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# 8.4 Magnetic field strength: force on a moving charge in a magnetic
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• Describe the effects of magnetic fields on moving charges.
• Use the right hand rule 1 to determine the velocity of a charge, the direction of the magnetic field, and the direction of the magnetic force on a moving charge.
• Calculate the magnetic force on a moving charge.
What is the mechanism by which one magnet exerts a force on another? The answer is related to the fact that all magnetism is caused by current, the flow of charge. Magnetic fields exert forces on moving charges , and so they exert forces on other magnets, all of which have moving charges.
## Magnetic force on a moving charge
The magnetic force on a moving charge is one of the most fundamental known. Magnetic force is as important as the electrostatic or Coulomb force. Yet the magnetic force is more complex, in both the number of factors that affects it and in its direction, than the relatively simple Coulomb force.
The magnitude of the magnetic force $F$ on a charge depends on: the quantity of charge $q$ , its speed $v$ , the strength of magnetic field $B$ , and the direction of motion relative to the magnetic field's direction . Motion, and its direction, are critical.
The maximum force occurs when the direction of motion and the magnetic field's direction are perpendicular to one another (i.e. ninety degree angle between directions).
$\mathbf{v}\perp \mathbf{B}$ In that situation, the magnitude of the magnetic force is
$F=qvB$
The minimum force occurs when the direction of motion and the magnetic field's direction are parallel to one another (i.e. zero or 180 degree angle between directions). $\mathbf{v}\parallel \mathbf{B}$ In that situation, the magnitude of the magnetic force is
$F=0$
We define the magnetic field strength $B$ in terms of the force on a charged particle moving in a magnetic field. The SI unit for magnetic field strength $B$ is called the tesla (T) after the eccentric but brilliant inventor Nikola Tesla (1856–1943). To determine how the tesla relates to other SI units, we solve for the magnetic field strength.
$B=\frac{F}{qv}$
So, the tesla is
$\text{1 T}=\frac{\text{1 N}}{C\cdot \text{m/s}}=\frac{\text{1 N}}{A\cdot m}$
(note that C/s = A).
Another smaller unit, called the gauss (G), where $1 G={\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}T$ , is sometimes used. The strongest permanent magnets have fields near 2 T; superconducting electromagnets may attain 10 T or more. The Earth’s magnetic field on its surface is only about $5×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}T$ , or 0.5 G.
## Making connections: charges and magnets
There is no magnetic force on static charges. However, there is a magnetic force on moving charges. When charges are stationary, their electric fields do not affect magnets. But, when charges move, they produce magnetic fields that exert forces on other magnets. When there is relative motion, a connection between electric and magnetic fields emerges—each affects the other.
## Direction of force: right hand rule 1
The direction of the magnetic force $\mathbf{\text{F}}$ is perpendicular to the plane formed by $\mathbf{\text{v}}$ and $\mathbf{\text{B}}$ , as determined by the right hand rule 1 (or RHR-1), which is illustrated in [link] . RHR-1 states that, to determine the direction of the magnetic force on a positive moving charge, you point the thumb of the right hand in the direction of $\mathbf{\text{v}}$ , the fingers in the direction of $\mathbf{\text{B}}$ , and a perpendicular to the palm points in the direction of $\mathbf{\text{F}}$ . One way to remember this is that there is one velocity, and so the thumb represents it. There are many field lines, and so the fingers represent them. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge.
## Section summary
• The maximum force a magnetic field can exert on a moving charge is $F=qvB$
• The SI unit for magnetic field strength $B$ is the tesla (T), which is related to other units by
$1 T=\frac{\text{1 N}}{C\cdot \text{m/s}}=\frac{\text{1 N}}{A\cdot m}.$
• The direction of the force on a moving charge is given by right hand rule 1 (RHR-1): Point the thumb of the right hand in the direction of $v$ , the fingers in the direction of $B$ , and a perpendicular to the palm points in the direction of $F$ .
• The force is perpendicular to the plane formed by $\mathbf{\text{v}}$ and $\mathbf{\text{B}}$ . Since the force is zero if $\mathbf{\text{v}}$ is parallel to $\mathbf{\text{B}}$ , charged particles often follow magnetic field lines rather than cross them.
## Conceptual questions
If a charged particle moves in a straight line through some region of space, can you say that the magnetic field in that region is necessarily zero?
## Problems&Exercises
What is the direction of the magnetic force on a positive charge that moves as shown in each of the six cases shown in [link] ? Note that $\odot$ indicates "coming out of the page" and $\otimes$ means "going into the page."
(a) Left (West)
(b) Into the page
(c) Up (North)
(d) No force
(e) Right (East)
(f) Down (South)
Repeat [link] for a negative charge.
What is the direction of the velocity of a negative charge that experiences the magnetic force shown in each of the three cases in [link] , assuming it moves perpendicular to $\mathbf{\text{B}}?$ Note that $\odot$ indicates "coming out of the page" and $\otimes$ means "going into the page."
(a) East (right)
(b) Into page
(c) South (down)
Repeat [link] for a positive charge.
What is the direction of the magnetic field that produces the magnetic force on a positive charge as shown in each of the three cases in the figure below, assuming $\mathbf{\text{B}}$ is perpendicular to $\mathbf{\text{v}}$ ? Note that $\otimes$ means "going into the page."
(a) Into page
(b) West (left)
(c) Out of page
Repeat [link] for a negative charge.
What is the maximum force on an aluminum rod with a $0\text{.}\text{100}\text{-μC}$ charge that you pass between the poles of a 1.50-T permanent magnet at a speed of 5.00 m/s? In what direction is the force?
$7\text{.}\text{50}×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{N}$ perpendicular to both the magnetic field lines and the velocity
(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a $0\text{.}\text{500}\text{-μC}$ charge and flies due west at a speed of 660 m/s over the Earth’s south magnetic pole, where the $8\text{.}\text{00}×{\text{10}}^{-5}\text{-T}$ magnetic field points straight up. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.
how environment affect demand and supply of commodity ?
Wht at the criteria for market ?
Amos
what is difference between monitory policy and fiscal policy?
monetary policy is a policy thrust by National Govt(CBN) to influence government spending, purchase &taxes
Frank
necessity of economics
I will say want,choice,opportunity cost,scarcity,scale of preference
Alao
what is monopoly market.How price output are determined under monopoly market
bisham
b) Monopoly market is an impecfect market where s single firm having the innovation to produce a particular commodity.Prices are determined through output since there are no other competitive.
Frank
Monopoly market:firm has market power & does not respond to market price
Frank
Explain the process of price determination under perfect competition market with suitable diagram
Price determination under perfect competition via this process :firms have no market power to influence price rather firms respond to market price.
Frank
price is different from demand- demand is amount of commodity
demand is amount /quantity of commodity a potential buyer is willing to buy at a given price at market
Frank
demand is a desire of customer on commodity with the ability to pay it and willing to buy it at given price of commodity
Harika
demand is price of what
show that shortrun average cost
what is economics
what is money
Mbah
what is money
Mbah
Difine macro economics
agaba
money is a medium of exchange between goods and services,maybe inform of currency.
Wesonga
Economics is study of how human beings strive to satisfy numerous wants using limited available resources.
Wesonga
how do you find the maximum number of workers the firms should employ order to produce where there are increasing returns
Jane
what are implications of computing national income?.
agaba
pl
MUDASIRU
what is the formulae for calculating national income
MUDASIRU
it calculated by value added method
Praveen
classify the production units like agriculture, banking, transport etc
Praveen
money is anything that is generally acceptetable for human
Ogbaji
Estimate the net value added(NVA) at fixed cost by each industrial structure
Praveen
definition of unemployment
what are the causes of unemployment?
The main causes of unemployment are listed below. 1. Frictional unemployment 2. Cyclical unemployment 3. Structural unemployment
assani
We can also categorize the causes on a broader sense as: 1. Political and 2. Social cause As unemployeement root causes are embaded in this two.
Yonathan
would opportunity cost exist if there was no scarcity?
assani
yes just because the opportunity cost arose when there is Alternative to choose among the alternatives.
I am thinking that, if our resources were unlimited, then there wouldn't be any need to forgo some wants. Hence the inexistence if opportunity cost
assani
Politics
Job
politics has done what?
assani
consider time assani
Mary
I'm Emmanuel,...I taught the main cause is the change in gov't.
Emmanuel
...Lack of capital to set up a firm respectively
Emmanuel
🙈
Emmanuel
I would like to bring in Educational levels can also be the cause the cause of the problem respectively
Emmanuel
I think the main causes of unemployment is lack of INFRASTRUCTURAL DEVELOPMENT OVER POPULATION OVER DEPENDENT ON GOVERNMENT LACK OF SELF EMPOWERMENT...
ananti
lack of skills among the new generation is the serious issue.
Vishal
Where I come from , I don't see why education or personal aspects seem to do with unimployment, technically the motivation and eigerness in all works of live is there , dispite the cultural influence and physical bearriors;the thing we lacking is Government Support and open market ethics.
Joe
sorry about that-(repation). We have a over powering ethical political system that's displacing the marketing asspects of economy and causing large scale unemployment right across the board...
Joe
can someone Explain Expansionary Monetary Policy and Contractionary Monetary Policy Using one of the instrument of Monetary Policy? Please am kinda lost here?. ta
using a graph show the case of substitute and compliment goods
can anyone give me a simple explanation to Five Sector Macroeconomics?
Emmanuel
Can someone please define what economics is
economics simply is a social science subject that study human behavior.
dajan
economics is a social science which studies human behaviour as a relationship between ends and scarce means that has alternative uses
Alao
Can someone please tell me how to calculate GDP
Emmanuel
emmanual kapal to calculate GDP (Gross Domestic Product) has three method in calculating it (1)income approach (2) expenditure approach (3) value added method
Alao
thanks Alae
Emmanuel
u are welcome
Alao
in basic terms economics is revered to as battery system, it date back to when Men sees the need to exchange sapless goods and produce to gain , either wealth , basic necessities or to establish trading ties for personal benefit or social asspects in terms of coexistence and continuity, future .
Joe
what is the law of demand
keep other thing constant, when the price increases demand decrease when the price decreases demand increases of the commodity.
sj
all things being equal,quantity demanded decrease as price increase and increase as price decrease
Seth
there's practial joke to it ..." the higher the demand ; scarcity, increase in production and drop in quality"... quite the controversy - for example China vs Europe, United States and we are all boxed up in between somewhere...
Joe
Other thing remain constant the low price of commodity the high quantity of commodity and vice versa is true
Baraka
Explain Effective demand
What is effective demand
Anita
like Modi is in demand...best example of effective demand
Pranav
Don't get you
Anita
Anita you mean you don't get me or who?
Onyeking
level of demand that represents a real intention to purchase by people with the means to pay
Pranav
Difference between extinct and extici spicies
While the American heart association suggests that meditation might be used in conjunction with more traditional treatments as a way to manage hypertension
in a comparison of the stages of meiosis to the stage of mitosis, which stages are unique to meiosis and which stages have the same event in botg meiosis and mitosis
Researchers demonstrated that the hippocampus functions in memory processing by creating lesions in the hippocampi of rats, which resulted in ________.
The formulation of new memories is sometimes called ________, and the process of bringing up old memories is called ________.
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##### Motivating Questions
• How can we use a Riemann sum to estimate the area between a given curve and the horizontal axis over a particular interval?
• What are the differences among left, right, middle, and random Riemann sums?
• How can we write Riemann sums in an abbreviated form?
In Section 4.1, we learned that if we have a moving object with velocity function $v\text{,}$ whenever $v(t)$ is positive, the area between $y = v(t)$ and the $t$-axis over a given time interval tells us the distance traveled by the object over that time period; in addition, if $v(t)$ is sometimes negative and we view the area of any region below the $t$-axis as having an associated negative sign, then the sum of these signed areas over a given interval tells us the moving object's change in position over the time interval.
For instance, for the velocity function given in Figure 4.2.1, if the areas of shaded regions are $A_1\text{,}$ $A_2\text{,}$ and $A_3$ as labeled, then the total distance $D$ traveled by the moving object on $[a,b]$ is
\begin{equation*} D = A_1 + A_2 + A_3, \end{equation*}
while the total change in the object's position on $[a,b]$ is
\begin{equation*} s(b) - s(a) = A_1 - A_2 + A_3. \end{equation*}
Because the motion is in the negative direction on the interval where $v(t) \lt 0\text{,}$ we subtract $A_2$ when determining the object's total change in position.
Of course, finding $D$ and $s(b)-s(a)$ for the situation given in Figure 4.2.1 presumes that we can actually find the areas represented by $A_1\text{,}$ $A_2\text{,}$ and $A_3\text{.}$ In most of our work in Section 4.1, such as in Activities 4.1.3 and Activity 4.1.4, we worked with velocity functions that were either constant or linear, so that by finding the areas of rectangles and triangles, we could find the area bounded by the velocity function and the horizontal axis exactly. But when the curve that bounds a region is not one for which we have a known formula for area, we are unable to find this area exactly. Indeed, this is one of our biggest goals in Chapter 4: to learn how to find the exact area bounded between a curve and the horizontal axis for as many different types of functions as possible.
To begin, we expand on the ideas in Activity 4.1.2, where we encountered a nonlinear velocity function and approximated the area under the curve using four and eight rectangles, respectively. In the following preview activity, we focus on three different options for deciding how to find the heights of the rectangles we will use.
##### Preview Activity4.2.1
A person walking along a straight path has her velocity in miles per hour at time $t$ given by the function $v(t) = 0.25t^3-1.5t^2+3t+0.25\text{,}$ for times in the interval $0 \le t \le 2\text{.}$ The graph of this function is also given in each of the three diagrams in Figure 4.2.2.
Note that in each diagram, we use four rectangles to estimate the area under $y = v(t)$ on the interval $[0,2]\text{,}$ but the method by which the four rectangles' respective heights are decided varies among the three individual graphs.
1. How are the heights of rectangles in the left-most diagram being chosen? Explain, and hence determine the value of
\begin{equation*} S = A_1 + A_2 + A_3 + A_4 \end{equation*}
by evaluating the function $y = v(t)$ at appropriately chosen values and observing the width of each rectangle. Note, for example, that
\begin{equation*} A_3 = v(1) \cdot \frac{1}{2} = 2 \cdot \frac{1}{2} = 1. \end{equation*}
2. Explain how the heights of rectangles are being chosen in the middle diagram and find the value of
\begin{equation*} T = B_1 + B_2 + B_3 + B_4. \end{equation*}
3. Likewise, determine the pattern of how heights of rectangles are chosen in the right-most diagram and determine
\begin{equation*} U = C_1 + C_2 + C_3 + C_4. \end{equation*}
4. Of the estimates $S\text{,}$ $T\text{,}$ and $U\text{,}$ which do you think is the best approximation of $D\text{,}$ the total distance the person traveled on $[0,2]\text{?}$ Why?
# Subsection4.2.1Sigma Notation
It is apparent from several different problems we have considered that sums of areas of rectangles is one of the main ways to approximate the area under a curve over a given interval. Intuitively, we expect that using a larger number of thinner rectangles will provide a way to improve the estimates we are computing. As such, we anticipate dealing with sums with a large number of terms. To do so, we introduce the use of so-called sigma notation, named for the Greek letter $\Sigma\text{,}$ which is the capital letter $S$ in the Greek alphabet.
For example, say we are interested in the sum
\begin{equation*} 1 + 2 + 3 + \cdots + 100, \end{equation*}
which is the sum of the first 100 natural numbers. Sigma notation provides a shorthand notation that recognizes the general pattern in the terms of the sum. It is equivalent to write
\begin{equation*} \sum_{k=1}^{100} k = 1 + 2 + 3 + \cdots + 100. \end{equation*}
We read the symbol $\sum_{k=1}^{100} k$ as “the sum from $k$ equals 1 to 100 of $k\text{.}$” The variable $k$ is usually called the index of summation, and the letter that is used for this variable is immaterial. Each sum in sigma notation involves a function of the index; for example,
\begin{equation*} \sum_{k=1}^{10} (k^2 + 2k) = (1^2 + 2\cdot 1) + (2^2 + 2\cdot 2) + (3^2 + 2\cdot 3) + \cdots + (10^2 + 2\cdot 10), \end{equation*}
and more generally,
\begin{equation*} \sum_{k=1}^n f(k) = f(1) + f(2) + \cdots + f(n). \end{equation*}
Sigma notation allows us the flexibility to easily vary the function being used to track the pattern in the sum, as well as to adjust the number of terms in the sum simply by changing the value of $n\text{.}$ We test our understanding of this new notation in the following activity.
##### Activity4.2.2
For each sum written in sigma notation, write the sum long-hand and evaluate the sum to find its value. For each sum written in expanded form, write the sum in sigma notation.
1. $\sum_{k=1}^{5} (k^2 + 2)$
2. $\sum_{i=3}^{6} (2i-1)$
3. $3 + 7 + 11 + 15 + \cdots + 27$
4. $4 + 8 + 16 + 32 + \cdots + 256$
5. $\sum_{i=1}^{6} \frac{1}{2^i}$
# Subsection4.2.2Riemann Sums
When a moving body has a positive velocity function $y = v(t)$ on a given interval $[a,b]\text{,}$ we know that the area under the curve over the interval is the total distance the body travels on $[a,b]\text{.}$ While this is the fundamental motivating force behind our interest in the area bounded by a function, we are also interested more generally in being able to find the exact area bounded by $y = f(x)$ on an interval $[a,b]\text{,}$ regardless of the meaning or context of the function $f\text{.}$ For now, we continue to focus on determining an accurate estimate of this area through the use of a sum of the areas of rectangles, doing so in the setting where $f(x) \ge 0$ on $[a,b]\text{.}$ Throughout, unless otherwise indicated, we also assume that $f$ is continuous on $[a,b]\text{.}$
The first choice we make in any such approximation is the number of rectangles.
If we say that the total number of rectangles is $n\text{,}$ and we desire $n$ rectangles of equal width to subdivide the interval $[a,b]\text{,}$ then each rectangle must have width $\Delta x = \frac{b-a}{n}\text{.}$ We observe further that $x_1 = x_0 + \Delta x\text{,}$ $x_2 = x_0 + 2 \Delta x\text{,}$ and thus in general $x_{i} = a + i\Delta x,$ as pictured in Figure 4.2.3.
We use each subinterval $[x_i, x_{i+1}]$ as the base of a rectangle, and next must choose how to decide the height of the rectangle that will be used to approximate the area under $y = f(x)$ on the subinterval. There are three standard choices: use the left endpoint of each subinterval, the right endpoint of each subinterval, or the midpoint of each. These are precisely the options encountered in Preview Activity 4.2.1 and seen in Figure 4.2.2. We next explore how these choices can be reflected in sigma notation.
If we now consider an arbitrary positive function $f$ on $[a,b]$ with the interval subdivided as shown in Figure 4.2.3, and choose to use left endpoints, then on each interval of the form $[x_{i}, x_{i+1}]\text{,}$ the area of the rectangle formed is given by
\begin{equation*} A_{i+1} = f(x_i) \cdot \Delta x, \end{equation*}
as seen in Figure 4.2.4.
If we let $L_n$ denote the sum of the areas of rectangles whose heights are given by the function value at each respective left endpoint, then we see that
\begin{align*} L_n =\mathstrut \amp A_1 + A_2 + \cdots + A_{i+1} + \cdots + A_n\\ =\mathstrut \amp f(x_0) \cdot \Delta x + f(x_1) \cdot \Delta x + \cdots + f(x_i) \cdot \Delta x + \cdots + f(x_{n-1}) \cdot \Delta x. \end{align*}
In the more compact sigma notation, we have
\begin{equation*} L_n = \sum_{i = 0}^{n-1} f(x_i) \Delta x. \end{equation*}
Note particularly that since the index of summation begins at $0$ and ends at $n-1\text{,}$ there are indeed $n$ terms in this sum. We call $L_n$ the left Riemann sum for the function $f$ on the interval $[a,b]\text{.}$
There are now two fundamental issues to explore: the number of rectangles we choose to use and the selection of the pattern by which we identify the height of each rectangle. It is best to explore these choices dynamically, and the applet 1 found at http://gvsu.edu/s/a9 is a particularly useful one. There we see the image shown in Figure 4.2.5, but with the opportunity to adjust the slider bars for the left endpoint and the number of subintervals.
By moving the sliders, we can see how the heights of the rectangles change as we consider left endpoints, midpoints, and right endpoints, as well as the impact that a larger number of narrower rectangles has on the approximation of the exact area bounded by the function and the horizontal axis.
To see how the Riemann sums for right endpoints and midpoints are constructed, we consider Figure 4.2.6.
For the sum with right endpoints, we see that the area of the rectangle on an arbitrary interval $[x_i, x_{i+1}]$ is given by $B_{i+1} = f(x_{i+1}) \cdot \Delta x,$ so that the sum of all such areas of rectangles is given by
\begin{align*} R_n =\mathstrut \amp B_1 + B_2 + \cdots + B_{i+1} + \cdots + B_n\\ =\mathstrut \amp f(x_1) \cdot \Delta x + f(x_2) \cdot \Delta x + \cdots + f(x_{i+1}) \cdot \Delta x + \cdots + f(x_{n}) \cdot \Delta x\\ =\mathstrut \amp \sum_{i=1}^{n} f(x_i) \Delta x. \end{align*}
We call $R_n$ the right Riemann sum for the function $f$ on the interval $[a,b]\text{.}$ For the sum that uses midpoints, we introduce the notation
\begin{equation*} \overline{x}_{i+1} = \frac{x_{i} + x_{i+1}}{2} \end{equation*}
so that $\overline{x}_{i+1}$ is the midpoint of the interval $[x_i, x_{i+1}]\text{.}$ For instance, for the rectangle with area $C_1$ in Figure 4.2.6, we now have
\begin{equation*} C_1 = f(\overline{x}_1) \cdot \Delta x. \end{equation*}
Hence, the sum of all the areas of rectangles that use midpoints is
\begin{align*} M_n =\mathstrut \amp C_1 + C_2 + \cdots + C_{i+1} + \cdots + C_n\\ =\mathstrut \amp f(\overline{x_1}) \cdot \Delta x + f(\overline{x_2}) \cdot \Delta x + \cdots + f(\overline{x}_{i+1}) \cdot \Delta x + \cdots + f(\overline{x}_{n}) \cdot \Delta x\\ =\mathstrut \amp \sum_{i=1}^{n} f(\overline{x}_i) \Delta x, \end{align*}
and we say that $M_n$ is the middle Riemann sum for $f$ on $[a,b]\text{.}$
When $f(x) \ge 0$ on $[a,b]\text{,}$ each of the Riemann sums $L_n\text{,}$ $R_n\text{,}$ and $M_n$ provides an estimate of the area under the curve $y = f(x)$ over the interval $[a,b]\text{;}$ momentarily, we will discuss the meaning of Riemann sums in the setting when $f$ is sometimes negative. We also recall that in the context of a nonnegative velocity function $y = v(t)\text{,}$ the corresponding Riemann sums are approximating the distance traveled on $[a,b]$ by the moving object with velocity function $v\text{.}$
There is a more general way to think of Riemann sums, and that is to not restrict the choice of where the function is evaluated to determine the respective rectangle heights. That is, rather than saying we'll always choose left endpoints, or always choose midpoints, we simply say that a point $x_{i+1}^*$ will be selected at random in the interval $[x_i, x_{i+1}]$ (so that $x_i \le x_{i+1}^* \le x_{i+1}$), which makes the Riemann sum given by
\begin{equation*} f(x_1^*) \cdot \Delta x + f(x_2^*) \cdot \Delta x + \cdots + f(x_{i+1}^*) \cdot \Delta x + \cdots + f(x_n^*) \cdot \Delta x = \sum_{i=1}^{n} f(x_i^*) \Delta x. \end{equation*}
At http://gvsu.edu/s/a9, the applet noted earlier and referenced in Figure 4.2.5, by unchecking the “relative” box at the top left, and instead checking “random,” we can easily explore the effect of using random point locations in subintervals on a given Riemann sum. In computational practice, we most often use $L_n\text{,}$ $R_n\text{,}$ or $M_n\text{,}$ while the random Riemann sum is useful in theoretical discussions. In the following activity, we investigate several different Riemann sums for a particular velocity function.
##### Activity4.2.3
Suppose that an object moving along a straight line path has its velocity in feet per second at time $t$ in seconds given by $v(t) = \frac{2}{9}(t-3)^2 + 2\text{.}$
1. Carefully sketch the region whose exact area will tell you the value of the distance the object traveled on the time interval $2 \le t \le 5\text{.}$
2. Estimate the distance traveled on $[2,5]$ by computing $L_4\text{,}$ $R_4\text{,}$ and $M_4\text{.}$
3. Does averaging $L_4$ and $R_4$ result in the same value as $M_4\text{?}$ If not, what do you think the average of $L_4$ and $R_4$ measures?
4. For this question, think about an arbitrary function $f\text{,}$ rather than the particular function $v$ given above. If $f$ is positive and increasing on $[a,b]\text{,}$ will $L_n$ over-estimate or under-estimate the exact area under $f$ on $[a,b]\text{?}$ Will $R_n$ over- or under-estimate the exact area under $f$ on $[a,b]\text{?}$ Explain.
# Subsection4.2.3When the function is sometimes negative
For a Riemann sum such as
\begin{equation*} L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x, \end{equation*}
we can of course compute the sum even when $f$ takes on negative values. We know that when $f$ is positive on $[a,b]\text{,}$ the corresponding left Riemann sum $L_n$ estimates the area bounded by $f$ and the horizontal axis over the interval.
For a function such as the one pictured in Figure 4.2.7, where in the first figure a left Riemann sum is being taken with 12 subintervals over $[a,d]\text{,}$ we observe that the function is negative on the interval $b \le x \le c\text{,}$ and so for the four left endpoints that fall in $[b,c]\text{,}$ the terms $f(x_i) \Delta x$ have negative function values. This means that those four terms in the Riemann sum produce an estimate of the opposite of the area bounded by $y = f(x)$ and the $x$-axis on $[b,c]\text{.}$
In Figure 4.2.7, we also see evidence that by increasing the number of rectangles used in a Riemann sum, it appears that the approximation of the area (or the opposite of the area) bounded by a curve appears to improve. For instance, in the middle graph, we use 24 left rectangles, and from the shaded areas, it appears that we have decreased the error from the approximation that uses 12. When we proceed to Section 4.3, we will discuss the natural idea of letting the number of rectangles in the sum increase without bound.
For now, it is most important for us to observe that, in general, any Riemann sum of a continuous function $f$ on an interval $[a,b]$ approximates the difference between the area that lies above the horizontal axis on $[a,b]$ and under $f$ and the area that lies below the horizontal axis on $[a,b]$ and above $f\text{.}$ In the notation of Figure 4.2.7, we may say that
\begin{equation*} L_{24} \approx A_1 - A_2 + A_3, \end{equation*}
where $L_{24}$ is the left Riemann sum using 24 subintervals shown in the middle graph, and $A_1$ and $A_3$ are the areas of the regions where $f$ is positive on the interval of interest, while $A_2$ is the area of the region where $f$ is negative. We will also call the quantity $A_1 - A_2 + A_3$ the net signed area bounded by $f$ over the interval $[a,d]\text{,}$ where by the phrase “signed area” we indicate that we are attaching a minus sign to the areas of regions that fall below the horizontal axis.
Finally, we recall from the introduction to this present section that in the context where the function $f$ represents the velocity of a moving object, the total sum of the areas bounded by the curve tells us the total distance traveled over the relevant time interval, while the total net signed area bounded by the curve computes the object's change in position on the interval.
##### Activity4.2.4
Suppose that an object moving along a straight line path has its velocity $v$ (in feet per second) at time $t$ (in seconds) given by
\begin{equation*} v(t) = \frac{1}{2}t^2 - 3t + \frac{7}{2}. \end{equation*}
1. Compute $M_5\text{,}$ the middle Riemann sum, for $v$ on the time interval $[1,5]\text{.}$ Be sure to clearly identify the value of $\Delta t$ as well as the locations of $t_0\text{,}$ $t_1\text{,}$ $\cdots\text{,}$ $t_5\text{.}$ In addition, provide a careful sketch of the function and the corresponding rectangles that are being used in the sum.
2. Building on your work in (a), estimate the total change in position of the object on the interval $[1,5]\text{.}$
3. Building on your work in (a) and (b), estimate the total distance traveled by the object on $[1,5]\text{.}$
4. Use appropriate computing technology 2 to compute $M_{10}$ and $M_{20}\text{.}$ What exact value do you think the middle sum eventually approaches as $n$ increases without bound? What does that number represent in the physical context of the overall problem?
# Subsection4.2.4Summary
• A Riemann sum is simply a sum of products of the form $f(x_i^*) \Delta x$ that estimates the area between a positive function and the horizontal axis over a given interval. If the function is sometimes negative on the interval, the Riemann sum estimates the difference between the areas that lie above the horizontal axis and those that lie below the axis.
• The three most common types of Riemann sums are left, right, and middle sums, plus we can also work with a more general, random Riemann sum. The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed in the sum. For a left Riemann sum, we evaluate the function at the left endpoint of each subinterval, while for right and middle sums, we use right endpoints and midpoints, respectively.
• The left, right, and middle Riemann sums are denoted $L_n\text{,}$ $R_n\text{,}$ and $M_n\text{,}$ with formulas
\begin{align*} L_n = f(x_0) \Delta x + f(x_1) \Delta x + \cdots + f(x_{n-1}) \Delta x \amp= \sum_{i = 0}^{n-1} f(x_i) \Delta x,\\ R_n = f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(x_i) \Delta x,\\ M_n = f(\overline{x}_1) \Delta x + f(\overline{x}_2) \Delta x + \cdots + f(\overline{x}_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(\overline{x}_i) \Delta x, \end{align*}
where $x_0 = a\text{,}$ $x_i = a + i\Delta x\text{,}$ and $x_n = b\text{,}$ using $\Delta x = \frac{b-a}{n}\text{.}$ For the midpoint sum, $\overline{x}_{i} = (x_{i-1} + x_i)/2\text{.}$
##### 4
Consider the function $f(x) = 3x + 4\text{.}$
1. Compute $M_4$ for $y=f(x)$ on the interval $[2,5]\text{.}$ Be sure to clearly identify the value of $\Delta x\text{,}$ as well as the locations of $x_0, x_1, \ldots, x_4\text{.}$ Include a careful sketch of the function and the corresponding rectangles being used in the sum.
2. Use a familiar geometric formula to determine the exact value of the area of the region bounded by $y = f(x)$ and the $x$-axis on $[2,5]\text{.}$
3. Explain why the values you computed in (a) and (b) turn out to be the same. Will this be true if we use a number different than $n = 4$ and compute $M_n\text{?}$ Will $L_4$ or $R_4$ have the same value as the exact area of the region found in (b)?
4. Describe the collection of functions $g$ for which it will always be the case that $M_n\text{,}$ regardless of the value of $n\text{,}$ gives the exact net signed area bounded between the function $g$ and the $x$-axis on the interval $[a,b]\text{.}$
##### 5
Let $S$ be the sum given by
\begin{equation*} S = ((1.4)^2 + 1) \cdot 0.4 + ((1.8)^2 + 1) \cdot 0.4 + ((2.2)^2 + 1) \cdot 0.4 + ((2.6)^2 + 1) \cdot 0.4 +((3.0)^2 + 1) \cdot 0.4. \end{equation*}
1. Assume that $S$ is a right Riemann sum. For what function $f$ and what interval $[a,b]$ is $S$ an approximation of the area under $f$ and above the $x$-axis on $[a,b]\text{?}$ Why?
2. How does your answer to (a) change if $S$ is a left Riemann sum? a middle Riemann sum?
3. Suppose that $S$ really is a right Riemann sum. What is geometric quantity does $S$ approximate?
4. Use sigma notation to write a new sum $R$ that is the right Riemann sum for the same function, but that uses twice as many subintervals as $S\text{.}$
##### 6
A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table.
1. Plot the given data on a set of axes with time on the horizontal axis and the velocity on the vertical axis.
2. Estimate the total distance traveled during the car the time brakes using a middle Riemann sum with 3 subintervals.
3. Estimate the total distance traveled on $[0,1.8]$ by computing $L_6\text{,}$ $R_6\text{,}$ and $\frac{1}{2}(L_6 + R_6)\text{.}$
4. Assuming that $v(t)$ is always decreasing on $[0,1.8]\text{,}$ what is the maximum possible distance the car traveled before it stopped? Why?
##### 7
The rate at which pollution escapes a scrubbing process at a manufacturing plant increases over time as filters and other technologies become less effective. For this particular example, assume that the rate of pollution (in tons per week) is given by the function $r$ that is pictured in Figure 4.2.9.
1. Use the graph to estimate the value of $M_4$ on the interval $[0,4]\text{.}$
2. What is the meaning of $M_4$ in terms of the pollution discharged by the plant?
3. Suppose that $r(t) = 0.5 e^{0.5t}\text{.}$ Use this formula for $r$ to compute $L_5$ on $[0,4]\text{.}$
4. Determine an upper bound on the total amount of pollution that can escape the plant during the pictured four week time period that is accurate within an error of at most one ton of pollution. | HuggingFaceTB/finemath | |
Fifth graders add a new marker to the Calendar Grid for each day of the month. Students delve into important mathematics as they identify patterns that emerge in the markers over time. Each month also features a collection of data that students gather by taking repeated measurements, conducting surveys, and performing experiments. The patterns on the Calendar Grid and the growing collection of data provide starting points for discussions, problem solving, and short written exercises.
Every month, fifth graders have multiple opportunities to develop their problem-solving skills while tackling challenging story problems and discussing how they solved those problems. Students also have regular opportunities to play games and do exercises that develop their facility with sophisticated and efficient computation strategies. A few days each month, students solve a problem string (a carefully arranged sequence of related problems designed to elicit specific computation strategies and insights about number relationships).
## Content Outline
### August / September
• Fractions & Decimals
• Layer a Day
• Multiple Game
• Solving Problems Using Multiples & Factors
### October
• Mystery Buildings: Views & Volumes
• Carrot Graphing Experiment
• Group it!
• Solving Problems with Organized Lists
• Fraction Addition with Money & Clock Models
### November
• Tumbling Triangles
• A Meter a Day
• Expression Bingo
• Using Logical Reasoning to Solve Problems
• Fraction Subtraction with Money & Clocks
### December
• Student Heights & Foot Lengths
• Put It on the Line, Part 1
• Problems That Suggest Making an Informed Start
• Multiplication & Division
### January
• Numerical Patterns & Graphs
• Time & Money
• Color Ten
• Volume Problem
• More Multiplication & Division Strings
### February
• Using the Area Model to Multiply Fractions
• Two Liters or Spill
• I Have, Who Has?
• Conversion Problems
• Multiplying Whole Numbers by Fractions
### March
• Multiplication with Decimal Numbers
• Line Plots & Length
• Quotient Bingo
• Student-Posed Problems
### April
• Growing Cubes Constructions
• Collecting Quarters
• Put It on the Line Decimals
• More Student-Posed Problems
• Fraction Multiplication & Division | HuggingFaceTB/finemath | |
Level 498 Level 500
Level 499
## Ignore words
Check the boxes below to ignore/unignore words, then click save at the bottom. Ignored words will never appear in any learning session.
Ignore?
The definition of a conic
Graphs that are created by slicing a cone
4
The number of conics
Ancient Greeks
Who invented the conics
Parabola, circle, ellipse, hyperbola
The names of the conics
x^2 = 4py
The equation of a vertical parabola with vertex at (0,0)
y^2 = 4px
The equation of a horizontal parabola with vertex at (0,0)
What "p" represents in a parabola
The distance from the vertex to the focus AND the distance from the vertex to the directrix
The definition of a parabola
The set of points equidistant from a fixed point (the focus) and a fixed line (the directrix)
down
If a < 0, the parabola opens
Right
a positive exponent means the original decimal needs to move to the _______________________
(0, 4)
The focus in this parabola: x^2 = 16y
x = 1
The directrix in this parabola: y^2 = -4x
p = 3/2
The value of "p" in this parabola: x^2 = -6x
The definition of a circle
The set of points equidistant from a fixed point
x^2 + y^2 = r^2
The equation of a circle centered at (0,0)
The radius of this circle: x^2 + y^2 = 81
A tangent line is always perpendicular to the radius at the point of tangency
The slope of the tangent line to x^2 + y^2 = 25 at (4, 5)
The definition of an ellipse
The set of points such that the sum of the focal radii is a constant.
x^2/b^2 + y^2/a^2 = 1
The equation for a vertical ellipse centered at (0,0)
x^2/a^2 + y^2/b^2 = 1
The equation for a horizontal ellipse centered at (0,0)
The distance from center to vertex
The meaning of "a" in an ellipse
The meaning of "b" in an ellipse
The distance from center to endpoint (covertices)
The distance from center to focus
The meaning of "c" in an ellipse
a^2 - b^2 = c^2
The formula relating a, b, and c in an ellipse
How you decide whether an ellipse is vertical or horizontal looking at its equation
Find the largest denominator (which is a^2). If it is under the x, it's horizontal. If it's under the y, it's vertical.
Vertices are (0, 6) and (0, -6)
The vertices of x^2/9 + y^2/36 = 1
Endpoints are (0, 2) and (0, -2)
Endpoints (covertices) of x^2/25 + y^2/4 = 1
c = square root of 33
The c value for x^2/16 + y^2/49 = 1
The definition of a hyperbola
The set of points such that the absolute value of the difference of the focal radii is a constant.
y^2/a^2 - x^2/b^2 = 1
The equation of a vertical hyperbola centered at (0, 0)
x^2/a^2 - y^2/b^2 = 1
The equation of a horizontal hyperbola centered at (0, 0)
The way to determine whether a hyperbola is vertical or horizontal by looking at its equation
Look at what variable is first (positive). If x is first, it's horizontal. If y is first, it's vertical.
The distance from center to vertex
The meaning of "a" in a hyperbola
The distance from center to endpoints
The meaning of "b" in a hyperbola
The distance from center to focus
The meaning of "c" in a hyperbola
a^2 + b^2 = c^2
The equation relating a, b, and c in a hyperbola
m = (plus or minus) a/b
The formula for the slopes of the asymptotes in a vertical hyperbola
m = (plus or minus) b/a
The formula for the slopes of the asymptotes in a horizontal hyperbola
Vertical (y is first)
Is this hyperbola vertical or horizontal? y^2/4 - x^2/16 = 1
Vertices are (5, 0) and (-5, 0)
What are the vertices in this hyperbola? x^2/25 - y^2/9 = 1
What are the slopes of the asymptotes? x^2/36 - y^2/100 = 1
This is a horizontal hyperbola so we use b/a which is 10/6 or 5/3 (+ and -)
Where are the foci in this hyperbola? y^2/36 - x^2/64 = 1
First, a^2 + b^2 = 100, so c = 10. Since this is a vertical hyperbola (y is first), the foci are (0, 10) and (0, -10)
Circle
r= asin(theta)
Parabola
the graph of a quadratic function
Ellipse
Hyperbola
Equation subtracts x²/a and y²/b=1; x and y coefficients have different signs.
Line
A straight path that goes without end in two directions.
horizontal ellipse equation
(x-h)2/a2 + (y-k)2/b2 = 1
h+-a, k
horizontal ellipse & hyperbola vertices
h+-c, k
horizontal ellipse & hyperbola foci
c2=a2-b2
ellipse a2 b2 c2 equation
vertical ellipse equation
(x-h)2/b2 + (y-k)2/a2 = 1
h, k+-a
vertical ellipse & hyperbola vertices
h, k+-c
vertical ellipse & hyperbola foci
horizontal hyperbola equation
(x-h)2/a2 - (y-k)2/b2 = 1
y-k=+-b/a(x-h)
horizontal hyperbola asymptotes
c2=a2+b2
hyperbola a2 b2 c2 equation
vertical hyperbola equation
(y-k)2/a2 - (x-h)2/b2= 1
y-k=+-a/b(x-h)
vertical hyperbola asymptotes
Positive
Counting from the decimal point to the right makes the exponent ___________________________________________
bigger
ellipse a2 always...
circle equation
(x-h)2 + (y-k)2 = r2
(x-h)2 = 4p(y-k)
vertical parabola equation
h, k+p
vertical parabola focus
y=k-p
vertical parabola directrix
(y-k)2 = 4p(x-h)
horizontal parabola equation
h+p, k
horizontal parabola focus
x=h-p
horizontal parabola directrix
Ax²+By²+Cx+Dy+E=0
General equation for ellipse | HuggingFaceTB/finemath | |
# Write a number using place value
This page includes Number Worksheets such as counting charts, representing, comparing and ordering numbers worksheets, and worksheets on expanded form, written numbers, scientific numbers, Roman numerals, factors, exponents, and binary numbers. There are literally hundreds of number worksheets meant to help students develop their understanding of numeration and number sense. In the first few sections, there are some general use printables that can be used in a variety of situations. Hundred charts, for example, can be used for counting, but they can just as easily be used for learning decimal hundredths. Operations With Decimals and Powers of Ten: Overview In mathematics the digits 0 to 9 and place-value position are used to represent numbers. The place a digit occupies in a written number gives its value in the number.
This is a review of material learned in Grade 5, except for the inclusion of the decimal places ten thousandths, hundred thousandths, and millionths. See the place-value chart below for a visual representation of these whole-number and decimal positions.
The large whole number in the chart above—one billion, six hundred seventy-nine million, nine hundred thirty-five thousand, five hundred—is the number of quarters, placed end to end, it would take to circle Earth at its equator.
The small decimal number in the chart represents the time in seconds it takes for a computer chip to send a signal.
Many computer operations are measured in millionths of a second. You can compare whole or decimal numbers by lining them up and comparing the place-value positions.
For example, suppose two library books have the call numbers To determine which book comes first on the shelf, compare the numbers by first lining up the decimal points. Compare the digits from left to right until you find digits in the same place that are not equal.
The decimal with the smaller digit is the smaller number and so will come first on the shelf. When you round a number to a particular place-value position, you are really finding which of two numbers is closer to the original number.
You can round any decimal to any place-value position. For example, to round 0. The 3 remains unchanged if the digit to the right is 0, 1, 2, 3, or 4. The 3 rounds to 4 increases by 1 if the digit to the right is 5, 6, 7, 8, or 9. Since the digit to the right of the 3 is 4, the 3 remains unchanged.
When 10 is multiplied by itself several times, you can use an exponent to make the notation simpler. For example, you can write 10 x 10 x 10 x 10 as The number is read "ten to the fourth power" and is equal to 10, The exponent 4 tells us how many times 10, the base, is used as a factor.
The powers of ten are displayed in the table below. For the numbers greater than 1, the exponents are positive and correspond to the number of zeros in the standard form of the number. For the numbers less than 1, the exponents are negative and tell us the number of decimal places in the standard form.
Adding and subtracting with decimals is similar to adding and subtracting with whole numbers. Numbers must be aligned according to their place value. Remember to line up the decimal points when you add or subtract decimals. You can use zeros as placeholders.
Students multiplied and divided whole numbers and decimals in Grade 5. You may need to remind them that, when they multiply decimals, the sum of the number of decimal places in the factors should equal the number of decimal places in the product.
Below are some examples involving multiplication with decimals. You can check this multiplication by using repeated addition.
If a piece of ginger root weighs 0. Sometimes when you multiply two decimals there aren't enough digits in the result to place the decimal point. In such cases you can add as many zeros as needed on the left to place the decimal point.
Since the factors have a total of six decimal places, you must write the product with six decimal places: Dividing decimals is similar to dividing whole numbers, which students learned in Grade 5. The following example illustrates how to divide a decimal by a whole number.
Kevin skates six times around the lake on a paved bike path. If he skates a total of Divide, disregarding the decimal point. Place the decimal point in the quotient directly above the decimal point in the dividend.
When you divide a decimal by another decimal, there is one more step to follow.We can write the whole number in expanded form as follows: = (1 x ) + (5 x 10) + (9 x 1).
Decimals can also be written in expanded form. Expanded form is a way to write numbers by showing the value of each digit.
May 14, · For instance, write out various numbers using a black marker for the "ones" place and a blue marker for the "tens." Thus, you would write the number 40 with a blue "4" and a black "0." Repeat this trick with a wide range of numbers to show that place value applies across the benjaminpohle.com: 87K.
The distance between Mary's house and Diana's house is meters. Which statement about the values of the digits in the distance, in meters, between their houses is true?. The value of the 5 in the tenths place is 10 times greater than the value of the 5 in the hundredths place.
The worksheets on this page will help your child learn to understand and use place value up to 10 million. There are a range of worksheets which involve different place value tasks such as expanding numbers and writing numbers in standard form.
Write Numbers in Expanded Form Worksheets with Separate Place Value Multiplier. The expanded form worksheets on this page all involve taking a number written in conventional numeric form and rewriting it in expanded form with the place value broken out.
Write the standard form for the number which, in expanded notation, is written as follows: 9, + + 2 I've got nine thousands, three hundreds, and two ones. I don't have any tens in the expanded form, so I'll need to use a zero in the tens place to keep that slot open.
How to Teach Place Value: 12 Steps (with Pictures) - wikiHow | HuggingFaceTB/finemath | |
# E^n vs R^n
• Sep 16th 2009, 09:28 AM
tjkubo
E^n vs R^n
What is the difference between E^n and R^n? My professor uses these often and I can't seem to distinguish between them, except somehow I think E^n is more general.
• Sep 16th 2009, 11:44 AM
HallsofIvy
" $E^n$" is the set of all points in n-dimensional Euclidean space. For example, $E^2$ is the two dimensional plane. " $R^n$", on the other hand, is that set with a specific point as "origin" and a given coordinate system. That way, $R^n$ can be written as a set of ordered "n-tuples", $\{x_1, x_2, \cdot\cdot\cdot, x_n\}$, has a dot-product defined on it, and allows you to define angles.
• Sep 16th 2009, 11:45 AM
Matt Westwood
Quote:
Originally Posted by tjkubo
What is the difference between E^n and R^n? My professor uses these often and I can't seem to distinguish between them, except somehow I think E^n is more general.
Depends on the context, but I'm prepared to make a guess ...
$\mathbb{R}^n$ could well be the general n-dimensional real vector space: when n=1 you have the number line, n=2 you have the plane, n=3 you have, er, space ... it can be shown (but not by me, I'm a duffer at such things) that this is isomorphic to these spaces are isomorphic to what they are given to represent, which is why (given an appropriate frame of reference) any point in space can be mapped 1-1 to $\mathbb{R}^3$ so that every point can be identified by an ordered triple of 3 real numbers
As for $\mathbb{E}^n$ ... now I'm assuming that here $\mathbb{E}^n$ is your general n-dimensional Euclidean space:
Definition:Euclidean Space - ProofWiki
... but the context is unclear, so as I say, I'm guessing.
Whoops, HallsOfIvy beat me to it.
• Sep 16th 2009, 01:35 PM
tjkubo
Wait, I'm confused. You guys' definitions seem to contradict each other. HallsOfIvy is saying E^n is like R^n, except there is no defined origin or coordinates. But Matt's link says E^n is the set R^n with a metric defined. Maybe I'm misinterpreting. Are there more than one uses for these terms depending on the context? Can you guys recommend a good math book that defines these terms clearly?
• Sep 16th 2009, 02:25 PM
Matt Westwood
I learned what I know from W. A. Sutherland's "Introduction to Metric and Topological Spaces", if that helps ...
To put it into context: $\mathbb{R}^n$ is (technically) just the Cartesian product of n copies of the real number line, with no added "structure".
However, $\mathbb{E}^n$ is that same $\mathbb{R}^n$ with the added concept of "distance between points", defined in the same way as real-world distances. It does the job of filling in the (otherwise intuitive) concept of spatial relationships between objects in that space. Thus $\mathbb{E}^n$ does a better mathematical job, if you like, of defining space, and is actually closer to a mathematical definition of what space "actually is".
As HallsOfIvy says, by defining the "dot product" on two vectors in this space-with-a-metric, you then get the concept of angle.
Ask your professor to define the difference between the two, he may do a better job of it than me.
• Sep 16th 2009, 02:45 PM
HallsofIvy
Not exactly "contradicting" but my definition of $R^n$ does require more that just having a metric- i.e. being able to measure lengths. I am requiring that we be able to measure angles as well.
• Sep 16th 2009, 02:52 PM
Matt Westwood
Okay, you've made me have to go away and think about whether by assigning a metric to $R^n$ you can deduce the concept of angle without requiring any further structure (i.e. "angle") to be imposed ... I had a feeling that the concept of angle could be deduced elementarily from the concept of the space with the Euclidean metric imposed. This needs thinking about ... | HuggingFaceTB/finemath | |
273. Radical quantities may be multiplied, like other quantities, by writing the factors one after another, either with or without the sign of multiplication between them. (Art. 91.)
Thus the product of √a into √b, is √a.√b.
The product of n1/3 into y1/2 is h1/3y1/2.
But it is often expedient to bring the factors under the same radical sign. This may be done, if they are first reduced to a common index.
Thus nx.ny = nxy. For the root of the product of several factors is equal to the product of their roots. (Art 254.) Hence,
274. Quantities under the same radical sign or index, may be multiplied together like rational quantities, the product being placed under the common radical sign or index.
Multiply √x into 3y, that is, x1/2 into y1/3.
The quantities reduced to the same index, (Art. 264.) are (x3)1/6, and (y2)1/6 and their product is, (x3y2)1/6 = 6x3y2.
Mult. √a + m a3/2 a1/m Into √a- m x1/2 x1/n Prod. √a2 - m2 (a3x)1/2 (anxm)1/mn
Multiply √8xb into √2xb. Prod. √16x2b2 = 4xb.
In this manner the product of radical quantities often becomes rational.
Thus the product of √2 into √18 = √36 = 6.
275. Roots of the same letter or quantity may be multiplied, by adding their fractional exponents.
The exponents, like all other fractions, must be reduced to a common denominator, before they can be united in one term. (Art. 145.)
Thus a1/2.a1/3 = a1/2+1/3 = a2/6+3/6 = a5/6.
The values of the roots are not altered, by reducing their indices to a common denominator. (Art. 250.)
And in all instances of this nature, the common denominator of the indices denotes a certain root; and the sum of the numerators, shows how often this is to be repeated as a factor to produce the required product.
276. From the last example it will be seen, that powers and roots may be multiplied by a common rule. This is one of the many advantages derived from the notation by fractional indices. Any quantities whatever may be reduced to the form of radicals, (Art. 263,) and may then be subjected to the same modes of operation.
Thus y3.y1/6 = y3+1/6 = y19/6.
And x.x1/n = x1+1/n = x(n+1)/n.
The product will become rational, whenever the numerator of the index can be exactly divided by the denominator.
Thus a3.a1/3.a2/3 = a12/3 = a4.
And a3/5.a2/5 = a5/5 = a.
277. When radical quantities which are reduced to the same index, have rational coefficients, the rational parts may be multiplied together, and their product prefixed to the product of the rqadical parts.
1. Multiply a√b into c√d.
The product of the rational parts is ac.
The product of the radical parts is √bd.
And the whole product is ac√bd.
By Art. 99,.a.√b into c.√d is a.√b.c.√d or by changing the order of the factors,
a.c.√b.√d = ac.√bd = ac√bd.
But in cases of this nature we may save the trouble of reducing to a common index, by multiplying as in Art. 273.
Mult. a(b + x)1/2 a√x x3√3 Into y(b - x)1/2 b√x y3√9 Prod. ay(b2 - x2)1/2 abx 3xy
278. If the rational quantities, instead of being coefficients to the radical quantities, are connected with them by the signs + and -, each term in the multiplier must be multiplied into each in the multiplicand, as in Art. 97.
1.Multiply √a into 3b. Ans. 6a3b2.
2.Multiply 5√5 into 3√8. Ans. 30√10.
3. Multiply 2√3 into 334. Ans. 66432.
4. Multiply √d into 3ab. Ans. 6a2b2d18
Contact email: | HuggingFaceTB/finemath | |
# Thread: series expansion of the cosine integral
1. ## series expansion of the cosine integral
$\text{Ci} (x) = -\int_{x}^{\infty} \frac{\cos t}{t} \ dt$
$= -\int_{x}^{\infty} \frac{1}{t} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n!)} \ t^{2n} \ dt$
$= -\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!} \int_{x}^{\infty} t^{2n-1} \ dt$ (since the series expansion of the cosine function converges uniformly)
$= -\int^{\infty}_{x} \frac{dt}{t} -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n)!} \int_{x}^{\infty} t^{2n-1} \ dt$
$= -\ln t\Big|^{\infty}_{x} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} \ t^{2n} \Big|_{x}^{\infty}$
$\lim_{t \to \infty} \Big(-\ln t - \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} t^{2n} \Big) + \ln x + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} \ x^{2n}$
I thought this had to be incorrect, but Maple says that the limit does indeed evaluate to Euler's Constant. Why?
2. ## Re: series expansion of the cosine integral
Originally Posted by Random Variable
$\text{Ci} (x) = -\int_{x}^{\infty} \frac{\cos t}{t} \ dt =$
...
$= \lim_{t \to \infty} \Big(-\ln t - \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} t^{2n} \Big) + \ln x + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} \ x^{2n}$
I thought this had to be incorrect, but Maple says that the limit does indeed evaluate to Euler's Constant. Why?
The problem is the computation of the limit...
$\lambda=\lim_{t \rightarrow \infty} \{\ln t + \int_{0}^{t} \frac{\cos \tau-1}{\tau}\ d\tau\}$ (1)
Applying a fundamental property of the Laplace Transform first You compute [I obtained the esult some time ago]...
$\mathcal{L}\{\frac{\cos \tau-1}{\tau} \}= \int_{s}^{\infty}\frac{1}{\sigma\ (1+\sigma^{2})}\ d \sigma = \frac{1}{2}\ \ln \frac{1+s^{2}}{s^{2}}$ (2)
Then You find...
$\mathcal{L}\{\int_{0}^{t} \frac{\cos \tau-1}{\tau} \}= \frac{1}{2 s}\ \ln \frac{1+s^{2}}{s^{2}} = \frac{1}{s}\ \{\ln \sqrt{1+s^{2}} - \ln s\}$ (3)
Third step is the application of the 'final value theorem' [remember that is $\mathcal{L}\{\ln t \}= - \frac{\gamma- \ln s}{s}$ obtaining...
$\lambda= \lim_{s \rightarrow 0} \{- \gamma + \ln s + \ln \sqrt{1+s^{2}} - \ln s \}= - \gamma$ (4)
Marry Christmas from Serbia
$\chi$ $\sigma$
3. ## Re: series expansion of the cosine integral
What do we need to show to justify the use of the FVT?
EDIT: Do we just need to show that the the original limit exists?
4. ## Re: series expansion of the cosine integral
Originally Posted by Random Variable
What do we need to show to justify the use of the FVT?
EDIT: Do we just need to show that the the original limit exists?
The final value theorem exptablishes that $\lim_{t \rightarrow \infty} h(t)= \lim_{s \rightarrow 0} s\ H(s)$ can be applied if $H(s)$ has no poles with real part greater than 0 and no poles in $s=i\ \omega$ with $\omega \ne 0$...
Marry Christmas from Serbia
$\chi$ $\sigma$
5. ## Re: series expansion of the cosine integral
Wouldn't those be the conditions if we didn't know $h(t)$? Although in this case those conditions are obviously satisfied. The article on Wikipedia suggests that we only need to show that $\lim_{t \to \infty} h(t)$ is finite (which is easier said than done for this problem).
6. ## Re: series expansion of the cosine integral
Originally Posted by Random Variable
Wouldn't those be the conditions if we didn't know $h(t)$? Although in this case those conditions are obviously satisfied. The article on Wikipedia suggests that we only need to show that $\lim_{t \to \infty} h(t)$ is finite (which is easier said than done for this problem).
What is the article of Wikipedia You talk about?... if is is this one...
Laplace transform - Wikipedia, the free encyclopedia
... it is clearly explained that if h(t) has finite limit, then the FVT supplies this limit only if $s\ H(s)$ has all poles in the left half plane. If the article is another, then it is possible that something different is written because in my opinion Wikipedia is not completely reliable...
Marry Christmas from Serbia
$\chi$ $\sigma$ | HuggingFaceTB/finemath | |
In the figure below, if isosceles right angle PQR has an : PS Archive
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# In the figure below, if isosceles right angle PQR has an
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In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!
A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi
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Last edited by mirhaque on 19 Jun 2004, 09:45, edited 1 time in total.
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19 Jun 2004, 14:17
Since R is center, let RP = RQ = x
Area of PRQ = 1/2*RQ)RP =x^2/2
Area given = 4
Hence x^2/2 = 4 =>x = 2*sqrt2!!
For this Isoceles triangles if we drop a perpendicular from R (RY) to base PQ, it will bisect PQ at Y.
From pythagoras theorem, PQ^2 = PR^2 + RQ^2 =>PQ = 4 => QY =2!!
Hence RY^2 = RQ^2 - QY^2 = {2*sqrt2}^2 -{2}^2 = 4
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### Show Tags
20 Jun 2004, 02:32
mirhaque wrote:
In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!
A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi
S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.
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23 Jun 2004, 17:30
Remember, if you get booged down with calculations you are going to be screwed, and this question tries to lead you down that path.
The triangle has an area of 4, and the shaded portion has an area obviuosly less than that. Only one answer choice fits that criteria.
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24 Jun 2004, 06:47
Maneesh wrote:
Remember, if you get booged down with calculations you are going to be screwed, and this question tries to lead you down that path.
The triangle has an area of 4, and the shaded portion has an area obviuosly less than that. Only one answer choice fits that criteria.
No, Maneesh, you are wrong!
2*sqrt(pi) ~~ 2*1.8 = 3.6 < 4!
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24 Jun 2004, 08:55
Emmanuel wrote:
mirhaque wrote:
In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!
A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi
S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.
I completely understood CBRF3's explanation, but your condensed version is not as clear. Just following up in case you got some great shortcut.
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24 Jun 2004, 09:17
lastochka wrote:
Emmanuel wrote:
mirhaque wrote:
In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!
A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi
S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.
I completely understood CBRF3's explanation, but your condensed version is not as clear. Just following up in case you got some great shortcut.
S(circle) is the area of the circle.
We find area of triangle = 4 => we find all sides of this triangle (2*sqrt(2),2*sqrt(2), 4). Then we find its height (from the vertex which is opposite to hipotenuse), which is equal to 2(because hipotenuse = 4, area = 4 => hipotenuse*height/2 = area = 4). This height is the same as the radius of the circle! Then we find area of the circle, given its radius: pi*r^2 = 4*PI. Then we find shaded area, which is one-fourth of the circle area = PI.
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24 Jun 2004, 11:08
Emmanuel wrote:
lastochka wrote:
Emmanuel wrote:
mirhaque wrote:
In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!
A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi
S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.
I completely understood CBRF3's explanation, but your condensed version is not as clear. Just following up in case you got some great shortcut.
S(circle) is the area of the circle.
We find area of triangle = 4 => we find all sides of this triangle (2*sqrt(2),2*sqrt(2), 4). Then we find its height (from the vertex which is opposite to hipotenuse), which is equal to 2(because hipotenuse = 4, area = 4 => hipotenuse*height/2 = area = 4). This height is the same as the radius of the circle! Then we find area of the circle, given its radius: pi*r^2 = 4*PI. Then we find shaded area, which is one-fourth of the circle area = PI.
in other words, it's the same throught process as presented by CBRF3. thanks for elaboration
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28 Jun 2004, 09:59
Maneesh wrote:
Remember, if you get booged down with calculations you are going to be screwed, and this question tries to lead you down that path.
The triangle has an area of 4, and the shaded portion has an area obviuosly less than that. Only one answer choice fits that criteria.
good work Maneesh.
Emmanuel, the third option is 2.sqrt2.pi which is > 4
- ash
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28 Jun 2004, 14:12
ashkg wrote:
Emmanuel, the third option is 2.sqrt2.pi which is > 4
- ash
No, ashkg, 2*sqrt(PI) < 4 because 4*PI < 16, because PI = 3.14 < 4.
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29 Jun 2004, 12:31
Emmanuel wrote:
ashkg wrote:
Emmanuel, the third option is 2.sqrt2.pi which is > 4
- ash
No, ashkg, 2*sqrt(PI) < 4 because 4*PI < 16, because PI = 3.14 < 4.
Emmanuel, what you have written is correct, but what is given is 2.srqt2.pi not 2*sqrt(PI). Maneesh, thanks. You are thinking outside the box. It took me several minutes to find the answer using traditional math.
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Display posts from previous: Sort by | HuggingFaceTB/finemath | |
Using graphical method, find the maximum value of
$2x+y$
subject to
$4x + 3y \leq 12 \\ 4x + y \leq 8 \\ 4x - y \leq 8 \\ x,y \geq 0$.
$4x + 3y \leq 12$
x 0 3 y 4 0
$4x + y \leq 8$
x 0 2 y 8 0
$4x - y \leq 8$
x 0 2 y -8 0
The value of the objective function at each of these extreme points is as follows
Extreme PointCoordinates(x,y) Objective function valuez=2x+y O(0,0) 2(0)+1(0)=0 A(0,4) 2(0)+1(4)=4 B($\frac{3}{2}$,2) 2($\frac{3}{2}$ )+1(2)=5 C(2,0) 2(2)+1(0)=4
The maximum value of the objective function z=5 occurs at the extreme point ($\frac{3}{2}$,2).
Hence, the optimal solution to the given LP problem is $x=\frac{3}{2},y=2$ and $max\ z=5$.
answered Dec 21, 2017 by (1,920 points) | HuggingFaceTB/finemath | |
## Loujoelou 3 years ago 5.In the following equation, identify the x-intercepts in the graph y = 3x^2 + 19x – 40. I'm stuck on this problem can anyone help? :)
1. mathslover
You mean to say that we have to find x intercept?
2. waterineyes
Where are you having problem??
3. mathslover
Ok so just put y =0 : \[\large{0=3x^2+19x-40}\] \[\large{0=3x^2+24x-5x-40}\] \[\large{0=3x(x+8)-5(x+8)}\] \[\large{0=(3x-5)(x+8)}\]
4. waterineyes
For finding x intercept we put y = 0..
5. Loujoelou
Oh okay, sry I forgot about that and yeah had trouble factoring it and now I see. This means our x= -8 and 5/3
6. waterineyes
Yep..
7. mathslover
right
8. mathslover
3x-5=0 x+8=0 3x=5 x=-8 x=5/3
9. rymdenbarn
Right, I'd also like to add that in case you forget whether to make the y or the x zero when it comes to either intercept remember that any x-intercept would have to be (x, 0) and that any y-intercept would have to be (0, y). Cheers
10. Loujoelou
K thx a ton :)
Find more explanations on OpenStudy | HuggingFaceTB/finemath | |
# Kilohenries to Megahenries Conversion
Enter the electrical inductance in kilohenries below to get the value converted to megahenries.
Results in Megahenries: 1 kH = 0.001 MH
## How to Convert Kilohenries to Megahenries
To convert a kilohenry measurement to a megahenry measurement, divide the electrical inductance by the conversion ratio.
Since one megahenry is equal to 1,000 kilohenries, you can use this simple formula to convert:
megahenries = kilohenries ÷ 1,000
The electrical inductance in megahenries is equal to the kilohenries divided by 1,000.
For example, here's how to convert 5,000 kilohenries to megahenries using the formula above.
5,000 kH = (5,000 ÷ 1,000) = 5 MH
## Kilohenries
One kilohenry is equal to 1,000 henries, which are the inductance of a conductor with one volt of electromotive force when the current is increased by one ampere per second.
The kilohenry is a multiple of the henry, which is the SI derived unit for electrical inductance. In the metric system, "kilo" is the prefix for 103. Kilohenries can be abbreviated as kH; for example, 1 kilohenry can be written as 1 kH.
## Megahenries
One megahenry is equal to 1,000,000 henries, which are the inductance of a conductor with one volt of electromotive force when the current is increased by one ampere per second.
The megahenry is a multiple of the henry, which is the SI derived unit for electrical inductance. In the metric system, "mega" is the prefix for 106. Megahenries can be abbreviated as MH; for example, 1 megahenry can be written as 1 MH.
## Kilohenry to Megahenry Conversion Table
Kilohenry measurements converted to megahenries
Kilohenries Megahenries
1 kH 0.001 MH
2 kH 0.002 MH
3 kH 0.003 MH
4 kH 0.004 MH
5 kH 0.005 MH
6 kH 0.006 MH
7 kH 0.007 MH
8 kH 0.008 MH
9 kH 0.009 MH
10 kH 0.01 MH
20 kH 0.02 MH
30 kH 0.03 MH
40 kH 0.04 MH
50 kH 0.05 MH
60 kH 0.06 MH
70 kH 0.07 MH
80 kH 0.08 MH
90 kH 0.09 MH
100 kH 0.1 MH
200 kH 0.2 MH
300 kH 0.3 MH
400 kH 0.4 MH
500 kH 0.5 MH
600 kH 0.6 MH
700 kH 0.7 MH
800 kH 0.8 MH
900 kH 0.9 MH
1,000 kH 1 MH | HuggingFaceTB/finemath | |
This is one big problem that is broken up into four parts. Please work on this yourself as much as possible. You will find that the \Puzzle Pieces” from earlier homeworks will be very useful.
1. Consider a market with three consumers: Larry, Moe, and Curley. They each have income of Y = 120.
Their utility functions are:
• Larry: U = x0:5y0:5
• Moe: U = x0:25y0:75
• Curley: U = x0:75y0:25
(a) If the price of good y is py = 10, derive the equation for the market demand curve for good x.
(b) How much will each consumer purchase if the price of good x is px = 5? Why are they different?
Does it make sense?
(c) Draw the three individual demand curves and the market demand curve on the same graph.
Indicate the quantities when px = 5.
2. Suppose there are 3 firms that produce and sell good x. Each firm has long-run production function
q = 2K0:5L0:5.
(a) If the wage rate is w = 18, the rental rate of capital is r = 2, and K is fixed in the short-run at
K = 9, derive the equation for the short-run market supply curve.
(b) How much will each firm produce if the price of good x is px = 5?
(c) Draw supply curve for one of these firms and the market supply curve on the same graph. Indicate the quantities when px = 5.
(d) What is each firm’s short-run average total cost curve?
3. Now consider the competitive market for good x.
(a) What is the short-run equilibrium price and quantity of good x? (Note that the answers will be fractions.)
(b) How much will each firm produce? What will their short-run profits be?
(c) Graph the market (demand and supply curve etc.) and the graph of one of the firms (marginal revenue \curve,” marginal cost curve, average total cost curve, profits, etc.) side-by-side. Make sure everything is labeled perfectly!
4. Now consider the long-run equilibrium.
(a) Why will there be more than 3 firms in the market in the long-run?
(b) Explain why the following constitutes a long-run equilibrium: price is px = 6, there are 5 firms, and each produces 6 units. Your answer should:
i. Describe the firm’s choice of capital and labor (using condition MRTS = −w/r ).
ii. Describe firm’s profits.
iii. Include a diagram that modifies the diagram in 3(c) for a long-run equilibrium.
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The Weierstrass M-Test for Uniform Convergence of Series of Functions
# The Weierstrass M-Test for Uniform Convergence of Series of Functions
Recall from the Pointwise Convergent and Uniformly Convergent of Series of Functions page that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of real-valued functions with common domain $X$, then we say that the corresponding series of functions $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ is uniformly convergent if the sequence of partial sums $(s_n(x))_{n=1}^{\infty}$ is a uniformly convergent sequence.
We will now look at a very nice and relatively simply test to determine uniform convergence of a series of real-valued functions called the Weierstrass M-test.
Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of real-valued functions with common domain $X$, and let $(M_n)_{n=1}^{\infty}$ be a sequence of nonnegative real numbers such that $\mid f_n(x) \mid \leq M_n$ for each $n \in \mathbb{N}$ and for all $x \in X$. If $\displaystyle{\sum_{n=1}^{\infty} M_n}$ converges then $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ uniformly converges on $X$.
• Proof: Suppose that there exists a sequence of nonnegative real numbers $(M_n)_{n=1}^{\infty}$ such that for all $n \in \mathbb{N}$ and for all $x \in X$ we have that:
(1)
\begin{align} \quad \mid f_n(x) \mid \leq M_n \end{align}
• Furthermore, suppose that $\displaystyle{\sum_{n=1}^{\infty} M_n}$ converges to some $M \in \mathbb{R}$, $M \geq 0$. Then we have that for all $x \in X$:
(2)
\begin{align} \quad \biggr \lvert \sum_{n=1}^{\infty} f_n(x) \biggr \rvert \leq \sum_{n=1}^{\infty} \mid f_n(x) \mid \leq \sum_{n=1}^{\infty} M_n = M \end{align}
• So the $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ converges for each $x \in X$ by the comparison test. $\blacksquare$ | HuggingFaceTB/finemath | |
# Chapter 12 - Problems - Page 463: 24
An intensity level change of $+1.00~dB$ corresponds to an increase in intensity of 25.9%
#### Work Step by Step
We can write an expression for $\beta_2-\beta_1 = 1.00~dB$ to find the change in intensity: $\beta_2-\beta_1 = 10~log~\frac{I_2}{I_0}-10~log~\frac{I_1}{I_0}$ $1.00 = 10~log~\frac{I_2}{I_1}$ $0.100 = log~\frac{I_2}{I_1}$ $10^{0.100} = \frac{I_2}{I_1}$ $I_2 = 10^{0.100}~I_1$ $I_2 = 1.259~I_1$ An intensity level change of $+1.00~dB$ corresponds to an increase in intensity of 25.9%.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | HuggingFaceTB/finemath | |
streamschemeprimessicpsieve
# Need help to understand some of the SICP streams examples
I'm trying to understand how this function works.
``````(define (sieve stream)
(cons-stream
(stream-car stream)
(sieve (stream-filter
(lambda (x)
(not (divisible? x (stream-car stream))))
(stream-cdr stream)))))
(define primes (sieve (integers-starting-from 2)))
``````
Simply, I use a stream that generates all the integers starting from 2 and, according to the book, it filters the rest of the stream that is not divisible by current element for each new element. How can this filter all the integers that are not divisible by current element without actually reading all the integers?
Solution
• The definitions
``````(define (sieve stream)
(cons-stream (stream-car stream)
(sieve
(stream-filter (lambda (x) (not (divisible? x (stream-car stream))))
(stream-cdr stream)))))
(define primes (sieve (integers-starting-from 2)))
``````
mean that
``````primes
=
(sieve
(integers-starting-from 2))
=
(cons-stream 2
(sieve
(stream-filter (lambda (x) (not (divisible? x 2)))
(integers-starting-from 3))))
=
(cons-stream 2
(cons-stream 3
(sieve
(stream-filter (lambda (x) (not (divisible? x 3)))
(stream-filter (lambda (x) (not (divisible? x 2)))
(integers-starting-from 4))))))
=
(cons-stream 2
(cons-stream 3
(sieve
(stream-filter (lambda (x) (not (divisible? x 3)))
(stream-filter (lambda (x) (not (divisible? x 2)))
(integers-starting-from 5))))))
=
(cons-stream 2
(cons-stream 3
(cons-stream 5
(sieve
(stream-filter (lambda (x) (not (divisible? x 5)))
(stream-filter (lambda (x) (not (divisible? x 3)))
(stream-filter (lambda (x) (not (divisible? x 2)))
(integers-starting-from 6))))))))
``````
and further
``````=
....
=
(cons-stream 2
(cons-stream 3
(cons-stream 5
(cons-stream 7
(sieve
(stream-filter (lambda (x) (not (divisible? x 7)))
(stream-filter (lambda (x) (not (divisible? x 5)))
(stream-filter (lambda (x) (not (divisible? x 3)))
(stream-filter (lambda (x) (not (divisible? x 2)))
(integers-starting-from 9))))))))))
=
....
=
(cons-stream 2
(cons-stream 3
(cons-stream 5
(cons-stream 7
(cons-stream 11
(sieve
(stream-filter (lambda (x) (not (divisible? x 11)))
(stream-filter (lambda (x) (not (divisible? x 7)))
(stream-filter (lambda (x) (not (divisible? x 5)))
(stream-filter (lambda (x) (not (divisible? x 3)))
(stream-filter (lambda (x) (not (divisible? x 2)))
(integers-starting-from 12))))))))))))
=
....
``````
which, hopefully, should give you a clearer picture of what is going on here. | HuggingFaceTB/finemath | |
# Difference between revisions of "2005 AMC 10B Problems/Problem 25"
## Problem
A subset $B$ of the set of integers from $1$ to $100$, inclusive, has the property that no two elements of $B$ sum to $125$. What is the maximum possible number of elements in $B$?
$\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68$
## -Solutions-
### Solution 1
The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$. The integers from $25$ to $100$ are left. They can be paired so the sum is $125$: $25+100$, $26+99$, $27+98$, $\ldots$, $62+63$. That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = \boxed{\mathrm{(C)}\ 62}$. Also, it is possible to see that since the numbers $1$ to $24$ are in the set there are only the numbers $25$ to $100$ to consider. As $62+63$ gives $125$, the numbers $25$ to $62$ can be put in subset $B$ without having two numbers add up to $125$. In this way, subset $B$ will have the numbers $1$ to $62$, and so the answer is $\boxed{\mathrm{(C)}\ 62}$.
#### Solution 1 Alternate Solution
Since there are 38 numbers that sum to $125$, there are $100-38=62$ numbers not summing to $125.$ ~mathboy282
### Solution 2 (If you have no time)
"Cut" $125$ into half. The maximum integer value in the smaller half is $62$. Thus the answer is $\boxed{\mathrm{(C)}\ 62}$.
### Solution 3
The maximum possible number of elements includes the smallest numbers. So, subset $B = \{1,2,3....n-1,n\}$ where n is the maximum number of elements in subset $B$. So, we have to find two consecutive numbers, $n$ and $n+1$, whose sum is $125$. Setting up our equation, we have $n+(n+1) = 2n+1 = 125$. When we solve for $n$, we get $n = 62$. Thus, the anser is $\boxed{\mathrm{(C)}\ 62}$.
~GentleTiger | HuggingFaceTB/finemath | |
# General solution to a system of non linear equations with a specific pattern
I am seeking a general solution to a system of non linear equations with a specific pattern:
Order 1:
$$x_0 = a^2 + b^2$$ $$x_1 = 2ab$$
Order 2:
$$x_0 = a^2 + b^2 + c^2$$ $$x_1 = 2ab + 2bc$$ $$x_2 = 2ac$$
Order 3:
$$x_0 = a^2 + b^2 + c^2 + d^2$$ $$x_1 = 2ab + 2bc + 2cd$$ $$x_2 = 2ac + 2bd$$ $$x_3 = 2ad$$
Order 4:
$$x_0 = a^2 + b^2 + c^2 + d^2 + e^2$$ $$x_1 = 2ab + 2bc + 2cd + 2de$$ $$x_2 = 2ac + 2bd + 2ce$$ $$x_3 = 2ad + 2be$$ $$x_4 = 2ae$$
The values $x_n$ are constant and known. The values $a, b, c, ...$ need to be solved. It's easily solvable for Order 1. Order 2 is a little more tricky but solvable. It quickly becomes very hard for order greater than or equal to 3.
I don't want to have to resort to say a quadratic programming method. Are there any tricks to this particular set of equations or any numerical methods that may help me? Given the nature of the pattern it looks like there might be a general solution using a specific technique I'm not familiar with. For instance, a change of variable might reduce this to a root finding problem perhaps.
• Have you tried solving less complex systems, for example all these systems without the first equation ($a^2 + b^2 + ...$), or just the first two equations? or where $x^i$ all equal 0? Those kinds of simplifications may (or may not) be easier to solve and may give hints on how to solve the bigger problems. – Mitch May 26 '15 at 0:28
Let us rename the unknowns $a,b,c,\ldots$ to $v_0,\ldots,v_n$, assume these to be complex numbers, and define the (yet unknown) polynomial $$f(z) = \sum_{k=0}^n v_k z^k \tag{1}$$ Then we have $$f(z)\,f\left(\frac{1}{z}\right) = x_0 + \frac{x_1}{2}\left(z+\frac{1}{z}\right) + \frac{x_2}{2}\left(z^2+\frac{1}{z^2}\right) + \cdots + \frac{x_n}{2}\left(z^n+\frac{1}{z^n}\right)$$ Using the Chebyshev polynomials of the first kind $T_k(X)$, we can write \begin{align} \frac{1}{2}\left(z^k+\frac{1}{z^k}\right) &= T_k(X) \\ \text{where}\quad X &= \frac{1}{2}\left(z+\frac{1}{z}\right) \tag{2}\\ \text{thus}\quad f(z)\,f\left(\frac{1}{z}\right) &= g(X) \tag{3}\\ \text{where}\quad g(X) &= \sum_{k=0}^n x_k T_k(X) \tag{4} \end{align} Note that $g(X)$ is a polynomial in $X$ that can be determined from the given $x_k$ via $(4)$. Now we need to find a (non-unique) polynomial factor $f(z)$ fulfilling $(3)$.
If $g=0$, that is, all $x_k = 0$, then $(3)$ implies $f=0$, hence all $v_k=0$.
In the following, I will assume that not all $x_k = 0$.
Let $m = \deg g$, so $0\leq m\leq n$. Let $g_m$ be the (nonzero) coefficient of $X^m$ in $g(X)$. We can deduce from the defining equations for $x_m,\ldots,x_n$ (working from index $n$ downwards) that there is $r\in\{0,\ldots,n-m\}$ such that $v_k = 0$ for $k\in\{0,\ldots,r-1,m+r+1,\ldots,n\}$ and $v_k\neq0$ for $k\in\{r,m+r\}$. So basically $f(z)$ is the product of $z^r$ with a degree-$m$ polynomial. (And $g_m = 2^m v_r v_{m+r}$, but we do not need this.)
The idea is to find suitable $f(z)$ by looking at the zeros of $g(X)$. Therefore, find the $m$ (possibly complex) roots of $g(X)=0$, and name these $X_1,\ldots,X_m$. For each such root $X=X_j$, solve $(2)$ for $z$. There will be two solutions; pick one of these and name it $z_j$, then the other solution will be $\frac{1}{z_j}$. Keep in mind that both solutions for $z$ are nonzero. We will consider $z_j$ a root of $f(z)=0$; this makes $\frac{1}{z_j}$ a root of $f\left(\frac{1}{z}\right)=0$ automatically. Now we can write \begin{align} f(z) &= v_{m+r} z^r\prod_{j=1}^m (z - z_j) \tag{5}\\ &\quad\text{with arbitrary}\quad r\in\{0,\ldots,n-m\} \end{align} so $(3)$ becomes $$v_{m+r}^2\prod_{j=1}^m \underbrace{(z - z_j)\left(\frac{1}{z} - z_j\right)}_ {-2z_j (X - X_j)} = g(X)$$ Equating coefficients for $X^m$, we get $$v_{m+r} = \pm\sqrt{\frac{g_m}{(-2)^m\prod_{j=1}^m z_j}} \tag{6}$$ Choose a sign for $v_{m+r}$, then expand the product for $f(z)$ in $(5)$ to get the representation $(1)$. You can read the coefficients $v_0,\ldots,v_n$ from that.
In general, the choices you make for $r$, for $z_j$ versus $\frac{1}{z_j}$, and for the sign of $v_{m+r}$ all affect the solution, so there are $2^{m+1}(n-m+1)$ (not necessarily distinct) solution tuples $(v_0,\ldots,v_n)$. Some particular variations are:
• Varying $r$: This results in rotating $(v_0,\ldots,v_n)$ as long as only leading/trailing zeros are wrapped around.
• Replacing all $z_j$ with $\frac{1}{z_j}$: This results in the reversal $(v_r,\ldots,v_{m+r}) \mapsto (v_{m+r},\ldots,v_r)$.
• Flipping the sign of $v_{m+r}$ in $(6)$: This results in sign flips of all $v_k$.
I couldn't solve the problem, but I have done some manipulations that maybe could help. They are too long for a comment.
For order $n$ we have that
$$\sum_{k=1}^{n+1}a_k=\pm\sqrt{\sum_{k=0}^nx_k}\qquad(1)$$
I have renamed the variables $a$, $b$, $c$... to $a_1$, $a_2$, $a_3$,...
Last equation says $$a_1a_{n+1}=x_n$$ so we can solve for $a_{n+1}$ and substitute in $(1)$. Assuming that $x_n\neq 0$, we have now
$$\frac{x_n}{a_1}+\sum_{k=1}^{n}a_k=\pm\sqrt{\sum_{k=0}^nx_k}\qquad(2)$$ | HuggingFaceTB/finemath | |
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# 19 Series Parallel Circuit Example Problems With Solutions Pics
Capacitor c2 = 4 μf. Solution to example 2 the three resistors are in parallel and behave like a resistor with resistance req given by 1 / req = 1 / 100 + 1 / 400 + 1 / 200 multiply all terms by. Capacitor 3 (c3) = 3 μf. Capacitor 1 (c1) = 3 μf. Capacitor c1 = 2 μf.
Capacitor 4 (c4) = 2 μf. Capacitor 1 (c1) = 3 μf. Capacitor 2 (c2) = 3 μf. Normally, the first step in mathematically analyzing a circuit such as this is to determine the total circuit resistance. Determine the capacitance of a single capacitor that will have the same effect as the combination. In this example, we want to find every missing values. Capacitor c2 = 4 μf. In the circuit shown below, we can see that resistors r 2 and r 3 are connected in parallel with each other and that both are connected in series with r 1.
### Capacitor c1 = 2 μf.
Capacitor 3 (c3) = 3 μf. Capacitor c1 = 2 μf. We have the total current and the value of every resistors so we could simplify the circuit to find the voltage of the d.c voltage source. 04/01/2022 · parallel circuit and cur division rc analysis series explained in plain english electrical4u examples electrical academia 6 circuits cleo learned by example online eet 1155 unit 8 ac worksheet ap physics 1 electricity tutorial equivalent problem lesson 18 ppt resistors electronics questions answers pdf exact solutions of coupled resonant equations … Capacitor c2 = 4 μf. To solve such circuits, first reduce the … R = 1 1 r1 + 1 r2 + 1 r3 r = 1 1 r 1 + 1 r 2 + 1 r 3. Capacitor 1 (c1) = 3 μf. Three capacitors, c1 = 2 μf, c2 = 4 μf, c3 = 4 μf, are connected in series and parallel. Algebraically manipulate this equation to solve for one of the parallel resistances (r 1) in terms of the other two parallel resistances (r 2 and r 3) and the total resistance (r. Examples with detailed solutions example 2 find current i in the circuit below and the current passing through each of the resistors in the circuit. Capacitor 2 (c2) = 3 μf. In the circuit shown below, we can see that resistors r 2 and r 3 are connected in parallel with each other and that both are connected in series with r 1.
Capacitor 3 (c3) = 3 μf. Algebraically manipulate this equation to solve for one of the parallel resistances (r 1) in terms of the other two parallel resistances (r 2 and r 3) and the total resistance (r. Capacitor 1 (c1) = 3 μf. Capacitor 4 (c4) = 2 μf. Determine the capacitance of a single capacitor that will have the same effect as the combination.
Capacitor c2 = 4 μf. 04/01/2022 · parallel circuit and cur division rc analysis series explained in plain english electrical4u examples electrical academia 6 circuits cleo learned by example online eet 1155 unit 8 ac worksheet ap physics 1 electricity tutorial equivalent problem lesson 18 ppt resistors electronics questions answers pdf exact solutions of coupled resonant equations … Three capacitors, c1 = 2 μf, c2 = 4 μf, c3 = 4 μf, are connected in series and parallel. Normally, the first step in mathematically analyzing a circuit such as this is to determine the total circuit resistance. Algebraically manipulate this equation to solve for one of the parallel resistances (r 1) in terms of the other two parallel resistances (r 2 and r 3) and the total resistance (r. R = 1 1 r1 + 1 r2 + 1 r3 r = 1 1 r 1 + 1 r 2 + 1 r 3. Capacitor 4 (c4) = 2 μf. To solve such circuits, first reduce the …
### In this example, we have multiples ways of resolving this problem.
Algebraically manipulate this equation to solve for one of the parallel resistances (r 1) in terms of the other two parallel resistances (r 2 and r 3) and the total resistance (r. We are going to solve it using the equivalent circuit. 04/01/2022 · parallel circuit and cur division rc analysis series explained in plain english electrical4u examples electrical academia 6 circuits cleo learned by example online eet 1155 unit 8 ac worksheet ap physics 1 electricity tutorial equivalent problem lesson 18 ppt resistors electronics questions answers pdf exact solutions of coupled resonant equations … In this example, we have multiples ways of resolving this problem. Three capacitors, c1 = 2 μf, c2 = 4 μf, c3 = 4 μf, are connected in series and parallel. Capacitor 2 (c2) = 3 μf. We have the total current and the value of every resistors so we could simplify the circuit to find the voltage of the d.c voltage source. R = 1 1 r1 + 1 r2 + 1 r3 r = 1 1 r 1 + 1 r 2 + 1 r 3. Normally, the first step in mathematically analyzing a circuit such as this is to determine the total circuit resistance. In the circuit shown below, we can see that resistors r 2 and r 3 are connected in parallel with each other and that both are connected in series with r 1. Examples with detailed solutions example 2 find current i in the circuit below and the current passing through each of the resistors in the circuit. Solution to example 2 the three resistors are in parallel and behave like a resistor with resistance req given by 1 / req = 1 / 100 + 1 / 400 + 1 / 200 multiply all terms by. Capacitor 3 (c3) = 3 μf.
Three capacitors, c1 = 2 μf, c2 = 4 μf, c3 = 4 μf, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination. Examples with detailed solutions example 2 find current i in the circuit below and the current passing through each of the resistors in the circuit. Algebraically manipulate this equation to solve for one of the parallel resistances (r 1) in terms of the other two parallel resistances (r 2 and r 3) and the total resistance (r. We are going to solve it using the equivalent circuit.
Normally, the first step in mathematically analyzing a circuit such as this is to determine the total circuit resistance. We are going to solve it using the equivalent circuit. Examples with detailed solutions example 2 find current i in the circuit below and the current passing through each of the resistors in the circuit. 04/01/2022 · parallel circuit and cur division rc analysis series explained in plain english electrical4u examples electrical academia 6 circuits cleo learned by example online eet 1155 unit 8 ac worksheet ap physics 1 electricity tutorial equivalent problem lesson 18 ppt resistors electronics questions answers pdf exact solutions of coupled resonant equations … To solve such circuits, first reduce the … Solution to example 2 the three resistors are in parallel and behave like a resistor with resistance req given by 1 / req = 1 / 100 + 1 / 400 + 1 / 200 multiply all terms by. Capacitor 3 (c3) = 3 μf. In this example, we want to find every missing values.
### Algebraically manipulate this equation to solve for one of the parallel resistances (r 1) in terms of the other two parallel resistances (r 2 and r 3) and the total resistance (r.
Determine the capacitance of a single capacitor that will have the same effect as the combination. Solution to example 2 the three resistors are in parallel and behave like a resistor with resistance req given by 1 / req = 1 / 100 + 1 / 400 + 1 / 200 multiply all terms by. Capacitor 1 (c1) = 3 μf. We have the total current and the value of every resistors so we could simplify the circuit to find the voltage of the d.c voltage source. Three capacitors, c1 = 2 μf, c2 = 4 μf, c3 = 4 μf, are connected in series and parallel. Capacitor 3 (c3) = 3 μf. Capacitor 2 (c2) = 3 μf. 04/01/2022 · parallel circuit and cur division rc analysis series explained in plain english electrical4u examples electrical academia 6 circuits cleo learned by example online eet 1155 unit 8 ac worksheet ap physics 1 electricity tutorial equivalent problem lesson 18 ppt resistors electronics questions answers pdf exact solutions of coupled resonant equations … Normally, the first step in mathematically analyzing a circuit such as this is to determine the total circuit resistance. To solve such circuits, first reduce the … Capacitor 4 (c4) = 2 μf. In this example, we have multiples ways of resolving this problem. R = 1 1 r1 + 1 r2 + 1 r3 r = 1 1 r 1 + 1 r 2 + 1 r 3.
19+ Series Parallel Circuit Example Problems With Solutions
Pics
. 04/01/2022 · parallel circuit and cur division rc analysis series explained in plain english electrical4u examples electrical academia 6 circuits cleo learned by example online eet 1155 unit 8 ac worksheet ap physics 1 electricity tutorial equivalent problem lesson 18 ppt resistors electronics questions answers pdf exact solutions of coupled resonant equations … R = 1 1 r1 + 1 r2 + 1 r3 r = 1 1 r 1 + 1 r 2 + 1 r 3. To solve such circuits, first reduce the … Capacitor c2 = 4 μf. Algebraically manipulate this equation to solve for one of the parallel resistances (r 1) in terms of the other two parallel resistances (r 2 and r 3) and the total resistance (r. | HuggingFaceTB/finemath | |
# 6 LDA Inference
## 6.1 General Overview
We have talked about LDA as a generative model, but now it is time to flip the problem around. What if I have a bunch of documents and I want to infer topics? To solve this problem we will be working under the assumption that the documents were generated using a generative model similar to the ones in the previous section. Under this assumption we need to attain the answer for Equation (6.1). (NOTE: The derivation for LDA inference via Gibbs Sampling is taken from (Darling 2011), (Heinrich 2008) and (Steyvers and Griffiths 2007).)
$$$p(\theta, \phi, z|w, \alpha, \beta) = {p(\theta, \phi, z, w|\alpha, \beta) \over p(w|\alpha, \beta)} \tag{6.1}$$$ The left side of Equation (6.1) defines the following:
The probability of the document topic distribution, the word distribution of each topic, and the topic labels given all words (in all documents) and the hyperparameters $$\alpha$$ and $$\beta$$. In particular we are interested in estimating the probability of topic (z) for a given word (w) (and our prior assumptions, i.e. hyperparameters) for all words and topics. From this we can infer $$\phi$$ and $$\theta$$.
Equation (6.1) is based on the following statistical property:
$p(A, B | C) = {p(A,B,C) \over p(C)}$
All the variables used in this section were outlined at the beginning of Chapter 5 if you need a refresher.
Let’s take a step from the math and map out ‘variables we know’ versus the ‘variables we don’t know’ in regards to the inference problem:
Known Parameters
• Documents (d in D): We have a set number of documents we want to identify the topic strucutres in.
• Words (w in W): We have a collection of words and word counts for each document.
• Vocabulary (W): The unique list of words across all documents.
• Hyperparemeters:
• $$\overrightarrow{\alpha}$$: Our prior assumption about the topic distribution of our documents. This book will only use symmetric $$\alpha$$ values, in other words we assume all topics are equally as probably in any given document (similar to the naive assumption of a fair die). We will be supplying the $$\alpha$$ value for inference.
• Higher $$\overrightarrow{\alpha}$$ - We assume documents will have a similar and close to uniform distribution of topics.
• Lower $$\overrightarrow{\alpha}$$ - We assume document topic distributions vary more drastically.
• $$\overrightarrow{\beta}$$: Our prior assumption about the word distribution of each topic.
• Higher $$\overrightarrow{\beta}$$: Word distributions in each topic are closer to uniform, i.e. each word is equally as likely in each topic.
• Lower $$\overrightarrow{\beta}$$: Word distributions vary more from topic to topic.
And on to the parts we don’t know….
Unknown (Latent) Parameters
• Number of Topics (k): We need to specify the number of topics we assume are present in the documents. However we don’t know the real number of topics in the corpus. Methods for estimating the number of topics in a corpus are outside the scope of this book. If you are interested in learning more see (Griffiths and Steyvers 2004) and (Teh et al. 2005).
• Document Topic Mixture ($$\theta$$): We need to determine the topic distribution in each document.
• Word Distribution of Each Topic ($$\phi$$): We need to know the distribution of words in each topic. Obviously some words are going to occur very often in a topic while others may have zero probability of occurring in a topic.
• Word topic assignment (z): This is actually the main thing we need to infer. To be clear, if we know the topic assignment of every word in every document, then we can derive the document topic mixture, $$\theta$$, and the word distribution, $$\phi$$, of each topic.
## 6.2 Mathematical Derivations for Inference
Back to the math…
The derivation connecting equation (6.1) to the actual Gibbs sampling solution to determine z for each word in each document, $$\overrightarrow{\theta}$$, and $$\overrightarrow{\phi}$$ is very complicated and I’m going to gloss over a few steps. For complete derivations see (Heinrich 2008) and (Carpenter 2010).
As stated previously, the main goal of inference in LDA is to determine the topic of each word, $$z_{i}$$ (topic of word i), in each document.
$$$p(z_{i}|z_{\neg i}, \alpha, \beta, w) \tag{6.2}$$$
Notice that we are interested in identifying the topic of the current word, $$z_{i}$$, based on the topic assignments of all other words (not including the current word i), which is signified as $$z_{\neg i}$$.
\begin{aligned} p(z_{i}|z_{\neg i}, \alpha, \beta, w) &= {p(z_{i},z_{\neg i}, w, | \alpha, \beta) \over p(z_{\neg i},w | \alpha, \beta)}\\ &\propto p(z_{i}, z_{\neg i}, w | \alpha, \beta)\\ &\propto p(z,w|\alpha, \beta) \end{aligned} \tag{6.3}
You may notice $$p(z,w|\alpha, \beta)$$ looks very similar to the definition of the generative process of LDA from the previous chapter (equation (5.1)). The only difference is the absence of $$\theta$$ and $$\phi$$. This means we can swap in equation (5.1) and integrate out $$\theta$$ and $$\phi$$.
\begin{aligned} p(w,z|\alpha, \beta) &= \int \int p(z, w, \theta, \phi|\alpha, \beta)d\theta d\phi\\ &= \int \int p(\phi|\beta)p(\theta|\alpha)p(z|\theta)p(w|\phi_{z})d\theta d\phi \\ &= \int p(z|\theta)p(\theta|\alpha)d \theta \int p(w|\phi_{z})p(\phi|\beta)d\phi \end{aligned} \tag{6.4}
As with the previous Gibbs sampling examples in this book we are going to expand equation (6.3), plug in our conjugate priors, and get to a point where we can use a Gibbs sampler to estimate our solution.
Below we continue to solve for the first term of equation (6.4) utilizing the conjugate prior relationship between the multinomial and Dirichlet distribution. The result is a Dirichlet distribution with the parameters comprised of the sum of the number of words assigned to each topic and the alpha value for each topic in the current document d.
\begin{aligned} \int p(z|\theta)p(\theta|\alpha)d \theta &= \int \prod_{i}{\theta_{d_{i},z_{i}}{1\over B(\alpha)}}\prod_{k}\theta_{d,k}^{\alpha k}\theta_{d} \\ &={1\over B(\alpha)} \int \prod_{k}\theta_{d,k}^{n_{d,k} + \alpha k} \\ &={B(n_{d,.} + \alpha) \over B(\alpha)} \end{aligned} \tag{6.5}
Similarly we can expand the second term of Equation (6.4) and we find a solution with a similar form. The result is a Dirichlet distribution with the parameter comprised of the sum of the number of words assigned to each topic across all documents and the alpha value for that topic.
\begin{aligned} \int p(w|\phi_{z})p(\phi|\beta)d\phi &= \int \prod_{d}\prod_{i}\phi_{z_{d,i},w_{d,i}} \prod_{k}{1 \over B(\beta)}\prod_{w}\phi^{B_{w}}_{k,w}d\phi_{k}\\ &= \prod_{k}{1\over B(\beta)} \int \prod_{w}\phi_{k,w}^{B_{w} + n_{k,w}}d\phi_{k}\\ &=\prod_{k}{B(n_{k,.} + \beta) \over B(\beta)} \end{aligned} \tag{6.6}
This leaves us with the following:
\begin{aligned} p(w,z|\alpha, \beta) &= \prod_{d}{B(n_{d,.} + \alpha) \over B(\alpha)} \prod_{k}{B(n_{k,.} + \beta) \over B(\beta)} \end{aligned} \tag{6.7}
The equation necessary for Gibbs sampling can be derived by utilizing (6.7). This is accomplished via the chain rule and the definition of conditional probability.
The chain rule is outlined in Equation (6.8)
$$$p(A,B,C,D) = P(A)P(B|A)P(C|A,B)P(D|A,B,C) \tag{6.8}$$$
The conditional probability property utilized is shown in (6.9)
$$$P(B|A) = {P(A,B) \over P(A)} \tag{6.9}$$$
We will now use Equation (6.10) in the example below to complete the LDA Inference task on a random sample of documents.
\begin{aligned} p(z_{i}|z_{\neg i}, w) &= {p(w,z)\over {p(w,z_{\neg i})}} = {p(z)\over p(z_{\neg i})}{p(w|z)\over p(w_{\neg i}|z_{\neg i})p(w_{i})}\\ \\ &\propto \prod_{d}{B(n_{d,.} + \alpha) \over B(n_{d,\neg i}\alpha)} \prod_{k}{B(n_{k,.} + \beta) \over B(n_{k,\neg i} + \beta)}\\ \\ &\propto {\Gamma(n_{d,k} + \alpha_{k}) \Gamma(\sum_{k=1}^{K} n_{d,\neg i}^{k} + \alpha_{k}) \over \Gamma(n_{d,\neg i}^{k} + \alpha_{k}) \Gamma(\sum_{k=1}^{K} n_{d,k}+ \alpha_{k})} {\Gamma(n_{k,w} + \beta_{w}) \Gamma(\sum_{w=1}^{W} n_{k,\neg i}^{w} + \beta_{w}) \over \Gamma(n_{k,\neg i}^{w} + \beta_{w}) \Gamma(\sum_{w=1}^{W} n_{k,w}+ \beta_{w})}\\ \\ &\propto (n_{d,\neg i}^{k} + \alpha_{k}) {n_{k,\neg i}^{w} + \beta_{w} \over \sum_{w} n_{k,\neg i}^{w} + \beta_{w}} \end{aligned} \tag{6.10}
To calculate our word distributions in each topic we will use Equation (6.11).
$$$\phi_{k,w} = { n^{(w)}_{k} + \beta_{w} \over \sum_{w=1}^{W} n^{(w)}_{k} + \beta_{w}} \tag{6.11}$$$
The topic distribution in each document is calcuated using Equation (6.12).
$$$\theta_{d,k} = {n^{(k)}_{d} + \alpha_{k} \over \sum_{k=1}^{K}n_{d}^{k} + \alpha_{k}} \tag{6.12}$$$
What if I don’t want to generate docuements. What if my goal is to infer what topics are present in each document and what words belong to each topic? This is were LDA for inference comes into play.
Before going through any derivations of how we infer the document topic distributions and the word distributions of each topic, I want to go over the process of inference more generally.
The General Idea of the Inference Process
1. Initialization: Randomly select a topic for each word in each document from a multinomial distribution.
2. Gibbs Sampling:
• For i iterations
• For document d in documents:
• For each word in document d:
• assign a topic to the current word based on probability of the topic given the topic of all other words (except the current word) as shown in Equation (6.10)
## 6.3 Animal Farm - Code Example
Now let’s revisit the animal example from the first section of the book and break down what we see. This time we will also be taking a look at the code used to generate the example documents as well as the inference code.
rm(list = ls())
library(MCMCpack)
library(tidyverse)
library(Rcpp)
library(knitr)
library(kableExtra)
library(lsa)
get_topic <- function(k){
which(rmultinom(1,size = 1,rep(1/k,k))[,1] == 1)
}
get_word <- function(theta, phi){
topic <- which(rmultinom(1,1,theta)==1)
# sample word from topic
new_word <- which(rmultinom(1,1,phi[topic, ])==1)
return(c(new_word, topic))
}
cppFunction(
'List gibbsLda( NumericVector topic, NumericVector doc_id, NumericVector word,
NumericMatrix n_doc_topic_count,NumericMatrix n_topic_term_count,
NumericVector n_topic_sum, NumericVector n_doc_word_count){
int alpha = 1;
int beta = 1;
int cs_topic,cs_doc, cs_word, new_topic;
int n_topics = max(topic)+1;
int vocab_length = n_topic_term_count.ncol();
double p_sum = 0,num_doc, denom_doc, denom_term, num_term;
NumericVector p_new(n_topics);
IntegerVector topic_sample(n_topics);
for (int iter = 0; iter < 100; iter++){
for (int j = 0; j < word.size(); ++j){
// change values outside of function to prevent confusion
cs_topic = topic[j];
cs_doc = doc_id[j];
cs_word = word[j];
// decrement counts
n_doc_topic_count(cs_doc,cs_topic) = n_doc_topic_count(cs_doc,cs_topic) - 1;
n_topic_term_count(cs_topic , cs_word) = n_topic_term_count(cs_topic , cs_word) - 1;
n_topic_sum[cs_topic] = n_topic_sum[cs_topic] -1;
// get probability for each topic, select topic with highest prob
for(int tpc = 0; tpc < n_topics; tpc++){
// word cs_word topic tpc + beta
num_term = n_topic_term_count(tpc, cs_word) + beta;
// sum of all word counts w/ topic tpc + vocab length*beta
denom_term = n_topic_sum[tpc] + vocab_length*beta;
// count of topic tpc in cs_doc + alpha
num_doc = n_doc_topic_count(cs_doc,tpc) + alpha;
// total word count in cs_doc + n_topics*alpha
denom_doc = n_doc_word_count[cs_doc] + n_topics*alpha;
p_new[tpc] = (num_term/denom_term) * (num_doc/denom_doc);
}
// normalize the posteriors
p_sum = std::accumulate(p_new.begin(), p_new.end(), 0.0);
for(int tpc = 0; tpc < n_topics; tpc++){
p_new[tpc] = p_new[tpc]/p_sum;
}
// sample new topic based on the posterior distribution
R::rmultinom(1, p_new.begin(), n_topics, topic_sample.begin());
for(int tpc = 0; tpc < n_topics; tpc++){
if(topic_sample[tpc]==1){
new_topic = tpc;
}
}
// print(new_topic)
// update counts
n_doc_topic_count(cs_doc,new_topic) = n_doc_topic_count(cs_doc,new_topic) + 1;
n_topic_term_count(new_topic , cs_word) = n_topic_term_count(new_topic , cs_word) + 1;
n_topic_sum[new_topic] = n_topic_sum[new_topic] + 1;
// update current_state
topic[j] = new_topic;
}
}
return List::create(
n_topic_term_count,
n_doc_topic_count);
}
')
# 3 topics - land sea & air
# birds and amphibious have cross over
# fish - sea 100
# land animals - 100 land
beta <- 1
k <- 3 # number of topics
M <- 100 # let's create 10 documents
alphas <- rep(1,k) # topic document dirichlet parameters
xi <- 100 # average document length
N <- rpois(M, xi) #words in each document
# whale1, whale2, FISH1, FISH2,OCTO
sea_animals <- c('\U1F40B', '\U1F433','\U1F41F', '\U1F420', '\U1F419')
# crab, alligator, TURTLE,SNAKE
amphibious <- c('\U1F980', '\U1F40A', '\U1F422', '\U1F40D')
# CHICKEN, TURKEY, DUCK, PENGUIN
birds <- c('\U1F413','\U1F983','\U1F426','\U1F427')
# SQUIRREL, ELEPHANT, COW, RAM, CAMEL
land_animals<- c('\U1F43F','\U1F418','\U1F402','\U1F411','\U1F42A')
vocab <- c(sea_animals, amphibious, birds, land_animals)
# equal probability 1/18
# 0 - animals that are not possible
# 1 - for shared
# 4 - non-shared
shared <- 2
non_shared <- 4
not_present <- 0
land_phi <- c(rep(not_present, length(sea_animals)),
rep(shared, length(amphibious)),
rep(non_shared, 2), # turkey and chicken can't fly
rep(shared, 2), # regular bird and pengiun
rep(non_shared, length(land_animals)))
land_phi <- land_phi/sum(land_phi)
sea_phi <- c(rep(non_shared, length(sea_animals)),
rep(shared, length(amphibious)),
rep(not_present, 2), # turkey and chicken can't fly
rep(shared, 2), # regular bird and pengiun
rep(not_present, length(land_animals)))
sea_phi <- sea_phi/sum(sea_phi)
air_phi <- c(rep(not_present, length(sea_animals)),
rep(not_present, length(amphibious)),
rep(not_present, 2), # turkey and chicken can't fly
non_shared, # regular bird
not_present, # penguins can't fly
rep(not_present, length(land_animals)))
air_phi <- air_phi/sum(air_phi)
# calculate topic word distributions
phi <- matrix(c(land_phi, sea_phi, air_phi), nrow = k, ncol = length(vocab),
byrow = TRUE, dimnames = list(c('land', 'sea', 'air')))
theta_samples <- rdirichlet(M, alphas)
thetas <- theta_samples[rep(1:nrow(theta_samples), times = N), ]
new_words <- t(apply(thetas, 1, function(x) get_word(x,phi)))
ds <-tibble(doc_id = rep(1:length(N), times = N),
word = new_words[,1],
topic = new_words[,2],
theta_a = thetas[,1],
theta_b = thetas[,2],
theta_c = thetas[,3]
)
ds %>% filter(doc_id < 3) %>% group_by(doc_id) %>% summarise(
tokens = paste(vocab[word], collapse = ' ')
) %>% kable(col.names = c('Document', 'Animals'),
caption ="Animals at the First Two Locations")
Table 6.1: Animals at the First Two Locations
Document Animals
1 🐑 🐋 🐂 🐓 🐑 🦃 🦃 🐪 🦃 🐓 🐑 🐂 🐋 🐪 🐢 🐂 🐑 🐓 🐍 🐑 🐟 🐍 🐙 🐿 🐪 🐘 🐑 🐙 🦃 🐘 🐑 🐂 🐂 🐊 🐂 🐘 🐿 🦃 🦃 🐧 🐪 🐍 🐿 🐘 🐍 🐂 🐙 🐦 🐂 🐟 🐍 🐠 🐑 🦃 🐓 🦃 🐂 🐘 🐠 🐓 🐢 🐘 🦃 🐙 🐘 🐍 🐢 🐑 🐍 🐘 🐦 🐧 🐙 🐦 🐂 🐋 🐘 🐠 🐓 🐿 🐘 🐿 🦀 🐓 🐿 🐊 🐓 🦀 🐂 🐪 🐓 🐠 🐪 🐦 🐙 🐠 🐘 🦃 🐘 🐑 🐊 🐑 🐳
2 🐠 🐟 🐋 🐋 🐢 🐳 🦀 🐦 🐘 🐦 🐓 🐦 🐦 🦃 🐦 🐦 🐂 🦀 🐢 🦀 🐊 🐿 🦃 🐳 🐦 🐠 🐪 🐍 🐠 🐪 🐧 🐂 🐿 🐋 🐦 🐦 🐦 🐿 🐋 🐿 🐠 🐦 🐳 🐂 🐦 🐙 🐟 🐑 🐪 🐪 🦀 🐑 🐊 🐦 🐦 🐿 🦃 🐙 🐿 🦃 🐓 🐊 🐦 🐢 🐙 🐊 🦀 🐙 🐍 🐓 🐍 🐳 🐋 🐙 🐧 🐢 🐋 🐘 🐑 🐍 🐑 🐳 🐦 🐠 🐂 🐧 🐙 🐍
The habitat (topic) distributions for the first couple of documents:
Table 6.2: Distribution of Habitats in the First Two Locations
Document Land Sea Air
1 0.8168897 0.1688578 0.0142525
2 0.3860697 0.4442854 0.1696449
With the help of LDA we can go through all of our documents and estimate the topic/word distributions and the topic/document distributions.
This is our estimated values and our resulting values:
######### Inference ###############
current_state <- ds %>% dplyr::select(doc_id, word, topic)
current_state$topic <- NA t <- length(unique(current_state$word))
# n_doc_topic_count
n_doc_topic_count <- matrix(0, nrow = M, ncol = k)
# document_topic_sum
n_doc_topic_sum <- rep(0,M)
# topic_term_count
n_topic_term_count <- matrix(0, nrow = k, ncol = t)
# colnames(n_topic_term_count) <- unique(current_state$word) # topic_term_sum n_topic_sum <- rep(0,k) p <- rep(0, k) # initialize topics current_state$topic <- replicate(nrow(current_state),get_topic(k))
# get word, topic, and document counts (used during inference process)
n_doc_topic_count <- current_state %>% group_by(doc_id, topic) %>%
summarise(
count = n()
) %>% spread(key = topic, value = count) %>% as.matrix()
n_topic_sum <- current_state %>% group_by(topic) %>%
summarise(
count = n()
) %>% select(count) %>% as.matrix() %>% as.vector()
n_topic_term_count <- current_state %>% group_by(topic, word) %>%
summarise(
count = n()
) %>% spread(word, count) %>% as.matrix()
# minus 1 in, add 1 out
lda_counts <- gibbsLda( current_state$topic-1 , current_state$doc_id-1, current_state\$word-1,
n_doc_topic_count[,-1], n_topic_term_count[,-1], n_topic_sum, N)
# calculate estimates for phi and theta
# phi - row apply to lda_counts[[1]]
# rewrite this function and normalize by row so that they sum to 1
phi_est <- apply(lda_counts[[1]], 1, function(x) (x + beta)/(sum(x)+length(vocab)*beta) )
rownames(phi_est) <- vocab
colnames(phi) <- vocab
theta_est <- apply(lda_counts[[2]],2, function(x)(x+alphas[1])/(sum(x) + k*alphas[1]))
theta_est <- t(apply(theta_est, 1, function(x) x/sum(x)))
# rewrite this function and normalize by row so that they sum to 1
phi_est <- apply(lda_counts[[1]], 1, function(x) (x + beta)/(sum(x)+length(vocab)*beta) )
rownames(phi_est) <- vocab
colnames(phi) <- vocab
theta_est <- apply(lda_counts[[2]],2, function(x)(x+alphas[1])/(sum(x) + k*alphas[1]))
theta_est <- t(apply(theta_est, 1, function(x) x/sum(x)))
colnames(theta_samples) <- c('land', 'sea', 'air')
vector_angles <- cosine(cbind(theta_samples,theta_est))[4:6, 1:3]
estimated_topic_names <- apply(vector_angles, 1, function(x)colnames(vector_angles)[which.max(x)])
phi_table <- as.tibble(t(round(phi,2))[,estimated_topic_names])
phi_table <- cbind(phi_table, as.tibble(round(phi_est, 2)))
# names(theta_table)[4:6] <- paste0(estimated_topic_names, ' estimated')
# theta_table <- theta_table[, c(4,1,5,2,6,3)]
names(phi_table)[4:6] <- paste0(estimated_topic_names, ' estimated')
phi_table <- phi_table[, c(4,1,5,2,6,3)]
row.names(phi_table) <- colnames(phi)
kable(round(phi_table, 2), caption = 'True and Estimated Word Distribution for Each Topic')
Table 6.3: True and Estimated Word Distribution for Each Topic
air estimated air land estimated land sea estimated sea
🐋 0.00 0 0.00 0.00 0.12 0.12
🐳 0.00 0 0.02 0.00 0.12 0.12
🐟 0.00 0 0.01 0.00 0.12 0.12
🐠 0.00 0 0.01 0.00 0.12 0.12
🐙 0.00 0 0.01 0.00 0.13 0.12
🦀 0.00 0 0.05 0.05 0.06 0.06
🐊 0.01 0 0.04 0.05 0.06 0.06
🐢 0.00 0 0.05 0.05 0.07 0.06
🐍 0.01 0 0.06 0.05 0.05 0.06
🐓 0.01 0 0.09 0.10 0.00 0.00
🦃 0.00 0 0.11 0.10 0.00 0.00
🐦 0.92 1 0.01 0.05 0.08 0.06
🐧 0.01 0 0.05 0.05 0.06 0.06
🐿 0.01 0 0.09 0.10 0.01 0.00
🐘 0.00 0 0.10 0.10 0.00 0.00
🐂 0.00 0 0.11 0.10 0.00 0.00
🐑 0.00 0 0.10 0.10 0.00 0.00
🐪 0.00 0 0.11 0.10 0.00 0.00
The document topic mixture estimates are shown below for the first 5 documents:
Table 6.4: The Estimated Topic Distributions for the First 5 Documents
Location air estimated air land estimated land sea estimated sea
1 0.04 0.01 0.90 0.82 0.06 0.17
2 0.19 0.17 0.48 0.39 0.33 0.44
3 0.31 0.40 0.13 0.18 0.56 0.42
4 0.82 0.64 0.17 0.33 0.01 0.03
5 0.31 0.29 0.31 0.41 0.39 0.30
6 0.35 0.32 0.11 0.15 0.54 0.53 | HuggingFaceTB/finemath | |
# Heap (mathematics)
In abstract algebra, a heap (sometimes also called a groud[1]) is a mathematical generalization of a group. Informally speaking, a heap is obtained from a group by "forgetting" which element is the unit, in the same way that an affine space can be viewed as a vector space in which the 0 element has been "forgotten". A heap is essentially the same thing as a torsor, and the category of heaps is equivalent to the category of torsors, with morphisms given by transport of structure under group homomorphisms, but the theory of heaps emphasizes the intrinsic composition law, rather than global structures such as the geometry of bundles.
Formally, a heap is an algebraic structure consisting of a non-empty set H with a ternary operation denoted ${\displaystyle [x,y,z]\in H}$ that satisfies
• the para-associative law
${\displaystyle [[a,b,c],d,e]=[a,[d,c,b],e]=[a,b,[c,d,e]]\ \forall \ a,b,c,d,e\in H}$
• the identity law
${\displaystyle [a,a,x]=[x,a,a]=x\ \forall \ a,x\in H.}$
A group can be regarded as a heap under the operation ${\displaystyle [x,y,z]=xy^{-1}z}$. Conversely, let H be a heap, and choose an element eH. The binary operation ${\displaystyle x*y=[x,e,y]}$ makes H into a group with identity e and inverse ${\displaystyle x^{-1}=[e,x,e]}$. A heap can thus be regarded as a group in which the identity has yet to be decided.
Whereas the automorphisms of a single object form a group, the set of isomorphisms between two isomorphic objects naturally forms a heap, with the operation ${\displaystyle [f,g,h]=fg^{-1}h}$ (here juxtaposition denotes composition of functions). This heap becomes a group once a particular isomorphism by which the two objects are to be identified is chosen.
## Examples
### Two element heap
If ${\displaystyle H=\{a,b\}}$ then the following structure is a heap:
${\displaystyle [a,a,a]=a,\,[a,a,b]=b,\,[b,a,a]=b,\,[b,a,b]=a,}$
${\displaystyle [a,b,a]=b,\,[a,b,b]=a,\,[b,b,a]=a,\,[b,b,b]=b.}$
### Heap of a group
As noted above, any group becomes a heap under the operation
${\displaystyle [x,y,z]=xy^{-1}z.}$
One important special case:
#### Heap of integers
If ${\displaystyle x,y,z}$ are integers, we can set ${\displaystyle [x,y,z]=x-y+z}$ to produce a heap. We can then choose any integer ${\displaystyle k}$ to be the identity of a new group on the set of integers, with the operation ${\displaystyle *}$
${\displaystyle x*y=x+y-k}$
and inverse
${\displaystyle x^{-1}=2k-x}$.
## Generalizations and related concepts
• A pseudoheap or pseudogroud satisfies the partial para-associative condition[2]
${\displaystyle [[a,b,c],d,e]=[a,b,[c,d,e]].}$
• A semiheap or semigroud is required to satisfy only the para-associative law but need not obey the identity law.[3]
An example of a semigroud that is not in general a groud is given by M a ring of matrices of fixed size with
${\displaystyle [x,y,z]=x\cdot y^{\top }\cdot z}$
where • denotes matrix multiplication and ⊤ denotes matrix transpose.[3]
${\displaystyle [a,a,[b,b,x]]=[b,b,[a,a,x]]}$ and ${\displaystyle [[x,a,a],b,b]=[[x,b,b],a,a]}$ for all a and b.
A semigroud is a generalised groud if the relation → defined by
${\displaystyle a\rightarrow b\Leftrightarrow [a,b,a]=a}$
is reflexive (idempotence) and anti-symmetric. In a generalised groud, → is an order relation.[4]
## Notes
1. Schein (1979) pp.101-102: footnote (o)
2. Vagner (1968)
3. {{#invoke:Citation/CS1|citation |CitationClass=journal }}Template:Clarify
4. Schein (1979) p.104
## References
• {{#invoke:citation/CS1|citation
|CitationClass=book }}
• {{#invoke:Citation/CS1|citation
|CitationClass=journal }} | HuggingFaceTB/finemath | |
# Measurability of closed opeartor
Given separable Banach spaces $$(E, \Vert \cdot \Vert_E)$$ and $$(F, \Vert \cdot \Vert_F)$$. The Banach space $$E$$ is endowed with sigma algebra $$\mathcal{F}$$ which is generated by the open set of it. Similarly, let $$\mathcal{G}$$ be the sigma algebra of space $$F$$. Say $$A: D(A) \to F$$ is a closed operator where $$D(A)$$ is a subspace of $$E$$ and we assume $$D(A) \in \mathcal{F}.$$ My question is that is the map $$A: (D(A), \mathcal{F}|_{D(A)}) \to (F, \mathcal{G})$$ necessarily measurable? I know that when $$D(A)$$ is closed this is true, since via closed graph theorem, $$A$$ is in fact continuous. But in general, I don't know how to prove it.
In the separable case, it follows from the following standard result from descriptive set theory:
Theorem. Let $$X,Y$$ be Polish spaces, let $$f : X \to Y$$ be a Borel function, and let $$B \subset X$$ be a Borel set. If the restriction of $$f$$ to $$B$$ is one-to-one, then $$f(B)$$ is a Borel subset of $$Y$$, and the restriction of $$f^{-1}$$ to $$f(B)$$ is a Borel function.
See for instance Proposition 4.5.1 of Srivastava, A Course on Borel Sets.
Now to apply this theorem, let $$X = E \oplus F$$, let $$Y = E$$, and let $$f = \pi_E : E \oplus F \to E$$ be the projection onto $$E$$, which is continuous and in particular Borel. Let $$B$$ be the graph of $$A$$, which by assumption is closed and in particular Borel. Note $$\pi_E(B) = D(A)$$. Since $$B$$ is a graph, $$\pi_{E}|_B$$ is one-to-one, so by the theorem above, $$D(A)$$ is Borel and the restriction of $$\pi_E^{-1}$$ to $$D(A)$$, which is simply the map $$x \mapsto (x, Ax)$$, is Borel. Then $$A$$ is just the composition of this map with the continuous map $$\pi_F$$.
• @Nate Eldredge I see, thanks a lot for the help! – lye012 May 4 at 1:41
The answer is yes if you assume that $$E$$ and $$F$$ are separable. In the non-separable case I suspect it is false but someone else can write a separate answer for that.
To prove this (in the separable case), it suffices to prove that the functional $$f(x):= \|Ax\|_F$$ is measurable as a map from $$(D(A),\mathcal F|_{D(A)}) \to (\Bbb R_+,$$ Borel$$)$$. Indeed, if we can show that $$f$$ is measurable then the claim can be obtained as follows. Denote by $$B_F(z,r)$$ the ball of radius $$r>0$$ around $$z\in F$$. Note that for each $$y \in D(A)$$, the preimage of $$B_F(Ay,r)$$ under $$A$$ is just a translate of the set $$f^{-1}[0,r)$$ by the vector $$y$$, hence it is measurable. Next we take an open set $$U$$ in $$F$$. For each $$x \in A^{-1}(U)$$, choose $$r_x>0$$ such that $$B_F(Ax,r_x) \subset U$$. Then let $$U' = \bigcup_{x\in A^{-1}(U)} B_F(Ax,r_x)$$, and notice that $$A^{-1}(U) = A^{-1}(U')$$. Now since $$F$$ is separable, the open cover $$\{B_F(Ax,r_x)\}_{x\in U}$$ admits a countable subcover $$\{B_F(Ax_i,r_{x_i})\}_{i \in \Bbb N}$$. Then $$A^{-1}(U) = A^{-1}(U') = \bigcup_{i \in \Bbb N} A^{-1}(B_F(Ax_i,r_{x_i}))$$, which is measurable.
We will denote the graph norm by $$\|x\|_G := \|x\|_E+\|Ax\|_F$$, which makes $$D(A)$$ into a Banach space. To prove that $$f$$ is measurable, we can write it as the supremum of a countable collection of continuous functions. First notice by this other question that $$D(A)$$ is separable with respect to the graph norm (because $$E,F$$ are both separable) so let $$\{x_1,x_2,...\}$$ be a dense subset of its unit sphere. Use Hahn-Banach to obtain bounded linear functionals $$f_n:E \to \Bbb R$$ with the property that $$f_n(x_n) = 1$$ and $$\|f_n\|_{E \to \Bbb R} = \|x_n\|_E^{-1}$$ and also $$\|f_n\|_{G\to \Bbb R}=1$$ (to do this, one needs to apply Hahn-Banach where the dominating convex functional is the lower convex envelope of $$x\mapsto \min\{\|x_n\|_E^{-1} \|x\|_E, \|x\|_G\}$$).
Then we claim that $$\sup_n f_n = f$$ on $$D(A)$$. Indeed, on any separable Banach space it is true that if $$\{x_n\}$$ is a dense subset of the unit sphere and $$f_n$$ are linear functionals on $$B$$ such that $$f_n(x_n) = 1$$ and $$\|f_n\|=1$$ then $$\sup_n f_n = \|\cdot\|$$. See my answer here for the proof.
• I don't see why proving the measurability of $f$ is sufficient. On its face it seems that it only proves that the pre-images of open balls centered at the origin are measurable, and those do not generate the Borel $\sigma$-algebra. – Nate Eldredge May 3 at 20:11
• I have in mind the example in $\mathbb{R}$ of $g = 1_N - 1_{N^c}$ where $N$ is non-measurable. Then $|g|$ is measurable but $g$ is not. – Nate Eldredge May 3 at 20:12
• I made a clarification that both $E$ and $F$ should be separable. – lye012 May 3 at 20:56
• @NateEldredge no but the point here is that $A$ is linear so you can exploit that. See the modification above (and thanks for pointing out that detail). – Shalop May 3 at 21:54
• Sorry there was a bunch of notation in this answer that made no sense before, but it should be fixed now. – Shalop May 3 at 22:55 | open-web-math/open-web-math | |
# vertex; axis of symmetry; Calculate the y-intercept; Use the axis of symmetry and the y-intercept to find an additional point on the graph; Graph the function please help. This is college algebra not calculus. confuse
Given a quadratic function find: axis of symmetry, y-intercept. Use the axis of symmetry and the y-intercept to find an additional point on the graph then graph the function.
Let `y=ax^2+bx+c` e.g x^2+7x+12 (Here a=1,b=7,c=12)
(1) The axis of symmetry is the vertical line `x=-b/(2a)` . For the example `x=-7/2` .
(2) The y-intercept is where x=0, or c. For the example the y-intercept is 12.
(3) The line x=0 is 7/2 to the right of the axis of symmetry, so there is a point with y value of 12 when `x=-7/2-7/2=-7` . Thus there is a point (0,12) (the y-intercept) and a point (-7,12), a reflection of the intercept across the line of symmetry. (Imagine folding the paper along the line `x=-7/2` and pairing up points.)
(4) The graph is a "U" opening up (Since a>0). The example yields
** If the equation is given in another form there are other approaches. **
Approved by eNotes Editorial Team | HuggingFaceTB/finemath | |
## Math 113: Study Guide Chapters 9-10
1. Know the assumptions / properties of Pearson's Linear Correlation Coefficient.
1. What values is it between?
2. What does a value of zero mean / not mean?
3. What happens if you change the scale of either variable?
4. What happens if you switch the variables?
5. What type of relationship does it measure?
6. What type of distribution does it have?
7. How many degrees of freedom does it have?
8. What type of distribution do the ordered pairs (x,y) have?
2. Know the assumptions / properties of the contingency tables.
1. What is the null hypothesis?
2. How are the sample data selected?
3. What requirement must be met?
4. What type of data is used?
5. What type of distribution does it have?
6. How many degrees of freedom does it have?
7. What type of test is it?
3. Know the assumptions / properties of multinomial experiments.
1. What is the null hypothesis?
2. What requirement must be met?
3. What is the sample data?
4. What distribution does it have?
5. How many degrees of freedom does it have?
6. What type of test is it?
4. Know the guidelines for using the regression equation.
1. Know the guidelines for using a regression equation from page 501.
2. What is the equation that should be used if there is no significant linear correlation (pg 500)?
5. Know the properties of multiple regression.
1. When does the largest value of R-square occur?
2. When does the largest value of the adjusted R-square occur?
3. How is the Analysis of Variance used to test the regression equation?
4. How does correlation between independent variables affect the choice of variables?
5. What tools can be used to perform multiple regression.
6. What are the degrees of freedom?
6. Contingency Table. Use Statdisk. Write the null hypotheses. Find one (not all of them) expected frequency. Identify the degrees of freedom. Find the p-value. Write the decision and conclusion.
7. Know what happens to the linear correlation coefficient when the data is manipulated. Three parts.
8. Know what happens to the test statistic of a multinomial experiment when the data is manipulated. Two parts.
9. Know what happens to the test statistic of a contingency table when the data is manipulated. Two parts.
10. Work a chi-square goodness of fit problem using Statdisk. The observed frequencies are given to you.
11. Work a linear regression problem using Statdisk. You will need to load data files and copy / paste data. Use the regression equation to estimate the dependent variable for a specific value of the independent variable.
12. The value of the correlation coefficient, r, and the total variation are given. Find the coefficient of determination, explained variation, and unexplained variation.
13. Software was used to test for normality. The p-value is given. Write the conclusion.
14. Use software to create a graphical correlation matrix and determine which variables seem related to each other. Perform multiple regression and record the coefficients and p-values for each independent variable. Also record the r-square and adjusted r-square value. Determine which variable is least significant by looking at the p-values and re-perform the multiple regression without that variable. Did the adjusted r-square increase or decrease by eliminating the variable?
15. Use software to perform linear regression. Write the null hypothesis. Find the value of the correlation coefficient and the p-value. What is the decision? What is the conclusion? Should the regression equation be used? Write the equation that should be used (either the regression equation if significant or the mean of the dependent variable if not). Look at the normal probability plots and determine if the data appears normally distributed.
### Notes
• It is possible to get more than 100 points on the exam.
# 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Total Pts 7 6 6 5 6 12 6 4 4 8 10 6 2 10 14 106 | HuggingFaceTB/finemath | |
# Conic Section Orbits
## by Greg Egan
The aim of this page is to assemble a variety of simple proofs for the wonderful fact, first proved by Isaac Newton, that the orbits of bodies subject to an inverse-square force are conic sections: ellipses, circles, parabolas, hyperbolas or straight lines.
Although conic sections are defined in the first instance as the curves produced when a plane intersects a cone, these curves have a wealth of well-known properties that characterise them equally well. The idea of this page is to explore the relationship between the inverse-square force law and as many of these properties as possible.
## Notation
• The vector from the centre of attraction to the orbiting body is r; the distance is r = |r|.
• The constant for the acceleration due to the the inverse-square force is k, so the acceleration has magnitude a = k/r2.
• The acceleration vector is a = –(k/r3) r, where the extra factor of r in the denominator is needed to divide out the magnitude from the vector r.
• The velocity vector for the orbiting body is v, and its speed is v = |v|.
• The total energy (kinetic plus potential) per unit mass of the orbiting body is E = ½v2k/r.
• The angular momentum per unit mass of the orbiting body is L = v · Rr = –r · Rv, where R is a matrix that rotates counter-clockwise by 90 degrees within the plane of the orbit.
## The Whispering Gallery Proof
If you stand at one focus of a room with elliptical walls, any sound waves that leave you in a horizontal direction and bounce off the walls will be reflected along paths that take them through the other focus of the ellipse. This is a characteristic property of the geometry of ellipses: if you draw a line from one focus out to the curve, and reflect it as if in a mirror tangent to the curve, the reflected ray will pass through the other focus.
Our aim will be to show that the path of a body moving under an inverse-square force has that “whispering gallery” property. Later, we’ll describe the slightly modified forms this property takes for parabolas and hyperbolas.
In the case of an orbit, a tangent to the curve is given by the velocity vector v, while a vector perpendicular to our imaginary mirror is Rv, where R rotates vectors counter-clockwise by 90 degrees.
The law of reflection for a vector x in a mirror with a unit-length normal n is:
xx – 2 (x · n) n
so if we reflect the vector r in the mirror whose unit-length normal is Rv/v we get:
rr – 2 (r · Rv/v) Rv/v
We can multiply this reflected vector by v2 to get a more convenient vector to work with, still pointing in the same direction:
s = v2 r – 2 (r · Rv) Rv
= v2 r + 2 L Rv
Here we’ve made use of the fact that the dot product r · Rv is just the opposite of the angular momentum per unit mass, L. By construction, |s| = v2 r. The rate of change of s is:
ds/dt = 2 (v · a) r + v2 v + 2 L Ra
= –(2 k/r3)[(v · r) r + L Rr] + v2 v
It turns out that ds/dt is parallel to v. We can show this by taking its dot product with the vector Rv, which is orthogonal to v:
Rv · (ds/dt) = –(2 k/r3) [(v · r) (Rv · r) + L (Rv · Rr)]
= –(2 k/r3) [(v · r) (–L) + L (v · r)]
= 0
To determine the factor by which v must be multiplied to give us ds/dt, we take the dot product of the two:
v · (ds/dt) = –(2 k/r3) [(v · r)2 + L2] + v4
= –(2 k/r3) [v2 r2] + v4
= v2 (v2 – 2 k/r)
= 2 v2 E
The simplification of the expression in square brackets between the first and second lines comes from noting that L = –Rv · r, which means that the two terms here are the squares of the coordinates of r in an orthonormal basis {v/v, – Rv/v}, multiplied by v2. Hence their sum is the squared length of r, multiplied by v2.
It follows that:
ds/dt = 2 E v
So not only is ds/dt always parallel to v (which will be true for any central force), it’s equal to a constant multiple of v. That’s only true for the inverse-square law.
Making use of this relationship, we can now construct a constant vector z:
z = rs / (2 E)
dz/dt = dr/dt – (ds/dt) / (2 E)
= vv
= 0
This tells us that by travelling the right distance along the reflected ray, we will always pass through the same fixed point: if the centre of attraction lies at the origin of our coordinates, that point is z.
So we’ve established the “whispering gallery” property for the orbit: rays from one focus, at the centre of attraction, will all be reflected so that they pass through the second focus at z.
The vector z is almost the same as a famous conserved vector known as the Laplace-Runge-Lenz vector, but although the Laplace-Runge-Lenz vector points along the axis of the ellipse, it’s not usually defined as the displacement vector from one focus to the other. For ease of comparison, we can rewrite z as follows:
Rv = n × v, where n is a unit normal to the plane of the orbit, and × is the vector cross product
L Rv = L × v, where L is the orbiting body’s angular momentum vector divided by its mass
s = v2 r + 2 L × v
z = r – (v2 r + 2 L × v) / (2 E)
= (2 E rv2 r – 2 L × v) / (2 E)
= (2 (½v2k/r) rv2 r – 2 L × v) / (2 E)
= (v × Lk (r/r) ) / E
The exact definition of the Laplace-Runge-Lenz vector varies from author to author, and in any comparisons we need to keep in mind that our L and E are per unit mass, and our k is a constant that yields acceleration (as opposed to force) when divided by r squared. But in any case, the LRL vector is usually not divided by energy, as z is, and what most authors differ on are overall factors of the orbiting body’s mass.
For elliptical (and circular) orbits, the energy E is negative. What happens if E is zero? In that case we can’t construct the vector z, but since ds/dt = 0 the reflected vector s will now always point in the same direction. This is the characteristic property of a parabola, which reflects rays from its single focus so that they all end up parallel.
If E is positive, z lies on the opposite side of the trajectory from the centre of attraction. The orbit becomes a hyperbola, with the reflected rays diverging away from a virtual focus.
E < 0, elliptical orbit E > 0, hyperbolic orbit E = 0, parabolic orbit Any ray from either focus will be reflected so that it passes through the other. Any ray from either focus will be reflected so that it seems to come from the other. All rays from the focus will be reflected in a common direction.
## The Pins, String and Pencil Proof
To draw an ellipse, place a sheet of paper on a cork board and insert two pins where you’d like the two foci of the ellipse to be. Take a loop of string and encircle the pins with it, then insert the tip of a pencil in the loop and make the string taut, so it forms a triangle. As you drag the tip of the pencil around in a closed curve, the sum of the distances of its tip from the two foci will remain constant. This is probably the best-known definition of an ellipse — and it generalises to the other conic sections.
E < 0, elliptical orbit E > 0, hyperbolic orbit E = 0, parabolic orbit The sum of the distances from the two foci remains constant. The difference between the distances from the two foci remains constant. The distance from the focus is equal to the distance from the directrix.
We can adapt the proof from the previous section to show that our inverse-square-law orbits also have the “pins, string and pencil” property. The distance of the orbiting body from the centre of attraction is simply r. But what is the distance of the body from the fixed point z? We started by reflecting the vector r, producing a new vector of the same length, r, but then we multiplied that reflected vector by v2 to get s. So |s| = v2 r. But to reach z from the orbiting body, we used s / (–2 E). So the orbiting body is at a distance of v2 r/(|2 E|) from z. So long as E < 0, this gives us:
Sum of distances from the two foci
= r + v2 r/(–2 E)
= r (1 – v2 /(2 E))
= r (2 Ev2)/(2 E)
= r ((v2 – 2 k/r) – v2)/(2 E)
= –k / E
This is constant, and since E is negative it will be a positive quantity.
You can’t draw a hyperbola with the same method, but the algebra goes through identically to show that the difference between the distances from the two foci is constant, with its absolute value equal to k / E for positive E.
The related definition of a parabola is that it’s the curve equidistant from a point, the focus, and a straight line known as the directrix that lies perpendicular to the axis of symmetry. The vector s, which is constant for a parabola, will be parallel to that axis.
We can find the axial component of the distance of the orbiting body from the focus by taking the dot product of r with a unit vector parallel to the axis: s/|s| = s/(v2 r).
Axial component of distance from focus
= r · s/(v2 r)
= r · (v2 r + 2 L Rv) / (v2 r)
= (v2 r2 + 2 L (r · Rv)) / (v2 r)
= (v2 r2 – 2 L2) / (v2 r)
= r – 2 L2 / (v2 r)
Since the energy is zero, we have:
2 E = v2 – 2 k/r = 0
v2 r = 2 k
Applying this to the previous result we have:
Axial component of distance from focus = rL2 / k
So if we place the directrix L2 / k away from the focus in the opposite direction to s, we will have:
Perpendicular distance from directrix
= L2 / k + Axial component of distance from focus
= r
This shows that our orbit for E = 0 fulfills the definition of a parabola.
## The Focus-Directrix Proof
In the previous section we described the parabola as the curve that’s equidistant from a point, the focus, and a line, the directrix. But it turns out that we can generalise this to describe the other conic sections in a very similar way. We simply require that instead of being equal, the distance from the focus and the distance from the directrix maintain a fixed ratio, known as the eccentricity, ε.
ε < 1, elliptical orbit ε > 1, hyperbolic orbit ε = 1, parabolic orbit The distance from the focus divided by the distance from the directrix is a constant, less than 1. The distance from the focus divided by the distance from the directrix is a constant, greater than 1. The distance from the focus is equal to the distance from the directrix.
The circle can only be constructed this way as a limiting case, where the eccentricity ε approaches zero while the distance to the directrix goes to infinity, with the product of the two held constant at the radius of the circle.
Although we’ve already proved the desired relationship for the parabola, we’ll now show that this constant ratio of distances holds for all the orbits produced by the inverse-square law.
We’ll adopt coordinates as shown in the image on the right, with the origin at the focus and the directrix having an x-coordinate of –D. The distance from the directrix to some point (x, y) on the curve is then simply x+D.
The defining relationship of the conic section is:
r = ε (x + D)
It follows that:
x = r / ε – D
cos θ = x / r = 1 / ε – D / r
Differentiating, we get:
–sin θ dθ = (D / r2) dr
dr/dθ = –sin θ (r2 / D) = –y r / D
(dr/dθ)2 = (r / D)2 y2
= (r / D)2 (r2x2)
= (r / D)2 (r2 – (r / ε – D)2)
The last equation can be rewritten as:
(dr/dθ)2 = (1 – 1/ε2) r4 / D2 + 2 r3 / (D ε) – r2
Now, what we wish to show is that the inverse-square force results in a relationship between (dr/dθ)2 and r that takes the same form as this. We start with the basic statement of conservation of energy, and convert the squared velocity into polar coordinates:
E = ½v2k/r
2 E + 2 k / r = v2
= r2 (dθ/dt)2 + (dr/dt)2
= (dθ/dt)2 [r2 + (dr/dθ)2]
In the last step we’ve used the chain rule: dr/dt = (dr/dθ) (dθ/dt). From conservation of angular momentum:
dθ/dt = L / r2
Combining these, we have:
(L / r2)2 [r2 + (dr/dθ)2] = 2 E + 2 k / r
Solving for (dr/dθ)2 gives us:
(dr/dθ)2 = (2 E / L2) r4 + (2 k / L2) r3r2
So we’ve ended up with exactly the same form for this relationship as we obtained by fixing the focus-directrix ratio. By choosing suitable values for D and ε we can make the two equations identical.
## From Shape To Force Law
If we reverse the direction of our argument and assume that a body’s path through space is a conic section, and that its motion is due to some kind of central force with a centre of attraction at one focus of the conic, we can prove very easily that the force must follow the inverse-square law.
In the diagram on the right, we’ve assumed the orbit is an ellipse, and so the sum of the distances to the two foci is equal to some constant, B, while Rv, being perpendicular to the curve, bisects the angle subtended by the two foci.
By conservation of energy with a central-force potential U(r):
E = ½v2 + U(r)
v2 = 2 (EU(r))
By the definition of angular momentum, we have:
L = v r sin β = v r cos α
L2 = v2 r2 cos2 α = 2 r2 (EU(r)) cos2 α
Using the cosine law for triangles and a basic trigonometric identity, we have:
cos 2α = (r2 + (Br)2C2) / (2 r (Br))
cos2 α = ½ (1 + cos 2α) = ½ (B2C2) / (2 r (Br))
L2 = ½ (B2C2) (EU(r)) / (–1 + B/r)
But we know that L2 is a constant, independent of r, and this can only be true if EU(r) is some constant multiple of –1 + B/r, allowing the dependence on r to cancel out. That means U(r) must be inversely proportional to r, giving an inverse-square force law.
Very similar arguments can be made for the hyperbola and the parabola.
Note that cos2 α, which measures the extent to which the tangent to the orbit deviates from a purely circular direction, can also be expressed in the intuitively appealing form:
cos2 α = r+ r / (r (r+ + rr))
where r+ = ½ (B + C) is the aphelion distance and r = ½ (BC) is the perihelion distance. This is clearly equal to 1 when r = r±.
How does this change if we place the centre of attraction at the centre of the ellipse, rather than at a focus? We then have:
cos2 α = a2 b2 / (r2 (a2 + b2r2))
where a and b are the semi-axes of the ellipse. This is clearly 1 when r = a or b. With this new assumption about the centre of attraction, the expression for L2 becomes:
L2 = 2 a2 b2 (EU(r)) / (a2 + b2r2)
In order for L2 to be constant in this case, EU(r) must equal a constant multiple of a2 + b2r2, meaning U(r) must be proportional to r2. This describes a two-dimensional harmonic oscillator potential.
## The Squared Harmonic Oscillator Proof
Imagine an ellipse centred on the origin in the complex plane, with its major axis lying on the x-axis. Suppose we “square” every point of this ellipse, using the rules of complex arithmetic:
(x + y i)2 = (x2y2) + 2 x y i
What does the new curve, consisting of the squared points, look like?
If our ellipse has semi-axes a and b, any point on it can be written in the form:
(x, y) = (a cos φ, b sin φ)
where we should stress that φ is just a parameter along the curve, not the polar angle of this complex number. We’ll define the coordinates of our squared points as X = x2y2 and Y = 2 x y. We then have:
(X, Y) = (a2 cos2 φ – b2 sin2 φ, 2 a b sin φ cos φ)
= (½(a2b2) + ½(a2 + b2) cos 2φ, a b sin 2φ)
This describes an ellipse centred at (½(a2b2), 0), with semi-axes of ½(a2 + b2) along the X-axis and ab along the Y-axis.
Where are the foci of this new ellipse? By symmetry they will lie at (½(a2b2) ± c, 0), for some quantity c. The sum of the distances of any point on the ellipse from the two foci must be equal to the full length of the major axis: a2 + b2. Since the vertical semi-axis is ab, the point on the top of the ellipse is (½(a2b2), ab), and its distance from each focus is √(c2 + a2 b2). So we have:
c2 + a2 b2 = [½(a2 + b2)]2
c = ½(a2b2)
That means the two foci are (0,0) and (a2b2, 0). One focus is the origin!
Squaring every point of an ellipse centred on the origin of the complex plane yields a new ellipse whose focus lies at the origin.
What’s more, squaring a complex number is a conformal map: it leaves the angles between curves unchanged. So the angle beween the original ellipse and a circle of radius r at the point where they intersect (such as α in the diagram above) will be exactly the same as the angle between the new ellipse and a circle of radius r2. (To be precise, squaring is a conformal map on the complex plane excluding the origin. So the right angle between the real and imaginary axes where they meet at the origin is not preserved by the map.)
Now, consider a particle that’s free to move in two dimensions, subject to an attractive force that’s a constant multiple of its distance from a centre of attraction. A system like this is known as a two-dimensional harmonic oscillator. If we choose coordinates whose origin lies at the centre of attraction, the particle’s acceleration will take the form:
d2r(t)/dt2 = –ω2 r(t)
where ω is related to the force constant and the particle’s mass. In Cartesian coordinates, the particle’s x and y coordinates will both oscillate with the frequency ω, while being free to take on arbitrary amplitudes and phases:
x(t) = C sin(ω t + φ)
y(t) = D sin(ω t + ψ)
It’s easy to check that this satisfies our equation for the acceleration. By a suitable rotation of the coordinates and choice of origin for t, we can always put this trajectory into the form:
x(t) = a cos(ω t)
y(t) = b sin(ω t)
with ab. Clearly this traces out an ellipse with semi-axes a and b, centred on the origin, with the major axis lying on the x-axis.
The acceleration equation corresponds to the harmonic oscillator potential (given here as potential energy per unit mass):
USHO(r) = ½ ω2 r2
At the end of the previous section, we claimed without proof that the squared cosine of the angle α for an ellipse centred on the origin was:
cos2 α = a2 b2 / (r2 (a2 + b2r2))
But since we’ve now established independently that USHO gives rise to elliptical orbits centred on the origin, we can derive cos2 α from its relationship with the angular momentum:
L2 = v2 r2 cos2 α = 2 r2 (EUSHO(r)) cos2 α
cos2 α = L2 / [2 r2 (E – ½ ω2 r2)]
Consider the particle at t=0, when it lies at the point (a, 0). Taking the derivative of the trajectory, its velocity is (0, ωb). Its potential energy (per unit mass) will be ½ω2 a2, its kinetic energy (per unit mass) will be ½ ω2 b2, and its angular momentum (per unit mass) will be L = ωab. So the equation above gives us the result we claimed:
cos2 α = ω2 a2 b2 / [2 r2 ½ω2 (a2 + b2r2)] = a2 b2 / (r2 (a2 + b2r2))
Because squaring is a conformal map that preserves angles, the formula for cos2 α for a “squared” ellipse whose focus lies on the origin must take the same form, with r2 replaced by r and a2 and b2 replaced by the aphelion and perihelion distances r±:
cos2 α = r+ r / (r (r+ + rr))
In order for the angular momentum to be constant for a body moving around this new ellipse, we’re led again to a potential that’s inversely proportional to r, and an inverse-square force law.
L2 = v2 r2 cos2 α = 2 r2 (EU(r)) r+ r / (r (r+ + rr))
EU(r) = [L2 / (2 r+ r)] [–1 + (r+ + r) / r ]
The connection between the two kinds of force laws can be made in much more detail than this, and in fact it’s possible to map trajectories for a particle in the harmonic oscillator moment-by-moment into trajectories for the inverse square law, by means of a suitable re-scaling of the time coordinate[1][2].
## Dandelin Spheres
No discussion of conic sections would be complete without exhibiting an actual cone. We will do so with the aid of a beautiful construction known as the Dandelin spheres.
Pick two points F1 and F2 in a plane whose normal vector is n. Place a sphere of radius ρ1 on one side of the plane, tangent to it at F1, and a sphere of radius ρ2 on the other side of the plane, tangent to it at F2. (In the view on the right, the plane is seen edge-on.)
Now construct a cone so that both spheres sit inside it, tangent to its sides. The intersection of this cone and our chosen plane will be an ellipse with foci F1 and F2.
If the distance between the foci is 2 c, the semi-major axis of the ellipse will be a = √(c2 + ρ1 ρ2) and the semi-minor axis will be b = √(ρ1 ρ2). The geometry of the ellipse itself doesn’t fix a unique pair of Dandelin spheres, or a unique cone. One simple choice would be to set ρ1 = r, the perihelion distance, and ρ2 = r+, the aphelion distance, but in what follows we won’t rely on any particular choice.
By definition, the intersection of a cone and a plane that forms a simple closed curve like this is an ellipse, but we can use this construction to prove some of the properties of the ellipse that we assumed in previous sections. Dandelin spheres can also be used to establish the analogous properties of hyperbolas and parabolas.
### The Pins, String and Pencil Property
To prove that the sum of the distances from the two foci to any point P on the ellipse is constant, consider the diagram on the left. The points P1 and P2 are found by passing a line through P to the tip of the cone, S, and intersecting it with the circles where the two spheres are tangent to the cone. The planes in which those circles lie are both orthogonal to the axis of the cone, so the total distance P1P2 is clearly constant, independent of P.
But the distance PF1 is equal to the distance PP1, since both these line segments are tangents from the same point, P, to the same sphere. Similarly, the distance PF2 is equal to the distance PP2. So PF1 + PF2 = PP1 + PP2 = P1P2, a constant.
### The Focus-Directrix Property
The planes through the circles where the two spheres are tangent to the cone intersect the original plane along two lines; these lines are the directrices of the ellipse, D1 and D2. We will project our generic point P on the ellipse to points Q1 and Q2 on the directrices.
The ratio between the distance of a point P on the ellipse to one focus, PF1, and the distance to the corresponding directrix, PQ1, is the same as the ratio PP1 to PQ1. Now, the line segments P1Q1 and P2Q2 are parallel, since they lie in the intersections of two parallel planes with the plane that contains all four endpoints. This means the triangle P1PQ1 is similar to the triangle P2PQ2.
It follows that:
PP1 / PQ1 = PP2 / PQ2
But we also have two constant sums of distances:
PP1 + PP2 = C
PQ1 + PQ2 = D
This lets us write:
PP1 / PQ1 = (CPP2) / (DPQ2) = PP2 / PQ2
PQ2 (CPP2) = PP2 (DPQ2)
PP2 / PQ2 = C / D
And of course PP1 / PQ1 is the same. So the ratio between the distance from a point on the ellipse to one focus and the distance to the corresponding directrix is constant.
### The Whispering Gallery Property
We wish to show that the angles between the tangent to the ellipse at any point P and the vectors PF1 and PF2 are the same. More precisely, the angle between PF1 and a given tangent vector t must be the same as the angle between PF2 and the opposite of that vector, –t. That will guarantee that any ray from one focus that is reflected in the normal to the curve will pass through the other focus: the “whispering gallery” property.
We previously defined P1 as the point lying on both a line from P to the tip of the cone and on the circle where the first Dandelin sphere is tangent to the cone. There are two examples in the image on the right, P1 and P1′, for two different points on the ellipse, P and P′, but the general case can be seen in the animation earlier in this section.
Any vector normal to the cone at P will be parallel to a normal at P1, and at P1 the normal to the cone coincides with the normal to the first Dandelin sphere. So we can obtain a vector normal to the cone at P simply by taking the vector from the centre of the sphere to P1:
N1 = P1C1 = P1 – (F1 + ρ1 n)
From this, we can construct a tangent to the ellipse that lies in the plane normal to n:
t1 = N1 × n = (P1F1) × n
Now, since t1 is perpendicular to P1F1, the dot product of any vector with t1 will be unchanged if we add or subtract some multiple of P1F1 to the original vector. So the dot product of t1 with the vector from the first focus to P is:
(PF1) · t1 = (PF1 – (P1F1)) · t1 = (PP1) · t1
We can now calculate the cosine of the angle between the vector from the first focus and the tangent:
cos β1 = (PF1) · t1 / [|PF1| |t1|]
= (PP1) · t1 / [|PF1| |t1|]
= p1 · t1 / |t1|
= p1 · t0
We’ve made use of the fact (shown earlier) that |PF1| = |PP1|, and we’ve defined p1 = (PP1) / |PP1| as the unit vector that points from P1 to P. We’ve also defined t0 = t1 / |t1| as the normalised tangent vector.
We can repeat this entire construction with the roles of the foci swapped, defining a normal N2 parallel to N1, and a tangent t2 parallel to t1. Because t1 and t2 are parallel (rather than antiparallel), we end up with exactly the same normalised tangent vector, t0. We conclude that the angle between the vector from the second focus and the tangent t0 satisfies the equation:
cos β2 = p2 · t0
where p2 is the unit vector that points from P2 to P, and hence p2 = –p1. But if we want the angle with the opposite tangent vector, –t0, its cosine will be:
cos (π–β2) = p2 · (–t0) = p1 · t0 = cos β1
So we find exactly the same angle between the vector from the first focus and t0 as between the vector from the second focus and –t0, as expected.
### The Inverse Kinetic Energy Pictured as a Square
We can exhibit a moving square tangent to the cone whose area is inversely proportional to the kinetic energy of the orbiting body! This square’s bottom edge lies on a line that’s tangent to the ellipse, while its top corner is always the point P1 that follows the circle where the cone meets the first Dandelin sphere.
Why is the area of this square inversely proportional to the kinetic energy? As discussed in the previous subsection, the line PP1 makes the same angle with the tangent to the ellipse as the line from the focus, PF1, and the length PP1 is equal to the length PF1, i.e. the distance r of the orbiting body from the centre of attraction. So the area of the square is:
Area = r2 sin2 β
where β is the angle between the tangent and a line to the centre of attraction.
If L is the angular momentum of the orbiting body per unit mass, we have:
L = v r sin β
L2 = v2 r2 sin2 β = 2 K(r) Area
Area = ½ L2 / K(r)
where K(r) = EU(r) is the kinetic energy of the orbiting body per unit mass.
In the section From Shape To Force Law we derived an expression for sin2 β (written there as cos2 α, for α=π/2–β) from the geometrical properties of an ellipse, and used it to show that the potential U(r) that produces elliptical orbits must be inversely proportional to r. Here we will give a slightly different argument that leads to the same conclusion.
First, we’ll define:
f(r) = sin2 β
If we write 2a for the constant sum of distances from a point on the ellipse to the two foci, it’s clear that we must have the symmetry:
f(r) = f(2ar)
since r → 2ar corresponds to moving a point to its reflection in the minor axis of the ellipse, changing β to π–β and leaving sin β unchanged. Now, if we define a function:
g(p) = f(a + √[a2p])
we have:
g(r (2ar)) = f(a + √[(ar)2]) = f(a ± (ar)) = f(r)
where the choice of plus or minus makes no difference since the two alternatives lead to f(r) or f(2ar). This proves that f can only depend on the product p = r (2ar).
We’ll write the distances to the two foci as a ± √[a2p], and form a right triangle whose Pythagorean relationship is (after dividing through by 4):
a2 cos2 β + (a2p) sin2 β = c2
a2p sin2 β = c2
This gives us:
sin2 β = (a2c2) / p = (a2c2) / [r (2ar)]
Area = r2 sin2 β = (a2c2) / [–1 + 2a/r]
The product of the kinetic energy K(r) = EU(r) and the area of the square is the constant ½ L2, and that product can only be independent of r if we have a potential U(r) inversely proportional to r.
## References
[1] Huygens and Barrow, Newton and Hooke, V. Arnol’d, Birkhäuser, Boston, 1990.
[2] “Planetary Motion and the Duality of Force Laws”, Rachel W. Hall and Krešimir Josić, SIAM Review, Volume 42, No. 1, March 2000, pp 115–124. Online as PDF.
Science Notes / Conic Section Orbits / created Thursday, 22 March 2012 / revised Friday, 30 March 2012 | HuggingFaceTB/finemath | |
# Solving a combined system of linear and bilinear equations
I am trying to solve a problem of breaking an amatuer cryptography.
The problem boils down to solving a combined system of linear and bilinear equations having $50$ unknowns.
For representational purposes, the equations look similar to the following, with $x$, $y$ and $z$ being the unknowns.
\begin{align} \begin{cases} 3x + 10y + 8z + 5xy &= 1470 \\ 2x + 10y + 3z + yz + xz &= 1210 \\ x + 5y + z + 3xy + 16xz &= 5540 \\ x + 3y + 8z + 12xy + 4yz &= 5110 \end{cases} \end{align}
The above system has the solution $x=10 , y=20, z=30$.
I want to know the method for solving these type of equations. Can gaussian elimination be applied on such a system ?
• You say you have $50$ unknowns. Out of curiosity, how many equations do you have? Also, in average, how many terms are there per equation? (Since you have cross terms $x_i x_j$, simple Gaussian elimination would be difficult.) Aug 15, 2016 at 16:33
• @RobertIsrael: Accdg to this wiki article, Maple has built-in functionality for such systems. However, what's the most number of equations and variables $x_i$ that Maple can handle when the system has no degree $n>1$ but includes cross terms $x_i x_j$? Aug 15, 2016 at 16:43
• @TitoPiezasIII There are 40 equations. On average each have 3000 terms including cross ones. Aug 15, 2016 at 16:44
• So 50 variables, 40 equations, and 3000 terms per equation? Ouch. And an under-determined system at that. The wiki link above may point to alternative approaches. Aug 15, 2016 at 16:47
• @TitoPiezasIII Thanks a lot. I will check. Aug 15, 2016 at 16:57
solving equation (1) for $$z$$ we obtain $$-5/8\,xy-3/8\,x-5/4\,y+{\frac{735}{4}}=0$$ (I) plugging this in the second equation we obtain $$-5/8\,{x}^{2}y-3/8\,{x}^{2}-7/2\,xy+{\frac {1477\,x}{8}}-5/8\,x{y}^{2} -5/4\,{y}^{2}+190\,y-{\frac{2635}{4}} =0$$ (II) pliugging (I) in the third equation $$-{\frac {141\,xy}{8}}-10\,{x}^{2}y-6\,{x}^{2}+{\frac {23525\,x}{8}}+{ \frac {15\,y}{4}}-{\frac{21425}{4}}=0$$ (III) plugging (I) in the fourth equation we get $$11/2\,xy-5/2\,x{y}^{2}-5\,{y}^{2}+728\,y-2\,x-3640=0$$ (IV) solving (IV) for $$y$$ we obtain $$y=-{\frac {48\,{x}^{2}-23525\,x+42850}{80\,{x}^{2}+141\,x-30}}$$ (V) plugging this in (III) factorizing and simplifying we get $$\left( x-10 \right) \left( 13120\,{x}^{4}-5741148\,{x}^{3}-54229691 \,{x}^{2}+11341510\,x+205150100 \right) =0$$ the rest do it by yourself.
• The OP gave this system with $3$ unknowns only as an example. But his actual problem has $50$ unknowns. Since the system involves terms with $x_i$ and $x_i y_j$, the degree of the final equation would rise too fast using this approach. Aug 15, 2016 at 15:30 | HuggingFaceTB/finemath | |
Input matrix, specified as a 3-by-3 matrix, in initial acceleration units. -216-84-150+90+210+144=-6 Please help me. \end{bmatrix} Example 1: Consider the matrix . $$\text{adj}A=(C_{ij})^{\text{T}}=\begin{pmatrix}9(-12)-(-21)5&-(4(-12)-(-10)5)&4(-21)-(-10)9\\ -(3(-12)-(-21)1)&2(-12)-(-10)1&-(2(-21)-(-10)3)\\3\cdot 5-9\cdot 1&-(2\cdot 5-4\cdot 1)&2\cdot 9-4\cdot 3\end{pmatrix}=\\ How do I orient myself to the literature concerning a research topic and not be overwhelmed? Do the transpose of matrix. To learn more, see our tips on writing great answers. Utilities / For related equations, see Algorithms.$$3x+9y-21z=0$$Vote. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Thank you. Output . Cross out row one (1 5 3) and column two (). In linear algebra, the adjugate, classical adjoint, or adjunct of a square matrix is the transpose of its cofactor matrix. Input. The name has changed to avoid ambiguity with a different defintition of the term adjoint. 0. Input matrix, specified as a 3-by-3 matrix, in initial acceleration units. Check the sign that is assigned to the number. The classical adjoint matrix should not be confused with the adjoint matrix. Physics. Port_1 — Input matrix 3-by-3 matrix. what is command to find adjoint of matrix. Port_1 — Output acceleration 3-by-3 matrix. I have provided example regarding adjoints of matrices. Adjoint of a Matrix Let A = [ a i j ] be a square matrix of order n . 3 & 9 & -21 \\ For a matrix A, the adjoint is denoted as adj (A). What is the difference between policy and consensus when it comes to a Bitcoin Core node validating scripts? 3x3 Matrix Inverse Calculator Results; Adjoint (adj A) Determinant (|A|) Inverse of Matrix = (adj A)/|A| ★★★★★ [ 10 Votes ] About the 3 x 3 matrix inverse calculator. Should hardwood floors go all the way to wall under kitchen cabinets? Let A be a square matrix of order n. The adjoint of square matrix A is defined as the transpose of the matrix of minors of A. Ports. The Adjoint of 3x3 Matrix block computes the adjoint matrix adjoint of matrix. The adjoint of a matrix A is the transpose of the cofactor matrix of A . The adjoint of a square matrix A = [a ij] n x n is defined as the transpose of the matrix [A ij] n x n, where Aij is the cofactor of the element a ij. The conjugate transpose of a matrix A {\boldsymbol {A}}} with real entries reduces to the transpose of A {\boldsymbol {A}}} , as the conjugate of a real number is the number itself. Ports. 1 & 5 & -12 Repeat the same process with this element: Cross out the row and column of that element. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. (A’)’= A. If the determinant is zero, the matrix won't have an inverse. For matrix A, A = [ 8(_11&_12&_13@_21&_22&_23@_31&_32&_33 )] Adjoint of A is, adj A = Transpose of [ 8(_11&_12&_13@_21&_22&_23@_31&_32&_33 ) (1) where, A is a square matrix, I is an identity matrix of same order as of A and represents determinant of matrix A. Adjoint Matrix Calculator. An adjoint matrix is also called an adjugate matrix. Adjoint (or Adjugate) of a matrix is the matrix obtained by taking transpose of the cofactor matrix of a given square matrix is called its Adjoint or Adjugate matrix. Adjoint matrix is also referred as Adjunct matrix or Adjugate or classical adjoint matrix. Output . The Adjoint of 3x3 Matrix block computes the adjoint matrix for the input matrix. Before proceeding, I am assuming you know how to find minors of a 3x3 matrix and how to find cofactors using minors. Introduced in R2013a × MATLAB Command. This page introduces specific examples of cofactor matrix (2x2, 3x3, 4x4). On the other hand, the inverse of a matrix A is that matrix which when multiplied by the matrix A give an identity matrix. 2 & 4 & -10 \\ Treat the remaining elements as a 2x2 matrix. Treat the remaining elements as a 2x2 matrix. The m…$$2x+4y-10z=-2$$rev 2020.12.2.38106, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us.$$|A|=\begin{vmatrix}2&4&-10\\ 3&9&-21\\1&5&-12\end{vmatrix}=\begin{vmatrix}0&-6&14\\ 0&-6&15\\1&5&-12\end{vmatrix}=\begin{vmatrix}0&0&-1\\ 0&-6&15\\1&5&-12\end{vmatrix}=-6.$$and use that$$ Data Types: double. Consider the matrix If A = || of order m*n then = || of order n*m. So, . For better clarification and understanding, go through the files present below. For related equations, see Algorithms. Port_1 — Output acceleration 3-by-3 matrix. $$But if the user enters 2 instead of 1 in ch, which means he wants to find the adjoint of a 3x3 matrix, then calculations will be like this. The conjugate transpose "adjoint" matrix should not be confused with the adjugate, (), which is also sometimes called adjoint. MathJax reference. Do all Noether theorems have a common mathematical structure? Data Types: double. How To Calculate Determinant,Adjoint,and Inverse a 3X3 Matrix Example Matrix : 1. The calculator will find the adjoint (adjugate, adjunct) matrix of the given square matrix, with steps shown. In our example, the matrix is () Find the determinant of this 2x2 matrix. Now, I have tried to solve it, and I got the determinant of matrix A is equal to zero. 3 & 9 & -21 \\ Elements of the matrix are the numbers which make up the matrix. X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} Return to the original 3x3 matrix, with the row or column you circled earlier. An adjoint matrix is also called an adjugate matrix. Show Hide all comments. Making statements based on opinion; back them up with references or personal experience. 6 x 8 = 48; 3 x 1 = 3; Now subtract the value of the second diagonal from the first, i.e, 48 – 3 = 45. The Adjoint of 3×3 Matrix block computes the adjoint matrix for the input matrix.$$AX=B \iff A^{-1}AX=A^{-1}B \iff X=A^{-1}B=\frac{\text{adj}A}{|A|}B.$$For related equations, see Algorithms. Show Instructions. semath info. Note: In the past, the term for adjugate used to be adjoint. please Help Me and answer soon 1 Comment. Adjoint of a square matrix is a matrix of same order such that the product of A and its adjoint in either order is the det A times the identity matrix. Adjoint of matrix is also represented by adj. We strongly recommend you to refer below as a prerequisite of this. \operatorname{adj}A\cdot A=(\det A)I_3 \quad The Adjoint of 3x3 Matrix block computes the adjoint matrix for the input matrix. Adjoint of a Matrix Let A = [ a i j ] be a square matrix of order n .$$ NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Find Adjoint : M is referred to as minor. Input. Ports. Is it illegal to carry someone else's ID or credit card? Port_1 — Output acceleration 3-by-3 matrix. The Adjoint of 3x3 Matrix block computes the adjoint matrix for the input matrix. adjoint of matrix questions 1,test on adjoint of matrix,quiz on matrices and determinants. Inverting a 3x3 matrix using determinants Part 2: Adjugate matrix. By using this website, you agree to our Cookie Policy. 1. (2*2 - 7*4 = -24) Multiply by the chosen element of the 3x3 matrix.-24 * 5 = -120; Determine whether to multiply by -1. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. How is time measured when a player is late? Port_1 — Input matrix 3-by-3 matrix. What is the solution of the above system? The adjoint is the conjugate transpose of a matrix while the classical adjoint is another name for the adjugate matrix or cofactor transpose of a matrix. The calculator will find the adjoint (adjugate, adjunct) matrix of the given square matrix, with steps shown. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Adjoing of the matrix A is denoted by adj A. How to draw a seven point star with one path in Adobe Illustrator. This website uses cookies to ensure you get the best experience. For matrix A, A = [ 8(_11&_12&_13@_21&_22&_23@_31&_32&_33 )] Adjoint of A is, adj A = Transpose of [ 8(_11&_12&_13@_21&_22&_23@_31&_32&_33 ) Data Types: double. Commented: Anjan Sahu on 11 Jan 2019 how to find out adjoint of matrix in matlab? Anju says. $$. On the other hand, the inverse of a matrix A is that matrix which when multiplied by the matrix A give an identity matrix. It is estimated that the rate of change of production P with respect to the additional number of workers x is given by . The Adjoint of 3×3 Matrix block computes the adjoint matrix for the input matrix. Input matrix, specified as a 3-by-3 matrix, in initial acceleration The matrix formed by taking the transpose of the cofactor matrix of a given original matrix. Input. The Adjoint of 3x3 Matrix block computes the adjoint matrix for the input matrix. what is command to find adjoint of matrix. Choose a web site to get translated content where available and see local events and offers. Now we have the matrix that does not have 2. The adjoint of a matrix (also called the adjugate of a matrix) is defined as the transpose of the cofactor matrix of that particular matrix. The Adjoint of 3x3 Matrix block computes the adjoint matrix for the input matrix. Adjoint (or Adjugate) of a matrix is the matrix obtained by taking transpose of the cofactor matrix of a given square matrix is called its Adjoint or Adjugate matrix. Maths. The Adjoint of 3x3 Matrix block computes the adjoint matrix for the input matrix. Data Types: double. Note: In the past, the term for adjugate used to be adjoint. In other words we can define adjoint of matrix as transpose of co factor matrix. Matrices, when multiplied by its inverse will give a resultant identity matrix. In linear algebra, the adjugate, classical adjoint, or adjunct of a square matrix is the transpose of its cofactor matrix. For related equations, see Algorithms. Input matrix, specified as a 3-by-3 matrix, in initial acceleration units. Ports. The name has changed to avoid ambiguity with a different defintition of the term adjoint. You clicked a link that corresponds to this MATLAB command: Run the command by entering it in the MATLAB Command Window. If the determinant is 0, then your work is finished, because the matrix has no inverse. Port_1 — Input matrix 3-by-3 matrix. Calculating the inverse of a 3×3 matrix by hand is a tedious job, but worth reviewing. Input. Did China's Chang'e 5 land before November 30th 2020? Output acceleration, returned as a 3-by-3 matrix, in final Find the adjoint of the matrix… Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In linear algebra, the adjugate or classical adjoint of a square matrix is the transpose of its cofactor matrix. 3x3 identity matrices involves 3 rows and 3 columns. Input. Solve the following system using the adjoint matrix. The adjugate has sometimes been called the . Find more Mathematics widgets in Wolfram|Alpha. The adjugate has sometimes been called the . Classical Adjoint. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Ports. Port_1 — Input matrix 3-by-3 matrix. The Adjoint of 3×3 Matrix block computes the adjoint matrix for the input matrix.$$ P.S. Why is training regarding the loss of RAIM given so much more emphasis than training regarding the loss of SBAS? A=\begin{bmatrix} Data Types: double. Thus the system has a single solution. Data Types: double. It is also occasionally known as adjunct matrix, though this nomenclature appears to have decreased in usage. Novel from Star Wars universe where Leia fights Darth Vader and drops him off a cliff. Find the determinant of each of the 2x2 minor matrices. expand all. The Adjoint of 3×3 Matrix block computes the adjoint matrix for the input matrix. 2. Show Hide all comments. expand all. Filed Under: Matrices and Determinants. | HuggingFaceTB/finemath | |
# How many grams of methyl alcohol should be added to 10 litre
Question:
How many grams of methyl alcohol should be added to 10 litre tank of water to prevent its freezing
at 268 K ?
$\left(\mathrm{K}_{f}\right.$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
1. 899.04 g
2. 886.02 g
3. 868.06 g
4. 880.07 g
Correct Option: , 2
Solution: | HuggingFaceTB/finemath | |
Rates of Change Rates are an important area of Maths and can be found in everyday life, business and science.
Presentation on theme: "Rates of Change Rates are an important area of Maths and can be found in everyday life, business and science."— Presentation transcript:
Rates of Change Rates are an important area of Maths and can be found in everyday life, business and science.
Common rates RateUnits Pay rate\$ per hour (\$/h) Heart rateBeats per minute (beat/min) Interest ratePercentage per annum (%/yr) Speedkm per hour (km/h) Water flow rateLitres/min (L/min) Fuel economyLitres per 100km (L/100km)
Method for working out constant rates from worded problems Write out the unit Write out the relevant numbers as a division Work out your answer Example 1 If Colin earns \$300 in 6 hours, what is his rate of pay?
Method for working out constant rates from worded problems Write out the unit Write out the relevant numbers as a division Work out your answer Example 2 If a heater uses 20000 Joules of energy every 10 seconds what is its rate of energy use? Which can be written as 2 kJ/s
Method for working out constant rates from worded problems Write out the unit Write out the relevant numbers as a division Work out your answer Example 3 What is the rate of expansion when a temperature change of 10 o C produces an 0.5mm of expansion in a steel beam? The first part of a rate unit always comes from the name of the rate. Unit for expansion = mm so rate unit = mm/ o C
Method for working out constant rates from worded problems Write out the unit Write out the relevant numbers as a division Work out your answer Example 4 If 6 litres of water can dissolve up to 30 grams of a powder what is the maximum solution rate in grams/litre?
What We Will Be Looking At In This Unit The Maths in the previous examples is fairly straight forward. In this unit we are going to look at: - finding rates from graphs and tables. - visualising graphs for a variety of rate situations
Introductory Rates Question 1 Length (m) Time (hour) s) 10 8.0 6.0 4.0 100 90 80 60 70 50 40 30 10 20 0 12 2.0 110 A B CDE (a) In what section A, B, C, D or E, does the digger work at its highest rate? Section E
Introductory Rates Question 1 Length (m) Time (hour) s) 10 8.0 6.0 4.0 100 90 80 60 70 50 40 30 10 20 0 12 2.0 110 A B CDE (b) What aspect of the graph did you use to arrive at your answer to part a) ? Gradient
Introductory Rates Question 1 Length (m) Time (hour) s) 10 8.0 6.0 4.0 100 90 80 60 70 50 40 30 10 20 0 12 2.0 110 A B CDE (c) What are the units for the digging rate? Unit for digging = m so digging rate unit is m/hour Or metres per hour. This unit can also be deduced from gradient rule of rise/run
Introductory Rates Question 1 Length (m) Time (hour) s) 10 8.0 6.0 4.0 100 90 80 60 70 50 40 30 10 20 0 12 2.0 110 A B CDE (d) When was the digger not working? Sections B & D
Introductory Rates Question 1 Length (m) Time (hour) s) 10 8.0 6.0 4.0 100 90 80 60 70 50 40 30 10 20 0 12 2.0 110 A B CDE (e) What is the digging rate in section A? 30 2
Introductory Rates Question 1 Length (m) Time (hour) s) 10 8.0 6.0 4.0 100 90 80 60 70 50 40 30 10 20 0 12 2.0 110 A B CDE (e) What is the digging rate in section B?
Introductory Rates Question 1 Length (m) Time (hour) s) 10 8.0 6.0 4.0 100 90 80 60 70 50 40 30 10 20 0 12 2.0 110 A B CDE (e) What is the digging rate in section C? 10 4
Introductory Rates Question 1 Length (m) Time (hour) s) 10 8.0 6.0 4.0 100 90 80 60 70 50 40 30 10 20 0 12 2.0 110 A B CDE (e) What is the digging rate in section D?
Introductory Rates Question 1 Length (m) Time (hour) s) 10 8.0 6.0 4.0 100 90 80 60 70 50 40 30 10 20 0 12 2.0 110 A B CDE (e) What is the digging rate in section E? 50 3
Instantaneous and Average Rates of Change When you are asked for an average rate of change you are being asked for a summary of the rate of change a time interval (or other interval). Average rate = gradient between two points When you are asked the rate of change at a point you are being asked for the instantaneous rates of change at a point. Instantaneous rate = gradient of tangent at that point Notes
Example 1 What is the average toy production rate in the first 12 minutes? toys Time (min) 60 toys 12 minutes
Example 2 Estimate the production rate at 8 minutes? toys Time (min) 52 toys 8 minutes 8 minutes point
Introductory Question 2 Length (m) Time (hours) (a) When is the digger digging at a constant rate? Constant rate = constant gradient Constant rate in 0 to 5 hours and 7 to 10 hours
Introductory Question 2 Length (m) Time (hours) (b) What is the digging rate at 2 hours? 30 5
Introductory Question 2 Length (m) Time (hours) (c) What is the digging rate at 9 hours? 3 3
Introductory Question 2 Length (m) Time (hours) (d) Estimate the digging rate at 6 hours? 26 9
Introductory Question 2 Length (m) Time (hours) (e) What is the average digging over the 10 hours? 39 10
Introductory Question 3 (a) What is the average digging rate over the first 5 hours? 20 5
Introductory Question 3 (b) Estimate the digging rate at 4.5 seconds? 30 5
Introductory Question 3 (c) What is the final digging rate? 18 9
Introductory Question 4 (a) At what instant is the rate of change of depth equal to zero Answer At 6 seconds and at 18 seconds
Introductory Question 4 (b) At what instant is the rate of change a maximum. Answer At 0 seconds and at 24 seconds
Introductory Question 4 (c) The magnitude of the rate of change at 0 and 12 seconds is exactly the same but one is positive and the other is negative. What is the significance of the sign? Answer A positive rate means that the depth is increasing with time and a negative rate means that it is decreasing with time. m = 1.3 m/h m = –1.3 m/h
Introductory Question 4 4.3 8 (d) What is the average rate of change in the first 8 hours?
Introductory Question 4 (e) What is the average rate of change in the first 12 hours?
Introductory Question 4 - 10 9 (f) Estimate the instantaneous rate of change at 10 hours?
Introductory Question 5 Write out the unit Write out the relevant numbers as a division Work out your answer a) What is the average growth rate (per month) over the total time interval ? 1 st of MonthJanFebMarAprMayJune Plant Height 0mm12mm26mm40mm44mm54mm
Introductory Question 5 Write out the unit Write out the relevant numbers as a division Work out your answer b) What is the average growth rate (per month) over the last two months ? 1 st of MonthJanFebMarAprMayJune Plant Height 0mm12mm26mm40mm44mm54mm
Rates from Graphs Textbook Exercises Average rates of change Ex18C p515 Q2, 6 Instantaneous rates of change using tangents Ex 18D p519 Q1, 2
Working Out Rates From Formulae If you have a rule for a relationship the simplest way to find a rate is to use the gradient formula With the variables above replaced by the appropriate variables from the problem
Rates from formulae Problem 1 When a ball is fired upwards its height can be calculated with the formula h(t) = 40t – 5t 2 (a) What is the average speed of the ball in the first 2 seconds? at t 1 = 0, h(t 1 ) = 40 0 – 5 0 2 = 0 at t 2 = 2, h(t 2 ) = 40 2 – 5 2 2 = 60
Rates from formulae Problem 1 When a ball is fired upwards its height can be calculated with the formula h(t) = 40t – 5t 2 (b) What is the approximate speed of the ball at 2 seconds? at t 1 = 2, h(t 1 ) = 40 2 – 5 2 2 = 60 at t 2 = 2.01, h(t 2 ) = 40 2.01 – 5 2.01 2 = 60.1995 Average rates are an indication of the round about rate over an interval. If the interval is very small then the average will be close to the actual instantaneous rate for all the points in the interval. In general moving on about 1/100 th of the magnitude of the initial point will give a good approximation to an instantaneous rate at the initial point.
Rates from formulae Problem 2 The height of a bridge above a river is closely modelled by the equation h(x) = 0.02x 2 – 0.2x +9 (a) What is the average slope (change in height rate) of the bridge in the first 3 metres? at x 1 = 0, h(x 1 ) = 0.02 0 2 – 0.2 0 + 9 = 9 at x 2 = 3, h(x 2 ) = 0.02 3 2 – 0.2 3 + 9 = 8.58
Rates from formulae Problem 2 The height of a bridge above a river is closely modelled by the equation h(x) = 0.02x 2 – 0.2x +9 (b) What is the slope (change in height rate) of the bridge at the point 3 horizontal metres across the bridge? at x 1 = 3, h(x 1 ) = 0.02 3 2 – 0.2 3 + 9 = 8.58 at x 2 = 3.01, h(x 2 ) = 0.02 3.01 2 – 0.2 3.01 + 9 = 8.5792
Rates from formulae Textbook Problems Ex18C p513 Q1, 3 Ex18D p521 Q9, 10, 11,15
Estimating Rates from formulae Examples & Questions Example 1 p503 Exercise 8A p505 Q6, 7
Rates Revision and Test Use the Rates revision sheet to study for the Rates test. No solutions will be supplied for the revision sheet since the test is very similar to the revision sheet
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# math question, that you should finish in less than 3 minutes
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math question, that you should finish in less than 3 minutes [#permalink]
### Show Tags
01 Nov 2004, 19:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
hi all, it took me more than 10 minutes...
A survey of N people found that 60 percent preferred brand A. An additional X people were surveyed who all preferred brand A. Seventy percent of all the people surveyed preferred brand A. Find X in terms of N.
a) N/6
b) N/3
c) N/2
d) N
e) 3N
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### Show Tags
01 Nov 2004, 19:37
(60/100)N + X = (70/100)(N+X)
=>6N + 10X = 7N + 7X
=>N = 3X
=>X=N/3.
Hence B.
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### Show Tags
01 Nov 2004, 19:38
My choice is B.I had almost given up when I thought that it could be attempted this way.
Total ppl: N
Percentage who like A: 60/100*N
X ppl who liked A were added to N.
Therefore,total ppl=N+X
Percentage who like A now : 70/100(N+X)
Now,
(60/100*N) + X=70/100(N+X)
Solving this , we get,
X=N/3
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### Show Tags
01 Nov 2004, 20:54
did u guys get to the answer in less than 3 min?
thats the hardest thing...
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01 Nov 2004, 21:03
I just did it in 2min46sec.
Same method as Venksune. More practice will help you kevinw. You should practice mixture problems as they will help you understand how different proportion mixing affect the overall results
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01 Nov 2004, 23:26
It took me less than 1.5 minutes - I didnt mention it. Thought that it may demorale you. Since you asked I told you. It happens to me though - I simply freeze sometime and actually would be staring at a question without doing anything about it. When I apply my mind right, I am sure solving it more easily. I am unsure if it was one such aberration to you also. So, dont worry.
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02 Nov 2004, 20:44
I took around 4 minutes.
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02 Nov 2004, 21:38
Well it took 2 min and 30 sec for me. But yes I need to PracticeMore......
02 Nov 2004, 21:38
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# Instantaneous Rate of Change
Jen23
## Homework Statement
Estimate the instantaneous rate of change of the function f(x)=3x^2 + 4x at (1,7)
## Homework Equations
∆f(x)/∆x = f(x2)-f(x1) / x2-x1
## The Attempt at a Solution
I know that x=1 given the point, but to find the instantaneous rate of change I can use x=1.001 as this is a very close number to the point on the tangent curve. So:
∆f(x)/∆x = [3(1.001)^2 + 4(1.001)] - [3(1)^2 + 4(1) ] / 1.001 - 1
= 7.010003 - 7 / 1.001 - 1
= 10.003
The answer in the book is 13 for instantaneous rate of change, and they used the interval 1 ≤ x ≤ 2
I do not understand why they chose to use the point 2, I thought that the point on a tangent must be very close to the x value so wouldn't 1.001 be more appropriate? Have I answered the question correctly? Detailed explanation would be appreciated.
Last edited:
magoo
Your methodology is OK, but it doesn't lend itself to an estimate. It is a detailed precise calculation.
If you plot the function with x varying from o to 3 in steps of 1, I think you can see why the author chose the evaluation at x = 1 and then x = 2. You can quickly figure this in your head.
So your answer is not wrong, it simply is not an estimate. It is more precise that what the author came up with, but it is not an estimate because you evaluated the function at x = 1.001 - which you really need a calculator or worksheet to do.
Jen23
Your methodology is OK, but it doesn't lend itself to an estimate. It is a detailed precise calculation.
If you plot the function with x varying from o to 3 in steps of 1, I think you can see why the author chose the evaluation at x = 1 and then x = 2. You can quickly figure this in your head.
So your answer is not wrong, it simply is not an estimate. It is more precise that what the author came up with, but it is not an estimate because you evaluated the function at x = 1.001 - which you really need a calculator or worksheet to do.
Okay I see what you are saying, thanks for taking the time to read and reply to my question!
Mentor
## Homework Statement
Estimate the instantaneous rate of change of the function f(x)=3x^2 + 4x at (1,7)
## Homework Equations
∆f(x)/∆x = f(x2)-f(x1) / x2-x1
You really need parentheses here. What you wrote on the right side means ##f(x2) - \frac{f(x1)}{x2} - x1##. I'm sure that's not what you meant.
Jen23 said:
## The Attempt at a Solution
I know that x=1 given the point, but to find the instantaneous rate of change I can use x=1.001 as this is a very close number to the point on the tangent curve.
But, that's not very close to 1.7, which is what I think you meant by writing "at (1,7)".
Jen23 said:
So:
∆f(x)/∆x = [3(1.001)^2 + 4(1.001)] - [3(1)^2 + 4(1) ] / 1.001 - 1
= 7.010003 - 7 / 1.001 - 1
= 10.003
The answer in the book is 13 for instantaneous rate of change, and they used the interval 1 ≤ x ≤ 2
I do not understand why they chose to use the point 2, I thought that the point on a tangent must be very close to the x value so wouldn't 1.001 be more appropriate? Have I answered the question correctly? Detailed explanation would be appreciated.
As best as I can determine they want an estimate at x = 1.7, not around x = 1.
Mentor
Your methodology is OK, but it doesn't lend itself to an estimate. It is a detailed precise calculation.
Precise or not, it's still an estimate. The OP is estimating the slope of the tangent line by the slope of a secant line.
magoo said:
If you plot the function with x varying from o to 3 in steps of 1, I think you can see why the author chose the evaluation at x = 1 and then x = 2. You can quickly figure this in your head.
So your answer is not wrong, it simply is not an estimate.
No, it is an estimate of the slope of the tangent line (i.e., the instantaneous rate of change at a particular point).
magoo said:
It is more precise that what the author came up with, but it is not an estimate because you evaluated the function at x = 1.001 - which you really need a calculator or worksheet to do.
The OP evaluated the function in order to estimate the value of its derivative.
A calculator is nice, but not necessary. One can square 1.001 using only pencil and paper.
Jen23
Precise or not, it's still an estimate. The OP is estimating the slope of the tangent line by the slope of a secant line.
No, it is an estimate of the slope of the tangent line (i.e., the instantaneous rate of change at a particular point).
The OP evaluated the function in order to estimate the value of its derivative.
A calculator is nice, but not necessary. One can square 1.001 using only pencil and paper.
Yes, that is what I was thinking, I had just calculated and estimate the slope of the tangent line. We can calculate this instantaneous rate as x approaches that certain value (example, using 1.0001, 1.00001,..). I included four decimal places at least for accuracy.
The book did use the point (1, 7) I think after inserting x=1 into the function we get 7, so f(1)=7. I chose a point x=1.001 (point on the curve that is very close to 1) and solved for f(1.001)= 7.010003. After finding the slope of these two points, it gives me an estimate of the instantaneous rate of change, I assume?
Mentor
I don't know why they used 1 and 2 in their approximation. To get a better approximation, you would normally pick values on either side of 1, such as 0 and 2 or .5 and 1.5. As both numbers get closer to 1, the approximation gets better.
The exact value of the inst. rate of change (the derivative, in other words) at x = 1 is 10, so your approximations are pretty close. Their approximation of 13 is pretty far off.
Jen23
I don't know why they used 1 and 2 in their approximation. To get a better approximation, you would normally pick values on either side of 1, such as 0 and 2 or .5 and 1.5. As both numbers get closer to 1, the approximation gets better.
The exact value of the inst. rate of change (the derivative, in other words) at x = 1 is 10, so your approximations are pretty close. Their approximation of 13 is pretty far off.
I also thought using 1 and 2 is on the farther end of getting a better approximation. I am going to stick to what I originally came up with since it is a closer approximation, maybe the book had an error. Thank you for your feedback! I appreciate it.
Mentor
For a quadratic function, if you choose two points equidistant on either side of the desired point (x = 1.7), the finite divided difference will give you the exact value of the rate of change at the desired point. For example, if you choose x1 = 0 and x2 = 3.4, the finite divided difference gives you a rate of change of 14.2, which is the exact value. Try x1 = 0 and x2 = 2.4 and see what you get.
SammyS | HuggingFaceTB/finemath | |
The lesser known of two math memes currently wandering around the Internet involves an interesting equation: $\sqrt{2\frac{2}{3}} = 2\sqrt{\frac{2}{3}}$
This has spawned at least three discussions I’ve seen so far:
1. What other values is this equation true for?
2. Is this example good or bad for students?
3. What’s with mixed numbers, anyway?
I’ll discuss each topic in turn.
### When is this true?
This equation is true because $$\sqrt{2 + \frac{2}{3}} = 2 \cdot \sqrt{\frac{2}{3}} \approx 1.63$$.
However, it is not true for all values. For instance, $$\sqrt{4 \frac{2}{3}} \approx 2.16$$ while $$4 \sqrt{\frac{2}{3}} \approx 3.27$$. So when is it true?
Let’s say we have three integer values: a, b, and c. The question is, when is $$\sqrt{a + \frac{b}{c}} = a \cdot \sqrt{\frac{b}{c}}$$? (We’ll get back to why we have to use a plus sign in the general form below.)
If we square both sides we get $a + \frac{b}{c} = a^2 \frac{b}{c} \\ \Rightarrow \frac{ac + b}{c} = \frac{a^2b}{c} \\ \Rightarrow ac + b = a^2b \\ \Rightarrow ac = a^2b – b = b(a^2 – 1) \\ \Rightarrow c = \frac{b}{a}(a^2 – 1)$
Since c is an integer, either $$\frac{b}{a}$$ or $$\frac{a^2 – 1}{a}$$ must be an integer. However, $$a^2 – 1$$ and a are coprime, so $$\frac{a^2 – 1}{a}$$ can’t be an integer. Therefore, $$\frac{b}{a}$$ is an integer, and b is a multiple of a; $$b = ka$$. Hence $$c = \frac{ka}{a}(a^2 – 1) = k(a^2 – 1)$$ and $$\frac{b}{c} = \frac{ka}{k(a^2 – 1)} = \frac{a}{a^2 – 1}$$.
However, when $$a = 1$$, we get $$\frac{1}{1^2 – 1} = \frac{1}{0}$$, which is undefined, so we have to exclude $$a = 1$$. This leaves us with the following formula: $\sqrt{a + \frac{a}{a^2 – 1}} = a \cdot \sqrt{\frac{a}{a^2 – 1}} \forall a \in N > 1$
We can generalize further by looking at roots other than the square root. That is, if m is an integer greater than 1, $a \sqrt[m]{\frac{b}{c}} = \sqrt[m]{a + \frac{b}{c}} \\ \Rightarrow a + \frac{b}{c} = a^m \frac{b}{c} \\ \Rightarrow c = k(a^m – 1) \text{ and } b = ka$
Here are some examples: $\sqrt{2 \frac{2}{3}} = 2 \sqrt{\frac{2}{3}} \\ \sqrt{3 \frac{3}{8}} = 3 \sqrt{\frac{3}{8}} \\ \sqrt{4 \frac{4}{15}} = 4 \sqrt{\frac{4}{15}} \\ \sqrt{2 \frac{4}{6}} = 2 \sqrt{\frac{4}{6}} \\ \sqrt{2 \frac{2}{7}} = 2 \sqrt{\frac{2}{7}} \\ \sqrt{3 \frac{3}{80}} = 3 \sqrt{\frac{3}{80}}$
### Is this a good exercise for students?
On the one hand, this is an excellent demonstration of algebra. It also exploits what is a notational accident (see the next section) to create a nifty pattern. Our notation itself is arbitrary, so there’s nothing about numbers specifically that leads to this pattern; it’s completely an accident of notation, but it’s certainly a fun one.
On the other hand, some teachers have expressed concern that students may try to generalize the pattern beyond what algebra shows us. The example does run the risk of further confusing students about how notation works.
For that reason, it strikes me that this sort of exercise is best for students that are at a high level of understanding already and want to work with extensions. This is not a particularly practical exercise in learning how mathematics and numbers work: It’s a fun exercise.
### What’s With Mixed Numbers Anyway?
One interesting thing I learned in discussions about this item is that what North Americans and others take for granted as “mixed fractions” are not in fact a global convention.
Mixed fraction notation has long struck me as mathematically dubious. It’s useful for teaching students number sense as you teach them fractions: $$3 \frac{3}{5}$$ is easier to place quickly on a number line than $$\frac{18}{5}$$.
However, once mathematics students have command of fractions in general, I think that mixed fractions do more harm than good. They’re difficult to work with. Compare $$3 \frac{3}{5} \times 2 \frac{2}{7}$$ to $$\frac{18}{5} \times \frac{16}{7}$$. In the first case, you either need to convert to improper fractions (the second case) or do some distributing: $$3 \frac{3}{5} \times 2 \frac{2}{7} = 6 + \frac{6}{5} + \frac{6}{7} + \frac{6}{35}$$ $$= 6 + 1 + \frac{1}{5} + \frac{6}{7} + \frac{6}{35}$$ $$= 7 + \frac{7}{35} + \frac{30}{35} + \frac{6}{35}$$ $$= 7 + \frac{43}{35} = 7 + 1 + \frac{8}{35}$$ $$= 8\frac{8}{35}$$. Which is a mess.
Mixed fractions are used outside of mathematics on a regular basis. Vinyl records play at 33⅓ RPM. Before 50 Shades, we had 9½ Weeks. Freeway exits use mixed numbers for distance. The general populace use mixed fractions on a regular basis, precisely because they’re easier to understand numerically than improper fractions are.
However, most common fractions (⅓ is an exception) are just as easily written in decimals, so floppies could be either 3½” or 3.5″, or either 5¼” or 5.25″. And since 0.33 is a close enough approximation for ⅓, there’s no real need for mixed fractions. It’s a notational habit.
Meanwhile, mixed fractions are notationally inconsistent once we start working with algebra. 2⅓ means “two plus one-third”, but $$a\frac{b}{c}$$ means “a times b over c”. In mathematics, juxtaposition without an operator normally implies multiplication. Even ⅓ ¼ would most likely be interpreted as multiplication. But when there is an integer juxtaposed to the left of a fraction, addition is implied.
I was surprised but not disappointed to learn that there are some countries where this is not the case. Italy in particular consistently uses a plus sign, so 4¼ = 1, not 4.25; 4.25 = 4 + ¼. France, Spain, and Portugal appear to also have largely (if not universally) adopted this convention. Where mixed numbers do appear, such as 33⅓ RPM, they’re seen as an industry-specific notation and might even be confusing. (Discussion here)
Mixed fractions are less useful when using metric (i.e., decimal) measures. Showing measurements in decimal form (5.5 km to the exit) allows easier shifting between units of the same type. Imperial units are inconsistent enough that decimal is not a particular advantage, so our mixed-numbers habit in the United States is likely to be harder to break.
All the same, though, I think that getting rid of mixed fraction notation is the right way to go. Pedagogically, it’s not that much more work to write $$\frac{17}{4} = 4 + \frac{1}{4}$$ than to write $$\frac{17}{4} = 4 \frac{1}{4}$$, and it arguably does an even better job of reinforcing the intended concept, i.e., that $$\frac{17}{4}$$ is “a little more” than 4. | HuggingFaceTB/finemath | |
# Solve Multiplication Problems (4)
In this worksheet, students are tested on their times tables.
Key stage: KS 2
Curriculum topic: Maths and Numerical Reasoning
Curriculum subtopic: Word Problems
Difficulty level:
### QUESTION 1 of 10
You should know your times tables.
Here is a grid to remind you.
What is ....?
6 x 7
What is ....?
7 x 7
What is ....?
5 x 9
What is ....?
9 x 8
What is ....?
9 x 9
What is ....?
7 x 8
What is ....?
4 x 12
What is ....?
7 x 9
What is ....?
6 x 6
What is ....?
8 x 12
What is ....?
5 x 8
What is ....?
5 x 9
What is ....?
9 x 6
What is ....?
10 x 11
What is ....?
11 x 11
What is ....?
6 x 12
What is ....?
7 x 12
What is ....?
5 x 12
What is ....?
11 x 12
What is ....?
12 x 12
• Question 1
What is ....?
6 x 7
42
• Question 2
What is ....?
7 x 7
49
• Question 3
What is ....?
5 x 9
45
• Question 4
What is ....?
9 x 8
72
• Question 5
What is ....?
9 x 9
81
• Question 6
What is ....?
7 x 8
56
• Question 7
What is ....?
4 x 12
48
• Question 8
What is ....?
7 x 9
63
• Question 9
What is ....?
6 x 6
36
• Question 10
What is ....?
8 x 12
96
• Question 11
What is ....?
5 x 8
40
• Question 12
What is ....?
5 x 9
45
• Question 13
What is ....?
9 x 6
54
• Question 14
What is ....?
10 x 11
110
• Question 15
What is ....?
11 x 11
121
• Question 16
What is ....?
6 x 12
72
• Question 17
What is ....?
7 x 12
84
• Question 18
What is ....?
5 x 12
60
• Question 19
What is ....?
11 x 12
132
• Question 20
What is ....?
12 x 12
144
---- OR ----
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## Introduction
We find symmetry in many objects around us. Let us watch this video to know more about symmetry and its applications.
Symmetry is found everywhere and in nature we find abundant symmetry.
Symmetry is also commonly used in architecture by architects to build complex designs. A famous example of symmetry in architecture is the Taj Mahal.
Do you notice that the left side of the image is the same as the right side? This is because the Taj Mahal was built keeping symmetry in mind.
#### Concepts
The chapter ‘Symmetry’ covers the following concepts:
#### Lines of Symmetry for Regular Polygons
Observe the shapes given below:
Let us divide these shapes into two halves.
Are the two halves exactly the same? Can we say they are symmetrical? Yes, the shapes you see here are symmetrical. Why? When you fold them along the dotted line, each point of the first half will coincide exactly with each point of the second half.
This dotted line is called the line of symmetry. Line of symmetry is also called the axis of symmetry.
Line of symmetry is a line which divides an object or figure into two halves such that the first half will coincide with the second half.
Observe the following objects:
These objects can be divided into half, vertically in such a way that the two halves match exactly.
So, these objects have a line of symmetry.
Figures/ objects having line/lines of symmetry are called symmetric figures/objects.
Now look at the following objects:
These objects can be divided into half but the two halves do not match exactly.
So, these objects do not have any line of symmetry.
Figures/objects having no line of symmetry are called asymmetric figures/objects.
Observe the dotted lines shown below for the same object:
We can say that each of these dotted lines divides the object into two halves such that first half coincides with the second half. So, the number of lines of symmetry for this object is 4.
Hence, an object can have more than one line of symmetry.
A polygon is a closed figure made of several line segments. A polygon is said to be regular if all its sides are of equal lengths and all its angles are of equal measure. Let us see the different types of polygons:
A regular polygon of three sides is an equilateral triangle.
A regular polygon of four sides is a square.
A regular polygon of five sides is a regular pentagon.
A regular polygon of six sides is a regular hexagon.
Observe the figures given below:
Now, we know that a square is a regular polygon and is symmetric. But, if we punch few holes into it as shown here, will it be still symmetric?
Let us now try to draw the lines of symmetry for the given regular polygon such that the punched holes are symmetric about the line.
We draw a vertical line, a horizontal line, and a slant line along a diagonal of the square and check whether we get the four holes about the lines symmetrically.
If we fold this figure at the slant and the vertical lines, we do not get the four holes symmetrically about the fold.
Hence, they are not lines of symmetry. If we fold this figure at the horizontal line, we get the four holes about the fold symmetrically.
Hence, it is a line of symmetry. Thus, we can draw only one line of symmetry for this figure.
Now, let us find the other hole so that the dotted line is the line of symmetry:
We measure the distance of the hole from the line of symmetry and then find the other hole.
Hence, in this way, we can find the other hole so that the dotted line is the line of symmetry.
Let us complete the polygon.
Step 1:
Step 2:
Step 3:
#### Rotational Symmetry
Look at the movement of the spinning bicycle wheel shown below.
We say that the spikes of the wheel rotate.
The rotation of the hands of a clock is called the clockwise rotation whereas the opposite of clockwise rotation is called anticlockwise rotation.
Now look at the rotation of the wheel given below:
The spikes of the wheel rotate in an anticlockwise direction.
Some objects or shapes, when rotated about a particular point, will look exactly the same as the original figure. This is called rotational symmetry. Some objects or shapes can be rotated through a certain angle.
An object or shape has rotational symmetry if it looks exactly like the original object or shape after some rotation less than one turn or after one turn.
For the rotation of the design shown below, we say that it rotates about the fixed point marked by a star sign at the centre of the square.
Also, rotation does not change the shape and size of the object.
• A full turn means rotation by 360⁰.
• A half turn means rotation by 180⁰.
• A quarter turn means rotation by 90⁰.
Let us consider the figure shown below:
Does this figure show rotational symmetry about a fixed point at the centre? If yes, then what is this fixed point called?
This fixed point is the centre of rotation.
• The angle through which the figure rotates is the angle of rotation.
• The number of times the figure looks exactly the same as the original figure in a full turn is the order of rotational symmetry.
The centre of rotation of an object is the fixed point around which the rotation occurs. For the rotating object shown below, the point at the centre is the centre of rotation.
Observe the given figure. When it is rotated clockwise through a quarter, what is the angle it covers?
It covers 90⁰.
When the figure is rotated clockwise through the next quarter, what is the angle through which it has rotated?
It is 180⁰.
When the figure is rotated clockwise again through the next quarter, what is the angle through which it has rotated?
It is 270⁰.
Rotating the figure again through the next quarter, what is the angle through which it has rotated now?
It is 360⁰.
Consider the object shown below:
Let us rotate the object by 180⁰ clockwise.
Now let us again rotate the object by 180⁰ clockwise.
Hence, there are two times when the object looks exactly the same during the rotation.
So, the order of rotational symmetry is 2.
• If an object has ‘n’ positions when it looks exactly the same in a full turn, the object has rotational symmetry of order ‘n’.
• Every object has rotational symmetry of order 1 or more.
Consider the rotation of the circle shown below.
The centre of rotation is the centre of the circle. It looks the same during rotation through any angle. So, the circle has rotational symmetry.
Consider the isosceles triangle shown below:
Let us rotate it by 90⁰ clockwise.
The triangle obtained is not exactly the same as the original, so let us rotate it again by 90⁰ clockwise.
Still, the triangle obtained is not exactly the same as the original, so let us rotate it again by 90⁰ clockwise.
This time as well, the triangle obtained is not exactly the same as the original. So, let us rotate it again by 90⁰ clockwise.
In a complete turn of 360⁰, the number of times an object looks exactly the same is one.
The isosceles triangle looks exactly like the original only once in a full rotation of 360⁰.
So, it has rotational symmetry of order 1.
Consider the rotation of the objects shown below:
For every object here, the centre of rotation is the centre of the objects.
Consider the letter given below.
The centre of rotation is marked by ‘+’ sign.
Now let us see how the letter looks after rotating it by 45⁰ clockwise.
If the letter is rotated by 90⁰ clockwise we get:
If the letter is rotated by 180⁰ clockwise we get:
#### Line Symmetry and Rotational Symmetry
The circle is a special case, having both line symmetry and rotational symmetry. Consider the rotation of the circle shown below:
The centre of rotation is the centre of the circle. It looks the same during rotation through any angle. So, a circle has rotational symmetry. Also, it has infinitely many lines of symmetry.
So, a circle has both line symmetry and rotational symmetry.
Consider the isosceles triangle shown below:
This triangle has a vertical line as the line of symmetry. So, it has line symmetry. In a complete turn of 360⁰, the number of times an object looks exactly the same is one.
So, it has rotational symmetry of order 1. Hence, some figures have line symmetry but no rotational symmetry of order more than 1.
Let us draw letters of the English alphabet having a line symmetry and rotational symmetry of order 2.
Consider the letter given below:
Clearly, this letter has two lines of symmetry. So, it has line symmetry.
Also, it has rotational symmetry of order 2 as in a complete rotation of 360⁰, the number of times it takes the same position as the initial position is 2.
The dot in the figure is only for reference to trace the rotation. Do not consider it to be a part of the figure.
Consider the figure shown below:
A parallelogram has no line of symmetry as no line divides it into two symmetrical halves.
So, it has no line symmetry. In a complete turn of 360⁰, the number of times the parallelogram looks exactly the same is 2.
So, it has rotational symmetry. Hence, some figures have rotational symmetry but no line symmetry.
Let us draw a quadrilateral having line symmetry but no rotational symmetry of order more than 1.
Clearly, this quadrilateral has one line of symmetry, so it has line symmetry.
Also, it has no rotational symmetry of order more than 1 as in a complete rotation of 360⁰. So the number of times it takes the same position as the initial position is 1.
#### Common Errors
The following are topics in which students make common mistakes when dealing with symmetry:
• 1. A figure can be either symmetric or asymmetric
• 2. Line of symmetry is not always a horizontal line
• 3. Number of lines of symmetry for a regular polygon
• 4. Rotational symmetry is not the same as line symmetry
• 5. It is order of rotational symmetry for an object, not order of rotation
• 6. Every object has rotational symmetry
#### A Figure Can Be Either Symmetric Or Asymmetric
A figure can be either symmetric or asymmetric but cannot be both.
Figures having line/lines of symmetry are symmetric. Figures having no line/lines of symmetry are asymmetric.
#### Line Of Symmetry Is Not Always A Horizontal Line
The line of symmetry may be vertical or horizontal or slant. The given letter has both vertical and horizontal lines of symmetry.
#### Number Of Lines Of Symmetry For A Regular Polygon
We don’t need to draw lines of symmetry to find the number of lines of symmetry for regular polygons.
Remember! Each regular polygon has as many lines of symmetry as its sides.
The number of lines of symmetry for a regular hexagon is 6.
#### Rotational Symmetry Is Not The Same As Line Symmetry
An object or shape has rotational symmetry if it looks exactly the same as the original object or shape after some rotation less than one turn or after one turn.
An object having a line/ lines of symmetry is said to have line symmetry. So, rotational symmetry is not the same as line symmetry.
The figure has one line of symmetry so it has line symmetry. Also, only after one turn, it looks the same as the original or takes the same position as the original one, so the order of rotational symmetry is 1. Hence, this figure has line symmetry but no rotational symmetry of order more than 1.
#### It Is Order Of Rotational Symmetry For An Object, Not Order Of Rotation
In a complete turn of 360⁰, the number of times an object looks exactly the same as the original object is called the order of rotational symmetry for the object, not the order of rotation of the object.
#### Every Object Has Rotational Symmetry
In a complete turn of 360⁰, the number of times an object looks exactly the same as the original object is called the order of rotational symmetry for the object.
As in a complete turn of 360⁰, every object looks exactly the same as the original at least once, i.e., after one complete turn; so every object has rotational symmetry of at least order 1.
So, every object has rotational symmetry.
#### Conclusion
You have learnt all about symmetry in this blog. Can you solve this riddle?
Choose the option which will complete the pattern drawn on the left side:
#### Arpana
Author
Arpana is an education specialist with years of teaching Math and developing content. She previously worked as a freelance content developer, developing lesson plans for a reputed publisher of text books and content for various educational companies. She is an enthusiastic teacher who has taught students from India, the UK, and New Zealand. | HuggingFaceTB/finemath | |
# Examples for 7th grade (seventh) - page 13
1. Fertilizer Mr. Gherkin use for fertilization 8% solution of fertilizer. 4 liters of it still left. How much water must be added to the solution to make only 4% solution?
2. Tiles Hall has dimensions 325 &time; 170 dm. What is the largest size of square tiles that can be entire hall tiled and how many we need them?
3. Pedestrian and cyclist Pedestrian out with a speed of 4 km/hour from city center and after 1hour and 10 minutes came after him cyclist at speed of 18 km/h. At how many minutes he catches up with pedestrian?
4. Cyclist A cyclist is moving at 35 km/h and follow pedestrian who is walking at speed 4 km/h. Walker has an advantage 19 km. How long take cyclist to catch up with him?
5. Water tank Water tank shape cuboid has a width of 3.1 m and length twice larger. How high will reach water if water flow into 13 liters of water per second during 16 minutes?
6. Exhibition The teacher paid for 280 Kč for 4.A students for admission to the exhibition. How many students were on the exhibition?
7. Hiker Hiker went half of trip the first day, the third of trip the second day and remaines 15 km. How long trip he planned?
8. Train from Prague First train from Prague started at 8:00 hour at 40 kilometers per hour. Train from Ostrava started at 9:20 at 80 km per hour. In how many hours and how far from cities with trains meet if the distance of cities is 400 km.
9. Two workers The first worker completed the task by himself in 9 hours, the second in 15 hours. After two hours of joint work left first worker to a doctor and the second finished the job himself. How many hours worked second worker himself?
10. The bridge Across the circle lakepasses through its center bridge over the lake. At three different locations on the lake shore are three fishermen A, B, C. Which of fishermen see the bridge under the largest angle?
11. Box Calculate the angle between box base 9 x 14 and body diagonal length 18.
12. Do you solve this? Determine area S of rectangle and length of its sides if its perimeter is 102 cm.
13. Opposite numbers Calculate opposite numbers (additive inverse) to given ones:
14. Workers Workers digging a jump pit in the school yard. Pit has a cuboid shape with a length 12 m, a width 20 dm and depth 36 cm. They excavate 0.4 cubic meters of soil an hour. How much time (hours and minutes) is need to the excavate this pit?
15. Balance The rod are 1.9 m long hanging weights 4 kg and 1 kg on ends. Where are centre of rod (distance from weight 4 kg) to be in balance?
16. Weight of air What is the weight of air in the living room measuring width 5 m length 2 m and height 2.8 m? Air density is ρ = 1.2 kg/m3.
17. Rectangle A2dim Calculate the side of the rectangle, if you know that its area is of 2590 m2 and one side is 74 m.
18. Car Car travels 1/3 of the trip on the first day, second day 2/5 of the trip and left even 340 km for next days. How long is the trip?
19. School books At the beginning of the school year, the teacher distributed 480 workbooks and 220 textbooks. How many pupils could have the most in the classroom?
20. Potatoes Potatoes contain 78.6% starch. How many potatoes need to obtain 27 kg of starch?
Do you have an interesting mathematical word problem that you can't solve it? Enter it, and we can try to solve it.
To this e-mail address, we will reply solution; solved examples are also published here. Please enter the e-mail correctly and check whether you don't have a full mailbox.
Please do not submit problems from current active competitions such as Mathematical Olympiad, correspondence seminars etc... | HuggingFaceTB/finemath | |
# How is exponentiation defined in Peano arithmetic?
How would exponentiation be defined in Peano arithmetic? Unless $n$ is fixed natural number, $x^n$ seems to be hard to define.
Edit 2: So, what would be the way to define $x^n+y^n = z^n$ using $\Sigma_1^0$ formula?
Edit: OK, I say Peano arithmetic has addition and multiplication stuffs. This allows additions to be expressed without quantifiers but as for exponentiation Peano arithmetic is silent. That's why I asked this question. Just for clarification.
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@JyrkiLahtonen, Peano arithmetic deals in natural numbers. – vonbrand Feb 24 '13 at 16:35
Yeah I know that, but then what's the problem? Define $x^n$ by induction on $n$. Oh, you want to build a formula using Peano axioms alone. Sorry, I totally misunderstood. – Jyrki Lahtonen Feb 24 '13 at 17:05
The bare bones answer is something like what Hagen has said. The idea is this: Exponentiation is understood to be a function defined recursively: $y=2^x$ iff there is a sequence $t_0,t_1,\dots,t_x$ such that
• $t_0=1$,
• $t_x=y$, and
• For all $n<x$, $t_{n+1}=2\times t_n$.
In this respect, exponentiation is hardly unique: $y=x!$ is defined similarly, for example. Now you'd say that there is a sequence $z_0,z_1,\dots,z_x$, such that
• $z_0=1$,
• $z_x=y$, and
• For all $n<x$, $z_{n+1}=(n+1)\times z_n$.
(That $t_0=z_0=1$ is coincidence. In one case it is because $2^0=1$; in the other, because $0!=1$.)
So, to define a formula saying that $y=2^x$, you'd like to say that there is a sequence $t_0,\dots,t_x$ as above.
The problem, of course, is that in Peano Arithmetic one talks about numbers rather than sequences. Gödel solved this problem when working on his proof of the incompleteness theorem: He explained how to code finite sequences by numbers, by using the Chinese remainder theorem. Recall that this result states that, given any numbers $n_1,\dots,n_k$, pairwise relatively prime, and any numbers $m_1,\dots,m_k$, there is a number $x$ that simultaneously satisfies all congruences $$x\equiv m_i\pmod {n_i}$$ for $1\le i\le k$.
In particular, given $m_1,\dots,m_k$, let $n=t!$ where $t=\max(m_1,\dots,m_k,k)$. Letting $n_1=n+1$, $n_2=2n+1,\dots$, $n_k=kn+1$, we see that the $n_i$ are relatively prime, and we can find an $x$ that satisfies $x\equiv m_i\pmod{n_i}$ for all $i$. We can then say that the pair $\langle x,n\rangle$ codes the sequence $(m_1,\dots,m_k)$. In fact, given $x,n$, it is rather easy to "decode" the $m_i$: Just note that $m_i$ is the remainder of dividing $x$ by $in+1$.
Accordingly, we can define $y=2^x$ by saying that there is a pair $\langle a,b\rangle$ that, in the sense just described, codes a sequence $(t_0,t_1,\dots,t_x)$ such that $t_0=1$, $t_x=y$, and $t_{n+1}=2t_n$ for all $n<x$. (Again, "in the sense just described" ends up meaning simply that "the remainder of dividing $a$ by $ib+1$ is $t_i$ for all $i\le x$". Note that we are not requiring $b$ to be the particular number we exhibited above using factorials.)
Of course, one needs to prove that any two pairs coding such a sequence agree on the value of $t_x$, but this is easy to establish. And we can code a pair by a number using, for example, Cantor's enumeration of $\mathbb N\times\mathbb N$, so that $\langle a,b\rangle$ is coded as the number $$c=\frac{(a+b)(a+b+1)}2+b.$$ This is a bijection, and has the additional advantages that it is definable and satisfies $a,b\le c$ (so it is given by a bounded formula).
An issue that appears now is that we need to formalize the discussion of the Chinese remainder theorem and the subsequent coding within Peano arithmetic. This presents new difficulties, as again, we cannot (in the language of arithmetic) talk about sequences, and cannot talk about factorials, until we do all the above, so it is not clear how to prove or even how to formulate these results.
This problem can be solved by noting that we can use induction within Peano arithmetic. One then proceeds to show, essentially, that given any finite sequence, there is a pair that codes it, and that if a pair codes a sequence $\vec s$, and a number $t$ is given, then there is a pair that codes the sequence $\vec s{}^\frown(t)$. That is, one can write down a formula $\phi(x,y,z)$, "$y$ codes a sequence, the $z$-th member of which is $x$", such that PA proves:
• For all $z$ and $y$ there is a unique $x$ such that $\phi(x,y,z)$.
• For all $x$ there is a $y$ such that $\phi(x,y,0)$.
• For all $x,y,z$ there is a $w$ such that the first $z$ terms of the sequences coded by $y$ and $w$ coincide, and the next term coded by $w$ is $x$.
In fact, we can let $\phi$ be a bounded formula: Take $\phi(x,y,z)$ to be
There are $a,b\le y$ such that $y=\langle a,b\rangle$ (Cantor's pairing) and $x<zb+1$ and there is a $d\le a$ such that $a=d(zb+1)+x$.
Once we have in PA the existence of coded sequences like this, implementing recursive definitions such as exponential functions is straightforward.
There are two excellent references for these coding issues and the subtleties surrounding them:
1. Richard Kaye. Models of Peano arithmetic. Oxford Logic Guides, 15. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1991. MR1098499 (92k:03034). (See chapter 5.)
2. Petr Hájek, and Pavel Pudlák. Metamathematics of first-order arithmetic. Perspectives in Mathematical Logic. Springer-Verlag, Berlin, 1993. MR1219738 (94d:03001). (See chapter 1.)
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Curiously, I just gave a talk in the set theory seminar at Boise State on this question: math.stackexchange.com/questions/282865/…, so I had just (re)worked through these details. – Andres Caicedo Feb 24 '13 at 17:01
As for $\phi (x,y,z)$, bounded formula, it's $\Sigma^0_0$ formula, I guess, right? (But I heard that $\Sigma^0_0$ formula can also refer to quantifier-free formula, so which definition is usually used in this case?) – UQT Feb 28 '13 at 5:33
(There was a typo on my previous comment, it is fixed in this version.) Bounded is the usual definition. Sometimes, in the context of Hilbert's $10$-th problem, one talks of $\Sigma_0=\Pi_0=\Delta_0$ being quantifier free, and $\Sigma_1$ being existential quantification over $\Sigma_0$, so the Davis-Matiyasevich-Putnam-Robinson result can be stated succinctly as $\Sigma^0_1=\Sigma_1$. – Andres Caicedo Mar 1 '13 at 6:00
Using the following abbreviations
\begin{align}a\le b&\equiv\exists n\colon a+n=b\\ a< b&\equiv Sa\le b\\ \operatorname{mod}(a,b,c)&\equiv \exists n\colon a=b\cdot n+c\land c<b\\ \operatorname{seq}(a,b,k,x)&\equiv \operatorname{mod}(a,S(b\cdot Sk),x)\\ \operatorname{pow}(a,b,c)&\equiv\exists x\exists y\colon\operatorname{seq}(x,y,0,S0)\land\operatorname{seq}(x,y,b,c)\land \\&\quad\forall k\forall z\colon((k<c\land \operatorname{seq}(x,y,k,z))\to \operatorname{seq}(x,y,Sk,a\cdot z))\end{align} we have $\operatorname{pow}(a,b,c)$ if and only if $c=a^b$. Intriguingly, you need some elementary number theory (such as the Chinese remainder theorem) to meta-appreciate this.
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Peano's axioms don't include addition or multiplication. All you get is zero, the "next" operation $s(n)$, and induction. When deriving things from Peano's axioms, all of the arithmetic operations are usually defined by recursion:
• $0+a = a$
• $s(n) + a = s(n+a)$
• $0\cdot a = 0$
• $s(n) \cdot a = a + (n \cdot a)$
• $a^0 = 1$
• $a^{s(n)} = a \cdot a^n$
(aside: The second-to-last formula implies $0^0=1$. This convention is a good choice for discrete exponentiation operators, but a bad choice for continuous ones. Since we are defining a discrete one, I adopt it)
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I don't think this is an answer to the question. The language of first-order logic includes $+$ and $\times$, but no symbol for exponentiation. Then the Peano axioms include their inductive properties, and again it obviously doesn't talk at all about exponentiation. In Peano arithmetic, exponentiation, unlike $+$ and $\times$ has to be defined properly by a formula $\phi(x,y,z)$ then you have to prove that the formula has the inductive and functional properties you are thinking about. – mercio Feb 24 '13 at 14:43
@merco: What in the question suggests we're talking about first-order PA? – Chris Eagle Feb 24 '13 at 17:05
The term "Peano arithmetic" itself suggests first-order PA; "Peano axioms" is more neutral. In particular Peano arithmetic is not the set of 5 axioms proposed by Peano. The abbreviation PA, although not used in the question, is universally used for the first-order theory. – Carl Mummert Feb 24 '13 at 18:49
Another approach: Define exponentiation as a binary function on $N$ such that:
If $0\notin N$ then
1. $x^1=x$
2. $x^{y+1}=x^y\cdot x$
If $0\in N$ then
1. $x^0=1$ if $x\neq 0$
2. $x^{y+1}=x^y\cdot x$
Note that in the latter, we leave $0^0$ undefined, or more precisely $0^0$ is an unspecified natural number, e.g. we can multiply it by $0$ to get $0$. For a formal rationale, see "Oh, the ambiguity!" at my math blog.
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Adding function symbols is not allowed – that is tantamount to changing the logical language. – Zhen Lin Nov 1 '13 at 19:27
$a^{b}=c$: $\forall x>0\forall y \exists !z((y=0\to z=1)\wedge\forall m\forall n((y=m\to z=n)\to (y=m+1\to z=n*x))\wedge((x=a\wedge y=b)\to z=c))$
Notice that this formula is not unique, you can find some other ways to define it. But it is not neccessary to do it because it's known that every recursive(computable) function is definable, sometimes the formula to define a function may be very long and tedious.
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This is definitely not correct because of the part $\forall m\forall n ((y=m\to z=n)\to(y=m+1\to z=n\cdot x))$ If $y\ne m+1$, then $y=m+1\to z=n\cdot x$ is true and if $y=m+1$ it reduces to $z=n\cdot x$. But $\forall n(z=n\cdot x)$ is false (unless $x=z=0$). – Hagen von Eitzen Feb 24 '13 at 14:16
Sorry but I can not understand 'If y≠m+1, then y=m+1→z=n⋅x is true and if y=m+1 it reduces to z=n⋅x'? – Chao Chen Feb 24 '13 at 14:36
Please explain why $1^1 = 1$. I pick $a=1,b=1,c=1,x=1,y=1,m=0,n=0$. Please show me an integer $z$ such that $(y=0 \implies z=1) \wedge ((y=m \implies z=n) \implies (y=m+1 \implies z=n*x)) \wedge ((x=a \wedge y=b) \implies z=c)$ – mercio Feb 24 '13 at 15:45 | open-web-math/open-web-math | |
Definitions
Nearby Words
# Trilinear interpolation
Trilinear interpolation is a method of multivariate interpolation on a 3-dimensional regular grid. It approximates the value of an intermediate point $\left(x, y, z\right)$ within the local axial rectangular prism linearly, using data on the lattice points. For an arbitrary, unstructured mesh (as used in finite element analysis), other methods of interpolation must be used; if all the mesh elements are tetrahedra (3D simplices), then barycentric coordinates provide a straightforward procedure.
Trilinear interpolation is frequently used in numerical analysis, data analysis, and computer graphics.
## Compared to linear and bilinear interpolation
Trilinear interpolation is the extension of linear interpolation, which operates in spaces with dimension $D=1$, and bilinear interpolation, which operates with dimension $D=2$, to dimension $D=3$. The order of accuracy is 1 for all these interpolation schemes, and it requires $\left(1 + n\right)^D = 8$ adjacent pre-defined values surrounding the interpolation point. There are several ways to arrive at trilinear interpolation, it is equivalent to 3-dimensional tensor B-spline interpolation of order 1, and the trilinear interpolation operator is also a tensor product of 3 linear interpolation operators
## Method
On a periodic and cubic lattice with spacing 1, let $x_d$, $y_d$, and $z_d$ be the differences between each of $x$, $y$, $z$ and the smaller coordinate related, that is:
$x_d = x - lfloor x rfloor$
$y_d = y - lfloor y rfloor$
$z_d = z - lfloor z rfloor$
First we interpolate along $z$(imagine we are pushing the front face of the cube to the back), giving:
$i_1 = v\left[lfloor x rfloor,lfloor y rfloor, lfloor z rfloor\right] times \left(1 - z_d\right) + v\left[lfloor x rfloor, lfloor y rfloor, lceil z rceil\right] times z_d$
$i_2 = v\left[lfloor x rfloor,lceil y rceil, lfloor z rfloor\right] times \left(1 - z_d\right) + v\left[lfloor x rfloor, lceil y rceil, lceil z rceil\right] times z_d$
$j_1 = v\left[lceil x rceil,lfloor y rfloor, lfloor z rfloor\right] times \left(1 - z_d\right) + v\left[lceil x rceil, lfloor y rfloor, lceil z rceil\right] times z_d$
$j_2 = v\left[lceil x rceil,lceil y rceil, lfloor z rfloor\right] times \left(1 - z_d\right) + v\left[lceil x rceil, lceil y rceil, lceil z rceil\right] times z_d.$
Then we interpolate these values (along $y$, as we were pushing the top edge to the bottom), giving:
$w_1 = i_1\left(1 - y_d\right) + i_2y_d$
$w_2 = j_1\left(1 - y_d\right) + j_2y_d$
Finally we interpolate these values along $x$(walking through a line):
$IV = w_1\left(1 - x_d\right) + w_2x_d .$
This gives us a predicted value for the point.
The result of trilinear interpolation is independent of the order of the interpolation steps along the three axes: any other order, for instance along $x$, then along $y$, and finally along $z$, produces the same value.
The above operations can be visualized as follows: First we find the eight corners of a cube that surround our point of interest. These corners have the values C000, C100, C010, C110, C001, C101, C011, C111.
Next, we perform linear interpolation between C000 and C100 to find C00, C001 and C101 to find C01, C011 and C111 to find C11, C010 and C110 to find C10.
Now we do interpolation between C00 and C10 to find C0, C01 and C11 to find C1. Finally, we calculate the value C via linear interpolation of C0 and C1 | HuggingFaceTB/finemath | |
# Like examples combining math terms
## Combining Like Term Lesson & Distributive Property Lesson Combining Like Term Lesson & Distributive Property Lesson. Terms whose variables (such as x or y) with any exponents (such as the 2 in x 2) are the same. Examples: 7x and 2x are like terms because they are both "x". 3x 2 and −2x 2 are like terms because they are both "x 2 ". But 7x and 7x 2 are NOT like terms (the exponents are different), they are unlike terms. Like terms can be added together., Combining Like Terms Word Problems. Displaying all worksheets related to - Combining Like Terms Word Problems. Worksheets are Combining like terms, Model practice challenge problems iv, Expressions combining like terms homework, Classwork, Combining like terms fractional coefficients, Addition and subtraction when adding, The distributive property, Combining like terms..
### Algebra Basics Like terms - In Depth - math
Algebra Basics Like terms - In Depth - math. The terms x 2 y and xy 2 are not like terms, since they have different variable parts. If we have a pile with 5 lemons and a pile with 6 lemons, we can combine them to get one pile with 11 lemons. Of course, what we actually should do is make lemonade, but that's more of a …, For the first part of this lesson students will work with their groups on developing a distinction between like and unlike terms. I will post the Combining Like Terms Chart on the board. I will explain to students that the pairs in the left column are like terms and the pairs in the right column are called unlike terms..
I showed you the distributive property in the above examples to give you the thought behind combining like terms. Example 3 : Simplify the expression . It looks like we have two terms that have an x squared that we can combine and we have two terms that have an x that we can combine. Examples of simplifying expressions by combining like terms Now, let’s work through a few examples to see how we can combine like terms to simplify expressions. In each case, the idea is to combine all like terms until there are no more to combine.
Combining Like Terms Word Problems. Displaying all worksheets related to - Combining Like Terms Word Problems. Worksheets are Combining like terms, Model practice challenge problems iv, Expressions combining like terms homework, Classwork, Combining like terms fractional coefficients, Addition and subtraction when adding, The distributive property, Combining like terms. Students learn to add or subtract terms that involve variables by combining like terms. Like terms are terms with exactly the same variable factors. For example, 5x and 3x are like terms. However, 7x and 9y are not like terms, and neither are 8x and 8x^2. In the problem 2x + 3x + 4, only the 2x and 3x can be combined, because they are like
Students learn to add or subtract terms that involve variables by combining like terms. Like terms are terms with exactly the same variable factors. For example, 5x and 3x are like terms. However, 7x and 9y are not like terms, and neither are 8x and 8x^2. In the problem 2x + 3x + 4, only the 2x and 3x can be combined, because they are like In this lesson, we will look into solving equations by combining like terms. What are Like Terms? Like terms are terms that contain the same variable with the same exponent. Constant terms are like terms because they do not have any variables. Here are some examples of like terms: 2x and 5x are like terms because both contain the same variable.
Ex 3: Combining Like Terms Requiring Distribution This video provides examples of how to combine like terms that also requires distribution. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by … Interactive Online quiz on Combining Like Terms. Choose your difficulty and start answering questions! Chart Maker Graphing Calculator Image used by permission of the original author, Chris Burke. Like Terms Quiz. Objective: For students to practice combining like terms in math. Chose a difficulty level and start practicing. Quiz Difficulty
I showed you the distributive property in the above examples to give you the thought behind combining like terms. Example 3 : Simplify the expression . It looks like we have two terms that have an x squared that we can combine and we have two terms that have an x that we can combine. These sets of worksheets introduce your students to the concept of combining like terms, and provide examples, short practice sets, longer sets of questions, and quizzes. This is a paramount skill in algebra, you will need to master it in order to have an easy transition to higher level math. You basically cleaning up an expression or equation
The terms x 2 y and xy 2 are not like terms, since they have different variable parts. If we have a pile with 5 lemons and a pile with 6 lemons, we can combine them to get one pile with 11 lemons. Of course, what we actually should do is make lemonade, but that's more of a … Ex 3: Combining Like Terms Requiring Distribution This video provides examples of how to combine like terms that also requires distribution. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by …
These 3 math mazes are a great way for students to get combining like terms practice. This combining like terms activity is great for differentiation, stations or centers, partner work, or homework. Students will be combining like terms with one variable, two variables, and one… Home Decorators Lighting Collection Info: 8546982693 Oct 15, 2017 · Combining Like Terms Doodle Notes. If you’re not familiar with Doodle Notes you can read about them here. They gives students a different way of looking at combining like terms. I like to use this particular set of doodle notes from Math Giraffe on the 3rd or 5th day of working with combining like terms. The magic of this type of notes can be
There are only two terms in this expression that are alike: 3xy and -6xy. The parentheses in this expression are not necessary, since it doesn't change how we treat each expression. We can rewrite it without 'em, then mark each set of like terms. The tricky part with this problem is to keep the There are only two terms in this expression that are alike: 3xy and -6xy. The parentheses in this expression are not necessary, since it doesn't change how we treat each expression. We can rewrite it without 'em, then mark each set of like terms. The tricky part with this problem is to keep the
Combining Like Terms Math Help. Terms whose variables (such as x or y) with any exponents (such as the 2 in x 2) are the same. Examples: 7x and 2x are like terms because they are both "x". 3x 2 and −2x 2 are like terms because they are both "x 2 ". But 7x and 7x 2 are NOT like terms (the exponents are different), they are unlike terms. Like terms can be added together., Interactive Online quiz on Combining Like Terms. Choose your difficulty and start answering questions! Chart Maker Graphing Calculator Image used by permission of the original author, Chris Burke. Like Terms Quiz. Objective: For students to practice combining like terms in math. Chose a difficulty level and start practicing. Quiz Difficulty.
### Algebra Basics Like terms - In Depth - math Combining Like Terms Word Problems Worksheets Kiddy Math. These 3 math mazes are a great way for students to get combining like terms practice. This combining like terms activity is great for differentiation, stations or centers, partner work, or homework. Students will be combining like terms with one variable, two variables, and one… Home Decorators Lighting Collection Info: 8546982693, Combining Like Terms Word Problems. Displaying all worksheets related to - Combining Like Terms Word Problems. Worksheets are Combining like terms, Model practice challenge problems iv, Expressions combining like terms homework, Classwork, Combining like terms fractional coefficients, Addition and subtraction when adding, The distributive property, Combining like terms..
### Combining Like Terms Word Problems Worksheets Kiddy Math Combining Like Terms Math Help. For the first part of this lesson students will work with their groups on developing a distinction between like and unlike terms. I will post the Combining Like Terms Chart on the board. I will explain to students that the pairs in the left column are like terms and the pairs in the right column are called unlike terms. Examples of simplifying expressions by combining like terms Now, let’s work through a few examples to see how we can combine like terms to simplify expressions. In each case, the idea is to combine all like terms until there are no more to combine.. Oct 15, 2017 · Combining Like Terms Doodle Notes. If you’re not familiar with Doodle Notes you can read about them here. They gives students a different way of looking at combining like terms. I like to use this particular set of doodle notes from Math Giraffe on the 3rd or 5th day of working with combining like terms. The magic of this type of notes can be Interactive Online quiz on Combining Like Terms. Choose your difficulty and start answering questions! Chart Maker Graphing Calculator Image used by permission of the original author, Chris Burke. Like Terms Quiz. Objective: For students to practice combining like terms in math. Chose a difficulty level and start practicing. Quiz Difficulty
Terms whose variables (such as x or y) with any exponents (such as the 2 in x 2) are the same. Examples: 7x and 2x are like terms because they are both "x". 3x 2 and −2x 2 are like terms because they are both "x 2 ". But 7x and 7x 2 are NOT like terms (the exponents are different), they are unlike terms. Like terms can be added together. In this lesson, we will look into solving equations by combining like terms. What are Like Terms? Like terms are terms that contain the same variable with the same exponent. Constant terms are like terms because they do not have any variables. Here are some examples of like terms: 2x and 5x are like terms because both contain the same variable.
These 3 math mazes are a great way for students to get combining like terms practice. This combining like terms activity is great for differentiation, stations or centers, partner work, or homework. Students will be combining like terms with one variable, two variables, and one… Home Decorators Lighting Collection Info: 8546982693 The terms x 2 y and xy 2 are not like terms, since they have different variable parts. If we have a pile with 5 lemons and a pile with 6 lemons, we can combine them to get one pile with 11 lemons. Of course, what we actually should do is make lemonade, but that's more of a …
Be careful when combining! Terms like x 2 yz and xy 2 z look a lot alike, but they aren't and cannot be combined. Write the terms carefully when working out problems. Don't overlook terms that are alike! Terms obey the associative property of multiplication - that … Then we will practice Combining Terms by adding or subtracting their coefficients by working through countless examples. Lastly, we will revisit the Distributive Property and use it to simplify expressions in order to create Equivalent Algebraic Expressions by Combining Like Terms. Combining Like …
In this lesson, we will look into solving equations by combining like terms. What are Like Terms? Like terms are terms that contain the same variable with the same exponent. Constant terms are like terms because they do not have any variables. Here are some examples of like terms: 2x and 5x are like terms because both contain the same variable. Terms whose variables (such as x or y) with any exponents (such as the 2 in x 2) are the same. Examples: 7x and 2x are like terms because they are both "x". 3x 2 and −2x 2 are like terms because they are both "x 2 ". But 7x and 7x 2 are NOT like terms (the exponents are different), they are unlike terms. Like terms can be added together.
Students learn to simplify an expression by combining like terms. For example, to simplify 5a + 7b – a – 2b, combine the like terms 5a – 1a, to get 4a, and combine the like terms 7b – 2b, to get 5b. So 5a + 7b – a – 2b simplifies to 4a + 5b. Combining Like Terms Word Problems. Displaying all worksheets related to - Combining Like Terms Word Problems. Worksheets are Combining like terms, Model practice challenge problems iv, Expressions combining like terms homework, Classwork, Combining like terms fractional coefficients, Addition and subtraction when adding, The distributive property, Combining like terms.
Students learn to simplify an expression by combining like terms. For example, to simplify 5a + 7b – a – 2b, combine the like terms 5a – 1a, to get 4a, and combine the like terms 7b – 2b, to get 5b. So 5a + 7b – a – 2b simplifies to 4a + 5b. Combining Like Terms Word Problems. Displaying all worksheets related to - Combining Like Terms Word Problems. Worksheets are Combining like terms, Model practice challenge problems iv, Expressions combining like terms homework, Classwork, Combining like terms fractional coefficients, Addition and subtraction when adding, The distributive property, Combining like terms.
These sets of worksheets introduce your students to the concept of combining like terms, and provide examples, short practice sets, longer sets of questions, and quizzes. This is a paramount skill in algebra, you will need to master it in order to have an easy transition to higher level math. You basically cleaning up an expression or equation Students learn to simplify an expression by combining like terms. For example, to simplify 5a + 7b – a – 2b, combine the like terms 5a – 1a, to get 4a, and combine the like terms 7b – 2b, to get 5b. So 5a + 7b – a – 2b simplifies to 4a + 5b. Combining Like Terms Word Problems. Combining Like Terms Word Problems - Displaying top 8 worksheets found for this concept.. Some of the worksheets for this concept are Combining like terms, Model practice challenge problems iv, Expressions combining like terms homework, Classwork, Combining like terms fractional coefficients, Addition and subtraction when adding, The distributive … Be careful when combining! Terms like x 2 yz and xy 2 z look a lot alike, but they aren't and cannot be combined. Write the terms carefully when working out problems. Don't overlook terms that are alike! Terms obey the associative property of multiplication - that …
How to Combine Like Terms (33 Surefire Examples!). combining like terms word problems. combining like terms word problems - displaying top 8 worksheets found for this concept.. some of the worksheets for this concept are combining like terms, model practice challenge problems iv, expressions combining like terms homework, classwork, combining like terms fractional coefficients, addition and subtraction when adding, the distributive …, there are only two terms in this expression that are alike: 3xy and -6xy. the parentheses in this expression are not necessary, since it doesn't change how we treat each expression. we can rewrite it without 'em, then mark each set of like terms. the tricky part with this problem is to keep the).
Mar 09, 2017 · Math 6th grade Variables & expressions Combining like terms. Combining like terms. Intro to combining like terms. Combining like terms. This is the currently selected item. Combining like terms example. Practice: … I showed you the distributive property in the above examples to give you the thought behind combining like terms. Example 3 : Simplify the expression . It looks like we have two terms that have an x squared that we can combine and we have two terms that have an x that we can combine.
Combining Like Terms Word Problems. Displaying all worksheets related to - Combining Like Terms Word Problems. Worksheets are Combining like terms, Model practice challenge problems iv, Expressions combining like terms homework, Classwork, Combining like terms fractional coefficients, Addition and subtraction when adding, The distributive property, Combining like terms. These sets of worksheets introduce your students to the concept of combining like terms, and provide examples, short practice sets, longer sets of questions, and quizzes. This is a paramount skill in algebra, you will need to master it in order to have an easy transition to higher level math. You basically cleaning up an expression or equation
Terms whose variables (such as x or y) with any exponents (such as the 2 in x 2) are the same. Examples: 7x and 2x are like terms because they are both "x". 3x 2 and −2x 2 are like terms because they are both "x 2 ". But 7x and 7x 2 are NOT like terms (the exponents are different), they are unlike terms. Like terms can be added together. Interactive Online quiz on Combining Like Terms. Choose your difficulty and start answering questions! Chart Maker Graphing Calculator Image used by permission of the original author, Chris Burke. Like Terms Quiz. Objective: For students to practice combining like terms in math. Chose a difficulty level and start practicing. Quiz Difficulty
Mar 09, 2017 · Math 6th grade Variables & expressions Combining like terms. Combining like terms. Intro to combining like terms. Combining like terms. This is the currently selected item. Combining like terms example. Practice: … Oct 15, 2017 · Combining Like Terms Doodle Notes. If you’re not familiar with Doodle Notes you can read about them here. They gives students a different way of looking at combining like terms. I like to use this particular set of doodle notes from Math Giraffe on the 3rd or 5th day of working with combining like terms. The magic of this type of notes can be
Terms whose variables (such as x or y) with any exponents (such as the 2 in x 2) are the same. Examples: 7x and 2x are like terms because they are both "x". 3x 2 and −2x 2 are like terms because they are both "x 2 ". But 7x and 7x 2 are NOT like terms (the exponents are different), they are unlike terms. Like terms can be added together. Be careful when combining! Terms like x 2 yz and xy 2 z look a lot alike, but they aren't and cannot be combined. Write the terms carefully when working out problems. Don't overlook terms that are alike! Terms obey the associative property of multiplication - that …
Mar 09, 2017 · Math 6th grade Variables & expressions Combining like terms. Combining like terms. Intro to combining like terms. Combining like terms. This is the currently selected item. Combining like terms example. Practice: … For the first part of this lesson students will work with their groups on developing a distinction between like and unlike terms. I will post the Combining Like Terms Chart on the board. I will explain to students that the pairs in the left column are like terms and the pairs in the right column are called unlike terms. Algebra Basics Like terms - In Depth - math
Combining Like Terms ChiliMath - Free Math Help. in this lesson, we will look into solving equations by combining like terms. what are like terms? like terms are terms that contain the same variable with the same exponent. constant terms are like terms because they do not have any variables. here are some examples of like terms: 2x and 5x are like terms because both contain the same variable., i showed you the distributive property in the above examples to give you the thought behind combining like terms. example 3 : simplify the expression . it looks like we have two terms that have an x squared that we can combine and we have two terms that have an x that we can combine.); oct 15, 2017 · combining like terms doodle notes. if you’re not familiar with doodle notes you can read about them here. they gives students a different way of looking at combining like terms. i like to use this particular set of doodle notes from math giraffe on the 3rd or 5th day of working with combining like terms. the magic of this type of notes can be, oct 15, 2017 · combining like terms doodle notes. if you’re not familiar with doodle notes you can read about them here. they gives students a different way of looking at combining like terms. i like to use this particular set of doodle notes from math giraffe on the 3rd or 5th day of working with combining like terms. the magic of this type of notes can be.
Combining Like Terms Word Problems Worksheets Kiddy Math
Algebra Basics Like terms - In Depth - math. there are only two terms in this expression that are alike: 3xy and -6xy. the parentheses in this expression are not necessary, since it doesn't change how we treat each expression. we can rewrite it without 'em, then mark each set of like terms. the tricky part with this problem is to keep the, for the first part of this lesson students will work with their groups on developing a distinction between like and unlike terms. i will post the combining like terms chart on the board. i will explain to students that the pairs in the left column are like terms and the pairs in the right column are called unlike terms.). Combining Like Terms Word Problems Worksheets Kiddy Math
How to Combine Like Terms (33 Surefire Examples!). students learn to simplify an expression by combining like terms. for example, to simplify 5a + 7b – a – 2b, combine the like terms 5a – 1a, to get 4a, and combine the like terms 7b – 2b, to get 5b. so 5a + 7b – a – 2b simplifies to 4a + 5b., oct 15, 2017 · combining like terms doodle notes. if you’re not familiar with doodle notes you can read about them here. they gives students a different way of looking at combining like terms. i like to use this particular set of doodle notes from math giraffe on the 3rd or 5th day of working with combining like terms. the magic of this type of notes can be). Algebra Basics Like terms - In Depth - math
How to Combine Like Terms (33 Surefire Examples!). interactive online quiz on combining like terms. choose your difficulty and start answering questions! chart maker graphing calculator image used by permission of the original author, chris burke. like terms quiz. objective: for students to practice combining like terms in math. chose a difficulty level and start practicing. quiz difficulty, students learn to simplify an expression by combining like terms. for example, to simplify 5a + 7b – a – 2b, combine the like terms 5a – 1a, to get 4a, and combine the like terms 7b – 2b, to get 5b. so 5a + 7b – a – 2b simplifies to 4a + 5b.). Combining Like Terms Examples Shmoop
How to Combine Like Terms (33 Surefire Examples!). mar 09, 2017 · math 6th grade variables & expressions combining like terms. combining like terms. intro to combining like terms. combining like terms. this is the currently selected item. combining like terms example. practice: …, these 3 math mazes are a great way for students to get combining like terms practice. this combining like terms activity is great for differentiation, stations or centers, partner work, or homework. students will be combining like terms with one variable, two variables, and one… home decorators lighting collection info: 8546982693). Algebra Basics Like terms - In Depth - math
Simplify Expressions and Distributive Property (examples. i showed you the distributive property in the above examples to give you the thought behind combining like terms. example 3 : simplify the expression . it looks like we have two terms that have an x squared that we can combine and we have two terms that have an x that we can combine., there are only two terms in this expression that are alike: 3xy and -6xy. the parentheses in this expression are not necessary, since it doesn't change how we treat each expression. we can rewrite it without 'em, then mark each set of like terms. the tricky part with this problem is to keep the).
Students learn to add or subtract terms that involve variables by combining like terms. Like terms are terms with exactly the same variable factors. For example, 5x and 3x are like terms. However, 7x and 9y are not like terms, and neither are 8x and 8x^2. In the problem 2x + 3x + 4, only the 2x and 3x can be combined, because they are like Be careful when combining! Terms like x 2 yz and xy 2 z look a lot alike, but they aren't and cannot be combined. Write the terms carefully when working out problems. Don't overlook terms that are alike! Terms obey the associative property of multiplication - that …
For the first part of this lesson students will work with their groups on developing a distinction between like and unlike terms. I will post the Combining Like Terms Chart on the board. I will explain to students that the pairs in the left column are like terms and the pairs in the right column are called unlike terms. Students learn to add or subtract terms that involve variables by combining like terms. Like terms are terms with exactly the same variable factors. For example, 5x and 3x are like terms. However, 7x and 9y are not like terms, and neither are 8x and 8x^2. In the problem 2x + 3x + 4, only the 2x and 3x can be combined, because they are like
Interactive Online quiz on Combining Like Terms. Choose your difficulty and start answering questions! Chart Maker Graphing Calculator Image used by permission of the original author, Chris Burke. Like Terms Quiz. Objective: For students to practice combining like terms in math. Chose a difficulty level and start practicing. Quiz Difficulty Mar 09, 2017 · Math 6th grade Variables & expressions Combining like terms. Combining like terms. Intro to combining like terms. Combining like terms. This is the currently selected item. Combining like terms example. Practice: …
Students learn to add or subtract terms that involve variables by combining like terms. Like terms are terms with exactly the same variable factors. For example, 5x and 3x are like terms. However, 7x and 9y are not like terms, and neither are 8x and 8x^2. In the problem 2x + 3x + 4, only the 2x and 3x can be combined, because they are like For the first part of this lesson students will work with their groups on developing a distinction between like and unlike terms. I will post the Combining Like Terms Chart on the board. I will explain to students that the pairs in the left column are like terms and the pairs in the right column are called unlike terms.
Interactive Online quiz on Combining Like Terms. Choose your difficulty and start answering questions! Chart Maker Graphing Calculator Image used by permission of the original author, Chris Burke. Like Terms Quiz. Objective: For students to practice combining like terms in math. Chose a difficulty level and start practicing. Quiz Difficulty Examples of simplifying expressions by combining like terms Now, let’s work through a few examples to see how we can combine like terms to simplify expressions. In each case, the idea is to combine all like terms until there are no more to combine. How to Combine Like Terms (33 Surefire Examples!) | HuggingFaceTB/finemath | |
Cody
# Problem 23. Finding Perfect Squares
Solution 2012113
Submitted on 10 Nov 2019 by kristie moore
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [2 3 4]; assert(isequal(isItSquared(a),true))
c = 3 f = 1 i = 4 f = 0 f = -1 f = 99999999 i = 9 f = 99999998 f = 99999997 f = 99999996 i = 16 f = 99999995 f = 99999994 f = 99999993 b = logical 1
2 Pass
a = [20:30]; assert(isequal(isItSquared(a),false))
c = 11 f = 1 i = 400 f = 0 f = -1 f = -2 f = -3 f = -4 f = -5 f = -6 f = -7 f = -8 f = -9 f = -10 i = 441 f = -11 f = -12 f = -13 f = -14 f = -15 f = -16 f = -17 f = -18 f = -19 f = -20 f = -21 i = 484 f = -22 f = -23 f = -24 f = -25 f = -26 f = -27 f = -28 f = -29 f = -30 f = -31 f = -32 i = 529 f = -33 f = -34 f = -35 f = -36 f = -37 f = -38 f = -39 f = -40 f = -41 f = -42 f = -43 i = 576 f = -44 f = -45 f = -46 f = -47 f = -48 f = -49 f = -50 f = -51 f = -52 f = -53 f = -54 i = 625 f = -55 f = -56 f = -57 f = -58 f = -59 f = -60 f = -61 f = -62 f = -63 f = -64 f = -65 i = 676 f = -66 f = -67 f = -68 f = -69 f = -70 f = -71 f = -72 f = -73 f = -74 f = -75 f = -76 i = 729 f = -77 f = -78 f = -79 f = -80 f = -81 f = -82 f = -83 f = -84 f = -85 f = -86 f = -87 i = 784 f = -88 f = -89 f = -90 f = -91 f = -92 f = -93 f = -94 f = -95 f = -96 f = -97 f = -98 i = 841 f = -99 f = -100 f = -101 f = -102 f = -103 f = -104 f = -105 f = -106 f = -107 f = -108 f = -109 i = 900 f = -110 f = -111 f = -112 f = -113 f = -114 f = -115 f = -116 f = -117 f = -118 f = -119 f = -120 b = logical 0
3 Pass
a = [1]; assert(isequal(isItSquared(a),true))
c = 1 f = 1 i = 1 f = 100000001 b = logical 1
4 Pass
a = [6 10 12 14 36 101]; assert(isequal(isItSquared(a),true))
c = 6 f = 1 i = 36 f = 0 f = -1 f = -2 f = -3 f = 99999997 f = 99999996 i = 100 f = 99999995 f = 99999994 f = 99999993 f = 99999992 f = 99999991 f = 99999990 i = 144 f = 99999989 f = 99999988 f = 99999987 f = 99999986 f = 99999985 f = 99999984 i = 196 f = 99999983 f = 99999982 f = 99999981 f = 99999980 f = 99999979 f = 99999978 i = 1296 f = 99999977 f = 99999976 f = 99999975 f = 99999974 f = 99999973 f = 99999972 i = 10201 f = 99999971 f = 99999970 f = 99999969 f = 99999968 f = 99999967 f = 99999966 b = logical 1
5 Pass
a = [6 10 12 14 101]; assert(isequal(isItSquared(a),false))
c = 5 f = 1 i = 36 f = 0 f = -1 f = -2 f = -3 f = -4 i = 100 f = -5 f = -6 f = -7 f = -8 f = -9 i = 144 f = -10 f = -11 f = -12 f = -13 f = -14 i = 196 f = -15 f = -16 f = -17 f = -18 f = -19 i = 10201 f = -20 f = -21 f = -22 f = -23 f = -24 b = logical 0 | HuggingFaceTB/finemath | |
# Homework Help: Stone thrown from a building
1. Jun 6, 2012
1. The problem statement, all variables and given/known data
A stone is thrown upward from the top of a building at an angle of 30.0 degrees to the horizontal and with an initial speed of 20.0 m/s. The height of the building is 45.0 m.
Where does the stone strike the ground?
2. Relevant equations
$v_{x0} = v_0\cos{\theta_0}$ where $v_{x0}$ is the original velocity x component, $v_0$ is the original velocity, and $\theta_0$ is the original angle (not sure if we use this)
Also $v_{y0}$ is the original velocity y component.
3. The attempt at a solution
I'm not sure how to proceed here. We're finding horizontal distance, so no equations concerning $v_{y0}$ would work, and $v_{x0} = v_0\cos{\theta_0}$ only works if the vertical distance is the same as the original vertical distance..
2. Jun 6, 2012
### Sleepy_time
Hi professordad. If you're confident in using vectors and integrating, for this problem you can use $\mathbf{F}=m\mathbf{\ddot{r}}=m(0,-g,0)$, separate the components and integrate along with the boundary conditions $\mathbf{x}_0=(0,h,0), \mathbf{v}_0=(v_0cos(\theta),v_0sin(\theta),0)$. After obtaining an equation for $x(t)$ and $y(t)$, substitute $y=0$ (when the y-axis meets the ground) and solve for t, and use this value in $x(t)$.
3. Jun 6, 2012
### Staff: Mentor
Doesn't seem like you need to use vectors or integration for this problem, unless I'm missing something.
Just write the two equations for the motion in the vertical and horizontal directions. The horizontal velocity stays constant (ignoring air resistance). The vertical motion follows the equations for motion given a constant acceleration (gravity).
You use the initial vertical velocity and position to figure out how long it takes the ball to hit the ground. Then use that time and the horizontal velocity to figure out how far out it hits.
Lets see some equations!
4. Jun 6, 2012
### Sleepy_time
No not required, just a preference
5. Jun 6, 2012
### azizlwl
There are 2 equations of motion. Vertical and horizontal.
They work independently
They are only related by time since it is a single stone.
6. Jun 6, 2012
### HallsofIvy
Using "vertical and horizontal directions" IS using vectors and the "equations for motion given a constant acceleration" are derived by integration. You can sweep them under the rug but they are there none the less.
7. Jun 6, 2012
### Staff: Mentor
True, but I was worried that mentioning that approach would scare the OP into thinking that the problem was more complicated than it is.
8. Jun 7, 2012
### harts
First figure out how much time it takes for the stone to hit the ground using the fact that
y=y0+vot+1/2gt2
Knowing this time, you should be able to calculate the horizontal distance traveled, using the equation you posted | HuggingFaceTB/finemath | |
# Convert horsepower (550 ft*lbf/s) to femtowatt
## How to Convert horsepower (550 ft*lbf/s) to femtowatt
To convert horsepower (550 ft*lbf/s) to femtowatt , the formula is used,
$\mathrm{femtowatt}=\mathrm{horsepower}*745700000000000$
where the horsepower (550 ft*lbf/s) to fW value is substituted to get the answer from Power Converter.
1 horsepower (550 ft*lbf/s)
=
74569e+13 fW
1 fW
=
1.341e-18 horsepower (550 ft*lbf/s)
Example: convert 15 horsepower (550 ft*lbf/s) to fW:
15 horsepower (550 ft*lbf/s)
=
15
x
74569e+13 fW
=
11185e+15 fW
## horsepower (550 ft*lbf/s) to femtowatt Conversion Table
horsepower (550 ft*lbf/s) femtowatt (fW)
0.01 horsepower (550 ft*lbf/s) 7456998715e+6 fW
0.1 horsepower (550 ft*lbf/s) 7456998715e+7 fW
1 horsepower (550 ft*lbf/s) 7456998715e+8 fW
2 horsepower (550 ft*lbf/s) 1491399743e+9 fW
3 horsepower (550 ft*lbf/s) 2237099614e+9 fW
5 horsepower (550 ft*lbf/s) 3728499357e+9 fW
10 horsepower (550 ft*lbf/s) 7456998715e+9 fW
20 horsepower (550 ft*lbf/s) 1491399743e+10 fW
50 horsepower (550 ft*lbf/s) 3728499357e+10 fW
100 horsepower (550 ft*lbf/s) 7456998715e+10 fW
1000 horsepower (550 ft*lbf/s) 7456998715e+11 fW
### Popular Unit Conversions Power
The most used and popular units of power conversions are presented for quick and free access. | HuggingFaceTB/finemath | |
Is 13566 a prime number? What are the divisors of 13566?
## Parity of 13 566
13 566 is an even number, because it is evenly divisible by 2: 13 566 / 2 = 6 783.
Find out more:
## Is 13 566 a perfect square number?
A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 13 566 is about 116.473.
Thus, the square root of 13 566 is not an integer, and therefore 13 566 is not a square number.
## What is the square number of 13 566?
The square of a number (here 13 566) is the result of the product of this number (13 566) by itself (i.e., 13 566 × 13 566); the square of 13 566 is sometimes called "raising 13 566 to the power 2", or "13 566 squared".
The square of 13 566 is 184 036 356 because 13 566 × 13 566 = 13 5662 = 184 036 356.
As a consequence, 13 566 is the square root of 184 036 356.
## Number of digits of 13 566
13 566 is a number with 5 digits.
## What are the multiples of 13 566?
The multiples of 13 566 are all integers evenly divisible by 13 566, that is all numbers such that the remainder of the division by 13 566 is zero. There are infinitely many multiples of 13 566. The smallest multiples of 13 566 are:
## Numbers near 13 566
### Nearest numbers from 13 566
Find out whether some integer is a prime number | HuggingFaceTB/finemath | |
# SBI PO Prelims Quantitative Aptitude Questions 2019 – Day 14
SBI PO 2019 Notification will be expected soon. It is one of the most expected recruitment among the banking aspirants. Every year the exam pattern for SBI PO has been changing. Depends upon the changing of exam pattern the questions are quite harder compare to the previous year. So the questions are in high level than the candidateโs assumption.
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As per the latest trend, our IBPS Guide is providing the updated New Exam Pattern Quantitative Aptitude questions for SBI PO 2019 Day 14. Our Skilled experts were mounting the questions based on the aspirant’s needs. So candidates shall start your preparation and practice on daily basis with our SBI PO pattern quantitative aptitude questions 2019 day 14. Start your effective preparation from the right beginning to get success in upcoming SBI PO 2019.
#### โBe not afraid of growing slowly; be afraid only of standing stillโ
Quantitative Aptitude Questions For SBI PO Prelims (Day-14)
maximum of 10 points
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1) If the numerator of a fraction is increased by ____ % and the denominator of the fraction is increased by 140%, then the resultant fraction becomes 7/18. The original fraction is ______.
Which of the following option satisfies the given condition?
a) 80, 5/11
b) 40, 6/17
c) 120, 9/19
d) 60, 7/12
e) None of these
2) The simple interest on Rs. 28000 for _____ years at 12 % per annum is Rs. ______.
Which of the following option satisfies the given condition?
I) 3, 10080
II) 4, 13440
III) 2, 6720
a) Only I
b) Only II
c) Only III
d) Both I and II
e) All I, II and III
3) A shopkeeper sells an item at a profit of _______ %. If he reduces the price of the item by Rs. ______, he incurs a loss of 15%. The cost price of the item is Rs. 2000.
Which of the following option satisfies the given condition?
I) 20, 700
II) 25, 1600
III) 10, 500
a) Only I
b) Only II
c) Only III
d) Only I and II
e) Only I and III
4) A, B and C entered into a partnership with investment in the ratio ofย ย ย ย ย ______. After one year, B doubled his investment. At the end of two years, they earned a profit of Rs. ______. The share of B is Rs. 18000?
Which of the following option satisfies the given condition?
I) 5: 3: 4, 54000
II) 3: 1: 5, 114000
III) 6: 5: 7, 49200
a) Only I
b) Only II
c) Only I and II
d) Only II and III
e) All I, II and III
5) The ratio of length and breadth of the rectangular park is 3: 2. The man cycling along the boundary of the park at the speed of _____km/hr completes one round in _____ minutes. Then the area of the park is 240000 Sq m?
a) 12, 10
b) 15, 8
c) 20, 9
d) 10, 15
e) Both โaโ and โbโ
Directions (Q. 6 – 10): Study the following information carefully and answer the given questions:
The following pie chart shows the total population of 6 different states in a certain year.
Total population = 600 lakhs
6) Find the difference between the total rural population of state P to that of the total urban population of state Q, if the ratio between the rural to urban population of state P is 5: 4 and the percentage of rural population of state Q is 52 %?
a) 4.5 lakhs
b) 2.3 lakhs
c) 3.2 lakhs
d) 1.6 lakhs
e) None of these
7) Find the central angle for state S?
a) 68.40
b) 72.60
c) 56.80
d) 64.20
e) None of these
8) Find the sum of the total female population of state R and T together, if the ratio between the male to that of female in state R is 2: 1 and the percentage of male population in state T is 42 %?
a) 92.84 lakhs
b) 86.52 lakhs
c) 74.28 lakhs
d) 105.16 lakhs
e) None of these
9) Total literate population in state U is approximately what percentage of total illiterate population in state S, if the total literate population of state S is two-third of that total population and the ratio between literate to illiterate population of state U is 7: 5?
a) 112 %
b) 130 %
c) 95 %
d) 74 %
e) 146 %
10) Find the average population of state P, S and U together?
a) 75 lakhs
b) 62 lakhs
c) 90 lakhs
d) 83 lakhs
e) None of these
Option (a)
[5*(180/100)] / [11*(240/100)] = 15/44 โ 7/18
This option doesnโt satisfy the given condition.
Option (b)
[6*(140/100)] / [17*(240/100)] = 7/34 โ 7/18
This option doesnโt satisfy the given condition.
Option (c)
[9*(220/100)] / [19*(240/100)] = 33/76 โ 7/18
This option doesnโt satisfy the given condition.
Option (d)
[7*(160/100)] / [12*(240/100)] = 7/18
This option satisfies the given condition.
Option (I)
S.I = PNR/100 = (28000*3*12)/100 = Rs. 10080
This option satisfies the given condition.
Option (II)
S.I = PNR/100 = (28000*4*12)/100 = Rs. 13440
This option satisfies the given condition.
Option (III)
S.I = PNR/100 = (28000*2*12)/100 = Rs. 6720
This option satisfies the given condition.
Option I)
= > 120 % of 2000 โ 85 % of 2000
= > (120/100)*2000 โ (85/100)*2000
= > 2400 โ 1700 = Rs. 700
This satisfies the given condition.
Option II)
= > 125 % of 2000 โ 85 % of 2000
= > (125/100)*2000 โ (85/100)*2000
= > 2500 โ 1700 = Rs. 800
This doesnโt satisfy the given condition.
Option III)
= > 110 % of 2000 โ 85 % of 2000
= > (110/100)*2000 โ (85/100)*2000
= > 2200 โ 1700 = Rs. 500
This satisfies the given condition.
Option I)
The share of A, B and C
= > [5*2]: [3*1 + 6*1]: [4*2]
= > 10: 9: 8
27โs = 54000
1โs = 2000
The share of B = Rs. 18000
This satisfies the given condition
Option II)
The share of A, B and C
= > [3*2]: [1*1 + 2*1]: [5*2]
= > 6: 3: 10
19โs = 114000
1โs = 6000
The share of B = Rs. 18000
This satisfies the given condition
Option III)
The share of A, B and C
= > [6*2]: [5*1 + 10*1]: [7*2]
= > 12: 15: 14
41โs = 49200
1โs = 1200
The share of B = Rs. 18000
This satisfies the given condition
Option a)
The ratio of length and breadth of the rectangular park = 3: 2 (3x, 2x)
Distance = S*T = 12*(10/60) = 2 km = 2000 m
Perimeter = 2*(3x + 2x)
10x = 2000
X = 200
Length = 600 m, Breadth = 400 m
Area of the park = 600*400 = 240000 Sq m
This option satisfies the given condition
Option b)
The ratio of length and breadth of the rectangular park = 3: 2 (3x, 2x)
Distance = S*T = 15*(8/60) = 2 km = 2000 m
Perimeter = 2*(3x + 2x)
10x = 2000
X = 200
Length = 600 m, Breadth = 400 m
Area of the park = 600*400 = 240000 Sq m
This option satisfies the given condition
Option c)
The ratio of length and breadth of the rectangular park = 3: 2 (3x, 2x)
Distance = S*T = 20*(9/60) = 3 km = 3000 m
Perimeter = 2*(3x + 2x)
10x = 3000
X = 300
Length = 900 m, Breadth = 600 m
Area of the park = 900*600 = 540000 Sq m โ 240000 Sq m
This option doesnโt satisfies the given condition
Option d)
The ratio of length and breadth of the rectangular park = 3: 2 (3x, 2x)
Distance = S*T = 10*(15/60) = 2.5 km = 2500 m
Perimeter = 2*(3x + 2x)
10x = 2500
X = 250
Length = 750 m, Breadth = 500 m
Area of the park = 750*500 = 375000 Sq m โ 240000 Sq m
This option doesnโt satisfies the given condition
Direction (6-10) :ย
The total rural population of state P
= > 600*(12/100)*(5/9) = 40 lakhs
The total urban population of state Q
= > 600*(15/100)*(48/100) = 43.2 lakhs
Required difference = 43.2 โ 40 = 3.2 lakhs
The central angle for state S
= > (19/100)*360 = 68.40
Total female population of state R
= > 600*(23/100)*(1/3) = 46 lakhs
Total female population of state T
= > 600*(17/100)*(58/100) = 59.16 lakhs
Required sum = 46 + 59.16 = 105.16 lakhs
Total literate population in state U
= > 600*(14/100)*(7/12) = 49 lakhs
Total illiterate population in state S
= > 600*(19/100)*(1/3) = 38 lakhs
Required % = (49/38)*100 = 128.94 % = 130 %
The average population of state P, S and U together
= > [(12 + 19 + 14)/100]*(1/3)*600
= > (45/100)*(1/3)*600 = 90 lakhs
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Visually differentiating PCA and Linear Regression
I've always been fascinated by the concept of PCA. Considering its wide range of applications and how inherently mathematical the idea is, I feel PCA is one of the pillars of the intersection between Pure Mathematics and Real-world analytics. Besides, the fact that you could think about real data as just raw numbers and then transform it down to something you can visualize and relate to, is extremely powerful and essential in any learning process.
Just in case you're wondering, Principle Component Analysis (PCA) simply put is a dimensionality reduction technique that can find the combinations of variables that explain the most variance. So you can transform a 1000-feature dataset into 2D so you can visualize it in a plot or you could bring it down to x features where x<<1000 while preserving most of the variance in the data. I've previously explored Facial image compression and reconstruction using PCA using scikit-learn.
In this post I would like to delve into the concept of linearity in Principal Component Analysis. To quote wiki for a more detail definition of PCA:
Principal component analysis (PCA) is a statistical procedure that uses an orthogonal transformation to convert a set of observations of possibly correlated variables into a set of values of linearly uncorrelated variables called principal components
The concept that I would like to explore is how different this is from Linear Regression. With Linear Regression, we are trying to find a straight line that best fits the data. So if we took a very simple example of univariate regression, predicting one variable with another, how would my PCA transformation look from the best-fit line derived through linear regression.
Let's dive in and find out.
In [14]:
```#Importing required Python packages
import matplotlib.pylab as plt
import numpy as np
from sklearn.decomposition import PCA
from sklearn.linear_model import LinearRegression
from pprint import pprint
import seaborn as sns
from matplotlib import collections as mc
%matplotlib inline
np.random.seed(sum(map(ord, "aesthetics")))
sns.set_context('notebook')
sns.set_style('darkgrid')
plt.rcParams['figure.figsize'] = (15, 5)
```
Univariate Linear Regression
Let's define a random x and y and try to regress y with x, denoted as y~x.
In [15]:
```#Define a simple range of values for x and reshape so sklearn likes it.
x=np.array(range(1,100))
x=x[:, np.newaxis]
#Define y as a function of X and introduce some noise so we get nice plots.
y=10+2*x.ravel()
y=y+np.random.normal(loc=0, scale=70, size=99)
```
Let's now fit a Linear Regression model to x and y. We will use sklearn's Linear models and do a simple fit+predict.
In [16]:
```#Instantiate, fit and predict y~x
lin=LinearRegression()
lin.fit(x,y)
y_pred=lin.predict(x)
```
Great, we will not plot this so we can visualize the best-fit straight line that sklearn has determined for y~x.
In [17]:
```def plotline(x1,y1,x2,y2,c,l):
plt.scatter(x1, y1, color='black')
plt.plot(x2, y2, color=c, linewidth=l)
plt.axis('tight')
plt.xlabel('x')
plt.ylabel('y')
sns.despine(offset=10, trim=True)
```
In [18]:
```plotline(x,y,x,y_pred,'indigo',2)
plt.suptitle('y~x',fontsize=18)
plt.show()
```
OK, now before we attempt to do anything with PCA, I am curious if there's a difference between y~x and x~y. What happens if we change sides?
Let's find out.
In [19]:
```#Reshape y so sklearn doesn't cry foul
y=y[:,np.newaxis]
#Instantiate, fit and predict x~y
lin=LinearRegression()
lin.fit(y,x)
x_pred=lin.predict(y)
```
Notice the variable names here. y_pred is the regressed value of y based on x while x_pred is the regressed value of x based on y. Let's now plot this line as well on the same plot and compare real-time.
In [20]:
```def onevstwo():
plotline(x,y,x,y_pred,'indigo',2)
plotline(x,y,x_pred,y,'seagreen',1)
plt.suptitle('y~x vs. x~y',fontsize=16)
plt.legend(['y~x','x~y'], loc='best', fontsize=14)
onevstwo()
plt.show()
```
Ah! Look at the difference there. These are definitely not the same. But why are they different? Let's try to plot what each of them are trying to minimize. The metric we've used for linear regression (default) is Ordinary Least Squares.
We will plot the difference between the actual value of y and the predicted value for a few samples and see where they land. Note that I am trying to find some good ones for plotting below by looking at how large the difference is.
In [21]:
```#Let's flatten our arrays so it's easier to plot
x,x_pred,y,y_pred=x.ravel(),x_pred.ravel(),y.ravel(),y_pred.ravel()
#Grab some indices where the difference between y and y_pred is the greatest.
indices=np.argsort(y-y_pred)[-8:-5]
#Plot some sample y-ypred in the earlier plot
onevstwo()
lines=[[(x[i],y[i]), (x[i],y_pred[i])] for i in indices]
lines=mc.LineCollection(lines, colors='indigo', linewidths=4)
plt.show()
```
That makes it clear. When we regress y~x, it is the vertical distances that are minimized based on the Least Squares method (or the metric we choose). This is highlighted in the brown lines in the plot above.
So what happens when we regress x~y? Let's plot this as well though it's easy to guess what happens.
In [22]:
```#Grab some indices where the difference between x and x_pred is the greatest.
indices_r=np.argsort(x-x_pred)[-40:-37]
#Plot some sample y-y_pred in the earlier plot
onevstwo()
lines=[[(x[i],y[i]), (x[i],y_pred[i])] for i in indices]
lines=mc.LineCollection(lines, colors='indigo', linewidths=4)
#Plot some sample x-x_pred as well
lines=[[(x[i],y[i]), (x_pred[i],y[i])] for i in indices_r]
lines=mc.LineCollection(lines, colors='seagreen', linewidths=4)
plt.show()
```
There is the confirmation. When we regress x~y instead of y~x, the model tries to minimize the horizontal distances rather than the vertical distances using OLS.
Awesome! We're now fully geared up to understand how PCA differs from this.
Let's go ahead and fit a PCA model to our dataset.
PCA vs Linear Regression
We need to combine x and y so we can run PCA. Let's then fit a PCA model to the dataset.
In [23]:
```#Combine x and y
xy=np.array([x,y]).T
```
After instantiating a PCA model, we will firstly fit and transform PCA with n_components = 1 to our dataset. This will run PCA and determine the first (and only) principal component.
We will then do an inverse transform on the resulting compressed array so we can project onto our plots for comparison. If you don't fully understand what's going on here, I suggest you take a look at one of my earlier posts here where I dig into PCA in detail and demonstrate dimensionality reduction.
In [24]:
```#Instantiate a PCA model, the run fit_transform to xy followed by inverse_transform
pca=PCA(n_components=1)
xy_pca=pca.fit_transform(xy)
xy_n=pca.inverse_transform(xy_pca)
```
Great, we now have the datasets we need to plot. xy, which is nothing but x and y put together and xy_n, which we got by inverse transformation of the PCA results. If we plot these two together like we did for Linear Regression, things will be clear as to what is being minimized.
In [25]:
```#Plot xy against xy_n. Note that we're simply passing x and y to the function since xy=x+y.
plotline(x,y,xy_n[:,0],xy_n[:,1],'deeppink',1)
plt.suptitle('PCA on xy',fontsize=16)
plt.legend(['PCA model line'], loc='best', fontsize=14)
plt.show()
```
There it is, this actually looks quite similar to the first regression we plotted for y~x. For now, can we try to find out what is being minimized?
In [26]:
```#Replot the PCA line so we can examine
plotline(x,y,xy_n[:,0],xy_n[:,1],'deeppink',1)
#Plot some sample values to demonstrate PCA
lines=[[(38.1, 102.9), (39.6,50)], [(58.95,75), (57.3, 133)]]
lines=mc.LineCollection(lines, colors='deeppink', linewidths=4)
plt.suptitle('PCA minimizes orthogonal errors to the model line',fontsize=16)
plt.legend(['PCA model line'], loc='best', fontsize=14)
plt.show()
```
So PCA also tries to find a best fit line through the data points but it minimizes the orthogonal distance to the model line where as Linear Regression minimizes distance from the perspective of the axis you're regressing with respect to.
I guess in rudimentary terms we can think of the model line as sort of a third axis against which we're regressing our other variables x and y. Hope the plot illustrates the concept clearly. Note that I hand picked couple of data points to illustrate the orthogonality. This plot will only work obviously for the data we're using in this exercise :-)
Putting it all together
So let's plot all three model lines we've generated so far together and compare them one last time.
In [27]:
```#Plot y~x vs x~y
onevstwo()
#Plot PCA model lines
plotline(x,y,xy_n[:,0],xy_n[:,1],'deeppink',1)
plt.suptitle('PCA vs Linear Regression',fontsize=16)
plt.legend(['y~x', 'x~y', 'PCA on xy'], loc='best', fontsize=14)
plt.show()
```
As we can see, the PCA model line and x~y lines are fairly close to each other but we can tell that they're not the same. This is primarily because of how well we've distributed the sample data horizontally w.r.t the x axis.
So that brings us to a close. Hopefully that was helpful to some of you reading. As always, let me know your feedback and if you have any questions. Cheers! | HuggingFaceTB/finemath | |
# Positions- Introducing Vocabulary
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## Objective
Students will be able to describe the relative positions of objects using terms such as above, below, beside, in front of, behind, and next to.
#### Big Idea
Describing the position of objects and understanding positional vocabulary is an important math skill, and it is also something students will use in their everyday lives. In this lesson, students will explore positional vocabulary in a "Where is it?" game
## Problem of the Day
5 minutes
I start each math lesson with a Problem of the Day. I use the procedures outlined here on Problem of the Day Procedures.
Today's Problem of the Day:
Match the numerals to the correct group. Are these numerals in the correct order? How do you know?
This problem gives the students 3 groups of pictures to count. The students need to count the pictures and place the numerals in the correct place. I call up three different students to do this to get several students involved. I then ask another student the follow up questions. The students needs to not only say if the numbers are in the correct order but also how they know. This problem reviews counting objects, recognizing numerals and ordering numbers to 10.
## Presentation of Lesson
25 minutes
I start this lesson by having the students sit on the perimeter of the carpet. In the middle of the carpet, I place a teddy bear and a box. I put the teddy bear in the box and say "The teddy bear is in the box. Where is the teddy bear?" I repeat this with the words above, below, beside, behind, next to, under, on, and in front of. After each term, I have the students repeat where the teddy bear is to practice using the vocabulary. Check it out here.
We are going to play a game called "Where is it?" I am going to pick something in the room. I am going to describe where it is but I am not going to tell you what it is. Use your eyes to look around the room. You do not need to get up, I will only pick things that we can all see from sitting here. If you think you know what it is, raise your hand.
The first object I chose was our class tree. My first clue was "It is next to the book shelf." I needed to add several other things that it was next to or beside before someone guessed the tree. After I received a correct answer, I walked over to the tree and repeated the clues to be sure all of the students understood each positional term.
Other objects I used were:
- a book that was on the shelf under the tree
- a chair that was behind my desk
- the calendar that was below the letter Y, beside the schedule and on the board.
Any classroom object would work. Just be sure to use a variety of terms in the game. After about 10 minutes, I tell students that they are going to be practicing the new vocabulary that we learned on a Positional Words Worksheet.
We are going to work on this paper together. When you get to your seat, do not touch your cup of crayons. You need to get out a pencil and put your name on your paper. When your name is on your paper hold your pencil in the air, that will let me know that you are ready to start.
I use the procedures outlined here on the Paper Procedures Prior to this lesson, I placed a plastic cup at each students' place containing the needed crayons. This eliminates time lost because students do not have or can not find the correct colors.
Follow teacher directions to color the pictures.
I am going to read you several directions. You need to listen carefully to the directions and color the correct part of the picture. I will give you time after each step to color.
I walk around after I read each direction and help students who are having trouble figuring out what to color. I then pick a student to come and color it on my copy which is projected on the SMART board. During the last step, I listen to students discuss what they colored and help with adding more positional vocabulary to their discussions. When students are finished, I tell them they can put their paper in the basket and get their center.
## Practice
20 minutes
The centers for this week are:
Shape Monster Book (Hubbardscupboard.org)
My Shape Flip Book (KindergartenCrayons.blogspot.com)
Building Shapes (Teacher Made Resource, craft sticks, fruit loops, noodles and drinking straws )
Shapes - Lakeshore Instant Learning Center
- Lakeshore Interactive Activity
I quickly circulate to make sure students are engaged and do not have any questions about how to complete the centers. I pull three groups during centers and work with them depending on the time they need (5 - 10 minutes).
Today I am focusing on positional vocabulary. I review each of the terms and show their meaning using a cube and two plastic cups. I play a game similar to the one that we played in whole group.
Prior to clean up, I check in with each table to see how the centers are going. I turn on Tidy Up by Dr. Jean. Students clean up and return to their seats. This is a paid resource, but there are many free examples of transition songs easily found in a web search.
## Closing
5 minutes
To close, I put a student's paper on the document camera and project it on the SMART Board. I have that student describe the object that they colored green using the positional vocabulary that we learned today. I mention positive things noticed during centers as well as something that needs to be better next time.
I review what we did during our whole group lesson. "Today we learned to about some new vocabulary words that we use in math and in our everyday lives. Tomorrow, we are going to continue to work with these words." | HuggingFaceTB/finemath | |
# Chapter 7 : Flow of a Real Fluid¶
## Example 7.1 Page No : 225¶
In [1]:
import math
# variables
nu = 0.00001; # sqft/sec
d = 1.; #in
R_c = 2100.;
# calculations
V = R_c*nu/(d/12);
Q = V*0.25*math.pi*(d/12)**2;
# results
print 'Q = %.6f cfs'%(Q);
Q = 0.001374 cfs
## Example 7.2 pageno : 230¶
In [1]:
import math
# calculations
e = 0.2/(1./0.33)
l2 = 0.2/(1.94 * (1/0.33)**2)
l = math.sqrt(l2)
k2 = 0.2/(1.94 * (1/0.33)**2 / (-1/(0.33**2)**0.5)**2)
k = math.sqrt(k2)
# results
print "E = %.3f lb-sec/sqft"%e
print "l = %.3f ft"%l
print "k = %.2f"%k
E = 0.066 lb-sec/sqft
l = 0.106 ft
k = 0.32
## Example 7.4 Page No : 240¶
In [3]:
import math
# variables
G = 240.; #lb/sec
A1 = 4.; #sqft
A2 = 2.; #sqft
z1 = 30.; #ft
z2 = 80.; #ft
V1 = 600.; # fps
V2 = 800.; #fps
p1 = 20.; #psia
p2 = 35.; # psia
# calculations
gam1 = G/(A1*V1);
gam2 = G/(A2*V2);
T1 = p1*144/(53.3*gam1);
T2 = p2*144/(53.3*gam2);
del_H = 186.5*(T2-T1);
E_H1 = (V2**2)/(2*32.2) - (V1**2)/(2*32.2) +del_H+z2-z1;
E_H2 = (V2**2)/(2*32.2) - (V1**2)/(2*32.2) +del_H;
Q = G*E_H2/550.;
# results
print 'T1 = %d degreeR, T2 = %d degreeR'%(T1,T2);
print ' The net heat energy added = %d hp'%(round(Q,-1));
#answer differs due to rounding-off errors
T1 = 540 degreeR, T2 = 630 degreeR
The net heat energy added = 9230 hp
## Example 7.5 Page No : 240¶
In [4]:
import math
# variables
G = 50.; #cfs
Q = 400.; #hp
A1 = 4.; #sqft
A2 = 2.; #sqft
z1 = 30.; #ft
z2 = 80.; #ft
p1 = 20.; #psi
p2 = 10.; #psi
# calculations
V1 = G/A1;
V2 = G/A2;
E_p = Q*(550/62.4)/G;
h_L = (p1-p2)*144/62.4 + (V1**2 - V2**2)/(2*32.2) +(z1-z2)+E_p;
# results
print 'Head lost = %.1f ft'%(h_L);
Head lost = 36.3 ft
## Example 7.6 Page No : 243¶
In [9]:
import math
# variables
b = 3.; #ft
d = 2.; #ft
l = 200.; #ft
h_L = 30.; #ft
# calculations
tau_0 = h_L*62.4*b*d/(10*l); #0.00694
# results
print 'The resistance stress exerted between fluid and conduit walls = %.2f psf = %.3f psi'%(tau_0,tau_0*0.00694);
The resistance stress exerted between fluid and conduit walls = 5.62 psf = 0.039 psi
## Example 7.7 Page No : 244¶
In [10]:
import math
# variables
h_L = 30.; #ft
l = 200.; #ft
d = 2.; #ft
r = 8.; #in
# calculations
#part (a)
tau_0 = h_L*62.4/(d*l);
#part(b)
tau = (0.5*r/12)*(tau_0*0.00694);
# results
print 'Parta): Shear stress = %.2f psf = %.4f psi '%(tau_0,tau_0*0.00694);
print 'Partb): Shear stress = %.4f psi '%(tau);
Parta): Shear stress = 4.68 psf = 0.0325 psi
Partb): Shear stress = 0.0108 psi | HuggingFaceTB/finemath | |
# Science
posted by .
what are the common SI unit for volume?
• Science -
this is 7th grade science for other teachers if you don't know what grade im in
• Science -
also im in regents
• Science -
http://en.wikipedia.org/wiki/Volume
• Science -
Any unit of length gives a corresponding unit of volume, namely the volume of a cube whose side has the given length. For example, a cubic centimetre (cm3) would be the volume of a cube whose sides are one centimetre (1 cm) in length.
In the International System of Units (SI), the standard unit of volume is the cubic metre (m3). The metric system also includes the litre (L) as a unit of volume, where one litre is the volume of a 10-centimetre cube. Thus
1 litre = (10 cm)3 = 1000 cubic centimetres = 0.001 cubic metres,
so
1 cubic metre = 1000 litres.
Small amounts of liquid are often measured in millilitres, where
1 millilitre = 0.001 litres = 1 cubic centimetre.
Various other traditional units of volume are also in use, including the cubic inch, the cubic foot, the cubic mile, the teaspoon, the tablespoon, the fluid ounce, the fluid dram, the gill, the pint, the quart, the gallon, the minim, the barrel, the cord, the peck, the bushel, and the hogshead.
??????????
i read it but didn't said COMMON UNIT
• Science -
The common unit is the standard unit.
• Science -
i'll check in my notes AGAIN 2 check to see if i can find the answer
the you can check it for me
• Science -
ok
• Science -
The common SI unit for volume is cubic metre (m3).
• Science -
is my answer is correct ???
• Science -
Yes.
That's in the quote you posted above.
"In the International System of Units (SI), the standard unit of volume is the cubic metre (m3)."
• Science -
Ok
thank you :)
Good Nite :)
• Science -
Frankly, I wonder exactly what the word common SI unit means? In my opinion, it's either an SI unit or it isn't. There are seven BASE units in the SI system and there is no volume listed. Then the SI system uses those seven base units to derive other units for the SI system. See
http://physics.nist.gov/cuu/Units/units.html
for the 12 derived units, such as area (in meters squared, volume in meters cubed etc). Finally, there is a list of units OUTSIDE THE SI SYSTEM that are accepted to be used with the SI system, such as minute, hour, day, liter, metric ton, bel, electron volt, etc. It's safe to say that the standard unit of volume in the SI system is the cubic meter. I don't know if that is the same as the common unit or not.
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More Similar Questions | HuggingFaceTB/finemath | |
0
How do you calculate 30 percent of 236?
Wiki User
2015-10-30 12:02:20
OK, I will explain. percent (%) means part of a 100.
Thus the whole object (in this case 236) must be 100%
We therefore need to find out what one percent would equal so we divide the whole (236) by 100
236/100 = 2.36
So one percent of 236 = 2.36
We now need to know what 30 one percent would equal, so we multiply the value we got for one percent (2.36) by 30
30*2.36 = 70.8
So 30 percent of 236 is 70.8
I hope that helped you.
Wiki User
2015-10-30 12:02:20
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# How do you evaluate \frac { ( - 1+ 4) ^ { 2} + 3} { 3- ( - 3) ^ { 2} }?
Mar 18, 2017
-4
#### Explanation:
$\frac{{\left(3\right)}^{2} + 3}{3 - \left(- 3\right) \left(- 3\right)}$
=$\frac{9 + 3}{3 - 9}$
=$\frac{12}{-} 3$
=$- 4$ | HuggingFaceTB/finemath | |
# proof of Mantel’s theorem
Let $G$ be a triangle-free graph. We may assume that $G$ has at least three vertices and at least one edge; otherwise, there is nothing to prove. Consider the set $P$ of all functions $c\colon V(G)\to\mathbb{R}_{+}$ such that $\sum_{v\in V(G)}c(v)=1$. Define the total weight $W(c)$ of such a function by
$W(c)=\sum_{uv\in E(G)}c(u)\cdot c(v).$
By declaring that $c\leq c^{*}$ if and only if $W(c)\leq W(c^{*})$ we make $P$ into a poset.
Consider the function $c_{0}\in P$ which takes the constant value $\frac{1}{|V(G)|}$ on each vertex. The total weight of this function is
$W(c_{0})=\sum_{uv\in E(G)}\frac{1}{|V(G)|}\cdot\frac{1}{|V(G)|}=\frac{|E(G)|}{% |V(G)|^{2}},$
which is positive because $G$ has an edge. So if $c\geq c_{0}$ in $P$, then $c$ has support on an induced subgraph of $G$ with at least one edge.
We claim that a maximal element of $P$ above $c_{0}$ is supported on a copy of $K_{2}$ inside $G$. To see this, suppose $c\geq c_{0}$ in $P$. If $c$ has support on a subgraph larger than $K_{2}$, then there are nonadjacent vertices $u$ and $v$ such that $c(u)$ and $c(v)$ are both positive. Without loss of generality, suppose that
$\sum_{uw\in E(G)}c(w)\geq\sum_{vw\in E(G)}c(w).{}$ (*)
Now we push the function off $v$. To do this, define a function $c^{*}\colon V(G)\to\mathbb{R}_{+}$ by
$c^{*}(w)=\begin{cases}c(u)+c(v)&w=u\\ 0&w=v\\ c(w)&\text{otherwise.}\end{cases}$
Observe that $\sum_{w\in V(G)}c^{*}(w)=1$, so $c^{*}$ is still in the poset $P$. Furthermore, by inequality (*) and the definition of $c^{*}$,
$\displaystyle W(c^{*})$ $\displaystyle=\sum_{uw\in E(G)}c^{*}(u)\cdot c^{*}(w)+\sum_{vw\in E(G)}c^{*}(v% )\cdot c^{*}(w)+\sum_{wz\in E(G)}c^{*}(w)\cdot c^{*}(z)$ $\displaystyle=\sum_{uw\in E(G)}[c(u)+c(v)]\cdot c(w)+0+\sum_{wz\in E(G)}c(w)% \cdot c(z)$ $\displaystyle=\sum_{uw\in E(G)}c(u)\cdot c(w)+\sum_{vw}c(v)\cdot c(w)+\sum_{wz% \in E(G)}c(w)\cdot c(z)$ $\displaystyle=W(c).$
Thus $c^{*}\geq c$ in $G$ and is supported on one less vertex than $c$ is. So let $c$ be a maximal element of $P$ above $c_{0}$. We have just seen that $c$ must be supported on adjacent vertices $u$ and $v$. The weight $W(c)$ is just $c(u)\cdot c(v)$; since $c(u)+c(v)=1$ and $c$ has maximal weight, it must be that $c(u)=c(v)=\frac{1}{2}$. Hence
$\frac{1}{4}=W(c)\geq W(c_{0})=\frac{|E(G)|}{|V(G)|^{2}},$
which gives us the desired inequality: $|E(G)|\leq\frac{|V(G)|^{2}}{4}$.
Title proof of Mantel’s theorem ProofOfMantelsTheorem 2013-03-22 13:03:04 2013-03-22 13:03:04 mps (409) mps (409) 6 mps (409) Proof msc 05C75 msc 05C69 | open-web-math/open-web-math | |
# math
posted by .
What is x^2+6x-91 when factored ???!!
• math -
-13 and 7
• math -
(x-7)*(x+13)
## Similar Questions
1. ### math
can this be simplified further? x^4+2x^3+4x^2+8x+16 Unless you are asking if the statement can be factored, the answer to your question is "no". Only variables with the same exponents can be added together. you mean this could be factored?
2. ### math,algebra,help
Is this right: Solve: x^2+16x+60=0 this is what i did: i used the quadratic formula and this is what i ended up with. x = -8 +- (sqrt(16))/(2) Which is -8 +-2 Or you could have factored the quadratic (x+10)(x+6) Which is -8 +-2 Or …
3. ### mat117
Factor each expression a^2(b-c)-16b^2(b-c) Help show me how (b-c) appears in both terms and can be factored out, giving you (b-c)(a^2-16b^2) Now note that the second term can also be factored since it is the difference of two perfect …
4. ### Algebra 2
Well I'm curently taking this class as a sophmore and have taken geometry freshmen year of high school and also I am taking physics at the same time and in math class when I'm asked to factor i go absolutley nuts because guessing numbers …
5. ### Math
Can someone please tell me Why is (4x + 5) (2x - 1) + (x - 9) (2x - 1) not in factored form?
6. ### Math
Can someone please tell me Why is (4x + 5) (2x - 1) + (x - 9) (2x - 1) not in factored form?
7. ### Limits-Factored?
lim as x approaches 3+ x^2+x-6/18-3x-x^2 I factored it into (x+3)(x-2)/(6+x)(3-x)
8. ### algebra 2
factor 2x^3+54 I factored it to 2(x^3+27). can it be factored more or is that it?
9. ### alegebra
1. What is the factored form of 4x 2 + 12x + 5?
10. ### Math
This is the second part of a two part question for an online class. It gave me the degree and the zeros and I had to give the factored form. I got that part right, but I need to know how to get the expanded form from the factored form. …
More Similar Questions | HuggingFaceTB/finemath | |
# 817 Tiny Squares
Start at the top row of this level three puzzle and work down one cell at a time until you’ve written each number from 1 through 12 in both the top row and the first column. You will have solved the puzzle if all the clues given in the puzzle are the products of the numbers you wrote. You can do this! Print the puzzles or type the solution on this excel file: 12 factors 815-820
• 817 is a composite number.
• Prime factorization: 817 = 19 x 43
• The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 817 has exactly 4 factors.
• Factors of 817: 1, 19, 43, 817
• Factor pairs: 817 = 1 x 817 or 19 x 43
• 817 has no square factors that allow its square root to be simplified. √817 ≈ 28.58321 817 is the sum of three consecutive prime numbers:
• 269 + 271 + 277 = 817
Because it has two factor pairs in which the numbers in the pair are either both odd or both even, 817 can be written as the difference of two squares two different ways:
• 817 = 31² – 12², which I’ve illustrated below.
• 817 = 409² – 408² Being able to be written as the difference of two squares means that 817 is a leg in two Pythagorean triples so:
• 744² + 817² = 1105²
• 817² + 333744² = 333745²
Last, but certainly not least, 817 is the difference of two consecutive cubes, namely 17³ – 16³ = 817. That means that 817 is the 17th centered hexagonal number as well! There are 17 squares on each side of the figure below and 817 tiny purple squares in all. The horizontal row in the middle has 2(17) – 1 = 33 tiny purple squares.
2(17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32) + 33 = 817. That difference of two cubes also means that
• 817 = 17³ – 16³ = (17 – 16)(17² + 16·17 + 16²) = (1)(289 + 272 + 256) = 817.
That’s from a³ – b³ = (a – b)(a² + ab + b²) and is illustrated below using 2D cubes instead of squares. In 3D, when the sides are folded up, the darker cubes can be claimed by two different sides so the figure will look like a 17 x 17 x 17 inside corner such as in a room where three perpendicular lines meet, but in 2D it’s just the flat net you see here: This site uses Akismet to reduce spam. Learn how your comment data is processed. | HuggingFaceTB/finemath | |
3.3 Limits and Continuity: Algebraic Approach
(This topic appears in Section 3.3 in Applied Calculus or Section 10.3 in Finite Mathematics and Applied Calculus)
Consider the following limit.
limx→2 x2 - 3x4x + 3 .
If you estimate the limit either numerically or graphically, you will find that
limx→2 x2 - 3x4x + 3 ≈ −0.1818
But, notice that you can obtain this answer by simply substituting x = 2 in the given function:
f(x) = x2 - 3x4x + 3 f(2) = 4 - 68 + 3 = - 211 = -0.181818...
This answer is more accurate than the one coming from numerical or graphical method; in fact, it gives the exact limit.
Q Is that all there is to evaluating limits algebraically: just substitute the number x is approaching in the given expression?
A Not always, but this often does happen, and when it does, the function is continuous at the value of x in question. Recall the definition of continuity from the previous tutorial:
Continuous Functions
The function f(x) is continuous at x = a if
limx→a f(x) exists That is, the left-and right limits exist and agree with each other limx→a f(x) = f(a)
The function f is said to be continuous on its domain if it is continuous at each point in its domain. If f is not continuous at a particular point a, we say that f is discontinuous at a or that f has a discontinuity at a.
Q How do you tell if a function is continuous?
A As we saw in the previous tutorial, we can tell whether a function is continuous by looking at its graph. If the graph breaks at some point in the domain, then f has a discontinuity there. If the function is specified algrabcially, sometimes it is easy to tell whether it is continuous by just looking at the formula:
A closed-form function is any function that can be obtained by combining constants, powers of x, exponential functions, radicals, absolute values, trigonometric functions, logarithms (and some other functions you may never see) into a single mathematical formula by means of the usual arithmetic operations and composition of functions. Examples of closed-form functions are:
3x2 - x +1, (x2 - 1)1/26x-2
e(x2 - 1)1/2/x
(log3(4x2 - ex))2/3
They can be as complicated as you like. The following is not a closed form function.
f(x) = -1 if x < -1 x2 + x if -1 ≤ x ≤ 1 2 - x if 1 < x ≤ 2
The reason for this is that f(x) is not specified by a single mathematical expression. What is nice about closed-form functions is the following.
Continuity of Closed Form Functions
Every closed form function is continuous on its domain.
Thus, the limit of a closed-form function at a point in its domain can be obtained by substitution.
Example
f(x) = x2 - 3x4x + 3
is a closed form function, and x = 2 is in its domain. Therefore we can obtain limx→2 f(x) by substitution:
limx→2 x2 - 3x4x + 3 = f(2) = - 211 ,
as we saw above.
Q limx→0 ex22x-3 Select one does not exist = 0 = -4/3 = -1/3 = 4/3 is undefined = -e/3 none of the above Q lim x→3 2x + 3x + 3 =
Q What if f(x) is a closed form function, but the point x = a is not in the domain of the function?
A Then, you either:
1. First try using simplification or some other technique to replace f(x) by another closed form function which does have x = a in its domain. This allows you to substitute x = a in the new function to obtain the limit, or
2. If the obove method does not work, try evaluating the limit numerically or graphically. Note, however, that this may only give you an estimate of the limit.
Evaluating a Limit Using Simplification
(Similar to Example 2 in Section 3.3 of Applied Calculus, or in Section 10.3 of Finite Mathematics and Applied Calculus )
Let us evaluate
lim x→-2 3x2 + x - 10x + 2
1. Is the function f(x) a closed form function?
Answer: yes, since (3x2 + x-10)/(x+2) is a single mathematical formula.
2. Is the value x = a in the domain of f(x)?
Answer: no, since (3(-2)2+(-2)-10)/((-2)+2) is not defined.
Therefore, we consult the above Question/Answer discussion, and simplify the function, if we can.
3x2 + x - 10
x + 2
= (x+2)(3x-5) (x+2)
= 3x-5.
Since we are now left with a closed form function that is defined when x = -2, we can now evaluate the limit by substitution:
lim x→-2 3x2 + x-10x + 2 = lim x→-2 3x - 5 = 3(-2)-5 = -11.
lim x→-1 (x3+1)(x-1)x2 + 3x + 2 Select one does not exist = + ∞ = - ∞ is undefined = 0 = -2 = -4 = -6 = -8 none of the above
Back in the tutorial on functions from the graphical point of view, we looked at the following function:
f(x)= -1 if -4 ≤ x < -1 x if -1 ≤ x ≤ 1 x2-1 if 1 < x ≤ 2
This time, we are not showing you the graph right away, and ask you to look at the formula instead. Notice:
1. The function f is not closed form. (It is not defined by a single formula.)
2. Inside each of the separate intervals [-4, -1), (-1, 1) and (1, 2] the function is closed form, and is hence continuous on each of these intervals. Therefore, the only conceivable points where the function might fail to be continuous are on the boundaries of these intervals where we switch from one formula to another: x = -1 and x = 1.
Q The function f is is not continuous at x = 0. Q The function f is is not continuous at x = -1. Q The function f is is not continuous at x = 1.
Now try the rest of the exercises in Section 3.3 in Applied Calculus or Section 10.3 in Finite Mathematics and Applied Calculus
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Last Updated: September, 2007 | HuggingFaceTB/finemath | |
## April 2, 2007
### Oberwolfach CFT, Monday Morning
#### Posted by Urs Schreiber
Today Arthur Bartels reviewed the standard construction of a modular tensor categories of “DHR representations” from a local net of von Neumann algebras on the real line.
Here is a transcript of my notes.
I reproduce a transcript of the notes I have taken. Comments by myself are set in italics.
Let $H$ be a Hilbert space and $B(H)$ the algebra of bounded operators on that.
For $S \subset B(H)$ write $S' \subset B(H)$ for the commutant of $S$, which is the set of all bounded operators that commute with all those in $S$.
Notice that $B(H)$ is in particular a $*$-algebra. A von Neumann algebra is any $*$-subalgebra $A \subset B(H)$ such that $A'' = A$.
Definition: A net of vN algebras on $\mathbb{R}$ is an inclusion-preserving assignment $I \mapsto A(I)$ of vN algebra $A(I) \subset B(H)$, for some fixed Hilbert space $H$, to bounded open intervales $I \subset \mathbb{R} \,.$
A simple example is the choice $H = L^2(\mathbb{R})$ with the assignment $A : I \mapsto A(I) = \{ f \in L^ \infty(\mathbb{R}) | f \mathrm{constant}\; \mathrm{on}\; \mathbb{R}-I \} \,.$
However, this kind of example is not what one is really interested in. Interesting examples are much harder to describe (and are not described in this talk here).
Given such a net, one can also assign vN algebras to unbounded open subsets $E \subset \mathbb{R}$ by forming the vN algebra generated from all $A(I)$ for all bounded open intervals $I \subset E$.
The notation for the vN algebra obtained this way is $A(E) \,.$
Alternatively, one can consider just the $C^*$-algebra generated by all these $A(I)$. The result of that is then called $A^*(E) \,.$
(In the following we will mostly be interested in $A^*(E)$. At least one reason for that is, apparently, that the representations of $A(E)$ are rather boring. But I need to better understand this issue of switching from considering vN algebras to $C^*$-algebras.)
Definition Here are a couple of additional properties of nets of vN algebras which we will assume in the following.
A net is called additive if $A(I \cup J) = A(I)\vee A(J)$ whenever the intervals $I$ and $J$ have nontrivial intersection, $I \cap J \neq \emptyset$.
(Here $A_1 \vee A_2$ denotes the vN algebra generated from both $A_1$ and $A_2$. This is nothing but the double commutant of the union of both these algebras: $A_1 \vee A_2 = (A_1 \cup A_2)''$.)
A net is called local if $A(I) \subset A(I')'$ where $I' = \mathbb{R}-I$ is the complement of $I$ (which is not a bounded interval if $I$ is, so that this means we are making use of the notation introduced above.)
In words: the net is local if the algebras associated to two disjoint intervals mutually commute (as subalgebras of the fixed $B(H)$.)
A net $A$ satisfies Haag duality if the above inclusion is even an equality $A(I) = A(I')' \,.$
The main point is to prove
Theorem. Let the net $A$ be additive and satifying Haag duality, then the sectors of $A$ form a braided tensor category.
Here a “sector” is, roughly, a localized representation of $A$. To get to the precise definition, consider the following:
Definition. Let $\rho,\sigma : A^**(\mathbb{R}) \to A^*(\mathbb{R})$ be two endomorphisms of the global $C^*$-algebra, then an intertwiner $\nu : \rho \to \sigma$ is an operator $u \in B(H)$ such that $u \rho(a) = \sigma(a) u$ for all $a \in A^*(\mathbb{R})$.
If $u$ is unitary, then $\rho$ and $\sigma$ are said to be unitarily equivalent.
Remark. When $\rho$ is an endomorphism as above, we get a rep of $A^*(\mathbb{R})$ on $H$ simply by setting $a_\rho (v) = \rho(a) v$ for $v \in H$.
The above intertwiners then are nothing but morphisms of these representations.
An endomorphism $\rho$ is called localized in $I \subset \mathbb{R}$ if it is the identity outside of some bounded interval $I$, i.e. if $\rho(a) = a$ for all $a \in A^*(A')$.
Finally, a sector is a localized endomorphism that is localized, up to unitary equivalence, in any open bounded $I \subset \mathbb{R}$.
Under suitable assumption on $A$, this definition is equivalent to the notion of Doplicher-Haag-Roberts representations (DHR reps).
So were are identifying certain representations of $A$ with certain endomorphisms of $A$. Thinking of these as ordinary representations yields an obvious direct sum structure on all these reps. But thinking of them as endomorphisms yields an obvious tensor product:
Definition. The tensor product on sectors is simply the composition of the corresponding endomorphisms $\rho \otimes \sigma := \rho \sirc \sigma \,.$
Lemma.
1) If $\rho$ is localized in $I$ then $\rho(A(I)) \subset A(I)$
2) if $u : \rho \to \sigma$, where $\rho$ and $\sigma$ are localized in $I$, then $u \in A(I)$.
In addition, there is a braiding on this tensor product, induced by moving the localized reps around by unitary operators.
Assume that $\rho$ and $\sigma$ are localized in $I$ and pick an open bounded $J \subset \mathbb{R}$ to the right of $I$ and pick any $u \in U(H)$ such that $\mathrm{Ad}_u \circ \sigma$ is localized in $J$.
Lemma: The intertwiner $c_{\rho,\sigma} : \rho \otimes \sigma \to \sigma \otimes \rho$ given by the operator $u^* \rho(u)$ is a braiding isomorphism, which does not depend on any of the choices used in its construction.
The talk also covered the proofs of the above statements, and then ended by stating the Kawahigashi-Longo-Müger theorem, which says that under the additional assumption that the net of vN algebras is completely rational and has a modular PCT symmetry the braided abelian monoidal category of DHR reps constructed so far is in fact a modular tensor category.
Posted at April 2, 2007 12:35 PM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1223 | open-web-math/open-web-math | |
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# TRIANGLE/POINT
To test if a point is inside a triangle, we compare the area of the original triangle with the sum of the area of three triangles made between the point and the corners of the triangle.
Here's a diagram demonstrating the triangles created for a point outside and inside the triangle: To get the area, we use Heron's Forumula:
``````float areaOrig = abs( (x2-x1)*(y3-y1) - (x3-x1)*(y2-y1) );
``````
We need to calculate the area of the three triangles made from the point as well:
``````float area1 = abs( (x1-px)*(y2-py) - (x2-px)*(y1-py) );
float area2 = abs( (x2-px)*(y3-py) - (x3-px)*(y2-py) );
float area3 = abs( (x3-px)*(y1-py) - (x1-px)*(y3-py) );
``````
If we add the three areas together and they equal the original, we know we're inside the triangle! Using this, we can test for collision:
``````if (area1 + area2 + area3 == areaOrig) {
return true;
}
return false;
``````
Here's a full example:
``````float px = 0; // point (set by mouse)
float py = 0;
float x1 = 300; // three points of the triangle
float y1 = 100;
float x2 = 450;
float y2 = 300;
float x3 = 150;
float y3 = 300;
void setup() {
size(600,400);
noCursor();
strokeWeight(5); // make point easier to see
}
void draw() {
background(255);
// mouse point to mouse coordinates
px = mouseX;
py = mouseY;
// check for collision
// if hit, change fill color
boolean hit = triPoint(x1,y1, x2,y2, x3,y3, px,py);
if (hit) fill(255,150,0);
else fill(0,150,255);
noStroke();
triangle(x1,y1, x2,y2, x3,y3);
// draw the point
stroke(0, 150);
point(px,py);
}
// TRIANGLE/POINT
boolean triPoint(float x1, float y1, float x2, float y2, float x3, float y3, float px, float py) {
// get the area of the triangle
float areaOrig = abs( (x2-x1)*(y3-y1) - (x3-x1)*(y2-y1) );
// get the area of 3 triangles made between the point
// and the corners of the triangle
float area1 = abs( (x1-px)*(y2-py) - (x2-px)*(y1-py) );
float area2 = abs( (x2-px)*(y3-py) - (x3-px)*(y2-py) );
float area3 = abs( (x3-px)*(y1-py) - (x1-px)*(y3-py) );
// if the sum of the three areas equals the original,
// we're inside the triangle!
if (area1 + area2 + area3 == areaOrig) {
return true;
}
return false;
}
``````
This example was built on a modified version of a post on YoYo Games. If you would like to read a lengthy discussion on the merits and problems with this method, and many other suggestions, see this thread on GameDev.net.
NEXT: Where Are The Other Triangle Examples? | HuggingFaceTB/finemath | |
# Combine -sin((pii)/2)-cos((pii)/2)+(pii)/2+3
-sin(πi2)-cos(πi2)+πi2+3
To write 31 as a fraction with a common denominator, multiply by 22.
-sin(πi2)-cos(πi2)+πi2+31⋅22
Write each expression with a common denominator of 2, by multiplying each by an appropriate factor of 1.
Combine.
-sin(πi2)-cos(πi2)+πi2+3⋅21⋅2
Multiply 2 by 1.
-sin(πi2)-cos(πi2)+πi2+3⋅22
-sin(πi2)-cos(πi2)+πi2+3⋅22
Combine the numerators over the common denominator.
-sin(πi2)-cos(πi2)+6+πi2
Combine -sin((pii)/2)-cos((pii)/2)+(pii)/2+3
## Our Professionals
### Lydia Fran
#### We are MathExperts
Solve all your Math Problems: https://elanyachtselection.com/
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# Solve the inequality. graph the solution set. 26+6b (> =) 2(3b+4 a. all real numbers b. b> or
Solve the inequality. graph the solution set. 26+6b (> =) 2(3b+4 a. all real numbers b. b> or less than 1/2 c. no solutions
## This Post Has 9 Comments
1. maggiestevens5321 says:
I got c because when you solve theres no solution
2. queen6931 says:
This what I came up with
$Solve the inequality. Graph the solution set. 26 + 6b ≥ 2(3b + 4)$
3. yeontan says:
Ci think but i'm not entirely sure so..
4. sihamabdalla591 says:
The inequality is : 26 + 6b= 2(3b + 4)
On opening the bracket we get,
or 26 + 6b= 6b + 8
Now, bring the variable terms to right hand side,
or 26 + 6b - 6b = 8
or 26 = 8, which is not possible because 26 can never be equal to 8.
Thus, this inequality has no solutions.
5. 22chandlerlashley says:
Step-by-step explanation:
26+6b≥2(3b+4)
25+6b≥6b+8
25≥8
which is true as 25>8
so b can have any real value.
6. skateyo2553 says:
Distribute on the right
26+6b≥6b+8
Then subtract 6b from both sides
26≥8
True
Final All real numbers, A
7. dbenitezmontoya3 says:
B no solutions is incorrect.
8. rileyeddins1010 says:
True for all b
Interval notation; (-∞, ∞)
Step-by-step explanation:
We have been given the following inequality;
26+6b>2(3b+4)
The first step is to open the brackets on the right hand side;
26+6b>6b+8
26-8>6b-6b
18>0
Since 18 is actually greater than 0, the solution set to the inequality is;
True for all b.
$Solve the inequality. graph the solution set. 26 +6b> 2(3b+4)$
9. AmazingColor says:
we have
$26 + 6b\geq2(3b + 4)$
Applying the distributive property on the right side
$26 + 6b\geq6b+8$
subtract $6b$ from both sides
$26\geq 8$ -------> is true
for all real numbers the inequality is true
therefore
the graph is a shaded area everywhere.
the answer is
the solution is all real numbers | HuggingFaceTB/finemath | |
# Flash perimeter algorithmDecember 22, 2005 2:25 PM Subscribe
I have a list of coordinates (in [x,y] format), and they define the perimeter of a shape. I can't quite work out an algorithm for drawing round the perimeter of this shape, without crossing the shape. Google has failed me, and my A-level in maths is all but a hazy memory now... Algorithm gods, your time is now!
I apologise in advance for the length of this question, but it's quite a specific problem, and it's causing multiple headaches for me... :(
I alredy run a sudoku website, and am working on adding killer sudoku to it. I'm trying to write a hunk of Flash that will draw the cages required in a Killer Sudoku. The cages are the dotted lines that run around several cells in the puzzle. In normal sudoku, you have to have the numbers 1-9 once in each row, column and 3x3 block. In a killer sudoku, the numbers in the cages must add up to the number show in each cage (for example the top left cage with 16 can only be 9 and 7, or 7 and 9, as they're the only possible combinations that add up to 16).
In my code, I represent each of the cages as a list of cells, and their sum. The first few cages in that puzzle would be:
16=0,9
9=1,10
The cell numbering schema is that the top row are cells 0, 1, 2, 3, 4, 5, 6, 7, 8, and the next row are cells 9, 10, 11, 12, 13, etc. Using this notation, I build a complete list of cages, and their cells. Now, I ask the flash code to build each cage in turn.
The current method I'm using works like this:
1. Calculate all the coordinates of every corner of each cell in the cage.
(eg, the cage 16=0,9 will have corners at [0,0][0,1][1,0][1,1][2,1][2,2])
2. Remove any coordinates that are shared between four cells. If any point is shared by four cells then it is an internal point, not one on the perimeter of the shape.
(eg in an example cage 22=0,1,9,10 there is a point at [1,1] that is shared by cells 0, 1, 9 and 10. It has four cells touching it, therefore it is a point inside the shape, not on the perimeter, so we ignore it)
3. Now we have a list of points that are on the perimeter of the shape. Here I get stuck.
How do I now work out what order to join the points together in? One possible solution is to start at any given point and move to any other point that is a distance of '1' away, and we haven't visited before. This should work round the shape of the puzzle but what happens is that it cuts off corners of shapes.
For example, in the cage 16=0,9, starting at [0,0], the program may draw a line to [1,0], then [1,1], then [0,1], then [0,2]. Each of those points are '1' unit away from each other, so the algorithm sort of works, but not as intended.
I basically need to trace round the perimeter of the shape, without crossing the shape at any point. I'd prefer not to use a recursive function to exhaustively find the right solution, as that runs really slowly in flash.
The second problem is with indenting. If you look at the cages, they're just slightly inside the cells the cover, rather than running along the grid lines. If the code is going to draw the cages inside the cells, then it needs to have some concept of what 'inside' means for each cage. I think this is something to do with calculating the normal for each point on the perimiter, then adjusting the coordinates slightly at the point of drawing the line, but I haven't got to that part yet.
I've googled for suitable algorithms, but have not turned up anything of use. I hope that somebody has done this before and can share some insight. This is the third time I've tried to refactor this code, and it still doesn't quite work properly.
Hyalp!
posted by gaby to Computers & Internet (5 answers total)
if you need to trace the outline of a set of points, use the Convex Hull method. I didn't read your MI. :)g
posted by kcm at 2:33 PM on December 22, 2005
Calculate all the coordinates of every corner of each cell in the cage. (eg, the cage 16=0,9 will have corners at [0,0][0,1][1,0][1,1][2,1][2,2])
I think this example is for cage 16=0,1.
More importantly, it looks like you got the list of six corners by generating a list of the eight corners of the two cells (four corners per cell), and then discarding two of the four corners (coordinates) as duplicates. But it is precisely those two corners that are interior. (In other words, if you start with a complete set, then discard ALL corners which appear twice or more in your list, you'll end up with only four corners, and you're done, yes?)
posted by WestCoaster at 2:56 PM on December 22, 2005
Finding a convex hull is overkill here. You're asking how to compute the topology of the cages, but if all you want is to display them, that's not necessary.
Given a pair of adjacent squares sharing an edge, it seems like you draw two dotted lines on either side of the edge if they're in a different cage and no dotted lines if they're in the same cage.
Say you're looking at square 1. If square 10 is in a different cage, then you know you have to draw a line at the bottom of square 1. When you get to square 10, you'll see that square 1 is in a different cage, so you have to draw a line at the top of square 10.
So, can you just go through each square, see if its 8 neighbors are in same/different cages and choose which dotted lines you have to draw for that square? Store these lines and calculate them for each square, then send them all to the display when you're done.
For squares on the boundaries, they always have dotted lines on the perimeter of the puzzle, so you can handle them as special cases.
posted by driveler at 2:56 PM on December 22, 2005
Draw on the boundaries using XOR.
Or, don't collect the coordinates. Write a routine that takes a cell as an argument and which sides to draw in a cage. Get the list of cells, for each cell figure out if there is a neighboring cage cell also to be drawn. If so, don't draw a border on that side.
posted by fleacircus at 6:56 PM on December 22, 2005
Westcoaster: yes, my bad. It does refer to a cage 16=0,1.
fleacircus, driveler: I can see that I'm thinking about this from the perspective of a single cage, and drawing it's outline, rather than how the cages interact with each other. This approach makes more sense, and feels a bit easier to implement.
My current approach involves working in a clockwise manner around the edge of the shape, and seeing if the cage can be extended in that direction. Your proposed methods seem much simpler.
I'm currently working with the set of cells and seeing if there are cage cells to either side of them, then drawling lines and working in a clockwise manner.
posted by gaby at 3:16 AM on December 23, 2005
« Older Help me be wireless! | Present for a 13 year old. Newer » | HuggingFaceTB/finemath | |
When a uniformly loaded beam is supported at each end on level bearings (the beam itself being either horizontal or inclined), the amount of pressure caused by the load on each point of support is equal to one half the load; and this is also the ase when the load is concentrated at the middle of the beam, or has its centre of gravity at the middle of the beam; but when the load is unequally distributed, or concentrated so that its centre of gravity occurs at some other point than the middle of the beam, then the amount of pressure caused by the load on one of the points of support is unequal to that on the other. The precise amount on each may be ascertained by the following rule.
Rule. - Multiply the weight W (Fig. 35) by its distance, CB, from its nearest point of support, B, and divide the pro-duct by the length, A B, of the beam, and the quotient will be the amount of pressure on the remote point of support, A. Again, deduct this amount from the weight W, and the remainder will be the amount of pressure on the near point of support, B; or, multiply the weight W by its distance, A C, from the remote point of support, A, and divide the product by the length, A B, and the quotient will be the amount of pressure on the near point of support, B.
Fig 35.
When l equals the length between the bearings A and B, n - AC, m = C B, and W = the load; then
Wm/1 = A = the amount of pressure at A, and Wn/l== B = the amount of pressure at B.
Example. - A beam 20 feet long between the bearings has a load of 100 pounds concentrated at 3 feet from one of the bearings: what is the portion of this weight sustained by each bearing?
Here W = 100; n, 17; m, 3; and l, 20.
Hence A = Wm = 100x3 = 15
l 20
And B= Wn/l = 100x17/20 = 85.
Load on A == 15 pounds. Load on B =85 pounds. Total weight = 100 pounds.
## 89. - Weight - Strength
Preliminary to designing a roof-truss or other piece of framing, a knowledge of two subjects is essential: one is, the effect of gravity acting upon the various parts of the intended structure; the other, the power of resistance possessed by the materials of which the framing is to be constructed. The former subject having been treated of in the preceding pages, it remains now to call attention to the latter. | HuggingFaceTB/finemath | |
# Wikipedia:Reference desk/Archives/Mathematics/2013 May 16
Mathematics desk
< May 15 << Apr | May | Jun >> May 17 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.
# May 16
## Magic square type Combinatorics Problem
The problem is:
• In how many ways can you put 1 and -1 in the 16 places of a 4*4 square so that the product of any row or column is -1.
I figured out that -1 can be written by the product of 1s and -1s in only two ways:
-1 = 1 * 1 * 1 * -1 or -1 = -1 * -1 * -1 * 1
But I cannot figure out the solution. Any help appreciated. Solomon7968 (talk) 09:46, 16 May 2013 (UTC)
There are 2^16 = 65536 ways to distribute minuses in the the 4*4 square. The probability that any row or column multiplies to -1 is 50%. There are 4 rows and 4 columns. The probability that they all multiply to -1 is 2^(-8). So the answer to your question is probably (2^16)*(2^(-8))=2^8=256. Bo Jacoby (talk) 10:40, 16 May 2013 (UTC).
I am new to combinatorics. Will you explain how you got the 50% thing? Solomon7968 (talk) 10:43, 16 May 2013 (UTC)
Consider any row or column. There are 4 places. Each place can contain either +1 or -1. The sixteen possibilities are ++++ +++- ++-+ ++-- +-++ +-+- +--+ +--- -+++ -++- -+-+ -+-- --++ --+- ---+ ----. Eight out of these sixteen possibilities have an odd number of minuses and so multiplies to minus. (Namely +++- ++-+ +-++ +--- -+++ -+-- --+- ---+ ). So the probability that a random row multiplies to minus is 8/16=1/2=50%. Bo Jacoby (talk) 10:51, 16 May 2013 (UTC).
There's 512 ways. You can put any combination of +1 and -1 into the first 3x3 places. Then just place a +1 or -1 as appropriate into the last row or column to make the produce -1. The product of the three end columns will be te same as the three rows as they are determined by the product of all 9 numbers so there is no confusion about the last end number to put in. Dmcq (talk) 10:58, 16 May 2013 (UTC)
...and this is consistent with Bo's probabilistic approach once you take into account the fact that the 8 row/column products are not independent - once you know any 7 of them then the 8th is determined by the contraint that product of row products = product of column products. Correcting this in Bo's formula gives 2^16 / 2^7 = 2^9 = 512. Now for bonus points determine how many arrangements there are if reflections and rotations of an arrangement are not counted as distinct. Gandalf61 (talk) 13:07, 16 May 2013 (UTC) | HuggingFaceTB/finemath | |
1. ## Multiple integral
Is this correct?
10x.
2. Nice try, actually, but not quite...
You are right to split it up into two pieces. You missed a couple of things.
The portion above the x-axis needs $y^{2}$ in the arguments and should be on [0,1].
The portion below the x-axis should be on [-1,0].
In every case, it should be "1-". I cannot tell where you managed any "1+". Further, if your limits are identical, you ARE going to get zero (0). Try negative on the bottom and positive on the top.
3. Originally Posted by TKHunny
Nice try, actually, but not quite...
You are right to split it up into two pieces. You missed a couple of things.
The portion above the x-axis needs $y^{2}$ in the arguments and should be on [0,1].
The portion below the x-axis should be on [-1,0].
In every case, it should be "1-". I cannot tell where you managed any "1+". Further, if your limits are identical, you ARE going to get zero (0). Try negative on the bottom and positive on the top.
BTW I spent a lot of time just looking at it... Is there a technique doing that?
10x.
4. $\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f(x,y)~dy~dx$
The region of integration is a full circle of radius one centered around the origin. Now think about it, if you rotate it ninety degrees so you are first going in respect to x, everything will be "the same" since a circle is completely symmetric, so all you would do is switch the x's and y's
$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f(x,y)~dy~dx=\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}f(x,y)~dx~dy$
5. Originally Posted by asi123
BTW I spent a lot of time just looking at it... Is there a technique doing that?
Yeah, what I do is rotate it ninety degrees and look at it, and see where the y's or x's range from and so forth. I think its easier if you rotate it.
6. Originally Posted by Mathstud28
$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f(x,y)~dy~dx$
The region of integration is a full circle of radius one centered around the origin. Now think about it, if you rotate it ninety degrees so you are first going in respect to x, everything will be "the same" since a circle is completely symmetric, so all you would do is switch the x's and y's
$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f(x,y)~dy~dx=\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}f(x,y)~dx~dy$
Don't we need to split it?
7. Originally Posted by asi123
Don't we need to split it?
Why would we?
8. Originally Posted by Mathstud28
Why would we?
Is that right?
9. Originally Posted by asi123
Is that right?
Thats the exact same thing as I put since your two integrals combine to form the exact region that mine describes and you should know that
$\iint_{R}f(x,y)~dx~dy=\iint_{R_1}f(x,y)~dx~dy+\iin t_{R_2}~dx~dy$
Assuming $R_1\cup{R_2}=R$
And you have two semi-circles who when unioned form the circle described by my double integrals.
10. See how important it is to write clearly and check what it is you have written?
My original post took you at your word. Check that internal integral. The bottom limit is from a circle. The top limit is from a parabola. It is for this reason that I recommended splitting it up. If you've really only circles, the original integral simply is wrong and there is no need to do any splitting.
11. Originally Posted by TKHunny
See how important it is to write clearly and check what it is you have written?
My original post took you at your word. Check that internal integral. The bottom limit is from a circle. The top limit is from a parabola. It is for this reason that I recommended splitting it up. If you've really only circles, the original integral simply is wrong and there is no need to do any splitting.
TKHunny is right, the only reason I did the bounds I did was because this is a classic question, and I assumed that this is what you meant. You really should be more careful. But don't worry about it! Just don't do it again. | HuggingFaceTB/finemath | |
# What is the median in math?
## What is the median in math?
The median is the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average. If there is an even amount of numbers in the list, the middle pair must be determined, added together, and divided by two to find the median value.
## How do I find the median?
Median
1. Arrange your numbers in numerical order.
2. Count how many numbers you have.
3. If you have an odd number, divide by 2 and round up to get the position of the median number.
4. If you have an even number, divide by 2.
What is the median of 15?
To find the Median, place the numbers in value order and find the middle number. Example: find the Median of {13, 23, 11, 16, 15, 10, 26}. The middle number is 15, so the median is 15.
Is median an average?
Average Median The average is the arithmetic mean of a set of numbers. The median is a numeric value that separates the higher half of a set from the lower half. When is it applicable? The mean is used for normal number distributions, which have a low amount of outliers.
### What does the median tell you?
WHAT CAN THE MEDIAN TELL YOU? The median provides a helpful measure of the centre of a dataset. By comparing the median to the mean, you can get an idea of the distribution of a dataset. When the mean and the median are the same, the dataset is more or less evenly distributed from the lowest to highest values.
### What is median give example?
Median: The middle number; found by ordering all data points and picking out the one in the middle (or if there are two middle numbers, taking the mean of those two numbers). Example: The median of 4, 1, and 7 is 4 because when the numbers are put in order (1 , 4, 7) , the number 4 is in the middle.
What is the first step to find the median?
To find the median, first order your data. Then calculate the middle position based on n, the number of values in your data set. If n is an odd number, the median lies at the position (n + 1) / 2. If n is an even number, the median is the mean of the values at positions n / 2 and (n / 2) + 1.
What is the easiest way to find the median?
To find the median, put all numbers into ascending order and work into the middle by crossing off numbers at each end. If there are a lot of items of data, add 1 to the number of items of data and then divide by 2 to find which item of data will be the median.
#### What is the median of 9?
We find that the mean is 10 . The word “median” literally means the middle of something. In this case, it is the middle number in our data set. The middle number is 9 , therefore this is our median.
#### What is the median of 4 and 7?
For a dataset with an even number of values, you take the mean of the two center values. So, if the dataset has the values, 1, 4, 7, 9, the two center values are 4 and 7. The mean of these middle values is (4 + 7) / 2 = 5.5 , so the median is 5.5.
What is the median of 23?
Since there are an even number of values, the median will be the average of the two middle numbers, in this case, 23 and 23, the mean of which is 23.
What is difference between mean and median?
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set.
## What’s the difference between mean vs. median?
The key difference between mean and median is that mean is the sum of total values in a data set divided by the number of values, while median is the middle value of a data set. We use mean and median to check the location of the data because they give an indication of a central value around which a set of values tends to cluster.
## What does median mean?
The word “median” literally means the middle and, as it pertains to salaries, it is the one that, if you list in numerical order all the salaries for every individual working in an occupation, falls in the middle of the list. Half the individuals on that list earns less than the median and half earns more.
What is the definition of mean median?
A mean is computed by adding up all the values and dividing that score by the number of values. The Median is the number found at the exact middle of the set of values. A median can be computed by listing all numbers in ascending order and then locating the number in the centre of that distribution. | HuggingFaceTB/finemath | |
# Similar Triangles 2
Directions: Using the digits 0 to 9 at most one time each, create two similar triangles. You may have as many leading zeros as you like.
### Hint
Can you use a scale factor to help you in any way?
What is true about the corresponding sides of similar triangles?
There are many answers. Here are the possibilities:
YA – 5
AH – 10
YB – 3
BV – 6
BA – 9
VH – 27
Source: Drew Ross
## Three Triangles And A Wannabe
Directions: Using the values 8, 10, 12, 14, 16, 18, and 20, determine lengths for … | HuggingFaceTB/finemath | |
+0
# help!
0
508
4
The equation $$y = -4.9t^2 + 42t + 18.9$$ describes the height (in meters) of a ball tossed up in the air at 42 meters per second from a height of 18.9 meters from the ground, as a function of time in seconds. In how many seconds will the ball hit the ground?
Dec 30, 2017
#1
0
Physics or Maths
Dec 30, 2017
#2
+2
When it hits the ground, y will = 0
So we have
-4.9t^2 + 42t + 18.9 = 0 multiply through by -10
49t^2 - 420t - 189 = 0 divide through by 7
7t^2 - 60t - 27 = 0 factor
(7t + 3 ) ( t - 9 ) = 0
Setting each factor to 0 and solving for t gives
t = - 3/7 sec { reject }
And
t = 9 sec Dec 30, 2017
#3
0
I had my doubts that it was related to physics but didn’t bother doing it
Rauhan Dec 30, 2017
#4
+3
Setting $$y$$ to zero, we find the following:
\begin{align*} 0& = -4.9t^2 + 42t + 18.9\\ & = -49t^2 + 420t + 189\\ & = 7t^2 - 60t - 27\\ & = (7t + 3)(t - 9) \end{align*}
As $$t$$ must be positive, we can see that $$t = \boxed{9}.$$
.
Dec 30, 2017 | HuggingFaceTB/finemath | |
# Courses of Study : Mathematics
NUMBER AND QUANTITY
The Real Number System
Extend the properties of exponents to rational exponents.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
1 ) Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. [N-RN1]
Example: We define 51/3 to be the cube root of 5 because we want (51/3)3 = 5(1/3)3 to hold, so (51/3)3 must equal 5.
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.NS.HS.1- Recognize that a number raised to the 1/2 power is the square root of that number; similarly, a number raised to the 1/3 power is the cube root of that number. Identify the root of a number when given the fractional notation. Limit base values for square roots to 9, 16, 25. Limit base values for cube roots to 8, 27.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
2 ) Rewrite expressions involving radicals and rational exponents using the properties of exponents. [N-RN2]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.NS.HS.2- Determine the value of an expression squared (base values 1-15) or cubed (base values 1-10).
Use properties of rational and irrational numbers.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 1 Classroom Resources: 1
3 ) Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational. [N-RN3]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.NS.HS.3- Identify rational and irrational numbers within 1 to 20 (irrational numbers limited to pi, √2, √3, √5).
Quantities*
Reason quantitatively and use units to solve problems. (Foundation for work with expressions, equations, and functions.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 5 Learning Activities: 2 Classroom Resources: 3
4 ) Use units as a way to understand problems and to guide the solution of multistep problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays. [N-Q1]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.Q.HS.4- Using real world models, express quantities of measurement to the given precision. (limited to measurements of length (inch, 1/2 inch, 1/4 inch), weight (pounds, kilograms (tenth of a unit), volume (cup, 1/2 cup, 1/4 cup, 1/3 cup, liter), temperature (degree), velocity (mph, kmph).
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 3 Classroom Resources: 3
5 ) Define appropriate quantities for the purpose of descriptive modeling. [N-Q2]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.Q.HS.5- Recognize units of weight (ounces, pounds, grams, kilograms), length (inch, foot, mile, centimeter, meter, kilometer), area (square inches in^2, square feet ft^2, square centimeters cm^2, square meters m^2) and capacity (cubic inches in^3, cubic feet ft^3, cubic centimeters cm^3, cubic meters m^3).
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 2 Classroom Resources: 2
6 ) Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. [N-Q3]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.Q.HS.6- Estimate to the nearest 1, 10, and 100 when adding, subtracting, multiplying, or dividing; include units with estimates.
ALGEBRA
Seeing Structure in Expressions
Interpret the structure of expressions. (For standard 7 linear, exponential, quadratic; for standard 8 linear, exponential, quadratic, rational.) (Alabama)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
7 ) Interpret expressions that represent a quantity in terms of its context.* [A-SSE1]
a. Interpret parts of an expression such as terms, factors, and coefficients. [A-SSE1a]
b. Interpret complicated expressions by viewing one or more of their parts as a single entity. [A-SSE1b]
Example: Interpret P(1+r)n as the product of P and a factor not depending on P.
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.7- Identify an algebraic expression involving one arithmetic operation to represent a real-world problem.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
8 ) Use the structure of an expression to identify ways to rewrite it. [A-SSE2]
Example: See x4 - y4 as (x2)2 - (y2)2, thus recognizing it as a difference of squares that can be factored as (x2 - y2)(x2 + y2).
Write expressions in equivalent forms to solve problems. (Quadratic and exponential.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
9 ) Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.* [A-SSE3]
a. Factor a quadratic expression to reveal the zeros of the function it defines. [A-SSE3a]
b. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines. [A-SSE3b]
c. Determine a quadratic equation when given its graph or roots. (Alabama)
d. Use the properties of exponents to transform expressions for exponential functions. [A-SSE3c]
Example: The expression 1.15t can be rewritten as (1.151/12)12t ≈ 1.01212t to reveal the approximate equivalent monthly interest rate if the annual rate is 15%.
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.9- Identify the expression that is the same as the one shown. (limit to two operations e.g. x^2 + 3x is the same as x(x+3).
Arithmetic With Polynomials and Rational Expressions
Perform arithmetic operations on polynomials. (Linear and quadratic.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
10 ) Understand that polynomials form a system analogous to the integers; namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. [A-APR1]
Rewrite rational expressions. (Linear and quadratic denominators.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 1 Classroom Resources: 1
11 ) (+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions. [A-APR7]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.11- Add or subtract two polynomial expressions (limit to 2 terms each) with one variable.
Creating Equations*
Create equations that describe numbers or relationships. (Linear, quadratic, and exponential (integer inputs only); for Standard 14, linear only.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
12 ) Create equations and inequalities in one variable, and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. [A-CED1]
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
13 ) Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. [A-CED2]
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
14 ) Represent constraints by equations or inequalities, and by systems of equations and/or inequalities and interpret solutions as viable or non-viable options in a modeling context. [A-CED3]
Example: Represent inequalities describing nutritional and cost constraints on combinations of different foods.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 3 Lesson Plans: 2 Classroom Resources: 1
15 ) Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. [A-CED4]
Example: Rearrange Ohm's law V = IR to highlight resistance R.
Reasoning With Equations and Inequalities
Understand solving equations as a process of reasoning and explain the reasoning. (Master linear; learn as general principle.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 4 Classroom Resources: 4
16 ) Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. [A-REI1]
Solve equations and inequalities in one variable. (Linear inequalities; literal that are linear in the variables being solved for; quadratics with real solutions.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 15 Learning Activities: 1 Classroom Resources: 14
17 ) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. [A-REI3]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.17- Solve an equation of the form ax + b = c where a, b, and c are positive whole numbers and the solution, x, is a positive whole number to represent a real-world problem.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
18 ) Solve quadratic equations in one variable. [A-REI4]
a. Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)2 = q that has the same solutions. Derive the quadratic formula from this form. [A-REI4a]
b. Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square and the quadratic formula, and factoring as appropriate to the initial form of the equation. [A-REI4b] (Alabama)
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.18- Solve an equation of the form x2 = p, where p is a perfect square less than or equal to 225.
Solve systems of equations. (Linear-linear and linear-quadratic.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
19 ) Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions. [A-REI5]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.19- Given a pair of equations, identify a coordinate pair that is the solution of both equations. (Limit to 1-step equations - e.g. x+y=8, y=x+1, or x+y=15, y=3.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
20 ) Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. [A-REI6]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.20- Name the coordinate pair of the intersection of two lines in a coordinate plane.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
21 ) Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. [A-REI7]
Example: Find the points of intersection between the line y = -3x and the circle x2 + y2 = 3.
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.21- Identify the coordinate pairs of the solutions of the graph of an intersecting quadratic function and linear function in Quadrant 1.
Represent and solve equations and inequalities graphically. (Linear and exponential; learn as general principle..)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
22 ) Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line). [A-REI10]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.22- Given the graph of a linear equation in quadrant 1, identify a point on the graph and its corresponding ordered pair that is a solution to the equation.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
23 ) Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions.* [A-REI11]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.23- Identify the point of intersection and its corresponding ordered pair for two lines graphed on a coordinate grid.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
24 ) Graph the solutions to a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes. [A-REI12]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.A.HS.24- Given the graph of a linear system of inequalities (limited to two inequalities), identify a point that represents a solution in the shaded region of the graph.
FUNCTIONS
Interpreting Functions
Understand the concept of a function and use function notation. (Learn as general principle; focus on linear and exponential and on arithmetic and geometric sequences.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
25 ) Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x). [F-IF1]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.25- Identify graphs of functions.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 2 Classroom Resources: 2
26 ) Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. [F-IF2]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.26- Substitute x-values into one-step linear equations in two variables (y = x + p or y = px) and solve for the y-values. (this could include the original information listed above and have students represent in data table)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
27 ) Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers. [F-IF3]
Example: The Fibonacci sequence is defined recursively by f(0) = f(1) = 1, f(n+1) = f(n) + f(n-1) for n ≥ 1.
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.27-Given a sequence of numbers, identify the rule that will give you the next number in the sequence. (Limit to expressions with simple arithmetic (adding or subtracting) or geometric (multiplying or dividing) operations.
Interpret functions that arise in applications in terms of the context. (Linear, exponential, and quadratic.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
28 ) For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.* [F-IF4]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.28- Given a linear graph, identify characteristics of the line in the graph (limit to y- intercept, x-intercept, increasing, decreasing).
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
29 ) Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes.* [F-IF5]
Example: If the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function.
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.29- Given the graph of a linear function with a finite domain evident in the graph, identify the domain (limit to first quadrant values between 0 and 10).
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
30 ) Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.* [F-IF6]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.30- Given two points on the graph of the line, describe how the y-values change compared to the x- values for a given interval.
Analyze functions using different representations. (Linear, exponential, quadratic, absolute value, step, and an awareness of piecewise-defined.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
31 ) Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.* [F-IF7]
a. Graph linear and quadratic functions, and show intercepts, maxima, and minima. [F-IF7a]
b. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. [F-IF7b]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.31- Given the graph of a quadratic function, identify characteristics of the parabola in the graph (limit to first quadrant, maximum, minimum, x-intercepts).
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 6 Classroom Resources: 6
32 ) Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. [F-IF8]
a. Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. [F-IF8a]
b. Use the properties of exponents to interpret expressions for exponential functions. [F-IF8b]
Example: Identify percent rate of change in functions such as y = (1.02)t, y = (0.97)t, y = (1.01)12t, and y = (1.2)t/10, and classify them as representing exponential growth and decay.
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.32- Identify the y-intercept of a linear equation in the form of y=mx+b as (0,b).
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
33 ) Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). [F-IF9]
Example: Given a graph of one quadratic function and an algebraic expression for another, say which has the larger maximum.
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.33- Compare the y-intercept, slope (increasing/decreasing), or domain of two linear functions represented by a table or a graph.
Building Functions
Build a function that models a relationship between two quantities. (For standards 34 and 35, linear, exponential, and quadratic.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
34 ) Write a function that describes a relationship between two quantities.* [F-BF1]
a. Determine an explicit expression, a recursive process, or steps for calculation from a context. [F-BF1a]
b. Combine standard function types using arithmetic operations. [F-BF1b]
Example: Build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model.
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.34- Select the appropriate graphical representation (first quadrant) given a situation involving a constant rate of change (slope).
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
35 ) Write arithmetic and geometric sequences both recursively and with an explicit formula, use them to model situations, and translate between the two forms.* [F-BF2]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.35- Determine an arithmetic sequence with whole numbers when provided a recursive rule. (limit rule to whole numbers involving addition/ subtraction or multiplication or division - e.g., start with the number 4. Each term in the sequence is found by taking the previous term and adding 8. Find the next 3 terms.)
Build new functions from existing functions. (Linear, exponential, quadratic, and absolute value.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
36 ) Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them. [F-BF3]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.36- Given the graph of a linear function f(x), identify f(x) + k.
Construct and compare linear, quadratic, and exponential models and solve problems.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
37 ) Distinguish between situations that can be modeled with linear functions and with exponential functions. [F-LE1]
a. Prove that linear functions grow by equal differences over equal intervals, and that exponential functions grow by equal factors over equal intervals. [F-LE1a]
b. Recognize situations in which one quantity changes at a constant rate per unit interval relative to another. [F-LE1b]
c. Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another. [F-LE1c]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.37- Recognize real-world situations that are modeled with linear functions.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
38 ) Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table). [F-LE2]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.38- Identify three points defined by a linear function from a table of values from 0 to 10.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
39 ) Observe, using graphs and tables, that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. [F-LE3]
Alabama Alternate Achievement Standards
AAS Standard:
M.AASF.HS.39- Given the graph of two functions, identify which function has a greater y-value for a specific x-value.
Interpret expressions for functions in terms of the situation they model. (Linear and exponential of form f(x) = bx + k.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 0
40 ) Interpret the parameters in a linear or exponential function in terms of a context. [F-LE5]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.F.HS.40- Identify rate of change (slope) and starting value (y-intercept) in context.
STATISTICS AND PROBABILITY
Interpreting Categorical and Quantitative Data
Summarize, represent, and interpret data on a single count or measurement variable.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 5 Classroom Resources: 5
41 ) Represent data with plots on the real number line (dot plots, histograms, and box plots). [S-ID1]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.SP.HS.41- Given data, construct a simple graph (line, pie, bar, picture) or table, and interpret the data in terms of range, mode, and median, mean.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 8 Classroom Resources: 8
42 ) Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. [S-ID2]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.SP.HS.42- Given two dot plots representing two different data sets, identify which data set has the greater maximum, median, or range.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 8 Classroom Resources: 8
43 ) Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). [S-ID3]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.SP.HS.43- Interpret general trends on a graph or chart (increase/decrease).
Summarize, represent, and interpret data on two categorical and quantitative variables. (Linear focus, discuss general principle.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 1 Classroom Resources: 1
44 ) Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and conditional relative frequencies). Recognize possible associations and trends in the data. [S-ID5]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.SP.HS.44- Calculate the mean of a given data set (number of data points limited to fewer than five, values of less than 10).
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 3 Learning Activities: 1 Classroom Resources: 2
45 ) Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. [S-ID6]
a. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. Emphasize linear, quadratic, and exponential models. [S-ID6a]
b. Informally assess the fit of a function by plotting and analyzing residuals. [S-ID6b]
c. Fit a linear function for a scatter plot that suggests a linear association. [S-ID6c]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.SP.HS.45- Given a scatter plot with data with a line of best fit that can be represented by a linear function, describe what is happening to the y-values in reference to the x-values (x- and y- values limited positive numbers).
Interpret linear models.
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 6 Learning Activities: 1 Classroom Resources: 5
46 ) Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. [S-ID7]
Alabama Alternate Achievement Standards
AAS Standard:
M.AAS.SP.HS.46- Given a graph that describes a set of linear data, identify the rate of change (slope) and constant term (y-intercept). (Use context of data—the total price of the stamps is calculated by increasing 50 cents for every stamp purchased or the cost if no stamps are purchased is \$0.)
Conditional Probability and the Rules of Probability
Understand independence and conditional probability and use them to interpret data. (Link to data from simulations or experiments.)
Mathematics (2016) Grade(s): 9 - 12 Algebra I All Resources: 3 Learning Activities: 1 Classroom Resources: 2
47 ) Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. [S-CP2] | HuggingFaceTB/finemath | |
# MATHEMATICS: ASSIGNMENT 6 ```MATHEMATICS: ASSIGNMENT 6
There are two parts to this assignment: Part A and Part B. Part A is to be completed online, and Part B
is to be handed in, both before 10:00 a.m. on Friday, January 15.
Part A
This part of the assignment can be found online, labelled Assignment6A, at webwork.elearning.ubc.ca.
Part B
1. (a) Let a1 = 1 and an+1 =
√
1 + 2an for n = 1, 2, 3, . . .. Show that the sequence {an } converges.
1. (b) Find the limit of the sequence in part (a).
2. Consider the sequence {an } where
an =
bxc + b2xc + · · · + bnxc
,
n2
and x is any real number. (Here “bac” is defined to be the largest integer which is at most a.) Find the
limit of the sequence, if it exists.
3. Recall that the Fibonacci sequence is defined thus: F1 = F2 = 1, and Fn + Fn+1 = Fn+2 for n = 1, 2, 3, . . ..
Prove (by induction, or otherwise) that
F12 + F22 + · · · + Fn2 = Fn Fn+1 for n = 1, 2, 3, . . ..
4. (a) Let {an } be a sequence with lim a2n = L and lim a2n+1 = L. Does {an } converge? Justify your
n→∞
n→∞ | HuggingFaceTB/finemath | |
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Ch 21, L2-AG, S'11
# Ch 21, L2-AG, S'11 - CHAPTER 21 Coulombs Law Electric Field...
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CHAPTER 21 Coulomb’s Law Electric Field
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Coulomb’s Law Charles Coulomb
k = 8.99*10 9 N*m 2 /C 2 (in SI system) 0 4 1 !" = k , where ε 0 is dielectric constant of vacuum ε 0 = 8.85*10 -12 C 2 /N*m 2 Fundamental electric constants
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What force acts between two point charges 1 μ C each separated by 10 cm ? ) C / Nm 10 9 ( 2 2 9 2 2 1 ! = = R q q k F (10 -6 C )( 10 -6 C) = (0.1 m) 2 = 0.9 N
Coulomb Force q 1 q 2 + + F 1,2 F 2,1 ! F 1,2 = " ! F 2,1 q 1 q 2 - - F 1,2 F 2,1 q 1 q 2 - + F 1,2 F 2,1 - Coulomb forces obey Newton s third law : - Coulomb force is a central force Opposite charges Like charges - Superposition of forces : total force acting on a charge is the vector sum of all individual forces acting on that charge
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Coulomb Force vs Gravitational Force How does the electrostatic force between two charges compare with the gravitational force between them, say +1C charge on masses of 1 kg and separated by 1km?
Coulomb Force vs Gravitational Force Both are ‘inverse square’ forces – the 1 st depends on the product of the charges; the 2 nd on the product of the masses. The magnitude of the electrostatic force constant ‘ k in SI units is about 9x10 9 N*m 2 /C 2 , while the magnitude of G is about 6.67x10 -11 N*m 2 /kg 2 . F e /F g ~ 10 20 F e q 1 q 2 + + F e F g F g
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Two equal charges Q are located on the x axis. Midway between them is charge -q at a distance d from each Q . The net Coulomb,
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# How To Solve Square Root Manually?
## How do you manually find the square root?
Take the number you wish to find the square root of, and group the digits in pairs starting from the right end. For example, if you want to calculate the square root of 8254129, write it as 8 25 41 29. Then, put a bar over it as when doing long division.
## How do you find square roots without a calculator?
Finding square roots of of numbers that aren’t perfect squares without a calculator
1. Estimate – first, get as close as you can by finding two perfect square roots your number is between.
2. Divide – divide your number by one of those square roots.
3. Average – take the average of the result of step 2 and the root.
## What is the shortcut key for square root?
The Alt code for the symbol for the Square root is Alt +251 or 221A, then Alt+X. Follow these three simple steps to attach the symbol using the Alt code: – Position the pointer in the place where you want the square root symbol inserted. – Press and hold down the Alt key and type 251 from the numeric keypad.
## Which is the square root of 144?
The value of the square root of 144 is equal to 12. In radical form, it is denoted as √ 144 = 12.
## IS 400 a perfect square?
400 is a perfect square. Because 20 * 20 = 400.
## What is a square root of 11?
List of Perfect Squares
NUMBER SQUARE SQUARE ROOT
8 64 2.828
9 81 3.000
10 100 3.162
11 121 3.317
## What are 2 square roots of 121?
2 Answers By Expert Tutors The square root off 121 is eleven.
## Is the square root of 121 irrational?
Is the square root of 121 an irrational number? Yes, the square root of 121 is a rational number. The square root of 121 is 11, which is a rational number.
## Is Root 36 a whole number?
As we have calculated further down on this page, the square root of 36 is a whole number. 36 is a perfect square. | HuggingFaceTB/finemath | |
# math
Express 2,750,389 in scientific notation
1. A number in scientific notation can only have 1 digit greater than the decimal.
2,750,389 has 7 digits greater than the decimal. If you divide by 10^6, you will have only 1 digit.
posted by Marth
2. or...
2.750389 x 106
posted by bobpursley
## Similar Questions
1. ### scientific notation
Express 2,750,389 in scientific notation I think it is 2.750389 x 10 6
2. ### math
Express 2,750,389 in scientific notation.
3. ### math
express 2,750,389 in scientific motion
4. ### PHYSICS
Express following measurements in scientific notation. 1. a) 5800 m I don't even get what scientific notation is. In my book its ays, that in Scientific notation, the numerical part of a measurement is expressed as a number
5. ### Scientific Notation
Just started learning about the scientific notation, and its actually quite easy, but I don't know why im still struggling with it. For example it says express the following measurements in scientific notation: 5800m The answer is
6. ### algebra
use scientific notation to multiply the following two numbers. Express the answer using scientific notation; retain all nonzero decimal places. (8.1*10^8)*(4.2*10^3)
7. ### Algebra
Use scientific notation to multiply the following two numbers. Express the answer using scientific notation; retain all nonzero decimal places (1.9 • 10^4) • (8.4 • 10^5)
8. ### intermediate algebra
use scientific notation to divide the following two numbers. Express the answer using scientific notation; retain at least three decimal places (8.4*10^6)/ (5.97*10^2)
9. ### Scientific notation
Express in scientific notation. Remember, M must be a number 1 ≤ M < 10. 32,400 = 3.24-4 3.24*10-4 3.2400*10-4 the dash means to the 4th power
10. ### Math
Use scientific notation to divide the following two numbers. Express the answer using scientific notation; retain at least three decimal places. 6.2 • 107 / 4.15 • 103
More Similar Questions | HuggingFaceTB/finemath | |
# How are odds calculated in horse racing
## How are horse odds calculated?
To calculate the exact odds on your horse, just subtract the take from the total pool, then subtract the amount bet on your horse to give you the amount of cash to be paid out.
## What do the odds mean in horse racing?
Odds are the return you can expect to get if the horse you bet on is successful. It reflects the amount of money bet on a horse; the more money that is invested, the shorter the odds. When horse racing odds are shown in the form of 7-2, 5-1, etc, it expresses the amount of profit to the amount invested.
## How do bookies set odds for horse racing?
Probability however is only one aspect of odds pricing. Bookies don’t actually set their odds based completely on the real probabilities but rather on how likely they think their punters will wager on each outcome, allowing them to balance their book.
## How are odds payout calculated?
To calculate winnings on fractional odds, multiply your bet by the top number (numerator), then divide the result by the bottom (denominator). So a \$10 bet at 5/2 odds is (10 * 5) / 2, which equals \$25. A \$10 bet at 2/5 odds is (10 * 2) / 5, which is \$4.
## What does 1 to 5 odds pay?
Standard Win Bets and PayoutsOdds\$ Payout\$2 Payout1/1\$4.00\$10.006/5\$4.40\$11.007/5\$4.80\$12.003/2\$5.00\$14.00
## What do Odds 1/3 mean?
This means that out of 4 possible outcomes, odds are that there will be 1 of one kind of outcome and 3 of another kind of outcome. For every 4, odds are that 1 will be a particular event and 3 will be another event.
You might be interested: How can i get rid of a uti without antibiotics?
## Can you make money from horse racing?
If you bet \$2 to win on each of those horses in every race you will have bet a total of \$10. … You will need to win 2 out of 5 races to show a profit. By being right 40% of the time you can make money betting on 2:1 shots at the racetrack. Of course things don’t always line up that simply in the real world.
## What is it called when you pick 4 horses?
Trifecta — Pick three horses. If they finish 1st, 2nd and 3rd, in exact order, you win. Superfecta — Pick four horses. If they finish 1st, 2nd, 3rd and 4th, in exact order, you win.
## What are true odds?
When you hear someone use the term “true odds” they are referring to the actual odds of something happening as opposed to what a linemaker or sportsbook would offer. The “true odds” are a better indication of the actual probability of something happening.
## How do you calculate odds?
The answer is the total number of outcomes. Probability can be expressed as 9/30 = 3/10 = 30% – the number of favorable outcomes over the number of total possible outcomes. A simple formula for calculating odds from probability is O = P / (1 – P). A formula for calculating probability from odds is P = O / (O + 1).
## What are dropping odds?
Playing on dropping odds means that you bet on a bookmaker that has yet to adjust the odds according to market movements. If the majority of the market offers 2.00 for over 2.5 goals on a match, while a single bookmaker offers 2.40, then you can assume this is a good bet and buy it at 2.40.
You might be interested: FAQ: How rare can you eat steak?
## What does 80 to 1 odds pay?
For every 81, odds are that 80 will be a particular event and 1 will be another event. There is a 98.77 percent probability of a particular outcome and 1.23 percent probability of another outcome. If you bet 1 on a game with 80 to 1 odds and you win, your total payout will be 81.00 which is your bet plus 80.00 profit.
## What do the odds 100 1 mean?
10/1 means the chances against it happening are higher than for 2/1. 100/1 means the chances against it happening are much higher still. Odds of 1/2, 1/3, 1/4, 1/5 and so on upwards are chances or odds on it happening, which means the chances are that it will happen. The higher the chances against ,the more you win.7 мая 2007 г.
2 years ago | HuggingFaceTB/finemath | |
# 480 is 209 Percent of what?
## 480 is 209 Percent of 229.67
%
480 is 209% of 229.67
Calculation steps:
480 ÷ ( 209 ÷ 100 ) = 229.67
### Calculate 480 is 209 Percent of what?
• F
Formula
480 ÷ ( 209 ÷ 100 )
• 1
Percent to decimal
209 ÷ 100 = 2.09
• 2
480 ÷ 2.09 = 229.67 So 480 is 209% of 229.67
Example | HuggingFaceTB/finemath | |
# Graph coloring
(Redirected from Distributed graph coloring)
A proper vertex coloring of the Petersen graph with 3 colors, the minimum number possible.
In graph theory, graph coloring is a special case of graph labeling; it is an assignment of labels traditionally called "colors" to elements of a graph subject to certain constraints. In its simplest form, it is a way of coloring the vertices of a graph such that no two adjacent vertices share the same color; this is called a vertex coloring. Similarly, an edge coloring assigns a color to each edge so that no two adjacent edges share the same color, and a face coloring of a planar graph assigns a color to each face or region so that no two faces that share a boundary have the same color.
Vertex coloring is the starting point of the subject, and other coloring problems can be transformed into a vertex version. For example, an edge coloring of a graph is just a vertex coloring of its line graph, and a face coloring of a plane graph is just a vertex coloring of its dual. However, non-vertex coloring problems are often stated and studied as is. That is partly for perspective, and partly because some problems are best studied in non-vertex form, as for instance is edge coloring.
The convention of using colors originates from coloring the countries of a map, where each face is literally colored. This was generalized to coloring the faces of a graph embedded in the plane. By planar duality it became coloring the vertices, and in this form it generalizes to all graphs. In mathematical and computer representations, it is typical to use the first few positive or nonnegative integers as the "colors". In general, one can use any finite set as the "color set". The nature of the coloring problem depends on the number of colors but not on what they are.
Graph coloring enjoys many practical applications as well as theoretical challenges. Beside the classical types of problems, different limitations can also be set on the graph, or on the way a color is assigned, or even on the color itself. It has even reached popularity with the general public in the form of the popular number puzzle Sudoku. Graph coloring is still a very active field of research.
Note: Many terms used in this article are defined in Glossary of graph theory.
## History
The first results about graph coloring deal almost exclusively with planar graphs in the form of the coloring of maps. While trying to color a map of the counties of England, Francis Guthrie postulated the four color conjecture, noting that four colors were sufficient to color the map so that no regions sharing a common border received the same color. Guthrie’s brother passed on the question to his mathematics teacher Augustus de Morgan at University College, who mentioned it in a letter to William Hamilton in 1852. Arthur Cayley raised the problem at a meeting of the London Mathematical Society in 1879. The same year, Alfred Kempe published a paper that claimed to establish the result, and for a decade the four color problem was considered solved. For his accomplishment Kempe was elected a Fellow of the Royal Society and later President of the London Mathematical Society.[1]
In 1890, Heawood pointed out that Kempe’s argument was wrong. However, in that paper he proved the five color theorem, saying that every planar map can be colored with no more than five colors, using ideas of Kempe. In the following century, a vast amount of work and theories were developed to reduce the number of colors to four, until the four color theorem was finally proved in 1976 by Kenneth Appel and Wolfgang Haken. The proof went back to the ideas of Heawood and Kempe and largely disregarded the intervening developments.[2] The proof of the four color theorem is also noteworthy for being the first major computer-aided proof.
In 1912, George David Birkhoff introduced the chromatic polynomial to study the coloring problems, which was generalised to the Tutte polynomial by Tutte, important structures in algebraic graph theory. Kempe had already drawn attention to the general, non-planar case in 1879,[3] and many results on generalisations of planar graph coloring to surfaces of higher order followed in the early 20th century.
In 1960, Claude Berge formulated another conjecture about graph coloring, the strong perfect graph conjecture, originally motivated by an information-theoretic concept called the zero-error capacity of a graph introduced by Shannon. The conjecture remained unresolved for 40 years, until it was established as the celebrated strong perfect graph theorem by Chudnovsky, Robertson, Seymour, and Thomas in 2002.
Graph coloring has been studied as an algorithmic problem since the early 1970s: the chromatic number problem is one of Karp’s 21 NP-complete problems from 1972, and at approximately the same time various exponential-time algorithms were developed based on backtracking and on the deletion-contraction recurrence of Zykov (1949). One of the major applications of graph coloring, register allocation in compilers, was introduced in 1981.
## Definition and terminology
This graph can be 3-colored in 12 different ways.
### Vertex coloring
When used without any qualification, a coloring of a graph is almost always a proper vertex coloring, namely a labeling of the graph’s vertices with colors such that no two vertices sharing the same edge have the same color. Since a vertex with a loop (i.e. a connection directly back to itself) could never be properly colored, it is understood that graphs in this context are loopless.
The terminology of using colors for vertex labels goes back to map coloring. Labels like red and blue are only used when the number of colors is small, and normally it is understood that the labels are drawn from the integers {1, 2, 3, ...}.
A coloring using at most k colors is called a (proper) k-coloring. The smallest number of colors needed to color a graph G is called its chromatic number, and is often denoted χ(G). Sometimes γ(G) is used, since χ(G) is also used to denote the Euler characteristic of a graph. A graph that can be assigned a (proper) k-coloring is k-colorable, and it is k-chromatic if its chromatic number is exactly k. A subset of vertices assigned to the same color is called a color class, every such class forms an independent set. Thus, a k-coloring is the same as a partition of the vertex set into k independent sets, and the terms k-partite and k-colorable have the same meaning.
### Chromatic polynomial
All non-isomorphic graphs on 3 vertices and their chromatic polynomials. The empty graph E3 (red) admits a 1-coloring, the others admit no such colorings. The green graph admits 12 colorings with 3 colors.
The chromatic polynomial counts the number of ways a graph can be colored using no more than a given number of colors. For example, using three colors, the graph in the adjacent image can be colored in 12 ways. With only two colors, it cannot be colored at all. With four colors, it can be colored in 24 + 4⋅12 = 72 ways: using all four colors, there are 4! = 24 valid colorings (every assignment of four colors to any 4-vertex graph is a proper coloring); and for every choice of three of the four colors, there are 12 valid 3-colorings. So, for the graph in the example, a table of the number of valid colorings would start like this:
Available colors 1 2 3 4 … Number of colorings 0 0 12 72 …
The chromatic polynomial is a function P(Gt) that counts the number of t-colorings of G. As the name indicates, for a given G the function is indeed a polynomial in t. For the example graph, P(Gt) = t(t − 1)2(t − 2), and indeed P(G, 4) = 72.
The chromatic polynomial includes at least as much information about the colorability of G as does the chromatic number. Indeed, χ is the smallest positive integer that is not a root of the chromatic polynomial
${\displaystyle \chi (G)=\min\{k\,\colon \,P(G,k)>0\}.}$
Triangle K3 ${\displaystyle t(t-1)(t-2)}$ Complete graph Kn ${\displaystyle t(t-1)(t-2)\cdots (t-(n-1))}$ Tree with n vertices ${\displaystyle t(t-1)^{n-1}}$ Cycle Cn ${\displaystyle (t-1)^{n}+(-1)^{n}(t-1)}$ Petersen graph ${\displaystyle t(t-1)(t-2)(t^{7}-12t^{6}+67t^{5}-230t^{4}+529t^{3}-814t^{2}+775t-352)}$
### Edge coloring
An edge coloring of a graph is a proper coloring of the edges, meaning an assignment of colors to edges so that no vertex is incident to two edges of the same color. An edge coloring with k colors is called a k-edge-coloring and is equivalent to the problem of partitioning the edge set into k matchings. The smallest number of colors needed for an edge coloring of a graph G is the chromatic index, or edge chromatic number, χ′(G). A Tait coloring is a 3-edge coloring of a cubic graph. The four color theorem is equivalent to the assertion that every planar cubic bridgeless graph admits a Tait coloring.
### Total coloring
Total coloring is a type of coloring on the vertices and edges of a graph. When used without any qualification, a total coloring is always assumed to be proper in the sense that no adjacent vertices, no adjacent edges, and no edge and its end-vertices are assigned the same color. The total chromatic number χ″(G) of a graph G is the fewest colors needed in any total coloring of G.
### Unlabeled coloring
An unlabeled coloring of a graph is an orbit of a coloring under the action of the automorphism group of the graph. If we interpret a coloring of a graph on ${\displaystyle d}$ vertices as a vector in ${\displaystyle \mathbb {Z} ^{d}}$, the action of an automorphism is a permutation of the coefficients of the coloring. There are analogues of the chromatic polynomials which count the number of unlabeled colorings of a graph from a given finite color set.
## Properties
### Bounds on the chromatic number
Assigning distinct colors to distinct vertices always yields a proper coloring, so
${\displaystyle 1\leq \chi (G)\leq n.}$
The only graphs that can be 1-colored are edgeless graphs. A complete graph ${\displaystyle K_{n}}$ of n vertices requires ${\displaystyle \chi (K_{n})=n}$ colors. In an optimal coloring there must be at least one of the graph’s m edges between every pair of color classes, so
${\displaystyle \chi (G)(\chi (G)-1)\leq 2m.}$
If G contains a clique of size k, then at least k colors are needed to color that clique; in other words, the chromatic number is at least the clique number:
${\displaystyle \chi (G)\geq \omega (G).}$
For perfect graphs this bound is tight.
The 2-colorable graphs are exactly the bipartite graphs, including trees and forests. By the four color theorem, every planar graph can be 4-colored.
A greedy coloring shows that every graph can be colored with one more color than the maximum vertex degree,
${\displaystyle \chi (G)\leq \Delta (G)+1.}$
Complete graphs have ${\displaystyle \chi (G)=n}$ and ${\displaystyle \Delta (G)=n-1}$, and odd cycles have ${\displaystyle \chi (G)=3}$ and ${\displaystyle \Delta (G)=2}$, so for these graphs this bound is best possible. In all other cases, the bound can be slightly improved; Brooks’ theorem[4] states that
Brooks’ theorem: ${\displaystyle \chi (G)\leq \Delta (G)}$ for a connected, simple graph G, unless G is a complete graph or an odd cycle.
### Lower bounds on the chromatic number
Several lower bounds for the chromatic bounds have been discovered over the years:
Hoffman's bound: Let ${\displaystyle W}$ be a real symmetric matrix such that ${\displaystyle W_{i,j}=0}$ whenever ${\displaystyle (i,j)}$ is not an edge in ${\displaystyle G}$. Define ${\displaystyle \chi _{W}(G)=1-{\tfrac {\lambda _{\max }(W)}{\lambda _{\min }(W)}}}$, where ${\displaystyle \lambda _{\max }(W),\lambda _{\min }(W)}$ are the largest and smallest eigenvalues of ${\displaystyle W}$. Define ${\displaystyle \chi _{H}(G)=\max _{W}\chi _{W}(G)}$, with ${\displaystyle W}$ as above. Then:
${\displaystyle \chi _{H}(G)\leq \chi (G)}$.
Vector chromatic number: Let ${\displaystyle W}$ be a positive semi-definite matrix such that ${\displaystyle W_{i,j}\leq -{\tfrac {1}{k-1}}}$ whenever ${\displaystyle (i,j)}$ is an edge in ${\displaystyle G}$. Define ${\displaystyle \chi _{V}(G)}$ to be the least k for which such a matrix ${\displaystyle W}$ exists. Then
${\displaystyle \chi _{V}(G)\leq \chi (G)}$.
Lovász number: The Lovász number of a complementary graph, is also a lower bound on the chromatic number:
${\displaystyle \vartheta ({\bar {G}})\leq \chi (G)}$.
Fractional chromatic number: The Fractional chromatic number of a graph, is a lower bound on the chromatic number as well:
${\displaystyle \chi _{f}(G)\leq \chi (G)}$.
These bounds are ordered as follows:
${\displaystyle \chi _{H}(G)\leq \chi _{V}(G)\leq \vartheta ({\bar {G}})\leq \chi _{f}(G)\leq \chi (G)}$.
### Graphs with high chromatic number
Graphs with large cliques have a high chromatic number, but the opposite is not true. The Grötzsch graph is an example of a 4-chromatic graph without a triangle, and the example can be generalised to the Mycielskians.
Mycielski’s Theorem (Alexander Zykov 1949, Jan Mycielski 1955): There exist triangle-free graphs with arbitrarily high chromatic number.
From Brooks’s theorem, graphs with high chromatic number must have high maximum degree. Another local property that leads to high chromatic number is the presence of a large clique. But colorability is not an entirely local phenomenon: A graph with high girth looks locally like a tree, because all cycles are long, but its chromatic number need not be 2:
Theorem (Erdős): There exist graphs of arbitrarily high girth and chromatic number.
### Bounds on the chromatic index
An edge coloring of G is a vertex coloring of its line graph ${\displaystyle L(G)}$, and vice versa. Thus,
${\displaystyle \chi '(G)=\chi (L(G)).}$
There is a strong relationship between edge colorability and the graph’s maximum degree ${\displaystyle \Delta (G)}$. Since all edges incident to the same vertex need their own color, we have
${\displaystyle \chi '(G)\geq \Delta (G).}$
Moreover,
König’s theorem: ${\displaystyle \chi '(G)=\Delta (G)}$ if G is bipartite.
In general, the relationship is even stronger than what Brooks’s theorem gives for vertex coloring:
Vizing’s Theorem: A graph of maximal degree ${\displaystyle \Delta }$ has edge-chromatic number ${\displaystyle \Delta }$ or ${\displaystyle \Delta +1}$.
### Other properties
A graph has a k-coloring if and only if it has an acyclic orientation for which the longest path has length at most k; this is the Gallai–Hasse–Roy–Vitaver theorem (Nešetřil & Ossona de Mendez 2012).
For planar graphs, vertex colorings are essentially dual to nowhere-zero flows.
About infinite graphs, much less is known. The following are two of the few results about infinite graph coloring:
### Open problems
As stated above, ${\displaystyle \omega (G)\leq \chi (G)\leq \Delta (G)+1.}$ A conjecture of Reed from 1998 is that the value is essentially closer to the lower bound, ${\displaystyle \chi (G)\leq \lceil {\frac {\omega (G)+\Delta (G)+1}{2}}\rceil .}$
The chromatic number of the plane, where two points are adjacent if they have unit distance, is unknown, although it is one of 4, 5, 6, or 7. Other open problems concerning the chromatic number of graphs include the Hadwiger conjecture stating that every graph with chromatic number k has a complete graph on k vertices as a minor, the Erdős–Faber–Lovász conjecture bounding the chromatic number of unions of complete graphs that have at exactly one vertex in common to each pair, and the Albertson conjecture that among k-chromatic graphs the complete graphs are the ones with smallest crossing number.
When Birkhoff and Lewis introduced the chromatic polynomial in their attack on the four-color theorem, they conjectured that for planar graphs G, the polynomial ${\displaystyle P(G,t)}$ has no zeros in the region ${\displaystyle [4,\infty )}$. Although it is known that such a chromatic polynomial has no zeros in the region ${\displaystyle [5,\infty )}$ and that ${\displaystyle P(G,4)\neq 0}$, their conjecture is still unresolved. It also remains an unsolved problem to characterize graphs which have the same chromatic polynomial and to determine which polynomials are chromatic.
## Algorithms
Graph coloring
Decision
Name Graph coloring, vertex coloring, k-coloring
Input Graph G with n vertices. Integer k
Output Does G admit a proper vertex coloring with k colors?
Running time O(2nn)[5]
Complexity NP-complete
Reduction from 3-Satisfiability
Garey–Johnson GT4
Optimisation
Name Chromatic number
Input Graph G with n vertices.
Output χ(G)
Complexity NP-hard
Approximability O(n (log n)−3(log log n)2)
Inapproximability O(n1−ε) unless P = NP
Counting problem
Name Chromatic polynomial
Input Graph G with n vertices. Integer k
Output The number P (G,k) of proper k-colorings of G
Running time O(2nn)
Complexity #P-complete
Approximability FPRAS for restricted cases
Inapproximability No PTAS unless P = NP
### Polynomial time
Determining if a graph can be colored with 2 colors is equivalent to determining whether or not the graph is bipartite, and thus computable in linear time using breadth-first search or depth-first search. More generally, the chromatic number and a corresponding coloring of perfect graphs can be computed in polynomial time using semidefinite programming. Closed formulas for chromatic polynomial are known for many classes of graphs, such as forests, chordal graphs, cycles, wheels, and ladders, so these can be evaluated in polynomial time.
If the graph is planar and has low branch-width (or is nonplanar but with a known branch decomposition), then it can be solved in polynomial time using dynamic programming. In general, the time required is polynomial in the graph size, but exponential in the branch-width.
### Exact algorithms
Brute-force search for a k-coloring considers each of the ${\displaystyle k^{n}}$ assignments of k colors to n vertices and checks for each if it is legal. To compute the chromatic number and the chromatic polynomial, this procedure is used for every ${\displaystyle k=1,\ldots ,n-1}$, impractical for all but the smallest input graphs.
Using dynamic programming and a bound on the number of maximal independent sets, k-colorability can be decided in time and space ${\displaystyle O(2.445^{n})}$.[6] Using the principle of inclusion–exclusion and Yates’s algorithm for the fast zeta transform, k-colorability can be decided in time ${\displaystyle O(2^{n}n)}$[5] for any k. Faster algorithms are known for 3- and 4-colorability, which can be decided in time ${\displaystyle O(1.3289^{n})}$[7] and ${\displaystyle O(1.7272^{n})}$,[8] respectively.
### Contraction
The contraction ${\displaystyle G/uv}$ of a graph G is the graph obtained by identifying the vertices u and v, and removing any edges between them. The remaining edges originally incident to u or v are now incident to their identification. This operation plays a major role in the analysis of graph coloring.
The chromatic number satisfies the recurrence relation:
${\displaystyle \chi (G)={\text{min}}\{\chi (G+uv),\chi (G/uv)\}}$
due to Zykov (1949), where u and v are non-adjacent vertices, and ${\displaystyle G+uv}$ is the graph with the edge uv added. Several algorithms are based on evaluating this recurrence and the resulting computation tree is sometimes called a Zykov tree. The running time is based on a heuristic for choosing the vertices u and v.
The chromatic polynomial satisfies the following recurrence relation
${\displaystyle P(G-uv,k)=P(G/uv,k)+P(G,k)}$
where u and v are adjacent vertices, and ${\displaystyle G-uv}$ is the graph with the edge uv removed. ${\displaystyle P(G-uv,k)}$ represents the number of possible proper colorings of the graph, where the vertices may have the same or different colors. Then the proper colorings arise from two different graphs. To explain, if the vertices u and v have different colors, then we might as well consider a graph where u and v are adjacent. If u and v have the same colors, we might as well consider a graph where u and v are contracted. Tutte’s curiosity about which other graph properties satisfied this recurrence led him to discover a bivariate generalization of the chromatic polynomial, the Tutte polynomial.
These expressions give rise to a recursive procedure called the deletion–contraction algorithm, which forms the basis of many algorithms for graph coloring. The running time satisfies the same recurrence relation as the Fibonacci numbers, so in the worst case the algorithm runs in time within a polynomial factor of ${\displaystyle \left({\tfrac {1+{\sqrt {5}}}{2}}\right)^{n+m}=O(1.6180^{n+m})}$ for n vertices and m edges.[9] The analysis can be improved to within a polynomial factor of the number ${\displaystyle t(G)}$ of spanning trees of the input graph.[10] In practice, branch and bound strategies and graph isomorphism rejection are employed to avoid some recursive calls. The running time depends on the heuristic used to pick the vertex pair.
### Greedy coloring
Two greedy colorings of the same graph using different vertex orders. The right example generalizes to 2-colorable graphs with n vertices, where the greedy algorithm expends ${\displaystyle n/2}$ colors.
The greedy algorithm considers the vertices in a specific order ${\displaystyle v_{1}}$,…,${\displaystyle v_{n}}$ and assigns to ${\displaystyle v_{i}}$ the smallest available color not used by ${\displaystyle v_{i}}$’s neighbours among ${\displaystyle v_{1}}$,…,${\displaystyle v_{i-1}}$, adding a fresh color if needed. The quality of the resulting coloring depends on the chosen ordering. There exists an ordering that leads to a greedy coloring with the optimal number of ${\displaystyle \chi (G)}$ colors. On the other hand, greedy colorings can be arbitrarily bad; for example, the crown graph on n vertices can be 2-colored, but has an ordering that leads to a greedy coloring with ${\displaystyle n/2}$ colors.
For chordal graphs, and for special cases of chordal graphs such as interval graphs and indifference graphs, the greedy coloring algorithm can be used to find optimal colorings in polynomial time, by choosing the vertex ordering to be the reverse of a perfect elimination ordering for the graph. The perfectly orderable graphs generalize this property, but it is NP-hard to find a perfect ordering of these graphs.
If the vertices are ordered according to their degrees, the resulting greedy coloring uses at most ${\displaystyle {\text{max}}_{i}{\text{ min}}\{d(x_{i})+1,i\}}$ colors, at most one more than the graph’s maximum degree. This heuristic is sometimes called the Welsh–Powell algorithm.[11] Another heuristic due to Brélaz establishes the ordering dynamically while the algorithm proceeds, choosing next the vertex adjacent to the largest number of different colors.[12] Many other graph coloring heuristics are similarly based on greedy coloring for a specific static or dynamic strategy of ordering the vertices, these algorithms are sometimes called sequential coloring algorithms.
The maximum (worst) number of colors that can be obtained by the greedy algorithm, by using a vertex ordering chosen to maximize this number, is called the Grundy number of a graph.
### Parallel and distributed algorithms
In the field of distributed algorithms, graph coloring is closely related to the problem of symmetry breaking. The current state-of-the-art randomized algorithms are faster for sufficiently large maximum degree Δ than deterministic algorithms. The fastest randomized algorithms employ the multi-trials technique by Schneider et al.[13]
In a symmetric graph, a deterministic distributed algorithm cannot find a proper vertex coloring. Some auxiliary information is needed in order to break symmetry. A standard assumption is that initially each node has a unique identifier, for example, from the set {1, 2, ..., n}. Put otherwise, we assume that we are given an n-coloring. The challenge is to reduce the number of colors from n to, e.g., Δ + 1. The more colors are employed, e.g. O(Δ) instead of Δ + 1, the fewer communication rounds are required.[13]
A straightforward distributed version of the greedy algorithm for (Δ + 1)-coloring requires Θ(n) communication rounds in the worst case − information may need to be propagated from one side of the network to another side.
The simplest interesting case is an n-cycle. Richard Cole and Uzi Vishkin[14] show that there is a distributed algorithm that reduces the number of colors from n to O(log n) in one synchronous communication step. By iterating the same procedure, it is possible to obtain a 3-coloring of an n-cycle in O(log* n) communication steps (assuming that we have unique node identifiers).
The function log*, iterated logarithm, is an extremely slowly growing function, "almost constant". Hence the result by Cole and Vishkin raised the question of whether there is a constant-time distributed algorithm for 3-coloring an n-cycle. Linial (1992) showed that this is not possible: any deterministic distributed algorithm requires Ω(log* n) communication steps to reduce an n-coloring to a 3-coloring in an n-cycle.
The technique by Cole and Vishkin can be applied in arbitrary bounded-degree graphs as well; the running time is poly(Δ) + O(log* n).[15] The technique was extended to unit disk graphs by Schneider et al.[16] The fastest deterministic algorithms for (Δ + 1)-coloring for small Δ are due to Leonid Barenboim, Michael Elkin and Fabian Kuhn.[17] The algorithm by Barenboim et al. runs in time O(Δ) + log*(n)/2, which is optimal in terms of n since the constant factor 1/2 cannot be improved due to Linial's lower bound. Panconesi & Srinivasan (1996) use network decompositions to compute a Δ+1 coloring in time ${\displaystyle 2^{O\left({\sqrt {\log n}}\right)}}$.
The problem of edge coloring has also been studied in the distributed model. Panconesi & Rizzi (2001) achieve a (2Δ − 1)-coloring in O(Δ + log* n) time in this model. The lower bound for distributed vertex coloring due to Linial (1992) applies to the distributed edge coloring problem as well.
### Decentralized algorithms
Decentralized algorithms are ones where no message passing is allowed (in contrast to distributed algorithms where local message passing takes places), and efficient decentralized algorithms exist that will color a graph if a proper coloring exists. These assume that a vertex is able to sense whether any of its neighbors are using the same color as the vertex i.e., whether a local conflict exists. This is a mild assumption in many applications e.g. in wireless channel allocation it is usually reasonable to assume that a station will be able to detect whether other interfering transmitters are using the same channel (e.g. by measuring the SINR). This sensing information is sufficient to allow algorithms based on learning automata to find a proper graph coloring with probability one.[18]
### Computational complexity
Graph coloring is computationally hard. It is NP-complete to decide if a given graph admits a k-coloring for a given k except for the cases k ∈ {0,1,2} . In particular, it is NP-hard to compute the chromatic number.[19] The 3-coloring problem remains NP-complete even on 4-regular planar graphs.[20] However, k-coloring of a planar graph is in P, for every k>3, since every planar graph has a 4-coloring (and thus, also a k-coloring, for every k≥4); see Four color theorem.
The best known approximation algorithm computes a coloring of size at most within a factor O(n(log log n)2(log n)-3) of the chromatic number.[21] For all ε > 0, approximating the chromatic number within n1−ε is NP-hard.[22]
It is also NP-hard to color a 3-colorable graph with 4 colors[23] and a k-colorable graph with k(log k ) / 25 colors for sufficiently large constant k.[24]
Computing the coefficients of the chromatic polynomial is #P-hard. In fact, even computing the value of ${\displaystyle \chi (G,k)}$ is #P-hard at any rational point k except for k = 1 and k = 2.[25] There is no FPRAS for evaluating the chromatic polynomial at any rational point k ≥ 1.5 except for k = 2 unless NP = RP.[26]
For edge coloring, the proof of Vizing’s result gives an algorithm that uses at most Δ+1 colors. However, deciding between the two candidate values for the edge chromatic number is NP-complete.[27] In terms of approximation algorithms, Vizing’s algorithm shows that the edge chromatic number can be approximated to within 4/3, and the hardness result shows that no (4/3 − ε )-algorithm exists for any ε > 0 unless P = NP. These are among the oldest results in the literature of approximation algorithms, even though neither paper makes explicit use of that notion.[28]
## Applications
### Scheduling
Vertex coloring models to a number of scheduling problems.[29] In the cleanest form, a given set of jobs need to be assigned to time slots, each job requires one such slot. Jobs can be scheduled in any order, but pairs of jobs may be in conflict in the sense that they may not be assigned to the same time slot, for example because they both rely on a shared resource. The corresponding graph contains a vertex for every job and an edge for every conflicting pair of jobs. The chromatic number of the graph is exactly the minimum makespan, the optimal time to finish all jobs without conflicts.
Details of the scheduling problem define the structure of the graph. For example, when assigning aircraft to flights, the resulting conflict graph is an interval graph, so the coloring problem can be solved efficiently. In bandwidth allocation to radio stations, the resulting conflict graph is a unit disk graph, so the coloring problem is 3-approximable.
### Register allocation
A compiler is a computer program that translates one computer language into another. To improve the execution time of the resulting code, one of the techniques of compiler optimization is register allocation, where the most frequently used values of the compiled program are kept in the fast processor registers. Ideally, values are assigned to registers so that they can all reside in the registers when they are used.
The textbook approach to this problem is to model it as a graph coloring problem.[30] The compiler constructs an interference graph, where vertices are variables and an edge connects two vertices if they are needed at the same time. If the graph can be colored with k colors then any set of variables needed at the same time can be stored in at most k registers.
### Other applications
The problem of coloring a graph arises in many practical areas such as pattern matching, sports scheduling, designing seating plans, exam timetabling, the scheduling of taxis, and solving Sudoku puzzles.[31]
## Other colorings
### Ramsey theory
An important class of improper coloring problems is studied in Ramsey theory, where the graph’s edges are assigned to colors, and there is no restriction on the colors of incident edges. A simple example is the friendship theorem, which states that in any coloring of the edges of ${\displaystyle K_{6}}$, the complete graph of six vertices, there will be a monochromatic triangle; often illustrated by saying that any group of six people either has three mutual strangers or three mutual acquaintances. Ramsey theory is concerned with generalisations of this idea to seek regularity amid disorder, finding general conditions for the existence of monochromatic subgraphs with given structure.
### Other colorings
Coloring can also be considered for signed graphs and gain graphs. | HuggingFaceTB/finemath | |
# Differential Slope Field Help
• Mar 27th 2010, 08:32 AM
rawkstar
Differential Slope Field Help
Consider the differential equation dy/dx=xy^2
A) on the axes provided sketch a slope field for the given differential equation at the nine points indicated [(-1,-1) (-1,0) (-1,1) (0,-1) (0,0) (0,1) (1,-1) (1,0) (1,1)]
B)find the general solution of the given differential equation in terms of aconstant C
C)Find the particular solution of the differential condition that satisfies the initial condition y(0)=1
D) For what values of the constant C will the solutions of the differential equation have on or more vertical asymptotes? Justify your answers
• Mar 28th 2010, 09:28 AM
rawkstar
anybody? i figured out part A
• Mar 28th 2010, 09:39 AM
ione
Quote:
Originally Posted by rawkstar
Consider the differential equation dy/dx=xy^2
A) on the axes provided sketch a slope field for the given differential equation at the nine points indicated [(-1,-1) (-1,0) (-1,1) (0,-1) (0,0) (0,1) (1,-1) (1,0) (1,1)]
For the first point (-1, -1) find dy/dx
$\displaystyle \frac{dy}{dx}=(-1)(-1)^2=-1$
Through the point (-1, -1) sketch a short segment with slope -1
Do the same for each of the other points
• Mar 28th 2010, 09:52 AM
rawkstar
I'm just still confused with B C & D although I did get a little hint
• Mar 28th 2010, 10:08 AM
ione
Quote:
Originally Posted by rawkstar
Consider the differential equation dy/dx=xy^2
B)find the general solution of the given differential equation in terms of aconstant C
$\displaystyle \frac{dy}{dx}=xy^2$
$\displaystyle \frac{dy}{y^2}=x\,dx$
$\displaystyle -\frac{1}{y}=\frac{1}{2}x^2+C$
$\displaystyle y=-\frac{2}{x^2+2C}$
• Mar 28th 2010, 10:21 AM
rawkstar
Thank you, I'm pretty sure I can figure out C from that.
I'm not sure how to do D though, I know you get a vertical asymptote when the denominator equals 0 or is undefined. However I'm not sure to do that when you don't know x and theyre asking for values of C
• Mar 28th 2010, 01:29 PM
HallsofIvy
Quote:
Originally Posted by rawkstar
Consider the differential equation dy/dx=xy^2
A) on the axes provided sketch a slope field for the given differential equation at the nine points indicated [(-1,-1) (-1,0) (-1,1) (0,-1) (0,0) (0,1) (1,-1) (1,0) (1,1)]
B)find the general solution of the given differential equation in terms of aconstant C
$\displaystyle \frac{dy}{dx}= xy^2$ is equivalent to $\displaystyle \frac{dy}{y^2}= \frac{x dx}$. Integrate both sides.
Quote:
C)Find the particular solution of the differential condition that satisfies the initial condition y(0)=1
Your answer to B will, of course, include a "constant of integration. Set x= 0, y= 1 and determine what that constant is.
Quote:
D) For what values of the constant C will the solutions of the differential equation have on or more vertical asymptotes? Justify your answers
Presumably your solution to B will be of the form y= a rational function with both x and C in the denominator. For what values of C is there some value of x that makes the denominator 0?
Quote: | HuggingFaceTB/finemath | |
# Volume of a 80 Lb Bag of Concrete in Cubic Yard & Cubic Foot
0
Table of Contents
## Volume of a 80 Lb Bag of Concrete in Cubic Yard & Cubic Foot:
The volume of an 80-pound bag of concrete in cubic yards and cubic feet. 80-pound bag of concrete cubic feet | How many cubic feet is in an 80-pound bag of concrete How much is an 80-pound bag of concrete cover? How much does 80lb of concrete make?
The volume of 80 lb of concrete in cubic yards to cubic feet: Each 80-pound bag of concrete mix will produce approximately 0.6 cubic feet. For 4 inches of topping on 5,500 square feet, you’ll need 1833-1/3 cubic feet (67.9 cubic yards) of concrete or 3056 80-pound bags. One bag of 80-pound concrete mix is about .60 cu. Ft.
So, one cubic yard of concrete would require 45 bags. Because one cubic foot of concrete equals 133 pounds, 80 pounds equals 133/80 or 1.66 weights, and 80 pounds of concrete is about 1.66 bags per cubic foot.
also read: How Many Bags of Concrete Do I Need?
## The Volume of an 80-Pound Bag of Concrete in Cubic Feet:
Concrete weighs about 133 pounds per cubic foot, 3600 pounds per cubic yard, 17.826lb/gallon, 21.36kN/m3, 2.136kg/liter, 1.234 oz/inch3, or about two in the dry state. An 80-pound (lb) bag of concrete is a form of concrete premixed with cement, sand, and gravel. The weight of concrete is determined by its density, which varies depending on the amount of aggregate, water, and air in the mix.
Mixing each 80-pound (36.3-kg) bag of concrete mix yields nearly 0.60 cubic feet (17 liters). One cubic foot of concrete weighs about 133 pounds, so the volume of an 80-pound bag of concrete = 80/133 = 0.60 cubic feet.
Therefore, an 80lb bag of concrete books is about 0.60 cubic feet. An 80 lb bag of premixed concrete holds about 0.60 cubic feet, 0.022 cubic yards or 17 liters, or about 4.5 gallons or 0.0169 cubic meters.
also read: How Many Bag Mix Is a 3000 PSI Concrete?
## The Volume of an 80-Pound Bag of Concrete in Cubic Yards:
80lb is the most popular bag size of Quikrete premixed concrete. You will need 45 bags of concrete mix to make a cubic concrete yard. One cubic yard requires about 45 bags of 80-pound concrete.
An 80-pound bag of concrete is a premix of cement, sand, and gravel concrete. A cubic yard of concrete weighs about 3600 pounds, so the volume of an 80-pound bag of concrete = 80/3600 = 0.0222 cubic yards.
Thus 45 packs of 80 lb concrete are similar to one cubic yard. Therefore, an 80lb bag of concrete book is approximately 0.0222 cubic yards. And an 80-pound bag of concrete contains 0.022 cubic yards.
An 80-pound bag (36.3 kg) of substantial mix yields roughly 0.60 cubic feet (17 liters) of concrete. There are 27 cubic feet of concrete in a cubic yard. If you divide 27 by .60, that equals 45 bags.
A cubic yard of concrete weighs about 3600 pounds, so the number of cubic feet in an 80-pound bag of concrete = concrete = 80÷3600 = 0.022 cubic yards.
Thus, an 80-pound bag of concrete contains 0.022 cubic yards. Therefore, one of Quirete’s 80lb bags of concrete gives a volume of 0.022 cubic yards.
A cubic yard of concrete weighs approximately 3600 lb, so the number of bags of 80lb concrete in one cubic yard = 3600/80 = 45 bags. Therefore, you will need 45 packs of 80-pound concrete to make one cubic yard. You will need 45 80-pound bags of concrete mixture to make a yard.
But how much concrete can 80-pound bags cover in a square foot? It depends on how thick your bag is. For example,
• The 80-pound bag of concrete, two ″ thick, will cover 3.6 square feet.
• The 80-pound bag of concrete, four ″ thick, will cover 1.8 square feet.
• The 80-pound bag of concrete, six ″ thick, will cover 1.2 square feet.
A cubic yard of concrete weighs roughly 3600 lb, so the number of bags of 80lb concrete in one cubic yard = 3600/80 = 45 bags. You will need about 45 bags of 80-pound concrete for one cubic yard. So, you’ll need 45 packs of 80-pound concrete to create a cubic yard covering about 108 square feet to a standard depth of 3 inches for your slab and patio.
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Frequently Asked Questions (FAQ)
### 80 Lb Bag of Concrete
An 80 lb bag of concrete weighs 80 pounds and contains a pre-mixed combination of cement, sand, and gravel. It is commonly used for small construction and DIY projects like setting fence posts or building walkways.
### High Strength Concrete Mix
While traditional concrete has a tensile strength of 400 – 700 psi, UHPC has a tensile strength of about 1400 psi. While UHPC’s strength is impressive, it’s durability further exceeds expectations. Durability is measured by how the material performs under extreme conditions.
### How Many Cu Ft in 80 Lb Bag of Concrete?
One 80 lb. bag of Quikrete Concrete Mix will yield approximately . 60 cu. ft.
### 60 Lb Bag of Concrete
approximately 0.45 cubic feet
Each 60-pound bag provides approximately 0.45 cubic feet, so the column will require 10 60-pound bags. Yields are approximate and will vary with waste, uneven subgrades, etc.
### Quikrete High Strength Concrete Mix
QUIKRETE 5000 High Early Strength Concrete Mix can be used for any application requiring concrete in a minimum thickness of 2 in (50 mm), such as slabs, footings, steps, columns, walls, and patios.
### How Many Cu Ft Per 80 Lb Bag Concrete?
Each 80-pound bag of QUIKRETE® Concrete Mix yields approximately 0.6 cubic feet when mixed, so for 12 square feet at 4” thick, you will need 4 cubic feet of concrete, or 7 80-pound bags of concrete. Yields are approximate and will vary with waste, uneven subgrades, etc.
### 80 Lb Bag of Concrete Cubic Feet
An 80 lb bag of concrete typically yields approximately 0.6 cubic feet of volume when mixed with water.
### 80 Pound Bag of Concrete Volume
bag provides 0.45 cubic feet of cured concrete. A 80 lb. bag provides 0.6 cubic feet of cured concrete. If you have concrete delivered it is sold by the yard (which is a cubic yard = 27 cubic feet).
### 80 Lb Bag of Concrete Price
The price of an 80 lb bag of concrete can vary depending on several factors, including the brand, location, and current market conditions. On average, you can expect to pay around \$5 to \$8 per bag. However, prices may differ in different regions or during periods of high demand. It’s always a good idea to check with local suppliers or home improvement stores for the most accurate and up-to-date pricing information.
### Volume of 80 Lb Bag of Concrete
The volume of an 80 lb bag of concrete can vary depending on the specific mix and manufacturer. However, as a rough estimate, an 80 lb bag of concrete typically yields around 0.6 cubic feet of volume when mixed with water. It’s important to note that the volume may change slightly based on the water-to-concrete ratio and how well the mixture is compacted.
### 80 Lb Concrete Volume
The volume of an 80 lb bag of concrete is typically around 0.6 cubic feet when mixed with water.
### Quikrete 80 Lb Bag Volume
45 80
To produce one yard of concrete, you’ll need to use about 45 80-pound bags (or 90 40-pound bags).
### Cu Ft in 80 Lb Bag of Concrete
An 80 lb bag of concrete typically yields approximately 0.6 cubic feet (or 0.0222 cubic meters) of volume when mixed with water.
Hey, I Am Joyce lincoln the Man Behind Civilstep.com I Started This Site to Spread Knowledge About Civil Engineering. I Am a Degree Holder in Civil Engineering. | HuggingFaceTB/finemath | |
# Evaluate the line integral, where C is the given curve $$\int\limits_{C}xy\,ds$$. $$C:x=t^2,\,\,y=2t,\,\,0\leq t\leq 5$$.
This question aims to find the given line integral using the parametric equations of the curve $C$.
A line integral represents the integration of a function along a curve. It can also be regarded as a path integral, curvilinear integral, or curve integral.
The line integrals are the extension of simple integrals (which helps in finding areas of flat and two-dimensional surfaces) and can be used to find the areas of the surfaces that curve out into three dimensions. It is integral that integrates a function along a curve in the coordinate system.
The function to be integrated can be defined as either a scalar or a vector field. Along a curve, we can integrate both scalar and vector-valued functions. The vector line integral can be calculated by adding the values of all the points on the vector field.
Since, $ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt$
Therefore, $\dfrac{dx}{dt}=2t$ and $\dfrac{dy}{dt}=2$
So, $ds=\sqrt{(2t)^2+\left(2\right)^2}\,dt$
$=\sqrt{4t^2+4}\,dt$
$=2\sqrt{t^2+1}\,dt$
And $\int\limits_{C}xy\,ds$ $=\int\limits_{0}^{5}(t^2)(2t)(2\sqrt{t^2+1})\,dt$
$=4\int\limits_{0}^{5} t^3\sqrt{1+t^2}\,dt$
Or, $\int\limits_{C}xy\,ds=2\int\limits_{0}^{5} t^2\sqrt{1+t^2}\cdot 2t\,dt$
Applying integration by substitution, let:
$1+t^2=u\implies t^2=u-1$
and $du=2t\,dt$
Also, when $t=0$, $u=1$
and when $t=5$, $u=26$
Therefore, $\int\limits_{C}xy\,ds=2\int\limits_{1}^{26} (u-1)\sqrt{u}\,du$
$=2\int\limits_{1}^{26} (u^{3/2}-u^{1/2})\,du$
$=2\left[\dfrac{u^{5/2}}{5/2}-\dfrac{u^{3/2}}{3/2}\right]_{1}^{26}$
$=4\left[\dfrac{u^{5/2}}{5}-\dfrac{u^{3/2}}{3}\right]_{1}^{26}$
$=4\left[\dfrac{(26)^{5/2}-(1)^{5/2}}{5}-\dfrac{(26)^{3/2}-(1)^{3/2}}{3}\right]$
$=4\left[\dfrac{(26)^2\sqrt{26}-1}{5}-\dfrac{26\sqrt{26}-1}{3}\right]$
$=4\left[\dfrac{676\sqrt{26}}{5}-\dfrac{1}{5}-\dfrac{26\sqrt{26}}{3}+\dfrac{1}{3}\right]$
$=4\left[\dfrac{(2028-130)\sqrt{26}}{15}+\dfrac{5-3}{15}\right]$
$\int\limits_{C}xy\,ds=\dfrac{4}{15}[1898\sqrt{26}+2]$ Graph of the given curve along with its surface area
## Example 1
Determine the line integral $\int\limits_{C}\left(\dfrac{y}{1+x^2}\right)\,ds$, where $C$ is a curve given by the parametric equations: $x=t,\,y=2+t$ for $0\leq t\leq 1$.
### Solution
Since, $ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt$
Therefore, $\dfrac{dx}{dt}=1$ and $\dfrac{dy}{dt}=1$
So, $ds=\sqrt{(1)^2+\left(1\right)^2}\,dt$
$=\sqrt{1+1}\,dt$
$=\sqrt{2}\,dt$
And $\int\limits_{C}\left(\dfrac{y}{1+x^2}\right)\,ds$ $=\int\limits_{0}^{1}\left(\dfrac{2+t}{1+t^2}\right)(\sqrt{2})\,dt$
$=\sqrt{2}\int\limits_{0}^{1} \left(\dfrac{2}{1+t^2}+\dfrac{t}{1+t^2}\right)\,dt$
$=\sqrt{2}\left[\int\limits_{0}^{1} \dfrac{2}{1+t^2}\,dt+\int\limits_{0}^{1} \dfrac{t}{1+t^2}\,dt\right]$
$=\sqrt{2}\left[2\tan^{-1}(t)+\dfrac{\ln(1+t^2)}{2}\right]_{0}^{1}$
Applying the limits of integration as:
$=\sqrt{2}\left(2\tan^{-1}(1)+\dfrac{\ln(1+(1)^2)}{2}\right)-\sqrt{2}\left(2\tan^{-1}(0)+\dfrac{\ln(1+(0)^2)}{2}\right)$
$=\sqrt{2}\left(2\cdot \dfrac{\pi}{4}+\dfrac{\ln(2)}{2}\right)-\sqrt{2}\left(0+0\right)$
$=\sqrt{2}\left(\dfrac{\pi}{2}+\dfrac{\ln(2)}{2}\right)$
$=\sqrt{2}\left(\dfrac{\pi+\ln(2)}{2}\right)$
Or $\int\limits_{C}\left(\dfrac{y}{1+x^2}\right)\,ds$ $=\dfrac{\pi+\ln(2)}{\sqrt{2}}$
## Example 2
Work out the line integral $\int\limits_{C}xy\,ds$, where $C$ is a curve defined by the parametric equations: $x=\cos t,\,y=\sin t$ for $0\leq t\leq \pi$.
### Solution
Since, $ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt$
Therefore, $\dfrac{dx}{dt}=-\sin t$ and $\dfrac{dy}{dt}=\cos t$
So, $ds=\sqrt{(-\sin t)^2+\left(\cos t\right)^2}\,dt$
$=\sqrt{\sin^2t+\cos^2t}\,dt$
$=\sqrt{1}\,dt$
So, $ds=1\cdot dt$
And $\int\limits_{C}xy\,ds$ $=\int\limits_{0}^{\pi}(\cos t)(\sin t)(1)\,dt$
$=\int\limits_{0}^{\pi} \cos t\sin t\,dt$
$=\int\limits_{0}^{\pi} \sin t (\cos t\,dt)$
Now, by using the power rule:
$=\left[\dfrac{\sin^2 t}{2}\right]_{0}^{\pi}$
Applying the limits of integration as:
$=\left[\dfrac{\sin^2 (\pi)}{2}-\dfrac{\sin^2 (0)}{2}\right]$
$=\left[\dfrac{0}{2}-\dfrac{0}{2}\right]$
Or $\int\limits_{C}xy\,ds=0$
Images/mathematical drawings are created with GeoGebra. | HuggingFaceTB/finemath | |
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# Lecture 6: Statics - PowerPoint PPT Presentation
Lecture 6: Statics. Equilibrium of Particles Free-body Diagram Equilibrium of Rigid Bodies. Chapter 12 Static Equilibrium. Units of Chapter 12. The Conditions for Equilibrium Solving Statics Problems. 12-1 The Conditions for Equilibrium.
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### Lecture 6: Statics
Equilibrium of Particles
Free-body Diagram
Equilibrium of Rigid Bodies
Chapter 12Static Equilibrium
• The Conditions for Equilibrium
• Solving Statics Problems
An object with forces acting on it, but with zero net force, is said to be in equilibrium.
The first condition for equilibrium:
Example 12-1: Chandelier cord tension.p312
Calculate the tensions A and B in the two cords that are connected to the vertical cord supporting the 200-kg chandelier shown. Ignore the mass of the cords.
The second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary.
### Second Condition of Equilibrium
= 0
= r F
(r F) = 0
(r F sin ) = 0
### Tips to carry out cross products
Put the tail of vector r & vector F at the same point.
Cross the four fingers from vector r to vector F through the smaller angle between the two vectors.
The thumb will give the direction of the cross product.
To be in equilibrium the magnitude of the cross products pointing in opposite directions must be equal.
• Choose one object at a time, and make a free-body diagram by showing all the forces on it and where they act.
• Choose a coordinate system and the origin and resolve the components of the forces.
• Write equilibrium equations for the forces.
• Write the torque equilibrium equation. A clever choice of origin can simplify the problem enormously.
• Solve.
Example 12-3: Balancing a seesaw.
A board of mass M = 2.0 kg serves as a seesaw for two children. Child A has a mass of 30 kg and sits 2.5 m from the pivot point, P (his center of gravity is 2.5 m from the pivot). At what distance x from the pivot must child B, of mass 25 kg, place herself to balance the seesaw? Assume the board is uniform and centered over the pivot.
If a force in your solution comes out negative (asA will here), it just means that it’s in the opposite direction from the one you chose. This is trivial to fix, so don’t worry about getting all the signs of the forces right before you start solving.
Example 12-5: Hinged beam and cable. p316
A uniform beam, 2.20 m long with mass m = 25.0 kg, is mounted by a small hinge on a wall. The beam is held in a horizontal position by a cable that makes an angle θ= 30.0°. The beam supports a sign of mass M = 28.0 kg suspended from its end. Determine the components of the force H that the (smooth) hinge exerts on the beam, and the tension FT in the supporting cable.
Example 12-6: Ladder. p317
A 5.0-m-long ladder leans against a smooth wall at a point 4.0 m above a cement floor. The ladder is uniform and has mass m = 12.0 kg. Assuming the wall is frictionless (but the floor is not), determine the forces exerted on the ladder by the floor and by the wall.
• An object at rest is in equilibrium; the study of such objects is called statics.
• In order for an object to be in equilibrium, there must be no net force on it along any coordinate, and there must be no net torque around any axis. | HuggingFaceTB/finemath | |
EASIER THAN YOU THINK...
# Pascal’s Triangle This amazing triangle is named after Blaise Pascal (1623-1662), the French mathematician who studied it at length and published what is probably the first mathematical treatise devoted specifically to an analysis of it. His famous work, Traité du triangle arithmétique (Treatise on the Arithmetic Triangle) was published posthumously in 1665 … although it was likely written in late 1654 (some say 1653) during the time he was corresponding with Pierre de Fermat concerning early probability theory. You will discover that this triangle is very useful in analysing probabilities.
What we have come to call Pascal’s Triangle has a very long history, however. It was known at least 1500 years before Pascal and is referred to as the Khayyam triangle in Iran, Yang Hui’s Triangle in China, and Tartaglia’s Triangle in Italy.
If you wish to learn about a wide range of properties that this wonderful triangle has, then I suggest you visit the Maths is Fun website for a comprehensive summary. In time, I will be discussing these properties on this page. I spent many very happy hours studying this triangle during my high school years!
A Brief History of Pascal's Triangle
### India
The triangle, and the means of constructing it (by adding the two terms above) were known by Pingala about two thousand years ago. He wrote the Chandaḥśāstra (also known as the Chandaḥsūtra), which was a study of metre in Sanskrit poetry. His discussion of light (laghu) and heavy (guru) syllables led him to a discussion of binary arithmetic. This corresponds to the binomial theorem upon which Pascal's Triangle is based.
Centuries later, the Indian astronomer and mathematician, Varāhamihira (505–587) also discussed the triangle and its additive properties. About 300 years later, about 850, the Jain mathematician Mahāvīra produced a means of generating the binomial coefficients by using multiplication. There are claims that his method is equivalent to the modern binomial formula which uses factorial notation. And, during the tenth century, another Indian mathematician, Halayudha, provides us with a description of this triangle which he calls Meru-prastaara (the Staircase of Mount Meru). It is through his commentary on Pingala's Chandaḥśāstra that we know of Pingala's use of Pascal's Triangle.
### Persia (Iran)
The Persian mathematician, Al-Karaji (953–1029) wrote about Pascal's Triangle and the binomial theorem. A century later, the poet-astronomer-mathematician Omar Khayyám (1048–1131) also discussed Pascal's Triangle in his famous Treatise on Demonstration of Problems of Algebra (1070). Consequently, the triangle is referred to as the Khayyam triangle in Iran.
### China It is believed that the Chinese mathematician Jia Xian (1010–1070) invented Pascal's Triangle independently and he certainly used it as a tool for extracting square and cube roots.
About 200 years later, Yang Hui (1238–1298) discussed Jia Xian's method (for finding roots using the triangle) at length and explicitly acknowledged Jia Xian as the source.
It is still called Yang Hui's Triangle (and occasionally, Jai Xian's Triangle) in China today. The illustration at right is from a publication by another Chinese mathematician, Zhu Shijie in his famous work, Jade Mirror of the Four Unknowns (1303).
### Europe
The Jewish-French mathematician, astronomer and philosopher, Gersonides (Levi ben Gershon, 1288–1344) calculated the binomial coefficients using the multiplicative formula. [As an aside, he was also the first person to experimentally falsify the Ptolemaic model of the universe (by observing the brightness of Mars, among other things) and appears to have been the first to argue that the stars were billions of times further away than others had calculated.]
In 1527, the German mathematician, astronomer and cartographer, Petrus Apianus (1495–1552) published a copy of the triangle on the frontispiece of his handbook of commercial arithmetic, Ein newe und wolgegründete underweisung aller Kauffmanns Rechnung in dreyen Büchern, mit schönen Regeln und fragstücken begriffen. Not long afterwards, in his Arithmetica integra (1544), Michael Stifel (1487-1567), a German monk, published a portion of the triangle (from the second to the middle column in each row), describing it as a table of figurate numbers.
Only twelve years later (1556), the Italian algebraist Niccolò Fontana Tartaglia (1500–1577) published six rows of the triangle in General Trattato di numeri, et misure. In Italy, Pascal's triangle is referred to as Tartaglia's triangle.
In 1570, the Italian mathematician Gerolamo Cardano published the triangle and explained the additive and multiplicative rules for constructing it ...
... which brings us to Blaise Pascal and his work!
Around 1654, Blaise Pascal (1623-1662) was corresponding with his fellow French mathematician, Pierre de Fermat (1601-1665) concerning some matters relating to probability theory and its application to resolving some problems in gambling. Because of the triangle's usefulness in binomial theory and, therefore, in probability, he wrote a detailed analysis of it, Traité du triangle arithmétique (Treatise on the Arithmetic Triangle). It was published posthumously, in 1665. What made this work stand out from all preceding texts was that Pascal's was the first treatise in history that was completely devoted to analysing the triangle.
In 1708, Pierre Raymond de Montmort (1678-1719) called the triangle the "Table de M. Pascal pour les combinaisons" (Table of Mr Pascal for Combinations) and in 1730, Abraham de Moivre (1667-1754) referred to it as "Triangulum Arithmeticum Pascalianum" (Latin for "Pascal's Arithmetic Triangle").
Since those days it has been known as Pascal's Triangle in the West.
Pascal's Traité du triangle arithmétique
The Traité du triangle arithmétique was written by Blaise Pascal in 1654 around the time that he was corresponding with another French mathematician, Pierre de Fermat. They were discussing problems in calculating the odds in games of chance and, basically, inventing the new mathematical discipline of probability. He did not publish this seminal work, however. It was found among his notes and published posthumously, in 1665.
To give you some historical context for this period let me share that, during 1665, the Great Plague was raging in England and, since Cambridge University had been closed because of the plague, Isaac Newton was living at his home in Woolsthorpe and developing his calculus, theories concerning optics, and laws of gravitation and motion. It was also a year before the Great Fire of London (1666).
The Traité du triangle arithmétique is just 36 pages long. It is divided into two parts. The first is the Treatise on the Arithmetical Triangle, consisting of a description of Pascal's Triangle and its structure. The second part, Uses of the Arithmetical Triangle, contains further analysis of the Triangle and its applications to probability.
The second part has four sections:
1. theory of figurate numbers
2. theory of combinations
3. how to divide the stakes in games of chance
4. a study concerning the powers of binomial expressions.
If you would like to read an English translation of this work, you may thank Richard Pulskamp from the Department of Mathematics and Computer Science at Xavier University, Cincinnati, Ohio who made his translation available (on the Internet) in 2009.
Thank you for this video. In a perfect world you would have been my trigonometry teacher in 1967.
Timothy Thomas (on a CCM YouTube video about How to Understand and Memorise the Four Trigonometric Identities)
See all Testimonials | HuggingFaceTB/finemath | |
# Exam P Practice Problem 110 – likelihood of auto accidents
Problem 110-A
An actuary studied the likelihood of accidents in a one-year period among a large group of insured drivers. The following table gives the results.
Age Group Percent of Drivers Probability of 0 Accidents Probability of 1 Accident
16-20 15% 0.20 0.25
21-30 25% 0.35 0.40
31-50 35% 0.60 0.30
51-70 20% 0.67 0.23
71+ 5% 0.50 0.35
Suppose that a randomly selected insured driver in the studied group had at least 2 accidents in the past year. Calculate the probability that the insured driver is in the age group 21-30.
$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 7$
$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 4$
$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 3 \bold 0$
$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 0$
$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5$
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Problem 110-B
An auto insurance company performed a study on the frequency of accidents of its insured drivers in a one-year period. The following table gives the results of the study.
Age Group Percent of Drivers Probability of At Least 1 Accident
16-20 10% 0.30
21-40 20% 0.20
41-65 35% 0.10
66+ 35% 0.12
A randomly selected insured driver from the study was found to have no accidents in the one-year period.
Calculate the probability that the insured driver is from the age group 16-20.
$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 0 \bold 8$
$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 2$
$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 5$
$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 9$
$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 0$
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exam P practice problem
probability exam P
actuarial exam
actuarial practice problem
math
Daniel Ma
mathematics
dan ma actuarial science
daniel ma actuarial science
Daniel Ma actuarial
$\copyright$ 2019 – Dan Ma
# Exam P Practice Problem 80 – Total Insurance Payment
Problem 80-A
An individual purchases an insurance policy to cover a random loss. If a random loss occurs during the year, the amount of loss is at least 1. Once a random loss occurs, the insurance payment to the insured is modeled by the random variable $X$ with the following density function
$\displaystyle f(x)=\frac{1}{x^2} \ \ \ \ \ 1
If there is a loss, there is only one loss in each year. In each year, the probability of a loss is 0.25. What is the probability that the annual amount paid to the policyholder under this policy is less than 2?
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.250$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.500$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.750$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.875$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.925$
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Problem 80-B
An individual purchases an insurance policy to cover a random loss. If a random loss occurs during the year, the loss amount is at least 1. Once a loss occurs, the insurance payment to the insured is modeled by the random variable $X$ with the following density function
$\displaystyle f(x)=\frac{1}{30} \ x(1+3x) \ \ \ \ \ 1
If there is a loss, there is only one loss in each year. In each year, the probability of a loss is 0.15. What is the probability that the annual amount paid to the policyholder under this policy is less than 2?
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.1500$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.2833$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.8500$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.8735$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.8925$
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# Exam P Practice Problem 51 – Expected Claim Payment
Problem 51-A
The probability that a property will not be damaged in the upcoming year is 0.80. Assume that there is at most one incidence of damage in a year.
When there is a damage to the property, the amount of damage (in thousands) has an exponential distribution with mean 20.
The property owner insured the property against damage by purchasing an insurance policy with a deductible of 5.
What is the probability that the insurer’s payment to the owner will exceed 17.5?
$\displaystyle (A) \ \ \ \ \ \ \ \ \ 0.0649$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ 0.0834$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ 0.3249$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ 0.4169$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ 0.8340$
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Problem 51-B
The probability that a property will not be damaged in the upcoming year is 0.80. Assume that there is at most one incidence of damage in a year.
When there is a damage to the property, the amount of damage (in thousands) has an exponential distribution with mean 20.
The property owner insured the property against damage by purchasing an insurance policy with a deductible of 5.
What is the expected payment made by the insurer to the owner of the property?
$\displaystyle (A) \ \ \ \ \ \ \ \ \ 20.00$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ 4.00$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ 3.89$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ 3.50$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ 3.12$
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# Exam P Practice Problem 40 – Total Claim Amount
Problem 40-A
The number of claims in a calendar year for an insured has a probability function indicated below.
$\displaystyle \begin{bmatrix} \text{Number of Claims}&\text{ }&\text{Probability} \\\text{ }&\text{ }&\text{ } \\ 0&\text{ }&\displaystyle \frac{27}{64} \\\text{ }&\text{ }&\text{ } \\ 1&\text{ }&\displaystyle \frac{27}{64} \\\text{ }&\text{ }&\text{ } \\ 2&\text{ }&\displaystyle \frac{9}{64} \\\text{ }&\text{ }&\text{ } \\ 3&\text{ }&\displaystyle \frac{1}{64} \end{bmatrix}$
When a claim occurs, the claim amount $X$, regardless of how many claims the insured will have in the calendar year, has probabilities $P(X=1)=0.8$ and $P(X=2)=0.2$. The claim amounts in a calendar year for this insured are independent.
Let $T$ be the total claim amount for this insured in a calendar year. Calculate $P(3 \le T \le 4)$.
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Problem 40-B
A bowl has 3 red balls and 6 white balls. Select two balls at random from this bowl with replacement. Let $N$ be the number of red balls found in the two selected balls. When $N=n$ where $n>0$, roll a fair die $n$ times.
Let $W$ be the sum of the rolls of the die. Calculate $P(4 \le W \le 5)$.
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$\copyright \ 2013$
# Exam P Practice Problem 10
Problem 10a
An individual is facing an outcome of an annual financial loss $X$ (in tens of thousands of dollars) whose probability density function is given by
$\displaystyle f(x)=0.003 x^2, \ \ \ \ 0
The probability of a loss in the next year is 0.08. If there is a loss, there is only one loss in any given year. An insurance policy is available to protect against the financial loss by paying in full when a loss occurs.
1. What is the probability that the insurer’s payment to the insured will exceed 50,000? 2. What is the mean payment made by the insurer to the insured? 3. What is the variance of the amount of payment made by the insurer? $\text{ }$ Problem 10b Suppose that instead of buying a policy that pays the loss in full, the individual buys a policy that has a 80/20 coinsurance provision, i.e., the insurance company pays 80% of the loss and the insured retains the remaining 20% of a loss. Answer the same three questions. $\text{ }$ Solution is found below. $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ Solution to Problem 10a Answers $\displaystyle 10a.1 \ \ \ \ \ 0.07$ $\displaystyle 10a.2 \ \ \ \ \ 0.6$ $\displaystyle 10a.3 \ \ \ \ \ 4.44$ $\text{ }$ Let $X$ be the loss variable as described in the problem. Then the following is the probability $P(X>t)$. \displaystyle \begin{aligned}P(X>t)&=\int_t^{10} 0.003 x^2 \ dx \\&=1-0.001 t^3 \end{aligned} One important thing to keep in mind is that the occurrence of a financial loss is not certain. So the answer to question #1 is not $P(X>5)$. Let $Y$ be the insurance payment to the insured. Note that $Y$ is conditional on the occurrence of a loss. If the loss does not happen, $Y=0$. If the loss does happen, $Y=X$. Likewise, $P(Y>t)=0$ in case of no loss and $P(Y>t)=P(X>t)$ in case of a loss. So we can use the law of total probability to obtain $P(Y>5)$. \displaystyle \begin{aligned}P(Y>5)&=0 \times 0.92+P(X>5) \times 0.08 \\&=(1-0.001 5^3) \times 0.08 \\&=0.07 \end{aligned} The answers to the other two questions can also be obtained by using the law of total probability. \displaystyle \begin{aligned}E(Y)&=0 \times 0.92+\int_0^{10} x 0.003 x^2 \ dx \times 0.08 \\&=\int_0^{10} 0.003 x^3 \ dx \times 0.08 \\&=7.5 \times 0.08 \\&=0.6 \\&=\6000 \end{aligned} \displaystyle \begin{aligned}E(Y^2)&=0 \times 0.92+\int_0^{10} x^2 0.003 x^2 \ dx \times 0.08 \\&=\int_0^{10} 0.003 x^4 \ dx \times 0.08 \\&=60 \times 0.08 \\&=4.8 \end{aligned} $\displaystyle Var(Y)=4.8-0.6^2=4.44$ $\text{ }$ $\text{ }$ Answers to Problem 10b $\displaystyle 10b.1 \ \ \ \ \ 0.06046875$ $\displaystyle 10b.2 \ \ \ \ \ 0.48$ $\displaystyle 10b.3 \ \ \ \ \ 2.8416$ # Exam P Practice Problem 7 Problem 7a The probability that a property will not be damaged in the upcoming year is 0.80. When there is a damage to the property, the probability density function of the amount of the damage (in thousands of dollars) is given by $\displaystyle f(x)=0.05 e^{-0.05x} \ \ \ \ \ x>0$ The property owner purchased an insurance policy that pays the amount of the damage in full during the next year. 1. What is the probability that the insurer’s payment to the owner will exceed17,500?
2. What is the mean payment made by the insurer to the owner of the property?
3. What is the variance of the amount of payment made by the insurer?
Problem 7b
The probability that a property will not be damaged in the upcoming year is 0.80. When there is a damage to the property, the probability density function of the amount of the damage (in thousands of dollars) is given by
$\displaystyle f(x)=0.05 e^{-0.05x} \ \ \ \ \ x>0$
The property owner purchased an insurance policy with a coinsurance provision that pays 80% of the amount of the damage during the next year. The remaining 20% of the amount of the damage is retained by the property owner.
1. What is the probability that the insurer’s payment to the owner will exceed \$17,500?
2. What is the mean payment made by the insurer to the owner of the property?
3. What is the variance of the amount of payment made by the insurer?
Solution is found below.
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Solution to Problem 7a
Let $X$ be the loss amount (i.e. the pdf is the one given in the problem). One important thing to keep in mind is that a loss to the property is not certain. So the answer for #1 is not $P(X>17.5)$. Let $Y$ be the payment made by the insurer to the property owner. The answer for #1 is $P(Y>17.5)$. One way to look at the problem is through the law of total probability.
\displaystyle \begin{aligned}P(Y>17.5)&=P(Y>17.5 \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +P(Y>17.5 \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+e^{-0.05 (17.5)} \times 0.2 \\&=0.2 \times e^{-0.875} \\&=0.08337 \end{aligned}
The following provides the answers to the rest of the problem:
\displaystyle \begin{aligned}E(Y)&=E(Y \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +E(Y \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+\frac{1}{0.05} \times 0.2 \\&=4 \\&=\4000 \end{aligned}
\displaystyle \begin{aligned}E(Y^2)&=E(Y^2 \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +E(Y^2 \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+\frac{2}{0.05^2} \times 0.2 \\&=160 \end{aligned}
\displaystyle \begin{aligned}Var(Y)&=E(Y^2)-E(Y)^2 \\&=160-4^2 \\&=144 \end{aligned}
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$\displaystyle 7b.1 \ \ \ \ \ \ 0.2 e^{-1.09375}=0.0669916086$
$\displaystyle 7b.2 \ \ \ \ \ \ \3200$
$\displaystyle 7b.3 \ \ \ \ \ \ 92.16$
# Exam P Practice Problem 6
Problem 6a
An auto insurer offers collison coverage to two large groups of policyholders, Group 1 and Group 2. On the basis of historical data, the insurer has determined that the loss due to collision for a policyholder in Group 1 has an exponential distribution with mean 5. On the other hand, the loss due to collision for a policyholder in Group 2 has an exponential distribution with mean 10.
Considering the two groups as one block, about 75% of the losses are from Group 1.
1. Given a randomly selected loss in this block, what is the probability that the loss is greater than 15?
2. If a randomly selected loss is greater than 15, what is the probability that it is a from a policyholder in Group 1?
Problem 6b
An auto insurer has two groups of policyholders – those considered good risks and those considered bad risks. On the basis of historical data, the insurer has determined that the number of car accidents during a policy year for a policyholder classified as good risk follows a binomial distribution with $n=2$ and $p=\frac{1}{10}$. The number of car accidents for a policyholder classified as bad risk follows a binomial distribution with $n=2$ and $p=\frac{3}{10}$. In this block of policies, 75% are classified as good risks and 25% are classified as bad risks. A new customer, whose risk class is not yet known with certainty, has just purchased a new policy.
1. What is the probability that this new policyholder is not accident-free in the upcoming policy year?
2. By the end of the policy year, it is found that this policyholder is not accident-free, what is the probability that the policyholder is a “good risk” policyholder?
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Soultion is found below.
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Solution to Problem 6a
Let $X$ be the loss amount of a randomly selected policyholder. The conditional probabilities of a loss greater than 7.5 are:
$\displaystyle P(X>15 \lvert \text{ Group 1 Policyholder})=e^{-\frac{15}{5}}$
$\displaystyle P(X>15 \lvert \text{ Group 2 Policyholder})=e^{-\frac{15}{10}}$
By the law of total probability, the unconditional probability is:
\displaystyle \begin{aligned}P(X>15)&=P(X>15 \lvert \text{ Group 1 Policyholder}) \times P(\text{ Group 1 Policyholder}) \\&\ \ \ +P(X>15 \lvert \text{ Group 2 Policyholder}) \times P(\text{ Group 2 Policyholder}) \\&=\frac{3}{4} \times e^{-\frac{15}{5}}+\frac{1}{4} \times e^{-\frac{15}{10}} \\&=\frac{3}{4} \times e^{-3}+\frac{1}{4} \times e^{-1.5} \\&=0.0931228413 \end{aligned}
The above calculation indicates that the unconditional probability is the weighted average of the conditional probabilities. The answer to the second question is obtained by applying the Bayes’ theorem:
\displaystyle \begin{aligned}P(\text{Group 1 Policyholder } \lvert X>15)&=\frac{P[(\text{Group 1 Policyholder}) \cap (X>15)]}{P(X>15)} \\&=\frac{\frac{3}{4} \times e^{-3}}{\frac{3}{4} \times e^{-3}+\frac{1}{4} \times e^{-1.5}} \\&=0.400978973 \end{aligned}
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$\displaystyle 6b.1 \ \ \ \ \frac{108}{400}=0.27$
$\displaystyle 6b.2 \ \ \ \ \frac{57}{108}=0.5278$ | HuggingFaceTB/finemath | |
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A271382 Least k with precisely n partitions k = x + y satisfying d(k) = d(x) + d(y), where d(k) is the number of divisors of k. 2
2, 14, 10, 26, 44, 45, 126, 68, 99, 104, 162, 117, 98, 124, 232, 164, 148, 200, 260, 333, 231, 244, 248, 297, 273, 284, 315, 406, 332, 345, 385, 430, 344, 399, 388, 436, 429, 488, 465, 495, 472, 525, 561, 555, 621, 556, 632, 604, 652, 712, 536, 693, 735, 675 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Paolo P. Lava, Table of n, a(n) for n = 1..1200 EXAMPLE d(10) = d(1) + d(9) = d(3) + d(7) = d(5) + d(5) = 4 and 10 is the least number with 3 partitions of two numbers with this property: therefore a(3) = 10; d(126) = d(21) + d(105) = d(22) + d(104) = d(28) + d(98) = d(38) + d(33) = d(40) + d(86) = d(50) + d(76) = d(63) + d(63) = 12 and 126 is the least number with 7 partitions of two numbers with this property: therefore a(7) = 126. MAPLE with(numtheory): P:=proc(q) local a, h, k, n; for h from 1 to q do for n from 2*h to q do a:=0; for k from 1 to trunc(n/2) do if tau(n)=tau(k)+tau(n-k) then a:=a+1; fi; od; if a=h then print(n); break; fi; od; od; end: P(10^6); MATHEMATICA nn = 10^3; Table[SelectFirst[Range@ nn, Function[k, With[{e = DivisorSigma[0, k]}, Count[Transpose@ {Range[k - 1, Ceiling[k/2], -1], Range@ Floor[k/2]}, x_ /; Total@ DivisorSigma[0, x] == e] == n]]], {n, 54}] (* Michael De Vlieger, Apr 06 2016 *) PROG (PARI) isok(k, n) = {my(nb = 0, tau = numdiv(k)); for (j=1, k\2, if (numdiv(j)+numdiv(k-j) == tau, nb++); if (nb > n, return (0)); ); nb == n; } a(n) = {k=2; while (!isok(k, n), k++); k; } \\ Michel Marcus, Apr 07 2016 CROSSREFS Cf. A000005, A211224, A271384. Sequence in context: A191995 A168207 A082609 * A158787 A174706 A072163 Adjacent sequences: A271379 A271380 A271381 * A271383 A271384 A271385 KEYWORD nonn,easy AUTHOR Paolo P. Lava, Apr 06 2016 STATUS approved
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Two to the power of m plus n divide by two to the power of n
QUESTION: Simplify 2^m+n ÷ 2^n.
2^(m+n) ÷ 2^n = 2^{(m+n) - n} (subtract the indices)
2^(m+n) ÷ 2^n = 2^m
by Level 11 User (80.5k points) | HuggingFaceTB/finemath | |
# need my work checked plz :)
• Oct 6th 2008, 09:35 PM
mathlovet
need my work checked plz :)
u= ln(sqrt(x^2+y^2))
u(x) = x / (2sqrt((x^2+y^2))
u(xx) = (1-2x^2)/(2(sqrt(x^2+Y^2))^3)
u(y)= y / (2sqrt((x^2+y^2))
u(yy)= (1-2y^2)/(2(sqrt(x^2+Y^2))^3)
i havent done derivatives for a while now (Doh).. not sure it is correct..
• Oct 6th 2008, 10:11 PM
lllll
you could use some of the properties of logarithms for this one:
$\displaystyle u = \ln(\sqrt{x^2+y^2}) = \frac{1}{2}\ln(x^2+y^2)$
so then:
$\displaystyle u(x) = \frac{1}{2} \times \frac{1}{x^2+y^2} \times 2x = \frac{x}{x^2+y^2}$
$\displaystyle u(xx) = \frac{(1)(x^2+y^2) - 2x(x)}{(x^2+y^2)^2} = \frac{-x^2+y^2}{(x^2+y^2)^2}$
$\displaystyle u(y) = \frac{1}{2} \times \frac{1}{x^2+y^2} \times 2y = \frac{y}{x^2+y^2}$
$\displaystyle u(yy) = \frac{(1)(x^2+y^2) - 2y(y)}{(x^2+y^2)^2} = \frac{x^2-y^2}{(x^2+y^2)^2}$ | HuggingFaceTB/finemath | |
# area of a parallelogram with 4 vertices calculator 3d
Determine the location of the vertices. The online calculator below calculates the area of a rectangle, given coordinates of its vertices. b. Notify administrators if there is objectionable content in this page. If x and y are matrices of the same size, then polyarea returns a row vector containing the areas of each polygon defined by the columnwise pairs in x and y.. (Be sure to answer the 2 questions that "follow".) Parallelogram calculator computes all properties of a parallelogram such as area, perimeter, sides and angles given a sufficient subset of these properties. Area of triangle using vectors: Advanced Algebra: Jun 25, 2014: Calculate the area of the triangle using vectors: Geometry: Feb 12, 2013: parallelogram area with 3D vectors: Calculus: Feb 2, 2009: 3D Vectors Parallelogram Area: Calculus: May 25, 2008 5 years ago. The quadrilateral displayed in the applet below is a PARALLELOGRAM. We note that the area of a triangle defined by two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. The online calculator below calculates the area of a rectangle, given coordinates of its vertices. $\begingroup$ The specific location of the vertices doesn’t affect the area, but it does determine where the middle of the parallelogram lies. Triangle area calculator by points. Please help! What is a 3D parallelogram called? It was created by user request. So area of parallelogram is simply length X breadth. By using this website, you agree to our Cookie Policy. By using this website, you agree to our Cookie Policy. Calculations include side lengths, corner angles, diagonals, height, perimeter and area of parallelograms. Use determinants to calculate the area of the parallelogram with vertices (1, 1), (−4, 5), (−2, 8), and (3, 4). Then click Calculate. Calculations include side lengths, corner angles, diagonals, height, perimeter and area of parallelograms. View/set parent page (used for creating breadcrumbs and structured layout). Plot the points on a graph and draw lines such that you have a parallelogram. Also deduce the condition for collinearity of the. This time you still need a vertex at (0,0), but instead of numbers for the other vertices, use letters ie. Vector area of parallelogram = a vector x b vector Male or Female ? A parallelepiped has equal opposite faces and edges. This calculator determines the area of a triangle using its vertex coordinates in the cartesian coordinate system. Area, Parallelogram The quadrilateral displayed in the applet below is a PARALLELOGRAM. From the details to the question: > Given points P,Q,R w/position vectors p(1,4,1), q(3,1,2), r(3,8,7). Name the vertices as A(2,5), B(5,10) ,C(10,15) , D(7,10) find the area of the triangle ABC and area of the triangle CDA and add their areas, [ since the parallelogram is 4 sided figure it can be taken as sum of the areas of two trianlges] Interactive Parallelogram. 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First, recall Lagrange's Identity: We can instantly make a substitution into Lagrange's formula as we have a convenient substitution for the dot product, that is $\vec{u} \cdot \vec{v} = \| \vec{u} \| \| \vec{v} \| \cos \theta$. How do you find the area of the parallelogram with vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3)? Triangle area calculator by points. (It will always remain a parallelogram, regardless of how much you choose to move one or more of its vertices around.) In fact, the calculation is quite generic, so it can also calculate the area of parallelogram, square, rhombus, trapezoid, kite, etc. The Area of a Triangle in 3-Space. Calculator solve the triangle specified by coordinates of three vertices in the plane (or in 3D space). Find the area of the triangle whose vertices are A(3, - 1, 2), B(1, - 1, - 3) and C(4, - 3, 1). Parallelogram. (It will always remain a parallelogram, regardless of how much you choose to move one or more of its vertices around.) Parallelogram Area Calculator. The online calculator below calculates the area of a rectangle, given coordinates of its vertices. We note that the area of a triangle defined by two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. Calculator solve the triangle specified by coordinates of three vertices in the plane (or in 3D space). Parallelogram Calculator. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. Example: Consider the parallelogram with vertices (0,0) (7,2) (5,9) (12,11) Area of Parallelogram = b×h Square units = 4 × 5 = 20 sq.cm. A parallelogram in three dimensions is found using the cross product. Parallelepiped Calculator. Learning about vertices and volumes [7] 2020/02/25 13:39 Male / Under 20 years old / - / Useful / Purpose of ... To improve this 'Volume of a tetrahedron and a parallelepiped Calculator', please fill in questionnaire. The shoelace formula or shoelace algorithm (also known as Gauss's area formula and the surveyor's formula) is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by their Cartesian coordinates in the plane. Area of a parallelogram = b × h = 8 × 11 cm 2 = 88 cm 2. Enter the three edge lengths and the three angles at one of the vertices and choose the number of decimal places. $\endgroup$ – amd Oct 31 '17 at 23:32 add a comment | So get difference of pairs of coordinates with one common coordinate. Change the name (also URL address, possibly the category) of the page. Arean av en parallellogram är lika med en sidas längd multiplicerat med det. Given two vectors $\vec{u} = (u_1, u_2, u_3)$ and $\vec{v} = (v_1, v_2, v_3)$, if we place $\vec{u}$ and $\vec{v}$ so that their initial points coincide, then a parallelogram is formed as illustrated: Calculating the area of this parallelogram in 3-space can be done with the formula $A= \| \vec{u} \| \| \vec{v} \| \sin \theta$. Learn about its shape, sides, angles and properties. Let the coordinates of fourth vertex be D (x, y) In a parallelogram, diagonals bisect each other. h = 11 cm. Solution: Calculate vector by initial and terminal points. Question 3 : Examine whether the given points. Hanna Ghorbani. If x and y are vectors of the same length, then polyarea returns the scalar area of the polygon defined by x and y.. the point B will have coordinates (a,b) and the … Thus we can give the area of a triangle with the following formula: The Areas of Parallelograms and Triangles in 3-Space, \begin{align} A = \| \vec{u} \| \| \vec{v} \| \sin \theta \\ \blacksquare \end{align}, \begin{align} \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}\|^2 \|\vec{v} \|^2 - (\vec{u} \cdot \vec{v})^2 \end{align}, \begin{align} \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}\|^2 \|\vec{v} \|^2 - (\| \vec{u} \| \| \vec{v} \| \cos\theta)^2 \\ \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}\|^2 \|\vec{v} \|^2 - \| \vec{u} \|^2 \| \vec{v} \|^2 \cos^2\theta \\ \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}\|^2 \|\vec{v} \|^2 (1 - \cos^2\theta) \\ \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}||^2 \|\vec{v} \|^2 \sin^2\theta \end{align}, \begin{align} \| \vec{u} \times \vec{v} \| = \|\vec{u}\| \|\vec{v}\| \sin \theta \end{align}, \begin{align} \: A = \frac{1}{2} \| \vec{u} \times \vec{v} \| = \frac{1}{2} \|\vec{u}\| \|\vec{v}\| \sin \theta \end{align}, Unless otherwise stated, the content of this page is licensed under. a. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. A parallelogram or rhomboid is a quadrilateral with parallel opposite sides of the same length and opposite angles of the same size. General Wikidot.com documentation and help section. ASAP i need to know what a parallelogram is called when its 3D. $A= \| \vec{u} \| \| \vec{v} \| \sin \theta$, $\mathrm{Area} = \| \vec{u} \| \| \vec{v} \| \sin \theta$, $\sin \theta = \frac{opposite}{hypotenuse}$, $\sin \theta = \frac{height}{\| \vec{u} \| }$, The Relationship of the Area of a Parallelogram to the Cross Product, $\vec{u} \cdot \vec{v} = \| \vec{u} \| \| \vec{v} \| \cos \theta$, $A = \| \vec{u} \times \vec{v} \| = \| \vec{u} \| \| \vec{v} \| \sin \theta$, $\mathrm{Area} = \frac{1}{2} \| \vec{u} \| \| \vec{v} \| \sin \theta$, Creative Commons Attribution-ShareAlike 3.0 License, Making appropriate substitutions, we see that the base of the parallelogram is the length of. Is equal to the determinant of your matrix squared. So, the given vertices will form a parallelogram. Click the "Customize" button above to learn more! In coordinate geometry, a parallelogram is similar to an ordinary parallelogram (See parallelogram definition ) with the addition that its position on the coordinate plane is known. Therefore area = LxB = 6x6 = 36 The shoelace formula or shoelace algorithm (also known as Gauss's area formula and the surveyor's formula) is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by their Cartesian coordinates in the plane. To find the area of a pallelogram-shaped surface requires information about its base and height. Click here to edit contents of this page. The task is simple - first, determine lengths of edges, then use the Heron formula to find the triangle area. View wiki source for this page without editing. Wikidot.com Terms of Service - what you can, what you should not etc. Parallelogram wiki article Watch headings for an "edit" link when available. Since the length/norm of a vector will always be positive and that $\sin \theta > 0$ for $0 ≤ \theta < \pi$, it follows that all parts under the square root are positive, therefore: Note that this is the same formula as the area of a parallelogram in 3-space, and thus it follows that $A = \| \vec{u} \times \vec{v} \| = \| \vec{u} \| \| \vec{v} \| \sin \theta$. Magnitude of the vector product of the vectors equals to the area of the parallelogram, build on corresponding vectors: Therefore, to calculate the area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. that is, the area of any convex quadrilateral. The area of a parallelogram determined by the vectors and is the length of the cross-product, that is, the area of a parallelogram is. So, the area of the given triangle is (1/2) â165 square units. Example: find the area of a parallelogram. are position vectors of the vertices A, B, C of a triangle ABC, show that the area of, . Say point A and B; they have 1 in common, so the length would be 4-(-2) = 4+2 = 6. Negate point A to get (-4, -2) and add to the other two points B and C. Add x's and y's so you have a new point (4, 2) (5, 4) Now use determinant to find the area. $\endgroup$ – amd Oct 31 '17 at 23:32 add a comment | Parallelogram is a polygon with four vertices (corners) and four edges (sides) that form two parallel pairs. Also deduce the condition for collinearity of the points A, B, and C. Area of triangle ABC = (1/2) |AB vector x AC vector|, = (1/2) |(b x c - b x a - a x c + a x a)|, = (1/2) |(b x c + a x b + c x a + 0 vector)|, If the points A, B and C are collinear, then. A parallelogram is a quadrilateral with opposite sides parallel. Solution : Length of AB : Here x 1 = -4, y 1 = -3, x 2 = 3 and y 2 = 1 = √(3-(-4)) 2 +(1-(-3)) 2 = √(3+4) 2 +(1+3) 2 = √7 2 +4 … In fact, the calculation is quite generic, so it can also calculate the area of parallelogram, square, rhombus, trapezoid, kite, etc. Calculations at a parallelepiped. Then click Calculate. Solution: Given, b = 8 cm. Each of the four vertices (corners) have known coordinates.From these coordinates, various … that is, the area of any convex quadrilateral. Solution : Let a vector = i vector + 2j vector + 3k vector. 0 0. Calculate the areas of each of these individually and add the results. In earlier classes, we have studied that the area of a triangle whose vertices are (x 1, y 1), (x 2, y 2) and (x 3, y 3), is given by the expression $$\frac{1}{2} [x 1 (y 2 –y 3) + x 2 (y 3 –y 1) + x 3 (y 1 –y 2)]$$.Now this expression can be written in … The point P = (0, 2, 3). Question: Find The Area Of The Parallelogram With Vertices:P(0,0,0), Q(-5,0,4), R(-5,1,2), S(-10,1,6). Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector − 2j vector + k vector. If you want to discuss contents of this page - this is the easiest way to do it. S_(ABCD) = 15*sqrt(6)~=36.742 First we should verify that the 4 points really are coplanar. The formula for area of a parallelogram is A = bh, where b is the base length and h is the height. Question 2: Find the area of a parallelogram whose breadth is 8 cm and height is 11 cm. In geometry, a parallelogram is a special type of the quadrilateral that has four vertices and the opposite sides are equal and parallel. Calculations at a parallelogram. The base of the parallelogram with vertices (-4, 2), (1, 6), (15, 6), and (10, 2) is 14 units, and the height is 4 units (see attachment). Vector area of parallelogram = a vector x b vector. Let's plug in our numbers and solve for the area. This will give you the area. A (-4, -3) and B (3, 1) and C (3, 6) and D (-4, 2) forms a parallelogram. How do I find the area of a parallelogram with the given vertices K (1, 3, 2) L (1, 4, 4) M (4, 9, 4) N (4, 8, 2) I have to use the cross product, but I'm pretty lost. Homework Statement Find the area of the parallelogram with vertices: P(0,0,0), Q(-3,0,-1), R(-3,1,0), S(-6,1,-1) Homework Equations A=BH The Attempt at a Solution I think I know why this is incorrect, but i dont know what else to try. For Heron formula, see Calculator of area … As we will soon see, the area of a parallelogram formed from two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$ can be seen as a geometric representation of the cross product $\vec{u} \times \vec{v}$. Answer to Calculate the area of the parallelogram with the given vertices. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how find area of triangle formed by vectors. The lengths of the four sides and two diagonals can be found by using the method described inDistance between two pointsto find the distance between point pairs. Or if you take the square root of both sides, you get the area is equal to the absolute value of the determinant of A. It was created by user request. Let a vector = i vector + 2j vector + 3k vector. Length of opposite sides are equal. If you want to customize the colors, size, and more to better fit your site, then pricing starts at just $29.99 for a one time purchase. 16-10 = … if you need any other stuff in math, please use our google custom search here. Linear Algebra Example Problems - Area Of A Parallelogram Also verify that the determinant approach to computing area yield the same answer obtained using "conventional" area computations. 1 Answer Massimiliano Feb 4, 2015 The answer is: #A=sqrt265#. 4 + x = 7 and and 3 + y = 8 and y = 5 Therefore, the fourth vertex, D is (3, 5). If you look at it, you can see that you should be able to create a square in the middle of the parallelogram, plus two right triangles on opposite side of the square. Calculate the perimeter and area of the quadrilateral formed by the points (0,0) and (5,10) and (10,15) and (5,5) Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Therefore, the area of a parallelogram = 20 cm 2. Area of a triangle (Heron's formula - given lengths of the three sides) Area of a triangle (By formula, given coordinates of vertices) Area of a triangle (Box method, given coordinates of vertices) Limitations The calculator will produce the wrong answer for crossed polygons, … Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how find area of parallelogram formed by vectors. Append content without editing the whole page source. For Heron formula, see Calculator of area … Something does not work as expected? Calculate certain variables of a parallelogram depending on the inputs provided. As you do, observe what happens. This free online calculator help you to find area of triangle formed by vectors. The user cross-multiplies corresponding coordinates to find the area encompassing the polygon, and subtracts it from the surrounding polygon … b vector = 3i vector − 2j vector + k vector. Check out how this page has evolved in the past. Learn how to find the area of a triangle when vectors in the form of (xi+yj+zk) of two adjacent sides are given along with solved examples. The parallelogram area calculator exactly as you see it above is 100% free for you to use. Uses Heron's formula and trigonometric functions to calculate area and other properties of a given triangle. The user cross-multiplies corresponding coordinates to find the area encompassing the polygon, and subtracts it from the surrounding polygon … For example, in the figure above click 'reset' and select "show diagonals' in the options menu.Using the method in Distance between two points, the diagonal AC isthe distance between the points A and C:AC=√(48−6)2+(26−7)2=46.1 Similarly the side AB can be found using the coordinates of the points A and B:AB=√(18−6)2+(26−7)2=22.5 Interactive Parallelogram. So the area of your parallelogram squared is equal to the determinant of the matrix whose column vectors construct that parallelogram. Can someone please help me? A parallelogram is formed in R3 (3-space/3D) by the vectors PA = (3, 2, –3) and PB = (4, 1, 5). b vector = 3i vector â 2j vector + k vector. The task is simple - first, determine lengths of edges, then use the Heron formula to find the triangle area. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. In Euclidean geometry, a parallelogram is a simple quadrilateral with two pairs of parallel sides. Area of a triangle (Heron's formula - given lengths of the three sides) Area of a triangle (By formula, given coordinates of vertices) Area of a triangle (Box method, given coordinates of vertices) Limitations The calculator will produce the wrong answer for crossed polygons, … View and manage file attachments for this page. 3D design 3D Parallelogram created by Cashew Man with Tinkerca . Find out what you can do. So your area-- this is exciting! c. Determine the length of the diagonals. AC vector = i vector - 2j vector - k vector, Area of triangle = (1/2) |AB vector x AC vector|. Male Female Age ... New coordinates by 3D rotation of points. See pages that link to and include this page. Determine the vectors representing the diagonals. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Consider the following vertices of a parallelogram: and.. Learn about its shape, sides, angles and properties.$\begingroup$The specific location of the vertices doesn’t affect the area, but it does determine where the middle of the parallelogram lies. Thus we can give the area of a triangle with the following formula: (5) If x and y are multidimensional arrays, then polyarea operates along the first dimension whose length is not equal to 1. A parallelogram is a quadrilateral with opposite sides parallel. Uses Heron's formula and trigonometric functions to calculate area and other properties of a given triangle. If a vector, b vector, c vector are position vectors of the vertices A, B, C of a triangle ABC, show that the area of the triangle ABC is (1/2) |a à b + b à c + c à a| vector. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. Given three of vertices of a parallelogram are A(1,2), B (4,3), C (6,6). A(3, - 1, 2), B(1, - 1, - 3) and C(4, - 3, 1). This problem has been solved! Once you have investigated the areas of the parallelograms above, take some graph paper and draw a new parallelogram. Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector − 2j vector + k vector. Enter the two side lengths and one angle and choose the number of decimal places. Area of Triangle Formula Using Determinants. Source(s): 3d parallelogram called: https://biturl.im/oeIbt. This calculator determines the area of a triangle using its vertex coordinates in the cartesian coordinate system. We will now begin to prove this. Please enter angles in … d. Find the area of the parallelogram. Calculate certain variables of a parallelogram depending on the inputs provided. It does not matter which side you take as base, as long as the height you use it perpendicular to it. This free online calculator help you to find area of parallelogram formed by vectors. Move the slider up top all the way to the right. Calculator Use. The objective is to find the area of this parallelogram. Point C and D; -2 is common; so 3-(-3) = 3+3 = 6. Online Calculator Area Of Parallelogram Formed By Vectors » Area Of Parallelogram Calculator Vertices Using vector values derived from the vertices, the product of a parallelogram's base and height is equal to the cross product of two of its adjacent sides. Click here to toggle editing of individual sections of the page (if possible). b vector = 3i vector − 2j vector + k vector. Making this substitution and the substitution that$\cos ^ \theta = 1 - \sin^2 \theta\$ we get that: The last step is to square root both sides of this equation. ... A C and the angle between the two sides are given by θ then the area of the parallelogram will be given by ... -5, 4) and C (3, 1, -4). Solution : Let a vector = i vector + 2j vector + 3k vector. Vector area of parallelogram = a vector x b vector See the answer Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector â 2j vector + k vector. Calculus Introduction to Integration Integration: the Area Problem. In this section, you will learn how to find the area of parallelogram formed by vectors. Has four vertices ( corners ) and four edges ( sides ) that form two parallel pairs quadrilateral. Or more of its vertices around. rhomboid is a special type the. Customize '' button above to learn more displayed in the plane ( or in 3D space ) med... Triangle formula using Determinants vector + 2j vector + 2j vector + k vector 3- ( )! Same size of coordinates with one common coordinate that is, the area of a triangle in.! Fourth vertex Be D ( x, y ) in a parallelogram is called when 3D! A given triangle is ( 1/2 ) â165 square units what you can, what should. Use our google custom search here that follow ''. by Cashew Man with Tinkerca ) that form parallel. View/Set parent page ( if possible ) numbers for the other vertices, use letters ie and properties a triangle... Points on a graph and draw lines such that you have a parallelogram as... Vector â 2j vector + k vector, area of your parallelogram squared is equal to determinant. Possible ) other stuff in math, please use our google custom search here any other stuff in,... # A=sqrt265 # link to and include this page formed by vectors with... Lengths, corner angles, diagonals, height, perimeter and area of parallelogram = 20 cm 2 regardless... = 20 cm 2 = 88 cm 2 common ; so 3- -3! Can, what you should not etc breadth is 8 cm and height such that you have investigated the of... Is found using the cross product three edge lengths and the opposite angles of a parallelogram a. Vector − 2j vector + k vector first dimension whose length is not equal to the determinant of same. Discuss contents of this page a new parallelogram triangle ABC, show that the area of a triangle!, what you can, what you can, what you should not etc height you use it perpendicular it! Parallelogram calculator computes all properties of a rectangle, given coordinates of its vertices,. Vector = 3i vector − 2j vector + 3k vector fourth vertex Be D x... Side lengths, corner angles, diagonals, height, perimeter and of... And the three angles at one of the given triangle is ( 1/2 area of a parallelogram with 4 vertices calculator 3d â165 square units investigated the of... -2 is common ; so 3- ( -3 ) = 3+3 = 6 choose. That the area of any convex quadrilateral ( also URL address, the! Quadrilateral that has four vertices and the opposite angles of the same.. It perpendicular to it is, the given vertices vectors of the vertices... Vertices around. coordinates of three vertices in the past 3 ) squared... Is exciting calculator of area … a parallelogram are of equal measure difference of pairs coordinates!, C of a parallelogram fourth vertex Be D ( x, y ) in parallelogram... '' button above to learn more opposite or facing sides of the page used! The vertices and choose the number of decimal places watch headings for an ''... Quadrilateral with opposite sides parallel trigonometric functions to calculate the areas of each of these individually and add the.. The easiest way to do it with one common coordinate a pallelogram-shaped surface requires information about its base height! … area of parallelogram = a vector x ac vector| = 88 cm 2 cm and is. Website, you agree to our Cookie Policy contents of this parallelogram structured layout.... The Heron formula to find the area of a parallelogram are a ( 1,2 ), b ( 4,3,...: calculate vector by initial and terminal points and opposite angles of a parallelogram, regardless how! Found using the cross product so your area -- this is exciting have parallelogram. The applet below is a polygon with four vertices ( corners ) and edges!: calculate vector by initial and terminal points Heron formula, see calculator of area a... Always remain a parallelogram is a quadrilateral with parallel opposite sides parallel other vertices use... By Cashew Man with Tinkerca content in this section, you will learn how to find the area decimal! = i vector + k vector, area of parallelogram = a =... Vertex coordinates in the plane ( or in 3D space ) parent page ( possible. This section, you agree to our Cookie Policy a new parallelogram calculator determines the of! Common ; so 3- ( -3 ) = 3+3 = 6 of each of the.! The coordinates of its vertices around. equal to the determinant of given! Let a vector = 3i vector − 2j vector - k vector you... The results click here to toggle area of a parallelogram with 4 vertices calculator 3d of individual sections of the a! Is objectionable content in this page has evolved in the applet below is a special type of vertices! Show that the area parallelogram squared is equal to the determinant of the quadrilateral that has four vertices choose... The area of a parallelogram are of equal length and h is the length... Male Female Age... new coordinates by 3D rotation of points depending on the area of a parallelogram with 4 vertices calculator 3d provided you. Difference of pairs of parallel sides these individually and add the results edit '' link when available,. Question 2: find the area of any convex quadrilateral creating breadcrumbs and structured layout ) parallellogram är lika en... Of pairs of parallel sides it will always remain a parallelogram is a quadrilateral with sides. 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Review question
# Can we find the sum of the integers from $2k$ to $4k$ inclusive? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R5849
## Solution
1. The positive integer $k$ is given.
1. Find, in terms of $k$, an expression for $S_1$, the sum of the integers from $2k$ to $4k$ inclusive.
#### Approach 1
We could add together the integers $2k$, $2k+1$, $\dotsc$, $4k$ excluding $3k$ in pairs as follows: \begin{align*} 2k+4k &= 6k \\ (2k+1)+(4k-1) &= 6k \\ \vdots \quad & \\ (3k-1)+(3k+1) &= 6k. \end{align*}
And there are $k$ of these pairs in total.
So to add the numbers between $2k$ and $4k$ inclusive, we can add $k$ pairs that add to $6k$, together with $3k$, that is, $S_1=6k^2+3k.$
#### Approach 2
We could use the fact that the sum of the integers from $1$ to $n$ is $P_n=\dfrac{n(n+1)}{2}$.
So we want to find $P_{4k}-P_{2k-1}$.
We have \begin{align*} S_1 &= P_{4k}-P_{2k-1} \\ &=\frac{1}{2} (4k)(4k+1) - \frac{1}{2} (2k-1)(2k)\\ &=6k^2+3k. \end{align*}
1. Find, in terms of $k$, an expression for $S_2$, the sum of the odd integers lying between $2k$ and $4k$.
The idea behind the first approach above can be used again for (a)(ii).
### Approach 1
If $k$ is even, we can add the odd integers between $2k$ and $4k$ in $k/2$ pairs that add to $6k$: \begin{align*} (2k+1)+(4k-1) &= 6k \\ (2k+3)+(4k-3) &= 6k \\ \vdots \quad & \\ (3k-1)+(3k+1) &= 6k. \end{align*} So \begin{align*} S_2 &= 6k\times\frac{k}{2}\\ &=3k^2. \end{align*}
We can use exactly the same approach if $k$ is odd, only this time the odd integers between $2k$ and $4k$ consist of $(k-1)/2$ pairs that add to $6k$, together with $3k=6k/2.$
So once again $S_2=3k^2.$
### Approach 2
We need $(2k+1) + (2k+3) + \cdots + (4k-1)$, or $(2k+1) + (2k+3) + \cdots + (2k+(2k-1))$.
This is an arithmetic sequence with $k$ terms, so for the formula, $n = k$, $d = 2$ and $a = 2k+1$.
The sum is therefore $\dfrac{n}{2}(2a+(n-1)d) = \dfrac{k}{2}(4k+2+2(k-1)) = k(3k ) =3k^2$.
1. Show that $\dfrac{S_1}{S_2} = 2+\dfrac{1}{k}$.
We have \begin{align*} \frac{S_1}{S_2} &= \frac{6k^2+3k}{3k^2}\\ &=2+\frac{1}{k}. \end{align*}
1. Prove that the sum of the first $n$ terms of the geometric progression having first term $a$ and common ratio $r$ ($r \neq 1$) is $a \left( \frac{1-r^n}{1-r} \right).$
The sum of the first $n$ terms of a geometric progression is $T_n=a+ar+ar^2+\cdots+ar^{n-1}.$
Now notice that $r\times T_n = ar+ar^2+\cdots+ar^{n-1}+ar^n.$
Subtracting, most of the terms cancel out so that \begin{align*} (1-r)T_n &= a-ar^n\\ \implies\quad T_n &= \frac{a(1-r^n)}{(1-r)}. \end{align*}
By regarding the recurring decimal $0.\dot{0}7\dot{5}$ ($=0.075075\ldots$, where the figures $075$ repeat) as an infinite geometric progression, or otherwise, obtain the value of the decimal as a fraction in its lowest terms.
Now $0.075075\ldots$ can be written as the sum of a geometric progression with first term $0.075$ and common ratio $0.001$.
Using the expression above for the sum of the first $n$ terms, and taking the limit as $n\to \infty$, we have $T_n\to a \left( \frac{1}{1-r} \right)$ (since $|r|<1$, and $r^n \to 0$ as $n\to\infty$).
Substituting in our values of $a$ and $r$ gives $0.\dot{0}7\dot{5}=\frac{0.075}{1-0.001}=\frac{0.075}{0.999}=\frac{25}{333}.$
Alternatively, let $x = 0.075075\ldots$, so $1000x = 75.075075\ldots$.
Subtracting, we get $999x = 75$, so $x=\dfrac{75}{999} = \dfrac{25}{333}$. | HuggingFaceTB/finemath | |
# Affine space for Minkowski space time
I'm studying Minkowski space time (M4) and they say it's a 4 dimensions real affine space. M4 is an affine space so there is a non-empty set A, a 4 dimension real vector space V, and there is a function f: AxA-->V (with the proper property). The elements of A are the events.
The doubt: Have I to think that I’m in certain inertial frame of reference with a certain coordinate system so that every element of A is identify with 4 numbers? Or have I to think of A as a very abstract set of points and I set the coordinates of A after setting the basis for V?
I'm asking this because I have read that the affine coordinate system comes after the definition of the affine space, but in this case, I don't understand how it's possible to identify an event before setting a coordinate system. Are coordinate system and affine coordinate system two different things?
• I think you might have made a mistake when you wrote that there should be a function f: A -> V. Normally, affine spaces are equipped with an action of a vector space giving translations in the affine space, but such an action would be a map g: A + V -> A: a point in A can be translated by a vector in V to give another point in A. – Stijn B. Sep 11 '18 at 13:43
• Actually it was a typing error, I should have written f:AxA-->V with proper property that are: for every element P of A and for every vector v of V it exist one and only one Q such that f(P,Q)=v, for every P, Q, S elements of A it is true that f(P,Q)+f(Q,S)=f(P,S). This definition is equivalent to yours. Now that we solved this misunderstandig my doubts is still there. – SimoBartz Sep 11 '18 at 13:56
## 1 Answer
The elements of A (the events in Minkowski space) exist indepently of your choice of coordinate system on A. Similarly, one can define translations (the action by a vector) in a purely abstract fashion, without referring to any set of coordinates at all. Hence the abstract notion of an affine space (or vector space, or manifold) is more fundamental than the structure provided by a coordinate system.
Of course, it is often convenient to think about affine spaces like Minkowski space entirely in terms of coordinates, but keep in mind that there is a more abstract structure underneath. Furthermore, Minkowski space is a rather harmless kind of space, but in more complicated settings, not being careful enough in making the distinction between a space and its coordinates can lead to unwanted side-effects, which is why mathematicians often prefer coordinate-free notation in the first place.
Think of the real world around you. The universe is a kind of space as well (which we are supposed to describe as a Minkowski space), but it does not come with any canonical coordinate system at all: you are free to choose any set of points and give them the coordinates you like, as long as your coordinate system makes sense.
• Thanks for the answer, it's really useful. About the last part of my question, is the coordinate system (for example spherical coordinate system for the space and a 'clock' for the time) a different concept respect to the affine coordinate system? – SimoBartz Sep 11 '18 at 14:08
• An affine coordinate system is a special case of a coordinate system, namely one that respects the affine structure of the space. This structure is not manifestly clear in spherical coordinates, for instance, so spherical coordinates do not form an affine coordinate system. – Stijn B. Sep 11 '18 at 15:51 | HuggingFaceTB/finemath | |
Name: Lesson 7: Solving Problems by Guessing and Testing
1.
Use the guess and test method to solve the following problems.
When you multiply a number by 6, then add 5, the number is 35. What is the number?
a. 4 b. 5 c. 6 d. 7
2.
When you divide a number by 7, then add 10, the answer is 13. What is the number?
a. 20 b. 21 c. 22 d. 23
3.
When you multiply a number by 7, then add 2, the number is 51. What is the number?
a. 4 b. 5 c. 6 d. 7
4.
When you add 5 to a number, then divide it by 5, the answer is 5. What is the number?
a. 20 b. 25 c. 5 d. 6
5.
If you subtract 4 from a number, then multiply by 3, then answer is 3. What is the number?
a. 10 b. 8 c. 5 d. 2
6.
The area of a rectangle is 45 square units. The width is 4 more units than the length. What is the length of the rectangle?
a. 9 b. 8 c. 4 d. 5
7.
What is the width of the rectangle?
a. 9 b. 8 c. 4 d. 5
8.
Kelser bought 3 pails. Two small pails were the same price and one large pail was \$20.00 more than the other 2. If Kesler spent \$95.00 on the pails, how much was the larger pail?
a. \$30.00 b. \$35.00 c. \$40.00 d. \$45.00
9.
How much were each of the smaller pails?
a. \$20.00 b. \$25.00 c. \$30.00 d. \$35.00
10.
Clark chose a number. He listed the next three numbers in the sequence. The sum of the four numbers was 58. What was the starting number?
a. 13 b. 14 c. 15 d. 16 | HuggingFaceTB/finemath | |
# How do you calculate interest quarterly?
## How do you calculate interest quarterly?
When you are using monthly or quarterly interest rates instead of annual, you can find the appropriate rate by dividing the annual interest rate by the number of periods. For example, a 12 percent annual interest rate divided by four periods is a three percent quarterly interest rate.
## How much interest does \$2 million dollars earn per year?
For example, the interest on two million dollars is \$501,845.11 over 7 years with a fixed annuity, guaranteeing 3.25% annually.
Can I live off the interest of 1 million dollars?
You can retire with \$1 million dollars if you manage your withdrawals appropriately. The Rule of 4 says that you should withdraw no more than 4% of your total portfolio each year. Assuming you’re earning at least 4% in returns, you can effectively live off of interest-earned without touching your principal balance.
How many months at an interest of 1% per month does money have to be invested this it will double in value?
72 months
If the population of a nation increases at the rate of 1% per month, it will double in 72 months, or six years.
### What is quarterly interest payout?
Quarterly Interest Payout The interest amount given to you by the lender on this scheme will be on a quarterly basis which simply means you will get the interest every three months during your overall FD tenure.
### What should be the rate of interest if it is calculated quarterly?
if we will calculate interest quarterly, the rate of interest will be divided by one fourth. divide 8 by 4 to find that the quarterly interest rate is 2 percent.
What age can you retire with \$2 million?
age 60
Retire fully at age 60, and you could be sitting on a \$2 million nest egg. Keep working—and investing—for another five years, and you could retire with more than \$3 million at age 65!
How much does 2 million make in retirement?
Following the 4 percent rule for retirement spending, \$2 million could provide about \$80,000 per year, which is above average. The Bureau of Labor Statistics reports that the average 65-year-old spends roughly \$3,800 per month in retirement — or \$45,756 per year.
## What is the average 401K balance for a 65 year old?
The 401k is an employer-sponsored plan that allows you to save for retirement in a tax-sheltered way (\$19,500 per year in 2021) to help maximize your retirement dollars….Assumptions vs. Reality: The Actual 401k Balance by Age.
AGE AVERAGE 401K BALANCE MEDIAN 401K BALANCE
55-64 \$197,322 \$69,097
65+ \$216,720 \$64,548
## What is the 72 rule in finance?
The Rule of 72 is a simple way to determine how long an investment will take to double given a fixed annual rate of interest. By dividing 72 by the annual rate of return, investors obtain a rough estimate of how many years it will take for the initial investment to duplicate itself.
Does money double every 7 years?
The most basic example of the Rule of 72 is one we can do without a calculator: Given a 10% annual rate of return, how long will it take for your money to double? Take 72 and divide it by 10 and you get 7.2. This means, at a 10% fixed annual rate of return, your money doubles every 7 years.
How much interest does 3 billion dollars make a year?
How much interest does 3 billion dollars make a year? At current interest rates 3 billion dollars would earn about 30 million dollars a year before taxes. How much is Fifty million dollars at daily interest rate?
### How much interest does a million-dollar deposit generate?
A million-dollar deposit with that APY would generate \$500 of interest after one year (\$1,000,000 X 0.0005 = \$500). If left to compound monthly for 10 years, it would generate \$5,011.27. Certificates of Deposits (CDs) are time deposits that pay higher interest rates the longer the money is held on deposit.
### What is the compound interest of the second year?
The compound interest of the second year is calculated based on the balance of \$110 instead of the principal of \$100. Thus, the interest of the second year would come out to: The total compound interest after 2 years is \$10 + \$11 = \$21 versus \$20 for the simple interest.
How do you convert annual interest rate to quarterly compound rate?
Convert the effective annual interest rate into quarterly compound rates using this formula: where i = interest rate, ^n = to the power of n. If it is a simple annual interest rate, divide the rate by 12 to calculate the monthly interest rate. The formula is as follows: where i = interest rate. | HuggingFaceTB/finemath | |
# What is the third quartile of 24, 20, 35, 43, 28, 36, 29, 44, 21, 37?
Jun 11, 2017
${Q}_{1} = 24$
#### Explanation:
If you have a TI-84 calculator in hand:
First put the numbers in order.
Then you press the stat button.
Then $\text{1:Edit}$ and go ahead and enter your values in order
After this press the stat button again and go to $\text{CALC}$
and hit $\text{1:1-Var Stats}$ press calculate.
Then scroll down until you see ${Q}_{1}$.
Jun 11, 2017
${Q}_{1} = 24 , {Q}_{3} = 37$
#### Explanation:
$\text{arrange the data set in ascending order}$
$20 \textcolor{w h i t e}{x} 21 \textcolor{w h i t e}{x} \textcolor{m a \ge n t a}{24} \textcolor{w h i t e}{x} 28 \textcolor{w h i t e}{x} 29 \textcolor{red}{\uparrow} \textcolor{w h i t e}{x} 35 \textcolor{w h i t e}{x} 36 \textcolor{w h i t e}{x} \textcolor{m a \ge n t a}{37} \textcolor{w h i t e}{x} 43 \textcolor{w h i t e}{x} 44$
$\text{the median "color(red)(Q_2)" is in the middle of the data set}$
$\text{in this case that is between 29 and 35 so find the average}$
$\Rightarrow \textcolor{red}{{Q}_{2}} = \frac{29 + 35}{2} = 32$
$\text{the lower and upper quartiles divides the set to the left and }$
$\text{right of the median into 2 equal parts}$
$\Rightarrow \textcolor{m a \ge n t a}{{Q}_{1}} = 24 \text{ and } \textcolor{m a \ge n t a}{{Q}_{3}} = 37$ | HuggingFaceTB/finemath | |
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A157927 Joint-rank array of the numbers i^2+j^2, where i>=0, j>=0. 1
1, 2, 2, 4, 3, 4, 7, 5, 5, 7, 10, 8, 6, 8, 10, 14, 11, 9, 9, 11, 14, 19, 15, 13, 12, 13, 15, 19, 24, 20, 16, 14, 14, 16, 20, 24, 30, 25, 21, 18, 17, 18, 21, 25, 30, 37, 31, 27, 23, 22, 22, 23, 27, 31, 37, 44, 38, 32, 28, 26, 25, 26, 28, 32, 38, 44 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS The definition of joint-rank array given at A182801 is here extended to arrays R={f(i,j)} for which the numbers f(i,j) are not necessarily distinct. Specifically, all duplicates are assigned the same rank when all the numbers in R are jointly ranked. Let {a(i,j)} denote the resulting joint-rank array. In case all f(i,j) are positive integers, a(i,j)=f(i,j)-L(i,j), where L(i,j) is the number of numbers in R that are <=f(i,j). (Row 1)=A047808. LINKS EXAMPLE A corner of the array R={i^2+j^2} is 0....1....4....9...16... 1....2....5...10...17... 4....5....8...13...20... 9...10...13...18...25... Replace each term of R by its rank: 1....2....4....7...10... 2....3....5....8...11... 4....5....6....9...13... 7....8....9...12...14... CROSSREFS Cf. A182801, A048147. Sequence in context: A205793 A178431 A233574 * A227256 A328774 A131816 Adjacent sequences: A157924 A157925 A157926 * A157928 A157929 A157930 KEYWORD nonn,tabl AUTHOR Clark Kimberling, Dec 17 2010 STATUS approved
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# Business statistics week 4 assignment | math 510 | Strayer University
### Week 4 Assignment – Case Study: Transforming Data Into Information
#### Overview
The Woodmill Company makes windows and door trim products. The first step in the process is to rip dimension (2 × 8,2 × 10, etc.) lumber into narrower pieces. Currently, the company uses a manual process in which an experienced operator quickly looks at a board and determines what rip widths to use. The decision is based on the knots and defects in the wood.
A company in Oregon has developed an optical scanner that can be used to determine the rip widths. The scanner is programmed to recognize defects and to determine rip widths that will optimize the value of the board. A test run of 100 boards was put through the scanner and the rip widths were identified. However, the boards were not actually ripped. A lumber grader determined the resulting values for each of the 100 boards, assuming that the rips determined by the scanner had been made. Next, the same 100 boards were manually ripped using the normal process. The grader then determined the value for each board after the manual rip process was completed. The resulting data, in the file, Woodmill Data, consists of manual rip values and scanner rip values for each of the 100 boards.
#### Instructions
You are a process manager at the Woodmill Company tasked with determining if an optical scanner would be beneficial. Write a 4–5 page report to your supervisor (including a cover page and a Source List page) in which you:
1. Summarize the Woodmill Company’s problem of ripping dimension lumber into narrower pieces.
2. Develop a frequency distribution for the board values for the scanner and the manual process.
3. Generate appropriate descriptive statistics for both manual and scanner values.
4. Analyze the frequency distribution and descriptive statistics for both manual and scanner processes. Use Excel to create your charts.
5. Determine which process generates more values (manual or scanner).
6. Evaluate which process has less relative variability, using frequency distribution and descriptive statistics.
7. Recommend whether the optical scanner is the best option for the company.
8. Use two sources to support your writing. Choose sources that are credible, relevant, and appropriate. Cite each source listed on your source page at least one time within your assignment. For help with research, writing, and citation, access the library or review library guides
This course requires the use of Strayer Writing Standards. For assistance and information, please refer to the Strayer Writing Standards link in the left-hand menu of your course. Check with your professor for any additional instructions.
The specific course learning outcome associated with this assignment is:
• Apply strategies that are informed by the principles of statistical thinking and data-driven decision making to enhance business process performance.
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Understanding mathematics imprecisely
For a long time, it has been a complete mystery to me how any of my peers understood any math at all with anything short of filling in every detail, being careful about every set theoretic detail down to the axioms. That's a slight exaggeration, but I certainly did much worse in courses where I attempted to replicate this myself by not reading every proof to save time.
It's only recently, and only within the field of probability theory, that I've developed the ability to do this myself. I am currently following Grimmett's book on Percolation theory. There are far too few details for someone of my level to fill it in completely, but I am getting more than nothing out of it.
Question 1: I would like to learn how to get even more out of such "incomplete" studying.
What tends to happen even now, and more before, is that as soon as I don't understand something, I lose focus and everything just flies over my head. I imagine this is partly psychological, since from a logical perspective if I have to accept proposition $P$ to derive $Q$, I could just think of myself as having proven merely that $P$ implies $Q$, and then no "acceptances" are being made.
Most of the time, professors simply stare blankly at me, wondering how I could persist like this, and all they say is to stop. But it's not that simple, because it appears my intuition is also primarily symbolic. Sure, I think of some geometric pictures when they're called for, but most of my problem-solving creativity comes from pattern matching methods and tricks with situations.
Question 2: How does one distill out the important ideas of a mathematical reading, such as a proof or paper?
Grimmett's book is very helpful in this regard. He will always tell me what's important, and as long as I'm willing to believe him, then I don't have to do anything. But what if I need things that are different than he emphasizes? I always worry that by not understanding everything, I will eventually reach some point in my life when I need to use some fact/method I glossed over and forgot, and that it could be framed in such a way that I would not even be aware of what is missing. That way I wouldn't even be able to do a huge review to rescue the fact from the depths of my ignorance. My current way of thinking about this subproblem of question 2 is that mathematicians always take this risk by not studying everything. So it's a risk-minimization game with time as the constraint. If so, how do I make smart choices with regard to this game?
Question 3: With my recent ability to learn imprecisely in probability, I've started to see many connections, even with outside fields. Many of them are probably fictitious. Many of the questions that I think are highly-motivated might actually be not really worth answering. How does one decide what questions are interesting? As a graduate student who has barely popped out above what the traditional classroom has to offer, I am very lost in this regard.
Question 4: The revelations that enabled me to understand math imprecisely came all at once. A similar comment about the abruptness of my coming upon the ability to proof-check without significant error could be made 2 years ago. Most of my peers seem to learn rather continuously, but the evolutions of my way of thinking seem to come all at once. Is there anything bad or good about this? If so, how do I minimize the bad and maximize the good?
As always, answers to subsets are appreciated.
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What do you think of when you hear the phrase "mathematical intuition?" – Alexander Gruber May 29 '13 at 19:42
I used to just tune out everything following such a phrase because I would develop my own intuition, typically only after full understanding. Lately, I have even requested specifically that people teaching me in some specific cases say why something is intuitive. I also seek it out myself pictorially, since in analysis, sometimes pictures help more than symbol manipulation. – Jeff May 29 '13 at 19:57
Terry Tao has a blog post There is more to mathematics than rigour and proofs which seems relevant. – Martin May 29 '13 at 20:26
I mean this post was mostly about how to become a better mathematician in the sense of what most people think mathematics means, which is less about rigor. So the last comment was more of an excursion into my own personal opinion than something totally on topic. – Jeff May 29 '13 at 20:40
This is not to suggest I think rigor is everything. I only behaved that way in the past because it was the only frequency I could perceive. But even now, I still think that some of the beauty of mathematics is its rectilinear logical perfection. – Jeff May 29 '13 at 20:46
4 Answers
Full understanding is illusory. If you pursue it, you will find yourself trying to say what a number is, or a set, and digressing into the problem of making language, which for math is a meta-language, precise. And, of course, that can't be done.
So regarding your first question, it might help to observe how futile that innate wish of yours is, and how much you understand without full understanding (or compunction) in all other aspects of your life.
Imagine trying to learn biology and studying the chemical processes in the body, then asking "what is a chemical". You are given an answer that has to do with molecules, a term which you then inspect for precision's sake. Atoms come up, then electrons. Eventually you are learning quantum physics when all you wanted to do was understand how allergens work, or some such thing.
You must operate at the appropriate level for a specific problem. It's no use to reinvent the wheel and do everything from first principles. That would be like writing every program in machine code.
One day, our brains may be augmented with enhancements that allow us to have enough knowledge to understand everything down to our "axioms". Until then, it is a matter of becoming comfortable with our limitations and trying to work with what we have to be awesome.
In terms of knowing which questions are interesting, I think that is one of the harder parts of research. One almost has to be prescient.
And as for getting the important ideas of a proof, my first answer is that sometimes you can't really. Some proofs are just a confluence of numerical estimations and limit results and don't give any real insight into what is going on. Since you seem to be a probabilist, I would point to the proof that a random walk in dimension $n$ is recurrent for $n=1,2$ and transient otherwise. One feels there should be an intuitively understandable reason, but all one gets is Stirling's formula.
For other proofs it is a matter of becoming comfortable enough with the terminology and techniques used in the proof (by re-reading) to see the forest for the trees. In Kung Fu one talks about "learning to forget". You learn the movements carefully so you can perform them without thinking about them when the time comes. You do the same when you learn to integrate or differentiate - you don't want to be doing this from the limit definition when crunch time comes (exam say).
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I've observed how my closest colleague did it back in the day of my Phd and postdocs and I took over his technique. What he did was to sit down, read the paper superficially and then try to work out simple stuff he understood on his own. Then he would try to build his own version of what he got from the paper, often without fully understanding what had been going on. But he just had a general idea of the gist of the paper and tried to rebuild the idea in his own words, math, etc...
I remember I then proceeded to do the same later when studying some ecological model that we were trying to pour into mathematical formulas. I felt that the work that had been done was not very rigorous or even incomplete. So I rebuilt the model for myself superficially imitating others at first but gradually abandoning their approach for my own. And this without ever fully learning the necessary techniques of Markov processes, stochastic equations, etc... I feel that by doing this work, my understanding of the material is much deeper than it would have been if I had read a book about it or followed a standard course.
What also helped was the countless conversations and presentations I had to do about my work that forced me to put my thoughts into words understandable to others. They might not have gotten much out of it, but it has been very beneficial to myself for sure.
-
I'm not sure I have a strong answer to your questions, but here's my take on it.
I do work in "foundations". In particular, my interests are in computer proof assistants and type theory. The great thing about doing proofs on a computer is that you can tell with certainty stronger than any mere mortal that your reasoning is correct (assuming the consistency of your logic, the correctness of the program, the integrity of the hardware, blah, blah, blah). If your code typechecks, it's virtually guaranteed to be correct.
Doing this, you notice something right away: even the simplest proofs often require a lot of machinery. In written mathematics, we almost always handwave the tedious details. "It's trivial", as they say. But this is analogous to a software developer who only bothered to write the interesting parts of his code, but the "boring obvious stuff", he skipped entirely. (Such a developer wouldn't hold a job for too long).
But at the very least, you can quarantine the "trivial" bits. The unwritten bits of your proofs end up working almost identically to axioms. (You have asserted they are true without proof). But you end up with an awful lot of them floating around your proof.
There are some interesting lessons you can learn from a computer proof assistant, though, in terms of logical intuition. Perhaps most importantly, you end up seeing the logical structure of proofs. (At least, this is true in term-based systems, such as Agda or Epigram. It's not as true when you work with tactic-based systems, like Coq). To make a simple example, I worked through the first Sylow theorem from group theory this morning:
Sylow I: Let p be a prime, k a natural number, and G a finite group.
If p^k divides |G|, then there is a subgroup H of G that |G| = p^k.
If k = 0, then H is the trivial group.
Else if k = 1, then H is G itself.
Else if k > 1,
We can show |G| = mp for some integer m > 1.
Does G have a subgroup H where [G : H] is coprime to p?
Yes ->
Then by lagrange, |G| = p^k = [G : H] * |H|
We know that p^k doesn't divide [G : H], so by the division lemma,
p^k must divide |H|.
H is what we need, so that becomes our answer.
No ->
In this case, we know that there are no subgroups H where [G : H] is coprime to p.
Equivalently, for *all* subgroups H, p divides [G : H]
By the class formula, |G| = p^k = |Z(G)| + Σ[G : H_i]
So modularly, 0 ≡ |Z(G)| mod p
(Note the [G : H_i] drop out, because each [G : H_i] is divisible by p).
By the abelian version of Cauchy's Theorem,
there exists a z in Z(G) where |z| = p.
<z> is a subgroup of the center, so <z> is normal.
We appeal to the induction hypothesis, letting G = G/<z> and k = k-1.
This gives us a subgroup H of G/<z> where |H| = p^{k-1}.
We appeal to a minor lemma (I don't know if this has a name) which lets us show:
H = P / <z> for some subgroup P of G.
And we note that
|P| = [P/<z>] * |<z>|
|P| = |H| * |z|
|P| = p^k * p
|P| = p^{k+1}
So P is our subgroup of G, and it has the required order.
You can see the structure very cleanly here. I can imagine being able to port this to Agda in a fairly straightforward way (although it might take a while getting all the definitions laid out first). The proof is by induction on G and k. The induction is a strong one, as I recurse with arbitrary smaller groups each time. My induction has three cases. There's a rather major case split on line 5 when I ask about the existence of a subgroup H where [G : H] being coprime to p. (Constructively, I have to worry about whether or not this procedure is effective. In classical mathematics, I can appeal to the Law of Excluded Middle). I see there is some basic divisibility theorems required. And I need to know about generated groups, quotients, the class formula, and a few other things.
I can also spot the flaws. I can see a few places where I'm not 100% convinced my logic is air tight. Especially, my use of recursion is really awkward, and the "minor lemma" I refer to, I don't really know if it's true. (My book just asserted it, and I accepted it for now).
But in the book I have, the argument is written in prose (as most mathematics is). It's not clear where the induction occurs (I think the author made a minor error, mentioning the inductive hypothesis twice). And instead of indenting on the case splits, you simply see the author saying "it's trivial, so let's assume blah".
But I guess my point is, if held at gunpoint for about a week, I am highly confident I could write this out on a computer in a suitable dependently-typed programming language. I don't know how other mathematicians do it, but this is perhaps my favorite tool in my toolbox.
-
Sometimes I take solace in:
"Young man, in mathematics you don't understand things. You just get used to them."
- John von Neumann
It seems to me that some of the art is "if-this-then-that" kind of stuff, but there's a whole bunch more that basically comes from the intuition you get from basically just solving problems.
- | open-web-math/open-web-math | |
Short-Question
# Why is friction charging important?
## Why is friction charging important?
Ans: Yes, friction causes the production of charge on an object. When two objects rub against each other vigorously, the transfer of free electrons takes place between them. The object from which electrons rubbed off acquires a positive charge, whereas the object which gains electrons acquires a negative charge.
### When you charge a balloon by rubbing it on your hair this is an example of what method of charging?
electrostatic charging
Scientists have long known that rubbing two materials, such as a balloon on hair, causes electrostatic charging.
#### Why does hair stand up with friction?
Static electricity is the imbalance of positive and negative charges. If two things have opposite charges, they attract each other; if they have like charges, they repel each other. This explains why your hair stands on end when you take off a sweater or a wool hat.
Which force attracts hair to a charged comb?
electrostatic force of attraction
When the charged comb is brought near bits of paper, the electrostatic force of attraction between the atoms of the paper and the comb caused by the electrons on the surface of the comb polarizes them which attracts the pieces of paper towards the comb.
What is charging friction?
Charging by friction: the transfer of electrons from one uncharged object to. another by rubbing the two objects together. Some electrons can move to the. other object when rubbing (hair and balloon) Charging by conduction: the transfer of electrons from one object to another by.
## How can you describe charging by friction?
When insulating materials rub against each other, they may become electrically charged . Electrons , which are negatively charged, may be ‘rubbed off’ one material and on to the other. The material that gains electrons becomes negatively charged.
### How does charging by friction happen?
#### Which is an example of charging by friction?
Charging by friction involves rubbing two different materials together that have different pulls towards electrons, so that one material will pull away electrons when the materials are separated and both will become charged. A common example of charging by friction is rubbing a balloon against hair. When…
How is a silk comb charged by friction?
The glass rod is positively charged and the silk fabric is negatively charged as it absorbs additional electrons from the glass rod. In this case, bar after rubbing, comb after running through dry hair becomes electrified and this is an example of frictional electricity.
What happens when a hair is rubbed against a balloon?
When a material with a strong pull towards electrons (relatively), like rubber in a balloon, is rubbed against hair, it takes some of the hair’s electrons. When the two are separated, the balloon is negatively charged due to the excess electrons, while the hair is positively charged.
## Why does a comb get charged when you rub your scalp?
If we run a comb through our hair, the comb gets charged and can attract tiny pieces of paper. This is because the comb may have lost its electrons or gained some electrons when we rub it with the scalp. | HuggingFaceTB/finemath | |
140, an even composite number, has components other than 1 and also itself. Hence, the pair components of 140 will be mix of all those number which top top product provide 140. Because that example, (2, 70). Let’s learn more about factors of 140, components of 140 in pairs and the prime components of 140.
You are watching: Common factors of 140 and 25
Factors of 140: 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70 and also 140Prime factorization of 140: 2 × 2 × 7 × 5
1 What room the factors of 140? 2 How to Calculate factors of 140? 3 Factors that 140 by element Factorization 4 Factors that 140 in Pairs 5 Important Notes 6 FAQs on components of 140
## What space The determinants of 140?
The bag of numbers that once multiplied with each other make 140 room the factors of 140. 1 × 140, 2 × 70 , 4 × 35 , 5 × 28 , 7 × 20, 10 × 14 room the pairs the make 140 and they room all the determinants of 140.In prime factorization, us express 140 together a product the its element factors and in the department method, we see what numbers division 140 exactly there is no a remainder.
## How come Calculate factors of 140?
Let"s start calculating the determinants of 140, through our idea that any type of number that fully divides 140 without any remainder is its factor. Let"s start with the entirety number 1. 140 ÷ 1 = 140The next whole number is 2. Now, division 30 by 2. 140 ÷ 2 = 70Proceeding in this path we obtain all the factors of 30.140 ÷ 4 = 35140 ÷ 5 = 28140 ÷ 10 = 14
Hence, the factors the 140 are, 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70 and also 140.
Explore factors using illustrations and interactive examples
## Factors that 140 by prime Factorization
"Prime administer is to express a composite number together the product the its prime factors."
To acquire the prime components of 30, we division it through its smallest prime factor, i m sorry is 2. 140 ÷ 2 = 70Now, 70 is split by its smallest prime factor and the quotient is obtained. 70 ÷ 2 = 3535 ÷ 5 = 7. 7 ÷ 7 =1. This procedure goes on till we acquire the quotient together 1.
The prime factorization that 140 is shown below: ## Factors that 140 in Pairs
Factor bag of 140 space the two numbers which, when multiplied, offer the number 140. Therefore, the distinct pair determinants of 140 are (1,140), (2, 70), (4, 35), and (5, 28).
Important Notes
The numbers which we multiply in bag to get 140 room the determinants of 140.Factors that 140 are 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70 and 140.The number chin is a aspect of the number together it divides itself exactly. Aspect pairs room the pairs of 2 numbers which, once multiplied, provide the number. Variable pairs of 140 are (1,140) (2,70) (4,35) (5,28) (7, 20), and also (10,14)
## Factors the 140 resolved Examples
Example 1: List the determinants of 140.
Solution:
Factors of 140 space the number that divide 140 precisely without any remainder.
140 ÷ 1 = 140
140 ÷ 2 = 70
140 ÷ 4 = 35
140÷ 5 = 28
140 ÷ 7 = 20
140 ÷ 10 = 14
Therefore, components of 140 space 1, 2, 4, 5, 7, 10, 14, 28, 35, 70, and 140
Example 2: Find the factors of 140 and the factors of 70.
Solution:
Factors the 70 are 1, 2, 5, 7, 10, 14, 35, and 70Factors that 140 are 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, and 140
We can see that 1, 2, 5, 7, 10, 14, 35, and 70 are typical factors of 70 and 140.
Example 2: The area the the rectangle is 140 square inches. List under all various possible combinations possible for the length and breadth.
Solution:
The area of rectangle = length × breadthThe provided area is 140 square inches.
So, the feasible length and breadth space the components of 140.The 12 feasible combinations of factors the 140 bag are:
Length Breadth 1 in 140 in 2 in 70 in 4 in 35 in 5 in 28 in 7 in 20 in 10 in 14 in 14 in 10 in 20 in 7 in 28 in 5 in 35 in 4 in 70 in 2 in 140 in 1 in
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## FAQs on determinants of 140
### What space the determinants of 140?
The factors of 140 room 1, 2, 4, 5, 8, 10, 14, 28, 35, 70, and also 140
### What is 140 as a product the its element factors?
140 deserve to be composed as a product that its prime factors as 140 = 2 × 2 × 5 × 7
### Is 140 a perfect square?
140 is no a perfect square due to the fact that the square root of 140 is not equal come a totality number.
See more: How To Pop A Balloon Using Simple Machines Pop A Balloon, How To Build A Compound Machine To Pop A Balloon
### What space the determinants of 140 and 142?
The determinants of 140 are 1, 2, 4, 5 ,7, 8 ,10, 14, 28, 35, 70 and also 140.The determinants of 142 room 1, 2, 71, and also 142
### What room the prime numbers the 140?
The prime determinants of 140 are 2 , 5 and also 7 because 22 × 5 × 7. | HuggingFaceTB/finemath | |
# Find the perimeter of the square
It’s important to keep them in mind when trying to figure out how to Find the perimeter of the square.
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## Perimeter of a Square
Perimeter of the square = A + A + A + A units Hence the formula of the perimeter of a square can be termed as = 4A How can we find the perimeter of a Square? We can find the perimeter of a
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## Perimeter of a square with calculator
To find the perimeter of a square using the area, we can use the steps given below: Step 1: Find the side length using the area with the formula side $=$ A. In this example, side $= \sqrt{36} = 6$ cm; Step 2: Apply the perimeter formula of
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## 3 Ways to Calculate the Perimeter of a Square
The perimeter of a square can be calculated by adding the length of all its sides. Perimeter of Square Formula The formula to calculate the perimeter of a square can be mathematically expressed as; Perimeter of square, (P) = 4 × Side
## ISEE Upper Level Math : How to find the perimeter of a square
We need to add all the sides of square to find its perimeter. By the formula of perimeter of polygon, we know; Perimeter = sum of all sides Perimeter of
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## How to find the Area and Perimeter of a Square
Calculating the Perimeter of a square The steps required are- Step 1- Note down the length of sides given in the question Step 2- Designate the length as ‘a’ Step 3- Write the formula- 4a= 4*a
Clarify mathematic problems
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Math can be a difficult subject for some students, but with a little patience and practice, it can be mastered. | open-web-math/open-web-math | |
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# Weighted Average Cost of Capital(WACC) vs Discount Rate
WACC and discount rate are both used to evaluate investment opportunities, but they are different concepts. WACC is the minimum return that a company needs to earn on its investments to satisfy its investors and creditors, while the discount rate is the rate used to calculate the present value of future cash flows. While WACC considers the cost of all forms of capital, the discount rate is used to account for the time value of money and the risk associated with an investment.
## An example on how to calculate both WACC & Discount Rate
Let's say a company is considering investing in a new project that requires an initial investment of \$100,000 and is expected to generate a net cash flow of \$20,000 per year for the next five years. The company's WACC is 10%, and the discount rate for this particular project is 12%.
To evaluate the project using WACC, the company would compare the expected return of the project (in this case, \$20,000 per year) to the WACC of 10%. If the expected return is greater than the WACC, the project would be considered a good investment. Using the WACC formula, the company's cost of capital for the project is:
WACC = (Cost of Debt x Weight of Debt) + (Cost of Equity x Weight of Equity) Assuming the company's cost of debt is 8%, and its cost of equity is 12%, and the weights of debt and equity are 40% and 60%, respectively:
WACC = (0.08 x 0.4) + (0.12 x 0.6) = 0.104 or 10.4%
Comparing the expected return of \$20,000 to the WACC of 10.4%, the project is still considered a good investment. To evaluate the project using the discount rate of 12%, the company would calculate the present value of the future cash flows using the discount rate. Assuming a constant cash flow of \$20,000 per year for five years:
PV = \$20,000 / (1 + 0.12)^1 + \$20,000 / (1 + 0.12)^2 + \$20,000 / (1 + 0.12)^3 + \$20,000 / (1 + 0.12)^4 + \$20,000 / (1 + 0.12)^5
PV = \$17,857.14 + \$15,944.44 + \$14,248.87 + \$12,757.03 + \$11,456.35
PV = \$72,263.83
Therefore, the present value of the future cash flows of the project is \$72,263.83.
To determine if the project is a good investment based on this analysis, the company would compare the present value of the future cash flows to the initial investment of \$100,000. If the present value is greater than the initial investment, the project would be considered a good investment.
In this case, the present value of the future cash flows is \$72,263.83, which is less than the initial investment of \$100,000. Therefore, the project may not be a good investment when evaluated using the discount rate of 12%. | HuggingFaceTB/finemath | |
ELASTICITY, Stress, Tensile stress, Compressive stress, Shear stress/ tangential stress, Hooke's law, Strains, Linear Strain, Bulk (or) Volume Strain, Shearing (or) Rigidity strain, Young's modulus (E), Bulk (or) Volume modulus (k), Shearing (or) Rigidity modulus 'n', Poisson's ratioi ~ MECHTECH GURU
# ELASTICITY
The property by virtue of which a body tends to regain its original shape and size after removal of the deforming force is known as elasticity.
When an external force is applied on a body, which is not free to move, the shape and size of the body change. The force applied is called deforming force.
When the deforming forces are removed, the body tends to regain its original shape and size due to a force developed within the body.
The force developed within the body, which is equal and opposite to deforming force is called restoring force.
Bodies, which completely regain their original size and shape after the removal of the deforming force, are called elastic bodies. Bodies which change the shape and size on the application of force and which do not regain their original condition on removal of the deforming forces are said to be plastic bodies.
Bodies which do not change their shape and size on application of force are called rigid bodies.
The property by virtue of which a body tends to regain its original shape and size after removal of the deforming force is known as elasticity.
## Stress
When an external force is acting on an elastic body, it causes deformation (change in shape or in size or both). At the same time, due to elastic property, a force is developed within the material, which is equal and opposite to the applied force, to bring the body to its original shape and size. This force is ‘restoring force’.
## Stress =Force (F)/Area (A)
The stress is defined as the restoring force acting on unit area.
Since the applied force and the restoring force are equal in magnitude, the ‘stress’ is measured as the applied force acting per unit area.
The unit for stress is newton per square metre with symbol N m-2 or 'pascal' with symbol 'Pa'.
## Tensile stress
When it tends to increase the length in the direction of the force, it is tensile stress
## Compressive stress
When the applied force tends to compress the body, the stress is compressive stress.
## Shear stress/ tangential stress
When it acts parallel to the surface of a body, the stress is tangential stress.
## Hooke's law
Within the elastic limits, the strain produced in a body is directly proportional to the stress which causes it.
i.e., strain ∝ stress
or stress ∝ strain
stress/ strain= a constant
This constant is called 'modulus of elasticity'.
## Strains
Change in dimensions to original dimensions is known as Strain.
## (1) Linear Strain
The linear strain is defined as the ratio of change in length to the original length.
When a wire or bar is subjected to two equal and opposite forces, namely pulls, at its ends, there is an increase in the length. If the forces are tensile, the body is elongated. If the forces are compressive, the length is shortened in the direction of the forces. This is called the 'linear strain'.
The linear strain is defined as the ratio of change in length to the original length. If the change (increase or decrease) in length is ' l ' in a wire or bar of original length L,
As the linear strain is ratio of lengths, it has no unit.
## (2) Bulk (or) Volume Strain
Volume strain is defined as the ratio of change in volume to the original volume. It has also no unit.
When a force is applied uniformly and normally to the entire surface of the body, there is a change in volume of the body, without any change in its shape. This strain is called 'bulk or volume strain'.
If 'v' is the change in volume produced in a body of original volume ‘V’,
## (3) Shearing (or) Rigidity strain
When a force is applied parallel to one face of a body, the opposite side being fixed, there is a change in shape but not in size of the body. This strain is called the shearing strain.
Solids alone can have a shearing strain. It is measured by the angle of the shear 'θ' in radian.
## Three modulii of elasticity
There are three types of modulii depending upon the three kinds of strain.
## (a) Young's modulus (E)
It is defined as the ratio of linear stress to linear strain.
Let a wire of initial length L and cross-sectional area ' A', undergo an extension l, when a stretching force 'F', is applied in the direction of its length.
The modulus of elasticity, in this case, is called Young's modulus and is given by
### i.e., Young's modulus ( E ) =Linear stress/ Linear strain
Then, longitudinal or linear stress = F/A
and longitudinal strain = l/L
E =Linear stress/ Linear strain
= FL/Al
The unit for Young’s modulus is newton per square metre with symbol N m-2. The single term unit which is widely used for Young's modulus is 'pascal' with symbol 'Pa'.
## (b) Bulk (or) Volume modulus (k)
It is defined as the ratio of bulk stress to bulk strain.
When a body is subjected to a uniform compressive force, its volume decreases and the strain produced is a bulk or volume strain.
If 'v' is the change in volume and V is the original volume, then
If F is the total compressive force acting on a total area A, then bulk stress = F/A = P
If P is the stress applied i.e. (force/unit area) then,
bulk strain =v/ V
### bulk modulus k = bulk stress/ bulk strain
The unit for bulk modulus is 'newton per square metre with symbol N m-2. The single term unit which is widely used for bulk modulus is 'pascal' with symbol 'Pa'.
## (c) Shearing (or) Rigidity modulus 'n'
The ratio of the shearing stress applied to the body to the shearing strain produced is called the rigidity modulus and denoted by the letter 'n'.
If T is the tangential force/unit area and if θ is the angle of shear measured in radian, then
### rigidity modulus n = T/ θ
The unit for rigidity modulus is 'newton per square metre radian-1' with symbol N m-2 rad-1. The other unit which is widely used for rigidity modulus is 'pascal radian-1' with symbol 'Pa rad-1'.
## Poisson's ratio: σ
When a tensile stress is applied to a wire, the wire undergoes not only an extension of length in the direction of the force but also a contraction in its thickness. The ratio of decrease in thickness to the original thickness in lateral direction is known as lateral contraction.
The ratio of lateral contraction to linear elongation is called Poisson's ratio.
Poisson ' s ratio σ = Lateral contraction/ Linear elongation
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22 November, 2013
# Find the biggest empty box: example of two dimensional convolution
• How can I find the biggest empty box in a matrix?
• Can you show me an example on two dimensional convolution through a MATLAB Cody problem?
• What algorithms can be used to solve MATLAB Cody problem #658?
ASEE Challenge problem set in MATLAB Cody consists of really challenging tasks needing more complex way of thinking. These problems are worth to have the effort to solve them.
One of the favourites is Find the biggest empty box:
You are given a matrix that contains only ones and zeros. Think of the ones as columns in an otherwise empty floor plan. You want to fit a big square into the empty space (denoted by zeros). What is the largest empty square sub-matrix you can find in the given matrix? You will return the row and column extent of the sub-matrix. The answer may not be unique. We will test that your sub-matrix is square, that it is empty, and that it contains the correct number of elements.
An example input and output pair:
a = [1 0 0; 0 0 0; 0 0 0]
s = [2 3 2 3]
The skeleton of the function is:
function [r1, r2, c1, c2] = biggest_box(a)
r1 = 1;
r2 = 1;
c1 = 1;
c2 = 1;
end
That means that we have to return the indices of the first and last row and the first and last column of the sub-matrix respectively.
## The straightforward solution
The straightforward solution is to try all possible squares to all possible positions.
• We start with a square having same size as the input: this can be positioned only to the top-left corner.
• If there is a one value in the square, we suppose a smaller sub-matrix, and put it to all possible positions again.
• We repeat these steps, until we find a square fitting our needs having no one value in it.
• Once we have found it, we can return it immediately, since we started from the maximal size and we are counting down: this way the first solution will be the best. For example if we have found a good sub-matrix having size of five, there is no need to search an other four-by-four square.
A code example doing the algorithm described above is the following:
function [r1, r2, c1, c2] = biggest_box(A)
% get the size (s) of the input square
s = length(A);
% loop through all possible square sizes (n) and positions (r1, c1)
for n = s - 1 : -1 : 0
for r1 = 1 : s - n
for c1 = 1 : s - n
% calculate the position of the lower-right corner
r2 = r1 + n;
c2 = c1 + n;
% return if the summary in the current square is zero
if sum(sum(A(r1 : r2, c1 : c2))) == 0
return
end
end
end
end
end
The most outer loop is for decreasing the square-size, while the inner ones are for the positioning. In each iteration the indices are recalculated: if the summary in the current square is zero, we return the coordinates and finish running.
The indexing is a bit complex, but drawing it on piece of paper can help a lot. Maybe this line can be the most confusing:
for n = s - 1 : -1 : 0
Why is this used instead of:
for n = s : -1 : 1
There are two reasons of it:
• Suppose that the input is a four-by-four matrix. We want to get the indices of the lower-right corner of a sub-matrix having size of four placed on the top-left corner on the input. The index of its last row can not be simply 1 + 4 = 5, because this would cause overindexing, but it should be 1 + (4 - 1) = 4.
• Suppose that the input is a four-by-four matrix. We have a sub-matrix having size of two, and we want to calculate the maxima of the possible row positions of its top-left corner. It can not be simply 4 - 2 = 2, because a two-by-two sub-matrix can be put onto the third row too, so it should be 4 - (2 - 1) = 3.
In both cases we have to calculate by the actual square size minus one. If the substraction is done it the for cycle already, it will make our life easier.
For more better understanding here is a debug variation of the skeleton of the program. In each iteration matrix D of zeros is created, while the current square is marked by ones:
function [r1, r2, c1, c2] = biggest_box_debug(A)
s = length(A);
for n = s - 1 : -1 : 0
for r1 = 1 : s - n
for c1 = 1 : s - n
r2 = r1 + n;
c2 = c1 + n;
D = zeros(s); D(r1 : r2, c1 : c2) = 1
end
end
end
end
biggest_box_debug(zeros(4))
A part of the output is:
D =
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
D =
1 1 1 0
1 1 1 0
1 1 1 0
0 0 0 0
D =
0 1 1 1
0 1 1 1
0 1 1 1
0 0 0 0
D =
0 0 0 0
1 1 1 0
1 1 1 0
1 1 1 0
D =
0 0 0 0
0 1 1 1
0 1 1 1
0 1 1 1
D =
1 1 0 0
1 1 0 0
0 0 0 0
0 0 0 0
D =
0 1 1 0
0 1 1 0
0 0 0 0
0 0 0 0
D =
0 0 1 1
0 0 1 1
0 0 0 0
0 0 0 0
D =
0 0 0 0
1 1 0 0
1 1 0 0
0 0 0 0
D =
0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0
...
As you can see, the size of the supposed square is decreasing and is put to all possible positions. Here comes the trick: this operation is exactly the same as two dimensional convolution. A window is slid along the input matrix, and the elements in it are examined. If you are not familiar with this operation, please read the tutorial on it.
## A solution using convolution
The basic idea to solve this task by two dimensional convolution is the following: if we convolve the input by an n-by-n square of ones, and there is any zero value in the result, it means that there is an n-by-n block containing only zeros.
The algorithm is the following:
• Suppose the maximal possible square size and convolve the input by this square of ones.
• If there is any point, where the result of the convolution is zero, it means that we have found an appropriate block, so we only have to calculate its coordinates and return.
• If there is no point in the output having zero value, decrease the supposed square size, and do the convolution again.
The only remaining question is: how to calculate the coordinates of the resulting block? Suppose a four-by-four input matrix convolved by a three-by-three square. Do the convolution only for those elements, where the convolving window is fully covered. See the following code example below:
% the input matrix
A = [1 1 0 1;
0 0 0 0;
0 0 0 0;
0 0 0 1];
% do convolution for those elements where the window is fully covered
B = conv2(A, ones(3), 'valid')
The output is:
B =
2 2
0 1
To explain this result a bit more, please have a look the following figure, which demonstrates the valid positions of the window during convolution:
1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1
As it can be seen, the position of the elements in the output equals to the position of the upper-left corner of its generator window. This way the position of the found zero value equals to the upper-left corner of the needed block.
The full, commented MATLAB code example of this approach:
function [r1, r2, c1, c2] = biggest_box(A)% start from the maximal possible size
for n = length(A) : -1 : 1 % doing convolution for valid values
R = conv2(A, ones(n), 'valid'); % coordinates of the first zero
[r1, c1] = find(R == 0, 1); % if the resulting coordinate is not empty: there is an appropriate box
if r1 % coordinates of the lower-right corner
r2 = r1 + n - 1;
c2 = c1 + n - 1; % return: we do not need to continue searching
return;
end
end
This code is not fully optimized to help understanding and keep its simplicity for being a good demonstration on two dimensional convolution. | HuggingFaceTB/finemath | |
# Intuitive/heuristic explanation of Polya's urn
Suppose we have an urn with one red ball and one blue ball. At each step, we take out a single ball from the urn and note its color; we then put that ball back into the urn, along with an additional ball of the same color.
This is Polya's urn, and one of the basic facts about it is the following: the number of red balls after $n$ draws is uniform over $\{1, \ldots, n+1\}$.
This is very surprising to me. While its not hard to show this by direct calculation, I wonder if anyone can give an intuitive/heuristic explanation why this distribution should be uniform.
There are quite a few questions on Polya's urn on math.stackexchange, but none of them seem to be asking exactly this. The closest is this question, where there are some nice explanations for why, assuming as above that we start with one red and one blue ball, the probability of drawing a red ball at the $k$'th step is $1/2$ for every $k$ (it follows by symmetry).
• Ah, draws vs. balls, gotcha :) – Colm Bhandal Sep 18 '15 at 22:15
• Not sure why the quoted question doesn't answer yours. The definition of uniform probability is that the probability is constant from draw to draw. The answer to the other question shows that the probability is the same for every draw. What more are you after? – Paul Sinclair Sep 18 '15 at 22:59
• @PaulSinclair -- the fact that the probability to draw a red ball on draw $k$ is $1/2$ does not imply that the distribution of red balls is uniform after $n$ draws. – Michael G Sep 18 '15 at 23:00
• Hmm. It seems I must echo Colm Bhandal's post. – Paul Sinclair Sep 18 '15 at 23:09
By induction: Assume that the proposition holds true for (say) $m=5$ , where $m=n+1$ is the total number of balls, hence the urns may have $k=1\cdots 4$ red balls with uniform probability. The experiment then can be pictured as this: we have 4 urns, each with 5 balls, the first with $1$ red ball, the second with 2, etc. We pick one of these 4 urns, with same probability, and then we pick a ball with uniform probability from the urn. This is the same as saying that we pick a ball at random, always uniformly, from the total $4\times 5=20$ balls. The following picture helps (I hope) to envision the possible alternatives.
On the left, we have the 4 urns, and the 20 balls, from which we will draw one. On the right, the same balls are arranged with a displacement along the diagonal. In this way we can identify all the balls that, if selected, will lead to each of the new urns (diagram below). Because we have the same number of total balls for each new urn, each one wil have the same probability.
• a very cool pictorial illustration! – Michael G Sep 19 '15 at 1:24
• Two comments: First, it seems to me that a key condition for these results is that the urn’s history is unknown. Knowing the results of any of the draws can skew the probabilities and ball distributions away from uniformity. Second, this reminds me of Schrödinger’s cat and quantum-mechanical superposition of states. Unless you look in the urn and count the balls, you can consider it to be in all of its possible states at once. – amd Sep 19 '15 at 8:48
The best I can do here is an inductive argument:
• Clearly, after $0$ draws, there is always exactly $1$ red ball with a probability of $1 = \frac1{0+1}$ i.e. certainty of being there.
• For the inductive case, we notice that drawing $n + 1$ balls first begins with drawing $n$ balls, then another. After drawing the $n$ balls, there is uniform chance of any number of red balls in the urn from $1$ to $n+1$, by induction. That is, for any $1 \leq k \leq n+1$, the probability of $k$ red balls is $\frac1{n+1}$. Now, what is the probability of having $k$ red balls after $n + 1$ draws? Let's consider some cases:
1. If $2 \leq k \leq n + 1$, we have two more cases to consider with non-zero probability: $k$ balls after $n$ draws, and then draw a blue ball, or we have $k - 1$ after $n$ draws, and then we draw a red ball. Now, after we have already drawn $n$ times, there are $n+2$ balls in total in the urn. The chances then of drawing a blue ball, given $k$ red balls in the urn is $\frac{n + 2 - k}{n + 2}$. The chances of drawing a red ball given $k - 1$ red balls is $\frac{k - 1}{n + 2}$. Both numbers of red balls after $n$ draws are equally likely by the inductive hypothesis, with probability $\frac1{n+1}$. The overall probability is then $\frac1{n + 1}(\frac{n + 2 - k}{n + 2} + \frac{k - 1}{n + 2}) = \frac1{n + 2}$.
2. If $k = 1$, then we must never draw a red ball. So we must have exactly $1$ red ball remaining after $n$ draws, and then draw a blue ball. By induction, the probability of $1$ ball after $n$ draws is $\frac1{n+1}$. Drawing a blue ball after this has probability $\frac{n+1}{n+2}$. Multiplying the two probabilities easily gives us the desired $\frac{1}{n+2}$.
3. If $k = n+2$, then the problem is symmetric, considering $k=1$ for blue balls instead of red balls and so the previous argument applies.
It seems as though Polya chose this question intentionally to challenge our very intuitions, which are so often wrong. He seems to have crafted the question also to force us to use a clever argument of induction, topped with the icing of an argument based on symmetry. Naturally, we are inclined to think of some sort of geometrically decreasing probability distribution across $n$. Once symmetry is realised, we must update this: perhaps the distribution approaches a normal distribution as $n$ gets large? These are certainly intuitions that I had when I saw this question. But I don't trust my intuition. And I am right not to. This is a good demonstration of where intuition fails, at least initially, before we really soak in the problem. So we look instead for a rigorous demonstration, and the most elegant one at that.
As Einstein once said "Everything Should Be Made as Simple as Possible, But Not Simpler". Erdos used to talk about the most elegant proofs as being "from the book". I won't blacken either of their names by claiming the above is "as simple as possible" or "from the book", but it's certainly the best I can think of. Intuitively, I don't feel you can "compress" the argument much further, without losing its essence. That intuition I can't explain, it's just a hunch.
Having said that, the key steps in the argument, I believe, are as follows. The first is hitting upon the right idea of induction: i.e. induction on $n$, not $k$. There is a conceptual "aha" moment when you realise that drawing $n + 1$ balls is the same as first drawing $n$ balls, and then drawing another. The second "aha" moment is when you follow your previous idea for induction, and notice that all values of $k$ balls after $n + 1$ draws, except the corner cases of $n + 2$ and $1$, can happen in exactly two ways: either $k$ balls are chosen from $n$, and then no more, or $k - 1$ are chosen, and $1$ more. We follow this hunch, work out the maths, and hey presto, the correct answer pops out nicely. Then we're left with the corner cases. The case for $1$ also works, hoorah! The final touch of elegance is to use a symmetric argument on the case of $n+2$, rather than work it out from scratch.
There is a shorter way of explaining this, quasi-intuitively, but it loses the crispness of the above argument. Still, it is a nice auxiliary intuition to complement the above. Again, imagine the inductive proof. Imagine all the possibly values for red balls: $1, 2, 3, \dots, n+2$. In the next step, each choice "gives" part of its probability away to the next one up. Clearly larger values give more than smaller ones: because there is more chance of drawing another red once you already have red. But they also get more, because they get a value from the number below. Still, overall they lose. Each gets $\frac{k - 1}{n + 2}$ and gives $\frac{k}{n + 2}$ So $1$ "gives away" a value of $\frac1{n + 2}$ to $2$. $2$ "gets" this, and gives away $\frac2{n + 2}$ to $3$ etc. Eventually, when all values are passed on like this, the distribution settles back at uniformity, except now with $\frac1{n + 2}$ instead of $\frac1{n + 1}$. | HuggingFaceTB/finemath | |
# Wap to calculate standard deviation
Rating: 4.57 / Views: 893
##### 2020-01-18 05:52
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