĂŶŝŶƚĞƌĂĐƚŝǀĞďŽĂƌĚĂůŽŶŐǁŝƚŚĂŶƵŵďĞƌůŝŶĞ͕ĂŶĚƐƚƵĚĞŶƚƐŽƌƚĞĂŵƐĐŽŵĞƵƉƚŽƚŚĞďŽĂƌĚĂŶĚƐůŝĚĞƚŚĞ ŶƵŵďĞƌƐŽŶƚŽƚŚĞŶƵŵďĞƌůŝŶĞŝŶƚŽƚŚĞĐŽƌƌĞĐƚŽƌĚĞƌ͘ůůŽǁƚŝŵĞĨŽƌƚŚĞĐůĂƐƐƚŽĐŽŵĞƚŽĂĐŽŶƐĞŶƐƵƐŽŶƚŚĞĐŽƌƌĞĐƚ ŽƌĚĞƌĂŶĚĨŽƌƐƚƵĚĞŶƚƐƚŽƐŚĂƌĞǁŝƚŚƚŚĞĐůĂƐƐƚŚĞŝƌƐƚƌĂƚĞŐŝĞƐĂŶĚƚŚŽƵŐŚƚƉƌŽĐĞƐƐĞƐ͘ dŚĞĨŽůůŽǁŝŶŐĂƌĞĞdžĂŵƉůĞƐŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŽƐŽƌƚĂŶĚŽƌĚĞƌ͗
Ͳ͕ െͶ ͕ ଵସ͕ െ ͳʹ͕ ͳ ͕ െ͵ ͵ͷ͕ ʹ ͕ െͶǤͳ ͕ െͲǤ ͕ ʹ͵ ͷ͕ ͕ െͳ ͕ ͶǤͷ ͕ െͷ ͕ ʹǤͳ
Solution:
െͷ , െͶǤͳ , െͶ , െ͵ ͵ͷ, െͳ ,െͲǤ , െ ͳʹ,Ͳ , ଵସ,ͳ ,ʹ ,ʹǤͳ ,ͶǤͷ , ʹ͵ ͷ ,
dŚĞĨŽůůŽǁŝŶŐůŝŶĞŽĨƋƵĞƐƚŝŽŶŝŶŐĐĂŶďĞƵƐĞĚƚŽĞůŝĐŝƚƐƚƵĚĞŶƚƌĞƐƉŽŶƐĞƐ͗
,ŽǁĚŝĚLJŽƵďĞŐŝŶƚŽƐŽƌƚĂŶĚŽƌĚĞƌƚŚĞŶƵŵďĞƌƐ͍tŚĂƚǁĂƐLJŽƵƌĨŝƌƐƚƐƚĞƉ͍
à Our group began by separating the numbers into two groups: negative numbers and positive numbers. Zero was not in either group, but we knew it fell in between the negative numbers and positive numbers.
tŚĂƚǁĂƐLJŽƵƌŶĞdžƚƐƚĞƉ͍tŚĂƚĚŝĚLJŽƵĚŽǁŝƚŚƚŚĞƚǁŽŐƌŽƵƉƐŽĨŶƵŵďĞƌƐ͍
à We ordered the positive whole numbers and then took the remaining positive numbers and determined which two whole numbers they fell in between.
Scaffolding:
ĚũƵƐƚƚŚĞŶƵŵďĞƌŽĨĐĂƌĚƐ ŐŝǀĞŶƚŽƐƚƵĚĞŶƚƐĚĞƉĞŶĚŝŶŐŽŶ ƚŚĞŝƌĂďŝůŝƚLJůĞǀĞů͘dŚĞƚLJƉĞƐŽĨ ƌĂƚŝŽŶĂůŶƵŵďĞƌƐŐŝǀĞŶƚŽĞĂĐŚ ŐƌŽƵƉŽĨƐƚƵĚĞŶƚƐŵĂLJĂůƐŽďĞ ĚŝĨĨĞƌĞŶƚŝĂƚĞĚ͘
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6ͻϯ Lesson 8 Lesson 8: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
,ŽǁĚŝĚLJŽƵŬŶŽǁǁŚĞƌĞƚŽƉůĂĐĞ ଵସĂŶĚ ଶଷ ହ͍
à Since ଵସ is less than a whole ( ͳ) but greater than zero, we knew the rational number was located between Ͳ and ͳ.
à We know that ଶଷ ହ is the same as Ͷ ͵ͷ, which is more than Ͷ but less than ͷ, so we knew the rational number was located between Ͷ and ͷ.
,ŽǁĚŝĚLJŽƵŽƌĚĞƌƚŚĞŶĞŐĂƚŝǀĞŶƵŵďĞƌƐ͍
à First, we started with the negative integers: െͷ ,െͶ , and െͳ . െͷ is the least because it is farthest left at ͷ units to the left of zero. Then came െͶ , and then came െͳ , which is only ͳ unit to the left of zero.
,ŽǁĚŝĚLJŽƵŽƌĚĞƌƚŚĞŶĞŐĂƚŝǀĞŶŽŶͲŝŶƚĞŐĞƌƐ͍
à We know െ ͳʹ is equivalent to െ ͷͳͲ , which is to the right of െͲǤ ;or െ ͳͲ Ϳ since െͲǤͷ is closer to zero than െͲǤ . Then, we ordered െͶǤͳ and െ͵ ͵ͷ. Both numbers are close to െͶ , but െͶǤͳ is to the left of
െͶ , and െ͵ ͵ͷ is to the right of െͶ and to the left of െ͵ ͘ Lastly, we put our ordered group of negative numbers to the left of zero and our ordered group of positive numbers to the right of zero and ended up with
െͷǡ െͶǤͳ ,െͶ ,െ͵ ͵ͷ,െͳ ,െͲǤ ,െ ͳʹ,Ͳ , ଵସ,ͳ ,ʹ ,ʹǤͳ ,ͶǤͷ ,ʹ͵ ͷ , .
Exercise 1 (8 minutes)
ϭ͘ ^ƚƵĚĞŶƚƐĂƌĞĞĂĐŚŐŝǀĞŶĨŽƵƌŝŶĚĞdžĐĂƌĚƐŽƌƐŵĂůůƐůŝƉƐŽĨƉĂƉĞƌ͘ ĂĐŚƐƚƵĚĞŶƚŵƵƐƚŝŶĚĞƉĞŶĚĞŶƚůLJĐŚŽŽƐĞĨŽƵƌŶŽŶͲ ŝŶƚĞŐĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐĂŶĚǁƌŝƚĞĞĂĐŚŽŶĞŽŶĂƐůŝƉŽĨƉĂƉĞƌ͘ƚůĞĂƐƚƚǁŽŽĨƚŚĞŶƵŵďĞƌƐŵƵƐƚďĞŶĞŐĂƚŝǀĞ͘ Ϯ͘ ^ƚƵĚĞŶƚƐŽƌĚĞƌƚŚĞŝƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚďLJƐůŝĚŝŶŐƚŚĞŝƌƐůŝƉƐŽĨƉĂƉĞƌŝŶƚŽƚŚĞĐŽƌƌĞĐƚŽƌĚĞƌ͘ dŚĞƚĞĂĐŚĞƌǁĂůŬƐĂƌŽƵŶĚƚŚĞƌŽŽŵƚŽĐŚĞĐŬĨŽƌƵŶĚĞƌƐƚĂŶĚŝŶŐĂŶĚƚŽƉƌŽǀŝĚĞŝŶĚŝǀŝĚƵĂůĂƐƐŝƐƚĂŶĐĞ͘^ƚƵĚĞŶƚƐŵĂLJ ƵƐĞƚŚĞŶƵŵďĞƌůŝŶĞŝŶƚŚĞŝƌƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐƚŽŚĞůƉĚĞƚĞƌŵŝŶĞƚŚĞŽƌĚĞƌ͘ ϯ͘ KŶĐĞĂůůƐƚƵĚĞŶƚƐŚĂǀĞĂƌƌĂŶŐĞĚƚŚĞŝƌŶƵŵďĞƌƐŝŶƚŽƚŚĞĐŽƌƌĞĐƚŽƌĚĞƌ͕ƚŚĞLJƐŚƵĨĨůĞƚŚĞŵĂŶĚƚŚĞŶƐǁŝƚĐŚǁŝƚŚ ĂŶŽƚŚĞƌƐƚƵĚĞŶƚ͘ ϰ͘ ^ƚƵĚĞŶƚƐĂƌƌĂŶŐĞƚŚĞŶĞǁƐĞƚŽĨĐĂƌĚƐƚŚĞLJƌĞĐĞŝǀĞŝŶƚŽƚŚĞĐŽƌƌĞĐƚŽƌĚĞƌĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͘ ϱ͘ dŚĞƉĂŝƌƐŽĨƐƚƵĚĞŶƚƐǁŚŽĞdžĐŚĂŶŐĞĚĐĂƌĚƐĚŝƐĐƵƐƐƚŚĞŝƌƐŽůƵƚŝŽŶƐĂŶĚĐŽŵĞƚŽĂĐŽŶƐĞŶƐƵƐ͘
Example 1 ;ϯ minutes): Ordering Rational Numbers from Least to Greatest
Example 1: Ordering Rational Numbers from Least to Greatest Sam has ̈́ Ǥ in the bank. He owes his friend Hank ̈́ Ǥ .He owes his sister ̈́ Ǥ ૠ . Consider the three rational numbers related to this story of Sam’s money. Write and order them from least to greatest.
െǤ ,െǤ ૠ ,Ǥ
Scaffolding:
WƌŽǀŝĚĞĂŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵĨŽƌǀŝƐƵĂůůĞĂƌŶĞƌƐƚŽŚĞůƉ ƚŚĞŵĚĞƚĞƌŵŝŶĞƚŚĞŶƵŵďĞƌƐ͛ŽƌĚĞƌ͘
A STORY OF RATIOS
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6ͻϯ Lesson 8 Lesson 8: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
džƉůĂŝŶƚŚĞƉƌŽĐĞƐƐLJŽƵƵƐĞĚƚŽĚĞƚĞƌŵŝŶĞƚŚĞŽƌĚĞƌŽĨƚŚĞŶƵŵďĞƌƐ͘
à There is only one positive number, ͳͲǤͲͲ , so I know that ͳͲǤͲͲ is the greatest. I know ʹǤʹͷ is farther to the right on the number line than ͳǤͷ ; therefore, its opposite, െʹǤʹͷ , will be farther to the left than the opposite of ͳǤͷ . This means െʹǤʹͷ is the least, and െͳǤͷ is between െʹǤʹͷ and ͳͲǤͲͲ .
,ŽǁǁŽƵůĚƚŚĞŽƌĚĞƌĐŚĂŶŐĞŝĨLJŽƵǁĞƌĞĂƐŬĞĚƚŽǁƌŝƚĞƚŚĞŶƵŵďĞƌƐĨƌŽŵŐƌĞĂƚĞƐƚƚŽůĞĂƐƚ͍
à The order would be reversed. I would list the numbers so that the number that comes first is the one farthest to the right on the number line, and the number that comes last is the one farthest to the left on the number line. The order would be ͳͲǤͲͲ (the greatest), followed by െͳǤͷ , and then followed by
െʹǤʹͷ (the least).
Exercises 2–4 (10 minutes)
ůůŽǁƚŝŵĞĨŽƌƐƚƵĚĞŶƚƐƚŽƐŚĂƌĞƚŚĞŝƌĂŶƐǁĞƌƐǁŝƚŚƚŚĞĐůĂƐƐĂŶĚĞdžƉůĂŝŶƚŚĞŝƌƌĞĂƐŽŶŝŶŐ͘
Exercises 2–4 For each problem, list the rational numbers that relate to each situation. Then, order them from least to greatest, and explain how you made your determination. 2. During their most recent visit to the optometrist (eye doctor), Kadijsha and her sister, Beth, had their vision tested. Kadijsha’s vision in her left eye was െǤ , and her vision in her right eye was the opposite number. Beth’s vision was െǤ in her left eye and Ǥ in her right eye.
െǤ , െǤ ,Ǥ ,Ǥ
The opposite of െǤ is Ǥ , and Ǥ is farthest right on the number line, so it is the greatest. െǤ is the same distance from zero but on the other side, so it is the least number. െǤ is to the right of െǤ , so it is greater than െǤ , and Ǥ is to the right of െǤ , so it is greater than െǤ . Finally, Ǥ is the greatest.
ϯ͘ There are three pieces of mail in Ms. Thomas’s mailbox: a bill from the phone company for ̈́ ૡǤ , a bill from the electric company for ̈́ ૠǤ , and a tax refund check for ̈́ Ǥ ૡૢ . (A bill is money that you owe, and a tax refund check is money that you receive.)
െૠǤ , െૡǤ , Ǥ ૡૢ
The change in Ms. Thomas’s money is represented by െૡǤ due to the phone bill, and െૠǤ represents the change in her money due to the electric bill. Since െૠǤ is farthest to the left on the number line, it is the least. Since െૡǤ is to the right of െૠǤ , it comes next. The check she has to deposit for ̈́ Ǥ ૡૢ can be represented by Ǥ ૡૢ , which is to the right of െૡǤ , and so it is the greatest number.
Monica, Jack, and Destiny measured their arm lengths for an experiment in science class. They compared their arm lengths to a standard length of inches. The listing below shows, in inches, how each student’s arm length compares to inches. Monica: െ ૡ
Jack:
Destiny: െ
െ , െ ૡ ,
I ordered the numbers on a number line, and െ was farthest to the left. To the right of that was െ ૡ. Lastly,
is to the right of െ ૡ, so is the greatest.
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6ͻϯ Lesson 8 Lesson 8: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Example 2 ;ϯ minutes): Ordering Rational Numbers from Greatest to Least
Example 2: Ordering Rational Numbers from Greatest to Least Jason is entering college and has opened a checking account, which he will use for college expenses. His parents gave him ̈́ Ǥ to deposit into the account. Jason wrote a check for ̈́ ૡǤ to pay for his calculus book and a check for ̈́ Ǥ to pay for miscellaneous school supplies. Write the three rational numbers related to the balance in Jason’s checking account in order from greatest to least.
Ǥ , െǤ ,െૡǤ
džƉůĂŝŶƚŚĞƉƌŽĐĞƐƐLJŽƵƵƐĞĚƚŽĚĞƚĞƌŵŝŶĞƚŚĞŽƌĚĞƌŽĨƚŚĞŶƵŵďĞƌƐ͘
à There was only one positive number, ʹͲͲǤͲͲ , so I know that ʹͲͲǤͲͲ is the greatest. I know ͺͷǤͲͲ is farther to the right on the number line than ʹͷǤ͵Ͷ , so its opposite, െͺͷǤͲͲ , will be farther to the left than the opposite of ʹͷǤ͵Ͷ . This means െͺͷǤͲͲ is the least, and െʹͷǤ͵Ͷ would be between െͺͷǤͲͲ
and ʹͲͲǤͲͲ .
Exercises 5–6 (6 minutes)
ůůŽǁƚŝŵĞĨŽƌƐƚƵĚĞŶƚƐƚŽƐŚĂƌĞƚŚĞŝƌĂŶƐǁĞƌƐǁŝƚŚƚŚĞĐůĂƐƐĂŶĚĞdžƉůĂŝŶƚŚĞŝƌƌĞĂƐŽŶŝŶŐ͘
Exercises 5–6 For each problem, list the rational numbers that relate to each situation in order from greatest to least. Explain how you arrived at the order. 5. The following are the current monthly bills that Mr. McGraw must pay: ̈́ Ǥ Cable and Internet ̈́ ૠǤ Gas and Electric ̈́ Ǥ Cell Phone
െǤ , െૠǤ ,െǤ
Because Mr. McGraw owes the money, I represented the amount of each bill as a negative number. Ordering them from greatest to least means I have to move from right to left on a number line. Since െǤ is farthest right, it is the greatest. To the left of that is െૠǤ , and to the left of that is െǤ , which means െǤ is the least.
െ , , െ , ૡ
ૡ , , െ , െ
I graphed them on the number line. Since I needed to order them from greatest to least, I moved from right to left to record the order. Farthest to the right is ૡ, so that is the greatest value. To the left of that number is . To the left of is െ , and the farthest left is െ , so that is the least.
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6ͻϯ Lesson 8 Lesson 8: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Closing ;ϯ minutes)
/ĨƚŚƌĞĞŶƵŵďĞƌƐĂƌĞŽƌĚĞƌĞĚĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚĂŶĚƚŚĞŽƌĚĞƌŝƐ ͕ܽ ܾ ͕ ͕ܿ ǁŚĂƚǁŽƵůĚƚŚĞŽƌĚĞƌďĞŝĨƚŚĞ ƐĂŵĞƚŚƌĞĞŶƵŵďĞƌƐǁĞƌĞĂƌƌĂŶŐĞĚŝŶŽƌĚĞƌĨƌŽŵŐƌĞĂƚĞƐƚƚŽůĞĂƐƚ͍,ŽǁĚŝĚLJŽƵĚĞƚĞƌŵŝŶĞƚŚĞŶĞǁŽƌĚĞƌ͍
àܿ ,ܾ ,ܽ
à This is the correct order because it has to be exactly the opposite order since we are now moving right to left on the number line, when originally we moved left to right.
,ŽǁĚŽĞƐŐƌĂƉŚŝŶŐŶƵŵďĞƌƐŽŶĂŶƵŵďĞƌůŝŶĞŚĞůƉƵƐĚĞƚĞƌŵŝŶĞƚŚĞŽƌĚĞƌǁŚĞŶĂƌƌĂŶŐŝŶŐƚŚĞŶƵŵďĞƌƐĨƌŽŵ ŐƌĞĂƚĞƐƚƚŽůĞĂƐƚŽƌůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͍
à Using a number line helps us order numbers because when numbers are placed on a number line, they are placed in order.
Exit Ticket (6 minutes)
Lesson Summary When we order rational numbers, their opposites are in the opposite order. For example, if ૠ is greater than , െૠ
is less than െ .
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6ͻϯ Lesson 8 Lesson 8: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
EĂŵĞ ĂƚĞ
Lesson 8: Ordering Integers and Other Rational Numbers Exit Ticket
KƌĚĞƌƚŚĞĨŽůůŽǁŝŶŐƐĞƚŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͕ĂŶĚĞdžƉůĂŝŶŚŽǁLJŽƵĚĞƚĞƌŵŝŶĞĚƚŚĞŽƌĚĞƌ͘
െ͵ ͕ Ͳ͕ െ ͳʹ͕ ͳ͕ െ͵ ͳ͵͕ ͕ ͷ͕ െͳ ͕ ଶଵ ହ͕ Ͷ
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6ͻϯ Lesson 8 Lesson 8: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Exit Ticket Sample Solutions Order the following set of rational numbers from least to greatest, and explain how you determined the order.
െ, , െ , , െ , , , െ , ,
െ , െ , െ , െ , , , , , ,
I drew a number line and started at zero. I located the positive numbers to the right and their opposites (the negative numbers) to the left of zero. The positive integers listed in order from left to right are , , , . And since is equal to
, I know that it is more than but less than . Therefore, I arrived at , , , , , . Next, I ordered the negative numbers. Since െ and െ are the opposites of and , they are unit and units from zero but to the left of zero. And
െ is even farther left, since it is units to the left of zero. The smallest number is farthest to the left, so I arrived at the following order: െ ,െ , െ ,െ , , , , , , .
Problem Set Sample Solutions 1.
a. In the table below, list each set of rational numbers from greatest to least. Then, in the appropriate column, state which number was farthest right and which number was farthest left on the number line. Column 1 Column 2 ŽůƵŵŶϯ Column 4 Rational Numbers Ordered from Greatest to Least Farthest Right on the Number Line Farthest Left on the Number Line
െǤ ૠ , െǤ െǤ ૠ ,െǤ െǤ ૠ െǤ
െૢ Ǥ ૠ ,െૢ െૢ ǡെૢ Ǥ ૠ െૢ െૢ Ǥ ૠ
, ,
െૠ , െ ૠ െૠ , െ ૠ െૠ െૠ
െ , െ െ , െ െ െ
, െ , െ െ
െૢૢ , െ , െૢૢ Ǥ െૢૢ , െૢૢ Ǥ , െ െૢૢ െ
Ǥ , Ǥ Ǥ , Ǥ Ǥ Ǥ
, െ , െ , െ , െ െ
െǤ , െǤ െǤ , െǤ െǤ െǤ
െ െ െ െ െ െ െૠ െૢ ૡૠ െ ૢ െૡ
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6ͻϯ Lesson 8 Lesson 8: KƌĚĞƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
b. For each row, describe the relationship between the number in Column ϯ and its order in Column 2. Why is this?
The number in Column 3 is the first number listed in Column 2. Since it is farthest right on the number line, it will be the greatest; therefore, it comes first when ordering the numbers from greatest to least.
c. For each row, describe the relationship between the number in Column 4 and its order in Column 2. Why is this?
The number in Column 4 is the last number listed in Column 2. Since it is farthest left on the number line, it will be the smallest; therefore, it comes last when ordering the numbers from greatest to least.
If two rational numbers, ࢇ and ࢈, are ordered such that ࢇ is less than ࢈, then what must be true about the order for their opposites: ࢇെ and ࢈െ ?
The order will be reversed for the opposites, which means ࢇെ is greater than ࢈െ .
ϯ͘ Read each statement, and then write a statement relating the opposites of each of the given numbers: a. ૠ is greater than .
െૠ is less than െ .
b. ૢ Ǥ is greater than .
െૢ Ǥ is less than െ .
c. െ is less than .
is greater than െ .
Order the following from least to greatest: െૡ , െૢ , , , .
െૢ , െ ૡ , , ,
Order the following from greatest to least: െ , , െૢ , , .
, , , െ , െ ૢ
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Lesson 9: Comparing Integers and Other Rational Numbers Student Outcomes
^ƚƵĚĞŶƚƐĐŽŵƉĂƌĞĂŶĚŝŶƚĞƌƉƌĞƚƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͛ŽƌĚĞƌŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͕ŵĂŬŝŶŐƐƚĂƚĞŵĞŶƚƐƚŚĂƚƌĞůĂƚĞƚŚĞ ŶƵŵďĞƌƐ͛ůŽĐĂƚŝŽŶŽŶƚŚĞŶƵŵďĞƌůŝŶĞƚŽƚŚĞŝƌŽƌĚĞƌ͘
^ƚƵĚĞŶƚƐĂƉƉůLJƚŚĞŝƌƉƌĞƌĞƋƵŝƐŝƚĞŬŶŽǁůĞĚŐĞŽĨƉůĂĐĞǀĂůƵĞ͕ĚĞĐŝŵĂůƐ͕ĂŶĚĨƌĂĐƚŝŽŶƐƚŽĐŽŵƉĂƌĞŝŶƚĞŐĞƌƐĂŶĚ ŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
^ƚƵĚĞŶƚƐƌĞůĂƚĞŝŶƚĞŐĞƌƐĂŶĚŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŽƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƐĂŶĚƉƌŽďůĞŵƐ͘
Lesson Notes
^ƚƵĚĞŶƚƐĐŽŵƉůĞƚĞĂŶĂĐƚŝǀŝƚLJĚƵƌŝŶŐ džĂŵƉůĞϮƚŚĂƚƌĞƋƵŝƌĞƐƉƌĞƉĂƌĂƚŝŽŶ͘dŚĞĐƚŝǀŝƚLJĂƌĚƐ;ĂƚƚĂĐŚĞĚƚŽƚŚĞĞŶĚŽĨƚŚĞ ůĞƐƐŽŶͿŶĞĞĚƚŽďĞƉƌĞƉĂƌĞĚďĞĨŽƌĞƚŚĞĚĞůŝǀĞƌLJŽĨƚŚŝƐůĞƐƐŽŶ͘
Classwork Example 1 ;ϯ minutes): Interpreting Number Line Models to Compare Numbers
ZĞĨĞƌƚŽƚŚĞŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵďĞůŽǁ͕ǁŚŝĐŚŝƐĂůƐŽůŽĐĂƚĞĚŝŶƚŚĞƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐ͘/ŶĂǁŚŽůĞͲŐƌŽƵƉĚŝƐĐƵƐƐŝŽŶ͕ ĐƌĞĂƚĞĂƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƚŚĂƚƌĞůĂƚĞƐƚŽƚŚĞŶƵŵďĞƌƐŐƌĂƉŚĞĚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘/ŶĐůƵĚĞĂŶĞdžƉůĂŶĂƚŝŽŶŽĨǁŚĂƚ njĞƌŽƌĞƉƌĞƐĞŶƚƐ͘^ƚƵĚĞŶƚƐƐŚŽƵůĚĐŽŶƚƌŝďƵƚĞƐƵŐŐĞƐƚŝŽŶƐƚŽŚĞůƉƚŚĞƐƚŽƌLJĞǀŽůǀĞĂŶĚĐŽŵĞƚŽĂĨŝŶĂůƐƚĂƚĞ͘^ƚƵĚĞŶƚƐ ǁƌŝƚĞƚŚĞƌĞůĂƚĞĚƐƚŽƌLJŝŶƚŚĞŝƌƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐ͘
Example 1: Interpreting Number Line Models to Compare Numbers
Answers may vary. Every August, the Boy Scouts go on an ૡ-day -mile hike. At the halfway point ( miles into the hike), there is a check-in station for Scouts to check in and register. Thomas and Evan are Scouts in different hiking groups. By Wednesday morning, Evan’s group has miles to go before it reaches the check-in station, and Thomas’s group is miles beyond the station. Zero on the number line represents the check-in station.
െ െ െ െ
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Exercise 1 (7 minutes)
ŝƐƉůĂLJƚŚĞĨŽůůŽǁŝŶŐǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞŵŽĚĞůŽŶƚŚĞďŽĂƌĚ͘^ƚƵĚĞŶƚƐĂƌĞƚŽ ŝŶĚĞƉĞŶĚĞŶƚůLJŝŶƚĞƌƉƌĞƚƚŚĞŶƵŵďĞƌůŝŶĞŵŽĚĞůƚŽĚĞƐĐƌŝďĞĂƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶŝŶǀŽůǀŝŶŐ ƚŚĞƐĞƚǁŽƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘ZĞŵŝŶĚƐƚƵĚĞŶƚƐƚŽĐŽŵƉĂƌĞƚŚĞŶƵŵďĞƌƐĂŶĚĚĞƐĐƌŝďĞƚŚĞŝƌ ŽƌĚĞƌŝŶƚŚĞŝƌǁƌŝƚĞͲƵƉƐ͘ĨƚĞƌĂůůŽǁŝŶŐĂĚĞƋƵĂƚĞƚŝŵĞĨŽƌƐƚƵĚĞŶƚƐƚŽǁƌŝƚĞƚŚĞŝƌ ƐŽůƵƚŝŽŶƐ͕ƐĞǀĞƌĂůƐƚƵĚĞŶƚƐƐŚĂƌĞǁŚĂƚƚŚĞLJǁƌŽƚĞǁŝƚŚƚŚĞĐůĂƐƐ͘^ƚƵĚĞŶƚƐŝŶƚŚĞĐůĂƐƐ ĚĞƚĞƌŵŝŶĞǁŚĞƚŚĞƌƚŚĞǁƌŝƚƚĞŶƌĞƐƉŽŶƐĞƐĐŽƌƌĞĐƚůLJƌĞůĂƚĞƚŽƚŚĞŶƵŵďĞƌůŝŶĞŵŽĚĞůƐ͘
Exercise 1 1. Create a real-world situation that relates to the points shown in the number line model. Be sure to describe the relationship between the values of the two points and how it relates to their order on the number line.
Answers will vary. Alvin lives in Canada and is recording the outside temperature each night before he goes to bed. On Monday night, he recorded a temperature of degrees Celsius. On Tuesday night, he recorded a temperature of െ degree Celsius. Tuesday night’s temperature was colder than Monday night’s temperature. െ is less than , so the associated point is below on a vertical number line.
Example 2 (10 minutes)
^ƚƵĚĞŶƚƐĂƌĞƐĞĂƚĞĚŝŶŐƌŽƵƉƐŽĨƚŚƌĞĞŽƌĨŽƵƌ͕ĂŶĚĞĂĐŚŐƌŽƵƉŝƐŐŝǀĞŶĂƐĞƚŽĨĐƚŝǀŝƚLJĂƌĚƐ͘&ŽƌĞĂĐŚŐƌŽƵƉ͕ ƉŚŽƚŽĐŽƉLJ͕ĐƵƚŽƵƚ͕ĂŶĚƐĐƌĂŵďůĞďŽƚŚƐŚĞĞƚƐŽĨĐƚŝǀŝƚLJĂƌĚƐƚŚĂƚĂƉƉĞĂƌĂƚƚŚĞĞŶĚŽĨƚŚŝƐůĞƐƐŽŶ͘
x ĂĐŚŐƌŽƵƉŽĨƐƚƵĚĞŶƚƐŵĂƚĐŚĞƐĞĂĐŚǁŽƌĚƐƚŽƌLJĐĂƌĚƚŽŝƚƐƌĞůĂƚĞĚŶƵŵďĞƌůŝŶĞĐĂƌĚ͘
x &ŽƌĞĂĐŚŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵ͕ƐƚƵĚĞŶƚƐŵƵƐƚǁƌŝƚĞĂƐƚĂƚĞŵĞŶƚƌĞůĂƚŝŶŐƚŚĞŶƵŵďĞƌƐ͛ƉůĂĐĞŵĞŶƚŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞƚŽƚŚĞŝƌŽƌĚĞƌ͘
x /ĨƚŝŵĞƉĞƌŵŝƚƐ͕ƚŚĞĐůĂƐƐŐŽĞƐŽǀĞƌĞĂĐŚĂŶƐǁĞƌĂƐĂǁŚŽůĞŐƌŽƵƉ͘&ŽƌĞĂĐŚŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵ͕ƐƚƵĚĞŶƚƐ ƉƌĞƐĞŶƚƚŚĞŝƌǁƌŝƚƚĞŶƐƚĂƚĞŵĞŶƚƐĂƐǀĞƌďĂůƐƚĂƚĞŵĞŶƚƐƚŽƚŚĞĐůĂƐƐ͘
x ŶĞdžĂŵƉůĞĨŽůůŽǁƐ͗
dŚĞEĂǀLJ^ĞĂůƐĂƌĞ ƉƌĂĐƚŝĐŝŶŐŶĞǁ ƚĞĐŚŶŝƋƵĞƐ͘dŚĞďůƵĞ ƐƵďŵĂƌŝŶĞŝƐ ͶͷͲǤ ďĞůŽǁƐĞĂůĞǀĞů͕ǁŚŝůĞ ƚŚĞƌĞĚƐƵďŵĂƌŝŶĞŝƐ
͵ͷǤ ďĞůŽǁƐĞĂůĞǀĞů͘
à The blue submarine is farther below sea level than the red submarine because െͶͷͲ is to the left of
െ͵ͷ on the number line; it is less than െ͵ͷ .
Scaffolding:
WƌŽǀŝĚĞĂƐƚŽƌLJƐƚĂƌƚĞƌĨŽƌ ƐƚƵĚĞŶƚƐǁŚŽĂƌĞƐƚƌƵŐŐůŝŶŐƚŽ ďĞŐŝŶƚŚĞǁƌŝƚŝŶŐƚĂƐŬ͘
െ
െ െ െ െ െ
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Exercises 2–8 (15 minutes)
^ƚƵĚĞŶƚƐƌĞĂĚĞĂĐŚŽĨƚŚĞĨŽůůŽǁŝŶŐƐĐĞŶĂƌŝŽƐĂŶĚĚĞĐŝĚĞǁŚĞƚŚĞƌƚŚĞLJĂŐƌĞĞŽƌĚŝƐĂŐƌĞĞ͘ dŚĞLJŵƵƐƚĚĞĨĞŶĚĂŶĚĞdžƉůĂŝŶƚŚĞŝƌƐƚĂŶĐĞŝŶǁƌŝƚŝŶŐ͘ůůŽǁƚŝŵĞĨŽƌƐƚƵĚĞŶƚƐƚŽƐŚĂƌĞ ƚŚĞŝƌĂŶƐǁĞƌƐǁŝƚŚƚŚĞĐůĂƐƐĂŶĚĞdžƉůĂŝŶƚŚĞŝƌƌĞĂƐŽŶŝŶŐ͘dŚĞĐůĂƐƐƐŚŽƵůĚĐŽŵĞƚŽĂ ĐŽŶƐĞŶƐƵƐĨŽƌĞĂĐŚŽŶĞ͘
Exercises 2–8 For each problem, determine if you agree or disagree with the representation. Then, defend your stance by citing specific details in your writing. 2. Felicia needs to write a story problem that relates to the order in which the numbers െ and െ are represented on a number line. She writes the following: “During a recent football game, our team lost yards on two different plays, one occurring in the first quarter of the game and the second occuring in the third quarter of the game. We lost yards on the pla y in the first quarter.
During the play in the third quarter, our quarterback was sacked for a -yard loss. On the number line, I represented this situation by first locating െ . I located the point by moving units to the left of zero. Then, I
graphed the second point by moving units to the left of .”
Agree. െ is less than െ since െ is to the left of െ on the number line. Since both numbers are negative, they indicate the team lost yards on both football plays, but they lost more yards on the second play.
ϯ͘ Manuel looks at a number line diagram that has the points െ and െ graphed. He writes the following related story: “I borrowed cents from my friend, Lester. I borrowed ૠ cents from my friend, Calvin. I owe Lester less than I owe Calvin.”
Agree. െ is equivalent to െǤ ૠ and െ is equivalent to െǤ . െǤ and െǤ ૠ both show that he owes money. But െǤ is farther to the right on a number line, so Manuel does not owe Lester as much as he owes Calvin.
Henry located and Ǥ on a number line. He wrote the following related story: “In gym class, both Jerry and I ran for minutes. Jerry ran miles, and I ran Ǥ miles. I ran a farther distance.”
Disagree. is greater than Ǥ since is equivalent to Ǥ . On the number line, the point associated with
Ǥ is to the right of Ǥ . Jerry ran a farther distance.
Sam looked at two points that were graphed on a vertical number line. He saw the points െ and Ǥ . He wrote the following description: “I am looking at a vertical number line that shows the location of two specific points. The first point is a negative number, so it is below zero. The second point is a positive number, so it is above zero. The negative number is െ .The positive number is unit more than the negative number.”
Disagree. Sam was right when he said the negative number is below zero and the positive number is above zero. But Ǥ is units above zero, and െ is units below zero. So, altogether, that means the positive number is
units more than െ .
Scaffolding:
WƌŽǀŝĚĞĂƐĞƚŽĨŚŽƌŝnjŽŶƚĂůĂŶĚ ǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞƐĨŽƌǀŝƐƵĂů ůĞĂƌŶĞƌƐƚŽĐƌĞĂƚĞĂŶƵŵďĞƌ ůŝŶĞŵŽĚĞůĨŽƌĞĂĐŚĞdžĞƌĐŝƐĞ͘
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Claire draws a vertical number line diagram and graphs two points: െ and . She writes the following related story: “These two locations represent different elevations. One location is feet above sea level, and one location is
feet below sea level. On a number line, feet above sea level is represented by graphing a point at , and
feet below sea level is represented by graphing a point at െ .”
Agree. Zero in this case represents sea level. Both locations are feet from zero but in opposite directions, so they are graphed on the number line at and െ .
Mrs. Kimble, the sixth-grade math teacher, asked the class to describe the relationship between two points on the number line, ૠ. and ૠ. , and to create a real-world scenario. Jackson writes the following story: “Two friends, Jackie and Jennie, each brought money to the fair. Jackie brought more than Jennie. Jackie brought
$ૠ. , and Jennie brought $ૠ. . Since ૠ. has more digits than ૠ. , it would come after ૠ. on the number line, or to the right, so it is a greater value.”
Disagree. Jackson is wrong by saying that ૠ. is to the right of ૠ. on the number line. ૠ. is the same as ૠ. ,and it is greater than ૠ. . When I count by hundredths starting at ૠ. , I would say ૠ. , ૠ. ૠ , ૠ. ૡ , ૠ. ૢ , and then ૠ. . So, ૠ. is greater than ૠ. , and the associated point falls to the right of the point associated with
ૠ. on the number line.
Justine graphs the points associated with the following numbers on a vertical number line: െ , െ , and . She then writes the following real-world scenario: “The nurse measured the height of three sixth-grade students and compared their heights to the height of a typical sixth grader. Two of the students’ heights are below the typical height, and one is above the typical height. The point whose coordinate is represents the student who has a height that is inch above the typical height. Given this information, Justine determined that the student represented by the point associated with െ is the shortest of the three students.”
Disagree. Justine was wrong when she said the point െ represents the shortest of the three students. If zero stands for no change from the typical height, then the point associated with െ is farther below zero than the point associated with െ . The greatest value is positive . Positive represents the tallest person. The shortest person is represented by െ .
Closing (4 minutes)
,ŽǁĐĂŶƵƐĞLJŽƵƵƐĞƚŚĞŶƵŵďĞƌůŝŶĞƚŽŽƌĚĞƌĂƐĞƚŽĨŶƵŵďĞƌƐ͍tŝůůŐƌĂƉŚŝŶŐƚŚĞŶƵŵďĞƌƐŽŶĂǀĞƌƚŝĐĂů ŶƵŵďĞƌůŝŶĞƌĂƚŚĞƌƚŚĂŶĂŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞĐŚĂŶŐĞƚŚŝƐƉƌŽĐĞƐƐ͍
à You can locate and graph the numbers on the number line to determine their order. If you use a vertical number line, their order is the same as it is on a horizontal number line, but instead of moving from left to right to go from least to greatest, you move from bottom to top. To determine the order of a set of numbers, the number that is farthest left (or farthest down on a vertical number line) is the smallest value. As you move right (or toward the top on a vertical number line), the numbers increase in value. So, the greatest number is graphed farthest right on a number line (or the highest one on a vertical number line).
/ĨƚǁŽƉŽŝŶƚƐĂƌĞŐƌĂƉŚĞĚŽŶĂŶƵŵďĞƌůŝŶĞ͕ǁŚĂƚĐĂŶLJŽƵƐĂLJĂďŽƵƚƚŚĞǀĂůƵĞŽĨƚŚĞŶƵŵďĞƌĂƐƐŽĐŝĂƚĞĚǁŝƚŚ ƚŚĞƉŽŝŶƚŽŶƚŚĞƌŝŐŚƚŝŶĐŽŵƉĂƌŝƐŽŶƚŽƚŚĞǀĂůƵĞŽĨƚŚĞŶƵŵďĞƌĂƐƐŽĐŝĂƚĞĚǁŝƚŚƚŚĞƉŽŝŶƚŽŶƚŚĞůĞĨƚ͍
à The number associated with the point on the right is greater than the number associated with the point on the left.
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
tŚŝĐŚŶƵŵďĞƌŝƐůĂƌŐĞƌ͗ െ͵ǤͶ Žƌ െ͵ ͳʹ͍,ŽǁǁŝůůŐƌĂƉŚŝŶŐƚŚĞƐĞŶƵŵďĞƌƐŽŶĂŶƵŵďĞƌůŝŶĞŚĞůƉLJŽƵŵĂŬĞƚŚŝƐ ĚĞƚĞƌŵŝŶĂƚŝŽŶ͍
à Whichever number is graphed farthest to the left (or below) is the smaller number. In this example,
െ͵ ͳʹ would be graphed to the left of െ͵ǤͶ , so it is the smaller number. You can compare the numbers to make sure they are graphed correctly by either representing them both as a decimal or both as a fraction. െ͵ ͳʹ is halfway between െ͵ and െͶ . So, if I divide the space into tenths, the associated point would be at െ͵Ǥͷ since െ͵ ͳʹ ൌ െ͵ ͷͳͲ . When I graph െ͵ǤͶ , it would be ͲǤͳ closer to െ͵ , so it would be to the right of െ͵ ͳʹ Ǥ This means െ͵ǤͶ is larger than െ͵ ͳʹ.
Exit Ticket (6 minutes)
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
EĂŵĞ ĂƚĞ
Lesson 9: Comparing Integers and Other Rational Numbers Exit Ticket
ϭ͘ /ŶƚĞƌƉƌĞƚƚŚĞŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵƐŚŽǁŶďĞůŽǁ͕ĂŶĚǁƌŝƚĞĂƐƚĂƚĞŵĞŶƚĂďŽƵƚƚŚĞƚĞŵƉĞƌĂƚƵƌĞĨŽƌdƵĞƐĚĂLJ ĐŽŵƉĂƌĞĚƚŽDŽŶĚĂLJĂƚϭϭ͗ϬϬƉ͘ŵ͘ Ϯ͘ /ĨƚŚĞƚĞŵƉĞƌĂƚƵƌĞĂƚϭϭ͗ϬϬƉ͘ŵ͘ŽŶtĞĚŶĞƐĚĂLJŝƐǁĂƌŵĞƌƚŚĂŶdƵĞƐĚĂLJ͛ƐƚĞŵƉĞƌĂƚƵƌĞďƵƚƐƚŝůůďĞůŽǁnjĞƌŽ͕ǁŚĂƚŝƐ ĂƉŽƐƐŝďůĞǀĂůƵĞĨŽƌƚŚĞƚĞŵƉĞƌĂƚƵƌĞĂƚϭϭ͗ϬϬƉ͘ŵ͘tĞĚŶĞƐĚĂLJ͍
DŽŶĚĂLJ͛ƐdĞŵƉĞƌĂƚƵƌĞ; ι ͿĂƚϭϭ͗ϬϬƉ͘ŵ͘
dƵĞƐĚĂLJ͛ƐdĞŵƉĞƌĂƚƵƌĞ; ι ͿĂƚϭϭ͗ϬϬƉ͘ŵ͘
െͷͲ
Ͳ
ͷͲ
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Exit Ticket Sample Solutions 1. Interpret the number line diagram shown below, and write a statement about the temperature for Tuesday compared to Monday at 11:00 p.m.
At 11:00 p.m. on Monday, the temperature was about degrees Fahrenheit, but at 11:00 p.m. on Tuesday, it was െ degrees Fahrenheit. Tuesday’s temperature of െ degrees is below zero, but degrees is above zero. It was much warmer on Monday at 11:00 p.m. than on Tuesday at that time.
If the temperature at 11:00 p.m. on Wednesday is warmer than Tuesday’s temperature but still below zero, what is a possible value for the temperature at 11:00 p.m. Wednesday?
Answers will vary but must be between and െ . A possible temperature for Wednesday at 11:00 p.m. is െ
degrees Fahrenheit because െ is less than zero and greater than െ .
Problem Set Sample Solutions Write a story related to the points shown in each graph. Be sure to include a statement relating the numbers graphed on the number line to their order. 1.
Answers will vary. Marcy earned no bonus points on her first math quiz. She earned bonus points on her second math quiz. Zero represents earning no bonus points, and represents earning bonus points. Zero is graphed to the left of on the number line. Zero is less than .
Answers will vary. My uncle’s investment lost ̈́ in May. In June, the investment gained ̈́ .The situation is represented by the points െ and on the vertical number line. Negative
is below zero, and is above zero. െ is less than .
Monday’s Temperature ( ι۴ ) at 11:00 p.m.
Tuesday’s Temperature ( ι۴ ) at 11:00 p.m.
െ
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
ϯ͘
Answers will vary. I gave my sister ̈́ Ǥ last week. This week, I gave her ̈́ Ǥ . The points െǤ and െǤ
represent the change to my money supply. We know that െǤ is to the left of െǤ on the number line; therefore, െǤ is greater than െǤ .
Answers will vary. A fish is swimming ૠ feet below the water’s surface. A turtle is swimming feet below the water’s surface. We know that െૠ is to the left of െ on the number line. This means െૠ is less than െ .
Answers will vary. I spent ̈́ ૡ on a CD last month. I earned ̈́ in allowance last month. െૡ and represent the changes to my money last month. െૡ is to the left of on a number line. െૡ is units farther away from zero than
, which means that I spent ̈́ more on the CD than I made in allowance.
Answers will vary. Skip, Mark, and Angelo were standing in line in gym class. Skip was the third person behind Mark. Angelo was the first person ahead of Mark. If Mark represents zero on the number line, then Skip is associated with the point at െ , and Angelo is associated with the point at . is unit to the right of zero, and െ
is units to the left of zero. െ is less than .
Answers will vary. I rode my bike miles on Saturday and miles on Sunday. On a vertical number line, and are both associated with points above zero, but is above . This means that is greater than .
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Activity Cards—Page 1
dŚĞEĂǀLJ^ĞĂůƐĂƌĞƉƌĂĐƚŝĐŝŶŐ ŶĞǁƚĞĐŚŶŝƋƵĞƐ͘dŚĞďůƵĞ ƐƵďŵĂƌŝŶĞŝƐ ͶͷͲǤ ďĞůŽǁ ƐĞĂůĞǀĞů͕ǁŚŝůĞƚŚĞƌĞĚ ƐƵďŵĂƌŝŶĞŝƐ ͵ͷǤ ďĞůŽǁ ƐĞĂůĞǀĞů͘ ŽůƉŚŝŶƐůŽǀĞƚŽũƵŵƉŽƵƚŽĨ ƚŚĞǁĂƚĞƌ͘ŽůůLJ͕ƚŚĞĚŽůƉŚŝŶ͕ ĐĂŶũƵŵƉ ͷŵĞƚĞƌƐĂďŽǀĞƚŚĞ ǁĂƚĞƌĂŶĚƐǁŝŵ ͶͷͲ ŵĞƚĞƌƐ ďĞůŽǁƚŚĞƐƵƌĨĂĐĞŽĨƚŚĞ ǁĂƚĞƌ͘ ŽůŽƌĂĚŽŝƐŬŶŽǁŶĨŽƌĚƌĂƐƚŝĐ ĐŚĂŶŐĞƐŝŶƚĞŵƉĞƌĂƚƵƌĞƐ͘ dƵĞƐĚĂLJŵŽƌŶŝŶŐƚŚĞ ƚĞŵƉĞƌĂƚƵƌĞǁĂƐ ͵ʹι ͕ ďƵƚ dƵĞƐĚĂLJŶŝŐŚƚƚŚĞ ƚĞŵƉĞƌĂƚƵƌĞǁĂƐ
െ͵ι ͘ dŚĞŚŝŐŚƐĐŚŽŽůĨŽŽƚďĂůůƚĞĂŵ ůŽƐƚ ͺLJĂƌĚƐŽŶĨŝƌƐƚĚŽǁŶ͘ KŶƐĞĐŽŶĚĚŽǁŶ͕ƚŚĞƚĞĂŵ ŐĂŝŶĞĚ ʹLJĂƌĚƐ͘ ,ŽůůLJƐŽůĚůĞŵŽŶĂĚĞƚǁŽĚĂLJƐ ŝŶĂƌŽǁ͘KŶ^ĂƚƵƌĚĂLJ͕,ŽůůLJ ĞĂƌŶĞĚ ̈́ ͷǤͷ ͘ KŶ^ƵŶĚĂLJ͕ ,ŽůůLJĞĂƌŶĞĚ ̈́ ͵Ǥʹͷ ͘ /ŶŐŽůĨ͕ƚŚĞůŽǁĞƐƚƐĐŽƌĞǁŝŶƐ͘ WĞƚĞ͛ƐĨŝŶĂůƐĐŽƌĞǁĂƐ െʹ ͕ ĂŶĚŶĚƌĞ͛ƐĨŝŶĂůƐĐŽƌĞǁĂƐ
െͷ ͘
െૡૡ
െૡ െ െ െ െ െૠૡ
െ െ െ െ
െૠ െ െ െ
െ െ െ െ െ
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6ͻϯ Lesson 9 Lesson 9: ŽŵƉĂƌŝŶŐ/ŶƚĞŐĞƌƐĂŶĚKƚŚĞƌZĂƚŝŽŶĂůEƵŵďĞƌƐ
Activity Cards—Page 2
dĞĂŐŽŶĞĂƌŶĞĚ ̈́ͶͷͲ ůĂƐƚ ŵŽŶƚŚĐƵƚƚŝŶŐŐƌĂƐƐ͘yĂǀŝĞƌ ƐƉĞŶƚ ̈́ ͵ͷ ŽŶĂŶĞǁ ĐŽŵƉƵƚĞƌ͘ :ĂLJĚĞŶŚĂƐĞĂƌŶĞĚ ͵ďŽŶƵƐ ƉŽŝŶƚƐĐŽŵƉůĞƚŝŶŐŵĂƚŚĞdžƚƌĂ ĐƌĞĚŝƚĂƐƐŝŐŶŵĞŶƚƐ͕ǁŚŝůĞ ^ŚŽŶƚĞůůĞŚĂƐĞĂƌŶĞĚ ͵ʹ ďŽŶƵƐ ƉŽŝŶƚƐ͘ <ŝŵĂŶĚŚĞƌĨƌŝĞŶĚ^ƚĂĐĞLJǁĞŶƚ ƚŽƚŚĞŬƐƚŽƌĞ͘^ƚĂĐĞLJƐƉĞŶƚ ̈́ͺ ŽŶŶŽƚĞŬƐ͘<ŝŵƐƉĞŶƚ ̈́ ͷ ŽŶƐŶĂĐŬƐĂŶĚƉĞŶĐŝůƐ͘ >ĂƐƚŵŽŶƚŚ͕ƚŚĞƐƚŽĐŬŵĂƌŬĞƚ ĚƌŽƉƉĞĚ ͷ ͵Ͷ ƉŽŝŶƚƐŽǀĞƌĂůů͘^Ž ĨĂƌƚŚŝƐŵŽŶƚŚ͕ƚŚĞƐƚŽĐŬ ŵĂƌŬĞƚƌŽƐĞ ͵ ͳͶƉŽŝŶƚƐ͘ ƚĂďĞĂĐŚŝŶĂůŝĨŽƌŶŝĂ͕ŝĨĂ ƉĞƌƐŽŶƐƚĂŶĚƐŝŶƚŚĞǁĂƚĞƌ͕ŚĞ ŝƐ ଵହ Ǥ ďĞůŽǁƐĞĂůĞǀĞů͘/ĨƚŚĞ ƉĞƌƐŽŶǁĂůŬƐŽŶƚŽƚŚĞďĞĂĐŚ͕ ŚĞŝƐ ଶହ Ǥ ĂďŽǀĞƐĞĂůĞǀĞů͘ ƌŝƚƚĂŶLJǁĞŶƚƚŽĂŶŽĨĨŝĐĞ ƐƵƉƉůLJƐƚŽƌĞƚǁŝĐĞůĂƐƚǁĞĞŬ͘ dŚĞĨŝƌƐƚƚŝŵĞ͕ƐŚĞŵĂĚĞ
ʹ ĐŽƉŝĞƐƚŚĂƚĐŽƐƚ ̈́ ͲǤʹͲ ĞĂĐŚ͘ dŚĞƐĞĐŽŶĚƚŝŵĞ͕ƐŚĞĚŝĚŶŽƚ ďƵLJĂŶLJƚŚŝŶŐďƵƚĨŽƵŶĚ ʹĚŝŵĞƐ ŝŶƚŚĞƉĂƌŬŝŶŐůŽƚ͘
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
Lesson 10: Writing and Interpreting Inequality Statements Involving Rational Numbers Student Outcomes
^ƚƵĚĞŶƚƐǁƌŝƚĞĂŶĚĞdžƉůĂŝŶŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƐŝŶǀŽůǀŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
^ƚƵĚĞŶƚƐũƵƐƚŝĨLJŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƐŝŶǀŽůǀŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
Lesson Notes
ĞƐƐŽŶƐϲ͕ϳ͕ϴ͕ĂŶĚϵŚĂǀĞƉƌĞƉĂƌĞĚƐƚƵĚĞŶƚƐĨŽƌƚŚŝƐůĞƐƐŽŶ͘^ƚƵĚĞŶƚƐŚĂǀĞĚĞǀĞůŽƉĞĚĂŶƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨŚŽǁƚŽ ƌĞƉƌĞƐĞŶƚ͕ŽƌĚĞƌ͕ĂŶĚĐŽŵƉĂƌĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐĂŶĚŶŽǁǁƌŝƚĞĂŶĚŝŶƚĞƌƉƌĞƚŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƐŝŶǀŽůǀŝŶŐƌĂƚŝŽŶĂů ŶƵŵďĞƌƐ͘/Ŷ'ƌĂĚĞϱDŽĚƵůĞϭ͕ƐƚƵĚĞŶƚƐǁƌŽƚĞŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƐŝŶǀŽůǀŝŶŐĚĞĐŝŵĂůƐ͕ĂŶĚƚŚĂƚĞdžƉĞƌŝĞŶĐĞƐĞƌǀĞƐĂƐ ĂĨŽƵŶĚĂƚŝŽŶĨŽƌƚŚŝƐůĞƐƐŽŶĂƐǁĞůů͘
Classwork Opening Exercise ;ϯ minutes)
ƐƐƚƵĚĞŶƚƐĞŶƚĞƌƚŚĞƌŽŽŵ͕ƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶŝƐƉŽƐƚĞĚŽŶƚŚĞďŽĂƌĚ;ĂŶĚŝƐŝŶƚŚĞƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐͿ͘͞ dŚĞĂŵŽƵŶƚŽĨŵŽŶĞLJ/ŚĂǀĞŝŶŵLJƉŽĐŬĞƚŝƐůĞƐƐƚŚĂŶ ̈́ ͷ ďƵƚŐƌĞĂƚĞƌƚŚĂŶ ̈́ Ͷ͘͟ ŝƌĞĐƚƐƚƵĚĞŶƚƐƚŽĚŝƐĐƵƐƐǁŝƚŚƚŚĞŝƌŐƌŽƵƉƐŽƌĞůďŽǁƉĂƌƚŶĞƌƐĂƉŽƐƐŝďůĞǀĂůƵĞĨŽƌƚŚĞ ĂŵŽƵŶƚŽĨŵŽŶĞLJŝŶLJŽƵƌƉŽĐŬĞƚ͕ĂŶĚƚŚĞŶĐŽŵƉůĞƚĞƚŚĞĨŽůůŽǁŝŶŐ͗
Opening Exercise “The amount of money I have in my pocket is less than ̈́ but greater than ̈́ .” a. One possible value for the amount of money in my pocket is ̈́ Ǥ ૠ .b. Write an inequality statement comparing the possible value of the money in my pocket to ̈́ .
Ǥ ൏ Ǥ ૠ
c. Write an inequality statement comparing the possible value of the money in my pocket to ̈́ .
Ǥ ૠ ൏ Ǥ
Discussion (5 minutes)
ůůŽǁƚŝŵĞĨŽƌƐƚƵĚĞŶƚƐƚŽƐŚĂƌĞǁŝƚŚƚŚĞĐůĂƐƐƚŚĞŝƌĂŶƐǁĞƌƐĨƌŽŵƉĂƌƚ;ĂͿŽĨƚŚĞKƉĞŶŝŶŐ džĞƌĐŝƐĞ͘ ŝƐĐƵƐƐĂŶƐǁĞƌƐƚŚĂƚǁŽƵůĚĂŶĚǁŽƵůĚŶŽƚĨĂůůďĞƚǁĞĞŶ ͶĂŶĚ ͷ͘ ůŝĐŝƚƐƚƵĚĞŶƚƌĞƐƉŽŶƐĞƐ͘ŝƐĐƵƐƐƉŽƚĞŶƚŝĂůĂŶƐǁĞƌƐ ƚŚĂƚǁŽƵůĚŽƌǁŽƵůĚŶŽƚŵĂŬĞƐĞŶƐĞŝŶƚŚĞĐŽŶƚĞdžƚŽĨƚŚĞƐŝƚƵĂƚŝŽŶ;Ğ͘Ő͕͘ ͶǤ͵Ͳͺ ĐĂŶŶŽƚďĞƌĞƉƌĞƐĞŶƚĞĚǁŝƚŚƉŚLJƐŝĐĂů ŵŽŶĞLJͿ͘
Scaffolding:
hƐĞƉůĂLJŵŽŶĞLJ;ŝŶĐůƵĚŝŶŐďŝůůƐ ĂŶĚĐŽŝŶƐͿƚŽƌĞƉƌĞƐĞŶƚ ̈́ Ͷ ĂŶĚ ̈́ͷ͘ ^ƚƵĚĞŶƚƐƚŚĞŶĐƌĞĂƚĞ ƌĞƉƌĞƐĞŶƚĂƚŝŽŶƐŝŶǀŽůǀŝŶŐ ̈́ Ͷ ĂŶĚƐŽŵĞĐŚĂŶŐĞƐŽĂƐŶŽƚƚŽ ĞdžĐĞĞĚ ̈́ ͷ͘
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
tŚĂƚĂƌĞƐŽŵĞƉŽƐƐŝďůĞǀĂůƵĞƐĨŽƌƚŚĞĂŵŽƵŶƚŽĨŵŽŶĞLJŝŶŵLJƉŽĐŬĞƚ͍
à ̈́ ͶǤʹͷǡ̈́ ͶǤ͵ͻǡ̈́ ͶǤʹǡ and ̈́ͶǤͻͻ
,ĂǀĞƐĞǀĞƌĂůƐƚƵĚĞŶƚƐǁƌŝƚĞƚŚĞŝƌŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƐŽŶƚŚĞďŽĂƌĚĨƌŽŵƉĂƌƚƐ;ďͿĂŶĚ;ĐͿ͘^ĞůĞĐƚĂƐƚƵĚĞŶƚ͛ƐĂŶƐǁĞƌ ;ŽƌĐƌĞĂƚĞĂŶŽƚŚĞƌĞdžĂŵƉůĞŽĨĂĐŽƌƌĞĐƚĂŶƐǁĞƌͿƚŽƵƐĞĂƐĂŵŽĚĞů͘tƌŝƚĞƚŚĞĂŶƐǁĞƌƐƚŽƉĂƌƚƐ;ďͿĂŶĚ;ĐͿƐŝĚĞďLJƐŝĚĞĂƐ ƐŚŽǁŶďĞůŽǁ͘
à ͶǤͲͲ ൏ ͶǤͷ and ͶǤͷ ൏ ͷǤͲͲ
ƌĞƚŚĞƌĞĂŶLJŝŶƚĞŐĞƌƐŽůƵƚŝŽŶƐ͍tŚLJŽƌǁŚLJŶŽƚ͍
à There are no integer solutions because there are no integers between Ͷ and ͷ because they are consecutive integers.
ƌĞƚŚĞƌĞĂŶLJŶƵŵďĞƌƐďĞƚǁĞĞŶ ͶĂŶĚ ͷƚŚĂƚĂƌĞŶŽƚƉŽƐƐŝďůĞǀĂůƵĞƐĨŽƌƚŚĞĂŵŽƵŶƚŽĨŵŽŶĞLJŝŶŵLJƉŽĐŬĞƚ͍ tŚLJŽƌǁŚLJŶŽƚ͍
à We are talking about money, so all possible answers should be rational numbers that terminate at the hundredths place. There are more possible answers between Ͷ and ͷ, but they would not be accurate in this situation.
/ƐƚŚĞƌĞĂǁĂLJƚŽǁƌŝƚĞŽŶĞŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƚŚĂƚĚĞƐĐƌŝďĞƐďŽƚŚŽĨƚŚĞƐĞƌĞůĂƚŝŽŶƐŚŝƉƐ͍
à Yes. ͶǤͲͲ ൏ ͶǤͷ ൏ ͷǤͲͲ
Exercises 1–4 (4 minutes)
^ƚƵĚĞŶƚƐƵƐĞĂŶƵŵďĞƌůŝŶĞŵŽĚĞůƚŽƌĞƉƌĞƐĞŶƚƚŚĞŽƌĚĞƌŽĨƚŚĞŶƵŵďĞƌƐƵƐĞĚŝŶƚŚĞŝƌKƉĞŶŝŶŐ džĞƌĐŝƐĞ;ŽƌŝŶƚŚĞ ĞdžĂŵƉůĞũƵƐƚĚŝƐĐƵƐƐĞĚĂƐĂǁŚŽůĞŐƌŽƵƉͿ͘^ƚƵĚĞŶƚƐƚŚĞŶŐƌĂƉŚƚŚƌĞĞƉŽŝŶƚƐ͗ Ͷ͕ ͷ͕ ĂŶĚĂǀĂůƵĞƚŚĂƚĨĂůůƐŝŶďĞƚǁĞĞŶ ͶĂŶĚ ͷ͘ ^ƚƵĚĞŶƚƐƐŚŽƵůĚƵƐĞƚŚĞŵŽĚĞůĂŶĚŶƵŵďĞƌůŝŶĞŽƌĚĞƌŝŶŐƚŽǁƌŝƚĞŽŶĞŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƌĞůĂƚŝŶŐƚŚĞƚŚƌĞĞ ŶƵŵďĞƌƐ͘
Exercises 1–4 1. Graph your answer from the Opening Exercise part (a) on the number line below. 2. Also, graph the points associated with and on the number line.
ϯ͘ Explain in words how the location of the three numbers on the number line supports the inequality statements you wrote in the Opening Exercise parts (b) and (c).
The numbers are ordered from least to greatest when I look at the number line from left to right. So, is less than
Ǥ ૠ , and Ǥ ૠ is less than .
4. Write one inequality statement that shows the relationship among all three numbers.
൏ Ǥ ૠ ൏
/ĨƐƚƵĚĞŶƚƐƐƚƌƵŐŐůĞǁŝƚŚ džĞƌĐŝƐĞϰ͕ƐƉĞŶĚĂĚĞƋƵĂƚĞƚŝŵĞĂƐĂǁŚŽůĞŐƌŽƵƉĚŽŝŶŐƐĞǀĞƌĂůĞdžĂŵƉůĞƐŽĨŽƌĚĞƌŝŶŐƐƚƵĚĞŶƚƐ͛ ƐĞƚƐŽĨŶƵŵďĞƌƐƵƐŝŶŐŽŶĞƐƚĂƚĞŵĞŶƚŽĨŝŶĞƋƵĂůŝƚLJ͘ dŚĞĨŽůůŽǁŝŶŐƚǁŽĞdžĂŵƉůĞƐƐŚŽƵůĚďĞĐŽŶĚƵĐƚĞĚĂƐǁŚŽůĞͲŐƌŽƵƉŝŶƐƚƌƵĐƚŝŽŶ͘
Ǥ ૠ
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
Example 1 (4 minutes): Writing Inequality Statements Involving Rational Numbers ^ƚƵĚĞŶƚƐƐŚŽƵůĚƌĞĐĂůůƵƐŝŶŐŝŶĞƋƵĂůŝƚLJƐLJŵďŽůƐƚŽĐŽŵƉĂƌĞĚĞĐŝŵĂůŶƵŵďĞƌƐĨƌŽŵ'ƌĂĚĞϱDŽĚƵůĞϭ͘ƐŶĞĞĚĞĚ͕ƌĞĨĞƌ ƚŽƚŚĞŶƵŵďĞƌůŝŶĞƌĞƉƌĞƐĞŶƚĂƚŝŽŶŽĨŶŽŶͲŝŶƚĞŐĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐĂŶĚƚŚĞŝƌŽƉƉŽƐŝƚĞƐĨƌŽŵ>ĞƐƐŽŶϲŽĨƚŚŝƐŵŽĚƵůĞ͕ ĂŶĚƌĞǀŝƐŝƚƚŚĞŽƌŝĞŶƚĂƚŝŽŶŽĨƚŚĞůĞƐƐƚŚĂŶƐLJŵďŽů; ൏ͿĂŶĚŐƌĞĂƚĞƌƚŚĂŶƐLJŵďŽů; Ϳ͘EŽƚĞƚŚĂƚƐƚƵĚĞŶƚƐŵĂLJĞƌƌŽŶĞŽƵƐůLJ ǁƌŝƚĞ ͺ ൏ ͳͲ ͳʹ ͻ ͕ ǁŚŝĐŚĚŽĞƐŶŽƚĂĐĐƵƌĂƚĞůLJĚĞƐĐƌŝďĞƚŚĞŽƌĚĞƌŽĨĂůůƚŚƌĞĞŶƵŵďĞƌƐ͘
Example 1: Writing Inequality Statements Involving Rational Numbers Write one inequality statement to show the relationship among the following shoe sizes: , ૡ, and ૢ .a. From least to greatest:
ૡ ൏ૢ ൏
b. From greatest to least:
ૢ ૡ
Example 2 (4 minutes): Interpreting Data and Writing Inequality Statements
Example 2: Interpreting Data and Writing Inequality Statements Mary is comparing the rainfall totals for May, June, and July. The data is reflected in the table below. Fill in the blanks below to create inequality statements that compare the Changes in Total Rainfall for each month (the right-most column of the table). Month This Year’s Total Rainfall (in inches) Last Year’s Total Rainfall (in inches) Change in Total Rainfall from Last Year to This Year (in inches) May Ǥ Ǥ ૠ െǤ
June Ǥ ૡ Ǥ Ǥ
July Ǥ ૠ Ǥ Ǥ
Write one inequality to order the Changes in Total Rainfall: െǤ ൏ Ǥ ൏ Ǥ Ǥ Ǥ െǤ
From least to greatest From greatest to least In this case, does the greatest number indicate the greatest change in rainfall? Explain.
No. In this situation, the greatest change is for the month of May since the average total rainfall went down from last year by Ǥ inches, but the greatest number in the inequality statement is Ǥ .
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
Exercises 5–8 (8 minutes)
^ƚƵĚĞŶƚƐǁŽƌŬŝŶĚĞƉĞŶĚĞŶƚůLJƚŽĂŶƐǁĞƌƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐ͘ůůŽǁƚŝŵĞĨŽƌƐƚƵĚĞŶƚƐƚŽƉƌĞƐĞŶƚƚŚĞŝƌĂŶƐǁĞƌƐĂŶĚ ƐŚĂƌĞƚŚĞŝƌƚŚŽƵŐŚƚƉƌŽĐĞƐƐĞƐƚŽƚŚĞĐůĂƐƐ͘hƐĞƚŚĞĨŽůůŽǁŝŶŐĂƐĂŶŽƉƚŝŽŶĂůƚĂƐŬ͗,ĂǀĞƐƚƵĚĞŶƚƐƚƌĂŶƐĨĞƌƚŚĞŝƌǁŽƌĚ ƉƌŽďůĞŵƐĨŽƌ džĞƌĐŝƐĞϴŽŶƚŽƉĂƉĞƌƵƐŝŶŐĐŽůŽƌĨƵůŵĂƌŬĞƌƐŽƌĐŽůŽƌĞĚƉĞŶĐŝůƐ͕ĂŶĚĚŝƐƉůĂLJƚŚĞŵŝŶƚŚĞĐůĂƐƐƌŽŽŵ͘
Exercises 5–8 5. Mark’s favorite football team lost yards on two back-to-back plays. They lost yards on the first play. They lost
yard on the second play. Write an inequality statement using integers to compare the forward progress made on each play.
െ ൏ െ
Sierra had to pay the school for two textbooks that she lost. One textbook cost ̈́ , and the other cost ̈́ ૠ . Her mother wrote two separate checks, one for each expense. Write two integers that represent the change to her mother’s checking account balance. Then, write an inequality statement that shows the relationship between these two numbers.
െ and െૠ ;െ െૠ
Jason ordered the numbers െૠ , െૡ , and െૡǤ from least to greatest by writing the following statement:
െૡ ൏ െૡǤ ൏ െૠǤ
Is this a true statement? Explain.
No, it is not a true statement because ૡ ൏ ૡǤ ൏ ૠ , so the opposites of these numbers are in the opposite order. The order should be െૠ ൏ െૡǤ ൏ ૡ .
Write a real-world situation that is represented by the following inequality: െૢ ൏ . Explain the position of the numbers on a number line.
The coldest temperature in January was െૢ degrees Fahrenheit, and the warmest temperature was degrees Fahrenheit. Since the point associated with is above zero on a vertical number line and െૢ is below zero, we know that is greater than െૢ . This means that degrees Fahrenheit is warmer than െૢ degrees Fahrenheit.
Sprint (5 minutes): Writing Inequalities
WŚŽƚŽĐŽƉLJƚŚĞĂƚƚĂĐŚĞĚƚǁŽͲƉĂŐĞ^ƉƌŝŶƚƐƐŽƚŚĂƚĞĂĐŚƐƚƵĚĞŶƚƌĞĐĞŝǀĞƐĂĐŽƉLJ͘dŝŵĞƐƚƵĚĞŶƚƐ͕ĂůůŽǁŝŶŐŽŶĞ ŵŝŶƵƚĞƚŽ ĐŽŵƉůĞƚĞ^ŝĚĞ͘ĞĨŽƌĞƐƚƵĚĞŶƚƐďĞŐŝŶ͕ŝŶĨŽƌŵƚŚĞŵƚŚĂƚƚŚĞLJŵĂLJŶŽƚƐŬŝƉŽǀĞƌƋƵĞƐƚŝŽŶƐĂŶĚƚŚĂƚƚŚĞLJŵƵƐƚŵŽǀĞŝŶ ŽƌĚĞƌ . ĨƚĞƌŽŶĞŵŝŶƵƚĞ͕ĚŝƐĐƵƐƐƚŚĞ ĂŶƐǁĞƌƐ͘ĞĨŽƌĞĂĚŵŝŶŝƐƚĞƌŝŶŐ^ŝĚĞ͕ĞůŝĐŝƚƐƚƌĂƚĞŐŝĞƐĨƌŽŵƚŚŽƐĞƐƚƵĚĞŶƚƐǁŚŽ ǁĞƌĞĂďůĞƚŽĂĐĐƵƌĂƚĞůLJĐŽŵƉůĞƚĞŵĂŶLJƉƌŽďůĞŵƐŽŶ^ŝĚĞ͘ĚŵŝŶŝƐƚĞƌ^ŝĚĞŝŶƚŚĞƐĂŵĞĨĂƐŚŝŽŶ͕ĂŶĚƌĞǀŝĞǁƚŚĞ ĂŶƐǁĞƌƐ͘ZĞĨĞƌƚŽƚŚĞ^ƉƌŝŶƚƐĂŶĚ^ƉƌŝŶƚĞůŝǀĞƌLJ^ĐƌŝƉƚƐĞĐƚŝŽŶƐŝŶƚŚĞDŽĚƵůĞKǀĞƌǀŝĞǁĨŽƌĚŝƌĞĐƚŝŽŶƐƚŽĂĚŵŝŶŝƐƚĞƌĂ ^ƉƌŝŶƚ͘
Exercise 9 (4 minutes): A Closer Look at the Sprint
^ƚƵĚĞŶƚƐĂƌĞĂƐŬĞĚƚŽůŽŽŬĐůŽƐĞůLJĂƚƚǁŽƌĞůĂƚĞĚĞdžĂŵƉůĞƐĨƌŽŵƚŚĞ^ƉƌŝŶƚĂŶĚĞdžƉůĂŝŶƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞ ŶƵŵďĞƌƐ͛ŽƌĚĞƌ͕ƚŚĞŝŶĞƋƵĂůŝƚLJƐLJŵďŽůƐ͕ĂŶĚƚŚĞŐƌĂƉŚƐŽĨƚŚĞŶƵŵďĞƌƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
Exercise 9: A Closer Look at the Sprint 9. Look at the following two examples from the Sprint.
൏ ൏
െΥǡ െǡ
െΥǡ െǡ
a. Fill in the numbers in the correct order.
െ ൏ െ ൏ and െ െ
b. Explain how the position of the numbers on the number line supports the inequality statements you created.
െ is the farthest left on the number line, so it is the least value. is farthest right, so it is the greatest value, and െ is in between.
c. Create a new pair of greater than and less than inequality statements using three other rational numbers.
Answers will vary. ૡ Ǥ െǤ ૡ and െǤ ૡ ൏ Ǥ ൏ ૡ
Closing ;ϯ minutes)
tŚĂƚĐĂŶLJŽƵĚŽďĞĨŽƌĞǁƌŝƚŝŶŐĂŶŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚŝŶǀŽůǀŝŶŐƚŚƌĞĞŶƵŵďĞƌƐƚŚĂƚŵĂŬĞƐŝƚĞĂƐŝĞƌƚŽǁƌŝƚĞ ƚŚĞŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚ͍&ŽƌĞdžĂŵƉůĞ͕ĞdžƉůĂŝŶƚŚĞƉƌŽĐĞƐƐĨŽƌǁƌŝƚŝŶŐŽŶĞŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚĐŽŵƉĂƌŝŶŐ
െ͵ ͕ ͺ ͕ ĂŶĚ െͳͲ ͘
à First, I would order the numbers, either from least to greatest or greatest to least.
/ĨLJŽƵŬŶŽǁƚŚĞŽƌĚĞƌŽĨĂƐĞƚŽĨŶƵŵďĞƌƐ͕ŚŽǁĐĂŶLJŽƵƌĞƉƌĞƐĞŶƚƚŚĞŽƌĚĞƌƵƐŝŶŐŝŶĞƋƵĂůŝƚLJƐLJŵďŽůƐ͍
à For example, if the numbers are ͳ, ʹ, and ͵, you can either write ͳ ൏ ʹ ൏ ͵ or ͵ ʹ ͳ .
/ĨƚǁŽŶĞŐĂƚŝǀĞŶƵŵďĞƌƐĂƌĞŽƌĚĞƌĞĚƵƐŝŶŐƚŚĞ ൏ƐLJŵďŽů͕ǁŚĂƚŵƵƐƚďĞƚƌƵĞĂďŽƵƚƚŚĞŝƌƉŽƐŝƚŝŽŶƐŽŶĂ ŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͍KŶĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͍
à The first number must be associated with a point to the left of the second number on a horizontal number line. The first number must be associated with a point below the second number on a vertical number line.
Exit Ticket (5 minutes)
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
EĂŵĞ ĂƚĞ
Lesson 10: Writing and Interpreting Inequality Statements Involving Rational Numbers Exit Ticket
<ĞŶĚƌĂĐŽůůĞĐƚĞĚĚĂƚĂĨŽƌŚĞƌƐĐŝĞŶĐĞƉƌŽũĞĐƚ͘^ŚĞƐƵƌǀĞLJĞĚƉĞŽƉůĞĂƐŬŝŶŐƚŚĞŵŚŽǁŵĂŶLJŚŽƵƌƐƚŚĞLJƐůĞĞƉĚƵƌŝŶŐĂ ƚLJƉŝĐĂůŶŝŐŚƚ͘dŚĞĐŚĂƌƚďĞůŽǁƐŚŽǁƐŚŽǁĞĂĐŚƉĞƌƐŽŶ͛ƐƌĞƐƉŽŶƐĞĐŽŵƉĂƌĞƐƚŽ ͺŚŽƵƌƐ;ǁŚŝĐŚŝƐƚŚĞĂŶƐǁĞƌƐŚĞ ĞdžƉĞĐƚĞĚŵŽƐƚƉĞŽƉůĞƚŽƐĂLJͿ͘
Name Number of Hours (usually slept each night) Compared to ૡ Hours
&ƌĂŶŬŝĞ ͺǤͷ ͲǤͷ
Dƌ͘&ŝĞůĚƐ െͳǤͲ
<ĂƌůĂ ͻǤͷ ͳǤͷ
ŽƵŝƐ ͺ Ͳ
dŝĨĨĂŶLJ ͵Ͷ െ ͳͶĂ͘ WůŽƚĂŶĚůĂďĞůĞĂĐŚŽĨƚŚĞŶƵŵďĞƌƐŝŶƚŚĞƌŝŐŚƚͲŵŽƐƚĐŽůƵŵŶŽĨƚŚĞƚĂďůĞĂďŽǀĞŽŶƚŚĞŶƵŵďĞƌůŝŶĞďĞůŽǁ͘ ď͘ >ŝƐƚƚŚĞŶƵŵďĞƌƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͘ Đ͘ hƐŝŶŐLJŽƵƌĂŶƐǁĞƌĨƌŽŵƉĂƌƚ;ďͿĂŶĚŝŶĞƋƵĂůŝƚLJƐLJŵďŽůƐ͕ǁƌŝƚĞŽŶĞƐƚĂƚĞŵĞŶƚƚŚĂƚƐŚŽǁƐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉ ĂŵŽŶŐĂůůŽĨƚŚĞŶƵŵďĞƌƐ͘
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
Exit Ticket Sample Solutions Kendra collected data for her science project. She surveyed people asking them how many hours they sleep during a typical night. The chart below shows how each person’s response compares to ૡ hours (which is the answer she expected most people to say). Name Number of Hours (usually slept each night) Compared to ૡ Hours Frankie ૡǤ Ǥ
Mr. Fields ૠ െǤ
Karla ૢ Ǥ Ǥ
Louis ૡ
Tiffany ૠ െ
a. Plot and label each of the numbers in the right-most column of the table above on the number line below.
b. List the numbers from least to greatest.
െǤ , െ , ,Ǥ ,Ǥ
c. Using your answer from part (b) and inequality symbols, write one statement that shows the relationship among all the numbers.
െǤ ൏ െ ൏ ൏ Ǥ ൏ Ǥ or Ǥ Ǥ െ െǤ
Problem Set Sample Solutions For each of the relationships described below, write an inequality that relates the rational numbers. 1. Seven feet below sea level is farther below sea level than feet below sea level.
െૠ ൏ െ
Sixteen degrees Celsius is warmer than zero degrees Celsius.
ϯ͘ Three and one-half yards of fabric is less than five and one-half yards of fabric.
൏
Ǥ Ǥ െǤ െǤ ૠ െǤ െǤ െ Ǥ Ǥ Ǥ ૠ Ǥ
െ
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
A loss of ̈́ in the stock market is worse than a gain of ̈́ in the stock market.
െ ൏
A test score of is worse than a test score of , and a test score of is worse than a test score of ૠ .
൏ ൏ ૠ
In December, the total snowfall was Ǥ inches, which is more than the total snowfall in October and November, which was Ǥ ૠ inches and Ǥ inches, respectively.
Ǥ Ǥ Ǥ ૠ
For each of the following, use the information given by the inequality to describe the relative position of the numbers on a horizontal number line. 7. െǤ ൏ െǤ
െǤ is to the left of െǤ , or െǤ is to the right of െǤ Ǥ
ૡ െૡ
ૡ is to the right of െૡ ǡ or െૡ is to the left of ૡ .
െ ൏ ൏
െ is to the left of zero and zero is to the left of , or is to the right of zero and zero is to the right of െ .
െૢૢ െ
െૢૢ is to the right of െ , or െ is to the left of െૢૢ .
െૠǤ ൏ െૠ ൏ െૠ
െૠǤ is to the left of െૠ and െૠ is to the left of െૠ ,or െૠ is to the right of െૠ and െૠ is to the right of
െૠǤ .
Fill in the blanks with numbers that correctly complete each of the statements. 12. Three integers between െ and െ ൏ െ ൏ െ
ϭϯ͘ Three rational numbers between and Ǥ ൏ Ǥ ൏ Ǥ ૠ Other answers are possible.
Three rational numbers between െ and െ െǤૢ ൏ െǤ ൏ െǤ Other answers are possible.
Three integers between and Ȃ െ ൏ ൏
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
Rational Numbers: Inequality Statements—Round 1 Directions: tŽƌŬŝŶŶƵŵĞƌŝĐĂůŽƌĚĞƌƚŽĂŶƐǁĞƌWƌŽďůĞŵƐϭʹϯϯ͘ƌƌĂŶŐĞĞĂĐŚƐĞƚŽĨŶƵŵďĞƌƐŝŶŽƌĚĞƌĂĐĐŽƌĚŝŶŐƚŽƚŚĞ ŝŶĞƋƵĂůŝƚLJƐLJŵďŽůƐ͘
ϭ͘ фф
ͳǡ െͳǡ Ͳ ϭϮ͘ хх
ǡ െǡ Ϯϯ͘ хх
ʹͷǡ Χǡ െΧ Ϯ͘ хх
ͳǡ െͳǡ Ͳ ϭϯ͘ хх
ͳǡ Ͷǡ ͳ Ϯϰ͘ фф
ʹͷǡ Χǡ െΧ ϯ͘ фф
͵Φǡ െ͵Φǡ Ͳ ϭϰ͘ фф
ͳǡ Ͷǡ ͳ Ϯϱ͘ хх
ʹǤʹǡ ʹǤ͵ǡ ʹǤͶ ϰ͘ хх
͵Φǡ െ͵Φǡ Ͳ ϭϱ͘ фф
Ͳǡ ͳʹǡ െͳͳ Ϯϲ͘ хх
ͳǤʹǡ ͳǤ͵ǡ ͳǤͶ ϱ͘ хх
ͳǡ െΦǡ Φ ϭϲ͘ хх
Ͳǡ ͳʹǡ െͳͳ Ϯϳ͘ хх
ͲǤʹǡ ͲǤ͵ǡ ͲǤͶ ϲ͘ фф
ͳǡ െΦǡ Φ ϭϳ͘ хх
ͳǡ Υǡ Φ Ϯϴ͘ хх
െͲǤͷǡ െͳǡ െͲǤ ϳ͘ фф
െ͵ǡ െͶǡ െͷ ϭϴ͘ фф
ͳǡ Υǡ Φ Ϯϵ͘ фф
െͲǤͷǡ െͳǡ െͲǤ ϴ͘ фф െͳ͵ǡ െͳͶǡ െͳͷ ϭϵ͘ фф
െΦǡ Φǡ Ͳ ϯϬ͘ фф
െͺǡ െͻǡ ͺ ϵ͘ хх
െͳ͵ǡ െͳͶǡ െͳͷ ϮϬ͘ хх
െΦǡ Φǡ Ͳ ϯϭ͘ фф
െͳͺǡ െͳͻǡ െʹ ϭϬ͘ фф
െΥǡ െͳǡ Ͳ Ϯϭ͘ фф
ͷͲǡ െͳͲǡ Ͳ ϯϮ͘ хх
െʹǡ െ͵ǡ ͳ ϭϭ͘ хх
െΥǡ െͳǡ Ͳ ϮϮ͘ фф
െͷͲǡ ͳͲǡ Ͳ ϯϯ͘ фф
െʹǡ െ͵ǡ ͳ
EƵŵďĞƌŽƌƌĞĐƚ͗ͺͺͺͺͺͺ
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
Rational Numbers: Inequality Statements—Round 1 [KEY] Directions: tŽƌŬŝŶŶƵŵĞƌŝĐĂůŽƌĚĞƌƚŽĂŶƐǁĞƌWƌŽďůĞŵƐϭʹϯϯ͘ƌƌĂŶŐĞĞĂĐŚƐĞƚŽĨŶƵŵďĞƌƐŝŶŽƌĚĞƌĂĐĐŽƌĚŝŶŐƚŽƚŚĞ ŝŶĞƋƵĂůŝƚLJƐLJŵďŽůƐ͘
ϭ͘ фф
ͳǡ െͳǡ Ͳ ϭϮ͘ хх
ǡ െǡ Ϯϯ͘ хх
ʹͷǡ Χǡ െΧ Ϯ͘ хх
ͳǡ െͳǡ Ͳ ϭϯ͘ хх
ͳǡ Ͷǡ ͳ Ϯϰ͘ фф
ʹͷǡ Χǡ െΧ ϯ͘ фф
͵Φǡ െ͵Φǡ Ͳ ϭϰ͘ фф
ͳǡ Ͷǡ ͳ Ϯϱ͘ хх
ʹǤʹǡ ʹǤ͵ǡ ʹǤͶ ϰ͘ хх
͵Φǡ െ͵Φǡ Ͳ ϭϱ͘ фф
Ͳǡ ͳʹǡ െͳͳ Ϯϲ͘ хх
ͳǤʹǡ ͳǤ͵ǡ ͳǤͶ ϱ͘ хх
ͳǡ െΦǡ Φ ϭϲ͘ хх
Ͳǡ ͳʹǡ െͳͳ Ϯϳ͘ хх
ͲǤʹǡ ͲǤ͵ǡ ͲǤͶ ϲ͘ фф
ͳǡ െΦǡ Φ ϭϳ͘ хх
ͳǡ Υǡ Φ Ϯϴ͘ хх
െͲǤͷǡ െͳǡ െͲǤ ϳ͘ фф
െ͵ǡ െͶǡ െͷ ϭϴ͘ фф
ͳǡ Υǡ Φ Ϯϵ͘ фф
െͲǤͷǡ െͳǡ െͲǤ ϴ͘ фф െͳ͵ǡ െͳͶǡ െͳͷ ϭϵ͘ фф
െΦǡ Φǡ Ͳ ϯϬ͘ фф
െͺǡ െͻǡ ͺ ϵ͘ хх
െͳ͵ǡ െͳͶǡ െͳͷ ϮϬ͘ хх
െΦǡ Φǡ Ͳ ϯϭ͘ фф
െͳͺǡ െͳͻǡ െʹ ϭϬ͘ фф
െΥǡ െͳǡ Ͳ Ϯϭ͘ фф
ͷͲǡ െͳͲǡ Ͳ ϯϮ͘ хх
െʹǡ െ͵ǡ ͳ ϭϭ͘ хх
െΥǡ െͳǡ Ͳ ϮϮ͘ фф
െͷͲǡ ͳͲǡ Ͳ ϯϯ͘ фф
െʹǡ െ͵ǡ ͳ
െ
ૠ െ Χ െΧ
െ ૠ െΧ
Χ
െΦ
Φ ૠ Ǥ Ǥ Ǥ
Φ
െΦ െ Ǥ Ǥ Ǥ
Φ െΦ
െ Ǥ Ǥ Ǥ
െΦ
Φ Φ Υ െǤ െǤ െ
െ
െ െ Υ Φ െ െǤ െǤ
െ
െ െ െΦ Φ െૢ െૡ ૡ
െ
െ െ Φ െΦ െૢ െૡ െ
െ
െΥ െ െ െ
െΥ െ െ െ െ
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
Rational Numbers: Inequality Statements—Round 2 Directions: tŽƌŬŝŶŶƵŵĞƌŝĐĂůŽƌĚĞƌƚŽĂŶƐǁĞƌWƌŽďůĞŵƐϭʹϯϯ͘ƌƌĂŶŐĞĞĂĐŚƐĞƚŽĨŶƵŵďĞƌƐŝŶŽƌĚĞƌĂĐĐŽƌĚŝŶŐƚŽƚŚĞ ŝŶĞƋƵĂůŝƚLJƐLJŵďŽůƐ͘
ϭ͘ фф
ͳȀǡ െͳȀǡ Ͳ ϭϮ͘ хх
ͳΥǡ ͳǡ ͳΦ Ϯϯ͘ хх ͳǡ ͳΧǡ െͳΧ Ϯ͘ хх
ͳȀǡ െͳȀǡ Ͳ ϭϯ͘ хх
ͳͳΥǡ ͳͳǡ ͳͳΦ Ϯϰ͘ фф ͳǡ ͳΧǡ െͳΧ ϯ͘ фф ͵Ȁǡ ʹȀǡ െͳȀ ϭϰ͘ фф
ͳͳΥǡ ͳͳǡ ͳͳΦ Ϯϱ͘ хх
െͺʹǡ െͻ͵ǡ െͳͲͶ ϰ͘ хх
͵Ȁǡ ʹȀǡ െͳȀ ϭϱ͘ фф
Ͳǡ ͲǤʹǡ െͲǤͳ Ϯϲ͘ фф
െͺʹǡ െͻ͵ǡ െͳͲͶ ϱ͘ хх
െͶȀͷǡ ͳȀͷǡ െͳȀͷ ϭϲ͘ хх
Ͳǡ ͲǤʹǡ െͲǤͳ Ϯϳ͘ хх
ͲǤͷǡ ͳǡ ͲǤ ϲ͘ фф
െͶȀͷǡ ͳȀͷǡ െͳȀͷ ϭϳ͘ хх
ͳǡ ͲǤǡ ͳȀͳͲ Ϯϴ͘ хх
െͲǤͷǡ െͳǡ െͲǤ ϳ͘ фф
െͺȀͻǡ ͷȀͻǡ ͳȀͻ ϭϴ͘ фф
ͳǡ ͲǤǡ ͳȀͳͲ Ϯϵ͘ фф
െͲǤͷǡ െͳǡ െͲǤ ϴ͘ хх
െͺȀͻǡ ͷȀͻǡ ͳȀͻ ϭϵ͘ фф
Ͳǡ െͳʹǡ െͳʹΦ ϯϬ͘ фф
ͳǡ ͺǡ ͻ ϵ͘ хх
െ͵Ͳǡ െͳͲǡ െͷͲ ϮϬ͘ хх
Ͳǡ െͳʹǡ െͳʹΦ ϯϭ͘ фф
െͳǡ െͺǡ െͻ ϭϬ͘ фф
െ͵Ͳǡ െͳͲǡ െͷͲ Ϯϭ͘ фф
ͷǡ െͳǡ Ͳ ϯϮ͘ хх
െʹǡ െ͵ǡ െͷ ϭϭ͘ хх
െͶͲǡ െʹͲǡ െͲ ϮϮ͘ фф
െͷǡ ͳǡ Ͳ ϯϯ͘ хх
ʹǡ ͵ǡ ͷ
EƵŵďĞƌŽƌƌĞĐƚ͗ͺͺͺͺͺͺ /ŵƉƌŽǀĞŵĞŶƚ͗ͺͺͺͺͺͺ
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6ͻϯ Lesson 10 Lesson 10: tƌŝƚŝŶŐĂŶĚ/ŶƚĞƌƉƌĞƚŝŶŐ/ŶĞƋƵĂůŝƚLJ^ƚĂƚĞŵĞŶƚƐ/ŶǀŽůǀŝŶŐZĂƚŝŽŶĂů EƵŵďĞƌƐ
Rational Numbers: Inequality Statements—Round 2 [KEY] Directions: tŽƌŬŝŶŶƵŵĞƌŝĐĂůŽƌĚĞƌƚŽĂŶƐǁĞƌWƌŽďůĞŵƐϭʹϯϯ͘ƌƌĂŶŐĞĞĂĐŚƐĞƚŽĨŶƵŵďĞƌƐŝŶŽƌĚĞƌĂĐĐŽƌĚŝŶŐƚŽƚŚĞ ŝŶĞƋƵĂůŝƚLJƐLJŵďŽůƐ͘
ϭ͘ фф
ͳȀǡ െͳȀǡ Ͳ ϭϮ͘ хх
ͳΥǡ ͳǡ ͳΦ Ϯϯ͘ хх ͳǡ ͳΧǡ െͳΧ Ϯ͘ хх
ͳȀǡ െͳȀǡ Ͳ ϭϯ͘ хх
ͳͳΥǡ ͳͳǡ ͳͳΦ Ϯϰ͘ фф ͳǡ ͳΧǡ െͳΧ ϯ͘ фф ͵Ȁǡ ʹȀǡ െͳȀ ϭϰ͘ фф
ͳͳΥǡ ͳͳǡ ͳͳΦ Ϯϱ͘ хх
െͺʹǡ െͻ͵ǡ െͳͲͶ ϰ͘ хх
͵Ȁǡ ʹȀǡ െͳȀ ϭϱ͘ фф
Ͳǡ ͲǤʹǡ െͲǤͳ Ϯϲ͘ фф
െͺʹǡ െͻ͵ǡ െͳͲͶ ϱ͘ хх
െͶȀͷǡ ͳȀͷǡ െͳȀͷ ϭϲ͘ хх
Ͳǡ ͲǤʹǡ െͲǤͳ Ϯϳ͘ хх
ͲǤͷǡ ͳǡ ͲǤ ϲ͘ фф
െͶȀͷǡ ͳȀͷǡ െͳȀͷ ϭϳ͘ хх
ͳǡ ͲǤǡ ͳȀͳͲ Ϯϴ͘ хх
െͲǤͷǡ െͳǡ െͲǤ ϳ͘ фф
െͺȀͻǡ ͷȀͻǡ ͳȀͻ ϭϴ͘ фф
ͳǡ ͲǤǡ ͳȀͳͲ Ϯϵ͘ фф
െͲǤͷǡ െͳǡ െͲǤ ϴ͘ хх
െͺȀͻǡ ͷȀͻǡ ͳȀͻ ϭϵ͘ фф
Ͳǡ െͳʹǡ െͳʹΦ ϯϬ͘ фф
ͳǡ ͺǡ ͻ ϵ͘ хх
െ͵Ͳǡ െͳͲǡ െͷͲ ϮϬ͘ хх
Ͳǡ െͳʹǡ െͳʹΦ ϯϭ͘ фф
െͳǡ െͺǡ െͻ ϭϬ͘ фф
െ͵Ͳǡ െͳͲǡ െͷͲ Ϯϭ͘ фф
ͷǡ െͳǡ Ͳ ϯϮ͘ хх
െʹǡ െ͵ǡ െͷ ϭϭ͘ хх
െͶͲǡ െʹͲǡ െͲ ϮϮ͘ фф
െͷǡ ͳǡ Ͳ ϯϯ͘ хх
ʹǡ ͵ǡ ͷ
െ ૠ
ૠ Φ Υ Χ െΧ
ૠ
െ ૠ Φ Υ െΧ Χ
െ ૠૠૠ
Υ Φ െૡ െૢ െ
ૠૠെ ૠ
െǤ Ǥ െ െૢ െૡ
െ െ
Ǥ െǤ Ǥ Ǥ
െ െ
Ǥ ૠ െǤ െǤ െ
െ ૡૢ ૢ ૢ
Ǥ ૠ െ െǤ െǤ
ૢ ૢ െ ૡૢ
െΦ െ ૡ ૢ
െ െ െ െ െΦ െૢ െૡ െ
െ െ െ െ െ െ െ
െ െ െ െ
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6ͻϯ Lesson 11 Lesson 11: ďƐŽůƵƚĞsĂůƵĞͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ
Lesson 11: Absolute Value—Magnitude and Distance Student Outcomes
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨĂŶƵŵďĞƌĂƐŝƚƐĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘
^ƚƵĚĞŶƚƐƵƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽĨŝŶĚƚŚĞŵĂŐŶŝƚƵĚĞŽĨĂƉŽƐŝƚŝǀĞŽƌŶĞŐĂƚŝǀĞƋƵĂŶƚŝƚLJŝŶĂƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶ͘
Classwork Opening Exercise (4 minutes)
&ŽƌƚŚŝƐǁĂƌŵͲƵƉĞdžĞƌĐŝƐĞ͕ƐƚƵĚĞŶƚƐǁŽƌŬŝŶĚŝǀŝĚƵĂůůLJƚŽƌĞĐŽƌĚƚǁŽĚŝĨĨĞƌĞŶƚƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŚĂƚĂƌĞƚŚĞƐĂŵĞ ĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽ͘^ƚƵĚĞŶƚƐĨŝŶĚĂƐŵĂŶLJĞdžĂŵƉůĞƐĂƐƉŽƐƐŝďůĞĂŶĚƌĞĂĐŚĂĐŽŶĐůƵƐŝŽŶĂďŽƵƚǁŚĂƚŵƵƐƚďĞƚƌƵĞĨŽƌ ĞǀĞƌLJƉĂŝƌŽĨŶƵŵďĞƌƐƚŚĂƚůŝĞƚŚĂƚƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽ͘
Opening Exercise
ĨƚĞƌƚǁŽŵŝŶƵƚĞƐ͗
tŚĂƚĂƌĞƐŽŵĞĞdžĂŵƉůĞƐLJŽƵĨŽƵŶĚ;ƉĂŝƌƐŽĨŶƵŵďĞƌƐƚŚĂƚĂƌĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽͿ͍
à െ ͳʹ and ଵଶ , ͺǤͲͳ and െͺǤͲͳ ,െ and .
tŚĂƚŝƐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶĞĂĐŚƉĂŝƌŽĨŶƵŵďĞƌƐ͍
à They are opposites.
,ŽǁĚŽĞƐĞĂĐŚƉĂŝƌŽĨŶƵŵďĞƌƐƌĞůĂƚĞƚŽnjĞƌŽ͍
à Both numbers in each pair are the same distance from zero.
Discussion ( ϯ minutes)
tĞũƵƐƚƐĂǁƚŚĂƚĞǀĞƌLJŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞĂƌĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵnjĞƌŽ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘dŚĞ absolute value ŽĨĂŶƵŵďĞƌŝƐƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƚŚĞ ŶƵŵďĞƌĂŶĚnjĞƌŽŽŶĂŶƵŵďĞƌůŝŶĞ͘
/ŶŽƚŚĞƌǁŽƌĚƐ͕ĂŶƵŵďĞƌĂŶĚŝƚƐŽƉƉŽƐŝƚĞŚĂǀĞƚŚĞƐĂŵĞ absolute value ͘
tŚĂƚŝƐƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨ ͷ͍ džƉůĂŝŶ͘
à The absolute value of ͷ is ͷ because it is ͷ units from zero.
tŚĂƚŝƐƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨ െͷ ͍
à The absolute value of െͷ is also ͷ because it is also ͷ units from zero.
ŽƚŚ ͷĂŶĚ െͷ ĂƌĞĨŝǀĞƵŶŝƚƐĨƌŽŵnjĞƌŽ͕ǁŚŝĐŚŵĂŬĞƐ ͷĂŶĚ െͷ ŽƉƉŽƐŝƚĞƐ͘
െ െ െ െ െ െ െૠ െૢ ૡૠ െ ૢ െૡ
Scaffolding:
WƌŽǀŝĚĞƐƚƵĚĞŶƚƐǁŝƚŚĂ ŶƵŵďĞƌůŝŶĞƐŽƚŚĞLJĐĂŶ ƉŚLJƐŝĐĂůůLJĐŽƵŶƚƚŚĞŶƵŵďĞƌŽĨ ƵŶŝƚƐďĞƚǁĞĞŶĂŶƵŵďĞƌĂŶĚ njĞƌŽ͘
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6ͻϯ Lesson 11 Lesson 11: ďƐŽůƵƚĞsĂůƵĞͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ
tŚĂƚŝƐƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨ െͳ ͍
à ͳ
tŚĂƚŽƚŚĞƌŶƵŵďĞƌŚĂƐĂŶĂďƐŽůƵƚĞǀĂůƵĞŽĨ ͳ͍ džƉůĂŝŶ͘
à ͳ also has an absolute value of ͳ because ͳ and െͳ are opposites, so they have the same absolute value.
tŚĂƚŝƐƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨ Ͳ͍
à Ͳ
Example 1 ;ϯ minutes): The Absolute Value of a Number Example 1: The Absolute Value of a Number The absolute value of ten is written as ȁȁ . On the number line, count the number of units from to . How many units is from ?
ȁȁ ൌ
What other number has an absolute value of ? Why?
ȁെȁ ൌ because െ is units from zero and െ and are opposites.
Exercises 1– ϯ (4 minutes) Exercises 1– ϯ
Complete the following chart. Number Absolute Value Number Line Diagram Different Number with the Same Absolute Value 1. െ ȁെȁ ൌ
ૡ ȁૡȁ ൌ ૡ െૡ
ϯ͘ െ ȁെȁ ൌ
െ െ െ െ െ െ െૠ െૢ ૡૠ െ ૢെૡ
െ െ െ െ െ െ െૠ െૢ ૡૠ െ ૢെૡ
െ െ െ െ െ െ െૠ െૢ ૡૠ െ ૢെૡ
െ െ െ െ െ െ െૠ െૢ ૡૠ െ ૢ െૡ
ૡૠ ૢ
The absolute value of a number is the distance between the number and zero on the number line.
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6ͻϯ Lesson 11 Lesson 11: ďƐŽůƵƚĞsĂůƵĞͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ
Example 2 ;ϯ minutes): Using Absolute Value to Find Magnitude Example 2: Using Absolute Value to Find Magnitude Mrs. Owens received a call from her bank because she had a checkbook balance of െ̈́ . What was the magnitude of the amount overdrawn?
ȁെȁ ൌ Mrs. Owens overdrew her checking account by ̈́ .
Exercises 4–8 (6 minutes) Exercises 4–8 For each scenario below, use absolute value to determine the magnitude of each quantity. 4. Maria was sick with the flu, and her weight change as a result of it is represented by െ pounds. How much weight did Maria lose?
ȁെȁ ൌ Maria lost pounds.
Jeffrey owes his friend ̈́ . How much is Jeffrey’s debt?
ȁെȁ ൌ Jeffrey has a ̈́ debt.
The elevation of Niagara Falls, which is located between Lake Erie and Lake Ontario, is feet. How far is this above sea level?
ȁȁ ൌ It is feet above sea level.
How far below zero is െ degrees Celsius?
ȁെȁ ൌ െι۱ is degrees below zero.
Frank received a monthly statement for his college savings account. It listed a deposit of ̈́ as Ǥ . It listed a withdrawal of ̈́ as െǤ . The statement showed an overall ending balance of ̈́ ૡǤ . How much money did Frank add to his account that month? How much did he take out? What is the total amount Frank has saved for college?
ȁȁ ൌ Frank added ̈́ to his account.
ȁെȁ ൌ Frank took ̈́ out of his account.
ȁૡǤ ȁ ൌ ૡǤ The total amount of Frank’s savings for college is ̈́ ૡǤ .
The magnitude of a measurement is the absolute value of its measure.
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6ͻϯ Lesson 11 Lesson 11: ďƐŽůƵƚĞsĂůƵĞͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ
Exercises 9–19 ( ϭϯ minutes)
^ƚƵĚĞŶƚƐǁŽƌŬŝŶĚĞƉĞŶĚĞŶƚůLJĨŽƌϴʹϭϬŵŝŶƵƚĞƐ͘ůůŽǁϯʹϱŵŝŶƵƚĞƐƚŽŐŽŽǀĞƌƚŚĞĂŶƐǁĞƌƐĂƐĂǁŚŽůĞŐƌŽƵƉ͘
Exercises 9–19 9. Meg is playing a card game with her friend, Iona. The cards have positive and negative numbers printed on them. Meg exclaims: “The absolute value of the number on my card equals ૡ.” What is the number on Meg’s card?
ȁെૡȁ ൌ ૡ or ȁૡȁ ൌ ૡ
Meg either has ૡ or െૡ on her card.
List a positive and negative number whose absolute value is greater than . Justify your answer using the number line.
Answers may vary. ȁെȁ ൌ and ȁૠȁ ൌ ૠ ; and ૠ . On a number line, the distance from zero to െ is
units. So, the absolute value of െ is . The number is to the right of on the number line, so is greater than .The distance from zero to ૠ on a number line is ૠ units, so the absolute value of ૠ is ૠ. Since ૠ is to the right of on the number line, ૠ is greater than .
Which of the following situations can be represented by the absolute value of ? Check all that apply. The temperature is degrees below zero. Express this as an integer. X Determine the size of Harold’s debt if he owes ̈́ .X Determine how far െ is from zero on a number line. X degrees is how many degrees above zero? 12. Julia used absolute value to find the distance between and on a number line. She then wrote a similar statement to represent the distance between and െ . Below is her work. Is it correct? Explain.
ȁȁ ൌ and ȁെȁ ൌ െ
No. The distance is units whether you go from to or to െ . So, the absolute value of െ should also be ,but Julia said it was െ .
ϭϯ͘ Use absolute value to represent the amount, in dollars, of a ̈́ ૡǤ profit.
ȁૡǤ ȁ ൌ ૡǤ
Judy lost pounds. Use absolute value to represent the number of pounds Judy lost.
ȁെȁ ൌ
In math class, Carl and Angela are debating about integers and absolute value. Carl said two integers can have the same absolute value, and Angela said one integer can have two absolute values. Who is right? Defend your answer.
Carl is right. An integer and its opposite are the same distance from zero. So, they have the same absolute values because absolute value is the distance between the number and zero.
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6ͻϯ Lesson 11 Lesson 11: ďƐŽůƵƚĞsĂůƵĞͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ
Jamie told his math teacher: “Give me any absolute value, and I can tell you two numbers that have that absolute value.” Is Jamie correct? For any given absolute value, will there always be two numbers that have that absolute value?
No, Jamie is not correct because zero is its own opposite. Only one number has an absolute value of , and that would be .
Use a number line to show why a number and its opposite have the same absolute value.
A number and its opposite are the same distance from zero but on opposite sides. An example is and െ . These numbers are both units from zero. Their distance is the same, so they have the same absolute value, .
A bank teller assisted two customers with transactions. One customer made a ̈́ withdrawal from a savings account. The other customer made a ̈́ deposit. Use absolute value to show the size of each transaction. Which transaction involved more money?
ȁെȁ ൌ and ȁȁ ൌ . The ̈́ withdrawal involved more money.
Which is farther from zero: െૠ or ૠ ? Use absolute value to defend your answer.
The number that is farther from is െૠ . This is because ቚെૠ ቚ ൌ ૠ and ቚૠ ቚ ൌ ૠ . Absolute value is a number’s distance from zero. I compared the absolute value of each number to determine which was farther from zero. The absolute value of െૠ is ૠ . The absolute value of ૠ is ૠ . We know that ૠ is greater than ૠ .Therefore, െૠ is farther from zero than ૠ .
Closing ;ϯ minutes)
/ĂŵƚŚŝŶŬŝŶŐŽĨƚǁŽŶƵŵďĞƌƐ͘ŽƚŚŶƵŵďĞƌƐŚĂǀĞƚŚĞƐĂŵĞĂďƐŽůƵƚĞǀĂůƵĞ͘tŚĂƚŵƵƐƚďĞƚƌƵĞĂďŽƵƚƚŚĞƚǁŽ ŶƵŵďĞƌƐ͍
à The numbers are opposites.
ĂŶƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨĂŶƵŵďĞƌĞǀĞƌďĞĂŶĞŐĂƚŝǀĞŶƵŵďĞƌ͍tŚLJŽƌǁŚLJŶŽƚ͍
à No. Absolute value is the distance a number is from zero. If you count the number of units from zero to the number, the number of units is its absolute value. You could be on the right or left side of zero, but the number of units you count represents the distance or absolute value, and that will always be a positive number.
,ŽǁĐĂŶǁĞƵƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽĚĞƚĞƌŵŝŶĞŵĂŐŶŝƚƵĚĞ͍&ŽƌŝŶƐƚĂŶĐĞ͕ŚŽǁĨĂƌďĞůŽǁnjĞƌŽŝƐ െͺ ĚĞŐƌĞĞƐ͍
à Absolute value represents magnitude. This means that െͺ degrees is ͺ units below zero.
Exit Ticket (6 minutes)
െ െ െ െ െ െ െૠ ૡૠെૡ
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6ͻϯ Lesson 11 Lesson 11: ďƐŽůƵƚĞsĂůƵĞͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ
EĂŵĞ ĂƚĞ
Lesson 11: Absolute Value—Magnitude and Distance Exit Ticket
:ĞƐƐŝĞĂŶĚŚŝƐĨĂŵŝůLJĚƌŽǀĞƵƉƚŽĂƉŝĐŶŝĐĂƌĞĂŽŶĂŵŽƵŶƚĂŝŶ͘/ŶƚŚĞŵŽƌŶŝŶŐ͕ƚŚĞLJĨŽůůŽǁĞĚĂƚƌĂŝůƚŚĂƚůĞĚƚŽƚŚĞ ŵŽƵŶƚĂŝŶƐƵŵŵŝƚ͕ǁŚŝĐŚǁĂƐ ʹǡͲͲͲ ĨĞĞƚĂďŽǀĞƚŚĞƉŝĐŶŝĐĂƌĞĂ͘dŚĞLJƚŚĞŶƌĞƚƵƌŶĞĚƚŽƚŚĞƉŝĐŶŝĐĂƌĞĂĨŽƌůƵŶĐŚ͘ĨƚĞƌ ůƵŶĐŚ͕ƚŚĞLJŚŝŬĞĚŽŶĂƚƌĂŝůƚŚĂƚůĞĚƚŽƚŚĞŵŽƵŶƚĂŝŶŽǀĞƌůŽŽŬ͕ǁŚŝĐŚǁĂƐ ͵ǡͷͲͲ ĨĞĞƚďĞůŽǁƚŚĞƉŝĐŶŝĐĂƌĞĂ͘ Ă͘ >ŽĐĂƚĞĂŶĚůĂďĞůƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞŵŽƵŶƚĂŝŶƐƵŵŵŝƚĂŶĚŵŽƵŶƚĂŝŶŽǀĞƌůŽŽŬŽŶĂǀĞƌƚŝĐĂů ŶƵŵďĞƌůŝŶĞ͘dŚĞƉŝĐŶŝĐĂƌĞĂƌĞƉƌĞƐĞŶƚƐnjĞƌŽ͘tƌŝƚĞĂƌĂƚŝŽŶĂůŶƵŵďĞƌƚŽƌĞƉƌĞƐĞŶƚĞĂĐŚ ůŽĐĂƚŝŽŶ͘ WŝĐŶŝĐĂƌĞĂ͗ Ͳ DŽƵŶƚĂŝŶƐƵŵŵŝƚ͗ DŽƵŶƚĂŝŶŽǀĞƌůŽŽŬ͗ ď͘ hƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽƌĞƉƌĞƐĞŶƚƚŚĞĚŝƐƚĂŶĐĞŽŶƚŚĞŶƵŵďĞƌůŝŶĞŽĨĞĂĐŚůŽĐĂƚŝŽŶĨƌŽŵƚŚĞ ƉŝĐŶŝĐĂƌĞĂ͘ ŝƐƚĂŶĐĞĨƌŽŵƚŚĞƉŝĐŶŝĐĂƌĞĂƚŽƚŚĞŵŽƵŶƚĂŝŶƐƵŵŵŝƚ͗ ŝƐƚĂŶĐĞĨƌŽŵƚŚĞƉŝĐŶŝĐĂƌĞĂƚŽƚŚĞŵŽƵŶƚĂŝŶŽǀĞƌůŽŽŬ͗ Đ͘ tŚĂƚŝƐƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƚŚĞĞůĞǀĂƚŝŽŶƐŽĨƚŚĞƐƵŵŵŝƚĂŶĚŽǀĞƌůŽŽŬ͍hƐĞĂďƐŽůƵƚĞǀĂůƵĞĂŶĚLJŽƵƌ ŶƵŵďĞƌůŝŶĞĨƌŽŵƉĂƌƚ;ĂͿƚŽĞdžƉůĂŝŶLJŽƵƌĂŶƐǁĞƌ͘
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6ͻϯ Lesson 11 Lesson 11: ďƐŽůƵƚĞsĂůƵĞͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ
Exit Ticket Sample Solutions Jessie and his family drove up to a picnic area on a mountain. In the morning, they followed a trail that led to the mountain summit, which was ǡ feet above the picnic area. They then returned to the picnic area for lunch. After lunch, they hiked on a trail that led to the mountain overlook, which was ǡ feet below the picnic area. a. Locate and label the elevation of the mountain summit and mountain overlook on a vertical number line. The picnic area represents zero. Write a rational number to represent each location. Picnic area:
Mountain summit: ǡ
Mountain overlook: െ
b. Use absolute value to represent the distance on the number line of each location from the picnic area. Distance from the picnic area to the mountain summit: ȁ ȁ ൌ
Distance from the picnic area to the mountain overlook: ȁെ ȁ ൌ
c. What is the distance between the elevations of the summit and overlook? Use absolute value and your number line from part (a) to explain your answer.
Summit to picnic area and picnic area to overlook: ǡ ǡ ൌ ǡ
There are ǡ units from zero to ǡ on the number line. There are ǡ units from zero to െǡ on the number line. Altogether, that equals ǡ units, which represents the distance on the number line between the two elevations. Therefore, the difference in elevations is ǡ feet.
Problem Set Sample Solutions For each of the following two quantities in Problems 1–4, which has the greater magnitude? (Use absolute value to defend your answers.) 1. dollars and െ dollars
ȁെȁ ൌ ȁȁ ൌ , so െ dollars has the greater magnitude.
െ feet and feet
ȁെȁ ൌ ȁȁ ൌ ൏ , so feet has the greater magnitude.
ǡ
ǡ
െǡ
െǡ
(Summit)
െǡ
(Overlook)
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6ͻϯ Lesson 11 Lesson 11: ďƐŽůƵƚĞsĂůƵĞͶDĂŐŶŝƚƵĚĞĂŶĚŝƐƚĂŶĐĞ
ϯ͘ െǤ pounds and െǤ ૡ pounds
ȁെǤ ȁ ൌ Ǥ ȁെǤ ૡȁ ൌ Ǥ ૡ Ǥ Ǥ ૡ , so െǤ pounds has the greater magnitude.
െ degrees and degrees
ቚെ ቚ ൌ ȁȁ ൌ , so െ degrees has the greater magnitude.
For Problems 5–7, answer true or false. If false, explain why. 5. The absolute value of a negative number will always be a positive number.
True
The absolute value of any number will always be a positive number.
False. Zero is the exception since the absolute value of zero is zero, and zero is not positive.
Positive numbers will always have a higher absolute value than negative numbers.
False. A number and its opposite have the same absolute value.
Write a word problem whose solution is ȁȁ ൌ .
Answers will vary. Kelli flew a kite feet above the ground. Determine the distance between the kite and the ground.
Write a word problem whose solution is ȁെૠȁ ൌ ૠ .
Answers will vary. Paul dug a hole in his yard ૠ inches deep to prepare for an in-ground swimming pool. Determine the distance between the ground and the bottom of the hole that Paul dug.
Look at the bank account transactions listed below, and determine which has the greatest impact on the account balance. Explain. a. A withdrawal of ̈́
b. A deposit of ̈́
c. A withdrawal of ̈́ ૡǤ
ȁെȁ ൌ ȁȁ ൌ ȁെૡǤ ȁ ൌ ૡǤ
ૡǤ , so a withdrawal of ̈́ has the greatest impact on the account balance.
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Lesson 12: dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ
6ͻϯ Lesson 12 Lesson 12: The Relationship Between Absolute Value and Order Student Outcomes
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĂƚƚŚĞŽƌĚĞƌŽĨƉŽƐŝƚŝǀĞŶƵŵďĞƌƐŝƐƚŚĞƐĂŵĞĂƐƚŚĞŽƌĚĞƌŽĨƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĂƚƚŚĞŽƌĚĞƌŽĨŶĞŐĂƚŝǀĞŶƵŵďĞƌƐŝƐƚŚĞŽƉƉŽƐŝƚĞŽƌĚĞƌŽĨƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĂƚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐĂƌĞĂůǁĂLJƐůĞƐƐƚŚĂŶƉŽƐŝƚŝǀĞŶƵŵďĞƌƐ͘
Lesson Notes
WƌŝŽƌƚŽƉƌĞƐĞŶƚŝŶŐƚŚĞůĞƐƐŽŶ͕ƉƌĞƉĂƌĞƐƚŝĐŬLJŶŽƚĞƐĐŽŶƚĂŝŶŝŶŐĂďĂůĂŶĐĞĚǀĂƌŝĞƚLJŽĨƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŝŶƚĞŐĞƌƐ ƌĂŶŐŝŶŐĨƌŽŵ െͷͲ ƚŽ ͷͲ ͘ ĂĐŚƉĂŝƌŽĨƐƚƵĚĞŶƚƐƌĞƋƵŝƌĞƐĂƐĞƚŽĨƚĞŶŝŶƚĞŐĞƌƐŝŶĐůƵĚŝŶŐĨŝǀĞŶĞŐĂƚŝǀĞǀĂůƵĞƐ͕njĞƌŽ͕ĂŶĚ ĨŽƵƌƉŽƐŝƚŝǀĞǀĂůƵĞƐ͘
Classwork Opening Exercise (5 minutes)
ŝǀŝĚĞƐƚƵĚĞŶƚƐŝŶƚŽƉĂŝƌƐ͘WƌŽǀŝĚĞĞĂĐŚƉĂŝƌǁŝƚŚĂƐĞƚŽĨƚĞŶŝŶƚĞŐĞƌƐŝŶĐůƵĚŝŶŐĨŝǀĞŶĞŐĂƚŝǀĞǀĂůƵĞƐ͕njĞƌŽ͕ĂŶĚĨŽƵƌ ƉŽƐŝƚŝǀĞǀĂůƵĞƐǁƌŝƚƚĞŶŽŶƐƚŝĐŬLJŶŽƚĞƐ͘/ŶƐƚƌƵĐƚƐƚƵĚĞŶƚŐƌŽƵƉƐƚŽŽƌĚĞƌƚŚĞŝƌŝŶƚĞŐĞƌƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚďĂƐĞĚŽŶ ƉƌŝŽƌŬŶŽǁůĞĚŐĞ͘tŚĞŶƚŚĞŝŶƚĞŐĞƌƐĂƌĞŝŶƚŚĞĐŽƌƌĞĐƚŽƌĚĞƌ͕ƐƚƵĚĞŶƚƐƌĞĐŽƌĚƚŚĞŝŶƚĞŐĞƌǀĂůƵĞƐŝŶŽƌĚĞƌŝŶƚŚĞŝƌƐƚƵĚĞŶƚ ŵĂƚĞƌŝĂůƐ͘
Opening Exercise Record your integer values in order from least to greatest in the space below.
Sample answer: െ ,െૢ ,െ ,െ ,െ , , , ,ૠ ,ૡ
,ĂǀĞŽŶĞƉĂŝƌŽĨƐƚƵĚĞŶƚƐƉŽƐƚƚŚĞŝƌƐƚŝĐŬLJŶŽƚĞƐƚŽƚŚĞǁĂůůŝŶƚŚĞƐƉĞĐŝĨŝĞĚŽƌĚĞƌ͘ƐŬĂŶŽƚŚĞƌƉĂŝƌŽĨƐƚƵĚĞŶƚƐ͗
džƉůĂŝŶƚŚĞƌĞĂƐŽŶŝŶŐĨŽƌƚŚĞŽƌĚĞƌ͘
à The integers are in the same order in which they would be found located from left (or bottom) to right (or top) on the number line.
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Lesson 12: dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ
6ͻϯ Lesson 12 Example 1 (8 minutes): Comparing Order of Integers to the Order of Their Absolute Values
^ƚƵĚĞŶƚƐƵƐĞƚŚĞŝŶƚĞŐĞƌǀĂůƵĞƐĨƌŽŵƚŚĞKƉĞŶŝŶŐ džĞƌĐŝƐĞƚŽĐŽŵƉĂƌĞƚŚĞŽƌĚĞƌŽĨŝŶƚĞŐĞƌƐƚŽƚŚĞŽƌĚĞƌŽĨƚŚĞŝƌ ĂďƐŽůƵƚĞǀĂůƵĞƐ͘
Example 1: Comparing Order of Integers to the Order of Their Absolute Values Write an inequality statement relating the ordered integers from the Opening Exercise. Below each integer, write its absolute value.
Sample answer: െ ൏ െૢ ൏ െ ൏ െ ൏ െ ൏ ൏ ൏ ൏ ૠ ൏ ૡ
ૢ ૠૡ
ƌĞƚŚĞĂďƐŽůƵƚĞǀĂůƵĞƐŽĨLJŽƵƌŝŶƚĞŐĞƌƐŝŶŽƌĚĞƌ͍ džƉůĂŝŶ͘
à No. The absolute values of the positive integers listed to the right of zero are still in ascending order, but the absolute values of the negative integers listed to the left of zero are now in descending order.
Circle the absolute values that are in increasing numerical order and their corresponding integers. Describe the circled values.
െ ൏ െૢ ൏ െ ൏ െ ൏ െ ൏ ൏ ൏ ൏ ૠ ൏ ૡ
ૢ ૠૡ
The circled integers are all positive values except zero. The positive integers and their absolute values have the same order.
Rewrite the integers that are not circled in the space below. How do these integers differ from the ones you circled?
െ , െૢ , െ , െ ,െ
They are all negative integers.
Rewrite the negative integers in ascending order and their absolute values in ascending order below them.
െ ൏ െૢ ൏ െ ൏ െ ൏ െ
ૢ
Describe how the order of the absolute values compares to the order of the negative integers.
The orders of the negative integers and their corresponding absolute values are opposite.
Example 2 (8 minutes): The Order of Negative Integers and Their Absolute Values
^ƚƵĚĞŶƚƐĞdžĂŵŝŶĞƚŚĞůĞŶŐƚŚƐŽĨĂƌƌŽǁƐĐŽƌƌĞƐƉŽŶĚŝŶŐǁŝƚŚƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŝŶƚĞŐĞƌƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞĂŶĚƵƐĞ ƚŚĞŝƌĂŶĂůLJƐŝƐƚŽƵŶĚĞƌƐƚĂŶĚǁŚLJƚŚĞŽƌĚĞƌŽĨŶĞŐĂƚŝǀĞŝŶƚĞŐĞƌƐŝƐŽƉƉŽƐŝƚĞƚŚĞŽƌĚĞƌŽĨƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘DŽŶŝƚŽƌ ƚŚĞƌŽŽŵ͕ĂŶĚƉƌŽǀŝĚĞŐƵŝĚĂŶĐĞĂƐŶĞĞĚĞĚ͕ĂŶĚƚŚĞŶŐƵŝĚĞĂǁŚŽůĞͲĐůĂƐƐĚŝƐĐƵƐƐŝŽŶǁŝƚŚƋƵĞƐƚŝŽŶƐ͘
Scaffolding:
&Žƌ ŶŐůŝƐŚ>ĂŶŐƵĂŐĞ>ĞĂƌŶĞƌƐ͗ /ŶƉůĂĐĞŽĨƚŚĞǁŽƌĚƐ ascending ĂŶĚ descending ͕ ĚĞƐĐƌŝďĞ ŶƵŵďĞƌƐĂƐŝŶĐƌĞĂƐŝŶŐĨƌŽŵůĞĨƚ ƚŽƌŝŐŚƚŽƌĚĞĐƌĞĂƐŝŶŐĨƌŽŵůĞĨƚ ƚŽƌŝŐŚƚ͘
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Lesson 12: dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ
6ͻϯ Lesson 12 Example 2: The Order of Negative Integers and Their Absolute Values Draw arrows starting at the dashed line (zero) to represent each of the integers shown on the number line below. The arrows that correspond with and have been modeled for you.
tŚŝĐŚŝŶƚĞŐĞƌ;ƐͿŝƐƌĞƉƌĞƐĞŶƚĞĚďLJƚŚĞůŽŶŐĞƐƚĂƌƌŽǁƚŚĂƚLJŽƵĚƌĞǁ͍tŚLJ͍
à ͷ and െͷ because, of the integers shown, they are farthest from zero on the number line.
tŚŝĐŚŶŽŶnjĞƌŽŝŶƚĞŐĞƌ;ƐͿŝƐƌĞƉƌĞƐĞŶƚĞĚďLJƚŚĞƐŚŽƌƚĞƐƚĂƌƌŽǁ͍tŚLJ͍
à ͳ and െͳ because, of the integers shown, they are closest to zero on the number line.
/ƐƚŚĞƌĞĂŶĂƌƌŽǁĨŽƌƚŚĞŝŶƚĞŐĞƌ Ͳ͍ džƉůĂŝŶ͘
à The length of such an arrow would be Ͳ͕ so we could not see the arrow. We could call it an arrow with zero length, but we could not draw it.
ŽƚŚĞƐĞ͞ĂƌƌŽǁƐ͟ƐƚĂƌƚĂƚƚŚĞƐĂŵĞƉůĂĐĞŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͍tŚLJŽƌǁŚLJŶŽƚ͍
à They all start at zero on the number line because Ͳ is the reference point for all numbers on the number line.
ŽĂůůĂƌƌŽǁƐƉŽŝŶƚŝŶƚŚĞƐĂŵĞĚŝƌĞĐƚŝŽŶ͍tŚLJŽƌǁŚLJŶŽƚ͍
à All arrows do not point in the same direction because some integers have opposite signs. The lengths of the arrows get shorter as you approach zero from the left or from the right, which means the absolute values decrease as you approach zero from the left or the right.
ŝƌĞĐƚƐƚƵĚĞŶƚƐƚŽĐŽŵƉůĞƚĞƚŚĞƐƚĂƚĞŵĞŶƚƐŝŶƚŚĞŝƌƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐďLJĨŝůůŝŶŐŝŶƚŚĞďůĂŶŬƐ͘
As you approach zero from the left on the number line, the integers
increase , but the absolute values of those integers decrease . This means that the order of negative integers is opposite the order of their absolute values.
െ െ െ െ െ
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Lesson 12: dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ
6ͻϯ Lesson 12 Discussion (2 minutes)
dŚŝŶŬŽĨĂƌĞĂůͲǁŽƌůĚĞdžĂŵƉůĞƚŚĂƚƐŚŽǁƐǁŚLJƚŚĞŽƌĚĞƌŽĨŶĞŐĂƚŝǀĞŝŶƚĞŐĞƌƐĂŶĚƚŚĞŽƌĚĞƌŽĨƚŚĞŝƌĂďƐŽůƵƚĞ ǀĂůƵĞƐĂƌĞŽƉƉŽƐŝƚĞ͘
à Alec, Benny, and Charlotte have cafeteria charge account balances of െ͵ ͕ െͶ , and െͷ dollars, respectively. The number that represents Alec’s cafeteria charge account balance is the greatest because he is the least in debt. Alec owes the least amount of money and is the closest to having a positive balance. His balance of െ͵ dollars is farthest right on the number line of the three balances.
Exercise 1 (5 minutes)
^ƚƵĚĞŶƚƐƐĞƉĂƌĂƚĞƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐĂŶĚŽƌĚĞƌƚŚĞŵĂĐĐŽƌĚŝŶŐƚŽƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘^ƚƵĚĞŶƚƐ ƚŚĞŶŽƌĚĞƌƚŚĞŐŝǀĞŶƐĞƚŽĨƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐƵƐŝŶŐƚŚĞŽƌĚĞƌƐŽĨƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘
Exercise 1 Complete the steps below to order these numbers:
൜Ǥ ǡ െ ǡ െǡ Ǥ ǡ െǤ ǡ ǡ Ǥૢ ǡ െǤ ǡ െǡ ǡ Ǥૢૢ ǡ െૢ ൠ
a. Separate the set of numbers into positive rational numbers, negative rational numbers, and zero in the top cells below (order does not matter). b. Write the absolute values of the rational numbers (order does not matter) in the bottom cells below. c. Order each subset of absolute values from least to greatest.
d. Order each subset of rational numbers from least to greatest.
Absolute Values
Ǥ Ǥ Ǥૢ
Ǥૢૢ
Absolute Values
Ǥ
ૢ Ǥ
Positive Rational Numbers
Ǥ Ǥ Ǥૢ
Ǥૢૢ
Negative Rational Numbers
െ െ െ Ǥ
െ െૢ െ Ǥ
Zero
Ǥ , , , , Ǥ ,ૢ Ǥ , Ǥ , , Ǥૢ , Ǥૢૢ
െૢ , െǤ , െ , െ , െ , െǤ Ǥ , Ǥ , , Ǥૢ , Ǥૢૢ
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Lesson 12: dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ
6ͻϯ Lesson 12 e. Order the whole given set of rational numbers from least to greatest.
Exercise 2 (8 minutes)
^ƚƵĚĞŶƚƐŝŶĚĞƉĞŶĚĞŶƚůLJĐŽŵƉůĞƚĞƚŚĞĨŽůůŽǁŝŶŐƉƌŽďůĞŵƐ͘DŽŶŝƚŽƌƐƚƵĚĞŶƚƉƌŽŐƌĞƐƐ͕ĂŶĚƉƌŽǀŝĚĞŐƵŝĚĂŶĐĞĂƐŶĞĞĚĞĚ͘
Exercise 2 a. Find a set of four integers such that their order and the order of their absolute values are the same.
Answers will vary. An example follows: , , ૡ ,
b. Find a set of four integers such that their order and the order of their absolute values are opposite.
Answers will vary. An example follows: െ ,െૡ , െ ,െ
c. Find a set of four non-integer rational numbers such that their order and the order of their absolute values are the same.
Answers will vary. An example follows: , , ,
d. Find a set of four non-integer rational numbers such that their order and the order of their absolute values are opposite.
Answers will vary. An example follows: െ , െ , െ , െ
e. Order all of your numbers from parts (a)–(d) in the space below. This means you should be ordering
numbers from least to greatest.
Answers will vary. An example follows:
െ , െ ૡ , െ , െ , െ , െ , െ , െ , , , , , , , ૡ ,
Closing (4 minutes)
ĞůŽǁĂƌĞƚŚĞĂďƐŽůƵƚĞǀĂůƵĞƐŽĨĂƐĞƚŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶŝŶĐƌĞĂƐŝŶŐŽƌĚĞƌ͘ ͲǤͶ ͕ ͳ͕ ʹ ͳʹ͕ Ͷ͕ ͶǤ͵ ͕ϳ͕ ͻǤͻ
ĂŶLJŽƵĚĞƚĞƌŵŝŶĞƚŚĞŽƌĚĞƌŽĨƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŚĞŵƐĞůǀĞƐ͍/ĨƐŽ͕ĞdžƉůĂŝŶǁŚLJ͕ĂŶĚǁƌŝƚĞƚŚĞŶƵŵďĞƌƐ ŝŶŝŶĐƌĞĂƐŝŶŐŽƌĚĞƌ͘/ĨŶŽƚ͕ĞdžƉůĂŝŶǁŚLJ͘
à It is not possible to determine the order of the rational numbers because we do not know the signs of the rational numbers.
െૢ , െǤ , െ , െ , െ , െǤ , , Ǥ , Ǥ , , Ǥૢ , Ǥૢૢ
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Lesson 12: dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ
6ͻϯ Lesson 12
/ĨLJŽƵĂƌĞƚŽůĚƚŚĂƚƚŚĞŶƵŵďĞƌƐĂďŽǀĞƌĞƉƌĞƐĞŶƚƚŚĞĂďƐŽůƵƚĞǀĂůƵĞƐŽĨƉŽƐŝƚŝǀĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͕ĐĂŶLJŽƵ ĚĞƚĞƌŵŝŶĞƚŚĞŽƌĚĞƌŽĨƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͍/ĨƐŽ͕ĞdžƉůĂŝŶǁŚLJ͕ĂŶĚǁƌŝƚĞƚŚĞŶƵŵďĞƌƐŝŶŝŶĐƌĞĂƐŝŶŐŽƌĚĞƌ͘/Ĩ ŶŽƚ͕ĞdžƉůĂŝŶǁŚLJŶŽƚ͘
à If the original numbers are all positive, we are able to order the rational numbers because we know their signs. The order of the original numbers will be the same as the order of their absolute values.
/ĨLJŽƵĂƌĞƚŽůĚƚŚĂƚƚŚĞŶƵŵďĞƌƐĂďŽǀĞƌĞƉƌĞƐĞŶƚƚŚĞĂďƐŽůƵƚĞǀĂůƵĞƐŽĨŶĞŐĂƚŝǀĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͕ĐĂŶLJŽƵ ĨŝŶĚƚŚĞŽƌĚĞƌŽĨƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͍/ĨƐŽ͕ĞdžƉůĂŝŶǁŚLJ͕ĂŶĚǁƌŝƚĞƚŚĞŶƵŵďĞƌƐŝŶŝŶĐƌĞĂƐŝŶŐŽƌĚĞƌ͘/ĨŶŽƚ͕ ĞdžƉůĂŝŶǁŚLJ͘
à If the original numbers are all negative, we are able to order the rational numbers because we know their signs. The order of the original numbers will be the opposite order of their absolute values.
Exit Ticket (5 minutes)
Lesson Summary The absolute values of positive numbers always have the same order as the positive numbers themselves. Negative numbers, however, have exactly the opposite order as their absolute values. The absolute values of numbers on the number line increase as you move away from zero in either direction.
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Lesson 12: dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ
6ͻϯ Lesson 12
EĂŵĞ ĂƚĞ
Lesson 12: The Relationship Between Absolute Value and Order Exit Ticket
ϭ͘ ĞƚŚĂŶLJǁƌŝƚĞƐĂƐĞƚŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶŝŶĐƌĞĂƐŝŶŐŽƌĚĞƌ͘,ĞƌƚĞĂĐŚĞƌĂƐŬƐŚĞƌƚŽǁƌŝƚĞƚŚĞĂďƐŽůƵƚĞǀĂůƵĞƐŽĨ ƚŚĞƐĞŶƵŵďĞƌƐŝŶŝŶĐƌĞĂƐŝŶŐŽƌĚĞƌ͘tŚĞŶŚĞƌƚĞĂĐŚĞƌĐŚĞĐŬƐĞƚŚĂŶLJ͛ƐǁŽƌŬ͕ƐŚĞŝƐƉůĞĂƐĞĚƚŽƐĞĞƚŚĂƚĞƚŚĂŶLJŚĂƐ ŶŽƚĐŚĂŶŐĞĚƚŚĞŽƌĚĞƌŽĨŚĞƌŶƵŵďĞƌƐ͘tŚLJŝƐƚŚŝƐ͍ Ϯ͘ DĂƐŽŶǁĂƐŽƌĚĞƌŝŶŐƚŚĞĨŽůůŽǁŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶŵĂƚŚĐůĂƐƐ͗ െ͵Ǥ͵ǡ െͳͷǡ െͺ ͺͻ͘ Ă͘ KƌĚĞƌƚŚĞŶƵŵďĞƌƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͘ ď͘ >ŝƐƚƚŚĞŽƌĚĞƌŽĨƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͘ Đ͘ džƉůĂŝŶǁŚLJƚŚĞŽƌĚĞƌŝŶŐƐŝŶƉĂƌƚƐ;ĂͿĂŶĚ;ďͿĂƌĞĚŝĨĨĞƌĞŶƚ͘
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Lesson 12: dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ
6ͻϯ Lesson 12 Exit Ticket Sample Solutions 1. Bethany writes a set of rational numbers in increasing order. Her teacher asks her to write the absolute values of these numbers in increasing order. When her teacher checks Bethany’s work, she is pleased to see that Bethany has not changed the order of her numbers. Why is this?
All of Bethany’s rational numbers are positive or . The positive rational numbers have the same order as their absolute values. If any of Bethany’s rational numbers are negative, then the order would be different.
Mason was ordering the following rational numbers in math class: െǤ ,െ ,െૡ ૡૢ .a. Order the numbers from least to greatest.
െ , െ ૡ ૡૢ , െ Ǥ
b. List the order of their absolute values from least to greatest.
Ǥ ,ૡ ૡૢ ,
c. Explain why the orderings in parts (a) and (b) are different.
Since these are all negative numbers, when I ordered them from least to greatest, the one farthest away from zero (farthest to the left on the number line) came first. This number is െ . Absolute value is the numbers’ distance from zero, and so the number farthest away from zero has the greatest absolute value, so will be greatest in the list of absolute values, and so on.
Problem Set Sample Solutions 1. Micah and Joel each have a set of five rational numbers. Although their sets are not the same, their sets of numbers have absolute values that are the same. Show an example of what Micah and Joel could have for numbers. Give the sets in order and the absolute values in order.
Examples may vary. If Micah had , , , , , then his order of absolute values would be the same: , , , ,
. If Joel had the numbers െ ,െ ,െ , െ ,െ , then his order of absolute values would also be , , , , .
Enrichment Extension: Show an example where Micah and Joel both have positive and negative numbers.
If Micah had the numbers: െ , െ ,െ , , , his order of absolute values would be , , , , . If Joel had the numbers െ ,െ , , , , then the order of his absolute values would also be , , , , .
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Lesson 12: dŚĞZĞůĂƚŝŽŶƐŚŝƉĞƚǁĞĞŶďƐŽůƵƚĞsĂůƵĞĂŶĚKƌĚĞƌ
6ͻϯ Lesson 12 2. For each pair of rational numbers below, place each number in the Venn diagram based on how it compares to the other. a. െ , െૡ
b. , ૡ
c. ૠ, െ
d. െૢ ,
e. ,
f. െ ,
g. െ ,
None of the Above
െ െ
Is the Greater Number
Has a Greater Absolute Value Is the Greater Number and Also the Greater Absolute Value
െૡ
െ
ૡ
ૠ
െૢ
െ
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6ͻϯ >ĞƐƐŽŶϭϯ
ĞƐƐŽŶϭϯ : ^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ
>ĞƐƐŽŶϭϯ : Statements of Order in the Real World Student Outcomes
^ƚƵĚĞŶƚƐĂƉƉůLJƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨŽƌĚĞƌĂŶĚĂďƐŽůƵƚĞǀĂůƵĞǁŚĞŶĞdžĂŵŝŶŝŶŐƌĞĂůͲǁŽƌůĚƐĐĞŶĂƌŝŽƐ͘^ƚƵĚĞŶƚƐ ƌĞĂůŝnjĞ͕ĨŽƌŝŶƐƚĂŶĐĞ͕ƚŚĂƚƚŚĞĚĞƉƚŚŽĨĂůŽĐĂƚŝŽŶďĞůŽǁƐĞĂůĞǀĞůŝƐƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨĂŶĞŐĂƚŝǀĞŶƵŵďĞƌ͕ ǁŚŝůĞƚŚĞŚĞŝŐŚƚŽĨĂŶŽďũĞĐƚĂďŽǀĞƐĞĂůĞǀĞůŝƐƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨĂƉŽƐŝƚŝǀĞŶƵŵďĞƌ͘
Classwork Opening Exercise (4 minutes)
^ƚƵĚĞŶƚƐĚŝƐĐƵƐƐƚŚĞŵĞĂŶŝŶŐŽĨƚŚĞƌĞƉŽƌƚďĞůŽǁ͕ǁƌŝƚĞĂƐƵŵŵĂƌLJŽĨƚŚĞŝƌĐŽŶĐůƵƐŝŽŶƐŝŶƚŚĞŝƌƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐ͕ĂŶĚ ƉƌŽǀŝĚĞƚŚĞŝƌĨĞĞĚďĂĐŬƚŽƚŚĞǁŚŽůĞŐƌŽƵƉ͘
Opening Exercise A radio disc jockey reports that the temperature outside his studio has changed degrees since he came on the air this morning. Discuss with your group what listeners can conclude from this report.
The report is not specific enough to be conclusive because degrees of change could mean an increase or a decrease in temperature. A listener might assume the report says an increase in temperature; however, the word “changed” is not specific enough to conclude a positive or negative change.
,ŽǁĐŽƵůĚLJŽƵĐŚĂŶŐĞƚŚĞƌĞƉŽƌƚƚŽŵĂŬĞŝƚŵŽƌĞŝŶĨŽƌŵĂƚŝǀĞ͍
à Using the words “increased” or “decreased” instead of “changed” would be much more informative.
/ŶƌĞĂůͲǁŽƌůĚĐŽŶƚĞdžƚƐ͕ĚĞƐĐƌŝƉƚŝǀĞǁŽƌĚƐƐƵĐŚĂƐ debt, credit, increase, ĂŶĚ decrease ŚĞůƉƵƐŝŶĚŝĐĂƚĞǁŚĞŶĂ ŐŝǀĞŶŵĂŐŶŝƚƵĚĞŝƐƌĞƉƌĞƐĞŶƚĂƚŝǀĞŽĨĂƉŽƐŝƚŝǀĞŽƌŶĞŐĂƚŝǀĞǀĂůƵĞ͘
Example 1 (4 minutes): Ordering Numbers in the Real World
^ƚƵĚĞŶƚƐƐĂǁƚŚĂƚĂďƐŽůƵƚĞǀĂůƵĞƌĞƉƌĞƐĞŶƚƐƚŚĞŵĂŐŶŝƚƵĚĞŽĨĂƉŽƐŝƚŝǀĞŽƌŶĞŐĂƚŝǀĞ ƋƵĂŶƚŝƚLJŝŶ>ĞƐƐŽŶϭϭ͘dŽŽƌĚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŐŝǀĞŶŝŶĂƌĞĂůͲǁŽƌůĚĐŽŶƚĞdžƚ͕ƐƚƵĚĞŶƚƐ ŶĞĞĚƚŽĐŽŶƐŝĚĞƌƚŚĞŵĞĂŶŝŶŐŽĨĚĞƐĐƌŝƉƚŽƌƐĂŶĚǁŚĂƚƚŚĞLJŝŶĚŝĐĂƚĞĂďŽƵƚĂŐŝǀĞŶ ƋƵĂŶƚŝƚLJ͘
Example 1: Ordering Numbers in the Real World A ̈́ credit and a ̈́ charge appear similar, yet they are very different. Describe what is similar about the two transactions.
The transactions look similar because they are described using the same number. Both transactions have the same magnitude (or absolute value) and, therefore, result in a change of ̈́ to an account balance.
Scaffolding:
ZĞǀŝĞǁƚŚĞĨŝŶĂŶĐŝĂůƚĞƌŵƐŝĨ ŶĞĐĞƐƐĂƌLJ͘,ĂǀĞƐƚƵĚĞŶƚƐ ĚĞǀĞůŽƉĂƉŽƐƚĞƌƚŽĐĂƚĞŐŽƌŝnjĞ ƚŚĞƚĞƌŵƐĂŶĚŝŶĐůƵĚĞ ĞdžĂŵƉůĞƐŽĨƚŚĞŝƌŵĞĂŶŝŶŐƐ͘
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6ͻϯ >ĞƐƐŽŶϭϯ
ĞƐƐŽŶϭϯ :^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ
How do the two transactions differ?
The credit would cause an increase to an account balance and, therefore, should be represented by , while the charge would instead decrease an account balance and should be represented by െ . The two transactions represent changes that are opposites.
Exercises 1–4 (22 minutes)
^ƚƵĚĞŶƚƐƵƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽƐŽůǀĞǀĂƌŝŽƵƐƌĞĂůͲǁŽƌůĚƉƌŽďůĞŵƐǁŝƚŚƉĂƌƚŶĞƌƐĂŶĚƚŚĞŶƐŚĂƌĞĂŶĚƐƵƉƉŽƌƚƚŚĞŝƌ ĐŽŶĐůƵƐŝŽŶƐǁŝƚŚƚŚĞǁŚŽůĞŐƌŽƵƉ͘ ůůŽǁƚǁŽŵŝŶƵƚĞƐĨŽƌƐĞƚƵƉ͘,ĂǀĞĞŝŐŚƚƐƚƵĚĞŶƚƐĂƌƌĂŶŐĞƚŚĞŝƌĚĞƐŬƐŝŶƚŽƚǁŽ ƌŽǁƐŽĨĨŽƵƌƐŽƚŚĂƚƚŚĞƌŽǁƐĂƌĞĨĂĐŝŶŐĞĂĐŚŽƚŚĞƌ͘ĚĚŝƚŝŽŶĂůŐƌŽƵƉƐŽĨĞŝŐŚƚ ƐƚƵĚĞŶƚƐƐŚŽƵůĚďĞĨŽƌŵĞĚƉĞƌŝŶĚŝǀŝĚƵĂůĐůĂƐƐƐŝnjĞ͘/ĨƵƐŝŶŐƚĂďůĞƐ͕ŚĂǀĞĨŽƵƌ ƐƚƵĚĞŶƚƐƐŝƚŽŶŽŶĞƐŝĚĞŽĨƚŚĞƚĂďůĞ;ƐͿĂŶĚƚŚĞŽƚŚĞƌĨŽƵƌƐƚƵĚĞŶƚƐƐŝƚŽƉƉŽƐŝƚĞ ƚŚĞŵ͘^ƚƵĚĞŶƚƐƌŽƚĂƚĞǁŚĞŶĞĂĐŚƉƌŽďůĞŵŝƐĐŽŵƉůĞƚĞĚ͘tŚĞŶƚŚĞƌŽƚĂƚŝŽŶ ŽĐĐƵƌƐ͕ƐƚƵĚĞŶƚƐǁŚŽĂƌĞŝŶƚŚĞƌŝŐŚƚͲŵŽƐƚƐĞĂƚŽĨĞĂĐŚƌŽǁƌŽƚĂƚĞƚŽĂƉŽƐŝƚŝŽŶŝŶ ƚŚĞŽƉƉŽƐŝƚĞƌŽǁ;ƐĞĞĚŝĂŐƌĂŵƚŽƚŚĞƌŝŐŚƚͿ͘,ĂǀŝŶŐƐƚƵĚĞŶƚƐŵŽǀĞŝŶŽƉƉŽƐŝƚĞ ĚŝƌĞĐƚŝŽŶƐĂůůŽǁƐĞĂĐŚƐƚƵĚĞŶƚƚŽǁŽƌŬǁŝƚŚĂĚŝĨĨĞƌĞŶƚƉĂƌƚŶĞƌŽŶĞĂĐŚŽĨƚŚĞ ĨŽƵƌƉƌŽďůĞŵƐ͘ ^ƚƵĚĞŶƚƐƐŚŽƵůĚǁŽƌŬĂƚƚŚĞŝƌƐƚĂƚŝŽŶƐĨŽƌϭϱŵŝŶƵƚĞƐ͕ĐŽŵƉůĞƚŝŶŐ džĞƌĐŝƐĞϭŝŶ ƚŚĞƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐǁŝƚŚƚŚĞŝƌĨŝƌƐƚƉĂƌƚŶĞƌ͕ džĞƌĐŝƐĞϮǁŝƚŚĂŶĞǁƉĂƌƚŶĞƌ͕ džĞƌĐŝƐĞϯǁŝƚŚĂĚŝĨĨĞƌĞŶƚŶĞǁƉĂƌƚŶĞƌ͕ĂŶĚĨŝŶĂůůLJ džĞƌĐŝƐĞϰǁŝƚŚĂĚŝĨĨĞƌĞŶƚŶĞǁ ƉĂƌƚŶĞƌ͘WĂƌƚŶĞƌƐĂƌĞŐŝǀĞŶƚŚƌĞĞŵŝŶƵƚĞƐƚŽĐŽŵƉůĞƚĞĞĂĐŚƉƌŽďůĞŵŝŶƚŚĞ ƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐĂŶĚƚŚĞŶŐŝǀĞŶŽŶĞŵŝŶƵƚĞƚŽƌŽƚĂƚĞƐĞĂƚƐ͘ ĂĐŚƉƌŽďůĞŵƌĞƋƵŝƌĞƐƚŚĂƚƐƚƵĚĞŶƚƐĚĞƚĞƌŵŝŶĞĂƉƉƌŽƉƌŝĂƚĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŽƌĞƉƌĞƐĞŶƚŐŝǀĞŶƋƵĂŶƚŝƚŝĞƐĂŶĚŽƌĚĞƌ ƚŚĞŶƵŵďĞƌƐĂƐƐƉĞĐŝĨŝĞĚ͘^ƚƵĚĞŶƚƐĂůƐŽƉƌŽǀŝĚĞƌĞĂƐŽŶŝŶŐĨŽƌƚŚĞŝƌĐŚŽŝĐĞƐŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐŝŶĞĂĐŚĐĂƐĞ͘
Exercises 1. Scientists are studying temperatures and weather patterns in the Northern Hemisphere. They recorded temperatures (in degrees Celsius) in the table below as reported in emails from various participants. Represent each reported temperature using a rational number. Order the rational numbers from least to greatest. Explain why the rational numbers that you chose appropriately represent the given temperatures. Temperatures as Reported
ૡbelow zero െ below zero above zero
below zero െ
Temperature
ሺι۱ሻ െૡ െ െ െ െ
െ ൏ െૡ ൏ െ ൏ െ ൏ െ ൏ ൏ ൏
The words “below zero ”refer to negative numbers because they are located below zero on a vertical number line.
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6ͻϯ >ĞƐƐŽŶϭϯ
ĞƐƐŽŶϭϯ : ^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ
Jami’s bank account statement shows the transactions below. Represent each transaction as a rational number describing how it changes Jami’s account balance. Then, order the rational numbers from greatest to least. Explain why the rational numbers that you chose appropriately reflect the given transactions. Listed Transactions Debit ̈́ Ǥ
Credit ̈́ Ǥ ૡ
Charge ̈́ Ǥ
Withdrawal ̈́ Ǥ
Deposit ̈́ Ǥ
Debit ̈́ Ǥૢ
Charge ̈́ Ǥ
Change to Jami’s Account െǤ Ǥ ૡ െǤ െ Ǥ െǤૢ െ
Ǥ Ǥ ૡ െǤ െ െǤૢ െǤ െ
The words “debit,” “charge,” and “withdrawal” all describe transactions in which money is taken out of Jami’s account, decreasing its balance. These transactions are represented by negative numbers. The words “credit” and “deposit” describe transactions that will put money into Jami’s account, increasing its balance. These transactions are represented by positive numbers.
ϯ͘ During the summer, Madison monitors the water level in her parents’ swimming pool to make sure it is not too far above or below normal. The table below shows the numbers she recorded in July and August to represent how the water levels compare to normal. Order the rational numbers from least to greatest. Explain why the rational numbers that you chose appropriately reflect the given water levels. Madison’s Readings
inch above normal
inch above normal
inch below normal
ૡ inch above normal
inches below normal
ૡ inch below normal
inch below normal Compared to Normal
െ
ૡ െ െ ૡ െ
െ ൏ െ ൏ െ ൏ െ ૡ ൏ ૡ ൏ ൏
The measurements are taken in reference to normal level, which is considered to be . The words “above normal” refer to the positive numbers located above zero on a vertical number line, and the words “below normal” refer to the negative numbers located below zero on a vertical number line.
Changes in the weather can be predicted by changes in the barometric pressure. Over several weeks, Stephanie recorded changes in barometric pressure seen on her barometer to compare to local weather forecasts. Her observations are recorded in the table below. Use rational numbers to record the indicated changes in the pressure in the second row of the table. Order the rational numbers from least to greatest. Explain why the rational numbers that you chose appropriately represent the given pressure changes. Barometric Pressure Change (Inches of Mercury) Rise Ǥ Fall Ǥ Rise Ǥ Fall Ǥ Rise Ǥ Fall Ǥ ૢ Fall Ǥ
Barometric Pressure Change (Inches of Mercury)
Ǥ െǤ Ǥ െǤ Ǥ െǤ ૢ െǤ
െǤ , െǤ ,െǤ ૢ ,െǤ , Ǥ ,Ǥ ,Ǥ
The records that include the word “rise ” refer to increases and are, therefore, represented by positive numbers. The records that include the word “fall” refer to decreases and are, therefore, represented by negative numbers.
ĨƚĞƌĐŽŵƉůĞƚŝŶŐĂůůƐƚĂƚŝŽŶƐ͕ĂƐŬƐƚƵĚĞŶƚƐƚŽƌĞƉŽƌƚƚŚĞŝƌĂŶƐǁĞƌƐĂŶĚƌĞĂƐŽŶŝŶŐĨŽƌƚŚĞŐŝǀĞŶƉƌŽďůĞŵƐ͘>ŽŽŬĨŽƌ ĚŝĨĨĞƌĞŶĐĞƐŝŶǀĂůŝĚƌĞĂƐŽŶŝŶŐ͕ĂŶĚĚŝƐĐƵƐƐƚŚŽƐĞĚŝĨĨĞƌĞŶĐĞƐǁŚĞƌĞĂƉƉƌŽƉƌŝĂƚĞ͘ ŶĐŽƵƌĂŐĞƐƚƵĚĞŶƚƐƚŽƉŽůŝƚĞůLJ ĐŚĂůůĞŶŐĞƚŚĞƌĞĂƐŽŶŝŶŐŽĨƚŚĞŝƌĐůĂƐƐŵĂƚĞƐŝĨĂƉƉůŝĐĂďůĞ͘dŚŝƐĂĐƚŝǀŝƚLJƐŚŽƵůĚƚĂŬĞĨŽƵƌŵŝŶƵƚĞƐ͘
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6ͻϯ >ĞƐƐŽŶϭϯ
ĞƐƐŽŶϭϯ : ^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ
Example 2 (5 minutes): Using Absolute Value to Solve Real-World Problems
^ƚƵĚĞŶƚƐƵƐĞƚŚĞĂďƐŽůƵƚĞǀĂůƵĞƐŽĨƉŽƐŝƚŝǀĞĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐƚŽƐŽůǀĞƉƌŽďůĞŵƐŝŶƌĞĂůͲǁŽƌůĚĐŽŶƚĞdžƚƐ͘^ƚƵĚĞŶƚƐ ŵĂLJĨŝŶĚŝƚŚĞůƉĨƵůƚŽĚƌĂǁĂƉŝĐƚƵƌĞĂƐĂƉƌŽďůĞŵͲƐŽůǀŝŶŐƐƚƌĂƚĞŐLJ͘'ƌŝĚƉĂƉĞƌŵĂLJďĞƉƌŽǀŝĚĞĚƐŽƚŚĂƚƚŚĞLJĐĂŶ ĂĐĐƵƌĂƚĞůLJĐŽŶƐƚƌƵĐƚĂƉŝĐƚƵƌĞĂŶĚŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵ͘
Example 2: Using Absolute Value to Solve Real-World Problems The captain of a fishing vessel is standing on the deck at feet above sea level. He holds a rope tied to his fishing net that is below him underwater at a depth of ૡ feet. Draw a diagram using a number line, and then use absolute value to compare the lengths of rope in and out of the water.
The captain is above the water, and the fishing net is below the water’s surface. Using the water level as reference point zero, I can draw the diagram using a vertical number line. The captain is located at , and the fishing net is located at െૡ .
ȁȁ ൌ and ȁെૡȁ ൌ ૡ , so there is more rope underwater than above.
ૡ െ ൌ
The length of rope below the water’s surface is feet longer than the rope above water.
ŝƐĐƵƐƐĂŶĚŵŽĚĞůŚŽǁƐƚƵĚĞŶƚƐĐĂŶĐŽŶƐƚƌƵĐƚĂŶƵŵďĞƌůŝŶĞĂŶĚƵƐĞŶƵŵďĞƌƐĞŶƐĞƚŽĨŝŶĚƚŚĞĂƉƉƌŽdžŝŵĂƚĞůŽĐĂƚŝŽŶŽĨ
ʹ͵ ĂŶĚ െ͵ͺ ǁŝƚŚŽƵƚŐƌŝĚƉĂƉĞƌ͘
Example ϯ (4 minutes): Making Sense of Absolute Value and Statements of Inequality
^ƚƵĚĞŶƚƐĞdžĂŵŝŶĞĂďƐŽůƵƚĞǀĂůƵĞƐŽĨŶĞŐĂƚŝǀĞŶƵŵďĞƌƐŝŶĂƌĞĂůͲǁŽƌůĚĐŽŶƚĞdžƚĂŶĚŵĂŬĞƐĞŶƐĞŽĨƐƚĂƚĞŵĞŶƚƐĂďŽƵƚ ŝŶĞƋƵĂůŝƚŝĞƐŝŶǀŽůǀŝŶŐƚŚŽƐĞǀĂůƵĞƐ͘
džĂŵƉůĞϯ͗DĂŬŝŶŐ^ĞŶƐĞŽĨďƐŽůƵƚĞsĂůƵĞĂŶĚ^ƚĂƚĞŵĞŶƚƐŽĨ/ŶĞƋƵĂůŝƚLJ
A recent television commercial asked viewers, “Do you have over ̈́ ǡ in credit card debt?” What types of numbers are associated with the word debt, and why? Write a number that represents the value from the television commercial.
Negative numbers; debt describes money that is owed; െǡ
Give one example of “over ̈́ ǡ in credit card debt.” Then, write a rational number that represents your example.
Answers will vary, but the number should have a value of less than െǡ . Credit card debt of ̈́ ǡ ; െǡ
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6ͻϯ >ĞƐƐŽŶϭϯ
ĞƐƐŽŶϭϯ : ^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ
How do the debts compare, and how do the rational numbers that describe them compare? Explain.
The example ̈́ ǡ is greater than ̈́ ǡ from the commercial; however, the rational numbers that represent these debt values have the opposite order because they are negative numbers. െǡ ൏ െǡ . The absolute values of negative numbers have the opposite order of the negative values themselves.
Closing ( ϯ minutes)
zŽƵƌĨƌŝĞŶĚ^ĂŵƵĞůƐĂLJƐŚĞŝƐ ͷͲ ĨĞĞƚĨƌŽŵƐĞĂůĞǀĞů͘tŚĂƚĂĚĚŝƚŝŽŶĂůŝŶĨŽƌŵĂƚŝŽŶƐŚŽƵůĚ^ĂŵƵĞůŐŝǀĞLJŽƵŝŶ ŽƌĚĞƌĨŽƌLJŽƵƚŽŝĚĞŶƚŝĨLJŚŝƐĞůĞǀĂƚŝŽŶ͍
à In order to know Samuel’s elevation, he would have to tell me if he is above or below sea level.
/ĚĞŶƚŝĨLJƚŚƌĞĞƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƐƚŚĂƚĂƌĞƌĞƉƌĞƐĞŶƚĞĚďLJŶĞŐĂƚŝǀĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
à The temperature is ͵Ԭ below zero. My mom was charged a ̈́ ͳͷ fee for missing a doctor appointment. Jason went scuba diving and was ʹͲ feet below sea level.
/ĚĞŶƚŝĨLJƚŚƌĞĞƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶƐƚŚĂƚĂƌĞƌĞƉƌĞƐĞŶƚĞĚďLJƉŽƐŝƚŝǀĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
à The temperature is ͵Ԭ above zero. My mom received a ̈́ ͳͷ credit for referring a friend to her Internet service. Jason went hiking and was ʹͲ feet above sea level.
Exit Ticket ( ϯ minutes)
Lesson Summary When comparing values in real-world situations, descriptive words help you to determine if the number represents a positive or negative number. Making this distinction is critical when solving problems in the real world. Also critical is to understand how an inequality statement about an absolute value compares to an inequality statement about the number itself.
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6ͻϯ >ĞƐƐŽŶϭϯ
ĞƐƐŽŶϭϯ : ^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ
EĂŵĞ ĂƚĞ
>ĞƐƐŽŶϭϯ : Statements of Order in the Real World Exit Ticket
ϭ͘ >ŽŶŝĂŶĚĂƌLJůĐĂůůĞĂĐŚŽƚŚĞƌĨƌŽŵĚŝĨĨĞƌĞŶƚƐŝĚĞƐŽĨtĂƚĞƌƚŽǁŶ͘dŚĞŝƌůŽĐĂƚŝŽŶƐĂƌĞƐŚŽǁŶŽŶƚŚĞŶƵŵďĞƌůŝŶĞ ďĞůŽǁƵƐŝŶŐŵŝůĞƐ͘hƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽĞdžƉůĂŝŶǁŚŽŝƐĂĨĂƌƚŚĞƌĚŝƐƚĂŶĐĞ;ŝŶŵŝůĞƐͿĨƌŽŵtĂƚĞƌƚŽǁŶ͘,ŽǁŵƵĐŚ ĐůŽƐĞƌŝƐŽŶĞƚŚĂŶƚŚĞŽƚŚĞƌ͍ Ϯ͘ ůĂƵĚĞƌĞĐĞŶƚůLJƌĞĂĚƚŚĂƚŶŽŽŶĞŚĂƐĞǀĞƌƐĐƵďĂĚŝǀĞĚŵŽƌĞƚŚĂŶ ͵͵Ͳ ŵĞƚĞƌƐďĞůŽǁƐĞĂůĞǀĞů͘ĞƐĐƌŝďĞǁŚĂƚƚŚŝƐ ŵĞĂŶƐŝŶƚĞƌŵƐŽĨĞůĞǀĂƚŝŽŶƵƐŝŶŐƐĞĂůĞǀĞůĂƐĂƌĞĨĞƌĞŶĐĞƉŽŝŶƚ͘
Ͳെʹ െͶ െ െͺ ʹ Ͷ ͺ
ŽŶŝ tĂƚĞƌƚŽǁŶ
ͳͲ െͳͲ
ĂƌLJů
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6ͻϯ >ĞƐƐŽŶϭϯ
ĞƐƐŽŶϭϯ : ^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ
Exit Ticket Sample Solutions 1. Loni and Daryl call each other from different sides of Watertown. Their locations are shown on the number line below using miles. Use absolute value to explain who is a farther distance (in miles) from Watertown. How much closer is one than the other?
Loni’s location is െ , and ȁെȁ ൌ because െ is units from on the number line. Daryl’s location is , and
ȁȁ ൌ because is units from on the number line. We know that , so Daryl is farther from Watertown than Loni.
െ ൌ ; Loni is miles closer to Watertown than Daryl.
Claude recently read that no one has ever scuba dived more than meters below sea level. Describe what this means in terms of elevation using sea level as a reference point.
meters below sea level is an elevation of െ meters. “More than meters below sea level” means that no diver has ever had more than meters between himself and sea level when he was below the water’s surface while scuba diving.
Problem Set Sample Solutions 1. Negative air pressure created by an air pump makes a vacuum cleaner able to collect air and dirt into a bag or other container. Below are several readings from a pressure gauge. Write rational numbers to represent each of the readings, and then order the rational numbers from least to greatest. Gauge Readings (pounds per square inch)
psi pressure
psi vacuum
Ǥ psi vacuum
ૠǤ ૡ psi vacuum
Ǥૢ psi vacuum
psi pressure
ૠǤ ૡ psi pressure Pressure Readings (pounds per square inch) െ െǤ െૠǤ ૡ െǤૢ ૠǤ ૡ
െ ൏ െૠǤ ૡ ൏ െǤ ൏ െǤૢ ൏ ൏ ૠǤ ૡ ൏
െ െ െ െૡ ૡ
Loni Watertown
െ
Daryl
miles miles
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6ͻϯ >ĞƐƐŽŶϭϯ
ĞƐƐŽŶϭϯ : ^ƚĂƚĞŵĞŶƚƐŽĨKƌĚĞƌŝŶƚŚĞZĞĂůtŽƌůĚ
The fuel gauge in Nic’s car says that he has miles to go until his tank is empty. He passed a fuel station ૢ miles ago, and a sign says there is a town only ૡ miles ahead. If he takes a chance and drives ahead to the town and there isn’t a fuel station there, does he have enough fuel to go back to the last station? Include a diagram along a number line, and use absolute value to find your answer.
No, he does not have enough fuel to drive to the town and then back to the fuel station. He needs ૡ miles’ worth of fuel to get to the town, which lowers his limit to ૡ miles. The total distance between the fuel station and the town is ૠ miles; ȁૡȁ ȁെૢ ȁ ൌ ૡ ૢ ൌ ૠ . Nic would be ૢ miles short on fuel. It would be safer to go back to the fuel station without going to the town first.
െૢ ૡ
Fuel Station Nic Town
A STORY OF RATIOS
125
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
ZŝǀĞƌtĂƚĞƌ
ͳǤͲ
ͲǤͷ
Ͳ
െʹǤͷ
െͲǤͷ
ͳǤͷ
െͳǤͲ
െͳǤͷ
െʹǤͲ
െ͵Ǥͷ
െ͵ǤͲ EĂŵĞ ĂƚĞ ϭ͘ dŚĞƉŝĐƚƵƌĞďĞůŽǁŝƐĂĨůŽŽĚŐĂƵŐĞƚŚĂƚŝƐƵƐĞĚƚŽŵĞĂƐƵƌĞŚŽǁĨĂƌ;ŝŶĨĞĞƚͿĂƌŝǀĞƌ͛ƐǁĂƚĞƌůĞǀĞůŝƐĂďŽǀĞ ŽƌďĞůŽǁŝƚƐŶŽƌŵĂůůĞǀĞů͘ Ă͘ džƉůĂŝŶǁŚĂƚƚŚĞŶƵŵďĞƌ ͲŽŶƚŚĞŐĂƵŐĞƌĞƉƌĞƐĞŶƚƐ͕ĂŶĚĞdžƉůĂŝŶǁŚĂƚƚŚĞŶƵŵďĞƌƐ ĂďŽǀĞĂŶĚďĞůŽǁ ͲƌĞƉƌĞƐĞŶƚ͘ ď͘ ĞƐĐƌŝďĞǁŚĂƚƚŚĞƉŝĐƚƵƌĞŝŶĚŝĐĂƚĞƐĂďŽƵƚƚŚĞƌŝǀĞƌ͛Ɛ ĐƵƌƌĞŶƚǁĂƚĞƌůĞǀĞů͘ Đ͘ tŚĂƚŶƵŵďĞƌƌĞƉƌĞƐĞŶƚƐƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞǁĂƚĞƌůĞǀĞůƐŚŽǁŶŝŶƚŚĞƉŝĐƚƵƌĞ͕ĂŶĚǁŚĞƌĞŝƐŝƚ ůŽĐĂƚĞĚŽŶƚŚĞŐĂƵŐĞ͍tŚĂƚǁŽƵůĚŝƚŵĞĂŶŝĨƚŚĞƌŝǀĞƌǁĂƚĞƌǁĂƐĂƚƚŚĂƚůĞǀĞů͍ Ě͘ /ĨŚĞĂǀLJƌĂŝŶŝƐŝŶƚŚĞĨŽƌĞĐĂƐƚĨŽƌƚŚĞĂƌĞĂĨŽƌƚŚĞŶĞdžƚ ʹͶ ŚŽƵƌƐ͕ǁŚĂƚƌĞĂĚŝŶŐŵŝŐŚƚLJŽƵĞdžƉĞĐƚƚŽ ƐĞĞŽŶƚŚŝƐŐĂƵŐĞƚŽŵŽƌƌŽǁ͍ džƉůĂŝŶLJŽƵƌƌĞĂƐŽŶŝŶŐ͘
A STORY OF RATIOS
126
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
Ϯ͘ /ƐĂĂĐŵĂĚĞĂŵŝƐƚĂŬĞŝŶŚŝƐĐŚĞĐŬŬ͘,ĞǁƌŽƚĞĂĐŚĞĐŬĨŽƌ ̈́ ͺǤͻͺ ƚŽƌĞŶƚĂǀŝĚĞŽŐĂŵĞďƵƚŵŝƐƚĂŬĞŶůLJ ƌĞĐŽƌĚĞĚŝƚŝŶŚŝƐĐŚĞĐŬŬĂƐĂŶ ̈́ͺǤͻͺ ĚĞƉŽƐŝƚ͘ Ă͘ ZĞƉƌĞƐĞŶƚĞĂĐŚƚƌĂŶƐĂĐƚŝŽŶǁŝƚŚĂƌĂƚŝŽŶĂůŶƵŵďĞƌ͕ĂŶĚĞdžƉůĂŝŶƚŚĞĚŝĨĨĞƌĞŶĐĞďĞƚǁĞĞŶƚŚĞ ƚƌĂŶƐĂĐƚŝŽŶƐ͘ ď͘ KŶƚŚĞŶƵŵďĞƌůŝŶĞďĞůŽǁ͕ůŽĐĂƚĞĂŶĚůĂďĞůƚŚĞƉŽŝŶƚƐƚŚĂƚƌĞƉƌĞƐĞŶƚƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐůŝƐƚĞĚŝŶ ƉĂƌƚ;ĂͿ͘ĞƐĐƌŝďĞƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞƐĞƚǁŽŶƵŵďĞƌƐ͘ĞƌŽŽŶƚŚĞŶƵŵďĞƌůŝŶĞƌĞƉƌĞƐĞŶƚƐ /ƐĂĂĐ͛ƐďĂůĂŶĐĞďĞĨŽƌĞƚŚĞŵŝƐƚĂŬĞǁĂƐŵĂĚĞ͘ Đ͘ hƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽĞdžƉůĂŝŶŚŽǁĂĚĞďŝƚŽĨ ̈́ ͺǤͻͺ ĂŶĚĂĐƌĞĚŝƚŽĨ ̈́ͺǤͻͺ ĂƌĞƐŝŵŝůĂƌ͘
Ͳ
A STORY OF RATIOS
127
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
ϯ͘ ůŽĐĂůƉĂƌŬ͛ƐƉƌŽŐƌĂŵƐĐŽŵŵŝƚƚĞĞŝƐƌĂŝƐŝŶŐŵŽŶĞLJďLJŚŽůĚŝŶŐŵŽƵŶƚĂŝŶďŝŬĞƌĂĐĞƐŽŶĂĐŽƵƌƐĞƚŚƌŽƵŐŚ ƚŚĞƉĂƌŬ͘ƵƌŝŶŐĞĂĐŚƌĂĐĞ͕ĂĐŽŵƉƵƚĞƌƚƌĂĐŬƐƚŚĞĐŽŵƉĞƚŝƚŽƌƐ͛ůŽĐĂƚŝŽŶƐŽŶƚŚĞĐŽƵƌƐĞƵƐŝŶŐ'W^ ƚƌĂĐŬŝŶŐ͘dŚĞƚĂďůĞƐŚŽǁƐŚŽǁĨĂƌĞĂĐŚĐŽŵƉĞƚŝƚŽƌŝƐĨƌŽŵĂĐŚĞĐŬƉŽŝŶƚ͘ EƵŵďĞƌ ŽŵƉĞƚŝƚŽƌEĂŵĞ ŝƐƚĂŶĐĞƚŽŚĞĐŬƉŽŝŶƚ
ʹʹ͵ &ůŽƌĞŶĐĞ ͲǤͳ ŵŝůĞƐďĞĨŽƌĞ
ʹ͵ͳ DĂƌLJ ଶହŵŝůĞƐƉĂƐƚ
ʹͶͲ ZĞďĞĐĐĂ ͲǤͷ ŵŝůĞƐďĞĨŽƌĞ
ʹͶͻ >ŝƚĂ ଵଶŵŝůĞƐƉĂƐƚ
ʹͷͷ EĂŶĐLJ ଶଵ ŵŝůĞƐďĞĨŽƌĞ Ă͘ dŚĞĐŚĞĐŬƉŽŝŶƚŝƐƌĞƉƌĞƐĞŶƚĞĚďLJ ͲŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘>ŽĐĂƚĞĂŶĚůĂďĞůƉŽŝŶƚƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ ĨŽƌƚŚĞƉŽƐŝƚŝŽŶƐŽĨĞĂĐŚůŝƐƚĞĚƉĂƌƚŝĐŝƉĂŶƚ͘>ĂďĞůƚŚĞƉŽŝŶƚƐƵƐŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘ ď͘ tŚŝĐŚŽĨƚŚĞĐŽŵƉĞƚŝƚŽƌƐŝƐĐůŽƐĞƐƚƚŽƚŚĞĐŚĞĐŬƉŽŝŶƚ͍ džƉůĂŝŶ͘ Đ͘ dǁŽĐŽŵƉĞƚŝƚŽƌƐĂƌĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵƚŚĞĐŚĞĐŬƉŽŝŶƚ͘ƌĞƚŚĞLJŝŶƚŚĞƐĂŵĞůŽĐĂƚŝŽŶ͍ džƉůĂŝŶ͘ Ě͘ tŚŽŝƐĐůŽƐĞƌƚŽĨŝŶŝƐŚŝŶŐƚŚĞƌĂĐĞ͕EĂŶĐLJŽƌ&ůŽƌĞŶĐĞ͍^ƵƉƉŽƌƚLJŽƵƌĂŶƐǁĞƌ͘
ͲŚĞĐŬƉŽŝŶƚ
A STORY OF RATIOS
128
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
ϰ͘ ŶĚƌĠĂĂŶĚDĂƌƚĂĂƌĞƚĞƐƚŝŶŐƚŚƌĞĞĚŝĨĨĞƌĞŶƚĐŽŽůĞƌƐƚŽƐĞĞǁŚŝĐŚŬĞĞƉƐƚŚĞĐŽůĚĞƐƚƚĞŵƉĞƌĂƚƵƌĞ͘dŚĞLJ ƉůĂĐĞĚĂďĂŐŽĨŝĐĞŝŶĞĂĐŚĐŽŽůĞƌ͕ĐůŽƐĞĚƚŚĞĐŽŽůĞƌƐ͕ĂŶĚƚŚĞŶŵĞĂƐƵƌĞĚƚŚĞĂŝƌƚĞŵƉĞƌĂƚƵƌĞŝŶƐŝĚĞĞĂĐŚ ĂĨƚĞƌ ͻͲ ŵŝŶƵƚĞƐ͘dŚĞƚĞŵƉĞƌĂƚƵƌĞƐĂƌĞƌĞĐŽƌĚĞĚŝŶƚŚĞƚĂďůĞďĞůŽǁ͗ ŽŽůĞƌ dĞŵƉĞƌĂƚƵƌĞ; ஈͿ െʹǤͻͳ ͷǤ െͶǤ͵
DĂƌƚĂǁƌŽƚĞƚŚĞĨŽůůŽǁŝŶŐŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚĂďŽƵƚƚŚĞƚĞŵƉĞƌĂƚƵƌĞƐ͗
െͶǤ͵ ൏ െʹǤͻͳ ൏ ͷǤ ͘ŶĚƌĠĂĐůĂŝŵƐƚŚĂƚDĂƌƚĂŵĂĚĞĂŵŝƐƚĂŬĞŝŶŚĞƌƐƚĂƚĞŵĞŶƚĂŶĚƚŚĂƚƚŚĞŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƐŚŽƵůĚďĞ ǁƌŝƚƚĞŶĂƐ
െʹǤͻͳ ൏ െͶǤ͵ ൏ ͷǤ ͘Ă͘ /ƐĞŝƚŚĞƌƐƚƵĚĞŶƚĐŽƌƌĞĐƚ͍ džƉůĂŝŶ͘ ď͘ dŚĞƐƚƵĚĞŶƚƐǁĂŶƚƚŽĨŝŶĚĂĐŽŽůĞƌƚŚĂƚŬĞĞƉƐƚŚĞƚĞŵƉĞƌĂƚƵƌĞŝŶƐŝĚĞƚŚĞĐŽŽůĞƌŵŽƌĞƚŚĂŶ
͵ĚĞŐƌĞĞƐďĞůŽǁƚŚĞĨƌĞĞnjŝŶŐƉŽŝŶƚŽĨǁĂƚĞƌ; Ͳᤪ ͿĂĨƚĞƌ ͻͲ ŵŝŶƵƚĞƐ͘/ŶĚŝĐĂƚĞǁŚŝĐŚŽĨƚŚĞƚĞƐƚĞĚ ĐŽŽůĞƌƐŵĞĞƚƐƚŚŝƐŐŽĂů͕ĂŶĚĞdžƉůĂŝŶǁŚLJ͘
A STORY OF RATIOS
129
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
ϱ͘ DĂƌLJŵĂŶĂŐĞƐĂĐŽŵƉĂŶLJƚŚĂƚŚĂƐďĞĞŶŚŝƌĞĚƚŽĨůĂƚƚĞŶĂƉůŽƚŽĨůĂŶĚ͘^ŚĞƚŽŽŬƐĞǀĞƌĂůĞůĞǀĂƚŝŽŶ ƐĂŵƉůĞƐĨƌŽŵƚŚĞůĂŶĚĂŶĚƌĞĐŽƌĚĞĚƚŚŽƐĞĞůĞǀĂƚŝŽŶƐďĞůŽǁ͗ ůĞǀĂƚŝŽŶ^ĂŵƉůĞ & ůĞǀĂƚŝŽŶ ;Ĩƚ͘ĂďŽǀĞƐĞĂůĞǀĞůͿ ͺʹǤͷ ͺ͵ͲǤʹ ͺ͵ʹǤͲ ͺ͵ͳǤͳ ͺʹͷǤͺ ͺʹǤͳ
Ă͘ dŚĞůĂŶĚŽǁŶĞƌǁĂŶƚƐƚŚĞůĂŶĚĨůĂƚĂŶĚĂƚƚŚĞƐĂŵĞůĞǀĞůĂƐƚŚĞƌŽĂĚƚŚĂƚƉĂƐƐĞƐŝŶĨƌŽŶƚŽĨŝƚ͘dŚĞ ƌŽĂĚ͛ƐĞůĞǀĂƚŝŽŶŝƐ ͺ͵Ͳ ĨĞĞƚĂďŽǀĞƐĞĂůĞǀĞů͘ĞƐĐƌŝďĞŝŶǁŽƌĚƐŚŽǁĞůĞǀĂƚŝŽŶƐĂŵƉůĞƐ͕͕ĂŶĚ ĐŽŵƉĂƌĞƚŽƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞƌŽĂĚ͘ ď͘ dŚĞƚĂďůĞďĞůŽǁƐŚŽǁƐŚŽǁƐŽŵĞŽƚŚĞƌĞůĞǀĂƚŝŽŶƐĂŵƉůĞƐĐŽŵƉĂƌĞƚŽƚŚĞůĞǀĞůŽĨƚŚĞƌŽĂĚ͗ ůĞǀĂƚŝŽŶ^ĂŵƉůĞ ' , / : < > ůĞǀĂƚŝŽŶ ;Ĩƚ͘ĨƌŽŵƚŚĞƌŽĂĚͿ ͵Ǥͳ െͲǤͷ ʹǤʹ ͳǤ͵ െͶǤͷ െͲǤͻ
tƌŝƚĞƚŚĞǀĂůƵĞƐŝŶƚŚĞƚĂďůĞŝŶŽƌĚĞƌĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͘ ͺͺͺͺͺͺͺͺͺ ൏ͺͺͺͺͺͺͺͺͺ ൏ͺͺͺͺͺͺͺͺͺ ൏ͺͺͺͺͺͺͺͺͺ ൏ͺͺͺͺͺͺͺͺͺ ൏ͺͺͺͺͺͺͺͺͺ Đ͘ /ŶĚŝĐĂƚĞǁŚŝĐŚŽĨƚŚĞǀĂůƵĞƐĨƌŽŵƚŚĞƚĂďůĞŝŶƉĂƌƚ;ďͿŝƐĨĂƌƚŚĞƐƚĨƌŽŵƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞƌŽĂĚ͘ hƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽĞdžƉůĂŝŶLJŽƵƌĂŶƐǁĞƌ͘
A STORY OF RATIOS
130
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task A Progression Toward Mastery Assessment Task Item STEP 1 Missing or incorrect answer and little evidence of reasoning or application of mathematics to solve the problem. STEP 2 Missing or incorrect answer but evidence of some reasoning or application of mathematics to solve the problem. STEP 3 A correct answer with some evidence of reasoning or application of mathematics to solve the problem, or an incorrect answer with substantial evidence of solid reasoning or application of mathematics to solve the problem. STEP 4 A correct answer supported by substantial evidence of solid reasoning or application of mathematics to solve the problem. 1 a
^ƚƵĚĞŶƚŝƐĂďůĞƚŽ ĚĞƚĞƌŵŝŶĞƚŚĂƚƚŚĞ ŐŝǀĞŶǁĂƚĞƌůĞǀĞůŝƐ ďĞůŽǁŶŽƌŵĂů;ŽƌůŽǁ ǁĂƚĞƌͿďƵƚĚŽĞƐŶŽƚ ŝŶĚŝĐĂƚĞĂĐůĞĂƌ ƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨnjĞƌŽ ŽƌƚŚĞŶƵŵďĞƌƐĂďŽǀĞ ĂŶĚďĞůŽǁnjĞƌŽŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞ͘ KZ ^ƚƵĚĞŶƚŝƐƵŶĂďůĞƚŽ ĚĞƚĞƌŵŝŶĞƚŚĂƚƚŚĞ ǁĂƚĞƌůĞǀĞůŝƐďĞůŽǁ ŶŽƌŵĂů͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚ ͲƌĞƉƌĞƐĞŶƚƐƚŚĞ ŶŽƌŵĂůǁĂƚĞƌůĞǀĞůďƵƚ ĚŽĞƐŶŽƚĐůĞĂƌůLJĚĞƐĐƌŝďĞ ƚŚĞŵĞĂŶŝŶŐƐŽĨŶƵŵďĞƌƐ ĂďŽǀĞĂŶĚďĞůŽǁnjĞƌŽŽŶ ƚŚĞŶƵŵďĞƌůŝŶĞ͘
^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚ ͲƌĞƉƌĞƐĞŶƚƐƚŚĞ ŶŽƌŵĂůǁĂƚĞƌůĞǀĞůĂŶĚ ƚŚĂƚĞŝƚŚĞƌƚŚĞŶƵŵďĞƌƐ ĂďŽǀĞnjĞƌŽƌĞƉƌĞƐĞŶƚ ĂďŽǀĞŶŽƌŵĂůǁĂƚĞƌ ůĞǀĞůƐŽƌƚŚĞŶƵŵďĞƌƐ ďĞůŽǁnjĞƌŽƌĞƉƌĞƐĞŶƚ ďĞůŽǁŶŽƌŵĂůǁĂƚĞƌ ůĞǀĞůƐďƵƚĚŽĞƐŶŽƚ ĐůĞĂƌůLJĚĞƐĐƌŝďĞďŽƚŚ͘
^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚ ͲƌĞƉƌĞƐĞŶƚƐŶŽƌŵĂů ǁĂƚĞƌůĞǀĞů͕ŶƵŵďĞƌƐ ĂďŽǀĞnjĞƌŽ;ŽƌƉŽƐŝƚŝǀĞ ŶƵŵďĞƌƐͿƌĞƉƌĞƐĞŶƚ ĂďŽǀĞŶŽƌŵĂůǁĂƚĞƌ ůĞǀĞůƐ͕ĂŶĚŶƵŵďĞƌƐ ďĞůŽǁnjĞƌŽ;ŽƌŶĞŐĂƚŝǀĞ ŶƵŵďĞƌƐͿƌĞƉƌĞƐĞŶƚ ďĞůŽǁŶŽƌŵĂůǁĂƚĞƌ ůĞǀĞůƐ͘
b ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐŽƌŝŶĐŽŵƉůĞƚĞ͘ &ŽƌŝŶƐƚĂŶĐĞ͕ƐƚƵĚĞŶƚ ŵĂŬĞƐĂŐĞŶĞƌĂů ƐƚĂƚĞŵĞŶƚƚŚĂƚƚŚĞ ƌŝǀĞƌ͛ƐǁĂƚĞƌůĞǀĞůŝƐ ďĞůŽǁŶŽƌŵĂůďƵƚĚŽĞƐ ŶŽƚƌĞĨĞƌƚŽŵĂŐŶŝƚƵĚĞ͕ ĚŝƌĞĐƚŝŽŶ͕ŽƌĂůŽĐĂƚŝŽŶ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽŵƉůĞƚĞďƵƚƐŚŽǁƐ ƐŽŵĞĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐƐƵĐŚĂƐ ƐƚĂƚŝŶŐƚŚĂƚƚŚĞǁĂƚĞƌ ůĞǀĞůŝƐĂƚ െʹ Žƌ െͳǤͻ Žƌ ƚŚĂƚƚŚĞǁĂƚĞƌůĞǀĞůŝƐ
ʹĨĞĞƚ;Žƌ ͳǤͻ ĨĞĞƚͿďƵƚ ǁŝƚŚŽƵƚĚĞƚĂŝůƐƐƵĐŚĂƐ ƵŶŝƚƐŽĨŵĞĂƐƵƌĞŵĞŶƚŽƌ ĚŝƌĞĐƚŝŽŶ͘
^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ŝŶƚĞƌƉƌĞƚƐƚŚĞƉŝĐƚƵƌĞƚŽ ŝŶĚŝĐĂƚĞƚŚĂƚƚŚĞĐƵƌƌĞŶƚ ǁĂƚĞƌůĞǀĞůŝƐďĞůŽǁ ŶŽƌŵĂůďƵƚƐƚĂƚĞƐŝƚŝƐ
exactly ʹĨĞĞƚďĞůŽǁ ŶŽƌŵĂůǁĂƚĞƌůĞǀĞů ƌĂƚŚĞƌƚŚĂŶ nearly ʹĨĞĞƚ ďĞůŽǁŶŽƌŵĂůǁĂƚĞƌ ůĞǀĞů͘
^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽƌƌĞĐƚĂŶĚĐŽŵƉůĞƚĞ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚƚŚĞ ƉŝĐƚƵƌĞŝŶĚŝĐĂƚĞƐƚŚĂƚƚŚĞ ƌŝǀĞƌ͛ƐĐƵƌƌĞŶƚǁĂƚĞƌ ůĞǀĞůŝƐ about ʹĨĞĞƚ ďĞůŽǁŶŽƌŵĂůǁĂƚĞƌ ůĞǀĞů͘
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©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task c
^ƚƵĚĞŶƚĂŶƐǁĞƌŝƐ ŝŶĐŽŵƉůĞƚĞŽƌŵŝƐƐŝŶŐ͘ džƉůĂŶĂƚŝŽŶƐŚŽǁƐůŝƚƚůĞ ŽƌŶŽĞǀŝĚĞŶĐĞŽĨŚŽǁ ƚŽĨŝŶĚŽƉƉŽƐŝƚĞƐŽŶĂ ŶƵŵďĞƌůŝŶĞ͘ ^ƚƵĚĞŶƚĐŽŵƉůĞƚĞƐƚŚĞ ĨŝƌƐƚƐƚĞƉƐƚĂƚŝŶŐƚŚĂƚƚŚĞ ŽƉƉŽƐŝƚĞŽĨ െʹ ŝƐ ʹ;Žƌ ƚŚĞŽƉƉŽƐŝƚĞŽĨ െͳǤͻ ŝƐ
ͳǤͻ ͕ĞƚĐ͘ͿďƵƚǁŝƚŚŶŽ ĨƵƌƚŚĞƌĚĞƚĂŝůƐŽƌĐŽƌƌĞĐƚ ƐƚĂƚĞŵĞŶƚƐ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚƚŚĞŽƉƉŽƐŝƚĞŽĨ െʹ ŝƐ ʹ;ŽƌƚŚĞŽƉƉŽƐŝƚĞŽĨ
െͳǤͻ ŝƐ ͳǤͻ ͕ĞƚĐ͘ͿĂŶĚ ƚŚĂƚŝŶƚŚĞŽƉƉŽƐŝƚĞ ƐŝƚƵĂƚŝŽŶƚŚĞƌŝǀĞƌ͛Ɛ ǁĂƚĞƌůĞǀĞůǁŽƵůĚďĞ ŚŝŐŚĞƌƚŚĂŶŶŽƌŵĂůďƵƚ ĚŽĞƐŶŽƚĐůĞĂƌůLJĚĞƐĐƌŝďĞ ŝƚƐůŽĐĂƚŝŽŶŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞͬŐĂƵŐĞ͘ KZ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ŝĚĞŶƚŝĨŝĞƐĂŶĚĐůĞĂƌůLJ ĚĞƐĐƌŝďĞƐƚŚĞůŽĐĂƚŝŽŶŽĨ ƚŚĞŽƉƉŽƐŝƚĞŶƵŵďĞƌŽŶ ƚŚĞŶƵŵďĞƌůŝŶĞͬŐĂƵŐĞ ďƵƚĚŽĞƐŶŽƚĂĚĚƌĞƐƐ ǁŚĂƚƚŚŝƐůĞǀĞůǁŽƵůĚ ŵĞĂŶŝŶƚŚĞĐŽŶƚĞdžƚŽĨ ƚŚĞƐŝƚƵĂƚŝŽŶ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ĂĚĚƌĞƐƐĞƐĂůůƉĂƌƚƐŽĨƚŚĞ ƋƵĞƐƚŝŽŶ͘^ƚƵĚĞŶƚ ĐŽƌƌĞĐƚůLJƐƚĂƚĞƐƚŚĂƚƚŚĞ ŽƉƉŽƐŝƚĞŽĨ െʹ ŝƐ ʹ;Žƌ ƚŚĞŽƉƉŽƐŝƚĞŽĨ െͳǤͻ ŝƐ
ͳǤͻ ͕ĞƚĐ͘ͿĂŶĚĞdžƉůĂŝŶƐ ǁŚĞƌĞƚŚĞƉŽƐŝƚŝǀĞ ŶƵŵďĞƌŝƐůŽĐĂƚĞĚ ƐƉĞĐŝĨLJŝŶŐƚŚĞŶƵŵďĞƌŽĨ ƵŶŝƚƐĂďŽǀĞ ͲŽƌŽŶƚŚĞ ŽƉƉŽƐŝƚĞƐŝĚĞŽĨnjĞƌŽ ĨƌŽŵƚŚĞŶĞŐĂƚŝǀĞǀĂůƵĞ͘ ^ƚƵĚĞŶƚĂůƐŽƐƚĂƚĞƐƚŚĂƚ ƚŚĞƉŽƐŝƚŝǀĞŶƵŵďĞƌ ǁŽƵůĚŵĞĂŶƚŚĞƌŝǀĞƌ͛Ɛ ǁĂƚĞƌůĞǀĞůŝƐƚŚĂƚŵĂŶLJ ĨĞĞƚŚŝŐŚĞƌƚŚĂŶƚŚĞ ŶŽƌŵĂůůĞǀĞů͘
d ^ƚƵĚĞŶƚĞdžƉůĂŶĂƚŝŽŶŝƐ ŵŝƐƐŝŶŐ͘ KZ dŚĞǁƌŝƚƚĞŶĞdžƉůĂŶĂƚŝŽŶ ĚĞŵŽŶƐƚƌĂƚĞƐůŝƚƚůĞŽƌ ŶŽĐŽƌƌĞĐƚ ŵĂƚŚĞŵĂƚŝĐĂů ŝŶƚĞƌƉƌĞƚĂƚŝŽŶŽĨƚŚĞ ƐŝƚƵĂƚŝŽŶ͕ƐƵĐŚĂƐ ĐůĂŝŵŝŶŐƚŚĂƚ ƚŽŵŽƌƌŽǁ͛ƐǁĂƚĞƌůĞǀĞů ǁŽƵůĚďĞďĞůŽǁƚŚĞ ĐƵƌƌĞŶƚůĞǀĞůƐŚŽǁŶ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚƚŚĞǁĂƚĞƌůĞǀĞů ǁŽƵůĚďĞŚŝŐŚĞƌƚŚĂŶƚŚĞ ůĞǀĞůƐŚŽǁŶďƵƚĚŽĞƐŶŽƚ ƉƌŽǀŝĚĞĂƐƉĞĐŝĨŝĐƌĞĂĚŝŶŐ ;ůĞǀĞůͿĂŶĚĚŽĞƐŶŽƚ ƉƌŽǀŝĚĞĂĚĞƋƵĂƚĞ ƌĞĂƐŽŶŝŶŐƚŽƐƵƉƉŽƌƚƚŚĞ ĐůĂŝŵ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚƚŚĞǁĂƚĞƌǁŽƵůĚ ƌŝƐĞƚŽĂƐƉĞĐŝĨŝĐůĞǀĞů ŚŝŐŚĞƌƚŚĂŶƚŚĞůĞǀĞů ƐŚŽǁŶĂŶĚŝĚĞŶƚŝĨŝĞƐƚŚĞ ŶĞǁůĞǀĞůďƵƚĨĂŝůƐƚŽ ƉƌŽǀŝĚĞĂĐůĞĂƌ ĞdžƉůĂŶĂƚŝŽŶƚŽƐƵƉƉŽƌƚ ƚŚĞĐůĂŝŵ͘ KZ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚƚŚĞǁĂƚĞƌůĞǀĞů ǁŽƵůĚďĞŚŝŐŚĞƌƚŚĂŶƚŚĞ ůĞǀĞůƐŚŽǁŶĂŶĚƉƌŽǀŝĚĞƐ ĂĚĞƋƵĂƚĞƌĞĂƐŽŶŝŶŐƚŽ ƐƵƉƉŽƌƚƚŚĞĐůĂŝŵďƵƚ ĚŽĞƐŶŽƚƉƌŽǀŝĚĞĂ ƐƉĞĐŝĨŝĐƌĞĂĚŝŶŐ;ůĞǀĞůͿ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽŵƉůĞƚĞĂŶĚĐŽƌƌĞĐƚ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚƚŚĞ ǁĂƚĞƌůĞǀĞůǁŽƵůĚƌŝƐĞƚŽ ĂƐƉĞĐŝĨŝĐůĞǀĞůŚŝŐŚĞƌ ƚŚĂŶƚŚĞůĞǀĞůƐŚŽǁŶ͕ ŝĚĞŶƚŝĨŝĞƐĂƐƉĞĐŝĨŝĐŶĞǁ ůĞǀĞů͕ĂŶĚƉƌŽǀŝĚĞƐĂ ĐůĞĂƌĞdžƉůĂŶĂƚŝŽŶƚŽ ƐƵƉƉŽƌƚƚŚĞĐůĂŝŵ͘
2 a
^ƚƵĚĞŶƚĂŶƐǁĞƌŝƐ ŝŶĐŽƌƌĞĐƚŽƌŵŝƐƐŝŶŐ͘ ^ƚƵĚĞŶƚŶĞŝƚŚĞƌĂƌƌŝǀĞƐ ĂƚďŽƚŚƌĂƚŝŽŶĂůŶƵŵďĞƌ ƌĞƉƌĞƐĞŶƚĂƚŝŽŶƐŶŽƌ ƉƌŽǀŝĚĞƐĂĐŽƌƌĞĐƚ ĞdžƉůĂŶĂƚŝŽŶŽĨƚŚĞ ĚŝĨĨĞƌĞŶĐĞŝŶƚŚĞƚǁŽ ƚƌĂŶƐĂĐƚŝŽŶƐ͕ĂůƚŚŽƵŐŚ ŽŶĞƚƌĂŶƐĂĐƚŝŽŶŵĂLJ ŚĂǀĞďĞĞŶƌĞƉƌĞƐĞŶƚĞĚ ǁŝƚŚĂĐŽƌƌĞĐƚƌĂƚŝŽŶĂů ŶƵŵďĞƌ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJĞdžƉůĂŝŶƐ ƚŚĞĚŝĨĨĞƌĞŶĐĞŝŶƚŚĞƚǁŽ ƚƌĂŶƐĂĐƚŝŽŶƐďƵƚŵĂŬĞƐ ĂŶĞƌƌŽƌŝŶƌĞƉƌĞƐĞŶƚŝŶŐ ŽŶĞŽƌďŽƚŚƚƌĂŶƐĂĐƚŝŽŶƐ ĂƐƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘ KZ ^ƚƵĚĞŶƚƵƐĞƐƚŚĞĐŽƌƌĞĐƚ ƚǁŽƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŽ ƌĞƉƌĞƐĞŶƚƚŚĞ ƚƌĂŶƐĂĐƚŝŽŶƐďƵƚĚŽĞƐŶŽƚ ĞdžƉůĂŝŶƚŚĞĚŝĨĨĞƌĞŶĐĞŝŶ ƚŚĞƚƌĂŶƐĂĐƚŝŽŶƐ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ƌĞƉƌĞƐĞŶƚƐƚŚĞĐŚĞĐŬĂƐ
െͺǤͻͺ ĂŶĚƚŚĞĚĞƉŽƐŝƚ ĂƐ ͺǤͻͺ ͘^ƚƵĚĞŶƚ ĞdžƉůĂŝŶƐƚŚĂƚƚŚĞĐŚĞĐŬ ĚĞĐƌĞĂƐĞƐƚŚĞĂĐĐŽƵŶƚ ďĂůĂŶĐĞŽƌƚŚĂƚƚŚĞ ĚĞƉŽƐŝƚŝŶĐƌĞĂƐĞƐƚŚĞ ĂĐĐŽƵŶƚďĂůĂŶĐĞďƵƚŶŽƚ ďŽƚŚ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽŵƉůĞƚĞĂŶĚĐŽƌƌĞĐƚ͘ dŚĞĐŚĞĐŬŝƐƌĞƉƌĞƐĞŶƚĞĚ ĂƐ െͺǤͻͺ ͕ĂŶĚƚŚĞ ĚĞƉŽƐŝƚŝƐƌĞƉƌĞƐĞŶƚĞĚĂƐ
ͺǤͻͺ ͘^ƚƵĚĞŶƚƉƌŽǀŝĚĞƐĂ ĐůĞĂƌĂŶĚĂĐĐƵƌĂƚĞ ĞdžƉůĂŶĂƚŝŽŶŽĨƚŚĞ ĚŝĨĨĞƌĞŶĐĞŝŶƚŚĞ ƚƌĂŶƐĂĐƚŝŽŶƐ͘
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©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task b ^ƚƵĚĞŶƚǁŽƌŬŝƐŵŝƐƐŝŶŐ ŽƌŝŶĐŽŵƉůĞƚĞ͘^ƚƵĚĞŶƚ ĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐĂŶĚͬŽƌ ůĂďĞůƐŽŶĞƉŽŝŶƚŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞďƵƚƐŚŽǁƐ ŶŽŽƚŚĞƌĂĐĐƵƌĂƚĞǁŽƌŬ͘ ^ƚƵĚĞŶƚƐŚŽǁƐŝŶƚĞŶƚƚŽ ŐƌĂƉŚ െͺǤͻͺ ĂŶĚ
ͺǤͻͺ ĐŽƌƌĞĐƚůLJŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞďƵƚĚŽĞƐŶŽƚ ĂĐĐƵƌĂƚĞůLJůŽĐĂƚĞďŽƚŚ ƉŽŝŶƚƐĂŶĚͬŽƌŚĂƐĂŶ ĞƌƌŽƌŝŶƚŚĞƐĐĂůĞ͘ ^ƚƵĚĞŶƚĚŽĞƐŶŽƚƐƚĂƚĞ ƚŚĂƚƚŚĞŶƵŵďĞƌƐĂƌĞ ŽƉƉŽƐŝƚĞƐ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ ĂŶĚůĂďĞůƐ െͺǤͻͺ ĂŶĚ
ͺǤͻͺ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ ďƵƚĚŽĞƐŶŽƚƐƚĂƚĞƚŚĂƚ ƚŚĞŶƵŵďĞƌƐĂƌĞ ŽƉƉŽƐŝƚĞƐ͘ KZ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ
െͺǤͻͺ ĂŶĚ ͺǤͻͺ ŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞĂŶĚƐƚĂƚĞƐ ƚŚĂƚƚŚĞŶƵŵďĞƌƐĂƌĞ ŽƉƉŽƐŝƚĞƐďƵƚĚŽĞƐŶŽƚ ůĂďĞůƚŚĞƉŽŝŶƚƐ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽƌƌĞĐƚĂŶĚĐŽŵƉůĞƚĞ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ ĂŶĚůĂďĞůƐ െͺǤͻͺ ĂŶĚ
ͺǤͻͺ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ ĂŶĚƐƚĂƚĞƐƚŚĂƚƚŚĞ ŶƵŵďĞƌƐĂƌĞŽƉƉŽƐŝƚĞƐ͘
c ^ƚƵĚĞŶƚĞdžƉůĂŶĂƚŝŽŶ ƐŚŽǁƐůŝƚƚůĞŽƌŶŽ ĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐƚŚĞ ĐŽŶĐĞƉƚŽĨĂďƐŽůƵƚĞ ǀĂůƵĞ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ŝŶĚŝĐĂƚĞƐƐŽŵĞĞǀŝĚĞŶĐĞ ŽĨĐŽƌƌĞĐƚƌĞĂƐŽŶŝŶŐƐƵĐŚ ĂƐƐƚĂƚŝŶŐƚŚĂƚƚŚĞ ƚƌĂŶƐĂĐƚŝŽŶƐĐŚĂŶŐĞƚŚĞ ĂĐĐŽƵŶƚďĂůĂŶĐĞďLJƚŚĞ ƐĂŵĞĂŵŽƵŶƚŽĨŵŽŶĞLJ͕ ďƵƚƚŚĞĞdžƉůĂŶĂƚŝŽŶŝƐ ŝŶĐŽŵƉůĞƚĞĂŶĚĚŽĞƐŶŽƚ ŝŶĐůƵĚĞĂĚŝƌĞĐƚ ƌĞĨĞƌĞŶĐĞƚŽĂďƐŽůƵƚĞ ǀĂůƵĞ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚĂ ĚĞďŝƚŽĨ ̈́ ͺǤͻͺ ĂŶĚĐƌĞĚŝƚ ŽĨ ̈́ͺǤͻͺ ĂƌĞƐŝŵŝůĂƌ ďĞĐĂƵƐĞďŽƚŚŚĂǀĞƚŚĞ ƐĂŵĞĂďƐŽůƵƚĞǀĂůƵĞ͕ďƵƚ ƚŚĞǁƌŝƚƚĞŶƌĞƐƉŽŶƐĞŝƐ ŶŽƚĐŽŵƉůĞƚĞĂŶĚĚŽĞƐ ŶŽƚĚĞŵŽŶƐƚƌĂƚĞ ĞǀŝĚĞŶĐĞŽĨƐŽůŝĚ ƌĞĂƐŽŶŝŶŐ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽƌƌĞĐƚĂŶĚĐŽŵƉůĞƚĞ͘ ^ƚƵĚĞŶƚĞdžƉůĂŝŶƐƚŚĂƚĂ ĚĞďŝƚŽĨ ̈́ ͺǤͻͺ ĂŶĚĂ ĐƌĞĚŝƚŽĨ ̈́ ͺǤͻͺ ĂƌĞ ƐŝŵŝůĂƌďĞĐĂƵƐĞƚŚĞƚǁŽ ƚƌĂŶƐĂĐƚŝŽŶƐ͕ǁŚŝĐŚĂƌĞ ƌĞƉƌĞƐĞŶƚĞĚďLJ െͺǤͻͺ ĂŶĚ ͺǤͻͺ ͕ŚĂǀĞƚŚĞƐĂŵĞ ĂďƐŽůƵƚĞǀĂůƵĞ͕ǁŚŝĐŚŝƐ
ͺǤͻͺ ͖ƐŽƚŚĞLJĐŚĂŶŐĞƚŚĞ ĂĐĐŽƵŶƚďĂůĂŶĐĞďLJƚŚĞ ƐĂŵĞĂŵŽƵŶƚŽĨŵŽŶĞLJ ďƵƚŝŶŽƉƉŽƐŝƚĞ ĚŝƌĞĐƚŝŽŶƐ͘
3 a
^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐĂŶĚůĂďĞůƐƚǁŽ ŽĨƚŚĞĨŝǀĞƉŽŝŶƚƐ͕Ăƚ ŵŽƐƚ͕ŽŶƚŚĞŶƵŵďĞƌ ůŝŶĞƵƐŝŶŐƌĂƚŝŽŶĂů ŶƵŵďĞƌƐ͘ ^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐĂŶĚůĂďĞůƐƚŚƌĞĞ ŽĨƚŚĞĨŝǀĞƉŽŝŶƚƐŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞƵƐŝŶŐ ƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘ ^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐĂŶĚůĂďĞůƐĨŽƵƌŽĨ ƚŚĞĨŝǀĞƉŽŝŶƚƐŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞƵƐŝŶŐ ƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘ ^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐĂŶĚůĂďĞůƐĂůůĨŝǀĞ ƉŽŝŶƚƐŽŶƚŚĞŶƵŵďĞƌ ůŝŶĞƵƐŝŶŐƌĂƚŝŽŶĂů ŶƵŵďĞƌƐ͘
b ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽŵƉůĞƚĞĂŶĚ ŝŶĐŽƌƌĞĐƚ͕ƐƵĐŚĂƐ ƐƚĂƚŝŶŐƚŚĂƚĂ ĐŽŵƉĞƚŝƚŽƌŽƚŚĞƌƚŚĂŶ &ůŽƌĞŶĐĞŝƐĐůŽƐĞƐƚƚŽ ƚŚĞĐŚĞĐŬƉŽŝŶƚǁŝƚŚŽƵƚ ĞdžƉůĂŝŶŝŶŐǁŚLJ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚ &ůŽƌĞŶĐĞŝƐĐůŽƐĞƐƚƚŽƚŚĞ ĐŚĞĐŬƉŽŝŶƚǁŝƚŚŽƵƚ ũƵƐƚŝĨŝĐĂƚŝŽŶ͘ KZ ^ƚƵĚĞŶƚƐƚĂƚĞƐĂŶŽƚŚĞƌ ĐŽŵƉĞƚŝƚŽƌ͛ƐŶĂŵĞĂŶĚ ĂƚƚĞŵƉƚƐƚŽũƵƐƚŝĨLJƚŚĞ ĂŶƐǁĞƌ͕ďƵƚƚŚĞ ĞdžƉůĂŶĂƚŝŽŶŝƐ ŝŶĐŽŵƉůĞƚĞ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚ &ůŽƌĞŶĐĞŝƐĐůŽƐĞƐƚƚŽƚŚĞ ĐŚĞĐŬƉŽŝŶƚ͕ďƵƚƚŚĞ ũƵƐƚŝĨŝĐĂƚŝŽŶĐŽŶƚĂŝŶƐĂŶ ĞƌƌŽƌ͘ KZ ^ƚƵĚĞŶƚƐƚĂƚĞƐĂŶŽƚŚĞƌ ĐŽŵƉĞƚŝƚŽƌ͛ƐŶĂŵĞĂŶĚ ũƵƐƚŝĨŝĞƐƚŚĞĂŶƐǁĞƌ ďĂƐĞĚŽŶƚŚĞƌĞƐƉŽŶƐĞƚŽ ƉĂƌƚ;ĂͿ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚ&ůŽƌĞŶĐĞŝƐĐůŽƐĞƐƚ ƚŽƚŚĞĐŚĞĐŬƉŽŝŶƚĂŶĚ ƉƌŽǀŝĚĞƐĐůĞĂƌĂŶĚ ĂĐĐƵƌĂƚĞũƵƐƚŝĨŝĐĂƚŝŽŶĨŽƌ ƚŚĞĐůĂŝŵ͘
A STORY OF RATIOS
133
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task c ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽŵƉůĞƚĞĂŶĚ ŝŶĐŽƌƌĞĐƚ͕ƐƵĐŚĂƐ ƐƚĂƚŝŶŐƚŚĂƚƚǁŽ ĐŽŵƉĞƚŝƚŽƌƐ other than ZĞďĞĐĐĂĂŶĚ>ŝƚĂĂƌĞ ƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵ ƚŚĞĐŚĞĐŬƉŽŝŶƚ͕ĂŶĚŶŽ ĨƵƌƚŚĞƌĞdžƉůĂŶĂƚŝŽŶŝƐ ƉƌŽǀŝĚĞĚ͘ ^ƚƵĚĞŶƚŝƐĂďůĞƚŽ ĚĞƚĞƌŵŝŶĞƚŚĂƚZĞďĞĐĐĂ ĂŶĚ>ŝƚĂĂƌĞƚŚĞƐĂŵĞ ĚŝƐƚĂŶĐĞĂǁĂLJĨƌŽŵƚŚĞ ĐŚĞĐŬƉŽŝŶƚ͕ďƵƚƚŚĞ ĞdžƉůĂŶĂƚŝŽŶĚŽĞƐŶŽƚ ĂĚĚƌĞƐƐǁŚĞƚŚĞƌŽƌŶŽƚ ƚŚĞĐŽŵƉĞƚŝƚŽƌƐĂƌĞŝŶ ƚŚĞƐĂŵĞůŽĐĂƚŝŽŶ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚ ZĞďĞĐĐĂĂŶĚ>ŝƚĂĂƌĞƚŚĞ ƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵƚŚĞ ĐŚĞĐŬƉŽŝŶƚďƵƚŽŶ ŽƉƉŽƐŝƚĞƐŝĚĞƐ͖ŚŽǁĞǀĞƌ͕ ƚŚĞĞdžƉůĂŶĂƚŝŽŶĚŽĞƐŶŽƚ ƐƉĞĐŝĨŝĐĂůůLJĂŶƐǁĞƌ ǁŚĞƚŚĞƌŽƌŶŽƚƚŚĞ ĐŽŵƉĞƚŝƚŽƌƐĂƌĞŝŶƚŚĞ ƐĂŵĞůŽĐĂƚŝŽŶ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ŝŶĚŝĐĂƚĞƐĂŶĚĞdžƉůĂŝŶƐ ƚŚĂƚZĞďĞĐĐĂĂŶĚ>ŝƚĂĂƌĞ ďŽƚŚ ͲǤͷ ŵŝůĞĨƌŽŵƚŚĞ ĐŚĞĐŬƉŽŝŶƚďƵƚƚŚĂƚƚŚĞLJ ĂƌĞƉŽƐŝƚŝŽŶĞĚŽŶ ŽƉƉŽƐŝƚĞƐŝĚĞƐŽĨƚŚĞ ĐŚĞĐŬƉŽŝŶƚ͕ĂŶĚƐŽƚŚĞLJ ĂƌĞŶŽƚŝŶƚŚĞƐĂŵĞ ůŽĐĂƚŝŽŶ͘
d ^ƚƵĚĞŶƚĞdžƉůĂŶĂƚŝŽŶ ƐŚŽǁƐůŝƚƚůĞŽƌŶŽ ĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐ͘&Žƌ ŝŶƐƚĂŶĐĞ͕ƐƚƵĚĞŶƚ ŝŶĐŽƌƌĞĐƚůLJƐƚĂƚĞƐEĂŶĐLJ ŝƐĐůŽƐĞƌƚŽĨŝŶŝƐŚŝŶŐƚŚĞ ƌĂĐĞǁŝƚŚŶŽ ĞdžƉůĂŶĂƚŝŽŶǁŚLJ͘ ^ƚƵĚĞŶƚŝŶĐŽƌƌĞĐƚůLJ ĚĞƚĞƌŵŝŶĞƐEĂŶĐLJŝƐ ĐůŽƐĞƌƚŽĨŝŶŝƐŚŝŶŐƚŚĞ ƌĂĐĞďƵƚƵƐĞƐĂǀĂůŝĚ ĂƌŐƵŵĞŶƚďĂƐĞĚŽŶ ĞĂƌůŝĞƌǁŽƌŬ͘ KZ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚ&ůŽƌĞŶĐĞŝƐĐůŽƐĞƌƚŽ ĨŝŶŝƐŚŝŶŐƚŚĞƌĂĐĞďƵƚ ǁŝƚŚŶŽĨƵƌƚŚĞƌ ĞdžƉůĂŶĂƚŝŽŶ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ &ůŽƌĞŶĐĞŝƐĐůŽƐĞƌƚŽ ĨŝŶŝƐŚŝŶŐƚŚĞƌĂĐĞ͕ďƵƚ ƚŚĞũƵƐƚŝĨŝĐĂƚŝŽŶĨŽƌƚŚĞ ĐůĂŝŵĐŽŶƚĂŝŶƐĂŶĞƌƌŽƌ ŝŶƌĞĂƐŽŶŝŶŐŽƌĂ ŵŝƐƌĞƉƌĞƐĞŶƚĂƚŝŽŶ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ĚĞƚĞƌŵŝŶĞƐĂŶĚƐƚĂƚĞƐ ƚŚĂƚ&ůŽƌĞŶĐĞŝƐĐůŽƐĞƌƚŽ ĨŝŶŝƐŚŝŶŐƚŚĞƌĂĐĞĂŶĚ ũƵƐƚŝĨŝĞƐƚŚĞĐůĂŝŵƵƐŝŶŐ ǀĂůŝĚĂŶĚĚĞƚĂŝůĞĚ ƌĞĂƐŽŶŝŶŐ͘
4 a
^ƚƵĚĞŶƚĞdžƉůĂŶĂƚŝŽŶ ƐŚŽǁƐůŝƚƚůĞŽƌŶŽ ĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐ͘&Žƌ ŝŶƐƚĂŶĐĞ͕ƐƚƵĚĞŶƚƐƚĂƚĞƐ ƚŚĂƚŶĞŝƚŚĞƌDĂƌƚĂŶŽƌ ŶĚƌĠĂŝƐĐŽƌƌĞĐƚ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚ DĂƌƚĂŝƐĐŽƌƌĞĐƚďƵƚĚŽĞƐ ŶŽƚƐƵƉƉŽƌƚƚŚĞĐůĂŝŵ͘ KZ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚ ŶĚƌĠĂŝƐĐŽƌƌĞĐƚ͕ďƵƚƚŚĞ ĞdžƉůĂŶĂƚŝŽŶŝŶĐůƵĚĞƐĂŶ ĞƌƌŽƌŝŶƌĞĂƐŽŶŝŶŐ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚDĂƌƚĂŝƐĐŽƌƌĞĐƚ͕ďƵƚ ƚŚĞĞdžƉůĂŶĂƚŝŽŶĐŽŶƚĂŝŶƐ ƌĞĂƐŽŶŝŶŐƚŚĂƚŝƐŶŽƚ ĐůĞĂƌĂŶĚĐŽŵƉůĞƚĞ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽƌƌĞĐƚĂŶĚĐŽŵƉůĞƚĞ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚ DĂƌƚĂŝƐĐŽƌƌĞĐƚ͕ ũƵƐƚŝĨLJŝŶŐƚŚĞĐůĂŝŵďLJ ĂĐĐƵƌĂƚĞůLJĚĞƐĐƌŝďŝŶŐƚŚĞ ŽƌĚĞƌŽĨƚŚĞƌĂƚŝŽŶĂů ŶƵŵďĞƌƐŽŶƚŚĞŶƵŵďĞƌ ůŝŶĞ͘
b ^ƚƵĚĞŶƚŝŶĐŽƌƌĞĐƚůLJ ƐƚĂƚĞƐĐŽŽůĞƌŽƌĂŶĚ ŵĞƚƚŚĞŐŽĂůĂŶĚ ƉƌŽǀŝĚĞƐŶŽũƵƐƚŝĨŝĐĂƚŝŽŶ ŽƌƉƌŽǀŝĚĞƐĂŶ ĞdžƉůĂŶĂƚŝŽŶƚŚĂƚ ĐŽŶƚĂŝŶƐŵƵůƚŝƉůĞĞƌƌŽƌƐ ŝŶƌĞĂƐŽŶŝŶŐ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚ ĐŽŽůĞƌŵĞƚƚŚĞŐŽĂůďƵƚ ƉƌŽǀŝĚĞƐŶŽũƵƐƚŝĨŝĐĂƚŝŽŶ ĨŽƌƚŚĞĐůĂŝŵ͘ KZ ^ƚƵĚĞŶƚĚĞƚĞƌŵŝŶĞƐƚŚĂƚ ĐŽŽůĞƌƐĂŶĚŵĞƚƚŚĞ ŐŽĂůĂŶĚŝŶĐůƵĚĞƐĂ ĐŽŵƉůĞƚĞĞdžƉůĂŶĂƚŝŽŶ ďƵƚĞƌƌŽŶĞŽƵƐůLJ ŝĚĞŶƚŝĨŝĞƐ െʹǤͻͳ ĚĞŐƌĞĞƐ ĂƐďĞŝŶŐŵŽƌĞƚŚĂŶ ͵ĚĞŐƌĞĞƐďĞůŽǁnjĞƌŽ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚĐŽŽůĞƌŵĞƚƚŚĞ ŐŽĂůĂŶĚũƵƐƚŝĨŝĞƐƚŚĞ ĐůĂŝŵ͕ďƵƚƚŚĞ ĞdžƉůĂŶĂƚŝŽŶĐŽŶƚĂŝŶƐĂ ƐůŝŐŚƚĞƌƌŽƌ͘&Žƌ ŝŶƐƚĂŶĐĞ͕ƐƚƵĚĞŶƚ ĚĞƐĐƌŝďĞƐƚŚĞŶƵŵďĞƌƐ ƚŽƚŚĞůĞĨƚŽĨ െ͵ ŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞĂƐďĞŝŶŐ ŵŽƌĞƚŚĂŶ െ͵ ƌĂƚŚĞƌ ƚŚĂŶůĞƐƐƚŚĂŶ െ͵ ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƚŚĂƚĐŽŽůĞƌŵĞƚƚŚĞ ŐŽĂůĂŶĚũƵƐƚŝĨŝĞƐƚŚĞ ĐůĂŝŵďLJĚĞƐĐƌŝďŝŶŐƚŚĂƚ͞ ŵŽƌĞƚŚĂŶ ͵ĚĞŐƌĞĞƐ ďĞůŽǁnjĞƌŽ͟ŝŶĚŝĐĂƚĞƐ ƚŚĞŶƵŵďĞƌƐŵƵƐƚďĞƚŽ ƚŚĞůĞĨƚŽĨ െ͵ ;ďĞůŽǁ
െ͵ ͿŽŶƚŚĞŶƵŵďĞƌůŝŶĞ ĂŶĚƚŚĂƚ െͶǤ͵ ŝƐƚŚĞŽŶůLJ ƉŝĞĐĞŽĨĚĂƚĂƚŚĂƚŵĞĞƚƐ ƚŚĂƚĐƌŝƚĞƌŝĂ͘
A STORY OF RATIOS
134
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task 5 a
^ƚƵĚĞŶƚĐŽŵƉĂƌŝƐŽŶŽĨ ƚŚĞĞůĞǀĂƚŝŽŶƐĂŵƉůĞƐ ƚŽƚŚĞůĞǀĞůŽĨƚŚĞƌŽĂĚ ŝƐŝŶĐŽƌƌĞĐƚ͘dŚĞ ǁƌŝƚƚĞŶǁŽƌŬƐŚŽǁƐ ůŝƚƚůĞŽƌŶŽ ƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨ ŽƌĚĞƌŝŶŐƌĂƚŝŽŶĂů ŶƵŵďĞƌƐ͘ ^ƚƵĚĞŶƚĐŽŵƉĂƌŝƐŽŶŽĨ ƚŚĞĞůĞǀĂƚŝŽŶƐĂŵƉůĞƐƚŽ ƚŚĞůĞǀĞůŽĨƚŚĞƌŽĂĚŝƐ ƉĂƌƚŝĂůůLJĐŽƌƌĞĐƚ͘^ƚƵĚĞŶƚ ĐŽƌƌĞĐƚůLJĐŽŵƉĂƌĞƐŽŶůLJ ŽŶĞŽƌƚǁŽŽĨƚŚĞ ƐĂŵƉůĞƐ;͕͕Žƌ ͿƚŽ ƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞƌŽĂĚ͘ ^ƚƵĚĞŶƚƐƚĂƚĞƐƚŚĂƚ ƐĂŵƉůĞŝƐŚŝŐŚĞƌƚŚĂŶ ƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞƌŽĂĚ ĂŶĚƐĂŵƉůĞ ŝƐůŽǁĞƌ ƚŚĂŶƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞ ƌŽĂĚĂŶĚƚŚĂƚƐĂŵƉůĞ ŝƐĂďŽƵƚůĞǀĞůǁŝƚŚƚŚĞ ƌŽĂĚďƵƚ does not distinguish ǁŚĞƚŚĞƌ ƐĂŵƉůĞ͛ƐĞůĞǀĂƚŝŽŶ ůĞǀĞůŝƐŚŝŐŚĞƌŽƌůŽǁĞƌ ƚŚĂŶ ͺ͵Ͳ ĨĞĞƚ͘ ^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ĚĞƐĐƌŝďĞƐĞĂĐŚƐĂŵƉůĞ͛Ɛ ƌĞůĂƚŝǀĞƉŽƐŝƚŝŽŶ ĐŽŵƉĂƌĞĚƚŽƚŚĞ ĞůĞǀĂƚŝŽŶŽĨƚŚĞƌŽĂĚ͕ ƐƚĂƚŝŶŐƚŚĂƚƐĂŵƉůĞƐ ĂŶĚĂƌĞŚŝŐŚĞƌƚŚĂŶ ƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞ ƌŽĂĚ͕ĂŶĚƐĂŵƉůĞ ŝƐ ůŽǁĞƌƚŚĂŶƚŚĞĞůĞǀĂƚŝŽŶ ŽĨƚŚĞƌŽĂĚ͘
b ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ƐŚŽǁƐůŝƚƚůĞŽƌŶŽ ĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐ͘ ^ƚƵĚĞŶƚŵĂLJƉůĂĐĞƚŚĞ ŶĞŐĂƚŝǀĞǀĂůƵĞƐůĞĨƚŽĨ ƚŚĞƉŽƐŝƚŝǀĞǀĂůƵĞƐďƵƚ ŵĂŬĞƐƐĞǀĞƌĂůĞƌƌŽƌƐŝŶ ŽƌĚĞƌ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞƐŚŽǁƐ ƐŽŵĞĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐ͘^ƚƵĚĞŶƚ ĐŽƌƌĞĐƚůLJŽƌĚĞƌƐĨŽƵƌŽĨ ƚŚĞƐŝdžǀĂůƵĞƐĨƌŽŵůĞĂƐƚ ƚŽŐƌĞĂƚĞƐƚ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJŽƌĚĞƌƐ ĂůůƐŝdžǀĂůƵĞƐĨƌŽŵůĞĂƐƚ ƚŽŐƌĞĂƚĞƐƚďƵƚĐŽƉŝĞƐ ŽŶĞŽĨƚŚĞǀĂůƵĞƐ ŝŶĐŽƌƌĞĐƚůLJ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJŽƌĚĞƌƐ ĂŶĚǁƌŝƚĞƐĂůůƐŝdžǀĂůƵĞƐ ĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ ;ŝ͘Ğ͕͘ െͶǤͷ ൏ െͲǤͻ ൏
െͲǤͷ ൏ ͳǤ͵ ൏ ʹǤʹ ൏ ͵Ǥͳ
Ϳ͘
c ^ƚƵĚĞŶƚŝŶĚŝĐĂƚĞƐ ƐĂŵƉůĞ<ďƵƚĚŽĞƐŶŽƚ ƉƌŽǀŝĚĞĂŶLJĨƵƌƚŚĞƌ ĚĞƚĂŝů͘ KZ ^ƚƵĚĞŶƚƐƚĂƚĞƐĂ ĚŝĨĨĞƌĞŶƚƐĂŵƉůĞƐƵĐŚ ĂƐ'ĂŶĚũƵƐƚŝĨŝĞƐƚŚĞ ĐŚŽŝĐĞƵƐŝŶŐĐůĞĂƌ ƌĞĂƐŽŶŝŶŐďƵƚĚŽĞƐŶŽƚ ĂĚĚƌĞƐƐĂďƐŽůƵƚĞǀĂůƵĞ͘ ^ƚƵĚĞŶƚŝŶĚŝĐĂƚĞƐƐĂŵƉůĞ <ĂŶĚƉƌŽǀŝĚĞƐĂǀĂůŝĚ ĞdžƉůĂŶĂƚŝŽŶďƵƚĚŽĞƐŶŽƚ ĂĚĚƌĞƐƐĂďƐŽůƵƚĞǀĂůƵĞŝŶ ƚŚĞĞdžƉůĂŶĂƚŝŽŶ͘ KZ ^ƚƵĚĞŶƚŝŶĐŽƌƌĞĐƚůLJ ƐƚĂƚĞƐƐĂŵƉůĞ'ĂŶĚ ũƵƐƚŝĨŝĞƐƚŚĞĐŚŽŝĐĞďLJ ĂĚĚƌĞƐƐŝŶŐƚŚĞŽƌĚĞƌŽĨ ƚŚĞƉŽƐŝƚŝǀĞŶƵŵďĞƌƐ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƐĂŵƉůĞ<ĂŶĚũƵƐƚŝĨŝĞƐ ƚŚĞƐƚĂƚĞŵĞŶƚƵƐŝŶŐ ĂďƐŽůƵƚĞǀĂůƵĞĐŽƌƌĞĐƚůLJ ŝŶƚŚĞĞdžƉůĂŶĂƚŝŽŶ͕ďƵƚ ƚŚĞĞdžƉůĂŶĂƚŝŽŶŝƐŶŽƚ ĐŽŵƉůĞƚĞ͘ KZ ^ƚƵĚĞŶƚŝŶĐŽƌƌĞĐƚůLJ ƐƚĂƚĞƐƐĂŵƉůĞ'ĂŶĚ ũƵƐƚŝĨŝĞƐƚŚĞĐŚŽŝĐĞƵƐŝŶŐ ĂďƐŽůƵƚĞǀĂůƵĞŝŶĂ ĐŽƌƌĞĐƚŵĂŶŶĞƌďƵƚ ǁŝƚŚŽƵƚĐŽŶƐŝĚĞƌŝŶŐƚŚĞ ĂďƐŽůƵƚĞǀĂůƵĞŽĨ െͶǤͷ ĨŽƌƐĂŵƉůĞ<͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ ƐĂŵƉůĞ<ŝƐƚŚĞĨĂƌƚŚĞƐƚ ĨƌŽŵƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞ ƌŽĂĚĂŶĚũƵƐƚŝĨŝĞƐƚŚĞ ƐƚĂƚĞŵĞŶƚďLJĐŽŵƉĂƌŝŶŐ ƚŚĞĂďƐŽůƵƚĞǀĂůƵĞƐŽĨ ƚŚĞƐĂŵƉůĞƐĨƌŽŵƚŚĞ ƚĂďůĞŝŶƉĂƌƚ;ďͿƵƐŝŶŐƚŚĞ ŽƌĚĞƌŽĨƌĂƚŝŽŶĂů ŶƵŵďĞƌƐƚŽƌĞĂĐŚƚŚĞ ĂŶƐǁĞƌ͘
A STORY OF RATIOS
135
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
ZŝǀĞƌtĂƚĞƌ
ͳǤͲ
ͲǤͷ
Ͳ
െʹǤͷ
െͲǤͷ
ͳǤͷ
െͳǤͲ
െͳǤͷ
െʹǤͲ
െ͵Ǥͷ
െ͵ǤͲ EĂŵĞ ĂƚĞ ϭ͘ dŚĞƉŝĐƚƵƌĞďĞůŽǁŝƐĂĨůŽŽĚŐĂƵŐĞƚŚĂƚŝƐƵƐĞĚƚŽŵĞĂƐƵƌĞŚŽǁĨĂƌ;ŝŶĨĞĞƚͿĂƌŝǀĞƌ͛ƐǁĂƚĞƌůĞǀĞůŝƐĂďŽǀĞ ŽƌďĞůŽǁŝƚƐŶŽƌŵĂůůĞǀĞů͘ Ă͘ džƉůĂŝŶǁŚĂƚƚŚĞŶƵŵďĞƌ ͲŽŶƚŚĞŐĂƵŐĞƌĞƉƌĞƐĞŶƚƐ͕ĂŶĚĞdžƉůĂŝŶǁŚĂƚƚŚĞŶƵŵďĞƌƐ ĂďŽǀĞĂŶĚďĞůŽǁ ͲƌĞƉƌĞƐĞŶƚ͘ ď͘ ĞƐĐƌŝďĞǁŚĂƚƚŚĞƉŝĐƚƵƌĞŝŶĚŝĐĂƚĞƐĂďŽƵƚƚŚĞ ƌŝǀĞƌ͛ƐĐƵƌƌĞŶƚǁĂƚĞƌůĞǀĞů͘ Đ͘ tŚĂƚŶƵŵďĞƌƌĞƉƌĞƐĞŶƚƐƚŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞǁĂƚĞƌůĞǀĞůƐŚŽǁŶŝŶƚŚĞƉŝĐƚƵƌĞ͕ĂŶĚǁŚĞƌĞŝƐŝƚ ůŽĐĂƚĞĚŽŶƚŚĞŐĂƵŐĞ͍tŚĂƚǁŽƵůĚŝƚŵĞĂŶŝĨƚŚĞƌŝǀĞƌǁĂƚĞƌǁĂƐĂƚƚŚĂƚůĞǀĞů͍ Ě͘ /ĨŚĞĂǀLJƌĂŝŶŝƐŝŶƚŚĞĨŽƌĞĐĂƐƚĨŽƌƚŚĞĂƌĞĂĨŽƌƚŚĞŶĞdžƚ ʹͶ ŚŽƵƌƐ͕ǁŚĂƚƌĞĂĚŝŶŐŵŝŐŚƚLJŽƵĞdžƉĞĐƚƚŽ ƐĞĞŽŶƚŚŝƐŐĂƵŐĞƚŽŵŽƌƌŽǁ͍ džƉůĂŝŶLJŽƵƌƌĞĂƐŽŶŝŶŐ͘
A STORY OF RATIOS
136
©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
Ϯ͘ /ƐĂĂĐŵĂĚĞĂŵŝƐƚĂŬĞŝŶŚŝƐĐŚĞĐŬŬ͘,ĞǁƌŽƚĞĂĐŚĞĐŬĨŽƌ ̈́ ͺǤͻͺ ƚŽƌĞŶƚĂǀŝĚĞŽŐĂŵĞďƵƚŵŝƐƚĂŬĞŶůLJ ƌĞĐŽƌĚĞĚŝƚŝŶŚŝƐĐŚĞĐŬŬĂƐĂŶ ̈́ͺǤͻͺ ĚĞƉŽƐŝƚ͘ Ă͘ ZĞƉƌĞƐĞŶƚĞĂĐŚƚƌĂŶƐĂĐƚŝŽŶǁŝƚŚĂƌĂƚŝŽŶĂůŶƵŵďĞƌ͕ĂŶĚĞdžƉůĂŝŶƚŚĞĚŝĨĨĞƌĞŶĐĞďĞƚǁĞĞŶƚŚĞ ƚƌĂŶƐĂĐƚŝŽŶƐ͘ ď͘ KŶƚŚĞŶƵŵďĞƌůŝŶĞďĞůŽǁ͕ůŽĐĂƚĞĂŶĚůĂďĞůƚŚĞƉŽŝŶƚƐƚŚĂƚƌĞƉƌĞƐĞŶƚƚŚĞƌĂƚŝŽŶĂůŶƵŵďĞƌƐůŝƐƚĞĚŝŶ ƉĂƌƚ;ĂͿ͘ĞƐĐƌŝďĞƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞƐĞƚǁŽŶƵŵďĞƌƐ͘ĞƌŽŽŶƚŚĞŶƵŵďĞƌůŝŶĞƌĞƉƌĞƐĞŶƚƐ /ƐĂĂĐ͛ƐďĂůĂŶĐĞďĞĨŽƌĞƚŚĞŵŝƐƚĂŬĞǁĂƐŵĂĚĞ͘ Đ͘ hƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽĞdžƉůĂŝŶŚŽǁĂĚĞďŝƚŽĨ ̈́ ͺǤͻͺ ĂŶĚĂĐƌĞĚŝƚŽĨ ̈́ͺǤͻͺ ĂƌĞƐŝŵŝůĂƌ͘
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Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
ϯ͘ ůŽĐĂůƉĂƌŬ͛ƐƉƌŽŐƌĂŵƐĐŽŵŵŝƚƚĞĞŝƐƌĂŝƐŝŶŐŵŽŶĞLJďLJŚŽůĚŝŶŐŵŽƵŶƚĂŝŶďŝŬĞƌĂĐĞƐŽŶĂĐŽƵƌƐĞƚŚƌŽƵŐŚ ƚŚĞƉĂƌŬ͘ƵƌŝŶŐĞĂĐŚƌĂĐĞ͕ĂĐŽŵƉƵƚĞƌƚƌĂĐŬƐƚŚĞĐŽŵƉĞƚŝƚŽƌƐ͛ůŽĐĂƚŝŽŶƐŽŶƚŚĞĐŽƵƌƐĞƵƐŝŶŐ'W^ ƚƌĂĐŬŝŶŐ͘dŚĞƚĂďůĞƐŚŽǁƐŚŽǁĨĂƌĞĂĐŚĐŽŵƉĞƚŝƚŽƌŝƐĨƌŽŵĂĐŚĞĐŬƉŽŝŶƚ͘ EƵŵďĞƌ ŽŵƉĞƚŝƚŽƌEĂŵĞ ŝƐƚĂŶĐĞƚŽŚĞĐŬƉŽŝŶƚ
ʹʹ͵ &ůŽƌĞŶĐĞ ͲǤͳ ŵŝůĞƐďĞĨŽƌĞ
ʹ͵ͳ DĂƌLJ ଶହŵŝůĞƐƉĂƐƚ
ʹͶͲ ZĞďĞĐĐĂ ͲǤͷ ŵŝůĞƐďĞĨŽƌĞ
ʹͶͻ >ŝƚĂ ଵଶŵŝůĞƐƉĂƐƚ
ʹͷͷ EĂŶĐLJ ଶଵ ŵŝůĞƐďĞĨŽƌĞ Ă͘ dŚĞĐŚĞĐŬƉŽŝŶƚŝƐƌĞƉƌĞƐĞŶƚĞĚďLJ ͲŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘>ŽĐĂƚĞĂŶĚůĂďĞůƉŽŝŶƚƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ ĨŽƌƚŚĞƉŽƐŝƚŝŽŶƐŽĨĞĂĐŚůŝƐƚĞĚƉĂƌƚŝĐŝƉĂŶƚ͘>ĂďĞůƚŚĞƉŽŝŶƚƐƵƐŝŶŐƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘ ď͘ tŚŝĐŚŽĨƚŚĞĐŽŵƉĞƚŝƚŽƌƐŝƐĐůŽƐĞƐƚƚŽƚŚĞĐŚĞĐŬƉŽŝŶƚ͍ džƉůĂŝŶ͘ Đ͘ dǁŽĐŽŵƉĞƚŝƚŽƌƐĂƌĞƚŚĞƐĂŵĞĚŝƐƚĂŶĐĞĨƌŽŵƚŚĞĐŚĞĐŬƉŽŝŶƚ͘ƌĞƚŚĞLJŝŶƚŚĞƐĂŵĞůŽĐĂƚŝŽŶ͍ džƉůĂŝŶ͘ Ě͘ tŚŽŝƐĐůŽƐĞƌƚŽĨŝŶŝƐŚŝŶŐƚŚĞƌĂĐĞ͕EĂŶĐLJŽƌ&ůŽƌĞŶĐĞ͍^ƵƉƉŽƌƚLJŽƵƌĂŶƐǁĞƌ͘
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©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
ϰ͘ ŶĚƌĠĂĂŶĚDĂƌƚĂĂƌĞƚĞƐƚŝŶŐƚŚƌĞĞĚŝĨĨĞƌĞŶƚĐŽŽůĞƌƐƚŽƐĞĞǁŚŝĐŚŬĞĞƉƐƚŚĞĐŽůĚĞƐƚƚĞŵƉĞƌĂƚƵƌĞ͘dŚĞLJ ƉůĂĐĞĚĂďĂŐŽĨŝĐĞŝŶĞĂĐŚĐŽŽůĞƌ͕ĐůŽƐĞĚƚŚĞĐŽŽůĞƌƐ͕ĂŶĚƚŚĞŶŵĞĂƐƵƌĞĚƚŚĞĂŝƌƚĞŵƉĞƌĂƚƵƌĞŝŶƐŝĚĞĞĂĐŚ ĂĨƚĞƌ ͻͲ ŵŝŶƵƚĞƐ͘dŚĞƚĞŵƉĞƌĂƚƵƌĞƐĂƌĞƌĞĐŽƌĚĞĚŝŶƚŚĞƚĂďůĞďĞůŽǁ͗ ŽŽůĞƌ dĞŵƉĞƌĂƚƵƌĞ; ஈͿ െʹǤͻͳ ͷǤ െͶǤ͵
DĂƌƚĂǁƌŽƚĞƚŚĞĨŽůůŽǁŝŶŐŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚĂďŽƵƚƚŚĞƚĞŵƉĞƌĂƚƵƌĞƐ͗
െͶǤ͵ ൏ െʹǤͻͳ ൏ ͷǤ ͘ŶĚƌĠĂĐůĂŝŵƐƚŚĂƚDĂƌƚĂŵĂĚĞĂŵŝƐƚĂŬĞŝŶŚĞƌƐƚĂƚĞŵĞŶƚĂŶĚƚŚĂƚƚŚĞŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƐŚŽƵůĚďĞ ǁƌŝƚƚĞŶĂƐ
െʹǤͻͳ ൏ െͶǤ͵ ൏ ͷǤ ͘Ă͘ /ƐĞŝƚŚĞƌƐƚƵĚĞŶƚĐŽƌƌĞĐƚ͍ džƉůĂŝŶ͘ ď͘ dŚĞƐƚƵĚĞŶƚƐǁĂŶƚƚŽĨŝŶĚĂĐŽŽůĞƌƚŚĂƚŬĞĞƉƐƚŚĞƚĞŵƉĞƌĂƚƵƌĞŝŶƐŝĚĞƚŚĞĐŽŽůĞƌŵŽƌĞƚŚĂŶ ͵ĚĞŐƌĞĞƐďĞůŽǁƚŚĞĨƌĞĞnjŝŶŐƉŽŝŶƚŽĨǁĂƚĞƌ; Ͳᤪ ͿĂĨƚĞƌ ͻͲ ŵŝŶƵƚĞƐ͘/ŶĚŝĐĂƚĞǁŚŝĐŚŽĨƚŚĞƚĞƐƚĞĚ ĐŽŽůĞƌƐŵĞĞƚƐƚŚŝƐŐŽĂů͕ĂŶĚĞdžƉůĂŝŶǁŚLJ͘
A STORY OF RATIOS
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©201 8Great Minds ®. eureka-math.org
Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
6ͻϯ Mid-Module Assessment Task
ϱ͘ DĂƌLJŵĂŶĂŐĞƐĂĐŽŵƉĂŶLJƚŚĂƚŚĂƐďĞĞŶŚŝƌĞĚƚŽĨůĂƚƚĞŶĂƉůŽƚŽĨůĂŶĚ͘^ŚĞƚŽŽŬƐĞǀĞƌĂůĞůĞǀĂƚŝŽŶ ƐĂŵƉůĞƐĨƌŽŵƚŚĞůĂŶĚĂŶĚƌĞĐŽƌĚĞĚƚŚŽƐĞĞůĞǀĂƚŝŽŶƐďĞůŽǁ͗ ůĞǀĂƚŝŽŶ^ĂŵƉůĞ & ůĞǀĂƚŝŽŶ ;Ĩƚ͘ĂďŽǀĞƐĞĂůĞǀĞůͿ ͺʹǤͷ ͺ͵ͲǤʹ ͺ͵ʹǤͲ ͺ͵ͳǤͳ ͺʹͷǤͺ ͺʹǤͳ
Ă͘ dŚĞůĂŶĚŽǁŶĞƌǁĂŶƚƐƚŚĞůĂŶĚĨůĂƚĂŶĚĂƚƚŚĞƐĂŵĞůĞǀĞůĂƐƚŚĞƌŽĂĚƚŚĂƚƉĂƐƐĞƐŝŶĨƌŽŶƚŽĨŝƚ͘dŚĞ ƌŽĂĚ͛ƐĞůĞǀĂƚŝŽŶŝƐ ͺ͵Ͳ ĨĞĞƚĂďŽǀĞƐĞĂůĞǀĞů͘ĞƐĐƌŝďĞŝŶǁŽƌĚƐŚŽǁĞůĞǀĂƚŝŽŶƐĂŵƉůĞƐ͕͕ĂŶĚ ĐŽŵƉĂƌĞƚŽƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞƌŽĂĚ͘ ď͘ dŚĞƚĂďůĞďĞůŽǁƐŚŽǁƐŚŽǁƐŽŵĞŽƚŚĞƌĞůĞǀĂƚŝŽŶƐĂŵƉůĞƐĐŽŵƉĂƌĞƚŽƚŚĞůĞǀĞůŽĨƚŚĞƌŽĂĚ͗ ůĞǀĂƚŝŽŶ^ĂŵƉůĞ ' , / : < > ůĞǀĂƚŝŽŶ ;Ĩƚ͘ĨƌŽŵƚŚĞƌŽĂĚͿ ͵Ǥͳ െͲǤͷ ʹǤʹ ͳǤ͵ െͶǤͷ െͲǤͻ
Đ͘ /ŶĚŝĐĂƚĞǁŚŝĐŚŽĨƚŚĞǀĂůƵĞƐĨƌŽŵƚŚĞƚĂďůĞŝŶƉĂƌƚ;ďͿŝƐĨĂƌƚŚĞƐƚĨƌŽŵƚŚĞĞůĞǀĂƚŝŽŶŽĨƚŚĞƌŽĂĚ͘hƐĞ ĂďƐŽůƵƚĞǀĂůƵĞƚŽĞdžƉůĂŝŶLJŽƵƌĂŶƐǁĞƌ͘
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GRADE 6 ͻDKh> 36G R A D E DĂƚŚĞŵĂƚŝĐƐƵƌƌŝĐƵůƵŵ
dŽƉŝĐ : ZĂƚŝŽŶĂůEƵŵďĞƌƐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ dŽƉŝĐ
ZĂƚŝŽŶĂůEƵŵďĞƌƐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
&ŽĐƵƐ^ƚĂŶĚĂƌĚƐ : hŶĚĞƌƐƚĂŶĚĂƌĂƚŝŽŶĂůŶƵŵďĞƌĂƐĂƉŽŝŶƚŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ džƚĞŶĚŶƵŵďĞƌ ůŝŶĞĚŝĂŐƌĂŵƐĂŶĚĐŽŽƌĚŝŶĂƚĞĂdžĞƐĨĂŵŝůŝĂƌĨƌŽŵƉƌĞǀŝŽƵƐŐƌĂĚĞƐƚŽƌĞƉƌĞƐĞŶƚ ƉŽŝŶƚƐŽŶƚŚĞůŝŶĞĂŶĚŝŶƚŚĞƉůĂŶĞǁŝƚŚŶĞŐĂƚŝǀĞŶƵŵďĞƌĐŽŽƌĚŝŶĂƚĞƐ͘ hŶĚĞƌƐƚĂŶĚƐŝŐŶƐŽĨŶƵŵďĞƌƐŝŶŽƌĚĞƌĞĚƉĂŝƌƐĂƐŝŶĚŝĐĂƚŝŶŐůŽĐĂƚŝŽŶƐŝŶ ƋƵĂĚƌĂŶƚƐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͖ƌĞĐŽŐŶŝnjĞƚŚĂƚǁŚĞŶƚǁŽŽƌĚĞƌĞĚ ƉĂŝƌƐĚŝĨĨĞƌŽŶůLJďLJƐŝŐŶƐ͕ƚŚĞůŽĐĂƚŝŽŶƐŽĨƚŚĞƉŽŝŶƚƐĂƌĞƌĞůĂƚĞĚďLJ ƌĞĨůĞĐƚŝŽŶƐĂĐƌŽƐƐŽŶĞŽƌďŽƚŚĂdžĞƐ͘ &ŝŶĚĂŶĚƉŽƐŝƚŝŽŶŝŶƚĞŐĞƌƐĂŶĚŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶĂŚŽƌŝnjŽŶƚĂůŽƌ ǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞĚŝĂŐƌĂŵ͖ĨŝŶĚĂŶĚƉŽƐŝƚŝŽŶƉĂŝƌƐŽĨŝŶƚĞŐĞƌƐĂŶĚ ŽƚŚĞƌƌĂƚŝŽŶĂůŶƵŵďĞƌƐŽŶĂĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘ ^ŽůǀĞƌĞĂůͲǁŽƌůĚĂŶĚŵĂƚŚĞŵĂƚŝĐĂůƉƌŽďůĞŵƐďLJŐƌĂƉŚŝŶŐƉŽŝŶƚƐŝŶĂůůĨŽƵƌ ƋƵĂĚƌĂŶƚƐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘/ŶĐůƵĚĞƵƐĞŽĨĐŽŽƌĚŝŶĂƚĞƐĂŶĚĂďƐŽůƵƚĞ ǀĂůƵĞƚŽĨŝŶĚĚŝƐƚĂŶĐĞƐďĞƚǁĞĞŶƉŽŝŶƚƐǁŝƚŚƚŚĞƐĂŵĞĨŝƌƐƚĐŽŽƌĚŝŶĂƚĞŽƌƚŚĞ ƐĂŵĞƐĞĐŽŶĚĐŽŽƌĚŝŶĂƚĞ͘
/ŶƐƚƌƵĐƚŝŽŶĂůĂLJƐ͗ ϲ
ĞƐƐŽŶϭ 4: KƌĚĞƌĞĚWĂŝƌƐ;WͿ ϭ
ĞƐƐŽŶ 15: >ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ;WͿ
ĞƐƐŽŶϭϲ : ^LJŵŵĞƚƌLJŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ;WͿ
ĞƐƐŽŶϭϳ : ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ;WͿ
ĞƐƐŽŶϭϴ͗ ŝƐƚĂŶĐĞŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ;WͿ
ĞƐƐŽŶϭϵ͗ WƌŽďůĞŵ^ŽůǀŝŶŐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ; Ϳ /ŶdŽƉŝĐ͕ƐƚƵĚĞŶƚƐƚƌĂŶƐŝƚŝŽŶĨƌŽŵƚŚĞŶƵŵďĞƌůŝŶĞŵŽĚĞůƚŽƌĞƉƌĞƐĞŶƚƉŽŝŶƚƐŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘dŚĞŝƌ
ĐŽŶĐĞƉƚƵĂůƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨƐLJŵŵĞƚƌLJĨƌŽŵ'ƌĂĚĞϰĂŶĚƚŚĞŝƌĞdžƉĞƌŝĞŶĐĞǁŝƚŚƚŚĞĨŝƌƐƚƋƵĂĚƌĂŶƚŽĨƚŚĞ
ĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŝŶ'ƌĂĚĞϱƐĞƌǀĞĂƐĂƐŝŐŶŝĨŝĐĂŶƚĨŽƵŶĚĂƚŝŽŶĂƐƚŚĞLJĞdžƚĞŶĚƚŚĞƉůĂŶĞƚŽĂůůĨŽƵƌƋƵĂĚƌĂŶƚƐ͘
/Ŷ>ĞƐƐŽŶϭϰ͕ƐƚƵĚĞŶƚƐƵƐĞŽƌĚĞƌĞĚƉĂŝƌƐŽĨƌĂƚŝŽŶĂůŶƵŵďĞƌƐƚŽ ϭ>ĞƐƐŽŶ^ƚƌƵĐƚƵƌĞ<ĞLJ͗ WͲWƌŽďůĞŵ^Ğƚ>ĞƐƐŽŶ͕ DͲDŽĚĞůŝŶŐLJĐůĞ>ĞƐƐŽŶ͕ EͲ džƉůŽƌĂƚŝŽŶ>ĞƐƐŽŶ͕ SͲ^ŽĐƌĂƚŝĐ>ĞƐƐŽŶ
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ŶĂŵĞƉŽŝŶƚƐŽŶĂŐƌŝĚ͕ ĂŶĚŐŝǀĞŶ ĂƉŽŝŶƚ Ɛ͛
©201 8Great Minds ®. eureka-math.org
6ͻϯ dŽƉŝĐ
dŽƉŝĐ :ZĂƚŝŽŶĂůEƵŵďĞƌƐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
A STORY OF RATIOS
142
loc ation, they identif y the f irst numb er in the ordered p air as the f irst c oordinate and the sec ond numb er as the sec ond c oordinate. I n L essons 15– 17 , students c onstruc t the p l ane; identif y the ax es, q uadrants, and origin; and grap h p oints in the p l ane, using an ap p rop riate sc al e on the ax es. S tudents rec ogniz e the rel ationship that ex ists b etw een p oints w hose c oordinates dif f er onl y b y signs ( as ref l ec tions ac ross one or both ax es) and l oc ate suc h p oints using the sy mmetry in the p l ane. F or instanc e, they rec ogniz e that the points (3, 4) and (3, −4) are b oth eq ual distanc e f rom the 𝑥𝑥 -ax is on the same v ertic al l ine, and so the p oints are ref l ec tions in the 𝑥𝑥 -ax is. I n L essons 18 and 19 , students grap h p oints in the c oordinate p l ane and use ab sol ute v al ue to f ind the l engths of v ertic al and horiz ontal segments to sol v e real -w orl d p rob l ems. ©201 8 Great Minds ®. eureka-math.org 6ͻϯ Lesson 14 Lesson 14: KƌĚĞƌĞĚWĂŝƌƐ
Lesson 14: Ordered Pairs Student Outcomes
^ƚƵĚĞŶƚƐƵƐĞŽƌĚĞƌĞĚƉĂŝƌƐƚŽŶĂŵĞƉŽŝŶƚƐŝŶĂŐƌŝĚĂŶĚƚŽůŽĐĂƚĞƉŽŝŶƚƐŽŶĂŵĂƉ͘
^ƚƵĚĞŶƚƐŝĚĞŶƚŝĨLJƚŚĞĨŝƌƐƚŶƵŵďĞƌŝŶĂŶŽƌĚĞƌĞĚƉĂŝƌĂƐƚŚĞ first coordinate ĂŶĚƚŚĞƐĞĐŽŶĚŶƵŵďĞƌĂƐƚŚĞ
second coordinate ͘
Lesson Notes
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĞƵƐĞŽĨŽƌĚĞƌĞĚƉĂŝƌƐŽĨŶƵŵďĞƌƐĂƐĚĞƐĐƌŝďŝŶŐƚŚĞůŽĐĂƚŝŽŶƐŽĨƉŽŝŶƚƐŽŶĂƉůĂŶĞŝŶǀĂƌŝŽƵƐ ƐŝƚƵĂƚŝŽŶƐ͘dŚĞLJƌĞĐŽŐŶŝnjĞƚŚĞƐŝŐŶŝĨŝĐĂŶĐĞŽĨƚŚĞŽƌĚĞƌŽĨŶƵŵďĞƌƐŝŶŽƌĚĞƌĞĚƉĂŝƌƐďLJůŽŽŬŝŶŐĂƚƚŚĞĚŝĨĨĞƌĞŶƚ ŝŶƚĞƌƉƌĞƚĂƚŝŽŶƐ͘ ĞĨŽƌĞƐƚĂƌƚŝŶŐƚŚĞůĞƐƐŽŶ͕ƚŚĞƚĞĂĐŚĞƌŵƵƐƚŵĂŬĞĐŽƉŝĞƐŽĨƚŚĞƐĐĞŶĂƌŝŽƐŝŶ džĂŵƉůĞϮĂŶĚĐƵƚƚŚĞŵĂƉĂƌƚŝŶŽƌĚĞƌƚŽ ƉƌŽǀŝĚĞĞĂĐŚŐƌŽƵƉǁŝƚŚĞdžĂĐƚůLJŽŶĞƐĐĞŶĂƌŝŽ͘
Classwork Opening Exercise (5 minutes)
ĞĨŽƌĞƐƚƵĚĞŶƚƐĂƌƌŝǀĞ͕ĂƌƌĂŶŐĞƚŚĞŝƌĚĞƐŬƐŝŶƚŽƐƚƌĂŝŐŚƚƌŽǁƐ͘ƐƐŝŐŶĂŶƵŵďĞƌ;ϭ͕Ϯ͕ϯ͕͙ͿƚŽĞĂĐŚƌŽǁĂŶĚĂůƐŽƚŽƚŚĞ ƐĞĂƚƐŝŶĞĂĐŚƌŽǁƐƚĂƌƚŝŶŐĂƚƚŚĞĨƌŽŶƚǁŝƚŚƐĞĂƚϭ͘ƐƐƚƵĚĞŶƚƐĞŶƚĞƌƚŚĞƌŽŽŵ͕ŐŝǀĞƚŚĞŵĂƐƚŝĐŬLJŶŽƚĞĐŽŶƚĂŝŶŝŶŐĂƉĂŝƌ ŽĨŶƵŵďĞƌƐĐŽƌƌĞƐƉŽŶĚŝŶŐǁŝƚŚƚŚĞƐĞĂƚŝŶŐůŽĐĂƚŝŽŶƐŝŶƚŚĞƌŽŽŵ͘/ŶƐƚƌƵĐƚƐƚƵĚĞŶƚƐƚŽĨŝŶĚƚŚĞƐĞĂƚĚĞƐĐƌŝďĞĚďLJƚŚĞŝƌ ƐƚŝĐŬLJŶŽƚĞ͕ĂƉƉůLJƚŚĞƐƚŝĐŬLJŶŽƚĞƚŽƚŚĞĚĞƐŬ͕ĂŶĚďĞƐĞĂƚĞĚ͘ DŽƐƚƐƚƵĚĞŶƚƐĂƌĞĐŽŶĨƵƐĞĚĂƐŽŶůLJƚŚŽƐĞǁŝƚŚŵĂƚĐŚŝŶŐŶƵŵďĞƌƐĂƌĞĂďůĞƚŽĨŝŶĚƚŚĞŝƌƐĞĂƚƐ͘DŽŶŝƚŽƌĐŽŶǀĞƌƐĂƚŝŽŶƐ ƚĂŬŝŶŐƉůĂĐĞďĞƚǁĞĞŶƐƚƵĚĞŶƚƐĂƐƚŚĞLJĂŐƌĞĞƵƉŽŶĂĐŽŶǀĞŶƚŝŽŶ;Ğ͘Ő͕͘ƚŚĂƚƚŚĞĨŝƌƐƚŶƵŵďĞƌƌĞƉƌĞƐĞŶƚƐƚŚĞƌŽǁ͕ĂŶĚƚŚĞ ƐĞĐŽŶĚŶƵŵďĞƌƌĞƉƌĞƐĞŶƚƐƚŚĞƐĞĂƚͿ͘
,ŽǁĚŝĚLJŽƵĨŝŶĚLJŽƵƌƐĞĂƚŝŶƚŚĞĐůĂƐƐƌŽŽŵ͍;ŶƐǁĞƌƐǁŝůůǀĂƌLJ͘Ϳ
ŝĚƚŚĞŽƌĚĞƌŽĨƚŚĞŶƵŵďĞƌƐŵĂƚƚĞƌ͍tŚLJŽƌǁŚLJŶŽƚ͍
à The order mattered since there are two different seats that involve the numbers ʹ and ͵. For instance, row 2, seat 3, and row 3, seat 2.
Example 1 (5 minutes): The Order in Ordered Pairs
/ŶƐƚƌƵĐƚƐƚƵĚĞŶƚƐƚŽƌŽƚĂƚĞƚŚĞŝƌĚĞƐŬƐ ͻͲ ĚĞŐƌĞĞƐŝŶŽŶĞĚŝƌĞĐƚŝŽŶ͘dŚŝƐĐŚĂŶŐĞƐƚŚĞŽƌŝĞŶƚĂƚŝŽŶŽĨƚŚĞƌŽǁƐƐŽƚŚĂƚ ƐƚƵĚĞŶƚƐĐĂŶďĞƚƚĞƌƐĞĞƚŚĞŵĞĂŶŝŶŐƐŽĨĞĂĐŚŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƐ͘^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĂƚƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨƚŚĞŝƌ ůŽĐĂƚŝŽŶĨƌŽŵƚŚĞKƉĞŶŝŶŐ džĞƌĐŝƐĞ;ŝŶŵŽƐƚĐĂƐĞƐͿĂƌĞĚŝĨĨĞƌĞŶƚŝŶ džĂŵƉůĞϭ͘&ŽƌĞdžĂŵƉůĞ͕ƚŚĞƐƚƵĚĞŶƚƐŝƚƚŝŶŐŝŶƌŽǁϭ͕ ƐĞĂƚϯ͕ĨŽƌƚŚĞKƉĞŶŝŶŐ džĞƌĐŝƐĞ͕ŝƐŶŽǁƐŝƚƚŝŶŐŝŶƌŽǁϯ͕ƐĞĂƚϭ͘
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©201 8Great Minds ®. eureka-math.org
6ͻϯ Lesson 14 Lesson 14: KƌĚĞƌĞĚWĂŝƌƐ
Example 1: The Order in Ordered Pairs The first number of an ordered pair is called the first coordinate .The second number of an ordered pair is called the second coordinate .
ĞĨŝŶĞƚŚĞĨŝƌƐƚĂŶĚƐĞĐŽŶĚĐŽŽƌĚŝŶĂƚĞƐŝŶƚŚŝƐĞdžĂŵƉůĞĂƐ;ƌŽǁŶƵŵďĞƌ͕ƐĞĂƚŶƵŵďĞƌͿ͘ƐŬĂůůƐƚƵĚĞŶƚƐŝŶƚŚĞĐůĂƐƐƌŽŽŵ ƚŽƐƚĂŶĚ͘ĂůůŽƵƚĂŶĂƉƉƌŽƉƌŝĂƚĞŽƌĚĞƌĞĚƉĂŝƌ͕ĂŶĚĂƐŬĨŽƌƚŚĞƐƚƵĚĞŶƚŝŶƚŚĂƚůŽĐĂƚŝŽŶƚŽƌĂŝƐĞŚŝƐŚĂŶĚ͕ďƌŝĞĨůLJĞdžƉůĂŝŶ ǁŚLJƚŚĞŽƌĚĞƌĞĚƉĂŝƌŽĨŶƵŵďĞƌƐĚĞƐĐƌŝďĞƐƚŚĂƚƐƚƵĚĞŶƚ͛ƐƉŽƐŝƚŝŽŶŝŶƚŚĞƌŽŽŵ͕ĂŶĚƚŚĞŶďĞƐĞĂƚĞĚ͘EŽǁŚĂǀĞƚŚĂƚ ƐƚƵĚĞŶƚĐĂůůŽƵƚĂĚŝĨĨĞƌĞŶƚŽƌĚĞƌĞĚƉĂŝƌƚŚĂƚĐŽƌƌĞƐƉŽŶĚƐǁŝƚŚƚŚĞůŽĐĂƚŝŽŶŽĨĂŶŽƚŚĞƌƐƚƵĚĞŶƚ͘ŽŶƚŝŶƵĞƚŚŝƐƉƌŽĐĞƐƐ ƵŶƚŝůĂůůƐƚƵĚĞŶƚƐŚĂǀĞƉĂƌƚŝĐŝƉĂƚĞĚ͘
Example 2 (10 minutes): Using Ordered Pairs to Name Locations
ŝǀŝĚĞƐƚƵĚĞŶƚƐŝŶƚŽƐŵĂůůŐƌŽƵƉƐ͕ĂŶĚƉƌŽǀŝĚĞĞĂĐŚŐƌŽƵƉǁŝƚŚŽŶĞŽĨƚŚĞŽƌĚĞƌĞĚƉĂŝƌ ƐĐĞŶĂƌŝŽƐďĞůŽǁ͘^ƚƵĚĞŶƚƐƌĞĂĚƚŚĞŝƌƐĐĞŶĂƌŝŽƐĂŶĚĚĞƐĐƌŝďĞŚŽǁƚŚĞŽƌĚĞƌĞĚƉĂŝƌŝƐďĞŝŶŐ ƵƐĞĚ͕ŝŶĚŝĐĂƚŝŶŐǁŚĂƚĚĞĨŝŶĞƐƚŚĞĨŝƌƐƚĐŽŽƌĚŝŶĂƚĞĂŶĚǁŚĂƚĚĞĨŝŶĞƐƚŚĞƐĞĐŽŶĚĐŽŽƌĚŝŶĂƚĞ͘ ůůŽǁŐƌŽƵƉƐϱŵŝŶƵƚĞƐƚŽƌĞĂĚĂŶĚĚŝƐĐƵƐƐƚŚĞƐĐĞŶĂƌŝŽĂŶĚƉƌĞƉĂƌĞĂƌĞƐƉŽŶƐĞƚŽƌĞƉŽƌƚ ŽƵƚƚŽƚŚĞĐůĂƐƐ͘
Example 2: Using Ordered Pairs to Name Locations Describe how the ordered pair is being used in your scenario. Indicate what defines the first coordinate and what defines the second coordinate in your scenario.
KƌĚĞƌĞĚƉĂŝƌƐĂƌĞůŝŬĞĂƐĞƚŽĨĚŝƌĞĐƚŝŽŶƐ͖ƚŚĞLJŝŶĚŝĐĂƚĞǁŚĞƌĞƚŽŐŽŝŶŽŶĞĚŝƌĞĐƚŝŽŶĂŶĚƚŚĞŶŝŶĚŝĐĂƚĞǁŚĞƌĞƚŽŐŽŝŶƚŚĞ ƐĞĐŽŶĚĚŝƌĞĐƚŝŽŶ͘
^ĐĞŶĂƌŝŽϭ͗dŚĞƐĞĂƚƐŝŶĂĐŽůůĞŐĞĨŽŽƚďĂůůƐƚĂĚŝƵŵĂƌĞĂƌƌĂŶŐĞĚŝŶƚŽ ʹͳͲ ƐĞĐƚŝŽŶƐ͕ǁŝƚŚ ͳͶͶ ƐĞĂƚƐŝŶĞĂĐŚ ƐĞĐƚŝŽŶ͘zŽƵƌƚŝĐŬĞƚƚŽƚŚĞŐĂŵĞŝŶĚŝĐĂƚĞƐƚŚĞůŽĐĂƚŝŽŶŽĨLJŽƵƌƐĞĂƚƵƐŝŶŐƚŚĞŽƌĚĞƌĞĚƉĂŝƌŽĨŶƵŵďĞƌƐ
ሺͳʹ͵ǡ ͵ሻ ͘ ĞƐĐƌŝďĞƚŚĞŵĞĂŶŝŶŐŽĨĞĂĐŚŶƵŵďĞƌŝŶƚŚĞŽƌĚĞƌĞĚƉĂŝƌĂŶĚŚŽǁLJŽƵǁŽƵůĚƵƐĞƚŚĞŵƚŽĨŝŶĚLJŽƵƌ ƐĞĂƚ͘
^ĐĞŶĂƌŝŽϮ͗ŝƌůŝŶĞƉŝůŽƚƐƵƐĞŵĞĂƐƵƌĞŵĞŶƚƐŽĨůŽŶŐŝƚƵĚĞĂŶĚůĂƚŝƚƵĚĞƚŽĚĞƚĞƌŵŝŶĞƚŚĞŝƌůŽĐĂƚŝŽŶĂŶĚƚŽĨŝŶĚ ĂŝƌƉŽƌƚƐĂƌŽƵŶĚƚŚĞǁŽƌůĚ͘>ŽŶŐŝƚƵĚĞŝƐŵĞĂƐƵƌĞĚĂƐ ͲʹͳͺͲᤪ ĞĂƐƚŽƌ ͲʹͳͺͲᤪ ǁĞƐƚŽĨĂůŝŶĞƐƚƌĞƚĐŚŝŶŐĨƌŽŵƚŚĞ EŽƌƚŚWŽůĞƚŽƚŚĞ^ŽƵƚŚWŽůĞƚŚƌŽƵŐŚ'ƌĞĞŶǁŝĐŚ͕ ŶŐůĂŶĚ͕ĐĂůůĞĚƚŚĞ prime meridian ͘ >ĂƚŝƚƵĚĞŝƐŵĞĂƐƵƌĞĚĂƐ
ͲʹͻͲᤪ ŶŽƌƚŚŽƌ ͲʹͻͲᤪ ƐŽƵƚŚŽĨƚŚĞĞĂƌƚŚ͛ƐĞƋƵĂƚŽƌ͘ƉŝůŽƚŚĂƐƚŚĞŽƌĚĞƌĞĚƉĂŝƌ; ͻͲᤪ ǁĞƐƚ͕ ͵Ͳᤪ ŶŽƌƚŚͿ͘tŚĂƚ ĚŽĞƐĞĂĐŚŶƵŵďĞƌŝŶƚŚĞŽƌĚĞƌĞĚƉĂŝƌĚĞƐĐƌŝďĞ͍,ŽǁǁŽƵůĚƚŚĞƉŝůŽƚůŽĐĂƚĞƚŚĞĂŝƌƉŽƌƚŽŶĂŵĂƉ͍tŽƵůĚ ƚŚĞƌĞďĞĂŶLJĐŽŶĨƵƐŝŽŶŝĨĂƉŝůŽƚǁĞƌĞŐŝǀĞŶƚŚĞŽƌĚĞƌĞĚƉĂŝƌ ሺͻͲιǡ ͵Ͳιሻ ͍ džƉůĂŝŶ͘
^ĐĞŶĂƌŝŽϯ͗ ĂĐŚƌŽŽŵŝŶĂƐĐŚŽŽůďƵŝůĚŝŶŐŝƐŶĂŵĞĚďLJĂŶŽƌĚĞƌĞĚƉĂŝƌŽĨŶƵŵďĞƌƐƚŚĂƚŝŶĚŝĐĂƚĞƐƚŚĞŶƵŵďĞƌ ŽĨƚŚĞĨůŽŽƌŽŶǁŚŝĐŚƚŚĞƌŽŽŵůŝĞƐ͕ĨŽůůŽǁĞĚďLJƚŚĞƐĞƋƵĞŶƚŝĂůŶƵŵďĞƌŽĨƚŚĞƌŽŽŵŽŶƚŚĞĨůŽŽƌĨƌŽŵƚŚĞŵĂŝŶ ƐƚĂŝƌĐĂƐĞ͘ŶĞǁƐƚƵĚĞŶƚĂƚƚŚĞƐĐŚŽŽůŝƐƚƌLJŝŶŐƚŽŐĞƚƚŽƐĐŝĞŶĐĞĐůĂƐƐŝŶƌŽŽŵϰʹϭϯ͘ĞƐĐƌŝďĞƚŽƚŚĞƐƚƵĚĞŶƚ ǁŚĂƚĞĂĐŚŶƵŵďĞƌŵĞĂŶƐĂŶĚŚŽǁƐŚĞƐŚŽƵůĚƵƐĞƚŚĞŶƵŵďĞƌƚŽĨŝŶĚŚĞƌĐůĂƐƐƌŽŽŵ͘^ƵƉƉŽƐĞƚŚĞƌĞĂƌĞ ĐůĂƐƐƌŽŽŵƐďĞůŽǁƚŚĞŵĂŝŶĨůŽŽƌ͘,ŽǁŵŝŐŚƚƚŚĞƐĞƌŽŽŵƐďĞĚĞƐĐƌŝďĞĚ͍ ƐŬƐƚƵĚĞŶƚŐƌŽƵƉƐƚŽƌĞƉŽƌƚƚŚĞŝƌĂŶƐǁĞƌƐƚŽƚŚĞƐĐĞŶĂƌŝŽƐĂůŽƵĚƚŽƚŚĞĐůĂƐƐ͘
Scaffolding:
WƌŽǀŝĚĞĞdžƚƌĂƉƌĂĐƚŝĐĞŝŶ ŶĂŵŝŶŐůŽĐĂƚŝŽŶƐƵƐŝŶŐŽƌĚĞƌĞĚ ƉĂŝƌƐďLJƉůĂLJŝŶŐĂŐĂŵĞŽŶƚŚĞ ĐŽŽƌĚŝŶĂƚĞƉůĂŶĞǁŚĞƌĞ ƐƚƵĚĞŶƚƐƚƌLJƚŽŐƵĞƐƐƚŚĞ ůŽĐĂƚŝŽŶƐŽĨƚŚĞŝƌŽƉƉŽŶĞŶƚƐ͛ ƉŽŝŶƚƐ .
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6ͻϯ Lesson 14 Lesson 14: KƌĚĞƌĞĚWĂŝƌƐ
Exercises 1–2 (12 minutes)
^ƚƵĚĞŶƚƐƵƐĞƚŚĞŐƌŝĚĚĞĚŵĂƉƐŝŶƚŚĞƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐƚŽŶĂŵĞƉŽŝŶƚƐƚŚĂƚĐŽƌƌĞƐƉŽŶĚ ǁŝƚŚƚŚĞŐŝǀĞŶŽƌĚĞƌĞĚƉĂŝƌƐ;ĂŶĚǀŝĐĞǀĞƌƐĂͿ͘dŚĞĨŝƌƐƚĐŽŽƌĚŝŶĂƚĞƐƌĞƉƌĞƐĞŶƚŶƵŵďĞƌƐŽŶ ƚŚĞůŝŶĞůĂďĞůĞĚ ݔ͕ ĂŶĚƚŚĞƐĞĐŽŶĚĐŽŽƌĚŝŶĂƚĞƐƌĞƉƌĞƐĞŶƚŶƵŵďĞƌƐŽŶƚŚĞůŝŶĞůĂďĞůĞĚ ݕ͘
Exercises The first coordinates of the ordered pairs represent the numbers on the line labeled ࢞ , and the second coordinates represent the numbers on the line labeled ࢟ .1. Name the letter from the grid below that corresponds with each ordered pair of numbers below. a. ሺǡ ሻ
Point ࡲ
b. ሺǡ ሻ
Point
c. ሺǡ ሻ
Point
d. ሺૡǤ ǡ ૡሻ
Point ࡸ
e. ሺǡ െሻ
Point ࡳ
f. ሺǡ Ǥ ሻ
Point ࡴ
g. ሺǡ െሻ
Point
h. ሺǡૢ ሻ
Point ࡱ
List the ordered pair of numbers that corresponds with each letter from the grid below. a. Point ࡹ
ሺǡ ૠሻ
b. Point ࡿ
ሺെǡ ሻ
c. Point ࡺ
ሺǡ ሻ
d. Point ࢀ
ሺെǡ ሻ
e. Point ࡼ
ሺǡ ሻ
f. Point ࢁ
ሺૠǡ ሻ
g. Point ࡽ
ሺǡ ሻ
h. Point ࢂ
ሺെǡ ሻ
i. Point ࡾ
ሺǡ ሻ
,ĂǀĞƐƚƵĚĞŶƚƐƉƌŽǀŝĚĞƚŚĞĐŽƌƌĞĐƚĂŶƐǁĞƌƐƚŽƚŚĞĞdžĞƌĐŝƐĞƐ͘
Scaffolding:
/ĨƐƚƵĚĞŶƚƐĚŽŶŽƚƵŶĚĞƌƐƚĂŶĚ ƚŚĞŶĞŐĂƚŝǀĞŶƵŵďĞƌƐŽŶƚŚĞ ǀĞƌƚŝĐĂůĂdžŝƐ͕ƌĞǀŝĞǁǁŝƚŚ ƐƚƵĚĞŶƚƐŚŽǁƚŚĞĨůŽŽƌƐďĞůŽǁ ŐƌŽƵŶĚůĞǀĞůŵŝŐŚƚďĞ ĚĞƐĐƌŝďĞĚŝŶ^ĐĞŶĂƌŝŽϯĨƌŽŵ džĂŵƉůĞϮ͘
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6ͻϯ Lesson 14 Lesson 14: KƌĚĞƌĞĚWĂŝƌƐ
Closing (5 minutes)
tŚLJĚŽĞƐŽƌĚĞƌŵĂƚƚĞƌǁŚĞŶƵƐŝŶŐŽƌĚĞƌĞĚƉĂŝƌƐŽĨŶƵŵďĞƌƐ͍
à The order is important because it provides one specific location in the coordinate plane.
ůĂLJŶĂƐĂLJƐƚŚĞŽƌĚĞƌŝŶǁŚŝĐŚƚŚĞǀĂůƵĞƐĂƌĞŐŝǀĞŶŝŶĂŶŽƌĚĞƌĞĚƉĂŝƌĚŽĞƐŶŽƚĂůǁĂLJƐŵĂƚƚĞƌ͘'ŝǀĞĂŶ ĞdžĂŵƉůĞŽĨǁŚĞŶƚŚĞŽƌĚĞƌĚŽĞƐŵĂƚƚĞƌĂŶĚĂŶĞdžĂŵƉůĞŽĨǁŚĞŶƚŚĞŽƌĚĞƌĚŽĞƐŶŽƚŵĂƚƚĞƌ͘
à The order does not matter if the first and second coordinates are the same number. For example, ሺ͵ǡ ͵ሻ
is the same location in the coordinate plane no matter which point is used as the first coordinate. However, order does matter when the two coordinates are not the same. For example, ሺͳǡ ͵ሻ has a different location in the coordinate plane than ሺ͵ǡ ͳሻǤ
džƉůĂŝŶŚŽǁƚŽůŽĐĂƚĞƉŽŝŶƚƐǁŚĞŶƉĂŝƌƐŽĨŝŶƚĞŐĞƌƐĂƌĞƵƐĞĚ͘
à The first coordinate describes the location of the point using the horizontal direction. Positive integers indicate moving to the right from zero, and we would move left for negative numbers. The second coordinate describes the location of the point using the vertical direction. Positive numbers indicate moving up from zero, and we would move down from zero for negative integers.
Exit Ticket (8 minutes)
Lesson Summary
The order of numbers in an ordered pair is important because the ordered pair should describe one location in the coordinate plane.
The first number (called the first coordinate ) describes a location using the horizontal direction.
The second number (called the second coordinate ) describes a location using the vertical direction.
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6ͻϯ Lesson 14 Lesson 14: KƌĚĞƌĞĚWĂŝƌƐ
EĂŵĞ ĂƚĞ
Lesson 14: Ordered Pairs Exit Ticket
ϭ͘ KŶƚŚĞŵĂƉďĞůŽǁ͕ƚŚĞĨŝƌĞĚĞƉĂƌƚŵĞŶƚĂŶĚƚŚĞŚŽƐƉŝƚĂůŚĂǀĞŽŶĞŵĂƚĐŚŝŶŐĐŽŽƌĚŝŶĂƚĞ͘ĞƚĞƌŵŝŶĞƚŚĞƉƌŽƉĞƌ ŽƌĚĞƌŽĨƚŚĞŽƌĚĞƌĞĚƉĂŝƌƐŝŶƚŚĞŵĂƉ͕ĂŶĚǁƌŝƚĞƚŚĞĐŽƌƌĞĐƚŽƌĚĞƌĞĚƉĂŝƌƐĨŽƌƚŚĞůŽĐĂƚŝŽŶƐŽĨƚŚĞĨŝƌĞĚĞƉĂƌƚŵĞŶƚ ĂŶĚŚŽƐƉŝƚĂů͘/ŶĚŝĐĂƚĞǁŚŝĐŚŽĨƚŚĞŝƌĐŽŽƌĚŝŶĂƚĞƐĂƌĞƚŚĞƐĂŵĞ͘ Ϯ͘ KŶƚŚĞŵĂƉĂďŽǀĞ͕ůŽĐĂƚĞĂŶĚůĂďĞůƚŚĞůŽĐĂƚŝŽŶƐŽĨĞĂĐŚĚĞƐĐƌŝƉƚŝŽŶďĞůŽǁ͗ Ă͘ dŚĞůŽĐĂůďĂŶŬŚĂƐƚŚĞƐĂŵĞĨŝƌƐƚĐŽŽƌĚŝŶĂƚĞĂƐƚŚĞĨŝƌĞĚĞƉĂƌƚŵĞŶƚ͕ďƵƚŝƚƐƐĞĐŽŶĚĐŽŽƌĚŝŶĂƚĞŝƐŚĂůĨŽĨƚŚĞĨŝƌĞ ĚĞƉĂƌƚŵĞŶƚ͛ƐƐĞĐŽŶĚĐŽŽƌĚŝŶĂƚĞ͘tŚĂƚŽƌĚĞƌĞĚƉĂŝƌĚĞƐĐƌŝďĞƐƚŚĞůŽĐĂƚŝŽŶŽĨƚŚĞďĂŶŬ͍>ŽĐĂƚĞĂŶĚůĂďĞůƚŚĞ ďĂŶŬŽŶƚŚĞŵĂƉƵƐŝŶŐƉŽŝŶƚ ܤ͘ ď͘ dŚĞsŝůůĂŐĞWŽůŝĐĞĞƉĂƌƚŵĞŶƚŚĂƐƚŚĞƐĂŵĞƐĞĐŽŶĚĐŽŽƌĚŝŶĂƚĞĂƐƚŚĞďĂŶŬ͕ďƵƚŝƚƐĨŝƌƐƚĐŽŽƌĚŝŶĂƚĞŝƐ െʹ ͘ tŚĂƚŽƌĚĞƌĞĚƉĂŝƌĚĞƐĐƌŝďĞƐƚŚĞůŽĐĂƚŝŽŶŽĨƚŚĞsŝůůĂŐĞWŽůŝĐĞĞƉĂƌƚŵĞŶƚ͍>ŽĐĂƚĞĂŶĚůĂďĞůƚŚĞsŝůůĂŐĞWŽůŝĐĞ ĞƉĂƌƚŵĞŶƚŽŶƚŚĞŵĂƉƵƐŝŶŐƉŽŝŶƚ ܲ͘
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6ͻϯ Lesson 14 Lesson 14: KƌĚĞƌĞĚWĂŝƌƐ
Exit Ticket Sample Solutions 1. On the map below, the fire department and the hospital have one matching coordinate. Determine the proper order of the ordered pairs in the map, and write the correct ordered pairs for the locations of the fire department and hospital. Indicate which of their coordinates are the same.
The order of the numbers is ሺ࢞ǡ࢟ ሻ ; fire department: ሺǡ ૠሻ and hospital: ሺǡ ૠሻ ; they have the same second coordinate.
On the map above, locate and label the location of each description below: a. The local bank has the same first coordinate as the fire department, but its second coordinate is half of the fire department’s second coordinate. What ordered pair describes the location of the bank? Locate and label the bank on the map using point .
ሺǡ Ǥ ሻ ; see the map image for the correct location of point .
b. The Village Police Department has the same second coordinate as the bank, but its first coordinate is െ .What ordered pair describes the location of the Village Police Department? Locate and label the Village Police Department on the map using point ࡼ.
ሺെǡ Ǥ ሻ ; see the map image for the correct location of point ࡼ.
Problem Set Sample Solutions 1. Use the set of ordered pairs below to answer each question.
ሼሺǡ ሻǡ ሺૡǡ ሻǡ ሺǡ ሻǡ ሺǡ ሻǡ ሺǡ ሻǡ ሺǡ ሻǡ ሺǡ ሻǡ ሺǡ ૡሻǡ ሺǡ ሻሽ
a. Write the ordered pair(s) whose first and second coordinate have a greatest common factor of .
ሺǡ ሻ and ሺǡ ሻ
b. Write the ordered pair(s) whose first coordinate is a factor of its second coordinate.
ሺǡ ሻ ,ሺǡ ሻ ,ሺǡ ሻ , and ሺǡ ૡሻ
ࡼ
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6ͻϯ Lesson 14 Lesson 14: KƌĚĞƌĞĚWĂŝƌƐ
c. Write the ordered pair(s) whose second coordinate is a prime number.
ሺǡ ሻ , ሺǡ ሻ , ሺǡ ሻ , and ሺǡ ሻ
Write ordered pairs that represent the location of points , , , and ࡰ, where the first coordinate represents the horizontal direction, and the second coordinate represents the vertical direction.
ሺǡ ሻ ; ሺǡ െሻ ; ሺǡ ሻ ; ࡰሺǡ ሻ
Extension:
ϯ͘ Write ordered pairs of integers that satisfy the criteria in each part below. Remember that the origin is the point whose coordinates are ሺǡ ሻ . When possible, give ordered pairs such that (i) both coordinates are positive, (ii) both coordinates are negative, and (iii) the coordinates have opposite signs in either order. a. These points’ vertical distance from the origin is twice their horizontal distance.
Answers will vary; examples are ሺǡ ሻ ,ሺെǡ ሻ , ሺെǡ െሻ , ሺǡ െሻ .
b. These points’ horizontal distance from the origin is two units more than the vertical distance.
Answers will vary; examples are ሺǡ ሻ , ሺെǡ ሻ , ሺെǡ െሻ , ሺǡ െሻ .
c. These points’ horizontal and vertical distances from the origin are equal, but only one coordinate is positive.
Answers will vary; examples are ሺǡ െሻ , ሺെૡǡ ૡሻ .
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Lesson 15: >ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
6ͻϯ Lesson 15 Lesson 15: Locating Ordered Pairs on the Coordinate Plane Student Outcomes
^ƚƵĚĞŶƚƐĞdžƚĞŶĚƚŚĞŝƌƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽŝŶĐůƵĚĞĂůůĨŽƵƌƋƵĂĚƌĂŶƚƐĂŶĚƌĞĐŽŐŶŝnjĞƚŚĂƚ ƚŚĞĂdžĞƐ;ŝĚĞŶƚŝĨŝĞĚĂƐƚŚĞ ݔͲĂdžŝƐĂŶĚ ݕͲĂdžŝƐͿŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞĚŝǀŝĚĞƚŚĞƉůĂŶĞŝŶƚŽĨŽƵƌƌĞŐŝŽŶƐĐĂůůĞĚ
quadrants ;ƚŚĂƚĂƌĞůĂďĞůĞĚĨƌŽŵĨŝƌƐƚƚŽĨŽƵƌƚŚĂŶĚĂƌĞĚĞŶŽƚĞĚďLJƌŽŵĂŶŶƵŵĞƌĂůƐͿ͘
^ƚƵĚĞŶƚƐŝĚĞŶƚŝĨLJƚŚĞŽƌŝŐŝŶĂŶĚůŽĐĂƚĞƉŽŝŶƚƐŽƚŚĞƌƚŚĂŶƚŚĞŽƌŝŐŝŶ͕ǁŚŝĐŚůŝĞŽŶĂŶĂdžŝƐ͘
^ƚƵĚĞŶƚƐůŽĐĂƚĞƉŽŝŶƚƐŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŚĂƚĐŽƌƌĞƐƉŽŶĚƚŽŐŝǀĞŶŽƌĚĞƌĞĚƉĂŝƌƐŽĨŝŶƚĞŐĞƌƐĂŶĚŽƚŚĞƌ ƌĂƚŝŽŶĂůŶƵŵďĞƌƐ͘
Classwork Opening Exercise (6 minutes)
,ĂŶŐƉŽƐƚĞƌƐŽŶƚŚĞǁĂůů͕ĞĂĐŚĐŽŶƚĂŝŶŝŶŐŽŶĞŽĨƚŚĞĨŽůůŽǁŝŶŐƚĞƌŵƐ͗ ݔͲĂdžŝƐ͕ ݕͲĂdžŝƐ͕ ݔͲĐŽŽƌĚŝŶĂƚĞ͕ ݕͲĐŽŽƌĚŝŶĂƚĞ͕ŽƌŝŐŝŶ͕ ĂŶĚĐŽŽƌĚŝŶĂƚĞƉĂŝƌ͘WĂŝƌƐƚƵĚĞŶƚƐƵƉ͕ĂŶĚŚĂǀĞƚŚĞŵĚŝƐĐƵƐƐƚŚĞƐĞǀŽĐĂďƵůĂƌLJƚĞƌŵƐĂŶĚǁŚĂƚƚŚĞLJƌĞŵĞŵďĞƌĂďŽƵƚ ƚŚĞƚĞƌŵƐĨƌŽŵ'ƌĂĚĞϱ͘^ƚƵĚĞŶƚƉĂŝƌƐƚŚĞŶǁƌŝƚĞǁŚĂƚƚŚĞLJĚŝƐĐƵƐƐĞĚŽŶƚŚĞƉŽƐƚĞƌƐǁŝƚŚƚŚĞĂƉƉƌŽƉƌŝĂƚĞǀŽĐĂďƵůĂƌLJ ƚĞƌŵ͘^ŽŵĞŝŵƉŽƌƚĂŶƚĂƐƉĞĐƚƐĨŽƌƐƚƵĚĞŶƚƐƚŽƌĞŵĞŵďĞƌŝŶĐůƵĚĞƚŚĞĨŽůůŽǁŝŶŐ͗
dŚĞ ݔͲĂdžŝƐŝƐĂŚŽƌŝnjŽŶƚĂůŶƵŵďĞƌůŝŶĞ͖ƚŚĞ ݕͲĂdžŝƐŝƐĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ͘
dŚĞĂdžĞƐŵĞĞƚĨŽƌŵŝŶŐĂ ͻͲι ĂŶŐůĞĂƚƚŚĞƉŽŝŶƚ ሺͲǡ Ͳሻ ĐĂůůĞĚƚŚĞ origin ͘
Example 1 (8 minutes): Extending the Axes Beyond Zero
^ƚƵĚĞŶƚƐƌĞĐŽŐŶŝnjĞƚŚĂƚƚŚĞĂdžĞƐĂƌĞŶƵŵďĞƌůŝŶĞƐĂŶĚ͕ƵƐŝŶŐƐƚƌĂŝŐŚƚĞĚŐĞƐ͕ĞdžƚĞŶĚƚŚĞĂdžĞƐŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽ ŝŶĐůƵĚĞŶĞŐĂƚŝǀĞŶƵŵďĞƌƐƌĞǀĞĂůŝŶŐƚŚĞƐĞĐŽŶĚ͕ƚŚŝƌĚ͕ĂŶĚĨŽƵƌƚŚƋƵĂĚƌĂŶƚƐ͘
ĞƐĐƌŝďĞƚŚĞ ݔͲĂdžŝƐ͘ŽŶƐŝĚĞƌŝŶŐǁŚĂƚǁĞŚĂǀĞƐĞĞŶŝŶƚŚŝƐŵŽĚƵůĞ͕ǁŚĂƚƚLJƉĞƐŽĨŶƵŵďĞƌƐƐŚŽƵůĚŝƚŝŶĐůƵĚĞ͍ tŚLJ͍
à The ݔ-axis is a horizontal number line that includes positive and negative numbers. The axis extends in both directions (left and right of zero) because signed numbers represent values or quantities that have opposite directions.
,ĂǀĞƐƚƵĚĞŶƚƵƐĞƚŚĞŝƌƉƌŝŽƌŬŶŽǁůĞĚŐĞƚŽĐŽŵƉůĞƚĞ džĂŵƉůĞϭŝŶĚŝǀŝĚƵĂůůLJ͘
Example 1: Extending the Axes Beyond Zero The point below represents zero on the number line. Draw a number line to the right starting at zero. Then, follow directions as provided by the teacher.
hƐĞĂƐƚƌĂŝŐŚƚĞĚŐĞƚŽĞdžƚĞŶĚƚŚĞ ݔͲĂdžŝƐƚŽƚŚĞůĞĨƚŽĨnjĞƌŽƚŽƌĞƉƌĞƐĞŶƚƚŚĞƌĞĂůŶƵŵďĞƌůŝŶĞŚŽƌŝnjŽŶƚĂůůLJ͕ĂŶĚ ĐŽŵƉůĞƚĞƚŚĞŶƵŵďĞƌůŝŶĞƵƐŝŶŐƚŚĞƐĂŵĞƐĐĂůĞĂƐŽŶƚŚĞƌŝŐŚƚƐŝĚĞŽĨnjĞƌŽ͘
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Lesson 15: >ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
6ͻϯ Lesson 15
ሺǡ ሻ
ሺǡ ሻ
ሺെǡ ሻ
ሺെǡ ሻ ሺെૡǡ ሻ
ሺǡ ૡሻ
ሺǡ െૡሻ
ሺǡ ሻ
ሺǡ ሻ
ሺǡ െሻ
ĞƐĐƌŝďĞƚŚĞ ݕͲĂdžŝƐ͘tŚĂƚƚLJƉĞƐŽĨŶƵŵďĞƌƐƐŚŽƵůĚŝƚŝŶĐůƵĚĞ͍
à The ݕ-axis is a vertical number line that includes numbers on both sides of zero (above and below), and so it includes both positive and negative numbers.
hƐĞĂƐƚƌĂŝŐŚƚĞĚŐĞƚŽĚƌĂǁĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞĂďŽǀĞnjĞƌŽ͘ WƌŽǀŝĚĞƐƚƵĚĞŶƚƐǁŝƚŚƚŝŵĞƚŽĚƌĂǁ͘
džƚĞŶĚƚŚĞ ݕͲĂdžŝƐďĞůŽǁnjĞƌŽƚŽƌĞƉƌĞƐĞŶƚƚŚĞƌĞĂůŶƵŵďĞƌůŝŶĞǀĞƌƚŝĐĂůůLJ͕ĂŶĚĐŽŵƉůĞƚĞƚŚĞŶƵŵďĞƌůŝŶĞƵƐŝŶŐ ƚŚĞƐĂŵĞƐĐĂůĞĂƐĂďŽǀĞnjĞƌŽ͘
Example 2 (4 minutes): Components of the Coordinate Plane
^ƚƵĚĞŶƚƐĞdžĂŵŝŶĞŚŽǁƚŽƵƐĞƚŚĞĂdžĞƐĂŶĚƚŚĞŽƌŝŐŝŶŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽĚĞƚĞƌŵŝŶĞ ŽƚŚĞƌůŽĐĂƚŝŽŶƐŝŶƚŚĞƉůĂŶĞ͘
Example 2: Components of the Coordinate Plane All points on the coordinate plane are described with reference to the origin. What is the origin, and what are its coordinates?
The origin is the point where the ࢞ - and ࢟ -axes intersect. The coordinates of the origin are ሺǡ ሻ .
dŚĞĂdžĞƐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŝŶƚĞƌƐĞĐƚĂƚƚŚĞŝƌnjĞƌŽĐŽŽƌĚŝŶĂƚĞƐ͕ǁŚŝĐŚŝƐĂƉŽŝŶƚĐĂůůĞĚƚŚĞ origin ͘ dŚĞ ŽƌŝŐŝŶŝƐƚŚĞƌĞĨĞƌĞŶĐĞƉŽŝŶƚĨƌŽŵǁŚŝĐŚĂůůƉŽŝŶƚƐŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞĂƌĞĚĞƐĐƌŝďĞĚ͘
To describe locations of points in the coordinate plane, we use ordered pairs of numbers. Order is important, so on the coordinate plane, we use the form ሺ࢟࢞ ሻ. The first coordinate represents the point’s location from zero on the ࢞ -axis, and the second coordinate represents the point’s location from zero on the ࢟ -axis.
Exercises 1– ϯ (8 minutes)
^ƚƵĚĞŶƚƐůŽĐĂƚĞĂŶĚůĂďĞůƉŽŝŶƚƐƚŚĂƚůŝĞŽŶƚŚĞĂdžĞƐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘
Exercises 1– ϯ
Use the coordinate plane below to answer parts (a)–(c). a. Graph at least five points on the ࢞ -axis, and label their coordinates.
Points will vary.
b. What do the coordinates of your points have in common?
Each point has a ࢟ -coordinate of .
c. What must be true about any point that lies on the ࢞ -axis? Explain.
If a point lies on the ࢞ -axis , its ࢟ -coordinate must be
because the point is located units above or below the ࢞
-axis. The ࢞ -axis intersects the ࢟ -axis at .
Scaffolding:
dŚĞƚĞƌŵ origin ŵĞĂŶƐ ƐƚĂƌƚŝŶŐƉŽŝŶƚ͘
ƉĞƌƐŽŶ͛ƐĐŽƵŶƚƌLJŽĨ ŽƌŝŐŝŶŝƐƚŚĞĐŽƵŶƚƌLJĨƌŽŵ ǁŚŝĐŚŚĞĐĂŵĞ͘
tŚĞŶƵƐŝŶŐĂŐůŽďĂů ƉŽƐŝƚŝŽŶŝŶŐƵŶŝƚ;'W^Ϳ ǁŚĞŶƚƌĂǀĞůŝŶŐ͕ƚŚĞ origin ŝƐǁŚĞƌĞƚŚĞƚƌŝƉďĞŐĂŶ͘
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Lesson 15: >ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
6ͻϯ Lesson 15 2. Use the coordinate plane to answer parts (a)–(c). a. Graph at least five points on the ࢟ -axis, and label their coordinates.
Points will vary.
b. What do the coordinates of your points have in common?
Each point has an ࢞ -coordinate of .
c. What must be true about any point that lies on the ࢟ -axis? Explain.
If a point lies on the ࢟ -axis, its ࢞ -coordinate must be because the point is located units left or right of the ࢟
-axis. The ࢟ -axis intersects on the ࢞ -axis.
ϯ͘ If the origin is the only point with for both coordinates, what must be true about the origin?
The origin is the only point that is on both the ࢞ -axis and the ࢟ -axis.
Example ϯ (6 minutes): Quadrants of the Coordinate Plane
^ƚƵĚĞŶƚƐĞdžĂŵŝŶĞƚŚĞĨŽƵƌƌĞŐŝŽŶƐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞĐƵƚďLJƚŚĞŝŶƚĞƌƐĞĐƚŝŶŐĂdžĞƐ͘
džĂŵƉůĞϯ͗YƵĂĚƌĂŶƚƐŽĨƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
dŚĞ ݔͲĂŶĚ ݕͲĂdžĞƐĚŝǀŝĚĞƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŝŶƚŽƌĞŐŝŽŶƐĐĂůůĞĚƋƵĂĚƌĂŶƚƐ͘tŚLJĂƌĞƚŚĞƌĞŐŝŽŶƐĐĂůůĞĚ ƋƵĂĚƌĂŶƚƐ͍
à The axes cut the plane into four regions. The prefix “quad” means four.
tŚŝĐŚŽĨƚŚĞĨŽƵƌƌĞŐŝŽŶƐĚŝĚLJŽƵǁŽƌŬǁŝƚŚŵŽƐƚŝŶ'ƌĂĚĞϱ͕ĂŶĚǁŚLJǁĂƐŝƚƚŚĞŽŶůLJƌĞŐŝŽŶLJŽƵƵƐĞĚ͍
à The region on the top right of the coordinate plane. We only used this region because we had not learned about negative numbers yet.
dŚĞĨŽƵƌƋƵĂĚƌĂŶƚƐĂƌĞŶƵŵďĞƌĞĚŽŶĞƚŚƌŽƵŐŚĨŽƵƌƵƐŝŶŐƌŽŵĂŶŶƵŵĞƌĂůƐ͘dŚĞƵƉƉĞƌͲƌŝŐŚƚƋƵĂĚƌĂŶƚŝƐ YƵĂĚƌĂŶƚ/͕ĂŶĚƚŚĞƌĞŵĂŝŶŝŶŐƋƵĂĚƌĂŶƚƐĂƌĞŶƵŵďĞƌĞĚŵŽǀŝŶŐĐŽƵŶƚĞƌĐůŽĐŬǁŝƐĞĨƌŽŵYƵĂĚƌĂŶƚ/͗YƵĂĚƌĂŶƚ //͕YƵĂĚƌĂŶƚ///͕ĂŶĚYƵĂĚƌĂŶƚ/s͘tŚĂƚǁĂƐƚŚĞĨŝƌƐƚĂdžŝƐƚŚĂƚǁĞĞdžƚĞŶĚĞĚŝŶ džĂŵƉůĞϭ͕ĂŶĚǁŚĂƚĚŝĚŝƚ ƌĞǀĞĂů͍
à We extended the ݔ-axis to the left beyond zero, and it revealed another region of the coordinate plane.
Scaffolding:
ZĞŵŝŶĚƐƚƵĚĞŶƚƐƚŚĂƚƚŚĞ ƉƌĞĨŝdž quad ŵĞĂŶƐĨŽƵƌ͕ ĂŶĚĐŝƚĞƐŽŵĞŽƚŚĞƌ ĞdžĂŵƉůĞƐǁŚĞƌĞƚŚĞƉƌĞĨŝdž ŝƐƵƐĞĚ͘
^ŽŵĞƐƚƵĚĞŶƚƐŵĂLJŶŽƚ ŚĂǀĞŬŶŽǁůĞĚŐĞŽĨƌŽŵĂŶ ŶƵŵĞƌĂůƐ͘ƌĞĂƚĞĂƚĂďůĞ ŝŶǁŚŝĐŚƐƚƵĚĞŶƚƐĐĂŶ ĐŽŵƉĂƌĞƚŚĞƐƚĂŶĚĂƌĚ ƐLJŵďŽůƐ ͳʹͺĂŶĚƚŚĞ ƌŽŵĂŶŶƵŵĞƌĂůƐ/ʹs///͘
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Lesson 15: >ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
6ͻϯ Lesson 15
dŚŝƐƚŽƉůĞĨƚƌĞŐŝŽŶŝƐĐĂůůĞĚYƵĂĚƌĂŶƚ//͘>ĂďĞůYƵĂĚƌĂŶƚ//ŝŶLJŽƵƌƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐ͘dŚĞƐĞƌĞŐŝŽŶƐŽŶůLJŵĂŬĞ ƵƉŚĂůĨŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘tŚĞƌĞĚŽĞƐƚŚĞƌĞŵĂŝŶŝŶŐŚĂůĨŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞĐŽŵĞĨƌŽŵ͍ džƉůĂŝŶ͘
à We need to extend the ݕ-axis down below zero to show its negative values. This reveals two other regions on the plane, one to the left of the ݕ-axis and one to the right of the ݕ-axis.
dŚĞƋƵĂĚƌĂŶƚƐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞĂƌĞŝŶĂĐŽƵŶƚĞƌĐůŽĐŬǁŝƐĞĚŝƌĞĐƚŝŽŶƐƚĂƌƚŝŶŐǁŝƚŚYƵĂĚƌĂŶƚ/͘>ĂďĞůƚŚĞ ƌĞŵĂŝŶŝŶŐƋƵĂĚƌĂŶƚƐŝŶLJŽƵƌƐƚƵĚĞŶƚŵĂƚĞƌŝĂůƐ͘
Exercises 4–6 (5 minutes)
^ƚƵĚĞŶƚƐůŽĐĂƚĞĂŶĚůĂďĞůƉŽŝŶƚƐƚŚĂƚůŝĞŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞĂŶĚŝŶĚŝĐĂƚĞŝŶǁŚŝĐŚŽĨƚŚĞĨŽƵƌƋƵĂĚƌĂŶƚƐƚŚĞƉŽŝŶƚƐ ůŝĞ͘
Exercises 4–6 4. Locate and label each point described by the ordered pairs below. Indicate which of the quadrants the points lie in. a. ሺૠǡ ሻ
Quadrant I
b. ሺǡ െሻ
Quadrant IV
c. ሺǡ െሻ
Quadrant IV
d. ሺെǡ ૡሻ
Quadrant II
e. ሺെǡ െሻ
Quadrant III
Write the coordinates of at least one other point in each of the four quadrants. a. Quadrant I
Answers will vary, but both numbers must be positive.
b. Quadrant II
Answers will vary, but the ࢞ -coordinate must be negative, and the ࢟ -coordinate must be positive.
c. Quadrant III
Answers will vary, but both numbers must be negative.
d. Quadrant IV
Answers will vary, but the ࢞ -coordinate must be positive, and the ࢟ -coordinate must be negative.
ሺૠǡ ሻ
ሺെǡ ૡሻ
ሺǡ െሻ
ሺǡ െሻ
ሺെǡ െሻ
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Lesson 15: >ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
6ͻϯ Lesson 15 6. Do you see any similarities in the points within each quadrant? Explain your reasoning.
The ordered pairs describing the points in Quadrant I contain both positive values. The ordered pairs describing the points in Quadrant III contain both negative values. The first coordinates of the ordered pairs describing the points in Quadrant II are negative values, but their second coordinates are positive values. The first coordinates of the ordered pairs describing the points in Quadrant IV are positive values, but their second coordinates are negative values.
Closing (4 minutes)
/ĨĂƉŽŝŶƚůŝĞƐŽŶĂŶĂdžŝƐ͕ǁŚĂƚŵƵƐƚďĞƚƌƵĞĂďŽƵƚŝƚƐĐŽŽƌĚŝŶĂƚĞƐ͍^ƉĞĐŝĨŝĐĂůůLJ͕ǁŚĂƚŝƐƚƌƵĞĨŽƌĂƉŽŝŶƚƚŚĂƚůŝĞƐ ŽŶƚŚĞ ݔͲĂdžŝƐ͍dŚĞ ݕͲĂdžŝƐ͍
à The ݕ-coordinate is always Ͳ if a point lies on the ݔ-axis. The ݔ-coordinate is always Ͳ if a point lies on the ݕ-axis.
tŚĂƚĚŽLJŽƵŬŶŽǁĂďŽƵƚƚŚĞůŽĐĂƚŝŽŶŽĨĂƉŽŝŶƚŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŝĨ͗
ŽƚŚĐŽŽƌĚŝŶĂƚĞƐĂƌĞƉŽƐŝƚŝǀĞ͍
à If both coordinates are positive, the point must be located in Quadrant I.
KŶůLJŽŶĞĐŽŽƌĚŝŶĂƚĞŝƐƉŽƐŝƚŝǀĞ͍
à If only one coordinate is positive, the point is either in Quadrant II or Quadrant IV. If only the first coordinate is positive, then the point is in Quadrant IV. If only the second coordinate is positive, then the point is in Quadrant II.
ŽƚŚĐŽŽƌĚŝŶĂƚĞƐĂƌĞŶĞŐĂƚŝǀĞ͍
à If both coordinates are negative, the point is located in Quadrant III.
KŶĞĐŽŽƌĚŝŶĂƚĞŝƐnjĞƌŽ͍
à If one coordinate is zero, then the point is located on the ݔ-axis or the ݕ -axis.
ŽƚŚĐŽŽƌĚŝŶĂƚĞƐĂƌĞnjĞƌŽ͍
à If both coordinates are zero, then the point represents the origin.
Exit Ticket (4 minutes) Lesson Summary
The ࢞ -axis and ࢟ -axis of the coordinate plane are number lines that intersect at zero on each number line.
The axes partition the coordinate plane into four quadrants.
Points in the coordinate plane lie either on an axis or in one of the four quadrants.
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Lesson 15: >ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
6ͻϯ Lesson 15
EĂŵĞ ĂƚĞ
Lesson 15: Locating Ordered Pairs on the Coordinate Plane Exit Ticket
ϭ͘ >ĂďĞůƚŚĞƐĞĐŽŶĚƋƵĂĚƌĂŶƚŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͕ĂŶĚ ƚŚĞŶĂŶƐǁĞƌƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐ͗ Ă͘ tƌŝƚĞƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨŽŶĞƉŽŝŶƚƚŚĂƚůŝĞƐŝŶƚŚĞ ƐĞĐŽŶĚƋƵĂĚƌĂŶƚŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘ ď͘ tŚĂƚŵƵƐƚďĞƚƌƵĞĂďŽƵƚƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨĂŶLJ ƉŽŝŶƚƚŚĂƚůŝĞƐŝŶƚŚĞƐĞĐŽŶĚƋƵĂĚƌĂŶƚ͍ Ϯ͘ >ĂďĞůƚŚĞƚŚŝƌĚƋƵĂĚƌĂŶƚŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͕ĂŶĚƚŚĞŶ ĂŶƐǁĞƌƚŚĞĨŽůůŽǁŝŶŐƋƵĞƐƚŝŽŶƐ͗ Ă͘ tƌŝƚĞƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨŽŶĞƉŽŝŶƚƚŚĂƚůŝĞƐŝŶƚŚĞƚŚŝƌĚƋƵĂĚƌĂŶƚŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘ ď͘ tŚĂƚŵƵƐƚďĞƚƌƵĞĂďŽƵƚƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨĂŶLJƉŽŝŶƚƚŚĂƚůŝĞƐŝŶƚŚĞƚŚŝƌĚƋƵĂĚƌĂŶƚ͍ ϯ͘ ŶŽƌĚĞƌĞĚƉĂŝƌŚĂƐĐŽŽƌĚŝŶĂƚĞƐƚŚĂƚŚĂǀĞƚŚĞƐĂŵĞƐŝŐŶ͘/ŶǁŚŝĐŚƋƵĂĚƌĂŶƚ;ƐͿĐŽƵůĚƚŚĞƉŽŝŶƚůŝĞ͍ džƉůĂŝŶ͘ ϰ͘ ŶŽƚŚĞƌŽƌĚĞƌĞĚƉĂŝƌŚĂƐĐŽŽƌĚŝŶĂƚĞƐƚŚĂƚĂƌĞŽƉƉŽƐŝƚĞƐ͘/ŶǁŚŝĐŚƋƵĂĚƌĂŶƚ;ƐͿĐŽƵůĚƚŚĞƉŽŝŶƚůŝĞ͍ džƉůĂŝŶ͘
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Lesson 15: >ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
6ͻϯ Lesson 15
Quadrant II
Quadrant III
Exit Ticket Sample Solutions 1. Label the second quadrant on the coordinate plane, and then answer the following questions: a. Write the coordinates of one point that lies in the second quadrant of the coordinate plane.
Answers will vary.
b. What must be true about the coordinates of any point that lies in the second quadrant?
The ࢞ -coordinate must be a negative value, and the ࢟
-coordinate must be a positive value.
Label the third quadrant on the coordinate plane, and then answer the following questions: a. Write the coordinates of one point that lies in the third quadrant of the coordinate plane.
Answers will vary.
b. What must be true about the coordinates of any point that lies in the third quadrant?
The ࢞ - and ࢟ -coordinates of any point in the third quadrant must both be negative values.
ϯ͘ An ordered pair has coordinates that have the same sign. In which quadrant(s) could the point lie? Explain.
The point would have to be located either in Quadrant I where both coordinates are positive values or in Quadrant III where both coordinates are negative values.
Another ordered pair has coordinates that are opposites. In which quadrant(s) could the point lie? Explain.
The point would have to be located in either Quadrant II or Quadrant IV because those are the two quadrants where the coordinates have opposite signs. The point could also be located at the origin ሺǡ ሻ since zero is its own opposite.
Problem Set Sample Solutions 1. Name the quadrant in which each of the points lies. If the point does not lie in a quadrant, specify which axis the point lies on. a. ሺെǡ ሻ
Quadrant II
b. ሺૡǡ െሻ
Quadrant IV
c. ሺെǡ െૡሻ
Quadrant III
d. ሺૢǤ ǡ ૠሻ
Quadrant I
e. ሺǡ െሻ
None; the point is not in a quadrant because it lies on the ࢟ -axis.
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Lesson 15: >ŽĐĂƚŝŶŐKƌĚĞƌĞĚWĂŝƌƐŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
6ͻϯ Lesson 15
Jackie claims that points with the same ࢞ - and ࢟ -coordinates must lie in Quadrant I or Quadrant III. Do you agree or disagree? Explain your answer.
Disagree; most points with the same ࢞ - and ࢟ -coordinates lie in Quadrant I or Quadrant III, but the origin ሺǡ ሻ is on the ࢞ - and ࢟ -axes, not in any quadrant.
ϯ͘ Locate and label each set of points on the coordinate plane. Describe similarities of the ordered pairs in each set, and describe the points on the plane. a. ሼሺെǡ ሻǡ ሺെǡ ሻǡ ሺെǡ ૠሻǡ ሺെǡ െሻǡ ሺെǡ െǤ ૡሻሽ
The ordered pairs all have ࢞ -coordinates of െ , and the points lie along a vertical line above and below
െ on the ࢞ -axis.
b. ሼሺെૢ ǡૢ ሻǡ ሺെǡ ሻǡ ሺെǡ ሻǡ ሺǡ െሻǡ ሺǡ െሻǡ ሺǡ ሻሽ
The ordered pairs each have opposite values for their ࢞ - and ࢟ -coordinates. The points in the plane line up diagonally through Quadrant II, the origin, and Quadrant IV.
c. ሼሺെૠǡ െૡሻǡ ሺǡ െૡሻǡ ሺǡ െૡሻǡ ሺǡ െૡሻǡ ሺെǡ െૡሻሽ
The ordered pairs all have ࢟ -coordinates of െૡ , and the points lie along a horizontal line to the left and right of െૡ on the ࢟ -axis.
Locate and label at least five points on the coordinate plane that have an ࢞ -coordinate of .a. What is true of the ࢟ -coordinates below the ࢞ -axis?
The ࢟ -coordinates are all negative values.
b. What is true of the ࢟ -coordinates above the ࢞ -axis?
The ࢟ -coordinates are all positive values.
c. What must be true of the ࢟ -coordinates on the ࢞
-axis?
The ࢟ -coordinates on the ࢞ -axis must be .
ሺǡ ሻ
ሺǡ ሻ
ሺǡ െሻ
ሺǡ െሻ
ሺǡ െሻ
ሺǡ െሻ
ሺǡ െሻ
ሺǡ െሻ
ሺǡ െૠሻ
ሺǡ ሻ
ሺǡ ሻ
ሺǡ ሻ
ሺǡ ሻ
ሺǡ ૠሻ
ሺǡ ૡሻ
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6ͻϯ Lesson 16 Lesson 16: ^LJŵŵĞƚƌLJŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
Lesson 16: Symmetry in the Coordinate Plane Student Outcomes
^ƚƵĚĞŶƚƐƵŶĚĞƌƐƚĂŶĚƚŚĂƚƚǁŽŶƵŵďĞƌƐĂƌĞƐĂŝĚƚŽĚŝĨĨĞƌŽŶůLJďLJƐŝŐŶƐŝĨƚŚĞLJĂƌĞŽƉƉŽƐŝƚĞƐŽĨĞĂĐŚŽƚŚĞƌ͘
^ƚƵĚĞŶƚƐƌĞĐŽŐŶŝnjĞƚŚĂƚǁŚĞŶƚǁŽŽƌĚĞƌĞĚƉĂŝƌƐĚŝĨĨĞƌŽŶůLJďLJƚŚĞƐŝŐŶŽĨŽŶĞŽƌďŽƚŚŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƐ͕ƚŚĞŶ ƚŚĞůŽĐĂƚŝŽŶƐŽĨƚŚĞƉŽŝŶƚƐĂƌĞƌĞůĂƚĞĚďLJƌĞĨůĞĐƚŝŽŶƐĂĐƌŽƐƐŽŶĞŽƌďŽƚŚĂdžĞƐ͘
Classwork Opening Exercise ;ϯ minutes) Opening Exercise Give an example of two opposite numbers, and describe where the numbers lie on the number line. How are opposite numbers similar, and how are they different?
Answers may vary. and െ are opposites because they are both units from zero on a number line but in opposite directions. Opposites are similar because they have the same absolute value, but they are different because opposites are on opposite sides of zero.
Example 1 (14 minutes): Extending Opposite Numbers to the Coordinate Plane
^ƚƵĚĞŶƚƐůŽĐĂƚĞĂŶĚůĂďĞůƉŽŝŶƚƐǁŚŽƐĞŽƌĚĞƌĞĚƉĂŝƌƐĚŝĨĨĞƌŽŶůLJďLJƚŚĞƐŝŐŶŽĨŽŶĞŽƌďŽƚŚĐŽŽƌĚŝŶĂƚĞƐ͘dŽŐĞƚŚĞƌ͕ ƐƚƵĚĞŶƚƐĂŶĚƚŚĞƚĞĂĐŚĞƌĞdžĂŵŝŶĞƚŚĞƌĞůĂƚŝŽŶƐŚŝƉƐŽĨƚŚĞƉŽŝŶƚƐŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞĂŶĚĞdžƉƌĞƐƐƚŚĞƐĞ ƌĞůĂƚŝŽŶƐŚŝƉƐŝŶĂŐƌĂƉŚŝĐŽƌŐĂŶŝnjĞƌ͘
x >ŽĐĂƚĞĂŶĚůĂďĞůƚŚĞƉŽŝŶƚƐ ሺ͵ǡ Ͷሻ ĂŶĚ ሺെ͵ǡ Ͷሻ ͘
x ZĞĐŽƌĚŽďƐĞƌǀĂƚŝŽŶƐŝŶƚŚĞĨŝƌƐƚĐŽůƵŵŶŽĨƚŚĞŐƌĂƉŚŝĐŽƌŐĂŶŝnjĞƌ͘ dŚĞĨŝƌƐƚĐŽůƵŵŶŽĨƚŚĞŐƌĂƉŚŝĐŽƌŐĂŶŝnjĞƌŝƐƚĞĂĐŚĞƌůĞĚƐŽƚŚĂƚƐƚƵĚĞŶƚƐĐĂŶƉĂLJƉĂƌƚŝĐƵůĂƌĂƚƚĞŶƚŝŽŶƚŽƚŚĞĂďƐŽůƵƚĞ ǀĂůƵĞƐŽĨĐŽŽƌĚŝŶĂƚĞƐĂŶĚƚŚĞŐĞŶĞƌĂůůŽĐĂƚŝŽŶƐŽĨƚŚĞĐŽƌƌĞƐƉŽŶĚŝŶŐƉŽŝŶƚƐǁŝƚŚƌĞŐĂƌĚƚŽĞĂĐŚĂdžŝƐ͘ŽůƵŵŶƐϮĂŶĚϯ ĂƌĞƉƌŝŵĂƌŝůLJƐƚƵĚĞŶƚůĞĚ͘
x >ŽĐĂƚĞĂŶĚůĂďĞůƚŚĞƉŽŝŶƚ ሺ͵ǡ െͶሻ ͘
x ZĞĐŽƌĚŽďƐĞƌǀĂƚŝŽŶƐŝŶƚŚĞƐĞĐŽŶĚĐŽůƵŵŶŽĨƚŚĞŐƌĂƉŚŝĐŽƌŐĂŶŝnjĞƌ͘
x >ŽĐĂƚĞĂŶĚůĂďĞůƚŚĞƉŽŝŶƚ ሺെ͵ǡ െͶሻ ͘
x ZĞĐŽƌĚŽďƐĞƌǀĂƚŝŽŶƐŝŶƚŚĞƚŚŝƌĚĐŽůƵŵŶŽĨƚŚĞŐƌĂƉŚŝĐŽƌŐĂŶŝnjĞƌ͘
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6ͻϯ Lesson 16 Lesson 16: ^LJŵŵĞƚƌLJŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
Example 1: Extending Opposite Numbers to the Coordinate Plane Extending Opposite Numbers to the Coordinates of Points on the Coordinate Plane Locate and label your points on the coordinate plane to the right. For each given pair of points in the table below, record your observations and conjectures in the appropriate cell. Pay attention to the absolute values of the coordinates and where the points lie in reference to each axis.
ሺǡ ሻ and ሺെǡ ሻ ሺǡ ሻ and ሺǡ െሻ ሺǡ ሻ and ሺെǡ െሻ
Similarities of Coordinates
Same ࢟ -coordinates The ࢞ -coordinates have the same absolute value. Same ࢞ -coordinates The ࢟ -coordinates have the same absolute value. The ࢞ -coordinates have the same absolute value. The ࢟ -coordinates have the same absolute value.
Differences of Coordinates
The ࢞ -coordinates are opposite numbers. The ࢟ -coordinates are opposite numbers. Both the ࢞ - and ࢟
-coordinates are opposite numbers.
Similarities in Location
Both points are units above the ࢞ -axis and units away from the ࢟ -axis. Both points are units to the right of the ࢟ -axis and
units away from the ࢞ -axis. Both points are units from the ࢟ -axis and units from the ࢞ -axis.
Differences in Location
One point is units to the right of the ࢟ -axis; the other is units to the left of the ࢟
-axis. One point is units above the ࢞ -axis; the other is
units below. One point is units right of the ࢟ -axis; the other is
units left. One point is
units above the ࢞ -axis; the other is units below.
Relationship Between Coordinates and Location on the Plane
The ࢞ -coordinates are opposite numbers, so the points lie on opposite sides of the ࢟ -axis. Because opposites have the same absolute value, both points lie the same distance from the ࢟ -axis. The points lie the same distance above the ࢞
-axis, so the points are symmetric about the ࢟ -axis. A reflection across the ࢟ -axis takes one point to the other. The ࢟ -coordinates are opposite numbers, so the points lie on opposite sides of the ࢞ -axis. Because opposites have the same absolute value, both points lie the same distance from the ࢞ -axis. The points lie the same distance right of the
࢟
-axis, so the points are symmetric about the ࢞ -axis. A reflection across the ࢞ -axis takes one point to the other. The points have opposite numbers for ࢞ - and ࢟ -coordinates, so the points must lie on opposite sides of each axis. Because the numbers are opposites and opposites have the same absolute values, each point must be the same distance from each axis. A reflection across one axis followed by a reflection across the other axis takes one point to the other.
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6ͻϯ Lesson 16 Lesson 16: ^LJŵŵĞƚƌLJŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
Exercises 1–2 (5 minutes) Exercises In each column, write the coordinates of the points that are related to the given point by the criteria listed in the first column of the table. Point ࡿሺǡ ሻ has been reflected over the ࢞ - and ࢟ -axes for you as a guide, and its images are shown on the coordinate plane. Use the coordinate grid to help you locate each point and its corresponding coordinates. Given Point: ࡿሺǡ ሻ ሺെǡ ሻ ሺǡ െሻ ሺെǡ െሻ
The given point is reflected across the ࢞
-axis.
ࡹሺǡ െሻ ሺെǡ െሻ ሺǡ ሻ ሺെǡ ሻ
The given point is reflected across the ࢟
-axis.
ࡸሺെǡ ሻ ሺǡ ሻ ሺെǡ െሻ ሺǡ െሻ
The given point is reflected first across the ࢞
-axis and then across the ࢟
-axis.
ሺെǡ െሻ
ሺǡ െሻ ሺെǡ ሻ ሺǡ ሻ
The given point is reflected first across the ࢟
-axis and then across the ࢞
-axis.
ሺെǡ െሻ
ሺǡ െሻ ሺെǡ ሻ ሺǡ ሻ
When the coordinates of two points are ࢟ǡ࢞ሺ ሻ and ሺെ ࢞ǡ࢟ ሻ , what line of symmetry do the points share? Explain.
They share the ࢟ -axis because the ࢟ -coordinates are the same and the ࢞ -coordinates are opposites, which means the points will be the same distance from the ࢟ -axis but on opposite sides.
When the coordinates of two points are ࢟ǡ࢞ሺ ሻ and ǡ࢞ሺ ሻ ࢟െ , what line of symmetry do the points share? Explain.
They share the ࢞ -axis because the ࢞ -coordinates are the same and the ࢟ -coordinates are opposites, which means the points will be the same distance from the ࢞ -axis but on opposite sides.
Example 2 (8 minutes): Navigating the Coordinate Plane Using Reflections
,ĂǀĞƐƚƵĚĞŶƚƐƵƐĞĂƉĞŶĐŝůĞƌĂƐĞƌŽƌĨŝŶŐĞƌƚŽŶĂǀŝŐĂƚĞƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŐŝǀĞŶǀĞƌďĂů ƉƌŽŵƉƚƐ͘dŚĞŶ͕ĐŝƌĐƵůĂƚĞƚŚĞƌŽŽŵĚƵƌŝŶŐƚŚĞĞdžĂŵƉůĞƚŽĂƐƐĞƐƐƐƚƵĚĞŶƚƐ͛ƵŶĚĞƌƐƚĂŶĚŝŶŐĂŶĚ ƉƌŽǀŝĚĞĂƐƐŝƐƚĂŶĐĞĂƐŶĞĞĚĞĚ͘
ĞŐŝŶĂƚ ሺǡ ʹሻǤ DŽǀĞ ͵ ƵŶŝƚƐĚŽǁŶ͕ĂŶĚƚŚĞŶƌĞĨůĞĐƚŽǀĞƌƚŚĞ ݕͲĂdžŝƐ͘tŚĞƌĞĂƌĞ LJŽƵ͍
à ሺെǡ െͳሻ
S
M
L
A࢞ ࢟
Scaffolding:
WƌŽũĞĐƚĞĂĐŚƉƌŽŵƉƚƐŽƚŚĂƚ ǀŝƐƵĂůůĞĂƌŶĞƌƐĐĂŶĨŽůůŽǁĂůŽŶŐ ǁŝƚŚƚŚĞƐƚĞƉƐ͘
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6ͻϯ Lesson 16 Lesson 16: ^LJŵŵĞƚƌLJŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
ĞŐŝŶĂƚ ሺͶǡ െͷሻ ͘ ZĞĨůĞĐƚŽǀĞƌƚŚĞ ݔͲĂdžŝƐ͕ĂŶĚƚŚĞŶŵŽǀĞ ƵŶŝƚƐĚŽǁŶĂŶĚƚŚĞŶƚŽƚŚĞƌŝŐŚƚ ʹƵŶŝƚƐ͘tŚĞƌĞ ĂƌĞLJŽƵ͍
à ሺǡ െʹሻ
ĞŐŝŶĂƚ ሺെ͵ǡ ͲሻǤ ZĞĨůĞĐƚŽǀĞƌƚŚĞ ݔͲĂdžŝƐ͕ĂŶĚƚŚĞŶŵŽǀĞ ƵŶŝƚƐƚŽƚŚĞƌŝŐŚƚ͘DŽǀĞƵƉƚǁŽƵŶŝƚƐ͕ĂŶĚƚŚĞŶ ƌĞĨůĞĐƚŽǀĞƌƚŚĞ ݔͲĂdžŝƐĂŐĂŝŶ͘tŚĞƌĞĂƌĞLJŽƵ͍
à ሺ͵ǡ െʹሻ
ĞŐŝŶĂƚ ሺെʹǡ ͺሻǤ ĞĐƌĞĂƐĞƚŚĞ ݕͲĐŽŽƌĚŝŶĂƚĞďLJ ƵŶŝƚƐ͘ZĞĨůĞĐƚŽǀĞƌƚŚĞ ݕͲĂdžŝƐ͕ĂŶĚƚŚĞŶŵŽǀĞĚŽǁŶ ͵ƵŶŝƚƐ͘ tŚĞƌĞĂƌĞLJŽƵ͍
à ሺʹǡ െͳሻ
ĞŐŝŶĂƚ ሺͷǡ െͳሻǤ ZĞĨůĞĐƚŽǀĞƌƚŚĞ ݔͲĂdžŝƐ͕ĂŶĚƚŚĞŶƌĞĨůĞĐƚŽǀĞƌƚŚĞ ݕͲĂdžŝƐ͘tŚĞƌĞĂƌĞLJŽƵ͍
à ሺെͷǡ ͳሻ
Examples 2– ϯ: Navigating the Coordinate Plane
Example ϯ (7 minutes): Describing How to Navigate the Coordinate Plane
'ŝǀĞŶĂƐƚĂƌƚŝŶŐƉŽŝŶƚĂŶĚĂŶĞŶĚŝŶŐƉŽŝŶƚ͕ƐƚƵĚĞŶƚƐĚĞƐĐƌŝďĞĂƐĞƋƵĞŶĐĞŽĨĚŝƌĞĐƚŝŽŶƐƵƐŝŶŐĂƚůĞĂƐƚŽŶĞƌĞĨůĞĐƚŝŽŶĂďŽƵƚ ĂŶĂdžŝƐƚŽŶĂǀŝŐĂƚĞĨƌŽŵƚŚĞƐƚĂƌƚŝŶŐƉŽŝŶƚƚŽƚŚĞĞŶĚŝŶŐƉŽŝŶƚ͘KŶĐĞƐƚƵĚĞŶƚƐŚĂǀĞĨŽƵŶĚĂƐĞƋƵĞŶĐĞ͕ŚĂǀĞƚŚĞŵĨŝŶĚ ĂŶŽƚŚĞƌƐĞƋƵĞŶĐĞǁŚŝůĞĐůĂƐƐŵĂƚĞƐĨŝŶŝƐŚƚŚĞƚĂƐŬ͘
ĞŐŝŶĂƚ ሺͻǡ െ͵ሻ ͕ ĂŶĚĞŶĚĂƚ ሺെͶǡ െ͵ሻ ͘ hƐĞĞdžĂĐƚůLJŽŶĞƌĞĨůĞĐƚŝŽŶ͘
à Possible answer: Reflect over the ݕ-axis, and then move ͷ units to the right.
ĞŐŝŶĂƚ ሺͲǡ Ͳሻ ͕ĂŶĚĞŶĚĂƚ ሺͷǡ െͳሻ ͘ hƐĞĞdžĂĐƚůLJŽŶĞƌĞĨůĞĐƚŝŽŶ͘
à Possible answer: Move ͷ units right, ͳ unit up, and then reflect over the ݔ-axis.
ĞŐŝŶĂƚ ሺͲǡ Ͳሻ ͕ĂŶĚĞŶĚĂƚ ሺെͳǡ െሻ ͘hƐĞĞdžĂĐƚůLJƚǁŽƌĞĨůĞĐƚŝŽŶƐ͘
à Possible answer: Move right ͳ unit, reflect over the ݕ-axis, up units, and then reflect over the ݔ -axis.
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6ͻϯ Lesson 16 Lesson 16: ^LJŵŵĞƚƌLJŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
Closing (4 minutes)
tŚĞŶƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨƚǁŽƉŽŝŶƚƐĚŝĨĨĞƌŽŶůLJďLJŽŶĞƐŝŐŶ͕ƐƵĐŚĂƐ ሺെͺǡ ʹሻ ĂŶĚ ሺͺǡ ʹሻ ͕ ǁŚĂƚĚŽƚŚĞƐŝŵŝůĂƌŝƚŝĞƐ ĂŶĚĚŝĨĨĞƌĞŶĐĞƐŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƐƚĞůůƵƐĂďŽƵƚƚŚĞŝƌƌĞůĂƚŝǀĞůŽĐĂƚŝŽŶƐŽŶƚŚĞƉůĂŶĞ͍
à The ݕ-coordinates are the same for both points, which means the points are on the same horizontal line. The ݔ-coordinates differ because they are opposites, which means the points are symmetric across the ݕ-axis.
tŚĂƚŝƐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶ ሺͷǡ ͳሻ ĂŶĚ ሺͷǡ െͳሻ ͍ 'ŝǀĞŶŽŶĞƉŽŝŶƚ͕ŚŽǁĐĂŶLJŽƵůŽĐĂƚĞƚŚĞŽƚŚĞƌ͍
à If you start at either point and reflect over the ݔ-axis, you will end at the other point.
Exit Ticket (4 minutes)
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6ͻϯ Lesson 16 Lesson 16: ^LJŵŵĞƚƌLJŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
EĂŵĞ ĂƚĞ
Lesson 16: Symmetry in the Coordinate Plane Exit Ticket
ϭ͘ ,ŽǁĂƌĞƚŚĞŽƌĚĞƌĞĚƉĂŝƌƐ ሺͶǡ ͻሻ ĂŶĚ ሺͶǡ െͻሻ ƐŝŵŝůĂƌ͕ĂŶĚŚŽǁĂƌĞƚŚĞLJĚŝĨĨĞƌĞŶƚ͍ƌĞƚŚĞƚǁŽƉŽŝŶƚƐƌĞůĂƚĞĚďLJĂ ƌĞĨůĞĐƚŝŽŶŽǀĞƌĂŶĂdžŝƐŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͍/ĨƐŽ͕ŝŶĚŝĐĂƚĞǁŚŝĐŚĂdžŝƐŝƐƚŚĞůŝŶĞŽĨƐLJŵŵĞƚƌLJďĞƚǁĞĞŶƚŚĞƉŽŝŶƚƐ͘ /ĨƚŚĞLJĂƌĞŶŽƚƌĞůĂƚĞĚďLJĂƌĞĨůĞĐƚŝŽŶŽǀĞƌĂŶĂdžŝƐŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͕ĞdžƉůĂŝŶŚŽǁLJŽƵŬŶŽǁ͘ Ϯ͘ 'ŝǀĞŶƚŚĞƉŽŝŶƚ ሺെͷǡ ʹሻ ͕ ǁƌŝƚĞƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨĂƉŽŝŶƚƚŚĂƚŝƐƌĞůĂƚĞĚďLJĂƌĞĨůĞĐƚŝŽŶŽǀĞƌƚŚĞ ݔͲŽƌ ݕͲĂdžŝƐ͘^ƉĞĐŝĨLJ ǁŚŝĐŚĂdžŝƐŝƐƚŚĞůŝŶĞŽĨƐLJŵŵĞƚƌLJ͘
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6ͻϯ Lesson 16 Lesson 16: ^LJŵŵĞƚƌLJŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
ሺǡ െሻ ሺെǡ െሻ
ሺെǡ ሻ ࡰሺǡ ሻ
Exit Ticket Sample Solutions 1. How are the ordered pairs ሺǡૢ ሻ and ሺǡ െૢ ሻ similar, and how are they different? Are the two points related by a reflection over an axis in the coordinate plane? If so, indicate which axis is the line of symmetry between the points. If they are not related by a reflection over an axis in the coordinate plane, explain how you know.
The ࢞ -coordinates are the same, but the ࢟ -coordinates are opposites, meaning they are the same distance from zero on the ࢞ -axis and the same distance but on opposite sides of zero on the ࢟ -axis. Reflecting about the ࢞ -axis interchanges these two points.
Given the point ሺെǡ ሻ , write the coordinates of a point that is related by a reflection over the ࢞ - or ࢟ -axis. Specify which axis is the line of symmetry.
Using the ࢞ -axis as a line of symmetry, ሺെǡ െሻ ; using the ࢟ -axis as a line of symmetry, ሺǡ ሻ
Problem Set Sample Solutions 1. Locate a point in Quadrant IV of the coordinate plane. Label the point , and write its ordered pair next to it.
Answers will vary; Quadrant IV ሺǡ െሻ
a. Reflect point over an axis so that its image is in Quadrant III. Label the image , and write its ordered pair next to it. Which axis did you reflect over? What is the only difference in the ordered pairs of points and ?
ሺെǡ െሻ ; reflected over the ࢟ -axis The ordered pairs differ only by the sign of their ࢞
-coordinates: ሺǡ െሻ and ሺെǡ െሻ .
b. Reflect point over an axis so that its image is in Quadrant II. Label the image , and write its ordered pair next to it. Which axis did you reflect over? What is the only difference in the ordered pairs of points and ? How does the ordered pair of point relate to the ordered pair of point ?
ሺെǡ ሻ ; reflected over the ࢞ -axis The ordered pairs differ only by the signs of their ࢟ -coordinates: ሺെǡ െሻ and ሺെǡ ሻ .The ordered pair for point differs from the ordered pair for point by the signs of both coordinates:
ሺǡ െሻ and ሺെǡ ሻ .
c. Reflect point over an axis so that its image is in Quadrant I. Label the image ࡰ, and write its ordered pair next to it. Which axis did you reflect over? How does the ordered pair for point ࡰ compare to the ordered pair for point ? How does the ordered pair for point ࡰ compare to points and ?
ࡰሺǡ ሻ ; reflected over the ࢟ -axis again Point ࡰ differs from point by only the sign of its ࢞ -coordinate: ࡰሺǡ ሻ and ሺെǡ ሻ .Point ࡰ differs from point by the signs of both coordinates: ࡰሺǡ ሻ and ሺെǡ െሻ .Point ࡰ differs from point by only the sign of the ࢟ -coordinate: ࡰሺǡ ሻ and ሺǡ െሻ .
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6ͻϯ Lesson 16 Lesson 16: ^LJŵŵĞƚƌLJŝŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
Bobbie listened to her teacher’s directions and navigated from the point ሺെǡ ሻ to ሺǡ െሻ . She knows that she has the correct answer, but she forgot part of the teacher’s directions. Her teacher’s directions included the following: “Move ૠ units down, reflect about the ? -axis, move up units, and then move right units.” Help Bobbie determine the missing axis in the directions, and explain your answer.
The missing line is a reflection over the ࢟ -axis. The first line would move the location to ሺെǡ െૠሻ . A reflection over the ࢟ -axis would move the location to ሺǡ െૠሻ in Quadrant IV, which is units left and units down from the end point ሺǡ െሻ .
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6ͻϯ Lesson 17 Lesson 17: ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ
Lesson 17: Drawing the Coordinate Plane and Points on the Plane Student Outcomes
^ƚƵĚĞŶƚƐĚƌĂǁĂĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŽŶŐƌĂƉŚƉĂƉĞƌŝŶƚǁŽƐƚĞƉƐ͗;ϭͿƌĂǁĂŶĚůĂďĞůƚŚĞŚŽƌŝnjŽŶƚĂůĂŶĚǀĞƌƚŝĐĂů ĂdžĞƐ͖;ϮͿDĂƌŬƚŚĞŶƵŵďĞƌƐĐĂůĞŽŶĞĂĐŚĂdžŝƐ͘
'ŝǀĞŶƐŽŵĞƉŽŝŶƚƐĂƐŽƌĚĞƌĞĚƉĂŝƌƐ͕ƐƚƵĚĞŶƚƐŵĂŬĞƌĞĂƐŽŶĂďůĞĐŚŽŝĐĞƐĨŽƌƐĐĂůĞƐŽŶďŽƚŚĂdžĞƐĂŶĚůŽĐĂƚĞĂŶĚ ůĂďĞůƚŚĞƉŽŝŶƚƐŽŶŐƌĂƉŚƉĂƉĞƌ͘
Classwork Opening Exercise (5 minutes)
/ŶƐƚƌƵĐƚƐƚƵĚĞŶƚƐƚŽĚƌĂǁĂůůŶĞĐĞƐƐĂƌLJĐŽŵƉŽŶĞŶƚƐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŽŶƚŚĞďůĂŶŬ ʹͲ ൈ ʹͲ ŐƌŝĚƉƌŽǀŝĚĞĚďĞůŽǁ͕ ƉůĂĐŝŶŐƚŚĞŽƌŝŐŝŶĂƚƚŚĞĐĞŶƚĞƌŽĨƚŚĞŐƌŝĚĂŶĚůĞƚƚŝŶŐĞĂĐŚŐƌŝĚůŝŶĞƌĞƉƌĞƐĞŶƚ ͳƵŶŝƚ͘KďƐĞƌǀĞƐƚƵĚĞŶƚƐĂƐƚŚĞLJĐŽŵƉůĞƚĞ ƚŚĞƚĂƐŬ͕ƵƐŝŶŐƚŚĞŝƌƉƌŝŽƌĞdžƉĞƌŝĞŶĐĞǁŝƚŚƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘
Opening Exercise Draw all necessary components of the coordinate plane on the blank ൈ grid provided below, placing the origin at the center of the grid and letting each grid line represent unit.
࢞࢟
െ െ
െ
െ
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6ͻϯ Lesson 17 Lesson 17: ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ
^ƚƵĚĞŶƚƐĂŶĚƚŚĞƚĞĂĐŚĞƌƚŽŐĞƚŚĞƌĚŝƐĐƵƐƐƚŚĞŶĞĞĚĨŽƌĞǀĞƌLJĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽŚĂǀĞƚŚĞĨŽůůŽǁŝŶŐ͗
x dŚĞ ݔ ͲĂŶĚ ݕͲĂdžĞƐĚƌĂǁŶƵƐŝŶŐĂƐƚƌĂŝŐŚƚĞĚŐĞ͕
x dŚĞŚŽƌŝnjŽŶƚĂůĂdžŝƐůĂďĞůĞĚ ݔ͕
x dŚĞǀĞƌƚŝĐĂůĂdžŝƐůĂďĞůĞĚ ݕ͕
x ĂĐŚĂdžŝƐůĂďĞůĞĚƵƐŝŶŐĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞĂƐĚŝĐƚĂƚĞĚďLJƚŚĞƉƌŽďůĞŵŽƌƐĞƚŽĨŽƌĚĞƌĞĚƉĂŝƌƐƚŽďĞŐƌĂƉŚĞĚ͘ ^ƚƵĚĞŶƚƐƐŚŽƵůĚĞƌĂƐĞĞƌƌŽƌƐĂŶĚŵĂŬĞĂŶLJŶĞĐĞƐƐĂƌLJĐŚĂŶŐĞƐďĞĨŽƌĞƉƌŽĐĞĞĚŝŶŐƚŽ džĂŵƉůĞϭ͘
Example 1 (8 minutes): Drawing the Coordinate Plane Using a ǣ Scale
/ƐƚŚĞƐŝnjĞŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞŐƌŝĚƚŚĂƚǁĞĚŝƐĐƵƐƐĞĚŝŶƚŚĞKƉĞŶŝŶŐ džĞƌĐŝƐĞƐƵĨĨŝĐŝĞŶƚƚŽŐƌĂƉŚƚŚĞƉŽŝŶƚƐŐŝǀĞŶ ŝŶƚŚĞƐĞƚŝŶ džĂŵƉůĞϭ͍
à Yes. All ݔ- and ݕ-coordinates are between െͳͲ and ͳͲ , and both axes on the grid range from
െͳͲ to ͳͲ .
Example 1: Drawing the Coordinate Plane Using a ǣ Scale Locate and label the points ሼሺǡ ሻǡ ሺૡǡ ሻǡ ሺെǡ ૡሻǡ ሺെǡ െૢ ሻǡ ሺǡ ሻǡ ሺെǡ െሻǡ ሺǡ െሻሽ on the grid below.
ĂŶLJŽƵŶĂŵĞĂƉŽŝŶƚƚŚĂƚĐŽƵůĚŶŽƚďĞůŽĐĂƚĞĚŽŶƚŚŝƐŐƌŝĚ͍ džƉůĂŝŶ͘
à The point ሺͳͺǡ ͷሻ could not be located on this grid because ͳͺ is greater than ͳͲ and, therefore, to the right of ͳͲ on the ݔ-axis. ͳͲ is the greatest number shown on this grid.
ŝƐĐƵƐƐǁĂLJƐŝŶǁŚŝĐŚƚŚĞƉŽŝŶƚ ሺͳͺǡ ͷሻ ĐŽƵůĚďĞŐƌĂƉŚĞĚǁŝƚŚŽƵƚĐŚĂŶŐŝŶŐƚŚĞƐŝnjĞŽĨƚŚĞŐƌŝĚ͘
à Changing the number of units that each grid line represents would allow us to fit greater numbers on the axes. Changing the number of units per grid line to ʹ units would allow a range of െʹͲ to ʹͲ on the ݔ-axis. ࢞ ࢟
െ െ
െ
െ
ሺെǡ ૡሻ
ሺǡ ሻ
ሺૡǡ ሻ
ሺെǡ െሻ
ሺǡ െሻ
ሺെǡ െૢ ሻ
ሺǡ ሻ
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6ͻϯ Lesson 17 Lesson 17: ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ
Example 2 (8 minutes): Drawing the Coordinate Plane Using an Increased Number Scale for One Axis
^ƚƵĚĞŶƚƐŝŶĐƌĞĂƐĞƚŚĞŶƵŵďĞƌŽĨƵŶŝƚƐƌĞƉƌĞƐĞŶƚĞĚďLJĞĂĐŚŐƌŝĚůŝŶĞŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŝŶŽƌĚĞƌƚŽŐƌĂƉŚĂŐŝǀĞŶƐĞƚ ŽĨŽƌĚĞƌĞĚƉĂŝƌƐ͘
džĂŵŝŶĞƚŚĞŐŝǀĞŶƉŽŝŶƚƐ͘tŚĂƚŝƐƚŚĞƌĂŶŐĞŽĨǀĂůƵĞƐƵƐĞĚĂƐ ݔͲĐŽŽƌĚŝŶĂƚĞƐ͍,ŽǁŵĂŶLJƵŶŝƚƐƐŚŽƵůĚǁĞ ĂƐƐŝŐŶƉĞƌŐƌŝĚůŝŶĞƚŽƐŚŽǁƚŚŝƐƌĂŶŐĞŽĨǀĂůƵĞƐ͍ džƉůĂŝŶ͘
à The ݔ -coordinates range from െͶ to ͻ, all within the range of െͳͲ to ͳͲ , so we will assign each grid line to represent ͳ unit.
tŚĂƚŝƐƚŚĞƌĂŶŐĞŽĨǀĂůƵĞƐƵƐĞĚĂƐ ݕͲĐŽŽƌĚŝŶĂƚĞƐ͍,ŽǁŵĂŶLJƵŶŝƚƐƐŚŽƵůĚǁĞĂƐƐŝŐŶƉĞƌŐƌŝĚůŝŶĞƚŽƐŚŽǁƚŚŝƐ ƌĂŶŐĞŽĨǀĂůƵĞƐ͍ džƉůĂŝŶ͘
à The ݕ-coordinates range from െͶͲ to ͵ͷ . If we let each grid line represent ͷ units, then the ݕ-axis will include the range െͷͲ to ͷͲ .
ƌĂǁĂŶĚůĂďĞůƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͕ĂŶĚƚŚĞŶůŽĐĂƚĞĂŶĚůĂďĞůƚŚĞƐĞƚŽĨƉŽŝŶƚƐ͘
Example 2: Drawing the Coordinate Plane Using an Increased Number Scale for One Axis Draw a coordinate plane on the grid below, and then locate and label the following points:
ሼሺെǡ ሻǡ ሺെǡ ሻǡ ሺǡ െሻǡ ሺǡ ሻǡ ሺૢǡ െሻሽ .
࢞࢟
െ െ
െ െ
ሺǡ ሻ
ሺെǡ ሻ
ሺെǡ ሻ
ሺǡ െሻ ሺૢ ǡ െሻ
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6ͻϯ Lesson 17 Lesson 17: ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ
Example ϯ (8 minutes): Drawing the Coordinate Plane Using a Decreased Number Scale for One Axis
^ƚƵĚĞŶƚƐĚŝǀŝĚĞƵŶŝƚƐĂŵŽŶŐŵƵůƚŝƉůĞŐƌŝĚůŝŶĞƐŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŝŶŽƌĚĞƌƚŽŐƌĂƉŚĂŐŝǀĞŶƐĞƚŽĨŽƌĚĞƌĞĚƉĂŝƌƐ͘
džĂŵŝŶĞƚŚĞŐŝǀĞŶƉŽŝŶƚƐ͘tŝůůĞŝƚŚĞƌƚŚĞ ݔͲŽƌ ݕͲĐŽŽƌĚŝŶĂƚĞƐƌĞƋƵŝƌĞĂĐŚĂŶŐĞŽĨƐĐĂůĞŝŶƚŚĞƉůĂŶĞ͍ džƉůĂŝŶ͘
à The ݔ-coordinates range from െͲǤ to ͲǤͺ . This means that if each grid line represents one unit, the points would all be very close to the ݕ-axis, making it difficult to interpret.
,ŽǁĐŽƵůĚǁĞĐŚĂŶŐĞƚŚĞŶƵŵďĞƌŽĨƵŶŝƚƐƌĞƉƌĞƐĞŶƚĞĚƉĞƌŐƌŝĚůŝŶĞƚŽďĞƚƚĞƌƐŚŽǁƚŚĞƉŽŝŶƚƐŝŶƚŚĞŐŝǀĞŶƐĞƚ͍
à Divide ͳ unit into tenths so that each grid line represents a tenth of a unit, and the ݔ-axis then ranges from െͳ to ͳ.
ƌĂǁĂŶĚůĂďĞůƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͕ĂŶĚƚŚĞŶůŽĐĂƚĞĂŶĚůĂďĞůƚŚĞƐĞƚŽĨƉŽŝŶƚƐ͘
džĂŵƉůĞϯ͗ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞhƐŝŶŐĂ Decreased Number Scale for One Axis Draw a coordinate plane on the grid below, and then locate and label the following points:
ሼሺǤ ǡ ሻǡ ሺǤ ǡ ૠሻǡ ሺെǤ ૠǡ െሻǡ ሺെǤ ǡ ሻǡ ሺǤ ૡǡ ሻሽ .
࢞࢟
െǤ െǤ Ǥ Ǥ
െ െ
ሺǤ ǡ ሻ
ሺǤ ǡ ૠሻ
ሺെǤ ǡ ሻ
ሺെǤ ૠǡ െሻ
ሺǤ ૡǡ ሻ
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6ͻϯ Lesson 17 Lesson 17: ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ
Example 4 (8 minutes): Drawing the Coordinate Plane Using a Different Number Scale for Both Axes
^ƚƵĚĞŶƚƐĂƉƉƌŽƉƌŝĂƚĞůLJƐĐĂůĞƚŚĞĂdžĞƐŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞŝŶŽƌĚĞƌƚŽŐƌĂƉŚĂŐŝǀĞŶƐĞƚŽĨŽƌĚĞƌĞĚƉĂŝƌƐ͘EŽƚĞƚŚĂƚ ƚŚĞƉƌŽǀŝĚĞĚŐƌŝĚŝƐ ͳ ൈ ͳ ǁŝƚŚĨĞǁĞƌŐƌŝĚůŝŶĞƐƚŚĂŶƚŚĞƉƌĞǀŝŽƵƐĞdžĂŵƉůĞƐ͘
Example 4: Drawing the Coordinate Plane Using a Different Number Scale for Both Axes Determine a scale for the ࢞ -axis that will allow all ࢞ -coordinates to be shown on your grid.
The grid is units wide, and the ࢞ -coordinates range from െ to . If I let each grid line represent units, then the ࢞
-axis will range from െ to .
Determine a scale for the ࢟ -axis that will allow all ࢟ -coordinates to be shown on your grid.
The grid is units high, and the ࢟ -coordinates range from െ to Ǥ . I could let each grid line represent one unit, but if I let each grid line represent of a unit, the points will be easier to graph.
Draw and label the coordinate plane, and then locate and label the set of points.
ሼሺെǡ ሻǡ ሺെǡ െǤ ሻǡ ሺǡ െǤ ሻǡ ሺǡ Ǥ ሻǡ ሺǡ Ǥ ሻǡ ሺെૡǡ െሻሽ
,ŽǁǁĂƐƚŚŝƐĞdžĂŵƉůĞĚŝĨĨĞƌĞŶƚĨƌŽŵƚŚĞĨŝƌƐƚƚŚƌĞĞĞdžĂŵƉůĞƐŝŶƚŚŝƐůĞƐƐŽŶ͍
à The given set of points caused me to change the scales on both axes, and the given grid had fewer grid lines.
ŝĚƚŚĞƐĞĚŝĨĨĞƌĞŶĐĞƐĂĨĨĞĐƚLJŽƵƌĚĞĐŝƐŝŽŶŵĂŬŝŶŐĂƐLJŽƵĐƌĞĂƚĞĚƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͍ džƉůĂŝŶ͘
à Shrinking the scale of the ݔ-axis allowed me to show a larger range of numbers, but fewer grid lines limited that range. ࢞࢟
െ െ െૡ െ ૡ
െ െ െ െ
ሺǡ Ǥ ሻ
ሺǡ െǤ ሻ
ሺെǡ ሻ
ሺǡ Ǥ ሻ
ሺെૡǡ െሻ
ሺെǡ െǤ ሻ
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6ͻϯ Lesson 17 Lesson 17: ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ
Closing (4 minutes)
tŚLJŝƐŝƚŝŵƉŽƌƚĂŶƚƚŽůĂďĞůƚŚĞĂdžĞƐǁŚĞŶƐĞƚƚŝŶŐƵƉĂĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͍
à It is important to label the axes when setting up a coordinate plane so that the person viewing the graph knows which axis represents which coordinate and what scale is being used. If a person does not know the scale being used, she will likely misinterpret the graph.
tŚLJƐŚŽƵůĚŶ͛ƚLJŽƵĚƌĂǁĂŶĚůĂďĞůƚŚĞĞŶƚŝƌĞĐŽŽƌĚŝŶĂƚĞŐƌŝĚďĞĨŽƌĞůŽŽŬŝŶŐĂƚƚŚĞƉŽŝŶƚƐƚŽďĞŐƌĂƉŚĞĚ͍
à Looking at the range of values in a given set of points allows you to determine an appropriate scale. If you set a scale before observing the given values, you will likely have to change the scale on your axes.
Exit Ticket (4 minutes)
Lesson Summary
The axes of the coordinate plane must be drawn using a straightedge and labeled ࢞ (horizontal axis) and ࢟ (vertical axis).
Before assigning a scale to the axes, it is important to assess the range of values found in a set of points as well as the number of grid lines available. This allows you to determine an appropriate scale so all points can be represented on the coordinate plane that you construct.
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6ͻϯ Lesson 17 Lesson 17: ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ
EĂŵĞ ĂƚĞ
Lesson 17: Drawing the Coordinate Plane and Points on the Plane Exit Ticket
ĞƚĞƌŵŝŶĞĂŶĂƉƉƌŽƉƌŝĂƚĞƐĐĂůĞĨŽƌƚŚĞƐĞƚŽĨƉŽŝŶƚƐŐŝǀĞŶďĞůŽǁ͘ƌĂǁĂŶĚůĂďĞůƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͕ĂŶĚƚŚĞŶůŽĐĂƚĞ ĂŶĚůĂďĞůƚŚĞƐĞƚŽĨƉŽŝŶƚƐ͘
ሼሺͳͲǡ ͲǤʹሻǡ ሺെʹͷǡ ͲǤͺሻǡ ሺͲǡ െͲǤͶሻǡ ሺʹͲǡ ͳሻǡ ሺെͷǡ െͲǤͺሻሽ
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6ͻϯ Lesson 17 Lesson 17: ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ
Exit Ticket Sample Solutions Determine an appropriate scale for the set of points given below. Draw and label the coordinate plane, and then locate and label the set of points.
ሼሺǡ Ǥ ሻǡ ሺെǡ Ǥ ૡሻǡ ሺǡ െǤ ሻǡ ሺǡ ሻǡ ሺെǡ െǤ ૡሻሽ
The ࢞ -coordinates range from െ to . The grid is units wide. If I let each grid line represent units, then the ࢞
-axis will range from െ to .The ࢟ -coordinates range from െǤ ૡ to . The grid is units high. If I let each grid line represent two-tenths of a unit, then the ࢟ -axis will range from െ to .
Problem Set Sample Solutions 1. Label the coordinate plane, and then locate and label the set of points below.
൜ሺǤ ǡ Ǥૢ ሻǡ ሺെǤ ǡ Ǥ ૠሻǡ ሺെǤ ǡ െǤ ሻǡ ሺെǤૢ ǡ Ǥ ሻǡ ሺǡ െǤ ሻ ൠ
െ െ
Ǥ ૡ
Ǥ
െǤ െǤ ૡ
ሺǡ Ǥ ሻ ࢞࢟
ሺെǡ Ǥ ૡሻ
ሺǡ െǤ ሻ
ሺǡ ሻ
ሺെǡ െǤ ૡሻ ࢞ ࢟
െǤ െǤ Ǥ Ǥ
Ǥ Ǥ
െǤ െǤ
ሺǡ െǤ ሻ
ሺെǤ ǡ െǤ ሻ
ሺെǤૢ ǡ Ǥ ሻ
ሺെǤ ǡ Ǥ ૠሻ
ሺǤ ǡ Ǥૢ ሻ
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6ͻϯ Lesson 17 Lesson 17: ƌĂǁŝŶŐƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞĂŶĚWŽŝŶƚƐŽŶƚŚĞWůĂŶĞ
Label the coordinate plane, and then locate and label the set of points below.
൜ሺૢ ǡૢ ሻǡ ሺെǡ െሻǡ ሺǡ ሻǡ ሺെǡ െሻǡ ሺെૡǡ െૡሻ ൠ
Extension:
ϯ͘ Describe the pattern you see in the coordinates in Problem 2 and the pattern you see in the points. Are these patterns consistent for other points too?
The ࢞ -coordinate for each of the given points is times its ࢟ -coordinate. When I graphed the points, they appear to make a straight line. I checked other ordered pairs with the same pattern, such as ሺെǡ െሻ , ሺǡ ሻ , and even ሺǡ ሻ , and it appears that these points are also on that line. ࢞ ࢟
െ െ
െ െ
ሺെૡǡ െૡሻ
ሺെǡ െሻ
ሺǡ ሻ
ሺૢ ǡૢ ሻ
ሺെǡ െሻ
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6ͻϯ Lesson 18 Lesson 18: ŝƐƚĂŶĐĞŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
Albertsville ՚ 8 mi. Blossville ՛ 3 mi. Cheyenne ՛ 12 mi. Dewey Falls ՜ 6 mi.
Lesson 18: Distance on the Coordinate Plane Student Outcomes
^ƚƵĚĞŶƚƐĐŽŵƉƵƚĞƚŚĞůĞŶŐƚŚŽĨŚŽƌŝnjŽŶƚĂůĂŶĚǀĞƌƚŝĐĂůůŝŶĞƐĞŐŵĞŶƚƐǁŝƚŚŝŶƚĞŐĞƌĐŽŽƌĚŝŶĂƚĞƐĨŽƌĞŶĚƉŽŝŶƚƐ ŝŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞďLJĐŽƵŶƚŝŶŐƚŚĞŶƵŵďĞƌŽĨƵŶŝƚƐďĞƚǁĞĞŶĞŶĚƉŽŝŶƚƐĂŶĚƵƐŝŶŐĂďƐŽůƵƚĞǀĂůƵĞ͘
Classwork Opening Exercise (5 minutes) Opening Exercise Four friends are touring on motorcycles. They come to an intersection of two roads; the road they are on continues straight, and the other is perpendicular to it. The sign at the intersection shows the distances to several towns. Draw a map/diagram of the roads, and use it and the information on the sign to answer the following questions: What is the distance between Albertsville and Dewey Falls?
Students draw and use their maps to answer. Albertsville is ૡ miles to the left, and Dewey Falls is miles to the right. Since the towns are in opposite directions from the intersection, their distances must be combined by addition, ૡ ൌ , so the distance between Albertsville and Dewey Falls is miles.
What is the distance between Blossville and Cheyenne?
Blossville and Cheyenne are both straight ahead from the intersection in the direction that they are going. Since they are on the same side of the intersection, Blossville is on the way to Cheyenne, so the distance to Cheyenne includes the miles to Blossville. To find the distance from Blossville to Cheyenne, I have to subtract; െ ൌૢ . So, the distance from Blossville to Cheyenne is ૢ miles.
On the coordinate plane, what represents the intersection of the two roads?
The intersection is represented by the origin.
Example 1 (6 minutes): The Distance Between Points on an Axis
^ƚƵĚĞŶƚƐĨŝŶĚƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƉŽŝŶƚƐŽŶƚŚĞ ݔͲĂdžŝƐďLJĨŝŶĚŝŶŐƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶŶƵŵďĞƌƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ dŚĞLJĨŝŶĚƚŚĞĂďƐŽůƵƚĞǀĂůƵĞƐŽĨƚŚĞ ݔͲĐŽŽƌĚŝŶĂƚĞƐĂŶĚĂĚĚŽƌƐƵďƚƌĂĐƚƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐƚŽĚĞƚĞƌŵŝŶĞƚŚĞĚŝƐƚĂŶĐĞ ďĞƚǁĞĞŶƚŚĞƉŽŝŶƚƐ͘
Example 1: The Distance Between Points on an Axis Consider the points ሺെǡ ሻ and ሺǡ ሻ .What do the ordered pairs have in common, and what does that mean about their location in the coordinate plane?
Both of their ࢟ -coordinates are zero, so each point lies on the ࢞ -axis, the horizontal number line.
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6ͻϯ Lesson 18 Lesson 18: ŝƐƚĂŶĐĞŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
Scaffolding:
^ƚƵĚĞŶƚƐŵĂLJŶĞĞĚƚŽĚƌĂǁĂŶ ĂƵdžŝůŝĂƌLJůŝŶĞƚŚƌŽƵŐŚƚŚĞĞŶĚ ƉŽŝŶƚƐƚŽŚĞůƉǀŝƐƵĂůŝnjĞĂ ŚŽƌŝnjŽŶƚĂůŽƌǀĞƌƚŝĐĂůŶƵŵďĞƌ ůŝŶĞ͘
How did we find the distance between two numbers on the number line?
We calculated the absolute values of the numbers, which told us how far the numbers were from zero. If the numbers were located on opposite sides of zero, then we added their absolute values together. If the numbers were located on the same side of zero, then we subtracted their absolute values.
Use the same method to find the distance between ሺെǡ ሻ and ሺǡ ሻ .
ȁെȁ ൌ and ȁȁ ൌ . The numbers are on opposite sides of zero, so the absolute values get combined: ൌૢ . The distance between ሺെǡ ሻ and ሺǡ ሻ is ૢ units.
Example 2 (5 minutes): The Length of a Line Segment on an Axis
^ƚƵĚĞŶƚƐĨŝŶĚƚŚĞůĞŶŐƚŚŽĨĂůŝŶĞƐĞŐŵĞŶƚƚŚĂƚůŝĞƐŽŶƚŚĞ ݕͲĂdžŝƐďLJĨŝŶĚŝŶŐƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶŝƚƐĞŶĚƉŽŝŶƚƐ͘
Example 2: The Length of a Line Segment on an Axis Consider the line segment with end points ሺǡ െሻ and ሺǡ െሻ .What do the ordered pairs of the end points have in common, and what does that mean about the line segment’s location in the coordinate plane?
The ࢞ -coordinates of both end points are zero, so the points lie on the ࢟ -axis, the vertical number line. If its end points lie on a vertical number line, then the line segment itself must also lie on the vertical line.
Find the length of the line segment described by finding the distance between its end points ሺǡ െሻ and ሺǡ െሻ .
ȁെȁ ൌ and ȁെȁ ൌ . The numbers are on the same side of zero, which means the longer distance contains the shorter distance, so the absolute values need to be subtracted: െ ൌ . The distance between ሺǡ െሻ and ሺǡ െሻ
is units, so the length of the line segment with end points ሺǡ െሻ and ሺǡ െሻ is units.
Example ϯ (10 minutes): Length of a Horizontal or Vertical Line Segment That Does Not Lie on an Axis
^ƚƵĚĞŶƚƐĨŝŶĚƚŚĞůĞŶŐƚŚŽĨĂǀĞƌƚŝĐĂůůŝŶĞƐĞŐŵĞŶƚƚŚĂƚĚŽĞƐŶŽƚůŝĞŽŶƚŚĞ ݕͲĂdžŝƐďLJĨŝŶĚŝŶŐ ƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶŝƚƐĞŶĚƉŽŝŶƚƐ͘
džĂŵƉůĞϯ͗>ĞŶŐƚŚŽĨĂ,ŽƌŝnjŽŶƚĂůŽƌsĞƌƚŝĐĂů>ŝŶ e Segment That Does Not Lie on an Axis Consider the line segment with end points ሺെǡ ሻ and ሺെǡ െሻ .What do the end points, which are represented by the ordered pairs, have in common? What does that tell us about the location of the line segment on the coordinate plane?
Both end points have ࢞ -coordinates of െ , so the points lie on the vertical line that intersects the ࢞ -axis at െ . This means that the end points of the line segment, and thus the line segment, lie on a vertical line.
Find the length of the line segment by finding the distance between its end points.
The end points are on the same vertical line, so we only need to find the distance between and െ on the number line.
ȁȁ ൌ and ȁെȁ ൌ , and the numbers are on opposite sides of zero, so the values must be added: ൌ ૡ . So, the distance between ሺെǡ ሻ and ሺെǡ െሻ is ૡ units.
Exercise (10 minutes)
^ƚƵĚĞŶƚƐĐĂůĐƵůĂƚĞƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƉĂŝƌƐŽĨƉŽŝŶƚƐƵƐŝŶŐĂďƐŽůƵƚĞǀĂůƵĞƐ͘
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6ͻϯ Lesson 18 Lesson 18: ŝƐƚĂŶĐĞŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
Exercise Find the lengths of the line segments whose end points are given below. Explain how you determined that the line segments are horizontal or vertical. a. ሺെǡ ሻ and ሺെǡૢ ሻ
Both end points have ࢞ -coordinates of െ , so the points lie on a vertical line that passes through െ on the ࢞
-axis. ȁȁ ൌ and ȁૢ ȁ ൌૢ , and the numbers are on the same side of zero. By subtraction, ૢ െ ൌ , so the length of the line segment with end points ሺെǡ ሻ and ሺെǡૢ ሻ is units.
b. ሺǡ െሻ and ሺെૡǡ െሻ
Both end points have ࢟ -coordinates of െ , so the points lie on a horizontal line that passes through െ on the ࢟
-axis. ȁȁ ൌ and ȁെૡȁ ൌ ૡ , and the numbers are on opposite sides of zero, so the absolute values must be added. By addition, ૡ ൌ , so the length of the line segment with end points ሺǡ െሻ and ሺെૡǡ െሻ is
units.
c. ሺെǡ െሻ and ሺെǡ ሻ
Both end points have ࢞ -coordinates of െ , so the points lie on a vertical line. ȁെȁ ൌ and ȁȁ ൌ , and the numbers are on opposite sides of zero, so the absolute values must be added. By addition, ൌ ૠ , so the length of the line segment with end points ሺെǡ െሻ and ሺെǡ ሻ is ૠ units.
d. ሺെૢ ǡ ሻ and ሺെǡ ሻ
Both end points have ࢟ -coordinates of , so the points lie on a horizontal line. ȁെૢ ȁ ൌૢ and ȁെȁ ൌ , and the numbers are on the same side of zero. By subtraction, ૢ െ ൌ , so the length of the line segment with end points ሺെૢ ǡ ሻ and ሺെǡ ሻ is units.
e. ሺǡ െሻ and ሺǡ ૡሻ
Both end points have ࢞ -coordinates of , so the points lie on the ࢟ -axis. ȁെȁ ൌ and ȁૡȁ ൌ ૡ , and the numbers are on opposite sides of zero, so their absolute values must be added. By addition, ૡ ൌ ૢ , so the length of the line segment with end points ሺǡ െሻ and ሺǡ ૡሻ is ૢ units.
Closing ;ϯ minutes)
tŚLJŝƐŝƚƉŽƐƐŝďůĞĨŽƌƵƐƚŽĨŝŶĚƚŚĞůĞŶŐƚŚŽĨĂŚŽƌŝnjŽŶƚĂůŽƌǀĞƌƚŝĐĂůůŝŶĞƐĞŐŵĞŶƚĞǀĞŶŝĨŝƚ͛ƐŶŽƚŽŶƚŚĞ ݔͲŽƌ
ݕͲĂdžŝƐ͍
à A line can still be a horizontal or vertical line even if it is not on the ݔ- or ݕ-axis; therefore, we can still use the same strategy.
ĂŶLJŽƵƚŚŝŶŬŽĨĂƌĞĂůͲǁŽƌůĚƐŝƚƵĂƚŝŽŶǁŚĞƌĞƚŚŝƐŵŝŐŚƚďĞƵƐĞĨƵů͍
à Finding the distance on a map
Exit Ticket (6 minutes) Lesson Summary To find the distance between points that lie on the same horizontal line or on the same vertical line, we can use the same strategy that we used to find the distance between points on the number line.
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6ͻϯ Lesson 18 Lesson 18: ŝƐƚĂŶĐĞŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
EĂŵĞ ĂƚĞ
Lesson 18: Distance on the Coordinate Plane Exit Ticket
ĞƚĞƌŵŝŶĞǁŚĞƚŚĞƌĞĂĐŚŐŝǀĞŶƉĂŝƌŽĨĞŶĚƉŽŝŶƚƐůŝĞƐŽŶƚŚĞƐĂŵĞŚŽƌŝnjŽŶƚĂůŽƌǀĞƌƚŝĐĂůůŝŶĞ͘/ĨƐŽ͕ĨŝŶĚƚŚĞůĞŶŐƚŚŽĨƚŚĞ ůŝŶĞƐĞŐŵĞŶƚƚŚĂƚũŽŝŶƐƚŚĞƉĂŝƌŽĨƉŽŝŶƚƐ͘/ĨŶŽƚ͕ĞdžƉůĂŝŶŚŽǁLJŽƵŬŶŽǁƚŚĞƉŽŝŶƚƐĂƌĞŶŽƚŽŶƚŚĞƐĂŵĞŚŽƌŝnjŽŶƚĂůŽƌ ǀĞƌƚŝĐĂůůŝŶĞ͘ Ă͘ ሺͲǡ െʹሻ ĂŶĚ ሺͲǡ ͻሻ ď͘ ሺͳͳǡ Ͷሻ ĂŶĚ ሺʹǡ ͳͳሻ Đ͘ ሺ͵ǡ െͺሻ ĂŶĚ ሺ͵ǡ െͳሻ Ě͘ ሺെͶǡ െͶሻ ĂŶĚ ሺͷǡ െͶሻ
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6ͻϯ Lesson 18 Lesson 18: ŝƐƚĂŶĐĞŽŶƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
Exit Ticket Sample Solutions Determine whether each given pair of end points lies on the same horizontal or vertical line. If so, find the length of the line segment that joins the pair of points. If not, explain how you know the points are not on the same horizontal or vertical line. a. ሺǡ െሻ and ሺǡૢ ሻ
The end points both have ࢞ -coordinates of , so they both lie on the ࢟ -axis, which is a vertical line. They lie on opposite sides of zero, so their absolute values have to be combined to get the total distance. ȁെȁ ൌ and
ȁૢ ȁ ൌૢ , so by addition, ૢ ൌ . The length of the line segment with end points ሺǡ െሻ and ሺǡૢ ሻ is
units.
b. ሺǡ ሻ and ሺǡ ሻ
The points do not lie on the same horizontal or vertical line because they do not share a common ࢞ - or ࢟
-coordinate.
c. ሺǡ െૡሻ and ሺǡ െሻ
The end points both have ࢞ -coordinates of , so the points lie on a vertical line that passes through on the
࢞
-axis. The ࢟ -coordinates lie on the same side of zero. The distance between the points is determined by subtracting their absolute values, ȁെૡȁ ൌ ૡ and ȁെȁ ൌ . So, by subtraction, ૡ െ ൌ ૠ . The length of the line segment with end points ሺǡ െૡሻ and ሺǡ െሻ is ૠ units.
d. ሺെǡ െሻ and ሺǡ െሻ
The end points have the same ࢟ -coordinate of െ , so they lie on a horizontal line that passes through െ on the ࢟ -axis. The numbers lie on opposite sides of zero on the number line, so their absolute values must be added to obtain the total distance, ȁെȁ ൌ and ȁȁ ൌ . So, by addition, ൌૢ . The length of the line segment with end points ሺെǡ െሻ and ሺǡ െሻ is ૢ units.
Problem Set Sample Solutions 1. Find the length of the line segment with end points ሺૠǡ ሻ and ሺെǡ ሻ , and explain how you arrived at your solution.
units. Both points have the same ࢟ -coordinate, so I knew they were on the same horizontal line. I found the distance between the ࢞ -coordinates by counting the number of units on a horizontal number line from െ to zero and then from zero to ૠ, and ૠ ൌ .or I found the distance between the ࢞ -coordinates by finding the absolute value of each coordinate. ȁૠȁ ൌ ૠ and
ȁെȁ ൌ . The coordinates lie on opposite sides of zero, so I found the length by adding the absolute values together. Therefore, the length of a line segment with end points ሺૠǡ ሻ and ሺെǡ ሻ is units.
Sarah and Jamal were learning partners in math class and were working independently. They each started at the point ሺെǡ ሻ and moved units vertically in the plane. Each student arrived at a different end point. How is this possible? Explain and list the two different end points.
It is possible because Sarah could have counted up and Jamal could have counted down or vice versa. Moving
units in either direction vertically would generate the following possible end points: ሺെǡ ૡሻ or ሺെǡ ሻ .
ϯ͘ The length of a line segment is units. One end point of the line segment is ሺെǡ ૠሻ . Find four points that could be the other end points of the line segment.
ሺെǡ ሻ , ሺെǡ െሻ , ሺെǡ ૠሻ or ሺǡ ૠሻ
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6ͻϯ Lesson 19 Lesson 19: Problem Solving and the Coordinate Plane Student Outcomes
^ƚƵĚĞŶƚƐƐŽůǀĞƉƌŽďůĞŵƐƌĞůĂƚĞĚƚŽƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƉŽŝŶƚƐƚŚĂƚůŝĞŽŶƚŚĞƐĂŵĞŚŽƌŝnjŽŶƚĂůŽƌǀĞƌƚŝĐĂůůŝŶĞ͘
^ƚƵĚĞŶƚƐƵƐĞƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽŐƌĂƉŚƉŽŝŶƚƐ͕ůŝŶĞƐĞŐŵĞŶƚƐ͕ĂŶĚŐĞŽŵĞƚƌŝĐƐŚĂƉĞƐŝŶƚŚĞǀĂƌŝŽƵƐ ƋƵĂĚƌĂŶƚƐĂŶĚƚŚĞŶƵƐĞƚŚĞĂďƐŽůƵƚĞǀĂůƵĞƚŽĨŝŶĚƚŚĞƌĞůĂƚĞĚĚŝƐƚĂŶĐĞƐ͘
Lesson Notes
dŚĞŐƌŝĚƉƌŽǀŝĚĞĚŝŶƚŚĞKƉĞŶŝŶŐ džĞƌĐŝƐĞŝƐĂůƐŽƵƐĞĚĨŽƌ džĞƌĐŝƐĞƐϭʹϲƐŝŶĐĞĞĂĐŚĞdžĞƌĐŝƐĞŝƐƐĞƋƵĞŶƚŝĂů͘^ƚƵĚĞŶƚƐ ĞdžƚĞŶĚƚŚĞŝƌŬŶŽǁůĞĚŐĞĂďŽƵƚĨŝŶĚŝŶŐĚŝƐƚĂŶĐĞƐďĞƚǁĞĞŶƉŽŝŶƚƐŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽƚŚĞĂƐƐŽĐŝĂƚĞĚůĞŶŐƚŚƐŽĨůŝŶĞ ƐĞŐŵĞŶƚƐĂŶĚƐŝĚĞƐŽĨŐĞŽŵĞƚƌŝĐĨŝŐƵƌĞƐ͘
Classwork Opening Exercise ;ϯ minutes) Opening Exercise In the coordinate plane, find the distance between the points using absolute value.
The distance between the points is ૡ units. The points have the same first coordinates and, therefore, lie on the same vertical line. |െ| = , and || = , and the numbers lie on opposite sides of , so their absolute values are added together; + = ૡ . We can check our answer by just counting the number of units between the two points.
|െ| =
|| =
Lesson 19: WƌŽďůĞŵ^ŽůǀŝŶŐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
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6ͻϯ Lesson 19 Exploratory Challenge Exercises 1–2 (8 minutes): The Length of a Line Segment Is the Distance Between Its End Points
^ƚƵĚĞŶƚƐƌĞůĂƚĞƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƚǁŽƉŽŝŶƚƐůLJŝŶŐŝŶĚŝĨĨĞƌĞŶƚƋƵĂĚƌĂŶƚƐŽĨƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞƚŽƚŚĞůĞŶŐƚŚŽĨĂ ůŝŶĞƐĞŐŵĞŶƚǁŝƚŚƚŚŽƐĞĞŶĚƉŽŝŶƚƐ͘^ƚƵĚĞŶƚƐƚŚĞŶƵƐĞƚŚŝƐƌĞůĂƚŝŽŶƐŚŝƉƚŽŐƌĂƉŚĂŚŽƌŝnjŽŶƚĂůŽƌǀĞƌƚŝĐĂůůŝŶĞƐĞŐŵĞŶƚ ƵƐŝŶŐĚŝƐƚĂŶĐĞƚŽĨŝŶĚƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨĞŶĚƉŽŝŶƚƐ͘
Exploratory Challenge Exercises 1–2: The Length of a Line Segment Is the Distance Between Its End Points 1. Locate and label (, ) and (, െ) . Draw the line segment between the end points given on the coordinate plane. How long is the line segment that you drew? Explain.
The length of the line segment is also ૡ units. I found that the distance between (, െ) and (, ) is ૡ units. Because the end points are on opposite sides of zero, I added the absolute values of the second coordinates together, so the distance from end to end is ૡ units.
Draw a horizontal line segment starting at (, െ) that has a length of ૢ
units. What are the possible coordinates of the other end point of the line segment? (There is more than one answer.)
(െ, െ) or (, െ)
Which point did you choose to be the other end point of the horizontal line segment? Explain how and why you chose that point. Locate and label the point on the coordinate grid.
The other end point of the horizontal line segment is (െ, െ) . I chose this point because the other option,
(, െ) , is located off of the given coordinate grid. Note: Students may choose the end point (, െ) , but they must change the number scale of the ࢞ -axis to do so.
Exercise ϯ (5 minutes): Extending Lengths of Line Segments to Sides of Geometric Figures
džĞƌĐŝƐĞϯ : Extending Lengths of Line Segments to Sides of Geometric Figures
ϯ͘ The two line segments that you have just drawn could be seen as two sides of a rectangle. Given this, the end points of the two line segments would be three of the vertices of this rectangle. a. Find the coordinates of the fourth vertex of the rectangle. Explain how you find the coordinates of the fourth vertex using absolute value.
The fourth vertex is (െ, ) . The opposite sides of a rectangle are the same length, so the length of the vertical side starting at (െ, െ) has to be ૡ units long. Also, the side from (െ, െ) to the remaining vertex is a vertical line, so the end points must have the same first coordinate.
|െ| = , and ૡ െ = , so the remaining vertex must be five units above the ࢞ -axis. Note: Students can use a similar argument using the length of the horizontal side starting at (, ) , knowing it has to be ૢ
units long.
(െ, െ)
(െ, )
(, െ)
(, )
Lesson 19: WƌŽďůĞŵ^ŽůǀŝŶŐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
(െ, െ)
ૡunits
(, െ)
(, )
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6ͻϯ Lesson 19 b. How does the fourth vertex that you found relate to each of the consecutive vertices in either direction? Explain.
The fourth vertex has the same first coordinate as (െ, െ) because they are the end points of a vertical line segment. The fourth vertex has the same second coordinate as (, ) since they are the end points of a horizontal line segment.
c. Draw the remaining sides of the rectangle. Exercises 4–6 (6 minutes): Using Lengths of Sides of Geometric Figures to Solve Problems Exercises 4–6: Using Lengths of Sides of Geometric Figures to Solve Problems 4. Using the vertices that you have found and the lengths of the line segments between them, find the perimeter of the rectangle.
ૡ +ૢ + ૡ +ૢ = ; the perimeter of the rectangle is units.
Find the area of the rectangle. ૢ × ૡ = ૠ ; the area of the rectangle is ૠ units 2.
Draw a diagonal line segment through the rectangle with opposite vertices for end points. What geometric figures are formed by this line segment? What are the areas of each of these figures? Explain.
The diagonal line segment cuts the rectangle into two right triangles. The areas of the triangles are units 2each because the triangles each make up half of the rectangle, and half of ૠ is .
Extension (If time allows): Line the edge of a piece of paper up to the diagonal in the rectangle. Mark the length of the diagonal on the edge of the paper. Align your marks horizontally or vertically on the grid, and estimate the length of the diagonal to the nearest integer. Use that estimation to now estimate the perimeter of the triangles.
The length of the diagonal is approximately units, and the perimeter of each triangle is approximately ૢ units.
Scaffolding:
^ƚƵĚĞŶƚƐŵĂLJŶĞĞĚƚŽƌĞǀŝĞǁ ĂŶĚĚŝƐĐƵƐƐƚŚĞĐŽŶĐĞƉƚƐŽĨ ƉĞƌŝŵĞƚĞƌĂŶĚĂƌĞĂĨƌŽŵ ĞĂƌůŝĞƌŐƌĂĚĞƐ͘
Lesson 19: WƌŽďůĞŵ^ŽůǀŝŶŐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
(െ, െ)
(െ, )
(, െ)
(, )
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6ͻϯ Lesson 19 Exercise 7 (8 minutes) Exercise 7 7. Construct a rectangle on the coordinate plane that satisfies each of the criteria listed below. Identify the coordinate of each of its vertices.
x Each of the vertices lies in a different quadrant.
x Its sides are either vertical or horizontal.
x The perimeter of the rectangle is ૡ units.
Answers will vary. The example to the right shows a rectangle with side lengths and units. The coordinates of the rectangle’s vertices are (െ, ) ,
(, ) , (, െ) , and (െ, െ) .
Using absolute value, show how the lengths of the sides of your rectangle provide a perimeter of ૡ units.
|െ| = , || = , and + = , so the width of my rectangle is units.
|| = , |െ| = , and + = , so the height of my rectangle is units.
+ + + = ૡ , so the perimeter of my rectangle is ૡ units.
Closing (5 minutes)
,ŽǁĚŽǁĞĚĞƚĞƌŵŝŶĞƚŚĞůĞŶŐƚŚŽĨĂŚŽƌŝnjŽŶƚĂůůŝŶĞƐĞŐŵĞŶƚǁŚŽƐĞĞŶĚƉŽŝŶƚƐůŝĞŝŶĚŝĨĨĞƌĞŶƚƋƵĂĚƌĂŶƚƐŽĨ ƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͍
à If the points are in different quadrants, then the ݔ-coordinates lie on opposite sides of zero. The distance between the ݔ-coordinates can be found by adding the absolute values of the ݔ-coordinates. (The ݕ-coordinates are the same and show that the points lie on a horizontal line.)
/ĨǁĞŬŶŽǁŽŶĞĞŶĚƉŽŝŶƚŽĨĂǀĞƌƚŝĐĂůůŝŶĞƐĞŐŵĞŶƚĂŶĚƚŚĞůĞŶŐƚŚŽĨƚŚĞůŝŶĞƐĞŐŵĞŶƚ͕ŚŽǁĚŽǁĞĨŝŶĚƚŚĞ ŽƚŚĞƌĞŶĚƉŽŝŶƚŽĨƚŚĞůŝŶĞƐĞŐŵĞŶƚ͍/ƐƚŚĞƉƌŽĐĞƐƐƚŚĞƐĂŵĞǁŝƚŚĂŚŽƌŝnjŽŶƚĂůůŝŶĞƐĞŐŵĞŶƚ͍
à If the line segment is vertical, then the other end point could be above or below the given end point. If we know the length of the line segment, then we can count up or down from the given end point to find the other end point. We can check our answer using the absolute values of the ݕ-coordinates. The process is similar with a horizontal line. If we know the length of the line segment, then we can count to the left or the right from the given end point to find the other end point.
Exit Ticket (10 minutes) Lesson Summary
The length of a line segment on the coordinate plane can be determined by finding the distance between its end points.
You can find the perimeter and area of figures such as rectangles and right triangles by finding the lengths of the line segments that make up their sides and then using the appropriate formula.
Lesson 19: WƌŽďůĞŵ^ŽůǀŝŶŐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
࢞࢟
(െ, )
(െ, െ)
(, )
(, െ)
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6ͻϯ Lesson 19
EĂŵĞ ĂƚĞ
Lesson 19: Problem Solving and the Coordinate Plane Exit Ticket
ϭ͘ dŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨŽŶĞĞŶĚƉŽŝŶƚŽĨĂůŝŶĞƐĞŐŵĞŶƚĂƌĞ (െ2, െ7) ͘ dŚĞůŝŶĞƐĞŐŵĞŶƚŝƐ 12 ƵŶŝƚƐůŽŶŐ͘'ŝǀĞƚŚƌĞĞ ƉŽƐƐŝďůĞĐŽŽƌĚŝŶĂƚĞƐŽĨƚŚĞůŝŶĞƐĞŐŵĞŶƚ͛ƐŽƚŚĞƌĞŶĚƉŽŝŶƚ͘ Ϯ͘ 'ƌĂƉŚĂƌĞĐƚĂŶŐůĞǁŝƚŚĂŶĂƌĞĂŽĨ 12 ƵŶŝƚƐ ϮƐƵĐŚƚŚĂƚŝƚƐǀĞƌƚŝĐĞƐůŝĞŝŶĂƚůĞĂƐƚƚǁŽŽĨƚŚĞĨŽƵƌƋƵĂĚƌĂŶƚƐŝŶƚŚĞ ĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͘^ƚĂƚĞƚŚĞůĞŶŐƚŚƐŽĨĞĂĐŚŽĨƚŚĞƐŝĚĞƐ͕ĂŶĚƵƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽƐŚŽǁŚŽǁLJŽƵĚĞƚĞƌŵŝŶĞĚƚŚĞ ůĞŶŐƚŚƐŽĨƚŚĞƐŝĚĞƐ͘ ࢞࢟
Lesson 19: WƌŽďůĞŵ^ŽůǀŝŶŐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
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6ͻϯ Lesson 19 Exit Ticket Sample Solutions 1. The coordinates of one end point of a line segment are (െ, െૠ) . The line segment is units long. Give three possible coordinates of the line segment’s other end point.
(, െૠ) ; (െ, െૠ) ; (െ, ) ; (െ, െૢ )
Graph a rectangle with an area of units 2 such that its vertices lie in at least two of the four quadrants in the coordinate plane. State the lengths of each of the sides, and use absolute value to show how you determined the lengths of the sides.
Answers will vary. The rectangle can have side lengths of and or and . A sample is provided on the grid on the right. × =
Problem Set Sample Solutions
WůĞĂƐĞƉƌŽǀŝĚĞƐƚƵĚĞŶƚƐǁŝƚŚƚŚƌĞĞĐŽŽƌĚŝŶĂƚĞŐƌŝĚƐƚŽƵƐĞŝŶĐŽŵƉůĞƚŝŶŐƚŚĞWƌŽďůĞŵ^Ğƚ͘
One end point of a line segment is (െ, െ) . The length of the line segment is ૠ units. Find four points that could serve as the other end point of the given line segment.
(െ, െ) ; (, െ) ; (െ, ) ; (െ, െ)
Two of the vertices of a rectangle are (, െ) and (െૡ, െ) . If the rectangle has a perimeter of units, what are the coordinates of its other two vertices?
(, െ) and (െૡ, െ) , or (, െ) and (െૡ, െ)
ϯ͘ A rectangle has a perimeter of ૡ units, an area of ૡ square units, and sides that are either horizontal or vertical. If one vertex is the point (െ, െૠ) and the origin is in the interior of the rectangle, find the vertex of the rectangle that is opposite (െ, െૠ) .
(, )
Lesson 19: WƌŽďůĞŵ^ŽůǀŝŶŐĂŶĚƚŚĞŽŽƌĚŝŶĂƚĞWůĂŶĞ
࢞࢟
unit unit
units
units
units
units
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6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
EĂŵĞ ĂƚĞ ϭ͘ Dƌ͘<ŝŶĚůĞŝŶǀĞƐƚĞĚƐŽŵĞŵŽŶĞLJŝŶƚŚĞƐƚŽĐŬŵĂƌŬĞƚ͘,ĞƚƌĂĐŬƐŚŝƐŐĂŝŶƐĂŶĚůŽƐƐĞƐƵƐŝŶŐĂĐŽŵƉƵƚĞƌ ƉƌŽŐƌĂŵ͘Dƌ͘<ŝŶĚůĞƌĞĐĞŝǀĞƐĂĚĂŝůLJĞŵĂŝůƚŚĂƚƵƉĚĂƚĞƐŚŝŵŽŶĂůůŚŝƐƚƌĂŶƐĂĐƚŝŽŶƐĨƌŽŵƚŚĞƉƌĞǀŝŽƵƐĚĂLJ͘ dŚŝƐŵŽƌŶŝŶŐ͕ŚŝƐĞŵĂŝůƌĞĂĚĂƐĨŽůůŽǁƐ͗
Good morning, Mr. Kindle, Yesterday’s investment activity included a loss of ̈́ ͺͲͲ , a gain of ̈́ ͻͲ , and another gain of ̈́ ʹ͵Ͳ . Log in now to see your current balance.
Ă͘ tƌŝƚĞĂŶŝŶƚĞŐĞƌƚŽƌĞƉƌĞƐĞŶƚĞĂĐŚŐĂŝŶĂŶĚůŽƐƐ͘
Description Integer Representation
ŽƐƐŽĨ ̈́ ͺͲͲ 'ĂŝŶŽĨ ̈́ ͻͲ 'ĂŝŶŽĨ ̈́ ʹ͵Ͳ ď͘ Dƌ͘<ŝŶĚůĞŶŽƚŝĐĞĚƚŚĂƚĂŶĞƌƌŽƌŚĂĚďĞĞŶŵĂĚĞŽŶŚŝƐĂĐĐŽƵŶƚ͘dŚĞ͞ůŽƐƐŽĨ ̈́ ͺͲͲ ͟ ƐŚŽƵůĚŚĂǀĞ ďĞĞŶĂ͞ŐĂŝŶŽĨ ̈́ ͺͲͲǤ ͟ >ŽĐĂƚĞĂŶĚůĂďĞůďŽƚŚƉŽŝŶƚƐƚŚĂƚƌĞƉƌĞƐĞŶƚ͞ĂůŽƐƐŽĨ ̈́ ͺͲͲ ͟ ĂŶĚ͞ĂŐĂŝŶŽĨ ̈́ͺͲͲ ͟ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞďĞůŽǁ͘ĞƐĐƌŝďĞƚŚĞƌĞůĂƚŝŽŶƐŚŝƉŽĨƚŚĞƐĞƚǁŽŶƵŵďĞƌƐǁŚĞŶnjĞƌŽ ƌĞƉƌĞƐĞŶƚƐŶŽĐŚĂŶŐĞ;ŐĂŝŶŽƌůŽƐƐͿ͘
Ͳ
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6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
Đ͘ Dƌ͘<ŝŶĚůĞǁĂŶƚĞĚƚŽĐŽƌƌĞĐƚƚŚĞĞƌƌŽƌ͕ƐŽŚĞĞŶƚĞƌĞĚ െሺെ̈́ͺͲͲሻ ŝŶƚŽƚŚĞƉƌŽŐƌĂŵ͘,ĞŵĂĚĞĂŶŽƚĞ ƚŚĂƚƌĞĂĚ͕͞dŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨ ̈́ ͺͲͲ ŝƐ ̈́ͺͲͲ ͘͟ /ƐŚŝƐƌĞĂƐŽŶŝŶŐĐŽƌƌĞĐƚ͍ džƉůĂŝŶ͘ Ϯ͘ ƚϲ͗ϬϬĂ͘ŵ͕͘ƵĨĨĂůŽ͕Ez͕ŚĂĚĂƚĞŵƉĞƌĂƚƵƌĞŽĨ ͳͲԬ ͘ƚŶŽŽŶ͕ƚŚĞƚĞŵƉĞƌĂƚƵƌĞǁĂƐ െͳͲԬ ͕ĂŶĚĂƚ ŵŝĚŶŝŐŚƚ͕ŝƚǁĂƐ െʹͲԬ ͘ Ă͘ tƌŝƚĞĂƐƚĂƚĞŵĞŶƚĐŽŵƉĂƌŝŶŐ െͳͲԬ ĂŶĚ െʹͲԬǤ ď͘ tƌŝƚĞĂŶŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƚŚĂƚƐŚŽǁƐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞƚŚƌĞĞƌĞĐŽƌĚĞĚ ƚĞŵƉĞƌĂƚƵƌĞƐ͘tŚŝĐŚƚĞŵƉĞƌĂƚƵƌĞŝƐƚŚĞǁĂƌŵĞƐƚ͍
Temperature in degrees Fahrenheit
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6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
Đ͘ džƉůĂŝŶŚŽǁƚŽƵƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽĨŝŶĚƚŚĞŶƵŵďĞƌŽĨĚĞŐƌĞĞƐďĞůŽǁnjĞƌŽƚŚĞƚĞŵƉĞƌĂƚƵƌĞǁĂƐĂƚ ŶŽŽŶ͘ Ě͘ /ŶWĞĞŬƐŬŝůů͕Ez͕ƚŚĞƚĞŵƉĞƌĂƚƵƌĞĂƚϲ͗ϬϬĂ͘ŵ͘ǁĂƐ െͳʹԬ ͘ƚŶŽŽŶ͕ƚŚĞƚĞŵƉĞƌĂƚƵƌĞǁĂƐƚŚĞĞdžĂĐƚ ŽƉƉŽƐŝƚĞŽĨƵĨĨĂůŽ͛ƐƚĞŵƉĞƌĂƚƵƌĞĂƚϲ͗ϬϬĂ͘ŵ͘ƚŵŝĚŶŝŐŚƚ͕ĂŵĞƚĞŽƌŽůŽŐŝƐƚƌĞĐŽƌĚĞĚƚŚĞ ƚĞŵƉĞƌĂƚƵƌĞĂƐ െԬ ŝŶWĞĞŬƐŬŝůů͘,ĞĐŽŶĐůƵĚĞĚƚŚĂƚ͞&ŽƌƚĞŵƉĞƌĂƚƵƌĞƐďĞůŽǁnjĞƌŽ͕ĂƐƚŚĞ ƚĞŵƉĞƌĂƚƵƌĞŝŶĐƌĞĂƐĞƐ͕ƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨƚŚĞƚĞŵƉĞƌĂƚƵƌĞĚĞĐƌĞĂƐĞƐ͘͟/ƐŚŝƐĐŽŶĐůƵƐŝŽŶǀĂůŝĚ͍ džƉůĂŝŶĂŶĚƵƐĞĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞƚŽƐƵƉƉŽƌƚLJŽƵƌĂŶƐǁĞƌ͘ ϯ͘ ŚŽŽƐĞĂŶŝŶƚĞŐĞƌďĞƚǁĞĞŶ ͲĂŶĚ െͷ ŽŶĂŶƵŵďĞƌůŝŶĞ͕ĂŶĚůĂďĞůƚŚĞƉŽŝŶƚ ܲ͘ >ŽĐĂƚĞĂŶĚůĂďĞůĞĂĐŚŽĨ ƚŚĞĨŽůůŽǁŝŶŐƉŽŝŶƚƐĂŶĚƚŚĞŝƌǀĂůƵĞƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ Ă͘ >ĂďĞůƉŽŝŶƚ ܣ͗ ƚŚĞŽƉƉŽƐŝƚĞ ŽĨƉŽŝŶƚ ܲ͘ ď͘ >ĂďĞůƉŽŝŶƚ ܤ͗ ĂŶƵŵďĞƌůĞƐƐƚŚĂŶƉŽŝŶƚ ܲ͘ Đ͘ >ĂďĞůƉŽŝŶƚ ܥ͗ ĂŶƵŵďĞƌŐƌĞĂƚĞƌƚŚĂŶƉŽŝŶƚ ܲ͘ Ě͘ >ĂďĞůƉŽŝŶƚ ܦ͗ ĂŶƵŵďĞƌŚĂůĨǁĂLJďĞƚǁĞĞŶƉŽŝŶƚ ܲ ĂŶĚƚŚĞŝŶƚĞŐĞƌƚŽƚŚĞƌŝŐŚƚŽĨƉŽŝŶƚ ܲ͘
െͳͲ െͻ െͺ െ െ െͷ െͶ െ͵ െʹ െͳ Ͳ ʹ ͵ Ͷ ͷ ͺ ͻ ͳͲ ͳ
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6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
ϰ͘ :ƵůŝĂŝƐůĞĂƌŶŝŶŐĂďŽƵƚĞůĞǀĂƚŝŽŶŝŶŵĂƚŚĐůĂƐƐ͘^ŚĞĚĞĐŝĚĞĚƚŽƌĞƐĞĂƌĐŚƐŽŵĞĨĂĐƚƐĂďŽƵƚEĞǁzŽƌŬ^ƚĂƚĞ ƚŽďĞƚƚĞƌƵŶĚĞƌƐƚĂŶĚƚŚĞĐŽŶĐĞƉƚ͘,ĞƌĞĂƌĞƐŽŵĞĨĂĐƚƐƚŚĂƚƐŚĞĨŽƵŶĚ͘
Mount Marcy is the highest point in New York State. It is ͷǡ͵Ͷ͵ feet above sea level.
Lake Erie is ʹͳͲ feet below sea level.
The elevation of Niagara Falls, NY, is ͳͶ feet above sea level.
The lobby of the Empire State Building is ͷͲ feet above sea level.
New York State borders the Atlantic Coast, which is at sea level.
The lowest point of Cayuga Lake is Ͷ͵ͷ feet below sea level.
Ă͘ tƌŝƚĞĂŶŝŶƚĞŐĞƌƚŚĂƚƌĞƉƌĞƐĞŶƚƐĞĂĐŚůŽĐĂƚŝŽŶŝŶƌĞůĂƚŝŽŶƐŚŝƉƚŽƐĞĂůĞǀĞů͘ DŽƵŶƚDĂƌĐLJ >ĂŬĞ ƌŝĞ EŝĂŐĂƌĂ&ĂůůƐ͕Ez ŵƉŝƌĞ^ƚĂƚĞƵŝůĚŝŶŐ ƚůĂŶƚŝĐŽĂƐƚ ĂLJƵŐĂ>ĂŬĞ ď͘ džƉůĂŝŶǁŚĂƚŶĞŐĂƚŝǀĞĂŶĚƉŽƐŝƚŝǀĞŶƵŵďĞƌƐƚĞůů:ƵůŝĂĂďŽƵƚĞůĞǀĂƚŝŽŶ͘
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6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
Đ͘ KƌĚĞƌƚŚĞĞůĞǀĂƚŝŽŶƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͕ĂŶĚƚŚĞŶƐƚĂƚĞƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘hƐĞƚŚĞĐŚĂƌƚ ďĞůŽǁƚŽƌĞĐŽƌĚLJŽƵƌǁŽƌŬ͘
Elevations Absolute Values of Elevations
Ě͘ ŝƌĐůĞƚŚĞƌŽǁŝŶƚŚĞƚĂďůĞƚŚĂƚƌĞƉƌĞƐĞŶƚƐƐĞĂůĞǀĞů͘ĞƐĐƌŝďĞŚŽǁƚŚĞŽƌĚĞƌŽĨƚŚĞĞůĞǀĂƚŝŽŶƐďĞůŽǁ ƐĞĂůĞǀĞůĐŽŵƉĂƌĞƐƚŽƚŚĞŽƌĚĞƌŽĨƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘ĞƐĐƌŝďĞŚŽǁƚŚĞŽƌĚĞƌŽĨƚŚĞĞůĞǀĂƚŝŽŶƐ ĂďŽǀĞƐĞĂůĞǀĞůĐŽŵƉĂƌĞƐƚŽƚŚĞŽƌĚĞƌŽĨƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘
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6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
ϱ͘ &ŽƌĐĞŶƚƵƌŝĞƐ͕ĂŵLJƐƚĞƌŝŽƵƐƐĞĂƐĞƌƉĞŶƚŚĂƐďĞĞŶƌƵŵŽƌĞĚƚŽůŝǀĞĂƚƚŚĞďŽƚƚŽŵŽĨDLJƐƚĞƌŝŽƵƐ>ĂŬĞ͘ ƚĞĂŵŽĨŚŝƐƚŽƌŝĂŶƐƵƐĞĚĂĐŽŵƉƵƚĞƌƉƌŽŐƌĂŵƚŽƉůŽƚƚŚĞůĂƐƚĨŝǀĞƉŽƐŝƚŝŽŶƐŽĨƚŚĞƐŝŐŚƚŝŶŐƐ͘ Ă͘ >ŽĐĂƚĞĂŶĚůĂďĞůƚŚĞůŽĐĂƚŝŽŶƐŽĨƚŚĞůĂƐƚĨŽƵƌƐŝŐŚƚŝŶŐƐ͗ ܣ ቀെͻ ͳʹ ǡ Ͳቁ ͕ ܤሺെ͵ǡ െͶǤͷሻ ͕ ܥሺͻǡ ʹሻ ͕ ĂŶĚ ܦሺͺǡ െʹǤͷሻ ͘ď͘ KǀĞƌƚŝŵĞ͕ŵŽƐƚŽĨƚŚĞƐŝŐŚƚŝŶŐƐŽĐĐƵƌƌĞĚŝŶYƵĂĚƌĂŶƚ///͘tƌŝƚĞƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨĂƉŽŝŶƚƚŚĂƚůŝĞƐ ŝŶYƵĂĚƌĂŶƚ///͘ Đ͘ tŚĂƚŝƐƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƉŽŝŶƚ ܣĂŶĚƚŚĞƉŽŝŶƚ ቀͻ ͳʹ ǡ Ͳቁ ͍ ^ŚŽǁLJŽƵƌǁŽƌŬƚŽƐƵƉƉŽƌƚLJŽƵƌ ĂŶƐǁĞƌ͘ Ě͘ tŚĂƚĂƌĞƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨƉŽŝŶƚ ܧŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͍ Ğ͘ WŽŝŶƚ ܨŝƐƌĞůĂƚĞĚƚŽƉŽŝŶƚ ܧ͘ /ƚƐ ݔͲĐŽŽƌĚŝŶĂƚĞŝƐƚŚĞƐĂŵĞĂƐƉŽŝŶƚ ܧ͛ Ɛ͕ďƵƚŝƚƐ ݕͲĐŽŽƌĚŝŶĂƚĞŝƐƚŚĞ ŽƉƉŽƐŝƚĞŽĨƉŽŝŶƚ ܧ͛ Ɛ͘>ŽĐĂƚĞĂŶĚůĂďĞůƉŽŝŶƚ ܨ͘ tŚĂƚĂƌĞƚŚĞĐŽŽƌĚŝŶĂƚĞƐ͍,ŽǁĨĂƌĂƉĂƌƚĂƌĞƉŽŝŶƚƐ
ܧĂŶĚ ܨ͍ džƉůĂŝŶŚŽǁLJŽƵĂƌƌŝǀĞĚĂƚLJŽƵƌĂŶƐǁĞƌ͘
E
A STORY OF RATIOS
191
©201 8Great Minds ®. eureka-math.org
6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
A Progression Toward Mastery Assessment Task Item STEP 1 Missing or incorrect answer and little evidence of reasoning or application of mathematics to solve the problem. STEP 2 Missing or incorrect answer but evidence of some reasoning or application of mathematics to solve the problem. STEP 3 A correct answer with some evidence of reasoning or application of mathematics to solve the problem, OR an incorrect answer with substantial evidence of solid reasoning or application of mathematics to solve the problem. STEP 4 A correct answer supported by substantial evidence of solid reasoning or application of mathematics to solve the problem. 1 a
^ƚƵĚĞŶƚŝƐƵŶĂďůĞƚŽ ĂŶƐǁĞƌƚŚĞƋƵĞƐƚŝŽŶ͘ EŽŶĞŽĨƚŚĞ ĚĞƐĐƌŝƉƚŝŽŶƐĂƌĞ ĐŽƌƌĞĐƚůLJƌĞƉƌĞƐĞŶƚĞĚ ǁŝƚŚĂŶŝŶƚĞŐĞƌ ĂůƚŚŽƵŐŚƐƚƵĚĞŶƚŵĂLJ ŵĂŬĞĂŶĞĨĨŽƌƚƚŽ ĂŶƐǁĞƌƚŚĞƋƵĞƐƚŝŽŶ .
^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ƌĞƉƌĞƐĞŶƚƐŽŶůLJŽŶĞŽĨ ƚŚĞƚŚƌĞĞĚĞƐĐƌŝƉƚŝŽŶƐ ǁŝƚŚĂŶŝŶƚĞŐĞƌ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ƌĞƉƌĞƐĞŶƚƐƚǁŽŽĨƚŚĞ ƚŚƌĞĞĚĞƐĐƌŝƉƚŝŽŶƐǁŝƚŚ ŝŶƚĞŐĞƌƐ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ƌĞƉƌĞƐĞŶƚƐĂůůƚŚƌĞĞ ĚĞƐĐƌŝƉƚŝŽŶƐǁŝƚŚ ŝŶƚĞŐĞƌƐ͗ െͺͲͲ ͕ ͻͲ ͕
ʹ͵Ͳ ͘
b ^ƚƵĚĞŶƚĚŽĞƐŶŽƚ ĂƚƚĞŵƉƚƚŽůŽĐĂƚĞĂŶĚ ůĂďĞů െͺͲͲ ĂŶĚ
ͺͲͲ ĂŶĚƉƌŽǀŝĚĞƐůŝƚƚůĞ ŽƌŶŽĞǀŝĚĞŶĐĞŽĨ ƌĞĂƐŽŶŝŶŐ͘ ^ƚƵĚĞŶƚĂƚƚĞŵƉƚƐƚŽ ůŽĐĂƚĞĂŶĚůĂďĞů െͺͲͲ ĂŶĚ ͺͲͲ ďƵƚŵĂŬĞƐĂŶ ĞƌƌŽƌ͘&ŽƌĞdžĂŵƉůĞ͕ďŽƚŚ ŝŶƚĞŐĞƌƐĂƌĞŶŽƚ ĞƋƵŝĚŝƐƚĂŶƚĨƌŽŵ Ͳ͘ ^ƚƵĚĞŶƚŵĂLJŽƌŵĂLJŶŽƚ ĐŽƌƌĞĐƚůLJŝĚĞŶƚŝĨLJƚŚĞ ƌĞůĂƚŝŽŶƐŚŝƉĂƐ ŽƉƉŽƐŝƚĞƐ .
^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐďƵƚ does not label െͺͲͲ ĂŶĚ ͺͲͲ ͖ƐƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ŝĚĞŶƚŝĨŝĞƐƚŚĞ ƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞ ŝŶƚĞŐĞƌƐĂƐŽƉƉŽƐŝƚĞƐ .
KZ
^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐĂŶĚůĂďĞůƐ െͺͲͲ ĂŶĚ ͺͲͲ ŽŶƚŚĞŶƵŵďĞƌ ůŝŶĞďƵƚĚŽĞƐŶŽƚŝĚĞŶƚŝĨLJ ƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶ ƚŚĞŝŶƚĞŐĞƌƐĂƐ ŽƉƉŽƐŝƚĞƐ͘
^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐĂŶĚůĂďĞůƐ
െͺͲͲ ĂŶĚ ͺͲͲ ŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞĂŶĚ ŝĚĞŶƚŝĨŝĞƐƚŚĞ ƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞ ŝŶƚĞŐĞƌƐĂƐŽƉƉŽƐŝƚĞƐ͘
A STORY OF RATIOS
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©201 8Great Minds ®. eureka-math.org
6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
c
^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽƌƌĞĐƚ͕ĂŶĚŶŽ ĞǀŝĚĞŶĐĞŽĨƌĞĂƐŽŶŝŶŐ͕ ƐƵĐŚĂƐĂŶĞdžƉůĂŶĂƚŝŽŶ ŽƌĂĚŝĂŐƌĂŵ͕ŝƐ ƉƌŽǀŝĚĞĚ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽƌƌĞĐƚ͕ďƵƚƐƚƵĚĞŶƚ ĂƚƚĞŵƉƚƐƚŽĂŶƐǁĞƌƚŚĞ ƋƵĞƐƚŝŽŶǁŝƚŚĂŶ ĞdžƉůĂŶĂƚŝŽŶĂŶĚͬŽƌ ĚŝĂŐƌĂŵƚŚĂƚ ĚĞŵŽŶƐƚƌĂƚĞƐĂŶ ƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨƚŚĞ ǁŽƌĚ opposite ĂůƚŚŽƵŐŚ ŝƚĚŽĞƐŶŽƚĂĚĚƌĞƐƐƚŚĞ ŵĞĂŶŝŶŐŽĨ͞ƚŚĞ ŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞ ŽĨ ̈́ͺͲͲ ͘͟ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ͕͞zĞƐ͕ Dƌ͘<ŝŶĚůĞ͛ƐƌĞĂƐŽŶŝŶŐŝƐ ĐŽƌƌĞĐƚ͘͟ƵƚƚŚĞ ĞdžƉůĂŶĂƚŝŽŶĂŶĚͬŽƌ ĚŝĂŐƌĂŵƉƌŽǀŝĚĞĚĚŽĞƐ ŶŽƚĐŽŵƉůĞƚĞůLJĞdžƉůĂŝŶ ǁŚLJDƌ͘<ŝŶĚůĞ͛Ɛ ƐƚĂƚĞŵĞŶƚŝƐĐŽƌƌĞĐƚ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ĐŽƌƌĞĐƚůLJƐƚĂƚĞƐ͕͞zĞƐ͕ Dƌ͘<ŝŶĚůĞ͛ƐƌĞĂƐŽŶŝŶŐŝƐ ĐŽƌƌĞĐƚ͘͟dŚĞƐƚĂŶĐĞŝƐ ƐƵƉƉŽƌƚĞĚǁŝƚŚĂǀĂůŝĚ ĞdžƉůĂŶĂƚŝŽŶƚŚĂƚ ĚĞŵŽŶƐƚƌĂƚĞƐĂƐŽůŝĚ ƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨƚŚĞ ĨĂĐƚƚŚĂƚƚŚĞŽƉƉŽƐŝƚĞŽĨ ƚŚĞŽƉƉŽƐŝƚĞŽĨĂ ŶƵŵďĞƌŝƐƚŚĞŶƵŵďĞƌ ŝƚƐĞůĨ͘
2 a ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ ^ƚƵĚĞŶƚƉƌŽǀŝĚĞƐĂŶ ŝŶĐŽƌƌĞĐƚƐƚĂƚĞŵĞŶƚďƵƚ ƉƌŽǀŝĚĞƐƐŽŵĞĞǀŝĚĞŶĐĞ ŽĨƵŶĚĞƌƐƚĂŶĚŝŶŐƚŚĞ ŽƌĚĞƌŝŶŐŽĨƌĂƚŝŽŶĂů ŶƵŵďĞƌƐŝŶƚŚĞǁƌŝƚƚĞŶ ǁŽƌŬ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ƉƌŽǀŝĚĞƐĂĐŽƌƌĞĐƚ ŽƌĚĞƌŝŶŐŽĨ െͳͲ ĂŶĚ
െʹͲ ďƵƚǁŝƚŚŽƵƚƵŶŝƚƐ ĂŶĚƌĞĨĞƌĞŶĐĞƚŽƚŚĞ ĐŽŶƚĞdžƚŽĨƚŚĞƐŝƚƵĂƚŝŽŶ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽƌƌĞĐƚ͘^ƚƵĚĞŶƚ ƉƌŽǀŝĚĞƐƚŚĞƐƚĂƚĞŵĞŶƚ͗
െͳͲԬ ŝƐǁĂƌŵĞƌƚŚĂŶ
െʹͲԬ Žƌ െʹͲԬ ŝƐ ĐŽůĚĞƌƚŚĂŶ െͳͲԬ ͘KZ ^ƚƵĚĞŶƚƉƌŽǀŝĚĞƐƐŽŵĞ ŽƚŚĞƌĞdžƉůĂŶĂƚŝŽŶƚŚĂƚ ĐŽŶƚĂŝŶƐĂǀĂůŝĚ ĐŽŵƉĂƌŝƐŽŶŽĨƚŚĞƚǁŽ ƚĞŵƉĞƌĂƚƵƌĞƐ͘
b ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ ^ƚƵĚĞŶƚĂƚƚĞŵƉƚƐƚŽ ǁƌŝƚĞĂŶŝŶĞƋƵĂůŝƚLJ ƐƚĂƚĞŵĞŶƚ͕ďƵƚƚŚĞ ƐƚĂƚĞŵĞŶƚŝƐŝŶĐŽƌƌĞĐƚ ĂŶĚĚŽĞƐŶŽƚŝŶĐůƵĚĞĂůů ƚŚƌĞĞŶƵŵďĞƌƐ͘ KZ dŚĞŝŶĐŽƌƌĞĐƚŝŶĞƋƵĂůŝƚLJ ƐƚĂƚĞŵĞŶƚůŝƐƚƐĂůůƚŚƌĞĞ ŶƵŵďĞƌƐďƵƚĚŽĞƐŶŽƚ ůŝƐƚ ͳͲ ĂƐƚŚĞŐƌĞĂƚĞƐƚ ǀĂůƵĞ͘ ^ƚƵĚĞŶƚǁƌŝƚĞƐĂŶ ŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚ ƚŚĂƚŽƌĚĞƌƐƚŚĞƚŚƌĞĞ ǀĂůƵĞƐǁŝƚŚ ͳͲ ĂƐƚŚĞ ŐƌĞĂƚĞƐƚŶƵŵďĞƌ͕ďƵƚƚŚĞ ƐƚĂƚĞŵĞŶƚĐŽŶƚĂŝŶƐĂŶ ĞƌƌŽƌ͘&ŽƌĞdžĂŵƉůĞ͕
െͳͲ ൏ െʹͲ ൏ ͳͲ ͘ dŚĞĐŽƌƌĞĐƚĂŶƐǁĞƌŝƐ ŐŝǀĞŶĂƐĂŶŝŶĞƋƵĂůŝƚLJ ƐƚĂƚĞŵĞŶƚŽĨ
െʹͲ ൏ െͳͲ ൏ ͳͲ Žƌ
ͳͲ െͳͲ െʹͲ ͕ĂŶĚ
ͳͲ ĚĞŐƌĞĞƐŝƐƚŚĞ ǁĂƌŵĞƐƚƚĞŵƉĞƌĂƚƵƌĞ͘
c ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ĞdžƉůĂŝŶƐŚŽǁƚŽƵƐĞĂ ŶƵŵďĞƌůŝŶĞƚŽĨŝŶĚƚŚĞ ŶƵŵďĞƌŽĨĚĞŐƌĞĞƐ ďĞůŽǁnjĞƌŽƚŚĞ ƚĞŵƉĞƌĂƚƵƌĞŝƐĂƚŶŽŽŶ͕ ďƵƚƚŚĞƵƐĞŽĨĂďƐŽůƵƚĞ ǀĂůƵĞŝƐŶŽƚŝŶĐůƵĚĞĚŝŶ ƚŚĞĞdžƉůĂŶĂƚŝŽŶ͕ŽƌŝƚŝƐ ƌĞĨĞƌĞŶĐĞĚŝŶĐŽƌƌĞĐƚůLJ͕ ƐƵĐŚĂƐ ȁ െ ͳͲȁ ൌ െͳͲ ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ŝŶĐůƵĚĞƐĂĐŽƌƌĞĐƚ ĞdžƉůĂŶĂƚŝŽŶĂŶĚ ƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨ ĂďƐŽůƵƚĞǀĂůƵĞ͗
ȁ െ ͳͲȁ ൌ ͳͲ ͕ďƵƚƚŚĞ ƚĞŵƉĞƌĂƚƵƌĞĂƚŶŽŽŶŝƐ ŝŶĐŽƌƌĞĐƚůLJƐƚĂƚĞĚĂƐ
െͳͲ ĚĞŐƌĞĞƐďĞůŽǁ Ͳ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ŝŶĐůƵĚĞƐĂĐŽƌƌĞĐƚ ĞdžƉůĂŶĂƚŝŽŶĂŶĚ ƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨ ĂďƐŽůƵƚĞǀĂůƵĞ͗
ȁ െ ͳͲȁ ൌ ͳͲ ͘ E dŚĞƚĞŵƉĞƌĂƚƵƌĞĂƚŶŽŽŶ ŝƐĐŽƌƌĞĐƚůLJƐƚĂƚĞĚĂƐ ͳͲ ĚĞŐƌĞĞƐďĞůŽǁ Ͳ͘
A STORY OF RATIOS
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©201 8Great Minds ®. eureka-math.org
6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
d ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐĂŶ ŝŶĐŽŵƉůĞƚĞƐƚĂƚĞŵĞŶƚ ƐƵƉƉŽƌƚĞĚďLJůŝƚƚůĞŽƌ ŶŽĞǀŝĚĞŶĐĞŽĨ ƌĞĂƐŽŶŝŶŐ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽƌƌĞĐƚďƵƚƐŚŽǁƐ ƐŽŵĞĞǀŝĚĞŶĐĞŽĨ ƌĞĂƐŽŶŝŶŐ͘,ŽǁĞǀĞƌ͕ƚŚĞ ĞdžƉůĂŶĂƚŝŽŶĚŽĞƐŶŽƚ ƐŚŽǁƚŚĂƚĂƐŶĞŐĂƚŝǀĞ ŶƵŵďĞƌƐŝŶĐƌĞĂƐĞ͕ƚŚĞŝƌ ĂďƐŽůƵƚĞǀĂůƵĞƐ ĚĞĐƌĞĂƐĞ͘^ƚƵĚĞŶƚ ĞdžƉůĂŶĂƚŝŽŶŵĂLJŽƌŵĂLJ ŶŽƚďĞƐƵƉƉŽƌƚĞĚǁŝƚŚ ĂŶĂĐĐƵƌĂƚĞǀĞƌƚŝĐĂů ŶƵŵďĞƌůŝŶĞŵŽĚĞů͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ŝŶĐůƵĚĞƐ͞zĞƐ͟ĂůŽŶŐǁŝƚŚ ĂǀĂůŝĚĞdžƉůĂŶĂƚŝŽŶƚŚĂƚ ŝŶĚŝĐĂƚĞƐƚŚĂƚĂƐ ŶĞŐĂƚŝǀĞŶƵŵďĞƌƐ ŝŶĐƌĞĂƐĞ͕ƚŚĞŝƌĂďƐŽůƵƚĞ ǀĂůƵĞƐĚĞĐƌĞĂƐĞ͘ƵƚĂ ǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞ ŵŽĚĞůŝƐŵŝƐƐŝŶŐŽƌ ĐŽŶƚĂŝŶƐĂŶĞƌƌŽƌ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ŝŶĐůƵĚĞƐ͞zĞƐ͟ĂůŽŶŐǁŝƚŚ ĂǀĂůŝĚĞdžƉůĂŶĂƚŝŽŶƚŚĂƚ ŝŶĚŝĐĂƚĞƐƚŚĂƚĂƐ ŶĞŐĂƚŝǀĞŶƵŵďĞƌƐ ŝŶĐƌĞĂƐĞ͕ƚŚĞŝƌĂďƐŽůƵƚĞ ǀĂůƵĞƐĚĞĐƌĞĂƐĞ͘dŚĞ ĂŶƐǁĞƌŝƐƐƵƉƉŽƌƚĞĚǁŝƚŚ ĂŶĂĐĐƵƌĂƚĞǀĞƌƚŝĐĂů ŶƵŵďĞƌůŝŶĞŵŽĚĞů ƌĞƉƌĞƐĞŶƚŝŶŐĂůůƚŚƌĞĞ ƚĞŵƉĞƌĂƚƵƌĞƐ͘
3 a ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ dŚĞƌĞŝƐůŝƚƚůĞŽƌŶŽ ĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐŝŶƚŚĞ ǁŽƌŬƐŚŽǁŶƚŽ ĚĞƚĞƌŵŝŶĞƚŚĞĐŽƌƌĞĐƚ ůŽĐĂƚŝŽŶĂŶĚǀĂůƵĞŽĨ ƉŽŝŶƚ ܣ͘ ^ƚƵĚĞŶƚŝŶĐŽƌƌĞĐƚůLJ ůŽĐĂƚĞƐƉŽŝŶƚ ܣ;ƚŚĞ ŽƉƉŽƐŝƚĞŽĨƉŽŝŶƚ ܲ ͿŽŶ ƚŚĞŶƵŵďĞƌůŝŶĞ͖ ŚŽǁĞǀĞƌ͕ƚŚĞůŽĐĂƚŝŽŶŽĨ ƉŽŝŶƚ ܣŝŶĚŝĐĂƚĞƐƐŽŵĞ ƵŶĚĞƌƐƚĂŶĚŝŶŐŽĨĂŶ ŝŶƚĞŐĞƌ͛ƐŽƉƉŽƐŝƚĞ͘ ^ƚƵĚĞŶƚůŽĐĂƚĞƐƚŚĞ ĐŽƌƌĞĐƚƉŽŝŶƚŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞĨŽƌƚŚĞ ŽƉƉŽƐŝƚĞ; ͳ͕ ʹ ͕ ͵ ͕Žƌ ͶͿ ďĂƐĞĚŽŶƚŚĞŝŶƚĞŐĞƌ ďĞƚǁĞĞŶ ͲĂŶĚ
െͷ ;െͳ ͕ െʹ ͕ െ͵ ͕ Žƌ െͶ Ϳ͘ ,ŽǁĞǀĞƌ͕ƚŚĞŽƉƉŽƐŝƚĞŝƐ ŶŽƚůĂďĞůĞĚŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞĂƐƉŽŝŶƚ ܣ͘ KZ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ ĂŶĚůĂďĞůƐƉŽŝŶƚ ܣ͕ ƚŚĞ ŽƉƉŽƐŝƚĞŽĨƉŽŝŶƚ ͕ܲ ďƵƚ ƉŽŝŶƚ ܲ ĚŽĞƐŶŽƚ ƌĞƉƌĞƐĞŶƚĂŶŝŶƚĞŐĞƌ ďĞƚǁĞĞŶ ͲĂŶĚ െͷ ͘ ĐŽƌƌĞĐƚĂŶƐǁĞƌŽĨƚŚĞ ŽƉƉŽƐŝƚĞ; ͳ͕ ʹ ͕ ͵ ͕ Žƌ Ͷ ͿŝƐ ŐŝǀĞŶďĂƐĞĚŽŶĐŽƌƌĞĐƚůLJ ĐŚŽŽƐŝŶŐĂŶŝŶƚĞŐĞƌ ďĞƚǁĞĞŶ ͲĂŶĚ
െͷ ;െͳ ͕ െʹ ͕ െ͵ ͕ Žƌ െͶ Ϳ ĂƐƉŽŝŶƚ ܲ͘ dŚĞŽƉƉŽƐŝƚĞ ŝƐĐŽƌƌĞĐƚůLJůŽĐĂƚĞĚŽŶ ƚŚĞŶƵŵďĞƌůŝŶĞĂŶĚ ůĂďĞůĞĚĂƐƉŽŝŶƚ ܣ͘
b ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ dŚĞƌĞŝƐůŝƚƚůĞŽƌŶŽ ĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐŝŶƚŚĞ ǁŽƌŬƐŚŽǁŶƚŽ ĚĞƚĞƌŵŝŶĞƚŚĞĐŽƌƌĞĐƚ ůŽĐĂƚŝŽŶĂŶĚǀĂůƵĞŽĨ ƉŽŝŶƚ ܤ͘ ^ƚƵĚĞŶƚŝŶĐŽƌƌĞĐƚůLJ ůŽĐĂƚĞƐƉŽŝŶƚ ܤŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞ͖ŚŽǁĞǀĞƌ͕ ƚŚĞůŽĐĂƚŝŽŶŽĨƉŽŝŶƚ
ܤ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞ ŝŶĚŝĐĂƚĞƐƚŚĂƚƉŽŝŶƚ ܤŝƐ ŶŽƚĞƋƵĂůƚŽƉŽŝŶƚ ܲ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ ĂƉŽŝŶƚŽŶƚŚĞŶƵŵďĞƌ ůŝŶĞƚŽƚŚĞůĞĨƚŽĨƉŽŝŶƚ ͖ܲŚŽǁĞǀĞƌ͕ƚŚĞƉŽŝŶƚŝƐ ŶŽƚůĂďĞůĞĚĂƐƉŽŝŶƚ ܤ͘ KZ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ ĂŶĚůĂďĞůƐƉŽŝŶƚ ܤĞǀĞŶ ƚŚŽƵŐŚƉŽŝŶƚ ܲ ĚŽĞƐŶŽƚ ƌĞƉƌĞƐĞŶƚĂŶŝŶƚĞŐĞƌ ďĞƚǁĞĞŶ Ͳ ĂŶĚ െͷ ͘ WŽŝŶƚ ܤŝƐĐŽƌƌĞĐƚůLJ ŐƌĂƉŚĞĚĂŶĚůĂďĞůĞĚŽŶ ƚŚĞŶƵŵďĞƌůŝŶĞ͘dŚĞ ƉŽŝŶƚŝƐƚŽƚŚĞůĞĨƚŽĨ ƉŽŝŶƚ ܲ ŽŶƚŚĞŶƵŵďĞƌ ůŝŶĞ͖ĨŽƌĞdžĂŵƉůĞ͕ŝĨƉŽŝŶƚ ܲ
ŝƐ െ͵ ͕ ƉŽŝŶƚ ܤĐŽƵůĚďĞ
െͷ ͘
A STORY OF RATIOS
194
©201 8Great Minds ®. eureka-math.org
6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
c ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ dŚĞƌĞŝƐůŝƚƚůĞŽƌŶŽ ĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐŝŶƚŚĞ ǁŽƌŬƐŚŽǁŶƚŽ ĚĞƚĞƌŵŝŶĞƚŚĞĐŽƌƌĞĐƚ ůŽĐĂƚŝŽŶĂŶĚǀĂůƵĞŽĨ ƉŽŝŶƚ ܥ͘ ^ƚƵĚĞŶƚŝŶĐŽƌƌĞĐƚůLJ ůŽĐĂƚĞƐƉŽŝŶƚ ܥŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞ͖ŚŽǁĞǀĞƌ͕ ƚŚĞůŽĐĂƚŝŽŶŽĨƉŽŝŶƚ ܥŽŶƚŚĞŶƵŵďĞƌůŝŶĞ ŝŶĚŝĐĂƚĞƐƚŚĂƚƉŽŝŶƚ ܥŝƐ ŶŽƚĞƋƵĂůƚŽƉŽŝŶƚ ܲ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ ĂƉŽŝŶƚŽŶƚŚĞŶƵŵďĞƌ ůŝŶĞƚŽƚŚĞƌŝŐŚƚŽĨƉŽŝŶƚ ͖ܲŚŽǁĞǀĞƌ͕ƚŚĞƉŽŝŶƚŝƐ ŶŽƚůĂďĞůĞĚĂƐƉŽŝŶƚ ܥ͘ KZ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ ĂŶĚůĂďĞůƐƉŽŝŶƚ ܥ͕ ĞǀĞŶ ƚŚŽƵŐŚƉŽŝŶƚ ܲ ĚŽĞƐŶŽƚ ƌĞƉƌĞƐĞŶƚĂŶŝŶƚĞŐĞƌ ďĞƚǁĞĞŶ ͲĂŶĚ െͷ ͘ WŽŝŶƚ ܥŝƐĐŽƌƌĞĐƚůLJ ŐƌĂƉŚĞĚĂŶĚůĂďĞůĞĚŽŶ ƚŚĞŶƵŵďĞƌůŝŶĞ͘dŚĞ ƉŽŝŶƚŝƐƚŽƚŚĞƌŝŐŚƚŽĨ ƉŽŝŶƚ ܲ ŽŶƚŚĞŶƵŵďĞƌ ůŝŶĞ͖ĨŽƌĞdžĂŵƉůĞ͕ŝĨƉŽŝŶƚ ܲ
ŝƐ െ͵ ͕ ƉŽŝŶƚ ܥĐŽƵůĚďĞ
Ͳ͘
d ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ dŚĞƌĞŝƐůŝƚƚůĞŽƌŶŽ ĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐŝŶƚŚĞ ǁŽƌŬƐŚŽǁŶƚŽ ĚĞƚĞƌŵŝŶĞƚŚĞĐŽƌƌĞĐƚ ůŽĐĂƚŝŽŶĂŶĚǀĂůƵĞŽĨ ƉŽŝŶƚ ܦ͘ ^ƚƵĚĞŶƚŝŶĐŽƌƌĞĐƚůLJ ůŽĐĂƚĞƐƉŽŝŶƚ ܦŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞ͖ŚŽǁĞǀĞƌ͕ ƚŚĞůŽĐĂƚŝŽŶŽĨƉŽŝŶƚ ܦŝƐ ƚŽƚŚĞƌŝŐŚƚŽĨƉŽŝŶƚ ܲ ĂůƚŚŽƵŐŚŶŽƚŚĂůĨǁĂLJ ďĞƚǁĞĞŶƚŚĞŝŶƚĞŐĞƌƚŽ ƚŚĞƌŝŐŚƚŽĨƉŽŝŶƚ ܲ ĂŶĚ ƉŽŝŶƚ ܲ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ ƚŚĞŶƵŵďĞƌƚŚĂƚŝƐ ŚĂůĨǁĂLJďĞƚǁĞĞŶƉŽŝŶƚ ܲ ĂŶĚƚŚĞŝŶƚĞŐĞƌƚŽƚŚĞ ƌŝŐŚƚŽĨƉŽŝŶƚ ͖ܲ ŚŽǁĞǀĞƌ͕ƚŚĞƉŽŝŶƚŝƐŶŽƚ ůĂďĞůĞĚĂƐƉŽŝŶƚ ܦ͘ KZ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJůŽĐĂƚĞƐ ĂŶĚůĂďĞůƐƉŽŝŶƚ ܦĞǀĞŶ ƚŚŽƵŐŚƉŽŝŶƚ ܲ ĚŽĞƐŶŽƚ ƌĞƉƌĞƐĞŶƚĂŶŝŶƚĞŐĞƌ ďĞƚǁĞĞŶ ͲĂŶĚ െͷ ͘ KZ ^ƚƵĚĞŶƚůŽĐĂƚĞƐĂŶĚ ůĂďĞůƐƉŽŝŶƚ ܦĂƐƚŚĞ ŶƵŵďĞƌƚŚĂƚŝƐŚĂůĨǁĂLJ ďĞƚǁĞĞŶƉŽŝŶƚ ܲ ĂŶĚƚŚĞ ŝŶƚĞŐĞƌƚŽƚŚĞ left ŽĨ ƉŽŝŶƚ ܲ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJŐƌĂƉŚƐ ĂŶĚůĂďĞůƐƉŽŝŶƚ ܦŽŶƚŚĞ ŶƵŵďĞƌůŝŶĞ͘dŚĞƉŽŝŶƚ ŝƐĞdžĂĐƚůLJŚĂůĨǁĂLJ ďĞƚǁĞĞŶƉŽŝŶƚ ܲ ĂŶĚƚŚĞ ŝŶƚĞŐĞƌƚŽƚŚĞƌŝŐŚƚŽĨ ƉŽŝŶƚ ܲ ŽŶƚŚĞŶƵŵďĞƌ ůŝŶĞ͖ĨŽƌĞdžĂŵƉůĞ͕ŝĨƉŽŝŶƚ ܲ
ŝƐ െ͵ ͕ ƉŽŝŶƚ ܦǁŽƵůĚ ďĞ െʹǤͷ ͘
4 a ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ ^ƚƵĚĞŶƚŵĂŬĞƐĂŶ ĞĨĨŽƌƚƚŽĂŶƐǁĞƌƚŚĞ ƋƵĞƐƚŝŽŶ͕ďƵƚŶŽŶĞŽĨ ƚŚĞƌĞƐƉŽŶƐĞƐĂƌĞ ĐŽƌƌĞĐƚ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ŝŶĐůƵĚĞƐ ͳ͕ ʹ ͕ ͵ ͕ŽƌĂƚ ŵŽƐƚ ͶůŽĐĂƚŝŽŶƐ ƌĞƉƌĞƐĞŶƚĞĚǁŝƚŚĐŽƌƌĞĐƚ ŝŶƚĞŐĞƌƐ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ŝŶĐůƵĚĞƐ ͷůŽĐĂƚŝŽŶƐ ƌĞƉƌĞƐĞŶƚĞĚǁŝƚŚĐŽƌƌĞĐƚ ŝŶƚĞŐĞƌƐ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ŝŶĐůƵĚĞƐĂůů ůŽĐĂƚŝŽŶƐ ƌĞƉƌĞƐĞŶƚĞĚǁŝƚŚƚŚĞ ĐŽƌƌĞĐƚŝŶƚĞŐĞƌƐ͗
ͷǡ͵Ͷ͵ ͕ െʹͳͲ ͕ ͳͶ ͕ ͷͲ ͕
Ͳ͕ െͶ͵ͷǤ
b ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ ^ƚƵĚĞŶƚŵĂŬĞƐĂŶ ĞĨĨŽƌƚƚŽĂŶƐǁĞƌƚŚĞ ƋƵĞƐƚŝŽŶ͕ďƵƚƚŚĞ ĞdžƉůĂŶĂƚŝŽŶĚŽĞƐŶŽƚ ƉƌŽǀŝĚĞĂŶLJĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐ͘ ^ƚƵĚĞŶƚĂƚƚĞŵƉƚƐƚŽ ƉƌŽǀŝĚĞĂŶĞdžƉůĂŶĂƚŝŽŶ͕ ĂŶĚƚŚĞĞdžƉůĂŶĂƚŝŽŶŝƐ ƐƵƉƉŽƌƚĞĚǁŝƚŚƐŽŵĞ ĞǀŝĚĞŶĐĞŽĨƌĞĂƐŽŶŝŶŐ͕ ďƵƚŝƚŝƐŝŶĐŽŵƉůĞƚĞ͘&Žƌ ĞdžĂŵƉůĞ͕͞WŽƐŝƚŝǀĞĂŶĚ ŶĞŐĂƚŝǀĞŶƵŵďĞƌƐƚĞůů :ƵůŝĂĂďŽƵƚƐĞĂůĞǀĞů͘͟ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ŝŶĐůƵĚĞƐĂŶĞdžƉůĂŶĂƚŝŽŶ ǁŝƚŚĞǀŝĚĞŶĐĞŽĨƐŽůŝĚ ƌĞĂƐŽŶŝŶŐ͕ďƵƚƚŚĞ ĞdžƉůĂŶĂƚŝŽŶůĂĐŬƐĚĞƚĂŝůƐ͘ &ŽƌĞdžĂŵƉůĞ͕͞WŽƐŝƚŝǀĞ ĂŶĚŶĞŐĂƚŝǀĞŶƵŵďĞƌƐ ƚĞůů:ƵůŝĂŚŽǁĨĂƌĨƌŽŵ ƐĞĂůĞǀĞůĂůŽĐĂƚŝŽŶŝƐ͘͟ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽƌƌĞĐƚ͘ŶĂĐĐƵƌĂƚĞĂŶĚ ĐŽŵƉůĞƚĞĞdžƉůĂŶĂƚŝŽŶŝƐ ŐŝǀĞŶ͕ƐƚĂƚŝŶŐƚŚĂƚĂ ƉŽƐŝƚŝǀĞŶƵŵďĞƌ ŝŶĚŝĐĂƚĞƐĂŶĞůĞǀĂƚŝŽŶ ĂďŽǀĞƐĞĂůĞǀĞů͕ĂŶĚ Ă ŶĞŐĂƚŝǀĞŶƵŵďĞƌ ŝŶĚŝĐĂƚĞƐĂŶĞůĞǀĂƚŝŽŶ ďĞůŽǁƐĞĂůĞǀĞů͘
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©201 8Great Minds ®. eureka-math.org
6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
c ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞƐĂƌĞ ŵŝƐƐŝŶŐ͕ĂŶĚͬŽƌƐƚƵĚĞŶƚ ŽŶůLJƉĂƌƚŝĂůůLJĨŝůůƐŝŶƚŚĞ ĐŚĂƌƚ͘ ^ƚƵĚĞŶƚĨŝůůƐŝŶƚŚĞĐŚĂƌƚ ĂƚƚĞŵƉƚŝŶŐƚŽŽƌĚĞƌƚŚĞ ĞůĞǀĂƚŝŽŶƐĂŶĚĨŝŶĚƚŚĞŝƌ ĂďƐŽůƵƚĞǀĂůƵĞƐ͕ďƵƚ ŵŽƌĞƚŚĂŶƚǁŽ ŶƵŵĞƌŝĐĂůĞƌƌŽƌƐĂƌĞ ŵĂĚĞ͘ KZ ^ƚƵĚĞŶƚĨŝůůƐŝŶƚŚĞĐŚĂƌƚ ĂŶĚĐŽƌƌĞĐƚůLJĨŝŶĚƐƚŚĞ ĂďƐŽůƵƚĞǀĂůƵĞŽĨĞĂĐŚ ŶƵŵďĞƌďƵƚĚŽĞƐŶŽƚ ŽƌĚĞƌƚŚĞĞůĞǀĂƚŝŽŶƐ ĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚŽƌ ĨƌŽŵŐƌĞĂƚĞƐƚƚŽůĞĂƐƚ͘ ^ƚƵĚĞŶƚĨŝůůƐŝŶƚŚĞĐŚĂƌƚ ŽƌĚĞƌŝŶŐƚŚĞĞůĞǀĂƚŝŽŶƐ ĂŶĚůŝƐƚŝŶŐƚŚĞŝƌĂďƐŽůƵƚĞ ǀĂůƵĞƐ͕ďƵƚŽŶĞŽƌƚǁŽ ŶƵŵďĞƌƐĂƌĞŝŶĐŽƌƌĞĐƚ͘ KZ ^ƚƵĚĞŶƚĨŝůůƐŝŶƚŚĞĐŚĂƌƚ ĂŶĚĐŽƌƌĞĐƚůLJĨŝŶĚƐƚŚĞ ĂďƐŽůƵƚĞǀĂůƵĞŽĨĞĂĐŚ ŶƵŵďĞƌ͖ŚŽǁĞǀĞƌ͕ƚŚĞ ĞůĞǀĂƚŝŽŶƐĂƌĞŽƌĚĞƌĞĚ ĨƌŽŵ greatest to least ƌĂƚŚĞƌƚŚĂŶůĞĂƐƚƚŽ ŐƌĞĂƚĞƐƚ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽƌƌĞĐƚĂŶĚĐŽŵƉůĞƚĞ͘ dŚĞĐŚĂƌƚŝƐĂĐĐƵƌĂƚĞůLJ ĐŽŵƉůĞƚĞĚǁŝƚŚ ĞůĞǀĂƚŝŽŶƐŽƌĚĞƌĞĚĨƌŽŵ ůĞĂƐƚƚŽŐƌĞĂƚĞƐƚĂŶĚ ƚŚĞŝƌƌĞƐƉĞĐƚŝǀĞĂďƐŽůƵƚĞ ǀĂůƵĞƐƌĞĐŽƌĚĞĚ͘
d ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞƐĂƌĞ ŵŝƐƐŝŶŐ͘ KZ ^ƚƵĚĞŶƚĐŝƌĐůĞƐƚŚĞƌŽǁ ǁŝƚŚnjĞƌŽƐŝŶƚŚĞĐŚĂƌƚ ƚŽƌĞƉƌĞƐĞŶƚƐĞĂůĞǀĞů ďƵƚƉƌŽǀŝĚĞƐŶŽĨƵƌƚŚĞƌ ĞdžƉůĂŶĂƚŝŽŶ͘ ^ƚƵĚĞŶƚĐŝƌĐůĞƐƚŚĞƌŽǁ ǁŝƚŚnjĞƌŽƐŝŶƚŚĞĐŚĂƌƚƚŽ ƌĞƉƌĞƐĞŶƚƐĞĂůĞǀĞůĂŶĚ ƉƌŽǀŝĚĞƐĂŶĞdžƉůĂŶĂƚŝŽŶ ƚŚĂƚĐŽŶƚĂŝŶƐƐŽŵĞ ĞǀŝĚĞŶĐĞŽĨƌĞĂƐŽŶŝŶŐ ĂůƚŚŽƵŐŚƚŚĞĞdžƉůĂŶĂƚŝŽŶ ŵĂLJďĞŝŶĐŽŵƉůĞƚĞŽƌ ĐŽŶƚĂŝŶŝŶĂĐĐƵƌĂƚĞ ƐƚĂƚĞŵĞŶƚƐ͘ ^ƚƵĚĞŶƚĐŝƌĐůĞƐƚŚĞƌŽǁ ǁŝƚŚnjĞƌŽƐŝŶƚŚĞĐŚĂƌƚƚŽ ƌĞƉƌĞƐĞŶƚƐĞĂůĞǀĞůE ƉƌŽǀŝĚĞƐĂǀĂůŝĚ ĞdžƉůĂŶĂƚŝŽŶ͕ďƵƚŝƚůĂĐŬƐ ĚĞƚĂŝůƐ͘/ƚŝƐƐƵƉƉŽƌƚĞĚ ǁŝƚŚƐŽŵĞĞǀŝĚĞŶĐĞŽĨ ƌĞĂƐŽŶŝŶŐƚŚŽƵŐŚŝƚŵĂLJ ďĞŐĞŶĞƌĂůŝŶŶĂƚƵƌĞ͘&Žƌ ĞdžĂŵƉůĞ͕͞ ůĞǀĂƚŝŽŶƐ ďĞůŽǁƐĞĂůĞǀĞůǁŝůůŚĂǀĞ ĚŝĨĨĞƌĞŶƚĂďƐŽůƵƚĞ ǀĂůƵĞƐ͘͟ ^ƚƵĚĞŶƚĐŝƌĐůĞƐƚŚĞƌŽǁ ǁŝƚŚnjĞƌŽƐŝŶƚŚĞĐŚĂƌƚƚŽ ƌĞƉƌĞƐĞŶƚƐĞĂůĞǀĞů͕E ĂŶĂĐĐƵƌĂƚĞĞdžƉůĂŶĂƚŝŽŶ ŝƐŐŝǀĞŶĂŶĚŝƐƐƵƉƉŽƌƚĞĚ ǁŝƚŚƐƵďƐƚĂŶƚŝĂů ĞǀŝĚĞŶĐĞƚŚĂƚƐĞĂůĞǀĞůƐ ďĞůŽǁnjĞƌŽŚĂǀĞ ŽƉƉŽƐŝƚĞĂďƐŽůƵƚĞǀĂůƵĞƐ ĂƐƚŚĞŝƌĞůĞǀĂƚŝŽŶƐ͕ĂŶĚ ƐĞĂůĞǀĞůƐĂďŽǀĞnjĞƌŽ ŚĂǀĞƚŚĞƐĂŵĞĂďƐŽůƵƚĞ ǀĂůƵĞƐĂƐƚŚĞŝƌ ĞůĞǀĂƚŝŽŶƐ͘
5 a ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ ůů ͶƉŽŝŶƚƐĂƌĞ ŝŶĂĐĐƵƌĂƚĞůLJůŽĐĂƚĞĚ͘ ^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐĂŶĚůĂďĞůƐ ͳʹʹƉŽŝŶƚƐ͘ ^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐĂŶĚůĂďĞůƐ ͵ƉŽŝŶƚƐ͘ ^ƚƵĚĞŶƚĂĐĐƵƌĂƚĞůLJ ůŽĐĂƚĞƐĂŶĚůĂďĞůƐĂůů ͶƉŽŝŶƚƐ͘
b ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽƌƌĞĐƚ͕EŶĞŝƚŚĞƌ ĐŽŽƌĚŝŶĂƚĞŝƐƐƚĂƚĞĚĂƐĂ ŶĞŐĂƚŝǀĞŶƵŵďĞƌ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽƌƌĞĐƚ͕ďƵƚŽŶĞŽĨƚŚĞ ĐŽŽƌĚŝŶĂƚĞƐŝƐĐŽƌƌĞĐƚ͘ &ŽƌĞdžĂŵƉůĞ͕ ሺെǡ ͵ሻ ŝƐ ƚŚĞƌĞƐƉŽŶƐĞ͕ĂŶĚƚŚĞ
ݔͲĐŽŽƌĚŝŶĂƚĞŝƐĐŽƌƌĞĐƚ͘ ^ƚƵĚĞŶƚƉƌŽǀŝĚĞƐĂ ĐŽƌƌĞĐƚĂŶƐǁĞƌ ĞdžƉƌĞƐƐĞĚĂƐĂŶŽƌĚĞƌĞĚ ƉĂŝƌǁŚĞƌĞďŽƚŚƚŚĞ ݔͲ ĂŶĚ ݕͲĐŽŽƌĚŝŶĂƚĞƐĂƌĞ ŶĞŐĂƚŝǀĞŶƵŵďĞƌƐ͘&Žƌ ĞdžĂŵƉůĞ͕ ሺെǡ െ͵ሻ ͘
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©201 8Great Minds ®. eureka-math.org
6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
c ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ ŶŝŶĐŽƌƌĞĐƚĂŶƐǁĞƌŝƐ ŐŝǀĞŶǁŝƚŚůŝƚƚůĞŽƌŶŽ ĂƉƉůŝĐĂƚŝŽŶŽĨ ŵĂƚŚĞŵĂƚŝĐƐƵƐĞĚƚŽ ƐŽůǀĞƚŚĞƉƌŽďůĞŵ͘ ^ƚƵĚĞŶƚƉƌŽǀŝĚĞƐĂŶ ŝŶĐŽƌƌĞĐƚĂŶƐǁĞƌĨŽƌƚŚĞ ĚŝƐƚĂŶĐĞďƵƚ ĚĞŵŽŶƐƚƌĂƚĞƐƐŽŵĞ ĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐŚŽǁƚŽ ĨŝŶĚƚŚĞĚŝƐƚĂŶĐĞ ďĞƚǁĞĞŶƚŚĞƉŽŝŶƚƐ ĂůƚŚŽƵŐŚĂƐŝŐŶŝĨŝĐĂŶƚ ĞƌƌŽƌǁĂƐŵĂĚĞ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞ ĐŽƌƌĞĐƚůLJƐƚĂƚĞƐĂ ĚŝƐƚĂŶĐĞŽĨ ͳͻ ƵŶŝƚƐ͕ďƵƚ ƚŚĞǁŽƌŬƐŚŽǁŶĚŽĞƐ ŶŽƚĂĚĞƋƵĂƚĞůLJƐƵƉƉŽƌƚ ƚŚĞĂŶƐǁĞƌ͘ KZ ŶŝŶĐŽƌƌĞĐƚĂŶƐǁĞƌĨŽƌ ƚŚĞĚŝƐƚĂŶĐĞŝƐŐŝǀĞŶ͕ďƵƚ ƚŚĞǁŽƌŬƐŚŽǁŶ ĚĞŵŽŶƐƚƌĂƚĞƐĂĐŽƌƌĞĐƚ ƉƌŽĐĞƐƐǁŝƚŚĂŵŝŶŽƌ ĞƌƌŽƌ͘&ŽƌĞdžĂŵƉůĞ͕ ƐƚƵĚĞŶƚŵĂĚĞĂŶĞƌƌŽƌŝŶ ƚŚĞĂĚĚŝƚŝŽŶŽƌ ŵŝƐĐŽƵŶƚĞĚǁŚĞŶƵƐŝŶŐ ƚŚĞŶƵŵďĞƌůŝŶĞ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽŵƉůĞƚĞĂŶĚĐŽƌƌĞĐƚ͘ dŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶ ƚŚĞƉŽŝŶƚƐŝƐĨŽƵŶĚƚŽďĞ
ͳͻ ƵŶŝƚƐ͕ĂŶĚĂŶ ĂĐĐƵƌĂƚĞĂŶĚĐŽŵƉůĞƚĞ ĞdžƉůĂŶĂƚŝŽŶ͕ƉƌŽĐĞƐƐ͕ ĂŶĚͬŽƌĚŝĂŐƌĂŵŝƐ ƉƌŽǀŝĚĞĚƚŽƐƵƉƉŽƌƚƚŚĞ ĂŶƐǁĞƌ͘
d ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽƌƌĞĐƚ͕ĂŶĚŶĞŝƚŚĞƌ ĐŽŽƌĚŝŶĂƚĞŝƐƐƚĂƚĞĚ ĐŽƌƌĞĐƚůLJ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŝŶĐŽƌƌĞĐƚ͕ďƵƚŽŶĞŽĨƚŚĞ ĐŽŽƌĚŝŶĂƚĞƐŝƐĐŽƌƌĞĐƚ͘ &ŽƌĞdžĂŵƉůĞ͕ ሺͷǡ െʹሻ ŝƐ ƚŚĞƌĞƐƉŽŶƐĞ͕ĂŶĚƚŚĞ
ݔͲĐŽŽƌĚŝŶĂƚĞŝƐĐŽƌƌĞĐƚ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ĐŽƌƌĞĐƚĂŶĚĐŽŵƉůĞƚĞ͘ WŽŝŶƚ ܧ͛ ƐĐŽŽƌĚŝŶĂƚĞƐĂƌĞ
ሺͷǡ ʹሻ ͘
e ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ŵŝƐƐŝŶŐ͘ KZ ^ƚƵĚĞŶƚŵĂŬĞƐĂŶ ĞĨĨŽƌƚƚŽĂŶƐǁĞƌƚŚĞ ƋƵĞƐƚŝŽŶ͕ďƵƚƚŚĞ ĂŶƐǁĞƌĂŶĚͬŽƌ ĞdžƉůĂŶĂƚŝŽŶĚŽĞƐŶŽƚ ƉƌŽǀŝĚĞĂŶLJĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐ͘ ^ƚƵĚĞŶƚĚŽĞƐŶŽƚĂƌƌŝǀĞ ĂƚƚŚĞĐŽƌƌĞĐƚ ĐŽŽƌĚŝŶĂƚĞƐĨŽƌƉŽŝŶƚ ܨĂŶĚŵĂLJŽƌŵĂLJŶŽƚ ĂƌƌŝǀĞĂƚƚŚĞĐŽƌƌĞĐƚ ĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƉŽŝŶƚƐ
ܧ ĂŶĚ ܨ͘ ƵƚƚŚĞƌĞŝƐ ƐŽŵĞĞǀŝĚĞŶĐĞŽĨ ƵŶĚĞƌƐƚĂŶĚŝŶŐŚŽǁƚŽ ůŽĐĂƚĞĂƉŽŝŶƚƌĞůĂƚĞĚƚŽ ƉŽŝŶƚ ܧĂŶĚͬŽƌŚŽǁƚŽ ĨŝŶĚƚŚĞĚŝƐƚĂŶĐĞ ďĞƚǁĞĞŶƚŚĞƚǁŽƉŽŝŶƚƐ͘ ^ƚƵĚĞŶƚƌĞƐƉŽŶƐĞŝƐ ƉĂƌƚŝĂůůLJĐŽƌƌĞĐƚ͘WŽŝŶƚ ܨŝƐĐŽƌƌĞĐƚůLJůŽĐĂƚĞĚĂŶĚ ůĂďĞůĞĚ͕ĂŶĚŝƚƐ ĐŽŽƌĚŝŶĂƚĞƐĂƌĞŐŝǀĞŶĂƐ
ሺͷǡ െʹሻ ͕ďƵƚƐƚƵĚĞŶƚŝƐ ƵŶĂďůĞƚŽĂƌƌŝǀĞĂƚƚŚĞ ĐŽƌƌĞĐƚĚŝƐƚĂŶĐĞ ďĞƚǁĞĞŶƉŽŝŶƚƐ ܧĂŶĚ ܨŽƌŝƐƵŶĂďůĞƚŽĞdžƉůĂŝŶ ƚŚĞƉƌŽĐĞƐƐĂĐĐƵƌĂƚĞůLJ͘ ^ƚƵĚĞŶƚĐŽƌƌĞĐƚůLJ ĐŽŵƉůĞƚĞƐĂůů ͵ƚĂƐŬƐ͘ WŽŝŶƚ ܨŝƐĐŽƌƌĞĐƚůLJ ůŽĐĂƚĞĚĂŶĚůĂďĞůĞĚŽŶ ƚŚĞĐŽŽƌĚŝŶĂƚĞŐƌŝĚ͕ĂŶĚ ŝƚƐĐŽŽƌĚŝŶĂƚĞƐĂƌĞŐŝǀĞŶ ĂƐ ሺͷǡ െʹሻ ͘dŚĞĚŝƐƚĂŶĐĞ ďĞƚǁĞĞŶƉŽŝŶƚƐ ܧ ĂŶĚ ܨŝƐ ͶƵŶŝƚƐĂŶĚŝƐ ƐƵƉƉŽƌƚĞĚǁŝƚŚ ƐƵďƐƚĂŶƚŝĂůĞǀŝĚĞŶĐĞŽĨ ƌĞĂƐŽŶŝŶŐ͘
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6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ
EĂŵĞ ĂƚĞ ϭ͘ Dƌ͘<ŝŶĚůĞŝŶǀĞƐƚĞĚƐŽŵĞŵŽŶĞLJŝŶƚŚĞƐƚŽĐŬŵĂƌŬĞƚ͘,ĞƚƌĂĐŬƐŚŝƐŐĂŝŶƐĂŶĚůŽƐƐĞƐƵƐŝŶŐĂĐŽŵƉƵƚĞƌ ƉƌŽŐƌĂŵ͘Dƌ͘<ŝŶĚůĞƌĞĐĞŝǀĞƐĂĚĂŝůLJĞŵĂŝůƚŚĂƚƵƉĚĂƚĞƐŚŝŵŽŶĂůůŚŝƐƚƌĂŶƐĂĐƚŝŽŶƐĨƌŽŵƚŚĞƉƌĞǀŝŽƵƐĚĂLJ͘ dŚŝƐŵŽƌŶŝŶŐ͕ŚŝƐĞŵĂŝůƌĞĂĚĂƐĨŽůůŽǁƐ͗
Good morning, Mr. Kindle, Yesterday’s investment activity included a loss of ̈́ ͺͲͲ , a gain of ̈́ ͻͲ , and another gain of ̈́ ʹ͵Ͳ . Log in now to see your current balance.
Ă͘ tƌŝƚĞĂŶŝŶƚĞŐĞƌƚŽƌĞƉƌĞƐĞŶƚĞĂĐŚŐĂŝŶĂŶĚůŽƐƐ͘
Description Integer Representation
ŽƐƐŽĨ ̈́ ͺͲͲ
-800
'ĂŝŶŽĨ ̈́ ͻͲ
960
'ĂŝŶŽĨ ̈́ ʹ͵Ͳ
230
ď͘ Dƌ͘<ŝŶĚůĞŶŽƚŝĐĞĚƚŚĂƚĂŶĞƌƌŽƌŚĂĚďĞĞŶŵĂĚĞŽŶŚŝƐĂĐĐŽƵŶƚ͘dŚĞ͞ůŽƐƐŽĨ ̈́ ͺͲͲ ͟ ƐŚŽƵůĚŚĂǀĞ ďĞĞŶĂ͞ŐĂŝŶŽĨ ̈́ ͺͲͲǤ ͟ >ŽĐĂƚĞĂŶĚůĂďĞůďŽƚŚƉŽŝŶƚƐƚŚĂƚƌĞƉƌĞƐĞŶƚ͞ĂůŽƐƐŽĨ ̈́ ͺͲͲ ͟ ĂŶĚ͞ĂŐĂŝŶŽĨ ̈́ͺͲͲ ͟ ŽŶƚŚĞŶƵŵďĞƌůŝŶĞďĞůŽǁ͘ĞƐĐƌŝďĞƚŚĞƌĞůĂƚŝŽŶƐŚŝƉŽĨƚŚĞƐĞƚǁŽŶƵŵďĞƌƐǁŚĞŶnjĞƌŽ ƌĞƉƌĞƐĞŶƚƐŶŽĐŚĂŶŐĞ;ŐĂŝŶŽƌůŽƐƐͿ͘
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Đ͘ Dƌ͘<ŝŶĚůĞǁĂŶƚĞĚƚŽĐŽƌƌĞĐƚƚŚĞĞƌƌŽƌ͕ƐŽŚĞĞŶƚĞƌĞĚ െሺെ̈́ͺͲͲሻ ŝŶƚŽƚŚĞƉƌŽŐƌĂŵ͘,ĞŵĂĚĞĂŶŽƚĞ ƚŚĂƚƌĞĂĚ͕͞dŚĞŽƉƉŽƐŝƚĞŽĨƚŚĞŽƉƉŽƐŝƚĞŽĨ ̈́ ͺͲͲ ŝƐ ̈́ͺͲͲ ͘͟ /ƐŚŝƐƌĞĂƐŽŶŝŶŐĐŽƌƌĞĐƚ͍ džƉůĂŝŶ͘ Ϯ͘ ƚϲ͗ϬϬĂ͘ŵ͕͘ƵĨĨĂůŽ͕Ez͕ŚĂĚĂƚĞŵƉĞƌĂƚƵƌĞŽĨ ͳͲԬ ͘ƚŶŽŽŶ͕ƚŚĞƚĞŵƉĞƌĂƚƵƌĞǁĂƐ െͳͲԬ ͕ĂŶĚĂƚ ŵŝĚŶŝŐŚƚ͕ŝƚǁĂƐ െʹͲԬ ͘ Ă͘ tƌŝƚĞĂƐƚĂƚĞŵĞŶƚĐŽŵƉĂƌŝŶŐ െͳͲԬ ĂŶĚ െʹͲԬǤ ď͘ tƌŝƚĞĂŶŝŶĞƋƵĂůŝƚLJƐƚĂƚĞŵĞŶƚƚŚĂƚƐŚŽǁƐƚŚĞƌĞůĂƚŝŽŶƐŚŝƉďĞƚǁĞĞŶƚŚĞƚŚƌĞĞƌĞĐŽƌĚĞĚ ƚĞŵƉĞƌĂƚƵƌĞƐ͘tŚŝĐŚƚĞŵƉĞƌĂƚƵƌĞŝƐƚŚĞǁĂƌŵĞƐƚ͍
Temperature in degrees Fahrenheit
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Đ͘ džƉůĂŝŶŚŽǁƚŽƵƐĞĂďƐŽůƵƚĞǀĂůƵĞƚŽĨŝŶĚƚŚĞŶƵŵďĞƌŽĨĚĞŐƌĞĞƐďĞůŽǁnjĞƌŽƚŚĞƚĞŵƉĞƌĂƚƵƌĞǁĂƐĂƚ ŶŽŽŶ͘ Ě͘ /ŶWĞĞŬƐŬŝůů͕Ez͕ƚŚĞƚĞŵƉĞƌĂƚƵƌĞĂƚϲ͗ϬϬĂ͘ŵ͘ǁĂƐ െͳʹԬ ͘ƚŶŽŽŶ͕ƚŚĞƚĞŵƉĞƌĂƚƵƌĞǁĂƐƚŚĞĞdžĂĐƚ ŽƉƉŽƐŝƚĞŽĨƵĨĨĂůŽ͛ƐƚĞŵƉĞƌĂƚƵƌĞĂƚϲ͗ϬϬĂ͘ŵ͘ƚŵŝĚŶŝŐŚƚ͕ĂŵĞƚĞŽƌŽůŽŐŝƐƚƌĞĐŽƌĚĞĚƚŚĞ ƚĞŵƉĞƌĂƚƵƌĞĂƐ െԬ ŝŶWĞĞŬƐŬŝůů͘,ĞĐŽŶĐůƵĚĞĚƚŚĂƚ͞&ŽƌƚĞŵƉĞƌĂƚƵƌĞƐďĞůŽǁnjĞƌŽ͕ĂƐƚŚĞ ƚĞŵƉĞƌĂƚƵƌĞŝŶĐƌĞĂƐĞƐ͕ƚŚĞĂďƐŽůƵƚĞǀĂůƵĞŽĨƚŚĞƚĞŵƉĞƌĂƚƵƌĞĚĞĐƌĞĂƐĞƐ͘͟/ƐŚŝƐĐŽŶĐůƵƐŝŽŶǀĂůŝĚ͍ džƉůĂŝŶĂŶĚƵƐĞĂǀĞƌƚŝĐĂůŶƵŵďĞƌůŝŶĞƚŽƐƵƉƉŽƌƚLJŽƵƌĂŶƐǁĞƌ͘
ϯ͘ ŚŽŽƐĞĂŶŝŶƚĞŐĞƌďĞƚǁĞĞŶ ͲĂŶĚ െͷ ŽŶĂŶƵŵďĞƌůŝŶĞ͕ĂŶĚůĂďĞůƚŚĞƉŽŝŶƚ ܲ͘ >ŽĐĂƚĞĂŶĚůĂďĞůĞĂĐŚŽĨ ƚŚĞĨŽůůŽǁŝŶŐƉŽŝŶƚƐĂŶĚƚŚĞŝƌǀĂůƵĞƐŽŶƚŚĞŶƵŵďĞƌůŝŶĞ͘ Ă͘ >ĂďĞůƉŽŝŶƚ ܣ͗ ƚŚĞŽƉƉŽƐŝƚĞ ŽĨƉŽŝŶƚ ܲ .ď͘ >ĂďĞůƉŽŝŶƚ ܤ͗ ĂŶƵŵďĞƌůĞƐƐƚŚĂŶƉŽŝŶƚ ܲ .Đ͘ >ĂďĞůƉŽŝŶƚ ܥ͗ ĂŶƵŵďĞƌŐƌĞĂƚĞƌƚŚĂŶƉŽŝŶƚ ܲ .Ě͘ >ĂďĞůƉŽŝŶƚ ܦ͗ ĂŶƵŵďĞƌŚĂůĨǁĂLJďĞƚǁĞĞŶ PĂŶĚƚŚĞŝŶƚĞŐĞƌƚŽƚŚĞƌŝŐŚƚŽĨƉŽŝŶƚ ܲ .
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ϰ͘ :ƵůŝĂŝƐůĞĂƌŶŝŶŐĂďŽƵƚĞůĞǀĂƚŝŽŶŝŶŵĂƚŚĐůĂƐƐ͘^ŚĞĚĞĐŝĚĞĚƚŽƌĞƐĞĂƌĐŚƐŽŵĞĨĂĐƚƐĂďŽƵƚEĞǁzŽƌŬ^ƚĂƚĞ ƚŽďĞƚƚĞƌƵŶĚĞƌƐƚĂŶĚƚŚĞĐŽŶĐĞƉƚ͘,ĞƌĞĂƌĞƐŽŵĞĨĂĐƚƐƚŚĂƚƐŚĞĨŽƵŶĚ͘
Mount Marcy is the highest point in New York State. It is ͷǡ͵Ͷ͵ feet above sea level.
Lake Erie is ʹͳͲ feet below sea level.
The elevation of Niagara Falls, NY, is ͳͶ feet above sea level.
The lobby of the Empire State Building is ͷͲ feet above sea level.
New York State borders the Atlantic Coast, which is at sea level.
The lowest point of Cayuga Lake is Ͷ͵ͷ feet below sea level.
Ă͘ tƌŝƚĞĂŶŝŶƚĞŐĞƌƚŚĂƚƌĞƉƌĞƐĞŶƚƐĞĂĐŚůŽĐĂƚŝŽŶŝŶƌĞůĂƚŝŽŶƐŚŝƉƚŽƐĞĂůĞǀĞů͘ ď͘ džƉůĂŝŶǁŚĂƚŶĞŐĂƚŝǀĞĂŶĚƉŽƐŝƚŝǀĞŶƵŵďĞƌƐƚĞůů:ƵůŝĂĂďŽƵƚĞůĞǀĂƚŝŽŶ͘
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Đ͘ KƌĚĞƌƚŚĞĞůĞǀĂƚŝŽŶƐĨƌŽŵůĞĂƐƚƚŽŐƌĞĂƚĞƐƚ͕ĂŶĚƚŚĞŶƐƚĂƚĞƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘hƐĞƚŚĞĐŚĂƌƚ ďĞůŽǁƚŽƌĞĐŽƌĚLJŽƵƌǁŽƌŬ͘
Ě͘ ŝƌĐůĞƚŚĞƌŽǁŝŶƚŚĞƚĂďůĞƚŚĂƚƌĞƉƌĞƐĞŶƚƐƐĞĂůĞǀĞů͘ĞƐĐƌŝďĞŚŽǁƚŚĞŽƌĚĞƌŽĨƚŚĞĞůĞǀĂƚŝŽŶƐďĞůŽǁ ƐĞĂůĞǀĞůĐŽŵƉĂƌĞƐƚŽƚŚĞŽƌĚĞƌŽĨƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘ĞƐĐƌŝďĞŚŽǁƚŚĞŽƌĚĞƌŽĨƚŚĞĞůĞǀĂƚŝŽŶƐ ĂďŽǀĞƐĞĂůĞǀĞůĐŽŵƉĂƌĞƐƚŽƚŚĞŽƌĚĞƌŽĨƚŚĞŝƌĂďƐŽůƵƚĞǀĂůƵĞƐ͘
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6ͻ3End-of-Module Assessment Task Module 3: ZĂƚŝŽŶĂůEƵŵďĞƌƐ ϱ͘ &ŽƌĐĞŶƚƵƌŝĞƐ͕ĂŵLJƐƚĞƌŝŽƵƐƐĞĂƐĞƌƉĞŶƚŚĂƐďĞĞŶƌƵŵŽƌĞĚƚŽůŝǀĞĂƚƚŚĞďŽƚƚŽŵŽĨDLJƐƚĞƌŝŽƵƐ>ĂŬĞ͘ ƚĞĂŵŽĨŚŝƐƚŽƌŝĂŶƐƵƐĞĚĂĐŽŵƉƵƚĞƌƉƌŽŐƌĂŵƚŽƉůŽƚƚŚĞůĂƐƚĨŝǀĞƉŽƐŝƚŝŽŶƐŽĨƚŚĞƐŝŐŚƚŝŶŐƐ͘ Ă͘ >ŽĐĂƚĞĂŶĚůĂďĞůƚŚĞůŽĐĂƚŝŽŶƐŽĨƚŚĞůĂƐƚĨŽƵƌƐŝŐŚƚŝŶŐƐ͗ ܣ ቀെͻ ͳʹ ǡ Ͳቁ ͕ ܤሺെ͵ǡ െͶǤͷሻ ͕ ܥሺͻǡ ʹሻ ͕
ĂŶĚ ܦሺͺǡ െʹǤͷሻ ͘ď͘ KǀĞƌƚŝŵĞ͕ŵŽƐƚŽĨƚŚĞƐŝŐŚƚŝŶŐƐŽĐĐƵƌƌĞĚŝŶYƵĂĚƌĂŶƚ///͘tƌŝƚĞƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨĂƉŽŝŶƚƚŚĂƚůŝĞƐ ŝŶYƵĂĚƌĂŶƚ///͘ Đ͘ tŚĂƚŝƐƚŚĞĚŝƐƚĂŶĐĞďĞƚǁĞĞŶƉŽŝŶƚ ܣĂŶĚƚŚĞƉŽŝŶƚ ቀͻ ͳʹ ǡ Ͳቁ ͍ ^ŚŽǁLJŽƵƌǁŽƌŬƚŽƐƵƉƉŽƌƚLJŽƵƌ ĂŶƐǁĞƌ͘ Ě͘ tŚĂƚĂƌĞƚŚĞĐŽŽƌĚŝŶĂƚĞƐŽĨƉŽŝŶƚ ܧŽŶƚŚĞĐŽŽƌĚŝŶĂƚĞƉůĂŶĞ͍ Ğ͘ WŽŝŶƚ ܨŝƐƌĞůĂƚĞĚƚŽƉŽŝŶƚ ܧ͘ /ƚƐ ݔͲĐŽŽƌĚŝŶĂƚĞŝƐƚŚĞƐĂŵĞĂƐƉŽŝŶƚ ܧ͛ Ɛ͕ďƵƚŝƚƐ ݕͲĐŽŽƌĚŝŶĂƚĞŝƐƚŚĞ ŽƉƉŽƐŝƚĞŽĨƉŽŝŶƚ ܧ͛ Ɛ͘>ŽĐĂƚĞĂŶĚůĂďĞůƉŽŝŶƚ ܨ͘ tŚĂƚĂƌĞƚŚĞĐŽŽƌĚŝŶĂƚĞƐ͍,ŽǁĨĂƌĂƉĂƌƚĂƌĞƉŽŝŶƚƐ
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8846 | https://mathoverflow.net/questions/48251/methods-for-solving-pells-equation | nt.number theory - Methods for solving Pell's equation? - MathOverflow
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Methods for solving Pell's equation?
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It is known that the minimum solution of Pell's equation x 2−d y 2=±1 x 2−d y 2=±1 can be found from the continued fraction expansion of d−−√d. Are there other methods for finding the minimum (or any other) solutions?
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edited Dec 6, 2010 at 12:24
Wadim Zudilin
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1 I've done some editing here, I hope I have captured the spirit. Gerry Myerson –Gerry Myerson 2010-12-04 05:40:01 +00:00 Commented Dec 4, 2010 at 5:40
I occasionally have to solve a Pell equation, and my method is almost invariably this: I fire up pari and just loop over the positive integers searching for some! The computer almost always finds two or three almost instantly, from which I can read off the fundamental unit and the degree 2 recurrence relation generating the solutions!Kevin Buzzard –Kevin Buzzard 2010-12-04 08:17:01 +00:00 Commented Dec 4, 2010 at 8:17
7 I guess my method would be called "Brute force and Ignorance". Given the speed of modern computers, there's a lot to be said for it.Kevin Buzzard –Kevin Buzzard 2010-12-04 09:51:07 +00:00 Commented Dec 4, 2010 at 9:51
1 Step 1. Build a scalable fault-tolerant quantum computer. Step 2. Use the algorithm of Hallgren: Polynomial-time quantum algorithms for Pell's equation and the principal ideal problem, STOC, 2002. Step 3: profit!Steve Huntsman –Steve Huntsman 2010-12-04 12:41:12 +00:00 Commented Dec 4, 2010 at 12:41
2 Just ran across this thread. I confess to using Kevin's usual method too on occasion, but if you're already firing up pari you should use its built-in function quadunit(x) (taking x=4d). Try telling gp for(d=1,100,if(!issquare(d),print([d,(quadunit(4d))]))) Noam D. Elkies –Noam D. Elkies 2011-05-28 04:30:17 +00:00 Commented May 28, 2011 at 4:30
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The basic and classical methods, apart from brute force, are
continued fraction expansions (regular, nearest integer, etc.) or, equivalently, some form of reduction theory for indefinite binary quadratic forms;
computing many elements of small norm in a quadratic number field, which often is a lot more effective; the technique used here is also used for factoring integers.
For a detailed algorithmic description see Jacobson & Williams (Solving the Pell Equation) or Buchmann & Vollmer (Binary Quadratic Forms).
In addition, you can compute a power of the fundamental unit from the class number formulas, which essentially consists in taking norms of suitable cyclotomic units. Kronecker has shown how to solve the Pell equation using elliptic and modular functions, and Girstmair (Kronecker's solution of the Pell equation on a computer {Kroneckers Lösung der Pellschen Gleichung auf dem Computer], Math. Semesterber. 53, 45-64 (2006)) has shown that it can be made to work in practice.
You can also imitate the theory of descent on elliptic curves; I have sketched connections with classical tricks in some preprints on higher descent on Pell conics.
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edited Dec 4, 2010 at 11:57
answered Dec 4, 2010 at 11:25
Franz LemmermeyerFranz Lemmermeyer
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Pell's equation can be reduced to the well-known quadratic equation and solved using the machinery of quadratic equations. The details can be found here. math.stackexchange.com/questions/3099910/…user25406 –user25406 2019-02-04 15:42:04 +00:00 Commented Feb 4, 2019 at 15:42
The link to "higher descent on Pell conics" is now dead. Sadly from wayback machine we don't get access to the original pdfs web.archive.org/web/20210506135403/ Ghoshal –Sidharth Ghoshal 2025-06-18 17:12:33 +00:00 Commented Jun 18 at 17:12
I'm working on putting these things on the archive now that my web page is gone.Franz Lemmermeyer –Franz Lemmermeyer 2025-06-19 09:24:45 +00:00 Commented Jun 19 at 9:24
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Another nice reference on this problem (and non-CF methods to solve it) is Lenstra's 2002 notices survey, "Solving the Pell Equation".
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answered Dec 4, 2010 at 5:59
David EppsteinDavid Eppstein
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Any algorithm for computing fundamental units of a real quadratic number field Q(D−−√)Q(D) can be used for solving Pell's equation. (You might have to do a bit of work to convert the result, but that can be done in polynomial time...)
See for example M. Jacobson, H. Williams: Solving the Pell Equation. Springer, 2009.
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answered Dec 4, 2010 at 5:55
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It is really the same method, but see my answer at
Upper bound of period length of continued fraction representation of very composite number square root
The one thing I would add here is that, if the main concern is the length of the period, the continued fraction digits are the absolute values of my δ δ's, so sometimes the continued fraction period is half the cycle length.
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edited Apr 13, 2017 at 12:58
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answered Dec 4, 2010 at 5:57
Will JagyWill Jagy
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Just to add another method to the collection: Let Δ=σ+4 m Δ=σ+4 m, be the fundamental discriminant of a quadratic field, where σ∈{0,1}σ∈{0,1}. Let (x/z,y/z)(x/z,y/z) be a rational solution of the Pell conic x 2+σ x y−m y 2=1 x 2+σ x y−m y 2=1, with gcd(x,y)=1 gcd(x,y)=1, and let β≡x⋅y−1(mod z)β≡x⋅y−1(mod z). Assume that q 1=(t 1,u 1)q 1=(t 1,u 1), q 2=(t 2,u 2)q 2=(t 2,u 2) satisfy z t 2+(2 β+σ)t u+β 2+σ β−m z u 2=1 z t 2+(2 β+σ)t u+β 2+σ β−m z u 2=1 in rational integers, then ν(q 2,q 1)=ν(q 2,q 1)= I am trying to write a matrix times a vector: (z t 1+(β+σ)u 1,β t 1+β 2+σ β−m z u 1,−u 1,t 1)(t 2,u 2)(z t 1+(β+σ)u 1,β t 1+β 2+σ β−m z u 1,−u 1,t 1)(t 2,u 2) satisfies the Pell conic in rational integers. I learned about the map ν ν from Franz Lemmermeyer, his articles and book `Binary quadratic forms'. There is a bijection between the integer points (t,u)(t,u) and the primitive integer points (T,U)(T,U) of z T 2+(2 x+σ y)T U+z U 2=y 2 z T 2+(2 x+σ y)T U+z U 2=y 2. Given a primitive integer point (T,U)(T,U), we also have a primitive integer point (U,T)(U,T). Using this bijection, given an integer point (t 1,u 1)(t 1,u 1), we obtain another point (t 2,u 2)=(κ t 1+κ′u 1,y t 1−κ u 1)(t 2,u 2)=(κ t 1+κ′u 1,y t 1−κ u 1), where κ=x−β y z κ=x−β y z, and κ′=(2 β+σ)x−(β 2+m)y z 2 κ′=(2 β+σ)x−(β 2+m)y z 2. The points (t 1,u 1),(t 2,u 2)(t 1,u 1),(t 2,u 2) are used with ν ν to obtain an integer point of the Pell conic. In particular, letting γ=β 2+σ β−m z γ=β 2+σ β−m z, (x t 2+(z κ′+σ κ+y γ)t u+((β+σ)κ′−κ γ)u 2,y t 2−2 κ t u−κ′u 2)(x t 2+(z κ′+σ κ+y γ)t u+((β+σ)κ′−κ γ)u 2,y t 2−2 κ t u−κ′u 2). One must check that T≠U T≠U, equivalently that y t≠(κ+1)u y t≠(κ+1)u, for otherwise this method will not work. This follows from "Arithmetic of Pell surfaces", Acta Arith., 146, (2011), no. 1, 1--12.
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edited Oct 7, 2011 at 5:42
answered Oct 7, 2011 at 5:37
Samuel HambletonSamuel Hambleton
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A quantum algorithm for solving Pell's equation exists, the algorithm is gives superpolynomial speedup. Given an equation, x 2−d y 2=1 x 2−d y 2=1, the goal to find integer solution to the equation. This problem can be solved by a Quantum Algorithm in which the algorithm for solving Pell's equation efficiently approximates the period of a periodic function with an irrational period. The quantum algorithm works by finding a regulator R=l n(x 1+y 1 d−−√)R=l n(x 1+y 1 d), and is closely related to finding the principal ideal.
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edited Mar 20, 2016 at 5:11
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8847 | https://math.stackexchange.com/questions/2634253/what-does-invariant-to-affine-transformations-mean | Skip to main content
What does invariant to affine transformations mean?
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I am researching multivariate medians. In one of my sources it is stated that "Liu showed that the simplicial median is invariant to affine transformations". I am a bit confused about what this means. I know that affine transformations preserve ratios of distances, collinearity but not necessarily angles or lengths. However, I don't know what it means for something to be invariant to affine transformations. Does it mean that it will always remain the same? Also I was wondering if there is a simple example to illustrate this to help improve my understanding.
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As long as I'm aware an affine transformation is a map ax+b for a≠0 in 1 dimension or Ax+b for an invertible matrix n in general. As I read in maa.org/sites/default/files/pdf/pubs/books/meg/meg_ch12.pdf bottom of page 9, it seems that a property is invariant under affine maps, if whenever the image of a function f satisfy the property implies that the image of f∘φ satisfy the property for every affine map φ
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It means that if you apply an affine transformation to the data, the median of the transformed data is the same as the affine transformation applied to the median of the original data. For example, if you rotate the data the median also gets rotated in exactly the same way.
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Affine transformation are a combination of a translation with a linear transformation
A(v)=Av+v0
which preserves points, straight lines and planes.
Ratio by segments are preserved and in particular mean points are preserved by linearity
indeed if
M=P+Q2⟹A(M)=A(P)+A(Q)2+v0
Affine transformations
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非物质文化名菜——澄城水盆羊肉
2022-08-23 来源：中国餐饮协会编辑：国家名厨编委会 阅读量：949
菜品文化
澄城水盆羊肉是渭北地区喜受大家欢迎的一道特色美食，源于商周，兴于唐宋，于 2016 年入选陕西省第五批非物质文化遗产名录，水盆羊肉又名羊肉泡，选用当地“同羊”，选料考究，山谷间茶食空间主理人王茜秉承祖传秘方，不忘初心，坚守传承人老父亲王学忠先生文化核心精神，匠心打造，不断钻研，以古法烹制，精选十六味调料，八道工序加工，配山泉之水，紫砂文火慢炖而成，汤色金黄，晶莹剔透，汤清肉香，长期食用药食同源，强身健体，营养丰富，驱寒补虚，调理气血，是养胃生阳的一道名小吃！
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菜点名：非物质文化名菜——澄城水盆羊肉
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8849 | https://stackoverflow.com/questions/45065160/find-exponential-function-given-two-points | solver - find exponential function given two points - Stack Overflow
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find exponential function given two points
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Asked 8 years, 2 months ago
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I know I can solve equations in maxima using the commands below but how do I solve for two different equations.
kill(all);
r:.5; a:1; b:-5.7; theta:theta; solve(ae^(btheta)=r,theta);
tex(''%);
I'm trying to get the equation of a exponetial function given two points. How do I go about doing this. Example point 1 is at (2,12) and point 2 is at (8,768)
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asked Jul 12, 2017 at 18:30
Rick TRick T
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Maxima needs some help to solve this problem, but it can be done. Start by expression the problem data.
(%i1) [x1,y1]:[2,12];
(%o1) [2, 12]
(%i2) [x2,y2]:[8, 768];
(%o2) [8, 768]
(%i3) eq1:y1 = aexp(bx1);
2 b
(%o3) 12 = a %e
(%i4) eq2:y2 = aexp(bx2);
8 b
(%o4) 768 = a %e
Now try to solve eq1 and eq2 for a and b.
(%i5) solve([eq1, eq2], [a, b]);
(%o5) []
Hmm, that's unsatisfying. I'm guessing that Maxima could solve it if we take logarithms which make it linear.
(%i6) log([eq1, eq2]);
2 b 8 b
(%o6) [log(12) = log(a %e ), log(768) = log(a %e )]
Apply the logexpand flag to simplify. Note that % means the previous result.
(%i7) %, logexpand;
2 b 8 b
(%o7) [log(12) = log(a %e ), log(768) = log(a %e )]
Hmm, that didn't do it. There are different forms of logexpand, try another.
(%i8) %, logexpand=super;
(%o8) [log(12) = 2 b + log(a), log(768) = 8 b + log(a)]
OK, good. Now try to solve it.
(%i9) solve (%, [a, b]);
(%o9) []
Well, that still didn't work. But I see it's linear in log(a) so solve for that instead.
(%i10) solve (%o8, [log(a), b]);
4 log(12) - log(768) log(12) - log(768)
(%o10)
3 6
Great. Here are the numerical values:
(%i11) float (%);
(%o11)
I'll try to simplify the exact values.
(%i12) %o10, logexpand=super;
4 log(12) - log(768) log(12) - log(768)
(%o12)
3 6
Hmm, that didn't work. I'll try another function:
(%i13) radcan(%);
(%o13)
OK, that was a little bit of work, but anyway maybe it helps.
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answered Jul 13, 2017 at 20:01
Robert DodierRobert Dodier
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8850 | https://www.expii.com/t/ionic-charge-overview-periodic-trend-11055 | Expii
Ionic Charge — Overview & Periodic Trend - Expii
The ionic charge trend in the periodic table is that generally elements on the left form cations and those on the right form anions. The charges are lower on the outer ends of the table and increase as you move in.
Explanations (2)
Cassie Gates
Text
2
Ionic Charges of the Periodic Table
An ionic charge is a positive or negative charge on an atom. The charge depends on the atom's electron configuration and the number of valence electrons. On the periodic table, elements in the same group have the same number of valence electrons. So, their ionic charge is usually the same. In general, atoms that form positive ions (cations) are found on the left of the periodic table. Atoms that form negative charges (anions) are found on the right. The lower charge values are on the outside of the periodic table and increase as you move inward.
Charges Among Periodic Groups
Let's look at each group on the periodic table.
Ionic charges by Group:
Group 1, the alkali metals: charge of +1
Group 2, the alkaline earth metals: charge of +2
Groups 3 - 12, the transition metals: charge can vary from negative to positive values. Some transition metals have been known to access up to a +7 charge.
Group 15 nonmetals (N, P): charge of -3
Group 16 nonmetals (O, S, Se): charge of -2
Group 17 halogens (F, Cl, Br, I): charge of -1
Group 18, the noble gases (He, Ne, Ar, Kr, Xn, Rn): charge of 0.
The noble gases already have a full octet. So, they don't have an ionic charge. There are a few Xenon compounds that are octet rule exceptions.
Determining Ionic Charges from the Electron Configuration
The trends of ionic charge on the periodic table are generally true and reliable. But have you ever wondered how the charge is determined? Atoms have the same number of electrons as their atomic number. The electrons fill the first shell or energy level of an atom with two electrons. The remaining electrons fill the next sets of shells with eight electrons. The valence electron shell is the outermost shell with electrons in it. What If there aren't eight electrons? The atom is unstable. So, electrons will fill or leave the valence shell. For example, sodium (Na) has one electron in its valence shell. It is more favorable to lose that electron than add seven more. When that electron is lost, the sodium atom has one more proton than it does electrons. So, there is a +1 charge associated with Na. This same charge and ion electron configuration are also true for the other element in group 1. Let's consider another example. Chlorine has seven electrons and needs one more for a complete octet or full valence shell. So chlorine will gain an electron and have a -1 charge. It has a large electron affinity. This is true for the other halogens.
Why Do We Care about Ionic Charge?
So, atoms can gain or lose valence electrons. Why do we care? The ionic charge is the basis of ionic bonding. For example, consider table salt. Do you know its chemical formula? It's NaCl, or sodium chloride. The sodium atom gave its one electron to chlorine. So, we have a cation and an anion. But, opposite charges experience electrostatic attraction. So, they bond together.
But, how does salt become a crystal? The atoms' charges actually radiate in every direction. So, each sodium is surrounded by six chlorine atoms. Two are above and below. The other four are in front and behind, and on the right and left. The same is true for each chlorine atom! Ionic bonds form a crystal lattice. They have repeating units of sodium surrounded by chlorine and vice versa. All of this happens because of the ionic charge!
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(Video) Finding Charges of Ions on Periodic Table
by MagnetsAndMotors (Dr. B's Other Channel)
In this video, Dr. Breslyn explains how to find the ionic charge. Thankfully, the periodic table's groups tell us exactly that! But, it only works for some elements. The alkali, alkaline earth, nonmetals, halogens, and noble gases are consistent. These atoms will either gain or lose electrons to create ions. So, if the electronegativity difference is extensive, they form ionic bonds.
What elements don't follow the trend? There are two answers. Let's start with the metalloids and carbon. They almost only form covalent bonds. They share valence electrons. So, they don't form ions and don't have an ionic charge. The other group is the transition metals. Why are they unique? Because they form multiple ions. The fancy term is that they have many oxidation states. Their charge depends on the other atoms they've bonded.
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8851 | https://www.edubull.com/Content/eRepository/SubjectContent/Exercises/05%20Ellipse_Exercise.pdf | MATHS FOR JEE MAIN & ADVANCED 278 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Ex. 1 Find the equation to the ellipse whose centre is origin, axes are the axes of co-ordinate and passes through the points (2, 2) and (3, 1).
Sol.
Let the equation to the ellipse is 2 2 x a + 2 2 y b = 1 Since it passes through the points (2, 2) and (3, 1) 2 4 a + 2 4 b = 1 ......(i) and 2 9 a + 2 1 b = 1 ......(ii) from (i) – 4 (ii), we get 2 4 36 a = 1– 4 a2 = 32 3 from (i), we get 2 1 b = 1 4 – 3 32 = 8 3 32 b2 = 32 5 Ellipse is 3x2 + 5y2 = 32 Ex. 2 If , are eccentric angles of end points of a focal chord of the ellipse 2 2 2 2 x y 1 a b , then tan /2. tan /2 is equal to -Sol.
Equation of line joining points ‘’ and ‘’ is x y cos sin cos a 2 b 2 2 If it is a focal chord, then it passes through focus (ae, 0), so e cos cos 2 2 cos e 2 1 cos 2 cos cos e 1 2 2 e 1 cos cos 2 2 2 sin / 2 sin / 2 e 1 2 cos / 2 cos / 2 e 1 tan e 1 tan 2 2 e 1 using (–ae, 0) , we get tan e 1 tan 2 2 e 1 SOLVED EXAMPLES ELLIPSE 279 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Ex. 3 From a point Q on the circle x2 + y2 = a2, perpendicular QM are drawn to x-axis, find the locus of point 'P' dividing QM in ratio 2 : 1.
Sol.
Let Q (a cos, a sin) M (a cos, 0) Let P (h, k) Q M P 1 2 h = a cos, k = asin 3 2 3k a + 2 h a = 1 Locus of P is 2 2 x a + 2 2 y (a / 3) = 1 Ex. 4 Find the equation of axes, directrix, co-ordinate of focii, centre, vertices, length of latus - rectum and eccentricity of an ellipse 2 (x 3) 25 + 2 (y 2) 16 = 1.
Sol.
Let x – 3 = X, y – 2 = Y, so equation of ellipse becomes as 2 2 X 5 + 2 2 Y 4 = 1.
equation of major axis is Y = 0 y = 2.
equation of minor axis is X = 0 x = 3.
centre (X = 0, Y = 0) x = 3, y = 2 C (3, 2) Length of semi-major axis a = 5 Length of major axis 2a = 10 Length of semi-minor axis b = 4 Length of minor axis = 2b = 8.
Let 'e' be eccentricity b2 = a2 (1 – e2) e = 2 2 2 a b a = 25 16 25 = 3 5 .
Length of latus rectum = LL = 2 2b a = 2 16 5 = 32 5 Co-ordinates focii are X = ± ae, Y = 0 S (X = 3, Y = 0) & S (X = –3, Y = 0) S (6, 2) & S (0, 2) Ea. 5 Find the distance from centre of the point P on the ellipse 2 2 x a + 2 2 y b = 1 whose radius makes angle with x-axis.
Sol.
Let P (a cos, b sin) m(op) = b a tan = tan tan = a b tan MATHS FOR JEE MAIN & ADVANCED 280 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 OP = 2 2 2 2 a cos b sin = 2 2 2 2 a b tan sec = 2 2 2 2 a b tan 1 tan = 2 2 2 2 2 2 2 2 a a b tan b a 1 tan b OP = 2 2 2 2 ab a sin b cos Ex. 6 The tangent at a point P on an ellipse intersects the major axis in T and N is the foot of the perpendicular from P to the same axis. Show that the circle drawn on NT as diameter intersects the auxiliary circle orthogonally.
Sol.
Let the equation of the ellipse be 2 2 2 2 x y 1 a b . Let P(acos, bsin) be a point on the ellipse. The equation of the tangent at P is xcos ysin 1 a b . It meets the major axis at T (a sec, 0).
The coordinates of N are (a cos, 0). The equation of the circle with NT as its diameter is (x – asec)(x – acos) + y2 = 0.
x2 + y2 – ax(sec + cos) + a2 = 0 It cuts the auxiliary circle x2 + y2 – a2 = 0 orthogonally if 2g . 0 + 2f . 0 = a2 – a2 = 0, which is true.
Ex. 7 If P() and P() are extremities of a focal chord of ellipse then prove that its eccentricity e = cos 2 cos 2 .
Sol.
Let the equation of ellipse is 2 2 x a + 2 2 y b = 1 equation of chord is x a cos 2 + y b sin 2 = cos 2 Since above chord is focal chord, it passes through focus (ae, 0) or (– ae, 0) ± e cos 2 = cos 2 e = cos 2 cos 2 ELLIPSE 281 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Ex. 8 If the normal at an end of a latus-rectum of an ellipse 2 2 2 2 x y 1 a b passes through one extremity of the minor axis, show that the eccentricity of the ellipse is given by 5 1 e 2 Sol.
The co-ordinates of an end of the latus-rectum are (ae, b2/a).
The equation of normal at P(ae, b2/a) is 2 2 2 2 2 a x b (y) a b ae b / a or 2 2 ax ay a b e It passes through one extremity of the minor axis whose co-ordinates are (0, –b) 0 + ab = a2 – b2 a2b2) = (a2 – b2)2 a2.a2(1 – e2) = (a2 e2)2 1 – e2 = e4 e4 + e2 – 1 = 0 (e2)2 + e2 – 1 = 0 2 1 1 4 e 2 5 1 e 2 (taking positive sign) Ex. 9 Find the angle between two diameters of the ellipse 2 2 x a + 2 2 y b = 1. Whose extremities have eccentricity angle and = + 2 .
Sol.
Let ellipse is 2 2 x a + 2 2 y b = 1 Slope of OP = m1 = bsin acos = b a tan Slope of OQ = m2 = bsin a cos = – b a cot given = + 2 tan = 1 2 1 2 m m 1 m m = 2 2 b (tan cot ) a b 1 a = 2 2 2ab (a b )sin 2 Ex. 10 Find the set of value(s) of '' for which the point P(, – ) lies inside the ellipse 2 x 16 + 2 y 9 = 1.
Sol.
If P(, – ) lies inside the ellipse S1 < 0 2 16 + 2 9 – 1 < 0 25 144 . 2 < 1 2 < 144 25 12 12 , 5 5 .
MATHS FOR JEE MAIN & ADVANCED 282 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Ex. 11 A tangent to the ellipse 2 2 x a + 2 2 y b = 1 touches at the point P on it in the first quadrant and meets the co-ordinate axes in A and B respectively. If P divides AB in the ratio 3 : 1, find the equation of the tangent.
Sol.
Let P ( a cos, b sin) P B A equation of tangent is x a cos + y b sin = 1 A (a sec, 0) B (0, b cosec) P divide AB internally in the ratio 3 : 1 a cos = asec 4 cos2 = 1 4 cos = 1 2 and b sin = 3bcosec 4 sin = 3 2 tangent is x 2a + 3y 2b = 1 bx + 3 ay = 2ab Ex. 12 A tangent to the ellipse x2 + 4y2 = 4 meets the ellipse x2 + 2y2 = 6 at P and Q. Prove that the tangents at P and Q of the ellipse x2 + 2y2 = 6 are at right angles.
Sol.
Given ellipse are 2 2 x y 1 4 1 ......(i) and, 2 2 x y 1 6 3 ......(ii) any tangent to (i) is xcos ysin 1 2 1 ......(iii) It cuts (ii) at P and Q, and suppose tangent at P and Q meet at (h, k) Then equation of chord of contact of (h, k) with respect to ellipse (ii) is hx ky 1 6 3 ......(iv) comparing (iii) and (iv), we get cos sin 1 h / 3 k / 3 h cos 3 and k sin 3 2 2 h k 9 locus of the point (h, k) is x2 + y2 = 9 x2 + y2 = 6 + 3 = a2 + b2 i.e. director circle of second ellipse. Hence the tangents are at right angles.
ELLIPSE 283 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Ex. 13 Prove that the locus of the point of intersection of tangents to an ellipse at two points whose eccentric angle differ by a constant is an ellipse.
Sol.
Let P (h, k) be the point of intersection of tangents at A() and B() to the ellipse.
h = acos 2 cos 2 & k = bsin 2 cos 2 2 h a + 2 k b = sec2 2 but given that – = locus is 2 2 2 x a sec 2 + 2 2 2 y b sec 2 = 1 Ex. 14 A point moves so that the sum of the squares of its distances from two intersecting straight lines is constant. Prove that its locus is an ellipse.
Sol.
Let two intersecting lines OA and OB, intersect at origin O and let both lines OA and OB makes equal angles with x axis.
i.e., XOA = XOB = .
Equations of straight lines OA and OB are y = x tan and y = –x tan or x sin – y cos = 0 ......... (i) and x sin + y cos = 0 ......... (ii) Let P() is the point whose locus is to be determine.
According to the example (PM)2 + (PN)2 = 22 (say) ( sin + cos)2 + ( sin – cos)2 = 22 22 sin2 + 22cos2 = 22 or 2 sin2 + 2cos2 = 2 2 2 2 2 2 2 1 cosec sec 2 2 2 2 1 ( cosec ) ( sec ) Hence required locus is 2 2 2 2 x y 1 ( cosec ) ( sec ) Ex. 15 Prove that, in an ellipse, the distance between the centre and any normal does not exceed the difference between the semi-axes of the ellipse.
Sol.
Let the equation of ellipse is 2 2 2 2 x y 1 a b MATHS FOR JEE MAIN & ADVANCED 284 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Equation of normal at P () is (a sec)x – (bcosec )y – a2 + b2 = 0 distance of normal from centre R O P = OR = 2 2 2 2 2 2 | a b | a b (a tan ) (bcot ) = 2 2 2 2 | a b | (a b) (a tan bcot ) (a + b)2 + (a tan – b cot)2 (a + b)2 or 2 2 2 | a b | (a b) |OR| (a – b) Ex. 16 Find the condition on 'a' and 'b' for which two distinct chords of the ellipse 2 2 2 2 x y 1 2a 2b passing through (a, –b) are bisected by the line x + y = b.
Sol.
Let (t, b – t) be a point on the line x + y = b.
Then equation of chord whose mid point (t, b – t) is 2 2 2 2 2 2 tx y(b t) t (b t) 1 1 2a 2b 2a 2b ......(i) (a, –b) lies on (i) then 2 2 2 2 2 2 ta b(b t) t (b t) 2a 2b 2a 2b t2(a2 + b2) – ab(3a + b)t + 2a2b2 = 0 Since t is real B2 – 4AC 0 a2b2(3a + b)2 – 4(a2 + b2)2a2b2 0 a2 + 6ab – 7b2 0 a2 + 6ab 7b2, which is the required condition.
Ex. 17 Find the locus of point of intersection of perpendicular tangents to the ellipse 2 2 2 2 x y a b = 1 Sol.
Let P(h, k) be the point of intersection of two perpendicular tangents equation of pair of tangents is SS1 = T2 2 2 2 2 x y 1 a b 2 2 2 2 h k 1 a b = 2 2 2 hx ky 1 a b 2 2 x a 2 2 k 1 b + 2 2 y b 2 2 h 1 a + ........ = 0 ......(i) Since equation (i) represents two perpendicular lines 2 1 a 2 2 k 1 b + 2 1 b 2 2 h 1 a = 0 k2 – b2 + h2 – a2 = 0 locus is x2 + y2 = a2 + b2 ELLIPSE 285 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Ex. 18 Any tangent to an ellipse is cut by the tangents at the ends of the major axis in T and T '. Prove that circle on TT' as diameter passes through foci.
Sol.
Let ellipse be 2 2 2 2 x y 1 a b and let P(acos, bsin) be any point on this ellipse Equation of tangent at P(acos, bsin) is x y cos sin 1 a b ......(i) The two tangents drawn at the ends of the major axis are x = a and x = –a Solving (i) and x = a we get b(1 cos ) T a, a,b tan sin 2 and solving (i) and x = – a we get b(1 cos ) T ' a, a, bcot sin 2 Equation of circle on TT' as diameter is (x – a)(x + a) + (y – b tan(/2))(y – b cot (/2)) = 0 or x2 + y2 – by (tan(/2) + cot(/2)) – a2 + b2 = 0 ......(ii) Now put x = ± ae and y = 0 in LHS of (ii), we get a2e2 + 0 – 0 – a2 + b2 = a2 – b2 – a2 + b2 = 0 = RHS Hence foci lie on this circle Ex. 19 Find the condition on 'a' and 'b' for which two distinct chords of the ellipse 2 2 x 2a + 2 2 y 2b = 1 passing through (a, – b) are bisected by the line x + y = b.
Sol.
Let the line x + y = b bisect the chord at P(, b – ) equation of chord whose mid-point is P(, b – ) 2 x 2a + 2 y(b ) 2b = 2 2 2a + 2 2 (b ) 2b Since it passes through (a, –b) 2a – (b ) 2b = 2 2 2a + 2 2 (b ) 2b 1 1 a b – 1 = 2 2 2 1 1 a b – 2 b + 1 2 2 2 1 1 a b – 3 1 b a + 2 = 0 since line bisect two chord above quadratic equation in must have two distinct real roots MATHS FOR JEE MAIN & ADVANCED 286 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 2 3 1 b a – 4 2 2 1 1 a b . 2 > 0 2 9 b + 2 1 a + 6 ab – 2 8 a – 2 8 b > 0 2 1 b – 2 7 a + 6 ab > 0 a2 – 7b2 + 6ab > 0 a2 > 7b2 – 6ab which is the required condition.
Ex. 20 A variable point P on an ellipse of eccentricity e, is joined to its foci S, S'. Prove that the locus of the incentre of the triangle PSS' is an ellipse whose eccentricity is 2e 1 e .
Sol.
Let the given ellipse be 2 2 2 2 x y 1 a b a(1+e cos ) a(1– e cos ) S' S P (a cos bsin ) (–ae, 0) (ae, 0) 2ae I(x, y) Let the co-ordinates of P are (a cos, b sin) By hypothesis b2 = a2(1 – e2) and S(ae, 0), S'(–ae, 0) SP = focal distance of the point P = a – ae cos and S'P = a + ae cos Also SS' = 2ae If (x, y) be the incentre of the PSS' then x = (2ae)a cos a( 1 e cos )( ae) a(1 e cos )ae 2ae a(1 e cos ) a( 1 e cos ) x = ae cos ....... (i) y = 2ae(bsin ) a(1 e cos ).0 a(1 e cos ).0 2ae a( 1 e cos ) a( 1 e cos ) ebsin y (e 1) ....... (ii) Eliminating from equations (i) and (ii), we get 2 2 2 2 2 x y 1 a e be e 1 which represents an ellipse.
Let e1 be its eccentricity.
2 2 2 2 1 2 b e a e ( 1 e ) (e 1) 2 2 1 2 2 b e 1 a (e 1) = 2 2 1 e 1 e 2e 1 1 1 e 1 e (e 1) 1 2e e 1 e ELLIPSE 287 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Exercise # 1 [Single Correct Type Questions] Choice 1.
The curve represented by x = 3(cost + sint), y = 4(cost – sint), is -(A) ellipse (B) parabola (C) hyperbola (D) circle 2.
Let 'E' be the ellipse 2 x 9 + 2 y 4 = 1 & 'C' be the circle x2 + y2 = 9. Let P & Q be the points (1 , 2) and (2, 1) respectively.
.
Then : (A) Q lies inside C but outside E (B) Q lies outside both C & E (C) P lies inside both C & E (D) P lies inside C but outside E.
3.
If PQR is an equilateral triangle inscribed in the auxiliary circle of the ellipse 2 2 2 2 x y a b = 1, (a > b), and P'Q'R' is the corresponding triangle inscribed withint the ellipse, then the centroid of triangle P'Q'R' lies at (A) center of ellipse (B) focus of ellipse (C) between focus and center on major axis (D) none of these 4.
An ellipse having foci at (3, 3) and (–4, 4) and passing through the origin has eccentricity equal to-(A) 3 7 (B) 2 7 (C) 5 7 (D) 3 5 5.
The length of the major axis of the ellipse (5x – 10)2 + (5y + 15)2 = 2 (3x 4y 7) 4 is (A) 10 (B) 20/3 (C) 20/7 (D) 4 6.
The normal at P() on the ellipse 2 2 2 2 x y 1 a b cuts the axis of x at G and PG is produced to Q so that GQ = 2PG then the locus of Q is given by (A) 2 2 2 2 2 2 2 a x y 1 (3b a ) b (B) 2 2 2 2 2 2 2 a x y 1 (a 3b ) 4b (C) 2 2 2 2 2 2 2 a x y 1 (a 3b ) 4b (D) 2 2 2 2 2 2 x y 1 (a 3b ) b 7.
A tangent having slope of – 4 3 to the ellipse 2 2 x y 1 18 32 intersects the major & minor axes in points A & B respectively. If C is the centre of the ellipse then the area of the triangle ABC is : (A) 12 sq. units (B) 24 sq. units (C) 36 sq. units (D) 48 sq. units 8.
The auxiliary circle of a family of ellipses passes through the origin and makes intercepts of 8 and 6 units on the x- and the y-axis, respectively. If the eccentricity of all such ellipse is 1/2, then the locus of the focus will be (A) 2 2 x y 16 9 = 25 (B) 4x2 + 4y2 – 32x – 24y + 75 = 0 (C) 2 2 x y 16 9 = 25 (D) none of these MATHS FOR JEE MAIN & ADVANCED 288 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 9.
x 2y + 4 = 0 is a common tangent to y2 = 4x & 2 2 2 x y 4 b = 1. Then the value of b and the other common tangent are given by : (A) b = 3 ; x + 2y + 4 = 0 (B) b = 3 ; x + 2y + 4 = 0 (C) b = 3 ; x + 2y 4 = 0 (D) b = 3 ; x 2y 4 = 0 10.
The equation to the locus of the middle point of the portion of the tangent to the ellipse 2 x 16 + 2 y 9 =1 included between the co-ordinate axes is the curve-(A) 9x2 + 16y2 = 4x2y2 (B) 16x2 + 9y2 = 4x2y2 (C) 3x2 + 4y2 = 4x2y2 (D) 9x2 + 16y2 = x2y2 11.
Tangents are drawn to the ellipse 2 2 2 2 x y a b = 1, (a > b) and the circle x2 + y2 = a2 at the points where a common ordinate cuts them (on the same side of the x-axis). Then the greatest acute angle between these tangents is given by (A) tan–1 a b 2 ab (B) tan–1 a b 2 ab (C) tan–1 2ab a b (D) tan–1 2ab a b 12.
The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is /4 is : (A) 2 2 2 2 a b ab a b (B) 2 2 2 2 a b a b ab (C) 2 2 2 2 a b ab a b (D) 2 2 2 2 a b a b ab 13.
Which of the following is the common tangent to the ellipses 2 2 2 2 2 x y 1 a b b & 2 2 2 2 2 x y 1 a a b ?
(A) ay = bx + 4 2 2 4 a a b b (B) by = ax – 4 2 2 4 a a b b (C) ay = bx – 4 2 2 4 a a b b (D) by = ax – 4 2 2 4 a a b b 14.
Let P be any point on a directrix of an ellipse of eccentricity e, S be the corresponding focus, and C the center of the ellipse. The line PC meets the ellipse at A. The angle between PS and tangent at A is , then is equal to (A) tan–1 e (B) /2 (C) tan–1(1 – e2) (D) none of these 15.
A conic passes through the point (2, 4) and is such that the segment of any of its tangents at any point contained between the co-ordinate axes is bisected at the point of tangency. Then the foci of the conic are (A) 2 2 , 0 & 2 2 , 0 (B) 2 2 , 2 2 & 2 2 , 2 2 (C) (4, 4) & ( 4, 4) (D) 4 2 , 4 2 & 4 2 , 4 2 16.
The point of intersection of the tangents at the point P on the ellipse 2 2 x a + 2 2 y b = 1, and its corresponding point Q on the auxiliary circle meet on the line -(A) x = a/e (B) x = 0 (C) y = 0 (D) none ELLIPSE 289 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 17.
If the ellipse 2 2 2 x y a 7 13 5a = 1 is inscribed in a square of side length 2 a, then a is equal to (A) 6/5 (B) (–, – 7 ) ( 7 , 13/5) (C) (–, – 7 ) (13/5, 7 ) (D) no such a exists 18.
Imagine that you have two thumbtacks placed at two points, A and B. If the ends of a fixed length of string are fastened to the thumbtacks and the string is drawn taut with a pencil, the path traced by the pencil will be an ellipse.
The best way to maximise the area surrounded by the ellipse with a fixed length of string occurs when I the two points A and B have the maximum distance between them.
I I two points A and B coincide.
III A and B are placed vertically.
IV The area is always same regardless of the location of A and B.
(A) I (B) II (C) III (D) IV 19.
The equation of the normal to the ellipse 2 2 2 2 x y 1 a b at the positive end of latus rectum is -(A) x + ey + e2a = 0 (B) x – ey – e3a = 0 (C) x – ey – e2a = 0 (D) none of these 20.
Let P be any point on any directrix of an ellipse. Then the chords of contact of point P with respect to the ellipse and its auxiliary circle intersect at (A) some point on the major axis depending upon the position of point P (B) the midpoint of the line segment joining the center to the corresponding focus (C) the corresponding focus (D) none of these 21.
PQ is a double ordinate of the ellipse x2 + 9y2 = 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR is (A) a circle (B) a parabola (C) an ellipse (D) a hyperbola 22.
PQ is a double ordinate of the ellipse x2 + 9y2 = 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR is -(A) a circle (B) a parabola (C) an ellipse (D) a hyperbola 23.
A parabola is drawn with focus at one of the foci of the ellipse 2 2 2 2 x y a b = 1, a > b, and directrix passing through the other focus and perpendicular to the major axis of the ellipse. If the latus rectum of the ellipse and that of the parabola are same, then the eccentricity of the ellipse is (A) 1 – 1 2 (B) 2 2 – 2 (C) 2 – 1 (D) none of these 24.
If & are the eccentric angles of the extremities of a focal chord of an standard ellipse, then the eccentricity of the ellipse is : (A) cos cos cos( ) (B) sin sin sin( ) (C) cos cos cos( ) (D) sin sin sin( ) MATHS FOR JEE MAIN & ADVANCED 290 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 25.
If F1 & F2 are the feet of the perpendiculars from the foci S1 & S2 of an ellipse 2 2 x y 1 5 3 on the tangent at any point P on the ellipse , then (S1F1) . (S2F2) is equal to (A) 2 (B) 3 (C) 4 (D) 5 26.
The set of values of m for which it is possible to draw the chord y = m x + 1 to the curve x2 + 2xy + (2 + sin2) y2 = 1, which subtends a right angle at the origin for some value of , is (A) [2, 3] (B) [0, 1] (C) [1, 3] (D) none of these 27.
An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that this point lies outside the ellipse is 2/3 then the eccentricity of the ellipse is : (A) 2 2 3 (B) 5 3 (C) 8 9 (D) 2 3 28.
The number of values of c such that the straight line y = 4x + c touches the curve (x2 / 4) + y2 = 1 is -(A) 0 (B) 1 (C) 2 (D) infinite 29.
An ellipse having foci at (3, 3) and (– 4, 4) and passing through the origin has eccentricity equal to (A) 3 7 (B) 2 7 (C) 5 7 (D) 3 5 30.
The point of intersection of the tangents at the point P on the ellipse 2 2 2 2 x y a b = 1 and its corresponding point Q on the auxiliary circle meet on the line : (A) x = a/e (B) x = 0 (C) y = 0 (D) None ELLIPSE 291 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Exercise # 2 Part # I [Multiple Correct Choice Type Questions] 1.
2 2 2 2 x y r r 6 r 6r 5 = 1 will represent an ellipse if r lies in the interval (A) (–, –2) (B) (3, ) (C) (5, ) (D) (1, ) 2.
On the x-y plane, the eccentricity of an ellipse is fixed (in size and position) by (A) both foci (B) both directrices (C) one focus and the corresponding directrix (D) the length of major axis 3.
The tangent at any point P on a standard ellipse with foci as S & S' meets the tangents at the vertices A & A' in the points V & V', then -(A) 2 AV). A'V') b l( l( (B) 2 AV .
A 'V' a l l (C) V'SV 90 (D) V'S' VS is a cyclic quadrilateral 4.
If the tangent at the point P() to the ellipse 16x2 + 11y2 = 256 is also a tangent to the circle x2 + y2 – 2x = 15, then = (A) 2/3 (B) 4/3 (C) 5/3 (D) /3 5.
If P is a point of the ellipse 2 2 2 2 x y 1 a b , whose foci are S and S'. Let PSS ' and PS 'S , then -(A) PS + PS' = 2a, if a > b (B) PS + PS' = 2b, if a < b (C) 1 e tan tan 2 2 1 e (D) 2 2 2 2 2 a b tan tan [a a b ] 2 2 b when a > b 6.
The coordinates (2, 3) and (1, 5) are the foci of an ellipse which passes through the origin. Then the equation of the (A) tangent at the origin is (3 2 – 5)x + (1 – 2 2 )y = 0 (B) tangent at the origin is (3 2 + 5)x + (1 + 2 2 y) = 0 (C) normal at the origin is (3 2 + 5)x – (2 2 + 1)y = 0 (D) normal at the origin is x(3 2 – 5) – y(1 – 2 2 ) = 0 7.
If the chord through the points whose eccentric angles are and on the ellipse 2 2 x y 25 9 = 1 passes through a focus, then the value of tan(/2) tan(/2) is (A) 1/9 (B) –9 (C) –1/9 (D) 9 8.
If a pair of variable straight line x2 + 4y2 + axy = 0 (where a is a real parameter) cuts the ellipse x2 + 4y2 = 4 at two points A and B, then the locus of the point of intersection of tangents at A and B is (A) x – 2y = 0 (B) 2x – y = 0 (C) x + 2y = 0 (D) 2x + y = 0 MATHS FOR JEE MAIN & ADVANCED 292 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 9.
Which of the following is/are true ?
(A) There are infinite positive integral values of a for which (13x – 1)2 + (13y – 2)2 = 2 5x 12y 1 a represents an ellipse.
(B) The minimum distance of a point (1, 2) from the ellipse 4x2 + 9y2 + 8x – 36y + 4 = 0 is 1.
(C) If from a point P(0, a) two normals other than the axes are drawn to the ellipse 2 2 x y 25 16 = 1, then |a| < 9/4.
(D) If the length of the latus rectum of an ellipse is one-third of its major axis, then its eccentricity is equal to 1/ 3 .
10.
Extremities of the latera recta of the ellipses 1 b y a x 2 2 2 2 (a > b) having a given major axis 2a lies on (A) x2 = a(a – y) (B) x2= a (a + y) (C) y2 = a(a + x) (D) y2 = a (a – x) 11.
If the chord through the points whose eccentric angles are on the ellipse, 2 2 2 2 x y 1 a b passes through the focus, then the value of tan (/2) tan (/2) is -(A) e 1 e 1 (B) e 1 e 1 (C) 1 e 1 e (D) 1 e 1 e 12.
Which of the following is/are true about the ellipse x2 + 4y2 – 2x – 16y + 13 = 0 ?
(A) The latus rectum of the ellipse is 1.
(B) The distance between the foci of the ellipse is 4 3 (C) The sum of the focal distances of a point P(x, y) on the ellipse is 4.
(D) Line y = 3 meets the tangents drawn at the vertices of the ellipse at points P and Q. Then PQ subtends a right angle of any of its foci.
13.
The locus of the image of the focus of the ellipse 2 2 x y 25 9 = 1, (a > b), with respect to any of the tangents to the ellipse is (A) (x + 4)2 + y2 = 100 (B) (x + 2)2 + y2 = 50 (C) (x – 4)2 + y2 = 100 (D) (x – 2)2 + y2 = 50 14.
Let E1 and E2, respectively, be two ellipse 2 2 x a + y2 = 1 and x2 + 2 2 y a = 1 (where a is a parameter). Then the locus of the points of intersection of the ellipses E1 and E2 is a set of curves comprising (A) two straight line (B) one straight line (C) one circle (D) one parabols 15.
If point P( + 1, ) lies between the ellipse 16x2 + 9y2 – 16x = 0 and its auxiliary circle, then -(A) [] = 0 (B) [] = –1 (C) no such real exist (D) [] = 1 where [.] denotes greatest integer function.
ELLIPSE 293 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 16.
Consider the ellipse 2 2 2 x y f(k 2k 5) f (k 11) = 1. If f(x) is a positive decreasing function, then (A) The set of values of k for which the major axis is the x-axis is (–3, 2) (B) The set of values of k for which the major axis is the y-axis is (–, 2) (C) The set of values of k for which the major axis is the y-axis is (–, –3) (2, ) (D) The set of values of k for which the major axis is the y-axis is (–3, ) 17.
Two concentric ellipses are such that the foci of one are on the other and their major axes are equal. Let e and e' be their eccentricities. Then, (A) The quadrilateral formed by joining the foci of the two ellipse is a parallelogram (B) The angle between their axes is given by = cos–1 2 2 2 2 1 1 1 e e' e e' (C) If e2 + e'2 = 1, then the angle between the axes of the two ellipse is 90°.
(D) none of these 18.
If latus rectum of an ellipse 2 2 x y 1 16 b {0< b < 4}, subtend angle 2 at farthest vertex such that cosec = 5 , then -(A) e = 1 2 (B) no such ellipse exist (C) b = 2 3 (D) area of formed by LR and nearest vertex is 6 sq. units 19.
If a number of ellipse be described having the same major axis 2a but a variable minor axis then the tangents at the ends of their latera recta pass through fixed points which can be (A) (0, a) (B) (0, 0) (C) (0, – a) (D) (a, a) 20.
If the tangent drawn at point (t2, 2t) on the parabola y2 = 4x is the same as the normal drawn at point ( 5 cos , 2sin ) on the ellipse 4x2 + 5y2 = 20, then (A) = cos–1 1 5 (B) = cos–1 1 5 (C) t = – 2 5 (D) t = – 1 5 MATHS FOR JEE MAIN & ADVANCED 294 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Part # II [Assertion & Reason Type Questions] These questions contains, Statement I (assertion) and Statement II (reason).
(A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
(B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I.
(C) Statement-I is true, Statement-II is false.
(D) Statement-I is false, Statement-II is true.
1.
Statement-I : Tangent drawn at a point P 4 5 , 2 3 on the ellipse 9x2 + 16y2 = 144 intersects a straight line x = 16 7 at M, then PM subtends a right angle at (– 7 , 0) Statement-II : The portion of the tangent to an ellipse between the point of contact and the directrix subtends a right angle at the corresponding focus.
2.
Statement-I : Tangents are drawn to the ellipse 2 2 x y 4 2 = 1 at the points where it is intersected by the line 2x + 3y =1. The point of intersection of these tangents is (8, 6) Statement-II : The equation of the chord of contact to the ellipse 2 2 2 2 x y a b = 1 from an external point is given by 1 1 2 2 xx yy a b – 1 = 0 3.
Statement-I : Any chord of the ellipse x2 + y2 + xy = 1 through (0, 0) is bisected at (0, 0) Statement-II : The centre of an ellipse is a point through which every chord is bisected.
4.
Statement-I : If a and b are real numbers and c > 0, then the locus represented by the equation |ay – bx| = c 2 2 (x a) (y b) is an ellipse.
Statement-II : An ellipse is the locus of a point which moves in a plane such that the ratio of its distances from a fixed point (i.e., focus) to that from the fixed line (i.e. directrix) is constant and less than 1.
5.
Statement-I : If P 3 3 , 1 2 is a point on the ellipse 4x2 + 9y2 = 36. Circle drawn AP as diameter touches another circle x2 + y2 = 9, where A (– 5 , 0) Statement-II : Circle drawn with focal radius as diameter touches the auxilliary circle.
ELLIPSE 295 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Exercise # 3 Part # I [Matrix Match Type Questions] Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II.
1.
Column - I Column - II (A) If the vertices of a rectangle of maximum area inscribed (p) 2 5 in the ellipse 2 2 2 2 x y 1 a b are extremities of latus rectum, then the eccentricity of the ellipse is (B) If the extremities of the diameter of the circle x2 + y2 = 16 (q) 1 2 are the foci of the ellipse, then the eccentricity of the ellipse, if its size is just sufficient to contain the circle, is (C) If the normal at point (6, 2) to the ellipse passes through (r) 1 3 its nearest focus (5, 2), having center at (4, 2), then its eccentricity is (D) It the extremities of the latus rectum of the parabola y2 = 24x (s) 1 2 are the foci of ellipse, and if the ellipse passes through the vertex of the parabola, then its eccentricity is 2.
Column - I Column - II (A) The minimum and maximum distance of a point (2, 6) from (p) 0 the ellipse are 9x2 + 8y2 – 36x – 16y – 28 = 0 (B) The minimum and maximum distance of a point 9 12 , 5 5 (q) 2 from the ellipse 4(3x + 4y)2 + 9(4x – 3y)2 = 900 are (C) If E : 2x2 + y2 = 2 and director circle of E is C1, director circle of C1 is C2 director circle of C2 is C3 and so on. If (r) 6 r1, r2, r3 .... are the radii of C1, C2, C3 ... respectively then G.M. of 2 1 r , 2 2 r , 2 3 r is (D) Minimum area of the triangle formed by any tangent to the (s) 8 ellipse x2 + 4y2 = 16 with coordinate axes is MATHS FOR JEE MAIN & ADVANCED 296 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Part # II [Comprehension Type Questions] Comprehension # 1 A coplanar beam of light emerging from a point source has the equation x – y + 2(1 + ) = 0, R. The rays of the beam strike an elliptical surface and get reflected. The reflected rays from another convergent beam having equation x – y + 2(1 – ) = 0, R. Further, it is found that the foot of the perpendicular from the point (2, 2) upon any tangent to the ellipse lies on the circle x2 + y2 – 4y – 5 = 0.
1. The eccentricity of the ellipse is equal to (A) 1/3 (B) 1/ 3 (C) 2/3 (D) 1/2 2.
The area of the largest triangle that an incident ray and the corresponding reflected ray can enclose with the axis of the ellipse is equal to (A) 4 5 (B) 2 5 (C) 5 (D) None of these 3.
The total distance travelled by an incident ray and the corresponding reflected ray is the least if the point of incidence coincides with (A) an end of the minor axis (B) an end of the major axis (C) an end of the latus rectum (D) none of these Comprehension # 2 An ellipse whose distance between foci S and S' is 4 units is inscribed in the triangle ABC touching the sides AB, AC and BC at P, Q and R. If centre of ellipse is at origin and major axis along x-axis, SP + S'P = 6.
On the basis of above information, answer the following questions : 1.
If BAC = 90°, then locus of point A is -(A) x2 + y2 = 12 (B) x2 + y2 = 4 (C) x2 + y2 = 14 (D) none of these 2.
If chord PQ subtends 90° angle at centre of ellipse, then locus of A is -(A) 25x2 + 81y2 = 620 (B) 25x2 + 81y2 = 630 (C) 9x2 + 16y2 = 25 (D) none of these 3.
If difference of eccentric angles of points P and Q is 60°, then locus of A is -(A) 16x2 + 9y2 = 144 (B) 16x2 + 45y2 = 576 (C) 5x2 + 9y2 = 60 (D) 5x2 + 9y2 = 15 ELLIPSE 297 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Exercise # 4 [Subjective Type Questions] 1.
A circle is concentric with the ellipse 2 2 2 2 x y 1 a b and passes through the foci F1 and F2 of the ellipse. Two curves intersect at four points. Let P be any point of intersection. If the major axis of the ellipse is 15 and the area of triangle PF1F2 is 26, then find the value of 4a2 – 4b2.
2.
Find the set of value(s) of for which the point 5 7 , 4 lies inside the ellipse 2 2 x y 1 25 16 .
3.
The tangent at any point P of a circle x2 + y2 = a2 meets the tangent at a fixed point A (a, 0) in T and T is joined to B , the other end of the diameter through A . Prove that the locus of the intersection of AP and BT is an ellipse whose eccentricity is 1 2 .
4.
The tangent at the point on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is (1 + sin²)1/2.
5.
Let P be a point on an ellipse with eccentricity 1/2, such that PS1S2 = , PS2S1 = , and S1PS2 = , where S and S2 are the foci of the ellipse. Then prove that cot(/2), cot(/2), and cot(/2) are in AP.
6.
The tangent at P 16 4 cos , sin 11 to the ellipse 16x2 + 11y2 = 256 is also a tangent to the circle x2 + y2 2x 15 = 0. Find . Find also the equation to the common tangent.
7.
ABC is an isosceles triangle with its base BC twice its altitude. A point P moves within the triangle such that the square of its distance from BC is half the area of rectangle contained by its distances from the two sides . Show that the locus of P is an ellipse with eccentricity 2 3 passing through B & C.
8.
Prove that the length of the focal chord of the ellipse 2 2 2 2 x y 1 a b which is inclined to the major axis at angle is 2 2 2 2 2 2 a b a sin b cos .
9.
‘O’ is the origin & also the centre of two concentric circles having radii of the inner & the outer circle as ‘a’ & ‘b’ respectively . A line OPQ is drawn to cut the inner circle in P & the outer circle in Q. PR is drawn parallel to the y-axis & QR is drawn parallel to the x axis . Prove that the locus of R is an ellipse touching the two circles. If the foci of this ellipse lie on the inner circle , find the ratio of inner : outer radii & find also the eccentricity of the ellipse.
10.
A tangent is drawn to the ellipse 2 2 2 2 x y 1 a b to cut the ellipse 2 2 2 2 x y 1 c d at the points P and Q. If the tangents at P and Q to the ellipse 2 2 2 2 x y 1 c d intersect at right angle, then prove that 2 2 2 2 a b 1 c d .
MATHS FOR JEE MAIN & ADVANCED 298 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 11.
If the normals at the points P, Q, R with eccentric angles , , on the ellipse 2 2 x a + 2 2 y b = 1 are concurrent, then show that sin cos sin 2 sin cos sin 2 sin cos sin 2 = 0.
12.
If tangent drawn at a point (t2, 2t ) on the parabola y2 = 4x is same as the normal drawn at a point 5 cos , 2 sin on the ellipse 4x2 + 5y2 = 20, then find the values of t & .
13.
Let d be the perpendicular distance from the centre of the ellipse 2 2 x a + 2 2 y b = 1 to the tangent drawn at a point P on the ellipse. If F1 & F2 are the two foci of the ellipse, then show that (PF1 – PF2)2 = 4a2 2 2 b 1 d 14.
The tangent and normal to the ellipse x2 + 4y2 = 4 at a point P() on it meet the major axis in Q and R respectively.
If QR = 2, show that the eccentric angle of P is given by cos = ± (2/3).
15.
If a triangle is inscribed in an ellipse and two of its sides are parallel to the given straight lines, then prove that the third side touches the fixed ellipse.
16.
Consider the family of circles, x2 + y2 = r2 , 2 < r < 5. If in the first quadrant, the common tangent to a circle of the family and the ellipse 4x2 + 25y2 = 100 meets the co-ordinate axes at A & B , then find the equation of the locus of the mid–point of AB.
17.
If the normal at a point P on the ellipse 2 2 2 2 x y 1 a b of semi axes a , b & centre C cuts the major & minor axes at G & g , show that a2. (CG)2 + b2. (Cg)2 = (a2 b2)2 . Also prove that CG = e2CN, where PN is the ordinate of P. (N is foot of perpendicular from P on its major axis.) 18.
Find the equation of the largest circle with centre (1, 0 ) that can be inscribed in the ellipse x2 + 4y2 = 16 .
19.
A ray emanating from the point (–4, 0) is incident on the ellipse 9x2 + 25y2 = 225 at the point P with abscissa 3. Find the equation of the reflected ray after first reflection.
20.
Tangents are drawn to the ellipse from the point 2 2 2 2 2 a / a b , a b . Prove that the tangents intercept on the ordinate througbh the nearer focus a distance equal to the major axis.
ELLIPSE 299 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Exercise # 5 Part # I [Previous Year Questions] [AIEEE/JEE-MAIN] 1.
If distance between the foci of an ellipse is equal to its minor axis, then eccentricity of the ellipse is-[AIEEE-2002] (1) e = 1 2 (2) e = 1 3 (3) e = 1 4 (4) e = 1 6 2.
The equation of an ellipse, whose major axis = 8 and eccentricity = 1/2 is- (a > b) [AIEEE-2002] (1) 3x2 + 4y2 = 12 (2) 3x2 + 4y2 = 48 (3) 4x2 + 3y2 = 48 (4) 3x2 + 9y2 = 12 3.
The eccentricity of an ellipse, with its centre at the origin, is 1/2. If one of the directirices is x = 4, then the equation of the ellipse is-[AIEEE-2004] (1) 3x2 + 4y2 = 1 (2) 3x2 + 4y2 = 12 (3) 4x2 + 3y2 = 12 (4) 4x2 + 3y2 = 1 4.
An ellipse has OB as semi minor axis, F and F' its focii and the angle FBF' is a right angle. Then the eccentricity of the ellipse is- [AIEEE-2005, IIT-1997] (1) 1 2 (2) 1 2 (3) 1 4 (4) 1 3 5.
In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is-[AIEEE-2006] (1) 1 2 (2) 4 5 (3) 1 5 (4) 3 5 6.
A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is 1/2. Then the length of the semi-major axis is-[AIEEE-2008] (1) 8/3 (2) 2/3 (3) 4/3 (4) 5/3 7.
The ellipse x2 + 4y2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is :-[AIEEE-2009] (1) 4x2 + 48y2 = 48 (2) 4x2 + 64y2 = 48 (3) x2 + 16y2 = 16 (4) x2 + 12y2 = 16 8.
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (–3, 1) and has eccentricity 2 / 5 is :-[AIEEE-2011] (1) 3x2 + 5y2 – 15 = 0 (2) 5x2 + 3y2 – 32 = 0 (3) 3x2 + 5y2 – 32 = 0 (4) 5x2 + 3y2 – 48 = 0 9.
An ellipse is drawn by taking a diameter of the circle (x – 1)2 + y2 = 1 as its semi-minor axis and a diameter of the circle x2 + (y – 2)2 = 4 as its semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is : [AIEEE-2012] (1) x2 + 4y2 = 16 (2) 4x2 + y2 = 4 (3) x2 + 4y2 = 8 (4) 4x2 + y2 = 8 10.
Statement–I : An equation of a common tangent to the parabola y2 = 16 3 x and the ellipse 2x2 + y2 = 4 is y = 2x + 2 3 .
Statement–II : If the line y = mx + 4 3 m , (m 0) is a common tangent to the parabola y2 = 16 3 x and the ellipse 2x2 + y2 = 4, then m satisfies m4 + 2m2 = 24.
[AIEEE-2012] MATHS FOR JEE MAIN & ADVANCED 300 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 (1) Statement–I is true, Statement–II is false.
(2) Statement–I is false, Statement–II is true.
(3) Statement–I is true, Statement–II is true ; Statement–II is a correct explanation for Statement–I.
(4) Statement–I is true, Statement–II is true ; Statement–II is not a correct explanation for Statement–I.
11.
The equation of the circle passing through the foci of the ellipse 2 2 x y 1 16 9 and having centre at (0, 3) is : [JEE (Main)-2013] (1) x2 + y2 – 6y – 7 = 0 (2) x2 + y2 – 6y + 7 = 0 (3) x2 + y2 – 6y – 5 = 0 (4) x2 + y2 – 6y + 5 = 0 12.
The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it is : [JEE Main 2014] (1) (x2 – y2) = 6x2 + 2y2 (2) (x2 – y2)2 = 6x2 – 2y2 (3) (x2 + y2)2 = 6x2 + 2y2 (4) (x2 + y2)2 = 6x2 – 2y2 13.
The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse 2 2 x y 1 9 5 , is [JEE Main 2015] (1) 27 2 (2) 27 (3) 27 4 (4) 18 Part # II [Previous Year Questions][IIT-JEE ADVANCED] 1.
Let ABC be an equilateral triangle inscribed in the circle x2 + y2 = a2.Suppose perpendiculars from A, B, C to the major axis of the ellipse, 2 2 2 2 x y 1 a b , (a > b) meet the ellipse respectively at P,Q,R so that P, Q,R lie on the same side of the major axis as A, B,C respectively . Prove that the normals to the ellipse drawn at the points P, Q and R are concurrent. [JEE 2000 (Mains)] 2.
Let C1 and C2 be two circles with C2 lying inside C1. A circle C lying inside C1 touches C1 internally and C2 externally.
Identify the locus of the centre of C. [JEE 2001 (Mains)] 3.
Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse to the point of contact meet on the corresponding directrix. [JEE 2002 (Mains)] 4.
Tangent is drawn to ellipse 2 2 x y 1 27 at (3 3 cos , sin ) (where (0, /2)). Then the value of , such that sum of intercepts on axes made by this tangent is least is - [JEE 2003 (Screening)] (A) 3 (B) 6 (C) 8 (D) 4 5.
The area of the quadrilateral formed by the tangents at the end points of the latus rectum of the ellipse 2 2 x y 1 9 5 , is - [JEE 2003 (Screening)] (A) 27/4 sq. units (B) 9 sq. units (C) 27/2 sq. units (D) 27 sq. units ELLIPSE 301 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 6.
Find a point on the curve x2 + 2y2 = 6 whose distance from the line x + y = 7, is as small as possible.
[JEE 2003 (Main)] 7.
Locus of the mid points of the segments which are tangents to the ellipse 2 2 1 x y 1 2 and which are intercepted between the coordinate axes is - [JEE 2004 (Screening)] (A) 2 2 1 1 x y 1 2 4 (B) 2 2 1 1 x y 1 4 2 (C) 2 2 1 1 1 3x 4 y (D) 2 2 1 1 1 2x 4 y 8.
The minimum area of triangle formed by tangent to the ellipse x a y b 2 2 2 2 1 and coordinate axes -(A) ab (B) a b 2 2 2 (C) ( ) a b 2 2 (D) a ab b 2 2 3 [JEE 2005 (Screening)] 9.
Find the equation of the common tangent in 1st quadrant to the circle x2 + y2 = 16 and the ellipse x y 2 2 25 4 1 . Also find the length of the intercept of the tangent between the coordinate axes. [JEE 2005 (Mains)] 10.
Let P(x1, y1) and Q(x2, y2), y1 < 0, y2 < 0, be the end points of the latus rectum of the ellipse x2+4y2=4. The equations of parabolas with latus rectum PQ are - [JEE 2008] (A) x2 + 2 3 y = 3 + 3 (B) x2 – 2 3 y = 3 + 3 (C) x2 + 2 3 y = 3 – 3 (D) x2 – 2 3 y = 3 – 3 11.
The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x2 + 9y2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and the origin O is :- [JEE 2009] (A) 31 10 (B) 29 10 (C) 21 10 (D) 27 10 12.
The normal at a point P on the ellipse x2 + 4y2 = 16 meets the x-axis at Q. If M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points -[JEE 2009] (A) 3 5 2 , 2 7 (B) 3 5 19 2 4 (C) 1 2 3, 7 (D) 4 3 2 3, 7 MATHS FOR JEE MAIN & ADVANCED 302 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 Paragraph for Question 13 to 15 [JEE 2010] Tangents are drawn from the point P(3, 4) to the ellipse 2 2 x y 1 9 4 touching the ellipse at points A and B.
13.
The coordinates of A and B are (A) (3, 0) and (0, 2) (B) 8 2 161 , 5 15 and 9 8 , 5 5 (C) 8 2 161 , 5 15 and (0, 2) (D) (3, 0) and 9 8 , 5 5 14.
The orthocenter of the triangle PAB is (A) 8 5, 7 (B) 7 25 , 5 8 (C) 11 8 , 5 5 (D) 8 7 , 25 5 15.
The equation of the locus of the point whose distances from the point P and the line AB are equal, is (A) 9x 2 + y 2 – 6xy – 54x – 62y + 241 = 0 (B) x 2 + 9y 2 + 6xy – 54x + 62y – 241 = 0 (C) 9x 2 + 9y 2 – 6xy – 54x – 62y – 241 = 0 (D) x 2 + y 2 – 2xy + 27x + 31y – 120 = 0 16.
The ellipse E1 : 2 2 x x 1 9 4 is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E2 passing through the point (0,4) circumscribes the rectangle R. The eccentricity of the ellipse E2 is - [JEE 2012] (A) 2 2 (B) 3 2 (C) 1 2 (D) 3 4 17.
A vertical line passing through the point (h,0) intersects the ellipse 2 2 x y 1 4 3 at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If (h) = area of the triangle PQR, 1 1 / 2 h 1 max (h) and 2 1 / 2 h 1 min (h), then 1 2 8 8 5 [JEE-Ad. 2013] 18.
Suppose that the foci of the ellipse 2 2 x y 9 5 = 1 are (f1, 0) and (f2, 0) where f1 > 0 and f2 < 0. Let P1 and P2 be two parabolas with a common vertex at (0, 0) and with foci at (f1, 0) and (2f2, 0), respectively. Let T1 be a tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through (f1, 0). If m1 is the slope of T1 and m2 is the slope of T2, then the value of 2 2 2 1 1 m m is [JEE-Ad. 2015] 19.
Let E1 and E2 be two ellipses whose centres are at origin. The major axes of E1 and E2 lie along the x-axis and the y-axis, respectively. Let S be the circle x2 + (y – 1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R, respectively. Suppose that PQ = 2 2 3 . If e1 and e2 are the eccentricites of E1 and E2, respectively, then the correct expression(s) is (are) [JEE-Ad. 2015] ELLIPSE 303 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 (A) 2 2 1 2 43 e e 40 (B) 1 2 7 e e 2 10 (C) 2 2 1 2 5 e e 8 (D) 1 2 3 e e 4 Comprehension Let F1(x1, 0) and F2(x2, 0) for x1 < 0 and x2 > 0, be the foci of the ellipse 2 2 x y 9 8 = 1. Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.
20.
The orthocentre of the triangle F1MN is (A) 9 ,0 10 (B) 2 ,0 3 (C) 9 ,0 10 (D) 2 , 6 3 21.
If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is (A) 3 : 4 (B) 4 : 5 (C) 5 : 8 (D) 2 : 3 MATHS FOR JEE MAIN & ADVANCED 304 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 M OCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE 1.
Find the locus of point of intersection of pair of tangents to the ellipse if the sum of the ordinates of the point of contact is b.
(A) 2 2 2 2 x y b a b 4y = 1 (B) 2 2 2 2 x y b a b 2y = 1 (C) 2 2 2 2 x y 2b a b y = 1 (D) 2 2 2 2 x y b 4 a b 2y 2.
If , are eccentric angles of the extremities of a focal chord of an ellipse, then eccentricity of the ellipse is (A) cos cos cos( ) (B) sin sin sin( ) (C) sec + sec (D) sin sin sin( ) 3.
If f(x) is a decreasing function then the set of values of ‘k’, for which the major axis of the ellipse 2 2 x f (k 2k 5) + 2 y f (k 11) = 1 is the x-axis, is: (A) k (–2, 3) (B) k (–3, 2) (C) k (–, –3) U (2, ) (D) k (–, –2) U (3, ) 4.
If circumcentre of an equilateral triangle inscribed in 2 2 x a + 2 2 y b = 1, with vertices having eccentric angles , , respectively is (x1, y1), then cos cos + sin sin = (A) 2 1 2 9x a + 2 1 2 9y b + 3 2 (B) 9x1 2 – 9y1 2 + a2 b2 (C) 2 1 9x a + 2 1 9y b + 3 (D) 2 1 2 9x 2a + 2 1 2 9y 2b – 3 2 5.
Q is a point on the auxiliary circle corresponding to the point P of the ellipse 2 2 2 2 x y a b = 1. If T is the foot of the perpendicular dropped from the focus S onto the tangent to the auxiliary circle at Q then the SPT is: (A) isosceles (B) equilateral (C) right angled (D) right isosceles 6.
The distance of the point ( 6cos , 2sin ) on the ellipse x2/6 + y2/2 = 1 from the centre of the ellipse is 2, if (A) = /3 (B) = /6 (C) = 5/4 (D) = 5/3 7.
If circumcentre of an equilateral triangle inscribed in 2 2 x a + 2 2 y b = 1, with vertices having eccentric angles , , respectively is (x1, y1), then cos cos + sin sin = (A) 2 1 2 9x a + 2 1 2 9y b + 3 2 (B) 9x1 2 – 9y1 2 + a2 b2 (C) 2 1 9x a + 2 1 9y b + 3 (D) 2 1 2 9x 2a + 2 1 2 9y 2b – 3 2 ELLIPSE 305 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 8.
The locus of extremities of latus rectum of the family of ellipse b2x2 + y2 = a2b2 where b is a parameter (b2 < 1), is-(A) x2 ± a2y2 = a2 (B) x2 ± ay = a2 (C) x ± ay2 = a2 (D) none of these 9.
S1 : If from a point P(0, ) two normals other than axes are drawn to ellipse 2 2 x y 25 16 = 1, where || k, then least value of k is 9 4 S2 : The minimum and maximum distances of a point (1, 2) from the ellipse 4x2 + 9y2 + 8x – 36y + 4 = 0 are L and G, then G – L is equal to 4 S3 : If the length of latus rectum of an ellipse is one-third of its major axis. Its eccentricity is equal to 2 3 S4 : The set of all positive values of a for which (13x – 1)2 + (13y – 2)2 = 2 5x 12y 1 a represents an ellipse is (1, 2) (A) TTFF (B) TTFT (C) TTTF (D) TFTF 10.
A series of concentric ellipses E1, E2, ......, En are drawn such that En touches the extremities of the major axis of En –1 and the foci of En coincide with the extremities of minor axis of En – 1. If the eccentricity of the ellipses is independent of n, then the value of the eccentricity, is (A) 5 3 (B) 5 1 2 (C) 5 1 2 (D) 1 5 SECTION - II : MULTIPLE CORRECT ANSWER TYPE 11.
If P is a point of the ellipse 2 2 x a + 2 2 y b = 1, whose focii are S and S. Let PSS = and PSS = , then (A) PS + PS = 2a, if a > b (B) PS + PS = 2b, if a < b (C) tan 2 tan 2 = 1 e 1 e (D) tan 2 tan 2 = 2 2 2 a b b [a – 2 2 a b ] when a > b 12.
The parametric angle , where – – , of the point on the ellipse 2 2 2 2 x y 1 a b at which the tangent drawn cuts the intercept of minimum length on the coordinate axes, is/are (A) tan–1 b a (B) –tan–1 b a (C) – tan–1 b a (D) + tan–1 b a 13.
Let F1, F2 be two focii of the ellipse and PT and PN be the tangent and the normal respectively to the ellipse at point P then (A) PN bisects F1 PF2 (B) PT bisects F1PF2 (C) PT bisects angle (180° – F1PF2) (D) None of these MATHS FOR JEE MAIN & ADVANCED 306 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 14.
Let F1, F2 be two focii of the ellipse and PT and PN be the tangent and the normal respectively to the ellipse at point P then (A) PN bisects F1 PF2 (B) PT bisects F1PF2 (C) PT bisects angle (180° – F1PF2) (D) None of these 15.
Let A() and B() be the extremeties of a chord of an ellipse . If the slope of AB is equal to the slope of the tangent at a point C() on the ellipse, then the value of , is (A) 2 (B) 2 (C) 2 + (D) 2 – SECTION - III : ASSERTION AND REASON TYPE 16.
Statement-I: If P 3 3 ,1 2 is a point on the ellipse 4x2 + 9y2 = 36. Circle drawn taking AP as diameter touches another circle x2 + y2 = 9, where A (– 5 , 0).
Statement-II : Circle drawn with focal radius as diameter touches the auxiliary circle.
(A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I.
(B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True 17.
Tangents are drawn from the point P 3, 2 to an ellipse 4x2 + y2 = 4.
Statement-I : The tangents are mutually perpendicular.
Statement-II : The locus of the points from which mutually perpendicular tangents can be drawn to given ellipse is x2 + y2 = 5.
(A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I.
(B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True 18.
Statement-I : Let tangent at a point P on the ellipse, which is not an extremity of major axis, meets a directrix at T. If circle drawn on PT as diameter cuts the directrix at Q, then PQ = ePS, where S is the focus corresponding to the directrix.
Statement-II : Let tangent at a point P on an ellipse, which not an extremity of major axis, meets the directrices at T and T. Then PT substends a right angle at the focus corresponding the directrix at which T lies.
(A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I.
(B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True ELLIPSE 307 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 19.
Statement-I: The ellipse 2 x 16 + 2 y 9 = 1 and 2 x 9 + 2 y 16 = 1 are congruent.
Statement-II: The ellipse 2 x 16 + 2 y 9 = 1 and 2 x 9 + 2 y 16 = 1 have the same eccentricity.
.
(A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I.
(B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True 20.
Statement-I : A triangle ABC right angled at A moves so that its perpendicular sides touch the curve 2 2 x a + 2 2 y b = 1 all the time. Then loci of the points A, B and C are circle.
Statement-II : Locus of point of intersection of two perpendicular tangents to the conic is director circle (A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I.
(B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True SECTION - IV : MATRIX - MATCH TYPE 21.
Column-I Column-II (A) The eccentricity of the ellipse which meets the straight line (p) 1 2 2x – 3y = 6 on the X-axis and the straight line 4x + 5y = 20 on the Y-axis and whose principal axes lie along the coordinate axes, is (B) A bar of length 20 units moves with its ends on two fixed straight lines (q) 1 2 at right angles. A point P marked on the bar at a distance of 8 units from one end describes a conic whose eccentricity is (C) If one extremity of the minor axis of the ellipse 2 2 2 2 x y 1 a b and the foci (r) 5 3 form an equilateral triangle, then its eccentricity, is (D) There are exactly two points on the ellipse 2 2 2 2 x y 1 a b (s) 7 4 whose distance from the centre of the ellipse are greatest and equal to 2 2 a 2b 2 . Eccentricity of this ellipse is equal to MATHS FOR JEE MAIN & ADVANCED 308 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 22.
Column – Column – (A) If the angle between the straight angle lines joining foci (p) 4 and one of the ends of the minor axis of the ellipse 2 2 2 2 x y a b = 1 is 90º. Find its eccentricity.
(B) For an ellipse 2 2 x y 1 9 4 with vertices A and A A, tangent (q) 2 drawn at the point P in the first quadrant meets the y-axis in Q and the chord AP meets the y-axis in M. If ‘O’ is the origin, then OQ2 – MQ2 equals to (C) The x-coordinate of points on the axis of the parabola (r) 1 2 4y2 – 32x + 4y + 65 = 0 from which all the three normals to the parabola are real is (D) The area of the parallelogram inscribed in the ellipse (s) 7 2 2 2 2 x y 2 (1/ 2) = 1 whose diagonals are the conjugate diameters of the ellipse is given by (t) 8 SECTION - V : COMPREHENSION TYPE 23.
Read the following comprehensions carefully and answer the questions.
Consider the ellipse 2 2 x y 9 4 = 1 and the parabola y2 = 2x. They intersect at P and Q in the first and fourth quadrants respectively. Tangents to the ellipse at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S.
1.
The ratio of the areas of the triangles PQS and PQR, is (A) 1 : 3 (B) 1 : 2 (C) 2 : 3 (D) 3 : 4 2.
The area of quadrilateral PRQS, is (A) 3 15 2 (B) 15 3 2 (C) 5 3 2 (D) 5 15 2 3.
The equation of circle touching the parabola at upper end of its latus rectum and passing through its vertex, is (A) 2x2 + 2y2 – x – 2y = 0 (B) 2x2 + 2y2 +4x – 9 2 y = 0 (C) 2x2 + 2y2 + x – 3y = 0 (D) 2x2 + 2y2 – 7x + y = 0 ELLIPSE 309 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 24.
Read the following comprehensions carefully and answer the questions.
Second degree equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents an ellipse if a h g h b f 0 g f c & h2 < ab. Intersection of major axis and minor axis gives centre of ellipse 1.
There are exactly ‘n’ integral values of for which equation x2 + xy + y2 = 1 represents an ellipse then ‘n’ must be (A) 0 (B) 1 (C) 2 (D) 3 2.
Length of the longest chord of the ellipse x2 + y2 + xy = 1 is __ (A) 2 (B) 1 2 (C) 2 2 (D) 1 3.
Length of the chord perpendiuclar to longest chord as in above question and pasing through centre of ellipse is __ (A) 1 2 (B) 3 2 (C) 2 2 3 (D) 1 3 25.
Read the following comprehensions carefully and answer the questions.
A bird flies on ellipse ax2 + by2 = 1 & z = 5 3 (b > a > 0) whose eccentricity is 1 2 . An observer stands at a point P(, , 0) where maximum and minimum angle of elevation of the bird are 60º and 30º when bird is at Q and R respectively on its path and Q and R are projection of Q and R on x- y plane, P, Q R are collinear & the distance between Q and R is maximum Let be the angle elevation of the bird when it is at a point on the arc of the ellipse exactly mid-way between Q & R. It is given that a2 + b2 – 1 > 0 1.
If > 0, then equation of the line along which minimum angle of elevation is observed, is (A) x 3 3 = z 3 1 (B) x 13 3 = z 3 1 , y = 0 (C) x 10 3 = z 3 1 , y = 0 (D) x 10 3 = y 0 = z 3 1 2.
Equation of plane which touches the ellipse at Q and passes through P ( > 0) is (A) – 3 x + y + z – 10 3 = 0 (B) 3 x + y + z – 10 3 = 0 (C) 3 x + z – 10 3 = 0 (D) 3 x + y – 10 3 = 0 3.
Value of tan , is (A) 2 3 (B) 3 2 (C) 2 3 (D) 6 5 MATHS FOR JEE MAIN & ADVANCED 310 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 SECTION - VI : INTEGER TYPE 26.
Number of distinct normal lines that can be drawn to ellipse 2 2 x y 169 25 = 1 from the point P (0, 6) is 27.
Find the number of integral values of parameter 'a' for which three chords of the ellipse 2 2 x 2a + 2 2 y a = 1 (other than its diameter) passing through the point 2 a P 11a, 4 are bisected by the parabola y2 = 4ax.
28.
Origin O is the centre of two concentric circles whose radii are a & b respectively, a < b. A line OPQ is drawn to cut the inner circle in P & the outer circle in Q. PR is drawn parallel to the y axis & QR is drawn parallel to the x axis. The locus of R is an ellipse touching the two circles. If the focii of this ellipse lie on the inner circle, if eccentricity is 2 , then find 29.
Consider two concentric circles S1 : | z | = 1 and S2 : | z | = 2 on the Argand plane. A variable parabola is drawn through the points where 'S1' meets the real axis and having arbitrary tangent drawn to 'S2' as its directrix. If the locus of the focus of the parabola is a conic C then find the area of the quadrilateral formed by the tangents at the ends of the latus-rectum of conic C.
30.
Number of points on the ellipse 2 x 50 + 2 y 20 = 1 from which pair of perpendicular tangents may be drawn to the ellipse 2 2 x y 16 9 = 1 is ELLIPSE 311 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 ANSWER KEY EXERCISE - 1 1.
A 2.
D 3.
A 4.
C 5.
B 6.
C 7.
B 8.
B 9.
A 10. A 11. A 12. A 13. B 14. B 15. C 16. C 17. D 18. B 19. B 20. C 21. C 22. C 23. C 24. D 25. B 26. A 27. A 28. C 29. C 30. C EXERCISE - 2 : PART # I 1.
AC 2.
AC 3.
ACD 4.
CD 5.
ABC 6.
AC 7.
CD 8.
AC 9.
ABC 10. AB 11. AB 12. ACD 13. AC 14. AC 15. AB 16. AC 17. ABC 18. ACD 19. AC 20. AD PART - II 1.
D 2.
A 3.
A 4.
D 5.
A EXERCISE - 3 : PART # I 1.
A q B q C s D p 2.
A q,s B p,r C r D s PART - II Comprehension # 1 : 1.
C 2.
B 3.
D Comprehension # 2 : 1.
C 2.
B 3.
C EXERCISE - 5 : PART # I 1.
1 2.
1 3.
2 4.
1 5.
4 6.
1 7.
4 8.
1 10. 3 11. 1 12. 3 13. 2 PART - II 2.
Locus is an ellipse with foci as the centres of the circles C1 and C2.
4.
B 5.
D 6.
(2,1)7.
D 8.
A 9.
2 7 y x 4 3 3 ; 14 3 10. B, C 11. D 12. C 13. D 14. C 15. A 16. C 17. 9 18. 4 19. AB 20. A 21. C MATHS FOR JEE MAIN & ADVANCED 312 Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 MOCK TEST 1.
B 2.
D 3.
B 4.
D 5.
A 6.
C 7.
D 8.
B 9.
A 10. B 11. A,B,C 12. A,B,C 13. A,C 14. A,C 15. A,C 16. A 17. A 18. D 19. B 20. D 21. A s B r C p D q 22. A r B p C s,t D q 23. 1.
C 2.
B 3.
D 24. 1.
D 2.
C 3.
C 25. 1.
B 2.
C 3.
C 26. 3 27. 2 28. 1/2 29. 16 30. 4 |
8852 | https://www.zhihu.com/question/321441965 | 如何证明不共线的三个点确定一个平面？ - 知乎
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将不共线的三点坐标代入到平面方程，得到待定系数方程组，证明该方程组对应矩阵的秩是3而未知数个数是4（ A x+B y+C z+D=0 Ax+By+Cz+D=0 ），从而证明基础解系只有一个向量，从而证明平面是唯一的。
具体过程，设点 M(x 1,y 1,z 1),N(x 2,y 2,z 2),P(x 3,y 3,z 3)M(x_1, y_1, z_1), N(x_2, y_2, z_2), P(x_3, y_3, z_3) ，将 M,N,P M, N, P 分别代入到 A x+B y+C z+D=0 Ax+By+Cz+D=0 ，得 (x 1 y 1 z 1 1 x 2 y 2 z 2 1 x 3 y 3 z 3 1)(A B C D)=(0 0 0)\left(\begin{matrix}x_1&y_1&z_1&1\x_2&y_2&z_2&1\x_3&y_3&z_3&1\end{matrix}\right)\left(\begin{matrix}A\B\C\D\end{matrix}\right)=\left(\begin{matrix}0\0\0\end{matrix}\right)\为方便起见，需对这个方程组进行改造，也就是对系数矩阵先进行一下初等列变换（交换列） (1 x 1 y 1 z 1 1 x 2 y 2 z 2 1 x 3 y 3 z 3)(D A B C)=(0 0 0)\left(\begin{matrix}1&x_1&y_1&z_1\1&x_2&y_2&z_2\1&x_3&y_3&z_3\end{matrix}\right)\left(\begin{matrix}D\A\B\C\end{matrix}\right)=\left(\begin{matrix}0\0\0\end{matrix}\right)\
对系数矩阵进行初等行变换，第二行减第一行，第三行减第一行，得(1 x 1 y 1 z 1 0 x 2−x 1 y 2−y 1 z 2−z 1 0 x 3−x 1 y 3−y 1 z 3−z 1)\left(\begin{matrix}1&x_1&y_1&z_1\0&x_2-x_1&y_2-y_1&z_2-z_1\0&x_3-x_1&y_3-y_1&z_3-z_1\end{matrix}\right)\按照矩阵的秩的定义，如果某矩阵存在一个行列式的值不为零的 N N 阶子矩阵，且任何 (N+1)(N+1)阶矩阵的行列式的值均为零，那么 该矩阵的秩就是 N N ．这里我们想证明上面这个矩阵的秩是 3 3 ，因而我们只要尝试通过计算找出一个行列式的值不为零的 3 3 阶子矩阵即可，这里我们选取如下三个 3 3 阶子矩阵 (1 x 1 y 1 0 x 2−x 1 y 2−y 1 0 x 3−x 1 y 3−y 1),(1 x 1 z 1 0 x 2−x 1 z 2−z 1 0 x 3−x 1 z 3−z 1),(1 y 1 z 1 0 y 2−y 1 z 2−z 1 0 y 3−y 1 z 3−z 1)\left(\begin{matrix}1&x_1&y_1\0&x_2-x_1&y_2-y_1\0&x_3-x_1&y_3-y_1\end{matrix}\right),\left(\begin{matrix}1&x_1&z_1\0&x_2-x_1&z_2-z_1\0&x_3-x_1&z_3-z_1\end{matrix}\right),\left(\begin{matrix}1&y_1&z_1\0&y_2-y_1&z_2-z_1\0&y_3-y_1&z_3-z_1\end{matrix}\right)\ 它们对应的行列式分别是（设这三个行列式的值分别是 \alpha, \beta, \gamma ） \alpha=\left|\begin{matrix}1&x_1&y_1\0&x_2-x_1&y_2-y_1\0&x_3-x_1&y_3-y_1\end{matrix}\right|,\beta=\left|\begin{matrix}1&x_1&z_1\0&x_2-x_1&z_2-z_1\0&x_3-x_1&z_3-z_1\end{matrix}\right|,\gamma=\left|\begin{matrix}1&y_1&z_1\0&y_2-y_1&z_2-z_1\0&y_3-y_1&z_3-z_1\end{matrix}\right|\按第一列展开，我们可以迅速得到 \alpha=\left|\begin{matrix}1&x_1&y_1\0&x_2-x_1&y_2-y_1\0&x_3-x_1&y_3-y_1\end{matrix}\right|=\left|\begin{matrix}x_2-x_1&y_2-y_1\x_3-x_1&y_3-y_1\end{matrix}\right|,\ \beta=\left|\begin{matrix}1&x_1&z_1\0&x_2-x_1&z_2-z_1\0&x_3-x_1&z_3-z_1\end{matrix}\right|=\left|\begin{matrix}x_2-x_1&z_2-z_1\x_3-x_1&z_3-z_1\end{matrix}\right|,\ \gamma=\left|\begin{matrix}1&y_1&z_1\0&y_2-y_1&z_2-z_1\0&y_3-y_1&z_3-z_1\end{matrix}\right|=\left|\begin{matrix}y_2-y_1&z_2-z_1\y_3-y_1&z_3-z_1\end{matrix}\right|\
因为 M, N, P 不共线，所以 \overrightarrow{MN} 和 \overrightarrow{MP} 线性无关，也就是说 \overrightarrow{MN} 和 \overrightarrow{MP} 不平行。一开始我们设的是点 M(x_1, y_1, z_1), N(x_2, y_2, z_2), P(x_3, y_3, z_3) ，这样我们有 \overrightarrow{MN}=(x_2-x_1, y_2-y_1, z_2-z_1), \overrightarrow{MP}=(x_3-x_1, y_3-y_1,z_3-z_1)
根据向量积的定义，我们有 \overrightarrow{MN}\times\overrightarrow{MP}=\left(\left|\begin{matrix}y_2-y_1&z_2-z_1\y_3-y_1&z_3-z_1\end{matrix}\right|,\left|\begin{matrix}z_2-z_1&x_2-x_1\z_3-z_1&x_3-x_1\end{matrix}\right|,\left|\begin{matrix}x_2-x_1&y_2-y_1\x_3-x_1&y_3-y_1\end{matrix}\right|\right)\
所以 \overrightarrow{MN}\times\overrightarrow{MP}=(\alpha, -\beta, \gamma) ，因为 \overrightarrow{MN} 和 \overrightarrow{MP} 不平行，所以 \overrightarrow{MN}\times\overrightarrow{MP}\neq\vec{0} ，所以 \alpha, \beta, \gamma 不全为零，即这三个行列式的值不全为零，由此证明，矩阵 \left(\begin{matrix}1&x_1&y_1&z_1\0&x_2-x_1&y_2-y_1&z_2-z_1\0&x_3-x_1&y_3-y_1&z_3-z_1\end{matrix}\right)\ 至少存在一个行列式的值不为零的 3 阶矩阵，即 \text{Rank}\left(\begin{matrix}1&x_1&y_1&z_1\0&x_2-x_1&y_2-y_1&z_2-z_1\0&x_3-x_1&y_3-y_1&z_3-z_1\end{matrix}\right)=3\所以，关于 \left(\begin{matrix}D\A\B\C\end{matrix}\right) 的方程组的基础解系只有一个向量，我们假设这个向量是 \left(\begin{matrix}d\a\b\c\end{matrix}\right) ，其中 a, b, c 不全为零，则有 \left(\begin{matrix}D\A\B\C\end{matrix}\right)=l\cdot\left(\begin{matrix}d\a\b\c\end{matrix}\right)\
综上所述，由不共线的三点 M, N, P 所确定的平面方程是 lax+lby+lcz+ld=0\其中，无论 l 取哪个非零值，这个方程都只表示一个平面，这就证明了不共线的三个点确定一个平面。
类似的，我们也可以证明：两点确定一条直线。
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"不共线的三个点确定一个平面."
在人教A版数学课本里，这叫基本事实！
也就是传说中的公理.
而公理，是人们在千百年的实践中总结出来的，是不需要证明的.
用数学语言讲，公理、公设，不证自明.
莫里斯·克莱因，在其名著《古今数学思想》一书中指出：
欧几里得在其《几何原本》里，很可能没有意识到数学的开头，必然有一些概念是未经定义的（现在称作不定义概念），因此常常不自觉地用物理概念来解释.
但他接受了亚里士多德的关于公理、公设正确性的观点.
"对公理提出证明，是对逻辑理解力的侮辱."
有兴趣的知友，可以从《古今数学思想》一书中受到更多启示，那里是数学家想象力和灵感的思想宝库.
广告 古今数学思想+数学世纪 全套4册 古今数学思想 第三册 天猫 ¥122.80 去购买
数学，需要底蕴！
文化底蕴，历史底蕴.
《古今数学思想》，是一部介绍从古代到20世纪初数学发展的最全面、最权威的著作！
没有任何一本其它的书，可以获得对数学史类似的理解！
您的书架上，值得永久珍藏.
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3 人赞同了该回答
反证法。既在平面a上又在平面b上的点，只能在两个面的交线上。与“不共线”矛盾。得证。
发布于 2019-04-27 08:26
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不共线的两个点可以确定几个平面? 3 个回答
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Holt McDougal Algebra 2: Online Textbook Help
How to Equate Two Complex Numbers
VideoQuizCourse
Instructions
Choose an answer and hit 'next'. You will receive your score and answers at the end.
1. Which of the following is a complex number?
a + 3b
3 - 3i
x + 7y
x - y
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Worksheet
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1. If 3a - 4i is the same complex number as 6 - bi, which of the following is true?
3a = 6
3a = b
4 = -b
4 = i
2. Solve 4a - bi = 8 + 2i
a = 2 and b = -2
a = 2 and b = 2
a = 1 and b = 1
a = 1 and b = -1
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About this quiz and worksheet
About This Quiz & Worksheet
You will quickly determine how much you know about complex numbers using this quiz and worksheet assessment. To do well on this quiz, you will need to identify complex numbers and solve sample problems.
Quiz & Worksheet Goals
For this quiz, you will be asked to do the following:
Identify a complex number from a given list
Select a true statement related to complex numbers based on given information
Solve a sample complex number problem
Note the real part of a given complex number equation
Recognize the imaginary part of complex number equation
Skills Practiced
Reading comprehension - ensure that you draw the most important information from the lesson, such as what a complex number looks like
Problem solving - use acquired knowledge to solve complex number practice problems
Knowledge application - use your knowledge to answer questions about the real and imaginary parts of a complex number equation
Additional Learning
Use the lesson called How to Equate Two Complex Numbers to broaden your knowledge about this subject. You will learn more about:
How a complex number is defined
What rule must be adhered to when solving complex number problems
How to solve for variables in a complex number problem
Practice Exams
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Holt McDougal Algebra 2: Online Textbook Help
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Holt McDougal Algebra 2: Online Textbook Help
14 chapters
233 quizzes
Chapter 1
Holt McDougal Algebra 2 Chapter 1: Foundations for Functions
Types of Numbers & Its Classifications QuizCardinality of a Set | Definition & Examples QuizSet Builder Notation | Definition, Symbols & Examples QuizInterval Notation | Definition, Rules & Examples QuizIntroduction to Groups and Sets in Algebra QuizCommutative Property | Definition, Examples & Applications QuizSquare Root | Definition, Formula & Examples QuizFactoring Radical Expressions QuizSimplifying Square Roots | Overview & Examples QuizAddition and Subtraction Using Radical Notation QuizTranslating Words into Algebraic Expressions | Phrases & Examples QuizEvaluating Algebraic Expressions | Rules & Examples QuizCombining Like Terms in Algebraic Expressions QuizSimplifying and Solving Exponential Expressions QuizProperties of Exponents | Formula & Examples QuizHow to Define a Zero and Negative Exponent QuizSimplifying Expressions with Exponents | Overview & Examples QuizScientific Notation | Definition, Conversion & Examples QuizFunctions: Identification, Notation & Practice Problems QuizDomain & Range of a Function | Definition, Equation & Examples QuizFunction in Math | Definition & Examples QuizTransformations: How to Shift Graphs on a Plane QuizReflection Rules in Math | Graph, Formula & Examples QuizVertical & Horizontal Compression of a Function QuizParent Function | Graphs, Types & Examples Quiz
Chapter 2
Holt McDougal Algebra 2 Chapter 2: Linear Functions
Solving Equations Using Both Addition and Multiplication Principles QuizSolving Equations Containing Parentheses QuizSolutions to Systems of Equations | Overview & Examples QuizProportion | Definition, Formula & Types QuizRatios & Rates | Differences & Examples QuizSimilar Triangles | Definition, Application Problems & Examples QuizLinear Equations | Definition, Formula & Solution QuizForms of a Linear Equation | Overview, Graphs & Conversion QuizUndefined & Zero Slope Graph | Definition & Examples QuizLinear Equation | Parts, Writing & Examples QuizEquation of a Line Using Point-Slope Formula QuizGraphing Inequalities | Definition, Rules & Examples QuizGraphing Inequalities | Overview & Examples QuizVertical & Horizontal Shifts | Definition & Equation QuizCreating & Interpreting Scatterplots: Process & Examples QuizUnderstanding Simple Linear Regression | Graphing & Examples QuizProblem Solving Using Linear Regression: Steps & Examples QuizSolving Absolute Value Functions & Equations | Rules & Examples QuizSolve & Graph an Absolute Value Inequality | Formula & Examples QuizAbsolute Value | Overview & Practice Problems QuizAbsolute Value | Graph & Transformations QuizGraphing Absolute Value Functions | Definition & Translation Quiz
Chapter 3
Holt McDougal Algebra 2 Chapter 3: Linear Systems
System of Equations in Algebra | Overview, Methods & Examples QuizClassifying Linear Systems in Math QuizHow Do I Use a System of Equations? QuizApplying Systems of Linear Equations to Breakeven Point: Steps & Example QuizInequality Notation | Overview & Examples QuizFeasible Region Definition & Graphs QuizObjective Function | Definition & Examples QuizGraphing Points & Lines in Three Dimensions QuizHow to Solve a Linear System in Three Variables With a Solution QuizParametric Equations | Definition & Examples QuizGraphs of Parametric Equations QuizParametric Equations in Applied Contexts QuizEvaluating Parametric Equations: Process & Examples Quiz
Chapter 4
Holt McDougal Algebra 2 Chapter 4: Matrices
Matrix in Math | Definition, Properties & Rules QuizScalars & Matrices: Properties & Application QuizSquare Matrix | Overview & Examples QuizIdentity Matrix | Definition, Properties & Examples QuizUsing Matrices to Complete Translations QuizUsing Matrices to Complete Rotations QuizUsing Matrices to Complete Reflections QuizFinding the Determinant of a Matrix | Properties, Rules & Formula QuizUsing Cramer's Rule with Inconsistent and Dependent Systems QuizInverse Matrix | Definition, Types & Example QuizMultiplicative Inverse of a Matrix | Overview & Examples QuizAugmented Matrix | Definition, Form & Examples QuizRow Operations & Reductions with Augmented Matrices Quiz
Chapter 5
Holt McDougal Algebra 2 Chapter 5: Quadratic Functions
Parabola | Definition & Parabolic Shape Equation QuizReflection Over X & Y Axis | Overview, Equation & Examples QuizTypes of Parabolas | Overview, Graphs & Examples QuizMaximum & Minimum Values of a Parabola | Overview & Formula QuizFactoring Quadratic Equations Using Reverse Foil Method QuizHow to Solve a Quadratic Equation by Factoring QuizHow to Complete the Square | Method & Examples QuizCompleting the Square Practice Problems QuizPrevious lesson Imaginary Numbers | Definition, History & Examples 8:40 minImaginary Numbers | Definition, History & Examples QuizCurrent lesson How to Equate Two Complex Numbers 5:54 minViewing now How to Equate Two Complex Numbers QuizUp next Quadratic Function | Formula, Equations & Examples 9:20 min Watch next lessonQuadratic Function | Formula, Equations & Examples QuizHow to Solve Quadratics with Complex Numbers as the Solution QuizHow to Solve & Graph Quadratic Inequalities QuizQuadratic Inequality | Solution Sets & Examples QuizQuadratic Function | Definition, Formula & Examples QuizHow to Add, Subtract and Multiply Complex Numbers QuizHow to Graph a Complex Number on the Complex Plane QuizComplex Numbers & Conjugates | Multiplication & Division QuizHow to Add Complex Numbers in the Complex Plane Quiz
Chapter 6
Holt McDougal Algebra 2 Chapter 6: Polynomial Functions
What are Polynomials, Binomials, and Quadratics? QuizAdding, Subtracting & Multiplying Polynomials | Steps & Examples QuizHow to Evaluate a Polynomial in Function Notation QuizWhat is the Binomial Theorem? QuizPolynomial Long Division | Overview & Examples QuizSynthetic Division of Polynomials | Method & Examples QuizDividing Polynomials with Long and Synthetic Division: Practice Problems QuizFactoring by Grouping | Definition, Steps & Examples QuizFactor & Remainder Theorem | Definition, Formula & Examples QuizHow to Solve Perfectly Cubed Equations QuizUsing the Rational Zeros Theorem to Find Rational Roots QuizIrrational Root Theorem Application & Examples QuizFundamental Theorem of Algebra | Definition, Examples & Proof QuizWriting a Polynomial Function With Given Zeros | Steps & Examples QuizCubic, Quartic & Quintic Equations | Graphs & Examples QuizBasic Transformations of Polynomial Graphs Quiz
Chapter 7
Holt McDougal Algebra 2 Chapter 7: Exponential and Logarithmic Functions
Exponential Function | Definition, Equation & Examples QuizExponential Growth & Decay | Formula, Function & Graphs QuizInverse Function | Graph & Examples QuizLogarithms | Overview, Process & Examples QuizEvaluating Logarithms | Properties & Examples QuizInverse of Log Functions | Definition & Examples QuizLogarithmic Properties | Product, Power & Quotient Properties QuizPractice Problems for Logarithmic Properties QuizChange-of-Base Formula for Logarithms | Rules & Examples QuizSolving Logarithmic Equations | Properties & Examples QuizExponential Equations | Definition, Solutions & Examples QuizSolving Exponential & Logarithmic Inequalities QuizRule of 72 Definition, Formula & Examples QuizExponentials, Logarithms & the Natural Log QuizNatural Base e | Overview & Importance QuizGraphing Logarithms | Overview, Transformations & Examples QuizGraphing Logarithmic Functions | Overview & Examples QuizStretching & Compression of Logarithmic Graphs Quiz
Chapter 8
Holt McDougal Algebra 2 Chapter 8: Rational and Radical Functions
Solving Direct Variation | Equation, Problems & Examples QuizSolving Equations of Inverse Variation QuizDirect & Inverse Variation | Equations, Relationships & Problems QuizHow to Multiply and Divide Rational Expressions QuizMultiplying and Dividing Rational Expressions: Practice Problems QuizAdding & Subtracting Rational Expressions | Overview & Examples QuizPractice Adding and Subtracting Rational Expressions QuizGraphing Rational Functions That Have Polynomials of Various Degrees: Steps & Examples QuizGraphing Rational Functions That Have Linear Polynomials: Steps & Examples QuizDomain & Range of Rational Functions | Definition & Graph QuizTransformations of the 1/x Function Graph | Definition & Types QuizRational Equations | Definition, Formula & Examples QuizRational Equations: Practice Problems QuizSolving Rational Inequalities | Steps & Examples QuizRational Exponents | Definition, Calculation & Examples QuizSimplifying Algebraic Expressions with Rational Exponent QuizRadical Functions | Graph, Equation & Examples QuizTransformations of Radical Functions QuizSolving Radical Equations | Overview & Examples QuizSolving Radical Equations with Two Radical Terms QuizSolving Radical Inequalities Quiz
Chapter 9
Holt McDougal Algebra 2 Chapter 9: Properties and Attributes of Functions
Graphing Basic Functions QuizPiecewise Function | Definition, Evaluation & Examples QuizPiecewise Functions | Graph & Examples QuizStep Function | Definition, Equation & Graph QuizTranslating Piecewise Functions QuizHow to Compose Functions QuizHow to Add, Subtract, Multiply and Divide Functions QuizApplying Function Operations Practice Problems QuizDomain & Range of Composite Functions | Steps & Examples QuizInverse Functions | Definition, Methods & Calculation QuizOne to One Function | Definition, Graph & Examples Quiz
Chapter 10
Holt McDougal Algebra 2 Chapter 10: Conic Sections
Conic Sections | Definition, Equations & Types QuizHow to Graph a Circle | Equation & Examples QuizTangent of a Circle | Definition, Formula & Examples QuizUnderstanding Circles with Inequalities QuizHow to Write the Equation of an Ellipse in Standard Form QuizEllipse Foci Formula & Calculations QuizHyperbola | Definition, Equation & Graphs QuizHyperbola Standard Form | Definition, Equations & Examples QuizHyperbola Equation | Foci Formula, Parts & Example QuizParabola | Equation, Formula & Examples QuizParabola | Definition, Formula & Examples QuizConic Sections | Overview, Equations & Types Quiz
Chapter 11
Holt McDougal Algebra 2 Chapter 11: Probability and Statistics
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8854 | https://math.stackexchange.com/questions/2289909/coefficients-and-derivatives-of-polynomials-and-solving-f-f | Skip to main content
Coefficients and Derivatives of Polynomials, and Solving f′=f
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Hi so I have been spending a long time trying to figure these two problems out,
1) Make up a few polynomials, express them as polynomials in x−2 and show that the coefficients are related to the derivatives at 2.
2) Show that the only functions f with f′=f are f(x)=ex
I think that for the first one I'm supposed to use some polynomials as an example to prove something, but I don't know how. And for the second one I believe I need to show constant multiples of ex
calculus
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edited May 21, 2017 at 4:15
Manuel Guillen
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Take f=0 everywhere then f′=f. But ex≢0. You need some initial conditions on f.
– Zain Patel
Commented
May 21, 2017 at 3:15
For the second item, have you learned Taylor series yet?
– d4rk_1nf1n1ty
Commented
May 21, 2017 at 3:16
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How does the l'Hopital's Rule in the title relate to the actual questions?
– dxiv
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May 21, 2017 at 3:42
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f(x)f′(x)f′′(x)=b0+==b1(x−2)+b1+b2(x−2)22b2(x−2)2b2+⋯++⋯++⋯+bn(x−2)nnbn(x−2)n−1n(n−1)bn(x−2)n−2f(2)f′(2)f′′(2)=b0=b1=2b2
The idea is, every time you differentiate, you get rid of the constant term, and every other term's degree is decremented by 1. When you plug in 2, all the non-constant terms evaluate to 0, so you are left with a constant term that is the result of the repeated differentiation of a single term. In general:
f(m)(2)=m!⋅bm
This is the essence behind Taylor polynomials. I recommend looking into it.
2
If we have a function f(x) such that:
f′(x)=f(x)
We see that:
ddx[f(x)e−x]=ddx[f(x)]⋅e−x+f(x)⋅ddxe−x=f′(x)e−x−f(x)e−x=e−x[f′(x)−f(x)]=e−x⋅0=0
Therefore, f(x)e−x is constant, meaning f(x)=cex for some constant c.
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answered May 21, 2017 at 3:35
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The 2nd part will be functions of the form cex for any real constant c.
It's equivalent to solving the differential equation dy/dx=y.
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answered May 21, 2017 at 3:29
Arpan1729Arpan1729
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Could be cex+k also.
– Ovi
Commented
May 21, 2017 at 3:37
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@ovi: cex+k is same thing as c′ex for suitable c′.
– Paramanand Singh
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8855 | https://www.bilibili.com/video/av494386224/ | 等比数列，从基础到进阶！_哔哩哔哩_bilibili
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赛事库课堂2021拜年纪 |
8856 | http://www.digimat.in/nptel/courses/video/108105153/L46.html | NPTEL : NOC:Electrical Measurement and Electronic Instruments (Electrical Engineering)
Videos
Lecture 1 - PMMC InstrumentsLecture 2 - Electrodynamic InstrumentLecture 3 - Demonstration of PMMC and Electrodynamic InstrumentsLecture 4 - Features of PMMC and Electrodynamic InstrumentsLecture 5 - Moving Iron InstrumentsLecture 6 - Demonstration of Moving Iron InstrumentLecture 7 - Electrostatic InstrumentLecture 8 - Derivation of Deflecting Torque in Electrodynamic, Electrostatic and Moving Iron InstrumentLecture 9 - Damping and Eddy Current DampingLecture 10 - Dynamics of the Moving Coil and DampingLecture 11 - Dynamics of the Moving Coil and Damping (Continued...)Lecture 12 - Ballistic GalvanometerLecture 13 - Ammeter - ILecture 14 - Ammeter - IILecture 15 - VoltmeterLecture 16 - Ohmmeters - ILecture 17 - Ohmmeters - IILecture 18 - Rectifier based Voltmeters and Ammeter - ILecture 19 - Rectifier based Voltmetesr and Ammeter - IILecture 20 - Resistance measurement with a Voltmeter and an AmmeterLecture 21 - Four-Terminal ResistanceLecture 22 - Problems: Four Terminal ResistanceLecture 23 - Error CalculationLecture 24 - Sensitivity, Accuracy, and Resolution of Wheatstone BridgeLecture 25 - Kelvin Double BridgeLecture 26 - High Resistance MeasurementLecture 27 - Wattmeter Connection and Compensated WattmeterLecture 28 - Single Phase Energy MeterLecture 29 - Demonstration: 1. Eddy Current Braking 2.Creation of Magnetic Field Without Moving ObjectsLecture 30 - Single Phase Energy Meter (Continued...)Lecture 31 - Connection of Energy Meter, Wattmeter, and Three Phase SupplyLecture 32 - DC PotentiometerLecture 33 - AC PotentiometerLecture 34 - Polar potentiometer and phase shifterLecture 35 - Polar potentiometerLecture 36 - Co-ordinate potentiometerLecture 37 - Kelvin-Varley potential dividerLecture 38 - Impedance measurementLecture 39 - AC bridges - ILecture 40 - AC bridges - IILecture 41 - AC bridges - IIILecture 42 - Current transformer and potential transformerLecture 43 - Review of transformer and magnetic circuitLecture 44 - Errrors in Instrument ransformerLecture 45 - Flux density measurement with Ballistic GalvanometerLecture 46 - Flux density measurement with Ballistic Galvanometer (Continued...)Lecture 47 - Background: From Flip Flop to Counters - ILecture 48 - Background: From Flip Flop to Counters - IILecture 49 - Background: Operational Amplifiers - ILecture 50 - Background: Operational Amplifiers - IILecture 51 - Background: Operational Amplifiers - IIILecture 52 - Background: Operational Amplifiers - IVLecture 53 - Inverting amplifier versus Schmitt TriggerLecture 54 - Non-inverting amplifier versus Schmitt TriggerLecture 55 - Difference amplifier - ILecture 56 - Difference amplifier - IILecture 57 - Difference amplifier - IIILecture 58 - Digital frequency meterLecture 59 - Digital frequency meter and Schmitt TriggerLecture 60 - Schmitt TriggerLecture 61 - Schmitt TriggerLecture 62 - Digital frequency meterLecture 63 - Linear ramp type digital voltmeterLecture 64 - Dual slope digital voltmeter - ILecture 65 - Dual slope digital voltmeter - IILecture 66 - Dual slope digital voltmeter and Integrator circuitLecture 67 - Digital ramp type voltmeterLecture 68 - Digital ramp type voltmeter and Successive approximation type voltmeterLecture 69 - ADC and DAC - ILecture 70 - ADC and DAC - IILecture 71 - Why we need electronic InstrumentsLecture 72 - Instruments with op-amp based amplifiers - ILecture 73 - Instruments with op-amp based amplifiers - IILecture 74 - Instruments with op-amp based amplifiers - IIILecture 75 - Instrumentation AmplifierLecture 76 - Function generatorLecture 77 - 555-Timer circuitLecture 78 - Astable and monostable oscillator circuitsLecture 79 - Pulse generatorLecture 80 - Oscilloscope - ILecture 81 - Oscilloscope - IILecture 82 - Emitter follower voltmeterLecture 83 - Linear ohmmeterLecture 84 - Ramp generator
PDF
Lecture 1 - PMMC InstrumentsLecture 2 - Electrodynamic InstrumentLecture 3 - Demonstration of PMMC and Electrodynamic InstrumentsLecture 4 - Features of PMMC and Electrodynamic InstrumentsLecture 5 - Moving Iron InstrumentsLecture 6 - Demonstration of Moving Iron InstrumentLecture 7 - Electrostatic InstrumentLecture 8 - Derivation of Deflecting Torque in Electrodynamic, Electrostatic and Moving Iron InstrumentLecture 9 - Damping and Eddy Current DampingLecture 10 - Dynamics of the Moving Coil and DampingLecture 11 - Dynamics of the Moving Coil and Damping (Continued...)Lecture 12 - Ballistic GalvanometerLecture 13 - Ammeter - ILecture 14 - Ammeter - IILecture 15 - VoltmeterLecture 16 - Ohmmeters - ILecture 17 - Ohmmeters - IILecture 18 - Rectifier based Voltmeters and Ammeter - ILecture 19 - Rectifier based Voltmetesr and Ammeter - IILecture 20 - Resistance measurement with a Voltmeter and an AmmeterLecture 21 - Four-Terminal ResistanceLecture 22 - Problems: Four Terminal ResistanceLecture 23 - Error CalculationLecture 24 - Sensitivity, Accuracy, and Resolution of Wheatstone BridgeLecture 25 - Kelvin Double BridgeLecture 26 - High Resistance MeasurementLecture 27 - Wattmeter Connection and Compensated WattmeterLecture 28 - Single Phase Energy MeterLecture 29 - Demonstration: 1. Eddy Current Braking 2.Creation of Magnetic Field Without Moving ObjectsLecture 30 - Single Phase Energy Meter (Continued...)Lecture 31 - Connection of Energy Meter, Wattmeter, and Three Phase SupplyLecture 32 - DC PotentiometerLecture 33 - AC PotentiometerLecture 34 - Polar potentiometer and phase shifterLecture 35 - Polar potentiometerLecture 36 - Co-ordinate potentiometerLecture 37 - Kelvin-Varley potential dividerLecture 38 - Impedance measurementLecture 39 - AC bridges - ILecture 40 - AC bridges - IILecture 41 - AC bridges - IIILecture 42 - Current transformer and potential transformerLecture 43 - Review of transformer and magnetic circuitLecture 44 - Errrors in Instrument ransformerLecture 45 - Flux density measurement with Ballistic GalvanometerLecture 46 - Flux density measurement with Ballistic Galvanometer (Continued...)Lecture 47 - Background: From Flip Flop to Counters - ILecture 48 - Background: From Flip Flop to Counters - IILecture 49 - Background: Operational Amplifiers - ILecture 50 - Background: Operational Amplifiers - IILecture 51 - Background: Operational Amplifiers - IIILecture 52 - Background: Operational Amplifiers - IVLecture 53 - Inverting amplifier versus Schmitt TriggerLecture 54 - Non-inverting amplifier versus Schmitt TriggerLecture 55 - Difference amplifier - ILecture 56 - Difference amplifier - IILecture 57 - Difference amplifier - IIILecture 58 - Digital frequency meterLecture 59 - Digital frequency meter and Schmitt TriggerLecture 60 - Schmitt TriggerLecture 61 - Schmitt TriggerLecture 62 - Digital frequency meterLecture 63 - Linear ramp type digital voltmeterLecture 64 - Dual slope digital voltmeter - ILecture 65 - Dual slope digital voltmeter - IILecture 66 - Dual slope digital voltmeter and Integrator circuitLecture 67 - Digital ramp type voltmeterLecture 68 - Digital ramp type voltmeter and Successive approximation type voltmeterLecture 69 - ADC and DAC - ILecture 70 - ADC and DAC - IILecture 71 - Why we need electronic InstrumentsLecture 72 - Instruments with op-amp based amplifiers - ILecture 73 - Instruments with op-amp based amplifiers - IILecture 74 - Instruments with op-amp based amplifiers - IIILecture 75 - Instrumentation AmplifierLecture 76 - Function generatorLecture 77 - 555-Timer circuitLecture 78 - Astable and monostable oscillator circuitsLecture 79 - Pulse generatorLecture 80 - Oscilloscope - ILecture 81 - Oscilloscope - IILecture 82 - Emitter follower voltmeterLecture 83 - Linear ohmmeterLecture 84 - Ramp generator
NPTEL Video Course : NOC:Electrical Measurement and Electronic Instruments
Lecture 46 - Flux density measurement with Ballistic Galvanometer (Continued...)
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8857 | https://www.apartmentlist.com/renter-life/largest-cities-in-california | Share this Article
10 Largest Cities in California
By: Julia Guerra Slater
June 25, 2025
Are you moving to the Golden State? Check out the 10 biggest cities in California. Compare lifestyles, rent, and amenities to find your next apartment.
As one of the largest and most populous states, California is home to several major cities that have a lot to offer. From the glamour of Los Angeles to the laid-back beach vibe of San Diego, each city offers distinct lifestyles, career opportunities, and housing markets worth exploring.
Meanwhile, cities like San Francisco and San Jose are known for their tech-driven economies and iconic landmarks. For families and young professionals, Sacramento offers more affordable living options and is in proximity to countless outdoor adventures.
Explore more insights about California’s largest cities, including their housing markets, transit systems, and neighborhood highlights. Dive deeper into the descriptions of each of these top 10 cities in California by population to find the perfect apartment that matches your needs and lifestyle.
All of the rental data included here comes from our internal database of rent estimates, which is available for download. Transit scores come from Walk Score, and census information comes directly from the U.S. Census Bureau website.
Largest Cities in California
Populations of the 10 Largest Cities in California
| Rank | City | Population |
---
| 1 | Los Angeles | 3,898,747 |
| 2 | San Diego | 1,386,932 |
| 3 | San Jose | 1,013,240 |
| 4 | San Francisco | 873,965 |
| 5 | Fresno | 542,107 |
| 6 | Sacramento | 524,943 |
| 7 | Long Beach | 466,742 |
| 8 | Oakland | 440,646 |
| 9 | Bakersfield | 403,455 |
| 10 | Anaheim | 346,824 |
Los Angeles
Median rent for a one-bedroom apartment: $2,435
Median rent for a two-bedroom apartment: $3,846
Population: 3,898,747
Median Household income: $80,366
Walk score: 69
Transit score: 53
Bike score: 59
Map of Los Angeles
Los Angeles is the biggest city in California and one that moves with energy and purpose. Life here is fast-paced, and while the traffic is legendary, so is the diversity that defines this vibrant metropolis.
Nowhere is that diversity more deliciously apparent than in LA’s incredible food scene. From health-focused cafés to a huge range of global cuisines, LA’s food scene is unmatched. Whether you're into street tacos or vegan sushi, there’s something for everyone in the largest city in California.
In addition to great food, you’ve also got easy access to beaches, amusement parks, national forests, and a nightlife that rarely slows down. Housing is expensive—averaging $2,435 for a one-bedroom—but many consider the lifestyle worth the price. Public transit is decent in some areas, but a car is often necessary. Luckily, there are plenty of backstreets to (hopefully) avoid the worst traffic.
San Diego
Median rent for a one-bedroom apartment: $1,973
Median rent for a two-bedroom apartment: $2,472
Population: 1,386,932
Median household income: $104,321
Walk score: 53
Transit score: 37
Bike score: 43
Map of
San Diego is the second-largest city in California. It’s a laid-back beach town with some of the best weather in the country—mild, sunny, and consistent year-round. Here, there’s always something to do outdoors, from surfing and hiking to visiting weekend farmers markets and taking coastal walks. The city has a strong military presence and a relaxed atmosphere that sets it apart from the hustle of Los Angeles. Culturally, it’s calmer but still highly active, with local breweries, art events, and diverse neighborhoods that each offer its own character.
With an average one-bedroom rent of $2,997, housing is expensive, especially since local salaries tend to lag behind the cost of living.
While public transit in San Diego is limited, options like buses and trolleys are available for those who prefer not to drive—though many residents opt for cars to get around more efficiently. San Diego is best for those who prioritize outdoor living, slower-paced days, and beach access over dense urban energy.
San Jose
Median rent for a one-bedroom apartment: $3,007
Median rent for a two-bedroom apartment: $3,668
Population: 1,013,240
Median household income: $141,565
Walk score: 51
Transit score: 40
Bike score: 62
Map of San Jose
San Jose is the heart of Silicon Valley, with close proximity to major tech employers like Google, Apple, and Adobe. While it’s a large city, it is serene, mostly residential, and well-suited for families. As such, the nightlife is limited, but the city hosts a steady stream of festivals, cultural events, concerts, and sports games. You'll also find quality museums, local farmers markets, and a growing calendar of community activities throughout the year. The food scene is solid, with diverse options that reflect the city’s international population.
Housing in San Jose is in high demand and expensive, with an average one-bedroom rent of $3,007. Public transit is limited, and many areas are car-dependent, despite the city’s moderate walk and bike scores. But if your budget matches, San Jose is a great fit for families or professionals who want a neighborly environment in the center of the tech world.
San Francisco
Median rent for a one-bedroom apartment: $2,969
Median rent for a two-bedroom apartment: $3,519
Population: 873,965
Median household income: $141,446
Walk score: 89
Transit score: 77
Bike score: 72
Map of San Francisco
San Francisco, a city of steep hills, coastal fog, and sweeping views, is one of the most populated cities in California. Its iconic Victorian homes, many beautifully restored and painted in bright colors, dot the city’s neighborhoods and contribute to its distinct charm.
Golden Gate Park is a central gem, offering 1,017 acres of green space and cultural institutions, while the city’s great location provides quick access to outdoor escapes like Mount Tamalpais, Point Reyes, and coastal beaches. With its cool, often foggy weather and dramatic vistas, San Francisco lives in a state of perpetual sweater weather, ideal for those who prefer mild climates and scenic backdrops.
Culturally, San Francisco punches far above its size. A longtime haven for countercultural movements, it has built a legacy of embracing diversity—from the Beat poets and the hippie movement to its vibrant LGBTQ+ community, which continues to thrive today.
It boasts world-class museums, theaters, symphonies, and a nightlife that spans jazz, salsa, and everything in between. The city also draws a large population of tech workers, many of whom live in San Francisco and commute to nearby offices in Silicon Valley or work remotely.
However, with an average one-bedroom rent of $3,459, affordability can be a real challenge, but those who brave the price often cite that its culture, progressive values, and outdoor access make it worthwhile.
Fresno
Median rent for a one-bedroom apartment: $1,093
Median rent for a two-bedroom apartment: $1,349
Population: 542,107
Median household income: $66,804
Walk score: 47
Transit score: 33
Bike score: 58
Map of Fresno
Fresno may be the fifth-largest city in California, but this large city feels more like a suburb. Situated just about halfway between San Francisco and Los Angeles, it boasts a central location that gives residents easy access to both mountains and beaches, making it ideal for outdoor enthusiasts.
The city offers a great mix of activities, whether it’s enjoying the lively bar scene in the Tower District or exploring the nearby Sierra Nevada mountains. Fresno’s affordability is what draws people in, with the median one-bedroom rent at $1,093, much lower than that of California’s coastal cities.
From a growing arts and culture scene to new dispensaries and a thriving food scene, Fresno has a little something for everyone. It’s best suited for people who want a quiet, suburban atmosphere while still having access to bigger cities.
Sacramento
Median rent for a one-bedroom apartment: $2,130
Median rent for a two-bedroom apartment: $2,518
Population: 524,943
Median household income: $83,753
Walk score: 49
Transit score: 34
Bike score: 67
Map of Sacramento
Known for its food scene, craft beer culture, and mild winters, Sacramento has plenty to offer year-round. The Delta breeze cools down the hot summer days, making outdoor activities like volleyball, hiking, or enjoying local parks a year-round option. Whether it’s a weekend trip to the mountains in Tahoe or wine tasting nearby in Lodi, Sacramento residents enjoy diverse recreational activities without straying too far from home.
Relative affordability makes Sacramento popular, with an average one-bedroom rent of $2,130, which is far more reasonable than Bay Area cities. With a transit score of 34, public transportation is limited, so a car is usually necessary for getting around. Sacramento is ideal for families, college students, and anyone seeking a more affordable, active lifestyle.
Long Beach
Median rent for a one-bedroom apartment: $1,578
Median rent for a two-bedroom apartment: $1,933
Population: 466,742
Median household income: $83,969
Walk score: 73
Transit score: 49
Bike score: 70
Map of Long Beach
Long Beach is a diverse, largely working-class city that offers a more low-key atmosphere than many parts of Los Angeles. While some neighborhoods are run-down, others feel upscale, especially in pockets near the coast.
Downtown has seen a noticeable boost in activity in recent years, with new restaurants, bars, and cultural events making it a much livelier place to spend time. Its location in Los Angeles County adds to its appeal, especially for those commuting or looking to stay connected to the city of Los Angeles and Orange County.
With a median one-bedroom rent of $1,578, housing isn’t cheap, but it’s still more approachable than many nearby coastal cities. Long Beach also offers waterfront parks, beaches, and bike-friendly streets, making it easy to stay active outdoors. A car is still the most practical way to get around, especially outside downtown. This city is best suited for people who like access to big-city activities in a place that feels rooted, local, and evolving.
Oakland
Median rent for a one-bedroom apartment: $1,830
Median rent for a two-bedroom apartment: $2,196
Population: 440,646
Median household income: $97,369
Walk score: 75
Transit score: 57
Bike score: 65
Map of Oakland
Oakland feels like a small town wrapped in a big city’s shell. It’s home to peaceful neighborhoods, a beautiful urban lake, and parks in the hills with sweeping Bay views. The waterfront at Jack London Square offers a relaxing spot to watch sailboats or grab a bite, while Lake Merritt is perfect for an afternoon walk or picnic.
Oakland also has a vibrant art scene, and its murals, independent galleries, and community-driven events reflect the city's deep cultural roots and history of activism. The local food scene is just as dynamic, ranging from food trucks to five-star restaurants, often influenced by the city’s incredible diversity.
Public transit is decent, with Bay Area Rapid Transit (BART) and bus lines connecting residents to nearby Berkeley and San Francisco. Some neighborhoods are very walkable, but safety and accessibility can vary depending on where you are, and many residents still rely on cars. Oakland offers strong value for those seeking culture, community, and access to both nature and city life.
Bakersfield
Starting rent for an apartment in Bakersfield: $700
Population: 403,455
Median household income: $77,397
Walk score: 37
Transit score: 25
Bike score: 44
Map of Bakersfield
Bakersfield is a Central Valley city renowned for its affordability and accessibility, particularly when compared to the high costs of coastal metros and other major cities in California. While it’s not a major tourist destination, it offers a strong restaurant scene, multiple concert venues, and a full range of national retailers. Residents enjoy easy access to cultural spots like the Bakersfield Museum of Art and the historic Fox Theater. The city’s location also puts it within driving distance of several national and state parks, making it a good base for weekend adventures.
Public transportation is limited in Bakersfield, so a car is essential for most residents. But with lower housing costs and a family-friendly pace of life, Bakersfield appeals to people looking for space, convenience, and a lower cost of living. It may not have the fast pace of Los Angeles or the coastal views of the Bay Area, but Bakersfield’s balance of affordability and accessibility gives it a distinct appeal.
Anaheim
Median rent for a one-bedroom apartment: $1,988
Median rent for a two-bedroom apartment: $2,453
Population: 346,824
Median household income: $90,583
Walk score: 56
Transit score: 34
Bike score: 52
Map of Anaheim
Known for being the home of Disneyland, Anaheim has a strong local identity beyond the theme park. Sports fans can catch a game at Angel Stadium or the Honda Center—also a popular concert venue—while beer lovers can explore a growing number of local breweries. Anaheim weather is typically mild and sunny, like much of Southern California, and the city benefits from being surrounded by neighboring areas with plenty to do.
While not as expensive as coastal cities, Anaheim housing costs are still high compared to national averages. Plus, public transit is limited, so having a car is a must for most residents. Even so, Anaheim is perfect for those who want suburban comfort within reach of beaches, sports, entertainment, and other major cities.
Need Help Finding an Apartment in California?
California's largest cities have something to offer every renter. Whether you prioritize affordability, outdoor adventure, access to culture, or a vibrant community atmosphere, California is waiting to welcome you.
Ready to find your perfect California apartment? Head over to our matching tool so you can narrow down your options and find an apartment in California that ticks all your boxes. With us, you’ll spend five minutes and save 50 hours searching. Happy hunting!
Frequently Asked Questions
What are the top three cities in California?
The top three California cities by population are Los Angeles, San Diego, and San Jose.
What are the five largest counties in California?
The five most populous counties in California, based on the 2020 census, are:
Los Angeles County: 10M+ (Los Angeles, Long Beach)
San Diego County: ~3.3M (San Diego)
Orange County: ~3.1M (Anaheim, Irvine)
Riverside County: ~2.4M (Riverside, Corona)
San Bernardino County: ~2.1M (San Bernardino, Fontana)
These counties are home to some of the state’s largest cities and key economic hubs.
What is the busiest city in California, and what is the largest city in California?
Los Angeles is both the largest and the busiest city in California, boasting a population of approximately 3.8 million residents.
What is the smallest city in California?
Amador City holds the distinction of being the smallest city in California, both in terms of population and land area. According to the 2020 United States Census, it has a population of 222 residents and encompasses a total area of 0.3 square miles.
What is the safest city in California?
Rancho Santa Margarita is recognized as the safest city in California for 2025. According to SafeWise, the city reported zero murders and zero rapes, with an overall violent crime rate of fewer than 2.2 incidents per 1,000 residents.
Which state has more counties, Texas or California?
Texas has more counties than California. Specifically, Texas is divided into 254 counties, the most of any U.S. state. In contrast, California comprises 58 counties.
What is the capital city in California?
Sacramento the capital city of California.
How many cities are in California?
California has 482 cities, towns, and villages and 2,949 special districts.
Share this Article
CONTENT EDITOR
Julia Guerra Slater
Julia Guerra Slater is a freelance writer for Apartment List, combining her passion for travel with her expertise in helping people find the perfect place to call home. Having explored cities and towns across the U. Read More
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Kinematics
Concepts and Principles
Kinematics is the formal language physicists use to describe motion. The need for a formal language is evidenced by a simple experiment: drop an object from about shoulder height and ask two people to independently describe the motion of the object. Chances are that the descriptions will not be in perfect agreement, even though both observers described the same motion. Obviously, a more formal way of describing motion is necessary to eliminate this type of descriptive ambiguity. Kinematics is the formal method of describing motion.
Three parameters are carefully defined and used by physicists to describe motion. Specifying these three parameters at all times forms a complete description of the motion of an object.
Position
The position of an object is its location relative to a well-defined coordinate system at a particular instant of time. Without a specified coordinate system, position is a meaningless concept. A coordinate system is comprised of a zero, a specified positive direction, and a scale.
For example, in the hypothetical experiment in which the object was dropped from shoulder height, a coordinate system could have been defined in which the zero position was at ground level, the positive direction was up, and the scale used was meters. Using this coordinate system, the position of the object could have been specified at any particular instant of time. Of course, choosing the zero at the location at which the object was dropped, the positive direction as down, and the scale in feet is also perfectly acceptable. It doesn’t matter what you choose as a coordinate system, only that you explicitly choose one. Depending on the coordinate system chosen, the position of an object can be positive, negative, or zero.
We will use the symbol r to designate position, and measure it in meters (m).
Velocity
Although the word velocity is often used loosely in everyday conversation, its meaning in physics is specific and well-defined. To physicists, the velocity is the rate at which the position is changing. The velocity can be specified at any particular instant of time.
For example, if the position is changing quickly the velocity is large and if the position is not changing at all the velocity is zero. A mathematical way to represent this definition is
Thus, velocity is measured in meters per second (m/s).
Actually, this is the definition of the average velocity of the object over the time interval Dt, but as the time interval becomes smaller and smaller the value of this expression becomes closer and closer to the actual rate at which the position is changing at one particular instant of time.
Since the final position of the object (rfinal) may be either positive, negative, or zero, and either larger, smaller, or the same as the initial position (rinitial), the velocity may be positive, negative, or zero. The sign of the velocity depends on the coordinate system chosen to define the position. A positive velocity simply means that the object is moving in the positive direction, as defined by the coordinate system, while a negative velocity means the object is traveling in the other direction.
Acceleration
Again, although the word acceleration is often used loosely in everyday conversation, its meaning in physics is specific and well-defined. To physicists, the acceleration is the rate at which the velocity is changing. Again, the acceleration can be specified at any particular instant of time.
For example, if the velocity is changing quickly the acceleration is large in magnitude, and if the velocity is not changing the acceleration is zero. If an object has non-zero acceleration, it does not mean that the object is speeding up. It simply means that the velocity is changing. Moreover, even if an object has a positive acceleration, it does not mean that the object is speeding up! A positive acceleration means that the change in the velocity points in the positive direction. (I can almost guarantee you will experience confusion about this. Take some time to think about the preceding statement right now.)
A mathematical way to represent acceleration is
Thus, acceleration is measured in meters per second per second (m/s2).
Again, this is actually the average acceleration of the object over the time interval Dt, but as the time interval becomes smaller and smaller, the value of this expression becomes closer and closer to the actual rate at which the velocity is changing at one particular instant of time.
Since vfinal may be either positive, negative, or zero, and either larger, smaller, or the same as vinitial, the acceleration may be positive, negative, or zero. The algebraic sign of the acceleration depends on the coordinate system chosen to define the position. A negative acceleration means that the change in the velocity points in the negative direction. For example, the velocity could be in the positive direction and the object slowing down or the velocity could be in the negative direction and the object speeding up. Both of these scenarios would result in a negative acceleration. Conversely, a positive acceleration means that the change in the velocity points in the positive direction.
Kinematics is the correct use of the parameters position, velocity, and acceleration to describe motion. Learning to use these three terms correctly can be made much easier by learning a few tricks of the trade. These tricks, or analysis tools, are detailed in the following section.
Analysis Tools
Drawing Motion Diagrams
The words used by physicists to describe the motion of objects are defined above. However learning to use these terms correctly is more difficult than simply memorizing definitions. An extremely useful tool for bridging the gap between a normal, conversational description of a situation and a physicists’ description is the motion diagram. A motion diagram is the first step in translating a verbal description of a phenomenon into a physicists’ description.
Start with the following verbal description of a physical situation:
The driver of an automobile traveling at 15 m/s, noticing a red-light 30 m ahead, applies the brakes of her car until she stops just short of the intersection.
Determining the position from a motion diagram
A motion diagram can be thought of as a multiple-exposure photograph of the physical situation, with the image of the object displayed at equal time intervals. For example, a multiple-exposure photograph of the situation described above would look something like this:
Note that the images of the automobile are getting closer together near the end of its motion because the car is traveling a smaller distance between the equally-timed exposures.
In general, in drawing motion diagrams it is better to represent the object as simply a dot, unless the actual shape of the object conveys some interesting information. Thus, a better motion diagram would be:
Since the purpose of the motion diagram is to help us describe the car’s motion, a coordinate system is necessary. Remember, to define a coordinate system you must choose a zero, define a positive direction, and select a scale. We will always use meters as our position scale in this course, so you must only select a zero and a positive direction. Remember, there is no correct answer. Any coordinate system is as correct as any other.
The choice below indicates that the initial position of the car is the origin, and positions to the right of that are positive. (In the text I’ll always use a little diamond to indicate the zero with an arrow pointing in the positive direction.)
We can now describe the position of the car. The car starts at position zero and then has positive, increasing positions throughout the remainder of its motion.
Determining the velocity from a motion diagram
Since velocity is the change in position of the car during a corresponding time interval, and we are free to select the time interval as the time interval between exposures on our multiple-exposure photograph, the velocity is simply the change in the position of the car “between dots.” Thus, the arrows (vectors) on the motion diagram below represent the velocity of the car.
We can now describe the velocity of the car. Since the velocity vectors always point in the positive direction, the velocity is always positive. The car starts with a large, positive velocity which gradually declines until the velocity of the car is zero at the end of its motion.
Determining the acceleration from a motion diagram
Since acceleration is the change in velocity of the car during a corresponding time interval, and we are free to select the time interval as the time interval between exposures on our multiple-exposure photograph, we can determine the acceleration by comparing two successive velocities. The change in these velocity vectors will represent the acceleration.
To determine the acceleration,
Comparing the first and second velocity vectors leads to the acceleration vector shown below:
Thus, the acceleration points to the left and is therefore negative. You could construct the acceleration vector at every point in time, but hopefully you can see that as long as the velocity vectors continue to point toward the right and decrease in magnitude, the acceleration will remain negative.
Thus, with the help of a motion diagram, you can extract lots of information about the position, velocity, and acceleration of an object. You are well on your way to a complete kinematic description.
Drawing Motion Graphs
Another useful way to describe the motion of an object is by constructing graphs of the object’s position, velocity, and acceleration vs. time. A graphical representation is a very effective means of presenting information concerning an object’s motion and, moreover, it is relatively easy to construct motion graphs if you have a correct motion diagram.
Examine the same situation as before:
The driver of an automobile traveling at 15 m/s, noticing a red-light 30 m ahead, applies the brakes of her car until she stops just short of the intersection.
The verbal representation of the situation has already been translated into a motion diagram. A careful reading of the motion diagram allows the construction of the motion graphs.
Drawing the position vs. time graph
We already know, from the motion diagram, that the car starts at position zero, then has positive, increasing positions throughout the remainder of its motion. This information can be transferred onto a position vs. time graph.
Notice that the position is zero when the time is equal to zero, the position is always positive, and the position increases as time increases. Also note that in each subsequent second, the car changes its position by a smaller amount. This leads to the graph of position vs. time gradually decreasing in slope until it achieves a slope of zero. Once the car stops, the position of the car should not change.
Drawing the velocity vs. time graph
From the motion diagram, we know that the velocity of the car is always positive, starts large in magnitude, and decreases until it is zero. This information can be transferred onto a velocity vs. time graph.
How do we know that the slope of the line is constant? The slope of the line represents the rate at which the velocity is changing, and the rate at which the velocity is changing is termed the acceleration. Since in this model of mechanics we will only consider particles undergoing constant acceleration, the slope of a line on a velocity vs. time graph must be constant.
Drawing the acceleration vs. time graph
From the motion diagram, the acceleration of the car can be determined to be negative at every point. Again, in this pass through mechanics we will only be investigating scenarios in which the acceleration is constant. Thus, a correct acceleration vs. time graph is shown below.
Tabulating Motion Information
After constructing the two qualitative representations of the motion (the motion diagram and the motion graphs), we are ready to tackle the quantitative aspects of the motion.
Utilizing the motion diagram,
you can now assign numerical values to several of the kinematic variables. A glance at the situation description should indicate that information is presented about the car at two distinct events. Information is available about the car at the instant the driver applies the brakes (the velocity is given), and the instant the driver stops (the position is given). Other information can also be determined by referencing the motion diagram. To tabulate this information, you should construct a motion table.
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| Event 1: The instant the driver first applies the brakes. t1 = 0 s r1 = 0 m v1 = +15 m/s a12 = | Event 2: The instant the car finally stops. t2 = r2 = +30 m v2 = 0 m/s |
t1 = 0 s
r1 = 0 m
v1 = +15 m/s
a12 =
t2 =
r2 = +30 m
v2 = 0 m/s
In addition to the information explicitly given, the velocity at the first event and the position at the second event, other information can be extracted from the problem statement and the motion diagram.
For example,
Since you are working under the assumption in this model that the acceleration is constant, the acceleration between the two instants in time is some unknown, constant value. To remind you that this assumption is in place, the acceleration is not labeled at the first instant, a1, or the second instant, a2, but rather as the acceleration between the two instants in time, a12.
You now have a complete tabulation of all the information presented, both explicitly and implicitly, in the situation description. Moreover, you now can easily see that the only kinematic information not known about the situation is the assumed constant acceleration of the auto and the time at which it finally stops. Thus, to complete a kinematic description of the situation these two quantities must be determined. What you may not know is that you have already been presented with the information needed to determine these two unknowns.
Doing the Math
In the concepts and principles portion of this unit, you were presented with two formal, mathematical relationships, the definitions of velocity and acceleration. In the example that you are working on, there are two unknown kinematic quantities. You should remember from algebra that two equations are sufficient to calculate two unknowns. Thus, by applying the two definitions you should be able to determine the acceleration of the car and the time at which it comes to rest.
Although you can simply apply the two definitions directly, normally the two definitions are rewritten, after some algebraic re-arranging, into two different relationships. This rearrangement is simply to make the algebra involved in solving for the unknowns easier. It is by no means necessary to solve the problem. In fact, the two definitions can be written in a large number of different ways, although this does not mean that there are a large number of different formulas you must memorize in order to analyze kinematic situations. There are only two independent kinematic relationships. The two kinematic relationships we will use when the acceleration is constant are:
To finish the analysis of this situation,
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| Event 1: The instant the driver first applies the brakes. t1 = 0 s r1 = 0 m v1 = +15 m/s a12 = | Event 2: The instant the car finally stops. t2 = r2 = +30 m v2 = 0 m/s |
t1 = 0 s
r1 = 0 m
v1 = +15 m/s
a12 =
t2 =
r2 = +30 m
v2 = 0 m/s
simply write the two kinematic relationships, input the known kinematic variables from the motion table, and solve the two relations for the two unknowns. (This process is not physics, it’s algebra.)
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| 2-300x216.png Now substitute this expression into the other equation: 41.png | 31-291x300.png Substitute this result back into the original equation: |
Now substitute this expression into the other equation:
Substitute this result back into the original equation:
Thus, the car must have accelerated at 3.8 m/s2 in the negative direction, and stopped after 4.0 seconds. The kinematic description of the situation is complete.
Analyzing a More Complex Motion
Let’s re-visit our scenario, although this time the light turns green while the car is slowing down:
The driver of an automobile traveling at 15 m/s, noticing a red-light 30 m ahead, applies the brakes of her car. When she is 10 m from the light, and traveling at 8.0 m/s, the light turns green. She instantly steps on the gas and is back at her original speed as she passes under the light.
Our first step in analyzing this motion should be to draw a motion diagram.
I’ve noted on the motion diagram the important events that take place during the motion. Notice that between the instant she hits the brakes and the instant she steps on the gas the acceleration is negative, while between the instant she steps on the gas and the instant she passes the light the acceleration is positive. Thus, in tabulating the motion information and applying the kinematic relations we will have to be careful not to confuse kinematic variables between these two intervals. Below is a tabulation of motion information using the coordinate system established in the motion diagram.
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| Event 1: She hits the brakes. t1 = 0 s r1 = 0 m v1 = +15 m/s a12 = | Event 2: She steps on the gas. t2 = r2 = +20 m v2 = +8.0 m/s a23 = | Event 3: She passes the light t3 = r3 = +30 m v3 = +15 m/s |
t1 = 0 s
r1 = 0 m
v1 = +15 m/s
a12 =
t2 =
r2 = +20 m
v2 = +8.0 m/s
a23 =
t3 =
r3 = +30 m
v3 = +15 m/s
First, notice that during the time interval between “hitting the brakes” and “stepping on the gas” there are two kinematic variables that are unknown. Recall that by using your two kinematic relations you should be able to determine these values. Second, notice that during the second time interval again two variables are unknown. Once again, the two kinematic relations will allow you to determine these values. Thus, before I actually begin to do the algebra I know the unknown variables can be determined!
First let’s examine the motion between hitting the brakes and stepping on the gas:
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| 61-300x206.png Now substitute this expression into the other equation: 81-300x186.png | 71-281x300.png Substitute this result back into the original equation: |
Now substitute this expression into the other equation:
Substitute this result back into the original equation:
Now, using these results, examine the kinematics between stepping on the gas and passing the light. Note that the initial values of the kinematic variables are denoted by ‘2’ and the final values by ‘3’, since we are examining the interval between event 2 and event 3.
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| 9-300x199.png Now substitute this expression into the other equation: 111.png | 101-282x300.png Substitute this result back into the original equation: |
Now substitute this expression into the other equation:
Substitute this result back into the original equation:
We now have a complete kinematic description of the motion.
Symbolic Analysis
Consider the following situation:
The driver of an automobile suddenly sees an obstacle blocking her lane. Determine the total distance the auto travels between seeing the obstacle and stopping (d) as a function of the initial velocity of the car (vi) and the magnitude of its acceleration while stopping (as).
As always, the first step in analyzing motion is to draw a motion diagram.
Rather than calculate the stopping distance for particular values of initial velocity and acceleration, the goal of this activity is to determine, in general, how the stopping distance depends on these two parameters. If we can construct this function we can then use the result to calculate the stopping distance for any car if we know its initial velocity and stopping acceleration.
Although this sounds like a different task from what we’ve done in the previous two examples we will approach this task exactly the same way, by tabulating what we know about the situation,
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| Event 1: The instant the driver hits the brakes. t1 = 0 s r1 = 0 m v1 = vi a12 = -as | Event 2: The instant the car stops. t2 = r2 = d v2 = 0 m/s |
t1 = 0 s
r1 = 0 m
v1 = vi
a12 = -as
t2 =
r2 = d
v2 = 0 m/s
and then applying our two kinematic relationships:
Since our goal is to determine d as a function of vi and as, we must eliminate t2. To do this, solve for t2 in the left equation and substitute this expression into the right equation.
Thus, the stopping distance appears to be proportional to the square of the initial velocity and inversely proportional to the stopping acceleration. Does this make sense?
To determine if a symbolic expression is sensible it is often useful to check limiting cases. A limiting case is when one of the variables in the expression takes on an extreme value, typically zero or infinity. For example, if the initial velocity of the car was zero the stopping distance would have to also be zero, since the car was never moving! Allowing vi to equal zero in the above expression results in a stopping distance of zero, so our expression “passes” this logical test.
Another limiting case would be setting the acceleration of the car equal to zero. With no acceleration, the car should never stop. In our expression, setting the acceleration equal to zero results in an infinite stopping distance, which again agrees with commonsense. If our expression didn’t give the correct results in these limiting cases we would know we made an error somewhere in the derivation (and, of course, we would then go back and find our mistake and fix it because we are good students …).
Hints and Suggestions
Algebraic Signs
Confusion about the meaning of algebraic signs is common among beginning physics students. The best way to clarify this confusion is to remember that algebraic signs are simply a mathematical way to describe direction. Instead of saying up and down, or east and west, physicists construct coordinate systems and translate the words east and west into the symbols ‘+’ and ‘-’, or even ‘-’ and ‘+’ if we choose a different coordinate system. The key to the translation is the coordinate system. A coordinate system is very similar to the English-French dictionary you might take with you on your first trip to France. When you see a ‘-’, use your coordinate system to translate it into a verbal description of direction.
Do not fall into the common habit of translating a ‘-’ into the word “decreasing”. A negative acceleration, for example, does NOT imply that the object is slowing down. It implies an acceleration that points in the negative direction. It is impossible to determine whether an object is speeding up or slowing down by looking at the sign of the acceleration! Conversely, the word “deceleration”, which does mean that an object is slowing down, does not give any information regarding the sign of the acceleration. I can decelerate in the positive direction as easily as I can decelerate in the negative direction.
Addendum
Deriving the kinematic relationships
Let’s construct the two independent kinematic relationships that you will use whenever the acceleration is constant. In a later chapter, we will return to the case in which the acceleration is not constant.
From the definition of acceleration:
The above relationship is our first kinematic relationship. The acceleration in this relationship is really the average acceleration. However, since the acceleration is constant in this model the average acceleration is the same as the acceleration at any instant between the initial and final state.
From the definition of velocity:
You must remember, however, that the velocity in this formula is really the average velocity of the object over the time interval selected. To keep you from having to remember this fact, we can rewrite the average velocity as the sum of the initial velocity and the final velocity divided by two:
Substituting in the first kinematic relationship for the final velocity yields:
The above relationship is our second kinematic relationship.
Although we could keep deriving new kinematic equations forever, it is impossible that any other derived equation could allow us to calculate some quantity that these equations do not allow us to calculate.
Activities
Construct motion diagrams for the motions described below on a separate piece of paper.
a. A subway train in Washington, D.C., starts from rest and accelerates at 2.0 m/s2 for 12 seconds.
b. The driver of a car traveling at 35m/s suddenly sees a police car. The driver attempts to reach the speed limit of 25 m/s by accelerating at 2.5 m/s2. The driver has a reaction time of 0.55 s. (The reaction time is the time between first seeing the police car and pressing the brake.)
c. A pole-vaulter, just before touching the cushion on which she lands after a jump, is falling downward at a speed of 10 m/s. The pole-vaulter sinks about 2.0 m into the cushion before stopping.
d. An elevator is moving downward at 4.0 m/s for 3.5 s before someone presses the emergency stop button. The elevator comes to rest after traveling 2.9 m.
Construct motion diagrams for the motions described below.
a. An automobile comes to rest after skidding 35 m. The car’s acceleration while skidding is known to be 6.0 m/s2.
b. A car, initially traveling at 20 m/s to the east, accelerates toward the west at 2.0 m/s2. At the same time that the car starts moving, a truck, 60 m west of the car and moving at 16 m/s toward the east, starts to move faster, accelerating at 1.0 m/s2. It’s a one-lane road and both drivers are too busy texting to notice each other.
c. A child is hanging from a rope by her hands. She exerts a burst of strength and 2.0 s later is traveling at 1.4 m/s up the rope.
d.
A two-stage rocket initially accelerates upward from rest at 13 m/s2 for 5.0s before the second stage initiates a 14 s long upward acceleration of 25 m/s2.
For each of the motion diagrams below, determine the algebraic sign (+, – or zero) of the position, velocity, and acceleration of the object at the location of the three open circles. Describe an actual motion that could be represented by each motion diagram.
a.
Description:
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| | 1 | 2 | 3 |
| r | | | |
| v | | | |
| a | | | |
b.
Description:
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--- --- |
| | 1 | 2 | 3 |
| r | | | |
| v | | | |
| a | | | |
c.
Description:
| | | | |
--- --- |
| | 1 | 2 | 3 |
| r | | | |
| v | | | |
| a | | | |
For each of the motion diagrams below, determine the algebraic sign (+, – or zero) of the position, velocity, and acceleration of the object at the location of the three open circles. Describe an actual motion that could be represented by each motion diagram.
a.
Description:
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--- --- |
| | 1 | 2 | 3 |
| r | | | |
| v | | | |
| a | | | |
b.
Description:
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--- --- |
| | 1 | 2 | 3 |
| r | | | |
| v | | | |
| a | | | |
c.
Description:
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--- --- |
| | 1 | 2 | 3 |
| r | | | |
| v | | | |
| a | | | |
For each of the position vs. time graphs below, construct a corresponding motion diagram and velocity vs. time graph.
For each of the position vs. time graphs below, construct a corresponding motion diagram and velocity vs. time graph.
For each of the graphs below, construct a corresponding graph and motion diagram.
For each of the graphs below, construct a corresponding graph and motion diagram.
For each of the motion diagrams below, construct the corresponding motion graphs.
For each of the motion diagrams below, construct the corresponding motion graphs.
An object’s motion is represented by the position vs. time graph above.
a. Rank the object’s position at the lettered times.
Largest Positive 1. E 2. D 3. C 4. B 5. A Largest Negative
b. Rank the object’s velocity at the lettered times.
Largest Positive 1. ABCDE 2. 3. 4. 5. Largest Negative
Constant slope means constant velocity (and zero acceleration)
c. Rank the object’s acceleration at the lettered times.
Largest Positive 1. 2. 3. ABCDE 4. 5. Largest Negative
An object’s motion is represented by the velocity vs. time graph above.
d. Rank the object’s position at the lettered times.
Largest Positive 1. E 2. A 3. D 4. B 5. C Largest Negative
The motion diagram for the object is sketched below. Notice that regardless of where the origin is located, the turn-around point (C) is the smallest position.
e. Rank the object’s velocity at the lettered times.
Largest Positive 1. E 2. D 3. C 4. B 5. A Largest Negative
f. Rank the object’s acceleration at the lettered times.
Largest Positive 1. ABCDE 2. 3. 4. 5. Largest Negative
Constant slope means constant acceleration.
An object’s motion is represented by the position vs. time graph above.
a. Rank the object’s position at the lettered times.
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative
b. Rank the object’s velocity at the lettered times.
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative
c. Rank the object’s acceleration at the lettered times.
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative
An object’s motion is represented by the velocity vs. time graph above.
a. Rank the object’s position at the lettered times.
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative
b. Rank the object’s velocity at the lettered times.
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative
c. Rank the object’s acceleration at the lettered times.
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative
An object’s motion is represented by the position vs. time graph at right.
a. Rank the object’s position at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
b. Rank the object’s velocity at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
c. Rank the object’s acceleration at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
An object’s motion is represented by the position vs. time graph at right.
d. Rank the object’s position at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
e. Rank the object’s velocity at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
f. Rank the object’s acceleration at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
An object’s motion is represented by the position vs. time graph at right.
a. Rank the object’s position at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
b. Rank the object’s velocity at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
c. Rank the object’s acceleration at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
An object’s motion is represented by the position vs. time graph at right.
d. Rank the object’s position at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
e. Rank the object’s velocity at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
f. Rank the object’s acceleration at the lettered times.
Largest Largest
Positive Negative
_____ 2. _____ 3. _____ 4. _____ 5. _____
Below are velocity vs. time graphs for six different objects.
Rank these graphs on the basis of the distance traveled by each object.
Largest 1. ___F___ 2. ___A___ 3. ___E___ 4. ___B D___ 5. ___C___ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
The faster you travel, the more distance you travel. The direction you are headed is not important. Therefore, F travels the largest distance because it is always moving the fastest, followed by A. E is consistently traveling faster than D, so it covers a larger distance than D. B and D always travel at the same speed (although in opposite directions) so they cover the same distance, therefore they are ranked as equal. C travels the slowest so it covers the least distance.
Below are position vs. time graphs for six different objects.
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
Below are velocity vs. time graphs for six different objects.
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
A pole-vaulter, just before touching the cushion on which she lands after a jump, is falling downward at a speed of 10 m/s. The pole-vaulter sinks about 2.0 m into the cushion before stopping.
A child is hanging from a rope by her hands. She exerts a burst of strength and 2.0 s later is traveling at 1.4 m/s up the rope.
Motion Diagram Motion Information
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| | Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
Mathematical Analysis
An elevator is moving downward at 4.0 m/s when someone presses the emergency stop button. The elevator comes to rest after traveling 2.9 m.
Motion Diagram Motion Information
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---
| | Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
Mathematical Analysis
The driver of a car traveling at 16 m/s suddenly sees a truck that has entered from a side street and blocks the car’s path. The car’s maximum magnitude acceleration while braking is 6.0 m/s2 and the driver has a reaction time of 0.75 s. (The reaction time is the time between first seeing the truck and pressing the brake.) The driver stops just in time to avoid an accident.
Motion Diagram
Motion Information
| | | |
---
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
Mathematical Analysis
The driver of a car traveling at 35 m/s suddenly sees a police car. The driver attempts to reach the speed limit of 25 m/s by accelerating at 2.5 m/s2. The driver has a reaction time of 0.55 s. (The reaction time is the time between first seeing the police car and pressing the brake.)
Motion Diagram
Motion Information
| | | |
---
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
Mathematical Analysis
An automobile, initially traveling at 15 m/s, begins to slow down as it approaches a red light. After traveling 15 m, and slowing to 3.0 m/s, the light turns green and the driver steps on the gas and accelerates for 2.4 seconds until she reaches her original speed.
Motion Diagram
Motion Information
| | | |
---
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
Mathematical Analysis
A two-stage rocket initially accelerates upward from rest at 13 m/s2 for 5.0s before the second stage ignites. The second stage “burns” for 14 s and results in the rocket achieving a speed of 415 m/s at the end of this stage.
Motion Diagram Motion Information
| | | | | | |
--- --- --- |
| | Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = | | |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
Mathematical Analysis
Rather than pushing the button for the correct floor a man prefers to hit the emergency stop button when an elevator approaches his floor, and then pry the doors apart. An elevator is moving at a constant speed of 2.8 m/s when it is 16 m from his floor. With uncanny timing, and an elevator that can slow at 3.5 m/s2, he makes the elevator stop precisely at his floor.
Motion Diagram Motion Information
| | | | | | |
--- --- --- |
| | Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = | | |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
Mathematical Analysis
A subway train in Washington, D.C., starts from rest and accelerates at 2.0 m/s2 for 12 s. The train travels at a constant speed for 65 s. The speed of the train then decreases for 25 s until it reaches the next station.
Motion Diagram
Motion Information
| | | | |
--- --- |
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = a34 = | Event 4: t4 = r4 = v4 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
a34 =
t4 =
r4 =
v4 =
Mathematical Analysis
A rocket ship is launched from rest from a space station. Its destination is 1.0 x 1011 m away. The ship is programmed to accelerate at 7.4 m/s2 for 12 hours. After 12 hours, the ship will travel at constant velocity until it comes within 1.0 x 1010 m of its destination. Then, it will fire its retrorockets to land safely.
Motion Diagram
Motion Information
| | | | |
--- --- |
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = a34 = | Event 4: t4 = r4 = v4 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
a34 =
t4 =
r4 =
v4 =
Mathematical Analysis
The driver of a car traveling at 16 m/s sees a truck 20 m ahead traveling at a constant speed of 12 m/s. The car starts without delay to accelerate at 4.0 m/s2 in an attempt to rear-end the truck. The truck driver is too busy talking on his cell phone to notice the car.
A car, initially at rest, accelerates toward the west at 2.0 m/s2. At the same time that the car starts, a truck, 350 m west of the car and moving at 16 m/s toward the east, starts to move slower, accelerating at 1.0 m/s2. The car and truck pass safely.
Motion Diagram
Motion Information
Object: Object:
| | | | |
--- --- |
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = | Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
Mathematical Analysis
A car, initially traveling at 20 m/s to the east, accelerates toward the west at 2.0 m/s2. At the same time that the car starts, a truck, 60 m west of the car and moving at 16 m/s toward the east, starts to move faster, accelerating at 1.0 m/s2. It’s a one-lane road and both drivers are too busy texting to notice each other.
Motion Diagram
Motion Information
Object: Object:
| | | | |
--- --- |
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = | Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
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Mathematical Analysis[xi]
A helium balloon begins to rise from rest with an acceleration of 2.0 m/s2 until it reaches a height of 50 m, and then continues upward at constant speed. However, a passenger in the balloon forgot their cell phone on the ground and threatens to jump from the balloon because of separation anxiety. Therefore a second balloon is launched 5.0 s after the first and accelerates upwards at 3.0 m/s2 to reunite the passenger and his phone.
Motion Diagram Motion Information
| | | | | | | |
--- --- ---
| | Object: Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = | Object: Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = |
Event 1:
t1 =
r1 =
v1 =
a12 =
Event 2:
t2 =
r2 =
v2 =
a23 =
Event 3:
t3 =
r3 =
v3 =
Event 1:
t1 =
r1 =
v1 =
a12 =
Event 2:
t2 =
r2 =
v2 =
a23 =
Event 3:
t3 =
r3 =
v3 =
Mathematical Analysis[xii]
A car traveling at 16 m/s sees a truck 20 m ahead traveling at a constant speed of 12 m/s. After thinking it over for 0.80 s, the car starts to accelerate at 4.0 m/s2 in an attempt to rear-end the truck. The truck driver has paid-up insurance so doesn’t care. The car driver’s motive is unknown.
Motion Diagram
Motion Information
Object: Object:
| | | | | | |
--- --- --- |
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = | Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
Mathematical Analysis[xiii]
A pole-vaulter lands on a cushion after a vault. Determine her acceleration as she sinks into the cushion (acushion) as a function of her velocity when she hits the cushion (vi) and the distance she sinks into the cushion (d).
Motion Diagram Motion Information
| | | |
---
| | Event 1: t1 = r1 = v1 = a12 = Mathematical Analysis | Event 2: t2 = r2 = v2 = |
t1 =
r1 =
v1 =
a12 =
Mathematical Analysis
t2 =
r2 =
v2 =
Questions
If vi = 0 m/s, what should acushion equal? Does your function agree with this observation?
If d = 0 m, what should acushion equal? Does your function agree with this observation?
What would result in a larger magnitude acceleration, hitting the cushion twice as fast or sinking one-half as far into the cushion?
The driver of an automobile suddenly sees an obstacle blocking her lane. Determine the total distance the car travels between seeing the obstacle and stopping (d) as a function of the initial velocity of the car (vi) and the magnitude of the car’s acceleration while stopping (as). Ignore the driver’s reaction time.
Motion Diagram
Motion Information
| | |
--- |
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
Mathematical Analysis
Questions
If vi = 0 m/s, what should d equal? Does your function agree with this observation?
If as = 0 m/s2, what should d equal? Does your function agree with this observation?
If the car was traveling twice as fast how much further would the car travel before stopping?
The driver of an automobile suddenly sees an obstacle blocking his lane. Determine the total distance the car travels between seeing the obstacle and stopping (d) as a function of his reaction time (tR), the initial velocity of the car (vi), and the magnitude of the car’s acceleration while stopping (as).
Motion Diagram
Motion Information
| | | |
---
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
Mathematical Analysis
Questions
If vi = 0 m/s, what should d equal? Does your function agree with this observation?
If as = 0 m/s2, what should d equal? Does your function agree with this observation?
Does the car travel a greater distance during the reaction phase or the braking phase? Estimate necessary quantities assuming a panic stop from highway speeds.
Two automobiles are involved in a race over level ground. Both cars begin at rest. The Audi accelerates at A for T seconds and then travels at constant velocity for the rest of the race. The Buick accelerates at A/2 for 2T seconds and then travels at constant velocity for the rest of the race. Determine the distance (d) between the two cars, at the instant t = 2T, as a function of A and T.
Motion Diagram
Motion Information
Object: Object:
| | | | | | |
--- --- --- |
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = | Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
Mathematical Analysis
Questions
If A = 0 m/s2, what should d equal? Does your function agree with this observation?
Assume the race takes longer than 2T to finish. What is the difference in speed between the two cars as they cross the finish line?
Assume the race takes longer than 2T to finish. What is the distance between the two cars as they cross the finish line?
Two automobiles are involved in a game of chicken. The two cars start from rest at opposite ends of a long, one-lane road. The Audi accelerates at A for T seconds and then travels at constant velocity. The Buick accelerates at A/2 for 2T seconds and then travels at constant velocity. Determine the elapsed time when the car’s collide (tcollide) as a function of the length of the road (d), A and T. The collision takes place after both cars have reached constant velocity.
Motion Diagram
Motion Information
Object: Object:
| | | | | | |
--- --- --- |
| Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = | Event 1: t1 = r1 = v1 = a12 = | Event 2: t2 = r2 = v2 = a23 = | Event 3: t3 = r3 = v3 = |
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
t1 =
r1 =
v1 =
a12 =
t2 =
r2 =
v2 =
a23 =
t3 =
r3 =
v3 =
Mathematical Analysis
Questions
If A = 0 m/s2, what should tcollide equal? Does your function agree with this observation?
If d = ∞, what should tcollide equal? Does your function agree with this observation?
What is the minimum length of road needed to guarantee the two cars collide at top speed?
[i] r2 = 1.4 m
[ii] t2 = 1.45 m
[iii] t3 = 3.4 s
[iv] t3 = 4.55 s
[v] r3 = 36.6 m
[vi] r3 = 3520 m
[vii] r2 = 14.9 m
[viii] r4 = 2000 m
[ix] t4 = 3.7 x 105 s
[x] t2 = 14.9 s
[xi] t2 = 7.8 s
[xii] t3 = 15.7 s
[xiii] t3 = 2.87 s
Homework 1 – Model 1: 25, 31, 32, 35, 39, 45, 47, 53, 55, and 59.
This page titled 1.1: Kinematics is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Lumen Learning.
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Algorithm: Triangle with two constraints, each corner on a given line
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Some time ago I asked a question on math.stackexchange and got an answer. I have difficulties deriving an algorithm from that answer because my background is in design and hope some of you can help me.
The original question with visual sketch and possible answer are here:
The question was: Given 3 3-dimensional lines (a, b and c) that coincide in a common point S and a given Point B on b, I'm looking for a point A on a and a point C on c where AB and BC have the same length and the angle ABC is 90 degrees.
I will have to implement this algorithm in an imperative language, any code in C++, Java, imperative pseudo-code or similar is fine.
Also, different approaches to this problem are equally welcome. Plus: Thanks for any hints, if the complete solution is indeed too time-consuming!
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edited Apr 13, 2017 at 12:19
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asked Apr 5, 2014 at 12:55
PatPat
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I tried to understand the answer :) I understand the first part of the answer but can't follow through with cubic equations and don't understand the whole second half with the conic intersection. I'm simply missing the mathematical background and decided to ask for help again.
– Pat
Commented
Apr 5, 2014 at 13:13
What I meant is, what have you tried so far codewise?
– wvdz
Commented
Apr 5, 2014 at 13:17
I wanted to understand the answer first. I can't implement an algorithm because I can't transform the answer into imperative steps because I don't understand the answer in the first place. Coding will (probably) not be the problem, because the problem is primarily a mathematical one, that's why I don't care about the language used in a solution here. As long as the solution is not too concisely expressed in a functional language.
– Pat
Commented
Apr 5, 2014 at 13:29
Seems a lot like you're asking us to write a program for you, rather than asking help on a program you're working on. In any case, I can't help you with this, cause I don't understand that mathematical answer either.
– wvdz
Commented
Apr 5, 2014 at 13:39
No, I'm not asking for a program. All I'm asking for is: If someone understands the mathematical answer, it's maybe not too time-consuming to sketch out an algorithm in pseudo-code. If it is, no problem, I don't ask here expecting everything/anything but hoping for any help or hints. And this problem is of course part of a larger program I'm working on.
– Pat
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Apr 5, 2014 at 13:43
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The two key formulas are
(I've replied the derivation for the formulas in the mathematics stack exchange site)
Substituting the first in the second gives in the end a 4th degree equation that is quite annoying to solve with a closed form. I've therefore used instead a trivial numerical solver in Python:
```
function to solve (we look for t such that f(t)=0)
def f(t):
s = (tcB - B2) / (tac - aB)
return ss - 2saB - tt + 2tcB
given f and an interval to search generates all solutions in the range
def solutions(f, x0, x1, n=100, eps=1E-10):
X = [x0 + i(x1 - x0)/(n - 1) for i in xrange(n)]
Y = map(f, X)
for i in xrange(n-1):
if (Y[i]<0 and Y[i+1]>=0 or Y[i+1]<0 and Y[i]>=0):
xa, xb = X[i], X[i+1]
ya, yb = Y[i], Y[i+1]
if (xb - xa) < eps:
# Linear interpolation
# 0 = ya + (x - xa)(yb - ya)/(xb - xa)
yield xa - ya (xb - xa) / (yb - ya)
else:
for x in solutions(f, xa, xb, n, eps):
yield x
```
The search algorithm samples the function in the interval and when it finds two adjacent samples that are crossing the f=0 line repeats the search recursively between those two samples (unless the interval size is below a specified limit, approximating the function with a line and computing the crossing point in that case).
I've tested the algorithm generating random problems and solving them with
```
from random import random as rnd
for test in xrange(1000):
a = normalize((rnd()-0.5, rnd()-0.5, rnd()-0.5))
b = normalize((rnd()-0.5, rnd()-0.5, rnd()-0.5))
c = normalize((rnd()-0.5, rnd()-0.5, rnd()-0.5))
L = rnd() 100
B = tuple(xL for x in b)
aB = dot(a, B)
cB = dot(c, B)
B2 = dot(B, B)
ac = dot(a, c)
sols = list(solutions(f, -1000., 1000.))
```
And there are cases in which the solutions are 0, 1, 2, 3 or 4. For example the problem
```
a = (-0.5900900304960981, 0.4717596600172049, 0.6551614908475357)
b = (-0.9831451620384042, -0.10306322574446096, 0.15100848274062748)
c = (-0.6250439408232388, 0.49902426033920616, -0.6002456660677057)
B = (-33.62793897729328, -3.5252208930692497, 5.165162011403056)
```
has four distinct solutions:
```
s = 57.3895941365 , t = -16.6969433689
A = (-33.865027354189415, 27.07409541837935, 37.59945205363035)
C = (10.436323283003153, -8.332179814593692, 10.022267893763457)
|A - B| = 44.5910029061
|C - B| = 44.5910029061
(A - B)·(C - B) = 1.70530256582e-13
s = 43.619078237 , t = 32.9673082734
A = (-25.739183207076163, 20.5777215193455, 28.577540327140607)
C = (-20.606016281518986, 16.45148662649085, -19.78848391300571)
|A - B| = 34.5155582156
|C - B| = 34.5155582156
(A - B)·(C - B) = 1.13686837722e-13
s = -47.5886624358 , t = 83.8222109697
A = (28.08159526800866, -22.450411211385674, -31.17825902887765)
C = (-52.39256507303229, 41.82931682916268, -50.313918854788845)
|A - B| = 74.0747844969
|C - B| = 74.0747844969
(A - B)·(C - B) = 4.54747350886e-13
s = 142.883074325 , t = 136.634726869
A = (-84.31387768560096, 67.4064705656035, 93.61148799140805)
C = (-85.40270813540043, 68.1840435123674, -82.01440263735996)
|A - B| = 124.189861967
|C - B| = 124.189861967
(A - B)·(C - B) = -9.09494701773e-13
```
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answered Apr 5, 2014 at 20:37
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Thanks a lot, I will try it out and let you know.
– Pat
Commented
Apr 6, 2014 at 17:30
For certain problems that should have a solution the algorithm doesn't give one... maybe a problem with the equation? See for example: a = (0.1836445, 0, 0.9829927); b = (0.0, 0.0, 1.0); c = (0, 0.3500095, 0.9367462); B = (0.0, 0.0, 8.3);
– Pat
Commented
Apr 20, 2014 at 15:26
@user479864: Checking the equation with these parameters indeed seems extremely regular but never crossing the zero (thus no real solutions). Why do you think this problem must have a solution? Can you provide one? Like it's said in the answer I've been able with random parameter sampling to create problems with 0, 1, 2, 3 or 4 distinct solutions.
– 6502
Commented
Apr 20, 2014 at 16:06
I'm displaying the solutions in a 3D environment and it seems/looks very strange that there is no solution for a = (0.3266604, 0, 0.9451418), c = (0, 0.3500095, 0.9367462) but one for a = (0.3266588, -0.001026195, 0.9451418), c = (0, 0.3500095, 0.9367462), both using b and B of the above example. In this example this problem always occurs when a) the angle between AB and BC is also 90 degrees when projecting the vectors onto a plane at B where b is its normal and b) |AB| != |BC|. I can't see a reason in my visualization why there shouldn't be a solution, but I don't have mathematical proof...
– Pat
Commented
Apr 20, 2014 at 18:15
@user479864: I'm not sure I understand. Even the equation xx - 1e-100 = 0 has two solutions but xx + 1e-100 = 0 has no solutions on the real line. The full equation has degree 4 and it can have between 0 and 4 solutions in general (IOW there are parameters for all 5 cases). Of course you can find values of the parameters for which a small variation can change the number of solutions. Note there are even infinite cases with infinite solutions each (e.g. a=(1,0,0), b=(sqrt(2)/2, sqrt(2)/2, 0), c=(0,1,0), B=b).
– 6502
Commented
Apr 21, 2014 at 16:40
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Write two quadratic equations for lambda, mu unknowns (just above matrix forms).
Solve this system with paper, pen and head, or with any mathematical software like Maple, Mathematica, Matlab, Derive etc. You will have 4th order equation. It has closed-form solution - apply Ferrari or Kardano method and get real roots, find mu, lambda, then point coordinates.
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answered Apr 5, 2014 at 15:53
MBoMBo
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Thanks for the steps, I will give them a try and let you know.
– Pat
Commented
Apr 6, 2014 at 17:30
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8860 | https://brainly.com/question/23839305 | [FREE] The largest 2-digit number is 99. How many 2-digit numbers must be in a set in order to apply the - brainly.com
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The largest 2-digit number is 99. How many 2-digit numbers must be in a set in order to apply the pigeonhole principle to conclude that there are two distinct subsets of the numbers whose elements sum to the same value?
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The solution can be defined as follows:
Explanation
→n=10
There are 210 = 1,024 sub-sets of 10 entries, however, the number of enters between the lower and upper limits total of 10 different submissions between 1 and 100 can be as low only as 901 (= 955 - 55 + 1). There has to be a minimum level of at least 1 subset with much more subgroups than possible, which equals at least two sets. Call for two main sub, S' and T' equivalent.
Let
C=S′∩T′
when
S=S′−C T=T′−C
ThenS an d T are disjoint
s u m(S)\t=s u m(S′−C)
=s u m(S′)−s u m(C)=s u m(T′)−s u m(C)=s u m(T′−C)=s u m(T)
Answered by codiepienagoya •6.5K answers•66.7M people helped
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To guarantee two distinct subsets of two-digit numbers that sum to the same value, we need at least 91 two-digit numbers in a set. The total number of two-digit numbers is 90, thus we must exceed this count by one to apply the pigeonhole principle effectively.
Explanation
To apply the pigeonhole principle in this situation, we need to consider the set of two-digit numbers. The largest two-digit number is 99, while the smallest is 10. This means the two-digit numbers range from 10 to 99, giving us a total of 90 two-digit numbers: 10, 11, 12, ..., 99.
When using the pigeonhole principle, we want to find how many of these numbers we need in a set to ensure that at least two distinct subsets of the numbers have the same sum.
First, we must calculate the maximum possible sum of the two-digit numbers. The sum of all two-digit numbers can be calculated using the formula for the sum of an arithmetic series:
S n=2 n×(a+l)
where n is the number of terms, a is the first term, and l is the last term.
Here, we have:
n=90 (the count of two-digit numbers)
a=10
l=99
Substituting these values into the formula:
S 90=2 90×(10+99)=45×109=4905
This means the sum of all two-digit numbers is 4905. Now, if we consider any subset of these numbers, the largest sum we can get from selecting different subsets will not exceed this total. However, when calculating the numbers of possible subsets, any subset can sum to various values.
The number of distinct sums of any subset from these numbers will range from 0 to 4905. However, the maximum number of distinct sums produced by subsets is smaller because not all integer values in this range can be achieved through distinct subsets. Specifically, when we take the subsets, the number of total subsets of a set with n elements is 2 n. Thus, with 90 two-digit numbers, there are a total of 2 90−1 possible non-empty subsets.
Given that there are many more possible distinct subsets than there are possible sums available (4906 total sums from 0 to 4905), the pigeonhole principle assures us that if we take more than 4905 two-digit numbers, at least two distinct subsets will have the same sum.
Therefore, we need to take at least 90+1=91 two-digit numbers in a set to ensure the existence of two distinct subsets whose elements sum to the same value.
Examples & Evidence
For example, if we have the numbers 10, 11, and 12, we could find subsets like {10} and {11, 12} that both sum to 10. Once we have all 90 numbers and add even one more, we can guarantee some overlapping sums among subsets.
The pigeonhole principle states that if we have n items to be distributed into m containers, and if n > m, then at least one container must contain more than one item. In this case, the subsets are our items and the sums are the containers.
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Applying the pigeonhole principle to this question, we can deduce that we need to select 199 numbers from a set of 2-digit numbers to ensure that there are two distinct subsets whose elements sum to the same value.
Explanation
The subject we're dealing with here is the pigeonhole principle. It is a concept in mathematics, more specifically, in combinatorics or the study of counting. In the realm of simple terms, the pigeonhole principle asserts that if 'n' items are put into 'm' pigeonholes with n > m, then at least one pigeonhole must contain more than one item.
When it comes to 2-digit numbers (from 10 to 99), the smallest possible sum of any two of them is 11 (1+10), and the largest possible sum is 198 (99+99). Therefore, there are 198 total possible sums we could achieve. With those 198 sums, there could be 90 2-digit numbers. According to the pigeonhole principle, if we select 199 numbers (198 possible sums plus 1 extra), we can be certain that there must be at least two different subsets of the numbers that will sum to the same total due to the nature of the principle.
Learn more about Pigeonhole Principle here:
brainly.com/question/34617354
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8861 | https://www.brighthubengineering.com/hydraulics-civil-engineering/67126-calculation-of-hydraulic-radius-for-uniform-open-channel-flow/ | Calculation of Open Channel Flow Hydraulic Radius: Calculate using Trapezoid Area
Introduction
One of the parameters needed in order to make use of the Manning equation for open channel flow calculations is the hydraulic radius of the channel cross section. Common shapes for open channel cross section include rectangle, trapezoid, triangle, and circle.
The use of hydraulic radius in Manning equation calculations is covered in the first article of this series, ‘Introduction to the Manning Equation for Open Channel Flow Calculations’ and the hydraulic radius and use of the Manning equation for a circular pipe are covered in ‘How to Use the Manning Equation for Storm Sewer Calculations.’
This article will cover rectangle, trapezoid, triangle, and circular shapes for an open channel cross section. Hydraulic radius is defined as the cross sectional area of flow divided by the wetted perimeter, so the calculation of rectangle and trapezoid area and triangle area will be included along with the perimeter for each.
Rectangular Cross Section
The simplest open channel flow cross section for calculation of hydraulic radius is a rectangle. The depth of flow is often represented by the symbol, y, and b is often used for the channel bottom width, as shown in the diagram at the left. From the hydraulic radius definition: RH = A/P, where A is the cross sectional area of flow and P is its wetted perimeter. From the diagram it is clear that A = by and P = 2y + b, so the hydraulic radius is: RH = by/(2y + b) for an open channel flow through a rectangular cross section.
Trapezoidal Cross Section
A trapezoid shape is sometimes used for manmade channels and the cross section of natural stream channels are often approximated by a trapezoid area. The diagram at the right shows a trapezoid and the parameters typically used for its shape and size in open channel flow calculations. Those parameters, which are used to calculate the trapezoid area and wetted perimeter, are y, the liquid depth; b, the bottom width; B the width of the liquid surface; λ, the wetted length measured along the sloped side; and α, the angle of the sloped side from vertical. The side slope is usually specified as horiz:vert = z:1.
The cross sectional area of flow is the trapezoid area: A = y(b + B)/2, or
A = (y/2)(b + b + 2zy), because B = b + 2zy, as can be seen from the diagram.
Simplifying, the trapezoid area is: A = by + zy2.
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
The use of hydraulic radius in Manning equation calculations is covered in the first article of this series, ‘Introduction to the Manning Equation for Open Channel Flow Calculations’ and the hydraulic radius and use of the Manning equation for a circular pipe are covered in ‘How to Use the Manning Equation for Storm Sewer Calculations.’
This article will cover rectangle, trapezoid, triangle, and circular shapes for an open channel cross section. Hydraulic radius is defined as the cross sectional area of flow divided by the wetted perimeter, so the calculation of rectangle and trapezoid area and triangle area will be included along with the perimeter for each.
Rectangular Cross Section
The simplest open channel flow cross section for calculation of hydraulic radius is a rectangle. The depth of flow is often represented by the symbol, y, and b is often used for the channel bottom width, as shown in the diagram at the left. From the hydraulic radius definition: RH = A/P, where A is the cross sectional area of flow and P is its wetted perimeter. From the diagram it is clear that A = by and P = 2y + b, so the hydraulic radius is: RH = by/(2y + b) for an open channel flow through a rectangular cross section.
Trapezoidal Cross Section
A trapezoid shape is sometimes used for manmade channels and the cross section of natural stream channels are often approximated by a trapezoid area. The diagram at the right shows a trapezoid and the parameters typically used for its shape and size in open channel flow calculations. Those parameters, which are used to calculate the trapezoid area and wetted perimeter, are y, the liquid depth; b, the bottom width; B the width of the liquid surface; λ, the wetted length measured along the sloped side; and α, the angle of the sloped side from vertical. The side slope is usually specified as horiz:vert = z:1.
The cross sectional area of flow is the trapezoid area: A = y(b + B)/2, or
A = (y/2)(b + b + 2zy), because B = b + 2zy, as can be seen from the diagram.
Simplifying, the trapezoid area is: A = by + zy2.
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
This article will cover rectangle, trapezoid, triangle, and circular shapes for an open channel cross section. Hydraulic radius is defined as the cross sectional area of flow divided by the wetted perimeter, so the calculation of rectangle and trapezoid area and triangle area will be included along with the perimeter for each.
Rectangular Cross Section
The simplest open channel flow cross section for calculation of hydraulic radius is a rectangle. The depth of flow is often represented by the symbol, y, and b is often used for the channel bottom width, as shown in the diagram at the left. From the hydraulic radius definition: RH = A/P, where A is the cross sectional area of flow and P is its wetted perimeter. From the diagram it is clear that A = by and P = 2y + b, so the hydraulic radius is: RH = by/(2y + b) for an open channel flow through a rectangular cross section.
Trapezoidal Cross Section
A trapezoid shape is sometimes used for manmade channels and the cross section of natural stream channels are often approximated by a trapezoid area. The diagram at the right shows a trapezoid and the parameters typically used for its shape and size in open channel flow calculations. Those parameters, which are used to calculate the trapezoid area and wetted perimeter, are y, the liquid depth; b, the bottom width; B the width of the liquid surface; λ, the wetted length measured along the sloped side; and α, the angle of the sloped side from vertical. The side slope is usually specified as horiz:vert = z:1.
The cross sectional area of flow is the trapezoid area: A = y(b + B)/2, or
A = (y/2)(b + b + 2zy), because B = b + 2zy, as can be seen from the diagram.
Simplifying, the trapezoid area is: A = by + zy2.
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
Rectangular Cross Section
The simplest open channel flow cross section for calculation of hydraulic radius is a rectangle. The depth of flow is often represented by the symbol, y, and b is often used for the channel bottom width, as shown in the diagram at the left. From the hydraulic radius definition: RH = A/P, where A is the cross sectional area of flow and P is its wetted perimeter. From the diagram it is clear that A = by and P = 2y + b, so the hydraulic radius is: RH = by/(2y + b) for an open channel flow through a rectangular cross section.
Trapezoidal Cross Section
A trapezoid shape is sometimes used for manmade channels and the cross section of natural stream channels are often approximated by a trapezoid area. The diagram at the right shows a trapezoid and the parameters typically used for its shape and size in open channel flow calculations. Those parameters, which are used to calculate the trapezoid area and wetted perimeter, are y, the liquid depth; b, the bottom width; B the width of the liquid surface; λ, the wetted length measured along the sloped side; and α, the angle of the sloped side from vertical. The side slope is usually specified as horiz:vert = z:1.
The cross sectional area of flow is the trapezoid area: A = y(b + B)/2, or
A = (y/2)(b + b + 2zy), because B = b + 2zy, as can be seen from the diagram.
Simplifying, the trapezoid area is: A = by + zy2.
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
Trapezoidal Cross Section
A trapezoid shape is sometimes used for manmade channels and the cross section of natural stream channels are often approximated by a trapezoid area. The diagram at the right shows a trapezoid and the parameters typically used for its shape and size in open channel flow calculations. Those parameters, which are used to calculate the trapezoid area and wetted perimeter, are y, the liquid depth; b, the bottom width; B the width of the liquid surface; λ, the wetted length measured along the sloped side; and α, the angle of the sloped side from vertical. The side slope is usually specified as horiz:vert = z:1.
The cross sectional area of flow is the trapezoid area: A = y(b + B)/2, or
A = (y/2)(b + b + 2zy), because B = b + 2zy, as can be seen from the diagram.
Simplifying, the trapezoid area is: A = by + zy2.
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
The cross sectional area of flow is the trapezoid area: A = y(b + B)/2, or
A = (y/2)(b + b + 2zy), because B = b + 2zy, as can be seen from the diagram.
Simplifying, the trapezoid area is: A = by + zy2.
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
A = (y/2)(b + b + 2zy), because B = b + 2zy, as can be seen from the diagram.
Simplifying, the trapezoid area is: A = by + zy2.
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
A = (y/2)(b + b + 2zy), because B = b + 2zy, as can be seen from the diagram.
Simplifying, the trapezoid area is: A = by + zy2.
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
Simplifying, the trapezoid area is: A = by + zy2.
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
The wetted perimeter is: P = b + 2λ, but by Pythagoras Theorem:
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
λ2 = y2 + (yz)2, or λ = [y2 + (yz)2]1/2, so the wetted perimeter is:
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
P = b + 2y(1 + z2)1/2, and the hydraulic radius for a trapezoid is:
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
RH = (by + zy2)/[b + 2y(1 + z2)1/2]
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
Triangular Cross Section
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
A triangular open channel cross section is shown in the diagram at the left. The diagram shows the typical case, where the two sides are sloped at the same angle. Fewer parameters are needed for the triangular area than for the trapezoid area. The parameters, as shown in the diagram are: B, the surface width of the liquid; λ, the sloped length of the triangle side; y, the liquid depth measured from the vertex of the triangle; and the side slope specification, horiz:vert = z:1.
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
The triangle area is: A = By/2, but the figure shows that B = 2yz, so the triangle area becomes simply: A = y2z.
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
The wetted perimeter is: P = 2λ with λ2 = y2 + (yz)2. This simplifies to: P = 2[y2(1 + z2)]1/2
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
The hydraulic radius is thus: RH = A/P = y2z/{2[y2(1 + z2)]1/2}
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
Circular Cross Section
The hydraulic radius for a circular pipe flowing full is easy to calculate. The cross-sectional area is A = πD2/4 and the wetted perimeter is P = πD. Substituting into the equation RH = A/P and simplifying the expression gives: RH = D/4. For calculation of the hydraulic radius for partially full pipe flow, see the article, “Use of Excel Spreadsheet Templates for Partially Full Pipe Flow.” Spreadsheet templates can be downloaded from this article to calculate hydraulic radius (and other parameters) for partially full pipe flow.
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
References and Image Credits
References for further information:
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
Bengtson, Harlan H., Open Channel Flow I - The Manning Equation and Uniform Flow, an online, continuing education course for PDH credit.
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002.
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
Images are all from reference #1
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
This post is part of the series: Uniform Open Channel Flow and the Manning Equation
The Manning equation is widely used for uniform open channel flow calculations with natural or man made channels. The Manning equation is used to relate parameters like river discharge and water flow velocity to hydraulic radius, and open channel slope, size, shape, and Manning roughness.
Introduction to the Manning Equation for Uniform Open Channel Flow Calculations- Calculation of Hydraulic Radius for Uniform Open Channel Flow- Use of the Manning Equation for Open Channel Flow in Natural Channels- Determining the Manning Roughness Coefficient for a Natural Channel- Calculating Uniform Open Channel Flow/Manning Equation Solutions
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8862 | https://www.quora.com/In-which-case-are-the-angle-of-incident-and-the-angle-of-refraction-the-same | In which case are the angle of incident and the angle of refraction the same? - Quora
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Physics
Index of Refraction
Geometrical Optics
Angle of Incidence
Light Refraction
Basic Optics
Angle of Reflection
Optics and Light
Angle of Refraction
5
In which case are the angle of incident and the angle of refraction the same?
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Harindra Kumar
Former Lecture in Physics in Science College · Author has 2K answers and 2.3M answer views
·6y
(1) when light falls normally on the separating surface of two media then angle of incidence and angle of refraction both are zero.
(2) when two media have same refractive index then angle of incidence and angle of reflection have same value. Example Canada Balsam and Crown glass both have refractive index n=1.55
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More answers below
When is the angle of an incident equal to the angle of a refraction?
What is the relation between angle of incidence and angle of refraction, if light travels from rarer to denser medium?
Can the angle of incidence be the same as the angle of refraction?
What is an incident angle when the refraction angle is 90?
Why is the angle of the incidence greater than angle of refraction?
Satya Parkash Sud
Former Professor at Himachal Pradesh University Shimla (1986–2002) · Author has 8.1K answers and 27.4M answer views
·7y
The angle of incidence and angle of refraction are the same, only when the incident rays enters the interface of two different media at right angle to the interface ie the angle of incidence is zero. In this case the angle of refraction is also zero.
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9 8
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Amit Kumar Yadav
Mathematics and Physics teacher at Genesis Global School · Author has 319 answers and 211.1K answer views
·3y
The angle of incidence equals angle of refraction in following case:
when the light ray is incident normally, the ray does not undergo refraction, it goes undeviated. In this case both the angle of incidence and angle of refraction is of 0 degrees.
Upvote ·
Gopal Menon
B Sc (Hons) in Mathematics, Indira Gandhi National Open University (IGNOU) (Graduated 2010) · Author has 10.2K answers and 15.2M answer views
·7y
Related
How can we say that the angle of incidence is equal to the angle of reflection?
The question is “How can we say that the angle of incidence is equal to the angle of reflection?”
This can be proved by the wave theory of light proposed by Huygens in 1678.
To explain the propagation of light waves, Huygens proposed the following:
Every point of a wave front behaves as a secondary source of light and sends secondary waves in all directions.
The secondary wavelets are more effective in the forward direction i.e in the direction of propagation of the wave.
The resultant wave front in any position is tangential to all the secondary wavelets at that point of time.
A plane wave front ap
Continue Reading
The question is “How can we say that the angle of incidence is equal to the angle of reflection?”
This can be proved by the wave theory of light proposed by Huygens in 1678.
To explain the propagation of light waves, Huygens proposed the following:
Every point of a wave front behaves as a secondary source of light and sends secondary waves in all directions.
The secondary wavelets are more effective in the forward direction i.e in the direction of propagation of the wave.
The resultant wave front in any position is tangential to all the secondary wavelets at that point of time.
A plane wave front appears as under:
Let A′B′A′B′ be the cross section of a wave front at any point of time. Some time back, the cross section of the wave front was A B A B and some time later, the wave front would be A′′B′′.A″B″.
Let us consider the cross section A′B′A′B′ as the primary wave front. Then, points P,Q,R,S,T P,Q,R,S,T act as secondary sources of light and create secondary spherical wavelets. These wavelets travel at the speed of light and after some time t would have a common tangential surface with cross section A′′B′′A″B″ which becomes the new wave front.
Now consider portion A B,A B, of a plane wave front bounded by two rays P A P A and P B P B perpendicular to it, as shown in the figure.
The wave front strikes the reflecting surface Y Z Y Z when it reaches the position A′B′.A′B′. At this point of time, when ray P A P A reaches A′,A′, it behaves as a secondary source of light and generates a secondary wavelet,. At this time, the ray Q B Q B has reached B′B′ and continues on its path until it strikes the reflecting surface at point C.C.
Let the time elapsed from the time ray P A P A reaches the reflecting surface to the time the ray Q B Q B reaches the reflecting surface be t.t. Then, the distance B′C=c t,B′C=c t, where c c is the speed of light. During this time, the secondary wavelet originating from A′A′ forms a hemisphere of radius c t,c t, whose cross section is shown in the figure as a semicircle. This becomes a secondary wave front. Draw a tangent to the secondary wave front from point C.C. Since B′D=c t=A D,B′D=c t=A D, the waves at points C C and D D are in phase and hence C D C D represents the reflected wave front bounded by the rays A′R A′R and C S.C S.
A′N A′N and C N′C N′ are normal to the reflecting surface. The angle of incidence is ∠A A′N=i∠A A′N=i and the angle of reflection is ∠N A′D=r.∠N A′D=r.
∠D A′C=90 o−r.∠D A′C=90 o−r.
⇒∠D C A′=90 o−D A′C=r,⇒∠D C A′=90 o−D A′C=r, since D C⊥A′R.D C⊥A′R.
Since ∠P A′B′=90 o,∠N A′B′=90 o−i.∠P A′B′=90 o,∠N A′B′=90 o−i.
⇒∠B′A′C=90 o−N A′B′=i⇒∠B′A′C=90 o−N A′B′=i
In △A′B′C△A′B′C and △A′D C,△A′D C,
∠A′B′C=∠A′D C=90 o.∠A′B′C=∠A′D C=90 o.
B′C=A′D=c t.B′C=A′D=c t.
A′C=A′C A′C=A′C as they are common sides.
⇒△A′B′C≅△A′D C⇒△A′B′C≅△A′D C (hypotenuse side test).
⇒∠B′A′C=∠D C A′.⇒∠B′A′C=∠D C A′.
We have already shown that ∠B′A′C=i∠B′A′C=i and ∠D C A′=r.∠D C A′=r.
⇒i=r.⇒i=r.
⇒⇒ The angle of incidence is equal to the angle of reflection.
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Related questions
More answers below
In which two cases taking place during refraction will the angle of incidence be equal to the angle of refraction?
If the angle of incidence is zero degrees, then what is the angle of refraction?
What is meant by that angle of incident, angle of refraction, and normal lie in the same plan?
If the angle between a plane mirror and an incident ray is 40 degrees, then what will be the angle of refraction?
What is the relation between an angle of incidence and an angle of refraction?
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·
1y
The angle of incidence and the angle of refraction are the same when light travels along the normal (perpendicular) to the boundary between two different media. In this situation, there is no bending of the light ray, and both angles are equal to 0 degrees.
In general, Snell's Law describes the relationship between the angles of incidence and refraction when light passes from one medium to another:
n 1 sin(θ 1)=n 2 sin(θ 2)n 1 sin(θ 1)=n 2 sin(θ 2)
Where:
n 1 n 1 and n 2 n 2 are the refractive indices of the two media,
θ 1 θ 1 is the angle of incidence,
θ 2 θ 2 is the angle of refraction.
When light strikes
Continue Reading
The angle of incidence and the angle of refraction are the same when light travels along the normal (perpendicular) to the boundary between two different media. In this situation, there is no bending of the light ray, and both angles are equal to 0 degrees.
In general, Snell's Law describes the relationship between the angles of incidence and refraction when light passes from one medium to another:
n 1 sin(θ 1)=n 2 sin(θ 2)n 1 sin(θ 1)=n 2 sin(θ 2)
Where:
n 1 n 1 and n 2 n 2 are the refractive indices of the two media,
θ 1 θ 1 is the angle of incidence,
θ 2 θ 2 is the angle of refraction.
When light strikes the boundary at 0 degrees (along the normal), both angles are 0, and thus they are equal.
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Richard Shagam
PhD in College of Optical Sciences, University of Arizona (Graduated 1980) · Author has 1.2K answers and 550.6K answer views
·2y
Originally Answered: When is the angle of an incident equal to the angle of a refraction? ·
Only if the refractive index is the same for both media, of if the angle of incidence is zero—-huh, you say? Look up Snell’s law and you should see why.
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Martin Bullen
BA in Mathematics, The Open University (Graduated 1987) · Author has 3.6K answers and 4.3M answer views
·7y
Originally Answered: Can the angle of incidence be the same as the angle of refraction? ·
The angle of incidence can be the same as the angle of refraction, but only so long as the medium has a refractive index of 1.0.
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Abhishek Singh
Chief Executive Officer (CEO) at Something Truth (2019–present)
·6y
According to snells law angle of incidence equal to angle of refraction only if relative reflective index is one. Means it is not possible for any two differebt medium, this kind of refraction occur every point in the path of light in a single medium.
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Wade Smeltzer
MBA in MBA in Accounting and Finance, Alumni of Colorado Technical University (Graduated 2015) · Author has 213 answers and 474.4K answer views
·6y
Related
How can we write the relationship between incidence angle and reflect angle?
Light is known to behave in a very predictable manner. If a ray of light could be observed approaching and reflecting off of a flat mirror, then the behavior of the light as it reflects would follow a predictable law known as the
law of reflection
. The diagram below illustrates the law of reflection.
In the diagram, the ray of light approaching the mirror is known as the
incident ray
(labeled I in the diagram).
The ray of light that leaves the mirror is known as the
reflected ray
(labeled R in the diagram).
At the point of incidence where the ray strikes the mirror, a line can be drawn perpendicular t
Continue Reading
Light is known to behave in a very predictable manner. If a ray of light could be observed approaching and reflecting off of a flat mirror, then the behavior of the light as it reflects would follow a predictable law known as the
law of reflection
. The diagram below illustrates the law of reflection.
In the diagram, the ray of light approaching the mirror is known as the
incident ray
(labeled I in the diagram).
The ray of light that leaves the mirror is known as the
reflected ray
(labeled R in the diagram).
At the point of incidence where the ray strikes the mirror, a line can be drawn perpendicular to the surface of the mirror. This line is known as a
normal line
(labeled N in the diagram)
. The normal line divides the angle between the incident ray and the reflected ray into two equal angles. The angle between the incident ray and the normal is known as the
angle of incidence
. The angle between the reflected ray and the normal is known as the
angle of reflection
. (These two angles are labeled with the Greek letter "theta" accompanied by a subscript; read as "theta-i" for angle of incidence and "theta-r" for angle of reflection.) The law of reflection states that when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection.
It is common to observe this law at work in a Physics lab such as the one described in the previous part of Lesson 1. To view an image of a pencil in a mirror, you must sight along a line at the image location. As you sight at the image, light travels to your eye along the path shown in the diagram below. The diagram shows that the light reflects off the mirror in such a manner that the angle of incidence is equal to the angle of reflection.
It just so happens that the light that travels along the line of sight to your eye follows the law of reflection. (The reason for this will be discussed later in Lesson 2). If you were to sight along a line at a different location than the image location, it would be impossible for a ray of light to come from the object, reflect off the mirror according to the law of reflection, and subsequently travel to your eye. Only when you sight at the image, does light from the object reflect off the mirror in accordance with the law of reflection and travel to your eye. This truth is depicted in the diagram below.
For example, in Diagram A above, the eye is sighting along a line at a position above the actual image location. For light from the object to reflect off the mirror and travel to the eye, the light would have to reflect in such a way that the angle of incidence is less than the angle of reflection. In Diagram B above, the eye is sighting along a line at a position below the actual image location. In this case, for light from the object to reflect off the mirror and travel to the eye, the light would have to reflect in such a way that the angle of incidence is more than the angle of reflection. Neither of these cases would follow the law of reflection. In fact, in each case, the image is not seen when sighting along the indicated line of sight. It is because of the law of reflection that an eye must sight at the image location in order to see the image of an object in a mirror.
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Nandan Umrikar
Studied Mechanical Engineering at Maharashtra Institute of Technology (MIT-Btech) ,Aurangabad (Graduated 2021) · Author has 171 answers and 678.6K answer views
·8y
Related
Why does refraction doesn't occur when incident angle is greater than critical angle?
Let us first discuss what is a critical angle. A critical angle is the angle beyond which the refracting surface acts as a reflecting surface. Why is it so?
It is because of the properties of the refraction. As per refraction theory, when a light ray incident on a transparent surface at some inclination, it gets refracted. That is it changes it's way as it enters into a different medium.
Now it can travel from denser medium to rarer medium or vice versa. Consider the case in which the light ray travels from denser medium to rarer medium. We know very well that as light ray enters into a rarer me
Continue Reading
Let us first discuss what is a critical angle. A critical angle is the angle beyond which the refracting surface acts as a reflecting surface. Why is it so?
It is because of the properties of the refraction. As per refraction theory, when a light ray incident on a transparent surface at some inclination, it gets refracted. That is it changes it's way as it enters into a different medium.
Now it can travel from denser medium to rarer medium or vice versa. Consider the case in which the light ray travels from denser medium to rarer medium. We know very well that as light ray enters into a rarer medium from a denser medium it moves away from the normal. As the incident angle increases, the refracted ray goes on moving away from the normal and at certain angle it becomes parallel to the refracting surface.
Now comes the interesting part. If we increase the incident angle a bit more than the one in the previous case what will happen to the refracted ray? Will it still be refracted to another medium? No. And this is where total internal reflection takes place.
So I hope your doubt is clear.
Cheers.
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Mayukh Chakrabarty
KVPY SA Fellow 2016 · Author has 126 answers and 261.3K answer views
·6y
When light is incident normally on the interface separating the two media
When the pair of media has equal refractive index
Upvote ·
Satya Parkash Sud
M.Sc. in Physics&Nuclear Physics, University of Delhi (Graduated 1962) · Author has 8.1K answers and 27.4M answer views
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When an incident angle is equal to a critical angle, then the angle of refraction will be what?
When light travels from a optically denser medium to an optically rarer medium, the refracted ray bends away from the normal. As we keep increasing the angle of incidence in the denser medium, the angle of refraction in the rarer medium keeps increasing. If we keep increasing the angle of incidence in the denser medium, a stage comes when the angle of refraction equals 90° ie the refracted ray mov
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When light travels from a optically denser medium to an optically rarer medium, the refracted ray bends away from the normal. As we keep increasing the angle of incidence in the denser medium, the angle of refraction in the rarer medium keeps increasing. If we keep increasing the angle of incidence in the denser medium, a stage comes when the angle of refraction equals 90° ie the refracted ray moves along the interface seperating the two media . The angle of incidence in the denser medium, for which the angle of refraction is 90°, is called critical angle.
When angle of incidence equals the critical angle, the angle of refraction is 90°.
If angle of incidence exceeds the critical angle, the incident ray is reflected back into the denser medium. This phenomenon is called Tota...
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Asima Chakraborty
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When the direction is same and there is no deviation then the angle of incidence is equal to angle of refraction
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Related questions
When is the angle of an incident equal to the angle of a refraction?
What is the relation between angle of incidence and angle of refraction, if light travels from rarer to denser medium?
Can the angle of incidence be the same as the angle of refraction?
What is an incident angle when the refraction angle is 90?
Why is the angle of the incidence greater than angle of refraction?
In which two cases taking place during refraction will the angle of incidence be equal to the angle of refraction?
If the angle of incidence is zero degrees, then what is the angle of refraction?
What is meant by that angle of incident, angle of refraction, and normal lie in the same plan?
If the angle between a plane mirror and an incident ray is 40 degrees, then what will be the angle of refraction?
What is the relation between an angle of incidence and an angle of refraction?
What will be the angle of refraction if angle of incidence is 90 degrees?
Does refraction take place more when the angle of refraction decreases or when the angle of refraction increases?
When is the angle of refraction greater than the angle of incidence?
When the angle of incidence is 0 degrees, what is the angle of refraction for the ray suffering refraction through a glass slab?
What is the angle of refraction if the angle of incidence is 60?
Related questions
When is the angle of an incident equal to the angle of a refraction?
What is the relation between angle of incidence and angle of refraction, if light travels from rarer to denser medium?
Can the angle of incidence be the same as the angle of refraction?
What is an incident angle when the refraction angle is 90?
Why is the angle of the incidence greater than angle of refraction?
In which two cases taking place during refraction will the angle of incidence be equal to the angle of refraction?
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8863 | https://theoraclesclassroom.com/wp-content/uploads/2019/11/1982-Scott-Fetzer-10K.pdf | SEC Fl.LE NO 1• 213 0.5 '"'03 scon & FETZER co NYS CA.Fi.:O 1 SJC.273 i:>t$Cl0$URE INC. WASl-llN.GtrON D. c. 20016 s 15 57 00 ,ooo iO·K 0 Cha1·tto Contents of SFC Filings ~J ""-0 .. E " 19-K . Registration .Statements c: .!? o3 .s a. c REPORT CONTENTS 10-K 20-F 10-Q 8-K 10-C 6-K ~ i< a..-1!>. r-·,...~A_ct--1 ~ e i _ f:10 8-A "S" a.. Ul 8-B Type AAS :!J: N-1R N-10 Auditor oName A A D Opinion 0 A A D Changes Compensation Plans D Equity DMonatary Company Information D Nature of Business A D History F D Organization and Change F Debt Structure A Depreciation & Other Schedules A Diiution Factors A Directors, Olllcars, Insiders D Identification D Background D Holdings D Compensation Earnings Par Shara Financial Information D Annual Audited D Interim Audited D Interim Unaudited Foreign Operations Labor Con1racts Legal Agreements Legal Counsel Loan Agreements Plants and Properties Portfolio Operation• F F A A A F A A.· F A A F ·F. A F F A ·A A A F. F A F F A A --· A A \ F A A A F ,.., F A F D Content (Listing of Securities) ,, ··•1-=,..,..,.--"'--~----'--'-+--+"--+--+--+--+--+--+--+--1-,,.4--4---l---1---1---1 D Management \:> A Product-Lina Breakout A A .A SeCurlties Structure A A A A A Subsidiaries A A A A A Underwriting A A A Unregistered Securities F F Block Movements A _ ,:.:i _ , _ , __ Legend A - a/ways included - Included - ii occurad or alflnllicanf F - frequently includ9d • spacial circumllancas only I TENDER OFFER/ ACQUISITION REPORTS ·/ 130 13 G 140'1 140-9 13E-3 13E·4 Name of Issuer (Subject Company) . A A A A A fl Fiiing Peraon (or Company) A A A A A A Amount of SharH Owned A A Percent of ClaM Oulltandlng . A A Financial Slltemenll of Bidders ., F F F PurpOM ol Tender Oller A A A A Source and Amount of Funda A A A Identity and Background Information A A A Par1on1 Retained Employed or to be Compan-.1 A A A A Exhlbill . F F F F F ©Copyright 1980 discLOS1JRE11tCORPORATEO/ . ' 5161 River Road o Washington, D.C. 20016 o 301/95"'1•1300 \ r; ·' 1) 0 /, • • a • • • • • ' \1 " 5 /:55700 ' SECURITIES AND EXCHANGE a:JMMISSIOO Washington, o. c. 20549 ,, FORM 10-K ANNUAL REPORI' PURSUANl' ro SECTION 13 or 15(d) 1• OF 'lllE SECURITIES EXCHANGE llCT OF 1934 For. the fiscal year ended Noverrber· 30, 1982 · Comnission File Nurrber 1-231 THE. SCOIT & FETZER COMPANY (Exact name of Registrant as spec1£1ea in its charter) : Eegistrant's Securities registered pursuant Title of each class comron shares Without Par Value ($1.25 Stated Value) ·~ 9-1/4% Notes Due 1985 of the Act: 0 Name of each exchange on which• 'registered New York Stock Exchange Midwest Stock Exchange Pacific Stock Exchange. l',jew York Stock Exchange Securities registerea pursuant ti:> Section 12(g) of the Act: None. Indicate bv check mark whether the registrant (1) has filed all reports required to be filea by _Section 13 or 15(d) of the Securities Exchange Act of 1934 during the J?receding 12 ironths (or for such shorter period that the registrant was , required to fife such reports), and (2) has been subject to such filing requirements for the past 90 days. Yes;.! No.. '!he aggregate market value of the voting stock held by .,non-affiliates of the . registrant as of January 31, 1983, excluding, for p.irposes of this a:imputation, only stock holdings of the registrant's Directors and Officers. $244,392,445 () ,, The number of Co!llron.Shares outstanding on.January 31, 1983, was 6,602,990. OOCUMENTS INCORPORJ\TID BY REFERENCE Portions of the annual shareholders' report for the year ended Noverrber 30, 1982, are incorporated by reference into Parts I and II. ' Portions of the anhual proxy statement for the year ended Novent>er 30, 1982, are incorporated by reference into Part III• ' 'file index to Exhibits is founa on page 32 of registrant's Form 10-K Annual Report as filed with the Securities anclEX'change Conrnission. \ ' () ""' \ • • PAR!' I I~ 1. Business 'rtle Scott & Fetzer company ("Scott Fetzer" or "company") is a diversified carpan;y which manufactures and sells products in the Cleaning Sy$tems & !k>usehold Products, E>:'lucation & Information Systems,oFluid Transn\ission, lJ Vehicular Products ;md Energy & Control segments. Tl«> of SCott Fetzer's principal product lines are vacuum cleaners and i;elated accessories primarily for home use sold under the Kir~ and other brand names, and encyclopedi<15 and related educational products s:> under the ~rld Book name. SCott Fetzer has 16 operating units located in 14 states, oost of Wh1dl wei:e independent ' 0 businesses acquired subseqlient to 1963. • ,, • • • • • • • " . Founded,Jn 1Jl4, ,Scott Fetzer was incorIX>rated under the laws of the State of Ohio on N:>ventier 30, l!U7. Business $egments 'rtle C11Dunt of net sales and other r:evenue, income before taxes and identifiable assets, for ,the past three years, attributable to each business segment is set forth in the tables on pages 11 and 12 of the 1982 Annual ReIX>rt to shareholders of Scott Fetzer. Additional information is noted in footnote 12 on page 17 of the 1982 Annual ReIX>rt. 'rtlese tables and footnote are inoorIX>rated herein by reference. segment data for 1980 have been rest\ited to reflect the. i:ealignmentoof qierating units initiated in 1981. SCott Fetzer's business segments ai:e based in part on the similarity of certain of its products and in part on the similarity of the markets in \tlich its products are sold, Such data reflect the allocation of certain expenses ·~and other arbitrary • • f) deternu.nat1ons. 0 ·:·! ,, " .. During the fiscal year ended Noventier 30, 1982, no single customer purchased products ftan the operating units of SCott Fetzer .. which in the aggregate accounted for nore than 10% of total sales ftan qierations fur such fiscal year. Scott Fetzer does not believe that the loss of arrt single custcrner would have a material a:nTerse effect on its total business. Cleaning Systemii & lk>usehold Products Scott Fetzer manufactures and distritllttes a wide ~ari~ty of vacuum cleaners and other floor maintenance equipnent and supplies for residential, 0 •• ;, industrial and institutional use. Scott Fetzer also manufactures and sells to other manufacturers certain CD111X>nent parts incorIX>rated in such equipnent • Floor maintenance equipnent for consumer use is sold primarily mder the Kirby name. In 1982, SCott Fetzer introduced oomnercial vacuurn cleaners Under ~Vac and Supervac names. Certain other floor maintenance equipirant is sold under both the private labels of CU!ll~rs and under certain conpany trade names. SCott Fetzel?', which entered the~household vacuurn cleaner field in 1919, manufactures and sells the Kirby uprigf;f vacuurn cleaner and·• i:elated floor care and other accessories, Kirby producti:i are sold by the direct sales methcla in the home through awroximately 8,00,!> independent dealers worldwide, bac.W;ed by some 942 factory and"area distributors. Ki~'s sales to distributors are substantially all for cash. Kirby has five maJor conpetitors in its conslll!ler 1 /) ii u • • • • • • • • • • • F/ \,. market and many m::>re in its comnercial market, although n:ine of the latter uses the direct sales method. The comnercial products are also sold by the direct sales method to end users and to janitorial supply rouses respectiyely. Kirby's sale prices to consumers are established individually by each aut:hOrized dealeri the ooriSumer product prices tend to be somewhat higher than tOOse of the oorrpetition, while the comnercial product prices generally a);Froximate those of the cxmpetition. Service for consumer products is provided by the iooependent dealers and the factory rebuild departmenti service for conmercial products is provided by the independent dealers and the growing network of authorized service centers. Kirby believes its service to be as good as, or better than, that of .the cx:q>etition. For both i;rQ(luct lines, Kirby's warranties generally afford m::>re C011Prehensive coverage for a lc.nger time than those of the oorrpetition. Both consurnrt- and a:.mnercial product performance appears equal to, or better than, that of tiie C011Petition. In fiscal 1982, m one distributor accounted for nore than 2% of Kirby sales. Danestic sales are fairly evenly ,, distributed throughout the country" based lipOn pop11latioo densities. , The sale of vacuum cleaners and related ~ssories pr:imarily for h:lme use under the Kirby and other names, as well as·private labels, accounted for approximately 13%, 14% and '15% of total revenues from continuing ~rations of Scott Fetzer for each of the fiscal years 1980 through 1982, respectively • As of Noveirber 30, 1982, united Retail Finance canpany, a \oilolly-owned subsidiary of tk>rld Book Finan~, Inc., which finances consumer installment accounts for Kirby distributord, had trade receivabl~s with a face cm:>unt totalling approximately $33 million. In 1982, the d:>q;>any established a new wholly-owned finance subsidiary, Scott Fetzer Financial Services Canpany (SFFS). SFFS has a wholly-owned consolidated subsidiary, United Acceptance Limited, (UAL) which pr:ovides funds to finance t:r:e installment receivables fran custaners of the Kirby Group distributors operating in the United Kingdan. As of November 30, 1982, UAL had trade receivables with a face cm:>unt totalling approximately $170,000. In addition to the Kirby products, Scott Fetzer manufactures and sells under the .American Lincoln name an~extensive line of power-driven industrial and institutional floor maintenance equipment and related supplies including industrial-type polishers, sweepers, wet floor scrubbers, tank-type vacuum cleaners and sweepers ranging in size from limited space vacuum cleaners up to, and including, tractor m::>unted scrubbers and sweepers~ The ~ also manufactures roller and flat brushes for vacuum cleaners, polishecs and other floor maintenance equipment, replacement vacuum cleaner parts, chemical cleaning products, and vacuum cleaner bags and replacement parts for h:lme vacuum cleaners. In addition to its floor care equipment, Scott Fetzer also manufactures and sells injection nolded plastic items, including containers, a household line of cutlery'-'and scissors and sterilized disposable medical supplies. In fiscal 1982, Scott Fetzer also began marketing, through direct sales by independent dealers, the HFAL'lll/mate, an innovative oral hygiene and skin care appliance. The Cleaning Systems & Household Products segment primarily sf1rves consumer, institutional, comnercial and industrial markets. rF 2 t ' I • • • • • • • F.ducation & Information Systems Scott Fetzer, through its i.bolly-owned subsidiary, w:>rld Book, Inc., ~blishes and sells The world Book Encyclopedia, other reference works and education and instruction material primarily l.l'lder the w:>rld Book and Childcraf t names. Domestic encyclopedia sales are principally l::iy the direct sales methOd in the txJme and to schools and libraries through approximately 8,850 full-and part-time independent. comnissioned sales representatives. world Book, through subsidiaries and branches having approximately 750 independent CXllllllissioned sales representatives, conducts. direq!:. selling operations in Canada, Australia, the British Isles and the caribbean. ·· 'n'lough w:>rld Book has six major oonpetitors, it remains the dominant firm in this market. Its prices are, generally, .lower than those of the oonpetition, yet w:>rld Book believes that its customer service program is nore extensive and its warranty CO\Terage nore conprehensive than thbse of the c:urpetition. w:>rld Book also markets, primarily l::iy direct mail, its annual encyclopedia sug>lements, other pUblications, insurance and other nerchandise th~h subsidiaries. r"I A large proportidi of encycloPeaia sales are made on a deferred consumer credit installment basis. '!tie domestic. and Caribbean installment acc6unts receivable are financed l::iy w:>rld Book through its wholly-oWned subsidiary, world Book Finance, Inc. At Novenber 30, 1982, the face aoourit of these receivables totaled ag>roximately $114 million. In 1982, oo ooe representative accounted for 11Dre than 3% of w:irld Book domestic sales. Domestic sales are fairly evenly distributed throughout the country based upon population densities. For fiscal years 1980, 1981, ancJCi982 sales of w:>rld Book products under the w:>rld Book and other names accounted for ag>roximately 42%, 40%, and 39%, respectively, of total revenues fran operations of Sci:>tt Fetzer. The F.ducation & Information Systems segmgnt primarily serves the consumer, educational and mass merchandiser markets • Fluid Transmission Scott Fetzer manufactures a variety of products involving the transmission of fluids, the major itefts of which are conplete, as well as conponent parts of, air carpressors, both reciprocating and single screw; spraying units, including small conpressors, for the spraying of paints and " other liquids; grey and ductile iron castings; air receivers; conpressed gas containers; and high pressure sprayers and washers. 'ltiese products are sold domestically and ab~ l::iy mass nerchandisers under private labels l::iy various ·retail stores under the Cc!r!fbell Hausfeld brand name and through independent distributors and sales representatives. Ccin;lbell Hausfeld has three main oonpetitors in the ooosumer market, and four in the oomnerci(I]. market. Its prices are fairly nents accounted for ag>roximately 12%, 13%, and 9% respectively, of total sales fran operations of Scott Fetzer. " In addition, Scott Fetzer assent>les and sells domestically under the wayne Home Equi~t name, p:Mer gas and oil burners and water circulating, sump and other Pll1'P3· ayr-e has twelve major oonpetitors in the punp market, and five in the burner market. Wayne's products are premi1.11t-priced1 its service, 3 • • • • • warranties, and product performance ai;:t>ear equal to, or better than, those of .··. the o::ilpetition. During fiscal 1982, Wayne Home Equipnent introduced a new · aui::lllersible utility punp called the Reliant One and developed a gasoline-powered mJlti-purpose centrifigal PJ111P called the Porta Purrp 11, to be marketed in early ~. . Also within the Fluid Transmission segment, Scott Fetzer manufactures connectors and fittings for oonpressed gas applications • The Fluid Transmission segment serves pt"imarily do-it-yourself hardware, home cent~r and mass merchandiser markets, but original equipnent and industrial markets represent growing areas of activity... 0 Vehicular Products Scott Fetzer manufactures and sells, pt"imarily through independent distributors, utility service truck bodies and related equipnent and suspension system OJnl?Ollel'lts for vehicles. In addition, Scott Fetzer manufactures and markets, pr1marily tt)rough independent distributors, electrical and mechanical winches for marine and qther applicationsi towing equipnent, including balls, ex>uplers and other related towin;i itemsi fan clutchesi oil coolersi antennas for the recreational vehicle marketi recreational vehicle awnin;JSi hydraulic cylindersi valvesi steering colunn oonponents for trucks and heavy equipnenti 1:rc and military tal'lk track links. There are a nunt>er of ng the leaders in the manufacture and sale of these productsi its pricing is ex>npetitive, .. and its !>ervice, warranties and product performance appear to be equal to, or better th~., those of the corrpetition. o The V~.hicular Products segment pi:imarily serves the recreational vehicle, heavy duty truck, construction, agricultural equipnent and military vehicular e equipnent markets. In 1982, Scott Fetzer established a new i.bolly-owned transportation subsidiary, SF'l Transportation, Inc., as a contract rotor carrier with pending application to become a licensed CXll'l'mJn carrier. e Energy & control scott Fetzer .is engaged in the manufacture and sale of explosion-proof fittings and junctiai boxes, notor efficiency control devices, instrument housin;is and control stations for electrical distribution systemsi a specialty line of armored cable connectors, and various other items used principally in e connection with high and low voltage electric cables, fittings, ex>uplers and liquid-tight conduit fittingsi various precision equipnent for use in processing ind_ustries for the measurement of pt"essure, vacuum and flow of liquidsi zinc and aluinim.111 die cast electrical fittin;JSi transformers and ballasts for indoor ~ outdoor electrical signs; ignition bystems for residential and industrial oil · furnaces, includin;i solid-state controli fractional horsepower rotors for ,, e electrical appliances and other productsi and timing devices for residential and OOlllllercial appliances. 'lbese products are }?rincipally sold by direct factory sales people·-~ independent manufacturer's representatives and distributed to and through original equipnent manufacturers and wholesale distributors. There • 4 • • • • • • • • arfi" a n.intJE!r of oorrpanies engaged in manufacturing each class of product within the energy and control business segement. Scott Fetzer believes it is ~ the leaders in the manufacture and sale of these products: its pricing is c:onpetitive, and its service, warranties and product performance a~ar to be equal to, or better than, those of the oonpetition. ·· Markets for 'the Energy & Control segment include construction, electrical sign, appliance and oil and gas industries • Acq\lisitions and Disp?Sitions Illlring the last quarter of 1982, Scott Fetzer sold its trailer hitch and manual winch business for cash • Backlog (; Scott Fetzer d::>es mt beli~e that the cbllar.aoount of backlog of orders and related information is material for an understanding of its business. A substantial portion of Scott Fetzer's cperations inVolve the recurring, mn-seasonal, production of standard items that are shi~ to the customer relatively shortly after an order is received by Scott Fetzer. 11 .• Raw Materials and Supplies Raw materials required for Scott Fetzer's various products are es mt depe~d on a single source of supply for its raw materials, oonp:>rient parts or suppli~s, and believes that its sources of supply are ~ate • Energy ,, , Scott Fetzer utilizes oil, gas and electricity as its principal energy sources. 'n'lere has been m material disruption of production at any of the Scott Fetzer plants because of energy shortages. However, there is m assurance that energy shortages in the future will not have an adverse effect on Sgott Fetzer either directly or indirectly by reason of their effect on OJstomers or suppliers. Environmental COntrols e Scott Fetzer believes its facilities are in stibstantial cmpliance with existing laws and regulations relating to control of air and water quality and waste disposal. Environmental cmpliance has mt had, and is mt expected to have, a material effect on the Coq>any's expenditures, earnings or c:onpetitive position. e Product Developnent, Patents and Trademarks • Scott Fetzer is oontinuously engaged'in the refinement and developnent of its various product lines, developnent of new applications for existing 5 = ,') • products, and the developnent of new products. SCOtt Fetzer's expenditures on o.:iiq;iany-sponsored research and developnent and on custaner-sponsored research t activities relating to the developnent of new products, services or techniques or the in1;>rovement of existing products, services or techniques during fiscal years 1982 and 1981 were rot material. ' Soott Fetzer a:intinues with research and developnent necessary to develop new products and applications for its various business segments through its · I Advanced Product Technology Center. • ' " • • • ,, • • • • Scott Fetzer uses in its rosiness various trademarks, trade names, patents, trade secrets and licenses. SCOtt .Fetzer dbes not consider that a material part of its rosiness is dependent on any one group of them, although the Kirby and world Book names are widely known and recognized. II ;: soott Fetzer's businesses are not generally affected by seasonal sales, fluctuations, alt;):!ough world Book's sales tend to increase during February and March because Of 'the sale Of anhual engyclopedia updates. ·1'.J E!!Eloyees ,, " . ..c .. {1· As of January 12, 1983, Scott. Fetzer f51i>loyed in. contim~ing qierations '' . Hi, 129 persons, of whan 12, 794 were salaried and 3,335 were hourly. A total of 2 1i,862 hourly erployees in 12 of SCOtt Fet~e,r's 48 facilities are represented by labor organizations. Soott Fetzer has enjoyed genercilly good relat;j,ons with its enployees. A total of 642 employees are C0\7ered by 3 labor oontr:acts ltlich are scheduled for renegotiatioo during fiscal 1983. ·· In addition. to erpl'?Yee'benefit programs M'lich include paid vacations, insurance, disability beneiiits, hospitalizatioo benefit$ and medical benefits, SCOtt Fetzer has in effect(~~ its divisions and subsidiaries various pension and retirement. plans for salaried and hourly personnel, including ron-contrirotory trusteed pension plans and profit-sharing retirement plans • See Note 9 of the Notes to Financial Statements oo page 16 of the 1982 Afuiual Report to shareholders of ~ canpany, M'lich rote is. incorporated herein by reference,. for information concerning contributions by SCOtt Fetzer under such plans and other data. I'l'91 2. Properties C' The following table sets forth the principal domestic plants and other materially inp>rtant domestic properties of the Conpany and its subsidiaries. The leased properties are irldicated with an asterisk. The nunt>er in parenthesis indicates nuntier of facility locations in each city • State Arkansas California Colorado llUTED Sl'ATES City Business Segment '':) Walnut Ridge (1) A I.odi (2) .. E Merced (1) E Broanfield (2) E 6 » ~ Plant Plant Plant Plant ~· Q ::: . (} \• :..;: " 'J • • • • • • • • • • ,, () Connecticut , Georgia Illioois Indiana Kentucky 0 Michigan New York Oklahana Te Mes see Texas Shelton (1) ""Valc??sta ( l) Chicago (2) Ft. wayne (2) KoltonD (1) Elkhart (1) Leitchfield (1) Shelbyville (1) Bronson (2) Watertown ( 1) Watertown (1) Akron (ll Avon Lake (1) Bedford Hts. (1) Bowling Green (1) Brookpark (1) cardingtoo (1) Chagrin Falls (ll ClevelarXI ( 5) Cleveland (2) Frem:>nt (1) Harrison (2) Harrisai (1) Lakewood (1) Twinsburg ( 1) westlake (1) WOoster (1) D.lrant (1) Fairview (2) Livingston (1) PortlarXI (1) Smithville (1) Andrews (1) A/E B D E E D E E c c A D c A A E A A/C 'i\ A D D F A A E E c c A c A Plant Plant Office/Warehouse Plant Plant Warehouse Plant Plant Plant Plant Plant Plant Plant Plant Plant Plant I/ Plant Plant Plant Plant Plant Plant Plant Off ice Plant Plant Plant Plant Plant Plant Plant Plant Plant (A) Cleaning Systems & Household' Products: (B) Education & Information Systems: (C) Energy & Control: (D) Fluid ':l'ransmission: (El Vehicular Products: (F) Corp:>rate Office • ·"" Scott Fetzer beli~ves that its pi:operties have been adequately maintained, are in good oonditioo generally, arXI are suitable arXI adequate for its business as p:esently oonducted. 'ttle extent of utilization of SCOtt Fetzer's properties varies am:xg its. plants ftaii time to time. 0 Scott Fetzer's various oontinuing operations are oonducted in 48 facilities in 33 locations in 14 states. Scott Fetzer maintains sales offices arXI warehouse facilities in various foreign countries and oonducts oeeitain manufacturing operatiohs in Whitby, Ontario. Many of Scott Fetzer's facilities are relatively new andul!Ddern, ~ile other facilities have been in operation for a s\lbstantial nunt>er of years. Management believes that the manufacturing capacity of S90tt i::et,zer's facilities is generally adequate at current levels of operation. V'at;ious of Scott Fetzer's facilities are leased, with options to purchase in sane cases. For additional information ooncerning the lease obligations of SCOtt Fetzer, see Note 7 of the Notes to Financial Statements bn ,., rr ,, • • • • • • ll • • • •" • " c ~ Page 15 of the 1982 Annual Report to shareholders of Soott Fetzer, ~ich oote is incorporated herein by reference. "" I'lDI 3. IA!gal Pt.oceedings In October, 1980, and January and August, 1981, respectively, the Attorneys General of Texas, New Mexioo and Iowa, oorrrnencaj C'ivil lawsuits in the Federal District Courts loeated in those states a;iainst the canpany, an independent distributor and certain independent divisional supeI'Visors, under state antitrust statutes and under federal antitrust statutes as parens patriae for citizens of their states. 'l'hese suits allege that the Conpany, through its Kirby Group operations, has in oonnection with the sale of Kirby vacuum cleaners and associated products, attempted to restrain trade, fix and maintain prices, boycott unauthorized purchasers, and restrict and allocate customers furl territories. Each state has requested.that the alleged violations be enjoined, \i,that civil penalties be i..np:ised and that nonetary damages, certain of ~ich '"''1ii0uld be trebled, be assessed. '!he Conpany believes that it has not ''7iolated the antitrust laws and has meritorious defenses ~ich it will vigorously assert. 'J.'le ultimate cost to the Cl:>npany of an unfavorable disposition of these cases is ?&>t rr::iw determinable. In the opinion of management, the canpany ~~- oot anticipate that the ultimate disposition of these lawsuits will have a material effect on the oonsolidated firianciai position and consolidated results of operations. I'lDI 4. Sulr.lission of Matters to a Vote of security lblders Proxies for Scott Fetzer's 1982 Annual Meeting were solicited under /' Regulation 14A of the securities and Exchan&e Comnission's proxy rules. 'l'her4 was no solicitation in ~sition to management's naninees listed in the proxy statement. All management 1YJ111inees listed in the proxy statement were elected • ,i::-.Cecutive Officers of SCOtt Fetzer a The follO\!ing is a schedule of names, a;ies and positions of officers as of February 1, 1983, of Scott Fetzer. There are no family relationships aroong these offi~~s. All executive offi,cers are elected annually by the Directors. ~ ~ Title John Bebbington John Bebbington has been ~loyed by SCOtt Fetzer for more than five years in such capacities as Group Vice President (7/19/71 to 9/1/78) and senior Vice President (9/1/78 to present) • J. F. Bradley ,jJ J, F. Bradley has been ~loyed by SCOtt Fetzer for'more than five years as Executive Vice President-J\dministration and Finance ( 3/1/77 to present) • 57 52 8 \1 senior Vice President Executive Vice President-.,, Administration & Finance 9i1 • • • • Gary A. Olildress Gary A. Childress has been 8rployed by Scott Fetzer as <' President and Chief Operating "' Officer ( 9/15/82 to present) • Fran 1977 to 1979 he was a Senior Vice President and Chief Operating Offieer of National can corporation, a manufa'bturer of metal, glass and plastic containers. Fran 1979 through 1981, he served as Chairman, President and Chief Executive Officer of ~ Warner ~y, a mineral and solid ~te management firm. During 1982, he served as Chief Executive Officer of the Qllf Resources & Chemical corporation, a mining and oil and gas exploratibert c. Weber has been employed by Soott Fetzer for 110re than five years in such capacities as Secretary1 General.Counsel (12/20/72 to 12/2/80) and vice President ~l.2/2/8.0 to present). C-' 58 46 52 Olairman and Chief Executive Officer Group vice President Vice President, General Counsel an9· Secretary ITEM 5. Market for the Registrant's Conm::>n Equity and Related Stockholder Matters t:xin1lcn Stock Market Price and Dividend. Information on page 1 of the Annual Repi:)rt to shareholders for the year ended November 30, 1982, is inoorporated herein by reference. 'll'le m.1mber of lx>lders of amron shares of Soott Fetzer as of January 31, 1983, ~as 6,239. , .. Im 6. Selected Financial Data · Selected . Financial Data oo ,{.page 1 of · the Annual Rep:>rt to shareholders for the year ended November 30, (f982, is inoorporated herein by reference • 10 ;: .. ,, • • • • ,, " • /, ,/ ·;.:, emeht 1s 0 Discussion and Anal sis of Financial condition Resu ts o rations Management's Discussion and Analysis of Financial COndition and Results of <:p!rations on pages 4 and 5 of the Annual Report to shareholders for the year ended Noveniler 30, 1982, is incorporated herein by reference. ITEM 8. Financial Statements and suwlementary Data '.l'he following CXl!lsolidated financial statements of the registrant and its subsidiaries, included on pages 7 through 18 in the Annual Report to shareholders for the fiscal year ended November 30, 1982, are incorporated herej.n by reference. " ITEM 9. consolidated. Statement of Income for the fiscal years ended November 30, 1982, 1981 and 1980. ,, :!? ::, ) , G Consolidated Balance Sheet as of Novent>er 30, 1982 and 1901. v ~ consolidated Statement of Changes in Financial Position for the fiscal years ended November 30, 1982, 1981 and.1980. COnsolidated St,atement of Shareholders' F.quity for the fiscal years ended Novent>er 30, 1982, 1981 and 1980. Notes to COnsolidated Financial St:atments • Disagreements bn Accounting and Financial Disclosure ·~ Not applicable. PAR!' III <> ITEM 10. Directors and Executive Officers of the Registrant Information as to the Direct:ori; of Scott Fetzer is CX>11tained on pages 2 ,;::::;:;;/ through 4 of Scott Fetzer'i;, .definitive Proxy Statement ("Proxy Statement") dated February 1B, 1983, which ii\:formation is incorporated herein by reference. ITEM 11. Management Renuneration and Transactions Information, under the captions "Rerruneration," "Pension Plans," and "Stock Options" on pages 6 through 10 of the Proxy Statement is incorporated herein by reference. ITEM 12., security Ownership of Certain Beneficial Owners and Management , '.l'he information CX>11cerning "Stock Ownership" on page 9 of Scott Fetzer's Pibxy Statement is incorporated herein by reference • () -t . 0 f PART r:v ;:·. ITEM 13. EAhibits'~ Financial Statement Schedules and Reports on Form 8-K (a) 1-2; 'lhe response to this portion of Item 13 is sul::mitted as a separate section of this report. . 3. Exhibits (3)(i) Articles of Incorporation (as Amended) previously sul::mitted as an Exhibit to Registrant's Form 10-K 1'.nnual Report for "' C:· 1981 and incorporated. herein by reference. I (ii) Regulations previously sul::mitted as an Exhibit to Registrant's Form 10-K Annual Report for 1981 and ,,) incorporated herein by reference. (4)(i) Indenture between The Scott & Fetzer Company and National City ·Bank dated May 15, 1975, for 30 million dollars of 9-1/4% Notes due May 15, 1985, i;reviously sul::mitted as ,.an Exhibit to Registrant's For#t 10"K Annual Report for 1981 arid incorporated herein by reference. ~ /,:;; J (ii) Note hjreement between The Scott & Fetzer c.onpany and '!he Prudential Insurance COnpany of America for 40 million I dollars of 9-1/2% Notes due in equal installments of 2.5 " million dollars 1983 through August, 1998, i;reviously /\'• ,, subnitted as an Exhibit to Registrant's Form. 10-K Annual · ,/;:,..""'-.. Report .for 1981 and incorporated herein by reference. ,, (iii) Note h;Jreement between world Book. Finance, Inc. , and The • Prudential Insurance COnpany of America for 50 million dollars of 10% Notes due in installments 1980 through August, 1998, i;reviously sul::mitted as an Exhibit to Registrant's Form lOK Annual Report for 1981 and incorporated herein by reference. "' • (iv) Note h;Jreement between world Book Finance, Inc., and The Prudential Insurance COnpany of America for a 25 million 0 dollar 14-3/4% Note due in equal installments June, 1985, \ through June, 1989, i;reviously sul::mitted as an Exhibit to Registrant's Form 10-K Annual Report for 1981 and incorporated herein by reference. ) I " (v) Term loans aggregating 25 million dollars between world Book Finance, Inc., and National City and Society National Banks of Cleveland at 15-1/4% due August 17, 1985~ thereafter_the (?-".:;:'; rate will be adjusted to equal the Banks' base rates pltis t~ percent, provided that oo time after maturity shall the • rate be less than 17-1/4% per annum. •C (10)(i) 'lhree-Party hjreement. dated June 15, 1981, involving the p.1rd1ase of land in Westlake, Ohio, for the building of the Co~rate off ices of '!he Scott & Fetzer ~ ~v ously sul::mitted as an Exhibit to • Registrant's Form ual Report for 1981 and incorporated hereiri by reference. " ~ '\'· 12" -~.!~ \ • {: ··:· ·"'.'.' "(ii) '!be Scott & Fetzer Conpany Stodt Option Plan adopted in 1973, as anended in 1977 and 1980, pt"eviously sul:rnitted as an Exhibit to Registrant's Form 10-K Annual Report for 1981 and incorporated herein by reference. ,, (iii) '!be Scott & Fetzer Conpany Stock Option Plan adopted in 1981. (iv) '!be Scott & Fetzer Company's Incentive COrrq;iensation Program previously sul:xnitted as an Exhibit to Registrant's Form 10-K Annual Report for 1981. ( v) Form of Deferl:'ea C.Orrpensation Agreement between '!be Scott & Fetzer Conpany and certain officers, directors and key enployees. (vi) Enployment Agreement Clated April. 7, 19821 between The Scott & Fetzer C.OJll>anY and. Mr. Rali;ti E; Schey. (vii) Enployment Agreement dated Septenber 15, 1982, between The Scott & ' reports on Form 8-K have been filed by the Company during the last quarter for the fiscal year e!Xled November 30, 1982. " 13 • • • • • • • • • • • \ SIGNATURES If (, 0 Pursuant to the requirements of Section 13 or lS(d) of the Securities Exchange Act of 1934, th~ Registrant has duly caused this report to be signed on its behalf by the undersigned, thereunto duly authorized • 0 • Bra ey ecutive Vice President-inance and Administration~ (Principal Financial Off,icer) Director FEB 25 1983 Date--------------Pursuant to the~:requirements of the",,.......,"'-report has been signed below by. the fo Registrant and in the capacities and ' .. 1 fl..,,( /= / 1:.. .. V~~<.#' R. E. Schey. Chairman and Chief Executive ·Officer (Principal Di recto •. G. President Officer Director/ /' W. A. Rajk:i Sen.~~r Vice Pr. es.ident DJJc7f :f~ J~~u.g eS\ D1 or ii J. E. Bradley utive Vice President-F ance and Administration (Principal Financial Officer) Director this /4 I • I • • • • • • • • 'nlE SCl)1'l' & mrzm CCMPANY AND SUBSIDIARY CCMPANIES INIElC '10 FIN!\N'.:IAL srATEMml'S AND SCHEOOI8i l& '!he consolidated financial statements and the report thereon of IndepE!Ddent Certified PUblic kcountants appear on pages 6 thro1J3h 18 of the attached 1982 Annual Report to shareholders, \hl.ch pages are incorporated by reference in this Fbrm 10-K Annual Report. With the e.'rrc:Mi.ngs X - 9.Jpplementary Incane statement Infonnation World B:>ok Finance, Inc. - Consolidated Financial statements Report of IOOependent Certified Public .!\coountants Balance Sheet, N::lvenber 30, 1982 and 1981 Statements of Incane ~,!Xi Retained Earnin:.:is, years eooed November 30, 1982, 1981 and 1980 Statement of Oian:.:ies in Financial Position, Nov6nber 30, 1982, 1981 and 1980 N:ltes to Financial statements Page NunU:lers Annual Report Form 'lb 10-K ,, Shareholders F-2 F-3 F-4 F-5 F-6 F-7 F-8 F-9 F-10 F-11 6 7 8 9 10 ll-18 F-12 - F-16 '!he individllal financial statemer,its of the registrant are anitted because the registrant is prirrarily an operatin;i cmpany. All subsidiaries inclooed in. the consolidated financial statements filed are Wholly-owned subsidiaries. Schedules other than th:>se listed abolle are anitted as the infonnation is not required, is either not pertinent or rot significant, or because the data is given in the consolidated financial statements or the notes thereto • F-1 . ,, ,, • • • • J '-""-Coopers &Lybrand certified public accounlanls REPORT ANO CONSENT OF INDEPENDENT CERTIFIED PUBLIC ACCOUNTANTS ·~..,,,, To The Shareholders and Board of Directors The Scott & Fet"zer Company Our report on the consolidated financial statements of The Scott & Fetzer Company has been incorporated by reference in this Form 10-K from the 1982 annual report to shareholders of The Scott 1f Fetzer Company and appears on page 6 therein. In connection with our examinations of such financial statements, we have also examined the related financial statement schedules listed in the index o.n Page F-1 of this Form 10-K. In our opinion, the financial statement schedules referred to above, when considered in relation to the basic financial statements., taken as a whole, present fairly the information required to be included therein in conformity with generally accepted accounting principles applied on a consistent basis. · We also consent to the incorporation by .. reference in the registration statement of The Scott & Fetzer Company on Form ,~-8 (File No. 2-58431) of our report dated January 25, 1983 on exam-inations of the consolidated financial sratements and financial statement schedules of The Scott & Fetze~ Company as of November 30, 1982 and 1981, and for the years ended November 30, 1982, 1981 and 1980, which report is included in this Annual Report on Form 10-K. \ . Cl~eland, Ohio (j ~?U.t" ....,. ~- u ..{ COOPERS & LYB~; 'I t January 25, 1983 f ·2. • 0 () O:>llltN A Year J;D:lel Novenber 30, Name of Debtor 1982 Robert H. King (r 1981 lbbert H. Kirg 1980 lbbert H. King • • 'lHE ocarr & FEI"LER CCMPl\NY /\ND SUBSIDIARY a:MPANIEH SCHEOOIE II - l>MXJNTS REX:E!VABIB FRCM RE£ATED PARl'IES /\ND UNDERWRITERS, PIO-IOl'ERS, AN:> n EMPIDYEES Ol'HER THAN REIATED PARl'IES o for the years ended Noveuber 30, 1982, 1981, and 1900 cblllt-U B rowmc Balance at Beginning of Period Additions $137,500 $187,500 $237,500 rowm o Deductions 1\mounts Collected $50,000 $50,000 $50,000f # .!\ncxmts written Off 0 • • Balance at En:i of Period Current Not Current $50,000 $' 37, 500 $50,000 $ 87,500 $50,000 $137,500 In Septeri:ler, 1979, the .CaJtlanY loaned $250,000 to lbbert H. King - Chainnan arrl 01.ief Executive Officer of 'W::>rld Ilcok, Inc. \1 " 'lbe :p:iey is to be repaid in c:ne of three ways: (a) in 60 installments of $4,166.66 per rrcnth, (b) in full 90 days frcrn date of tennlnation, either voluntary or for cause, q.r (cl 180 days fran. date of any other tenninatic:n of enploirment. 'lbe loan is noninterest bearing. J F",-3 ,, -0 -l • • • • • • • • ,, ·-· 'lliE OCulT'' & FEI'ZER Cvember 30, 1982 W::>rld Boak Finance, Inc. Short-term notes ColtlllTI B Title of Issue of Eadt Class of Securities Guaranteed 7.875% notes payable Field Enterprises Internaticnal, Inc. ( tmconsolidated subsidiary of W::>rld lb:>k, Inc.) il -• • • • Colunns D, E, am G are not presented as the infonnation l.S mt applicable. · Colum C 'lbtal J\rrotmt Guaranteed and Outstanding §.,:: 690 ,685 • .... Colum F Nature of Guarantee Principal ').' -'~-- ',~:t At Nov~r 30, 1?82, 'Ille. ~t & Fetzer ~ an:l it;; .. c<.msolidatoo subsidiaries maintainoo guarantees of ~;4. 306, 702 for a consolidated foreign subsidiary. ' · · F-4 • ..,., • • • • • • • .-.-, ' -\V 'llfE scarr & FE'l'ZER CcMPANY AND SUBSIDIARY CCMPANIFB SCHEDUIE VII£ -VALUATICN AND QUALIFYIN3 ACxxxlNTs ~or the years ended Novenber 30, 1982, 1981, ~~ ~~ ""' "'.'.,::::, ColUllll A Descflption Year Eh:led Noysnber 30, . 1982 AllCMance for D:>ubtful Acrounts Year Eh:led lbrsnber 30, 1981 AllCMance for D:>llhtful Acoounts ,, Year Eh:led t-lollember 30, 1980 · AllCMance for D:>ubtful Accbl.mts Col\llll1 B ,; 0 Balance at !3e$Jinning of Period $12,022,051 $12,124,043 $10, 724, tX,7 (A) Write-Oft oZ~rollectible acoounts less recCJ11eries. . ,, F-5 Col\llll1 C Additions ~a-ia~rg=ed'"''to rosts ~:am expenses ] $ 7,788,122 $ 7,840,030 $ 8, 586, 540 Colurm D " Deductions $ 7,265,482 (A) $ 7,942,022 (A) $ 7~ 187, 164· (A) 0 • Colurm E Balanee Eni, of"J?'briod $12,544,891 $12,022,051 di $12,124,043 -• • • • At Novanber 301 1982, 1981 and 1980, the Cl:tlpany naintaine:'I short-tenn debt balances doo only to foreign banks. 0 (A) 'nle average amouni() outstanding during the period was corrputed by dividing the total of rronth-end out.Sta~j.!1')g principal balances by 12 • F-6 • • • • • • • • • • (.Z . Ti'IB scorr Ii FETZER CXMPANY AND SUB.SIDIARY CG!PANIFS SCHE001E X - SUPPUMENI'ARY IN'.XME STATEMENI' INFOR>IATIOO for the years' erded No\Tenber 30, 1982, l 98f:l and l 98Q Colurm A Item ;year Fndoo No.Yenber 3o, 1982 Maintenance and Repairs ·o J\dvertising Year Fndoo No\Tenber 30, 1981 Maintenance and Repairs J\dvertising Year Fndoo November 30, 1980 Maintenance and Repairs J\dvertising Colurm B <::::~....,.,__-....-,----Cllargoo to Costs and Expenses $ 7,656,002 $17,686,891 $ 8,104,136 $12,554,566 $ 6,214,507 $10,660,431 (\ Amounts. for itens other than those reported have been excltrled because they 111Dunt to less than 1% of net sales. 0 , ,, " ' \ F-7 •• • • • • • • 0 • ,; . "' 0 ,'·' ,, . WORLD BOOK FINANCE, INC. AND ITS SUBSIDIARY 0 REPORT.ON EXAMINATION OF CONSOLIDATED FINANCIAL STATEMENTS " \'t ' for the years ended November 30, 19 182 and 1981 ,, . ~ ' c ·~, ' ' . , o" 0 0 b fl,, . Cooper.;; &Lyorand. C8rtlfled· futiilc-1\ccOuntant_s 22. 0 • a 'Coopers !&Lybrand certified public accountants 222 South Riverulde Plaza . Chicago, Illinois 60606 telephone (312) 559-5500 twx (910) 221-5211 cables Colybrand To the Board of Directors and Shareholder World Book Finance, Inc. In principal areas ol tho world We have ~mined the consolidated balance sheet of World Book Finance, Inc. and its subsidiary at Nove_mber 30, 1982 and 1981, and the related consolidated statements of income and retained earn-ings, and changes in financial pos'i ti on for the years ended November 30, 1982, 1981 and 1980. Our examinations were made in ac-cordance with generally accepted auditinwfstandards and, accordingly, included such. tests of the accounting records and such other auditing procedures as we considered necessary in the circumstances. In our opinion, the financial statements referred to above present fairly the consolidated financial position of World Book Finance, Inc. and its subsidiary at November 30, 1982 and 1981, and the consolidated results of their operations and changes in tbeir financial position for the years ended November 30, 1982, 1981 and 1980, in conformity with generally accepted accounting principles applied on a consistent basis. Chicago, Illinois January 25, 1983 F-8 '23 \l " • • • • • • • • WORLD BOOK FINANCE, INC. AND ITS SUBSIDIARY CONSOLIDATED BALANCE SHEET November 30, 1982 and 1981 (In Thousands) ASSETS Cash Commercial paper and certificates of 0 deposit Finance receivables, net Due from World Book, Inc. Accrued interest receivable Prepaid expenses Furniture and fixtures, net of accumulated depreciation of $12 in 1982 and $6 in 1981 Total assets LIABILITIES Current portion of long-term debt Accrued interest and other expenses Distributors' reserves Unearned acquisition fees Due to World Book, Inc. Long-term debt Total liabilities SHAREHOLDER'S EQUITY Common stock - authorized 1,000 shares, without par value, 261 shares issued and outstanding Additional capital Retained earnings Total shareholder's equity Total liabilities and shareholder's equity The accompanying notes are an integral part of the financial statements. 2 . 1982 $ 5 29,850 101,078 2,357 419 26 155 $133,890 $ 2,625 3,933 2,504 524 89,500 99,086 1 26,000 8 1803 34,804 $133,890 (-' 1981 ~' $ 1,385 40,955 89,518 1,575 r.> 64 $133,497 $ 2,625 3,189 1,446 282 1,697 92,125 101, 364 1 26,000 6,132 32,133 $133,497 • • I I • • • • • • WORLD BOOK FINANCE, INC. AND ITS SUBSIDIARY CONSOLIDATED STATE~ENT OF INCOME AND RETAINED EARNINGS t ' for the years er:!f.ed .November 30, 1982, 1981 and 1980 0 "(In Thousands) 1982 Revenue Facilitating fee --;:, $12,102 Interest income and acquisition fees 5,337 Investment income 4,989 22,428 Expenses Provision for credit losses 2,067 Operating expenses 3,828 Interest expense 11,022 16,917 Income bei'ore income taxes 5,511 Provision for income taxes Federal 2,275 State and local 565 2,840 Net income 2,671 (• Beginning retained earnings 6,132 Ending retained earnings $ 8,803 \ The accompanying notes are an integral part of uhe financial statements. 3 f• /O 1981 $ 7,650 1,156 6,333 15,139 322 2,178 8,426 10,926 4,213 1,759 390 2,149 2,064 4,068 $ 6,132 $ $ 25 1980 8,275 44 3,231 11,550 813 7,15!! 7,971 3,579 1,528 258 1,786 1,793 2,275 4,068 ' ,I: () {) c WORLD BOOK FINANCE, IN9. AND ITS SUBSIDIARY CONSOLIDATED STA1f EMENT OF CHANGES IN FINANCIAL POSITION for the years ended November 30, 1982, 1981 and 1980 ' (In Thousands) Sources of funds From operations Net income Add (deduct) items not requiring funds: Increase (decrease) in operations portion of int~rcompDnY receivable Decrease (increase) in accrued interest receivable Depreciation Increase (decrease) in accrued interest and other expenses Increase in unearned acquisition fees (Increase) decrease in prepaid expenses Total from operations Decrease (increase) in cash and \ short-term investments Increase in distribute~~· reserves \ Application of funds Increase (decrease) in finance receivables, net Increase (decrease) in non-operation~ portion of intercompany receivables Decrease (increase} in long-term debts Purchase of furITT.ture and fixtures $ 2,671 $ 2,064 $ 1,793 (2,142) 1,302 4,067 1,156 (1,298) (278) 6 6 744 1,911 (175) 242 193 89 (26) 45 (158) --~-2,651 1,058 4,223 (14,292) 1,242 5,338 (1,590) 204 $16,194 $ (8,827) $ 3,952 $11,560 1,912 2,625 97 $ 13, 805 \ (327) Q,22,375) 70 $ (822) 2,149 2, 625 $16,194 $ (8,827) $ 3,952 The accompanying notes are an integral part of the financial statements. D 4 f· II 0 \ • I 1. :~· I WORLD BOOK FINANCE, INC. AND ITS SUBSIDIARY NOTES TO CONSOLIDATED FINANCIAL STATEMENTS Accounting Policies Principles of Consolidation - The consolidated financial statements include the accounts of World Book Finance, Inc. and its wholly-owned subsidiary, United Retail Finance Company which was organized April 11, 1980. Intercomp~ny balances and transactions have been elimina\ed. · · Relationshi with Parent Com an - Or anizi~on - World Book Finance, Inc. the Company is a wholly..;owned sub-sidiary of World Book, Inc. (World Book), which is a wholly-owned subsidiary of The Scott & Fetzer Company (Scott Fetzer). The Company purchases receivables from World Book on a nonrecourse basis under the provisions of an operating agreement, as amend .. d, dated August 31, 1978. The agreement calls for World Book to make facilitating fee payments to the Company as required to cause net earnings available for fixed charges as defined for each accounting period of three months or for each fiscal year to be 1-1/2 times the "fixed charges" for such period. 27 United Ret~.il Finance Company (United Retail) purchases receivables with recourse, under the provisions of an operating agreement dated June 30, 1981, from the independent-contractor-distributors of the Kirby Group of Scott Fetzer.. United Retail is required by this operating agreement to ma~ntain a provision for credit losses of not less than 21 of its finance receivables. These receivables have unearned finance charges and acquisition fees, which are recognized over the term of the receivable on an effective yield basis. The recourse receivabi!es in the financial statements are stated net of the unearned fin~nce charges and provision for credit losses. " 1 Short-Term Investments - Commercial paper and certificates of deposit are carried at cost, which approximates market value. Interest income is recognized as .earned. Contract R~serve - Upon the purchase of nonrecourse receivables, an amount equal to not less than 111 of the receivable balance 0,is withheld, which is credited to a contract reserve. Accounts that are determined to be uncollectible are charged to the reserve. Any amount in the contract reserve at the end of the fiscal year in excess of the sum of 51 of the aggregate amount of all receivables plus certain defined past-due accounts can be remitted to World Book. The non-recourse receivables in the financial statements are stated net of the contract reserve. 5 f· l'l I I • • • • • • • • • WORLD BOOK FINANCE, INC. AND ITS SUBSIDIARY NOTES TO CONSOLIDATED FINANCIAL STATEMENTS, Continued 0 ~. Accounting Policies, continued Distributors' Reserves - Upon the purchase of recourse receivables, an amount equal to 10% of the gross receivable balance is withheld, which is credited to a distributors' reserve account. Defaulted accounts are charged against this reserve if they are not repurchased by the distributors. Amounts in the Clistributors' reser-vell,<;at the end of each quarter in excess of 15% of a distributor's total receivables, but nnt less than $5,000, are remitted to that distributor. Under certain circumstances, the entire recourse receivable may not be purchased. Any such amount not purchased is established as an additional deferred payable and is included as a part of distributors' reserves. Such deferred payables are eligible for release to the distributor if the receivable i~ current at the midpoint in the financing term. Income Taxes - The Company's taxable income is included in the consolidated tax return o:f Scott Fetzer. The Federal tax provision is calculated on a separate return basis, without any benefit for the surtax exemption • Furniture and Fixtures -Furn~ture and fixtures are stated at cost. Depreciation is charged to operations based on the straight-line method over the useful lives of the assets . () [l 28" C> • • ,, WORLD BOOK FINANCE, INC. AND ITS SUBSIDIARY NOTES TO CONSOLIDATED FINANCIAL STATEMENTS, Continued "' 2. ~inance Receivables The maturities and an analysis of the amounts deducted from finance receivables at November 30, 1982 and 1981 are as follows (in thousands) : Due in one year or less Due after one but not more than two years Due after two years Finance receivables Contract reserves Unearned finance charges Allowance for credit losses Finance receivables, net Total nonrecourse and recourse finance receivables, net 1982 1981 Nonrecourse Recourse Nonrecourse Recourse $ 42,128 45,866 25,899 113,893 (39,850) ----74,043 $ 6,466 26,887 33,353 £,4' 773) lb545) 27,035 $101,078 $ 38,555 48,950 30,073 117,578 (38,333) l ·79,245 $ 2,329 9,930 12,259 (1,6,~4) '>" (32;} 10,273 $89,518 The change in the con~ract reserve from December 1, 1980 through November 30, 1982 is is follows (in thousands): (l Balance, beginning of year $ 38 ,333 1981 ~.-$ 36,378 ·-~~ (~ :::"-~ -: . __ --c;///' ', Amount withheld on purchase of obligations /I I I/ Defaulted obligations Reserves returned per operating agreement Ba~ance, end of year 15,903 (13, 787) (599) 1,517 $ 39,850 7 17,223 (14,618) ( 650) 1,955 $ 38,333 t • • • • . " • WORLD BOOK FINANCE, INC. AND ITS SUB~IDIARY )""'·, •'/: ~v, NOTES TO CONSOLIDATED FINANCIAL STATEM~TS, tontinued 3"/J 3. ~istributor Reserves The following reflects the changes in the distributor reserves ·~· from December 1, 1980 through November 30, 1982 (in thousands): 4. Balance, beginning of.". year Amounts withheld on purchases Repurchases () Refunds to distributors Balance, end of year ,, Long-'Il!3rm Debt Long-term debt at November 30, 1§82 the following (in thousands0: ~romissory notes (10%) Promissory note (14.751) Term loan (8.751) Term loans (15.25%) $ I: $ " 1982 1981 1,446 $ 188 3,873 1,276 (2,752) ( 12) (63) ( 6) 2,504 ·0 $1,446 ~~ consisted {f 19~2 $39,500 251,0.00~· 25,000 1981 $42,125 25,000 25,000 Total long-term debt J$89, 50°-$92, 125 c=~' Under the provisions of the debt agreements, the Company is required to c'bmply with its operating agreements with World Book and the Kirby Group and to meet ·certain covenants, the major restrictions;of which are: 1) net eligible assets as defined will riot be less than 1101 of total indebtedness (less the aggregate of senior and junior subordinated indebtedness held by World Book or Scott Fetzer); 2) net earnings available for fixed charges will not be les~s than 150% of fixed charges for any quarter; 3) the senior indebtedness of the Company shall not exceed 4001 of shareholders' equity; 4) shareholders' equity will no~./1"!lVhan $25, 000, 000; 5) the aggregate indebtedness of the Company plus deferred tax obligations srrall not exceed 6001 0 of shareholders' equity and 6) the Company will not pay or declare any dividend, redeem, purchase or otherwise acquire any shares of its stock unless the cumulative amount of' all such payments does not exceed net earningG subsequent to November 30, 1979. At November 30, 1982 retained earnings in the amount of $6,528,000 were available for the payment of cl,ividends. 0 ~ 8 f-IS I • • 4. WORLD BOOK FINANCE, INC. AND ITS SUBSIDIARY NOTES TO CONSOLIDATED FINANCIAL STATEMENTS, Continued Long-Term Debt, , ,, ="""\ cdtitinued \1 The 101 promissory notes $2,625,000 to 1997, with 1998. :r--~ -. are payable in annual instal~ments of a final payment of $2,750,000~on August "·-3/ 31, The 14.751 promissory note represents financing to be used for the purchase of receivables from qualifying \Jistributon,s of the Kirby Group of Scott E.etzer. The note is payable in five annual installments of $5,000,000, commencing June 30, 1985 . The 15.251 term loans maturepn August 17, 1985. Upon maturity, the in~~rest rate is subject to adjustment on a defined basis, the amount ol which is not to be less than 17.251. Aggregate maturities of long-term debt during the five-year period November 30, 1983 through November 30, 1987 are $2,625,000, $2,625,000, $32,625,000, $7,625,000 and $7,625,000, respectively. 5. Contingent Liabilities t As of No'vember 30, 1982, the. Company has guaranteed $690,685 of debt obligations of Field Enterprises International, Inc., an unconsolidated foreign finance subsidiary of World Book. It ' ! • • • 6. Reclassifications Certain amounts reflected in the prior year financial statements have been reclassified to conform to the 1981 presentation. 9 ,;:\ EXHIBI'r INDEX () 'l'O FOR-! 10-K THE SO'.Jl'l' & FETZER cn-!PANY "' For The Fiscal Year Ended N:>veni:ler 30, 1982 Exhibit P!!2e Number " ITEM 13: (4)(v) Term loans cggregating 25 million 33 - 48 dollars between ~rld Book Finance, Inc. I ard National City ard society • National Banks of Clevelard ~-(lO)(iii) The Scott & Fetzer Company Stock 49 - 50 Option Plan adopted in 1981 ~ (v) Form of Deferred compensation 51 - 63 • Agreement (vi) Mr. Ralph E. Schey enployment 64 - 69 agr~n~ ' (vii) Mr. Gary A. Childress employment 70 - 73 • agreement (ix) IDng-term disability ard death 74 - 99 benefit agreements ( 13) 1982 Annual Report to shareholders 100 - 123 (22) Subsidiaries of The Scott & Fetzer 124 Company ,- '-=-'"/ :::, I I • • \ • ,, • • • • EJtHIBiT (If) (v) AGREEMENT WORLD BOOR FINANCE, INC. NATIONAL CITY BANR and / SOCIETY NATIONAL BANK Dated: August 17, 1982 Term loans aggregating $25,000,00Q payable on August ...!.!.• 19,85 ' \ 33 • • l. 2. • 3. 4. 5. 6. ,, 7. TABLE OF CONTENTS Cross-Reference • • • • • • • • • • • • • • • • • • Term Loans • 2.1 Amounts • • • • • • • • • • • • • • • • • • • • • • • • • • 2.2 Maturity • • • • • • • • • • • • • 2.3 Interest 2.4 Prepayment 2.5 Term Notes • • • • • • •• • • • • • . . . . . . . . Prohibition • • • • • • • • • • • • 0 Prudential Covenants • • • • • • • • • • Covenants • • • • • • • • Opening 4.1 4.2 4.3 4.4 4.5 • • • • • • Resolutions . . •: . . . . '"~ . . . . Legal Opinion ..... -,·~-----7 ~~~,. Prudential Agreements • • ~c • • • • Operating Agreements • Lett.er Agreement • • • • • • • • • • • • • • Warranties • • • • • • • • • • • • • • 5.1 5". 2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 Events 6.1 6.2 6.3 6.4 6.5 6.6 Banks' 7.1 7.2 of Existence • • • • • • Right to Act • • • • • Litigation • • • • • Taxes • • • • • • • • • Title • • • • • ••• Lawful Operations • • ERISA Compliance • • "Non-Purpose" Borrowing Financial Statements Defaults ••••••• ·-, 0 Default • • • • • • • Payments • • • • • Warranties • • • • • • Covenants • • • • Cross-Default • • • Subsidiary's Solvency • Borrower's Solvency •• • Rights Upon, Default Optional Defaults • Automatic Defaults • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • () Page 1 1 1 1 l 2 2 2 3 3 3 3 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 3.Lf? : TABLE OF CONTENTS - Continued t Page 8. Interpretation • • • • • • . • . • • • • • • • • • 7 9. Notice • • • • • • • • • • • • • • • • • • • • B 10. Definitions I) B • • • • • • • • • • • • • • • • • • • . accumulated funding deficiency • • • • • 8 Bank: Banks • • • • • • • • • • • • • • • B Borrower • • • • • • • • • • • • • • • B company • • • • • • • • • • • • • • • • B default under this Agreement • • • . • B I employee benefit plan . • • . • • • • • • 8 ER ISA • • • • • • • • • • • • • • • • • 8 event of default • • • • • • • • • • • • 9 First Operating Agreement • • • • • • • • q funded indebtedness • • • • • • • • • • • 9 insolvency action • • • • • • • • • • • 9 I Kirby Operating Agreement • • • . • • • • 9 lien • • • • • • • • • • • • • • . • • • 9 NCB . • • • • • • • • • • • • • • • • . • 9 .NCB's base rate • • . • • • • • • • . 10 overdue interest • • • • • • • • • • • • 10 1981 Prudential Agreement • • • • • • 10 • 1978 Prudential Agreeme.nt • • • • • • • • 10 person . • • • • • • • • • • • • • • • • 10 Prudential • • • . • • • • • . 10 Prudl')ntial Agreements • . • • • • • • • • 10 Prudential note n 10 • • • • • • • • • .. • • • related writing • • • . • . • • • • 10 • reportable event • • • • • • • 10 SNB . 10 • • • • • • • • • • • • . • • SNB's prime rate • • • . • • • • • • 10 Subsidiary • • • • . • • • • • . • • . • 11 ''>--,,;j term loan: term loan'3 • • • • • • • • • 11 ~~i term notei term notes 11 ' .-~.-• • • • '~I definitions • • • • • • • • . • • . • 11 11. Investment Purpose • • • • • • • • • • • • • • 11 (Signatures) • Exhibit A: Form of Ter.m Note (s. 2.5) Exhibit B't Form of Te\:ln Note (s. 2. 5) • •• • • t • • • •• AGREEMENT Agreement made as of August 17, 1982 by and among WORLD BOOK FINANCE, INC. ("Borrower"), NATIONAL CITY BANK ("NCB•), and SOCIETY NATIONAL BANK ("SNB") (NCB and SNB hereinafter some-times referred to individually as "Bank" and collectively as the •Banks•): 1. (CROSS-REFERENCE) Certain terms used in this Agreement are defined in section 10. 2. (TERM LOANS) This Agreement contemplates that each Bank will, upon the terms and subject to the conditions of this Agreement, grant Borrower a term loan • I~ 2.1 (AMOUNTS) The amount of the term loan granted by NCB to Borrower shall be fifteen million dollars ($15 ,000 ,000). (:) The amount of the term loan granted by SNB to Borrower shall be ten million dollars ($10,000,000), 2. 2 (MATURITY) . The term loans shall mature on August 1; , 1985, unless either or both of the Banks accelerate the~ time at which their term loans shall be due and payable in full pursuant to this Agreement. 2.3 (INTEREST) Interest shall accrue and be payable on" both of the term loans as follows: (a) Prior to maturity, interest shall accrue on the principal amount of and any overdue interest on each term loan at the fixed rate of fifteen and one-quarter percent (15-1/4%) per annum. (b) After maturity, interest shall accrue on the unpaid principal of and any overdue interest (1) on NCB's term loan at a rate equal to two percent (2%) per annum above NCB's base rate from time to time in effect and (2) on SNB's term loan at a rate eaual to two percent (2%) p~r annum above SNB's prime rate from time to time in effect, with each change in NCB's base rate or·SNB's prime rate respectively, automatically and immediately · changing the rate thereafter applicable to that Bank's term loan, provided that at no time after maturfty shall the rate applicable to either term loan be. less than seventeen and one-quarter percent (17-1/4%) per annum • /) • • • • • ,, (c) Interest shall be computed, on the basis of a 36S-day year for the actual number of days elapsed • (d) Interest shall be paid in arrears by Bor-rower directly to each Bank begi,nning on December 1, 1982 and thereafter on the first day of each March, June, September, and December, and at maturity • 2.4 (PREPAYMENT PROHIBITION) No prepayment Of the prin-cipal of any term loan shall be permitted. 2.S (TERM NOTES) To eviaence the term loans, Borrower shall execute and deliver to NCB a term no'te in the form and substance of Exhibit A and shall execute and deliver to SNB a term note in the form and substance of Exhibit, B. 3. (PRUDENTIAL COVENANTS) Borrower agrees that until the principal of and interest on both of the term loans shall have been paid in full, Borrower will perform and observe, and will cause the Subsidiary to perform and observe their respective obligations under (a) each pro,vision of paragraphs SA, SB, SD and SE of each of the Prudential Agreements, provided that each reference to Prudential in these para-graphs shall be deemed to be, for purposes of this Agreement, a reference to both NCB and SN~ (thus entitling each Bank to the same rights an'd privileges afforded to Prudential by such provi-sions) and all references to a •oefault• or an •Event of Default" in paragraph SA shall be deemed to include •events of default• and "defaults under this Agreement• as those terms are defined in section 10 hereof: and (b) each provision of paragraphs 6A, 6B, 6C, 6D and 6E in each of the Prudential Agreements, without regard to any payment in full, satisfaction or other cancellation of any Prudential note and as any such provisions may be affected by any waiver, consent, amend-ment, supplement or any other modification of any kind by Prudential to, or affecting the provisions of, the Pruden-tial Agreements set forth in subsections (a) and (b) above, provided that, in the case of each such waiver, consent, amendment, supplement or other modification of, any kind, Borrower shall have given prior written notice thereof to each Bank and both Banks shall have given written consent or have no reasonable basis for withholding such consent with respect thereto. · \ -2,... I ' t. 4. (OPENING COVENANTS) Prior to or at the execution and deliv-ery of this Agreement Borrower shall have complied or caused compliance with the following: 4.1 (RESOLUTIONS) Borrower's secrP.tary or assistant secretary shall have certified to each Bank a copy of reso-lutions duly adopted by Borrower's board of airectors or its Executive Committee in respect of this Agreement and the certificate (if required) referred to in the first sentence of section 5. 4.2 (LEGAL OPINION) Borrower's outside counsel shall have rendered to the Banks their written opinion in respect o,f' the matters referred to in subsections 5.1 and 5.2, which opinion may be subject to such qualifications and excepl tions, if any, as shall be satifactory to the Banks. ,\ 4.3 (PRUDENTIAL AGREEMENTS) Borrower's secretary or assistant secretary shall have. certified to each Bank a true and complete copy of each of the Prudential Agreements. 4.4 (OPERATING AGREEMENTS) Borrower's secretary or assis-tant secretary shall have certified to each Bank a true and complete copy of each of the First Operating Agreement and the Kirby Operating Agreement. 4.5 (LETTER AGREEMENT) Borrower shall have deliv.ered to each Bank a letter agreement among Borrower, The Scott-Fetzer Company, World Book, Inc., World Book Encyclopedia, Inc., and United Reta,il Finance Company in respect of the First Operating Agreement and the Kirby Operating Agreement. 5. (WARRANTIES) Subject only-to such exceptions, if any, as may be set forth in a certificate delivered to each Bank by Borrower • .. s secretary or assista11t secretary prior to the execu-tion and delivery of this Agreement, Borrower represents and warrants as follows: . 5.1 (EXISTENCE) Borrower and the Subsidiary are each corporations duly organized and validly existing in good standing un.der the laws of the State of Delaware. Each company is duly qualified as a foreign corporation in each state or other jurisdiction in which the nature of the business conducted by it makes such qualification necessary or where failure to qualify would have a material adverse effect upon the companies' financial conditions and their ability to transact business. 5.2 (RIGHT TO ACT) Borrower has requisite corporate power and authority to enter into this AgreemP.l'lt and to obtain -3-·~ • I • • • • • • • the term loans. No registration with or approval of any governmental agency of any kind is requireu,on the part of Borrower for the due execution and delivery or for its per-formance of this Agreement or the term notes. The officer executing and delivering this Agreement and the term notes on behalf of Borrower has) been duly authorized to do so. Neither the execution and delivery of this Agreement and ·the term notes by Borrower nor the performance and obser-vance of the respective provisions thereof by any company will violate. any existing provision in the certificates of incorporation or by-laws ~f any company or any applicable .. law or cause any company t:o violate or otherwise become in default under any contract or other obligation now existing and binding upon it. Qpon the execution and delivery thereof, this Agreement and the term notes will each become a valid and binding obligation of Borrower. 5. 3 (LITIGATIO!O No ii tigation or proceeding is pending against any company before any court or any administrative agency which in the opinion of Borrower's officers might, if successful, have a material, adverse effect on the.com-panies. viewed on a consolidated basis. 5.4 (TAXES) 'Each company has filed all federal, state and local tax returns which are required to be filed by it and paid all taxes due as shown thereon (except to the extent, if any, that any tax is being contested in good faith and by timely and appropriate proceedings). The Internal Revenue Service has not alleged any material default by Borrower in the payment of any tax material in amour.t or threatened to make any assessment in. respect thereof which has not been reflecteq in the financial statements referred to in subsection 5.8. · 5.5 (TITLE) The companies have good and marketable title to their resp~ctive real properties, and good title to all of their othe1: respective properties and assets reflected in their consolidated balanca;sheet as at November 30, 1981 (except for changes resulting from transactions in the ordinary course of business) subject to no lien of any kind except those permitted by paragraph 6C(l) in each of the Prudential Agreements. 5.6 (LAWFUL OPERATIONS) Each company's operations are in substantial compliance with all material requirements imposed by applicable law, whether federal or state. 5.7 (ERISA COMPLIANCE) No material accumulated funding ··deficiency exists in respect of any of the companies' employee benefit plans. No reportable event has occurred -4-• • • • • • I ~ • I) in respect of any such plan which is continuing and which constitutes grounds either for termination of the plan or for court. appointment of a trus.t'\~e for the administration thereof, ·· 5.8 (•No~-PURPOSE" BORROWING) Borrower is net engaged in the business of extending credit for the purpose of pur-chasing or carrying any margin stock (as defined in Regu-lation U of the Board of Governors of the Federal Reserve System>,,. No proceeds of the term loans will be used for the purpose of purchasing or carrying any margin stock, to extend credit to others for any such purpose, or to reduce or re,tire any Indebtedness incurred for any such purpose. 5.9 (FINANCIAL STA~EMENTS) Borrower's annual audit report, prepared as at: November 30, 1981 and certified by Coopers & Lybrand, and the companies' interim financial statements, prepared as of May 31, 1982, each of which has been here-tofore furnished by Borrower to Bank, are true and correct in all material respects (subject, as to interim state-ments, to routine changes resulting from audits and year-end adjustments) and have been prepared in accordance with ·generally accepted accounting principles consistently followed throughout the periods involved, except as noted therein, and fairly present their financial condition. as of thee.respective dates thereof (including a full disclosure of material contingent liabilities, if any, required to be shown in accordance with such accounting principles) and the results of their operations for the' respective fiscal periods then ending, all on a consolidated baS!iS. There .has been no material adverse change in the financial con-dition, properties or business of· the companies since May 31, 1982, nor has there been any material change in their accounting procedures since November 30 ,. 1981. , __ , 5.10 (DEFAULTS) No default under this Agreement exists, nor will any exist immediately after the execution and delivery of this Agreement and the term notes • 6. (EVENTS OF DEFAULT) Each of the following shall constitute an event of default hereunder: 6.1 (PAYMENTS) If Borrower fails to pay fully \ (a) any principal of either term note when the same shall become due; or (b) any interest on either term note when the same shall become due and which rema'oins unpaid· for ten (10) days after the due date~ · · -5-0 • • ' ·~", ==~~~\ 6.2 (WARRANTIES) If any representation, warranty or statement made in this Agreement or any related writing or any other material information furnished by or on behalf of any company to the Banks in writing shall be erroneous in any material respect on the date as of which made. )\6.3 (COVF;NANTS) If any company shall fail or omit to "-=''perform ai-ld observe an~ covenant or agreement which it is obligated''to per~orm or observe under the terms of this Agreemeri't (other than those referred to in subsection 6 .1) and that failure or omission shall not have been fully corrected within thirty (30) days after the Banks have given written notice to Borrower of such failure or omission. 6.4 (CROSS-DEFAULT) If the maturity of any indebtedness for borrowed money or of any funded indebtedness now owing or hereafter incurred by any company shall be accelerated pursuant to any indenture, agreement or other instrument which evidences or secures any such indebtedness or pur-suant to which any such indebtedness is issued or assumed; or if any indebtedness which has matured by lapse of time (and not by acceleration) or which was payable on demand shall not have been paid in full within thirty (30) con-secutive days after the same became due and payable or after demand for payment was made; provided that this sub-section 6.4 shali not apply to any indebtedness of less than two hundred fifty thousand dollars ($250,000). 6.5 (SUBSIDIARY'S SOLVENCY) If (a) Subsidiary shall com-mence any insolvency acti.on of any kind or admit (by answer, 'default or otherwise) the material allegations of, .or con-sent to any relie~ requested in, any insolvency action of any kind commenc.ta against Subsidiary by its creditors or any thereof, or (b) Subsidiary's creditors or any thereof shall commence against Subsidiary any insolvency action of any kind which shall remain in effect (neither dismissed nor stayed) for sixty (60) consecutive days. 6.6 (BORROWER'S SOLVENCY) If (a) Borrower shall discon-tinue substantially all of i.ts operations, or (b) Borrower shall commence any insolvency action of any kind or admit (by answer, default or otherwise) the material allegations of, pr consent to any relief requested in, any insolvency a:;tion of any kind commenced against Borrower by its cred-itor:s or any thereof, or (c) Bor;:ower's creditors or any ttlbreof shall commence against Borrower any insolvency action of any kind which shall remain in effect (neither dismissed nor stayed) for sixty (60) consecutive days. 7. (BANKS' RIGHTS UPON DEFAULT) Notwithstanding any con~rary provisi!on or inference in this Agreement or in any related writing, -6-,.t..f , • • .. • '( ; t •' t· 0 ;:1 ~"'· \,~ .... ,, . 7.1 (OPTIONAirDEFAULTS) if .any event of default referred to in subsections 6.1 through 6.5, both inclusive, shall occur and be.continuing and shall not have been theretofore remedied 9r waived, NCB and. SNB shall each have the right in its sole discretion, by giving written notice to Bor-rower, to accel~rate the matuQ'.iity of the term notes which shall thereuponc'become and thereafter be immediately due and payable in full ~ithout any presentment or demand and without any further or other notice of any kind, all of which are hereby waivedi -"• \? 7.2 (AUTOMATIC DEFAULTS) if any event of default referred to in subsection 6.6 shall occur, the term notes shall thereup.on become and therea.fter be immediately due and pay-able in full, all withou.t any presentment, demand or notice of any kind, all of which are hereby waivedi a. (INTERPRETkT·ION) No course of dealing bY either Bank in respect of, nor\ any omission or delay by either Bank in the exercise of, any right, power or privilege referred to in sub-section 7.3 or elsewhere in this Agreement or in any related writing shall operate as a waiver thereof, nor shall any single or partial exercise thereof preclude any further or other exet.-cise thereof or of any other, as each such c!'~i9ht, power or pr i-vilege may be exercised either independently or concurrently with others and as of ten and in such order as such Bank may deem expedient. Thill Agreement may be amended by written action of the parti,es and the Banks may from time to time in their dis-, ct·e€ion g,~ant Bprrower waivers ''arid consents in respect of this ·>' Agreemenf".or any related writing, but no such waiver o.r consent sh'al1;1bin4, the Banks ur,iless specifically grari;ted by the Banks ~in writing''· which writing shall be str ict 0ly construed. Each right, powe/r· or privilege specified or referred to in this Agreement or any related writing is in addition to and not in limitation of anYc, other rights, powers and pi\i,,vileges that either Bank may otherwise have or acquire by operation of law, by other contract' Q.r othe.i:wise. The provisions of this Agree-: ment and the related wri,t,ings shall bind and benefit Borrower and each Bank and their respective successors and assigns, including. each subsequent holder, if.• any, of, either of the term notes. Ttle several captions to different sections• and sub-~J~c'tions of this Agreement are inserted for convenience only., "aitd shall be ignored in interpreting the provisions thereof. . This Agreement may be executed simul):aneously in twq\or IJlpre counterparts, each of which shall be··aeemed an original !:Yut which togi:?ther. constitute one agreement. T.his Agreement .,.and the related writings and the re.spective rights and obligations o·f the parties hereto shall be construed in accordance with and governed by Ohio law. \ / \~ <;•, ,, -7-t t • I I • • • • • <·I.\''' 9. (NOTICE) A notice to or request of Borrower shall be deemed to J"i'ave been given or made hereunder when a writing to that effect shall have been delivered to an officer of Borrower or five (5) days after a writing to that effect shall have been deposited in the United States mail and sent, 'With postage pre-paid, by registered or certified mail to Borrower at the address set forth. opposite Borrower's signature below (or to such other address as Borrower may hereafter furnish to each Bank in 'writing for that purpose), irrespective of whether the writing i~ctually received by Borrower. No other method of.giving actual notice tc. or making a request of Borrower is hereby pre-cluded. Every notice required to be given to NCB or SNB pur-suant to this Agreement shall be delivered to a loan officer at the address set forth opposite such Bank's signature below (or such other address as s~ch Bank may furnish to Borrower in !} writing for that purpose) • · 10. (DEFINITIONS) As used in. this Agreement and in the related writings, accumulated funding deficiency sh!1111 ha.ve the ll)eaning ascribed thereto in section 30'2 (a) (2) of ERISA; Bank refers to either NCB or SNB,.as the case may be; Banks refers to both NCB and SNB; · · \ Borrower means World Book Finance,. Inc., a Delaware corpo-ration having its principal place of business in Chicago, Illinois; company refers to Borrower or to ~ubsidiary or to both, as· the case may be; " default under: this Agreement means an event, condition or thing which ~as occurred or exists that constftutes or which with.<-.the lapse of any applicable grace period or the giving of notice or both would constitute an event of default referred to in section 6 and whi,f"h has not been appropriately waived in writing in accordance with this Agreemenb or corrected to the Banks' full satiSfaction1 (I . employee benefit plan shall have the meaning ascribed there-to in section 3(3) of ERISA and shall mean such a plan of one or more or all of the companies; f_J ERISA me'fins the Employee Retirement Inc6me Security Act of l974 as"amended from time to time; and in the event of any amendment affecting any sec,tion thereof referred to in this Agreement, that refere.nce tl:)~reafter shall be a refe.rence to that section as amended, supplemented, replaced cir other-wise modified; ' • -a-43 • " • • • • • •• " • event of default is defined in section 61 First Operating Agreement. shall,. have the same meaning as given to it in paragraph 10,AA of the 1981 Prudential Agree-·"" ment; o funded indebtedness means indebtedness which matures or which (including each renewal or extension, if any, in whole or in part) remains unpaid for more than twelve ., months a·fter the date originally incurred, it being under-· stood that in the case of any indebtedness payable in installments or evidenced by serial notes or calling for sinking fund payments, those payments maturing within twelve months after the date of determination shall be con-sidered current indebtedness r~ther than f~nded indebted-ness; insolvency action means either (a) a pleading of any kind filed by a company to seek rel,ief from its creditors, or filed by that company's credit'brs or any thereof to seek relief of any kind agains't that company, in anir court or other tribunal pursuant to any law (whether fe/~eral, state or other) relating generally to the rights of creditors or the relief of debtors or both, or (1p) any other action of any kind commenced by tl!iat company qr its creditors or any thereof for the purpose of marshalling that company's assets and liabilities for the benefit of its creditprsi and "insolvency action" includes (without limitation) a petition com.~encing a case pursuant to any chapter of the federal bankruptcy code, any application for the appointment of a .r•eceiver, trustee,, liquidator or custodian for a debtor or any substantial part of its assets, and any assignment by a debtor for the general benefit of its creditors; ,, Kirbv Operating A•;j'reement shall have the sa!!lfo' meaning as given to it in paragraph 3G of the 1981 Pruci\ential Agree-ment; . ,. a lien means any mortgage, pledge, security interest, encumbrance, lien or charge .of any kind (including any agreement to give any of the foregoing, any conditional sale or other t:itle retention agreement, any lease in the nature thereof, and the filing of an agreement to give any financing statement under the Uniform Commercial Code of any jurisdictio~.); ~ NCB means National City Bank, a national banking association organized under the laws of the United States and having its maip office in Cleveland, Ohioi. , -'\' (f • • • • • • • • • • ·~ NCB's base rate means the fluctuating rate, as in effect at the time in question, that is publicly announced by NCB from time to time in Cleveland, Ohio, as being its base rate thereafter in effecti overdue interest means interest which at the time in question shall have become due in accordance with the terms of subsection 2.3(d) of this Agreement and which has not been paidi 1981 Prudential Aareement means the Note Agreement between Borrower and Prudential dated June 30, 1981, as in effect on the date .of this Agreementi 1978 Prudential Agreement means the Note Agreement between Borrower and Prudential dated August 31, 1978, as amended by a letter from Borrower to Prudential dated June 30, 1981, and in effect on the date of this Agreementi ~.§.2.!l means any individual, partnership, jiint venture, corporation, trust, estate, or other association or entity, or any governmental unit or agencyi Prudential means The Prudential Insurance Company of America, a New Jersey corporation having its principal office in Newark, New Jersey; Prudential Agreements means the 1978 Prudential ~greement and the 1981 Prudential Agreement; .Prudential note means any promissory note issued under the 1978 Prudential '.Agreement or under the 1981 Prude,ntici1 Agreement; (/' related writing means any notice, note, financial statement, audit report, request or other writing of any kind which is executed by a company, or certified or signed by one or more of the officers, auditors or counsel of the companies, and is delivered to either Bank pursuant to this Agreement; reportable event shall be defined in accordance w~th ERISA; SNB means Society National Bank, a national banking asso-ciation organized under the laws of the United States and having its main Qffice in Cleveland, Ohio; SNB's prime rate means the fluctuating rate, as in effect at the time in question, that is publicly announced by SNB from time to time in Cleveland, Ohio, as being its prime rate thereafter in effect; '-'-' ~ -10-.-,. ,, •, ,, .. t I • • . • • • () .. D Subsidiary shall mean United Retail Finar:ce Company, a Delaware corporation and wholly-owned subsidiary of Borrower: term loan means the loan made by each Bank under this Agreement: term loans means the term loan made by NCB and the term loan made by SNB: · term note means a promissory note being either in the form of Exhibit A or Exhibit B and executed and delivered by Borrower to NCB or SNB to evidence each Bank's term loan: term notes means the term note executed and delivered to NCB and the 'term note executed .. and delivered to SNB: the foregoing definitions shall be applicable to the respective plurals of the foregoing defined terms. ,. 11. (INVESTMENT PURPOSE) Each Bank represents and warrants to Borrower that .it is familiar with the' Sectirities Act of 1933 as amended and that it is not acquiring its' term note for the pur-pose of distribution or resale:. provide•:!, that each Bank at all times shall retain full control over the disposition of its assets. Address: Suite 500 Merchandise Mart Plaza Chicago, Illinois 60654 Address: 1900 East Ninth Stre.et .- Cleveland, Ohio 4411& Attention: Metropolitan Division Address: 127 Public Cle"ieland, Attehtion: Square Ohio 44114 Metropolitan Division WORLD BOOK FINANCE, INC. NATIONAL CITY BANK By ...,t......:.l:.~~AJ'd:;:::.:'.-:'C_:.':.......,. __ ',_,(- /,~~ 6~i SOCIETY NATIONAL BANK By k7f;z;r~fli Title Ucua- /l.tt•"· . .f -11-" '-' • • • $15,000,000 Cleveland, Ohio August j], 1982 On August 17 , 1985, the undersigned, ~ BCXl< FillANcE, INC. (a Delaware c:orporiatioil)for value received promises to pay to the order of NATIOOAL CITY BANK ("Bank") at Bank's main office in Cleveland, Ohio, the principal sum of FIF'l'EEN MILLIOO RID ID/100 -- OOLLARS together with interest as provided below. The principal sum, of this note shall bear interest payable oo December 1, 1982, and oo the first day of each March, June, september and December thereafter and at the maturity hereof, and conputed oo a basis of a 365-day year for the actual nllli>er of.,days elapsed as followsi prior to maturity, at the rate of fifteen and ooe-cjuarter";~rcent (15-1/4%) per annum; and after maturity, at a · fluct:uatil'll3 rate equal to two percent (2%) per anm111 above Bank's base rate from time to time in effect, with each change in Bank's base rate automatically and imnediately ~h~il'll3 the rate there~fter a~licable to this note after maturity, provided that in no event after maturity shall this nQte bear interest at a l.esser rate than seventeen and one-quarter percent (17-1/4%) per annum. No prepayment of the principal sum of this note may be made. e As. used in this note, the term "base rate" means the fluctuating rate, as in effect at the time in .question, that is publicly announced by Bank from ti.rre to time in Cleveland, Ohio, as being its base rate thereafter in effect. · This. note is ooe of the term notes referred to in a certain Agreement of e11en date herewith by and between Bank.~ the undersigned, which Agreement contains e provisions for the acce.leration of the maturity hereof upon the happening of certain events. This oote shall be governed by and construed in accordance with ... Ohio law. • • • Address: Suite 500 Merchandise Mart Plaza Olicago, Illinois 60654 K>RID BX>K f!NANCE, INC. !) .. I I • • • • • • • • • $10,000,000 Cleveland, Ohio August .u_, 1982. Q\ August 17 1 1985, the undersigned, K>RID BOO< FINANCE, INC. (a Delaware corporatioO'i"'for value received promises to pay to the order of SOCIETY Nl'.TION.l\L BANK (•Bank") at Bank's main office in Cleveland, Ohio, the principal sum of C• TEN MILLIOO AND N'.>/100 -- OOLLARS together with interest as provided below. The principal sum of this note shall bear interest payable on December 1, 1982, and oo the first day of each March, June, September and December thereafter and at the maturity hereof, and o:inputed on a basis of a 365-day year for the actual nlnlber of days elapsed as follows: prior to maturity, at the rate of fifteen and one-quarter percent (15-1/4%) per annumi and after maturity, at a fluctuating rate equal to two percent (2%) per annlJll above Bank's prim=,,rate fran time to time in effect with each change. in Bank's i:rime rate automatically and imnediately changing the rate tl'iereafter applicable to this note after maturity, provided that in no event after maturity shall this note bear interest at a lesser rate than seventeen and one-quarter percent (17-1/4%) per annum. No {repayment of the principal sum of this note may be made • As used in this note, the term "prime rate" means the fluctuating rate, as in effect at the time in question, that is ?Jblicly announced by Bank from time to time in Cleveland, Ohio, as being its prime rate thereafter in effect. This note is one of the term notes referred to in a certain Agreement of even date herewith by and between Bank and the undersigned, which Agreement contains pravisions for the acceleration of the maturity hereof upon the happening of certain events. This note shall be governed by and constru/00 in accordaneoe with Ohio law. Address: Suite 500 Merchandise Mart Plaza Cllicago, Illinois 60654 ~RLD ln>K FINANCE, INC. •I " • • • • • • • • • • •• II !/ .--.t: v 0 EXHIBIT TitE scon • FETZER COMPANY 1•1 Stock Option Plln I. 1be total number of lhua wbicb may be luued lllCI IOld under opcioas pntc4 punUllll ID dlil Stock Opcioa 1'1111 lhlll DOI eu.eed 500,000 lbarel of tbe Compuy'• Common Stock without pc Yllue, w:cpt to tbe ealall of ldjustments IUlboriz.ed by the lut 1entence of Pmppb 5 of Chis Stocll: Opciaa Pllll. Sucb lhua may be treuu:y Ibara or lbua of aria· 11111 iuue or 1 combinalioa of the fcnaoiaa. 2. 1be Boud of Directan of the Comp111y may, ftom time to time and upoa sucb tenm and coaditioas u it may delcrmiDe, autMrlzC •the pntina to omcerl (incJudina oftic:en who lie memben.oftbe Boud of Directan) and to atber key employees of tbe Company or any of iU IUbsidiariea of opcioas IO .. buy from the Cml· puy ab~\ot Common Stocll:. and may fil tbe 9!1D1ber of ~s to be eovcnd by cacb 1UCb opcioa, provided, bowever, 11111 die agreptc flir market value (detcrmined,,u of tbe time tbe opcioa is pied) of tbe ICOCk for wbicb any employee may be panted opcioas in any Cllen· dar year wbicb lie intended ID qualify u "IDcentlve Stock Opcioas" UDder Seclion 422(A) of the lntcrnal Revenue cOde (UDder all plua of tbe Coinpuy and ill parent and IUb-lidiary carpontioas, if any) aba!I DOI eu.eed SJ00,000 plus any unused •limit cmyover to IUCb year (sucb unused limit carryover ID be determined u provide in llUCb Section 422(A)). Succaaive opcions may .be panted to tbe ume pen.111 wbltber or DOI tbe opcioa or opciooa fint (10) (iii) panted to llUCb ~ ftlDlill uncun:iled. Tbe Directan may, wida the CGDCUrlellCC of tbe If. fecteci opcioaeea, cuccl uy opcioD panted un· der this Stock OptioD Pllll llld may 111tboriz.e tbc pntina ol 1 -opcioD or opcioas, ID buy from tbe Company lhua of tbc C-puy's Common Stocll: in tbe ume number or 1 differ· eat number u provided by tbe cucclled option. 3. Optloas panted Ulll!cr dlil Stock Option Pllll may be (i) opcioas wbicb are intended to qualify uader penicular provisioas of tbe Inter· 1111 Revenue Code, u ill ~Kt from time to time, (ii) opcioDs wbicb 1R DOI intlCDded IO ID qualify under tbc ~~Revenue Code, or(iii) cambia•tioaa of dic~foreaoinl. No opcioD aba1J IUD for llllft than ICll ,s from tbe date pant• ed. No opcion lba1J be ll'llllfcnble by tbe op-tiallee Glhawile than by will or the Jaws of descellt ..: cliluibutloa. 4. 1be opcioD price 111111 DOI be leu dlln tbc fair mmtet Yllue of tbe abana eovcnd by tbc ' opcioD II the time the opcioD ii pnted. 1be opcioD pri=e 111111 be payable in cub or, in wbole or in plft, in lbua of the Compufs Cmln!O!! Sliocli:· Yllued II flir IDllket Yllue. 1be Directan may llllboriz.c, upoa llUCb conditions llld Jimiwioas u Ibey deem ldvislble, the eurreadel of.• opcioa, or ~Y poniOD tbereof, pnted, lmdet dais Stoc:ll: Ojicion Pllll ud the delivery in evb'DF damfor of: (1) abares of tbc Compuy.'•'f'--' S10Ct uvina 111e flir mmUt value OD tbe dale of auncuder oqual IO the W:CU of the fair mmUt value II tbe date of IWicuder af'the lbua CO\cnil by tbe opcioa, or padioD thereof, eumadered over tbe aaarea•tc opcioD price of IUCb abana, (b) cub in ID ~ equivalent ID 11111 determined under () t • • • • • • • • dlu1e (1), or: (c) plltly ill cub and pmtJy ill lbms ol lhe Compuiy'a 0•1111- SIOCt u die Direclon may delMnine ill !heir discretioa. 'l1lc Common Shares coveted by any apcion, ar par· 1ioo lhereof, surrendered u ben:ia provided abalJ aot be deemed ID have been issued and told under chis Stock Opcioa Plan, e1cep1, llld ID die u1e111 ol, lhlres of die Compuiy'a C-DMI!! Stock issued ar delivencl ill respect ol IUCb auacudtr. 5. 'l1lc Boud ol Directon may mate ar pro-vide far aucb adju11men11 ia ll:le apcion price llld ID die number ar tiad of abarea of die Com-pany's Common SIOCt ar alher leCUritiea COY• end by outstandia1 apci011111 aucb Board ill iu 11>le disaetioa, eacn:ised il1 SoCJd flilb, may f~ is equitably required ID prevent dilu-("'tiCJll ar enlqemeat of die riabls of apcionecs dw would ocherwise mull from (1) any llOct dividend, ltoct split, combiaalion of abarea, re-capilllizatioa ar alher c:baaae ill die capital tlnlclUre of die Compuiy, ar (b) aay mcracr •. COlllOJidation, aepmtica, recqanization or par-lial ar complete liquidation, ar (c) aay alher corporate tnallClion ar eveatbavilla an drect llimil1r ID aay of die fore1oilla. 'l1lc Board of Directon may aho mate ar provide far IUCb adj1111mea11 in die number ar kind ol lbarel of die Compuiy'a Common Stock ar alher llCUri-liea wbicb may be told under Ibis Stock Opcioa Plan U IUCb Board ill its IOle dillCl'Clion, Ger· ciled ill aood flilb, may delamille ii ljJjiiiljHi· lie ID reflect lay event of die type dacribod ill dlule. <1> o1111e P"""1iaa -"'"· 6. Each Stock Optioa Aireemeat 1ba11 con-taill IUCb alher lmlll and conditions fiDI iacon-\ lillent berewilb II aba1J be appiOved by die Board ofDirecton ar by tbe commiaee !hereof referred ID ill Par.graph 7 ol Ibis Stock Opcioa Plan, 7. 1liis Stock Option Plan aball be admiais-leled by die Board of Directon wbicb may from lime IO lime dele11re all or aay pan of ill aulbor-ily under Ibis Stock Option Plan ID I commiaee ol not leu lbaa diree Directon appointed by die Board of Directon. 'l1lc majority of the commit-tie lbaJI COlllli!ute 1 quonam, llld tbe actioa of tbe "'"Dbers ol tbe commiaee praeat 11 any meetiJla II which I quon&lll is pmeat, ar ICIS lllllDil!JOUaJy approved ill wfitin;, abalJ be die ac11 oflhe c:ommiaee. No apcion lWll be pant-ed by tbe camaiaee ID any DIClllbCl''6r die com-miaee Hcepl ID die Htent specifically aulbo-riial by die Board of Directon. I. nu. Sloct Option Pim may be amended from lime ID lime by tbe Board of Dinc:lors, but without .runbcr approval by die abarebolders of · die Compuiy Ibis Stock Option Plan may aot be ' uamded ID iDcreue die agrepte number of abarea of die Compuiy'a Commoa'Stoct wilb-Cllll par value dw may be iuued llld sold under tbis Stock Option Plan (e1cep1 dw adjustments llltborized' by tbe lit!, ICf.llellce of Puqraph S abalJ not. be ., limited) ar c:baaae die designa-tion ill Parq:npb 2 of die clus of employees elipble ID be .gnated apciOlll. <' SD (10) (v) t . . . I I • ri~ AN EXECUTIVE, UNFUNDED, NONQUALIFIED DEr£RRED COMPENSATION AGREEMENT BETWEEN THE • SCOTT FETZER COMPANY AND • • ,, • • ,, • • • • This· Agreement made as of -------• 19, by and ,, between·The Scott' Fetzer Company, an Ohio corporation with executive offices at 14600 Detroit Avenue., Lakewood, Ohio 44107 (the ..52. •company•), and -------- residing at. ____ _ --------- (the •Executive•). Witnesseth That: In consideration of the premises, arid the mutual promises • and agreements herein contained the parties hereto agree as follows: • • 1. As used in this Agreement, the following words. and phrases shall have. the meanings as indicated: (a) •Executive's Compensat.ion" - fo:r purposes of this, Agreement, Ex.ecutive • s Ce>mpensation" shall mean •Annual Bas.e Salary" and "Executive Bonus" as defined immediately herein. '.:::· 11 ( 1) "Annual Base Salary" - Tile annual rate of salary of Executive in effect at any time, excluding all fringe. benefits such as but not ~imited to, pension, profit sharing, bonuses, group life insurance and supplementary benefits ordinarily available to executives of the Company. ( 21 "Executive Bonusn - Added compensation that may be made to an Executive under any Company incentive plan for services rend.ered during the Company fiscal year. {' (bl "Beneficiary" - The individual or entity designat~.d • in Schedule B annexed hereto. If the Executive has failed to • • ,, designate a Beneficiary, or if Executive's designation fails by' virtue of the designated Beneficiary predeceasingoExecutive, then the Beneficiary shall be the Executive's estate • 1 .. -., • • • • • • (I (c) •committee" ~ The Committee appointed by the Chief Executive Officer of the Company to administer this Agreement. (d) •oeferred Compensation" -deferred pursuant to Paragraph The amount of Executive's l~mpensation "' (e) ~Deferred described in Paragraph 3 o.f 2 of this Agreement • Compensation Accolmt• ,, - The account this Agreement • (f) "Retirement• - As used in this Agreement, "Retirement" shall have the same meaning as it does under the Company's applicable qualified pension plan • ,, (g) "Retirement Plan• - The Company•s·applicnble pension plan meeting the requirements for qualification under Section 40l(a) 0 of the Internal Revenue Code of 1954, as amended • 1 1 2. The Company and Executive agree to irrevocably defer a portion of Executive's Annual Base SalaJ;Y otherwise payable for "' "' services rendered after the date of this Agreement in the amount of e .. ' $ per month for the year 19, and $ per mqnth for each succeeding yea.r. The Company and Executive further agree to irrevocably defer , of any cash Executive Bonus or cash allotment · under any incentive compensation plan of the Company for services e rendered for Fiscal Year 19 and , for:. each succeeding year. Executive may amend in writing (using Schedule A for such purposes), no later than the last day of the Calendar Year preceding the Calendar Year in which Executive receives his Annual Base Salary or " e no later than the last day of the Fiscal Year preced.ing the Fiscal Year in which the Executive Bonus is accrued by the Company subject to the approval of the. Committee. Executive understands that he may elect only to defer compensation prospectively; that is, he can e irrevocably defer a portion of future • 2 • IJ 0 • • • • • • !I 0 Annual Base Salary in a stipulated amount pet month commencing the first month after the date of this Agreement and be can irrevocably elect to defer all or part of Executive Bonus payal:l)e for services after the date of the election to defer said Executive Bonus. 0 = 0 , 0 3. (a) The Company" shall establish a Deferr~d Compen-sation Account fot: the F.xeC!.Jtive, wbich shall be credited with all amounts of Deferred Compensation as of the. day such Def{irred Compensation would have otherwise be'1n payable •. 0 [I (b) At thf! end of each calendar quarter~ the balance of such Deferred Compensation Account, shall be credited with earnings, for,the period comme7cing the date on which the Deferred Compensation Account was credite~, which earnings shall be equal to an amount derived' by applying a percentage rate of return to the account balance as of the last day of each month during the quarter, such percentage rate of return being the higher of the six-month $100,000 Certificate of Deposit rate or the prime rate of National City Bank of Cleveland, in effect on the last day of the quarter. (c) The value of the shall be determined in accGrdance ,, Deferred Compensation Account with Paragraph 3 at the time of any payment required by paragraphs 4 or 6 of this Agreell)et1t. the determination. of the amount of the Deferred Compensation Account ""' shall be made by crediting the balance of the Deferred Compensation Account with all interest which has accrued since the last quarterly adjustment ~s required by paragraph 3(b) •.. In determining the interest that has accrued since the last quarterly adustment, the last phrase of Paragraph 3(b) should be applied by substituting the word month for quarter. Also any accrual of inJ;erest for a period of less than a full month should use the rate in effect on the last day of the previous month. The dlite of determination shall be a date fixed by the Committee and be no more t.han thirty (30) days prior to the first payment to be made to the Executive. 3 " ..
0 G • (d) The Company ahall adviae the Executive of the amount credited to hi• Deferred Compensation,Account upon his ,, reaaonable requeat and shall at least annually give the Executive a atatement of the amount in auch Account. 0 . 4. (a) (1) Unless the Executive h!'s elected otherwise in accordance with Paragraph 4(a)(2) of this Agx:eement, conunencing not later than six 0(6) months following the date that Executive's c-·employment is terminated with the Company, the Company shall pay to Executive compensation consisting of ten (10) annual. installments of the unpaid balance of 0 the Deferred Compensation Account, each payment to be determined by multiplying the unpaid balance by" a fraction, the numerator Of which is. one and the denominator of which is the. ,number of instal~ments remaining to be paid. (2) By election on Schedule A, annexed hereto and made a part of thi.s Agreement, Executive may elect to have sueh. 0 . _ 1 a compensation commen~f!at the time he reaches the age apecified in Schedule A. lf he does so elect, commencing not later than sixty (60) days after Executive reaches the age apecified in Schedule A, the Company shall pay to Executive ten (10) annual installments of the unpaid balance of the Deferred Compensation Account, .eaeh payment to .be determined by multiplying the unpaid balance by a fraction, the numerator of which ia one is the number·of installments remaining ,,1, r; and the denominator to be paid. of which., (3) The Balance in the Deferred Compens'ation Ac'countQ as of the date of Executive'• lletirement ot termination of employment or death, or date apecified in 4{a)(2) ahall be vested one hundred per,cent (100'1.) to the Executive's Deferred Compensation Account. At the end of each calendar quarter during the ten-year period of distribution of the balance in the Deferred Compensation Account, the undistribu~ed balance ahall be credited with •arnings, which ahall be derived by applying a percent11ge rate of return to the uniliatributed balance, auch percentage rate of return as indi· cated in paragraph 3(b) or 3(c) if appropriate. 0 ·. I ,, • • • ) • • • • (4) Executive, before or after the time otherwise fixed for commencement of payments, may request in writing from the Board Of Directors eai:lier or lat:r installments, more or less JI '' frequent i·nstallments and/or installments of greater oi lesser amount•. The.Board of Directors, Jn its sole discretion, after giving full consideration to the Executi~~·s intereilt will ''make a determination as to changing the pattern of the distribution, including lump sum distribution lf, in its judgment any is to be made, and communicate its decision to the Executive withi~ sixty (6.0) days • (5) (A) If Executive dies either before distribution star.ts or after it has commenced, but has not yet been completed, Bene~iciary assllmes the place of Executive ··as to the remaining installments to be .Pai,d. However, in the case of the Executive's death, the Committee m.Jy be .petitioned in writing·by the Beneficiary for Committee approval of earlier or later installments, more or less frequent installments and/or installments of greater ··or lesser amounts. The Committee, in its sole discretion, wili decide whether a change in the pattern of distribution, includi~g lump sum distribution, is in the interest of the Beneficiary, and 'wi,11 respond in writing 0to the ,petition within sixty (60) days, informing. the Beneficiary of its decision and, if it has decided to·ch~l'lge the pattern of distribution, informing the Beneficiary of the type of change and when it is to take effect. (B) If both the. Executive af!d his designated Beneficiary should die before a tctal of 10 annual payments are made-:1 • by the Company, then the remaining value of the Deferred Comp'ensation Account shall be determine.d as of the date of the death of the designated. Beneficiary and shall be paid as promptly as possible in one lump sum to the estate of such designated Beneficiary • . !/ J,~-" // 5 " I) \ ' ' 0 ' ' it ~· ' • (b) For the purpoaea of.this Paragraph 4~ Executive shall not be deemed to have terminated hia employment if he is n transferred from the employ of fhe Company to the employ of a CG'rporation in which the Company owna more than fifty percent (SO'l) of the equity intereat or one which,directly or inditectly baa own-,,. ,,-rahip in common vi th that of the Company of more than fifty per.-, ; cent (SO'l.). s1 S. Neither Executive nor hia Beneficiary shall have any right to commute,z;.ell, aasign, tranafer, or otherwiae convey or encumber the rights to receive any payments hereunder, which payments ,, .:;;· (1 and all the rights thereto are expreaaly declared to be nona11ignable and nontransferable. 6. any time. If satisfaction terminated, Account (as Agreement); The Board of Direc~ors may terminate this Agreement at the Board,...terminates this Agreement,. the Company in of its obl\~ation hereunder, uat its option shall: (a) pay Executive"' on the date this Agreement is an imlount equal· to the value of hia Deferred Compensation determined in accordance with Paragraph 3 of this or· = 0 0 (b) pay Executive, commencing on the date this Agree-ment ia terminated or upon termination of hia employment, an "amount equal to the value of hia Deferred Compensation Account (as I dete~mined in accordanc.e with paragraph 3 of this Aireelnent) valued aa of the. date this Agreement ia terminated in no more than. ten (10) annual installments, calculated in the aame manner as payments under Paragraph 4(•)(1), with interest on auch amount from the date .~ '· I thia agreement is terminated to the date of auch payment at a rate ',./ to be determined in accordance with paragraph 3(b) or 3(c) if . Q •.rlpropriate. ';ii ,:;:.. • l'F' ~ 6 • ·. Q I I \ .58 7. lf Employee, on termination of employment, 1• entitled to any benefit• under the Retirement Plan, the Company shall pay Employee a auppl1mentary benefit (at the aame time, and in the same manner, as the benefit to which he i• entitled under the Retirement 0 -Plan) equal to the excess of the amount computed in (a) below over ~he amount computed in (b) below: ,, -(a) The benefit to which Employee would have been entitled if hta Annual Earnings, as defined in the Retirement Plan, for each year had included all compeneation earned in such year which would have been included in hi• annual earnings if this Agreement were not in effect. ( (b) The actual benefit to which he is entitled under the Ret 0irement Plan. 8. Nothing contained in this Agreement and no action taken pursuant to the provisions of this Agreement shall create \Ir be construed to create a trust of any kind, or a fiduciary relation--ship between the Company and the Executive, hi• designated Bene-~ ficiary or any other person. Any fund• which may be invested ~nder the provisions of this Agreement shall continue for all purposes to be a part of the general funds of the Company and no pe~aon other than the Company shall by virtue of the provisions of this Agreement have any interest in such funds. To the extent that any person acquires a right to receive payment• from the Company under this Agreement, such right shall be no greater than the right of any unsecured general creditor ... of the Company. 9. Thi• Agreement does not constitute a contract for the continued employment of the-Executive by the Company. The Company reserves the right to modify Executive'• compensation at any time and i) ~ from time to time as it consider• appropriate and to terminate his 0 employment for any reason at any time notwithstanding this Agreement, unless there ar.e restriction• on such termination outside this Agreement. 7 10. Thl1 Agreement may be changed, modified, or amended' 0 by written Agreement between the partie1. 11 •. Any notice to Executive hereunder may be given either by delivering it to Executive or by depo1iting it in the United State• mall, po1tage prepaid, addre11ed to hl1 la1t known address. ~ 0 Failure to in1i1t upon 1trlct compliance with any of the terms, covenant•, or condition• hereof 1he'll not be deemed a waiver of 1uch term, covenant, or condition, nor 1hall any waiver or relinquishment of any right or power hereunder at one or more times b• deemed a waiver or relinqui1hment of 1uch right or power at any other time or time1.
0 The invalidity or unenforceability of any provi1ion hereof shall in no way a£fect the validity or enforceability of any other provi1ion. 12. Except as otherwi1e provided herein, this Agreement 1hall inure to the benefit of and )?e binding upon tll,e Company, its 1uccessors and a1signs, including but not limited to any Company which may acquire all or substantially all of the Company's assets 0 and busine1s or with or into which the Company may be consolidated or merged. 13. Thl1. Agreement 1hall be governed by the laws of the !itate of Ohio. 14. Executive acknowledge• that he has read all parts of this Agreement and has 1ought and obtained 1ati1fac~ory answer(s) to any question(•) he had as to his rlght1, obligations and potential 0 liabilities under this Agreement prior to affixing hi• 1ignature to any part of this Agreement. 8 'I I . " ·' ' & ' 11 IM WITNESS WHEREOF the. parties hereto have executed this ' " 0 Asreement as of the day and year first above written. VlTMESS u 0 9 ·. THE SCOTT FETZER COMPANY '00 By ..... ,.-·· -----0 Executfve () 0 0 <' 0 . i i • I I I 11 •·a I , I i 'I . I i I ,, I !1 I I I I , 11 • • • • • • • Section 1. ROTE: () ,, ,, Schedule A Elections in Connection With Deferral of Annual Base Salary and/or Executive •onus Change in Election aa to Compensation to be Deferred. THI~. SECTION IS TO BE USED ONLY IF YOU WANT TO MAKE A CHANGE IN THE ELECTION THAT YOU PREVIOUSLY HADE UNDER PARAGRAPH 2 OF THIS AGREEMENT. ANY CHANGE IN ELECTION MUST BE MADE NO LATER THAN THE LAST DAY OF THE CALENDAR YEAR PRECEDING THE CALENDAR YEAR IN WHICH _YOU RECEIVE1 v YOUR ANNUAL BASE SALARY OR NO LATER THAN. THt:J LAST DAY .. OF THE FISCAL YEAR PRECEDING THE FISCAL YEAR. IN WHICH YOUR EX~CUTIVE BONUS IS ACCRUED B~ THE COMPANY. . 0 . I hereby elect to irrev~c;ably def~r ... until my retirement '· \/ .• \;:;, or other termination of employment., $ co 11 month of my Annual, Base Salary, for the,,19 _ ~.•lender ,Year lind $ -----a month for eaich succeeding Calendar Year, and further, I elect to 1"1 -irrevocably defer\ 1 of Executive Bonus for services rendered for the lS~~iscal Year and 1 for each succeeding fiscal year. "' Section 2. 0 Election aa to Commencement of,. Distribution From Deferred·Compensation Account. 0 I hereby irrevocably elect to have annual installment payments commence: December , 19 Witness Six (6) months after cessation of employment, Within sixty (60) days after reach1ing the age , b which will occur on-----------Executive 0 • • • • • • • 0 • • • () Schedule B ,, Deaign•tion of Benefici•ry(iea) In the event of my de•th. I hereby design•te the following individ-u•la, in their own right or in their representative c::ap•city, in 0 the proportions .and in the priority of interest deaign•ted, to be the beneficiaries of any benefits owing to me under the foregoing Executive Unfunded, Nonqu•lified Deferred Compensation Agreement between The Scott Fetzer Company •nd myself, dated » 19 • 0 PRIMARY BENEFICIARIES - (The following ini:Uvidual' or individuals shall r•ceive all benefit• payable under this Ag~eelna:irlt in the •! ' proportions designated hereunder. If •ny one of the primary bene-ficiaries designated hereunder shall predecease aie, then his or her. share ah•ll .be divided equally among the then living primary beneficiaries.) NAME AND PRESENT ADDRESS OF PRIMARY BENEFICIARIES Q () PROPORTIONATE INTEREST OF PRIMARY BENEFICl~RY(IES) Page 1 of 2 - Schedule B Dealgnatlo~ of Beneflciary(les) RELATIONSHIP TO EXECUTIVE , i • • • • I [) I 0 . 0 ' ~. I • • .o . . / 0 ,, SECONDARY IENEFiCIARIES - (The following •h•ll receive •11 benefit• p•y•ble under thi• Agreement in the proportion• designeted hereunder only if •11 of myoprimery bene-,flci•rie• heve predece••ed me. If the primery beneflc!ery or beneflci•ries·' h•ve pred~ceased me •nd if •ny one of the eecondary 0 -. •. benefici•rie• design•ted hereunder •h•ll predecease me, then his or her •here •h•ll be divided equ•lly among, the then li,Jing sec-ondary beneficieries.) c NAME AND, PRESENT ADDRESS OF SECONDARY BENEFICIARIES PROPORTIONATE INTEREST OF SECONDARY .BENEFICIARY(IES) RELATIONSHIP TO EXECUTIVE " ___ 'I. ~/ --"'---.-.· 0 ESTATE - I hereby ecknowledge th•t if I heve declined to designate • benefici•rY hereunder or if •11 of the benef iclarie• that I have design•ted predece••e meo, then my e•t~te •h•ll be the beneficiary of •11 benefit• p•y•ble. under the Executive Unfunded, Nonqualified Defer,red Compens•tion Agreement between myself •nd THE SCOTT FETZER " COMPANY deted ----------· ,, " D•te: 0 :;::. '" ·:;Cl WITNESS EX£~UTIVE'S SIGNATURE I P•ge 2 of 2 • Schedule 8 Design•tion of lenefici•ry(ies) .~l ,I . I •\ E1tHl8lT {lo) (v~) II 1 ! '1'he Scott r. Fetzer ~y, al aiio wtpocatien (•Scott retzer•H and Ralph Sc:hey (•Schey") beret¥ lqCf!e ~ -Of ~il .J_, 1982 ~ follcws: <? 1. 'l'O induce Schey to aintinue to- serve • chief executive Officer of 10Q Scott Fetzer for i:he piri:iod camencing •il ..:J_, 1982 through August 31, 1987, Q ,, . (•) Scott Fetzer shall pay Schey for his setvices in that ' -capacity m~annual salary Of It leZlllt $150,000 ri' shall pra'lide him II .. o o \fith such other benefits • may be ;"8taury ri -,iitable for a chi:f t~ecutive officer Of a cc:mpiny such • Scott Fetzer ri at least the ' Q \ . •. ~tuivalent of benefits pra'lided other senior officers Of Scott Fetzer; \ 0 (b) ~t Fetzer shall pay Schey a retiranent benefit fu the ·< llllOUl\t ri mnier, for the duration, Sid subject to the conditions pra'lided in parllJraphs 3 n1 4; aid (cl in the event that Schey's ~ • chief executive " officer of Scott Fetzer is teminated by Scott Fetzer for 1SfJ reason, at my time during the aforementioned perilJd ending August 31, 1987, Scott Fetzer l!hall: (i) aintinue to ~ Schey at his rate of salary then? in effect for the bal.mce of the calendar year in Wiich . , D his enployment is t~minated, (ii) pay Schey $150,000 per year in -periodic installments for each of the five (5) calendar years, followi113 the te'llllination year, Sid (iii) pra'lide Schey the equivalent of all insuRd benefits made availr,ble by Scott Fetzer to Schey at the time his t!ll(>loyment is 1teminated for the fU;il perlod Specified in clmses u> Sid uu. !I r; Payments Sid benefits pursuant to c:lmses (i), (ii) Sid (iii) shall 0continm only imtil Schey reaches age 65 at Wiich time he becalles. entitled to 0 (\ 0 0 i) '· the retb:ement benefit en1111erated in paragriiph 4. Notwithstanding the foregoing, if .llnj of the various events enl.llW!rated in par113raph 3 occur during e Schi!y's tenure as chief executive officer of Scott Fetzer, payments Sid benefits • • • • • ,, :'· pursuant to clauses (i), (ii) and (iii), shali continue even if beycrx1 113~ 65. In the ~ent that SChey' s enployment is terminated Sid the payments speeified in clllllleS ( i) Sid (ii) cXllllnence, and Schey subsequenUy dies before all payments have been mme, ;payments pursuant to clauses ( i) Sid (ii.) shall . . •. . . I) I\ cease and oo further paymems will be made. ':' In the event that Schey's enployment is terminated on a basis that he becanes entiUed to payments under Scott Fetzez{~,;disability incane plans; such ., pa}'lllents shall offset any payment under clause (c) on a cbllar-for-dollar basis 1" the pericx1 they are made. " 2. Por purposes of clause (c) of paragraph 1, Schey's enployment :. 'J shall not be ·deemed to have been tetminated by Scott Fetzer if he voluntarily Ii resigns, voluntarily retires, voluntarily takes another position requiring a substantial portiat of his time, or dies. 3. In 1;he event of (a) the change in control of Scott Fetzer, or (bl the signin;J of .llnJ a,ireement providin:J (i) for the merger ex c:onsolidatiat of , .----- -. --B, Scotti Fetzer into or with ...Other corporation/or (ii) for the sale of substantially all the assets of Scott,J'gtzer to another wtporation, or (iii) e for Scott Fet:i;er to becXime a subsidiary of another corporation, "~here the ccns~ion of llnJ sucti 113reement would rl!sult in a chan:Je of control of Scott Fetzer, any option held by Schey to purchase c::amon shares of Scott Fetzer shall ,.,"•, e .. becxiie inmediately exercisable notwithstandin:J 1!1nJ Other provisions of art/I stoc:k • option agreement, Sid, at the election of Schiy, my other benefit payable to him in th€ future shall becale ilm)ediately payable, inclUdin:J the retirenent fr , . II • •' • • • • . --' ,, benefit ccntenplated by paragraph 4 actuarially reduced for payment ~ior to 11ge 65, but excluding lfl'tlJ benefit payable fran a qualified pension plm unless allowed by its terms, and excluding my benefit subject to a oondition other thlrl continued enployment, such • a death or disability benefit. FurthetnPre, ~ if ,...y of the various events en1.111erated in this paragraph 3 occur during Schey' s tenure • dlief executive officer of Scx>tt Fetzer and Schey voluntarily leaves ·-· the enployment of Scx>tt Fetzer (or my successor entity) within three ll'D!lths after such event, Scx>tt Fetzer shall make the payments and fulfill its other obltgations described in paragraph 1 notwithstanding my other prcwisions in this ltgreement: pro17ided, however, that if Schey continues in his position as ·C:::) dlief executive officer of Scx>tt Fetzer (or my successor entity) beyond sudl 0 c~ three m:>nthsr he shall continue to enjCJ all the rights pro17ided to him by this ltgreement to the extent not theretofqre exercised .• 4. 'l'he retirenent benefit referenced in clause (b) of par11graEh 1 shall be a life. annuity, w11oencing the first day of the Jlalth following /~le ' . J later of Schey's termination of eiployment with Scx>tt Fetzer (whether b'ecause of death or t111y other reasorij or the date on whidl he readies or would have readied 11ge 65, based on Sdley's length of service at termination of enployment as follow9•. (',) (a) If Schey has 15 years of service at the time his eiployment terminates, he shall be entiUed to m S1Dunt equal to what he would,;;~' have. been entitled to Wlder 'lhe Scx>tt ' Fetzer Oolrpany General Pension Plan (•General Plan") in effect at that time in a straight life foi:m of payment had he termil'iated enployment at 11ge 65 with 25 years of service. (b) If Schey has less tht111 15 but 10 or nme years of service at the t~ his enployment terminates, he shall be entitled ~ an atDunt equal to the SIDUl\t calculated in (a) above times a percentage equal CJ . ' "'"' I ,I I .. • I I I I I I • to (y) div.idea by (x) where (y) represents, his years of service at n e . teminaticn Of enpl~!nt mi (x) equals ~5 years. ' (c) If Schey has less.than 10 years of service at the time. his eaployment teminates, he shall be entitled to a straight life annual llllOUnt of $100, ooo. (d) ·'Diere shall be deducted frcrn my lllDunts payafile·pursuant to (a) or (bl aboue ill B!Dunts payable to Schey pursulllt to the provis:l.ons of the General Plan. Solely for the purpose of determining I the lllDll1t to be deducted frail the retiranent benefit pursuant to (d) . re , above, the smie form of payment elected for payment of the retirement benefit hereunder will be asslllllld for the General Plm benefit. (e) Schey may designate beneficiaries .md make elections with !'J (I respect to the fom am manner of payments made pursuant to either (a) , (bl or ( c) alxwe in the .m111ner md to the extent such elections are pemitted under the General Plan in effect at that time. If an annuity form of ~t is elected by Schey, the retirement benefit will be paid DDnthly. " (f) 'Dle 111Dunts in (.a), (bl llld ( c) alxwe represent straight life. annuities. lbWelrer, if Schey does not elect another fo1111 of payment (e.g. l1J11P 81111 or joint md survivor) md dies before 120 monthly payments hlM! been made, the payments will be made to Schey's beneficiary until 111 aggregate of 120 11Dnthly payments have been mi!de to Schey an:i his beneficiary. '1tle nDnthly 111Dun1; payable to such if beneficiary will be equal to the lllDl.lllt payable to Sc:hey pursuant to rta), (b) or (c), less lln'J lllDUDts payable to lln'J beneficiary of Schey Wider say other benefit plans providing death benefits, M\ather i.JISijf'ed or ~inSured, of Sc:>tt Fetzer but not inclUding the General c • ,, • • • •• • • • • • • .. Plan. SChey may designate a beneficiary in writing to Scott Fetzer. , If no audl designatim is mllde or the designatim made is ineffective, Scott Fetzer may make payment to such relative as it may select or to Sc:hey's estate. (g) Notwithstanding the foregoing, Scott Fetzer is mt obligated to set aside fums, maintain a special accoW'lt, or take lDJ similar actipn with respect to the obligation under this paragraph 4 to pay a retirement benefit to Schey. S. ~ring the period !that Schey receives CXJnllensation, payments or other benefits un3er this Agreement, Schey shall not: (a) curpete with Scott Fetzer in the mil'lufacture, design, sale or distribution of products of the general type manufactured or sold by Scott Fetzer at sum time in the areas '~ ' where Scott Fetzer is «bing business, or (b) except as an enployee. of Scott Fetzer, eng or participate directly or indirectly in the business or .,1 businesses then a:>ndu~.6d by Scott Fetzer • 6. Any pa:r"llent or delivery required under this Agreement shall tie subject to all requirements of the la.r with regatd to withholding, fililYi!r making of reports and. the .like, and Scott Fetzer shall use its best ef~orts to satisfy,. prarptly all such requirements. 7. Any dispute or claim ex>ncerning this Agreement or the terms Cl'ld CX>nditions of enplo~nt, incltx'lilYi! \tlether such claim is arbitrable, shall be settled by arbitration. '!be arbitration shall be a:>nduct~ uiider the Ccrnnercial Arbitration ail.es of the American Arbitration Associaticn in effect at the time a demand for arbitration under the rules is made. '!be decision of the arbitrator, incltx'ling determinatim of the 11110unt of ltrrJ danages suffered, shall be exclusive, final and binding on both Scott Fetzer and Schey, their heirs, executorsj. administrators, ~uccessors, am assigtis, If necessary, the decision .
' .. ~ ... . .. . ..... of the arbitrator may be entered • a j~rt; in Sly a:>urt of the State of Olio or el..nere. 'i )/ ,, t 8. 'l'his Jlgreement supersedes the Jlg~E!eml!l\t of August 10, 1979 and the Deferred Calpensaton Jlgreement of October 23, 1979 between SChey ri Scott Fetzer with respect to similar matters. Furthernore, the aca>unts maintained t pursuant to prior deferred caipensation a:iceernents will tie closed .m mf fW'lds returned to the general •sets of Scott Fetzer. 'l'his Jlgreement shall be bindin13 upon Scott Fetzer and its successors ~ operation of lar or otherwiR, a shall t inure to the benefit of. Schey, or his personal representative and heirs. • Neither SChey nor mf other I>erson, includi"iJ mf beneficim:y described in paragraph 4, shall ~e my right to make my msigment of Sly rights under this Jlgreement wit:OOut th'e' expcess a:>nsent of the Directors of Scott Peter or a cx:mnittee thereof havini3 11.1thority to grant such a ainsent, and Sly attempt to make such assigment without such., consent stiall be null, void a of n:> effect: t mch rights shall rot be subject to gamistnent, attachnent, execution or levy of mf kind. Executed at Lakel«XX3, Cliio as of the date hereinabove written • • • • • E1Cttl8lT (ao) {vii) 70 • f • • • • 'l'h~s l!MPLOYMml' IC~ is effective the 15th day of Septentier, 1982, between 'l'HE sa:nT ' FETZER CXK'ANY, an C»lio corporaticn (the •caipany•t and Gary A. Olildress (the "&tployee•) • Section 1. b>loyment. Upon and subject to the terms and conditions hereinafter set forth;"'thiOiipany hereby enploys the &lployee durin;J the term of this ~reement as President and Chief Operating Officer of 'lbe Ccmpany and .usigns him the responsibility for such duties as are generally performed in such a position and such other duties as may frail time to time be assigned to him by the Oonpeny. 'lbe &tployee will report to the Qiief Executive Officer of the Q)npany and will have such position, ait.'lority, duties and responsibility as lllOUld noi:mally be associated with such position. 'lbe &lployee hereby accepts such enployment and agrees to devote his full business and (XOfessional time and energy to the Qinpany, c•rull-Time &nployment•) and to perfom such duties and responsibilities in an efficient, trJstworthy and businesslike manner. Without limiting the generality of the fore:ioing, the &tployee shall.not, without the written ~val of the Oonpeny, ~nder 111.y services of a business, professional, or CCl!lllercial nature to any other person, fim or corporation, whether for o:inpensation or otherwise, and, so cbing shall constitute the renderin;J of less than Full-Time &lployment ·hereunder. Section 2. CclnJ?ensation • 2.1 Sl'J.ary. 'Die &tployee shall be entitled to receive frail the Oonpeny aSi fixedWary (•Base Pay") 000)erlSat-ion at the rate of $250,000.00 per year, payable in llDl'lthly installments of $20,833.34 each. Said annual Base Pay shall be reviewed on Noveni)er 30, 1983 and 1984 and should artJ increase be I granted it shall be effective January 1, thereafter and shall be related to the performan~ of the o:>npany and increases granted to other off i~rs of the Carpany. 2.2 Experuies. 'lbe carpany shall reinburse the &lployee for all ·· authorizea-and approved expenses incurred and paid by· him in the 'course of the I performance of his dutiesnpursuant to this ~reement and consistent with policies and rules of the o:>npany relatin;J to reinblrsement of such business expenses. Such expenses shall include mentiership in The union Club, Cleveland, C»lio. 2.3 Benefits. 'lbe Oonpeny will provide fringe and supplemental I benefits (such as Ille and health insurance and sick leave) to the &tployee oonparable_to the benefits.aist:anarily provided all other cxinpany officers. ~ 1111>loyee will be eligible for the benefits. custanarily provided to enployees in the Carpany's Executive Group A subject only to such changes • may be made fran time to time by the O:x!pany and effective with respect to all auch enployees, and will be eligible to participate in the carpany•s Stock ~ion Plan. As I added incentive to render profitable the activities of the Oonpeny, &rployee shall be granted options to purchase 10,000 shares of the Cclrpany's axrm:m stock at its closing price en the New York Stock Exchange on Sept:eni)er 21, 1982. Ii . ,• • • • • • • • ··c-JJ • • • '11 2 : 2.4 Bonus. In addition to the annual Base Pll'J, &lployee shall be eligible iC>receive a bonus (•eonus•) for Ccnpany's fiscal years ending 0 ltM!nt>er 30, 1982, 1983, 1984 and 1985 (~Eb'lus Period"). &lployee shall recei•Je a bonus for the fiscal year eliding ~r 30, 1982 of ro less than $10,417.00, Md for the fiscal year ending Novenber 30, 1983 of ro less than $50,000.00. 'lhe determiniiticn of all bonuses shall be based en the fot11Ula used for other · Olllpany officers. i/ Should &lployee 's aiploym!nt terminate for' any reason prior to any of said dates, or should .he be . rendering less thll'I Full-'l'ime &lployment hereunder on 111y of said dates, &lployee shall rot be eligible for a Bonus for the fiscal year preceding such date or arry portion thereof whatsoever unless nutually agreed to at the time of such termination. In addition to other forms of terminaticn of enployment, for the purposes of this Paragraph 2.4 any enployment with the,,Oonpany that is less than Full-'l'ime &lployment, as defined in this Agreement, &hall constitute a terminaticn of enployment herewxler. 2.5 Retirement 9!'P;nsation. &lployee. shall participate in The Soott ' Fetzer 5i1>aiiY General Pension Plan • Section 3. Term and Termination. 3.1 ~l~nt Term. The term of this Ag~nt shall o::mnence oo Septentier ~2~ unless extended by nutual ag~t of the parties hereto, or sooner terminated or cancelled pursuant to Section 3.2 hereof or otherwise hereunder, shall terminate am expire on 1'bYerlb!r 30, 1985 • 3.2 Cause, Illness, Incapacity, and Termination Pay• The Enployee shall recelVi fuii oonpensat1cn for the services to se performed hereunder during 111y period of illness or ineapacity during the term hereof. ~er, the Chief Executive Officer of the OJnpany shall have the right to terminate this Agreement inmediately by written rotice for to'¥; following causes: . (a) CXll'lduct of the &lployee materially detrimental to the Conpany's reputation or business operations or its ability to secure and, r:enew future contractsi or ' (b) gross or habitual neglect of duty or misCXll'lduct of !:he Enployee in discharging his duties hereunderi or (c) prolonged absence ftQ!I his duties without the CXX1Sent of the Calpanyi or 11 (d) failure or refusal of the &lployee to faiJ;hfully and diligently perform the provisions of this Agreement, policies of the Conpany, or the usual Md aJ&tanary dutie5 of his aiploymenti or (e) failure or refusal of the &lployee to ccnply materially with the policies, standards and. regulations of the Oonpany as fran time to time may be lilade lcram to Enployeei or (f) rendering less than Full-'l'ime Enployment ISlder this Agreement • ,, < •• (? 72. 0 3 In m:lition, the catpany ahall have the right to terminate this Agreement in the event the Enployee suffers an illness or inc:apacity of such o chara~er • to aublltantially diHble '1im fran perfoming his duties hereunder fer: a jleriod of 11>re thin sixty (60) consecutive days in iny one year upcn giving at leut thirty (30) days'. written notice of intention to d:> ao. If the llployee shall ~his duties hereunder within thirty (30) days follow~ the receipt of aicb notice ll1d shall perfom aich cllties for sixty (60) consecutive business days thereafter, this Agreement shall continue in full force or ·validity. Notwithstanding any other pro11ision of this· Agreelll!nt, this Agreement forthwith shall teminate upcn the death of the Enploy~. ,. In m:lition, the Oonpany may teminate this Agreement at my tine, without cause, upcn thirty (30) days prior written notice to Enployee. Should Enployee's enployment hereunder be ao teminated pr:ior to Septent:ler 15, 1983, " Ellployee shall be paid an mount ~al to his annual Bue pay on the date of such temination. Should such temination occ:ur on er: after Septent:ler 15, 1983, Enployee shall (be paid a s1111 ~al to one and one half times his annual Base pay on the date of such temination. In either case, such aiount shall be paid in ~ installments monthly on the last day of the ncnth until the final payment which lhall be paid on Noventier 30, 1985. -For a period of blo (2) years after the te?lllination of enployment of the Enployee with the Ccnpilny (for aey reason), the Enployee shall not, except with the written CXX1Sent of the Ccnpilny: · (a) engage, in the United States, directly or indirectly, in any business or entity which axipetes with the ~or aey of its operati11:1 units or subsidiaries: or (b) engage in any activity or axiduct.olllich damages. the business or welfare of the Ccnpilny. --'l'he Enployee shall be deemed to be a:inpeting with the Ccnpilny if. he ahall engage in the businesses described directly er: .indirectly, whether for his own account cc for that of my other perscn, fim or m~rat~on, ll1d lllether as stockholder, principal, agent, investor, proprietor, officer,· director, enployee or axusultant,. er: in any other capacity, pro11ided, !Pilever, that this paragr:aph shall not be interpreted to prohibit Enployee fran purchasi":J securities in any publicly held caipany for investment purposes. · Section 5. Notice. Any notice or other mmunication required or pemitted to be given to the parties hereto shall be deemed to have been given al the day personally delivered, er: if not ao personally delivered, al the third business day following the day on which mailed 1:7;,r certified or registered mail, return receipt· ~ted, first class postage prepaid, m:lressed as follows: .-1 (a)( If to the Enployee: Gary A. Childress '!be Scott ' Fetzer Ccnpilny 14600 Detroit Avenue Lllltelilood, Ohio 44107 I~· • •• • • • • ~ • • • • • • (b) ;-/--' If to thelf~y: '1be SCX>tt ' Fetze~ Claipmy 14600 Detroit Avenue Lalcewcoc!, Cllio 44107 Attention: Ral.i;:ti Schey, Oiairman 13 4 •. , 0 0 or 1 'at llUCh other addresses • the parties hereto may fran time to time designate in writing. " IN WI'l:NESS NIEREX)F, the parties hereto have mused this Agreement to be duly executed Md delivered in Cleveland, Cllio, on the date first above written • i} I/ I •1 • • ( (lo J (be) \j () DISABILITY BENEFIT AGREEMENT " BETWEEN THE SCOTT & FETZER COMPANY 0 AND KEY EXECUTIVES l".:· 0 ·-·:~· I n 0 ,, 15 . ( (i " THIS AGREEMENT made as of , .by and . ' • • ( D between the SCOTT & FETZER COMPANY, an Ohio Corporation in behalf of itself and its wholly owned subsidiaries (the "Company"), with . . {· ;;.. . executive offices at 14600 Detroit Avenue, Lakewood, Ohio, and (the "Executive") residing at ------------------0 u WHEREAS, it is to the beneficial interest of the stock-' holders of the Company to retain the services of its key and highly compensated executives; and \, WHEREAS, it is to the beneficial interest of these same c::;':r key and highly compensated executives to secure certain disability benefits. ..-. NOW, THEREFORE, the Company and the Executive have agreed to enter into this Agreement. 1. · .. , Definitions. As used in this Agreement, the follow-ing words and phrases shall have the meanings indicated: (a) "Committee" - The Committee appointed by the Chief Executive Officer of the Company to administer this Agree-ment. (b) "Total Disability" - Total Disability shall mean tpe definition of total disability as defined in the dis-ability income insurance policies purchased by the Company on the Executive . Executive's Initials Page 1 of 5 I 0 •· • I: I I • I • II 0 0 0 2. Obligations of the Committee. .~ (a) The Commit~ee any question that may arise under shall be the final arbiter of this Agreement. The Executive 1b should communicate with the Committee in writing as to any question ~ or petition said Executive should have arisi~g out of this Agree-'.' I/ ment. ) " (b) insurance policies Company under this 0 The'' Committee shall purchase disabifl.ty income on the Executive to fund the obligations of the Agreement. (c) Upon reasonable request by th~ Executive, the Committee shall notify the Executive of the amount of benefits payable under this Agreement. Any disability income insurance policy o~ed by the Company on the Ex~cutive sh~ff~··made avail-able to the Executive for review upon reasonable request by the c /1 Executive . .1 C/ L;:/ 3. Benefits Under This Agreement. (a) Amount of Benefit. In the event that the Executive suffers a Total Disability such that benefits are payab~e to the Company.under any disability income\policies owned by the Company on the Executive, the Executive will be entitled to receive a benefit equal to twice the amount payable to.the Company I:'• urider the policies owned b'y the companl'. .on the Executive except as limited by Paragraph 3(c). (b) Benefit PaymentFreguency. The benefits pay-able to the Executive by the Company shall be payable when the income is received by the Company from the disability income ' policies on the Executive owned by the Company. Executive's Initials Page 2 of 5 l) ;J I iJ 'I • • • • • • • i • ':Ci (c) Duration of Benef~ts. Benefits will continue for as long as the Company receives benefits under any disability '1'1 . \} income p~lic~es owned by"lme Company on ~: ~xecutive exc~pt~~at~~ no benefit will be·payable to the Executive if the Executive i-s ~ employed by other than the Company even though ~e Company may still be receiving benefits under the disability income policies \l.,__, owned by the. Company on the ~;xecuti ve. · ~· . ~ ~ : ~ 4. Payments Not Transferable. The' Execu'ffi.ve shall have no right to commute, sell, assign, transfer or otherwise convey or encumber the rights to receive any payments hereunder, which pay-ments and all the rights, thereto are exp1:essly declared to be non-assignable; 5. Not a Contract for Continued Employment. This l\greeme~.t does not constitute a contract for the continued employ-ment of Executive by the Company. The Company resexves the right to ~odify Executive's compensation at any time and from time to time as it consigers appropriate and to terminate his employment for any reason at any time notwithstanding this Agreement. ~ k' 7~~. Termination .. The Company, in its sole discretion, may tE!:rm .. Ji~te this Agreement at any time by gi v,ing written notice to Executive, subject however, to the provision that the Company may not terminate this Agreement for any Executive currently receiving or in the process of applying for benefits under this' ,, .Agreement. 7. Notice. Any notice given in connection with this Agreemen_t .. shall be in 'l<irl ting and delivered in person or by de-/J..~ positing it in,the United States mail, postage prepa:i:•l/ addressed ... to the last-known address. -~-"-!] 8. Waiver. Failure to insist upon strict compliance with any of the terms, covenants, or conditions hereof sh~ll not be . ~ deemed a waiver of such term, covenant, or condition, nor shall" a'ny Executive's Initials Page t ~·. , • • • • • • • • • • • I ' Q, waiver or relinquishment of any right or power hereunder at any one or more times be deemed a waiver or relinquishment of such right or power at any other time or times. 9. Effect of Invalidity of Any Part of the Agreement Upon the Whole Agreement. The invalidity or unenforceabiiity of any provision hereof sh~ll in no way affect the validity or en-forceability of any other provision. 10. Consolidation. or Merger. provided,, this Agreement shall inure to Except as otherwise herein the benefit of and be binding upon the Company, its successors and assigns, including but not limited to any corporation which may acquire all or substan-tially all of the Company's assets and business or with or into which the Company may be consolidated or merged. 11. State Laws Governing This. Agreement. This Agreement shall be governed bf::the laws of the State of Ohio . 12. Agreement supersedes Prior Agreement. Thts Agreement supersedes all prior agreements, which provide for disability bene-fits, between the Executive and the Company. Any prior agreerlients providing disability benefits to the Executive are null and void as of February 1, 1979. 13. Counterparts of This Agreement and Executive Ac-knowledgment That He Has Read and Understands All Parts of This Agreement. This. Agreement has been executed in several counter-parts~ each of which shall be an original, but such counterparts shall together constitute but one instrument. Executive aclo:i.owl-edges that he has read all parts of this Agreement and has sought and obtained, !satisfactory answer(s) to any question(s) he had as to his rights, obligations and potential liabilities under this Agreement prior to affixing his signature and initials to any part "'.") of this Agreement. Executive's Initials Page 4 of 5 Q • • Q •1 t • • • "· . 0 •. ·~ . (\ f th·. . h' . t Signature~ o . e Parties to T is Agreemen IN WITNESS WHEREOF. the parties hereto have executed this Agreement as of the day and year first above written. ' (SEAL) , I L ,·1 -~· THE SCOTT & FETZER COMPANY WITNESS '(; Executive G o Executive's Initials 0 • 1·' ~-~ ·:::) • • • • C· DEATH BENEFIT AGREEMENT • BETWEEN THE SCOTT & FETZER COMPANY AND KEY EXECUTIVES • • 0 • • Q • ii,·,• ,/,Ji:';.. ' ~ .. • •• • • "THIS AGREEMENT made as of by and between THE SCQTT & FETZER COMPANY, an Ohio corporation (the "Company") wl.th executive offices tt 14600 Detroit Avenue, Lakewoo4, Ohio, and (the "Execativfiii") residing at ., (....-: • wHEREAS, it is to the beneficial interest of the stock-holders of the Company t~ retain the services of its key and highly compensated executives; and ;/' • WHEREAS, it is to the beneficial interest of these same • • • • • • • ,, -key and highly compensated executives to secure certain death benefits, NOW, THEREFORE, the Company and the Executive have agreed to enter into this Agreement. 1. Definitions. As used in this Agreement, t.~e follow-ing words and phrases shall have the meanings indicated: (a) "Base Salary" -- Base Salary shall mean the Executive's annualized base salary as defined in the most recently effective Schedule A, which is attached hereto and is specifically incorporated into be amended by the this Agreement Company. and which may, from time to time, (b) "Beneficiary" The person or persons desig-0 1, nated by Executive on Schedule B, which is attached hereto and is specifically incorporated into this Agreement, to receive .any benefits payable by the Company under this Agree~ent on account of Executive's death. If Executive has failed to designate a Benefi-ciary, or if Executive's designation fails by virtue of the desig-nated Beneficiary predeceasing Executive, then the Committee shall have the right to designate the Beneficiary of .. ail benefits payable under this Agreement. Solely at the discretion of the Executive, ·< . \1 1 of 10 Executive's Initials.: 0 • • • • • S~hedule B may be amended from time to time. Two copies of the amended Schedule B shall be executed by the Executive. One copy shall be retained by the Executive, and the second copy shall be forwarded to the Company and shall be attached hereto and specif-ically incorporated into this Agreement. (c) "Committee" -- The Chief Executive Officer of the Company ment. Committee appointed by th~ to administer this Agre;;-(d) "Death Benefit" -- Death Benefit shall mean the amount defined in Paragraph 2 or 3 . ) ~ . . (e "Gross Compensation" Gross Compensation shall mean the sum of the Executive's annualized base salary as of the Executive's date of death,· disability or Retirement plus the amount of the last annual bonus awarde.d to the Executive. ~ b (f) "Retirement" -- Retirement shall mean an Executive retiring from the employment of the Company under conditions whereby he is entitled to receive pension benefits under the Company's General Pension Plan in the form of an Early Re.tirement Pension, a Normal Retirement Pension, or a Disability Pension as defined in the Company's General Pension Plan. Retire-ment shall not mean an Executive retiring under the Company's General Pension Plan who is eligible for and elects a Def erred Vested Pension as defined in the Company's General Pension Plan. (g) "AcciC!ental Death Benefit" -- Accidental Death Benefit shall mean the amount indicated 1n the most recently effec-tive Schedule C, which is attached hereto and specifically incor-porated into.this Agreement . (h) "Basic Death Benefit" •- Basic Death Benefit shall mean the amount indicated in the most recently effective 2 of 10 Executive's Initials: • • a '• • Schedule C, which is attached hereto and specifically incorporated into this Agreement. 2. Pre-Retirement,I:itiath Benefit. -11 (a) If death should occur to the Executive due to natural or accidental causes while actively employed by the Company, or while the Executive is disabled and eligible to receive dis-ability benefits under any of the Company's group long-term dis-ability plans, the Company shall pay an annual Death Benefit to the Executive's Beneficiary equal to the Executive's Base Salary, which shall equal the amount of the Base Salary as indicated on the most recently effective Schedule A, which is attached hereto and is specifically incorporated into this Agreement. The Base Salary (~ amount shall be paid to the Beneficiary for a period of years equal to the greater of ten years or the number of years from the date of the Executive's death to the Executive's sixty-fifth birthday. (b) T~e payment of the Death Benefit defined in Paragraph 2(a) is subject to the limitations as defined in Paragraph 5. (c) Solely at the discretion of the Company, the Base Salary amount may be increased from time to time. Any such increase shall be accomplished by completing an amended Schedule A having the effective date and amount of the new Base Salary. Two copies of the amended Schedule A shall be executed by the ~om pany. One copy shall be attached hereto and specifically incorpo-rated into this Agreement. The second copy shall be forwarded to the Executive as notification of the new amount of'the Executive's Base Salary for the purpose of computing the Executive's pre-retirement Death Benefit. 3 of 10 Executive's Initials: I . '. 3. Post-Retirement Death Benefit. (a) If at any point in time the Executive enters into the employ of or in any way renders either direct or in-direct service to any comp,Jtitor of the company or any of the Company's subsidiaries o~~affiliates without the prior writteg approval of the Committee, the Executive shall not be eligible for and the Company shall not be liable for any Death Benefit sub-sequent to his Retirement. (b) If the Executive enters Retirement prior to his sixty-second birthday, the Company shall not be liable for and the Executive shall not be eligible for the Death Benefit defined in Paragraphs 3(c), 3(d), and 3(e). However, if the Executive retires prior to age sixty-two (62), the Executive may request of the Com-mittee that he be eligible for the Death Benefit as defined in Paragraphs 3(c), 3(d) and 3(e). If the,Executive receives written approval from the Committee designating the Executive to be eligible for post-retirement Death''Benefits, the Company shall be liable for and the Executive. shall be eligible for the Death Benefits as de-0 fined in Paragraphs 3(c), 3(d) and 3(e). (c) In the event of the Executive's death due to natural causes subsequent to his Retirement but prior to the Executive attaining age seventy (70), the Company shall pay a Death. Benefit to the Executive's Beneficiary equal to the lesser of two aiia one-half (2.5) times the Executive•s·Gross Compensa-tion less fifty thousand ($50,000.00) dollars or the Basic Death Benefit. (d) In the event of the Executive's death due to accidental c~uses subsequent to his Retirement but prior to the Executive attaining age seventy (70), the Company shall pay a Death Benefit to the Executive's Beneficiary equal to th~ lesser .. l\·· r:-:::.::-« 4 of 10 Executive's Initials: " . • '·' o.f five (5) times the Executive's Gross compensation less one hundred thousand ($100,000.00) dollars or the Accidental Death Benefit. SS I . I • • • • (1 • • • • • (e) In the event of the Executive's death due to natural or accidental causes subsequent to his Retirement and the attainment of age seventy (70) but prior to his attaining age eighty (80), the Company shall pay a Death Benefit to the Exec-utive's Beneficiary equal to one half (~) of the Death Benefit defined in Paragraph 3(~). (~' (f) In the event of the Executive's death subse-quent to his Retirement and his attaining age eighty (80), the Company shall not be liable for the payment of any Death. Benefit under this Agreement • .(g) The payment of the Death Benefit defined in :Paragraphs 3(c), 3(d), and 3(e), is subject to the limitations as defined in Paragraphs 3(a), 3(b), and 5 • (h) Solei'.y at the discr~tion o/¥fhe company, the Basic Death Benefit and A&cidental Death Benefit may be increased from time to time. Any su~h increase shall be accomplished by completing an amended Schedule C showing the effective date and amounts of the new Basic Death Benefit and Accidental Death Bene-fit. Two copies of the amended Schedule c shall be executed by the Company. One copy shall be attached hereto and specifically incorporated into this Agreement. The second copy shall be for-warded to the Executive as notification of the new amount of the Executive's Basic Death Benefit and Accidental Death Benefit for the purpose of computing the Executive's post-retirement Death Benefit . ::-· s .. of 10 Executive's Initials: 0 \ .. J.1 t " 4. Payment of Death Benefit. (a) The pre-retirement Death Benefit, as defined in Paragraph 2, shall be payable to the Executive's Beneficiary in quarterly installments commencing not later than six (6) months subseqlient to the Executive's date of death. (b) The applicable post-retirement Death Benefit, as defined in Paragraph 3, shall be payable to th_e\~Executi ve' s Beneficiary in forty (40) quarterly installments b~inning no later than six (6) months subsequent to the Executive's date of death. Each installment shall equal two and one-half percent (2.5%) of the gross Death !'"nefit, as defined in Paragraph 3, plus interest. Interest shall be credited to each payment based on the unpaid balance of the gross Death Benefit and shall equal the average quarterly prime rate during the preceding quarter as determined by the Citibank of New York, but shall not be less than one and three-qua~~ers (1.75%) percent. (e) Executive's Beneficiary may request a change " in the timing and/or dollar amount and/or number of benefit pay-\~ ments under this Agreement. Such request shall be in writing and .;::/ shall be directed to the Committee. The Committee, in its sole ;/ discretion, may grant such requests. In the event,;that the Committee shall grant a request which would accelerate. the timing of the payment of the· Death Benefit to the Beneficiary, a penalty, in the form of a reduced Death Benefit, may be imposed by the Committee. 5. Limitation C·on Death Benefits. (a) The payment of any Death Benefit as defined in Paragraphs 2 and 3 s.hall be contingent upon evidence of the Execu-tive's death as reqUired by the Committee . • 6 of 10 Executive's Initials: • • • • • • • • • • • (b) In the event that Executive's death is the result of suicide while the Executive is either sane or insane, the Company shall not be liable for any Death Benefit as defined in Paragraphs 2 and 3. EVen though the Company has no contractual obligation to pay any Death Bdnefit under this Agreement in the event that the Executive's death is due to suicide, the Executive's Beneficiary may petition the Company for the payment of a Death Benef'Jt. Any such petition by Executive's Beneficiary shall be.· directed to the Committee. The Committee, in its sole discretion, shall determine whether·or not the Company will pay any such Death Benefit. c-. (c) If the Company should terminate this Agree-ment pursuant to Paragraph 9 hereinbelow, the Company's obliga-tion to pay any benefits under this Agreement shall likewise terminate, provided however, that the Company may not terminate this Agreement subsequent to the Executive's death or in the event that the Executive is disabled and is eligible to receive disa-bility benefits under any of the Company's group long-term disa-bility plans . 6. Obligations of Company. In the performance and administration of this Agreement": the Company shall be under no obligation to purchase any insurance policies or to segregate any assets for Executive's account or benefit. Should the Comp~ny pur- 0 chase any insurance policies, any insurance proceeds, together with any and all assets resulting from any investment (if anyd of such amounts or earnings thereon, shall b~ the property and assets of the Company, and neither Executive nor the Beneficiary shall have any interest therein. The interest of Executive or the Beneficiary is limited to that of a general creditor of the Company and is limited to the rights to receive payments pursuant to the terms of this Agreement . 0 7 of 10 Executive.' s Initials: . '• .. 0 7. PaY!!!ents Hot Transferable. Heither Executive nor • (I the Beneficiary shall have any right to commute, sell, assign, . tr'an~fer or otherwise convey or enc\Jmber the rights to rec~ive '~~ny payments hereunder, which payments and all the rights thereto are expressly declared to be nonassignable. 8. H~£· a Contract for Continued Employment. This Agreement does not constitute a contract for the continued em-ployment of Executive by the Company. The company reserves ~he right to modify E...>tecutive's compensation at any time and from ,time to time as it considers appropriate and to tr.rminate his employment for any reason at any time notwithstanding this Agreement. 0 9. Termination of Agreement at Company's Discretion. The Company, in its sole discretion, may terminate 0 this Agreelilent at any to the time by giving written notice to·' ·Executive,0 subject however, CJ . provision of Paragraph S(c) herein. {i_ 10. Waiver of Group Life Insurance Benefits. In con-sideration of the benefits payable under this Agree~ent, Executive waives participation in and entitlement to any b~~efits under' the Company's group life insurance plans in excess of fifty.thousand " gg dollars ($50,000.00) in the event of a natural death and one o hundred thousand dollars , 0($100,000.00) in the event of an acci-dental death. ~ 11. Function of the Committee. The Committee shall be final arbiter of any question that may arise undei'\:;~1is Agree-ment. Executive or his Beneficiary should communic~te with the Committee\in ~riting as to any question or petition said.Execu-tive or Benefi'C"iary should have arising out of this Ag~eement. -~ . "' 12. Notice. Any notice given in connection with this Agreement shall be in writing and delivered in personzor by de-positing it in the United States mail, postage prepaid, addressed to the last-known. address. 8 of 10 Executive's Initials: • . " • • • • • • • ' ' ' 13. Waiver. Failure to insist upon strict compliance with any of the terms, covenants, or conditions hereof shall not be deemed a waiver of such term, covenant, or cond~.fr~~n, nor shall any waiver or relinquishment of any· right or power /Aereilnder at any one or more times be deemed a wa~ver or relinquishment of such right or ~ower at any other time or times. 14. Effect of Invalidity of Any Part of The Agreement U~on The Whole Agreement. The invalidity or unenforceability of any provision hereof shall in no way affect the validity or en-forceability of any other provision. JI '~, .. '-;< 15 ... ~ility to Amend The Agreement. This Agreement may be changed, modified, or amended only in writing by both parties except for amendments to Schedules A? B, and c. Schedules A and C,may be amenged.solely by the Company as indicated in Paragraphs 2(c) and 3{h). Schedule B may be amended solely by th~, Executive as indicated in Paragraph l(b). (! 16. consolidation or Merger. Except as herein provided, this Agreement shall inure to the otherwis{ . I benefit of and be binding upon the Company, its successors· and assigns, #including but not l,imited to any corporation which may acquire all.or substan-,..t.ially all, of the Company's assets and busin~.ss or with •ox:--into \ .~./ ,',; -· ··which the Company may be consolidated or merged. 17. State shall be governed .by Laws Governing This Agreemen~f=T~greement the laws of the State of Ohio. 18. Counterparts of This Agreement And Executive Ac-ltnowledqrnent That He Has Read and Understands All Parts of This Agreement. ~his. Agreement has been execu,ted in several counter-parts, each of which shall be an original, but such counterparts 111hall together-constitute but one instrument. Executive acknowl- .,-. ~~-.....~-' ·edges that he has read al1'· parts of this Agreement•·and has sought and obtained satisfactory answer(s) to any questions(s) he had as 9 of 10 Executive 1'f Initials: • . " • • • • • • • • • • to his rights; obligations and potential liabilities under this Agreement prior to affixing his signature and initials to any part of this Agreement . 19. Agreement Supersedes Prior Agreements. Executive acknowledges that this Agree~ent supersedei;; any and all prior Estate or Death Benefit ;~gre~en~~ s) between the Company and the Executive and that any and all prior Estate or Death Benef\tt . i Agreements shall become null and void upon the execution of{_his Agreement. '. Signatures of The Parties to This Agreement . IN WITNESS WHEREOF the parties hereto have executed this Agreement as of the day and year first above written . (SEAL) THE SCOTT & FETZER COMPANY WITNESS: .Executive 10 Of 10 • • • • • I • II II \". •. 0 SCHEDULE A BASE SALARY (; ,Under the Death Benefit Agreement dated between THE SCOT!' & FETZER COMPANY and. thp_e_E_x_e_c_u_t~i-v_e_n_am_e~d~b-e~iow, the Executive's Base Salary for the purpose of computing the Pre-Retirement Death ~enefit as defined in Paragraph 2 of the Death Benefit Agreement'shall be the amount defined below and shall be effective as of the effective date stated below. Executive: Base Salary: Effective Date: The Base Salary and Effective Date indicated above shall supersede a11y Base Salary amount previously awarded to the Executive named abo.ve urider the Death Benefit Agreement dated between hTHE SCOT!' & FETZER COMPANY and the Exe-cu-t-:-1..-·v--e __ n_am_e_d...--ab...-o_v_e-. ~ ' . 1 \ , THE SCOT!' & FETZER COMP ANY Date ~{ • • • • • • • • • .. . ' . " SCHEDULE B DESIGNATION OF BENEFICIARY(IES) In.the event of my death, I hereby designate the following indi-viduals, in tpeir own right or in their representative capacity, in the proportions and in the priority of interest designated, . to be the beneficiaries of any benefits owing to me, under the Agreement betw,een myself and THE SCOTT & FETZER COMPANY dated PRIMARY BENEFICIARIES - (The following individual or indivi4uals shall receive all benefits payable under this Agreement in the proportions designated hereunder. If any one of ,,,the primary beneficiaries designated hereunder shall predecease me, then his or her share shall be divided equally among the then living primary beneficiaries.) NAME AND PRESENT ADDRESS OF PRIMARY BENEFICIARIES ,.l,..,.,.----------11 ''·' ii ,~,-----------~ ~.~------------.~.,-:;; ...• · PROPORTIONATE INTEREST OF.PRIMARY RELATIONSHIP BENEFICIARY{IES) TO EMPLOYEE (,· Executive's Initials: l of 2 - Schedule B • • ... 1.· • • • • • • • • • • • t) SECONDARY BENEFICIARIES - (The following individual or individuals shall receive all benefits payable under this Agreement in the proportions designated hereunder ~ly if all of my primary benefi-ciaries have predeceased me. If e primary beneficiary or bene-ficiaries have predeceased me and if any one of the secondary beneficiaries designated hereunder shall]>redecease me, then his or her share shall be divided equally among the then living secondary beneficiaries.) PROPORTIONATE · INTEREST OF NAME AND PRESENT ADDRESS OF SECONDARY. RELATIONSHIP SECONDARY BENEFICIARIES BENEFICIARY( IES) " TO EMPLOYEE J \1 _J _J.~~·----_J r ESTATE - I hereby acknowledge that if I t1ave declined to designate a beneficiary hereunder or if all of the beneficiaries that I have designated predecease me, then all benefits payable under this Agreement will be payable to my estate • Date: WITNESS EXECUTIVE 1S SIGNATURE c i-', (J 2 of 2 - Schedule B • ... • I > • • • • • • • . 41 • SCHEDULE C POST-RETIREMENT DEATH BENEFITS Under the Death Benefit Agreement dated , between THE SCOTT & FETZER COMPANY and the Executive named below, the Executive's Basic Death Benefit and Accidental Death Benefit for the purpose of computing the Executive's Post-Retirement Death Benefit as defined in Paragraph 3 of the Death Benefit Agreement, shall be the amounts stated below and shall be effective as of the effective date stated below. Executive: Basic Death Benefit: Acc:ldental Death Benefit: Effective Date: The Basic Death Benefit and Accidental Death Benefit amounts stated above shall supersede any Basic Death Benefit and Accidental Death Benefit previously awarded to the Executive named above under the Death Benefit Agreement dated , between THE SCOTT & FETZER COMPANY.:and the Executive named above. THE SCOTT & FETZER COMPANY Date 0 • • • fl • • • • !his ltmenlhent to Deat.'l Benefit Agreement mde • c1 this day of -;;...... __ , l::lit' and between b Scott .. ' Fetzer ws married at his death, however, the Estate Tax Burden with respect tc the United I States federal estate tax shall be the lesser of the following llllJUl'lts: (a) 34• ~ the part, if any, of the federal estate tax value of the Death Benefit which ia mt CJlalified for the federal estate tax mrital deduction; •(b) 17• of the total of the Death Benefit • valued ~er federal estate tu puzposes, and (c) I the Onited States estate tax actually payable with nspect to the Executive's 0 •tate. b Olapmty'• determinaticn of the Estate Tax ~ shall be final and oancluaive upon the Executive's Estate and Beneficiaries.• I 2. b following paragrapi •s(A) Estate Tu Loan• shall be added m.ldiately following paragraph S(c): : • 1. !xecutive'• Initials .. •• • • • • • • • • • • •s. (A) Estate Tax Loan • In the event that the beneficiaries of an Executive are entitled to an annual benefit under paragraph 2, then the Conpany shall loan the Estate of the Executive (or such other person \Ille> bears the turden of the estate am inheritance taxes involved) the amount of the Estate Tax Burden at the tine the taxes involved become payable, but a'lly upon ixesentation of sufficient inforniation to OOllpJte the Estate Tax Burden. 'Die loan shall be payable in equal installment:3 CNer the period cllring which Death Benefit payments are made, without interest, or, at the discretioo of the Calpany, may be withheld from payments which ,'°1ld otherwise be made. to the Beneficiary." In all other respects the Death Benefit Agreement remains in effect am unchanged • Pf WI'!NESS \tiEREDF, Calpany and Executi ~ have executed this Amenanent to Death Benefit Agreement as of the day .and year first above written • '1'llE 9Dl'l' ' FETZER aK'ANY Witness as to Executive: Executive 2 Ii • .. '!bis Amerdnent to Death Benefit Agreement lll!lde as of this .3J.£tdaY of January,·, 198~, ~ and between 'lbe Sclott ' Fetzer caipany (the •Cla!peny") and ---------------------- (the •Executive•). WHEREAS, (bnpany and Executive entered into a Death Benefit Agn:~t dated -of ' 19, and '.( ',', ~. the CCflpany desires to EEOVide to the Estate am Beneficiaries .. of the Executive the opportunity to obtain an Estate Tu loan to assist in the Jl c payment of estate am inheritance taxes ~ they bea:ate cileJ and WHEREAS, the Executive desires to obtain the opportunity fer his Estate and Beneficiaries to obtain such a loan; and WHEREAS, the (bnpany desires to place a ceiling oo the mount of the Death Benefit to which the Executive is entitled; and .. WHERF.AS, the Executive is willing to have such a ceiling plac:ed oo the 111DUnt of the Death Benefit to which tie is entitled; and WHERF.AS, (bnpany and Executive desire that the Death Benefit Agreement be supplemented am mnended as hereinafter 11et fbrth; tl:M, '111EREfORE, in ccnsideratioo of the premises the parties hereby mutually agree that the Death Benefit Agreement be anended as follows: 1. Paragraph 2(a) Pre-Retirement Death Benefit shall be deleted in its entirety am the .following paragraph 2(a) substituted in its place: •) •2(a) If death should occur to the Executive due to natural or accidental causes while ctively 91P~\ ~ the Caipany, er while . the Executive is disabled md eligible to receive disability '\:: 1. Bleaitive'a Initials: • • • () benefits under any of the <:atpany's group long-tetm disability plans, the <:atpany shall pay an annual Death Benefit to the Executive's Beneficiary equal to the lesser of, (a) the Executive's Base Salary, which shall be equal to the arount of the Base Salary indicated on the nost recently effective Schedule A, which is attached hereto and is specifically inex>rporated into this Agreement, or (b) the aliount of Fifty ibousand Dollars ($50,000.00). Such Death Benefit shall be paid to the Beneficiary for a period of years equal to the greater of (a) ten years or (b) the nuntier of years fran t.'le date of the Executive's death to the Executive's sixty-fifth birthday.• 2. The following ~ragraph (i) .shall 'be added to paragraph 1 Definitions: . •1. (i) Estate Tax Burden. "Estate Tax Burden" shall rrean the excess, if any, of. the cm::>unt of the United States and state estate or inheritance taxes inp:>sed upoo the Estate. or Beneficiaries of the Executive CNer the arount of such taxes which would have been inposed if the Death Benefit provided ~ this Agreement had not been included in the Executive's Estate. In the case of an Executive who was married at his death, In.lever, the Estate Tax Burden with respect to the United States federal estate tax shal1 be the lesser of the following arounts: (a) 34% of the part, if any, of the federal estafe tax value of the Death Benefit which is mt cpalified for the federal estate tax marital deduction; (b) 17% of the total of the Death Benefit as valued for federal estate tax purposes; and (c) the United States estate tax actually payable with respect to the Executive's estate. The <:atpany's determination 9f the. Estate Tax Burden shall be final and conclusive upon the Executive's Estate and Beneficiaries.• 3. The following paragraph "S(A) Estate Tax Loan" shall be added imnediately following paragraph S(c): 2. Executive's Initials: \:> I •s. (A) Estate Tax IDan. In the event that the beneficiaries of an Executive are entitled to an annual benefit under paragraph 2, then the c:oopany shall loan the Estate of the Executive (or such other person ..tio bears the l::urden of the estate and inheritance taxes involved) the anoont of the Estate Tax Burden at the tine the taxes involved becane payable, l::ut mly ~ p:esentation of sufficient informatim to CXlll'plte the Estate Tax Burden. The loan shall be payable in equal installments 011er the period chring which Death Benefit paJllllE!nts are made, without interest, or, at the discretim of the Conpany, l1li¥ be withheld from paJllllE!nts which would,,otherwise be made. to the Beneficiary.• In all other respects the Death Benefit Agreement remains in effect and unchanged. ?· J),. IN WI'lWFSS tii!EREOF, O:lnpany and Executive have executed. this Amenanent to Death Benefit Agreement as of the day and year first above written. 'mE scorr & FETZER QJMPANY By: . ...~~~~~~~~~ Witness as to Executive: Executive 3. Executive's Initials: ',, '\ tl .: About th,~ Company) ,- ,, -'\~~-,, \ ~~::-) Scott Fetzer, founded in 1914, is a diversified manufacturer '/ and marketer of high quality produds for the home, family and industry. The Comp'any's businesses are divid-ecj into five major operating segments: Cleaning Systems & Household Products, Education & lnfon;nation Systems, Energy & Control, Fluid 'lransmission, andVehicul;lr · Products. Headquartered in Lakewood, Ohio, the Company and its subsidiaries employ more'lhan 15,000 persons world-wide. Its decentralized management str,ncture allows each operating unit to operate as a" separate blisiness. Market-ing is accomplished through direct, in-the-home, mass mercfiandislhg, distributor and mail order sales·:· The largest and'best known among Scott Fetzer's product lines are Kirby home care systems, World Book •· encyclopedias, Campbell Hausfeld air compressors and Wayne burners and water pumps. . '\1 \ About this Report As you can see, this is an unusual annual report for Scott Fetzer. It is designerl justJo tell our shareholders and <',other interested parties liow the Company performed in 1982. In the next few months, another publication will be issued lbat will more clearly describe how the Company is being positioned for growth and how it plans to meet the challenges of the fut.ure. Together, these two docu-ments will provide a c1!(arer pictur\; of Scott Fetzer's preseQ.t firiantial condition and its plans for the future. - ---\ y----0 0 ! 0 {) Contents Qi'/ 1 Selected Financial Data • '' 2 Lett~r to Shareholders 3 Chairman's Perspective 4 Financial Review "'6 Management and Auditors' Reports "7 Consolidated Statement of Income 8, Consolidated Balani:e Sheet 9 Consolidated Statement of Changes in Financial Position 10 Consolidated Statement of Shareholders' Equity 11 Business Segment fnformation 13 Notes to Consolidated Financial Statements :. 19 Directors 20 Corporate Management/Information .. 21 Operating Units C--. 0 Selected Financial Data (Dollars in thousands except per share data) 0 ~) FjscalcYear Ended November 30 1982 1981 1980 879 1978 ' OPEMI'.JONS Net Sale5 and otherrevenue ...........•...... $600,008 $656,386 $632,398 $697,401 $478,222 Income before income taxes .................. 48,112 56,927 37,338 63,060 60,807 Net income ....• , .......•................. 27,092 29,067 23,069 34,096 30,247 ~ 4.5% 4.4% 3.6% 4.9% 6.3% Return on net sales ar.d other revenue .........• c PER SHARE DATA 0 Earnings .....••.......................... 4.04 4.01 3.12 4.62 4.11 Dividends paid .................... , , ....... 1.80 1.80 1.80 1.70 1.50 Book value ............................... 29.40 27.13 25.04 23.76 20.85 Oiviilends as a percent of net income ........... 44.6% 44.9% " 57.7% 36.8% 36.5% Market price range •........................ 36%-25.'2 303/.i-21 Vs 0 253/a-17 l-1' 30%-22 36Vii-23 Price/ earnings ratio ........................ 9-6 8-5 8-6 7.5 9-6 BALANCE SHEET DATA Current assets ............................ ' •·' 252,654 2zy,153 263,095 271,268 229,573 Current liabilities .........•................ 122,651 133, 104 117,459 128,067 131,036 Wi k;i . I 130,003 140,049 . 145,636 143,201 9~,537 or 1hg capita ......•.•................... ·;:,: Current ratio .•.. '. .....•................... 2.1 2.1 2.2 2.1 1.8 Property, plant'and equipment, net ............ 81,970 \ 82,827 83,609 78,651 62,236 'lbtal assets ..• ;• .............•............ ·: . I) ,:.. '" 403,790 405,794 394,043 395,376 337,977 Long-term debt ........•................... 72,786 76,901 79,595 82,346 46,036 Sharerlders' equity ......... ,, .............. 194,085 184,000 185,079 174,968 153,014 " Returl on shareholders' equity ................ 14.0% 15.8% 12.5% 19.5% 19.8% ~ "".· YEAR-ENO DATA t} ,, Shares outstanding (OOO's) ................... , 6,601 6,781 7,390 7,364 7,338 Average shares outstanding (OOO's) ....... 'J ... 6,708 7,252 7,388 ·7,373 7,354 Number of shareholders of record ..... <0.·ft·· •••• 6,393 7,105 7,937 8,119 8,439 Nu,mber of employees ...................... , 15,439 16,618 16,225 17,934 17,425 ~ I Common Stock Market Price and Dividend Information " Market Price of Common Stock Dividends Per Share 1982 1981 1982 . 1981 Fiscal Quarter, , ,,y High Low High Low First ................. : ............•....... $29.'s $27.'s $25V, $21 Vii $.45 $.45 Second ............ , ....... .' ............... . '30 253 /f 30Vii 23% .45 .45 Third .................................... . Fourth ................................... . Closing pric~ on December 31.~. : .................. . 31.'2 '25.'2 303 /i 275/s .45 .45 36% . ,30.'s 30 24V, .45 .45 > . , S28V, $35% "'\ 2. Letter to Shareholders Dear Shareholder: In September, 1982, I joined Scott Fetzer as president and chief operating officer. Since that time, I have visited most of the Company's operations, lear~ing about our businesses and meeting with management people in the. field. Witq,out exception, I hav_e been 0 impressed by the operating unit managements' dedication and enthu-siasm, and believe that with our ,, help, their commitment to growing their individual businesses is one of Scott Fetzer's most valuable assets. It is the strength of this commit-ment by our employees that enabled Scott Fetzer to weather the stormy eeonomic climate that battered the financial performance of many com-panies inJ982. Indeed, the economy had a negative effect on 11s as well, but it could have been much worse. In fiscal 1982, consolidated sales and other revenues were $600 mil-· lion, compared.with $656.4 million diiSt year. Net income was $27.1 mil-lion in 1982, moderately below the $29.1 million earned in 1981. Earn-ings per share increased slightly to $4.04 from $4.0l in the prior year. Obviously, as managers we are not pleased with this performance, and indeed have already taken additional steps to improve profitability in fiscal 1983. On the other hand, it is important to recog-nize that Scott Fetzer di.d well compared wiih most industrial companies. Operations Review The EDUCATION & INFORMA-TION SYSTEMS segment's sales were moderately lower while earn-ings were substantially higher. The drop in s~les reflected lower unif volume a't World Book and the absence in 1982 of the Japanese operations which were sold at the end of 1981. These decreases were partially offset by stronger results from I.he direct response business of World Book. , , It is important to note that 1981 earnings were depressed reflecting the provision of additional reserves' "" A. v to cover costs associated with the disposition of the Japanese opera-tions of World Book. As far as the future is concerned, we are currently looking at several new electronics-related products for World Book that look quite promis-ing, as well as other possibly interesting new ventures. The CLEANING SYSTEMS & HOUSEHOLD PRODUCTS seg-ment's lower results in 1982 were due to decreased unit volume at Kirby and continuing costs associ-ated with the start-up of two consumer finance companies and a new line of commercial vacuum cleaners. We believe that,]982 was the peak of start-up costHor the finance companies which will offer reliable consumer financing for sales of our Kir,by home care sys-temi} in the U.S. and the United """'. Kingdom. Many positive steps w~re comi;ileted in 1982. which are ex- \ pecte'ndltlon and Reawtll of Operations D Reawtll of Operatlona-1982 Compared with 1981 Scott FelZers consolidated sales and other revenue for 1982 totalled $600 million, compared with $656~4 million in 1981. The lower ' sales\yolume for 1982 reHected re-duced customer demand due to the continuing economic rece5sion and. the sale..ol \brld Books Japan operations~ . . . !';let income in 1982 feU6.8% to $27J m111ion from $29.1 million in the prioryear. Earnings per share for 1982 were $4.04 per share, up slightly from last years $4.0l per share.· . Earnings per share for 1982 were ''favorably affected by Scott Fetzer's common share repurch<!Sll pro-gram which began in May, 1981. Since then the Company has on an i.msolicited basis repur~hased " 790,425 shares for $22,2 million which has been charged to share-holders' equity. The average number of common and common equivalent shares outstandin'g h~ decreased 7.5% since N0Ve1nber " D 30, 1981. No shares were purchased .. • For the fourth quarter' of 1982, during the fourth quarter ol 1982; sales. were.$141. 7 million compared 0 The moderate decline in net in· with $163.5 million for the same come for 1982 was primarily due to the period in 1981. Net income was $6 decrease in sales volume. lower net " million, down from $6.4 million in interest income, and continuing start· the year earlier quarter. Earnings up costs a.5Sf)Ciated with two consumer per sharti were 91 cents &impared finance subsidiaries, partially offset by with 94 cents reported in the same a lower effective l!l" rate, Earnings in -. period a year ago. ' 1~81 wer~ ~~versely affected by a spe-Uquldlty and Capital c1al provision for closedown costs Resources associated with World Books interna· Ll 'd"ty • 11 d f ed Jionaloperations, particularly in · qui 1 ,is gt;~era Y e in as a .Japan'. Interest income in 1982 de-comp.anys ab1hty ~o genera~e oish ·dined to $12.1 million from $16.8 suflicte~t to,c?ver its pperatmg . million in 1981 'reflecting the overall .. needs (including capital expend!· ' decrease in .interest rates arl\'l a lower %. tl!r~) repay ~ebt, and pay cash level of cash as a rest:il! ofth~ sh'are d1v1dends to its shareholders. In repurchase progra)n andjn~'.reased ad· each of the past three years, ~OU vances.to unconsolftlated'§ubsidiaries. Fl)\Zer has.ger:ie.r~ted ful1ds to sup-The effective inc&rre. tax rate for 1982 port th~ acttvities entirely from • 7 % . . .\ ed 1 · . h· · · its operations. was 4~. compaR wit 48.9\! for Cash and cash equivalents the pnor year. T~e .• o.wer tax rate in . decreased by $16 million in 1982, 198~ r~ulted pnma~1ly from t~e eff~t follo\'ing a $)0.2 million increase of foreign taxes and increased invest· . in 1981. The decline in cash during ment. tax ~nd resear~h ~nd develop- . . . 1982 resulted primarily from th.e ment cred1~. The principal f~ctors that cinvestment in and loari to a n affectl!CI net income and earnings per· h. II ·. · ed'li . . b -d~w share for'1982 are summarized below· 1N ~ y-own •nanc.e s.u st tary, · · · Untied Acceptance Llm1ted, the Net . Earnings ,, Income' Per ($0005) Share ·common sharerepurchase pro-gram referred to above, and an investment in tax benefit transferc, leij.Ses. , · , . 1981-asreported .....•.. , .............. :·.· ..... ) •.. " $ 29,067 $ 4.0l ~ 'Working capital (current assel$ . Jess current liabilities).was $130 mil-.. lion at year-end 1982, representing (Decrease)Increase in 1982 from: . · · o Operation~· .. .. ·:·. '· ............... -. ......•.. ~· .,.,. .. ~ Interest income, net. 1 ...... ......•.• ; .....•. ? •••.•••• . Lower special provisidn for international operations ......•........ : , , •. : , ..... lntome tax changes: · Effect of foreign taxes ; ..................... ; ..... ., 'investment tax and research •· & deyelopment credits . : ............ : ..•........ Lower "QISC'' export credit , ................... , .. . Tu.~exempt interest'~ •... , . • .•...... , . . . . . . . . .. . Lower state and lo(al income taxes . , : .... , . . . . . ... . All other .... .";., ... ;,.· ....•....... , ...•.......... Net Change~ ...... -...... -.: .. _ . .'~~ ... ·.-;, ... ~· .. 1 •••• Decrease in'avei·age number of shares outst~nding ...................... , ... . ..... ; 1982 .... asreported , ..... , .. , , .......•. ·"' . .' . .' ...... . . :~ '( I, t :-> [) . ' -(4,013) (2,633) 2,028 1;647 0 355 "· (480) 724 , 191 206 (1,975) (.55) (.36) .23 .04 (.07J .10 .03 .03 (.27} (",·." -"·"' .30 $ 27,092 ' $ 4.04 ' ' a current ratio of 2.1, unchanged from the ptjor year. Working capital·• per dollar ofsales improved slightly in 1982. Accounts receivable were below the 1981 year.end level and inventories at year-end were slightly higher. The notes payable on the bal• . ance sheet represent the short term debt of foreign subsidiaries, whi.ch increased slighily from the prior . yearo(!nd . .The Company has aggre-gate credit facilities in various foreign countries of $7.9 million. o . 0 . . Long-term debt as a percentage . of total capital (including long-term debt) was 27;3% at theend of 1982, a decrease. from the 1981 level of 29.5 % • Current maturities of $4.6 . million will be met by funds 11rovided from operations. · Capital expenditures in 1~82 were $11.9 million •. only slightly more than Jhe $11.5 million expend-ed in 1981. · · Supplemental information.re-garding the impact of inflation is presented on pages 17.andl8 of this report. GeperaUy,. the Company strives to offset cost increases . through illlprove<l pr~uctivity, on-going cost reduction programs and selling price increases .. RQults of 0pe..;.ti0m.-. · 1981Compareclwltb 1980" Consolidated' sales and other reve- · . nue for 1981 totalled $656.4 miilio0, 4 % above the prior year total of, . $632.4 million. Sales volume for 1981 was affected by the economic downturn, the sale of the.Stream-way division, and the curtailment. of World Book's international busines5es. · , Net·income amounted to $29.1 million,26% above.the $23.1 mil-lion earned in 1980. Earnings pe~ share increased 29.% to $4.01 per shl!re compa[e<i'wi!h $3.12 per · · share in 1980. • The increase in net income for .1981 over the prior year was at-,... tributable to the 17 % improvement in earnings from operations, signifi· cantly higher net interest income, • · and a decrease .in special provisions associated with the closedown of . some o!World Book's international operations. The effective income tax rate was 48.9%"compared with 38.2 % in 1980.The lower tax rate ... In 1980 W'I$ prim1uily the. result. of · . significant tax cre<lits for worthless o stock deductions and the effect of · foreign tax credits. '6.' Business Seg.nents -1982 Compared with 1981 . · Results by business segment for · J982 and the two prior years are presented on pages 11 and 12 of this report. · . Sales and other revenue for 1982 decreased in four of the five busi-ness segments, with the most significant decline occurring in the. ,fluid Transmission segment which was most affected by the recession. Income before taxes was down in all bt!siness segments except for' E',ducation & Information Systems; · Cleanln11Sy1tem1 & Ho11se-bold Products ·Sales in 1982 were $119.6 million compared with ~)18 million in 1981. .Income before takes. was $11.3 mil-lion.in'1982, a decrease from the $15.5 m.illion in 1981. Results in this category were primarily affe<:ted by start'up costs associated with the finance subsidiaries and the intro-duction of a new line of commercial vacuum cleaners. Education.& Information Sy1teD11 ' Income before taxes in J982 amounted to $17.4 milhon, nearly ., 70% abbve the $10.3 milli·on re-.,, ported in 1981,,.Earnirigs in 1981 , · · were reduced by provisions for an· ticipated losses. associated With the closedown of World Book's Japan operations. Sales and other reve-nues declined to. $235.6 million ·compared with $259.8 milliop in 1981. The decline primarily re-fiected the absence in 1982 of the " o " 0sales o!Jhe.Japcln op~rations, slightly·Jowef'aomestic and other international sales, partially offset by higher direct response business. Energy & Control Sales for 1982 declined 6 % to $85.1 million from a record $90.3 million in 1981. Income before taxes was $10.3 million in 1982 compared with $13.2 .million in the prior year. Sales volume declined at the Adalet and Halex divisions, while volume at the France and Northland divi-sions showed moderate increases. Sales. were ti~ld down by reduced commercial and .industrial con· struction ;md a drop.off in the mining and oil and gas industries. The earnings decline reflected the drop in sales volume at the Adalet and Halex divisions and substantial new product development costs at the Northland division. Fluid 'lhlnsnllssion Sales fi;>r 1982. were $100.3 million, a decrease of about' 20 % from the $124.6 rriilliori last year. Income before taxes was approximately $7.2 million, about 32% below the $10.6 million earned in 1981. The sales decline primarily occurred at the Campbell Hausfeld division reflecting general economic condi-tions. Sales for the Wayne division in 1982were10% above the 1981 •· level. ~hicular Products Sales declined 7 % to $7 4.8 million in 1982, compared with last year's $80 million. Income before taxes was $10 million in 1982 compaJed with $10.4 million a year ago. Sales declined for all divisions in this segment with the exception of the Carefree division, )Yhich showed a ., 17% increase over 1981. The.Care-17 free, Douglas, Powerwinch and · __.,,, Sta. hi di.vi.sioml'reported. increasedV'·· ·· ;.:;::::-· ~. · earnings, while the Valley division07 experienced a sharp earnings decline. , 0 0 '•.<::.:: 5 n'. -; 6 Report of Management The management of The Scott & Fetzer Company has prepared the consolidated financial statements and related information included in this Annual Report and is respon· sible for their integrit)(, and object. ivity. The siatements have been prepared in co°liformity with generally accepted ac-counting principles consistently applied and as such in· elude amounts based on· estimates and judgments by management. Management is also re5ponsible for maintaining a sys-tem of internal accounting controls'which is designed to provide reasonable assurance at reasonable cost that assets are safeguarded against loss or unauthqrized use and that financial records. are adequate and can be relied upon to produce financial statements iii' accordance 'h;ith generally accepted accounting principles, The system ts supported by written policies and guidelines, by careful selection and training of financial management personnel . and by an · internal audit staff which. coordinates its activities with the Compar,iys indeiiendent certified public accountants .. ,,, Coopers & Lybrand, independent certified public ac· countants, are retained to examine the Company's financial statements, Their .examination is conducted i.n 0 Report of Independent · Certified Public Acc:Ountants To theShareholdel'S and Board of Directors The Scott & Fetzer Company: We have examined the consolidated balance sheet of The 'Scott & Fetzer Company a,nd subsidiary companies at November 30, 1982 and.1981, and the related consolidated .. statements of income, shareholders' equity, and changes in finanCial position for the years ended November 30, 1982, 1981 and 1.980; Our examinations were made. in accordance with generally accepted auditing standards and, accordingly, included such tests of the accounting records and such other auditing procedures as we consid· ~ ered,.necessaryin the circumstances. " •) accordance with generally accepted auditing standards and provides an independent ass«i!ssment that helps e.nsure fair presentation of the Company's financial Position, results of operations and changes in financial position. The Audit Committee of the Board of D.irectors is com-" posed entirely of outside directors and meets periodically with management, internal auditors and the independent ·certified public accountants. These meetings include dis-cussions of internal aceounting controls and the quality of financial reporting. Financial management, as well as the internal auditors and the independent certified public accountants, have full and free access to the Audit Committee. ' -R.E.Schey Chairman and , Chief Executive Officer Ii J. F:Bradley Executive Vice. President· Administration and Finance fl In our opinion, the finar,i,i:ial state!Ilents . referred. to above p~esent fairly the con5olidated flnancial position of The Scott & Fetzer Company and subsilliary companies at November 30, 1982 and 1981, and the consolidated, results of their operations and changes in financial position for the years ended November 30, 1982, 1981 and 1980, in con· formity.with generally accepted accounting principles ap-0 pUodoo•~ii:..~;~t41 Cleveland, Ohio --.,..--"1f""-r-.., January 25, 1983 0 . " 0 G Ji' CoD110llclaW Statement of Income {) : iii-c The Scott & Felzef Company and Subsidiary Companies (Dollars In thousands except per share data) ' . NETSALF.SAND OTHER REVE~UE •••••....•.••..••••. ; .......... . Cost of goods 5olense .... • _ ....... · ...... , ......... _ ....... · ........ _ ... . Charges for services of finance subsidiary (Nole 4) ..... , .... · ....•... · Income from unconsolidated subsidiaries and joint ventures • , ....•.. Other, net .. .- . ....... , ........ _ ........... _ ... _ ................•.... ' . . . ~' Jncomeffefore incometaxes ... •:•.·.'.., •••.. n; .............. • ••••••••• :,. Income taxesJNofe JO) .....•.•..••••....•..•••......•.•.. : ....• ' . NET INCOME •• ·: •••• 0 ••••••••••• ·~ •• ,,, •• ; ••• : ••••••• "''" ••••• . EARNING.$ PER SHARE •...• :· ••••••••• ' . : ..••••..•••.•. : ....••.••• ' 0 ' -:X) ,(; -(1 DIVIDENDS PER SHARE .•••.• , ................................. ... Average number of common and . , common equivalent shates outstanding (0005) ..•...........•....... • 0 Th~ accompan)i"8 noleS are an ln•al part of the financial Slalements. ., i\ _,. . 0 Year Ended November 30 1982 1981 1980 $600,008 $656,386 $632,398 327,899 366,691 355,485. 272,109 289,695 276,913 219,916 235,592 232,321 52,193 54,103 44,592 ' .· 12,132 16,808 8,119 (9.~) (9.~6) {9,858) (12,102) {7,650) (8,275) ,, :g:" 4,244 4,434 6,690-:J) (1,157)-· (1,452) (1,674) (4,081) 2,824 ,,(7,254) 4'112 '· 56,927 . 37,338 21,020 27,860 14,269 . ' 0 $. 27,092 $ 29,067 $ 23,069 -,~'' $ 4,04 $ 4.01 $ 3,12 $ 1.80 $. 1.80 $ 1.80 ~.708 7,252 " 7,388. " a "' ( ' 0 " 0 7 '!! 0 Conaolldated BaJance Sheet 0 The Scott & Fetzer CQmpany and Subsidiary Companies 0 (Dollars In thousands except peishare data) •.l ASSETS . Current Assets Ca5h .••....• •; ................... · • · · · · · · · · •. · · ·· · · · · · · · · ·" · · • · · · · · · · Interest-bearing deposits, .........• ; .. ,, ............................•...... -'-Short-term investments ................... ; ............................ . ~eivables(Note2) .• ':' .. . , .......................... ·; •........ '. ........ . Inventories (Note 3) •..•..........•.. "·' .......... , :' . . ,"', :· .......•........ Prepaid expenses •.... , .... , ....................... , ... ; .............. . TOTAL CURRENT A§SETS ...•.• , .............................. . ~ = ' 0 ' Investments in and advances.to unconsolidated subsidiarie5 arid joint ventures (Note 4) .....•......... , .................... ,, . Property, plant andequipment(Note5).;., ...•.•..• , ...•............. '; .......... . Ex ' .f. '' f ' ts . ed ' ' 0 ' ' ' . c~ cost over air v~,J.Je o asse acqu1r •.................... _ . , ......... ; .. . Other assets.: .• ,. 1~ •• •...• .' ..................... .· ...... ; .•.....•..•..• , ..• ., 1UTALASSETS· .. . _ ...... ,\ .................. -.. ·--·. ~---· .. -.. , ........ _ ·-·· ~- .-,, . UAB~ ~ ,, Current Llabilitie5 · 0 ' ' Notes payable (Note 6) ......•...... :; •.•......•. ; .............. , ..•. , .•... , AccountS payable .•..•.•............• ; , ... , : .. , ................ ; ... ; .. . Accrued employment costs . ., ........... o • ••••••••••••••••••••••••• '. •••• " Accrued liabilities .........................•....... , •................•... , . ; Income taxes .••. ,,. .• ............... , .•. , .................................. . Deferred inl:ome taxes (Notes 4'&10) ..•.................•...... ; ..•.... ,;, .. CJ Current portion of long-term debt.. " . , ......•.. ; ................... ~" ....... . TOTAL CURRENT LIABILl'IJES •....•..• ; ............ ; ....... "'· •.. Deferred income taxes(Note 10) ............................. ' .•.......•.. '' ...•.. Other deferred credits .... , .... ~'. .. 1 , •• , ••••••••••.••••• •· •••• , ••••••••••• , ••••• :·. . Long-term debt (Notes 6 & 7), ., .... , ......... , ... ··: ...... ,,. ........ '. ..•• , •....... TOT AL LIABILITIES . ; ..•..... '!' ••••• , •. " .••••..•••••• ,, •••••••••. fl . SHAREHOLDERS' EQUITV . , 0 Serial preferenc~stock · · 0 . . . . ' • 1uthoriz~ 1~000 shares without par value; none issued Commonstock ·.. . . . . Authorized 15,000;000 shares without par value; stated value , . . o of outstanding shares: $1.25 per share {Note 8) .... ,, ....•..•.....• "' ... ; ..... Ad .· ·d·t· a1· . 't I . · • , . · · "' " 1 ion. cap1 a ....•. , ......•.. , . , ....................................... . "Retained earnings(Note 6) ..... ,,, .. , .....•.. , • " .•..... ~ .. , ' •.•. , "'" ....... .' ...• . . , , ~TAL SHAREHOLD~RS' EQUITY .• ~ .... , ... ,· : ...... ·: .•.. '. ...... , • 0 · .· TOTALLIABILITIES AND SHAREHOLDERS' EQUITY ...•••.•...•• , .. . ,- -. ~ - ---' The acro.,mpaijy!ng r,ol;,. il{e an Integral pan Ot the financial ~~t~ments. ' ' ~· ' ' .''' ' ,, 0 n " 0, Q ,,, 0 Q " iJ " November30 1982 1981 $ 2,651 $ 8,924 46,163 . 72,246 ,. 37;504 21, 179 80,483 86, 117 ' 81,045 "'·"''" 79,611 ' 4,808 . 5,076 ' ' 252,654 273, 153 56'883 '--Ji 81,t\70 ' 2,62'& '9,658 $403,790 " $ 4,971 31,1,95 21,472 ' .33,449 6,1S6 20,549 4,559 122,651 8,444 5,824 72,786 209,705 i} 8,25.1 6,169 ··179,665 194,085 $403,790 0 40,702 > 82,827 2,642 q,47(., (} ~-· $405,794 !> •· ... f 4;505 34,~94 0 24,810 25,979 12,?44 28,269 2,602 133,104 7,480 4,309 76,901 221,794 8,476. 6,316 169,208 184,000 ---$405,794 11 Statement of Changes In Financial l\isltlon :Ompany and Subsidiary Companies . G Is) )PERATIONS . •. · .............. ,• ·-.... · ....... -. ............ ·' ........ ·-· .... . not affecting funds: on· .... _ .. ~~-; .. •.-~- .. ·.:~,, ..... -........... ., ....... -· ........ . 1t deferred income taxes .. : •............. "" .. ; ......•... ~m unconsolidated subsidiaries and joint vel)t~res .•.•........ . ' ,• -. -. . . . ·-........ , ................................... . .......... . !,(used) bychaliges in certain ciment'assets and liabilities: ~ .. :-.... ~,,~ ...... --. •.•. -........ ' ~ ...... ; . -· ·- .... ·-· .......... . ~. ' .. - ·- ...... ·-~·-. .--; .. ~ ·-·. ·-· .- -· ............ - . .. _ ... ·- •::- ...... . :penses ............ ; ..............•........ : .....•.. payable •....... , .... ,. • ..•...•....... , ..... / ...... , .. mployment costS, accrued liabilities and income taxes ... ; ... . nco.me taxes ... , .................... , .... / .......... . funds provided from oiierations .......................•.. ,. OTHER ~CES, EXCWDING FINANCING . rMENT ACTIVITIES disp0sitiqn qi division •..•••....•.•.........•. 1' ..•....... property, pli!nl and.equipment, net .... ; , .... : ........... . er deferred credits . "" ....•......... ; . . . . . . . . . . . . . . • . . . . · . · -, -.. ' . ' • ' , •- -, ~ o; • • ·•- ' • -• ' •- • • • • • • • • • • ' • ' ;, .• - • • • • • • • ' • ' • ·, ·- • • ·' • - '- • • • funds provided before 1ihancing and im 1estment activities .. :' .. EXCWDING FINANCING AND INVESTMENT ACTIVITIES -. --Q ' . opetty, plant and equipment. , •.•.... "'' •................. ·pa"1d · " · ~ ' ' -... ;. -. -..... ·-. __ ... -; '·~ .... '· ....... -. -· . •-. -..... " ...... _ ... . . ets of division .sold ....... , . , .•... , •.............. '· ; •..... . ,,:- .. ,~.·-··· .. : ... . ............................... -.......... . funds used before finand11g and investment activities .....•..• ~ . '· 0 • . • ROVIDED BEFORE FINANCING AND ITACTIVITIES .. ·•· •... : .' ... , ..............••. · ............. , DED FROM (USED IN) FJNANCING AND IT ACTIVITIES ~. . cllrrent maturities oflong-term debt ...... ; ....•..• , •....... n st~k under stock opti1ms, ...... , .......... , .. :' : .. : . ' .• !ase) 1~ho~t-term debt , ... , . , , ..... ,, ........ ; ... ~ , .... . easucyiShares .... , : ....... ·, ......•........ , ........... . tax leases netol current benefit .•......•................... , and advances to unconsolidated . . arid joint ven,tures, ........ , . ; ... , .... , ............ ·. , •. ,'.'. I funds used in financing and investment activities ..... : ... , .. .. ~ ....... ···.. . . IN CASH AND CASH EQUIVALENTS .. : ..•..•.... . ·: ...... . ·~-'·~ • --. '•1 •• ,-·· ·--,-• .-oot~_clr¢~ i~l~-P,cirt_of i_h~. fi~ariciaJ St~temeilt~~ . , . Year Ended November 30 1982 1981 1980 $ 27,092 $ 2~j067 $ 23,069 10,455 10,053 9,022 964 164' 1,768 (3,333)' (I,451) (386) 929 420 594 5,634 (845) 16,422 ' (1,434) 534 20,450 268 355 293 (3,099) . 7,083 (9,046) (2,056) 6,795 2,525 (7,720) 4,702 (3,244) 27,700 urchases of treasury sha1es~ ...... . Cash dividends-\ $1.80 per share ...... . . h -.Balances at November 30, 1982 " ...... '' . , . --,, e -•'iiea..ury shares are c:uried •I staled value. 7,576,924 $9,471 " The accompanying notes...; an Integral part ol the flnaJlcial statements. 0 . " ,:·~ .... ~~-.. 'Hf/• I (i (1,000) 180,900 I (226) 976,296 $(1,220) 0 " Additional Capital $6,573 311 ,6,884 (5!)8) 0 ·. 6,316 21 (168) $6,169 0 • Retained Earnings $159,190 23,069 (13.302). 0 168,957 29,067 .. (15,813) .. (13;003) 16~.208 27,092 (4,661) (11,9?4) . $179,665 IJ o· : \ . 'lblal $.174,968 23,069 ;,344 (13,302) ,, · .. 185,079 ' 29,067 (17,143) o. ,; (13,003)<~ .. 'n ' 184,000 .27,092 . '· 22. (5~55) • (l l,974) .\f'i,~ - ti -~ 'i; ' " ;·.ti ' $194,085 -"" r.-Bualneu Seplent lnformaUon by Industry ' " Tue.SCott & Fetzer Company and Subsidiary CQmpanlcs (Dolla~·in thousands) " Cr 0 Year EndedNovcmbcr 30 NET SALES AmfoTHER REVENUE (I Cleaning Systems & Household Products .....•...........• ,, ...•.. Education & lnformatiilri Systems .. v· .................. · ...... . :: · Energy&Conlro1 .. .... ~ ........ ~ ....... ~ ...... , .... ,,,', ... _ .... -.. -.. . Fluid 'lransmission ..... ·; .•.... "' ......... , .................. . Vehicular Products .••.•.......... 0 ••••••••••••••.••••••••••••• · · lntersegment Sales .••.... · .............•..• , •..... , .•... , . , . : . INCOME BEF<;i,IIB INCOME TA}(ES . . . .:. , ··. Cleaningii]stems &Hou'khold Products, .... , .•...... , ........... -Uncon~Hll!lted FinancecSubsidiaries ...•.... , ..........•...... ;k-" 0 0 0 Education& Information Systems .•.....•..• • •..... , .... , .. , .... . Uncc>nSolidatecl Finance Subsidiaries and. " · Joint Ventures. " " .................................... . (!,, E . &Co. 1· . ·. . ' ··' ''l:l . nergy . ntro ...... , .....•..........••................. _ •. Fluid 'Jtansmission ...............• , .... ·: ................... . .~ . Vehicular Products •.. , " ................ · .....•....... :. ....... " .•. ~ ,j, Operating Earnings ......•..•..•.•••............ ' ............ · Corporate ~pl!nses'and Net Interest , .. , ...•............. : ....•. IDENTIFIABLE ASSETS CleanillgSystems&Household Products.,, ........... •i ......... ,'; Unconsolidated'Finance Subsidiaries .. :· ..•... ,1, ••• ; ••••••••••••• . 0 . Educat.ion & Information Systems ... ~~-.... ,., .... : .....•........ Um:onsolidatecl Fin<!nce Subsidiaries and ' . Joint Ventures .•... : ............. ., .. · .......•.. , ...••.... =·· "' . . . ~ .. .Energy & Control .................................... 6 : .••••• · · Fluid Transmission ....................•..•.......• , .. · .. , . : ..... Vehicular Products .•.... : . o· ..................... " ............• • • • '-' >l Corporate and Other ..............•.... .' ....... : ............ . 1982 $119,583 235,559 r.1 I 85,055 100,310 74,791 (15,290) . . " $600,008 . $ 14,547 (3,226) ll,321 " . 10,107 7,31~, 17,423 10,300 7,170 9,969 \ 56,183 (8,071) $ 48,,12 (;, '" $ 41,5!!4. 15,25.0 0 °56~784 .c 107,276 r. ", P \F;.{ I>; tl:;.~ "\ 29,5~ 13,,782 37,00«l " " 62,055 31,939 79,230 l.0> ' $403,790 ) Q --· -: ~ ,. 1981 1980 ,, $118,020 $110,583 259,801 262,965 90,324 79,390 124,646 116,428 80,024 73,259 {16,429) (10,227) $656,386 $632,398 $ 16,951 $ 15,394 (1,486}"' (192) 15,465 15,202 5,898 2,742 I 4,370 ... 3,823 10,268 6,565. . 11,802" 0 13,156 10,557 4,016 10,446 8,836 59,892 46,421 "'(2,965) (9,083) $(56,927 $ 37,338 $ 45,631 $ 39,741 (905) (103) ' 44,726 39,638 .112,304. 108,103 ·~ 31,374 '28,747 143,678 136,850 37,513 33,345 63,283 67,646 34,455 28,605 ·82,1,39 . 87,959 ' $405,794 "$394,043 0 n : ,,~ ? (• 0 0 -- n: ~ -·- . ., ' c {• c . " " II l) .:~· ;~·-: . c ,, .. • ..•.. Year Ended Novernbe~, 30 J...... . . ~c • " . c.~J982 1981 @ · 1980 .1:>.·1·· ---·' ' $ J,67.7 c2,329 . 2 •. 379 3,203 . 1,235 3.828. ·. $s'12,l48 S611,331 s579,680 · .·s~;s8s · 57,337 63,390 O:i,ozs}' ·. · . ,J1.2,252~ • 1.ro.672> $600;@8 . $656,386. $632,398, • . -_Ii . ·<:!: ii>.· .. ·.. w ·o \· ~~~:~,-~;~~:~:~::~ \~· '·.:L·t:L~.'.:-:·.'-~~~~:~~~~~:.~ .. ~~-:~~~ .. :-. __ ~ .. ~-~.~~;,.~~~,;,:.i Noteslc) Consolidated Financial Statement8 ·• · (DOUa(S in tho~sands except per share data) " •1'·· ·"·. · :: ,. ·/r . -,,- ·-·: " . . . _ -· .· . !Ji 2. R~Vllhies . , t.~1gn1ftciintACclluntinlf Policies .. ~~ . . . . · Recltissification; ~. In 1982, the~olllpany !ldopt~ the cash ... · .. ·· fl?~·.c?pcept ()f. tlie.·S\atl!Iilent of,Chang;SJ~. Financicil --· ' ,. -__ .· ·--- !i' ' . Receivables at November 30; 1982and1981 consisted of the · following: . · . f'Os1,1i;in, Tile 198},and 1980 Statell)en~ have heel! re: •. . dassifi¥ to.con!Oiinwlththis concept: .. ·. .. • ; · .•.. ··.·· .•.• . . · '!lade receivables, .less altow~'ce . . .... Piina. 11.les.'.•·.o. r .. o. o. ~~o .. l.ldati···.on. "· Tl! ... '. e .. corisofo:i.a1ec1.·· fi.1.nancicil · · 10• doubtful ·=~ 01 • .·· . · • · · . . · · . · · · . . P!l82-$5;871; 1981-$5,551); .......... .. statements i.ncludeth~ac~ouilts !'>fall doJTlesticaildforeign lristallinenl receivables, less. . Q 1982 198!. . $54,490 . $57,292 sullSidiaries, except for the finance and insurance subsidi-alloWance !Or doubtful accounts ,, arles and j()intventuresiri .,.,.hichbwnershlp is5Q% or less, O!ll!2-S5,772; t98t ,-S6,005) .••.... ,.. . . . 13,679 16,861 ' wliichare tamed on the equify basis;;IDtercQmpany bcil-Otherret:eiVables, net ....... ··" .......... • '\ s~:!~ 0 s~:: , . aiices and tr3nsactiolis are elim,inatetl'in conSQliclation. . . • v Short-Term investri\ehts .~ ShorHerm l'rivestments, prln~ · Iri accordance with industry practi~e. total installment clpally co111me~i:iaI i:>apef!lnd governmeritsecurities, are receivables are included iri cun:ent ass~ts. The portions of ~ . '.· carried at ci>st,:whkh approximates market vcilue. \ ~.µch accounts due alter one year from~~;,he bcilance she.et" ·. ' inst(JlimentS<lles - Profits Oil instalinlbnt sales are creel~ . date a'!lounted to $5, 080 and $6, 452 at November 30, 1982 . 0 . ited tQ income at the liinebf sa!e.<Monihly finance charges arid 19"81, respectively. levied on domestic instcillinent accounts are ci'edited to inoome over the lives of the contracts, after dedui:ting a .· . prcivlsiori for estimated tincollectible clial"ges. ln/JentoJfes ~ lnyentories ares!ated at the lower of<:ost or marke.t CoSt has been deterinlned on a Jasf'iil. first.,Out . , (LIFO)basis for approxilllaiely 79% and 84% of totcil i~ventories in 1982 andJ981, respectlvelY. CQSt for the. , rem~ning inventory has been d~iermiried on a first-in, I ' · fiPro1rs~~(Fl[,O)tb~isd: .·· . .,, .. · .. ' .. • "t. ·, .. ·p··· .. ·. ·.rt· . I '·t··· · .. d. , ··. };·-.'. .,. .,. '.(ln an , .,qu,1~(!1en ::-.. rope y, ,,r ~ ~n ·. ·. "'=' eqmpment1s stated at cost Deprec1at1on and amortization 0 · are.c{)rnput!!d by the stiaight:iine methOcl using esth'!lated . . . useful lives of the'.~ts. . . .. . . . '\ . . . . . .. - --. - - -- ---- .-'· :--' -,· --.- -, .-. .- __ ---, - . -/ _ - - -' : . EJcceSS. Cost Over Faif Valuf!t df .Assets Acquired -'- The · 3. lilv'entorles Inventories at November 30, 1982 and 1981, consisted of the following: · ~ ~~ierials, .. ~ .. , .• ~ ~-. '. .... ·-· .. ~·· .... b ..... · \brk 1n-p_rocess ••••••••••••• ~ •.••••• -•• : •••• FiniShed gOOds ••••• ; •.•••.••• · •••••.• , ••••• 0 1982 . $22,919 3!,tl77 27,049 s81;045 1981 $26,067 27,830 . 25,714 $79,611 ·' --~ . . . . ..::;.:: ln1980, theCompjiny adopted theJast-in, first-out (LIFO) . ; . ~c~ C()St ()f lrivl)Stine11ts over.asset'S,acquired is being .amortized oh a,:sir.aight-Iine basis, principally over a'"40 . Q year-_pen_od1 :-· .. -::::~:'.~~~iZ: __ ::· - -:,: ,.::- ___ --=:--,;··'.·- -- --~ .. _ 11 - - · -·.1?~~;,ie.f.~~::·0Qe~.·1.ruces.o.n.i~i:6irte at~Pfoyid~ .. for methooAffoventocy vciluation for severcil operatiRg units that had previously been rep<irted under the first-in, first-out (FIFO) vciluati<>n !lletliC>CI. This change resulted in a $1,385 feduction Ofl980 netincomeor19 cents per share . The Company believes that the use of the LIFO inethoo · reduces the effeet <>I inflationary cost increases iri invento-0 ry and t}µs r~ults in,,a better m~tchipg of. costs against " · current Sciles. The. totcil impact of. LIFO on net income a.nd -· ' -' - .·-· . ~ l!mmg. differences. between finlll)l:•cil and tax·. reporting; ... a,risjng prilllaril}' from acceleratiid depreciatiori; .• infome .. . recogni.zed fro ht insirulinellt <,accounts recei$able,, pur-chased tax benefits, ~nd !ltcrued 'expenses nofcutrently • tax deductible. · · · • ·.' · · v . · ... · · ·· ·. . · · •. ' )~vestment tax credits are applied. to reduce the pfovi'; ..... ·· Si(Jti f()rfederal incollle taxes in ilie ye11r the credits arise. · · ·. • · faensions.-'- Ailnuai . pension costs charged t1>. incotrie in-. ,cltide actuarially°estiritated. curr~nt service, costs ~d provi-csi!)ns f()r \jripmvided, or o.vei'J>rQvlded acttlaficilllabilities, . . . ~hich are ainortiZedOv~\ ~ri~ of fifteen to. forty years. · Thl!·chmpanys general p()icy is 1 to fund the minimum · amount required 1,1ridef ERISA'. · • net income ~r share is presented in the following table. . GiQss UFO eifect · 11 Am0unt..; : •...... , ...... ... Per share:'., .. ," .. ,.: •... - Less incie~-ln.nft_ incoine related_ to i_riVeniorY JfQuMatlonSof UFO quaniiiies ati 0 If FIFO costs had been used for inventories presently • . • ' ' • .. ' • '. -' . !) valued usmg the,UFO method, mventones wouldbave been $27 ,023 and$27,354 higher than re parted at Novem· ber 30, 1982 and 1981, w.spectively. ' · .· -, .-.--. __ >"-~::=~.--',. --·--· - ·:-' :'[) .-. ,, .. 4, lnftlltmellts ln·andAdvBD.ces to Unconsolidated .subeldiarlea and®mt\ea~ " :· ---- - .-. - -- -. -- , . -- -. N '111e inve5tments in ancl advances to wholly-owned uncoli· solidated subsidiari~ and jlljnt ventures in whicll:owner, ship is ~0% or less are carried on the equity basis and at' November30, 1982and1981, were as follows: o () f"maOCesubsidiaries. ~ ........ _ •. .-......... . 1982 $44,516 12,126 1981 $30,614 10,233 c . \brld BOok Uie lrisurance ....•• ; ••••••••••. --,,. .IOini vefitu,~ .. ~ ...............•......... ill $56,883 ~·· $40.702 ',.,-,' _cy The consolidated balance sheet included net amountS ·· receivable from unconsolidated subsidiaries of $3,314 and $2,118 at November 30, 1982 and 1981, respectively. II) 1982, the Company advance9 $9;107 and established . '. ··. ' ,'' .. ·~· .. • '·. . a new wholly-owned finance subsidiary, Scott FetZer Financial Services Company (SFFS). SFFS ha5 awholly-owned consolidated subsidiary, United Acceptance Lim· . ited, which provide5 funds to finance the installment r~ ceivables. from custom.ers of tile Kirby Group clistributors operating in th~ United Kingdom .. World Book Fffiance, Inc. (WBFI) provides funds prin-cipally to finance the domestic installment receivabies of \brld Book, Inc. WBFI has a wholly-owned co~lidated subsidiary, United ~etail Finance Company, wfifch. pro-. vides funds to linan~e the domestic installment receivables " . from customers of Kirby Group distribui01~, The Company· 0 15 obligated under. an operaiing agreemerii lo make availc ,, · able. to WBFI amounts stiffiCient so that e;irnings, as de, fined, are at least 150 % of fixed cha!'ges, prlmanly interest. The amo1,mts ptovidedw~re.$12,102in1982; $7,650 inl981 and $8,275 in 19Bo. The';;urrent liability for deferred in-come taxes in the consolidated balance sheet includes amounts related to installment receivables financed by WBFL Summarized financial s\atemenis of WBFI follow. \ 14 ,, \ D !I , . " ,, A-ta · Cash and c;i,ib•equivaletils., •••••...•.•• ; ••• Finance ieceivObles, net ol allowance -~o~c;:~l=: :: :-: : :~:'-~::~-:-·:':::::::::: :· Tu1al AssetS ...................... , •••••• Llabilltles ud F.qaity Aa:oun~ payable and other liabilities ..••. ;.: .• · Payables to affiliated company ........... : .. . cuirent portion ol lotig-tenn debt ••••••••••••• I.Dog.term debt ••••••••••••••••• • ••••••••• Scott Fetzer equity •••..•• : ...••••• , ••.•••... Toia! Uabililies and F.quity .. , , •••••••.•• ' November30 1982 1981 $ 29,855 - ,,( • $ 42;340 0 10i,078 c 89,518 2,957 ~ $133.890 $133.497 $ 6,961 $ 4,917 r.697 · 2;625 2,625 89,500 92,125 34,804 ' 32.133 $133,890 $133,497 Year Ended November 30 1982 1981 1980 Jbtal revenue: ................. . $22,428 .•• $15,139 $11,550 interest expenS. •..••••.•••••• Other expenses .•• , ..••••....• lncornetaxeS ...... . _ •. -.......• 11.022 ' 8,426 07;158 5,89J " 2,500 813 2,840 2,149 1,786 Nel income . .. : .............. ,. $ 2,671, $ 2,064 $.1,793 s. ~ i>llmtuilEqulpment . Property, plant and e<fuipment at NovemSer 30, 1982 and i98!, consisted of ihe'1 following: · d~~d 18nd irllproVementS~ ~ ~ ~ . ~ .......• ~. Buildings •.•••.•.••..••.•••••••••.•..•••• Maichinery and equipment ...... ; ....••••.••• Capitalized leases .•••.•....•••..••••• ,. . ; • · 1982 $ 5,199 ,26,738 99,375 14,370 145,682 63,712 $ 81-970 '1981 $ 4,436 25,546 94,862 -~ 138,932 . 56.105 $ 82,827 6. Notes Pll~le and LOng-1.erfu Debt ~ The Co!Dpally hall borroWini!s of $4,971 and $4,506 at November 30, lg82 and 0l9Sl under bank credit ilnesavail' . able under' ccintractual arrangements -iri yanouftoreign counttj~. These.~redit lines totalled $1,857 and,$10,498 at these dates and. required commitment fees up to J. ~ .% per ;\ annum on the umised portion. . . - . . ., ' A stimniary oflong-term debt, at November 30, l982 and' · 1981, excltidirig the current p0rtion;folloW5: 9Y.i% Notes, payablein annual lnstallmebiS of $2,500beginning1983; due 1998 ..•••..•• 9~% Notes,due1985, ....•••••••••.... : •.. Other long.term debt with interest tales from 8 % toll% due In . , . installments l~ni 1987 ................... . Capitalized leases With interest () ratesfrom,,t:I; tQ16%.dueln · installmentS lhru 1989 ••••••.•...•..•.•••. ,.,., .. , 1982 $37,500 30;000 lio 816 " .4,470 $72,786 Q 1981 .(l <l '$40,00o 30,0oo n 0 0 {J 871 6,030 $~6,901 D <J The 9~% and 914 % ·ribte agreements coiltain cove. mints regarding increases ln debt, minimum asset~to-debt ratios and fue extent to which div~dends may be paid. Under the most restrictive of the debt c~venahts, retained earnings available for the payment of dividends amounted to $44,828 ai November 30; 1982. · · . · · · Aggregate maturities of long.term""'debt are $4,559, $4,063, $34,008, $3,497, $3,248 for the years ending 8 .. Stoek Options 0 The C\?mpanys 1973 and 1981 Stock Option Plans provide for the granting of options to key E!Xecutive employees to purchase shares oUhe Company\; common stock at not less .than 100% of its fair market value on the date of grant Options l'Jsued under the 1981 flan, which was approvedo by shareholders in March 1982, and certain .converted Ncvember30, 1983 through.1987, respectively.· 0 o options issu!!(I under the 1973 Plan are intended to qualify 0 as "lncentlv~ Stock Options""'~ defined in the'\intemal D " 7. Jeeinea &~) The Company, leases certain equipment, offices, ware-houses <ind piOductioh facilities. ~ ~evenue ~Code. All outstanding options .are exercisable one-fourtn each year commencing one year after date of grant and expire 10 years after date of grant. No 0 charge to income,ismade with respect to options granted. Q . _ _ o , ·I~ -• ·• , Both plans permit the granting of stock appreciall<m rights to ORtionees.As of November 30,198.2 no s.•1ch rights Lease terms for capitalized facilities and machinery and equipment are generally 15 years and 5 to 8 years, res~ ~V~f~ !f1e Company has options to purchase ~any of the cap1c(iiirul assets and can renew or purchase many I.eased ~currently classified as operating leases. . · had been granted. ·.· • ·· . . , . ·· . Changes ih options outstanding under these Plans dut' . ing the years 1980, 1981and1982 follow. Capitai leases, Which are accounted for.and aihortized as 0 Company-owned assets. at Nov~rriber 30, fJ2 and 1~81, 1173 CTIV'll Ol"l10N PLAN were as follows: · · . 0 ~ • ......,, · · · e 4 rJovember 30.1979 .•••.•.••••• ---..---~----,,..---· 1=982=·--.-~198=1.,.. Granted •.•••••• : •••••••••••• . Land and lahd improvemen!S ........ ,,. ••• ·• , • s 333 s. 333 · ':?Exercised "" .•.•••..••••.•••• 60 Buildings ••••••••••••••••••••• ,.......... 9,517 9 9,5qo, 0 "' Cancelled ••••••..••••••••••• Machineiy and equipment •..•.••. , • . . • . • • • • . 4,520 •. 0 4,255 0 November 30, 1980 ••••.•••• , .. . 14,370 14,088 Granted .• • .• ' •••••• ; ........ . Act11mulated amortization • • • • • • • . • • • • • . • • • • 8.fil!. 7.622 Cancelled •••••• ; •••••••.••.. • • . . $ 5,699 $ 6.466 November 30, 1981.. • ti • ..••.. . " 07: . . •. = (]ranted .••••••••••••••••• :~.' Future minimum lease p11yment5 Ullde~ noricancelable Exercised ••••••••••••••••••• capital and operatii;ig leases are: ~ 0 0 0 . o eance11.d • • • • · • • • • • • • · • • ., • · Capital!zfil Operating November 30, 1982 ........... . Year Ending November30 !.eos 7,657 $24.075 ··ss 1,679 CancellOd .•••••.• , .. • • . • • • • • @.650) 28 Novembei30, 1982............ 129.060 28-351/2 At Novemb!!r 30, 1982, 1~81, and 19J:,options exeri:isa-ble under the 1973 Stock Option Phm were 171,581, 139,264 and 79,869 respectively, and 5).3,213, 514,213, and 514,213 shares were re5erved for issuance. Shares available for fliture grants under the 1973 Plan.at the end of each' ofihe Prcs<.it valueolminumum lease three yei1r5 ~ere 254,023, 249,3~ and 242,338, . . · .. · =~~~t';!, 5 J,;~ ~-.............. .s 5• 880 1 ' Options. ei .. -~;.,,.,.;.-:. ;,~"'(.') o ~. •O-• ...... -.·, " 0 0 c 16 '.' " 0 0 0 9. Pension and Retirement Plarui ' The Company has various pension, profit'sharing and thrift plans, mostly non-contributory, covering substan-'~ially all of its en1ployees. It also has accrued .deferred compensatiOn, non~ of which has been funded, pur5uant to agl;eelnents with certain officers and other senior man-llgj!ment employees. o o Total pension eitpense for the year5 ended November · · 30, 1982, 1981, and 1980 was,$7,467, $6,587 and $7,363, respectively. During 1982, the Company changed certain actuaJial assumptions used for pension expense determin-ation and funding purposes to more closely approximate actual experience of the plans. These changes primarily consisted of increases in the; as5umed rate of return, pro-jected salary scales and fut~'fe Social Security benefits. The decrea5e in pension cost in 1981 was due principalty to an increase in the assumed rate of return on plan assets. A ~compari~n~ ofcaccumulated plan benefits and net assets for the domestic. defined benefit plans as of the most recent valuation dates, principally December 1, 1981 and 1980, is as follo~s: Actuarial present value of accumulated plan benefits: ::;Vested ••••••.•••••••••••••••••••..•••• Ncn·vesled ••••••••••••••• , ••••••••••.•• Total value ••.••••••••••••••••••••••.• Net assets, at market.value, available for benefits. 1982 1981 $19.596;> 3.710 $23.306 $32.976 $19,947 2,871 $22.818,~. $30,345 0 1) The rate of return Used in determining the actuarihl present value of accumulated plan benefits was 7~% for both years. The Company has a .limited number of employees wh~ participate in noncompany sponsored multi-emp!Oyer plans. IO. Income 18xes The provisions for income taxes were as follows: 0 Year Ended November 30 1982 1981 -r: '1980 -Current: Federal •.••• '" ........... . $14,910 $19,798° si4,I!J9 Stale and focal •••••••••••••• 2,986 3,984 2,343 Foreign., ................ . ~ 2,190 ___rrw 18,531 25,972 15,74S --,, 0 Deferred: -(3,241) Current •••..•••..••. t;;'I ••••• Noncvrrent ..... , ......•... 0 (2,848) S,S37 2.489 $21.020 " ,; 0 D 788 l.100 1,768 . 1.688 0 . (1,476) $27.860 $14.269 Cl 0 0 The deferred income tax provisions resulted from the follqwing: 0 .-Year Ended November 30 1982 1981 1980 Accelerated depreciation ••••••• $ 709 S !,OSI $ 1.892 Expenses deductible for lax .. {') .purpose; when paid., "0; • ., • ' · (845) (1,366) (4.4~) Tuxes defeiled on installmenl sales ..................... (2,346). ,, 2,078 3M Purchased tax benefits .• , •••••• 4,526 State and local income/franchise c tax timing differfnces .. ~ ..... 445 J!it) ~ 338 Other. net •••••..••• , •••••..• @ 383 0 ' $ 2.489 S 1.888 $(1.476) The effective income tax rate is reconciled to the statu-tory federal in,come tax r~te as f9il9w~:· ., a Year Ended November 30 1982 1981 1980 Statuiory federal tax rate •••••••• 46.0% 46.0%" 46.0% State and local income taxes net of federal income tax benefit •••• 3.3 3.8 3.3' Investment tax credit and ~ 0 research &development credit ' (2.5) (1.5) (3.4) DISC income not taxed ••••••••• (1.0) (1.7) (1.6)' Foreign operations t~ effect .·~ .. lB> 2.8 (1.2) \lifthless stock deduttlons -foreign operations. .......... ' (5.1) 0 ....ML Other, net ................... l:fil_ __ .2_ 43.7% 48.9% 38.2% During 1982, the Company entered into safe ha!'bor leasing agreements involving proper1Y with a cost basis of $26, 781. The $7,893 paid as consideration for the proper1Y has been included in Other Assets and is being amortized over the period during which the. corresponding tax bene: fits are to be realized. These transactions have had no effe1;t on net income in 1982 and will not have a material effect oil net income in any future period. Wfiile nOt reducing the provision for income taxes, cun:ent tax payments have been reduced by the $4,526 in tax benefits. realized from these transactions. 0 IL DllpOea1 of Operatlona Effective December 1, 1981, the Company,el\tered into an agreement to dispo5e of es..c;entially. all of its '\brld Book, Inc. operations in Japan. · · ' During 1981and1980, lhe Company charged operations for $4,000 and $6,000, respectively, to cover anticipated co5ts relating to this disposal and other aspects of '\brld · Book~ i11ternational operations. These charges priniarlly represent estimated future l0sses on the rtialization of net asset vatues. · · 0 . 0 , __ '?. ... '-0 ~ ~ 0 ,,, -1'\ I' ' ·iz. Bllalnee1 Segment Information Information with respect to the Company's. busines5 seg-ments is contained on page5 11 and 12 or this report. Segment data for 1980 has been restated to reflect the realignment of operatirigunits initiated in l98L Addi· tionally, certain financial data of unconsolidated finance . subsidiaries and joint ventures have been included with the segments with which they are vertically integrated. • · \brld Book Finance, Inc. (WBFI) is a wholly-owned unconsolidated domestic subsidiary of \brld-Book, Inc. (WBI). WBFI has a wholly-owned dOll\E!§tic fOnsolidated . subsidiary, United Retail Finance !;ompany (URFC), which was organized in 1980. URFC is vertically integrated with ·the operations of the Cleaning Systems & Household Prod-ucts segment. " Scott Fetzer Financial Services Company (SFFS} is a wholly-owned unconsolidated domestic subsidiary or the C<imi:iany. SFFS has a wholly-owned foreign consolidated subsidiary, United Acceptance Limited (UAL), which was organized in 1982. UAL is vertically ;ntegrated with the operations of the Cleaning Systems & Household Products segment. Wiii has. a whoily-owned unconsolidated foreign finance subsidiary i!lld also maintains two domestfo joint ventures on the equity basis. These .entities and WBFI, excluding URFC, are vertically integrated with the Education & Infor-mation Systems segment. •lntersegment and interarea sales ate accounted for at prices which generally approximate fair market value. Operating earnings are total revenue less operating ex· penses, excluding interest, ·general corporate expenses, and earnings of the unconsolidated life insurance subsidiary. ~ 13. Contingent LlabWtles The Company is a defendant in several lawsuits and other claims which, in the opinion of management;' will not have a material effect on the consolidated financial position and consolidated results ofoperations. 0 .· .. il ) 0 -14. Quarterly Information (Unaudited) Q 1982 Fiscal Year b~ guarler First Second Third Fourth Net sales and other revenue ..•....• 5142,801 5172,010 $143,505 $141,692 Gross p!tlfit ............ 61,825 78,164 . 66,285 65,835 Net income. ~ . -. ...... , . 5,375 8,895 6,i80 6,042 F.amings pei' share •••••• .79 e ).:i2 1.02 .91 '1981 Fiscal Year !!l: guarter First Second· Third Fourth Net sales and 0 other revenue, .•....• $153,639 5175,581 $163,687 $163,479 Gross profit ............ 68,870 74,424 67,596 78,805 Nerini:ome~······-····· 5,548 9;068 8,052 6,399 Earnings per share •••••• .75 1.23 ;o l,~09 .94 . .... 15. lmpactof Innatlon on the BualneSI (Unaudited) The Company's financial statements are prepared on a historical cost basis. This basis does not account for the effeets of inllation, either general or specific, on the results of operations or changes in financial position~The follow-ing supplemental financial i!iformation has beell prepared in accordance with the experimental techniques of Sta@ ment No~ 33 issued by The Financial Accounting Stand- ' ards Board. This statementrn<1Uires computation ofcertain v supplemeJllary informatioil' utilizing two methods, con-stant dollar and current cost. Both of these 11.1ethods inherently involve the u~ of assumptions, estiAfates and subjective judgments and should not.be viewed as precise indi~to!s of the effects of inflation on the Company. " .. The constant dollaf'method was computed by adjusting historical costs for changes in the purchasing power of the dollar as measured by the consumer Price Index for All Urban Consumers. The current cost of invent~ries was based on the 'first-in, first4>ut assumption as to inventory usage. The. current cast of property, plant and equipment was determined by application of the, Engineering News. Record Index :ror buildings and selected Prbducer Price Indices for other fixed assets. The CCl1J1pany's use of the last-in, first-out method of inventory valuation for most locations rµinimized the. adjustment to cost of sales under the current cost method. Depreciation appliCllble to both methods was COll\Puted with the same asset lives an.d in the same ipanner as used in the primary statements. Since inflation adjustments are not deductible for tax purposes, no adjustment was made to incomf,! tax expense=!W resulting in an effective tax rate of 57 .5 % for i:onst{int dollat~1:;,; 0 basis net income and 55.2 96 for current cost 'basis llet " income, compared with 43. 7 ~ in the primary statements. D ~-C) 0 D n. ) o. 0 0 ' 17 . 0 0 0 0 0 0 " u, . . . . . . . 0 . ·. Ii · . . .. ~·swnaeat~liicoaaeAd,lla.ledtorb.nihi& ~(Unaudited) FortheyearendedNovember30,1982 . 0 o As!l.ported " (Dollm in lhoosailds except~shareda!a} , !nl'rinWy a ~-Net!iletinddber~ ............ ~ ..... : ..... ·-·(;;-"~ ...... ..--~ ':": .......... ·i-~··~,;·-· •.... Costof fJiodsdd .. ltf, •• ~ •• ~ .-~ ...... _ ................ ·- •.• ........... \ •••• ·.,, ~. -••••• -~ •. ~ ••• ·~ .. Gro&s profit . 0 >I . -' ,,,· -··· -:0..·~,.···-·····:-···-···-·······-···············~~·::a··., ... . .. ::;-···:,••••t•••• Stl!1118; ~lnd~"'l"ll"'S •.••• ' .............. "' ....... '." ....•.... ~lnco.tie ~ ... :;):-... ~ ~: .. -..,. ... ···,~ .. _ .. ~.·~ ;' ... ~ .... -...... -... ~-............ ., .. . Otllrnt .•• "tr'·-· ••• • .••••••• , •• _ •••• , ................. '• •••• ·-;, • ~ ,t;:. •• -. ··~-· ...... •.• •••.•• -. . -. - . -a . ~ ::..:::e~~:;,:: :~ :: :::::: :: : : : : : :: :: : : ::::~: ::.:;::::: :.: : :::: :: : : :: 0 . "$600,008 0 327;899 272,109 2\9~916 " 52.193 0 ,, • .• " lll.975' (9,644) (12:102) .. 6,690 48.112 . 21.020, • 27,092 · 'lo455 --""-""" ......:.Ml! 0 ' 0 Adjusledlw Geneial !nftllion . (Comtant Dollar) c,~ $000,008 337.761 262.247 22t,6ill 40.~' 0 J0,97S .\9,6441 (12.102) 6.tiSo i1;j6,558 ,, Q . 21,020 D il5.S31l 'Ii ~ !g '>) $ .i.322 {) Qi' $600,008:. "" 336~ /i) 263,s.53 ' 221,fi6.'I• D 42,190·' 10,975. (9,644} (12,102) .6,690 38;109 21,020 !.:ill!!!l!.' . 'i6,0sl 2.55 S5.2\ll · • 155,3cl1 ~ . 11.538 ~ 10,373 ~ 1981 . 'iSllO 1979 1978 >terSalesaiilc:tbd~. ;: .Q~ : .... •. .-: .- ~:;, •· ............ ~ .: •••• -••• ~-•• ,.~:; •• ;,'D-o ~tdQUari~: · _ ·' 11 . -· • •• .
11 • (_I() C Netsale$andci..'ler~1Je. •• .,.::.,, ~~~. • • .,,., • •• .~~ •• • •~•·"'~"""'' }', •·Y• o Netiocorne. ·-• .. a· .... :~,.:~.·~-.• ,~.~.~··•· ••. ····•. .. ·"'·••··· ..... ····-~ ... :.~ ~ inc:orntptrahafe.. • • ~ • ;;r~ • • ..... <:!<." "' .~ .• • • • • • • • • ~ • ~ • .-. • • ,., ,' .-. • ~ • .·.. Neta:s:w:tsiil.yeai-elid ., .. ~··~ .... ~ ...... -. ..... .,. ........ ~ .....••.... •" .. .. I) -' ' ., 'Olrrent-illlormlllo!>! -·;. " . . q 0 ' <;J .Net~ ... '!. ·····~·t;o ............ · .•.•. : ............... , •.......... ;. " '~~~ .. ~.~·····•'"'""'"J~:~····· .. ·•················ -''i· ,, ·~~ 1~ .• .., ...... ,. ••••••••••• -!S·.;··----·-...................... f • . Eiooestolmci.....inCW11"1taJtlt . . . ~ ;:," • -· •· z:...a.: __ -· ·'' , ':f tN'tf l~ 1n general UPJill.MI, ,-: • .. , , , , , • , •, • , • , •. • ,, •, • ,.. , , , •" • .• ,. • 00...inlonnatiori: -Gain r~dedlne in puithaslngpow.r olnet_atnoUnts.a.witd ••••.•.. -. ...•...•.••..•..... ' .................... -..... '-0..\1 dividends per share-. 0 As~ ..... ······ ·····-·······--·········n···· .......... ' ..... -~ ..... . ",, Aveaget982doUari •••••••••... ••••• , •••••• ~ .............. ·,,. ............ · ... .. 1: 0 CommoriWlemamtpmat~-. . . ~.;............, . .. . . . . ' ,, . \~,~-·-· ................. ··.····-· .-... .,".. .... ,..,..,., ..... ~ ... ,., ...... •.•.• ..... -, AYefll! 1!)82dolJMs, .•• -.,., •.. , .•... • ••. ··•, ••. o, ~" ..... ...-,. ••. ,., , • ; , ; • ·-· -' .1 " ...... _ ..., ·-~ . . ' 0 n.veflle~l3Uluc:rr1a;e~~'•-.•-,.... ..................... , ... ~····•··~···•·• _ · ,., ,,,,:'."-. ~~··· 600.008 15,538 2.32 258,021 . lt.069 . 2.SS" -: 2$,307 655 1,:122 l.ilo 1.80 34.12 33.41 288.3 ., 0 6!'il.519 . 715,992' 1&,747 !l.40l . ..;µt. .27 24$;701 236,698 20.001 \,9l4 2.80 UI 24t.06l!n 232,949· (4,389) " (4,178) 2,\40 Q -:,3.742, i.80 1.80 1.112 2.i2 2825 23.~ 29.GI 26.44 270.5 :144.4 ' > 0 0 0 0 1.7!} 2-2&' 22.75 28.83 215.2 $478,222 -710,6" D "' .50 2.23 25.81· 36,92 i!H.O 0 " \j Q ,, 0 v \ (• (! -~. Directors ROBERf W. BJORK Managing Director of Stuyvesant capital Management Corp., Investment advisory firm CompenS<!tion Committee; Planning and Organization Committee; Nominating Committee, chairman J. F. .BRADLEY Exeeutive Vice President -Administration and Finance r1 GARY A. CHILDRESS President and Chief Operating Officer Planning and Organization Committee; Executive Committee STEPHEN H. FULLER, Ph.D. Professor. Harvard• Business School Audit,,Committee;o "Compensation Committee; Planning and Organization Committee JAMES A. HUGHES 0 Chairman, First Union P,eal Estate . .. Investments, Real· estaleinvestment tr·ust Nominating Committee; · · Compensatim1.Committee, chairman; Planning and Organization Committee; Investment and Pension Committee \'; \~ 0 D c 0 II /! I) LAWRENCE C. JONES Chairman and President, c Van Dorn Company Manufacturer of special purpose containers, plastic injection molding machinery, and heat treating of steel Audit Committee; Investment and Pension Committee, chairman; Planning and Organization Committee WALTER A. RAJKI Senior Vice President Investment and Pension Committee; Executive Committee RALPH SCHEY Chairman and Chief Executive Officer o Executive Committee, chai!t;"1an; Nominating Committee; ~ Investment and Pension Committee; Plar:min!l.and Organization Committee, chauma\1 · , c " ROBERf L. SWIGGETI Chairman and Chief Executive Officer, Kollmorgen Corporation Manufacturer of printed circuits, direct current motors and control systems, ccilor and. photometry instruments and electro-0ptical systems ' Investment and Pension Committee; Audit CommitteE!, chairman; · Planning and Organization Committee 0 JJ 0 jg ' c Corpor:ate Management " Corporate Information ?~' :~·" 0 .·,·~~; :'.' r'.'.:p ~ 0 0 0 0 .:;. ~ I·:!;; ~o ') -,':::;::, , ,- ... -~-··~ c ' RALPH SCHEY 0 KENNETH D. HUGHES Corporate omce .+. ' " .Chairman and Vice President arid 'iteasurer 146QQ Detroit Avenue Chief Executive Officer G Lakewood, Ohio 44107 IDBERlli(,:. WEBER Telephone: 216/228-6200 GARY A. CHILDRF.$ Vice President, . Telexf·~.to I 421-8128 "'\1 President and General Counsel, and Sec.<etary AnlltJll! Meeting r Chief Operating Officer c i;, The annual meeting of shareholders "' oO b THEODORE C'. BU$ " will be held on Tuesday, March 22, ' ~ J. F. BRADLEY Assistant 'll'easurer 1983, at 10:30 a.m. at the "71 Executive Vice President-Westlake Holiday Inn, Administration and Finance RICHARDE.HERfHNECK 1100 Crocker Road, " Westlake, Ohio 44145. " WALTF.FiA. RAIKI Assistant Secretary ~~ and Associate Counsel Form to.K Report . Senior Vice President B" " Copies of Scott Fetzer's Form 10.K KENNETH A. HOOK \. c ,report, flied with the Securities and D JOHN·BEBBINGTON Assistant Secretary ' " Exchange Commission, are available <'.' ,without chargP. uplln written request to Senior Vice President _if! and Associate Counsel '" )__, Robert C. Weber, Secretary of the " 0 0 •· .Company. KEARNEY K. KIER' CORRINE E. )IBYNOLDS· Z'I'' Group Vice President "'" Assistant Secretary 'lhw8ler Agent and Registrar &nd Associate Counsel Arneriliust Company KENNETH J. SEMEISBERGER "o ~ P. o. Bqi.t6fG .. Group Vice President " Clevelafl'il,·ohio .44101 ; . " ~ .· ~ Common Stock setiit Fetzer common shares are traded on .the, New York Stock Excha11ge, the Midwest Stock Exchange, and the Pacific Stock Exchange. The ticker symbol for the shares is SFZ . . ~;~;.: •.Q ~ __ ,,.-'" ~ , I G ':r-0 " 0 t}' C•$ ·~· ,5, 0 •if,"" ' ' ~~ t;',• 0 0 0 0 () v ,. [~ "' .I ,. a ~ 0 20 lb r;· " & " ,~. !:! " D ·~ ""' 9 \', , /'-r...-~ " ~' ' Operating Units n .-;-P I~ \ 0 ~ "' Adalet Division Klevac Division "') 4801 Wesft50th Street 2021 Mi~y:Drive " Cleveland, Ohio 44135 '!Winsoifqj; Ohio 44087 216/267-9000 216/ 425-3371 •" - -American Uncoln Divisim1 , Northland Division 1100 Haskins Road 96S Bradley Street J:!owling Green, Ohio 43402 Wdtertown, New York 13601 ,0 419/352-7511 3151782'.2350 0 Campbell Hausfeld, Division Powerwinch/Jason/Quikut!MedCare Division " , 100 Production Drive 217 Long Hill Cross R!lad 0 Hamson, Ohio 45030 Shelton, Connecticut Oti484 0 " 513/367-4811 203/929-5371 ~' Carefree of Colorado Division Stahl,Division c ;::;.-.r " 2760 Industrial Lane I . '0 3201 West Uncoln Way Broomfield, Colorado 80020 \boster, Ohio 44691 t3 303/ 469-3324 216/264-7441 c; Douglas Division Valley Industries Qjvision 141 Railroad Street 1313 South Stockton Street Bronson, Michigan 49028 Lodi, California 95240 517 /369-2315 209/368-8881 France Division Wdyne Home Equipment Division 726 Fairview Boulevard, West 801 Glasgow Avenue Fairview, Tennessee 37062 Fort Wayne, Indiana 46803 / 6151799-0551 219/ 426-4000 Halex Division Western Enter'i>rises Division 23901 Aurora Road "' .. 33672 Pin Oak Parkway ,, . ., Bedford Heights, Ohio ~4146 Avon Lake, Ohio 44012 2161439-1616 216/933-2171 ./ Kirby Group \brld llook, Inc. (a subsidiary) 1920West114th Street Merchandise Mart Plaza 0 Cleveland, Ohio 44102 Chicago, Illinois 60654 0 216/228-2400 312/245-3456 0 " ,, ,, 0 21 ! • •• " • • •• r \ ' 0 • Q (( Exhibit 22 0 There are oo parents of the Conpany. Its principal subsidiaries are as'' follows: a ,, Subsidiaries of Scott & Fetzer A. The Kirby Sales c.onpany, Inc. ~ A. Melben Products Co., Inc. A. SFZ International, Inc. ,, 0 0 A. SCOtt & Fetzer , (Canada) Ltd. A. 'Die Scott & Fetzer International Conpany 0::1 A. w:>rld Book, Inc. A. SFZ Transportation, Inc • 0 B. w:>r ld Book Finance, Inc. C. United R~tail Finance Conpany D. Scott,Fetzer Financial services ' Conpany 0 " c " 0 E. United l\cceptance Limited State of Incorporation Ohio :.:(,) ·l'llio ,. Canada Ontario Ohio () 0 Delaware Ohio Delaware Delaware ,, Delaware Delaware ;; .p % 6'f Stock Owned 100% 100% ' "' 100% ,:;,: 100% 100% 100% 100% ),1 100% 100% 100% 100% ·•-· A. Included in the Consblidated Financial Statements contained in the 1982 Annual Report to1;zhareholders of Scott Fetzer. ,, ,, B· Unconsolidated subsidiary of w:>rld Book, Inc., for 'Which separate financial statements are filed on pages F-8 through F~l6 of the 1982 Fonn l~K Report. c. Consolidated subsidiary of w:>rld Book Finance, Inc. D. Unconsolidated subsidiary of Scott Fetzer. 0 E. eonsolidat'ed subsidiary of Scott Fetzer Financial Services c.anpany. 12 "f 0 |
8864 | https://math.stackexchange.com/questions/1840075/the-intuition-behind-conditional-probability-and-independence-in-the-case-of-dif | The intuition behind conditional probability and independence in the case of different sample space - Mathematics Stack Exchange
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The intuition behind conditional probability and independence in the case of different sample space
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I came up with this question when doing this problem:
In throwing a pair of dice, let A be the event that "the first die turns up odd", B the event that "the second die turns up odd", and C the event that "the total number of spots is odd". Then since P(C|A)=P(C), we proved that A and C are independent (B and C are similar).
Normally, we can use Venn diagram to explain conditional probability. And the intuition for P(X|Y) is we can first denote the number of outcomes for XY and Y as n(XY) and n(Y) (graphically, the corresponding area), then both divided by the number of trials n, we can finally use (n(XY)/n)/(n(Y)/n) to approximate "in n trials, the probability that X happens given Y happens." Note that intuitively, we believe relative frequency converges to probability.
But in our case, the thing gets a bit complicated. The sample space for C is {2,3,...,12} and the sample space for A is {1,...,6} , they are different sample space! We can't explain in the traditional way of doing an experiment and analyzing the intersection of events. So I wonder the true intuition behind the definition of conditional probability (P(X|Y)=P(XY)/P(Y)), and the definition of independence (P(XY)=P(X)P(Y) or P(X|Y)=P(X))
FYI, try compare with a normal question like this one: Let A be the event that a card picked at random from a full deck is a spade, and B the event that it is a queen. Then A and B are independent.
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edited Jun 26, 2016 at 10:21
user14972
asked Jun 26, 2016 at 10:11
NicholasNicholas
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The sample space here is simply all possible rolls of two (distinguishable) dice.
There are various random functions on the sample space: e.g. the function
The value of the first die
The value of the second die
The sum of the two values
Your mistake is confusing the ranges of these random functions as being sample spaces.
While "the number is odd" is a proposition one can ask about integers, that is not what event C C is. Event C C is the application of that proposition to the value of the third random variable mentioned above, and is still an event on the sample space of all outcomes of two die rolls.
For example, event B B is true on the outcomes
{(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}{(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}
Usually, all of the random variables and events you consider at one time will all be defined on the same sample space.
In those circumstances where you had considered two separate problems, and then for some reason decided to consider them both again at the same time, you would combine the sample spaces together by taking the Cartesian product.
e.g. if you studied the first die roll on its own with sample space {1,2,3,4,5,6}{1,2,3,4,5,6}, then studied the second die roll on its own with sample {1,2,3,4,5,6}{1,2,3,4,5,6}, and then decided to study them both together, you'd switch everything over to the sample space consisting of all pairs of rolls.
e.g. if you had an event E={2,3}E={2,3} when studying the first die roll, when you switch over to the combined sample space, you would extend the event by just considering the first coordinate: the extension of E E would be {(2,1),(2,2),…,(2,6),(3,1),…,(3,6)}{(2,1),(2,2),…,(2,6),(3,1),…,(3,6)}.
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edited Jun 26, 2016 at 10:31
answered Jun 26, 2016 at 10:21
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8865 | https://www.khoury.northeastern.edu/home/viola/papers/bivsmod.pdf | Bounded Independence versus Symmetric Tests∗ Ravi Boppana† Johan H˚ astad‡ Chin Ho Lee§ Emanuele Viola§ June 1, 2018 Abstract For a test T ⊆{0, 1}n define k∗to be the maximum k such that there exists a k-wise uniform distribution over {0, 1}n whose support is a subset of T.
For T = {x ∈{0, 1}n : |P i xi −n/2| ≤t} we prove k∗= Θ(t2/n + 1).
For T = {x ∈{0, 1}n : P i xi ≡c (mod m)} we prove that k∗= Θ(n/m2 + 1). For some k = O(n/m) we also show that any k-wise uniform distribution puts probability mass at most 1/m + 1/100 over T. Finally, for any fixed odd m we show that there is an integer k = (1 −Ω(1))n such that any k-wise uniform distribution lands in T with probability exponentially close to |T|/2n; and this result is false for any even m.
∗A preliminary version of this paper appeared in RANDOM 2016 [BHLV16] †Research Affiliate, Department of Mathematics, Massachusetts Institute of Technology.
‡KTH-Royal Institute of Technology, work done in part while visiting the Simons Institute. Supported by the Swedish Research Council.
§College of Computer and Information Science, Northeastern University. Supported by NSF grant CCF-1319206. Work done in part while visiting Harvard University, with support from Salil Vadhan’s Simons Investigator grant, and in part while at the Simons Institute for the Theory of Computing.
1 Introduction and our results A distribution on {0, 1}n is k-wise uniform, aka k-wise independent, if any k bits are uniform in {0, 1}k. The study of k-wise uniformity has been central to theoretical computer science since at least the seminal work [CW79] by Carter and Wegman. A specific direction has been to show that k-wise uniformity “looks random” to several classes of tests.
These classes include combinatorial rectangles [EGL+92, CRS00] (for an exposition see e.g. Lecture 1 in [Vio17]), bounded-depth circuits, aka AC0, [Baz09, Raz09, Bra10, Tal17, HS16] (see e.g. Lectures 2-3 in [Vio17]), and halfspaces [RS10, DGJ+10, GOWZ10, DKN10].
More recently a series of works considers smoothed versions of the first two classes, where the input is perturbed with noise, and gives improved bounds [AW89, GMR+12, HLV, LV17a]. These results have in turn found many applications. For example, the recent exciting constructions of 2-source extractors for polylogarithmic min-entropy rely on [Bra10, DGJ+10].
In this work we extend this direction by giving new bounds for two classes of tests, both symmetric. First we consider the class of mod m tests.
Definition 1. For an input length n, and integers m and c, we define the set Sm,c := {x ∈ {0, 1}n : P i xi ≡c (mod m)}.
These tests have been intensely studied at least since circuit complexity theory “hit the wall” of circuits with mod m gates for composite m in the 80’s. However, the effect of k-wise uniformity on mod m tests does not seem to have been known before this paper. We study for what values of k does there exist a k-wise uniform distribution over {0, 1}n supported on Sm,c. Our first main result gives tight bounds on the maximum value of such a k, establishing k = Θ(n/m2 + 1). The constants hidden in the O, Ω, and Θ notation are absolute. The “+1” is there in case n/m2 is smaller than any constant.
Theorem 2. For all integers m ≥2 and c, there exists an integer k ≥n/32m2 and a k-wise uniform distribution on {0, 1}n that is supported on Sm,c.
Theorem 3. For all integers m ≥2, c, and k ≥140n/m2+4, no k-wise uniform distribution on {0, 1}n can be supported on Sm,c.
Theorem 2 is trivial for m = 2, as the uniform distribution over S2,0 is (n −1)-wise uniform. But already for m = 3 the result is not trivial.
Theorem 3 is equivalent to saying that when k ≥100n/m2 +4 then every k-wise uniform distribution must land in Sm,c with non-zero probability.
For motivation, recall from above the line of works [Baz09, Raz09, Bra10, Tal17, HS16] showing that k-wise uniform distributions fool AC0 circuits. Specifically, these works show k = poly log n suffices to fool AC0 circuits on n bits of size poly(n) and depth O(1). It is natural to ask whether the same distribution also fools AC0 circuits with mod m gates, a “frontier” class for which we have exponential lower bounds [Raz87, Smo87] (when m is prime) but not good pseudorandom generators. A positive answer might have looked plausible, given that for example the parity function is hard even with mod 3 gates [Smo87].
But in fact Theorem 2 gives a strong negative answer, showing that k = Ω(n) is necessary even for a single mod 3 gate.
1 Theorem 2 proves a conjecture in [LV17b] where this question is also raised.
Their motivation was a study of the “mod 3” rank of k-wise uniform distributions, started in [MZ09], which is the dimension of the space spanned by the support of the distribution over F3. [LV17b] shows that achieving 100 log n-wise uniformity with dimension ≤n0.49 would have applications to pseudorandomness. It also exhibits a distribution with dimension n0.72 and uniformity k = 2. Theorem 2 yields a distribution with dimension n −1 and Ω(n)-wise uniformity.
We then prove another theorem which is in the same spirit of Theorem 3 but gives different information.
First Theorem 4 (a) shows that the largest possible value of k in Theorem 2 is k ≤2(n + 1)/m + 2. Compared to Theorem 3, this result is asymptotically inferior, but gives better constants and has a simpler proof. Theorem 4 (b) shows that when m is odd, if k is larger than (1 −γ)n for a positive constant γ depending only on m then k-wise uniformity fools Sm,c with exponentially small error. The proof of Theorem 4 (b) however does not carry to the setting of k < n/2, for any m. So we establish Theorem 4 (c), which gives a worse error bound but allows for k to become smaller for larger m, specifically, k = O(n/m) for constant error. The error bound in Theorem 4 (c) and the density of Sm,c are such that it only provides a meaningful upper bound on the probability that the k-wise uniform distribution lands in Sm,c, but not a lower bound. In fact, we conjecture that no lower bound is possible in the sense that there is a constant C > 0 such that for every m there is a Cn-wise uniform distribution supported on the complement of Sm,c.
Theorem 4. Let m and c be two integers.
(a) For k ≥2n/m + 2, a k-wise uniform distribution over {0, 1}n cannot be supported on Sm,c.
(b) If m is odd, then there is a γ > 0 depending only on m such that for any (1 −γ)n-wise uniform distribution D over {0, 1}n, we have |Pr[D ∈Sm,c] −|Sm,c|/2n| ≤2−γn.
(c) There exists a universal constant C such that for every ε > 0, n ≥Cm2 log(m/ε), and any C(n/m)(1/ε)2-wise uniform distribution D over {0, 1}n, Pr[D ∈Sm,c] ≤ |Sm,c|/2n + ε.
We note that Theorem 4 (b) is false when m is even because the uniform distribution on S2,0 has uniformity k = n −1 but puts about 2/m mass on Sm,0, a set which as we shall see later (cf. Remark 1) has density about 1/m. The latter density bound, in combination with Theorem 4 (b) and Theorem 4 (c) implies that for some k = min{O(n/m), (1−Ω(1))n}, every k-wise uniform distribution puts probability mass at most 1/m + 1/100 over Sm,c for odd m and any integer c.
We then consider another class of tests which can be written as the intersection of two halfspaces.
Definition 5. For an input length n, and an integer t, we define the set Ht := {x ∈{0, 1}n : |P i xi −n/2| ≤t}.
Again, we ask for what values of k does there exist a k-wise uniform distribution over {0, 1}n supported on Ht. We obtain tight bounds for k up to a constant factor, showing that the maximum value of such a k is Θ(t2/n + 1).
2 Theorem 6. For every integer t, there exists an integer k ≥t2/50n and a k-wise uniform distribution over {0, 1}n that is supported on Ht, Theorem 7. For all integers t and k ≥36t2/n + 3, no k-wise uniform distribution over {0, 1}n can be supported on Ht.
One motivation for these results is to understand for which tests the smoothed version of the test obtained by perturbing coordinates with random noise is fooled by k-wise uniformity.
As mentioned earlier, this understanding underlies recent, state-of-the-art pseudorandom generators [AW89, GMR+12, HLV, LV17a]. See also [LV17b]. Using Theorem 6 we prove that independence Ω(log n) is necessary to fool read-once DNF on n bits, even with constant noise. Note that independence O(log n) is sufficient, even without noise [EGL+92].
Theorem 8. There exists a read-once DNF d : {0, 1}n →{0, 1}, a constant α, and an α log n-uniform distribution D such that | Pr[d(U) = 1] −Pr[d(D + Nα) = 1]| ≥Ω(1), where U is uniform over {0, 1}n, Nα is the distribution over {0, 1}n whose bits are independent and are set to uniform with probability α and 0 otherwise, and ‘+’ is bit-wise XOR.
Proof. Let d be the Tribes DNF with width w = log n −log ln n + on(1), see e.g. [O’D14].
We have |Pr[d(U) = 1] −1/2| = o(1).
Partition the n bits into n/w blocks of size w.
The distribution D has i.i.d. blocks. The projection of each block is an αw-wise uniform distribution with Hamming weight ≤2w/3. The probability that d(D + Nα) = 1 is at most the probability that there exists a block where the noise vector Nα has Hamming weight ≥w/3. This probability is at most (n/w)2w(α/2)w/3 ≤1/3, for a sufficiently small α.
1.1 Techniques We give two related approaches to proving Theorem 2. At a high level, both approaches are similar to the work of Alon, Goldreich, and Mansour [AGM03], which shows that one can apply a small perturbation to the probability masses of every almost k-wise uniform distribution on {0, 1}n to make it k-wise uniform, showing that every ε-almost k-wise uniform distribution on {0, 1}n is nO(k)ε-close to a k-wise uniform distribution, in statistical distance.
However, in their setting there is no constraint on the support.
This makes our proof significantly more technical.
Our first approach uses the following equivalent definition for a distribution on {0, 1}n to be k-wise uniform: the distribution is unbiased under any parity test on at most k bits.
To construct our distribution, we first start with the uniform distribution over the set Sm,c, and show that the bias under each of these parity tests is small enough, so that they can be made zero by a small perturbation of the probability masses of the distribution. Our goal is then to show that the change in the probability on each weight is no more than the probability we start with, so that it remains non-negative after the perturbation. In the conference version of this paper [BHLV16], we use this approach to prove a slightly weaker version of Theorem 2. We refer the interested readers to [BHLV16] for details.
3 We now give an overview of the second approach, which is developed in this paper. Instead of looking at the biases of parity tests, we consider another equivalent characterization of k-wise uniform distributions that are symmetric. To simplify the calculations, we will switch to {−1, 1} and consider distributions supported on S′ m,c := {y ∈{−1, 1}n : P i yi ≡c (mod m)}. One can then translate results for {−1, 1}n back to {0, 1}n (See Fact 12). A symmetric distribution is k-wise uniform on {−1, 1}n if and only if the first k moments of the sum of its n bits match the ones of the uniform bits. Similar to the first approach, we start with the uniform distribution on S′ m,c, and show that the first k moments of the sum of the bits are close to the ones of the uniform bits. Then, we perturb the probabilities on k +1 of the sums P i yi of the distribution to match these moments exactly. Once again, our goal is to show that the amount of correction is small enough so that the adjusted probabilities remain non-negative. Note that in this approach we work with distributions over the integers instead of {0, 1}n.
While the two approaches seem similar to each other, the latter allows us to perform a more refined analysis on the tests we consider in this paper, and obtain the tight lower bounds for both modular and threshold tests.
Organization.
We begin with Theorem 4 in Section 2 because it is simpler.
We use the second approach to prove the tight lower bounds for Sm,c and Ht in Sections 3 and 4, respectively. The proof of Theorem 2 involves a somewhat technical lemma which we defer to Section 5. Finally, we prove our tight upper bounds for Sm,c and Ht in Sections 6 and 7, respectively.
2 Proof of Theorem 4 In this section we prove Theorem 4. We start with the following theorem which will give Theorem 4 (a) as a corollary.
Theorem 9. Let I ⊆{0, 1, . . . , n} be a subset of size |I| ≤n/2. There does not exist a 2|I|-wise uniform distribution on {0, 1}n that is supported on S := {x ∈{0, 1}n : P i xi ∈I}.
Proof. Suppose there exists such a distribution D. Define the n-variate nonzero real poly-nomial p by p(x) := Y i∈I −i + n X j=1 xj .
Note that p(x) = 0 when x ∈S. And so E[p2(D)] = 0 in particular. However, since p2 has degree at most 2|I|, we have E[p2(D)] = E[p2(U)] > 0, where U is the uniform distribution over {0, 1}n, a contradiction.
Proof of Theorem 4 (a). When I corresponds to the mod m test Sm,c, |I| ≤n/m + 1.
We now move to Theorem 4 (b). First we prove a lemma giving a useful estimate of X x∈Sm,c (−1) Pk i=1 xi.
4 Similar bounds have been established elsewhere, cf. e.g. Theorem 2.9 in [VW08], but we do not know of a reference with an explicit dependence on m, which will be used in the next section. Theorem 4 (b) follows from bounding above the tail of the Fourier coefficients of the indicator function of Sm,c.
Lemma 10. For any 1 ≤k ≤n −1 and any 0 ≤c ≤m −1, we have X x∈Sm,c (−1) Pk i=1 xi ≤2n cos π 2m n , while for k = 0, we have X x∈Sm,c (−1) Pk i=1 xi −2n/m ≤2n cos π 2m n .
For odd m the first bound also holds for k = n.
Proof. Consider an expansion of p(y) = (1 −y)k(1 + y)n−k into 2n terms indexed by x ∈{0, 1}n where xi = 0 indicates that we take the term 1 from the ith factor. It is easy to see that the coefficient of yd is P |x|=d(−1) Pk i=1 xi, where |x| denotes the number of occurrences of 1 in x. Denote ζ := e2πi/m as the mth root of unity. Recall the identity 1 m m−1 X j=0 ζj(d−c) = ( 1 if d ≡c (mod m) 0 otherwise.
Thus the sum we want to bound is equal to 1 m m−1 X j=0 ζ−jcp(ζj).
Note that p(ζ0) = p(1) = 0 for k ̸= 0 while for k = 0, p(ζ0) = 2n. For the other terms we have the following bound.
Claim 11. For 1 ≤j ≤m −1, |p(ζj)| ≤2n cos π 2m k cos π m n−k.
Proof. As |1 + eiθ| = 2|cos(θ/2)| and |1 −eiθ| = 2|sin(θ/2)| we have |p(ζj)| = |1 −ζj|k|1 + ζj|n−k = 2n sin jπ m k cos jπ m n−k ≤2n cos π 2m k cos π m n−k , where the last inequality holds for odd m because (1) sin jπ m is largest when j = m−1 2 or j = m+1 2 , (2) sin( π 2 −x) = cos x, and (3) cos jπ m is largest when j = 1 or j = m −1. For even m the term with j = m/2 is 0, as in this case we are assuming that k < n, and the bounds for odd m are valid for the other terms.
5 Therefore, for k ̸= 0 we have X x∈Sm,c (−1) Pk i=1 xi = m −1 m · 2n cos π 2m k cos π m n−k ≤2n cos π 2m k cos π m n−k , and we complete the proof using the fact that cos(π/m) ≤cos(π/2m). For k = 0 we also need to include the term p(1) = 2n which divided by m gives the term 2n/m.
Remark 1. Clearly the lemma for k = 0 simply is the well-known fact that the cardinality of Sm,c is very close to 2n/m. Equivalently, if x is uniform in {0, 1}n then the probability that P i xi ≡c (mod m) is very close to 1/m.
Proof of Theorem 4 (b). Let f : {0, 1}n →{0, 1} be the characteristic function of Sm,c. We first bound above the nonzero Fourier coefficients of f. By Lemma 10, we have for any β with |β| = k > 0, | ˆ fβ| = 2−n X x∈Sm,c (−1) Pk i=1 xi ≤ cos π 2m n ≤2−αn, where α = −ln cos(π/2m) depends only on m. Thus, if D is k-wise uniform, |E[f(D)]−E[f(U)]| ≤ X |β|>k | ˆ fβ|· Ex∼D h (−1) P xiβii ≤ X |β|>k | ˆ fβ| ≤2−αn n X t=k+1 n t = 2−αn n−k−1 X t=0 n t .
For k ≥(1 −δ)n, we have an upper bound of 2n(H(δ)−α).
Pick δ small enough so that H(δ) ≤α/2. The result follows by setting γ := min{α/2, δ}.
Note that the above proof fails when m is even as we cannot handle the term with |β| = n.
Finally, we prove Theorem 4 (c). We use approximation theory.
Proof of Theorem 4 (c). Let f : {0, 1}n →{0, 1} be the characteristic function of Sm,c. The proof amounts to exhibiting a real polynomial p in n variables of degree d = C(n/m)(1/ε)2 such that f(x) ≤p(x) for every x ∈{0, 1}n, and E[p(U)] ≤ε for U uniform over {0, 1}n.
To see that this suffices, note that E[p(U)] = E[p(D)] for any distribution D that is d-wise uniform. Using this and the fact that f is non-negative, we can write 0 ≤E[f(U)] ≤E[p(U)] ≤ε and 0 ≤E[f(D)] ≤E[p(D)] ≤ε Hence, |E[f(U)]−E[f(D)]| ≤ε. This is the method of sandwiching polynomials from [Baz09].
Let us write f = g(P i xi/n), for g: {0, 1/n, . . . , 1} →{0, 1}. We exhibit a univariate polynomial q of degree d such that g(x) ≤q(x) for every x, and the expectation of q under the binomial distribution is at most ε. The polynomial p is then q(P i xi/n).
Consider the continuous, piecewise linear function s : [−1, 1] →[0, 1] defined as follows.
The function is always 0, except at intervals of radius a/n around the inputs x where g equals 1, i.e., inputs x such that nx is congruent to c modulo m. In those intervals it goes up and down like a ‘Λ’, reaching the value of 1 at x. We set a = εm/10.
6 By Jackson’s theorem (see e.g., [Car, Theorem 7.4] or [Che66]), for d = O(nε−1a−1) = O(nε−2m−1), there exists a univariate polynomial q′ of degree d that approximates s with pointwise error ε/10. Our polynomial q is defined as q := q′ + ε/10.
It is clear that g(x) ≤q(x) for every x ∈{0, 1/n, . . . , 1}. It remains to estimate E[q(U)].
As q′ is a good approximation of s we have E[q(U)] ≤2ε/10 + E[s(U)]. We noted in Remark 1 that the remainder modulo m of P xi is δ-close to uniform for δ = cos(π/2m)n = e−Ω(n/m2). Now the function s, as a function of P xi, is a periodic function with period m and if we feed the uniform distribution over {0, 1/n, . . . , m/n} into s we have E[s] ≤ε/10.
It follows that if n is at least a large constant times m2(log(m/ε)), we have E[s(U)] ≤2ε/10 and we conclude that E[q(U)] ≤4ε/10.
3 Tight lower bound on k-wise uniformity vs. mod m In this section we prove Theorem 2. For convenience, from now on we will consider the space {−1, 1}n instead of {0, 1}n. In particular, we will consider strings x ∈{−1, 1}n that satisfy P i xi ≡c (mod m). One can translate results stated for {0, 1}n to results for {−1, 1}n and vice versa using the following fact.
Fact 12. Let x ∈{0, 1}n and y ∈{−1, 1}n be the string obtained by replacing each xi by yi = 1 −2xi. Then X i yi = n −2 X i xi (mod m), and conversely, X i xi ≡ ( 2−1(n −P i yi) (mod m) if m odd, (n −P i yi)/2 (mod m 2 ) if m and n are even.
Let n be a positive integer. Let X1, X2, . . . , Xn be independent random variables chosen uniformly from {−1, 1}. Let B be the sum of all the Xj. The distribution of B is a shifted binomial distribution. Note that B has the same parity as n.
Theorem 13. Let m and n be positive integers and c be an integer. Suppose that m is odd or n and c have the same parity. Let k be a positive integer such that k ≤ n 8m2. Then there is a probability distribution on the c mod m integers that matches the first k moments of B.
Furthermore, the support of the probability distribution is a subset of the support of B.
Theorem 2 follows from applying Fact 12 to Theorem 13.
Our goal is to come up with a distribution supported on c mod m so that its first k moments match the moments of B. We first start with a measure q on the c mod m integers.
Here q may not be a probability measure — its values may not sum to 1. However, we will show that we can turn q into a probability measure p by a small adjustment ∆on k + 1 appropriately chosen positive values of q(x).
7 3.1 Defining Cm,c(x) Let m be a positive integer (the modulus).
Let c be an integer (the residue).
We will assume that either m is odd or n and c have the same parity. We will use Iverson bracket notation: JtrueK = 1 and JfalseK = 0. Define the comb function Cm,c on the integers by Cm,c(x) = mJx ≡c (mod m)K if m is odd and Cm,c(x) = m 2 Jx ≡c (mod m)K if m is even.
3.2 Defining q(x) Define the function q on the integers by q(x) = Cm,c(x) Pr[B = x]. Note that q is nonnegative.
Also if q(x) ̸= 0, then x is c (mod m) and in the support of B.
Lemma 14. If f is a function on the integers, then X x q(x)f(x) = E[Cm,c(B)f(B)].
Proof. By the definition of expected value, we have X x q(x)f(x) = X x Pr[B = x]Cm,c(x)f(x) = E[Cm,c(B)f(B)].
3.3 Defining Lagrange polynomials Let k be a positive integer. Let a0, a1, . . . , ak be k + 1 distinct integers that are c mod m, n mod 2, and as close to 0 as possible. Because they are as close to 0 as possible, we have |aj| ≤(k + 1)m ≤2km. In our application, 2km will be at most n. So each aj will be in the support of B.
Given an integer v such that 0 ≤v ≤k, define the Lagrange polynomial Lv as follows: Lv(x) = Y 0≤j≤k j̸=v (x −aj).
Note that Lv(aw) = 0 if and only if v ̸= w. It’s well known that L0, L1, . . . , Lk form a basis (the Lagrange basis) of the vector space of polynomials of degree at most k.
3.4 Defining ∆(x) Define the function ∆on the integers as follows. If x equals av (for some v), then ∆(av) = E[Cm,c(B)Lv(B)] −E[Lv(B)] Lv(av) .
For x ̸= aw for any w, then ∆(x) = 0.
Lemma 15. If f is a polynomial of degree at most k, then X x ∆(x)f(x) = E[Cm,c(B)f(B)] −E[f(B)].
8 Proof. We will first prove the claim when f is a Lagrange polynomial Lv. If ∆(x) ̸= 0, then x is of the form aw for some w. But if Lv(aw) ̸= 0, then v = w. So the sum has at most one nonzero term, corresponding to x = av. And the equation is true in this case by the definition of ∆.
We have proved the claim for Lagrange polynomials. But every polynomial of degree at most k is a linear combination of the Lagrange polynomials. This completes the proof.
3.5 Defining p(x) Define the function p on the integers by p(x) = q(x) −∆(x). Note that if p(x) ̸= 0, then x is c mod m and (assuming 2km ≤n) in the support of B.
Lemma 16. If f is a polynomial of degree at most k, then X x p(x)f(x) = E[f(B)].
Proof. By Lemmas 14 and 15, we have X x p(x)f(x) = X x q(x)f(x) − X x ∆(x)f(x) = E[Cm,c(B)f(B)] − E[Cm,c(B)f(B)] −E[f(B)] = E[f(B)].
To show that p is nonnegative, we will show that |∆(x)| ≤1 2q(x). First we bound above |∆(x)|. Then we bound below q(x).
Recall that ∆(x) = E[Cm,c(B)Lv(B)]−E[Lv(B)] Lv(av) if x equals av for some v (and 0 otherwise).
Lemmas 20 and 22 below give upper and lower bounds on the numerator and denominator, respectively.
3.6 Bounding above |E[Lv(B)Cm,c(B)] −E[Lv(B)]| We start this section by proving a few lemmas that will be useful to obtaining the upper bound.
Lemma 17. If θ is a real number such that |θ| < π 2, then |tan θ| ≥|θ|.
Proof. By symmetry, we may assume that θ ≥0. Recall that sec x is 1/ cos x. The derivative of tan is sec2. Because cos x ≤1, it follows that sec2 x ≥1. Integrating both sides (from 0 to θ) gives tan θ ≥θ.
Lemma 18. If θ is a real number such that |θ| ≤π 2, then cos θ ≤e−θ2/2.
Proof. By symmetry, we may assume that θ ≥0. The case θ = π/2 is trivial, so we will assume that θ < π/2. The derivative of ln cos is −tan. By Lemma 17, we have tan x ≥x for 0 ≤x < π 2. Integrating both sides (from 0 to θ) gives −ln cos θ ≥θ2/2. Exponentiating gives the desired inequality.
9 Lemma 19. Let r be an integer such that 0 ≤r ≤ n 8.
Let m be an integer such that 1 ≤m ≤√n. Suppose that m is odd or n and c have the same parity. Then E[BrCm,c(B)] −E[Br] ≤8(3rm)re−2n/m2.
Proof. Let m′ be m if m is odd and m/2 if m is even. Let α be π/m′. For now, we will assume that n and c have the same parity. At the end, we will show how to adjust the proof when n and c have the opposite parity.
Because B and n have the same parity, B −c is even. So we have the identity m′−1 X j=1 eijα(B−c) = Cm,c(B) −1.
Hence, by the triangle inequality and then Lemma 38, whose proof we defer to Section 5, we have (Lemma 38 is the second inequality) E[BrCm,c(B)] −E[Br] = E h Br m′−1 X j=1 eijα(B−c)i = m′−1 X j=1 e−ijαc E[BreijαB] ≤ m′−1 X j=1 |E[BreijαB]| ≤ m′−1 X j=1 2(8r|cot jα|)r|cos jα|n/2 = 2(8r)r m′−1 X j=1 |cot jα|r|cos jα|n/2.
The sum is symmetric: the terms corresponding to j = ℓand j = m′ −ℓare equal. So we can double its first half: E[BrCm,c(B)] −E[Br] ≤4(8r)r ⌊m′/2⌋ X j=1 |cot jα|r|cos jα|n/2.
10 Therefore, by Lemmas 17 and 18, we have E[BrCm,c(B)] −E[Br] ≤4(8r)r ⌊m′/2⌋ X j=1 1 jα r e−j2α2n/4 ≤4 8r α r ⌊m′/2⌋ X j=1 e−j2α2n/4 ≤4 8rm π r ⌊m′/2⌋ X j=1 e−j2π2n/(4m2) ≤4(3rm)r ⌊m′/2⌋ X j=1 e−2j2n/m2 ≤4(3rm)r ⌊m′/2⌋ X j=1 e−2jn/m2.
The sum is a geometric series whose common ratio is less than 1 2, so we can bound it by twice its first term: E[BrCm,c(B)] −E[Br] ≤8(3rm)re−2n/m2.
The proof above assumed that n and c have the same parity. When n and c have the opposite parity, we can adjust the proof as follows.
From the parity hypothesis in the theorem, we know that m is odd. In particular, m′ = m. Because B and n have the same parity, B −c is odd. So we have the identity m′−1 X j=1 (−1)jeijα(B−c) = Cm,c(B) −1.
It’s the same identity as before except for the factor of (−1)j. We can now continue with the remainder of the proof. The factor of (−1)j goes away as soon as we apply the triangle inequality. Hence we obtain the same bound.
Lemma 20. Let m be an integer such that 1 ≤m ≤√n. Suppose that m is odd or n and c have the same parity. Suppose that k ≤n 8. If v is an integer such that 0 ≤v ≤k, then E[Lv(B)Cm,c(B)] −E[Lv(B)] ≤8(5km)ke−2n/m2.
Proof. Given a subset A of {1, 2, . . . , n}, define XA to be the product of the Xj for which j is in A.
By expanding the product, we have Lv(B) = Y j̸=v (B −aj) = X A (−1)|A|aABk−|A|, 11 where A ranges over every subset of {0, 1, . . . , k}−{v}. Therefore, by the triangle inequality, Lemma 19, and the binomial theorem, we have |E[Lv(B)Cm,c(B)] −E[Lv(B)]| = E h [Cm,c(B) −1]Lv(B) i = E h [Cm,c(B) −1] X A (−1)|A|aABk−|A|i = X A (−1)|A|aA E h [Cm,c(B) −1]Bk−|A|i ≤ X A aA E h [Cm,c(B) −1]Bk−|A|i ≤ X A (2km)|A| E h [Cm,c(B) −1]Bk−|A|i ≤ X A (2km)|A| · 8(3(k −|A|)m)k−|A|e−2n/m2 = 8e−2n/m2 X A (2km)|A|(3(k −|A|)m)k−|A| ≤8e−2n/m2 X A (2km)|A|(3km)k−|A| = 8e−2n/m2(5km)k.
3.7 Bounding below |Lv(av)| Our lower bound on |Lv(av)| follows from the following claim.
Claim 21. Let a0 < · · · < ak be k + 1 points such that for every j ∈{1, . . . , k} we have aj −aj−1 ≥d. Then for any integer t such that 0 ≤t ≤k, Y j̸=t |at −aj| ≥ kd 2e k .
Proof. First observe that we have |at −aj| ≥d|t −j|. So we have Y j̸=t |at −aj| ≥ Y j̸=t d|t −j| = dkt!(k −t)!.
We will use Stirling’s formula in the form x! ≥ef(x), where f(x) = x ln x e for x > 0 and f(0) = 0. Note that f is convex on the interval [0, ∞). Hence we have Y j̸=t |at −aj| ≥dkt!(k −t)! ≥dkef(t)+f(k−t) ≥dke2f(k/2) = kd 2e k .
Lemma 22. If v is an integer such that 0 ≤v ≤k, then |Lv(av)| ≥ km 6 k .
Proof. Without loss of generality, assume that the aj are in sorted order. Then this lemma follows from Claim 21 with d replaced by m.
12 3.8 Conclude upper bound on |∆(x)| Lemma 23. Let m be an integer such that 1 ≤m ≤√n. Suppose that m is odd or n and c have the same parity. Suppose that k ≤n 8. If x is an integer, then |∆(x)| ≤8(30)ke−2n/m2.
Proof. If x is different from av for every v, then ∆(x) = 0. So we may assume that x = av for some v. By Lemmas 20 and 22, we have |∆(x)| = |E[Cm,c(B)Lv(B)] −E[Lv(B)]| |Lv(av)| ≤8(5km)ke−2n/m2 (km/6)k = 8(30)ke−2n/m2.
3.9 Bounding below q(x) Lemma 24. If a is an integer such that |a| ≤n and a ≡n (mod 2), then Pr[B = a] ≥2−a2/n 1 2√n .
Proof. The event B = a is equivalent to n+a 2 of the Xj being 1 and the other n−a 2 being −1.
Hence Pr[B = a] = 1 2n n (n + a)/2 .
If a = n or a = −n, then the desired inequality is easy to check. So we will assume that |a| < n. We will use Stirling’s formula in the form xxe−x√ 2πx ≤x! ≤xxe−xe√x , where x is a positive integer. We have n (n + a)/2 = n!
n+a 2 ! n−a 2 !
≥ nn√ 2πn ( n+a 2 )(n+a)/2( n−a 2 )(n−a)/2e2p (n + a)/2 p (n −a)/2 = 2H( 1 2 + a 2n )n 2 √ 2πn e2√ n2 −a2 , where H is the binary entropy function H(x) = −x log2 x −(1 −x) log2(1 −x). It’s known 13 that H(x) ≥4x(1 −x). That means H( 1 2 + a 2n) ≥1 −a2 n2. So we have n (n + a)/2 ≥2n−a2/n 2 √ 2πn e2√ n2 −a2 ≥2n−a2/n2 √ 2π e2√n ≥2n−a2/n 1 2√n .
Dividing by 2n gives the desired inequality.
3.10 Conclude |∆(x)| ≤q(x) Lemma 25. Let m and n be positive integers and c be an integer. Suppose that m is odd or n and c have the same parity. Let k be a positive integer such that k ≤ n 8m2. If x is an integer, then |∆(x)| ≤1 2q(x).
Proof. If x is different from av for every v, then ∆(x) = 0. So we may assume that x = av for some v. By Lemma 23, we have |∆(x)| ≤8(30)ke−2n/m2 ≤1 4e8ke−2n/m2 ≤1 4en/m2e−2n/m2 = 1 4e−n/m2.
By the definition of q and Lemma 24, we have q(x) = Cm,0(x) Pr[B = x] ≥m 2 Pr[B = x] ≥2−x2/n m 4√n .
We know that |x| = |av| ≤2km ≤n 4m .
So q(x) ≥2−n/(16m2) m 4√n ≥e−n/(16m2) m 4√n .
Applying the inequality x ≤ex/e (to x = 4n/m2), we have 4n m2 ≤e4n/(em2) ≤e3n/(2m2).
Thus m 4√n = 1 2 4n m2 −1/2 ≥1 2e−3n/(4m2).
Hence q(x) ≥1 2e−3n/(4m2)e−n/(16m2) ≥1 2e−n/m2.
Comparing our bounds for ∆and q, we see that |∆(x)| ≤1 2q(x).
14 3.11 Conclude lower bound Proof of Theorem 13. Recall the function p from Lemma 16. We will show that p is the desired probability distribution. From the definition of p and Lemma 25, we get p(x) ≥1 2q(x); in particular, p is nonnegative. Applying Lemma 16 to the constant function 1 (the zeroth moment), we see that the sum of the p(x) is 1. In other words, p is indeed a probability distribution. Applying Lemma 16 to the other monomials (namely x, x2, . . . , xk), we see that p matches the first k moments of B. This completes the proof.
4 Tight lower bound on k-wise uniformity vs. threshold In this section we prove Theorem 6. Like the last section, we will work with {−1, 1}n and translate the results back to {0, 1}n using the following fact.
Fact 26. Let x ∈{0, 1}n and y ∈{−1, 1}n be the string obtained by replacing each xi by yi = 1 −2xi. Then |P i xi −n/2| ≤t if and only if |P i yi| ≤2t.
Let n be a positive integer. Let X1, X2, . . . , Xn be independent random variables chosen uniformly from {−1, 1}. Let B be the sum of all the Xj. The distribution of B is a shifted binomial distribution. Note that B has the same parity as n.
Theorem 27. Let n and t be positive integers such that t ≤n. Let k be a positive integer such that k ≤ t2 200n. Then there is a probability distribution on the integers with absolute value at most t that matches the first k moments of B. Furthermore, the support of the probability distribution is a subset of the support of B.
Theorem 6 follows from applying Fact 26 to Theorem 27.
Let m be an odd integer between n 3t and n t . In Theorem 13, we constructed a probability distribution (call it p′) on the 0 mod m integers that matches the first t2 8n moments of B.
Furthermore, the support of p′ is a subset of the support of B. Looking at the proof, we see that p′(x) ≥1 2Cm,0(x) Pr[B = x] for all x. Let C′(x) be p′(x)/ Pr[B = x] if Pr[B = x] > 0 and Cm,0(x) otherwise. We have p′(x) = C′(x) Pr[B = x] for all x. Because p′ matches the first few moments of B, we have E[C′(B)f(B)] = E[f(B)] for every polynomial f of degree at most t2 8n. Also, C′(x) ≥1 2Cm,0(x) for all x.
Given an integer x, let T(x) be J |x| ≤t KC′(x).
4.1 Defining q(x) Given an integer x, define q(x) to be T(x) Pr[B = x]. Note that q is nonnegative. Also if q(x) ̸= 0, then |x| ≤t and x is in the support of B.
Lemma 28. If f is a function on the integers, then X x q(x)f(x) = E[T(B)f(B)].
Proof. By the definition of expected value, we have X x q(x)f(x) = X x Pr[B = x]T(x)f(x) = E[T(B)f(B)].
15 4.2 Defining Lagrange polynomials Let a0, a1, . . . , ak be k + 1 distinct integers that are 0 mod m, n mod 2, and as close to 0 as possible. Because they are as close to 0 as possible, we have |aj| ≤(k + 1)m ≤2km.
Because k ≤ t2 200n and m ≤n t , we have 2km ≤t ≤n. So |aj| ≤t and aj is in the support of B.
Given an integer v such that 0 ≤v ≤k, define the Lagrange polynomial Lv as follows: Lv(x) = Y 0≤j≤k j̸=v (x −aj).
Note that Lv(aw) = 0 if and only if v ̸= w. It’s well known that L0, L1, . . . , Lk form a basis (the Lagrange basis) of the vector space of polynomials of degree at most k.
4.3 Defining ∆(x) Define the function ∆on the integers as follows. If x equals av (for some v), then ∆(av) = E[T(B)Lv(B)] −E[Lv(B)] Lv(av) .
For x ̸= aw for any w, then ∆(x) = 0.
Lemma 29. If f is a polynomial of degree at most k, then X x ∆(x)f(x) = E[T(B)f(B)] −E[f(B)].
Proof. We will first prove the claim when f is a Lagrange polynomial Lv. If ∆(x) ̸= 0, then x is of the form aw for some w. But if Lv(aw) ̸= 0, then v = w. So the sum has at most one nonzero term, corresponding to x = av. And the equation is true in this case by the definition of ∆.
We have proved the claim for Lagrange polynomials. But every polynomial of degree at most k is a linear combination of the Lagrange polynomials. This completes the proof.
4.4 Defining p(x) Define the function p on the integers by p(x) = q(x) −∆(x). Note that if p(x) ̸= 0, then |x| ≤t and x is in the support of B.
Lemma 30. If f is a polynomial of degree at most k, then X x p(x)f(x) = E[f(B)].
Proof. By Lemmas 28 and 29, we have X x p(x)f(x) = X x q(x)f(x) − X x ∆(x)f(x) = E[T(B)f(B)] − E[T(B)f(B)] −E[f(B)] = E[f(B)].
16 We will show that |∆(x)| ≤1 2q(x). First we bound above |∆(x)|. Then we bound below q(x). Recall that ∆(x) = E[Lv(B)T(B)]−E[Lv(B)] Lv(av) if x equals av for some v and 0 otherwise.
Lemmas 34 and 35 below give upper and lower bounds on the numerator and denominator, respectively.
4.5 Bounding above |E[Lv(B)T(B)] −E[Lv(B)]| We start this section by proving the following lemma.
Lemma 31. Let d be a nonnegative integer. Then the (2d)th moment E[B2d] is at most (2d)!
2dd! nd.
Proof. The odd moments of each Xj are all 0. The even moments of Xj are all 1. Let g1, g2, . . . , gn be independent standard Gaussians (with mean zero and unit variance). The odd moments of gj are all 0. If c is a nonnegative integer, then the (2c)th moment of gj is known to be (2c)!
2cc! , the product of the positive odd integers less than 2c. In particular, the even moments of gj are all at least 1. So the moments of Xj are at most the corresponding moments of gj.
Let G be the sum of the gj. When we expand B2d and G2d, each gives a sum of n2d terms.
By the previous paragraph, the expectation of each term of B2d is at most the expectation of the corresponding term of G2d. Hence the (2d)th moment of B is at most the (2d)th moment of G. But G is a Gaussian with mean zero and variance n. In particular, G/√n is a standard Gaussian. So we have E[B2d] ≤E[G2d] = nd E[(G/√n ) 2d] = (2d)!
2dd! nd.
Lemma 32. If d is a nonnegative integer, then E[B2d] is at most √ 2 (2dn/e)d.
Proof. The case d = 0 is trivial (interpreting 00 as 1), so we will assume that d is positive.
Lemma 31 says that E[B2d] ≤(2d)!
2dd! nd.
To bound the factorials, we will use the following precise form of Stirling’s formula due to Robbins [Rob55]: if x is a positive integer, then xxe−x√ 2πx e1/(12x+1) < x! < xxe−x√ 2πx e1/(12x).
Hence we have (2d)!
2dd! < (2d)2de−2d√ 4πd e1/(24d) 2ddde−d√ 2πd e1/(12d+1) = √ 2 2d e d e1/(24d)−1/(12d+1) < √ 2 2d e d .
Lemma 33. Let n and t be positive integers such that t ≤n and t2 ≥200n. Let r be an integer such that 0 ≤r ≤t2 9n. Then E[BrT(B)] −E[Br] ≤2tre−t2/(6n).
17 Proof. Let s be a nonnegative integer such that r + s is an even number between t2 9n and t2 8n.
(Because t2 ≥200n, there is such an s.) We have J |B| > t K ≤t−s|B|s.
Hence, by the moment-matching property of C′, the definition of T, the triangle inequality, and Lemma 32, we have E[Br] −E[BrT(B)] = E[BrC′(B)] −E[BrT(B)] = E BrC′(B)J|B| > tK ≤E |B|rC′(B)J|B| > tK ≤t−s E |B|r+sC′(B) = t−s E Br+sC′(B) = t−s E Br+s ≤2t−s(r + s)n e (r+s)/2 = 2tr(r + s)n et2 (r+s)/2 ≤2tr 1 8e t2/(18n) ≤2tre−t2/(6n).
Lemma 34. Let n and t be positive integers such that t ≤n. Let k be a positive integer such that k ≤ t2 200n. If v is an integer such that 0 ≤v ≤k, then E[Lv(B)T(B)] −E[Lv(B)] ≤2(2t)ke−t2/(6n).
Proof. Recall that for a subset A of {1, 2, . . . , n}, we define XA to be the product of the Xj for which j is in A.
By expanding the product, we have Lv(B) = Y j̸=v (B −aj) = X A (−1)|A|aABk−|A|, where A ranges over every subset of {0, 1, . . . , k}−{v}. Therefore, by the triangle inequality, 18 Lemma 33, and the binomial theorem, we have |E[Lv(B)T(B)] −E[Lv(B)]| = E h [T(B) −1]Lv(B) i = E h [T(B) −1] X A (−1)|A|aABk−|A|i = X A (−1)|A|aA E h [T(B) −1]Bk−|A|i ≤ X A aA E h [T(B) −1]Bk−|A|i ≤ X A t|A| E h [T(B) −1]Bk−|A|i ≤ X A t|A| · 2tk−|A|e−t2/(6n) = 2e−t2/(6n) X A tk = 2e−t2/(6n)(2t)k.
4.6 Bounding below |Lv(av)| Lemma 35. If v is an integer such that 0 ≤v ≤k, then |Lv(av)| ≥ kn 9t k .
Proof. Without loss of generality, assume that the aj are in sorted order. Then this lemma follows from Claim 21 (with d replaced by 2m) and the bound m ≥n 3t.
4.7 Conclude upper bound on |∆(x)| Lemma 36. Let n and t be positive integers such that t ≤n. Let k be a positive integer such that k ≤ t2 200n. If x is an integer, then |∆(x)| ≤1 50e−t2/(12n).
Proof. If x is different from av for every v, then ∆(x) = 0. So we may assume that x = av for some v. By Lemmas 34 and 35, we have |∆(x)| = |E[T(B)Lv(B)] −E[Lv(B)]| |Lv(av)| ≤2(2t)ke−t2/(6n) (kn/(9t))k = 2 18t2 kn k e−t2/(6n) ≤1 50 1800t2 kn k e−t2/(6n).
19 The expression ( 1800t2 kn )k is an increasing function of k on the interval (0, 1800t2 en ]. Because k ≤ t2 200n, we have 1800t2 kn k ≤(1800 · 200)t2/(200n) ≤et2/(12n).
Plugging this bound into our previous inequality for |∆(x)| completes the proof.
4.8 Conclude |∆(x)| ≤q(x) Lemma 37. Let n and t be positive integers such that t ≤n. Let k be a positive integer such that k ≤ t2 200n. If x is an integer, then |∆(x)| ≤1 2q(x).
Proof. If x is different from av for every v, then ∆(x) = 0. So we may assume that x = av for some v. By Lemma 36, we have |∆(x)| ≤ 1 50e−t2/(12n). By the definition of q and Lemma 24, we have q(x) = C′(x) Pr[B = x] ≥1 2Cm,0(x) Pr[B = x] ≥n 6t Pr[B = x] ≥2−x2/n √n 12t .
We know that |x| = |av| ≤2km ≤2 · t2 200n · n t = t 100 .
So q(x) ≥2−t2/(10000n) √n 12t ≥e−t2/(10000n) √n 12t .
Applying the inequality x ≤ex/e (to x = t2 4n), we have t2 4n ≤et2/(4en) ≤et2/(10n).
Thus √n 12t = 1 24 t2 4n −1/2 ≥1 24e−t2/(20n).
Hence q(x) ≥1 24e−t2/(20n)e−t2/(10000n) ≥1 24e−t2/(12n).
Comparing our bounds for ∆and q, we see that |∆(x)| ≤1 2q(x).
4.9 Conclude lower bound Proof of Theorem 27. Recall the function p from Lemma 30. We will show that p is the desired probability distribution. From the definition of p and Lemma 37, we get p(x) ≥1 2q(x); in particular, p is nonnegative. Applying Lemma 30 to the constant function 1 (the zeroth moment), we see that the sum of the p(x) is 1. In other words, p is indeed a probability distribution. Applying Lemma 30 to the other monomials (namely x, x2, . . . , xk), we see that p matches the first k moments of B. This completes the proof.
20 5 Proof of Lemma 38 In this section, we will prove the following lemma that was used in the proof of Lemma 19.
Recall that X1, X2, . . . , Xn are independent random variables chosen uniformly from {−1, 1}, and B is the sum of all the Xj.
Lemma 38. Let r be an integer such that 0 ≤r ≤n 8. Let θ be a real number such that sin θ ̸= 0. Then |E[BreiθB]| ≤2(8r|cot θ|)r|cos θ|n/2.
Remark 2. By the triangle inequality and Lemma 31, we have |E[B2reiθB]| ≤(2r)!
2rr! nr, whether sin θ is zero or not.
Recall that for a subset A of {1, 2, . . . , n}, we define XA to be the product of the Xj for which j is in A.
We will need the hyperbolic functions. Recall that cosh z is (ez + e−z)/2, sinh z is (ez − e−z)/2, tanh z is sinh z/ cosh z, coth z is cosh z/ sinh z, and sech z is 1/ cosh z.
Lemma 39. If A is a subset of {1, 2, . . . , n} and z is a complex number, then E[XAezB] = (sinh z)|A|(cosh z)n−|A|.
Proof. Because B is the sum of the Xj, we have E[XAezB] = E hY j∈A Xj n Y j=1 ezXji = E hY j∈A XjezXj Y j / ∈A ezXji = Y j∈A E[XjezXj] Y j / ∈A E[ezXj] = Y j∈A sinh z Y j / ∈A cosh z = (sinh z)|A|(cosh z)n−|A|.
Lemma 40. Let A be a subset of {1, 2, . . . , n}. Let θ be a real number. If |sin θ| < |cos θ|, then |E[XAeiθB]| = (cos 2θ)n/2 E[XAeλB], where λ is 1 2 ln 1+|tan θ| 1−|tan θ|.
Proof. Because |sin θ| < |cos θ|, we have |tan θ| < 1, so λ is well defined. From our choice of λ, we have tanh λ = e2λ −1 e2λ + 1 = (1 + |tan θ|) −(1 −|tan θ|) (1 + |tan θ|) + (1 −|tan θ|) = |tan θ|.
It follows that cosh λ = 1 p 1 −tanh2 λ = 1 √ 1 −tan2 θ = |cos θ| p cos2 θ −sin2 θ = |cos θ| √ cos 2θ .
21 Hence sinh λ = tanh λ cosh λ = |tan θ| |cos θ| √ cos 2θ = |sin θ| √ cos 2θ .
Therefore, applying Lemma 39 twice, we have |E[XAeiθB]| = |sinh iθ||A||cosh iθ|n−|A| = |sin θ||A||cos θ|n−|A| = (cos 2θ)n/2(sinh λ)|A|(cosh λ)n−|A| = (cos 2θ)n/2 E[XAeλB].
Let r be a nonnegative integer. Let f be a function from {1, 2, . . . , r} to {1, 2, . . . , n}.
Define odd(f), the odd image of f, to be the set of j in {1, 2, . . . , n} such that |f −1(j)| is odd. Note that |odd(f)| ≤r and |odd(f)| ≤n.
Lemma 41. If r is a nonnegative integer, then Br = P f Xodd(f), where the sum is over every function f from {1, . . . , r} to {1, . . . , n}.
Proof. By expanding Br and exploiting the constraint that each Xj is ±1, we have Br = X f Xf(1) · · · Xf(r) = X f n Y j=1 X|f−1(j)| j = X f n Y j=1 X|f−1(j)| mod 2 j = X f Y j∈odd(f) Xj = X f Xodd(f).
Lemma 42. Let r be a nonnegative integer. Let θ be a real number.
(a) If r ≤n, then |E[BreiθB]| ≤nr|cos θ|n−r.
(b) If |sin θ| < |cos θ|, then |E[BreiθB]| ≤(cos 2θ)n/2 E[BreλB], where λ is 1 2 ln 1+|tan θ| 1−|tan θ|.
Proof. By Lemma 41 and the triangle inequality, we have |E[BreiθB]| = E hX f Xodd(f)eiθBi = X f E[Xodd(f)eiθB] ≤ X f |E[Xodd(f)eiθB]|.
We will use this inequality in both parts.
22 (a) By Lemma 39, we have |E[BreiθB]| ≤ X f |E[Xodd(f)eiθB]| = X f |sin θ||odd(f)||cos θ|n−|odd(f)| = |cos θ|n−r X f |sin θ||odd(f)||cos θ|r−|odd(f)| ≤|cos θ|n−r X f 1 = nr|cos θ|n−r.
(b) By Lemmas 40 and 41, we have |E[BreiθB]| ≤ X f |E[Xodd(f)eiθB]| = X f (cos 2θ)n/2 E[Xodd(f)eλB] = (cos 2θ)n/2 X f E[Xodd(f)eλB] = (cos 2θ)n/2 E[BreλB].
Lemma 43. If r ≥0 and λ is a nonzero real number, then E[|B|reλB] ≤2 4r |λ| r cosh 11 10λ n .
Proof. First we will bound |B|r by an exponential. Let a be 10r e|λ|. Let’s temporarily assume that r ̸= 0. Applying the inequality x ≤ex/e (to x = |B|/a), we have |B| ≤ae|B|/(ea).
Raising both sides to the rth power gives |B|r ≤arer|B|/(ea). Plugging in the definition of a, we have |B|r ≤are|λB|/10. This inequality is true for r = 0 too.
Now we are ready to prove the desired inequality. By Lemma 39 (with A = ∅), we have E[|B|reλB] ≤E[|B|re|λB|] ≤ar E[e|λB|/10e|λB|] = ar E[e11|λB|/10] ≤ar E[e11λB/10 + e−11λB/10] = 2ar cosh 11 10λ n = 2 10r e|λ| r cosh 11 10λ n ≤2 4r |λ| r cosh 11 10λ n .
23 Lemma 44. If λ is a real number, then |tanh λ| ≤|λ|.
Proof. By symmetry, we may assume that λ ≥0. The derivative of tanh is sech2. Because cosh x ≥1, it follows that sech2 x ≤1. Integrating both sides (from 0 to λ) gives tanh λ ≤ λ.
Lemma 45. If λ is a real number and c ≥1, then cosh cλ ≤(cosh λ)c2.
Proof. We will first prove that tanh cx ≤c tanh x for every x ≥0. The derivative of tanh is sech2. Because cosh is increasing on [0, ∞), it follows that sech2 ct ≤sech2 t for every t ≥0. Integrating both sides (from 0 to x) gives 1 c tanh cx ≤tanh x. Multiplying by c gives tanh cx ≤c tanh x.
Next we will prove the cosh inequality. By symmetry, we may assume that λ ≥0. The derivative of ln cosh is tanh. By the previous paragraph, we have tanh cx ≤c tanh x for every x ≥0. Integrating both sides (from 0 to λ) gives 1 c ln cosh cλ ≤c ln cosh λ. Multiplying by c and exponentiating gives the desired inequality.
Lemma 46. If θ is a real number such that cos2 θ ≥ √ 5−1 2 , then cos 2θ ≥cos6 θ.
Proof. From the hypothesis, we have cos2 θ + cos4 θ = cos2 θ(1 + cos2 θ) ≥ √ 5 −1 2 · √ 5 + 1 2 = 1.
Therefore, we have cos6 θ = 1 −(1 −cos2 θ)(1 + cos2 θ + cos4 θ) = 1 −sin2 θ(1 + cos2 θ + cos4 θ) ≤1 −2 sin2 θ = cos 2θ.
5.1 Main lemma, bounding E[BreiθB] Lemma 47 (Lemma 38 restated). Let r be an integer such that 0 ≤r ≤n 8. Let θ be a real number such that sin θ ̸= 0. Then |E[BreiθB]| ≤2(8r|cot θ|)r|cos θ|n/2.
Proof. We will consider two cases: cos2 θ ≤e−1/e and cos2 θ ≥ √ 5−1 2 . Because √ 5−1 2 < e−1/e, these two cases cover all possible θ.
Case 1: cos2 θ ≤e−1/e. Applying the inequality xx ≥e−1/e for x ≥0 (to x = 8r/n), we have cos2 θ ≤e−1/e ≤ 8r n 8r/n .
24 Hence, by Lemma 42(a), we have |E[BreiθB]| ≤nr|cos θ|n−r ≤nr|cos θ|3n/4+r = nr(cos2 θ)n/8|cos θ|n/2+r ≤nr8r n r |cos θ|n/2+r = (8r|cos θ|)r|cos θ|n/2 ≤(8r|cot θ|)r|cos θ|n/2 ≤2(8r|cot θ|)r|cos θ|n/2.
Case 2: cos2 θ ≥ √ 5−1 2 . In particular, |sin θ| < |cos θ|. Let λ be 1 2 ln 1+|tan θ| 1−|tan θ|. In the proof of Lemma 40, we showed that tanh λ is |tan θ| and cosh λ is |cos θ|/ √ cos 2θ. By Lemma 44, we have |λ| ≥|tanh λ| = |tan θ|. Hence, by Lemmas 42(b), 43, 45, and 46, we have |E[BreiθB]| ≤(cos 2θ)n/2 E[BreλB] ≤(cos 2θ)n/2 E[|B|reλB] ≤2(cos 2θ)n/2 4r |λ| r cosh 11 10λ n ≤2(cos 2θ)n/2(4r|cot θ|)r cosh 11 10λ n ≤2(cos 2θ)n/2(4r|cot θ|)r(cosh λ)121n/100 ≤2(cos 2θ)n/2(4r|cot θ|)r(cosh λ)5n/4 = 2(4r|cot θ|)r(cos 2θ)n/2 |cos θ| √ cos 2θ 5n/4 = 2(4r|cot θ|)r |cos θ|5n/4 (cos 2θ)n/8 ≤2(4r|cot θ|)r |cos θ|5n/4 |cos θ|3n/4 = 2(4r|cot θ|)r|cos θ|n/2 ≤2(8r|cot θ|)r|cos θ|n/2.
6 Tight upper bound on k-wise uniformity vs. mod m In this section we prove Theorem 3. It follows from Theorem 48 below by translating the statement for {−1, 1}n back to {0, 1}n using Fact 12. Recall that S′ m,c = {y ∈{−1, 1}n : P i yi ≡c (mod m)}.
Theorem 48. Let m be a positive integer and let c be an integer.
Let k be an integer greater than or equal to 4. Suppose there is a k-wise uniform distribution on {−1, 1}n that is supported on S′ m,c. Then k ≤140n m2 .
25 We can restate the theorem using moments as follows. Let B be the sum of n independent random bits chosen uniformly from {−1, 1}. Let Y be a random variable that is always c mod m. Suppose the first k moments of Y match those of B. Then k ≤140n m2 .
The high-level idea of the proof is as follows. We compare |E[e2πiY/m]| and |E[e2πiB/m]|.
The first is 1 because Y is always c mod m. The second we show is less than 1 2. We then take the Taylor approximations of the exponentials. The first k −1 terms are equal by the moment-matching property of Y . The error is given by the kth term, which gives us an upper bound on k.
Proof of Theorem 48. If m < 4, then k ≤n ≤16n m2 . So we may assume that m ≥4. Let α = 2π/m. Because m ≥4, we have 0 < α ≤π/2. For now, we will assume that k is even.
At the end, we will handle the odd case.
Because Y is always c mod m, we have E[eiαY ] = eiαc = 1.
By Lemma 39 (with A = ∅) and Lemma 18, we have E[eiαB] = |cos α|n ≤e−α2n/2.
By Taylor’s theorem, for every real θ we have eiθ − k−1 X j=0 (iθ)j j! ≤θk k! .
Hence by the triangle inequality we have E[eiαY ] − k−1 X j=0 (iα)j j!
E[Y j] ≤E h eiαY − k−1 X j=0 (iαY )j j! i ≤αk k! E[Y k].
Similarly we have E[eiαB] − k−1 X j=0 (iα)j j!
E[Bj] ≤αk k! E[Bk].
Because the first k moments of Y match those of B, we get a ton of cancellation: E[eiαY ] −E[eiαB] ≤αk k! E[Y k] + αk k! E[Bk] = 2αk k! E[Bk].
In particular, we have 1 = E[eiαY ] ≤ E[eiαB] + 2αk k! E[Bk] ≤e−α2n/2 + 2αk k! E[Bk].
Hence, using the moment bound of Lemma 31, we have 1 ≤e−α2n/2 + 2 αk 2k/2(k/2)!nk/2 ≤e−α2n/2 + 2 (k/2)!
α2n 2 k/2 .
26 Let f be the function defined by f(x) = e−x + 2 (k/2)!xk/2. Then the inequality above simplifies to f(α2n/2) ≥1. Note that f(0) = 1. Also f is convex on the interval [0, ∞).
We claim that f(k √ 2/8) < 1. To prove it, we will show that the first term of f is less than 1 2 and the second term is at most 1 2. Because k ≥4, the first term indeed satisfies e−k √ 2/8 ≤e− √ 2/2 < 1 2 .
The second term is 2 (k/2)!(k √ 2/8)k/2. If k = 4, then this term is exactly 1 2. Otherwise k ≥6, and so by Stirling’s formula we have 2 (k/2)!
k √ 2 8 k/2 ≤ 2 (k/2)k/2e−k/2√ πk k √ 2 8 k/2 = 2 √ πk e √ 2 4 k/2 < 2 √ πk < 1 2 .
In either case, the second term is at most 1 2. Hence f(k √ 2/8) < 1.
To summarize, f is convex on [0, ∞), f(0) = 1, and f(k √ 2/8) < 1. It follows that f is less than 1 on the interval (0, k √ 2/8]. Because f(α2n/2) ≥1, we have α2n/2 > k √ 2/8.
Solving for k gives k < 2 √ 2α2n = 2 √ 2 2π m 2 n < 112n m2 .
So far, we assumed that k is even. Now suppose that k is odd. We can apply the proof above to k −1, which gives the bound k −1 < 112n m2 . Because k ≥5, we have k ≤5 4(k −1) < 5 4 · 112n m2 = 140n m2 .
7 Tight upper bound on k-wise uniformity vs. thresh-old In this section we prove Theorem 6, which follows from Theorem 50 below by translating the statement for {−1, 1}n to {0, 1}n using Fact 26.
We will show that for any k ≥3, any k-wise uniform distribution over {−1, 1}n must put nonzero probability masses on strings x whose sums Pn i=1 xi are −Ω( √ nk) and Ω( √ nk) away from 0. This result shows that the lower bound we obtain in Theorem 6 is tight. We note that this is not true for k = 2, as when n is odd, there exists a pairwise uniform distribution supported on the all −1 vector and vectors with (n + 1)/2 ones.
Let X1, X2, . . . , Xn be independent random variables chosen uniformly from {−1, 1}.
Let B be the sum of all the Xj. The distribution of B is a shifted binomial distribution.
First we give a lower bound on the dth moment of B.
Claim 49. Let d be a nonnegative integer. Then E[|B|d] ≥( n(d−1) e2 ) d/2.
Proof. We first prove the claim assuming d is even. Let d = 2r. Consider expanding B2r, which gives us a sum of n2r terms. If for some index i, the variable Xi appears an odd number of times in a term, then this term has expectation zero. So the terms with nonzero expectation are the ones in which each Xi appears an even number of times. In particular, 27 each such term has expectation 1. It suffices to consider the terms in which either each Xi appears exactly twice or does not appear at all. There are n r ways of choosing the indices that appear twice in a term, and each term appears (2r)!/2r number of times in the n2r terms. Hence we have E[B2r] ≥ n r (2r)!
2r .
Using the inequality n r ≥(n/r)r and a crude form of Stirling’s formula, n! ≥(n/e)n, we have n r (2r)!
2r ≥ n r r2r e 2r 1 2r = 2nr e2 r , proving the claim for even d.
For odd d, let d = 2r+1. Then by Jensen’s inequality, we have E[|B|2r+1] ≥E[B2r] 2r+1 2r ≥ ( 2nr e2 ) r+1 2 .
Theorem 50. Let t+ and t−be two positive integers. Let Y be a random variable that is supported on {−1, 1}n so that P i Yi ≥−t−and P i Yi ≤t+. Let k be a positive integer.
Suppose that the (2k + 1)th moment of Y is equal to the (2k + 1)th moment of B. Then min{t−, t+} ≥ √ nk/3.
Remark 3. The conclusion is false when Y only matches the first two moments of B.
When n is odd, there exists a pairwise uniform distribution supported on the all −1 vector and vectors with (n + 1)/2 ones.
Proof. Let p+ and p−denote Pr[Y ≥0] and Pr[Y < 0] respectively. Note that E[Y 2k+1] = E[B2k+1] = 0. Together with Claim 49, we have p+ E[|Y |2k+1 | Y ≥0] −p−E[|Y |2k+1 | Y < 0] = E[Y 2k+1] = 0 p+ E[|Y |2k+1 | Y ≥0] + p−E[|Y |2k+1 | Y < 0] = E[|Y |2k+1] ≥ 2nk e2 k .
Summing the two relations, we have 2p+ E[|Y |2k+1 | Y ≥0] ≥( 2nk e2 ) 2k+1 2 . Hence, there must be a point y in Y such that y2k+1 ≥(nk/9) 2k+1 2k , and so y ≥ √ nk/3. By symmetry, there is another point y′ in Y such that y′ ≤− √ nk/3.
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Below are links to additional reading sources from the Congressional Research Service as well as select print secondary sources in the Library of Congress' collections on constitutional law utilized by CRS attorneys.
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Beyond the Constitution Annotated: Table of Additional Resources
A searchable table of CRS reports relevant to constitutional research and case law.
CRS Reports
This collection provides the public with access to research products produced by the Congressional Research Service (CRS) for the United States Congress. These reports were created for the sole purpose of supporting Congress in its legislative, oversight, and representational duties.
Books
Aspen Treatise for Federal Jurisdiction by Erwin Chemerinsky
Call Number: KF8858
ISBN: 9781543821116
Published/Created: 2020-11-16
In the Eighth Edition of Federal Jurisdiction, luminary author Erwin Chemerinsky unpacks the black letter law and underlying policy issues of his subject with the clarity and penetrating insight for which he is renowned.
Century of Struggle by Eleanor Flexner; Ellen Fitzpatrick
Call Number: HQ1410 .F6 1996
ISBN: 0674106539
Published/Created: 1996-03-01
Century of Struggle tells the story of one of the great social movements in American history. The struggle for women's voting rights was one of the longest, most successful, and in some respects most radical challenges ever posed to the American system of electoral politics.
Commentaries on the Constitution of the United States by Joseph Story
Call Number: KF4541 .S7 2005
ISBN: 9781584775157
Published/Created: 2005-07-01
Reprint of the second edition, with additions by his son, W.W. Story 1819-1895]. Originally published: Boston: Little, Brown and Company, 1851. Two volumes. xxxiii, 734; 632 pp. First published in 1833, this work is generally considered to be the most important work written on the American Constitution before the Civil War, and it remains an important work. Dedicated to John Marshall, it presents a strongly Nationalist interpretation. It is divided into three books. Book I contains a history of the colonies and discussion of their charters. Book II discusses the Continental Congress and analyzes the flaws that crippled the Articles of Confederation. Book III begins with a history of the Constitution and its ratification. This is followed by a brilliant line-by-line exposition of each of its articles and amendments.
The Constitution in Congress: Descent into the Maelstrom, 1829-1861 by David P. Currie
Call Number: KF4541 .C833 2005
ISBN: 0226129160
Published/Created: 2006-01-15
This acclaimed series serves as a biography of the U.S. Constitution, offering an indispensable survey of the congressional history behind its development. In a rare examination of the role that both the legislative and executive branches have played in the development of constitutional interpretation, The Constitution in Congress shows how the actions and proceedings of these branches reveal perhaps even more about constitutional disputes than Supreme Court decisions of the time.
The Constitution in Congress: the Federalist Period, 1789-1801 by David P. Currie
Call Number: KF4541 .C834 1997
ISBN: 0226131157
Published/Created: 1999-02-15
In the most thorough examination to date, David P. Currie analyzes from a legal perspective the work of the first six congresses and of the executive branch during the Federalist era, with a view to its significance for constitutional interpretation. He concludes that the original understanding of the Constitution was forged not so much in the courts as in the legislative and executive branches, an argument of crucial importance for scholars in constitutional law, history, and government.
The Constitution in the Supreme Court by David P. Currie
Call Number: KF4550 .C87 1985
ISBN: 0226131084
Published/Created: 1995-06-29
Currie's masterful synthesis of legal analysis and narrative history, gives us a sophisticated and much-needed evaluation of the Supreme Court's first hundred years.
The Creation of the American Republic, 1776-1787 by Gordon S. Wood
Call Number: JA84.U5 W6
ISBN: 9780807824221
Published/Created: 1998-04-06
The Creation of the Presidency, 1775-1789 by Charles C. Thach
Call Number: JK511 .T52 2007
ISBN: 9781330817124
Published/Created: 2015-07-06
Democracy and Distrust by John H. Ely
Call Number: KF4575 .E4
ISBN: 0674196368
Published/Created: 1980-02-19
Ely criticizes the two prevailing legal approaches to the Constitution and sets forth a proposal for determining the role of the Supreme Court today based on the view that the Court should assure majority governance while protecting minority rights.
Everyman's Constitution by Howard Jay Graham
Call Number: KF4558 14th .G7
ISBN: 9780870206351
Published/Created: 2013-05-31
In 1938, Howard Jay Graham, a deaf law librarian, successfully argued that the authors of the Fourteenth Amendment--ratified after the American Civil War to establish equal protection under the law for all American citizens regardless of race--were motivated by abolitionist fervor, debunking the notion of a corporate conspiracy at the heart of the amendment's wording. For over half a century, the amendment had been used to endow corporations with rights as individuals and thus protect them from state legislation.
For the People by Akhil R. Amar; Alan Hirsch
Call Number: KF4550.Z9 A43 1998
ISBN: 0684826941
Published/Created: 1998-02-02
Certain to provoke legal debates and seize the attention of readers and commentators across the political spectrum, For the People presents a dramatically populist interpretation of the Constitution which reveals the numerous, often overlooked powers it reserves for all citizens.
Impeachment by Raoul Berger
Call Number: KF4958 .B46
ISBN: 0674444752
Published/Created: 1973-01-01
The little understood yet volcanic power of impeachment lodged in the Congress is dissected through history by the nation's leading legal scholar on the subject. Berger offers authoritative insight into high crimes and misdemeanors. He sheds new light on whether impeachment is limited to indictable crimes, on whether there is jurisdiction to impeach for misconduct outside of office, and on whether impeachment must precede indictment. In an addition to the book, Berger finds firm footing in contesting the views of one-time Judge Robert Bork and President Nixon's lawyer, James St. Clair.
Law of Federal Courts by Charles Alan Wright; Mary Kay Kane
Call Number: KF8840 .W7 1983b
ISBN: 9780314927071
Published/Created: 2011-06-15
This work offers practical guidance and comprehensive coverage of all aspects of federal court jurisdiction and litigation procedure, as well as the relationship between the state and federal courts. The text reviews the federal judicial system; judicial power of the United States; diversity of citizenship; venue; law applied in federal courts; pleadings, trials, and judgments; and appellate court and Supreme Court jurisdiction.
No State Shall Abridge by Michael Kent Curtis; Willian Van Alstyne (Introduction by)
Call Number: KF4757 .C87 1986
ISBN: 0822305992
Published/Created: 1986-05-19
Religion and the American Constitutional Experiment by John Witte; Joel A. Nichols
Call Number: KF4783 .W58 2016
ISBN: 9780190459420
Published/Created: 2016-04-13
This accessible introduction tells the American story of religious liberty from its colonial beginnings to the latest Supreme Court cases. The authors provide extensive analysis of the formation of the First Amendment religion clauses and the plausible original intent or understanding of the founders. They describe the enduring principles of American religious freedom--liberty of conscience, free exercise of religion, religious equality, religious pluralism, separation of church and state, and no establishment of religion--as those principles were developed by the founders and applied by the Supreme Court.
The Second Amendment in Law and History by Carl T. Bogus (Editor); Michael A. Bellesiles (Editor)
Call Number: KF3941 .S43 2000
ISBN: 1565846990
Published/Created: 2002-01-01
Few issues of public policy are so misunderstood, so oversimplified--and so crucially important to the health and welfare of all Americans. The gun lobby and its proponents would have us believe that the constitutional issue is moot, and that the regulation of firearms is beyond the reach of legislation. But as the contributors to this important anthology demonstrate, both the historical and constitutional arguments are very much alive--and in fact weigh heavily in favor of those who would restrict gun ownership.
To Keep and Bear Arms by Joyce L. Malcolm
Call Number: KF4558 2nd .M35 1994
ISBN: 0674893069
Published/Created: 1994-02-01
Joyce Malcolm illuminates the historical facts underlying the current passionate debate about gun-related violence, the Brady Bill, and the NRA, revealing the original meaning and intentions behind the individual right to "bear arms." few on either side of the Atlantic realize that this extraordinary, controversial, and least understood liberty was a direct legacy of English law. This book explains how the Englishmen's hazardous duty evolved into a right, and how it was transferred to America and transformed into the Second Amendment.
War and Responsibility by John Hart Ely
Call Number: KF5060 .E58 1993
ISBN: 9780691025520
Published/Created: 1995-03-26
Twenty years after the signing of the Paris Accords, the constitutional ambiguities of American involvement in the Vietnam War remain unresolved. John Hart Ely examines the overall constitutionality of America's role in Vietnam; and shows that Congress authorized each new phase of American involvement without committing itself to the stated aims of intervention.
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A block is sitting on a frictionless table and has a rope attached that goes over a massless and frictionless pulley on the edge of the table connecting to another block hanging below. What will happen in the system? O A. The blocks will remain motionless because of friction. OB. Unable to determine what will happen because we do not know the masses of the
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Question: A block is sitting on a frictionless table and has a rope attached that goes over a massless and frictionless pulley on the edge of the table connecting to another block hanging below. What will happen in the system? O A. The blocks will remain motionless because of friction. OB. Unable to determine what will happen because we do not know the masses of the
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A block is sitting on a frictionless table and has a rope attached that goes over a massless and frictionless pulley on the edge of the table connecting to another block hanging below. What will happen in the system? O A. The blocks will remain motionless because of friction. OB. Unable to determine what will happen because we do not know the masses of the blocks. OC Unable to determine what will happen because we do not know the mass and size of the pulley OD. The hanging mass will drop as the mass on the table slides with an acceleration based upon the ratio of the two masses because of no friction Two objects are in a styble orbit around a planet and have equal radil from the center of the planet. If the mass of one object is ten times the mass of the other, what is the speed at which the heavier mass rotates about the planet compared to the first? O A Less than the first. OB. The same as the first OC. Depends on the planet's mass. OD. Greater than the first How is the gravitational force between two uniform and symmetric objects directed? A. Always up. B. Always down. OC. Along a random path between the two objects' centers. OD. Along a straight line connecting the two objects' centers. Suppose that a box was floating in space, far away from any other object. Is there any possible way that two forces, not in direct opposition with each other, could cause the net force to be zero? A. Only if the forces are perpendicular to one another. OB. Only if the forces are both large. OC. Only if the forces are incredibly small OD. There is no possible way The centripetal force on an object is: A. the force that causes uniform circular motion. B. the force that is created when moving in circular motion. OC. centrifugal force. OD. the combination of forces necessary to cause an object to move in uniform circular motion. Normal force is: A. the force a surface pushes on an object that is directed perpendicularly away from the surface. B. always the weight of the object. OC. Always directed against gravity. OD. only present when the surface is a table or the ground. The gravitational force between two objects is said to be inversely proportional to the square of the distance between them. What does this mean? O A. As the distance between the two objects increases, the gravitational force decreases. OB. It doesn't mean anything because the acceleration due to gravity between two objects is always a constant. OC. As the distance between the two objects increases, the gravitational force increases. OD. As the distance between the two objects decreases, the gravitational force decreases. Over the years, humans have placed objects into orbit around the planet. How is this possible? O A. Orbiting is possible as long as the orbiting object has enough tangential speed. OB. Orbiting is possible as long as the orbiting object is constantly rotating about itself. OC. Orbiting is possible as long as the orbiting object is small enough. OD. Orbiting is possible as long as the orbiting object is large enough. Question 9 If the forces necessary to cause an object to move in uniform circular motion were to suddenly cease, the object will: O A. move off on a path tangent to the circle at the point where the forces ceased. O B. keep moving in a uniform circle while slowing down. O C. change its direction of motion, D. stop moving because there is no more force to push it. Save 10 points Suppose a car is moving down a slope and through a vertical shaped part of the road that causes the car to go from moving down a hill to up another. At the bottom, will the force on the passengers be greater than, less than or equal to their weight? A. Equal to their weight O B. Less than their weight OC. Depends on the weight of the passengers. OD. Greater than their weight,
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8868 | https://www.youtube.com/playlist?list=PLduM7bkxBdOd85vOyZAK71FTXX_qYrVsd | Number System - YouTube
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number system binary to decimal decimal to binary hexadecimal decimal to hexadecimal hexadecimal to binary binary to hexadecimal octal to decimal decimal to octal binary number binary to hex international number system binary to octal octal to binary class 9 maths chapter 1 octal base 10 binary number system international system of numeration hexadecimal number system c in roman numerals ncert solutions for class 9 maths chapter 1 octal to hexadecimal number system class 9 octal number system decimal number system 10 in binary number system in computer real number system number system in maths decimal system numeration number to binary 9 in roman number binary digit greek numerals binary to denary the base of binary number system is binary to number binary numbers list denary to binary hexadecimal number 0 in roman numerals 9 roman number types of number system octal number adding binary numbers 10101 binary to decimal base 10 to base 2 binary number system in computer roman numeral numbers base 8 hex to octal class 9 chapter 1 numeral system roman numeral d base 2 to base 10 base 10 number system roman number 4 class 9 maths chapter 1 solutions 9th class math solution chapter 1 converting decimal to binary denary to hexadecimal binary code numbers converting binary to decimal the base of hexadecimal number system is international place value system zero in roman numerals the real number system babylonian number system signed binary to decimal binary to base 10 class 9 math chapter 1 roman number 6 ncert class 9 maths chapter 1 ex 1.6 class 9 positional number system egyptian numerals 60 roman number binary math 1011 binary to decimal the base of decimal number system is base 16 to base 10 define number system 11111 binary to decimal class 9 chapter 1 maths base of binary number system is exercise 1.6 class 9 maths hexadecimal system class9 maths ch1 base 8 to base 10 base 2 number system international number system example international place value 16 in hexadecimal hexadecimal to denary from decimal to binary from binary to decimal egyptian number system binary counting
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8869 | https://www2.math.uconn.edu/~stein/math103/Slides/math103-08.pdf | Number Theory Divisibility and Primes Definition. If a and b are integers and there is some integer c such that a = b · c, then we say that b divides a or is a factor or divisor of a and write b|a.
Definition (Prime Number). A prime number is an integer greater than 1 whose only positive divisors are itself and 1. A non-prime number greater than 1 is called a composite number.
Theorem (The Fundamental Theorem of Arithmetic). Every positive integer greater than 1 may be expressed as a product of primes and this representation is unique up to the order in which the factors are written.
Theorem. There are infinitely many prime num-bers.
Proof. Suppose otherwise. Then there would be a finite number n of primes, which we may denote by p1, p2, p3, . . . , pn. Consider x = p1·p2· p3 . . . pn + 1. There must be some prime num-ber greater than 1 which divides x, but clearly x is not divisible by any of p1, p2, p3, . . . , pn.
This contradicts the assumption that there is are finitely many primes, proving there are in-finitely many.
Sieve of Erastothenes The Sieve of Erastothenes is a technique which may be used to determine all the prime num-bers up to a certain size. One writes down all the integers up to that size. One then crosses out all the multiples of 2 (the even numbers) greater than 2.
At each step, one takes the smallest number left whose multiples haven’t been crossed out and crosses out all its multi-ples. One is ultimately left only with the prime numbers.
Test for Primality One may check every integer less than the number’s square root.
If none are divisors, then the integer is prime.
This may be seen by recognizing that if an integer n is not prime, there must be integers p ≤q both dividing n. But then p2 ≤pq ≤n, so p ≤√n. So every non-prime number must have a divisor no greater than its square root.
There are much more sophisticated tests for primality.
Goldbach’s Conjecture Goldbach’s Conjecture is that every even inte-ger greater than 4 may be written as a sum of two odd primes.
Goldbach’s Conjecture has been shown to hold for all even integers up to 400 trillion, but has not yet been proven in general. Hence, it re-mains a conjecture rather than a theorem.
Theorem (The Division Algorithm). If a, b are integers with b > 0, then there exist unique integers q, r such that a = q · b + r with 0 ≤r < b. q is called the quotient and r is called the remainder.
Note: The Division Algorithm is not an algo-rithm!
Note: Any number which divides both a and b also divides both b and r and visa versa.
Definition (Greatest Common Divisor). The greatest common divisor of integers a and b is the largest positive integer which divides both a and b. We denote the greatest common di-visor by gcd(a, b) or simply (a, b).
The Euclidean Algorithm gives a method (an algorithm!) for finding the greatest common divisor of any two positive integers: Given a, b, we apply the Euclidean Algorithm and find (a, b) = (b, r).
We then apply the Euclidean Algorithm to the pair b, r. We keep repeating the process, each time getting a new pair of numbers with the same gcd as a, b, until we get two numbers such that one divides the other.
That divisor is the gcd we’re looking for.
Modular Arithmetic Definition (mod). If a is an integer and n is a positive integer, then a mod n is the remain-der obtained when we divide a by n using the Euclidean Algorithm.
Definition (congruence). If n is a positive in-teger, two integers a, b are said to be congruent modulo n if they both have the same remainder when divided by n. We write a ≡b mod n.
Corollary. a ≡b mod n if and only if n|(a −b).
Modular arithmetic has many of the same prop-erties as ordinary arithmetic.
We may define addition, subtraction and multiplication mod-ulo n because it is easily seen that if a ≡b mod n and c ≡d mod n, then: 1. a + c ≡b + d mod n 2. a −c ≡b −d mod n 3. a · b ≡b · d mod n Divisibility Tests Modular arithmetic may be used to show the validity of a number of common divisibility tests.
Casting Out Nines A test for divisibility is called Casting Out Nines: Theorem. A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9.
Proof. Since 10 ≡1 mod 9, it follows that 10n ≡1 mod 9 for any positive integer n. Given any integer N, we may write N = am · 10m + am−1·10m−1+am−2·10m−2+. . . a0·100, where a0, a1, a2, . . . am are the digits in N.
But then N ≡am · 1 + am−1 · 1 + · · · + a0 mod 9.
Essentially the same reasoning shows: Theorem. A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
A variation gives a method called Casting out Elevens for testing divisibility by 11. It’s based on the fact that 10 ≡−1 mod 11, so 10n ≡ (−1)n mod 11.
Theorem (Casting Out Elevens). A positive integer is divisible by 11 if and only if the al-ternating sum of its digits is divisible by 11.
Proof. Since 10 ≡−1 mod 9, it follows that 10n ≡(−1)n mod 11 for any positive integer n. Given any integer N, we may write N = am· 10m+am−1·10m−1+am−2·10m−2+. . . a0·100, where a0, a1, a2, . . . am are the digits in N. But then N ≡am·(−1)m+am−1·(−1)m−1+· · ·+a0 mod 11 ≡a0 −a1 + a2 −a3 + · · · + (−1)mam mod 11.
Other Tests • Divisibility By 2 – The units digit must be even.
• Divisibility By 4 – The number formed by its last two digits must be divisible by 4.
• Divisibility By 5 – The units digit must be 0 or 5.
• Divisibility By 6 – It must be even and di-visible by 3.
• Divisibility By 7 – When the units digit is doubled and subtracted from the number formed by the remaining digits, the result-ing number must be divisible by 7. (To ver-ify, write the original number in the form 10a + b ≡3a + b mod 7, so the resulting number is a −2b, and check the possible ways for 3a + b to be divisible by 7.) • Divisibility By 8 – The number formed by its last three digits must be divisible by 8.
• Divisibility By 10 – Its last digit must be 0.
Check Digits Definition (Check Digit). A check digit is an extra digit tacked onto a number which is mathematically related in some way to the other digits.
Example: Airline Tickets – The check digit is the main part mod 7.
Example: U.S. Postal Service Money Orders – The check digit is the main part mod 9.
These do not catch all single-digit errors nor do they catch transposition errors.
Bank Identification Number Check Digit For-mula: Every bank has a nine digit identification number of the form a8a7a6a5a4a3a2a1a0 where a0 = (7a8+3a7+9a6+7a5+3a4+9a3+7a2+ 3a1) mod 10.
UPC Number Check Digit Formula: a0 is cho-sen so (3a11+a10+3a9+a8+3a7+a6+3a5+ a4 + 3a3 + a2 + 3a1) ≡0 mod 10.
ISBN Check Digit Formula: a0 ≡(a9 + 2a8 + 3a7+4a6+5a5+6a4+7a3+8a2+9a1) mod 11.
There is a check digit method that detects all single-digit and transposition errors and only generates 0 through 9 as a check digit.
Tournament Scheduling Problem: How do we schedule the teams play-ing in a round-robin tournament?
Solution: Let N be the number of teams in the tourna-ment and number the teams 1, 2, 3, . . . , N.
Let Tm,r be the team which Team m plays in Round r.
If there is an odd number of teams, we let Tm,r be the unique integer between 1 and N such that Tm,r ≡r −m mod N.
If Tm,r = m, then Team m gets a bye.
If there is an even number of teams, we sched-ule the teams as if there was one fewer team and let the team that would otherwise get a bye play the last team.
Cryptology Definition (Cryptology). Cryptology is the dis-cipline of encoding and decoding messages.
Cryptology is critical in everyday life today.
Our banking system, including the ability to use ATM’s and to do online banking, would collapse without the ability to securely trans-mit financial information over public networks.
Cryptology has played a crucial role in history.
Many believe that World War II was shortened by several years because the Allies were able to crack the secret codes used by the Axis powers.
Definition (Cipher). A cipher is a method for encoding messages.
Definition (Plaintext). Plaintext refers to the original text that is being encoded.
Definition (Ciphertext). Ciphertext refers to the encoded message.
Definition (Enciphering, Encryption). The process of encoding a message is sometimes referred to as enciphering or encryption.
Definition (Deciphering, Decryption). The process of Decoding a message is sometimes referred to as deciphering or decryption.
The Caesar Cipher The Caesar Cipher is one of the earliest known ciphers and was used by Julius Casar. Each let-ter in a message is simply replaced by the letter coming three letters after it in the alphabet.
Obvious problem: What about x, y and z?
Obvious solution: Replace them with a, b and c.
We may make this somewhat quantitative by assigning a numerical value to each letter: 0 to A, 1 to B, 2 to C, . . . , 25 to Z. If we let P represent the numerical value of a given letter in plaintext and C represent the number it is replaced by in ciphertext, we have C ≡(P + 3) mod 26.
To decode, we have P ≡(C −3) mod 26.
More generally, we might use a shift other than 3 and let C ≡(P + b) mod 26 for some other integer b.
We might mix things up a little more and let C ≡(aP + b) mod 26 (1) for some choice of integers a and b.
Such a cipher would be called an affine cipher.
To decode, we could try to solve the congru-ence (1) for P in terms of C. We might pro-ceed as follows: C ≡aP + b mod 26, aP ≡C −b mod 26, P ≡a−1(C −b) mod 26.
Question: What is a−1, the multiplicative in-verse of a?
It would have to be a number such that a·a−1 ≡ 1 mod 26.
It turns out that not all integers have such an inverse. For example, no even numbers could have inverses mod 26 since any multiple of an even number would be even and could only be congruent to another even integer and thus could not possibly be congruent to 1.
Theorem 1. An integer a has a multiplicative inverse mod n if and only if a and n have no factors in common other than 1, in other words, if (a, n) = 1.
The Hill Cipher With the Hill Cipher, blocks of letters are en-coded simultaneously rather than encoding let-ters separately in a method similar to an affine cipher. For a block of two letters, P1, P2, the corresponding encoded letters C1, C2 would be determined by the formulas C1 ≡(aP1 + bP2) mod 26 C2 ≡(cP1 + dP2) mod 26, where a, b, c, d are integers. It turns out we will need (ad −bc, 26) = 1 in order to be able to decipher the encoded message.
To see this, we may try to solve for P1 and P2 in terms of C1 and C2 the same way we would if we were dealing with ordinary equations–by matching coefficients and using elimination.
We might match the coefficients of P2 by mul-tiplying the first equation by d and the second by b to get dC1 ≡(adP1 + bdP2) mod 26 bC2 ≡(bcP1 + bdP2) mod 26, .
Subtracting, we get dC1 −bC2 ≡(ad −bc)P1 mod 26.
We need ad −bc to have an inverse mod 26 in order to solve: P1 ≡(ad −bc)−1(dC1 −bC2) mod 26.
Similarly, we may find P2 ≡(ad −bc)−1(aC2 −cC1) mod 26.
Variations may be used for larger blocks of let-ters.
The RSA Public Key System In public key systems, the encoding method is made public, so that anyone may send an encoded message, but the decoding method is known only to the recipient. This depends on the difficulty of determining the decoding method even if the encoding method is known.
The RSA Public Key System was created by Ron Rivest, Adi Shamir and Len Adelman in 1975. It is probably the most widely used sys-tem today.
The Method Two large primes, p, q, are determined along with their product n = pq and a positive integer r relatively prime to both p −1 and q −1. The values of n and r are published.
A block of letters is then encoded by letting C ≡P r mod n.
To decode the message, the recipient deter-mines the multiplicative inverse k = r−1 mod (p− 1)(q −1) and calculates P ≡Ck mod n. |
8870 | https://store.lonestarlearning.com/shop/target-vocabulary-pictures-for-math/ | Math Vocabulary Pictures™
You asked and we listened! Our award-winning cards are now aligned with the updated TEKS and available in grade level sets! Help your students quickly learn new math terms with these colorful vocabulary cards, each illustrated to look just like its meaning! Buy one set, or the full bundle for extra products and savings!
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| Grade | Choose an optionKindergarten1st Grade2nd Grade3rd Grade4th Grade5th Grade6th - 8th Grade |
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Description
Math lessons tend to be vocabulary-dense. A lesson on fractions, for example, has to discuss numerators and denominators and may also go over equivalency, proper and improper fractions, and many other terms. These terms can be difficult to keep straight for students, so how can you help them? Quick-reference hand-outs that are costly to photocopy? Explaining terms every lesson, using valuable class time?
Instead, let Math Vocabulary Pictures™ vocabulary cards handle the explaining for you! Each brightly-colored card features a different math term on the face, illustrated to look like its meaning, while the reverse has the definition in black text on a white background. These cards are wonderful for your math word wall for general review, as well as for your visual learners, ESL students, and struggling readers, since the illustrations quickly and easily create a visual connection between the term and its meaning.
All of the words are TEKS-specific terms. Each bundle contains both Math Vocabulary Pictures™ card sets for the grade, plus either a set of our Operation Options™ posters (Kindergarten and First grade) or a set of Measurement Benchmarks™ cards. Operations Options™ are visual reminders to help give your students clues for the various ways to both add and subtract, while Measurement Benchmarks™ are illustrated cards to help your students become familiar with the most common units in both the metric and customary systems.
Math Vocabulary Pictures™ vocabulary cards are printed in full color on 11″ x 8.5″ coated cardstock, durable and ready for display.
These cards are printed with pride in the USA.
Word Lists
Kindergarten (64 cards): add, addend/sum, altogether, capacity, circle, classify, coins, compare, cone, contain, corner, count, cube, cylinder, decreasing, difference, digit, dime, each, edge, equal, fewer than, five frame, height, income, increasing, irregular, join, length, less than, minuend/subtrahend, minus, money, more than, need, nickel, numeral, one, pattern, penny, pictograph, plus, polygon, quarter, rectangle, regular, shape, side, solid, sort, sphere, square, subtract, take away, tally, ten, ten frame, total, triangle, vertex, want, weight, whole number, width
1st Grade Set 1 (65 cards): one digit, two digits, add, addend, addition, backward, classify, combined, compare, complete, contain, decreasing, difference, double, dozen, equal, equal amount, equation, even, expanded form, expression, fact family, fair shares, fewer than, forward, greater than, greatest, halves, hundred, increasing, join, least, less than, minuend/subtrahend, minus, more than, non-example, number sentence, number line, numeral, odd, one, one-fourth, operations, order, pair, period, place value, plus, point, separate, skip count, sort, strategies, subtract, sum, symbol, ten, ten frame, total, twice, unknown, whole, whole numbers
1st Grade Set 2 (64 cards): 2-dimensional, 3-dimensional, analog, bar graph, categorical data, cent, circle, clockwise, coins, colon, cone, cube, curved surface, cylinder, diagonal, digital, dime, distance, dollar, earn, edge, exact, exchange, face, figure, flat surface, hexagon, horizontal, hour, income, irregular, key, left, length, measure, minute, need, nickel, penny, pictograph, plane figure, polygon, quarter, rectangle, rectangular prism, regular, rhombus, right, rotation, save, side, spend, sphere, square, square corner, T chart, table, tally, triangle, triangular prism, vertex, vertical, want, weight
2nd Grade Set 1 (64 cards): one digit, addition, addend, bar graph, categorical data, classify, colon, column/row, combined, comparison, compose, decimal, decompose, decreasing, difference, division, dot plot, double, eighths, equal, equal amount, equation, exact, expanded form, expression, even, fact family, fair shares, fractional parts, frequency, greater than, greatest to least, halves, increasing, interval, join, least to greatest, less than, minuend/subtrahend, minus, non-example, number line, numeral, odd, one-fourth, operations, order, partitions, period, pictograph, point, predict, quotient, repeated addition, rotation, separate, skip count, sort, strategies, subtract, sum, total, unknown, whole numbers
2nd Grade Set 2 (64 cards): 2-dimensional, 3-dimensional, area, array, balance scale, borrow, capacity, cent, charity, circle, cone, producer/consumer, cost, counting numbers, cube, cup, curved surface, cylinder, decagon, deposit, distance, dodecagon, dollar, edge, exchange, face, height, heptagon, hexagon, horizontal, hour, income, interest, irregular, labor, lend, length, measuring tape, minute, octagon, parallel, parallelogram, pentagon, place value, polygon, property, quadrilateral, quarter, rectangle, regular, resources, rhombus, ruler, save, spend, sphere, square, square unit, trapezoid, triangle, triangular prism, vertex, vertical, withdrawal
3rd Grade Set 1 (64 cards): addend/sum, array, axis, bar graph, categorical data, column/row, comparison, compose, decimal, decompose, decreasing, difference, dividend, division, divisor, dot plot, double, eighths, equal, equation, equivalent fractions, even, expanded form, expression, fact family, factor, fewer than, fraction, fractional parts, frequency table, greater than, greatest to least, halves, increasing, increment, input/ output, interval, key, least to greatest, less than, minuend/subtrahend, multiples, number line, numerator/ denominator, odd, operations, order, partitions, period, pictograph, point, predict, product, quotient, repeated addition, skip count, strategies, subtract, symbol, tally, twice, unit fraction, unknown, whole
3rd Grade Set 2 (64 cards): 2-dimensional, 3-dimensional, analog, area, balance scale, base, borrow, capacity, charity, classify, cone, congruent, credit, cube, cup, cylinder, decagon, digital, distance, edge, equilateral, expenses, face, height, hexagon, horizontal, hour, income, interest, irregular, isosceles, labor, lend, length, line, line segment, measuring tape, metric, minute, octagon, parallel, parallelogram, pentagon, perimeter, plane figure, polygon, quadrilateral, rectangle, rectangular prism, regular, resources, rhombus, ruler, save, scalene, scarcity, spend, sphere, trapezoid, triangular prism, vertex, vertical, weight, width
4th Grade Set 1 (64 cards): analog, area, array, axes, capacity, classify, column/row, consecutive, convert, cup, decimal, decompose, decreasing, numerator/denominator, difference, digital, dimension, distance, dividend, divisor, equation, equivalent, expanded form, expenses, expression, factors, greater than, height, improper fraction, increasing, increment, input/output, intersection, interval, length, less than, line of reflection, mixed number, multiples, number line, numeral, operations, perimeter, period, place value, product, proper fraction, protractor, quotient, relative size, remainder, repeated addition, round, rule, sequential, side congruency marks, sum/addend, trailing zeroes, unit fraction, unknown, volume, weight, whole numbers, width
4th Grade Set 2 (64 cards): acute angle, adjacent angles, angle, balance scale, bar graph, categorical data, complementary angles, compose, composite figures, congruent, congruent angles, credit, decagon, degree, diagonal, dot plot, earn, edge, equilateral triangle, fixed expenses, formula, frequency table, hexagon, horizontal, income, interest, intersecting lines, irregular, isosceles triangle, key, line, line segment, nonagon, obtuse angle, octagon, parallel, parallelogram, pentagon, perpendicular, polygon, profit, quadrilateral, range, ray, rectangle, rectangular, regular, rhombus, right angle, right triangle, rotation, scale drawing, scalene triangle, spend, stem and leaf, supplementary angles, table, tally, trapezoid, triangle, variable, variable expenses, vertex, vertical
5th Grade Set 1 (64 cards): additive pattern, approximately, axes, bracket, categorical data, column/row, compose, composite number, convert, coordinates, decimal, decompose, decreasing, numerator/denominator, difference, continuous data/ discrete data, dividend, divisor, equation, equivalent, equivalent fractions, expanded form, expression, factors , frequency table, greater than, gross income/net income, hierarchy, income tax, increasing, increment, input/output, interval, least common denominator, less than, mixed number, more than, multiples, multiplicative pattern, operations, ordered pair, parenthesis, place value, predict, prime numbers, proper fraction, property tax, quantity, quotient, ratio table, rational numbers, reciprocal, relative size, remainder, sequential, simplify fraction, stem and leaf, trailing zeros, unit fraction, unknown, whole numbers, withdrawal, x-axis, y-axis
5th Grade Set 2 (64 cards): 2-dimensional, 3-dimensional, acute angle, area, area of base, balance scale, bar graph, budget, classify, composite figures, congruent, congruent angles, consecutive, coordinate plane, counting numbers, credit card, debit card, dot plot, edge, equilateral triangle, face, fixed expenses, formula, height, horizontal, improper fraction, inequality, intersection, irregular, isosceles triangle, length, line segment, obtuse angle, origin, parallel, parallelogram, perimeter, perpendicular, polygon, prime factorization, product, quadrant, quadrilateral, range, ray, rectangular prism, regular, rhombus, right angle, right triangle, round, rule, scale drawing, scalene triangle, scatter plot, side congruency marks, transaction, triangular prism, variable, variable expenses, vertex, vertical, volume, width
6th-8th Grade Set 1 (72 cards): 529 plan, acute angle, area, area of base, arithmetic sequence, ascending, complementary angles, composite figures, composite numbers, cone, congruent, continuous data, discrete data, coordinate grid, coordinates, coupon, cylinder, descending, dimension, direct deposit, dividend, divisor, dot plot, edge, equation, equilateral triangle, equivalent, expression, face, factors, fixed expenses, formula, greater than, heptagon, hexagon, histogram, hypotenuse, improper fraction, interest, isosceles triangle, legs, less than, mixed number, numerator/denominator, obtuse angle, pentagon, percent, percentage bar graph, perimeter, prime factorization, prime numbers, product, proportional, quadrant, qualitative data, quantitative data, quotient, range, rebate, rectangular prism, right triangle, scalene triangle, scholarship, simplify fraction, stem and leaf, supplementary angles, symmetrical data distribution, triangular prism, variable expenses, volume, wage, x-axis, y-axis
6th-8th Grade Set 2 (72 cards): absolute value, additive inverse, algebraic expression, average, bivariate data, box & whiskers plot, categorical data, circle graph, circumference, coefficient, collateral, compound event, compound interest, constant, corresponding angles, diameter, dilation, exponent, exterior angles, function, identity element of multiplication, income tax, inequality, inflation, integers, intercept, interquartile range, inverse operation, investment, irrational, isosceles trapezoid, lateral surface area, like terms, line, linear, mean, median, negative, net, ordered pair, origin, outlier, pi, positive, possible outcome, principal/interest, probability, pyramid, Pythagorean theorem, radius, rate, rational numbers, reciprocal, reduction, reflection, relative frequency table, rotation, scale factor, scatter plot, semi-circle, similar, simple interest, skewed data, slant height, slope, spread, square root, translation, transversal, variable, vertex, vertical angles
10 reviews for Math Vocabulary Pictures™
Kimberly Carter
– 2019-07-17
I absolutely love the vocabulary cards. I started using these with ESL (3rd-8th) a few years ago. Now I use them with everyone.
Becky Teague
– 2019-04-15
Love it! Great way to express the meaning of the vocabulary word in a visual way.
7th grade math teacher
– 2019-04-05
I like to use the picture vocabulary on my word wall. I refer back to it every time we use the vocabulary word. It is really helpful for my ESL students.
Learning Magazine® Teachers’ Choice Award® product evaluator
– 2019-02-07
I love these! Tutors and basic skills teachers will want these in their inventory of teacher tools. Fun to use and don’t take much time for kids to learn. They make a great review for tests and quizzes, too!
Learning Magazine® Teachers’ Choice Award® product evaluator
– 2019-02-07
I recommend this product to every teacher I know! This was by far my favorite product to evaluate this year. Both my students and I love these
Jeanette Mercer, ESL program director, Aldine ISD
– 2018-04-26
These Math Vocabulary Picture Cards really help our English learners to remember the academic language. The teachers love them too. Vocabulary understanding is essential in tests. This product helps!
Sheila Moore
– 2017-08-10
I use these in my 8th grade classroom and love the graphics and color.
Laura Holmes
– 2017-07-18
These cards are the best. They brighten up the class. They catch the students attention.
NCTM 2017 attendee
– 2017-04-07
I’m buying these for my teacher friend for teacher appreciation week! She loves mine that I put around my room!
Claudia Perez
– 2017-04-07
Your cards are up in my room! My kids love them and are so smart!
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8871 | https://www.allvuesystems.com/resources/what-is-private-debt/ | Private Equity
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What Is Private Debt?
By: Allvue Team
July 14, 2023
It may not be the largest private capital strategy, but private debt has seen an exciting jump in popularity in the last few years. In 2022, private debt funds raised $200.4 billion, up from $184.8 billion in 2019.
So what is private debt, and what’s behind its growth in the institutional investing space? Let’s break down private debt’s definition, fund structure, recent growth, and more.
Private debt definition
Private debt – also known as private credit – is a private capital strategy in which investment managers and institutions invest by making private, non-bank loans to companies. These loans are not available via publicly traded markets, and they generate returns for investment managers and their private debt fund investors via interest payments.
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Table of Contents
1 Private debt definition
2 The evolution of private debt as an investment strategy
3 How do private debt funds operate vs. private equity funds?
4 Diverse strategies and capital sources in private debt
5 Risk and return in private debt strategies
6 Private debt and the capital structure
7 What are common types of private debt?
8 What are some real-world examples of private debt?
9 How has private debt grown?
10 Why investors are turning to private debt
11 Industry trends in the private debt space
12 What do private debt managers need to succeed?
The evolution of private debt as an investment strategy
Private debt emerged as a prominent investment strategy in the aftermath of the Global Financial Crisis in 2008. This period marked a significant shift in the lending landscape, with traditional banks tightening their lending standards and reducing exposure to riskier loans. This gap in the market paved the way for private debt funds to become a key player in providing alternative sources of lending. Private debt has now established itself as a distinct asset class, offering a diverse range of lending options and attracting a substantial investor base.
How do private debt funds operate vs. private equity funds?
Private debt funds have much in common with other private capital strategies such as private equity and venture capital. For example, it’s a highly illiquid investment with long lockup times for investors’ capital. However, private debt as a strategy is seen as slightly less risky than equity investments, partially due to capital structure and due to the ability to highly customize the risk profile of a private debt investment.
The lifecycle of a private debt fund is quite similar to private equity, but fundamental differences do exist. In the fundraising stage, private debt managers typically target institutions like pension funds, university endowments, or family offices. However, some high-net-worth individual investors are also drawn to the asset class. The average-sized private debt fund in 2022 raised $964 million.
Once fundraising has closed and the fund enters the investment period, the private debt manager seeks out investment opportunities and calls investor capital when it finds the right deals. Rather than buying equity in the portfolio companies they target, private debt funds instead make an investment by setting up a loan.
Just as with a private equity fund, a private debt fund’s lifespan averages around seven to 10 years, and investors can count on the high illiquidity that accompanies such a timeframe. In the final few years of the fund, winddown begins, and the private debt manager will stop seeking new investments while letting their various investment repayment periods run out. At the closing of the fund, final returns are paid out and investors recover the full liquidity of their investment.
Diverse strategies and capital sources in private debt
Private debt funds adopt a range of investment strategies to cater to various investor needs. These strategies include direct lending, venture debt, and special situations. Each strategy offers unique risk and return profiles, aligning with different investor objectives. The capital for these funds is sourced from various channels, including Collateralized Debt Obligations (CDOs), Business Development Companies (BDCs), and hedge funds. These sources provide the necessary liquidity and financial backing, allowing private debt funds to offer tailored lending solutions to their clients.
Risk and return in private debt strategies
The allure of private debt investments lies in their diverse risk/return profiles. Strategies like direct lending generally offer lower risk but also lower returns, making them attractive for conservative investors. On the other hand, ventures like mezzanine debt carry higher risk, potentially leading to higher returns. This variety allows investors to choose strategies that align with their risk tolerance and investment goals. Understanding these profiles is crucial for investors to navigate the private debt landscape effectively and make informed investment decisions.
Private debt and the capital structure
Debt sits above equity in the private capital structure. This means that if a company were to declare bankruptcy, its debts are paid out before equity, making it less risky of an investment overall. Within each level of debt in the capital structure, there are a variety of debt vehicles and different levels of distress/risk as well as different loan terms based on the borrower’s status.
Private debt in the capital structure, visualized
Senior debt
The term “senior debt” describes loans that are prioritized as first to be repaid in the event of the borrower’s bankruptcy. Since senior debt has the highest priority, it is therefore the lowest risk. As a result, senior debt typically offers lower interest rates, aka a lower return profile for the investor – but also a lower risk profile since the borrower is seen as favorable. Senior debt is most often secured, meaning that it’s tied to certain assets as collateral in the case that the company cannot make repayment.
Mezzanine debt (or junior debt)
While still sitting above equity, mezzanine or junior debt describes credit investments that have a higher risk profile than its senior debt counterpart. In the event of the borrower’s bankruptcy, it is paid out after senior debt. Because of this, its higher risk could mean a higher reward for the lender in the form of higher interest rates and more stringent loan terms. Mezzanine financing is increasingly popular with mid-marketing companies in recent years.
What are common types of private debt?
Private debt loans can take many forms, but a few of the most common loan structures are as follows.
Term loan – A term loan provides a borrower with a one-time lump sum of cash up front attached to certain borrowing terms, including a set repayment schedule and either a fixed or floating interest rate.
Revolving credit facility – A revolving credit facility allows the borrower to borrow up to a set limit and pay back that money on a continuous basis.
Convertible debt – A convertible debt arrangement stipulates that the money the borrower receives from the lender will be repaid in the form equity. Hence the name, the investment is ultimately converted from debt to equity.
What are some real-world examples of private debt?
Part of the reason private debt is experiencing a surge in popularity over the last several years likely can be attributed at least in part to its varying classifications and types. With many options to choose from, there is a private debt niche that works best for every institutional investor.
With the many structures and vehicles of private debt that are available, private debt managers and their investors have many different applications for carrying out their private debt investments. Here are some of the more common private debt scenarios that market participants rely on.
Leveraged buyouts (LBO) – A leveraged buyout involves the acquisition of a company with borrowed money. The acquiree’s assets are then designated as collateral for the loan. Some of the largest leveraged buyout deals include H.J. Heinz in 2015, Hilton Worldwide in 2007, and R.J.R. Nabisco in 1988.
Collateralized loan obligations (CLOs) – CLOs describe a collection of separate loans that are pooled together and packaged into a security for purchase to investors by CLO managers. The loans are often poorly rated, but investors can benefit from the diversification across the full pool of the loans.
Venture debt – Venture debt describes loans made to small, early-stage businesses, usually in tandem with equity arrangements. Often, venture loans are constructed with specific benefits to the portfolio company in mind, such as by protecting the equity ratio as the company grows or by helping the young company establish a credit track record.
Real estate debt – A real estate debt fund is typically set up by a private debt manager and backed by their investors. The fund then makes loans to real estate developers for the purpose of building or purchasing real estate properties. The property itself serves as collateral to a senior debt asset.
Infrastructure debt – An infrastructure debt fund is set up with the intention of loaning money to finance the building of essential societal and economic needs, such as roads, bridges, airports, power plants and beyond. Infrastructure debt is seen as a relatively stable and lower-risk form of private debt, with lower default rates than corporate debt.
How has private debt grown?
Private debt as an asset class has exploded since 2000. But market factors have shifted significantly over the past few years – thanks to Covid-fueled economic instability, global conflicts, climbing interest rates and beyond. Through all of it, private debt, just like private capital in general, has seen its impressive growth sustain. There are several factors behind its continued growth.
Why investors are turning to private debt
Private debt has become increasingly popular among investors for several compelling reasons. Its reputation as a low-risk investment relative to other alternative asset classes makes it a preferred choice for many. Additionally, private debt offers attractive risk-adjusted returns, especially in low-interest-rate environments. Investors value the portfolio diversification it provides, along with its low correlation to public markets. These characteristics, combined with the potential for predictable and contractual returns, make private debt a lucrative and strategic investment option in the diverse world of private capital.
Larger private debt deal sizes
While the number of private debt deals carried out has shrunk, the average size of deals has grown considerably. The U.S. has seen over $50 billion in unitranche loans sized at $1 billion or larger since September 2019.
Customizable nature of private debt
Private debt hit its stride as we entered a nearly 15-year period of extremely low interest rates and investors sought investments that would provide the returns they needed at an acceptable risk level. Even as interest rates are climbing again, private debt’s flexibility and differing levels of risk and return by class of debt offers permanent appeal. There’s always a debt strategy that fits the market conditions, whatever they may be.
Interest rate flexibility
As rates rise, interest in private debt doesn’t necessarily dissipate since managers can turn to a floating interest rate. While this raises costs for borrowers, it does raise the potential for higher returns for investors.
Continued private debt adoption in Europe and Asia
Private debt first took hold in the U.S., but has been seeing increased adoption in Europe and Asia recently. One-fifth of all private debt deal activity since 2019 has occurred in Europe, and APAC’s private debt AUM has grown 195% over the last five years.
Industry trends in the private debt space
With all the growth the private debt space has seen in the last few decades, exciting developments are on the horizon for this industry. What are some of the key private debt trends to watch for?
Less market share for banks
Tightened regulations are keeping banks away from leveraged loans, pushing companies of all sizes to private debt as an attractive alternative to bank financing. Between 1994 and 2020, banks saw an 80% drop in loan market participation.
Larger deals and funds
Private debt has encroached on the bank dominated broadly syndicated loan market, with several headline-grabbing $1 billion+ deals throughout 2021. And as deal sizes balloon, fund sizes are growing alongside them, with the average fund size surpassing $1 billion in the first quarter of 2021.
Rising popularity for venture debt
Venture debt emerged as a major alternative source of financing for high-growth startups that have traditionally opted to solely finance through equity VC. More than $20 billion has been loaned to VC-backed companies in the US during each of the past three years.
What do private debt managers need to succeed?
As the private debt and direct lending industries continue to expand, many fund managers find themselves lacking the proper infrastructure and processes to scale their businesses. Allvue offers an award-winning suite of solutions meant to support a private debt manager at every turn of the investment lifecycle, including the following challenges:
Delivering Meaningful Portfolio Data
Reducing Your Quarterly Reporting Cycle
Optimizing Deal Management
Automating Financial and Covenant Tracking
Managing Loan Principal Activity
For more on Allvue’s private debt solutions, reach out for a live demo below.
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8872 | https://www.youtube.com/watch?v=-GC2bhP6DuQ | Show that the set of all polynomials over X with constant term zero is a prime ideal of Z[x] but not
MATHS A TO Z
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Posted: 11 Feb 2024
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Rajkishor Patra (M.Sc Jadavpur University)
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show that set of all polynomials over Z with constant term zero is a prime ideal of Z [x] but not maximal ideal.
Or
Prove that the ideal (x) of Z[x] is prime but not maximal.[Let I={f(x)€Z[x]:f(0)=0} is an ideal andI= (x)]
If in a ring R, x^3=x for all x€R, then show that R is commutative.(Solution)
Proof of 'Principal ideal (x^2+1) is not a prime ideal of Z _2[x]'
If R is a ring with non zero nilpotent elements then show that for any idempotent element e,ex=xe for all x€R and thus e€Z(R).
Let J={a+ib€Z[i]:a,b€7Z}. Show that J is an ideal of Z[i].Is J a maximal ideal? Also find the number of elements of Z[i]/J.(Bankura University -2022 based on maximal ideal).
This video contains an solution of a problem of ring theory of Bankura University -2019
Two important exercises solution of ring theory of S K Mapa Higher Algebra .
This video contains Statement and proof of third Isomorphism theorem of ring with explanation.
This video contains Statement and proof of second isomorphism theorem of ring.
This video contains statment and proof of First Isomorphism theorem of ring(Fundamental theorem of ring homomorphism)
This video contains definition of quotient ring or factor ring, important example.Statment and proof of such Theory which gives the relationship between quotient ring and prime Ideal and with maximal ideal.
This video contains definition and important example of maximal ideal in a ring.The condition that a subring pZ is maximal ideal together with the proof.
This video contains definition, example and property of prime ideal.The condition that an ideal of ring of integer be prime.
This video contains definition and important example of principal ideal, principal ideal ring and principal ideal domain
This video contains definition of Ideal different property, theory and important example.simple ring
This video contents definition of integral domain, it's important example property, related theory and their proof.
(This is the class:01 of ring Theory
This video contents definition of ring and different property , theory example etc )
(This is Class:02 of Ring Theory .This video contents important definition, property, theory of ring Theory.This is very helpful for beginners )
Transcript:
सो दैट द सेट ऑफ ऑल पॉलिनो मिल्स ओवर जड ज इ द सेट ऑफ इंजर थ कांस्टेंट ट जीरो इज ए प्राइम आइडियल ऑफ जऑफ एक्स बट नॉट मैक्सिमम दैट इज नॉट मैक्सिम आइडियल सो वी हैव टू प्रूव दिस इज ए प्राइम आइडियल बट नॉट मैक्सिम आइ ओके देखो थरम थरम कि नट प्र आइडियल और द प्रिंसिपल आइ जेनरेटेड बाय एक ऑफ ज इ प्राइम आडिल बट न मैसम आ हम प्रिंसिपल आ बा बो म जीरो तो न आल ल प्र की न मैक्सम आ तो ि हम जानी आइडियल आई आल आ ए बी बिल टू आ इ ए माइनस बी ब आ आ फ ए ब आ r बिंग्स टू r हेयर r इज z x इंप्ला a आ और आ बिंग्स टू आ ठीक है सेकंड लो आइडियल डेफिनेशन ब हम प्राइम आइ प्राइम आइडियल डेफिनेशन की बोल ए बिंग्स टू आ इंप्ला आइर आइर a बिंग्स टू आ और ओके डेफिनेशन ऑफ प्राइम मैक्सम मैक्सम आ की आ आ मैक्सम आ बो र अनदर आल ज स आई सबसे ज सबसेट ज एक ओके आइर आ इ ज और ज इ ज एकस [संगीत] आम बस तो 0 प् ए1 प् ए एक्स स् प् ए एक् पावर बि टू ज एकने सेट ऑ इजरर तो बोट ल प थ का टम जी [संगीत] [संगीत] [संगीत] जी तो प् बी एक्स स् प् बी पावर बल टूस ब ू आ डिगी डिग्री ी ब ज नाउ माइन एव माइनस बीव इन एक्स प्स ए2 माइनस बीट इन एक्स स् प्स ए माइनस बी ए इन एक्स पावर प्स ब n प्व एक् पा प्व ब ए एक् पावर ए और उ हम ए ग ए न कि की ए माइनस बी ए एक्स टू द पावर ए प्लस ए प्व एक्सू पावर ए प् बो सब माइनस माइनस जा प्र इंजर ए ब ब द बिल टू आ हो बिल टू आ सो हम बोलते प की f माइज एक बिल टू आ सो आइडियल फर्स्ट प्रॉपर्टी हम प्रू आडिल फर्प ब देखो h प प् एक प् स एक्स स् प् सी एक ब ज इन म स सी [संगीत] प्स स ए2 प् सव एक्स प्रोड म बज की ना जी बि टू आट इ होलो ना सेकंड प्रॉपर्टी प्रप फ राइट साइड प्रोडक्टर नहीं बिकॉज की प्रोडक्ट सब मल्ट क्लियर ो प्रथम स्टेप से स्टेप ब देखो ने बोले [संगीत] आइडिया आडिल के हम प्रिंसिपल आइडियल जेनरेटेड बाय जेनरेटेड बाय x ख खाने ज हम देख f 0 दिस मीस f इ a1 एक a2 एक्स स् ए एक पावर टा की बि टू आ सो दिस इ x इन a1 प् a2 एक प् a3 एक स् प् एंड दिस एलिमेंट बल टू ज एक सो राइट दिस ब टू प्रिंसिपल आ बा ओके आट ने बो आ सब प्रिंसिपल आल जटे बा एमली [संगीत] सेम टाइप ऑस इन समथ समथ माने रि हो रि आ सो व गेट दिस सब ऑ सो फम दिस टू ल् गेट आ इ प्रिंसिपल आइ जड बा एक सो वन राइट दिस इ ओके दिस इ द फर्स्ट स्प ऑफस थम एंड नाउ गो टू द सेकंड स्ट प्र दिस ए प्राइम आ प्राइम प्र की द एमेंट तो एक एमेंट ब ट एंड लेट fxxx.pro [संगीत] एंड g ऑफ x दिस इज इक्वल टू b0 प् b1x प् b2x स् प् b ए एकू पावर ए नाउ इफ व मल्टीप्लाई f g एक वी गेट द रिजल्ट एज a0 बी0 प्स a0 b1 प् a1 बी0 इन x प्स a0 बीट प्लस बी0 a2 प्स a1 b1 इन x स् एंड सो ऑन नाउ दिस दिस प्रोडक्ट बिलंग टू आ सो व मस्ट हैव f0 इन g0 दिस इ इ इफ टेक दिस रिजल्ट एंड देन इट मस्ट बी इ 0 बिकॉज इनस सेट सो व गेट ए0 बी इ प्र टम टम प्रम जीरो 0 सो a0 बी0 इ 0 हम की जा औरी इंजर आ सेट ऑफ इंटी जर है नो डिवाइजर ऑफ जीरो सो वी मस्ट हैव आइर आइर a0 = 0 और बी0 इ इ ब देखो इ a0 इ 0 a0 इ जीरो तन हम देखते f f जीरो थाक ना f चले जा आ बिलंग टू आ और जन जद v0 इ 0 और आर आइर और आ ओके सो आ सेक ट प्र आई प्रा थड स्टेप ट आई इ नॉ ए मैक्सिमम आ थ आ इ न द मैक्सम आडिल आडिल कडर ज हम ज बो आडर ज प् और द पमल तो प्रथम देख र आल प्रमा की प्रमा प् x f1 प् 2 इन g1x बोथ एलिमेंट ऑफ ज नाउ सबट कर x न f माइ f1 प् 2 नज एक माइज सबट करले की देर सो दिस इज द सेम एस द गिवन सो दिस इ बिलंग टू ज आल्सो हम h बोले एमेंट ् और ने एमेंट ज ो बो की h एक् इन एक f प् नजस इ एकन प् h सो दिस इ आल्सो ऑ द फॉमस सो [संगीत] म ब आल आल आ आ सब ज हम ज एमेंट एमेंट ऑ आन न फम प् र ऑफ एक दिस इ जी तो जी न दिस इ दिस बिलंग टू प्रिंसिपल आ जड बा एक एंडस आ सो ट आ इ सबसेट ऑ ज एटलीस्ट नमेंट आ आ न इने दिस 0 और [संगीत] एक प्ज एक x0 ए न प् एक् प् एक् स् प् डस नट बल टूट ऑ बज इ दिस इन आ देन मस्ट ब का जीरो सो प्र आ सबसे ज ब हम प्रूफ करते दिस इज सबसेट ऑफ z x z x बट ज नॉ इ z ज सबसेट ऑफ z बिकॉज की बिकॉज z ज इज आइडियल ऑफ z प्रमान सबसेट प्रमान प्र न इ z तो लो करो z x ज तो य तो केट इ पॉसिबल न दिस न बिल टू आरी न बिल टू ज ट इ न दिस इल टू एक्स ट एक् प्सन बज द एलिमेंट ऑफ ज इ ऑफ फम प् [संगीत] प् प् [संगीत] प् ना ट ल ऑ ए व गेट द रिजल्ट ए x इन a0 प् a1 एकट ए एकू पावर प् 2 इन g एक मीस बी0 प् b1 एक प् ब एक पावर ए एंड दिस इ इक्ट 1 नाउ कंपरिंग द कोफ ऑफ इल पावर व गेट र ली काटम 2 v 2 v0 दिस इ इ 1 दैट इज v0 बी0 दिस इ इक्वल टू हाफ बट हाफ इज नॉट एन इंजर सो दिस इ ड नॉट बिल टू ज ओके सो ड नॉट बिल टू z दिस मीस g ड नॉ बिल टू z एंड दिस गिव्स ए कांट्रडिक्शन सो वी मस्ट हैव वन डज नॉट बिलंग टू ज क्लियर सो वी से दैट j इज नॉट इक्वल टू z ले देखो हमरा की पम आ आ के कंटेन ज ज के कंटेन जऑफ x किंतु दे गलो देखा गलो आ आ सा ज समान अब ज किंतु z समान खा बोलते आ इ नॉ ए मैक्सिमम मैक्सिम आर न न लास नसमस वेरी वेरी इंटरेस्टिंग ए पोर्टेंट प्रॉब्लम और थम बो आ आस इक्वल ट बिल ू ज स ए इन इर इन ओके इट इर जन बो प्रिंसिपल आडिल जटे बा प्रिंसिपल आ बाटल आ [संगीत] क यक |
8873 | https://www.healthline.com/health/sexually-transmitted-diseases/chlamydia-vs-gonorrhea | What’s the Difference Between Chlamydia and Gonorrhea?
Medically reviewed by Holly Ernst, PA-C — Written by Tim Jewell — Updated on May 5, 2023
People of any anatomy can contract chlamydia or gonorrhea and never develop any symptoms. When symptoms do occur, there are a few telltale signs differentiating the two conditions.
Chlamydia and gonorrhea are both sexually transmitted infections (STIs) caused by bacteria. They can be contracted through oral, genital, or anal sex.
The symptoms of these two STIs overlap, so if you have one of these conditions, it’s sometimes hard to be sure which one it is without having a diagnostic test at a doctor’s office.
Some people with chlamydia or gonorrhea may have no symptoms. But when symptoms occur, there are some similarities, such as an abnormal, bad-smelling discharge from the penis or vagina, or a burning feeling when you pee.
Chlamydia is more common than gonorrhea. According to a 2017 report, over 1.7 million cases of chlamydia were reported in the United States, while just over 550,000 cases of gonorrhea were documented.
Read on to learn about how these two STIs are different, how they’re similar, and how you can reduce your risk for these infections.
How do the symptoms compare?
People of any anatomy can contract chlamydia or gonorrhea and never develop any symptoms.
With chlamydia, symptoms may not appear for a few weeks after you’ve contracted the infection. And with gonorrhea, people who have female anatomy may never experience any symptoms at all or may only show mild symptoms, while people who have male anatomy are more likely to have symptoms that are more severe.
A couple of the most telltale symptoms of these STIs overlap between the two, such as:
burning when you pee
abnormal, discolored discharge from the penis or vagina
abnormal discharge from the rectum
pain in the rectum
bleeding from the rectum
With both gonorrhea and chlamydia, people with male anatomy may also experience abnormal swelling in their testicles and scrotum, and pain when they ejaculate.
You may also develop symptoms that affect your throat if you engage in oral sex with someone who has one of these conditions. This can cause mouth and throat symptoms, including sore throat and a cough.
Chlamydia symptoms
With chlamydia, people with female anatomy may experience more severe symptoms if the infection moves upward to the uterus and fallopian tubes. This can cause pelvic inflammatory disease (PID).
PID can cause symptoms such as:
fever
feeling sick
vaginal bleeding, even if you’re not having a period
intense pain in your pelvic area
Seek emergency medical help if you think you may have PID.
Gonorrhea symptoms
With gonorrhea, you may also notice rectal symptoms like itching, soreness, and pain when you defecate.
People with female anatomy may also notice heavier bleeding during their periods and pain during sex.
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What causes each condition?
Both conditions are caused by an overgrowth of bacteria. Chlamydia is caused by an overgrowth of the bacteria Chlamydia trachomatis.
Gonorrhea is caused by an overgrowth of bacteria called Neisseria gonorrhoeae.
How is each condition transmitted?
Both STIs are caused by bacterial infections that are transmitted through unprotected sexual contact, meaning sex without using a condom, dental dam, or another protective barrier between you and your partner during vaginal, anal, or oral sex.
It’s also possible to contract the infection through sexual contact that doesn’t involve penetration. For example, if your genitals come into contact with the genitals of someone who’s contracted the infection, it’s possible to develop the condition.
Both STIs can also be contracted through protected sex with a condom or other barrier if you don’t use protection properly, or if the barrier breaks.
Either STI can be contracted even if you aren’t showing visible symptoms. Both STIs can also be transmitted to a child at birth if the mother has either condition.
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Who’s at increased risk for these conditions?
You’re at increased risk for developing these and other STIs if you:
have multiple sexual partners at one time
don’t properly use protection, such as condoms, female condoms, or dental dams
regularly use douches which can irritate your vagina, killing healthy vaginal bacteria
have contracted an STI before
Sexual assault can also increase your risk of both chlamydia or gonorrhea.
Get tested for STIs as soon as possible if you’ve recently been forced to have non-consensual oral, genital, or anal sex. If you’re in the United States, you can also call the Rape, Abuse, and Incest National Network (RAINN) for support from people who can help without revealing any of your personal information or details of your experience.
How is each condition diagnosed?
Both STIs can be diagnosed using similar diagnostic methods. Your doctor may use one or more of these tests to ensure that the diagnosis is accurate and that the right treatment is given:
physical examination to look for symptoms of an STI and determine your overall health
urine test to test your urine for the bacteria that cause chlamydia or gonorrhea
blood test to test for signs of bacterial infection
swab culture to take a sample of discharge from your penis, vagina, or anus to test for signs of infection
How is each condition treated?
Both STIs are curable and can be treated with antibiotics, but you’re more likely to contract the infection again if you’ve had either STI before.
Treatment for chlamydia
Chlamydia is usually treated with a dose of azithromycin (Zithromax, Z-Pak) taken either all at once or over a period of a week or so (typically about five days).
Chlamydia can also be treated with doxycycline (Oracea, Monodox). This antibiotic is usually given as a twice-daily oral tablet that you need to take for about a week.
Follow your doctor’s dosage instructions carefully. It’s important to take the full dosage for the prescribed number of days so that the antibiotics can clear the infection. Not completing the round of antibiotics can cause you to become resistant to that antibiotic. This can be dangerous if you contract the infection again.
If you’re experiencing symptoms, they should begin to fade a few days after you start treatment.
Avoid sex until your doctor tells you that the infection has fully been cleared by the antibiotics. It can take two weeks or more for the infection to clear up, and during that time, you can still transmit the infection.
Treatment for gonorrhea
Your doctor will likely prescribe ceftriaxone (Rocephin) in the form of an injection into your buttock.
The CDC previously recommended ceftriaxone plus azithromycin, but the guidelines were changed because the bacteria causing gonorrhea are becoming increasingly resistant to azithromycin.
Using both antibiotics helps clear the infection better than using only one treatment alone.
As with chlamydia, don’t have sex until the infection clears, and be sure to take your entire dose.
Gonorrhea is more likely than chlamydia to become resistant to antibiotics. If you contract the infection with a resistant strain, you’ll need treatment with alternative antibiotics, which your doctor will recommend.
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What complications are possible for each condition?
Some complications of these STIs can happen to anyone. Others are unique to each sex due to differences in sexual anatomy.
Gonorrhea has more severe possible complications and is more likely to cause long-term problems like infertility.
In both males and females
Complications that may be seen in anyone include:
Other STIs. Chlamydia and gonorrhea both make you more susceptible to other STIs, including human immunodeficiency virus (HIV). Having chlamydia can also increase your risk of developing gonorrhea, and vice versa.
Reactive arthritis (chlamydia only). Also called Reiter’s syndrome, this condition results from an infection in your urinary tract (your urethra, bladder, kidneys, and ureters — the tubes that connect the kidneys to your bladder) or intestines. Symptoms of this condition cause pain, swelling, or tightness in your joints and eyes, and a variety of other symptoms.
Infertility. Damage to reproductive organs or to sperm can make it more challenging or, in some cases, impossible to become pregnant or to impregnate your partner.
In males
Testicular infection (epididymitis). Chlamydia or gonorrhea bacteria can spread to the tubes next to each of your testicles, resulting in infection and inflammation of testicle tissue. This can make your testicles swollen or painful.
Prostate gland infection (prostatitis). Bacteria from both STIs can spread to your prostate gland, which adds fluid to your semen when you ejaculate. This can make ejaculation or peeing painful, and cause fevers or pain in your lower back.
In females
Pelvic inflammatory disease (PID). PID happens when your uterus or fallopian tubes contain an infection from the bacteria. PID requires immediate medical attention in order to prevent damage to your reproductive organs.
Infections in newborns. Both STIs can be transmitted to a baby during birth from vaginal tissue that contains an infection from the bacteria. This can result in complications like eye infections or pneumonia.
Ectopic pregnancy. These STIs can cause a fertilized egg to become attached to tissue outside the uterus. This type of pregnancy won’t last until birth and can also threaten the mother’s life and future fertility if it’s not treated.
What measures can I take to prevent these conditions?
The only way that you can completely prevent yourself from catching chlamydia, gonorrhea, or another STI is by abstaining from sexual activity.
But there are also plenty of ways you can reduce your risk of contracting or transmitting these infections:
Use protection. Both male and female condoms are effective in helping to reduce your risk from infection by either bacteria. Using proper protection during oral or anal sex can also reduce your risk of infection.
Limit your sexual partners. The more sex partners you have, the more you risk exposing yourself to an infection. And because these STIs may not cause noticeable symptoms, sex partners may not know they have the condition.
Get regularly tested. Whether you’re having sex with multiple people or not, regular STI tests can help you remain aware of your sexual health and ensure that you’re not unknowingly transmitting an infection to others. Regular testing can also help you identify an infection even if you’re not experiencing any symptoms.
Don’t use products that affect your vaginal bacteria. Healthy bacteria in the vagina (called vaginal flora) helps fight off infections. Using products like douches or scented odor-reduction products can upset the balance of vaginal flora and make you more susceptible to infection.
The takeaway
Both chlamydia and gonorrhea can be transmitted in the same ways, and both can easily be treated using antibiotics.
Both are also preventable if you take precautions during sex, such as using protection and limiting the number of people you have unsafe sex with at any given time.
Regular STI testing, for both you and your sexual partners, can also help reduce the risk of transmitting an infection if you or a sexual partner develop an STI.
If you suspect an STI or have been diagnosed with one, stop all sexual activity and get treatment as soon as possible. If you’re diagnosed, tell anyone you’ve had sex with to get tested just in case.
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How we reviewed this article:
Healthline has strict sourcing guidelines and relies on peer-reviewed studies, academic research institutions, and medical journals and associations. We only use quality, credible sources to ensure content accuracy and integrity. You can learn more about how we ensure our content is accurate and current by reading our editorial policy.
Chlamydia [fact sheet]. (2017).
Gonococcal infections. (2018).
Gonorrhea [fact sheet]. (2017).
Table 1. sexually transmitted diseases — reported cases and rates of reported cases per 100,000 population, United States, 1941-2017. (2018).
Whiley DM, et al. (2012). The ticking time bomb: Escalating antibiotic resistance in Neisseria gonorrhoeae is a public health disaster in waiting.
Young K. (2020). CDC guidelines recommend ceftriaxone monotherapy for uncomplicated gonorrhea.
Our experts continually monitor the health and wellness space, and we update our articles when new information becomes available.
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8874 | https://www.grc.nasa.gov/www/k-12/airplane/vectpart.html | Vector Components
+ Text Only Site + Non-Flash Version + Contact Glenn Math and science were invented by humans to describe and understand the world around us. We observe that there are some quantities and processes in our world that depend on the direction in which they occur, and there are some quantities that do not depend on direction. Mathematicians and scientists call a quantity which depends on direction a vector quantity. A quantity which does not depend on direction is called a scalar quantity. A vector quantity has two characteristics, a magnitude and a direction. When comparing two vector quantities of the same type, you have to compare both the magnitude and the direction. On this slide we describe a mathematical concept which is unique to vectors; vector components. Vector components allow us to break a single vector quantity into two (or more) scalar quantities with which we have more mathematical experience. Vector components are used in vector algebra to add, subtract, and multiply vectors. Vectors are usually denoted on figures by an arrow. The length of the arrow indicates the magnitude of the vector and the tip of the arrow indicates the direction. The vector is labeled with an alphabetical letter with a line over the top to distinguish it from a scalar. We will denote the magnitude of the vector by the symbol |a|. The direction will be measured by an angle phi relative to a coordinate axis x. The coordinate axis y is perpendicular to x. Note: The coordinate axes x and y are themselves vectors! They have a magnitude and a direction. You first encounter coordinates axes when you learn to graph. So, you have been using vectors for some time without even knowing it! If we construct a dashed line from the tip of the vector a running parallel to the x-axis, it cuts the y-axis at a location we label ay. Similarly, a line from the tip of the vector parallel to the y-axis cuts the x-axis at ax. Using the sine and cosine relations from trigonometry: ay = |a| sin(phi) ax = |a| cos(phi) We call ax the x-component of a, and ay the y-component of a. The component equations are scalar equations; |a| and the trigonometric functions are just scalars. Any algebra involved with these quantities will be scalar algebra, not vector algebra. We have in essence replaced the single vector quantity a with two scalar quantities ax and ay. Looking very closely at these two equations, we notice that they completely define the vector quantity a; they specify both the magnitude and the direction of a. We can find the magnitude of the vector by using the Pythagorean Theorem. The components form two sides of a right triangle. To determine the length of the hypotenuse of the triangle: |a|^2 = ax^2 + ay^2 |a|^2 = |a|^2 sin^2(phi) + |a|^2 cos^2(phi) |a|^2 = |a|^2 (sin^2(phi) + cos^2(phi) ) |a|^2 = |a|^2 |a| = sqrt(|a|^2) Why go to all this trouble? Because, in aerospace, we are often dealing with forces and forces are vectors. Breaking a single vector force into several components allows us to study the resulting motion much more easily. Note: On this slide, for simplicity, we have developed the components in only two dimensions; there are two coordinate axes. In reality, there are three spatial dimensions and three components of all forces. This is important in our derivation of the general equations of motion for flight trajectories and for the Navier-Stokes and Euler equations which describe the forces and resulting motion of fluids in the engine. We can break very complex, three-dimensional, vector problems into only three scalar equations. Activities: Guided Tours Navigation .. Beginner's Guide Home Page + Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility CertificationEditor: Nancy Hall NASA Official: Nancy Hall Last Updated: May 13 2021 + Contact Glenn |
8875 | https://fiveable.me/hs-honors-geometry/unit-6/special-parallelograms-rectangles-rhombuses-squares/study-guide/QXQE9xnmaKAR3HX7 | Special parallelograms: rectangles, rhombuses, and squares | Honors Geometry Class Notes | Fiveable | Fiveable
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🔷Honors Geometry Unit 6 Review
6.3 Special parallelograms: rectangles, rhombuses, and squares
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Topic: 6.3
🔷Honors Geometry Unit 6 Review
6.3 Special parallelograms: rectangles, rhombuses, and squares
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🔷Honors Geometry
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Foundations of Geometry
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Congruent Triangles
Relationships in Triangles
Quadrilaterals
6.1 Classification and properties of quadrilaterals
6.2 Parallelograms and their properties
6.3 Special parallelograms: rectangles, rhombuses, and squares
6.4 Trapezoids and kites
Proportions and Similarity
Right Triangles and Trigonometry
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Special parallelograms are unique quadrilaterals with specific properties. Rectangles have right angles and congruent diagonals, while rhombuses have all sides equal and perpendicular diagonals. Squares combine both, making them extra special.
These shapes pop up everywhere in real life. Knowing their properties helps solve problems involving angles, side lengths, and diagonals. We can even construct these shapes using basic tools, which is pretty cool for design and architecture.
Properties and Theorems of Special Parallelograms
Properties of special quadrilaterals
Rectangles
Opposite sides parallel and congruent (same length)
All angles right angles (90°)
Diagonals congruent and bisect each other (divide into two equal parts)
Rhombuses
All sides congruent (same length)
Opposite sides parallel
Opposite angles congruent
Diagonals perpendicular and bisect each other
Squares
All properties of rectangles and rhombuses combined
All sides congruent and parallel
All angles right angles (90°)
Diagonals congruent, perpendicular, and bisect each other
Theorems for special parallelograms
Diagonals of a rectangle are congruent
Prove using the SSS (Side-Side-Side) congruence criterion
Opposite sides of a rectangle congruent
Diagonals bisect each other, creating four congruent triangles
Diagonals of a rhombus are perpendicular
Prove using the properties of isosceles triangles
Diagonals of a rhombus bisect each other, creating four congruent isosceles triangles
Base angles of isosceles triangles congruent
Sum of the base angles 90°, proving the diagonals perpendicular
Problem-solving with special quadrilaterals
Use the properties of special parallelograms to solve for missing angles, side lengths, and diagonal lengths
Given a rhombus with a side length of 5 units and one diagonal length of 8 units, find the length of the other diagonal
Use the Pythagorean theorem to solve for the missing diagonal length: $a^2 + b^2 = c^2$
$5^2 + 5^2 = 8^2 + d^2$
$50 = 64 + d^2$
$d^2 = 14$
$d = \sqrt{14}$
Construction of special parallelograms
Construct a rectangle given the lengths of two adjacent sides
Draw a segment with the length of one side
Construct perpendicular lines at each endpoint of the segment
Mark the length of the other side on one of the perpendicular lines
Connect the endpoints to form the rectangle
Construct a rhombus given the lengths of the diagonals
Draw the diagonals intersecting at their midpoints
Connect the endpoints of the diagonals to form the rhombus
Construct a square given the length of one side
Draw a segment with the given side length
Construct perpendicular lines at each endpoint of the segment
Mark the same side length on each perpendicular line
Connect the endpoints to form the square
Applying Special Parallelogram Properties
Problem-solving with special quadrilaterals
Utilize the unique properties of each special parallelogram to find missing measurements
In a square with a diagonal length of 10 units, find the side length
Diagonals of a square bisect each other, so each half of the diagonal is 5 units
Use the Pythagorean theorem to solve for the side length: $a^2 + a^2 = 5^2$
$2a^2 = 25$
$a^2 = 12.5$
$a = \sqrt{12.5}$
Apply the properties of special parallelograms in real-world contexts
A rectangular garden has a perimeter of 60 meters and a length that is 5 meters longer than its width. Find the dimensions of the garden.
Let the width be $x$ meters
The length is $x + 5$ meters
Perimeter formula for a rectangle: $P = 2l + 2w$
$60 = 2(x + 5) + 2x$
$60 = 2x + 10 + 2x$
$60 = 4x + 10$
$50 = 4x$
$x = 12.5$
The width is 12.5 meters, and the length is 17.5 meters
6.2 BackNext 6.4
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8876 | https://www.khanacademy.org/math/cc-fifth-grade-math/powers-of-ten/imp-multiplying-and-dividing-decimals-by-10-100-and-1000/a/multiplying-and-dividing-by-powers-of-10 | Dividing decimals by 10, 100, and 1000 (article) | Khan Academy
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Skip to lesson content
5th grade math
Course: 5th grade math > Unit 10
Lesson 2: Multiplying and dividing decimals by 10, 100, and 1000
Multiplying and dividing decimals by 10
Multiply and divide decimals by 10
Multiplying and dividing decimals by 10, 100, 1000
Multiply and divide decimals by 10, 100, and 1000
Multiplying decimals by 10, 100, and 1000
Dividing decimals by 10, 100, and 1000
Math>
5th grade math>
Powers of ten>
Multiplying and dividing decimals by 10, 100, and 1000
© 2025 Khan Academy
Terms of usePrivacy PolicyCookie NoticeAccessibility Statement
Dividing decimals by 10, 100, and 1000
AZ.Math: 5.NBT.A.2
Google Classroom
Microsoft Teams
Practice dividing decimals by 10, 100, and 1000.
Part 1: Dividing decimals by 10
Key idea: Dividing by 10 moves every digit one place to the right.
Let's visualize it! Drag the dot all the way to the right.
1 5 4 5.14 5.14÷10
5.14÷10\=
Check
Show solution
5.14÷10\=0.514
Let's try a few more dividing by 10 :
44.2÷10\=
Check
145.18÷10\=
Check
0.678÷10\=
Check
Show solutions
Dividing by 10 moves every digit one place to the right.
44.2÷10\=4.42
145.18÷10\=14.518
0.678÷10\=0.0678
And one multiplying by 10 (for review):
0.933×10\=
Check
Show solution
Multiplying by 10 moves every digit one place to the left.
0.933×10\=9.33
Part 2: Dividing decimals by 100
Key idea: Dividing by 100 moves every digit two places to the right.
Let's visualize it! Drag the dot all the way to the right.
Notice that we add zeros to fill the empty place values.
8 2 2.8 2.8÷100
2.8÷100\=
Check
Show solution
2.8÷100\=0.028
Let's try a few more dividing by 100 :
6.33÷100\=
Check
71.005÷100\=
Check
189.2÷100\=
Check
Show solutions
Dividing by 100 moves every digit two places to the right.
6.33÷100\=0.0633
71.005÷100\=0.71005
189.2÷100\=1.892
And one multiplying by 100 (for review):
5.4×100\=
Check
Show solution
Multiplying by 100 moves every digit two places to the left.
5.4×100\=540
Part 3: Dividing decimals by 1000
Key idea: Dividing by 1000 moves every digit three places to the right.
Let's visualize it! Drag the dot all the way to the right.
Notice that we add zeros to fill the empty place values.
0 6 0.6 0.6÷1000
0.6÷1000\=
Check
Show solution
0.6÷1000\=0.0006
Let's try a few more dividing by 1000 :
7500.9÷1000\=
Check
0.1÷1000\=
Check
623.88÷1000\=
Check
Show solutions
Dividing by 1000 moves every digit three places to the right.
7500.9÷1000\=7.5009
0.1÷1000\=0.0001
623.88÷1000\=0.62388
And one multiplying by 1000 (for review):
0.0043×1000\=
Check
Show solution
Multiplying by 1000 moves every digit three places to the left.
0.0043×1000\=4.3
Part 4: Let's look at the pattern.
Dividing by 10 moves every digit 1 place to the right.
Dividing by 100 moves every digit 2 places to the right.
Dividing by 1000 moves every digit 3 places to the right.
Dividing by 100,000 moves every digit
places to the right.
Dividing by 1,000,000 moves every digit
places to the right.
Check
Show solutions
The number of zeros tells us how many times to move the digits.
Dividing by 1,000,000 moves every digit 6 places to the right.
Dividing by 100,000 moves every digit 5 places to the right.
Part 5: Challenge time!
Use the pattern above to answer the following questions.
333.4÷100,000\=
Check
29.7÷1,000,000\=
Check
Show solutions
Dividing by 100,000 moves every digit five places to the right.
333.4÷100,000\=0.003334
Dividing by 1,000,000 moves every digit six places to the right.
29.7÷1,000,000\=0.0000297
And a multiplication challenge:
0.055×10,000\=
Check
Show solution
Multiplying by 10,000 moves every digit four places to the left.
0.055×10,000\=550
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gets hard once it gets to 100
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janakba Rathod
5 years ago Posted 5 years ago. Direct link to janakba Rathod's post “wanna tip? Just remember ...”
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wanna tip? Just remember when dividing with decimals just slide decimal to the right acoordingto how many zeros in the number
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4 years ago Posted 4 years ago. Direct link to Caden Florencio's post “gets hard on 100,000 and ...”
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gets hard on 100,000 and 1m
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4 years ago Posted 4 years ago. Direct link to MukhtarTuraani's post “think about like this: if...”
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think about like this: if you want to divide shift the decimal point to the left based on the numbers of zeros. for example if you want to divide 4.25 ÷ 10 you will shift the decimal point one time to the left (because there is one zero in the ten). if you want to divide 4.35 ÷ 100 you will shift the decimal point two times to the left because there is two zeros in the 100.
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Ridhima ghiya
4 years ago Posted 4 years ago. Direct link to Ridhima ghiya's post “i always forget where to ...”
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i always forget where to move the point while dividing and multiplying
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Ian Pulizzotto
4 years ago Posted 4 years ago. Direct link to Ian Pulizzotto's post “When multiplying by a pow...”
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When multiplying by a power of 10 (such as 10, 100, 1000, etc), move the decimal point the same number of places to the right as there are zeros in the power of 10.
When dividing by a power of 10, move the decimal point the same number of places to the left as there are zeros in the power of 10.
Have a blessed, wonderful new year!
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i thought it was neat and a good website for learners
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5 years ago Posted 5 years ago. Direct link to Jaeda's post “Can we get hints, and the...”
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Can we get hints, and then when we are really stuck, we can get the answer.
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bo wang
9 months ago Posted 9 months ago. Direct link to bo wang's post “move the numbers left, or...”
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move the numbers left, or decimal point left
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am I the only one who is doing this?
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2 years ago Posted 2 years ago. Direct link to alapankozhapn's post “Nah bro you good. you goo...”
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Nah bro you good. you good
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I can say its easy because I used a calculater cause my teacher don't care but if you didn't get to use a calculater then it would probably be hard...
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bo wang
9 months ago Posted 9 months ago. Direct link to bo wang's post “move the numbers left, or...”
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move the numbers left, or decimal point left(if you can't/don't want use a calculator)
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its a little difficlet when you go to diviseion
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9 months ago Posted 9 months ago. Direct link to bo wang's post “move the numbers left, or...”
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move the numbers left, or decimal point right
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never gonna give u up
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there is so much questions):
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never gonna live u downn
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kekuacara
a month ago Posted a month ago. Direct link to kekuacara's post “How ever many zeros are i...”
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How ever many zeros are in the number after the multiplication or division sign, that's how many times you move the decimal point.
EXAMPLE: 145.28 x 100 =14528.0
EXAMPLE: 145.28➗100 =1.4528
ps. Multiplication=move it right
Division=move it left. I'm only in 5th grade. Hope this helped! :)
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8877 | https://www.ck12.org/section/exploring-geometric-sequences-%3A%3Aof%3A%3A-te-infinite-series---ti/ | Exploring Geometric Sequences | CK-12 Foundation
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Exploring Geometric Sequences
Difficulty Level: Basic | Created by: CK-12
Last Modified: Nov 04, 2014
Read Resources Details Attributions
This activity is intended to supplement Calculus, Chapter 8, Lesson 1.
ID: 10992
Time required: 20 minutes
Activity Overview
In this activity, students will explore geometric series. They will consider the effect of the value for the common ratio and first term using the TRANSFRM APP for the TI-84. Students will graphically analyze geometric series using graphs.
Topic: Sequences and Series
Explore geometric sequences
Teacher Preparation and Notes
This activity serves as a nice introduction to geometric series. Students will need the TRANSFRM APP on their TI-84. The accompanying worksheet can add depth to the questions and helpful key press explanation.
An extension would be to solve infinite geometric series that converge using sigma notation and the limit of the partial sum formula.
Students should make sure to turn off all Stat Plots before beginning the activity.
To exit the Transformation App, go to the App menu, press enter on Transfrm and then select Uninstall.
To download the Transformation Graphing Application, go to and select Transformation Graphing from the Applications header.
Associated Materials
Student Worksheet: Exploring Geometric Sequences
Transformation Graphing Application
Example of Geometric Sequence
Students are shown the path of a ball that is bouncing. Show that the common ratio of the heights is approximately the same.
2.8 4.0=0.7 2.0 2.8≈0.71 1.4 2.0=0.7
[Figure 1]
Changing the Common Ratio
For the first part of this activity, students explore geometric sequences graphically by varying the value of r, the common ratio. Pressing the right or left arrows will change the value of r and updates the graph each time. Note that students will not see a graph for A values less than zero.
They should see that r−values between 0 and 1 could model the heights of the ball in the example.
Inquiry questions
Why do you think the r−value is called the common ratio? [Answer: The ratios of consecutive terms are the same; They have that in common.]
What would happen if you added all the terms of this sequence? For what common ratio conditions do you think the sum will diverge (get larger, and not converge to some number)?
[Figure 2]
[Figure 3]
Changing the Initial Valueandthe Common Ratio
Students will now change the equation in the Transfrm App to explore the changes in the graph of the general series, a n=a 1⋅r n−1. Students are use the up and down arrow to move from one variable to another and use the left and right arrows to change the value of the variable. They should be sure to try negative and positive values for B for various values of A. (Y 1=B∗A∧(X−1)
Inquiry question
Which variable seems to have a more profound effect on the sequence? Explain.
Extension – Deriving and Applying the Partial Sum Formula
On the worksheet, students are shown the derivation of the formula for the sum of a finite geometric series. You may need to explain in detail the substitution in the third line. For example, a 2=r⋅a 1, so then a 3=r⋅a 2=r⋅(r⋅a 1)=r 2⋅a 1. In the fourth line, r is multiplied by both sides, changing r n−1 to r n.
Students can apply the formula to find the sum of the series given on the worksheet.
Inquiry questions:
If you cut a piece of paper in half, then cut one of the halves in half and repeated the process, what is the sum of all these fractional pieces? [Answer: one whole]
However, if you gave your teacher one penny on the first day of class, 2 on the next, then 4, 8, 16, 32, ... would this number converge?
How would the formula for a finite geometric sequence look if r is a fractional number and n is increasing?
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Difficulty Level
Basic
Tags
geometric sequences,sequences,infinite series,AP exam,transfrm application,Taylor and Maclaurin series,TI.MAT.ENG.TE.1.Calculus.9, (4 more)
Subjects
mathematics
Grades
11,12
Standards Correlations
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License
CC BY NC
Language
English
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Date Created
Feb 24, 2012
Last Modified
Nov 04, 2014
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| | [Figure 2] | License:CC BY-NC |
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8878 | https://www.lvnhm.org/post/the-days-are-getting-longer-it-s-recorded-in-the-fossils | The Days are Getting Longer: It’s Recorded in the Fossils
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The Days are Getting Longer: It’s Recorded in the Fossils
Las Vegas Natural History Museum
Jan 1, 2021
2 min read
Dr. Rowland back for our first Fossil Friday of 2021!
Does it seem like the years are flying by, with each year going faster than the one before? Well, the years are actually going by more slowly. As we celebrate the beginning of a new year, it is a good time to explore how some fossils record celestial dynamics. Let me start by saying that the length of time it takes for each trip around the sun does not change. What changes is Earth’s spin rate; it is slowing down. Fluids sloshing around on Earth―the oceans and the fluid outer core―are a drag on Earth’s spin rate. So, Earth spins slightly slower each year than it did the previous year. “How much slower?” you might ask. So, I’ll tell you: 0.0018 seconds per year slower! Okay, it’s a tiny amount, but it adds up over geologic time.
Over the past several hundreds of millions of years, as the spin rate has slowed, the number of days per year has been decreasing. And the results are recorded in fossil corals. The first image shows a rugose coral (also called a horn coral) that is approximately 400 million years old, from the Devonian Period. Notice the conspicuous ridges, or ‘rugosities,’ on the shell. The presence of these rugosities is what gives these corals their name rugose corals. Each ridge represents one year of growth.
The second image schematically shows tiny growth lines within a rugose coral, seen by cutting open the fossil and looking at it with a microscope. These microscopic growth lines record daily increments of shell growth. You see where I’m going with this: the presence of daily growth lines, together with annual ridges, allow paleontologists to determine how many days were in a year when this coral was growing its shell. The results indicate that each year was about 400 days long in the Devonian Period, 400 million years ago.
‘Happy New Year’ from the Las Vegas Natural History Museum. This year you will have 180,000 more nanoseconds to visit the museum!
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1996 AHSME Problems/Problem 26
Problem
An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:
(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(d) the selection of one marble of each color.
What is the smallest number of marbles satisfying the given condition?
Solution
Let the bag contain marbles total, with representing the number of red, white, blue, and green marbles, respectively. Note that .
The number of ways to select four red marbles out of the set of marbles without replacement is:
The number of ways to select one white and three red marbles is:
The number of ways to select one white, one blue, and two red marbles is:
The number of ways to select one marble of each colors is:
Setting the first and second statements equal, we find:
Setting the first and third statements equal, we find:
Setting the last two statements equal, we find:
These are all the "linking equations" that are needed; the transitive property of equality makes the other three equalities unnecessary.
From the first equation, we know that must be more than a multiple of , or that
Putting the first equation into the second equation, we find . Therefore, . Using the Chinese Remainder Theorem, we find that .
The third equation gives no new restrictions on ; it is already odd by the first equation.
Thus, the minimal positive value of is . This requires by the third equation, and by the first equation. Finally, the second equation gives .
The minimal total number of marbles is , which is option .
See also
1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25Followed by
Problem 27
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25•26•27•28•29•30
All AHSME Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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2022 AIME I Problems/Problem 6
Contents
1 Problem
2 Solution 1
3 Solution 2 (Rigorous)
4 Solution 3
5 Solution 4
6 See Also
Problem
Find the number of ordered pairs of integers such that the sequenceis strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
Solution 1
Since and cannot be an arithmetic progression, or can never be . Since and cannot be an arithmetic progression, and can never be . Since , there are ways to choose and with these two restrictions in mind.
However, there are still specific invalid cases counted in these pairs . Since cannot form an arithmetic progression, . cannot be an arithmetic progression, so ; however, since this pair was not counted in our , we do not need to subtract it off. cannot form an arithmetic progression, so . cannot form an arithmetic progression, so . cannot form an arithmetic progression, ; however, since this pair was not counted in our (since we disallowed or to be ), we do not to subtract it off.
Also, the sequences , , , , and will never be arithmetic, since that would require and to be non-integers.
So, we need to subtract off progressions from the we counted, to get our final answer of .
~ ihatemath123
Solution 2 (Rigorous)
We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50. (Note if we take only 1 at a time, there will have to be 3 of , which is impossible.)
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.
Finally, we count the sequences that are something like (one of 3,4,5,), , (one of 30, 40, 50). If this is to be the case, then let be the starting value in the sequence. The sequence will be ; We see that if we subtract the largest term by the smallest term we have , so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are . Of these, only the last is invalid because it gives , larger than our bounds . Therefore, we subtract from this case.
Our final answer is
~KingRavi
Solution 3
Denote .
Denote by a subset of , such that there exists an arithmetic sequence that has 4 terms and includes but not .
Denote by a subset of , such that there exists an arithmetic sequence that has 4 terms and includes but not .
Hence, is a subset of , such that there exists an arithmetic sequence that has 4 terms and includes both and .
Hence, this problem asks us to compute
First, we compute .
We have .
Second, we compute .
: .
We have . Thus, the number of solutions is 22.
: .
We have . Thus, the number of solutions is 9.
Thus, .
Third, we compute .
In , we have . However, because , we have . Thus, .
This implies . Note that belongs in .
Thus, .
Fourth, we compute .
: In the arithmetic sequence, the two numbers beyond and are on the same side of and .
Hence, . Therefore, the number solutions in this case is 3.
: In the arithmetic sequence, the two numbers beyond and are on the opposite sides of and .
: The arithmetic sequence is .
Hence, .
: The arithmetic sequence is .
Hence, .
: The arithmetic sequence is .
Hence, . However, the sequence is not strictly increasing.
Putting two cases together,
Therefore,
~Steven Chen (www.professorchenedu.com)
Solution 4
divide cases into .(Notice that can't be equal to , that's why I divide them into two parts. There are three cases that arithmetic sequence forms: .(NOTICE that IS NOT A VALID SEQUENCE!) So when , there are possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)
When , there are ways.
In all, there are possible sequences.
~bluesoul
See Also
2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5Followed by
Problem 7
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
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8881 | https://www.pearson.com/channels/organic-chemistry/asset/6839ffe5/do-the-sp2-carbons-and-the-indicated-sp3-carbons-lie-in-the-same-plane | Organic Chemistry
Improve your experience by picking them
Predict the hybridization, geometry, and bond angles for the central atoms in
b. CH3CH=NH
Predict the hybridization, geometry, and bond angles for the central atoms in
a. but-2-ene, CH3CH=CHCH3
a. Which of the species have bond angles of 109.5°?
b. Which of the species have bond angles of 120°?
H2O H3O+ +CH3 BF3
For each of the following molecules, indicate the hybridization of each carbon and give the approximate values of all the bond angles:
a. CH3C≡CH
Predict the hybridization and geometry of the carbon and nitrogen atoms in the following molecules and ions. (Hint: Resonance.)
a.
b.
c.
In most amines, the nitrogen atom is sp3 hybridized, with a pyramidal structure and bond angles close to 109°. In urea, both nitrogen atoms are found to be planar, with bond angles close to 120°. Explain this surprising finding. (Hint: Consider resonance forms and the overlap needed in them.)
Predict the hybridization, geometry, and bond angles for the carbon and nitrogen atoms in acetonitrile (CH3–C≡N:). |
8882 | https://users.wpi.edu/~bservat/birigid.pdf | BIRIGIDITY IN THE PLANE BRIGITTE SERVATIUS Abstract. We consider the 2-dimensional generic rigidity matroid R(G) of a graph G.
The notions of vertex and edge birigidity are introduced.
We prove that vertex birigidity of G implies the connectivity of R(G) and that the connectivity of R(G) implies the edge birigidity of G. These implications are not equivalences.
A class of minimal vertex birigid graphs is exhibited and used to show that R(G) is not representable over any finite field.
1. Introduction and Basic Definitions Let G = (V, E) be a simple graph on the edge set E, vertex set V . We define the support σ(F) of a subset F of E to be the set of endpoints of edges in F.
We define a subset F of E to be independent if |F ′| ≤2|σ(F ′)| −3 holds for all subsets F ′ of F. It is well known, see and , that these independent edge sets are the independent sets of a matroid, the so-called 2-dimensional generic rigidity matroid, R(G), of the graph G. The closure operator and rank function of this matroid will be denoted by c and r respectively. The term circuit will always refer to a circuit in R(G). Some properties of circuits are discussed in . Note that R(G) may be considered as a restriction of the rigidity matroid of a sufficiently large complete graph.
G = (V, E) is called rigid if r(E) = 2n −3, where |V | = n. G is called edge birigid, if r(E −e) = 2n −3 for every e ∈E. G is called vertex birigid, if G is rigid and r(E −star(v)) = 2(n −1) −3 = 2n −5 for every v ∈V , where star(v) denotes the set of edges adjacent to v. We will henceforth abbreviate E −star(v) with E −v.
To simplify notation and language we will not distinguish between sets of edges and the subgraphs they induce. Some simple examples of graphs with specified rigidity properties are given in figure 1.
The following observations are immediate consequences of the definitions. The union of two graphs G1 and G2 having at most one vertex in common is not rigid, and c(G1 T G2) = c(G1) T c(G2).
If two rigid graphs intersect in two or more vertices, their union is rigid.
Let us call two edges of G related if they are both contained in a rigid subgraph of G. Clearly the so defined relation is symmetric. An edge constitutes a rigid subgraph, which shows reflexivity. For transitivity; if edges e and f are contained in a rigid subgraph H1 of G and f and h are contained in a rigid subgraph H2 of G, then H1 and H2 intersect in at least two vertices, namely the endpoints of f, so their union is a rigid graph containing e and h.
Thus rigidity induces an equivalence relation the edge set of G. The equivalence classes are called r-components. It follows that r-components have at most one vertex in common and 1 2 BRIGITTE SERVATIUS that birigid graphs are at least 3-connected. Moreover, R(G) can be written as the direct sum over the r-components of G. This follows from the observation that circuits are rigid, in fact edge birigid, see .
We shall often use the following property of R(G): Assume the edge set F induces a subgraph of G containing a vertex, v, of valence three. Then F is independent if and only if there is an edge e connecting neighbors of v such that e is not contained in F and F −v+e is independent. We say R(G) satisfies the 1-extendability property, see . Note that e need not be contained in G.
If the vertices of G are ”generically” embedded in the plane, see , and the edges of G are replaced by rigid bars, which are pin-jointed at their endpoints, the resulting structure will be rigid if and only if G is rigid in the sense defined above.
See .
If the vertices of G are restricted to a line, and the edges of G are again replaced by rigid bars, the resulting structure will be rigid if and only if G is connected, and we may characterize the 1-dimensional generic rigidity matroid M(G), of the graph G as follows: a subset F of E is independent if and only if |F ′| ≤|σ(F ′)| −1, holds for all subsets F ′ of F, i.e., the independent sets in this matroid are simply the edge sets of subforests of G. M(G), is called the cycle matroid M(G), of G, see .
Observe that M(G), and R(G) are matroids defined on the edge set of G, and that the vertex set of G is used only via the support function, to define independent sets. Consequently, there is no property of M(G), or R(G) that corresponds directly to the connectivity of G. Whitney, , calls a matroid M on S connected if r(A) + r(S−A) > r(S) holds for every non-empty proper subset a of S. With this definition M(G), is connected if and only if G is biconnected. It is natural to ask for relations between the connectivity of R(G) and the rigidity of G. This will be done in section II.
Every pair of edges in a biconnected graph is contained in a cycle. A cycle is an edge-minimal vertex-biconnected graph. Note that any cycle has exactly one edge more than it needs to be connected. A biconnected graph can simply be thought of as a union of sufficiently intersecting cycles.
It is natural to look for a rigid analogue: Given a birigid graph, can we write it as a union of birigid graphs of minimal excess, where the excess of a rigid graph G = (V, E) is defined to be |E| −r(E). Observe that the only birigid graph of excess one is the complete graph on four vertices, since the average valence in a birigid graph of excess one on n vertices is greater than or equal to 4 −(4/n).
Therefore, a birigid graph on more than four vertices contains a vertex of valence at least four. The removal of a vertex of valence four decreases the excess by two, therefore a birigid graph on more than four vertices has to have excess at least two.
In section 3 we show that there are infinitely many birigid graphs of excess two.
We give an inductive procedure to construct them all. We also show that they do not, unfortunately, fullfill the role of universal building blocks of birigid graphs.
2. Birigidity of G and Connectivity of R(G) Theorem 1. If G has no isolated vertices and more than one edge, and R(G) is connected, then G is edge birigid, but not conversely.
Proof. G = (V, E) is rigid, otherwise R(G) could be written as the direct sum over the rigid components of G. Hence r(E) = 2|V | −3.
BIRIGIDITY IN THE PLANE 3 Assume that there is an edge, e, such that G −e is not rigid. Then r(E −e) = 2|V |−4 and r(E −e)+r(e) = r(E). The last equation contradicts the connectivity of R(G).
The converse is not true: Let Go be minimally rigid, having no vertices and 2no −3 edges. We attach to each edge ei a circuit Ci, 1 ≤i ≤(2no −3), Ci having ni vertices, by identifying one edge of each Ci with one edge of Go. Then the resulting graph is clearly rigid and hence has rank 2n −3, where n = n0 + 2no−3 X i=1 (ni −2).
So Pn i=1 ni = n + 3n0 −6. The rank if each Ci is 2ni −3.
If we sum over the ranks, we get 2no−3 X i=1 r(Ci) = 2no−3 X i=1 (2ni −3) = −3(2no −3) + 2 2no−3 X i=1 ni = −6no + 9 + 2n + 6no −12 = 2n −3 = r(G).
So M(G)is not connected. On the other hand, G is clearly edge birigid.
□ An example with n0 = 3 is drawn in figure 1(vi).
Theorem 2. If G = (V, E) is birigid and |V | > 3, then R(G) is connected but not conversely.
Proof. Assume that G is birigid and that R(G) is not connected.
Consider the connected components Ri of R(G). Then there is a partition of E, E = E1 S E2 S · · · S Ek, such that R(G) = R1 + R2 + ... + Rk, where Ri = R(Gi), with Gi = (σ(Ei), Ei). Every Gi is rigid, so it follows that (1) 2|V | −3 = r(G) = k X i=1 r(Gi) = k X i=1 (2|σ(Ei)| −3).
Let ni be the number of vertices in the support of Ei which are also contained in the support of some Ej, i ̸= j and let Ni be the number of vertices contained only in the support of Ei. Denote by N the number of vertices of G which are contained in exactly one of the σ(Ei)’s, and by n the number of vertices which occur in more than one of these supports. Ni, N, ni, and n satisfy the following equations: i. Ni = |σ(Ei) − [ j̸=i σ(Ej)| ii. |σ(Ei)| = ni + Ni iii. N = k X i=1 Ni iv. |V | = n + N.
So (2) n ≤1 2 k X i=1 ni.
4 BRIGITTE SERVATIUS Rewriting 1 in this new notation we obtain 2n + 2N −3 = k X i=1 (2(ni + Ni) −3) or (3) 2n = 3(1 −k) + k X i=1 2ni so that 2 and 3 give [ k X i=1 2ni] −3(k −1) ≤ k X i=1 ni, or (4) k X i=1 ni ≤3(k −1).
Furthermore, since every cutset in a birigid graph has cardinality at least 3, we have that |σ(Ei) \ i̸=j σ(Ej)| ≥3, which implies that ni ≥3 for all i. This combined with 4 gives 3k ≤ k X i=1 ni ≤3(k −1), a contradiction.
If R(G) is connected, G need not be birigid: If G is a wheel, R(G) consists of a single circuit and hence is connected.
□ But the removal of the center vertex leaves a non-rigid graph if the number of spokes is larger than 3.
3. Birigid Graphs of Excess Two G is called edge minimally birigid if G is birigid but G −e is not birigid for all e ∈E(V ).
In this section we will restrict our attention to an edge minimal vertex birigid graph G = (V, E), which has exactly two edges more than it needs to be rigid, i.e.
|E| = 2|V | −1, r(E) = 2|V | −3.
We first list some elementary properties of G.
Proposition 1. Let G be a birigid graph of excess two. Then (1) G contains at least five vertices, (2) If e ∈E(G), then G −e is not birigid, and (3) G has exactly two vertices of valence three and the remaining vertices each have valence four.
Proof.
(1) Simple graphs on less than five vertices do not contain enough edges to satisfy |E| = 2|V | −1.
BIRIGIDITY IN THE PLANE 5 (2) G −e is not a complete graph. G −e has excess one. Since the only birigid graph of excess one is K4, G −e is not birigid.
(3) Since G is rigid, it contains no vertex of valence less than two. Suppose that G had a vertex v of valence two. Let w be adjacent to v. Then G −w contains a vertex of valence one and is not rigid. Now suppose G has a vertex v of valence k. Then G −v has n −1 vertices and 2(n −1) −(k −1) edges. Since G −v is rigid, k −1 < 4, which implies that k < 5. Finally, if there are m vertices of valence three, we have 3m + 4(n −m) = 2(2n −1), which gives m = 2.
□ The simplest birigid graph of excess two can be obtained from K5 by deleting an edge. This graph contains two copies of K4 as subgraphs. By “attaching” to K4 two adjacent vertices of valence three, we obtain a birigid graph on six vertices.
We remark that birigid graphs on more than 6 vertices do not contain a birigid subgraph of positive excess.
Next, we examine the circuit structure of R(G): Theorem 3. A graph on n vertices with 2n −1 edges is birigid if and only if there is a partition of the edge set E of G, E = E1 T E2 T · · · T Ek such that E −Ei is a circuit in R(G) for all i, and either (1) Ei is an edge for 3 ≤i ≤k and E1 and E2 are stars of two vertices of valence three, or, (2) Ei is an edge for 2 ≤i ≤k and E1 is the union of stars of two adjacent vertices of valence three.
Proof. Assume that there exists such a partition. Consider a class containing ex-actly one element e. Then E −e is a circuit of R(G), so G −e is a graph with minimum valence at least three, and e has endpoints of valence at least four in G.
Condition i) or ii) imply that G has 2 vertices of valence three and we conclude by a simple counting argument that all other vertices are of valence four.
Depending on whether or not the two vertices of valence three are adjacent in G, conditions i) or ii) imply that the removal of a vertex of valence three of G results in a circuit or in a circuit with a vertex of valence two attached, a rigid graph in both cases.
Consider a vertex v of valence four in G. Remove an edge e of star(v) with endpoints of valence four. E −e is a circuit by assumption, and v has valence three in this circuit. Recall that a circuit is edge birigid. By deleting an edge in star v, we therefore obtain a rigid graph in which v has valence two. The removal of a vertex of valence two does not destroy the rigidity of a graph, so E −v is rigid.
Conversely, assume that G is edge birigid on n vertices and 2n −1 edges. Since r(E) = |E|−2, and every edge is contained in a circuit, E is the union of two distinct circuits, and can be partitioned into a collection of sets Ei such that E −Ei is a circuit for each i, and |E −Ei| = 2|σ(E −Ei)| −2, see for example or .
Subtracting this equation from |E| = 2|σ(E)| −1 gives ∗|Ei| = 2|σ(E) −σ(E −Ei)| + 1.
If E and E −Ei have the same support then Ei is a single edge. If σ(E)−σ(E − Ei) = 1, then Ei contains all edges of the star of a vertex in G. The equation gives |Ei| = 3. Since every vertex in G has valence at least three, Ei must be a 6 BRIGITTE SERVATIUS star of a vertex of valence three, and the two vertices of valence three in G are not adjacent because E −Ei is a circuit.
If |σ(E)−σ(E−Ei)| = 2, then Ei contains all edges of the stars of two vertices of G. The equation gives |Ei| = 5, so Ei must be the union of two adjacent vertices of valence three in G.
If |σ(E)−σ(E −Ei)| > 2, then Ei contains all edges of the star of 3 vertices of G.
One of these must be of valence four. But the removal of a vertex of valence four leaves an independent set since G is birigid. The desired partition is so established and the proof of the theorem is complete.
□ Examples of graphs with a partition of type i) and ii) are given in the figure below.
Clearly we can ”string together” as many triangles as we wish to obtain birigid graphs of excess two of arbitrarily large size. Also the number of classes in the partition described in theorem 1 is unbounded. From a theorem of Tutte , we know that, if M is a matroid representable over a finite field k of order n and S is the union of two cycles of M with r(S) = |S| −2 and S1, · · · , Sm is a partition of S such that S −Si is a cycle of M, then m is bounded by n + 1. Hence we have proved Corollary 1. There is no finite field k such that R(G) is representable over k for all G.
Consider an edge minimal birigid graph G = (V, E). For every edge e in E there exists a nonempty set Ve of vertices of G such that E −e −v is nonrigid for all v ∈Ve. Elements of Ve are called essential vertices for the edge e.
From a given birigid graph of excess two, we want to construct a larger birigid graph of excess two, by attaching a vertex of valence three and removing one edge from the given graph.
To formalize this idea, we introduce some notation: Let T be a graph on four vertices and three edges, where one vertex is of valence three and construct a graph G + T by identifying vertices a, b, c of G with the vertices of valence one in T.
We can now prove Theorem 4. Let G be a birigid graph of excess two, and let T and {a, b, c} be as described above. Then: (1) G + T is birigid, (2) a necessary and sufficient condition for G + T to be edge minimally birigid is that the set {a, b, c} not be contained in V −Ve for any edge e of G; (3) if G + T is not edge minimally birigid, then there is an edge e such that G + T −e is birigid of excess two; and (4) there is always a choice of {a, b, c} such that G −T is not edge minimal.
Proof.
(1) The removal of T results in a birigid, and hence rigid graph, and the removal of any vertex v ∈G from G + T removes at most one edge from T, and since G −v is rigid, so is G + T −v.
(2) Sufficiency:: Let e be any edge of G. Since the intersection of Ve with {a, b, c} is nonempty, the removal of e and any vertex v in this inter-section leaves a nonrigid graph, G −e −v, which has the same rigidity properties as G + T −e −v.
BIRIGIDITY IN THE PLANE 7 Necessity:: Assume the existence of an edge e of G such that {a, b, c} is contained in V −Ve. Observe that all vertices of valence four of G which are not endpoints of e are elements of Ve. Therefore least one vertex in the set {a, b, c} is an endpoint of e.
There are two cases.
(a) a and b are endpoints of e and c is of valence three in G.
Theorem 3.1 implies that a vertex v of valence three is essential for a nonempty set of edges only if the two vertices of valence three in G are adjacent. In this case v is essential for the two edges not contained in a circuit in G −v. It follows that c is not adjacent to a possible endpoint of e of valence three and all essential vertices for e are of valence four. This means that E−v is rigid of zero excess, i.e. independent for all v ∈Ve. Therefore E −v −e is independent and e is not in its closure. By the 1-extendability property E +T −e−v is independent and hence rigid.
(b) e has endpoints of valence four in G, one of them being a, and b and c are of valence three. Remove e and star(v) for some v ∈Ve. Repeating the argument in (a) we show that E −e −v is independent and non-rigid. Consider the r −components of E −e−v, and assume that a, b, and c are contained in the same r-component.
This component is independent, and we count that exactly 3 edges of G −e are incident with it, contradicting the fact that G −e is a circuit by theorem 1. So, {a, b, c} is not contained in one r- component of E−e−v and the 1-extendability property implies that E + T −e −v is independent and hence rigid for all v in Ve, so G + T −e is birigid.
(3) If G + T is not edge minimally birigid, then there is an edge e in G + T such that G + T −e is birigid. G + T −e has excess two.
(4) For an edge e with endpoints a and b, both of valence four, a vertex c of valence three is not essential by theorem 1.
The proof of the theorem is now complete.
□ Given an edge minimal vertex birigid graph of excess two on n vertices, we can get an edge minimal vertex birigid graph on n + 1 vertices by choosing an edge e in G with |V −Ve| ≥3 and forming (G −e) + T by identifying three vertices of V −Ve with the endpoints of T of valence one. In fact, we obtain all birigid graphs of excess two by this process.
Theorem 5. Let G be a birigid graph of excess two with |V | > 5.
Let v be one of its vertices of valence three, T = star(v) and let x, y, and z denote the vertices adjacent to v. Then there is an edge e with endpoints in {x, y, z} such that e is not an edge of G and G −T + e is birigid.
Proof. : |V | > 5 insures that G −T is not complete. There are two cases.
(1) The two vertices of valence three in G are adjacent. By Theorem 1, the removal of v leaves a circuit, C, with a vertex, x, of valence two attached.
Assume x and y are in the same rigid component of C + x −w, where w is a vertex of valence four in C. We count that exactly three edges leave this component, contradicting the fact that a cutset of C has cardinality greater 8 BRIGITTE SERVATIUS than three. Observe that x is not adjacent to y or z, this would contradict the birigidity of G. So x and y are never in the same rigid component of C + x −w, and neither are y and z, therefore C + x −w + e is rigid if e is one of (x, y), (x, z) respectively.
(2) the two vertices of valence three in G are not adjacent.
By Theorem 1, if we remove v, we are left with a circuit, C. Let w be a vertex of valence four in C. C −w + T is rigid of zero excess, hence independent and consequently C −w is independent and nonrigid. By the 1-extendability property there exists an edge e with endpoints in {x, y, z} such that C −w + e is rigid. However, the choice of e depends on w, and we have to find an e that achieves rigidity independently from the choice of the removed vertex w.
If C contains already two of the possible three edges with endpoints in x, y, z we are done. Assume now that C does not contain e = (x, z) and f = (y, z) and there is a vertex w of valence four in C such that x and z are in the same rigid component A of C −w, but C −w + f is rigid and that there is a vertex u of C such that y and z are in the same rigid component B of C −u and C −u + e is rigid. A and B intersect in at least one edge, since z is of valence three in C, and their union is not equal to C. a contains at least two vertices which are not in B, so there are at least three edges of A −B incident with vertices of B, and by symmetry, three edges of B −A are incident with vertices of A. We count that exactly four edges leave each of A and B. So |C −(A S B)| ≤2, contradicting the fact that C −(A S B) contains a vertex.
Therefore, we can always find an edge e with endpoints in {x, y, z} such that C −w + e is rigid for all vertices w in C, i.e., G −T + e is birigid.
□ We have now found all birigid graphs of excess two, and we have seen that they are not only edge minimally birigid, but also minimal in the sense that they do not, with the exception of the ones on five and six vertices, contain any birigid subgraph of positive excess. Now we ask if every birigid graph on more than six vertices contains a birigid graph of excess two. The answer is no: The graph in figure 3 is birigid, has excess three and is minimal. The question if there are minimally birigid graphs of arbitrary excess is still open.
BIRIGIDITY IN THE PLANE 9 s s s s s s s s s s s s s s s s s s @ @ @ @ @ PPPP P @ @ @ A A A A A A B B B B B B B B B @ @ @ A A A A A A @ @ @ @ @ @ A A A A A A B B B B B B B B B @ @ @ A A A A A A @ @ @ @ @ @ @ @ PPPP P Figure 1. This is our favorite figure Acknowledgement:: I wish to thank Jack Graver for introducing me to R(G), and for his guidance throughout this research.
References J. E. Graver, a Combinatorial Approach to Infinitesimal Rigidity, Technical Report M-10, Department of Mathematics, Syracuse University, August, 1984.
G. Laman, On Graphs and Rigidity of Plane Skeletal Structures, J. Engrg. Math. 4(1970) 331-340.
L. Lovasz & Y. Yemini, On Generic Rigidity in the Plane, SIAM J. Alg. Disc. Methods 3(1982) 91-98.
K. Sugihara, On Redundant Bracing in Plane Skeletal Structures, Bulletin of the Electrotech-nical Laboratory Vol 44 Nos. 5 & 6 (1980) 78-88.
W. T. Tutte, Lectures on Matroids. J. Res. Nat. Bur. Stand. 69B (1965), 1-48.
D. J. A. Welsh, Matroid Theory, Academic Press, London (1976).
H. Whitney, On the Abstract Properties of Linear Independence. Amer. J. Math. 57(1935), 509-533.
Dept. of Math. Sciences, Worcester Polytechnic Institute, Worcester, Mass. 01609 |
8883 | https://people.csail.mit.edu/madhu/ST12/scribe/lect10.pdf | 6.S897 Algebra and Computation March 12, 2012 Lecture 10 Lecturer: Madhu Sudan Scribe: Rachel Lin Today, we will continue our approach to factoring bivariate polynomials. We will first focus on the tool of Hensel’s Lifting; and then describe how we perform the factoring.
1 Hensel’s Lifting Suppose f(x, y) = g(x)h(x) (mod y). We wanted a factorization for higher powers of y: f(x, y) = ˜ g(x, y)˜ h(x, y) (mod y2k). Hensel’s Lifting says we can obtain this if g and h are “relatively prime,” and that the ˜ g and ˜ h we obtain are essentially unique. Formally: Lemma 1 (Hensel’s Lifting) If R is a ring, I ⊆R is an ideal, and there exist f, g, h, a, b in R such that (H1) f = gh (mod I) (H2) ag + bh = 1 (mod I).
Then, for every positive interger s that is a power of 2, there exist ˜ g, ˜ h and ˜ a,˜ b in R, such that (C1) f = ˜ g˜ h (mod Is) (C2) ˜ g˜ a + ˜ b˜ h = 1 (mod Is) (C3) g = ˜ g (mod I) and h = ˜ h (mod I) Furthermore, the solution satisfying the above three conditions is “unique” in the following sense: Uniqueness: We say that an ideal J is special if for all all interger k, and a, b such that ab ∈Jk, there is an integer l such that a ∈Jl and b ∈Jk−l.
Assume that I is special. Then, for every two solutions g1, h1 and g2, h2 satisfying contidions C1 to C3, there exists u ∈It, such that: g2 = g1(1 + u) (mod Is), h2 = h1(1 −u) (mod Is) As we will see later, towards factoring a bivariate polynomial f ∈Fq[x, y], we will apply the Hensel’s lifting to R = Fq[x, y], I = (y) and a factorization of f mod I. When I = (y), f (mod y) is simply a univariate polynomial on x over Fq, which we know how to factor from previous lectures. Next we proceed to prove Lemma 1.
Proof We prove this lemma by induction.
Base Case s = 2: We first show that there exists ˜ g, ˜ h and ˜ a,˜ b satisfying C1 to C3, and then establish the uniqueness of the solution. By condition H1 and H2, we have f = gh + q, for some q ∈I ag + bh = 1 + r for some r ∈I Write ˜ g = g + g1, ˜ h = h + h1, for some g1, h1 ∈I to be set. Then ˜ g˜ h = gh + g1h + h1g + h1g1 10-1 Since h1g1 ∈I2, in order to satisfy condition C1, we want g1h + h1g + h1 = q (mod I2). To satisfy this, set g1 = bq, h1 = aq, and get that g1h + h1g = q(bh + ag) = q(1 + r), which equals to q modulo I2 as required. By construction ˜ g = g (mod I) and ˜ h = h (mod I), satisfying condition C3. To show that ˜ g and ˜ h are also relatively prime, observe that a˜ g + b˜ h = ag + bh + r′ = 1 + r + r′, for some r′ ∈I. Let r′′ = r + r′ ∈I. Now we can take ˜ a = a(1 −r′′) and ˜ b = b(1 −r′′), and get that: ˜ a˜ g + ˜ b˜ h = (1 −r′′)(a˜ g + b˜ h) = (1 −r′′)(1 + r′′) = 1 −r′′2 = 1 (mod I2) Now it remains to show that ˜ g, ˜ h is the unique solution satisfying C1 to C3. That is, if g∗, h∗is a different solution satisfying C1 to C3, then there is u ∈I such that g∗= ˜ g(1+u) and h∗= ˜ h(1−u).
By contidion C3, we have g∗= ˜ g + g2 and h∗= ˜ h + h2 for some g2, h2 ∈I (because, modulo I, we know that g∗= g = ˜ g and h∗= h = ˜ h). Therefore, we have: g∗h∗= ˜ g˜ h + g2˜ h + h2˜ g + g2h2 By condition C1, we know that g∗h∗= f = ˜ g˜ h (mod I2). Thus, the above equation modulo I2 gives, g2˜ h + h2˜ g = 0 (mod I2) By condition C2, we have ˜ a˜ g + ˜ b˜ h = 1 (mod I2). Therefore, ˜ b(g2˜ h + h2˜ g) = 0 (mod I2) g2˜ b˜ h + ˜ bh2˜ g = 0 (mod I2) g2(1 −˜ a˜ g) + ˜ bh2˜ g = 0 (mod I2) g2 = (˜ ag2 −˜ bh2)˜ g (mod I2) Let u = ˜ ag2 −˜ bh2. Since g2 and h2 are all elements in I, so is u. Furthermore we have g∗= ˜ g +g2 = ˜ g(1 + u). Similarly, by symmetry, we obtain that h2 = (˜ bh2 −˜ ag2)˜ h (mod I2) Therefore, h∗= ˜ h(1 −u). This concludes the proof for the base case.
Induction Step: Assume that for the case of s = t, there exist g0, h0 ∈R[x] and a0, b0 ∈It satisfying conditions C1 to C3, and the solution to the three conditions is “unique”. We show that for the case of s = 2t, we can construct g1, h1 ∈R[x] and a1, b1 ∈I2t satisfying conditions C1 to C3, and the solution is also unique.
The existence of g1, h1, a1, b1 satisfying conditions C1 to C3 follows exactly the same proof as in the base case. Therefore, we focus on the proof of uniqueness. Let g1, h2 be a different solution from g1, h1. Then both g1 (mod It), h1 (mod It) and g2 (mod It), h2 (mod It) are solutions satisfying C1 to C3 for the case of s = t. Then by the induction hypothesis, we have that there is a u0 ∈It/2 such that, (g2 (mod It)) = (g1 (mod It))(1 + u0) (mod It) (h2 (mod It)) = (h1 (mod It))(1 −u0) (mod It) This implies that g2 = g1(1 + u0) (mod It) h2 = h1(1 −u0) (mod It) Notice that this is differerent from the condition in the base case where any two solutions must equal modulo I. Nevertheless, following the same argument, we can derive that there is an element u ∈It such that, g2 = g1(1 + u0)(1 + u) (mod I2t) 10-2 h2 = h1(1 −u0)(1 −u) (mod I2t) Below we show that u0 is in fact an element in It, then g2 = g1(1 + u0 + u + u0u) = g1(1 + u0 + u) (mod I2t) h2 = h1(1 −u0 −u + u0u) = h1(1 −u0 −u) (mod I2t) Thus g2 = g1(1 + u′) and h2 = h1(1 −u′) for u′ = u0 + u ∈It as desired.
To show that u0 ∈It, consider: g2h2 = g1h1(1 −u2 0)(1 −u2) (mod I2t) g2h2 = g1h1(1 −u2 0) (mod I2t) [as u ∈It] Since g1 and h1 are not elements in I, for the last euqation to hold, it must be the case that u0 ∈It.
Therefore, we conclude the lemma.
2 Outline of Factoring, revisited We now give a more complete outline for factoring bivariate polynomials.
Given a monic f(x, y) ∈Fq[x, y], with total degree d, the factoring algorithm SPLIT proceeds as follows: 1. If g = gcd(f, ∂f ∂x) ̸= 1, then output (g, f/g) and stop. Otherwise, continue the following steps.
2. Find y0 ∈F such that f(x, y0) has no repeated factors.
This can be done by computing Res f, ∂f ∂x , and pluging in y0 = 1, 2, . . . until we find one that makes the resultant non-zero.
We claim that this will terminiate in at most d2 iterations, as Res f, ∂f ∂x is a polynomial in y with degree at most d2. Futhermore, the first step ensures that f does not have repeated roots; therefore, Res f, ∂f ∂x is not a zero polynomial. Hence it has at most d2 roots.
3. Put fy0(x) = f(x, y) (mod (y −y0)) = f(x, y0) and factor it. This can be done by using the factoring algorithm for univariate polynomial over F. Let g be an irreducable factor of fy0(x), and h such that f = gh (mod (y −y0)).
4. Now we apply Hensel’s Lifting to obtain f = g1h1 (mod (y −y0)t) for a t ≈d2 5. Next, from g1 we ask if we can find a nontrivial factor ˜ g of f. This is done through the “Jump” step, which tries to find polynomials ˜ g and ˜ h such that ˜ g = g1˜ h (mod (y −y0)t), and ˜ g has small degrees in y (smaller than d) and minimal degree in x.
6. Finally, return ˜ g and f/˜ g.
10-3 |
8884 | https://math.stackexchange.com/questions/4001113/calculate-the-intersection-point-of-m-with-ac | geometry - Calculate the intersection point of $m$ with $AC$. - Mathematics Stack Exchange
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Calculate the intersection point of m m with A C A C.
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Let A B C A B C be the triangle with A=(0,0)A=(0,0), B=(16,0)B=(16,0) and C=(4,16)C=(4,16).
Let B′B′ be the midpoint of the side A C¯¯¯¯¯¯¯¯A C¯, let C′C′ be the midpoint of the side A B¯¯¯¯¯¯¯¯A B¯ and let I I be the point (2 17−−√−2,9−17−−√)(2 17−2,9−17).
Calculate the line that goes through B B and I I.
The angle bisector m m at B B of the triangle A B C A B C divides the opposite A C¯¯¯¯¯¯¯¯A C¯ proportional to the adjacent sides. Calculate the intersection point of m m with A C A C. Show that this intersection point is on the line B I B I.
Show that I I has the same distance to A B A B as to A C A C.
Conclude that I I is the center of the inscribed circle of the triangle A B C A B C and calculate the line that goes through I I and the contact point of the inscribed circle and the side B C¯¯¯¯¯¯¯¯B C¯.
For question 1 we have:
y−0=9−17−−√−0 2 17−−√−2−16(x−16)⇒y=−1 2(x−16)y−0=9−17−0 2 17−2−16(x−16)⇒y=−1 2(x−16)
For question 2 we have:
We have to calculate the side lengths |A B||A B| and |B C||B C|, right?
We have that |A B|=16|A B|=16 and |B C|=12 2+16 2−−−−−−−−√=20|B C|=12 2+16 2=20. Therefore m m divides the side A C¯¯¯¯¯¯¯¯A C¯ in relation 20 16=5 4 20 16=5 4, right? But how can we calculate the coordinates of that point?
For question 3 we have:
Here we just use the formula for distances and calculate both and see that these are equal, right?
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asked Jan 26, 2021 at 21:51
Mary StarMary Star
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What you need to do for part (2) is to compute the intersection point of the line you got from part (1), with the line that passes through A A and C C, which you have not yet calculated. This is simply
y=16 4 x=4 x,y=16 4 x=4 x,
consequently the intersection point, which we will call P P, satisfies the system
y y=−x 2+8=4 x.y=−x 2+8 y=4 x.
Then compute the distances |P A||P A| and |P C||P C|, and show that because
|P A||A B|=|P C||C B|,|P A||A B|=|P C||C B|,
point P P obeys the proportionality criterion, thus P P lies on the angle bisector of ∠B∠B. Conclude that, since this line is also the line passing through B I B I from part (1), that I I lies on the angle bisector of ∠B∠B.
For part (3), simply perform the distance calculation. The distance of I I to A B A B is easy. However, the distance of I I to A C A C requires us to find the line perpendicular to A C A C passing through I I, computing its intersection point Q Q with A C A C, then computing the distance |Q I||Q I|.
For part (4), observe from parts (2) and (3) that if I I lies on the angle bisector of ∠B∠B, then the distance of I I to B C B C equals the distance of I I to A B A B; therefore, I I is equidistant to all three sides of △A B C△A B C, hence I I is the incenter.
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answered Jan 27, 2021 at 1:35
heropupheropup
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1 @MaryStar The purpose of part (2) is to show that I I lies on the angle bisector. You can do this two ways: one is to locate the point P P on A C A C that satisfies the proportionality condition (thus is on the bisector), then show I I is collinear with P B P B because the equation for line I B I B passes through P P. The other way is to do what I suggested in the answer, which is to find the intersection of I B I B with A C A C, call this P P, and then show that the proportionality condition |P A|/|A B|=|P C|/|C B||P A|/|A B|=|P C|/|C B| is true by calculating the distances.heropup –heropup 2021-01-27 18:59:05 +00:00 Commented Jan 27, 2021 at 18:59
1 For part (3), yes you absolutely can. In fact, the formula you cite can be derived by the process I described in my solution. Since I did not know that you were aware of this formula, I did not assume you knew to use it, and instead described an approach using first principles.heropup –heropup 2021-01-27 19:00:42 +00:00 Commented Jan 27, 2021 at 19:00
1 As for part (4), your proposal is unnecessary. The reason is because in part (2), you showed that I I lies on the angle bisector of ∠B∠B, therefore the distance from I I to side A B A B is the same as the distance from I I to side B C B C. In other words, my explanation in part (4) is more or less a complete proof in itself. No additional calculation is required.heropup –heropup 2021-01-27 19:02:56 +00:00 Commented Jan 27, 2021 at 19:02
1 @MaryStar For part (2), that is precisely correct. For part (4), the key idea is that the distance from a point I I to a line m m is equal to the radius of a circle whose center is I I and is tangent to m m. So if you can show that I I is equidistant from lines A B A B and B C B C, then the circle with center at I I and radius equal to this common distance is tangent to A B A B and B C B C. You do not need to calculate an equation for the circle.heropup –heropup 2021-01-27 19:16:42 +00:00 Commented Jan 27, 2021 at 19:16
1 @MaryStar I apologize, I forgot about that part of the question. All it is asking you to do is compute the equation of the line through I I and the point of tangency of the incircle to B C B C. One way to get the point of tangency is to calculate the equation of line B C B C, then for a point R=(x,y)R=(x,y) on B C B C, find the value of (x,y)(x,y) such that |R I||R I| is the inradius. Or you can reflect the point (2 17−−√−2,0)(2 17−2,0) about the line B I B I, if you know such a formula. Or you can solve the system of equations for the incircle I I and the line B C B C. There are multiple ways to do it.heropup –heropup 2021-01-27 19:28:26 +00:00 Commented Jan 27, 2021 at 19:28
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Why irrational implies having an infinite decimal expansion?
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Why irrational means having an infinite decimal expansion?
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edited Apr 19, 2014 at 14:29
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1 Do you mean "infinite without a repeating pattern"?naslundx –naslundx 2014-04-19 14:28:28 +00:00 Commented Apr 19, 2014 at 14:28
@naslundxs I mean that an irrational number have an end. You count and count then you get to an end, you know all the digits of that number.user144243 –user144243 2014-04-19 14:31:02 +00:00 Commented Apr 19, 2014 at 14:31
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If the decimal was finite, then it can be represented as 0.d 1 d 2 d 3…d n where 0≤d i<10 are the digits of the decimal expansion and n is finite. This can be represented as a fraction, namely d 1 d 2 d 3…d n 10 n (d 1 d 2 d 3…d n is a number in base 10, not the product of all d i)
If the decimal was infinite but repeating, say it is 0.b 1 b 2 b 3…b n¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯. Then it can be represented as a fraction, namely b 1 b 2 b 3…b n 10 n−1
Since they can be represented as fractions of integers, they are not irrational.
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answered Apr 19, 2014 at 14:31
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Your claim in the title is different from what you have in the question.
The following claim below is true.
Irrational⟹infinite decimal expansion
This is equivalent to the contrapositive, i.e.,
Not infinite decimal expansion⟹not irrational
which (after getting rid of double negative) is same as
Finite decimal expansion⟹rational
I assume you can prove this statement.
Let the finite decimal expanion have k digits after the decimal point, i.e., the number is a+0.a 1 a 2…a k, where a is an integer and a i∈{0,1,…,9}. This can be written as
a+a 1 a 2…a k 10 k=10 k a+a 1 a 2…a k 10 k
However, note that any number with infinite decimal expansion need not be irrational. For instance, the number 0.3¯ is the infinite decimal expansion for the rational number 1/3.
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edited Apr 19, 2014 at 14:33
answered Apr 19, 2014 at 14:28
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What you are saying is correct. And as Git Gud says you can look at what happens when the decimal expansion is finite. suppose then the number in decimal looks like a m a m−1…a 1.d 1 d 2…d n Then it is equivalent to a m a m−1…a 1+d 1 d 2…d n 10 n which is rational.
However not all numbers with infinite expansion are irrational. look at .9¯¯¯ which is 1.
see here also.
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edited Apr 13, 2017 at 12:20
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answered Apr 19, 2014 at 14:32
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1Convergence of an infinite product of a quotient of an irrational number and its decimal expansion
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0Decimal Representation vs Decimal Expansion
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8886 | https://satprepteacher.com/wp-content/uploads/2020/09/All-Things-Equation-of-a-Circle-Worksheet.pdf | SAT “Equation of a Circle” Practice Problems Worksheet Over the past few years, the College Board has transitioned from practically never asking about the equation for graphing a circle to now asking it on just about every test (I would say “every test,” but there is always that possibility it may not be on one test as a rare exception). For reference, here’s the equation for graphing a circle: For a circle with center (h, k), (𝑥−ℎ)2 + (𝑦−𝑘)2 = 𝑅𝑎𝑑𝑖𝑢𝑠2 The good news is that the questions they use to test your knowledge of this concept are not varied; in fact, there are only three ways I have seen them actually test this concept, and the three question types are seen in questions 1-3 (here’s the equation, now identify the center and radius), 4-5 (here’s the center and radius, now identify the equation), and 6-8 (reformat the given equation to look like the equation for graphing a circle by completing the square, then identify the center or radius). If you can do these problems, then you’re good for this question type. 1. (𝑥−8)2 + (𝑦−6)2 = 36 For the equation above, what is the coordinate point for the center of the circle as well as the circle’s radius? A) Center: (-8, -6), Radius: 36 B) Center: (8, 6), Radius: 36 C) Center: (-8, -6), Radius: 6 D) Center: (8, 6), Radius: 6 2. (𝑥+ 14)2 + (𝑦+ 2)2 = 64 For the equation above, what is the coordinate point for the center of the circle as well as the circle’s radius? A) Center: (14, 2), Radius: 8 B) Center: (14, 2), Radius: 64 C) Center: (-14, -2), Radius: 8 D) Center: (-14, -2), Radius: 64 3. (𝑥−11)2 + (𝑦+ 4)2 = 49 For the equation above, what is the coordinate point for the center of the circle as well as the circle’s radius? A) Center: (11, -4), Radius: 7 B) Center: (11, -4), Radius: 49 C) Center: (-11, 4), Radius: 49 D) Center: (-11, 4), Radius: 7 4. Find the equation of a circle with center (12, -3) and a radius of 3: A) (𝑥−12)2 + (𝑦+ 3)2 = 3 B) (𝑥−12)2 + (𝑦+ 3)2 = 9 C) (𝑥+ 12)2 + (𝑦−3)2 = 3 D) (𝑥+ 12)2 + (𝑦−3)2 = 9 5. Find the equation of a circle with center (-7, 11) and a radius of 9: A) (𝑥+ 7)2 + (𝑦−11)2 = 81 B) (𝑥+ 7)2 + (𝑦+ 11)2 = 9 C) (𝑥−7)2 + (𝑦−11)2 = 81 D) (𝑥−7)2 + (𝑦+ 11)2 = 9 𝑥2 + 10x + 𝑦2 – 6y = -18 6. The graph of the equation shown above is a circle. What is the radius of the circle? A) 3 B) 4 C) 5 D) 9 𝑥2 + 18x + 𝑦2 – 8y = -48 7. The graph of the equation shown above is a circle. What is the radius of the circle? A) 4 B) 5 C) 6 D) 7 𝑥2 -4x + 𝑦2 +6y = 87 8. The graph of the equation shown above is a circle. What is the coordinate point of the center of the circle? A) (13, 10) B) (4, 13) C) (-4, 6) D) (2, -3) Answer Key: 1. D 2. C 3. A 4. B 5. A 6. B 7. D 8. D |
8887 | https://opendsa-server.cs.vt.edu/ODSA/Books/CS3/html/BinaryTree.html | | About
« 7.1. Binary Trees Chapter Introduction
::
Contents
::
7.3. Binary Tree as a Recursive Data Structure »
7.2. Binary Trees¶
7.2.1. Definitions and Properties¶
A binary tree is made up of a finite set of elements
called nodes.
This set either is empty or consists of a node called the
root together with two binary trees, called the left and
right
subtrees, which are disjoint from each other and
from the root.
(Disjoint means that they have no nodes in common.)
The roots of these subtrees are
children of the root.
There is an edge from a node to each of its children,
and a node is said to be the parent of its children.
If n1,n2,...,nk
is a sequence of nodes in the tree such
that ni is the parent of
ni+1 for 1≤i<k,
then this sequence is called a path from
n1 to nk.
The length of the path is k−1.
If there is a path from node R to node M,
then R is an ancestor of M, and
M is a descendant of R.
Thus, all nodes in the tree are descendants of the root of the tree,
while the root is the ancestor of all nodes.
The depth of a node M in the tree is the length
of the path from the root of the tree to M.
The height of a tree is the depth of the
deepest node in the tree.
All nodes of depth d are at
level d in the tree.
The root is the only node at level 0, and its depth is 0.
A leaf node is any node that has two empty children.
An internal node is
any node that has at least one non-empty child.
A
B
C
D
E
F
G
H
I
A
B
A
B
(a)
(b)
A
B
EMPTY
A
EMPTY
B
(c)
(d)
Figure 7.2.1 illustrates the various terms used to
identify parts of a binary tree.
Figure 7.2.2 illustrates an important point regarding
the structure of binary trees.
Because all binary tree nodes have two children
(one or both of which might be empty), the two binary
trees of Figure 7.2.2 are not the same.
Two restricted forms of binary tree are sufficiently
important to warrant special names.
Each node in a full binary tree
is either (1) an internal node with exactly two non-empty children or
(2) a leaf.
A complete binary tree has a restricted shape obtained by
starting at the root and filling the tree by levels from left to
right.
In a complete binary tree of height d, all levels
except possibly level d are completely full.
The bottom level has its nodes filled in from the left side.
(a)
(b)
Figure 7.2.3 illustrates the differences between
full and complete binary trees.
There is no particular relationship between these two tree shapes;
that is, the tree of Figure 7.2.3 (a) is full
but not complete while the tree of Figure 7.2.3
(b) is
complete but not full.
The heap data structure is an example
of a complete binary tree.
The Huffman coding tree
is an example of a full binary tree.
7.2.2. Practice Questions¶
Privacy | | License
« 7.1. Binary Trees Chapter Introduction
::
Contents
::
7.3. Binary Tree as a Recursive Data Structure » |
8888 | https://www.ajog.org/article/S0002-9378(22)00461-6/fulltext | The “30-minute rule” for expedited delivery: fact or fiction? - American Journal of Obstetrics & Gynecology
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Clinical OpinionVolume 228, Issue 5, SupplementS1110-S1116 May 2023
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The “30-minute rule” for expedited delivery: fact or fiction?
Tracy Caroline Bank, MD
Tracy Caroline Bank, MD
Correspondence
Corresponding author: Tracy Caroline Bank, MD.
Tracy.C.Bank@christianacare.org
Affiliations
Department of Obstetrics and Gynecology, ChristianaCare, Newark, DE
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aTracy.C.Bank@christianacare.org ∙ George Macones, MD, MSCE
George Macones, MD, MSCE
Affiliations
Department of Obstetrics and Gynecology, The University of Texas at Austin, Austin, TX
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b ∙ Anthony Sciscione, DO
Anthony Sciscione, DO
Affiliations
Department of Obstetrics and Gynecology, ChristianaCare, Newark, DE
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a
Affiliations & Notes Article Info
a Department of Obstetrics and Gynecology, ChristianaCare, Newark, DE
b Department of Obstetrics and Gynecology, The University of Texas at Austin, Austin, TX
Publication History:
Received January 27, 2022; Revised June 12, 2022; Accepted June 13, 2022; Published online March 16, 2023
Footnotes:
The authors report no conflict of interest.
This manuscript was completed without the benefit of external support.
DOI: 10.1016/j.ajog.2022.06.015 External LinkAlso available on ScienceDirect External Link
Copyright: © 2022 Elsevier Inc. All rights reserved.
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Limitations of existing data
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Abstract
Initially developed from hospital feasibility data from the 1980s, the “30-minute rule” has perpetuated the belief that the decision-to-incision time in an emergency cesarean delivery should be <30 minutes to preserve favorable neonatal outcomes. Through a review of the history, available data on delivery timing and associated outcomes, and consideration of feasibility across several hospital systems, the use and applicability of this “rule” are explored, and its reconsideration is called for. Moreover, we have advocated for balanced consideration of maternal safety with rapidity of delivery, encouraged process-based approaches, and proposed standardization of terminology regarding delivery urgency. Furthermore, a standardized 4-tier classification system for delivery urgency, from class I, for a perceived threat to maternal or fetal life, to class IV, a scheduled delivery, and a call for further research with a standardized structure to facilitate comparison have been proposed.
Key words
birth trauma
cesarean delivery
decision-to-delivery
decision-to-incision
fetal acidemia
fetal hypoxia
fetal monitoring
intrauterine resuscitation
maternal morbidity
maternal mortality
maternal outcomes
neonatal morbidity
neonatal mortality
neonatal outcomes
obstetrical emergency
Introduction
Does the time from the decision to perform an expedited cesarean delivery to the first incision, the so-called “decision-to-incision” time, affect neonatal outcomes? If so, how long is too long? These are questions that bewilder many providers, hospital systems, patients, and juries. Optimal delivery timing in an emergent and nonemergent situation must balance the availability of resources and maternal safety with evidence-based optimization of neonatal outcomes. The so-called 30-minute rule describes a commonly held belief that the best neonatal outcomes occur when the decision-to-incision time of an “emergency” cesarean delivery is fewer than 30 minutes. This has been promulgated as a “standard of care,” and thus, failure to adhere is associated with legal and quality of care implications, despite several calls to reconsider its application.1
1.
Boehm, F.H.
Decision to incision: time to reconsider
Am J Obstet Gynecol. 2012; 206:97-98
Full Text
Full Text (PDF)
Scopus (20)
PubMed
Google Scholar
AJOG at a Glance
Why was this study conducted?
This study was conducted to investigate the evidence behind the “30 Minute Rule” for expedited delivery and to develop recommendations for rapid delivery.
Key findings
There is insufficient evidence to establish an optimal clear decision-to-incision or decision-to-delivery interval for urgent or emergent deliveries.
What does this add to what is known?
This study calls for reconsideration of a commonly held “30 Minute Rule” quality metric. It proposes a 4-tier classification system to standardize urgency of delivery.
Here, the current data and supporting evidence for or against this “rule” are reviewed. Furthermore we call for the use of consistent nomenclature and research to develop new quality measures to ensure that “emergency” or “stat” deliveries are carried out in the best, evidence-based approach, supporting maternal safety, improving neonatal outcomes and resource allocation, and maintaining feasibility across multiple hospital systems.
Background
History
In 1972, Myers et al2
2.
Myers, R.E.
Two patterns of perinatal brain damage and their conditions of occurrence
Am J Obstet Gynecol. 1972; 112:246-276
Abstract
Full Text (PDF)
PubMed
Google Scholar
demonstrated that when term monkey fetuses underwent total asphyxia, via head envelopment and cord occlusion at the time of surgical delivery, the first evidence of brain damage, as evidenced by necrosis on autopsy, occurred after 10 minutes. This was followed by widespread brain injury after 16 to 18 minutes and death after 25 minutes. Based, in part, on this work, the 1982 publication Standards for Obstetrics and Gynecology suggested 15 minutes as the time for the initiation of emergency cesarean delivery for any “obstetrical service that generally cares for high-risk patients.”3
3.
American College of Obstetricians and Gynecologists Committee on Professional Standards
Standards for obstetric-gynecologic services
The Committee, Washington, DC, 1982
Google Scholar
In 1986, Brann et al4
4.
Brann, Jr., A.W.
Hypoxic ischemic encephalopathy (asphyxia)
Pediatr Clin North Am. 1986; 33:451-464
Crossref
Scopus (31)
PubMed
Google Scholar
described the pathogenesis of hypoxic-ischemic encephalopathy and its sequelae based on 2 monkey models. The first was one of prolonged partial asphyxia (like an incomplete placental abruption), in which 1 to 2 hours of partial asphyxia was associated with seizures in 50% of monkey neonates. In this group, a wide variety of cerebral lesions were found, including those similar to lesions seen in human infants who die after birth and those who are diagnosed with cerebral palsy. The second model was one of acute total asphyxia (like a cord prolapse), in which a period of 8 to 10 minutes was associated with long-term neurologic dysfunction consistent with spastic quadriplegia. Once more, this implied that delivery within the studied period would improve neonatal outcomes.
The advent of the 30-minute rule
In 1987, a landmark hospital survey was released, suggesting that most hospitals in the United States were able to initiate a cesarean delivery within 30 minutes. This survey was sent to a random sample of hospitals that performed ≥5 deliveries per year. Sampling rates based on the number of births per year were used to control for differences in volume. Of note, 538 hospitals were selected, and questionnaires were sent to the “Chief of Obstetrics” or another individual who is thoroughly familiar with the obstetrics service; after several mailings and telephone calls, an 87% response rate was achieved. Of note, 88% of hospitals reported the capability to prepare for an emergency cesarean delivery within 30 minutes. It should be noted that the purpose of this survey was not to describe the time frame for the performance of “emergency” cesarean deliveries but rather to explore patterns in rates of cesarean delivery and trial of labor after previous cesarean delivery. The ability to prepare for a cesarean delivery within 30 minutes was assessed, along with the availability of technology to monitor fetal heart rate and contractions and whether this was a potential limiting factor to a hospital’s decision to offer trial of labor after a previous cesarean delivery. Neither the characteristics of the hospitals among that 88% nor the distribution of time frames considered feasible by various hospitals were included in this report.5
5.
Shiono, P.H. ∙ Fielden, J.G. ∙ McNellis, D. ...
Recent trends in cesarean birth and trial of labor rates in the United States
JAMA. 1987; 257:494-497
Crossref
Scopus (94)
PubMed
Google Scholar
After the release of this survey, the previous 15-minute suggestion was expanded to 30 minutes; this was put forward as a national guideline through the 1988 publication of the Standards for Obstetric Services3
3.
American College of Obstetricians and Gynecologists Committee on Professional Standards
Standards for obstetric-gynecologic services
The Committee, Washington, DC, 1982
Google Scholar
and the second edition of the Guidelines for Perinatal Care.6
6.
American Academy of Pediatrics, American College of Obstetricians and Gynecologists
Guidelines for perinatal care
American Academy of Pediatrics, Elk Grove Village, IL, 1988
Google Scholar
It was often implemented as a “standard of care” among hospitals throughout the United States, without substantial medical evidence to support its association with improved outcomes.1
1.
Boehm, F.H.
Decision to incision: time to reconsider
Am J Obstet Gynecol. 2012; 206:97-98
Full Text
Full Text (PDF)
Scopus (20)
PubMed
Google Scholar
National guidelines
Before the early 2010s, the American College of Obstetricians and Gynecologists (ACOG) and the American Association of Pediatrics (AAP) held the position that hospitals should have the ability to initiate an emergency cesarean delivery within 30 minutes.6
6.
American Academy of Pediatrics, American College of Obstetricians and Gynecologists
Guidelines for perinatal care
American Academy of Pediatrics, Elk Grove Village, IL, 1988
Google Scholar
,7
7.
Racine, A.D.
Standards of Obstetric-Gynecologic Services
Washington (DC): ACOG, 1989
American College of Obstetricians and Gynecologists
Google Scholar
However, these guidelines have changed.
The eighth edition of the Guidelines for Perinatal Care, developed by the AAP and the ACOG and published in 2017, states the following:
“Historically, the consensus has been that hospitals need to have the capability of beginning a cesarean delivery within 30 minutes of the decision to operate. However, the scientific evidence to support this threshold is lacking. The decision-to-incision interval should be based on the timing that best incorporates maternal and fetal risks and benefits […] it is reasonable to tailor the time to delivery to local circumstances and logistics.”6
6.
American Academy of Pediatrics, American College of Obstetricians and Gynecologists
Guidelines for perinatal care
American Academy of Pediatrics, Elk Grove Village, IL, 1988
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In the United Kingdom, as of 2021, the National Health Service (NHS) recommended dividing cesarean deliveries into 4 categories, as defined by the clinical concern: category 1 for the presence of an immediate threat to maternal or fetal life (with examples of suspected umbilical cord prolapse, persistent fetal bradycardia, major placental abruption, or concern for uterine rupture), category 2 for non–life-threatening maternal or fetal compromise, category 3 for the requirement of delivery in the absence of maternal or fetal compromise, and category 4 for a birth that can be scheduled as desired by the mother and provider. The NHS suggests a decision-to-delivery interval of fewer than 30 minutes for a category 1 and 75 minutes for a category 2 cesarean delivery but cautions that rapid delivery bears risks in certain situations.8
8.
National Institute for Health and Clinical Excellence
Caesarean birth
Available at:
Date: 2021
Accessed June 30, 2022
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Although the recommendations of the NHS are more temporally restrictive than those of the AAP and ACOG, both new sets of guidelines do allow more flexibility for a medical team to tailor the time frame to the clinical scenario. Despite these statements, the “30-minute rule” is often considered the legal standard9
9.
Miller, L.A.
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and is used as part of the causation argument, that is, that a perceived delay in delivery led to a poor neonatal outcome, typically neurologic.10
10.
Veltman, L.
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Feasibility of rapid delivery
As described above, the 30-minute rule arose from a 1987 study on feasibility, which found that 30 minutes was considered an achievable time frame in most hospitals surveyed.5
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It should be noted that response times, at least internationally, often are quite rapid when there is a concern for a “true” clinical emergency. A Norwegian study found that although overall decision-to-delivery interval was >50 minutes among all emergency deliveries, the mean decision-to-delivery interval was 11.8 minutes among those cesarean deliveries thought to require delivery “as quickly as at all possible.”11
11.
Kolås, T. ∙ Hofoss, D. ∙ Øian, P.
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A Singaporean report found that the mean decision-to-delivery interval was 7.7 minutes for those deliveries labeled as “extremely urgent.”12
12.
Lim, Y. ∙ Shah, M.K. ∙ Tan, H.M.
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Lastly, in Germany, the median decision-to-delivery time has been as short as 10 minutes in settings of severe maternal or fetal distress (specifically cord prolapse, abruption, and bradycardia).13
13.
Hillemanns, P. ∙ Hasbargen, U. ∙ Strauss, A. ...
Maternal and neonatal morbidity of emergency caesarean sections with a decision-to-delivery interval under 30 minutes: evidence from 10 years
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Response times this fast may not be feasible in a broader range of hospital systems and birthing centers, some of which do not have in-house obstetrical or anesthetic teams. Furthermore, it has been suggested that because of challenges in positioning and anesthetizing women with higher body mass indices, it is more difficult to achieve a 30-minute time interval in this population.14
14.
Lucas, D.N.
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As obesity rates increase, a response time of <10 minutes may not be safe or feasible for certain patients, even in highly efficient hospital settings.
Association between time to delivery and outcomes
Neonatal outcomes
Over the last 30 years, we have learned more about the neonatal and maternal outcomes associated with time to delivery. Of note, it is challenging to determine the exact effects of delivery timing on neonatal outcomes. Although neurologic outcomes, including encephalopathy and cerebral palsy, are of concern, these are currently thought to occur through several causal pathways, spanning gestation and the perinatal period.15
15.
American College of Obstetricians and Gynecologists
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Among neonatal outcomes, there also seems to be a significant difference related to the reversibility of a potential stressor, for example, a reversible bradycardia related to tachysystole vs an irreversible bradycardia related to complete abruption. Leung et al16
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retrospectively characterized a cohort of 235 women who delivered via “extremely urgent cesarean delivery” into those delivered for “reversible (n=22),” “irreversible (n=39),” and “unknown (n=174)” causes. Reversible causes included tachysystole, hypotension after epidural, and bradycardia after external cephalic version in the absence of placental abruption. Irreversible causes included placental abruption, uterine rupture, preeclampsia, cord prolapse, and failed instrumented delivery. Among the group categorized as potentially reversible or unknown, there was no correlation between bradycardia-to-delivery interval and neonatal pH. Among the group categorized as having irreversible bradycardia, pH fell at a rate of 0.011 per minute, suggesting a time-associated decrease. There was no correlation made between time to delivery and neonatal outcomes.
In some cases, 30 minutes may be too long a decision-to-incision time, such as umbilical cord prolapse. As suggested in the initial animal study by Myers2
2.
Myers, R.E.
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described earlier, at >15 minutes of total asphyxia via cord occlusion, widespread brain injury was seen. There have been several human studies analyzing acidemia in the setting of fetal bradycardia; one found that metabolic acidosis was significantly (P<.001) higher after 15 minutes.17
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Katz, M. ∙ Shani, N. ∙ Meizner, I. ...
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A study assessing the effects of maintaining an in-house operating team found a significant risk of intrauterine fetal demise associated with a time frame >20 minutes.18
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reviewing the outcomes of 39,291 neonates delivered via emergency cesarean delivery found decreased odds of low Apgar scores with a decision-to-delivery cutoff of 20 minutes, with no difference noted at a cutoff of 30 minutes. Finally, in a study of 19 deliveries complicated by sustained fetal bradycardia (related to cord prolapse, abruption, uterine rupture, maternal respiratory failure, among other causes), which followed the children involved to the age of 2, Kamoshita et al20
20.
Kamoshita, E. ∙ Amano, K. ∙ Kanai, Y. ...
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found that a bradycardia-to-delivery interval of fewer than 25 minutes was associated with normal neonatal neurologic development.
Conversely, several studies of neonates delivered emergently have shown worse neonatal outcomes among those delivered more quickly. Outcomes noted included increased acidemia,21–23
21.
Holcroft, C.J. ∙ Graham, E.M. ∙ Aina-Mumuney, A. ...
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lower Apgar scores,13
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an increased incidence of umbilical artery pH of <7.10,25
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and a higher rate of delivery room intubation among those delivered within 30 minutes than those delivered after.26
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There may be a bias that fetuses delivered more quickly are in greater jeopardy or that those in most jeopardy do not have a modifiable outcome by speed of delivery. Alternatively, improved outcomes among those delivered with less haste may be attributable to fetal recovery after a maternal catecholamine surge, which often occurs when the decision is made to proceed with emergency delivery.24
24.
Schauberger, C.W. ∙ Rooney, B.L. ∙ Beguin, E.A. ...
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Other studies have shown no difference in neonatal outcomes, including low Apgar scores, neonatal intensive care unit admission, acidosis, seizures, and fetal or neonatal death, in time intervals of 30,27
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or even 180 minutes.30
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In a review of the available literature in 2013, Leung and Lao31
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concluded that in certain life-threatening situations, the more rapid the delivery, the better for fetal well-being. They emphasized the importance of identifying and addressing potential causes of hypoxia and acknowledged 30 minutes as an achievable standard, if not one supported by outcome data. Although it is the decision-to-incision time that is most commonly debated, the total duration and severity of hypoxia seem to correlate more closely with adverse neonatal outcomes.16
16.
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Although prompt action to deliver a compromised infant is admirable, this relies on timely and reliable recognition of irreversible hypoxia and treatment of reversible hypoxia. In addition, the notion that shorter decision-to-incision times improve outcomes implies that most fetal injuries occur in the intrapartum period and are thus preventable, which is known to not be the case; according to the ACOG, 70% of neonatal encephalopathy cases are likely attributable to prelabor events, including genetic differences, abnormal fetal growth, maternal infection, and placental lesions.15
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Maternal outcomes
There is substantial evidence to suggest that emergency cesarean deliveries are associated with higher rates of surgical complications than scheduled cesarean deliveries. These include increased risk of surgical site infection, urinary tract infection, wound dehiscence, and disseminated intravascular coagulation.32–34
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Nielsen, T.F. ∙ Hökegård, K.H.
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However, It must be noted that the nature of these associations is confounded by the indication for emergency cesarean delivery; emergency delivery for nonreassuring fetal heart tracing in the setting of chorioamnionitis may be more associated with postpartum infectious complications, although emergency delivery for a significant abruption could confer risk of disseminated intravascular coagulation. As many as 75% of bladder injuries during cesarean deliveries occurred during emergency cesarean deliveries,35
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and although less common, bowel injury is also believed to occur more frequently in emergency deliveries.36
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In a study of 19,416 cesarean deliveries, Schneid-Kofman et al37
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found that emergency cesarean delivery independently increased the rate of early wound infections (diagnosed before discharge) with an odds ratio of 1.3 (95% confidence interval, 1.1–1.5), although this was not seen in a later study of 1605 cesarean deliveries by Olsen et al,38
38.
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also examining risk factors for surgical site infection.
Maternal concerns with rapid delivery
Traditionally, within the category of emergency cesarean deliveries, maternal complications have been thought to be independent of delivery speed.26
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However, more recent literature has suggested several risks worthy of consideration. Moroz et al39
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found significantly higher odds of transfusion, broad ligament hematoma, and uterine artery ligation among those with an incision-to-delivery time of <2 minutes during emergency repeat cesarean delivery.
In addition, although general anesthesia has been associated with shorter decision-to-incision times and has been proposed to be necessary to achieve time intervals <20 minutes, there seems to be a significant risk associated with the use of general anesthesia.11
11.
Kolås, T. ∙ Hofoss, D. ∙ Øian, P.
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Exposure to general anesthesia has been associated with increased rates of postpartum hemorrhage,42
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along with an increased risk of severe postpartum depression necessitating hospitalization, self-injury, and suicidal ideation.43
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In pursuit of an arbitrary time goal, the potentially safer option of regional anesthesia may not be offered. However, it is highly challenging to separate the impact of the indication for general anesthesia from the anesthesia itself. Emergency cesarean delivery has been associated with posttraumatic stress disorder; however, the same study found that subjective distress, including fear for one’s life or the life of one’s child, was the greatest predictor of maternal posttraumatic stress disorder44
44.
Andersen, L.B. ∙ Melvaer, L.B. ∙ Videbech, P. ...
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; it is quite possible that it is not the emergent delivery, but the situation necessitating emergent delivery that affects maternal mental health most.
Limitations of existing data
Designation of delivery urgency
Based on our review of the medical literature, there is no standard definition for expedited delivery. Terms, such as “stat,” “emergency,” and “urgent,” are used frequently without consistent definitions. Consistent terms must be established so that outcomes can be tracked across hospitals and systems. Heterogeneity of terms complicates the interpretation of data, including the use of “decision-to-incision” times and “decision-to-delivery” times, along with various terms used to categorize urgency of delivery. The Table summarizes the time frames and outcomes studied, illustrating this variety. By establishing consistent nomenclature, we hope to facilitate a comparison of future literature as it emerges.
| Study | n | Time interval assessed | Outcomes assessed | Key findings |
--- ---
| Heller et al,19 19. Heller, G. ∙ Bauer, E. ∙ Schill, S. ... Decision-to-delivery time and perinatal complications in emergency cesarean section Dtsch Arztebl Int. 2017; 114:589-596 PubMed Google Scholar 2017 | 39,291 | Decision-to-delivery time | 5- and 10-min Apgar scores, arterial cord blood pH, inhospital neonatal death | Low Apgar scores (<4) less common with decision-to-delivery time within 10 or 20 min |
| Kamoshita et al,20 20. Kamoshita, E. ∙ Amano, K. ∙ Kanai, Y. ... Effect of the interval between onset of sustained fetal bradycardia and cesarean delivery on long-term neonatal neurologic prognosis Int J Gynaecol Obstet. 2010; 111:23-27 Crossref Scopus (25) PubMed Google Scholar 2010 | 19 | Bradycardia-to-delivery time | Umbilical arterial pH, death, neurologic assessment at 2 y | All normal neurologic development when bradycardia-to-delivery time is <25 min |
| Holcroft et al,21 21. Holcroft, C.J. ∙ Graham, E.M. ∙ Aina-Mumuney, A. ... Cord gas analysis, decision-to-delivery interval, and the 30-minute rule for emergency cesareans J Perinatol. 2005; 25:229-235 Crossref Scopus (40) PubMed Google Scholar 2005 | 117 | Decision-to-incision time | Umbilical artery pH | Increasing decision-to-incision interval correlated with increased umbilical artery pH and base excess |
| MacKenzie and Cooke,22 22. MacKenzie, I.Z. ∙ Cooke, I. What is a reasonable time from decision-to-delivery by caesarean section? Evidence from 415 deliveries BJOG. 2002; 109:498-504 Crossref Scopus (113) PubMed Google Scholar 2002 | 415 | Decision-to-delivery time | 1- and 5-min Apgar scores, cord arterial pH values, immediate neonatal care provided | Trend toward less acidemia with increasing decision-to-delivery interval |
| Bloom et al,26 26. Bloom, S.L. ∙ Leveno, K.J. ∙ Spong, C.Y. ... Decision-to-incision times and maternal and infant outcomes Obstet Gynecol. 2006; 108:6-11 Crossref Scopus (138) PubMed Google Scholar 2006 | 2808 | Decision-to-incision ≤30 min vs >30 min | Endometritis, wound complication, operative injury, 5-min Apgar score of ≤3, umbilical artery pH<7.0, intubation in the delivery room, CPR, HIE, fetal death in labor, neonatal death | Umbilical artery pH<7, neonatal intubation rate higher when decision-to-incision times is <30 min, no difference in maternal complication rates |
| Grobman et al, 27 27. Grobman, W.A. ∙ Bailit, J. ∙ Sandoval, G. ... The association of decision-to-incision time for cesarean delivery with maternal and neonatal outcomes Am J Perinatol. 2018; 35:247-253 Crossref Scopus (13) PubMed Google Scholar 2018 | 3,482 | Decision-to-incision ≤15 min, 16-30 min, >30 min | Composite neonatal outcome (pH<7.0, Apgar score<5 at 5 min, HIE, seizures, death, NICU admission, ventilator use, hypotension with pressor support), composite maternal outcome (EBL>1000 mL, postpartum hemorrhage, blood transfusion, endometritis, wound complication, operative injury, hysterectomy) | Decision-to-incision interval of >30 min not associated with worse composite maternal or neonatal outcomes |
| Lurie et al,28 28. Lurie, S. ∙ Sulema, V. ∙ Kohen-Sacher, B. ... The decision to delivery interval in emergency and non-urgent cesarean sections Eur J Obstet Gynecol Reprod Biol. 2004; 113:182-185 Full Text Full Text (PDF) Scopus (29) PubMed Google Scholar 2003 | 155 | Decision-to-delivery time | Umbilical artery pH, 1- and 5-min Apgar scores | No correlation between decision-to-delivery time and neonatal outcomes |
| Thomas et al,29 29. Thomas, J. ∙ Paranjothy, S. ∙ James, D. National cross sectional survey to determine whether the decision to delivery interval is critical in emergency caesarean section BMJ. 2004; 328:665 Crossref PubMed Google Scholar 2004 | 17,780 | Decision-to delivery ≤15 min, 16-75 min, >75 min | Apgar scores of <7 and <4 at 5 min, stillbirth, maternal requirement for special care additional to routine care after cesarean delivery | No difference in outcomes in the ≤15-min and 16- to 75-min groups, higher adverse outcomes if >75 min |
| Onah et al,30 30. Onah, H.E. ∙ Ibeziako, N. ∙ Umezulike, A.C. ... Decision - delivery interval and perinatal outcome in emergency caesarean sections J Obstet Gynaecol. 2005; 25:342-346 Crossref Scopus (46) PubMed Google Scholar 2005 | 224 | Decision-to-delivery time | Indication for cesarean delivery, 1- and 5-min Apgar scores, newborn admission to special care, perinatal death, reasons for delay in decision-to-delivery interval of >30 min | No correlation between decision-to-delivery time and adverse perinatal outcomes |
| Moroz et al,39 39. Moroz, L. ∙ DiNapoli, M. ∙ D’Alton, M. ... Surgical speed and risk for maternal operative morbidity in emergent repeat cesarean delivery Am J Obstet Gynecol. 2015; 213:584.e1-584.e6 Full Text Full Text (PDF) Scopus (17) PubMed Google Scholar 2015 | 793 | Decision-to-delivery ≤2 min or >2 min | Composite morbidity, intraoperative events (atony, transfusion, uterine artery ligation, broad ligament hematoma, cystotomy, exploratory laparotomy or hysterectomy), postoperative events (postpartum transfusion, DVT, endometritis, wound complication, postoperative ileus, ICU admission), uterine artery pH | Decision-to-delivery interval of ≤2 min associated with increased adverse maternal outcomes, no difference in neonatal acidemia |
| Pallasmaa et al,33 33. Pallasmaa, N. ∙ Ekblad, U. ∙ Aitokallio-Tallberg, A. ... Cesarean delivery in Finland: maternal complications and obstetric risk factors Acta Obstet Gynecol Scand. 2010; 89:896-902 Crossref Scopus (104) PubMed Google Scholar 2010 | 2,496 | Elective vs emergency vs “crash-emergency” cesarean deliveries | Maternal complications (hemorrhage of >1500 mL, need for blood transfusion, intraoperative complications, complications related to anesthesia, hysterectomy and other reoperations, septicemia, endometritis, wound infection, UTI, pneumonia, pulmonary edema, DVT, bowel obstruction) | Increased urgency of cesarean delivery correlated with increased rate of maternal complications |
Table
Time interval and outcomes assessed
CPR, cardiopulmonary resuscitation; DVT, deep vein thrombosis; EBL, estimated blood loss; HIE, hypoxic-ischemic encephalopathy; ICU, intensive care unit; NICU, neonatal intensive care unit; UTI, urinary tract infection.
Bank. Thirty-minute rule for expedited delivery. Am J Obstet Gynecol 2023.
Open table in a new tab
The following definitions for request of a cesarean delivery, similar to those proposed by others, have been proposed8
8.
National Institute for Health and Clinical Excellence
Caesarean birth
Available at:
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Accessed June 30, 2022
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,45
45.
Lucas, D.N. ∙ Yentis, S.M. ∙ Kinsella, S.M. ...
Urgency of caesarean section: a new classification
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:
Class I—A cesarean delivery requested because of an immediate threat to the life of the mother or fetus, necessitating rapid delivery. Examples of this include umbilical cord prolapse, persistent fetal bradycardia, complete placental abruption, and resuscitative cesarean delivery, in cases of maternal cardiac arrest or impending maternal arrest.
Class II—A cesarean delivery requested because of a concern for the fetus or mother, but the timing of which may be delayed as felt to be clinically appropriate, allowing for additional measures, such as further intrauterine resuscitation, the arrival of additional personnel, cross-matching of blood if need is anticipated, or administration of regional anesthetic. Examples of this include nonbradycardic category III fetal heart tracing, worsening preeclampsia remote from delivery, or development of other concerning, although not immediately life-threatening, maternal or fetal conditions.
Class III—A cesarean delivery requested in the intrapartum period, for which delivery is recommended but there is no immediate perceived threat to the mother or fetus. Examples of this include labor arrest disorders with reassuring maternal and fetal status.
Class IV—A cesarean delivery that could be scheduled to best accommodate the patient and the delivery team. Examples of this include elective cesarean delivery, malpresentation, nonbleeding placenta previa, or previous cesarean deliveries in a nonlaboring patient.
Recommendations
A change of focus
Through the focus on decision-to-incision time, we have learned the importance of establishing clear clinical pathways and when feasible maintaining an in-house operating team,18
18.
Korhonen, J. ∙ Kariniemi, V.
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minimizing the distance between the patient and the delivery room,46
46.
Chauhan, S.P. ∙ Magann, E.F. ∙ Scott, J.R. ...
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the staff, and the operating room accessibility.47
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Spencer, M.K. ∙ MacLennan, A.H.
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We encourage investigation of process measures and research to better enable the prediction of fetal and maternal adverse outcomes within the context of intrapartum expedited delivery within different clinical settings.
In the cardiology literature, where the “door-to-balloon” time for percutaneous coronary intervention has been studied extensively, more recent research has raised questions of whether adherence to a time goal is an appropriate quality metric and whether the focus on time has distracted from therapeutic innovation, which may be more crucial to improving outcomes.48
48.
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Is door-to-balloon time a misleading metric? American College of Cardiology
Available at:
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Further research is needed to better understand accurate signs of impending fetal and maternal compromise rather than on how to achieve a specific time.
Of note, 1 hospital, after an adverse outcome, implemented a process flowchart for emergency cesarean deliveries; after the implementation, they experienced both a decrease in the number of cesarean deliveries labeled as emergent and an improvement in the timing among those that were. Their project included ensuring adequate supplies were available in the operating room, designating nursing duties, and conducting STAT pathway simulations.49
49.
Mooney, S.E. ∙ Ogrinc, G. ∙ Steadman, W.
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We encourage a focus on the creation of clear pathways and evidence-based safety measures to reduce maternal and neonatal morbidity and mortality during cesarean delivery, including the administration of appropriate antibiotic prophylaxis and neonatal resuscitation interventions.
Conclusions and future directions
Although a concrete time cutoff serves as a useful metric for ensuring standardization of care and historically has helped ensure adequate administrative support to enable prompt initiation of cesarean deliveries with a perceived need,1
1.
Boehm, F.H.
Decision to incision: time to reconsider
Am J Obstet Gynecol. 2012; 206:97-98
Full Text
Full Text (PDF)
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we recommend abandoning the “30-minute rule” as a quality measure, as a best practice guideline, or most importantly as a “standard of care,” given the lack of evidence that it is associated with desired outcomes for both the mother and the neonate.
It has been suggested that treatment within a specified time frame improves clinical outcomes.21
21.
Holcroft, C.J. ∙ Graham, E.M. ∙ Aina-Mumuney, A. ...
Cord gas analysis, decision-to-delivery interval, and the 30-minute rule for emergency cesareans
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However, through our review of the literature, there remains insufficient evidence to support a time frame for the decision-to-delivery interval.
For class I cases, delivery can be performed as expeditiously as possible while ensuring maternal safety. Class II cases, in which the fetal heart tracing is of concern (ie, nonbradycardic category III tracings), should first be managed conservatively. In light of available data, apparent fetal compromise (eg, fetal bradycardia or persistent category III fetal heart rate tracing)50
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should be identified and evidenced-based actions should be taken to address potentially reversible causes of hypoxia within 15 minutes. This includes the following: addressing maternal hypotension, maternal repositioning, discontinuation of uterine stimulants and/or administration of terbutaline in the setting of tachysystole, and correction of maternal acidemia or hypoxia.51
51.
American College of Obstetricians and Gynecologists
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,52
52.
Simpson, K.R.
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Close monitoring and individualization of delivery planning throughout resuscitative efforts, especially as these may vary in the time to recovery, should be continued. In the cases where fetal recovery is perceived to be unlikely, such as umbilical cord prolapse, severe placental abruption, and uterine rupture, efforts to move toward delivery start within this time. This would include mobilization of a delivery team, preparation of an operating room, and patient counseling regarding potential cesarean or operative vaginal delivery.
In addition, a focus is recommended on process improvement, eliminating unnecessary inefficiencies at each step along the way, a strategy some have used in the past.53
53.
de Regt, R.H. ∙ Marks, K. ∙ Joseph, D.L. ...
Time from decision to incision for cesarean deliveries at a community hospital
Obstet Gynecol. 2009; 113:625-629
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This study aimed to direct hospital attention and resources to develop evidence-based improvement in neonatal and maternal outcomes rather than on the direct shortening of decision-to-incision times. For example, increasing education and simulations of strategies for in utero fetal resuscitation, including prompt recognition and treatment of tachysystole, maternal hypotension, positional cord compression, or other causes of fetal hypoxia, may yield better neonatal outcomes than increasing speed of delivery. This is an area in which we hope to see more literature forthcoming.
In summary (as presented in “The ‘30 Minute Rule’ for Expedited Delivery: Fact or Fiction?” Video), there is scant and unreliable evidence that there is a dependable time goal for cesarean delivery that improves neonatal or maternal outcomes. The “30-minute rule” is not based on sound scientific evidence and should not be used as a quality measure or a proxy for improved neonatal outcomes. Furthermore, the importance of balancing maternal safety with the perceived need for speed of a delivery cannot be overstated. Finally, a call has been made for more research on intrapartum predictors of both short-term and, importantly, long-term neonatal outcomes, enabling a better assessment of delivery urgency.
Supplementary Data (1)
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Video (24.53 MB)
Video 1
Bank. Thirty-minute rule for expedited delivery. Am J Obstet Gynecol 2023 .
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8889 | https://onlinelibrary.wiley.com/doi/10.1111/j.1467-9469.2012.00808.x | On Sampling with Prescribed Second‐order Inclusion Probabilities - BONDESSON - 2012 - Scandinavian Journal of Statistics - Wiley Online Library
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Scandinavian Journal of Statistics
Volume 39, Issue 4 pp. 813-829
On Sampling with Prescribed Second-order Inclusion Probabilities
LENNART BONDESSON,
LENNART BONDESSON
Department of Mathematics and Mathematical Statistics, Umeå University
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LENNART BONDESSON,
LENNART BONDESSON
Department of Mathematics and Mathematical Statistics, Umeå University
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First published: 31 August 2012
Citations: 2
Lennart Bondesson, Department of Mathematics and Mathematical Statistics, Umeå University, SE-90187, Sweden.
E-mail: Lennart.Bondesson@math.umu.se
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Abstract
Abstract. Methods to perform fixed size sampling with prescribed second-order inclusion probabilities are presented. The focus is on a conditional Poisson design of order 2, a CP(2) design. It is an exponential design of quadratic type and it is carefully studied. In particular, methods to find the suitable values of the parameters and methods to sample are described. Small examples illustrate.
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References ----------
Related -------
Information -----------
Recommended
Pareto Sampling versus Sampford and Conditional Poisson Sampling
LENNART BONDESSON,IMBI TRAAT,ANDERS LUNDQVIST, Scandinavian Journal of Statistics
An Extension of Sampford's Method for Unequal Probability Sampling
LENNART BONDESSON,ANTON GRAFSTRÖM, Scandinavian Journal of Statistics
Conditional and Restricted Pareto Sampling: Two New Methods for Unequal Probability Sampling
LENNART BONDESSON, Scandinavian Journal of Statistics
Negative Dependence in Sampling
PETTER BRÄNDÉN,JOHAN JONASSON, Scandinavian Journal of Statistics
Order Sampling Design with Prescribed Inclusion Probabilities
NIBIA AIRES,JOHAN JONASSON,OLLE NERMAN, Scandinavian Journal of Statistics
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Citations: 2
Details
© 2012 Board of the Foundation of the Scandinavian Journal of Statistics
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Keywords
complementary sampling
conditional Poisson sampling
CP(2) sampling
entropy
Gibbs sampling
inclusion probabilities
linear programming
Pareto sampling
Sampford sampling
Sinha's method
Publication History
Issue Online: 21 November 2012
Version of Record online: 31 August 2012
Editorial history: Received January 2011, in final form March 2012
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Foci of an ellipse from equation
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The equation of an ellipse is given below.
(x−3)2 25+(y+1)2 169=1
What are the foci of this ellipse?
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Choose 1 answer:
(Choice A) (−3,13) and (−3,−11) A (−3,13) and (−3,−11)
(Choice B) (3,−13) and (3,11) B (3,−13) and (3,11)
(Choice C) (−12,−1) and (15,−1) C (−12,−1) and (15,−1)
(Choice D) (9,1) and (−15,1) D (9,1) and (−15,1)
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Is an odd/even function multiplied by i i still an odd/ even function?
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Let's say I have a function that I want to integrate an even function multiplied by i sin(x)i sin(x) between −1−1 and 1 1 and the function is 1−|x|1−|x| then does this integral become zero because integrating an odd function over a symmetric domain is zero or is i sin(x)i sin(x) not odd anymore when multiplied by i i.
even-and-odd-functions
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edited Nov 9, 2019 at 19:39
Allawonder
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You can pull constants out of an integral ...Rushabh Mehta –Rushabh Mehta 2019-11-09 17:52:45 +00:00 Commented Nov 9, 2019 at 17:52
I always thought of even and odd functions as functions with domain and range being subsets of R R. So, to me, i sin(x)i sin(x) would be i i times an odd function.Steven Alexis Gregory –Steven Alexis Gregory 2020-03-31 09:39:00 +00:00 Commented Mar 31, 2020 at 9:39
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Multiplying an odd or even function by a constant doesn't change the parity of the function.
The symbol i i sometimes represents the imaginary unit, which is a number, a constant.
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answered Nov 9, 2019 at 18:03
AllawonderAllawonder
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8893 | https://zhuanlan.zhihu.com/p/176512369 | 数据的集中位置：众数、中位数和平均数 - 知乎
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数据的集中位置：众数、中位数和平均数
首发于超智星球·从浅入深统计学
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数据的集中位置：众数、中位数和平均数
超级老乔
Good good study，day day up！
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如果你的老板让你汇报上个季度的销量表现，而他赶着开一个会，让你只用一个指标来概括。那你需要一个合适的数字，反映了上个季度的销量的集中位置，来代替整体水平。
本文，我们就来讲讲，有哪些关键的数值，可以代表整体的水平。
从猪肉价格上涨说起
最近国内猪肉价格飙升，不知道无肉不欢的朋友们可还能撑得住啊，就连小餐馆的炒饼炒肉都涨了好几块钱。。。然而，最近的微博有条热搜，某集团老板曹某旺谈猪肉涨价：应大胆、勇敢的接受涨价，不涨价才不合理！！！！！
具体言辞如下“1978年的猪肉两块多钱一斤，工人的工资才30块钱一个月。现在工资平均要7000块，涨了200倍。而农民养猪，猪肉就按照今天60块来算，也才涨30倍。我们应该大胆的、勇敢的接受这些农副产品涨价。可能涨价是合理的，不涨价才是不合理的”。
听完他的话，你是不是感觉貌似有点道理但又哪里不对的样子，其实这里他就让我们进入到了一个误区“平均”。
在贫富两极化的比较严重的情况下，平均工资根本不能代表普通人工资的水平；而且经济越发达，基本消费品的价格占总收入的比例应该越小才对。所以，下次再看到某媒体拿平均数来说事要谨慎了，我和马云的平均年收入为10亿，能说我年赚5亿吗，我倒是想呢。
这里出现了第一个知识点：平均数容易受异常数值影响，有时不能表现一组数据真实水平。
既然平均数不能表示真实工资水平，那么什么指标能表示工资水平呢？咱们本期谈谈代表数据集中数值的三个数：众数、中位数、平均数。
众数、中位数和平均数
上文讲到，针对数值型数据，我们可以进行分类统计、数据排序、数值计算三个层级的操作。
根据分组统计，我们可以得到众数
根据数据排序，我们可以得到中位数
根据数值计算，我们可以得到平均数
众数、中位数和平均数是集中趋势的三个主要测度值，它们具有不同的特点和应用场合。掌握它们的特点，有助于在实际应用中选择合理的测度值来描述数据的集中趋势。
众数众数是一组数据分布的峰值，不受数据极端值的影响。比如，B站刚创办时的用户主要是二次元爱好者，这说的就是众数。众数的缺点是具有不唯一性，一组数据可能有一个众数，也可能有两个或多个众数，也可能没有众数。众数只有在数据量较多时才有意义，当数据量较少时，不宜使用众数。众数适合作为分类数据的集中趋势测度值。
中位数是一组数据中间位置上的值，不受数据极端值的影响。举个栗子，房间里有5人，收入分别为「10万,11万,12万,13万」，此时的中位数为11.5万。即使此时马云加入，「10万,11万,12万,13万，马云」,中位数变为12万，仍然可以反映真实水平。当一组数据的分布偏斜程度较大时，使用中位数也许是一个好的选择。中位数适合作为顺序数据的集中趋势测度值。
平均数是针对数值型数据计算的，而且利用了全部数据信息，它是应用最广泛的集中趋势测度值。当数据呈对称分布或接近对称分布时，三个代表值相等或接近相等，这时则应选择平均数作为集中趋势的代表值。但平均数的主要缺点是易受数据极端值的影响，对于偏态分布的数据，平均数的代表性较差。因此，当数据为偏态分布，特别是偏斜程度较大时，可以考虑选择中位数或众数，这时它们的代表性要比平均数好。
从分布的角度看，众数始终是一组数据分布的最高峰值，中位数是处于一组数据中间位置上的值，而平均数则是全部数据的算术平均。因此，对于具有单峰分布的大多数数据而言，众数、中位数和平均数之间具有以下关系：
如果数据的分布是对称的，众数、中位数和平均数必定相等;
如果数据是左偏分布，说明数据存在极小值，必然拉动平均数向极小值一方靠，而众数和中位数由于是位置代表值，不受极值的影响，因此三者之间的关系表现为：平均数<中位数<众数；
如果数据是右偏分布，说明数据存在极大值，必然拉动平均数向极大值一方靠，因此，众数<中位数＜平均数。
上述关系如图所示：
“平均数有三种算法”
虽然平均数有缺点，但平均数在统计学中具有极其重要的地位，是集中趋势的最主要的测度值。所以我们要重点讲一下平均数。关键的是，平均数的计算方法。
有同学就要问了，平均数怎么算初中就已经学过了，还用的着教么？
哼，孔乙己说:“回字有四样写法”，老乔说：“平均数有三种算法”。在统计学中，根据所掌握数据的不同，平均数有不同的计算形式和计算公式，分别为：简单平均数、加权平均数、几何平均数。
这里有两点，咱扯远些。
首先，孔乙己的说法不对。
回字其实只有三种写法，即：回、囘、囬。这里是鲁迅先生描画孔乙己迂腐形象的神来之笔，说明孔乙己醉心于钻研无用的知识，自以为博学，其实是错误的。
其次，老乔的说法也不对。
数学世界里，除了上述三种，还有一种“调和平均数（harmonic mean）”。在数学中调和平均数与算术平均数都是独立的自成体系的。计算结果两者不相同且前者恒小于等于后者。因而数学调和平均数定义为：数值倒数的平均数的倒数。
但统计和数学不同，统计学中，统计加权调和平均数是加权算术平均数的变形，附属于算术平均数，不能单独成立体系。它主要是用来解决在无法掌握总体单位数（频数）的情况下，只有每组的变量值和相应的标志总量，而需要求得平均数的情况下使用的一种数据方法，公式：n/(1/A1+1/A2+...+1/An)。其计算结果与加权算术平均数完全相等。
没听懂没关系，为了这事，行业里也有学者专门研究这问题：
陈正谟. 统计学中不应有调和平均数[J]. 统计工作, 1957(7).
常乐. 论取消调和平均数[J]. 贵州财经学院学报, 1996.
姜长文. 关于取消"调和平均数"统计概念的论证分析[J]. 金融理论与教学, 2007(第2期):51-53.
咱截图看看其中一篇的文风，很有那个时代的特色。老乔我是文科生，对这个瞬间勾起无数上学时的知识回忆。
好啦，还是回正题，咱讲统计学的平均数三种算法：简单平均数、加权平均数、几何平均数。
简单平均数
根据未经分组数据计算的平均数称为简单平均数(simple mean)。
设一组样本数据为x1,x2...xn,样本量为n,简单样本平均数用$\bar{x}$表示，计算公式为：
\bar{x}=\frac{\sum_{i=1}^n x_{i}}{n}
我们可以利用Excel中的AVERAGE函数可以计算一组数值型数据的平均数。
加权平均数
根据分组数据计算的平均数称为加权平均数(weighted mean)。
设原始数据被分成k组，各组的组中值分别用M1,M2,…，Mk表示，各组变量值出现的频数分别用f1,f2,…，fk表示，则样本加权平均数的计算公式为：
\bar{x}=\frac{\sum_{i=1}^k M_{i}f_{i}}{n}
要注意，这里其实是用各组中值代表各组的实际数据。如果各组数据在组内是均匀分布的，那计算的结果是比较准确的，否则误差就会较大。
几何平均数
几何平均数(geometric mean)是n个变量值乘积的n次方根，用G表示。计算公式为：
G_{m}=\sqrt[n] {\prod_{i=1}^n {(1+x_{i})}}-1
当数据中出现零值或负值时、不宜计算几何平均数。
利用Excel中的GEOMEAN函数可以计算—组数值型数据的几何平均数。
几何平均数是适用于特殊数据的一种平均数，它主要用于计算平均比率。当所掌握的变量值本身是比率形式时，采用几何平均法计算平均比率更为合理。在实际应用中，几何平均数主要用于计算现象的平均增长率。
一位投资者持有一种股票。连续4年的收益率分别为4.5%,2.1%,25.5%,1.9%。要求计算该投资者在这4年内的平均收益率。
平均收益率，几何平均数为8.0787%；平均收益为13644.57(元)
平均收益率，算数平均数为8.5%；平均收益为13858.59(元)
两者相差214.02元，而这部分收益投资者没有获得。这说明，对于比率数据的平均采用几何平均要比算术平均更合理。
并且，当所平均的各比率数值差别不大时，算术平均和几何平均的结果相差不大；如果各比率的数值相差较大时，二者的差别就更加明显。
谨慎对待数字
还有讲一个故事，关于如何选择集中数字：
2010年至2013年间，美国家庭的平均工资从8.4万美元增加到超过8.7万美元，增长了4%。
但并不是每个人都在欢呼，年内收入中位数下降了5%，家庭收入中位数从4.9万美元下降到略高于4.65万美元。
那些不道德的政客，口口声声说“平均收入在增长”———人民大众切切实时的感受着收入的下降。平均收入增长的同时，收入中位数下降。这是金融危机后的真实情况。因为家庭在收入分配上不平等，部分人赚了更多的钱，而其他许多家庭的收入却减少了。
这指向了关于统计学一个非常重要的问题，在这个系列中一次又一次强调的，统计数字是会“说谎”的。统计学除了学习知识本身，更重要的一个重要部分是理解。统计数字可以帮助我们做决定，但我们用我们的常识谨慎的接纳这些数字。
总结
参考文献：
孙静娟主编.《统计学》.清华大学出版社.2015：18-20.
袁卫,刘超.《统计学--思想、方法与应用》(第二版).中国人民大学出版社.2011.
贾俊平,何晓群,金勇进.《统计学》(第七版).中国人民大学出版社.2018.
陈正谟. 统计学中不应有调和平均数[J]. 统计工作, 1957(7).
发布于 2020-08-09 06:44
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现在工资平均7000块？我一个月工资才三四千，7000块钱叫平均啊！
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8894 | https://www.wikidoc.org/index.php/Gibbs_free_energy | Gibbs free energy - wikidoc
Gibbs free energy
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Template:Statistical mechanics In thermodynamics, the Gibbs free energy (IUPAC recommended name: Gibbs energy or Gibbs function) is a thermodynamic potential which measures the "useful" or process-initiating work obtainable from an isothermal, isobaricthermodynamic system. Technically, the Gibbs free energy is the maximum amount of non-expansion work which can be extracted from a closed system or this maximum can be attained only in a completely reversible process. When a system changes from a well-defined initial state to a well-defined final state, the Gibbs free energy ΔG equals the work exchanged by the system with its surroundings, less the work of the pressure forces, during a reversible transformation of the system from the same initial state to the same final state.
Gibbs energy is also the chemical potential that is minimized when a system reaches equilibrium at constant pressure and temperature. As such, it is a convenient criterion of spontaneity for processes with constant pressure and temperature.
The Gibbs free energy, originally called available energy, was developed in the 1870s by the American mathematical physicist Willard Gibbs. In 1873, in a footnote, Gibbs defined what he called the “available energy” of a body as such:
“The greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition.”
The initial state of the body, according to Gibbs, is supposed to be such that "the body can be made to pass from it to states of dissipated energy by reversible processes". In his 1876 magnum opusOn the Equilibrium of Heterogeneous Substances, a graphical analysis of multi-phase chemical systems, he engaged his thoughts on chemical free energy in full.
[x]
Contents
1 Definitions
2 Derivation
3 Overview
4 History
5 What does the term ‘free’ mean?
6 Free energy of reactions
7 Useful identities
8 Standard change of formation
9 Table of Selected Substances
10 See also
11 References
12 External links
Definitions
File:Gibbs-plot.jpg
Willard Gibbs’ 1873 available energy (free energy) graph, which shows a plane perpendicular to the axis of v (volume) and passing through point A, which represents the initial state of the body. MN is the section of the surface of dissipated energy. Qε and Qη are sections of the planes η = 0 and ε = 0, and therefore parallel to the axes of ε (internal energy) and η (entropy) respectively. AD and AE are the energy and entropy of the body in its initial state, AB and AC its available energy (Gibbs free energy) and its capacity for entropy (the amount by which the entropy of the body can be increased without changing the energy of the body or increasing its volume) respectively.
The Gibbs free energy is defined as:
G = U+pV-TS \,
which is the same as:
G = H-TS \,
where:
U is the internal energy (SI unit: joule)
p is pressure (SI unit: pascal)
V is volume (SI unit: m 3)
T is the temperature (SI unit: kelvin)
S is the entropy (SI unit: joule per kelvin)
H is the enthalpy (SI unit: joule)
The expression for the infinitesimal reversible change in the Gibbs free energy, for an open system, subjected to the operation of external forces X i, which cause the external parameters of the system a i to change by an amount da i, is given by:
\mathrm{d}G =V\mathrm{d}p-S\mathrm{d}T+\sum_{i=1}^k \mu_i \,\mathrm{d}N_i + \sum_{i=1}^n X_i \,\mathrm{d}a_i + \ldots \,
T\mathrm{d}S=\mathrm{d}q=\mathrm{d}U+p\mathrm{d}V,
where:
\mu_i is the chemical potential of the i-th chemical component. (SI unit: joules per particle or joules per mol)
N_i is the number of particles (or number of moles) composing the i-th chemical component.
This is one form of Gibbs fundamental equation. In the infinitesimal expression, the term involving the chemical potential accounts for changes in Gibbs free energy resulting from an influx or outflux of particles. In other words, it holds for an open system. For a closed system, this term may be dropped.
Any number of extra terms may be added, depending on the particular system being considered. Aside from mechanical work, a system may in addition perform numerous other types of work. For example, in the infinitesimal expression, the contractile work energy associated with a thermodynamic system that is a contractile fiber which shortens by an amount -dl under a force f would result in a term fdl being added. If a quantity of charge -de is acquired by a system at an electrical potential Ψ, the electrical work associated with this is -Ψ de, which would be included in the infinitesimal expression. Other work terms are added on per system requirements.
Each quantity in the equations above can be divided by the amount of substance, measured in moles, to form molar Gibbs free energy. The Gibbs free energy is one of the most important thermodynamic functions for the characterization of a system. It is a factor in determining outcomes such as the voltage of an electrochemical cell, and the equilibrium constant for a reversible reaction. In isothermal, isobaric systems, Gibbs free energy can be thought of as a "dynamic" quantity, in that it is a representative measure of the competing effects of the enthalpic and entropic driving forces involved in a thermodynamic process.
The temperature dependence of the Gibbs energy for an ideal gas is given by the Gibbs-Helmholtz equation and its pressure dependence is given by:
\frac{G}{N} = \frac{G}{N}^\circ + kT\ln \frac{p}Template:P^\circ")
if the volume is known rather than pressure then it becomes:
\frac{G}{N} = \frac{G}{N}^\circ + kT\ln \frac{V^\circ}Template:V")
or more conveniently as its chemical potential:
\frac{G}{N} = \mu = \mu^\circ + RT\ln \frac{p}Template:P^\circ")
In non-ideal systems, fugacity comes into play.
Derivation
The Gibbs free energy total differential in terms of its natural variables may be derived via Legendre transforms of the internal energy. For a system undergoing an internally reversible process that is allowed to exchange matter, heat and work with its surroundings, the differential of the internal energy is given from the first law of thermodynamics as
\mathrm{d}U = T\mathrm{d}S - p \,\mathrm{d}V + \sum_i \mu_i \,\mathrm{d} N_i\,
.
Because S, V, and {N_i} are extensive variables, Euler's homogeneous function theorem allows easy integration of \mathrm{d}U:
U = T S - p V + \sum_i \mu_i N_i\,
.
The definition of G from above is
G = U + p V - T S\,
.
Taking the total differential, we have
\mathrm{d} G = \mathrm{d}U + p\,\mathrm{d}V + V\mathrm{d}p - T\mathrm{d}S - S\mathrm{d}T\,
.
Replacing \mathrm{d}U with the result from the first law gives
\mathrm{d} G = T\mathrm{d}S - p\,\mathrm{d}V + \sum_i \mu_i \,\mathrm{d} N_i + p \,\mathrm{d}V + V\mathrm{d}p - T\mathrm{d}S - S\mathrm{d}T\,
\mathrm{d} G = V\mathrm{d}p - S\mathrm{d}T + \sum_i \mu_i \,\mathrm{d} N_i \,
.
The natural variables of G are then p, T, and \left { N_i \right }. Because some of the natural variables are intensive, d G may not be integrated using Euler integrals as is the case with internal energy. However, simply substituting the result for U into the definition of G gives a standard expression for G:
G = T S - p V + \sum_i \mu_i N_i + p V - T S\,
G = \sum_i \mu_i N_i\,
.
Overview
In a simple manner, with respect to STP reacting systems, a general rule of thumb is:
“Every system seeks to achieve a minimum of free energy.”
Hence, out of this general natural tendency, a quantitative measure as to how near or far a potential reaction is from this minimum is when the calculated energetics of the process indicate that the change in Gibbs free energy ΔG is negative. Essentially, this means that such a reaction will be favored and will release energy. The energy released equals the maximum amount of work that can be performed as a result of the chemical reaction. Conversely, if conditions indicated a positive ΔG, then energy--in the form of work--would have to be added to the reacting system to make the reaction go.
History
The quantity called "free energy" is essentially a more advanced and accurate replacement for the outdated term “affinity”, which was used by chemists in previous years to describe the “force” that caused chemical reactions. The term affinity, as used in chemical relation, dates back to at least the time of Albertus Magnus in 1250.
From the 1998 textbook Modern Thermodynamics by Nobel Laureate and chemical engineering professor Ilya Prigogine we find: "As motion was explained by the Newtonian concept of force, chemists wanted a similar concept of ‘driving force’ for chemical change? Why do chemical reactions occur, and why do they stop at certain points? Chemists called the ‘force’ that caused chemical reactions affinity, but it lacked a clear definition."
During the entire 18th century, the dominant view in regards to heat and light was that put forward by Isaac Newton, called the “Newtonian hypothesis”, which stated that light and heat are forms of matter attracted or repelled by other forms of matter, with forces analogous to gravitation or to chemical affinity.
In the 19th century, the French chemist Marcellin Berthelot and the Danish chemist Julius Thomsen had attempted to quantify affinity using heats of reaction. In 1875, after quantifying the heats of reaction for a large number of compounds, Berthelot proposed the “principle of maximum work” in which all chemical changes occurring without intervention of outside energy tend toward the production of bodies or of a system of bodies which liberate heat.
In addition to this, in 1780 Antoine Lavoisier and Pierre-Simon Laplace laid the foundations of thermochemistry by showing that the heat given out in a reaction is equal to the heat absorbed in the reverse reaction. They also investigated the specific heat and latent heat of a number of substances, and amounts of heat given out in combustion. Similarly, in 1840 Swiss chemist Germain Hess formulated the principle that the evolution of heat in a reaction is the same whether the process is accomplished in one-step or in a number of stages. This is known as Hess' law. With the advent of the mechanical theory of heat in the early 19th century, Hess’s law came to be viewed as a consequence of the law of conservation of energy.
Based on these and other ideas, Berthelot and Thomsen, as well as others, considered the heat given out in the formation of a compound as a measure of the affinity, or the work done by the chemical forces. This view, however, was not entirely correct. In 1847, the English physicist James Joule showed that he could raise the temperature of water by turning a paddle wheel in it, thus showing that heat and mechanical work were equivalent or proportional to each other, i.e. approximately, dW \propto dQ. This statement came to be known as the mechanical equivalent of heat and was a precursory form of the first law of thermodynamics.
By 1865 the German physicist Rudolf Clausius had shown that this equivalence principle needed amendment. That is, one can use the heat derived from a combustion reaction in a coal furnace to boil water, and use this heat to vaporize steam, and then use the enhanced high pressure energy of the vaporized steam to push a piston. Thus, we might naively reason that one can entirely convert the initial combustion heat of the chemical reaction into the work of pushing the piston. Clausius showed, however, that we need to take into account the work that the molecules of the working body, i.e. the water molecules in the cylinder, do on each other as they pass or transform from one step of or state of the engine cycle to the next, e.g. from (P1,V1) to (P2,V2). Clausius originally called this the “transformation content” of the body, and then later changed the name to entropy. Thus, the heat used to transform the working body of molecules from one state to the next cannot be used to do external work, e.g. to push the piston. Clausius defined this transformation heat as dQ = TdS.
In 1873, Willard Gibbs published A Method of Geometrical Representation of the Thermodynamic Properties of Substances by Means of Surfaces in which he introduced the preliminary outline of the principles of his new equation able to predict or estimate the tendencies of various natural processes to ensue when bodies or systems are brought into contact. By studying the interactions of homogeneous substances in contact, i.e. bodies, being in composition part solid, part liquid, and part vapor, and by using a three-dimensional volume-entropy-internal energy graph, Gibbs was able to determine three states of equilibrium, i.e. "necessarily stable", "neutral", and "unstable", and whether or not changes will ensue. In 1876, Gibbs built on this framework by introducing the concept of chemical potential so to take into account chemical reactions and states of bodies which are chemically different from each other. In his own words, to summarize his results in 1873, Gibbs states:
If we wish to express in a single equation the necessary and sufficient condition of thermodynamic equilibrium for a substance when surrounded by a medium of constant pressurep and temperatureT, this equation may be written:
\delta (\epsilon - T\eta + p\nu) = 0
when \delta refers to the variation produced by any variations in the state of the parts of the body, and (when different parts of the body are in different states) in the proportion in which the body is divided between the different states. The condition of stable equilibrium is that the value of the expression in the parenthesis shall be a minimum.
In this description, as used by Gibbs, ε refers to the internal energy of the body, η refers to the entropy of the body, and ν is the volume of the body.
Hence, in 1882, after the introduction of these arguments by Clausius and Gibbs, the German scientist Hermann von Helmholtz stated, in opposition to Berthelot and Thomas’ hypothesis that chemical affinity is a measure of the heat of reaction of chemical reaction as based on the principle of maximal work, that affinity is not the heat given out in the formation of a compound but rather it is the largest quantity of work which can be gained when the reaction is carried out in a reversible manner, e.g. electrical work in a reversible cell. The maximum work is thus regarded as the diminution of the free, or available, energy of the system (Gibbs free energy G at T = constant, P = constant or Helmholtz free energy F at T = constant, V = constant), whilst the heat given out is usually a measure of the diminution of the total energy of the system (Internal energy). Thus, G or F is the amount of energy “free” for work under the given conditions.
Up until this point, the general view had been such that: “all chemical reactions drive the system to a state of equilibrium in which the affinities of the reactions vanish”. Over the next 60 years, the term affinity came to be replaced with the term free energy. According to chemistry historian Henry Leicester, the influential 1923 textbook Thermodynamics and the Free Energy of Chemical Reactions by Gilbert N. Lewis and Merle Randall led to the replacement of the term “affinity” by the term “free energy” in much of the English-speaking world.
What does the term ‘free’ mean?
In the 18th and 19th centuries, the theory of heat, i.e. that heat is a form of energy having relation to vibratory motion, was beginning to supplant both the caloric theory, i.e. that heat is a fluid, and the four element theory in which heat was the lightest of the four elements. Many textbooks and teaching articles during these centuries presented these theories side by side. Similarly, during these years, heat was beginning to be distinguished into different classification categories, such as “free heat”, “combined heat”, “radiant heat”, specific heat, heat capacity, “absolute heat”, “latent caloric”, “free” or “perceptible” caloric (calorique sensible), among others.
In 1780, for example, Laplace and Lavoisier stated: “In general, one can change the first hypothesis into the second by changing the words ‘free heat, combined heat, and heat released’ into ‘vis viva, loss of vis viva, and increase of vis viva.’” In this manner, the total mass of caloric in a body, called absolute heat, was regarded as a mixture of two components; the free or perceptible caloric could affect a thermometer while the other component, the latent caloric, could not. The use of the words “latent heat” implied a similarity to latent heat in the more usual sense; it was regarded as chemically bound to the molecules of the body. In the adiabaticcompression of a gas the absolute heat remained constant by the observed rise of temperature, indicating that some latent caloric had become “free” or perceptible.
During the early 19th century, the concept of perceptible or free caloric began to be referred to as “free heat” or heat set free. In 1824, for example, the French physicist Sadi Carnot, in his famous “Reflections on the Motive Power of Fire”, speaks of quantities of heat ‘absorbed or set free’ in different transformations. In 1882, the German physicist and physiologist Hermann von Helmholtz coined the phrase ‘free energy’ for the expression E – TS, in which the change in F (or G) determines the amount of energy ‘free’ for work under the given conditions.
Thus, in traditional use, the term “free” was attached to Gibbs free energy, i.e. for systems at constant pressure and temperature, or to Helmholtz free energy, i.e. for systems at constant volume and temperature, to mean ‘available in the form of useful work.’ With reference to the Gibbs free energy, we add the qualification that it is the energy free for non-volume work.
An increasing number of books and journal articles do not include the attachment “free”, referring to G as simply Gibbs energy (and likewise for the Helmholtz energy). This is the result of a 1988 IUPAC meeting to set unified terminologies for the international scientific community, in which the adjective ‘free’ was supposedly banished. This standard, however, has not yet been universally adopted, and many published articles and books still include the descriptive ‘free’.
Free energy of reactions
To derive the Gibbs free energy equation for an isolated system, let S tot be the total entropy of the isolated system, that is, a system which cannot exchange heat or mass with its surroundings. According to the second law of thermodynamics:
\Delta S_{tot} \ge 0 \,
and if \Delta S_{tot} = 0 \, then the process is reversible. The heat transfer \Delta Q vanishes for an adiabatic system. Any adiabatic process that is also reversible is called an isentropic \left( {\Delta Q\over T} = \Delta S = 0 \right) \, process.
Now consider diabatic systems, having internal entropy S int. Such a system is thermally connected to its surroundings, which have entropy S ext. The entropy form of the second law does not apply directly to the diabatic system, it only applies to the closed system formed by both the system and its surroundings. Therefore a process is possible if
\Delta S_{int} + \Delta S_{ext} \ge 0 \,
.
We will try to express the left side of this equation entirely in terms of state functions. ΔS ext is defined as:
\Delta S_{ext} = - {\Delta Q\over T} \,
Temperature T is the same for two systems in thermal equilibrium. By the zeroth law of thermodynamics, if a system is in thermal equilibrium with a second and a third system, the latter two are in equilibrium as well. Also, \Delta Q is heat transferred to the system, so -\Delta Q is heat transferred to the surroundings, and −ΔQ/T is entropy gained by the surroundings. We now have:
\Delta S_{int} - {\Delta Q\over T} \ge 0 \,
Multiply both sides by T:
T \Delta S_{int} - \Delta Q\ge 0 \,
ΔQ is heat transferred to the system; if the process is now assumed to be isobaric, then ΔQ p = ΔH:
T \Delta S_{int} - \Delta H \ge 0\,
ΔH is the enthalpy change of reaction (for a chemical reaction at constant pressure and temperature). Then
\Delta H - T \Delta S_{int} \le 0 \,
for a possible process. Let the change Δ G in Gibbs free energy be defined as
\Delta G = \Delta H - T \Delta S_{int} \,
(eq.1)
Notice that it is not defined in terms of any external state functions, such as Δ S ext or Δ S tot. Then the second law becomes, which also tells us about the spontaneity of the reaction:
\Delta G < 0 \,
favored reaction (Spontaneous)
\Delta G = 0 \,
Neither the forward nor the reverse reaction prevails (Equilibrium)
\Delta G > 0 \,
disfavored reaction (Nonspontaneous)
Gibbs free energy G itself is defined as
G = H - T S_{int} \,
(eq.2)
but notice that to obtain equation (2) from equation (1) we must assume that T is constant. Thus, Gibbs free energy is most useful for thermochemical processes at constant temperature and pressure: both isothermal and isobaric. Such processes don't move on a P-T diagram, such as phase change of a pure substance, which takes place at the saturation pressure and temperature. Chemical reactions, however, do undergo changes in chemical potential, which is a state function. Thus, thermodynamic processes are not confined to the two dimensional P-V diagram. There is a third dimension for n, the quantity of gas. Naturally for the study of explosive chemicals, the processes are not necessarily isothermal and isobaric. For these studies, Helmholtz free energy is used.
If a closed system (Δ Q = 0) is at constant pressure (Δ Q = Δ H), then
\Delta H = 0 \,
Therefore the Gibbs free energy of a closed system is:
\Delta G = -T \Delta S \,
and if \Delta G \le 0 \, then this implies that \Delta S \ge 0 \,, back to where we started the derivation of Δ G.
Useful identities
\Delta G = \Delta H - T \Delta S \,
for constant temperature
\Delta G^\circ = -R T \ln K \,
\Delta G = \Delta G^\circ + R T \ln Q \,
\Delta G = -nF \Delta E \,
and rearranging gives
nF\Delta E^\circ = RT \ln K \,
nF\Delta E = nF\Delta E^\circ - R T \ln Q \, \,
\Delta E = \Delta E^\circ - \frac{R T}{n F} \ln Q \, \,
which relates the electrical potential of a reaction to the equilibrium coefficient for that reaction.
where
Δ G = change in Gibbs free energy, Δ H = change in enthalpy, T = absolute temperature, Δ S = change in entropy, R = gas constant, ln = natural logarithm, K = equilibrium constant, Q = reaction quotient, n = number of electrons per mole product, F = Faraday constant (coulombs per mole), and Δ E = electrical potential of the reaction. Moreover, we also have:
K_{eq}=e^{- \frac{\Delta G^\circ}{RT}}
\Delta G^\circ = -RT(\ln K_{eq}) = -2.303RT(\log K_{eq})
which relates the equilibrium constant with Gibbs free energy.
Standard change of formation
The standard Gibbs free energy of formation of a compound is the change of Gibbs free energy that accompanies the formation of 1 mole of that substance from its component elements, at their standard states (the most stable form of the element at 25 degrees Celsius and 100 kilopascals). Its symbol is Δ G f~~O~~.
All elements in their standard states (oxygen gas, graphite, etc.) have 0 standard Gibbs free energy change of formation, as there is no change involved.
Δ G = Δ G˚ + RT ln Q
At equilibrium, Δ G=0 and Q = K so the equation becomes Δ G˚= −RT ln K
Table of Selected Substances
| Substance | State | Δ G˚ (cal/mol) |
---
| NH 3 | g | -3.976 |
| H 2 O | lq | -56.69 |
| H 2 O | g | -54.64 |
| CO 2 | g | -94.26 |
| CO | g | -32.81 |
| CH 4 | g | -12.14 |
| C 2 H 6 | g | -7.86 |
| C 3 H 8 | g | -5.614 |
| C 8 H 18 | g | 4.14 |
| C 10 H 22 | g | 8.23 |
See also
Enthalpy
Entropy
Free energy
Helmholtz free energy
Free entropy
Thermodynamics
Willard Gibbs
Calphad
Grand potential
References
↑ 1.01.11.2Perrot, Pierre (1998). A to Z of Thermodynamics. Oxford University Press. ISBN 0-19-856552-6.
↑Chemical Potential - IUPAC Gold Book
↑Müller, Ingo (2007). A History of Thermodynamics - the Doctrine of Energy and Entropy. Springer. ISBN 978-3-540-46226-2.
↑Katchalsky, A. (1965). Nonequilibrium Thermodynamics in Biophysics. Harvard University Press. CCN 65-22045.Unknown parameter |coauthors= ignored (help)
↑ 5.05.15.2Salzman, William R. (2001-08-21). "Open Systems". Chemical Thermodynamics. University of Arizona. Archived from the original on 2007-07-07. Retrieved 2007-10-11.
↑Mendoza, E. (1988). Reflections on the Motive Power of Fire – and other Papers on the Second Law of Thermodynamics by E. Clapeyron and R. Carnot. Dover Publications, Inc. ISBN 0-486-44641-7.
↑Baierlein, Ralph (2003). Thermal Physics. Cambridge University Press. ISBN 0-521-65838-1.
↑Reiss, Howard (1965). Methods of Thermodynamics. Dover Publications. ISBN 0-486-69445-3.
↑International Union of Pure and Applied Chemistry Commission on Atomspheric Chemistry (1990). "Glossary of Atmospheric Chemistry Terms (Recommendations 1990)"(PDF). Pure Appl. Chem.62: 2167&ndash, 2219. Retrieved 2006-12-28.International Union of Pure and Applied Chemistry Commission on Physicochemical Symbols Terminology and Units (1993). Quantities, Units and Symbols in Physical Chemistry (2nd Edition)(PDF). Oxford: Blackwell Scientific Publications. p.48. ISBN 0-632-03583-8. Retrieved 2006-12-28.International Union of Pure and Applied Chemistry Commission on Quantities and Units in Clinical Chemistry (1996). "Glossary of Terms in Quantities and Units in Clinical Chemistry (IUPAC-IFCC Recommendations 1996)"(PDF). Pure Appl. Chem.68: 957&ndash, 1000. Retrieved 2006-12-28.Unknown parameter |coauthors= ignored (help)
↑Handbook of chemistry and physics, 1960, p.1882-1915, p.1919-1921, 42nd ed., Harrison
External links
IUPAC definition (Gibbs energy)
Gibbs energy - Florida State University
Gibbs Free Energy - Eric Weissteins World of Physics
Gibbs Free Energy - Chemistry Gateway
Entropy and Gibbs Free Energy - www.2ndlaw.com
Gibbs Free Energy - Georgia State University
Gibbs Free Energy Java Applet - University of California, Berkeley
Gibbs Free Energy - Illinois State University
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8895 | https://www.101computing.net/random-odd-and-even-number-functions/ | Published Time: 2021-10-20T21:03:23+00:00
Random Odd and Even Numbers - 101 Computing
Random Odd and Even Numbers - 101 Computing
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Random Odd and Even Numbers
Posted on October 20, 2021 by Administrator Posted in Computer Science, Python - Intermediate, Python Challenges, Solved Challenges
In this challenge we will focus on using the random library to generate random numbers in Python.
The randint() function is used to generate a random integer (whole number) within a given range.
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import random
number = randint(1,100)
import random number = randint(1,100)
import random
number = randint(1,100)
The above code would generate a random number between 1 and 100.
The aim of this challenge is to write our own function called randomOddNumber() to generate a random odd number within a given range and a another function called randomEvenNumber() to generate a random even number. Both functions will take two parameters, the lower and higher values of the range.
Random Odd Number Functions
We will investigate three different algorithms to implement the randomOddNumber() function:
Method #1: Sequencing Method #2: Iteration Method #3: Selection
The first method is based on a sequencing algorithm.
Here is the approach we will use:
To generate an odd number, let’s say between 0 and 100, we will first generate a random number between 0 and 49, then multiply this number by 2 (this will hence become an even number between 0 and 98) and finally we will add 1 to make it an odd number (between 1 and 99).
Python Code:
The second algorithm is based on a while loop (a form of iteration).
With this approach we are generating a random number between 1 and 100. We will repeat this process as long as the number generated is even (not odd). A number is even if the remainder of dividing this number by two is zero.
Python Code:
Our third approach will be based on using an IF statement, a form of selection.
We will generate a random number, check if this number is odd or even by checking whether the remainder of dividing this number by two is zero (for an even number) or 1 (for an odd number). If the number is even, we will add 1 to this number to change it to an odd number.
Python Code:
Random Even Number Functions
Your task consists of adapting the above three functions to create three additional functions, this time to generate random even numbers using all three methods/programming constructs: sequencing, selection and iteration.
Extension Task: Random Prime Number Function
For this more complex extension task, you will need to write a function called randomPrimeNumber() to generate a random prime number within a given range. This function will take two parameters: the lower and higher values of the range.
Solution...
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8896 | https://cs.uwaterloo.ca/~smann/PGABLE/PGAtutorial.pdf | A Tutorial for Plane-based Geometric Algebra Version 2.0.0 Zachary Leger and Stephen Mann March 27, 2025 Abstract In this tutorial we give an introduction to geometric algebra, using our Matlab package PGABLE (Plane-based Geometric AlgeBra Learning Environment). We begin with an introduction to OGA (Orig-inal Geometric Algebra), where we graphically demonstrate the ideas of the geometric product, the outer product, and the inner product, and the geometric operators that may be formed from them. We give several demonstrations of computations you can do using the geometric algebra, including projection and rejection, orthogonalization, interpolation of rotations, and intersection of linear offset spaces such as lines and planes. We emphasize the importance of blades as representations of subspaces, and the use of meet and join to manipulate them. We then give a tutorial for PGA (Plane-based Geometric Algebra), showing how to construct planes, lines, and points, and how to manipulate, query, and transform these objects. Finally, we give a similar tutorial for CGA (Conformal Geometric Algebra), which has circles and spheres in addition to point,s lines, and planes.
1 Introduction This is an introduction to geometric algebra, which is a structural way to think about and calculate with geometry. It differs in its nature from linear algebra, constructive Euclidean geometry, differential geometry, vector calculus and other ways you may have learned before; yet it encompasses them all in a natural manner, including some extra things like complex numbers and quaternions.
To help understand and visualize the geometry, we have implemented PGABLE (Plane-based Geometric AlgeBra Learning Environment) in Matlab, from which we created a tutorial to illustrate our examples.
This tutorial is heavily based on the tutorial for GABLE [MDB99a, MDB99b, MDB01], although that later parts of this tutorial look at PGA rather than the homogeneous model that was covered in the GABLE tutorial. PGABLE itself is heavily based on GABLE, although significant changes have been made to update the software to a more modern version of Matlab.
There are several reasons why geometric algebra is convenient to work with in geometry, and they all involve the capability to talk constructively about important geometrical concepts, which are all embedded as elementary objects in the algebra. Having those available will change your thinking in a strangely pow-erful way, and geometric algebra then provides the computational tools to implement that new thinking.
Obviously, you can’t change your thinking overnight, and so we will demonstrate some of it in the tutorial to give you a flavor.
You may find in this tutorial many things that are familiar, because a lot of this work has been invented before in other contexts than geometric algebra. It is only recently that we understand how it all fits together into one framework, and how important that is for the computer sciences. It now becomes possible to write one book (see, for example, [DMF07]), and one computer program, which contains all the geometry we might ever need. Gone would be the transitions between parts of the real-world problem that are solved by linear algebra, vector calculus, or differential geometry, with the accompanying efficiency and sensitivity to bugs.
All would just be done within the framework of geometric algebra.
In this tutorial, we introduce terms gradually, and give you a geometric intuition of their meaning as we go along. We limit almost all of our explanations to three geometric algebras of Euclidean 3-dimensional 1 space, which are denoted R3,0,0 (OGA), R3,0,1 (PGA), and R3,0,2 (CGA). Other geometric algebras are important as well, but this tutorial is already long enough.
If you are primarily interested in PGA, you should read (and run the Matlab examples) the rest of this section (Sections 1.2 and 1.3), as well Section 2 through Sections 2.5. You may then skip the remainder of Section 2 and move straight to Section 3, although you should probably read Section 2.6.3 on reflections and Section 2.7 on rotations.
Similarly, if you are primarily interested in CGA, you should read (and run the Matlab examples) the rest of this section (Sections 1.2 and 1.3), as well Section 2 through Sections 2.5. You may then skip the remainder of Section 2 and move straight to Section 4, although you should probably read Section 2.6.3 on reflections and Section 2.7 on rotations.
1.1 Release Notes This version of the tutorial matches PGABLE version 2.0.0. Note that if you use an older version of Matlab (e.g., pre-2024), you will need to add the function clip.m to your PGABLE folder.
The big change in PGABLE 2.0.0 from earlier versions is support for CGA, the Conformal Geometric algebra. In addition to adding CGA to PGABLE, there is a new section in this tutorial for CGA. We have also fixed DrawBivector and DrawTrivector, and added AnimateBivector, all of which are OGA routines, as well as made some other minor corrections.
1.2 Getting started To get the most out of reading this tutorial, you should read it while running Matlab, and try the sample code and exercises. This tutorial is not a tutorial on Matlab, and you should read an introduction to Matlab before using our PGABLE package. Yet you should be able to run most of the demos described in this tutorial with no supporting documentation.
You can download the software from the following web page: The instructions there will tell you how to set up the software. You will have to install Matlab separately.
Assuming you install PGABLE in a directory called PGABLE, then to use it, add to your Matlab path with addpath(genpath(’.../PGABLE’)); where ... is the directory path to the PGABLE directory. You may also wish to put this command in your initialization file. See the Matlab documentation for details on both subjects. You can also add PGABLE to your Matlab path using the Setpath button in the Matlab GUI, although be sure to use the “add with subfolders” option. The location of the Setpath button depends on which version of Matlab you’re running.
You can get a quick introduction to the basics of geometric algebra by running the PGABLE command GAdemo. This demonstration routine will show you vectors, bivectors, and trivectors, as well as introduce the three products of geometric algebra. However, GAdemo is not a substitute for reading this tutorial, as the interpretations and description of how to use the geometric algebra is too involved for the demo script.
Thus, after running GAdemo you will need to read the remainder of this tutorial We intend for the reader to read this tutorial while running Matlab. We have put Matlab commands for you to type, which will result on objects being displayed on the screen. To get a better feel for the geometry, you should use the mouse to adjust the view. In Figure 1.2, we have labeled the controls to select for translation and rotation.
If you wish to record your session (i.e., commands and text output), use the Matlab diary command. If you wish to save a figure in the figure window, you will either need to take a screenshot or use the Matlab print command, which can print to a file in various formats, such as PDF, JPEG, or PNG. Type help diary for more details.
2 R T Figure 1: Controls for Rotation (R) and Translation (T), labeled in red. Note that you need to move your mouse into the white drawing area for the controls to appear.
1.3 Notation In this tutorial, we will use standard, italic math fonts for our equations. When giving Matlab code, or specifying Matlab variables in our text, we will use typewriter font. We will elaborate on some further parts of our notation as we introduce it. Further, in our Matlab code samples, unless otherwise specified, we will assume that you clear the graphics window (using clf) before running each code fragment. If we have a running example (i.e., where the sample code is separated with explanatory text), the later code fragments will begin with >> %...
to indicate that this is a continuation of the previous code fragment (you should not type the ‘...’ in Matlab!).
We may denote the variables you need to continue from the previous segment. E.g., >> %... needs X means that the example is continued from the previous fragment, but that you only need the variable X from that fragment. Occasionally, we will put comments in our code fragments to indicate what the code is doing; such comments look like >> % === words You do not need to type such comments into Matlab, and in general you do not need to type anything following a percent sign on a line. Sometimes an illustration involves a lot of typing. To save you typing, we have put this code in a routine called GAblock. Any code sequence that appears in a GAblock will be prefaced by >> %GAblock(N) where N denotes the appropriate section within GAblock. To run this sequence, type everything after the ’%’ sign on that line. The running of such a code sequence will stop on any line with a ‘%%’. At such times, we want you to see something, and give you a special prompt: GAblock >> At this prompt, you may type Matlab/PGABLE commands; when done, just press return and the block of code will resume running. For a continued code fragment (i.e., one that starts with %...’), you will be prompted to read the tutorial before continuing. To quit a GAblock sequence early, type ^C. While you can access the variables defined in a block at the GAblock >> prompt, once a GAblock ends (and the regular >> prompt returns), the variables defined in the block are forgotten by Matlab.
3 Table 1: Table summarizing OGA Matlab commands.
Command Arguments Result e1,e2,e3 Basis vectors for generating the geometric algebra e12,e23,e31 Bivectors for the algebra e123 The trivector for the algebra +,-,,/,ˆ,. multivectors The operations of our geometric algebra meet, join join and meet dual (multivector) Compute the dual of a multivector gexp (multivector) The geometric product exponential, emv glog (multivector) Returns the geometric logarithm norm (multivector) Returns the 2-norm or Euclidean norm of the multivector normalize (multivector) Returns the normalized multivector grade (multivector,n) Return the portion of the multivector that is of grade n isgrade (blade,n) Return true (1) if grade of the blade is n (0 if a multivector) draw (blade) Draw the given blade Note that we insert the GAblock prompts for a reason: either you should be understanding something in the picture, or you need to understand a result on the screen. At these prompts, you can and should type Matlab/PGABLE code to test things, to rotate the view on the screen, etc., until you understand what is being illustrated.
Finally, at times you may need to convert floating point numbers from Matlab to the appropriate ge-ometry package and back again. For example, rather than type glog(2), you will need type glog(OGA(2)) or glog(PGA(2)) depending on which geometry you’re using. If the floating point number is the result of GA operation, then you won’t need to do the conversion.
E.g., glog(inner(e1,e1)) will work just fine. Likewise, if you want to call standard Matlab functions on scalar values computed by GA routines, you will need to convert back to double; e.g., rather than acos(inner(e1,e1)), you will need to type acos(double(inner(e1,e1))).
2 Ordinary Geometric Algebra (OGA) This section introduces the basics of the geometric algebra, and gives the Matlab commands for performing the operations. Most of these operations work in both OGA and PGA, although the discussion in this section is focused on OGA.
Many of the objects have a graphical representation, and can be displayed on the screen using the draw command. Table 1 summarizes the OGA Matlab commands described in this section and in the rest of this tutorial. We will introduce these routines gradually, as we need them; the table is here just for reference.
Some additional commands and details on some of the commands in this table can be found in the appendices.
You can also get a summary of commands using the Matlab ‘help(OGA)’ command.
2.1 Scalar product The defining elements of geometric algebra are the vectors of a linear space of vectors over scalars. In our 3D package, we use an orthonormal basis to represent this linear space, with vectors e1, e2, and e3, and we will always use integer or real numbers as our scalars.1 You can scale and add these vectors in the usual manner. For example, to create the vector a = 3e1 + 2e2, you would type >> a = 3e1+2e3 1Frames are a necessary crutch for input and output; most of our computations will be independent of the frame of repre-sentation, in the sense that our equations can be expressed in a coordinate-free manner. For a further discussion of frames, see Appendix B.
4 3 2 0 1 0.5 1 0.5 1 1.5 0 0 2 Figure 2: 3e1+2e3, e2 a = 3e1 + 2e3 Do not forget the ‘’ for the product with a scalar, or Matlab will interpret ‘3e1’ as 3 × 101 = 30.
The norm of a vector is its length (in the standard metric in Euclidean space): >> %... needs a >> norm(a) ans = 3.6056 You can draw a vector (and may of the geometric objects) using the command draw. To draw the above vector, type >> draw(a) In the graphics window, you will see a line with an arrow head. If we draw a second vector, >> draw(e2) we see the space get rescaled so that both vectors appear in the same plot. Your screen should look something like Figure 2. Note that both vectors start at the origin and have their arrow head at the end of the line segment away from the origin. Rotate the scene if you would like to understand the spatial relationship better.
2.2 The outer product Definition Geometric algebra has an outer product, often called a wedge product. The outer product has the properties of anti-symmetry, linearity and associativity. For vectors u, v, w we have: • v ∧w = −w ∧v so that v ∧v = 0.
• u ∧(v + w) = u ∧v + u ∧w • u ∧(v ∧w) = (u ∧v) ∧w 5 The outer product of a vector v with a scalar α, or of two scalars α. and β, we define to be equal to the scalar product α ∧β = αβ and α ∧v = αv, and then use associativity to extend that to the evaluation of more involved terms.
(The properties of the outer product of two vectors are similar to the properties of the cross product for two vectors in 3D. Yet it results in a different geometrical object, as we will soon see. We will discuss the correspondence between the two in Section 2.4.3).
2.2.1 Bivectors In Matlab, the circumflex symbol or hat (ˆ) is used to take the outer product of objects. E.g., the outer product of e1 and e2 is formed by e1ˆe2, which will be displayed as e12 (i.e., e12=e1ˆe2); e23 and e31 should be interpreted in a similar manner. Here are some to try, and you may want to verify the results on paper using the definition: >> 2^e2 >> e1^e2 >> e2^e1 >> e1^(e1+e2) >> (e1+e2)^(e1-e2) The outcome of the wedge product of two vectors thus contains terms like α(e1 ∧e2), etc., which can not be simplified further to vectors or scalars. This outcome is therefore a new kind of object in our geometric algebra, called a bivector.
As you try more combinations, you find that any bivector can be expressed as a scalar-weighted linear combination of the standard bivectors e12, e23, e13 (which are e1 ∧e2, e2 ∧e3, e3 ∧e1, when written in terms of vectors). These standard bivectors are formed by outer product between the vectors in the vector basis.
This follows easily from the linearity and anti-symmetry properties in the definition of the outer product.
These bivectors thus form a bivector basis (but as in the case of a basis for vectors, other bases may serve just as well). Algebraically, the set of bivectors in 3-dimensional space is therefore in itself a 3-dimensional linear space.
You can view a bivector as a directed area element, directed both in the sense of specifying a plane and an orientation in that plane. In general, with ϕ denoting the angle from u to v and with i the unit directed area of the (u, v)-plane, we can write: u ∧v = |u||v| sin(ϕ) i.
You recognize that |u||v| sin(ϕ) is the directed area spanned by u and v; and as u and v get more parallel, this quantity becomes smaller. As you make the angle negative, the bivector becomes negative (in agreement with the anti-symmetry of the outer product since now u and v have switched roles); this is what we mean by a directed area element. The i indicates the plane in which this takes place; it is therefore a geometric ‘unit direction of dimension 2’ of what is measured.
Graphically we represent the bivector in our package as a directed circle in the bivector plane, with arrows along its border to indicate the orientation. The area of the circle is the magnitude of the bivector.
For example, let us draw some vectors, and then draw a bivector: >> clf; draw(2e1+e3); draw(e2); >> draw(e1^e2) You should see in the graphics window something like Figure 3. Rotate the scene to appreciate the spatial relationships better: note that the circle lies in the plane containing both e1 and e2. Note that if you type an expression without assigning it to a variable, Matlab will assign the object to the built-in variable ans, so the next time you enter a command that changes ans, the previous object will be lost.
6 2 0 1 0.5 1 1 0 0 -1 -1 Figure 3: 2e1+e3, e2, e1^e2.
Warning: In one case, our implementation of the outer product gives the wrong answer, namely if you perform the outer product of two scalars: 2^3 gives 8 (wrong!) rather than 6 (correct!), since we were not able to overload the exponentiation operator in Matlab for scalars. So when you want to multiply scalars, you will have to use , as in 23.
2.2.2 Trivectors Taking the outer product of three vectors yields yet another object, which is naturally called a trivector. It is a directed volume element. In 3-dimensional space, all such elements must be multiples of the unit directed volume element, which we denote by I3. (In other words, algebraically the trivectors of a 3-dimensional vector space form a 1-dimensional linear space with basis I3.) In an orthonormal basis e1, e2, e3 for our Euclidean 3-space, we equate it with the volume spanned by the ‘right-handed’ frame: I3 ≡e1∧e2∧e3 which we also write as e123. The unit directed volume I3 is often called the (unit) pseudoscalar of 3-dimensional Euclidean space.
Verify the outcome of the following expressions by hand, to get some dexterity in manipulations with the wedge product on the basis of its definition: >> e1^(e2^e3) >> (e1^e2)^e3 >> e1^e2^e3 >> e3^e2^e1 >> e1^(e1+2e2)^e3 >> e1^(e1+2e2)^e1 >> norm(e1^e2^e3) Notice that the trivector e3 ∧e2 ∧e1 equals −I3: these vectors in this order form a ‘left-handed’ frame. The terms denoting ‘handedness’ are therefore not explicit conventions anymore, they have become part of the computational framework of geometric algebra as the signs of trivectors. The norm of a trivector is absolute value of the volume; if you need a consistently signed scalar denoting the volume of a trivector T, use T/I3.
Conceptually, a trivector represents an oriented volume. Graphically, we represent it by a sphere, which we render as a line drawing. The magnitude of the trivector is represented by the volume of the sphere. To indicate the orientation, we draw line segments emanating from the points on the sphere; the orientation is indicated by whether these line segments go into or out of the sphere.
Try >> draw(I3,'g') >> draw(-0.5I3,'r') 7 Note that it is unfortunately difficult to visualize more than one pseudoscalar using this representation as it makes the drawing too cluttered. Also note that although we represent pseudoscalars with a line drawing of the surface of the sphere, the pseudoscalar is actually a solid volume element.
2.2.3 Quadvectors?
If you try taking some outer products of four or more vectors, you will find that these are all zero. You may understand why this should be: since only three vectors in 3-space can be independent, any fourth must be writable as a weighted sum of the other three; and then the anti-symmetry of the outer product kills any term in the expansion. For instance: e1 ∧e2 ∧e3 ∧(e1 + e2) = e1 ∧e2 ∧e3 ∧e1 + e1 ∧e2 ∧e3 ∧e2 = (e1 ∧e1) ∧e2 ∧e3 −e1 ∧(e2 ∧e2) ∧e3 = 0.
So the highest order object that can exist in a 3-dimensional space is a trivector. But you can also see that this is not a limitation of geometric algebra in general: if the space had more dimensions, the outer product would create the appropriate hyper-volumes.
If all we are interested in is planar geometry, then all vectors can be written as the linear combination of two basis vectors, such as e1 and e2; in that case, the highest order object would be a bivector. We would then call I2 = e1 ∧e2 a pseudoscalar of that 2-dimensional space. In n-dimensional space, the pseudoscalar is the highest dimensional object in the space. It received this rather strange name because it is ‘dual’ to a scalar, as we will see in Section 2.4.3.
2.2.4 0-vectors In the same vein of the interpretation of k-vectors as geometrical k-dimensional subspaces based in the origin, we can reinterpret the scalars geometrically. Since these are 0-vectors, they should represent a 0-dimensional subspace at the origin, i.e., geometrically, a scalar is a weighted point at the origin. This is a fully admissible geometric object, and therefore it should not be surprising that it is a member of the basis {1, e1, e2, e3, , e1 ∧e2, e2 ∧e3, e3 ∧e1, e1 ∧e2 ∧e3} of the geometric algebra of 3-dimensional space.
2.2.5 Use: parallelness and spanning subspaces The outer product of two vectors u and v forms a bivector.
When you keep u constant but make v increasingly more parallel to u (by turning it in the (u, v)-plane), you find that the bivector remains in the same plane, but becomes smaller, for the area spanned by the vectors decreases. When the vectors are parallel, the bivector is zero; when they move beyond parallel (v turning to the ‘other side’ of u) the bivector reappears with opposite magnitude. Try this: >> e1^e2 >> e1^(e1+e2)/norm(e1+e2) >> e1^(10e1+e2)/norm(10e1+e2) >> e1^(1000e1+e2)/norm(1000e1+e2) >> e1^e1 >> e1^(1000e1-e2)/norm(1000e1-e2) >> e1^(10e1-e2)/norm(10e1-e2) A bivector may therefore be used as a measure of parallelness: in Euclidean space u ∧v equals zero if and only if u and v are parallel, i.e., lie on the same 1-dimensional subspace. Note that this even holds for 1-dimensional space.
Similarly, a trivector is zero if and only if the three vectors that compose it lie in the same plane (2-dimensional subspace). We then do not call them ‘parallel’; the customary expression is ‘linearly dependent’, but the geometric intuition is the same. If the vectors are ‘almost’ in the same plane, they span a ‘small’ trivector (relative to their norms times the unit pseudoscalar).
8 In fact, we can use a bivector B to represent a plane through the origin: vector x in plane of B ⇐ ⇒x ∧B = 0.
And this B even represents a directed plane, for we can say that a point y is at the ‘positive side’ of the plane if y ∧B is a positive volume, i.e., a positive multiple of the unit pseudoscalar I3. We will come back to this powerful way of representing planes later (Section 3.7); but for now you understand why we like to think of a bivector as a directed plane element.
2.2.6 Blades and grades We now have all the basic elements for our geometric algebra of 3-space: scalars, vectors, bivectors, and trivectors (pseudoscalars).
We have constructed each of these from vectors and scalars using the outer product. There are some useful terms to describe this construction.
A blade is an object that can be written as the outer product of vectors. For instance, e1, or e1∧(e1+2e2), or I3 ≡e1 ∧e2 ∧e3.
The grade of a blade is the dimension of the subspace that the blade represents. So for instance grade(e1∧ (e1 +e2 +e3))=2 and grade(I3)=3. As you see, the outer product of an object with a vector raises the grade of the object by one (or gives 0).
We can make general objects in our algebra by taking scalar-weighted sums of blades such as 1+e1+e2∧e3.
Such objects are called multivectors. In this construction, a blade is called an m-vector, with m the grade of the blade. In that sense, a scalar is a 0-vector. Often such a multivector is of mixed grade (do not worry about its geometrical interpretation yet).
In PGABLE, we have a routine grade that when given an object, returns the grade of that object. If the object is of mixed grade, grade returns −1. When invoked with a geometric object and an integer, grade returns the portion of the geometric object of that grade. For example, >> grade(e1+I3,1) ans = e1 >> grade(e1+e1^e2^e3,3) ans = e1^e2^e3 To test if an object is of a particular grade, use the isgrade command: >> isgrade(e1^e2,1) ans = logical 0 >> isgrade(e1^e2,2) ans = logical 1 In this context of blades and grades, there is a peculiarity of 3-dimensional space that does not carry over to higher dimensions: in 3-space (or 2-space, or 1-space), any multivector that is not of mixed grade can be factored into a blade. For example, we can rewrite e1 ∧e2 + e2 ∧e3 as (e1 −e3) ∧e2. The former is the sum of two bivectors (and thus not in blade form), while the latter is the outer product of two vectors (and is therefore obviously a blade). In 4-dimensional space, this fails: e1 ∧e2 + ee ∧e4 cannot be rewritten as the outer product of two vectors.
2.2.7 Other ways of visualizing the outer product The interpretation of the bivector as a directed circle, which we have used so far, is not what everyone uses to visualize them. The standard interpretation works directly with the outer product. If you have e1∧(e1+e2), then for the standard interpretation, we construct the parallelogram having e1 and e1 +e2 as two of its sides.
Graphically, we would draw the vector (e1 +e2) starting from the head of e1. The area of this parallelogram is the area of the bivector, and the two vectors give the orientation.
You can see this interpretation of the bivector in Matlab by typing 9 >> a = DrawBivector(e1,e1+e2) The vector e1 is drawn in blue, the vector e1+e2 is offset by e1 and drawn in green, and the area is shaded in yellow. However, note that the particular vectors used to construct the bivector are not unique. Any two vectors in this plane that form a parallelogram of the same directed area give the same bivector. For example, >> %...
b = DrawBivector(e1,e2) will draw a second parallelogram (a square) that overlaps the first parallelogram. Although these are two different parallelograms, they are coplanar and have the same directed area, therefore they represent the same bivector. (If you are having trouble seeing e1^e2, you may wish to run clf and rerun the second DrawBivector command.) Using Matlab, you can test that these represent the same bivector by using the equality operator: >> e1^(e1+e2) == e1^e2 ans = logical 1 In Matlab, a result of ‘1’ for a Boolean test means “true” and a result of ‘0’ means “false”. In our example, this means the geometric objects are the same, which we can also prove algebraically: e2 ∧(e1 + e2) = e1 ∧e1 + e1 ∧e2 = e1 ∧e2.
Note that the area is oriented. In particular, if we reverse the order of the vectors in the outer product, we get a different result: >> e1^e2 == e2^e1 ans = logical 0 However, they only differ by a sign: >> e1^e2 == -1e2^e1 ans = logical 1 This is a result of the bivector representing a directed area: if we reverse the order of the arguments, then we get the opposite direction.
More generally, we can think of the bivector as representing any directed area within some simple, closed, directed curve. (To prove this we would need to develop a calculus; we will not do so in this introduction, so please just accept this statement.) We used a circle for our representation since we often will not have the creating vectors for a bivector. Indeed, depending on how we constructed the bivector such specific vectors may not exist, for example when we take the dual (Section 2.4.3) of a vector. While we may use any closed curve of the appropriate area, the circle is the closed curve with perfect symmetry. In our rendering of the bivector, the area of the circle indicates the area of the bivector; to indicate the orientation of the area, we draw arrows along its rim.
Yet another way to visualize a bivector is the “area swept out as one vector moves along the other vector.” Type AnimateBivector(e1,e2) to see an animation of this visualization, where e2 is swept along e1.
In a similar manner, we can view a trivector as a parallelepiped. Clear the screen, then run >> T = DrawTrivector(e1,e1+e2,e3) and rotate the scene to see the wireframe of the parallelepiped constructed for a trivector. If you specify the optional argument ‘‘Solid’’, then the parallelepiped will be drawn with opaque faces: >> T = DrawTrivector(e1,e1+e2,e3,"Solid") 10 2.2.8 Summary In this section we have seen the outer product, which combines elements of geometric algebra to form higher dimensional elements. In particular, the outer product of two vectors is a directed area element that spans the space containing those two vectors. When applied to other blades of our space, we get higher dimensional directed volume elements.
If the vectors we combine with the outer product are linearly dependent, then the result of the outer product will be zero; if they are ‘almost linearly dependent’ in the sense of almost aligned, the outer product will be small. Therefore the outer product provides a quantitative and computational way to treat linear dependence.
Exercises In this section we have seen the outer product, which combines elements of geometric algebra to form higher dimensional elements. In particular, the outer product of two vectors is a directed area element that spans the space containing those two vectors. When applied to other blades of our space, we get higher dimensional directed volume elements. If the vectors we combine with the outer product are linearly dependent, then the result of the outer product will be zero; if they are ‘almost linearly dependent’ in the sense of almost aligned, the outer product will be small. Therefore the outer product provides a quantitative and computational way to treat linear dependence.
1. Draw (using draw) in different colors the bivectors e1 ∧e2, e2 ∧e3, and e3 ∧e1.
2. Redraw the bivectors of the previous exercise using DrawBivector. Based on this and the previous exercise, do you have a feeling that these bivectors are orthogonal?
3. Draw using DrawBivector the bivectors e1 ∧e2 and e1 ∧(e2 + 3e1). From this picture, do you get a feeling whether or not these two bivectors are equal? Test their equality by comparing them with the == operator.
4. Draw the bivectors e1 ∧e2 and e2 ∧e1. Are these two bivectors the same? Hint: Notice how the arrows point in opposite directions. Repeat this exercise using DrawBivector and comment on the results.
5. Draw with and without using DrawTrivector the trivector e1 ∧e2 ∧e3. Now clear the screen and draw the trivector e1 ∧e3 ∧e2 with both drawing routines. Which graphical representation gives you a better feel for orientation?
2.3 The inner product Definition In a Euclidean vector space, you may have used an inner product, often called the dot product, to quantify lengths and angles. Geometric algebra also has an inner product, and in our specific geometric algebra, R3,0,0, the inner product of two vectors is the same as the Euclidean dot product of two vectors.
On vectors, the inner product of our geometric algebra has the standard properties of symmetry and linearity: • u · v = v · u • (αu + βv) · w = α(u · w) + β(v · w), for α, β scalars • In Euclidean spaces: u · u > 0 if u is not zero, and u · u = 0 if and only if u is zero.
In geometric algebra, the inner product can be applied to any elements of the geometry. Its definition for such arbitrary elements is rather complicated (for instance, it is neither associative nor symmetric, though it is linear), and we defer its definition to Section 2.5, although we will use it before then in some examples to develop a feeling for its meaning.
11 In our mathematical formulas, we will use ‘·’ to represent the inner product. In PGABLE, it is the function inner(), which takes two arguments: >> inner(2e1+3e3, e1) ans = 2 Alternatively, you may use . to take the inner product.
The inner product between a scalar α and a vector u is defined to be αu2: α · u = u · α = αu. You can also use the inner product on bivectors and trivectors. Try some: >> 2 . e1 >> e1 . 2 >> e1.(e1^e2) >> (e1^e2).(e1^e2) >> (e1^e2).(-e1^e2^e3) Note the space between the ‘2’ and the ‘.’ in the first example; this space is needed for parsing reasons.
While you would get the same answer without the space, a different function would be called to do the computation.
We need to interpret these intriguing results geometrically, and see how we can put this extended inner product to practical use.
2.3.1 Interpretation: perpendicularity For two vectors u and v, the inner product u·v is a scalar. From linear algebra, we know how to interpret this scalar: if two vectors are of unit length, then the inner product is the length of the perpendicular projection of each vector on to the other, which is equal to the cosine of the angle between the two vectors. If either vector is not of unit length, then the cosine is scaled by the lengths of the vectors. So if the angle between vectors u and v is ϕ, then u · v = |u||v| cos(ϕ).
Thus the inner product is a measure of perpendicularity: if we keep u constant and turn v to become more and more perpendicular to u, the inner product gets smaller, becomes zero when they are precisely perpendicular, and changes sign as v moves ‘beyond’ perpendicular.
The inner product keeps this interpretation of perpendicularity when applied to bivectors, but becomes much more specific geometrically: for instance x · B is a vector in the B-plane perpendicular to x.
Let us visualize this: >> B = e1^e2; >> x = e1+e3; >> xiB = inner(x,B) >> clf; draw(x,'b'); draw(B,'g'); draw(xiB,'r') >> GAview([0 90]) Rotate this figure around to get a better feel of the resulting geometry.
Note that the result is also perpendicular to x and in the B-plane if the vector x was in the B-plane to begin with.
2The reader who knows geometric algebra can see that we are using the modified Hestenes inner product here. ???However, the choice of inner product can be changed; we will get back to that.
12 >> B = e1^e2; >> b = e1; >> biB = inner(b,B) >> clf; draw(b,'b'); draw(B,'g'); draw(biB,'r') >> GAview([0 90]) So in a sense, the operation ‘·B’, applied to a vector b in the B-plane (so that b ∧B = 0), produces the perpendicular to b in that plane, the complement to b in the B-plane. This generalizes, as we will see later (Section 4.2).
Even with a trivector, the interpretation of the inner product as producing a perpendicular result remains: >> x = e1+e3; >> clf; draw(x,'b'); draw(inner(x,I3),'r') >> GAorbiter The inner product x · I3 is now the bivector representing the plane perpendicular to x. Conversely, the inner product of a bivector with a trivector produces a vector perpendicular to the plane of the bivector: >> B = (e1+e2)^e3; >> clf; draw(B,'b'); draw(inner(B,I3),'r') >> GAorbiter You begin to see how conveniently the inner and outer product work together to produce simple expres-sions for such constructions.
In general, for blades of different grades, their inner product is a blade whose grade is the difference in grades of the two objects, lies in the subspace of the object of higher grade, and is perpendicular to the object of lower grade. The inner product is thus grade decreasing. However, if the first argument has a larger grade than the second, our inner product is zero (because of this grade-reducing property it is also known as a contraction, and you cannot contract something bigger onto something smaller).3 But what happens if we take the inner product of two blades of the same grade? We already know that the inner product of two vectors yields a scalar. Try taking the inner product of two bivectors: >> inner(e1^e2, e2^e3) >> inner(e1^e2, e1^e2) In both cases, the result is a scalar; in the first example, the result is 0, while in the second example, it is −1. Likewise, if we take the inner product of the pseudoscalar with itself the result is a scalar. In general, the inner product of two blades of the same grade results in a scalar.
2.3.2 Summary The inner product is a generalization of the dot product, and may be applied to any two elements of our space. When applied to vectors, it is the familiar dot product. More generally, the inner product is associated with perpendicularity.
2.4 The geometric product Definition We have seen how the inner and outer product of two vectors specify aspects of their perpendicularity and parallelness, but neither gives the complete relationship between the two vectors.
So it makes sense to combine the two products in a new product. This is the geometric product, and it is amazingly powerful.
3This is the commonly used ‘Hestenes inner product’; other forms of the inner product will be discussed later.
13 We denote the geometric product of objects by writing them next to each other leaving the multiplication symbol understood. For vectors u and v, we define: u v = u ∧v + u · v.
(1) This is therefore an object of mixed grade: it has a scalar part u · v and a bivector part u ∧v. This mixed grade is not a problem, as geometric algebra spans a linear space in which such scalar-weighted combinations of blades are perfectly permissible.
Changing the order of u and v gives: v u = v · u + u · v = −u ∧v + u · v so the geometric product is neither fully symmetric, nor fully anti-symmetric.
With the geometric product defined, we can use it as the basis of geometric algebra, and view the inner and outer products as secondary, derived constructions. For instance, for vectors we we can retrieve the inner and outer products as its symmetric and anti-symmetric parts, respectively: u · v = 1 2(u v + vu) (2) u ∧v = 1 2(u v −vu) (3) and these formulas can be extended to arbitrary multivectors (see Section 2.5).
Although the products algebraically have this relationship to each other (with the geometric product being the more fundamental one), yet we will show that geometrically it is convenient to think of them as three basic products, each with their own geometric annotations and usage. Inner product and outer product are indeed highly meaningful ‘macros’ for geometric algebra.
Unfortunately, in PGABLE we cannot just omit the multiplication sign, and we have chosen to use ‘’ for the geometric product. Play around with it – but check the outcomes by hand to familiarize yourself with the computations (the “arrow” will be explained later.) >> e1(e1+e2) >> (e1+e2)(e1+e3) >> (e1+e2)(e1+e2) >> e1e2 >> e1e1 The geometric product is extended by linearity and associativity to general multivectors (Section 2.5), which is how we have implemented it. A geometric product of general multivectors may produce multivector with many different grades: >> (1+e1)(e1e2 + e1^e2^e3 + inner(e1,e2^e3)) ans = e2 + e1^e2 + e2^e3 + e1^e2^e3 The geometric interpretation of the geometric product is more difficult than the geometric interpretations of vectors, bivectors, and trivectors. By Equation 1, e1 (e1+2e2) results in the scalar e1·(e1+2e2) = e1·e1 = 1 and a bivector e1∧(e1+2e2) = e1∧e2, and it is hard to understand what that means. We will soon recognize that the geometric product produces a geometric operator rather than a geometric object, and that therefore we had better visualize it through its effect on objects, rather than by itself.
2.4.1 Invertibility of the geometric product The inner product and outer product each specify the relationships between vectors incompletely, so they cannot be inverted (e.g., knowing the value of the inner product of an unknown vector x with a known vector u does not tell you what x is). However, the geometric product is invertible. This is extremely powerful in computations.
14 Still, not all multivectors have inverses in general geometric algebras.
Fortunately, in the geometric algebra of Euclidean space, subspaces (represented as blades, i.e., multivectors of a single grade) do. Thus, for each blade A ̸= 0 in Euclidean space, we can find A−1 such that A A−1 = 1.
Let us first take a linear subspace, characterized by a vector v. (Why does a vector characterize a linear subspace? Because x ∧v = 0 characterizes all vectors x in this subspace.) What is its inverse? Think about this, then ask PGABLE: >> v = 2e1 >> iv = inverse(v) >> viv >> ivv So, as you thought, the inverse of a vector is parallel to the vector, but differs by a scalar factor: v−1 = v v · v.
This is easily verified: v(v/(v · v)) = (vv)/(v · v) = (v · v + v ∧v)/(v · v) = (v · v + 0)/(v · v) = 1.
Now consider a two-dimensional subspace, characterized by a bivector B.
For instance, what is the inverse of 2e1 ∧e2, where the basis {e1, e2, e3} is orthonormal? In such a basis we have: e1 e1 = e1 · e1 = 1, etc. and e1 e2 = e1 ∧e2, etc.
(4) With this, we observe that (e1 ∧e2)(e1 ∧e2) = (e1e2)(e1e2) = e1(e2e1)e2 = −(e1e1)(e2e2) = −1, (5) so, in the sense of the geometric product: the square of a bivector is negative. Then the inverse is simple to determine: (2e1 ∧e2)−1 = −1 2(e1 ∧e2) = 1 2e2 ∧e1. In general, in R3,0,0 B−1 = B B · B.
The inverse of a pseudoscalar αI3 is also easy. Observe that I3I3 = (e1 ∧e2 ∧e3)(e1 ∧e2 ∧e3) = e1 e2 e3 e1 e2 e3 = −e2 e1 e3 e1 e2 e3 = e2 e3 e1 e1 e2 e3 = e2 e3 e2 e3 = −e3 e2 e2 e3 = −e3 e3 = −1 Thus, in our algebra the inverse of αI3 is: (αI3)−1 = I3/α.
In PGABLE, use inverse() to compute the inverse of a geometric object. As a short-hand for Binverse(A), you may write B/A. Note that the geometric product is not in general commutative, so we would write inverse(A)B as (1/A)B, which is rarely equal to B/A.
2.4.2 Duality The dual of an element A of our geometric algebra is defined to be dualA ≡A/I3 = −AI3, (6) and in OGA the command dual returns the dual of an object. The dual of a blade representing a subspace is a blade representing the orthogonal complement of that subspace, i.e., the space of all vectors perpendicular to it. This is a common construction, and it is great to have it in such a simple form: just divide by I3.
For example, type the following: 15 >> clf >> draw(e1) >> draw(dual(e1)) The blue arrow represents e1; the green circle represents the dual of e1, which is −e2 ∧e3 as shown by the following derivation: dual(e1) = e1I−1 3 = e1(e1 ∧e2 ∧e3) = e1(e1e2e3)−1 = −e1e1e2e3 = −e2e3 = −e2 ∧e3.
The construction is more striking for more arbitrary vectors, of course (see exercises).
Note that for a blade U we have grade(dual(U)) = 3−grade(U), so that the dual of a scalar is a pseudoscalar and vice versa: dual(1) = −I3 and dual(I3) = 1 (this is true in any space and partly explains the name ‘pseudoscalar’ for the n-dimensional volume element).
As a consequence of this rule on grades, the dual relationship between vectors and bivectors is only valid in 3-dimensional space. In 3-space, we may characterize a plane through the origin (which is really a bivector) dually by a vector; that vector is commonly called the normal vector of the plane. We now see that it is the dual of the bivector. Indeed, both of the following two equations characterize the same plane B: x ∧B = 0 and x · dual(B) = 0.
We recognize the latter as the ‘normal equation’ of the plane B, the inner product of a vector x with the normal vector n ≡dual(B) = B/I3.
This is an example of a duality relationship between the outer product and inner product. Since the outer product produces an element of geometric algebra, we can take its dual. One can then prove (nice exercise, try it for vectors) dual(u ∧v) = u · dual(v) and dual(u · v) = u ∧dual(v) (7) for any multivectors u and v from the geometric algebra of the space with the pseudoscalar I3 used to define the dual. This is a good moment to explain how the 3-dimensional cross product of vectors fits into geometric algebra. The cross product obeys u × v = dual(u ∧v).
You can see this with the following PGABLE commands: >> a = e1+0.5e2; b = e1+e2+e3; >> B = a^b; >> draw(a);draw(b); >> draw(B,'g'); >> draw(dual(B),'r'); >> GAorbiter Here we see that the red vector is perpendicular to both of the blue vectors. To check that the dual matches the cross-product, print the dual of B and then use the Matlab cross command to compute the cross-product of the two vectors.
So we could define the cross product using the above equation. However, we will not do so, for two reasons. Firstly, the cross product is too particular for 3-dimensional space; in no other space is the dual of a ‘span of vectors’ a vector of that space. And secondly, its only use is to characterize planes and rotations.
Geometric algebra offers a much more convenient way to characterize those, directly through bivectors. We have seen this for planes and we will see it for rotations soon. Since these bivector characterizations are valid in arbitrary dimensions, we prefer them to any specific, 3D-only construction. So: we will not need the cross product.
16 Exercises 1. Determine the subspace perpendicular to the vector e1 + 02e3.
2. Determine the subspace perpendicular to the plane spanned by the vectors e1 + 0.3e3 and e2 + 0.5e3, in one line of PGABLE.
3. Prove Equation 7.
4. We used the geometric product to define the dual. We can also make the dual using the inner product (thus directly conveying the intuition of ‘orthogonal complement’). Give such a formulation of the dual in a way that is equivalent to the geometric product formulation of the dual and show this equivalence.
2.4.3 Summary The geometric product is a third product of our geometric algebra. Unlike the inner and outer products, the geometric product is invertible, which is useful when doing algebraic manipulations. In Section 3, we will see geometric interpretations of the geometric product.
2.5 Extension of the products to general multivectors We have stated that the inner, outer, and geometric products can be generalized to arbitrary multivectors. In this section, we will first show how to extend the definition of the geometric product to general multivectors.
Once we have that, it is easy to extend the inner and outer products.
For general objects of geometric algebra, the geometric product can be defined as follows (this is not the only way, but it is the most easy to understand). In the n-dimensional vector space considered, introduce an orthogonal basis {e1, e2, ..., en}. Use the outer product to extend this to a basis for the whole geometric algebra (the one containing e1 ∧e2, etcetera). Any multivector can be written as a weighted sum of basis elements on this extended basis. The geometric product is defined to be linear and associative in its ar-guments, and distributive over +, so it is sufficient to define what the result is of combining two arbitrary elements of the extended basis. We first observe that the desired compatibility with Equation 1 combined with the orthogonality of the basis vectors leads to ei ej = −ej ei if i ̸= j (8) because on the orthogonal basis, effectively ei ej equals ei ∧ej if i ̸= j. If i = j, Equation 1 gives the scalar result ei · ei, which in our Euclidean space equals 1.4 So we have eiei = 1.
(9) This is now enough to define the geometric product of any elements in the geometric algebra. For instance (2+e1∧e2)(e1+e2∧e3) is expanded by distributivity over + to 2e1+(e1∧e2)e1+2(e2∧e3)+(e1∧e2)(e2∧e3).
The term (e1 ∧e2)e1 in this equals (e1 e2)e1, and by associativity this equals e1e2e1. We apply Equation 8 to get −e1e1e2, and then Equation 9 to get −e2. The other terms are computed in a similar way. Note that this definition is heavily based on the introduction of an orthogonal basis (which is somewhat inelegant); other definitions manage to avoid that. You may also think that all these expansions make the geometric product an expensive operation. But the above was just to show how those minimal definitions actually define the outcome mathematically; a more alternative computation scheme using matrices on the extended basis is what we used to make PGABLE (such details may be found in [12, 13]).
Now that we have defined the general geometric product, it is easy to generalize both the inner and outer products. Both products are linear in their arguments, and so are sufficiently specified when we say what they 4To get a general geometric algebra, of a space with a quadratic form (‘metric’) Q, this is set to some specified scalar Q(ei), usually taken to be +1, −1, or 0. The sign of Q(ei) is called the signature of ei.
17 do on blades. For instance, if we would want to know the outcome of (A1 +A2)·(B1 +B2) (where the index denotes the grade of the blades involved), then this can be written out as A1·B1+A1·B2+A2·B1+A2·B2.
For a blade U of grade r, and a blade V of grade s, the definitions for inner and outer products are: U · V = grade(U V, |s −r|) U ∧V = grade(U V, s + r).
Note that the inner product lowers the grade, and the outer product increases the grade.
For a vector u and an s-blade V, these formulas can be shown to produce: u ∧V = 1 2(u V + b Vu) (10) u · V = 1 2(u V −b Vu) (11) where we used b V as a shorthand for (−1)sV (it is sometimes called the grade involution). Compare this to equations (3) and (2).
Beware: it is not generally true that U V = U · V + U ∧V; that equality is only so if U is a vector. For instance, compute the geometric product of two specific bivectors: >> (e1^e2)((e1+e2)^e3) ans = e1^e3 + -e2^e3 >> inner(e1^e2,(e1+e2)^e3) ans = 0 >> (e1^e2)^((e1+e2)^e3) ans = 0 In general, the geometric product of an r-vector and an s-vector contains vectors of grade |r −s|, |r − s| + 2, ...r + s −2; r + s; the inner and outer product specify only two terms of this sequence, and are therefore only a partial representation of the geometric product (which contains all geometric relationships between its arguments). For objects other than vectors, there is much more to geometric algebra than just perpendicularity (inner product) and spanning (outer product), but in this tutorial we focus on those.
There are some useful formulas permitting the computation of the inner product of multivectors made using the geometric product or the outer product. We state them without proof, for vectors u and pure blades U, V and W (the general case then follows by linearity).
u · (V W) = (u · V)W + b V(u · W) (12) u · (V ∧W) = (u · V) ∧W + b V ∧(u · W) (13) (U ∧V) · W = U · (V · W) (14) 2.6 Geometry In this section we will show how the products of geometric algebra can be used to perform many geometrical tasks.
2.6.1 Projection, rejection Given a subspace and a vector, one operation we commonly need to perform is to find the part of the vector that lies in the subspace and the part of the vector that lies outside the subspace. These operations are called projection and rejection respectively. Both are easy to perform with geometric algebra.
Let us begin with a vector v and the desire to write it as v⊥+ v|| relative to a subspace characterized by a blade M, where v⊥is the component of v perpendicular to M, and v|| the parallel component. Therefore v⊥and v|| need to satisfy v⊥· M = 0 and v|| ∧M = 0.
18 Thus v⊥M = v⊥· M + v⊥∧M = v⊥∧M = v⊥∧M + v|| ∧M = v ∧M.
But we can divide by the blade M, on both sides, so we obtain: v⊥= (v ∧M)/M (15) A similar derivation shows that v|| = (v · M)/M.
(16) These are the general projection and rejection formulas for a vector v relative to any subspace with invertible blade M, in any number of dimensions.
It is possible to visualize what is going on geometrically. Take M to be a 2-blade, determining a plane.
Then v ∧M is a volume spanned by v with that plane. It is a ‘reshapable’ volume: any vector v that has its endpoint on the plane parallel to M, through v’s endpoint, spans the same trivector. The division by M in the formula for the projection demands the factoring of this volume into a component M, and therefore returns what is left: the unique vector perpendicular to M that spans the volume. This is a general property: Dividing a space B by a subspace A produces the orthogonal complement to A in B.
If A is not a proper subspace, other things happen that we’ll cover later.
These relationships can be seen in PGABLE. To begin, clear the graphics screen (clf) and draw a bivector. Next, draw a vector that lies outside the plane of this bivector: >> clf; B = DrawBivector(e1,e1+e2) >> v = 1.5e1+e2/3+e3; >> draw(v) To get a better feel for the 3D relationships, rotate the scene around a bit.
We can now type our formulas for the perpendicular and parallel parts of v directly into Matlab and draw the resulting perpendicular and parallel components of v relative to B: >> %... needs v,B >> vpar = inner(v,B)/B; draw(vpar); >> vperp = (v^B)/B; draw(vperp,'r'); We stated that this computation works for any subspace B. In particular, we can set B to be a vector, and the same computations for vpar and vperp work. Try this!
As we will see later (but you could try it now), you may project a bivector A onto a bivector B using the ‘same’ formula: A|| = (A · B)/B.
However, the rejection of the bivector is now not obtained by A⊥= (A ∧B)/B, (which is zero since quadvectors do not exist in 3D), but simply by A⊥= A −A|| (a formula that also works in the previous case where A is a vector). More about the relationships of subspaces in Section 4.
Exercises 1. Redo the example of projection and rejection using the same v as above, but with B = e1 + e2. You will need rotate the scene a bit to see the perpendicular relationship, although it can also be tested non-graphically using the inner product.
19 2. Let v = −3e1 −2e2 and let B = 2e2 · (e3 + 3e1). Compute the projection and rejection of v by B.
Draw v, B, and this projection and rejection. Try this exercise once using DrawBivector to draw B and a second time using draw to draw B.
3. With the same v and B as the previous exercise, study how the rejection formula works: first draw v · B using DrawTrivector; as decomposition use v, v · B and the projection P((v), B). Then draw that trivector again in a decomposition that uses the rejection.
4. Prove the formula for projection: v|| = (v · M)/M.
5. (Not easy!) Using Equations 12, 13, 14, show that ((a ∧b) · M)/M = a|| ∧b|| (still a useful exercise!). This gives two ways to compute the projection of a bivector a ∧b relative to a blade M. Which do you prefer and why?
2.6.2 Orthogonalization Geometric algebra does not require the representation of its elements in terms of a particular basis of vectors.
Therefore the specific treatment of issues like Gram-Schmidt orthogonalization are much less necessary. Yet it is sometimes convenient to have an orthogonal basis, and they are simple to construct using our products.
Suppose we have a set of three vectors u, v, w, and would like to form them into an orthogonal basis.
We arbitrarily keep u as one of the vectors of the perpendicularized frame, which will have vectors denoted by primes: u′ ≡u.
Then we form the rejection of v by u′, which is perpendicular to u′: v′ ≡(v ∧u′)/u′.
Now we take the rejection of w by u′ ∧v′, which is perpendicular to both u′ and v′: w′ ≡(w ∧u′ ∧v′)/(u′ ∧v′) and we are done. (This is the Gram-Schmidt orthogonalization procedure, rewritten in geometric algebra.) Here’s an example (no need to type it, use GAblock(1), see Section 1.3): >> %GAblock(1) >> % ORTHOGONALIZATION >> clf; >> u = e1+e2; >> v = 0.3e1 + 0.6e2 - 0.8e3; >> w = e1 -0.2e2 + 0.5e3; >> up = u; >> vp = (v^up)/up; >> wp = (w^up^vp)/(up^vp); >> draw(u); draw(v); draw(w); %% The original vectors ...
draw(up,’r’); draw(vp,’r’); draw(wp,’r’); %% ... and orthognalized >> GAorbiter You might want to draw the duals to show the perpendicularity of the resulting basis more clearly (see Exercise 1).
Note that in this construction, v′ ∧u′ = v′ u′ = v ∧u, and w′ ∧v′ ∧u′ = w′ (v′u′) = w ∧v ∧u, so that it preserves the trivector spanned by the basis, in magnitude and orientation. Check this in PGABLE: >> u^v^w == up^vp^wp ans = 1 As an exercise, you might want to give the algorithm for n-dimensional orthogonalization.
20 Figure 4: Reflecting v through m.
Exercises 1. Convince yourself that up, vp, and wp are orthogonal, using graphics routines to explore their con-struction. You might for instance draw dual(up), and in that plane study vp and wp.
2.6.3 Reflection Suppose we wish to reflect a vector v through some subspace (a vector, a plane, whatever). In a geometric algebra, this subspace is naturally described by a blade, so we will look at reflecting v through a unit blade M. If we write v as v = v⊥+ v|| where v⊥is the part of v perpendicular to M and v|| is the part parallel to M (we derived formulas for both in the Section 3.1), then r, the reflection of v through M, is given by r = v|| −v⊥.
Using our formulas for the parallel and perpendicular parts of v relative to M, we see that r = v|| −v⊥ = = (v · M) M−1 −(v ∧M) M−1 = 1 2(v M −c M v)M−1 −1 2(v M + c M v) M−1 = −c M v M−1.
This is an interestingly simple expression for such arbitrary reflections.
Let us see what the formula yields for specific choices of the blade we reflect in. In 3D, there are four possibilities for the blades.
• scalar: M = 1 The scalar, viewed as a subspace, is a point at the origin (see Section 2.2.5). And indeed, our reflection formula yields v →−v, which is clearly a point reflection in the origin.
• vector: M = u Let the blade M characterize the reflecting subspace be a unit vector u; then the reflection is relative to a line in direction u. This gives r = u v u.
An attractive formula for a basic process!
21 • bivector M = B For a plane in 3D characterized by a unit bivector B (so that B2 = −1, and therefore B−1 = −B), we obtain r = B v B.
We could also write this in terms of the dual to the plane, i.e., the normal vector n defined as n = B I−1 3 : r = B v B = n I3v n I3 = n I2 3 v n = −n v n.
(It is a good exercise to prove r = −n v n for reflection in a hyperplane dual to n in arbitrary-dimensional space.) You can see this in PGABLE with the following sequence of commands: >> v = e1+2e3 >> M = e1^e2 >> draw(v) >> draw(M,'g') >> draw(MvM,'r') • volume M = I3 Since b I3 = −I3 and I3 commutes with v, this gives v →−v. Reflection in the containing space is the identity, since v⊥= 0.
To develop the formula for the reflection of an arbitrary blade of grade k in the blade M of grade m, first see what happens for the geometric product of the reflection of k vectors: (−c M u1 M−1) (−c M u2 M−1) · · · (−c M uk M−1) = = (−1)kc M u1 M−1c M u2 M−1 · · · c M um M−1 = (1)k+kmM u1 M−1 M u2 M−1 · · · M uk M−1 = (−1)k(m+1)M u1 u2 · · · uk M−1.
Therefore: X even grade: X →M X M−1 X odd grade: X →−c M X M−1 A bit complicated; but that is how reflections are for such subspaces.
By the way, you see from this what happens if you reflect a plane through the origin characterized by a bivector B, relative to the origin: it is preserved. However, its normal would reflect, and that is undesirable.
In the past, when people could only characterize planes by normal vectors, they therefore had to make two kinds of vectors: position vectors which do reflect in the origin, and ‘axial’ vectors which don’t, and realize that a normal vector is an axial vector. Using bivectors directly to characterize planes, we have no need of such a strange distinction: the algebra and its semantics simplifies by admitting objects of higher grade!
Exercises 1. Reflect the bivector B = (e1^e2+2/3e2^e3+e3^e1/3) through the plane spanned by the bivector M = e1^e2. Note carefully what happens to the orientation of the bivector. Is that what you expected?
2. Reflect the pseudoscalar I3=e1^e2^e3 through a bivector. Note the orientation.
3. Do orientation and magnitude of the reflecting blade M matter to the reflection result?
22 2.7 Rotations 2.7.1 Rotations in a plane Consider a vector a, and a vector b that is ‘a rotated version of a’. We can try to construct b as the multiplication of a with some element R of our algebra: b = R a, and since the geometric product with a is invertible we find simply that R = b/a = b a a · a This is a bit too general to describe a rotation, since we have nowhere demanded explicitly that a and b should have the same norm. Therefore b a contains both a rotation and a dilation (‘stretch’) of the vector a. This is then a possible interpretation for the geometric product: b a is the operator that maps a−1 to b.
(Or, alternatively, b/a is the operator that transforms a into b.) For the pure rotation we wished to consider, the length of the rotated vector (and equivalently, the norm of the vector) does not change. To create an R that does not depend on the length of the vectors we used to construct it, we should construct R using vectors of the same norm in the direction of a and b; for instance, both unit vectors. In Euclidean space the inverse of a unit vector is the vector itself. So then we get: a rotation operator is the geometric product of two unit vectors (and vice versa).
Try this in PGABLE using the following sequence of commands: >> a = e1 >> b = (e1+e2)/norm(e1+e2) >> R = b/a >> clf; draw(a); draw(b); At this point, we see the vectors a and b drawn in red. By construction, they are separated by a 45 degree angle. If you now type: >> %... needs a,R >> draw(Ra,'r') we see a blue vector drawn overtop b, since this is a rotated to become b (we’ve scaled the blue vector slightly so that it can be seen overtop of b). And we finally see the meaning of the “arrow” inside the red disk representing R: the arrow represents the direction of rotation. Rotating the vector a a second time and looking from above >> %... needs a,R >> draw(RRa,'g'); >> GAview([0 90]); we see that we have rotated a by the angle between a and b (45 degrees) twice, giving rotation of a by 90 degrees (drawn in green).
You can use this rotation to rotate any vector in the a ∧b-plane; for instance, if you wish to rotate c = 2a + b, then you are looking for a vector d such that d is to c as b is to a. In formula: d/c = b/a, and therefore d = (b/a) c = R c. Let us try that: >> %... needs a,b >> c=2a+b; >> draw(c); draw(Rc,'c') 23 Writing things in their components of fixed grade, the expression for R is (taking a and b unit vectors) R = b a = b · a + b ∧a = cos ϕ −i sin ϕ.
where i is the bivector denoting the oriented plane containing a and b, oriented from a to b, and ϕ is the angle between them in that oriented plane, so from a to b. Note that if we would orient the plane from b to a, then i would change sign, and R would only be the same if ϕ changes sign as well. So the orientation of i specifies how to measure the angle.
If you have had complex numbers or Taylor series in your math courses, you may realize that we can write the above as an exponential: cos ϕ −i sin ϕ = e−iϕ This is based on i2 = −1, which is correct since i is a unit bivector in R3,0,0 (5). If you have not had those subjects, just consider this as a definition of a convenient shorthand notation in terms of ‘abstract’ exponentials. In fact, there are several equivalent ways of writing the relationship between a and b, and hence the rotation over the ‘bivector angle’ iϕ: b = e−iϕ a = aeiϕ.
The sign-change in the exponent is due to the non-commutative properties of the geometric product. So i is not really a complex number: then the order would not have mattered.
As we showed before when rotating the vector c = 2a + b, you can use this formula to rotate arbitrary vectors in the i-plane over an angle ϕ as R x = e−iϕx.
Let us try this, using the geometric exponent function exp(): >> i = e1^e2; >> R = gexp(-ipi/2); >> a = e1; >> clf; draw(a); draw(Ra,'r'); draw(aR,'g'); Note that R a indeed rotates a over π/2 in the positive orientation direction of the plane e1 ∧e2 (from e1 to e2), and a R is the opposite orientation (it rotates over −π/2). If you print R, you will notice that there are some numerical issues: R is not the pure bivector it should have been, but this affects the result R a but little.
However, if x is not in the i-plane, the formula e−iϕ does not produce a pure vector. You might try this, for instance with b=e1+e3; you will see something a bit surprising. So the above rotation formulas (R a and a R) only work for vectors a in the plane in which R is defined; i.e., we do not yet have the formula for general rotations in 3-space, which is the topic of the next section.
2.7.2 Rotations as spinors There is a better representation of those rotations that keep the i-plane invariant, and which will work on any vector, whether in the i-plane or not. We construct it by realizing that a rotation can be made by two reflections. We saw in Section 3.3 that the reflection of a vector x in a vector a can be written as: a x a−1.
Following this by a reflection in a vector b we obtain (take a and b as unit vectors): b (a x a)b−1 = (b a) x (b a) = e−iϕ/2xeiϕ/2, where i is the plane of a and b (so proportional to a∧b) and ϕ/2 the angle between a and b. This produces a rotation over ϕ, as you may find by inspecting Figure 5 or writing out the shorthand explicitly. In doing this, 24 Figure 5: A rotation represented as two reflections.
it is convenient to split x in components perpendicular and contained in i, and to use that x⊥i = i x⊥(since x⊥· i = 0) and x|| i = −i x|| (since x|| ∧i = 0): e−iϕ/2xeiϕ/2 = (cos(ϕ/2) −i sin(ϕ/2))x(cos(ϕ/2) + i sin(ϕ/2)) = (cos(ϕ/2) −i sin(ϕ/2))(x⊥+ x||)(cos(ϕ/2) + i sin(ϕ/2)) = (cos2(ϕ/2) −i2 sin2(ϕ/2))x⊥+ x||(cos2(ϕ/2) −sin2(ϕ/2) + 2 sin(ϕ/2) cos(ϕ/2)i) = x⊥+ x||(cos(ϕ) + sin(ϕ)i) = x⊥+ x||eiϕ.
So the perpendicular component is unchanged, and the parallel component rotates over ϕ. Therefore the rotation of x over ϕ in the i-plane is given by the formula x →e−iϕ/2xeiϕ/2.
This formula represents the desired rotation. Let us try the last example of the previous section (a 90 degree rotation in the e1 ∧e2-plane) again, now properly using the new formula: >> i = e1^e2; >> R = gexp(-ipi/2/2); % note the half angle!
a = e1; >> b=e1+e3; >> Ra = Ra/R; >> Rb = Rb/R; >> clf; draw(a,'b'); draw(Ra,'c'); >> draw(b,'b'); draw(a^b,'b'); draw(R(a^b)/R,'c'); The new formula easily extends to arbitrary multivectors, as follows. Suppose we want to rotate c d. We can perform this rotation by rotating c and d independently and multiplying the results. But this is simply e−iϕ/2ceiϕ/2) e−iϕ/2deiϕ/2) = e−iϕ/2c deiϕ/2) Linearity of the rotation permits us to rotate sums of geometric products using this formula; and since that is all that inner products and outer products are (see Equations 10 and 11), the formula applies to those as well. Therefore, characterizing a rotation R by R = e−iϕ/2, we can use it as X →RXR−1 25 to produce the rotated version of X, whatever X is.
%... needs a,b,R >> draw(a^b,'g'); draw(R(a^b)/R,'c'); >> draw(b^Rb,'r'); draw(R(b^Rb)/R,'m'); The object R = e−iϕ/2 is called a spinor (or sometimes a rotor). It is all we need to characterize a rotation in n dimensions.
Composition of rotations (independent of what we are going to rotate) is then just a multiplication of their spinors. For rotating X first by R1 and then by R2 gives: R2(R1XR−1 1 )R−1 2 = (R2R1)X(R2R1)−1, which is a new rotation using the spinor R = R2R1. Be careful, though: spinors do not commute, so you cannot simply say e−i2ϕ2e−i1ϕ1 = e−(i2ϕ2+i1ϕ1).
Exercises 1. On paper and on the computer, make an operator that implements a rotation of 90 degrees around the e1 axis, followed by 90 degrees around the e2 axis. After you have constructed it, write it in exponential form again (on paper only) – this should give you the effective total rotation, in terms of a rotation plane and an angle. (Use GAview([135 -45]) to bring the view more or less along the effective rotation axis.) 2. Repeat the same exercise using rotation matrices; especially the final step of retrieving axis and angle is quite a bit more work!
3. Draw, rotate and draw a bivector, and make sure your intuition keeps up with these results!
4. What happens when you rotate a trivector? Can you prove that?
2.7.3 Rotations around an axis In the previous section, we took a rotation in the plane and extended it to a rotation in 3-space. However, we can also characterize rotations in 3-dimensions by an axis and an angle. You may wonder how this 3-dimensional rotation is related to the above rotation.
The answer: it is simply related by duality. We can rewrite the bivector iϕ/2 (the rotation plane and angle) as iϕ/2 = (iϕ/2)I−1 3 I3 = (dual(i)ϕ/2)I3 = (uϕ/2)I3 = I3uϕ/2, with u ≡dual(i) the unit vector dual to i. The vector u is the rotation axis. Thus a rotation in 3D may be characterized by a vector of length ϕ/2 along the axis u: R = e−iϕ/2 = e−I3uϕ/2.
A very handy formula, all you need is uϕ to get a computable representation of the rotation! This is natural, much more so than rotation matrices in which the axis is hidden as the eigenvector of eigenvalue 1, and the angle in the complex eigenvalues or the trace. (Beware: in higher dimensions, this rewriting does not work, and you will have to use the bivectors. For instance in 4-dimensional space the dual of the bivector characterizing the rotation plane is a bivector; rotations in 4D do not have an axis).
Obviously, all this is similar to the cross product trick, which is also valid in 3D only. We have seen that a × b = dual(a ∧b), so that u = dual(i) can be written as the cross product of two vectors in the i-plane.
However, the characterization of rotation by a bivector works in arbitrary dimensions, and is therefore a good habit even in 3D.
To move back and forth between the spinor in exponential form, and the argument of the exponential, it is convenient to have an inverse of the exponential, i.e., a logarithm. This inverse is not unique, since if 26 R = exp(iϕ/2), then also exp(iϕ/2+2πi) = R; so the logarithm of a spinor has multiple values (those of you familiar with complex analysis may recognize the phenomenon). In the implementation for the logarithm of a spinor we choose the value of the bivector between −πi and πi;this is the function glog(). You see glog used in the exercise below.5 Exercises 1. Verify that the following code is an implementation of the first exercise of the previous section. Note especially how we retrieve the direction of the effective rotation axis.
%GAblock(2) >> % ROTATION EXERCISE >> clf >> R1 = gexp(-I3e1pi/2/2); >> R2 = gexp(-I3e2pi/2/2); >> R = R2R1; >> a = e1+e2; >> Ra = Ra/R; >> RRa = RRa/R; >> draw(a); draw(Ra,'m'); draw(RRa,'r'); %% Draw the objects >> axisR = normalize( -GAZ(glog(R))/I3 ); >> draw(axisR,'g'); %% Draw the axis of rotation You may be able to visualize things a little better using the bivector of the rotation plane: >> %... needs axisR >> draw(dual(axisR),'g'); 2.8 Orientations in 3-space When a vector, bivector, or other geometric element is rotated relative to another element, we can use that to encode orientation. This representation is itself an element of the geometric algebra. Here is how that works, in a Euclidean space.
2.8.1 Interpolation of orientations Interpolation between orientations in 3-space is a notorious problem; often required, but not easy to do tidily in many a classical representation. In geometric algebra, the problem is fairly easily solved – about as easily as it is stated. First, we characterize orientations in terms of rotation operators, in a coordinate-free manner. The idea is straightforward: when you consider an object to have a certain orientation, what you mean is that relative to its standard orientation (whatever that may be), it has undergone some rotation R.
So its orientation may be characterized by this rotation R. If we have two orientations, then these may be characterized by two rotations RA and RB. A smooth interpolation between these two orientations may then be achieved by n intermediate identical rotation operators R, to be applied to RA, so that the subsequent rotations are R0 = RA; Ri+1 = RRi; Rn = RB.
The intermediate orientations of whatever object X we wanted oriented are then RiXR−1 i , for i = 1, ...n.
With this specification of the manner of rotation (n identical operators) the problem can be solved in a fairly straightforward manner.
5If you use glog on a non-unit spinor S, then there is an additive scalar term equal to log(jSj), in agreement with the usual behavior of the logarithm on products.
27 The total rotation to be made by the n operators R is the rotation from RA to RB, which is RBR−1 A .
This should equal Rn, so we have Rn = RBR−1 A = e−I3aϕ/2, where a and ϕ follow from RB and RA (see below for an example). The solution is then R = e−I3aϕ/21/n = e−I3aϕ/(2n).
Taking the n-th root of a spinor can be done by taking its logarithm using glog (giving the argument of the exponential), dividing by n, and then exponentiating.
The following example shows the computations to implement this. Let RA be the orientation achieved from some standard orientation by a rotation over Ie1π/2, and let RB be the rotation Ie2π/2. Let us say that the aim is to get from RA to RB in eight intermediate steps. We compute: >> %GAblock(3) >> % INTERPOLATION OF ORIENTATIONS >> clf; >> RA = gexp(-I3e1pi/2/2); >> RB = gexp(-I3e2pi/2/2); >> Rtot = RB/RA ; >> n=8; % we rotate in 8 steps >> R = gexp(glog(Rtot)/n); %% To demonstrate the correctness, we need an object that shows the orientations. So we choose to draw some bivector u^v in orientation RA and RB, and show the intermediate orientations. Caution: unfortunately, for loops don’t work quite right from the console, so you can only run this as a GAblock and can’t type it in on the console. Also note the use of an array in the loop. PGABLE’s support of arrays in minimal; they can’t be passed to functions, for example.
%... needs n, R, RA, RB >> u = e1+e2-e3; >> v = e1+e3; >> view = [-0.6 2.5 -1 1.16 -2 1.1]; % select the view >> % === initial orientation: >> DrawBivector(RA/RA, RAv/RA,'b'); axis(view); GAview([30 30]); %% >> % === final orientation: >> DrawBivector(RBu/RB, RBv/RB), 'g'); axis(view); %% >> axisR = unit(GAZ(-glog(R)/I3)); % reorientation axis: >> % === display of the 7 intermediate orientations >> Ri = RA; >> for i=1:n-1 >> Ri = RRi; >> ui = Riu/Ri; >> vi = Riv/Ri; >> DrawBivector(ui,vi); >> end >> GAorbiter(190); This gives a result similar to Figure 6. Make sure you understand the reason behind these steps. If you have learned about quaternions before, you may realize that the above geometric algebra approach to interpolating orientation is equivalent to unit quaternions (see also next section). For most people, those are an isolated technique; you may now appreciate how they are a natural part of the algebraic structure of the 3-dimensional space.
28 Figure 6: Interpolation of orientations for a bivector.
Exercises 1. Verify the axis and angle in the example above by hand.
2. What happens in the example if one of the rotations, say RB, was over 180 degrees, so ϕ = π. Is the solution still well-behaved?
3. What happens if RA and RB occur in the same plane?
2.9 Complex numbers and quaternions: subsumed So rotations are characterized by the plane in which they occur; those planes are characterized by bivectors; and combinations of rotations are represented by geometric products of these bivectors. If you have heard of complex numbers and quaternions, you know that those may also be used to do rotations, in the plane and in space, respectively. How does this compare to the geometric algebra method? Surprisingly, the answer is: it is exactly the same (but in a richer context)!
Algebraically, we may observe that the linear space C↕3,0 has several interesting subspaces with corre-sponding sub-algebras. In particular, consider the subspace spanned by {1, e1 ∧e2} on an orthonormal basis e1, e2, e3. The product of e1 ∧e2 with itself is (e1 ∧e2) (e1 ∧e2) = e1 e2 e1 e2 = −e1 e1 e2 e2 = −1.
Thus in this subspace, the bivector e1 ∧e2 has similar properties to the complex number i, and we see that indeed we can use this linear subspace of our geometric algebra spanned by 1 and e1 ∧e2 as the complex numbers, with their product represented by the geometric product (at least if we are careful about the order, due to difference in the commutative properties of the two algebras). Within geometric algebra, these can be seen as operators, which can act on geometrical subspaces represented by vectors, bivectors, trivectors or scalars. Complex numbers can only multiply each other, not ‘operate’ on other objects such as vectors.
Now consider the subalgebra spanned by {1, e1 ∧e2, e2 ∧e3, e1 ∧e3}. The analysis of the previous paragraph shows that (e1 ∧e2)2(e2 ∧e3)2(e1 ∧e3)2 = −1.
29 Figure 7: A line described by offset p and tangent v, with moment V.
Further, we have (e1 ∧e2)(e2 ∧e3) = e1 ∧e3 (e2 ∧e3)(e1 ∧e3) = e1 ∧e2 (e1 ∧e3)(e1 ∧e2) = e2 ∧e3 and changing the order in which we multiply these bivectors changes the sign of the result. In other words, with the identification i2 = e1 ∧e2, j = e2 ∧e3, and k = e1 ∧e3 we get: i2 = j2 = k2 = −1, ij = k, ji = i, ki = j.
You may recognize this: they are the defining equations of quaternions, usually introduced as a special ‘number system’, somehow mysteriously handy for computations with rotations.
Now we see that the basic ‘complex numbers’ occurring in quaternions are actually a basis of bivectors in the geometric algebra C↕3,0. Those seemingly imaginary numbers are actually real spatial planes in disguise, and their product is just the geometric product. Not a big surprise, then, that quaternions are so useful in describing orientations: quaternions are actually disguised spinors representing the rotation planes, and their products therefore represent successive rotations. Within the same framework, we can now also represent the objects on which the rotations need to act, and that is a significant improvement over the quaternion representation.
2.10 Blades and subspace relationships Blades are the fundamental objects in geometric algebra. We have seen how a k-blade is formed by taking the outer product of k independent vectors. We have also shown that a blade represents a k-dimensional subspace containing the origin. In the next section, we present a trick to make blades represent subspaces off-set from the origin. That obviously makes blades very relevant objects in geometric algebra, and their geometry worthy of a separate study.
In this section, we determine the operations required to assess the relative positions of blades: how they project onto each other, how they can be decomposed relative to each other, and how they intersect. This intersection will necessitate the definition of a new kind of product in geometric algebra, the meet, and a related product, the join, as geometric union.
2.10.1 Subspaces as blades Picture, in your mind, two blades; they can be a vector and a bivector, two vectors, two bivectors, or they may even be scalars (blades of grade 0). As you visualize them in the way we have done, you actually make each of those k-blades characterize a k-dimensional subspace, through the equivalence 30 Figure 8: Subspace A and its geometric relationship relative to a subspace B, for several relative grades and positions.
vector x in subspace determined by A ⇔ x ∧A = 0.
This interpretation of blades is not specific on sign or magnitude of A, so by themselves, subspaces have no significant sign or magnitude. For now, we can think of subspaces as ‘blades modulo a multiplicative scalar’ – but we’ll keep in mind that geometric algebra would allow us to be more quantitative about subspaces than we may have been used to. By the way, note that a scalar blade A yields the equation x = 0, so that it represents a point at the origin.
We would like to encode the following relationships of subspaces into geometric algebra: • projection of subspaces onto each other, and its counterpart: rejection of subspaces by each other • intersection of subspaces, including how ‘orthogonal. this intersection is • ‘union’ of subspaces: common span, or common superspace, with some measure of their degree of parallelness • the ‘sides’ of a space split by a subspace • a distance measure between subspaces We will see that a limited set of operators can express all these concepts, and that they are composed of the products we have defined so far.
2.11 Projection, rejection, orthogonal complement Consider the examples of the subspaces sketched in Figure 8. It depicts subspaces A and B of various grades, and some subspaces derived from their geometric relationships. We will now find the corresponding formulas.
We have already mentioned before, in Section 3.1, that the projection of a blade A onto a blade B is given by the formula proj(A, B) ≡(A · B)/B.
We interpret this as the projection formula for subspaces: it gives the component of A entirely in B. This projection is the inner product divided by B, which is like the dual in the B-plane of the inner product.
The inner product A · B by itself therefore also has geometric significance: it is the orthogonal complement, within B, of the projection of A. We will denote it by comp(A, B) if we mean to imply this meaning. You may want to think of it as ‘the part of B not contributing to A’, The ‘perpendicular’ part of A to B is the rejection, it is the component of A not in B: rej(A, B) = A −proj(A, B) = (AB −A · B)/B (17) Only when A is a vector a can we write this as rej(a, B) = a −proj(a, B) = (a B −a · B)/B = (a ∧B)/B, and even then there are problems in 1-dimensional space; it is better to use the universal Equation 18.
So the relationship of two subspaces A and B split into three parts: 31 projection rejection ‘complement’ proj(A, B) rej(A, A) comp(A, B) description = (A · B)/B = A −(A · B)/B A · B scalar 0 ℓ-blade A is a scalar k-blade 0 (ℓ−k)-blade A lies in B k-blade 0 scalar A and B coincide k-blade k-blade (ℓ−k)-blade ℓ̸= k: in general position k-blade k-blade scalar non-trivial intersection 0 k-blade 0 A perpendicular to B Table 2: Relationship between two subspaces A (of grade k) and B (of grade ℓ−k).
• The projection proj(A, B) of A onto B, which equals (A · B)/B.
• The rejection rej(A, B) of A by B, which is A −proj(A, B); it projects to zero, which we may express by saying it is the part of A perpendicular to B.
• The complement comp(A, B), which is the part of B orthogonal to A and its projection; this is the subspace A · B. This is a new concept; you see how it is an atom in the construction of the projection.
(Proof of orthogonality to A is easy using Equation 14: A · (A · B) = (A ∧A) · B = 0 · B = 0.) Depending on how they relate, some of these subspaces may be zero, or identical to A. Such events define the various relative situations of A and B, see Table 2 and Figure 8.
Let us play with this in 3 dimensions: >> A = e1-e2/2+e3; % a line >> B = e1^(e2+e3/2); % a plane >> cAB = inner(A,B); >> pAB = cAB/B; >> rAB = A - pAB; >> clf; draw(A,'b'); draw(B,'g'); >> draw(cAB,'r'); draw(pAB,'c'); draw(rAB,'m'); (use GAorbiter for interpretation!). Now replace B by the line e1/2+e2+e3/3 (or a line of your choice) and run the same commands. Note that projection and rejection are as expected, and that the orthogonal complement to A in B is now a scalar.
The point at the origin is indeed the only subspace of B and ‘orthogonal’ to both A and its projection.
Now do the same on two non-identical planes, for instance A = e2^(e1+e3) and B = e1^(e2+e3/2), and interpret the results. (As you may see, the magnitudes of the bivectors are somewhat difficult to interpret as circles. We’ll soon use factored bivector instead, but then we first need to find the intersection of the planes, to span them nicely.) This subspace interpretation casts an interesting new light on the inner product of vectors. We have seen that a scalar is, geometrically, the representation of a point at the origin. The inner product u · v of two vectors u and v is the subspace of v that is the orthogonal complement of u; and geometrically, that is the point at the origin, on the line determined by v. So the inner product of two vectors is a weighted point...
If you want to stretch your intuition for generalization to new limits, use the space itself for B, so take B = e1^e2^e3 and take A to be a line, and then a plane. Verify what you see by computation! Why does the dual subspace appear? Can you make yourself see and feel that the rejection of A from 3-space is 0, which is ‘not even a point’?
The same formulas and interpretation are still applicable when A and B are coincident spaces: play around with B = 2A, choosing a line, plane or I3 for A. What is the projection now?
32 What you may learn from all this is that geometric algebra has no special cases for computing such geometric relationships. It is not always easy to find the characterization in words of what is computed since it may not be a classically recognized concept (our notion of ‘orthogonal complement of A in B’ is a bit of a stretch), but we have found it rewarding to add the outcomes of straightforward computations as a elementary concepts to our intuition.
2.12 Angles and distances A vector x has several relevant relationships with respect to a planar or linear subspace A in 3-space such as its incidence angle, the distance of its tip to the subspace, and the ‘side’ it lies on. With projection and rejection, these are easy to compute.
• perpendicular distance The perpendicular distance is obviously the length of the rejection |rej(x, A)| = |x −(x · A)/A| = |(x ∧A)/A|.
• side The quantity x ∧A is the space spanned by x and A; we can compare this to the blade we might already have for this subspace to determine whether x lies on the ‘positive’ or ‘negative’ side of A. If A is a 2-blade, then x∧A is a 3-blade (or zero, in which case x lies in A). So we can compute the side as the sign of the scalar (x ∧A)/I3. If A is one-dimensional, there is no natural orientation to give to the 2-blade x ∧A: so a line in 3-space has no sides (for a point – it does induce a sense in lines).
Note that the side is closely related to the perpendicular distance: for the plane we can write I3 = nA, with n the normal vector of A (so that n · A = 0). Then (x ∧A)I−1 3 = (x ∧A)/A/n = rej(x, A)/n.
Since the rejection and n are parallel, this is a scalar, expressing the perpendicular distance in units of n (if you take x = n, you get 1). So for a point relative to a plane, we obtain a signed distance, and that is often more useful than just the absolute value of the perpendicular distance.
• angle The angle of a vector relative to a line or plane is characterized by the ratio between its rejection and projection, which is like the tangent function from trigonometry: rej(x, A) proj(x, A) = x ∧A x · A .
The magnitudes of x and A cancel, so it is properly independent of scale. Note that we have not taken norms, so that the result is a bivector (it is the ratio of a (k + 1)-blade and a (k −1)-blade, so a 2-blade). We may write the outcome as i tan ϕ. This explicitly gives us the plane i in which the angle is to be measured, and it cancels the bothersome ambiguity of sign which the use of a tangent always involves. The above expression computes i tan ϕ and if you wished to interpret the angle in the oppositely oriented plane −i, the value of the tangent function (and therefore the angleϕ) automatically changes sign. As we have seen when treating rotations, angles are essentially bivectors! (It is even possible to extend the trigonometric functions naturally to geometric algebra; then the formula may be said to compute tan(iϕ), see ).
Let us try this out: >> % GAblock(5) >> % PROJECTION, REJECTION >> x = e1 + e2/2+e3; >> A = e2 + e3/3; % a linear subspace >> xA = geoall(x,A); >> clf; draw(x,'b'); draw(A,'g'); %% Draw A and x >> DrawPolyline({xA.rej,xA.rej+xA.proj,xA.proj},'k'); 33 >> draw(xA.rej,'m'); draw(xA.proj,'m'); %% Draw rej and proj >> distxA = norm(xA.rej) >> tanglexA = xA.rej/xA.proj >> anglexA = atan(norm(tanglexA))180/pi %% >> B = e1^A; % A planar subspace (containing A) >> xB = geoall(x,B); >> draw(B,'y'); %% >> DrawPolyline({xB.rej,xB.rej+xB.proj,xB.proj},'k'); >> draw(xB.rej,'r'); draw(xB.proj,'r'); %% rej and proj for B >> distxB = norm(xB.rej) >> tanglexB = xB.rej/xB.proj >> anglexB = atan(norm(tanglexB))180/pi >> side = sign(double((x^B)/I3)) Note the use of double to convert an OGA scalar into a double when calling the Matlab function sign.
By the way, the PGABLE code OGA(0) creates the geometric zero object.
Exercises 1. What do you get if you compute these quantities relative to a 0-dimensional subspace A = 2, or relative to the total 3-space A = I3? Can you interpret those outcomes sensibly?
2. If you add ‘1’ to the bivector we computed above, you get an operator which (under post-multiplication) rotates and scales x to be in the subspace A. Why? Check this for the subspaces in the PGABLE example above.
3. In 3-dimensional space, the relative orientation of two planes is also determined by an angle. Can you give a formula for the value of its tangent function?
2.13 Intersection and union of subspaces: meet and join In the example of the two planes: >> % GAblock(6) >> % MEET, JOIN >> A = e2^(e1+e3); >> B = e1^(e2+e3/2); >> cAB = inner(A,B); pAB = cAB/B; rAB = A - pAB; >> clf; draw(A,'b'); draw(B,'g'); %% >> draw(cAB,'r'); draw(pAB,'c'); draw(rAB,'m'); we see that the formulas so far do not determine their line of intersection. Such an intersection between subspaces is obviously something we often need; so we need to make it computational.
First, how do we determine that a subspace A has a non-trivial intersection with subspace B? In such a case, there is at least a 1-dimensional subspace c in common. Rewriting the blades in the form A = A′ ∧c and B = c ∧B′, we see that A ∧B = (A′ ∧c) ∧(c ∧B′) = A′ ∧(c ∧c) ∧B′ = A′ ∧0 ∧B′ = 0. So we find: A and B have non-trivial intersection ⇔ A ∧B = 0.
Of course the fact that A ∧B = 0 may mean that there are several independent 1-dimensional subspaces in common; A and B might even coincide completely. Let us suppose that C is a largest common subspace of A and B (in the sense of: the highest dimensionality, so the highest grade). Then we can write A = A′ ∧C and B = C ∧B′, so we can ‘factor out’ the common subspace C. (The weird order is to get better signs later on.) The common subspace C is the ‘intersection. of the two subspaces A and B. We would like to define an operation in geometric algebra that produces it; let us call that the meet operation. It is thus defined by: 34 If A and B can be factored using a highest grade sub-blade C as A = A′ ∧C and B = C ∧B′, then their meet is meet(A, B) = C.
We have indeed implemented such an operation: >> %... needs A,B >> mAB = meet(A,B) >> clf; draw(A,’b’); draw(B,’g’); draw(mAB,’y’); However, you should be somewhat careful with using its outcome. The problem is that the factorization is not unique, and so neither is the meet. Since C is a blade indicating a subspace, any multiple γC is just as valid, and that would produce a meet of γC. We could demand |C| = 1 (at least in Euclidean spaces), but that would still leave the sign undetermined. The outcome of the meet is therefore a strange ‘object’: it has no well-defined sign. Its meaning as a subspace is perfectly clear (x is in C iff x ∧C = 0), but as a blade it is ambiguous: it has no unique orientation that can be established on the basis of A and B alone.
If we consider C in the context of the superspace spanned by A and B, we can be more specific on its orientation. This common superspace is like a union of the spaces A and B. It is often called the join of A and B, and we define it through the factorization we had before: If A and B can be factored using a highest grade sub-blade C as A = A′ ∧C and B = C ∧B′, then their join is: join(A, B) = A′ ∧C ∧B′ (= A′ ∧B = A ∧B′).
This we have also implemented: >> %... needs A,B >> jAB = join(A,B) >> draw(jAB,'k'); Obviously, in this case of two distinct 2-blades in 3-space, it is proportional to I3. If A and B would be disjoint, then join(A, B) is proportional to A ∧B.
Again, the problem is that the factorization is not unique, and replacing C by γC now changes the join by a factor of 1 = 1/γ. Thus we have the problem of assigning a magnitude to the join, as we did for the meet. Although we have no mechanism in geometric algebra to fix the magnitudes uniquely (and this is a fundamental impossibility, since it is the ambiguity of the factorization that causes it), we can at least make a meet and join that are consistent, in the sense that they are based on the same factorization. This can be done based on the observation that join and meet are related by join(A, B) = A meet(A, B) ∧B.
(A derivation may be found in [MDB99b, MDB01].) It can be rewritten to the more commonly seen form meet(A, B) = B join(A, B) · A.
We have moreover scaled the meet to have unit norm. This fixes meet and join up to one shared ambiguous sign.
2.13.1 Combining subspaces: examples With projection, rejection, orthogonal complement, meet and join, we have a powerful set of terms to describe many (all?) relationships between subspaces. And they are all computationally related, we can use their properties both synthetically to construct new objects reflecting the relationships, or analytically in derivations.
Let us play around with these new tools.
It is convenient to have a function that computes all the quantities, so that you can execute it whenever you have defined new objects (using cut-and-paste, or the scroll in PGABLE, or as a .m-file), and a function to draw them all in standard colors. So we use what we had before, but make it into a function that computes the things in a single structure: 35 function r = geoall(A,B) r.obj1 = A; r.obj2 = B; r.comp = inner(A,B); r.proj = r.comp/B; r.rej = A - r.proj; r.meet = meet(A,B); r.join = join(A,B); function drawall(r) clf; draw(r.obj1,’b’); draw(r.obj2,’g’); draw(r.comp,’r’); draw(r.proj,’c’); draw(r.rej,’m’); draw(r.meet,’k’); draw(r.join,’y’); These functions have been precoded for your convenience (in the actual code, we have included some argument protection – testing for blades – to make them more robust). Note the colors used to draw objects, and realize that if objects overlap exactly, only the last one drawn is visible. It is therefore a good idea to use non-unit-norm blades for A and B, so that the various objects are more likely to have different magnitudes.
Let us first continue the example of the two intersecting planes A and B of Section 4.2. Remember how it was hard to see whether the projection was done correctly when we drew the bivectors using draw? We can use the above functions to redo that quickly: >> clf; >> A = e2^(e1+e3); >> B = e1^(e2+e3/2); >> clf; drawall(geoall(A,B)); Now that we have the meet operation, we can use the meet as a common vector in the two planes, and this permits us to use factored bivector to show the various blades more clearly in a factored form.
% GAblock(7) >> % MEET AND JOIN DECOMPOSED >> A = e2^(e1+e3); >> B = e1^(e2+e3/2); >> aAB = geoall(A,B); >> M = aAB.meet; >> clf; >> DrawBivector(A/M,M,’b’); % draw A, decomposed >> DrawBivector(M,1/MB,’g’); %% draw B, decomposed >> DrawBivector(aAB.proj/M,M,’c’); % draw proj, decomposed >> DrawBivector(aAB.rej/M,M,’m’); %% draw rej, decomposed >> draw(aAB.comp,’r’); >> draw(aAB.meet,’k’); >> draw(aAB.join,’y’); (use GAorbiter(30,2) repeatedly to interpret the result – note how this partially shows up the projection in cyan). You now see how the magnitudes of projection and rejection obviously give the original A as their sum, which was hard to appreciate in the circular way of drawing the bivectors.
What we have defined is general for blades of any dimension. For instance, let’s take a 1-blade and a 2-blade.
A = e1-e2/2+e3; >> B = e1^(e2+e3/2); >> drawall(geoall(A,B)); 36 The black caption scalar = 1 shows that the intersection is now the point at the origin.
Or take a 0-blade and a 1-blade, a rather degenerate case: >> A = 3; >> B = 2(e2+e3); >> drawall(geoall(A,B)); No matter how degenerate the situation, our relationship operators have no exceptions, and the results are always sensibly interpretable. These functions and operators are extremely stable to any blades you might use them on, and they can be used to implement all of classical geometry. One more step is needed: to show how blades can represent off-set subspaces such as lines and planes in general position. We do that in the next section.
Exercises 1. The vector p = e1 + e2 lies in the plane A = (e1 + e2) ∧e3 (as you may verify). How far does its endpoint lie from the intersection of A with the plane B = 2(e1 ∧(e2 + e3)? Draw this!
2. What is the angle between p and the intersection of A and B?
3 Plane-based Geometric Algebra (PGA) For many applications, we wish to treat points and vectors as different objects, where points have a location in space, while vectors have direction and magnitude but no fixed position. This distinction makes points and vectors transform differently under translation: vectors remain unchanged while points “move”. The standard way to achieve this difference between points and vectors is to add an extra dimension to your space, and use this extra dimension to encode the difference between points and vectors, where the extra dimension typically relates to an origin for the space.
In Plane-based Geometric Algebra (PGA), planes are the basic element of the algebra, and are represented as vectors (grade 1 objects). The outer product operator acts as the intersection of planes, and we will intersect two planes to get a line, and intersect three planes to get a point (although both lines and points can be constructed directly if desired). Thus, lines will be bivectors (grade 2 objects) and points will be trivectors (grade 3 objects).
In this trivector representation of points, we will have three trivectors corresponding to the x, y, z coordinates, and a fourth basis element that is the origin of the space. The points are weighted points in the sense that the coefficient of this extra coordinate can be different than 1. However, there are no “vectors” in PGA, at least not in the standard sense of an object having direction and magnitude but no location.
Instead, if the coefficient of the origin of a point is 0, then we have a point at infinity.
While it may seem odd to have points represented as trivectors, realize that our space is spanned by 24 = 16 basis vectors (scalars, vectors, bivectors, trivectors, and quadvectors), and we have just chosen to use the trivector basis elements to represent points rather than the vector basis elements. Computationally, for example, there is no difference in which we choose (vectors or trivectors) for our representation of points.
With this model of planes, lines, and points, we will then use the PGA planes to construct transforma-tions via reflections, similar to the way transformations were constructed earlier in this tutorial, and similar to the OGA part of the tutorial, we will also look at an exponential method for constructing transforma-tions. Regardless, using reflections, we will be able to generate all rigid body transformations. For a quick introduction to PGA, run PGAdemo. The rest of the tutorial will expand on the ideas in this demo.
Table 3 summarizes the PGA Matlab commands in PGABLE; for a complete list of PGA Matlab com-mands, type help PGA. Note that PGA geometry has three Euclidean vectors basis vectors, e1, e2, e3, where ei · ej = δij, and an additional basis, e0, where ei · e0 = 0 and e0 · e0 = 0. The vector e0 may seem strange to you (a non-zero vector whose inner product with itself is 0); such a vector is known as a null vector.
We can test these inner products in PGABLE. First, you must switch PGABLE to the PGA mode: 37 Table 3: Table summarizing PGA Matlab commands.
Command Arguments Result e0,e1,e2,e3 Basis vectors for generating the geometric algebra e12,e23,e13 Bivectors for the Euclidean subalgebra e123 The trivector for the Euclidean subalgebra e01,e02,e03 e0 bivectors e012,e023,e013 e0 trivectors e0123 The quadvector for the space gapoint (x,y,z) Compute the PGA point with coordinates x, y, z.
galine (l1,l2,l3, p1,p2,p3) Compute the PGA line in direction l=(l1,l2,l3) (l, p) passing through point p=(p1,p2,p3) +,-,,/,ˆ,. multivectors The operations of our geometric algebra inverse Compute the inverse join, meet (multivector,multivector) join and meet of two multivectors dual (multivector) Compute the dual of a multivector poincaredual, pdual (multivector) Compute the Poincar` e dual of a multivector gexp (multivector) The geometric product exponential, emv glog (multivector) The geometric logarithm grade (blade) Return the grade of a blade (−1 if a multivector) (multivector,n) Return the portion of the multivector that is of grade n isgrade (multivector, g) To determine if a multivector is of grade g euclidean (multivector) Return the Euclidean portion of a multivector noneuclidean (multivector) Return the e0 coefficient of a multivector getx, gety, getz (point) Get the x, y, z coordinates of a point draw (blade) Draw the given blade pclf Clear all objects (including vanishing objects) from the screen 38 >> GA.model(PGA) Now check the following: >> inner(e1,e1) >> inner(e1,e2) >> inner(e1,e0) >> inner(e0,e0) The last one is of particular interest, since the inner product e0 · e0 = 0. However, if you test >> e0==0 ans = logical 0 we see that e0 is not the zero vector.
If you wish to switch back to the Ordinary Geometric Algebra (OGA), type >> GA.model(OGA) We present the material as a mix of PGABLE examples, some mathematics, and exercises. The theoretical material in this section is heavily based on [DdK]. Many of the exercises discuss topics beyond what is discussed in the tutorial, and thus should not be skipped when reading the tutorial. We limit the math mostly to parts that should be understandable at the undergraduate level, leaving the more advanced derivations, etc., to the references. The reader less interested in the mathematics may wish to only skim through the longer derivations.
Note that geometric entities will be drawn differently in PGA than they were in OGA. For example, e1 is drawn as an arrow in OGA, while it will be drawn as a square (representing a plane) in PGA. Also, note that the routine to run blocks of PGA code is called PGAblock rather than GAblock. Finally, for those readers who skipped the OGA part of this tutorial and are just interested in the PGA part, if you haven’t read the introduction (Section 1) and Section 2 through Sections 2.5 of this tutorial, you should read those earlier sections before proceeding.
3.1 Planes In PGA, we use the extra dimension e0 to carry the distance between a plane and the origin. In particular, we represent a plane with unit normal n = n1e1 + n2e2 + n3e3 at a signed distance δ from the origin as n = n −δe0.
The sense in which n represents a plane will become more apparent once we’ve seen points, where a point p is on plane n if and only if p ∧n = 0.
In PGABLE we can draw planes with the draw command: >> n1 = e3-.5e0 >> draw(n1); You should see something like Figure 9 (a). The arrows indicate the orientation of the plane, with the arrows pointing direction of the normal to the plane. Since Matlab rescales everything to fit in its display box, by default we draw our planes as unit squares.6 Now draw a second plane: >> %...
n2 = e1-0.25e0 >> draw(n2,'b'); You should see something like Figure 9 (b). The planes are drawn semi-transparent so that you can better see overlapping planes. You should rotate this figure to get a better sense of the 3D geometry of the two planes.
6If you want your plane drawn as a larger square, just scale the plane; e.g., draw(4n1).
39 0 0.1 0.2 1 1 0.5 0.5 0 0 -0.5 -0.5 -1 -1 (a) (b) Figure 9: PGA planes. (a) A single plane; (b) two planes 3.1.1 Exercises • In the code above, we drew the plane n1=e3-0.5e0. Clear the screen and redraw this plane, n1.
Now draw the planes -e3-0.5e0, -e3+0.5e0, and e3+0.5e0. You may wish to draw the planes in different colors to better match which plane matches which formula. Notice that the sign of the normal (e3) in the equation determines the direction of the normal in the plane, and that the offset from the origin is in the direction of the normal. To see things better, you may wish to clear the screen and draw the planes e3-0.5e0, e3+0.45e0, -e3-0.5e0, -e3+0.45e0, again in different colors.
3.2 Lines In PGA, the outer product of two planes represents the line they have in common. We can intersect the two planes we’ve created so far: >> %...
L = n1^n2 L = -0.5e0^e1 + 0.25e0^e3 + e3^e1 We see that L has a “Euclidean” part (i.e., terms that do not contain e0) of e3 ∧e1, and an e0 part, e0 ∧(−0.5e1 +0.25e3). The vector (−0.5e1 +0.25e3) is the vector that the line is offset from the origin. And the e3 ∧e1 part is the plane perpendicular to the line; i.e., it describes the direction of the line.
We can show that in general the outer product of two planes n1 = n1 −δ1e0 and n2 = n2 −δ2e0 has a directional component and an offset component: L = n1 ∧n2 = (n1 −δ1e0) ∧(n2 −δ2e0) = n1 ∧n2 −e0 ∧(δ1n2 −δ2n1) = n1 ∧n2 −e0 ∧ (δ1n2 −δ2n1)/(n1 ∧n2) (n1 ∧n2) = n1 ∧n2 −e0d(n1 ∧n2) = (1 −e0d)(n1 ∧n2), (18) where d = (δ1n2 −δ2n1)/(n1 ∧n2) embodies the offset of the line from the origin, and n1 ∧n2 is a plane perpendicular to the direction of the line. In Section 3.6, we will show how to extract the direction of the line, as well as how to extract the point on the line closest to the origin, and we’ll see another way to construct a line from a direction and the closest point on the line.
If we draw the line, 40 (a) (b) Figure 10: PGA planes.
(a) Two intersecting planes and their line of intersection (b) just the line of intersection.
%...
draw(L,'r') we see something lying at the intersection of the two planes (Figure 10 (a)). To see this line of intersection better, clear the screen and draw it again: >> %...
clf; draw(L,'r') What we in Figure 10 (b) see is a line with “hairs”. Our lines are directed lines, where the hairs indicate the direction of the line.
3.2.1 Exercises 1. Try draw(-L,’r’). The two lines, L and -L will overlap; what do you notice about the “hairs” of the two lines?
3.3 Points Since the outer product of planes is the intersection of the planes, taking the outer product of three planes will give us a point: >> %PGAblock(1) >> % POINTS >> n1 = e3-.5e0; >> n2 = e1-0.25e0; >> n3 = e2+0.75e0; >> clf; draw(n1); draw(n2,'r'); draw(n3,'b'); %% >> P = n1^n2^n3 P = 0.25e0^e3^e2 + -0.75e0^e1^e3 + 0.5e0^e2^e1 + e1^e2^e3 >> draw(P) You should see something like in Figure 11 (a).
We observe several things. First, points in PGA are trivectors. Second, notice that p has an e0 term of e0 ∧(−0.25e2 ∧e3 + 0.75e3 ∧e1 −0.5e1 ∧e2)—the coordinates of the point of intersection are embedded in the e0 term, with each coordinate corresponding to the missing basis vector in a term. Third, you may have 41 (a) (b) Figure 11: Intersection of three PGA planes (red, green, and blue), with the intersection point drawn in yellow. (a) The planes are orthogonal; (b) One plane is not orthogonal to the other two.
also noticed that we switched the order of e3 ∧e1; this “switched” order is needed to have the correct sign for the e2 coordinate. Fourth, there is an e1 ∧e2 ∧e3 term. This is similar to the “extra coordinate” used in other geometries that represent points and vectors as separate objects. We will discuss the e1 ∧e2 ∧e3 term shortly.
Finally, notice that we have drawn the point as a small octahedron (although if you clear the screen and just draw P, it won’t look so small due to Matlab’s rescaling). We used an octahedron to have 3D shape for points that protrudes through planes, making points-on-a-plane more visible.
Returning to the e1 ∧e2 ∧e3 term, try intersecting non-orthogonal planes: >> %PGAblock(2) >> % INTERSECTING NON-ORTHOGONAL PLANES >> n1 = (e1+e2)/norm(e1+e2) -0.25e0; >> n2 = e2 +0.5e0; >> n3 = e3 -0.25e0; >> clf; draw(n1); draw(n2,'r'); draw(n3,'b'); >> P = n1^n2^n3 P = 0.60355e0^e3^e2 + -0.35355e0^e1^e3 + 0.17678e0^e2^e1 + 0.70711e1^e2^e3 >> draw(P) >> GAview([65 30]) %% You should see something like in Figure 11 (b).
You’ll note that the coefficient of e1 ∧e2 ∧e3 is not 1. Further, the coefficients of the e0 terms are not the coordinates of the point of intersection (we know that the intersection point’s e2 coordinate should be −0.5 and its e3 coordinates should be 0.25, since all points on the intersection of n2 and n3 have those coordinates). If we normalize by the e1 ∧e2 ∧e3 coefficient, >> P/(sqrt(2)/2) ans = 0.85355e0^e3^e2 + -0.5e0^e1^e3 + 0.25e0^e2^e1 + 1e1^e2^e3 we see the appropriate e2 and e3 coordinates. Thus, in PGA we may have unnormalized (or weighted) points, depending on how we construct a point. However, we won’t use weighted points very much in this tutorial.
It may seem awkward to construct points as the intersection of planes. Instead, you can construct a point directly. For a point with coordinates [x1, x2, x3], we construct three coordinate planes pi = ei −xie0 42 (i = 1, 2, 3), and expand the intersection of these three planes: P = p1 ∧p2 ∧p3 = (e1 −x1e0) ∧(e2 −x2e0) ∧(e3 −x3e0) = e1 ∧e2 ∧e3 −e0 ∧(x1e2 ∧e3 + x2e3 ∧e1 + x3e1 ∧e2) = I3 −e0 ∧(x · I3) = e123 + x (e0 I3) (19) where x = xe1 + ye2 + ze3 and (x, y, z) are the desired coordinates of your point and I3 = e123. From this formula, we see that e123 is the point at the origin (i.e., it’s the result of intersecting the three planes created with x1 = x2 = x3 = 0). In general, we will try to use e123 when referring to the point at the origin, and I3 when taking a form of a “dual,” with duals being something that we will discuss more later.
Using (19), we can directly construct and display a point in PGABLE: >> x = -0.5e1 - 0.5e2 + 0.25e3; >> P = e123+x(e0I3) P = -0.5e0^e3^e2 + -0.5e0^e1^e3 + 0.25e0^e2^e1 + e1^e2^e3 >> draw(P) This method of creating a point is only slightly better than creating and intersecting three coordinate planes, so we have written a routine, point(x,y,z), that creates the point whose position vector is xe1 +ye2 +ze3.
You can also create a point by giving the gapoint routine a Euclidean vector: gapoint( xe1+ye2+ze3 ).
3.3.1 Exercises 1. Construct a point with coordinates (1.5, 0.5, 1.25) in PGABLE three different ways, with three different names: P1 by using the gapoint command; P2 by intersecting three orthogonal planes; and P3 using (19). The expressions for P1, P2, and P3 should all be the same.
2. Construct the line with direction e1 + 2e2 + e3 whose closest point to the origin is (0.2, 0.4, −1) using (18). You will need to compute the Euclidean dual of the line direction to use (18), which you get by dividing the direction by I3. Draw your line, the origin (e123) and the closest point.
3.4 Positioning and Resizing Planes and Lines By default, a line is draw to be two units in length with its center being the closest point to the origin. And a plane is draw to be four units in area, with its center being the closest point to the origin. However, this size is based on lines and planes of unit norm. To draw a bigger (or smaller) square for a plane, or a longer line segment for a line, just scale the plane or line, and the result will be scaled.
Additionally, we provide an option argument to the draw command to position the line segment drawn for a line and the square drawn for a plane. This optional argument is a point P, where the drawn line segment and the square are positioned so that their center is the closest point to P. The expectation is that the point P will lie on the drawn line of plane, but our draw command will accept any point P.
As an example, let’s create a point and a plane passing through the point: >> P = gapoint(3,1,1); draw(P); >> n = e1-3e0; draw(n); When we draw both with the default draw command, the point is positioned at the corner of the plane.
Using the optional positional argument to draw and rescaling the plane, >> %...
draw(2n,P,'k') we see now that we get a larger square, centered at P. See Figure 12.
43 -1 -0.5 0 2 0.5 1 1.5 1 2 0 3.4 -1 3 Figure 12: The default plane (shown in yellow) and a scaled offset plane (shown in black) centered at the yellow point.
P1 P2 P3 P4 Figure 13: Points P1 and P2 lie on the plane; points P3, P4 do not.
3.5 Points on Planes We can test if a point lies on a plane by computing the outer product of the point with the plane. As an example, we use PGAblock to create a plane, two points on the plane and two points not on the plane: >> %PGAblock(3) >> % POINTS ON PLANES >> n = e3-0.5e0; >> clf; draw(n); >> x1=0.5e3; P1 = (1-e0x1)e123; >> x2=e1+0.5e3; P2 = (1-e0x2)e123; >> draw(P1); draw(P2); view([-58, 12]); %% >> x3=e3; P3 = (1-e0x3)e123; >> x4=0; P4 = (1-e0x4)e123; >> draw(P3); draw(P4); view([-58, 12]); %% See Figure 13. Before exiting the demo, try computing the outer product of each point with the plane: >> outer(P1,n) 44 Plane δ e0xI3 = e0(xe2e3 + x2e3e1 + x3e1e2) n ∧e0xI3 = (n1x1 + n2x2 + n3x3)e0e1e2e3 n = n1e1 + n2e2 + n3e3 Figure 14: Testing for point on plane. The plane is at a distance δ in direction n; the point has coordinates x = x1e1 + x2e2 + xee3.
outer(P2,n) >> outer(P3,n) >> outer(P4,n) We see that the outer product is 0 for the points on the plane and non-zero for the points not on the plane.
Mathematically, the plane n = n −δe0 is represented by its unit normal n = n1e1 + n2e2 + n3e3 and the distance from the plane to the origin. The point is represented as P = (1 −e0x)I3 = I3 −e0 x I3 = I3 −e0(x1e2e3 + x2e3e1 + x3e1e2), where x = x1e1 + x2e2 + x3e3 is the coordinates of the point.
The outer product n ∧P is essentially projecting the point onto the normal, and computing the length of this projection, and subtracting δ from it (see Figure 14). Thus, if the point lies on the plane, this distance is 0, otherwise it is non-zero.
Algebraically, n ∧p = (n −δe0) ∧(I3 −e0 x I3) = n ∧I3 −n ∧e0 x I3 −δe0 ∧I3 + δe0 e0 x I3 The n ∧I3 and δe0 ∧e0 x I3 terms are 0; the δe0 I3 term carries the distance to the origin (as a coefficient to I4), while the n ∧e0 x I3 term computes the inner product of n and x, again as a coefficient to I4 (if this is unclear, expand x I3 and take its outer product with n). Thus, if the coefficient of I4 is 0, then the point lies on the plane; otherwise, the coefficient of I4 is the signed distance from the point to the plane.
3.6 More on Lines Now that we have seen points, we can further investigate lines. In PGABLE, we can extract the direction of the line from (18) by using the euclidean command. First, recreate the line L that we created in Section 3.2: >> n1 = e3-0.5e0; >> n2 = e1-0.25e0; >> L = n1^n2; >> draw(n1); draw(n2); draw(L); 45 We can extract the direction of the line using euclidean: >> eL = euclidean(L); >> draw(eL,'b'); draw(e123); eL is a line parallel to L through the origin (e123=e1^e2^e3 is the origin, as we saw in Section 3.3). To convert L to a direction vector, divide eL by I3 (dividing by I3 gives the Euclidean dual): >> dir = eL/I3 We can now extract the closest point on the line to the origin as >> CP = L/dir >> draw(CP) Rotate this around until you’ve convinced yourself the CP is the closest point on L to the origin.
Mathematically, from equation (18) in Section 3.2 L = (1 −e0d)(n1 ∧n2), it’s clear that the Euclidean part of L is the plane perpendicular to the line. Dividing L by dir gives L/dir = (1 −e0d)(n1 ∧n2)/dir = (1 −e0d)I3, which we recognize from (19) as the point offset from the origin by d, the orthogonal offset of the line from the origin.
And this works the other way around: if we know the direction of the line v and the closest point P on the line (to the origin), then the line is given by L = P v.
(20) Those familiar with Pl¨ ucker coordinates will notice a similarity in constructing a line from its direction and moment.
While (20) is algebraically nice (i.e., if we know any two of L, P, or v, we can compute the third), it’s an inconvenient way to construct a line. I.e., we would rather construct a line from the direction of the line and any point on the line, not one special point. If we switch from the geometric product to the outer product, L = P · v, (21) then the equation works for any point P on the line L. In PGABLE, you can construct a line with the function galine(l1,l2,l3, p1,p2,p3), where l1,l2,l3 is the direction of the line and p1,p2,p3 is a point on the line.
At times, we want a parameterized line or ray. For example, in ray tracing, we want a ray r(t) = P + tv for a point P and vector v. We can construct such a ray in PGA as follows. Let L = n1 ∧n2 be the plane perpendicular to v (note that v = (n1 ∧n2)/I3). Then the ray is r(t) = P −t(e0L) = (1 −e0x)I3 −t e0(1 −e0d)(n1 ∧n2) = (1 −e0x)I3 −t e0(n1 ∧n2) = (1 −e0x)I3 −(t e0(n1 ∧n2)/I3) I3 = (1 −e0x)I3 −(t e0v) I3 = e123 + (x + tv) e0 I3 which we see from (19) is a point that is the translation of P by tv We can test this formula with the line L that we have constructed: 46 v x1 x2 e123 d Figure 15: Point on line. The line is in direction v. Consider two points, one offset from the origin by x1, a point on the line; the other offset from the origin by x2, a point not on the line. x1 −d is parallel to v, while x1 −d isn’t parallel to v.
%... needs L, CP >> clf; draw(L); draw(CP); >> CP1 = CP - e0L; draw(CP1); >> CP2 = CP - 2e0L; draw(CP2); Another useful operation is to test if a point lies on a line. In PGA, a point Pt is on the line L if and only if >> PtL has a zero e0, trivector component. Try this with L, CP, CP2, and e123: >> %... needs L, CP, CP2 >> CPL >> CP2L >> e123L To prove that a point P = (1 −e0x)I3 is on a line L = (1 −e0d)(n1 ∧n2) if and only if the e0, grade 3 term is 0, we note that geometrically, if v = dual(n1 ∧n2), then the point P is on line L when (x −d) · v = 0.
See Figure 15. Now, mathematically, we can expand “point ∗line” as (1 −e0x)I3 ∗(1 −e0d)(n1 ∧n2) = (I3 −e0xI3) ∗((n1 ∧n2) −e0d(n1 ∧n2)).
Since we’re only interested in the e0 terms (i.e., we’re not interested in the purely Euclidean term) and since e2 0 = 0, this simplifies to −e0xI3(n1 ∧n2) −I3e0d(n1 ∧n2).
I3 is commutative with a Euclidean vector (e.g., I3e1 = e1I3), but anti-commutative with e0 (e.g., I3e0 = −e0I3), so we have −e0xI3(n1 ∧n2) −I3e0d(n1 ∧n2) = −e0xI3(n1 ∧n2) + e0dI3(n1 ∧n2) = e0(d −x)I3(n1 ∧n2).
The term (d −x)I3 is the Euclidean dual of (d −x) (i.e., a bivector); if (d −x) is parallel to v, then (d −x)I3(n1 ∧n2) = c(n1 ∧n2)(n1 ∧n2) is a scalar (see Section 2.9) and e0(d −x)I3(n1 ∧n2) is a scalar factor of e0, otherwise (if (d −x) is not parallel to v) this is a bivector and e0(d −x)I3(n1 ∧n2) is a trivector. Thus, the e0 term is a trivector if and only if (d −x) is not parallel to v (and thus, P is on L if and only if the e0 trivector term is 0). See Figure 15.
47 3.6.1 Exercises 1. In exercise 2 in Section 3.3.1, you constructed a line using (18). Now construct this same line (in direction d = e1 + 2e2 + e3 through the point P = (0.2, 0.4, −1), the closest point on the line to the origin) using (20). Be sure to draw the line, the origin, and the closest point.
2. Suppose you try to use (20) to construct a line using a point other than the closest point on the line through the origin. E.g., using the same direction as in the previous question, d = e1 + 2e2 + e3, use the point Q = (0.2 + 1, 0.4 + 2, −1 + 1) = (1.2, 2.4, 0). What happens when you use (20) using d and Q?
3. Now use (21) to construct a line using Q from the previous question and d from Exercise 1.
3.7 Degenerate Objects PGA contains a variety of degenerate objects, many of which are useful. We will look at vanishing lines, vanishing points, and mention the vanishing plane.
In general, the intersection of two planes is a line. But if the planes are parallel, this line of intersection contains no location, but only the common normal vector of the two planes: (n −δ1e0) ∧(n −δ2e0) = (δ2 −δ1)e0 ∧n, where n is a Euclidean vector. This line is commonly thought of as a line at infinity. We will follow the lead of [DdK] and refer to this line as a vanishing line. e0 ∧e1 is an example of a vanishing line.
When drawing in Matlab, Matlab finds a bounding box for all the drawn objects, and rescales things so that this box fits on the screen, and all drawn objects are visible on the screen. Thus, in Matlab, we can’t draw things “at infinity”, or even very far away, since Matlab will rescale things to be in a box that fits on the screen.
In PGABLE, to draw vanishing lines (i.e., lines at infinity), we will draw the projection of this line at infinity onto Matlab’s box. However, we need a box to draw on, i.e., you must draw something else before drawing a vanishing line. So as an example, to show the vanishing line e0 ∧e1, we first draw the point e123 and the plane e2, and then the vanishing line e0 ∧e1: >> draw(e123); draw(e2); draw(e3); >> draw(e0^e1); view([66 30]) You should see something like Figure 16(a).
If we intersect a vanishing line with a plane (not parallel to the vanishing line), then get get a vanishing point: (e0 ∧n1) ∧(n2 −δe0) = e0 ∧(n1 ∧n2), which is a point with no positional aspect (i.e., no e1 ∧e2 ∧e3 term; just a direction, the Euclidean dual of n1 ∧n2; see Section 3.3 for a review of points). In PGABLE, we draw a vanishing point as a star on the boundary of Matlab’s drawing box, with a short line indicating the direction: >>%...
draw(e0^e1^e3); >> draw(e0^e1^(e2+2e3)); view([130 10]) You should see something like Figure 16(b). You may wish to also draw the plane e2+2e3.
Note that as expected, the vanishing points lie on the vanishing line. In these examples, the vanishing points also lies on the plane intersected with the vanishing line. However, when you intersect the vanishing line e0^e1 with any plane parallel to e3 or e2+2e3 (in our examples), you will get the same vanishing points.
Thus, in general, the vanishing point (constructed as the intersection of a vanishing line with a plane) will lie on the vanishing line, but the vanishing point (in our figures) won’t lie on the plane of intersection; however, the direction line of the vanishing point will be parallel to this plane of intersection.
48 (a) (b) Figure 16: (a) The vanishing line e0 ∧e1 is drawn as the blue dashed line around the edge of the Matlab drawing box. (b) Vanishing points are drawn as stars with a line segment indicating their direction.
Further note that there is some ambiguity as to where to draw the vanishing points (and vanishing lines), since where they appear should depend on the location of the viewer. Since we don’t really know where the viewer is, we have chosen to place the vanishing points on the Matlab box as if the viewer were positioned at the center of the box. Note also that if you draw an additional object that changes the size of Matlab’s drawing box, then the location of the vanishing lines (and vanishing points) will move, since they are redrawn relative to the center of the new box.
We conclude this section with two more degenerate objects. The equation for a plane in PGA is n = n −δe0. Degenerate planes include • δ = 0. This is a plane passing through the origin. While in some sense, these planes are not special, proofs are often easier for planes through the origin than they are for general planes.
• n = −δe0. This plane contains all the vanishing points. Classical names include the ideal plane and the improper plane; again, we will follow the lead of [DdK] and refer to it as the vanishing plane. In PGABLE, we don’t draw the vanishing plane, since it would cover the entire Matlab drawing box, obscuring everything inside the box.
Special Note: Vanishing lines and vanishing points are treated different with regards to drawing than other items, since their location in the scene changes based on the objects in the scene. Thus, clf doesn’t clear our vanishing objects. Instead, you should call pclf to clear these (and all) items from the scene.
3.8 Duality Duality typically occurs when there is a correspondence between one set of objects in a space to another set of objects in the space. In geometric algebras, duality is typically defined as dual A ≡A/In, where In is the pseudoscalar for the algebra. This sets up a correspondence between scalars and the pseu-doscalar, vectors and grade n −1 blades, etc. However, this definition of duality (with the division by the 49 pseudoscalar) is problematic in PGA, since the pseudoscalar, I = e0e1e2e3 is a null vector, and has no inverse (i.e., you can’t divide by it). An alternative dual, the Poincar´ e dual, will be used, which in our algebra makes grade 1 objects dual to grade 3 objects (i.e., planes dual to points). Unfortunately, this dual is dependent on the choice of basis, and the point that is dual to any particular plane will be dependent on the choice of origin. Details on the Poincar´ e dual are beyond this scope of this tutorial; for our purposes, the Poincar´ e dual creates a mapping between points and planes. See [DdK] for the math behind the Poincar´ e dual7.
Mathematically, the Poincar´ e dual is represented by ⋆.
In PGA, you can either call the function poincaredual or pdual to compute the Poincar´ e dual of an object. The Poincar´ e dual is its own inverse, so applying it twice to an object returns that object; try computing pdual(e1) and pdual(pdual(e1)) in PGABLE.
We can explore the Poincar´ e dual in PGABLE by constructing three planes and taking their duals: >> n1 = e1-0.5e0; n2 = e1-e0; n3 = e1-2e0; >> pclf; draw(n1); draw(n2,'r'); draw(n3,'m'); >> draw(pdual(n1),'g'); draw(pdual(n2),'r'); draw(pdual(n3),'m'); >> draw(e123); view([9 25]) What we see are three planes and their duals (three points). In addition to the three planes and their dual three points, we have also drawn the origin. In the figure, we see that as the planes move further from the origin, their dual points get closer to the origin. Conversely, we would expect that as the planes get closer to the origin, their dual points move towards infinity. And if we construct a plane passing through the origin and compute its dual, >> %...
n4 = e1; draw(n4,'y'); draw(pdual(n4),'y'); view([25 6]) we see that the dual of a plane through the origin is indeed a point at infinity (the yellow star in the figure).
This relative placement of each dual point based on the distance of the plane from the origin shows the dependence of the Poincar´ e dual on the origin: if we chose to construct the Poincar´ e dual relative to a different point, each plane in this example would have a different dual point.
3.8.1 Exercises 1. In the sample code in this section, we computed the Poincar´ e dual of several planes, getting points.
Now construct the points (1, 0, 0) and (1, 1, 1) and compute the Poincar´ e dual of these points, drawing both points and their duals. You probably also want to draw the origin, e123 (drawing each point and its dual in their own color will make it easier to see the correspondence between a point and its dual).
What do you observe about the relative placement of point (1, 1, 1), its dual, and the origin?
2. In the code above, when we computed the Poincar´ e dual of a plane through the origin, we got a point at infinity. What do you get if you compute the Poincar´ e dual of the origin, e123?
3. Compute the planes n1 = e1 −e0 and n2 = e2 −e0, and compute (and draw) the line L = n1 ∧n2.
Now compute (and draw) the Poincar´ e dual of this line. The result is a second line; what relationship to you observe about the two lines, and about the two lines and the origin (which you may want to draw)?
4. Draw the line e1 ∧e2, which passes through the origin (you probably also want to draw the origin).
Next, draw the line that is the Poincar´ e dual of e1 ∧e2. What can you say about this dual line?
7They actually discuss the Hodge dual, which is closely related to the Poincar´ e dual.
50 3.9 Meet and Join As in OGA, in PGA there are also meet and join operations. The meet operation in PGA is the outer product: two planes meet at a line; three planes meet at a point. The join operator is more complicated to derive. Using ∧for the meet and ∨for the join, the join operator is defined as A ∨B = ⋆−1(⋆A ∧⋆B).
See [DdK] for details on the derivation of this equation.
Our interest in the join operation is that we can use it as a kind of dual to the meet to construct lines and planes from points. We can test this in PGABLE by constructing two points and computing their join: >> P1 = gapoint(0.5,1,0); P2 = gapoint(1,1,1); >> L = join(P1,P2); >> clf; draw(P1); draw(P2); draw(L); Joining the line L to a third point gives us a plane: >> %...
P3=gapoint(1,0,0); >> n = join(L,P3); >> draw(P3); draw(n); 3.9.1 Exercises 1. We have used (18) and (20) to construct a line. This line passes through points P = (0.2, 0.4, −1) and Q = (1.2, 2, 4, 0). Use the join operation to create this line in a third way.
2. We used the join of two points to create a line, and the join of this line with a third point to create a plane. Verify that the three points used to create this plane actually lie in the plane (see Section 3.5 if you’re unsure how to test if a point lies in a plane).
3. Explore other cases of join, in particular (a) Compute the join of a point with itself.
(b) Compute the join of two planes.
(c) Compute the join of two lines.
4. e031 and e012 are points at infinity. Draw these two points, as well as the origin (e123) and the plane e2. Now draw the join of these two points at infinity: join(e031,e012). Comment on the result of the join, as well as the location of the two points at infinity relative to the join of these two points at infinity.
3.10 Transformations We will construct transformations in PGA as sandwiching operators. Similar to OGA, we will use a plane n = n −δe0 as an mirror operator via the sandwiching operator. Composing multiple mirror reflections will give us the remaining transformations. The mapping with this operator n to an arbitrary element X is X 7→n b Xn−1, (22) where b X = (−1)grade(X)X, introducing a minus sign for odd grade elements (such as planes and points) and no sign for even grade elements (such as lines).
51 -1 1.5 1 0.5 0 0 -0.5 1 -1 1 0.5 0 -0.5 n2 L n1 n2r pt1r Lr pt1 Figure 17: Reflecting a point pt1, a line L, and a plane n2 through a plane n1.
We can try this in PGABLE: >> %PGAblock(4) >> % REFLECTION IN A PLANE >> n1 = e1-0.5e0; % construct a plane to reflect through >> n2 = (e1+e2+e3)/norm(e1+e2+e3)-0.5e0; % construct a plane to reflect >> Pt1 = n2^e1^e2; % construct point on plane n2 >> Pt1 = -Pt1/inner(Pt1,e1^e2^e3); % normalize the point >> Pt1r = -n1Pt1n1; % compute its reflection >> clf; draw(n1); draw(Pt1); draw(Pt1r,'r'); GAview([-15 21]); %% >> n2r = n1n2n1; % reflect plane n2 through plane n1 >> draw(n2,'b'); draw(n2r,'m'); GAview([-5,50]); %% >> L = e1^e2; % construct a line >> Lr = n1Ln1; % reflect line L through plane n1 >> draw(L); draw(Lr); GAview([-5,50]); %% You likely need to rotate the diagram a bit to see the reflections. See also Figure 17.
In this example, you may have noticed that when we applied n1 and n2 as reflection operators, we multi-plied by n1 and n2 on the right, rather than inverse(n1) or inverse(n2). That’s because we constructed n1 and n2 to have unit norm, making them their own inverses. If we had constructed n2 to have normal e1+e2+e3, then when we used n2 as a reflection operator, we would have had to multiply by inverse(n2) on the right. However, constructing n2 with normal e1+e2+e3 would have changed the coefficient of e0 in this equation to account for the non-unit normal.
Further, if you look at Pt1r, you’ll notice that it has a negative e1^e2^e3 component. Although we won’t use them here, at times the concept of negatives points is useful. Also notice that the reflected line Lr has the opposite orientation to L.
The point pt1r should clearly look like the reflection of pt1 in the plane n1. However, the plane n2r may look more like the reflection of n2 in a plane perpendicular to n1 through the intersection of n1 and n2. But realize that we’ve only drawn a square portion of each plane. The point pt1 was chosen to lie on the plane n2 so that you could see where its reflection maps to. While pt1 (the yellow point) lies in the plane n2 (the blue square), the reflection of pt1 doesn’t lie in the magenta square. But you should be able to convince yourself that pt1r lies in the plane containing the magenta square. And it’s the infinite plane that contains the blue square that is being reflected through n1, even though our diagrams only show square subportions 52 of these planes.
3.11 Reflection in Two Parallel Planes: Translation We can compose two reflections with the geometric product to get a new transformation. Given planes n1 = n1 −δ1e0 and n2 = n2 −δ2e0. Their product is n2 n1 = (n1 −δ1e0) (n2 −δ2e0) = (n2 · n1) + n2 ∧n1 + e0(δ1n2 −δ2n1).
(23) If the two planes are parallel, with n1 = n2 = n, with |n| = 1 then n2 n1 = 1 + (δ1 −δ2)e0 ∧n ≡1 −e0t/2, (24) where t = 2(δ2 −δ1)n. To use 1 −e0t/2 as an operator as in (22), we also need its inverse: 1 + e0t/2.
Applying this operator to a point (1 −e0x)I3 and expanding gives (dropping all terms with e2 0 = 0 and observing that I3ett/2 = −ettI3/2) (1 −e0t/2) (1 −e0x)I3 (1 + e0t/2) = (1 −e0t/2) (1 −e0x)(I3 + I3e0t/2) = (1 −e0t/2)(I3 −e0xI3 + I3e0t/2) = I3 −e0tI3/2 −e0xI3 + I3e0t/2 = e123 + (t + x) e0 I3 we see that Tt ≡n2n1 is a translation by t. Of course, we can directly construction Tt from (24) without having to construct two planes of reflection.
We can visualize all of this in PGABLE: >> %PGAblock(5) >> % TRANSLATION >> n1=(e1+e2+e3)/norm(e1+e2+e3)-0.5e0; >> n2=(e1+e2+e3)/norm(e1+e2+e3)-1.5e0; >> Pt = (1-e0(e1+e3))e123; >> clf; draw(n1); draw(n2,'b'); draw(Pt); GAview([30,36]) %% >> Ptr1 = n1Ptn1; draw(Ptr1,'m'); GAview([30,36]) %% First reflection >> Ptr2 = n2Ptr1n2; draw(Ptr2,'r'); GAview([30,36]) %% Second reflection In the figure, the red point is the translation of the yellow point, and that the red point has been translated by twice the distance between the two planes, as suggested by the derivation above. We can construct the translation as either the reflection in two parallel planes, or we can construct the translation directly from Equation 24: >> %...
Tt = n2n1 >> t = 2(1.5-0.5)(e1+e2+e3)/norm(e1+e2+e3); >> Tt = 1-e0t/2 %% Observe the two Tt's are the same In the text window, observe that Tt is the same, whether constructed as a reflection in two planes or constructed directly via (24). We can also translate lines with the same operator: >> %...
L = e1^e2; draw(L); GAview([30,36]); %% A line >> Lt = TtLinverse(Tt); draw(Lt); GAview([30,36]); %% The line translated 53 Because of our visualization of lines as line segments, it can be hard to tell from the figure that the line L has translated by t; to better see the translation, try drawing the point at the origin (which lies on L), and then drawing the translation of the origin (using Tt).
You can also translate planes with Tt, although depending on the plane you translate, it can be difficult to see the translation in the graphics window.
Note that unlike reflection, we did not have to adjust the sign of the line after transforming it. That’s because a translation is a composition of two reflections, and the minus sign cancels out. A similar thing will happen with rotations in the next section.
3.11.1 Exercises 1. In PGABLE, construct a translation T1, which translates by e1, and a second translation T2, which translates by e2. Translate the origin by T1 and translate the origin by T2. Now construct to composite transformation Tc = T2T1, and translate the origin by Tc. Did the composed transformation perform as you expected?
2. You are going to construct an operator T to translate by t = e1 + 0.5e2 + 0.25e3 (a) Construct T in two ways: directly, using (24) and by reflecting through two planes, n1 = ˆ t −δ1e0 and n2 = ˆ t −δ2e0, where ˆ t = t/norm(t), δ1 = 0 and δ2 = √ 12 + 0.52 + 0.252/2.
i. Are the two formulates of T the same? Why was δ2 set to √ 12 + 0.52 + 0.252/2?
ii. The translation operator we made can translate points, lines, and planes, as T XT−1, where X is the object to translate. Draw the origin (e123), the points P1 = (1, 0, 0) and P2 = (0, 1, 0), the line L = join(P1, P2), and the plane n = t. Now apply T to P1, P2, L, and n, and draw each of these transformed objects. Does it look like each object was translated by t?
Try using view([-3.5, 63]) to get a view that should show that at least P1 and P2 were translated in the direction of t.
3.12 Reflection in Two Non-Parallel Planes: Rotations Reflection in two non-parallel planes is a rotation, as is readily seen for two planes through the origin (e.g., n1 = n1 −δ1e0 = n1 and n2 = n2 −δ2e0 = n2, since δ1 = δ2 = 0 for planes through the origin): Tr = n1 n1 = n2 · n1 + n2 ∧n1 = cos(ϕ/2) −sin(ϕ/2)I (25) where I is the unit bivector spanning n2 ∧n1, which we saw in Section 2.7.1 is a rotation by ϕ around the line of intersection between n1 and n2. However, if we reflect in arbitrary planes, we still get a rotation around the line of intersection of the two planes, as shown in this PGABLE example: >> %PGAblock(6) >> % ROTATIONS >> n1=e1-0.25e0; >> n2=(5e1+e2+e3)/norm(5e1+e2+e3)-0.25e0; >> Tr = n2n1; >> Pt = (1-e0(e1+e3))e123; >> clf; draw(n1); draw(n2,'b'); draw(Pt); %% >> for i=1:10 >> Pt = zeroepsilons(TrPtinverse(Tr)); >> draw(Pt,'r'); >> end; >> %% See Figure 18.
54 Figure 18: Rotation as reflection in two planes. We rotate the yellow point 10 times to get each of the 10 red points.
In this example, we first drew the two planes and a point, and constructed a rotation operator Tr from the two planes; we rotated the point 10 times with Tr, drawing the rotated point after each rotation. By rotating the figure, you should be able to see that the point has been rotated around the intersecting line between the two planes. While you can construct a rotation about an axis through the origin directly from (25) rather than as a reflection through two planes, constructing a rotation around an arbitrary axis (not necessarily through the origin) requires more work (see the exercises at the end of this section).
Note also the use of zeroepsilons; at times the computations will generate terms with coefficients close to machine precision; you can see an example of these terms if you type Trptinverse(Tr) before exiting the demo. zeroepsilons sets these terms to 0.
By reflecting in three planes in 3D, you can get glide reflections and rotoreflection. By reflecting in four planes, you get screw motions. See the Dorst-DeKeninck report [DdK] for details on such motions, as well as a discussion on the exponential form of these motions.
3.12.1 Exercises 1. We want to construct a rotation by 45◦around an axis a = e1 + e2 passing through the point (1, 0, 1).
Construct this transformation in PGABLE by composing three transformations: Ti, which translates the point (1, 0, 1) to the origin using (24); R, which rotates around a by 45◦using (25), and then T, which translates the origin back to (1, 0, 1), again using (24). Your composed transformation would be C = T R Ti. Apply your transformation to several points to convince yourself that it works, and repeatedly transform the origin to see it form a circle of 8 points.
Tips: • You may wish to construct and draw the line through (1, 0, 1) in direction a with either (18) or (20). Note that (1, 0, 1) is not the closest point on the line to the origin. Rather than compute the closest point on the line, you may wish to compute the line through the origin in direction a and draw the translation of that line by T.
• You may wish to normalize a. You will also want to compute the Euclidean dual of a to use (25).
• Note that Ti and T are each others inverses.
55 3.13 Euclidean Motions Characterized as Exponentials In addition to constructing linear transformations as multiple reflections in planes, we can also construct them via exponentiation. To begin, recall the Taylor expansion of the exponential: ex = X i=0 ∞xi i! .
Now, if we have a Euclidean translation vector t, we see that (e0t)2 = e0 t e0 t = −t2e2 0 = 0 (since e2 0 = 0), and if we exponentiate −e0t, we get e−e0t/2 = 1 −e0t/2 + 1 2(e0t/2)2 −. . .
= 1 −e0t/2, which we see is the equation for a translation operator (24). Further, if we have a line L = n1 ∧n2 for two orthonormal planes n1 and n2 then L2 = −1 (see the exercises) and e−ϕL/2 = 1 −ϕL/2 + 1 2(ϕL/2)2 −1 3(ϕL/2)3 + . . .
= cos(ϕ/2) −sin(ϕ/2)L, which we see is the equation for rotation (25) (where you’ll need to review Taylor series expansions of trig functions to follow that last step). While the example above is for a line through the origin, the idea generalizes to arbitrary lines (see the exercises).
3.13.1 Exercises 1. In PGABLE, for the Euclidean vector t = e1 + 0.25e2 + 0.5e2, construct the translation versor pair e−t/2, et/2 using gexp. Draw points P1 = e1 and P2 = e1 + e3, as well as e123. Apply the versor you created to P1 and P2, drawing the results in a different color than you draw P1 and P2.
2. Prove that for L = n1 ∧n2, L2 = −1, where n2 1 = n2 2 = 1 and n1 · n2 = 0.
3. In PGABLE, construct points Q1 = 2e1 + e2 and Q2 = e + 2e2 + e3, and a line L = join(Q1, Q2).
(a) Is L unit length? If not, normalize it using the norm function. Check that L2 = −1, where L is unit length.
(b) Draw Q1, Q2, e123 and L.
(c) Construct the rotor Lr = e−ϕpi/12L/2, L−1 r = eϕpi/12L/2 using the gexp function, and apply your rotor to Q3 = 2e1 + e2 giving Q′ 3, and apply your rotor to Q′ 3 giving Q′′ 3. Draw Q3, Q′ 3, and Q′′ 3, giving each a different color.
4. Repeat exercise 1 of Section 3.12.1 using the exponential of a line to compute the versor. You may wish to construct the line as the join of the given point plus the translation of this point by the direction of the line. However you construct your line, be sure to normalize it before constructing the versor.
3.14 General Euclidean Motions The most general Euclidean motion is a screw motion, which can be computed as the reflection in four planes. However, a screw motion is a translation along a line composed with a rotation around that line (or vice versa–the operations commute). Thus, if we have a line L in direction t, we can compute a screw motion as the product of the translation versor of (24) with the rotation by ϕ around L, e−ϕL/2, where we could also use the exponential form of the translation, e−e0t/2.
56 We can construct and visualize these transformations in PGABLE. We begin by constructing a line L from a direction tv and a point P=gapoint(1,0,1): >> %PGAblock(7) >> % SCREWS >> tv = (e1+e2+0.5e3)/5; >> P = gapoint(1,0,1); >> L = P.tv/norm(tv); >> draw(P); draw(e123); draw(L); We next construct a point Q and a translation versor T,Ti to translate in the direction of the line L: >> %...
Q = gapoint(0,1,0); >> draw(Q,'b'); >> T = gexp(-e0tv/2); >> Ti = gexp(e0tv/2); We now repeatedly apply this translation versor to the point Q: >> %...
for i=1:5 >> Q = TQTi; draw(Q,'r'); >> end This gives us Figure 19(a). Now we reset the point Q, and construct a rotation versor RL,RLi to rotate around the line L: >> %...
Q = gapoint(0,1,0); >> RL = gexp(-pi/12L/2); RLi = gexp(pi/12L/2); We now repeatedly apply this rotation versor to Q: >> %...
for i=1:23 >> Q = RLQRLi; draw(Q,'g'); >> end This gives us Figure 19(b). Note that we have abbreviated the text window output so that you only see iterations i = 1, 2. You may wish to rotate the figure to better see the rotation of Q around L.
Finally, we construct a new initial point R and construct the screw versor Sc,Sci. Note that Sc is the product of the versors we constructed earlier, T and R: >> %...
R = gapoint(0,0,0); >> Sc = TRL; Sci = RLiTi; We now repeatedly apply this screw versor to R: >> %...
for i=1:30 >> Q = ScQSci; draw(Q,'m'); >> end This gives us Figure 19(c). Continuing the PGAblock provides a view looking down the line, so that L appears as a point and you can see that both the points computed by the rotation and by the screw motion rotate around this line. Concluding the PGAblock is a GAorbiter of the figure.
57 0 1.5 0.5 1 1 0.5 0 1 -0.5 0.5 -1 0 -0.5 0 0.5 1 1 1.5 2 2 0 2.5 1 -1 0 0 2 1 2 1 2 0 1 -1 0 (a) (b) (c) Figure 19: Constructing a screw motion. (a) A translation; (b) a rotation; (c) a screw motion that is the composition of the translation and rotation 3.14.1 Exercises 1. In PGAblock(7) note the normalization of the direction tv when building the line L. Rebuild tv, then construct L without this normalization. Next, rebuild the rotor RL and the screw motion Sc, and apply repeatedly (as in the PGAblock(7)) to Q=gapoint(0,1,0) and R=gapoint(0,0,0) to see the effect of this normalization.
2. In PGAblock(7), we constructed the screw versor as Sc=TRL. What if we commute T and RL? Compute this (RLT) in PGABLE and compare it to the original form, TRL.
3. Rather than use exponentials to construct the versors T, RL, we could construct them by intersecting planes. Construct T and RL each as the intersection of two planes. Verify your result by comparing the versors you constructed as the intersection of planes to the T and RL computed in PGAblock(7).
Notes: (i) the solution is not unique; (ii) the hard part of this exercise is getting the distance between the translation planes and the angle between the rotations planes correct.
3.15 Geometric Meaning of Operations We have seen a variety of operators in this tutorial.
Most of these operators have geometric meaning, although the precise meaning depends on the particular space and geometric interpretation one has of that space. For our model of PGA, we note the following geometric interpretation of some of the operators: • Outer product (a ∧b): On blades, the outer product acts as an intersection of spaces.
• Inner product (a · b): projection • Geometric product (a b): The geometric product of two vectors constructs an operator (such as reflec-tion or rotation), to be used in the sandwich product.
• Regressive product: join • Sandwich product (R a R): As in most geometric algebras, the sandwich product in PGA applies a linear transformation (R) to an object (a).
58 Table 4: Inner product table for the conformal model.
o e1 e2 e3 ∞ o 0 0 0 0 −1 e1 0 1 0 0 0 e2 0 0 1 0 0 e3 0 0 0 1 0 ∞ −1 0 0 0 0 Table 5: Table summarizing CGA Matlab commands.
Command Arguments Result no,e1,e2,e3,ni Basis vectors for generating the geometric algebra e12,e23,e13 Bivectors for the Euclidean subalgebra e123 The trivector for the Euclidean subalgebra I3 The pseudo-scalar for the Euclidean subspace I5 The pseudo-scalar gapoint (x,y,z) Compute the PGA point with coordinates x, y, z.
(l, p) passing through point p=(p1,p2,p3) +,-,,/,ˆ,. multivectors The operations of our geometric algebra inverse Compute the inverse join, meet (multivector,multivector) join and meet of two multivectors dual (multivector) Compute the dual of a multivector gexp (multivector) The geometric product exponential, emv glog (multivector) The geometric logarithm grade (blade) Return the grade of a blade (−1 if a multivector) (multivector,n) Return the portion of the multivector that is of grade n isgrade (multivector, g) To determine if a multivector is of grade g euclidean (multivector) Return the Euclidean portion of a multivector noneuclidean (multivector) Return the e0 coefficient of a multivector getx, gety, getz (point) Get the x, y, z coordinates of a point draw (blade) Draw the given blade pclf Clear all objects (including vanishing objects) from the screen 4 The Conformal Model The 3D Conformal Geometric Algebra has representations for points, lines, circles, planes and spheres, and among other things, we can intersect these objects using the products of the geometry. Further, we can represent conformal transformation (angle preserving) as versors in the Conformal model.
Unlike GABLE, PGABLE supports the Conformal Model . The Conformal Model has three Euclidean vectors, e1, e2, e3, and two null vector o and ∞, which represent the points at the origin and at infinity, where o · ∞= −1. Table 4 gives the inner product table for these elements of the conformal model. Table 5 gives a summary of the CGA commands in PGABLE.
A point p in the conformal model are represented as p = o + p + 1 2p2∞, (26) where p is a Euclidean vector (i.e, a linear sum of e1, e2, e3). In this model, the inner product between two points is proportional to the distance squared between the points. In particular, if we have p = o+p+ 1 2p2∞ and q = o + q + 1 2q2∞, then p · q = −1 2(q −p)2.
(27) 59 To use the conformal model in PGABLE, first type >> GA.model(CGA) You can construct points in the conformal model in PGABLE using (26), >> v = e1+e2 >> pt1 = no + v + 0.5(vv)ni >> draw(pt1) but there are also routines to assist you in building points: >> ...
w = e2+e3 >> pt2 = gapoint(w) >> pt3 = gapoint(-1,1,0) % construct the point for -e1 + e2 >> draw(pt2); draw(pt3) We can now compute the distance between two of these points using the inner product: >> ...
sqrt(double(-2pt1.pt2)) In the conformal model, in addition to points, lines, and planes, you can construct circles and spheres. A sphere can be constructed as the outer product of four points. Let’s construct a fourth point and then construct a sphere in PGABLE: >> ...
pt4 = gapoint(0,2,0); draw(pt4) >> sp1 = pt1^pt2^pt3^pt4 sp1 = -2no^e2^e3^ni + 6no^e3^e1^ni + 6no^e1^e2^ni + 4no^e1^e2^e3 >> draw(sp1) In PGABLE, a sphere is drawn in wireframe so that you can see the constructing points. We have selected the points so that sp1 is a unit sphere centered at e2. Try drawing the center point gapoint(0,1,0) in red (i.e., with the optional argument to draw, ’r’).
This is an outer product null space object; i.e., if you take the outer of the sphere with a point P, then the result is 0 if and only P lies on the sphere. We can try this with several points: >> pt1^sp1 >> pt2^sp1 >> gapoint(0,1,0)^sp1 >> gapoint(0,3,0)^sp1 You’ll notice that the result of the first two operations is 0, while the last two are non-zero and of opposite sign. Our sphere is an oriented sphere; taking the outer product of the sphere with points on the inside of the sphere gives the opposite sign as taking the outer product with points on the outside of the sphere. The line segments in the figure indicate the orientation of the sphere. If you swap the order of the points used the create the sphere, >> sp1n = pt2^pt1^pt3^pt4 then you’ll noticed that sp1 and sp1n have opposite signs, and thus the outer product with points on the inside/outside of these sphere also have opposite signs. When you draw a sphere in PGABLE, if the sign of the o term is positive, the normals point to the “outside” of the sphere, and if the sign of o is negative, the normals point to the “inside” of the sphere.
You can also construct a sphere using the center of the sphere c and the radius r: s = c −1 2r2∞.
(28) Let’s try using this formula in PGABLE to construct the dual of sp1 60 -1 -0.5 0 0.5 1 1.5 1 0 0 -1 1 2 Figure 20: Four points, two spheres, a line, a plane, and a circle in the Conformal Model, with view view([-116,14]).
...
ptC = gapoint(0,1,0) >> sp2d = ptC - 0.511ni sp1d = no + e1 You may have noticed that the formulas for sp1 and sp2d are significantly different from one another.
This difference is because these two spheres are in dual representations. If you type dual(sp1), then you’ll find dual(sp1)’s representation is the same as sp1d (up to sign).
To get a plane in PGABLE, just construct a sphere with one of the four points being ni, the point at infinity: >> ...
pi1 = pt1^pt2^pt3^ni >> draw(pi1) To get a circle, we take the outer product of three points: >> ...
circ1 = pt1^pt2^pt3 >> draw(circ1) And a line is the outer product of three points, where one of the points is the point at infinity: >> ...
line1 = pt1^pt2^ni >> draw(line1) Figure 20 shows a view of the objects we’ve created.
Exercises The first three exercises are meant to be drawn on top of the example in this section as shown in Figure 20.
You should regenerate that figure for each of these exercises, although you may wish to adjust the view to better see the new object drawn in the exercise.
1. A third way to construct a sphere is from its center, C, and a point P on the sphere: s = P · (C ∧∞).
61 In PGABLE, construct the unit sphere centered at (0, 1, 0) and see that it is the same sphere (up to sign) as the spheres constructed in the examples above.
2. In PGABLE, construct and draw the plane through pt2, pt3, and pt4.
3. We can also construct a line as the outer product of a point, a direction, and the point at infinity. In PGABLE, construct and draw any line tangent to the sphere sph at pt1.
4. Expand the inner product of two points p and q to prove (27).
4.1 Intersection of Objects Suppose we have two dual spheres, one sphere of radius 1 centered at (1, 0, 0) and a second sphere of radius 2 centered at (2.5, 0, 0): >> clf; >> s1 = gapoint(1,0,0) - 0.5(11)ni >> s2 = gapoint(2.5,0,0) - 0.5(22)ni >> draw(s1); draw(s2); view([-6,18]) We can intersect s1 and s2 by taking their outer product: >> %...
ic = s1^s2 >> draw(ic); view([-6,18]) We can move s2 further along the x-axis, >> %...
s2=gapoint(3,0,0) - 0.5(22)ni >> draw(s2,'m'); >> ic = s1^s2 >> draw(ic,'g'); view([-6,18]) and the intersection moves as expected. If we continue to move s2 until the s1 and s2 are tangent, >> %...
pclf; >> s2=gapoint(4,0,0) - 0.5(22)ni >> draw(s1); draw(s2); >> ic = s1^s2 >> draw(ic); view([-6,18]) the intersection ic is a circle of radius 0 and isn’t visible in the picture. And if we continue to move s2, the two spheres s1 and s2 no longer intersect. However, s1^s2 still computes something. What we find looking more closely at the intersection is that we get an imaginary circle, a circle whose radius is an imaginary number. In PGABLE, we draw this circle as a “dashed circle”: >> %...
pclf; >> s2=gapoint(5,0,0) - 0.5(22)ni >> draw(s1); draw(s2); >> ic = s1^s2 >> draw(ic,'c'); view([ -13, 19]) While an imaginary circle may seem to have little to no use, realize that the expression s1^s2 is an algebraic equation that always has a solution, and at times, we will want to test the result, etc.
See Figure 21 for screenshots of these examples.
62 -2 -1.5 -1 -0.5 0 2 0.5 1 1.5 2 0 -2 5 4 3 2 1 0 -2 -1.5 -1 -0.5 0 2 0.5 1 1.5 2 0 -2 6 5 4 3 2 1 0 -2 -1 0 2 1 2 0 7 6 5 4 3 -2 2 1 0 (a) (b) (c) Figure 21: Intersection of spheres. (a) Intersection small sphere with each of two large spheres; (b) two spheres tangent; (c) Imaginary circle of intersection.
Exercises 1. For a point P and normal n, we can construct the dual representation of the plane through P perpen-dicular to n as Π = −n −(P · n) ∞.
Using s1 as above (s1 = gapoint(1,0,0) - 0.5(11)ni), construct the dual representation of the plane Pi through the center of s1 with normal n. Draw s1, Pi, and the intersection of s1 and Pi, s1^Pi. When drawing Pi, be sure to offset it so that it is centered at the center of s1.
4.2 Transformations In CGA, we can represent transformations as versors. And some objects, such as planes and spheres, act as operators. For example, given a plane Π, we can reflect an object X (point, line, circle, plane, sphere, all of which are blades in CGA) in Π as Π e X Π−1, where e X is the reversion of X; basically, you reverse the order of the vectors whose outer product is X. The result of the reversion of X is e X = (−1)k(k−1)/2X; i.e., for some grades, the reversion of X is X, while for other grades, the reversion of X is −X.
Constructing a point, a line, a sphere and a plane, >> P = gapoint(1,0,0); >> L = P^gapoint(2,0,0)^ni; >> sph = gapoint(0,2,0) - 0.5(0.50.5)ni; >> n = e1+e2; >> pi1 = -n-(gapoint(0,0,1).n)ni; >> draw(P); draw(L); draw(sph); draw(pi1) we can now reflect P, L, and sph through pi1: >> Pr = pi1P/pi1; >> Lr = -pi1L/pi1; >> sphR = pi1sph/pi1; >> draw(Pr,'m'); draw(Lr,'c'); draw(sphR); view([34 56]) Note the minus sign when reflecting a line, since the line is of grade 3, and (−1)3∗2/2 = −1. Figure 22 shows these objects and their reflections.
63 2 -1 1 -2 -1 0 0 0 -1 1 1 Figure 22: Reflection of a point, line, and sphere in a plane.
We could construct translation and rotation via multiple reflections.
Instead, we will construct the exponential form of these transformations. If we specify our translation with a Euclidean vector ⃗ t, then the corresponding translation versor is given by T = e−⃗ t ∞/2 = 1 −0.5⃗ t ∞.
As in Section 3.13 for PGA, you’ll need to review Taylor expansions to see the last step (where (⃗ t ∞)2 = 0 keeps the Taylor series finite).
In PGABLE, we construct a translation versor and apply it to the original point, line and sphere con-structed above: >> tv = 3e3; >> Ttv = gexp(-0.5tvni); >> Pt = TtvP/Ttv; >> Lt = -TtvL/Ttv; >> sphT = Ttvsph/Ttv; >> draw(Pt,'g'); draw(Lt,'g'); draw(sphT) Similarly, we can construct a versor for a rotation around a line L by an angle ϕ as e−L∗ϕ/2, where L∗= L/I5 is the dual of L. In PGABLE, we’ll construct a new point (since the point in our running example lies on our line), and a versor to rotated around L by π/6 radians. We’ll rotate our new point around L twice and align the view along L to better see the rotation: >> P2 = gapoint(0,1,0); draw(P2) >> Trot = gexp(-dual(L)pi/6/2); >> P2rot = TrotP2/Trot; draw(P2rot,'r'); >> P2rot = TrotP2rot/Trot; draw(P2rot,'b'); view([85,0]) We can also rotate sph and pi1 around L: 64 >> sphRot = Trotsph/Trot; draw(sphRot); >> pi1rot = Trotpi1/Trot; draw(pi1rot,'b'); view([85,0]) While the rotation of the sphere should be clear in the Matlab figure, the rotation of the plane is harder to see, although if you rotate the image to look down the intersection line of the two planes, you should see that the angle between the planes is π/6.
Exercises 1. Just as we used a plane as an operator for reflection, we can also use a sphere as an operator for “reflection” through the sphere. However, the operation we get when using a sphere as a versor isn’t reflection, but spherical inversion. To understand spherical inversion, it’s a little easier to start with circular inversion. Consider a unit circle centered at the origin. Points on the boundary of the circle are stationary under spherical inversion. Consider a ray ℓstarting on the origin, and a point P a distance r from the origin that lies on ℓ. Under spherical (circular) inversion, P gets mapped to the point on ℓthat is a distance 1/r from the origin.
Use PGABLE to visualize spherical (circular) inversion. Construct and draw a unit circle c centered at the origin that lies in the xy-plane. Construct the point P=gapoint(1/2,0,0), and map P through c: Ps=cP/c. Draw P, Ps, and the center of the circle, no. To see things better, use view([0 90]).
Repeat with points gapoint(1/3,0,0) and gapoint(1/4,0,0).
While we did this example for a unit circle centered at the origin, spherical inversion also works for circles/spheres of arbitrary radius centered at an arbitrary point.
2. We can apply spherical inversion to lines. Again, construct the unit circle c centered at the origin. Now construct the lines l=gapoint(0.5,0,0)^gapoint(0.5,1,0)^ni and map l through c: ls = cl/c.
Draw c, l, ls (you may wish to draw ls a different color than c). You’ll see that while l is a line, ls is a circle; the points on l that were inside of c were mapped to the points on ls that are outside of c, and the points on l outside of c were mapped to points on the circle ls that are inside of c.
Note that the line extends to infinity, or more precisely, to ∞. Given a sphere s, one way to find the center of s is to map ∞through s: c = s ∞/s.
3. Screw motions are the most general Euclidean motion. We can construct a screw motion as the compo-sition of a translation and a rotation. Construct two points P=gapoint(1,0,1) and Q=gapoint(2,0,1, and a line l=P^Q^ni. Now construct a translation T in the direction of l but scaled by a factor of 1/5 (e.g., in direction e1/5). Now construct a rotation R around l by ϕ = π/10. Our versor Sc = R ∗T.
To visualize the screw motion, draw l, no. Now compute Psc = Scno/Sc; draw(Psc), and then repeatedly call Psc = ScPsc/Sc; draw(Psc) (for ten or more times).
You should see the point translating and rotating around the line l, although you’ll likely want to rotate the figure to see it better.
4.3 Duality You likely noticed that objects in CGA have both a primal and dual form. Objects in the primal form are mostly constructed using the outer product of points, and in the dual form tend to be constructed from “pieces” (e.g., the center and radius of a sphere; the normal to and a point on a plane). You can convert between the primal and dual representations by dividing by the pseudo-scalar, I5, or by calling the function dual.
Table 6 gives various ways to construct objects in CGA.
65 Table 6: Construction primal and dual objects in CGA. P, Pi, C are points. ⃗ v is a Euclidean vector, and ˆ n is a unit length Euclidean vector. r is a positive scalar. For the sphere, C is the center of the sphere, and P is a point on the surface of the sphere. For the circle, ˆ n is perpendicular to the plane of the circle.
Primal Dual Point o + ⃗ v + 1 2 ⃗ v 2∞ Sphere P1 ∧P2 ∧P3 ∧P4 C −1 2 r2 ∞ P · (C ∧∞) Plane P1 ∧P2 ∧P3 ∧∞ −ˆ n −(P · ˆ n) ∞ Circle P1 ∧P2 ∧P3 (C −1 2r2∞) ∧ˆ n Line P1 ∧P2 ∧∞ 5 Is this all there is?
Our tutorial stops here. We have familiarized you with the basic operators that geometric algebra offers, and some of their geometrical significance. The intuition acquired should enable you to read some of the other texts which are available, and get you into the right mode of thinking in geometrical concepts.
There are some things we left out of the tutorial. Since they are rather elementary, you may run into them soon after trying other texts, so a few words may be appropriate.
• Addendum to A Tutorial for Plane-based Geometric Algebra is a follow-up document to this one, containing a few constructions (ratios of lines) that weren’t included in this tutorial; some material that was removed from the GABLE tutorial; and material on polynomial curves and surfaces [MZ25].
• The Conformal Model. Our treatment of the Conformal Model was very brief. In particular, there are other objects of interest in the conformal model, such as point pairs (which arise when intersecting a line with a sphere). For a more in-depth discussion of the conformal model, see, for example, [DMF07].
• signatures 3-dimensional space can be given different signatures, so that the unit vectors do not square to 1, but to another scalar such as −1. We have touched on this matter when discussion PGA (where e2 0 = 0), and CGA (where ∞2 = o2 = 0), but see any textbook on Geometric Algebra for a more in-depth discussion of signature. And a variety of software packages exist for using geometries with arbitrary signatures; for example, see SUGAR for a Matlab package that allows you to construct other geometries [VZDM24].
• inner products There are several inner products around; the one we have chosen is by no means standard, though it has the most straightforward geometric interpretation.
• higher dimensions At the moment, our implementation is only for 3-dimensional spaces. A variety of other packages exist that allow for higher dimensional geometric algebras. Again, SUGAR is a Matlab package that allows for geometries of arbitrary dimension and signature (although there are limits on how high of a dimension is practical in SUGAR): • differential geometry A major subject we have not treated is differential geometry, and geometric calculus. The objects in 66 geometric algebra can be differentiated, in new and powerful ways, and this is beginning to expand the scope of all fields using differential geometry. Matlab is less suited to show this flexibly.
Further reading There is a growing body of literature on geometric algebra. Unfortunately much of the more readable writing is not very accessible, being found in specialized books rather than general journals. We would recommend the following as natural follow-ups on this paper: • The material on PGA is based on [DdK], which goes deeper into PGA than we covered in this tutorial.
• Our book [DMF07], although while it covers OGA and CGA, it doesn’t cover PGA.
• The first part of the two part CG&A tutorial [DM02, MD02] is mostly covered by this tutorial; the second part goes into more advanced topics. Again, this CG&A tutorial doesn’t cover PGA.
• The introductory chapters of ‘New Foundations of Classical Mechanics [Hestnes00].
• An introductory course intended for physicists [DL99].
Read them in approximately this order.
You can also find a large amount of additional literature from the usual sources. However, one warning about the literature: some people make no distinction between ‘geometric algebra’ and ‘Clifford algebra’ so you may want to check both; but you will find that the literature on the latter contains little basic geometry (it tends to be about spinors). Your best keyword for initial search is ‘geometric algebra’. You will find that most of the literature in the field aims at an audience familiar with physics; yet the introductory chapters are often also readable to non-physicists.
6 Acknowledgements The OGA portion of this tutorial was jointly developed by Leo Dorst, Stephen Mann, and Tim Bouma [MDB99a, MDB99b]. The PGA portion of this tutorial was based on the paper of Leo Dorst and Steven DeKenick [DdK].
This work was funded in part by NSERC, The Natural Sciences and Engineering Research Council of Canada.
A Matlab details In this appendix, we discuss some Matlab implementation details that are relevant to using our geometric algebra package.
In many ways, Matlab eased the implementation of this geometric algebra. It was convenient to not have to write the matrix routines, etc. Also, Matlab objects and operator overloading nicely encapsulated many of the ideas we wanted to show while hiding many of the distracting details of the implementation.
On the other hand, using Matlab caused us to introduce several warts. in GABLE. Some of these are straight-forward, such as Matlab objects using standard arithmetic precedence for arithmetic on objects.
This means that ‘^’ has a higher precedence than ‘’, while for our system, we would want the two operators to have equal precedence. The standard arithmetic precedence may necessitate the use of parentheses at times.
Other problems are more complex and subtle. This appendix describes some of these problems, how we addressed them, and what problems the user of GABLE may need to work around. This appendix also discusses several features not mentioned in the tutorial that users may find useful.
67 A.1 Modes PGABLE has three modes: OGA, PGA, and CGA. As mentioned in the tutorial, to change OGA mode, type GA.model(OGA); to change to PGA mode, type GA.model(PGA); and to change to CGA mode, type GA.model(CGA). However, you can force the creation of an element relative to one geometry even if you’ve switched the mode to the other geometry. For example, >> GA.model(PGA) >> v = e1(OGA) >> draw(v) draws an arrow for v even though you’re in PGA mode. The example illustrates why you might want to do this: when in PGA mode, at times, you want to draw a vector (or bivector) as an arrow rooted at the origin (or disk through the origin) to check perpendicularity, etc.
You can also convert OGA elements into PGA elements, and you can convert PGA elements into OGA elements.
There are several ways to do this.
First, can convert an object from OGA to PGA without considering any geometric properties? E.g., converting e1 from OGA into the vector e1 in PGA. In this case, you can use PGAcast. For example, PGAcast(e1(OGA)) returns the vector e1 as a PGA object.
You can also convert the other way, i.e., from PGA to OGA. However, the question arises: what to do about any e0 components? The function OGAcast does this conversion, dropping any terms with a e0 component. So OGAcast(PGA(e0)+PGA(e1)) will return the vector e1 as an OGA object (with the e0 term being omitted).
However, you may wish to convert a geometric object in one algebra to a similar geometric object in the other algebra. Since all OGA objects are rooted at the origin, we provide a function geoPGA to convert an OGA vector to a PGA line, and to convert an OGA bivector to a PGA plane. We currently do not provide a mechanism to convert from PGA objects to OGA objects geometrically.
A.2 Objects at Infinity Displaying objects at infinity is problematic for several reasons. In particular, the location of an object at infinity depends on the location of the viewer. Additionally, objects at infinity behind the viewer would not be seen if these objects were displayed relative to the viewer.
What we chose to do to display objects at infinity was to make use of the Matlab “bounding box”. Matlab computes a bounding box for all the objects it draws, and rescales the scene so that all objects appear in a box that’s visible on the screen. We chose to draw our objects at infinity on the surface of this box as if the viewer was located at the center of the box. This choice has several implications. First, objects behind the viewer will now be visible on the surface of the box. And second, we have to do a 2-level rendering, first drawing objects not at infinity using Matlab in the usual way, then fixing the bounding box, and then drawing the objects at infinity on this box.
Such as approach requires us to keep a separate list of objects. You can access this list of objects with GAScene.displayitems(). Each object is given a number, as well as its GA representation. You can clear an individual object from the scene with GAScene.clearitems(2), where instead of “2” you would put the number of the object that you wish to delete.
Note that to clear the screen, you really should call pclf rather than clf. In addition to clearing the list of vanishing objects, pclf clears the list of stored items that are accessed with GAScene.displayitems() and GAScene.clearitems(); if you only have still objects, then calling clf will clear Matlab’s copy of the objects, but not the ones in GAScene.
See Appendix A.4 for more details on drawing and on details on how we handle objects at infinity.
A.3 Code Structure PGABLE is written using Matlab objects.
There is an abstract class, GA.m, that gets instantiated for particular algebras. Methods common to most algebras are included as function prototypes, giving a common 68 base set of functions to all algebras.
Currently there are three algebras, OGA.m, PGA.m, and CGA.m, all implementing “3D” Euclidean geometry, although the dimension of the base vector space of PGA and CGA are 4 and 5.
Some functions in GA.m are not part of some geometries, necessitating the use of a dummy function that just prints an error message and returns.
To implement a new algebra, one would copy the existing algebra most similar to the new algebra and update the functions to reflect this new algebra. The hardest part in this process is updating the various products, since products are computing by expanding one of the elements of the product into a square matrix, and multiplying the column matrix of the other element by this square matrix. Generating these matrices is non-trivial.
PGABLE has an epsilon that it uses when calls are made to zeroepsilons(); any coefficient that is less than this epsilon gets set to 0. When calling draw routines, the draw routing calls zeroepsilon() on the object being drawn to reduce the burden on the (novice) user. The default epsilon is 10−15; the value of epsilon may be queried and set by calling GA.epsilon tolerance(). In one place in CGA.m, a relative epsilon is used, scaling the GA epsilon by the norm of the object under consideration. Such an approach might be needed elsewhere in the code.
The default format of grade 2 and higher elements is to write them as an outer product of the basis elements. If you prefer to see e12 rather than e1^e2, then use the function compact notation to query and set whether compact notation is used. Similarly, if you prefer to see I4 rather than e0^e1^e2^e3, the function compact pseudoscalar() can query and set the pseudo-scalar notion. Regardless, both formats can always be read as input.
GAScene.m manages the still and dynamic objects (with dynamic objects being those at infinity), along with GASceneDynamicItem.m and GESceneStillItem.m. PGABLEDraw.m has most of the drawing routines; PGABLEDraw.m tries to use functions like getx() to allow the drawing routines to work on any geometry.
A.3.1 Basis Elements as Functions As with GABLE, we chose to implement the basis elements of the algebra as functions. This approach has its advantages and disadvantages. In particular, when the user type e1, we can check what the current geometry is, and return the appropriate “e1”. The functions for the basis elements are in the subfolder /elements.
A.4 Drawing Routines An attempt was made to keep all calls to Matlab drawing routines in PGABLEDraw.m. In PGABLEDraw.m are routines to draw arrows, polygons, lines, hairy lines, planes, spheres, points (as octahedrons), and other objects. If you wish to draw things yourself (rather than using the draw command), you can use the routines in PGABLEDraw.m. However, you should be aware of a few things to effectively use these routines.
First, at the start of your code (or at least, before the first call that draws things, whether it is to Matlab or to a PGABLEDraw.m routine), you should call GAScene.usefigure(). This ensures that your drawing appears on the correct Matlab display. Next, you may also want to call pclf to clear the screen.
Further, PGABLE keeps track of all objects drawn, both static (standard Matlab) and dynamic (objects at infinity). When Matlab draws an object, it returns a handle for that object. You can use this handle to adjust the properties of the object. PGABLE maintains a table of these handles and uses them to delete objects. If you do your own drawing, then you should make a list of these handles and add your object to PGABLE’s list of objects.
In particular, your code would look something like this: h = []; ht = plot3(...); h = [h ht]; ...
GAScene.addstillitem(GASceneStillItem(A,h)); 69 where A is the PGABLE object you’re drawing (e.g., A=e1^e2). Note that the draw routines in PGABLEDrawing.m also return these drawing handles. See the code for DrawBivector() for an example.
One other thing to be aware of is that many of the PGABLEDraw.m routines are PGA based, and you’ll need to convert some of the things you’re drawing to PGA in order to use these routines, even if you’re using OGA or CGA.
B Glossary This glossary is meant to convey the geometric flavor of terms, rather than their exact definitions – for those the reader is referred to the appropriate sections.
• Bivector – A directed area element, a 2-dimensional direction. Formed by the outer product of two independent vectors, it determines a plane through the origin. (Section 2.2.1) • Blade – a subspace through the origin; a k-blade is constructed as the outer product of k vectors.
(Section 2.2.6) • Dual – The orthogonal complement of a multivector (usually a blade); made through division by the unit volume. (Section 2.4.2) • Geometric product – The product to scale, rotate and orthogonalize multivectors; it provides a geo-metric operator. (Section 2.4) • Grade involution – take the k-blade A and multiply it by (−1)k; for multivectors, do this for each of its blades.
• Inner product – the product used when orthogonal complements are involved. (Section 2.3) • Inverse – The inverse under the geometric product; denoted by A−1 (as object) or by /A (as right-side operator). (Section 2.4.1) • Multivector – a sum of blades, the most general element of geometric algebra. (Section 2.2.6) • Outer product – The product used for spanning subspaces. (Section 2.2) • Projection – The component of a blade totally contained in another blade; or the operation producing this. (Section 2.6.1) • Pseudoscalar – The outer product of all the vectors in a basis for our vector space; the ‘volume-element.
of the space. (Section 2.2.2) • Rejection – The component of a blade totally outside another blade; or the operation producing this.
(Section 2.6.1) • Spinor – a product of vectors, to be used as the operator S in the spinor product S×S−1. (Section 2.7.2) • Trivector – A directed volume element. Formed by the outer product of three independent vectors.
(Section 2.2.2) • Vector – The basic 1-dimensional elements of a linear space spanning a geometric space. A basis for this vector space generates the linear space of the geometric algebra of the geometric space.
• Wedge product – Another name for the outer product. (Section 2.2) 70 References [DLG96] C Doran, A Lasenby, and S Gull. Linear algebra. In WE Bulls, editor, Clifford (Geometric) Algebras with Applications in Physics, Mathematics and Engineering, chapter 6. Birkhauser, 1996.
[DL99] Chris Doran and Anthony Lasenby. Physical applications of geometric algebra. Available at clifford/ptIIIcourse/, 1999.
[DdK] Leo Dorst and Steven De Keninck. A Guided Tour to the Plane-Based Geometric Algebra PGA.
Available at
[DM02] Leo Dorst and Stephen Mann. Geometric algebra: a computation framework for geometrical appli-cations: Part i. Computer Graphics and Applications, 22(3):24–31, May/June 2002.
[DMF07] L. Dorst, S. Mann, and D. Fontijne. Geometric Algebra for Computer Science. Morgan-Kaufmann, 2007.
[Hestenes91] D. Hestenes, The design of linear algebra and geometry, Acta Applicandae Mathematicae 23: 65-93, 1991.
[Hestnes00] D. Hestenes, New foundations for classical mechanics, 2nd edition, D. Reidel, Dordrecht, 2000.
[LF98] J. Lasenby, W. J. Fitzgerald, C. J. L. Doran and A. N. Lasenby. New Geometric Methods for Computer Vision, Int. J. Comp. Vision 36(3), p. 191-213 (1998).
[LHR01] Li, H., Hestenes, D., Rockwood, A. Generalized Homogeneous Coordinates for Computational Ge-ometry. In: Sommer, G. (eds) Geometric Computing with Clifford Algebras. Springer, Berlin, Heidel-berg. 2, 2001.
[MD02] Stephen Mann and Leo Dorst. Geometric algebra: a computation framework for geometrical appli-cations: Part ii.
Computer Graphics and Applications, 22(4):58–67, July/August 2002.
[MDB99a] Leo Dorst, Stephen Mann, and Tim Bouma. GABLE: A matlab tutorial for geometric algebra.
Available at 1999.
[MDB99b] Stephen Mann, Leo Dorst, and Tim Bouma. The making of a geometric algebra package in Matlab.
Technical Report CS-99-27, University of Waterloo, December 1999. Available as https: //cs.uwaterloo.ca/research/tr/1999/27/CS-99-27.pdf.
[MDB01] Stephen Mann, Leo Dorst, and Tim Bouma. The making of GABLE, a geometric algebra learning environment in Matlab.
In E. Bayro-Corrochano and G. Sobczyk, editors, Geometric Algebra with Applications in Science and Engineering, pages 491–511. Birkhauser, 2001.
[MZ25] Stephen Mann and Zachary Leger. Addendum to A Tutorial for Plane-based Geometric Algebra online 2025.
[VZDM24] Manel Velasco, Isiah Zaplana, Arnau D´ oria-Cerezo, Pau Mart´ ı. Symbolic and User-friendly Ge-ometric Algebra Routines (SUGAR) for Computations in Matlab. arXiv preprint abs/2403.16634 71 |
8897 | https://sites.ualberta.ca/~gingrich/courses/phys395/notes/node119.html | Binary, Octal and Hexadecimal Numbers
Next:Number RepresentationUp:Number SystemsPrevious:Number Systems
Binary, Octal and Hexadecimal Numbers
Consider a decimal number with digits a b c. We can write abc as
Similarly, in the binary system a number with digits a b c can be written as
Each digit is known as a bit and can take on only two values: 0 or 1. The left most bit is the highest-order bit and represents the most significant bit (MSB), while the lowest-order bit is the least significant bit (LSB).
Conversion from binary to decimal can be done using a set of rules, but it is much easier to use a calculator or tables (table7.1).
Table 7.1:Decimal, binary, hexadecimal and octal equivalents.
The eight octal numbers are represented with the symbols , while the 16 hexadecimal numbers use .
In the octal system a number with digits a b c can be written as
while one in the hexadecimal system is written as
A binary number is converted to octal by grouping the bits in groups of three, and converted to hexadecimal by grouping the bits in groups of four. Octal to hexadecimal conversion, or visa versa, is most easily performed by first converting to binary.
Example: Convert the binary number 1001 1110 to hexadecimal and to decimal.
_> _Example: Convert the octal number to hexadecimal.
_
_> \_Example: Convert the number 146 to binary by repeated subtraction of the largest power of 2 contained in the remaining number.
\_
\_> \_Example: Devise a method similar to that used in the previous problem and convert 785 to hexadecimal by subtracting powers of 16.
\_\_
_Doug Gingrich
Tue Jul 13 16:55:15 EDT 1999_ |
8898 | https://chess.stackexchange.com/questions/32043/wrongly-placed-chessboard-what-happens | Skip to main content
Wrongly placed chessboard - what happens?
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What happens if the chessboard is kept with black square on right hand side instead of white square? I think that the notations will change.
In the new positon, the White king will be positioned on a white square and the Black king will be positioned on a black square. The White queen will be positioned on a black square and the Black queen will be positioned on a white square.
Why this is considered a wrong placing of the chess board? Addtionally, does White or Black have an advantage, and will strategies and tactics change?
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edited Aug 31, 2020 at 18:26
Rewan Demontay
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asked Aug 31, 2020 at 17:35
Prashant AkerkarPrashant Akerkar
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This plays out like any other chess game, provided black goes first. Everything you've done to the game equates to swapping the colors. Let black go first and it's the same game as always. Otherwise it's a different game.
candied_orange
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candied_orange
09/02/2020 01:59:38
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Sep 2, 2020 at 1:59
Game play won't change for computers at all. But for human, it might have psychological effects. A french player might trade the light bishop off carelessly.
jf328
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jf328
09/02/2020 08:31:50
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Sep 2, 2020 at 8:31
What do you mean "what happens?" Nothing happens. This isn't allowed by the rules of chess, so it would never happen. Bit of a strange question...
user91988
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user91988
09/02/2020 18:23:44
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@candied_orange Why would black going first result in it playing like other chess games?
Acccumulation
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Acccumulation
09/02/2020 20:42:08
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Sep 2, 2020 at 20:42
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@Acccumulation because the queen goes on her own color. Make her go on the other color AND play the other colors turn and you get the same game as always. Just with the colors swapped.
candied_orange
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candied_orange
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It's wrong because the rules say so:
2.1 The chessboard is composed of an 8 x 8 grid of 64 equal squares alternately light (the ‘white’ squares) and dark (the ‘black’ squares).
The chessboard is placed between the players in such a way that the near corner square to the right of the player is white.
Of course, those rules are based on what has been common practice for a very long time. Apparently, an incorrectly placed board happens often enough to warrant a rule for what to do when that happens:
7.2.2 If during a game it is found that the chessboard has been placed contrary to Article 2.1, the game shall continue but the position reached must be transferred to a correctly placed chessboard.
Since the pieces have the same position relative to each other, it doesn't affect strategy and tactics. Notation won't change; the column of the starting positions of the queens is still column 'd'. If you have a board with numbers/letters, it's even easier to spot the problem.
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answered Aug 31, 2020 at 18:13
Glorfindel♦Glorfindel
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If White still makes the first move, there's no difference at all. That's why rule 7.2.2 says "hey, you put the board wrong, but you can still continue with the game, since it doesn't influence the gameplay".
Glorfindel
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Glorfindel
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08/31/2020 18:21:24
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Aug 31, 2020 at 18:21
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Thank you. To summarise, this chessboard location is not going to change the tactics or strategies of both the players neither the notations.
Prashant Akerkar
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Prashant Akerkar
08/31/2020 18:36:38
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Aug 31, 2020 at 18:36
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@PrashantAkerkar Correct. The colour of the tiles is completely irrelevant to the game. If you decide to paint the white tiles red and the black ones green, the movements of the pieces don't change. You could even make all tiles the same colour, assuming you have marked the borders of each tile so you can still find the tiles. The colours are an aid, but for informal games you can play on anything vaguely resembling a chess board. It just won't be a regulatory board. As long as it has 8x8 tiles, you can play (informal) chess on it.
Mast
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Mast
09/01/2020 05:47:01
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Sep 1, 2020 at 5:47
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@Mast I'm now thinking of a way to combine a 9x9 go game with chess on the same board. Years ago, me and some friends already developed a variation on Bughouse where one board started playing chess and the other one draughts, so you could drop draughtsmen your teammate captured on the chessboard and vice versa.
Glorfindel
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Glorfindel
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09/01/2020 05:53:13
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Sep 1, 2020 at 5:53
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@Glorfindel : Randall's done it (go+chess): xkcd.com/1287
Eric Towers
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Eric Towers
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I don't think there's anything in the game that's strictly tied to the colors of the squares. You can still number the ranks and files from the bottom-left corner and e.g. Bishops would start on opposite colors than usual, but the diagonal moves still work the same.
Though, if you set the board up using the mnemonic that the queens are placed on squares matching their color, i.e. white queen on a white square, black queen on a black square, then the setup would be different and the game along with it. But not by much, what you'd end up is this:
That's just the regular starting position flipped. Since all moves are also symmetrical with regard to left and right, it still technically makes no difference other than that now you'd have to number the squares starting from the lower right to match the notation of a regular chess game. (Counting from lower right, the kings here are at e1 and e8 as in a normal game.)
(Castling is the one move which works differently to the left than to the right, but it's defined by the sides where the king/queen start at, and anyway basically reduces down to "king takes two steps and the rook jumps to the square the king passed" which is direction-independent.)
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edited Sep 2, 2020 at 7:56
answered Sep 1, 2020 at 14:26
ilkkachuilkkachu
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I don't think it's true to say "it makes no difference". It's theoretically the same game, but flipped. But in practice, both players will have their play strongly affected by where they expect pieces to be, patterns learnt etc. Like, by habit protecting F2 more than C2, even though in this variant, that's the wrong way around.
Steve Bennett
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Steve Bennett
09/01/2020 23:24:26
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@SteveBennett I think it's unlikely that players with enough experience to have those habits could possibly make the mistake of putting their kings on the wrong side, no matter what the mnemonic says.
Misha Lavrov
–
Misha Lavrov
09/02/2020 00:22:54
Commented
Sep 2, 2020 at 0:22
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Yeah, me too. Or more likely, as has happened with me OTB, the K and Q are flipped, you realise after a few moves, and flip them back. No one ever continues with "oh I guess my K is on D1 now" :)
Steve Bennett
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Steve Bennett
09/02/2020 03:05:53
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Sep 2, 2020 at 3:05
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@SteveBennett, yes, you're right; edited to say "technically no difference". But I also don't think it's very likely to happen except with beginners.
ilkkachu
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ilkkachu
09/02/2020 08:07:53
Commented
Sep 2, 2020 at 8:07
We had one where only one side's king was placed incorrectly.
Joshua
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Joshua
09/02/2020 18:20:57
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Sep 2, 2020 at 18:20
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It makes no difference at all in any way, even if the kings and queens are reversed left to right, or if Black goes first. It's just convention, the thought process, how to play, etc. It's identical other than the callout above about psychologically. Which bishop means more in the French is the example cited, but that's just because that player has that "rule" memorized The reasoning is no different, however.
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8899 | https://web.stanford.edu/group/ctr/ResBriefs/2024/19_Cabrera-Booman.pdf | Center for Turbulence Research Annual Research Briefs 2024 High-Reynolds-number and low-Mach-number wind tunnel at Stanford University By F. Cabrera-Booman, J. M. O. Massey AND B. J. McKeon This study presents the recommissioning and modernization of the High Pressure Wind Tunnel at Stanford University, originally constructed by Prof. Eaton’s research group (DeGraaff 1999). The facility generates high-Reynolds-number flows with Reθ ≤31, 000 while maintaining a low Mach number by pressurizing air up to 8 atmospheres. At these Reynolds numbers, viscous scales reduce to as small as 3.5 µm, necessitating the develop-ment of careful measurement techniques. This brief introduces the issues being addressed by this device, which are part of the SAPPHIRE (Shear-stress And Propagating Pressure measurements in HIgh-Reynolds-number Experiments) project and can be summarized as the study of low-wave-number, or subconvective, wall pressure and shear-stress fluc-tuations in high-Reynolds-number turbulent boundary layers.
1. Introduction Due to the limitations of current computational power, direct numerical simulation of Navier–Stokes equations is not feasible for turbulent boundary layers at high-Reynolds-numbers. As a result, turbulence models are necessary for such calculations and are developed to match experimental results. However, large-scale experimental facilities are costly, and accurately measuring high-Reynolds-number flows presents significant chal-lenges (DeGraaf & Eaton 2000, Marusic et al. 2010, Bodenschatz et al. 2014). Conse-quently, most available data are from low-Reynolds-number flows. Since many engineering applications operate at much higher Reynolds numbers, understanding how turbulence behaves at high-Reynolds-numbers in situations closer to those in applications is neces-sary.
Three variables need to be tuned to experimentally generate high-Reynolds-number turbulence: freestream velocity U∞, development region L, and fluid kinematic viscosity ν, which relate to Reynolds number as Re = U∞L/ν. It is inconvenient and expensive to build a real-size wind tunnel to test, say, an airplane. However, if a reduced-size model is used, the freestream velocity would need to be increased by the same factor, which in most cases would make the flow supersonic. The final parameter to vary is the kinematic viscosity ν: Samuels (1993) and Donnelly et al. (1990) have proposed using superfluids as the working fluid for a wind tunnel. In those scenarios, the kinematic viscosity is 2 orders of magnitude lower than that of air; therefore Reynolds number would increase by a factor of 100, in the case of helium above the lambda point (i.e., temperature at which helium becomes a superfluid). However, such a wind tunnel would be very costly and difficult to operate as, for instance, the liquid helium needs to be maintained at approximately 4 K. The present facility achieves high-Reynolds-numbers by pressurizing the air that is used as the working fluid to increase its density, and, in turn, increase Reynolds number while keeping a low Mach number.
The High-Pressure Wind Tunnel at Stanford University, originally constructed by Prof.
Eaton’s research group (DeGraaff 1999), consists of a laboratory-sized wind tunnel placed 203 Cabrera-Booman, Massey, McKeon.
Reθ (1430, 31000) ν/uτ (3.5, 58) µm U max ∞ 19 m/s δ 35 mm pmax 8 bar νmin 1.9 × 10−6 m2/s Table 1. Relevant parameters of the setup and turbulent boundary layer: Reynolds number based on momentum thickness Reθ = U∞θ/ν; the viscous length scale ν/uτ estimated by a log-law fit (De Graaff & Eaton (2000)); the freestream velocity measured at the center of the test section U∞; the boundary layer thickness δ; the maximum air pressure inside the vessel pmax; and the lowest kinematic viscosity, achieved at pmax, νmin.
inside a pressure vessel. The device achieves high-Reynolds-numbers using pressurization of up to 8 times atmospheric pressure (with associated density increase), and a factor of 3 increase in the freestream velocity, thus permitting an increase in the Reynolds number by a factor of approximately 24 for a given model. This setup allows for the generation of high-Reynolds-number turbulent boundary layers with parameters in the ranges presented in Table 1, which were obtained for the first time by Prof. Eaton’s research group under similar conditions (DeGraaff 1999, DeGraaff & Eaton 2000).
The High-Pressure Wind Tunnel is currently being used for the SAPPHIRE project (Cabrera-Booman et al. 2024) which aims to elucidate the physics driving low-wave-number unsteady wall pressure and wall shear-stress fluctuations in high-Reynolds-number, low-Mach-number turbulent boundary layer flows. The project is in collaboration with Princeton University and the University of Melbourne with the goal of constructing a data set of high-Reynolds-number measurements of wall pressure and shear-stress across multiple facilities (McKeon et al. 2014; Marusic et al. 2010) to mitigate the effects of sensor noise and spatial resolution. In parallel, another goal is to evaluate resolvent anal-ysis as a tool to predict the effect of very-large-scale motions (VLSMs) on pressure and shear-stress fluctuations.
This brief is organized as follows: First, experimental details on the High Pressure Wind Tunnel are detailed in Section 2. The diagnostic techniques used to characterize the turbulent boundary layer and to measure wall pressure and shear-stress are pre-sented in Section 3. Detailed information about the goals of the SAPPHIRE project are explained in Section 4. Finally, a summary of the High Pressure Wind Tunnel capabilities, diagnostics, and its use in the SAPPHIRE project is presented in Section 5.
2. High-Pressure-Wind-Tunnel 2.1. Wind tunnel The wind tunnel structure inside the pressure vessel is composed of plexiglass, wood, and galvanized steel. The dimensions of the plexiglass working section are 3.0 × 0.15 × 0.71 m. The test section is a clear 7/16-inch-thick plexiglass removable plate at the lower wall with an area of 0.917 × 0.308 m2 and holes of 3/4 inch in diameter to insert plugs that house shear-stress sensors and microphones. The wind is produced by an Eaton Eddy current drive motor (Model AS-272-504-01, 25 HP, 0-1690 RPM), which is controlled via an Eaton Model 4000 controller. Additional components include a furnace motor starter (Model ESP100) and a centrifugal blower (size 3, class III, 18-inch-diameter wheel) from Pacific Fan and Blower.
204 High-Reynolds-Number Wind Tunnel at Stanford University Figure 1. Airflow path. Taken from DeGraaff (1999).
Figure 1, taken from DeGraaff (1999), shows the airflow, which begins at a rectangular inlet (762 × 711 mm2) on the lower section of the vessel door (right bell-shaped end of the vessel), then passes through heat exchangers and a flow contraction. The flow moves through a flexible duct around the motor and then enters the blower. A Hexcel honeycomb (3/16-inch cell size, 75-mm length) helps align the flow as it exits the duct. After passing through the blower, the flow turns 180◦and passes through a series of conditioning devices: three grids (50% blockage), two additional Hexcel honeycombs (3/16-inch cell size, 75-mm length), and three Howard Wire Cloth screens (34 mesh, 0.165-mm wire).
The flow is then funneled through a contraction that uses a fifth-order polynomial design to reduce the height from 762 mm to 152 mm, achieving a 5:1 contraction ratio while maintaining width, which allows for an uniform flow. See DeGraaff (1999) for more details.
At the contraction exit, a boundary layer trip strip 1.6 mm tall and 6.4 mm wide is positioned 150 mm downstream, followed by a development section about 920 times the height of the trip or 9.8 times the channel height. The tunnel walls have V-tape (1-mm-height “V” shapes produced with Dymo tape) along the sides and ceiling located 120 mm downstream. A straight-walled diffuser at the end expands the height from 152 mm to 279 mm over a length of 762 mm. Afterward, the flow passes through two heat transfer coils in the tank door before cycling back to the inlet.
Both the motor and blower are mounted on a steel frame supported by four Gimbal Piston isolators (Technical Manufacturing Corp, Model 64-25765-01) filled with pressur-ized air from the building supply, which minimizes vibration transfer between the motor and the wind tunnel.
2.2. Pressure vessel & pressurization The pressure vessel where the wind tunnel is installed is 8 feet in diameter and 19 feet in length with a wall thickness of 1/2 inch (fabricated by Melco Steel in Azusa, California).
The tank can be accessed by a Harris quick opening door at one end. The vessel is rated for a maximum working pressure of 120 psig at 38◦C. To reach ambient pressure of 8 atmospheres, the original compressor used by Prof. Eaton’s research group (see DeGraaff 1999) has been upgraded to an Ingersoll Rand compressor (Model SSR-XF-400), which pressurizes the tank. Air fed to the compressor is conditioned via an Ingersoll Rand 205 Cabrera-Booman, Massey, McKeon.
dryer (Model DA4000NVCA4HG000) to maintain a constant temperature and humidity below 30%. The rate of pressurization and depressurization is slower than 500 Pa/s to protect electronic components inside the pressure vessel. A Thermo Fisher distilled-water chiller (Model TF5000) was installed, alongside the corresponding valve control system, to circulate cold water through two heat exchangers placed inside the pressure vessel with the goal of maintaining a constant temperature against the heat caused by the pressurization procedure and the motor.
The pressure vessel is opaque and the door needs to be closed to operate the wind tunnel and close the airflow loop. This requires the experiments to be performed and supervised remotely, which was achieved by the installation of a monitoring camera and high-pressure-rated plugs to allow cables to pass through the vessel.
2.3. Improvements to the original version Several elements from the original design (DeGraaff 1999) were upgraded or replaced.
Inside the pressure vessel, the motor, blower, piston isolators, and contractions were recycled from the original experiment. The plexiglass working and test sections were redesigned and rebuilt. Moreover, the heat exchangers were replaced, a new monitoring camera was installed, and plugs were reconditioned. Outside the pressure vessel: the air compressor was upgraded, and the dryer, chiller, and auxiliary piping were fully replaced.
All of the diagnostic techniques and tools described in the next section were developed entirely from the ground up.
3. Diagnostic Techniques 3.1. Boundary layer characterization To measure the turbulent boundary layer, a pitot tube and hot wire are attached to a sting that penetrates the test section from below and through a linear bearing. The sting is then displaced away from the wall by an XSlide positioning device (Velmex), which has a precision of 2.5 µm. This is needed to characterize output parameters of the turbulent boundary layer, such as Reynolds number and viscous length scales, for different input parameters, e.g., wind speed and air pressure. In particular, an estimation of the friction velocity (uτ) is central. This velocity can be estimated from a mean velocity profile. In principle, if measurements near the wall can be obtained, the friction velocity can be obtained from the linear relation for the mean velocity profile in the viscous sublayer (y+ ≤5). This is rarely technically possible, much less so in the current setup where the viscous length scale is between 3.5 and 58 µm (see Table 1), which would require mean velocity measurements at distances from the wall y ≤17.5 µm, for the highest-Reynolds-number case. Note that regular hot wire diameters range from 1 to 5 µm (Dantec Dynamics). Therefore, in the present case where only measurements further from the wall can be accessed, the Clauser method (Clauser (1954)) can be used. This method is based on the idea that there is an overlap region of the mean velocity profile, which follows a logarithmic law. More sophisticated versions of such a fit are available (e.g., see Rodriguez-L´ opez 2015).
3.2. Wall pressure and shear-stress The mean and fluctuating wall shear-stress are measured using DirectShear sensors from IC2, which can be placed directly in the removable plate that makes the test section.
To measure wall pressure fluctuations, an infrasonic microphone (Hottinger Br¨ uel & 206 High-Reynolds-Number Wind Tunnel at Stanford University 27 (a) (b) ϕ L h Microphone Plug Figure 2. (a) A CAD of the microphone inside a plug with a pinhole and (b) a photograph of the microphone inside one of the plugs.
Kjær 4964) is used alongside a preamplifier (Hottinger Br¨ uel & Kjær 2669). This setup can measure low frequencies of 0.7 Hz up to 20 kHz (±3 dB) with a dynamic range of (1.1 × 10−4, 4.0 × 102) Pa and a sensitivity of 50 mV/Pa. The microphone has been calibrated using a constant-frequency calibrator (Hottinger Br¨ uel & Kjær 4231), which produces a sine wave at 1 kHz with an accuracy of ±0.2 dB.
Contrary to the shear-stress sensor, which can be placed directly on the wall, a pinhole and a resonating cavity are often needed to acquire wall pressure measurements with a microphone (Gibeau & Ghaemi 2021). This is for the following reasons: (i) The pinhole helps create a localized, more accurate measurement of the wall pressure at specific points. This is particularly important for the measurement of multipoint correlations.
(ii) The pinhole acts as a low-pass filter, reducing higher-frequency noise or vibrations from the environment. Figure 2(a) presents a CAD of the microphone inside a plug with a pinhole, and Figure 2(b) shows a photograph of the microphone inside one of the plugs. The microphone is fixed to the plug by a thread, which is originally intended to fix a protective casing for the diaphragm (where the setup is resting in Figure 2(b)).
The plugs were 3D printed by the Additive Manufacturing and Prototyping Facility at Stanford University using a Carbon L1 3D printer.
The dimensions of the cavity and pinhole, ϕ, L, and h, as depicted in Figure 2, de-termine what frequencies are resolved. The work of Shaw (1960) proposes that twice the length of the pinhole, 2 × L, must be larger than its diameter, ϕ. Moreover, to avoid attenuation of the high frequencies, the maximum pinhole diameter must respect 12 < ϕ+ < 18, as discussed by Gravante et al. (1998). Finally, Tsuji et al. (2007) indi-cated that the pinhole diameter needs to be small enough compared to the flow length scale ϕ+ < 20. Building on these studies, the following plug dimensions are needed in this experimental setup: ϕlowest Re ≤1.2 mm and ϕhighest Re ≤71 µm; Llowest Re = 4.4 mm and Lhighest Re = 262 µm. The cavity height is h ≥190 µm and is determined by the height of the lip around the microphone diaphragm, which can be seen in Figure 2(a). The resonant frequency of this Helmholtz resonator can then be found as: fres = c 2π q A L×V , with A the cross-sectional area of the pinhole, V the volume of the cavity, and c the speed of sound.
207 Cabrera-Booman, Massey, McKeon.
Figure 3. The region of interest is between the convective ridge and the acoustic ridge. The acoustic ridge is O(104) below the convective ridge. Figure adapted from Liu et al. (2024).
4. The SAPPHIRE project The SAPPHIRE project focuses on the subconvective region, which lies between the convective and acoustic peaks (highlighted in Figure 3). This region is of particular inter-est because low-wave-number fluctuations have been shown to excite structural modes and significantly contribute to noise generation in wall-bounded flows (Lighthill 1952, 1954). These fluctuations are critical for practical applications such as reducing skin-friction drag on vehicles and mitigating flow-induced vibrations that affect structural stability.
As Reynolds numbers increase, the nature of turbulence changes. One defining char-acteristic of this change is the separation between inner and outer scales (Klewicki et al.
2008). Hutchins & Marusic (2007) demonstrated that large-scale structures maintain a near-wall footprint and actively modulate near-wall turbulence production. Rather than merely imposing a low-wave-number shift, these large-scale motions redistribute fine-scale turbulent energy across the boundary layer. This interaction has significant implications for wall pressure and wall shear-stress fluctuations, which remain understudied at high-Reynolds-numbers.
However, studying wall quantities at high-Reynolds-numbers presents substantial chal-lenges due to limitations in spatial resolution arising from the small size of viscous scales.
These restrictions are exacerbated by sensor design constraints and poor signal-to-noise ratios in experimental measurements (Abtahi et al. 2024).
4.1. Key objectives The following objectives guide this research: (a) Characterizing Large-Scale Effects on Subconvective Pressure and shear-stress: The project will focus on understanding how large-scale turbulent structures, particu-larly VLSMs, influence subconvective pressure and wall shear-stress at high-Reynolds-numbers. VLSMs dominate at high-Reynolds-numbers, with streamwise lengths up to 10 times the boundary layer thickness. These structures overlap with the subconvective wave number range, which is crucial for understanding flow-induced noise and vibrations.
(b) Nonlinear Interactions Driving Wall Fluctuations: We aim to identify the nonlin-208 High-Reynolds-Number Wind Tunnel at Stanford University ear interactions responsible for driving wall pressure and shear-stress fluctuations. This involves isolating the spatial locations and spatiotemporal scales that contribute to these nonlinear excitations.
(c) Impact of Surface Treatments: The study will investigate how changes in sur-face conditions—such as roughness, spatial inhomogeneity, and bumps—affect low-wave-number contributions to wall pressure and shear-stress. These surface modifications are expected to alter the energy distribution across different wave numbers, particularly in the subconvective range.
(d) Evaluation of Predictive Models: By integrating experimental data with resolvent analysis the project aims to evaluate predictive models for wall-bounded turbulence at high-Reynolds-numbers. These models will be compared against ground truth measure-ments obtained from experiments conducted in unique facilities.
5. Summary In this brief, we outline the recommissioning and modernization of the High Pres-sure Wind Tunnel at Stanford University, originally constructed by Prof. Eaton’s re-search group (DeGraaff 1999), which is now poised to contribute to a multi facility cam-paign. The updates made to the facility ensure its capability to generate high-Reynolds-number flows while maintaining a low Mach number, a critical feature for studying turbu-lent boundary layers in certain conditions. In particular, this device produces turbulent boundary layers with Reθ ≤31, 000 while maintaining a Mach number below 0.02 by pressurizing air up to 8 bars. At these Reynolds numbers, the viscous scales reduce to ν/uτ ≈3.5 µm, which presents a measurement challenge. The diagnostic and measure-ment techniques implemented in the setup are delineated, especially regarding the direct measurement of wall pressure and wall shear-stress.
The High Pressure Wind Tunnel is currently being used in the context of the SAP-PHIRE project. This campaign is aimed at capturing the physics of low-wave-number, or subconvective, wall pressure and shear-stress fluctuations in high-Reynolds-number turbulent boundary layers. The goal is to advance both experimental techniques and theoretical modeling approaches for studying wall-bounded turbulence at high-Reynolds-numbers by providing data in flow regimes currently limited by spatial resolution and low sensor signal-to-noise ratio, and by investigating and modeling the nonlinear interactions driving these fluctuating quantities.
Acknowledgments The support of DARPA under award number HR0011-24-9-0465 is gratefully acknowl-edged. The authors acknowledge the help from Dr. MA Saccone in the 3D printing or the microphone plugs.
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