text stringlengths 22 1.01M |
|---|
How do you solve x+y=9, x-y=-1 by graphing and classify the system?
May 30, 2017
Solution is $x = 4$ and $y = 5$.
Explanation:
We select a few points satisfying the first equation i.e. $x + y = 9$. Let these be $\left(- 3 , 12\right)$, $\left(0 , 9\right)$ and $\left(5 , 4\right)$. Joining them gives the graph as follows:
graph{(x+y-9)((x+3)^2+(y-12)^2-0.03)(x^2+(y-9)^2-0.03)((x-5)^2+(y-4)^2-0.03)=0 [-9.57, 10.43, 2.64, 12.64]}
Similarly for second equation $x - y = - 1$, let some of the points satisfying the equation be $\left(- 6 , - 5\right)$, $\left(- 1 , 0\right)$ and $\left(5 , 6\right)$ and graph appears as follows.
graph{(x-y+1)((x+6)^2+(y+5)^2-0.05)((x+1)^2+y^2-0.05)((x-5)^2+(y-6)^2-0.05)=0 [-21.28, 18.72, -9, 11]}
If we draw the two graphs on te same Cartesian plane they intersect each other at $\left(4 , 5\right)$ asshown below.
graph{(x+y-9)(x-y+1)((x-4)^2+(y-5)^2-0.05)=0 [-18.11, 21.89, -5.32, 14.68]}
Hence, solution is $x = 4$ and $y = 5$.
Note $-$ In case you find that lines are parallel and do not intersect, we have #no solution. Some times two lines may coincide.. This means the two lines are essentially same. In such cases you wil find that the one line is just a multiple of anther line. |
# Section B: Practice Problems Reason about Attributes to Solve Problems
## Section Summary
Details
In this section, we used the attributes such as side lengths, angles, lines of symmetry, and parallel sides to solve problems about perimeter of shapes.
We learned that, if a shape has certain attributes, we can use them to find its perimeter, even if we don’t have all of its side lengths. Or, if we know the perimeter of a shape, we can find its side lengths if there is enough information about their attributes.
For example, here are two shapes:
If we know the perimeter of each shape is 48 units and the dashed line through shape A is a line of symmetry, we can find the missing side lengths.
Shape B doesn’t have a line of symmetry, but if we were told that its opposite sides have equal lengths, then we can also reason about the three missing side lengths.
## Problem 1 (Lesson 7)
1. What is the perimeter of the rhombus? Explain or show your reasoning.
2. Diego says he can find the area of this rectangle because he knows two side lengths.
Do you agree with Diego? Explain your reasoning.
## Problem 2 (Lesson 8)
1. Draw the lines of symmetry for the windmill blades.
2. Each blade is 5 feet long and feet wide. What is the perimeter of the windmill? Explain or show your reasoning.
## Problem 3 (Lesson 9)
Here is a rectangle R.
1. What shape can be folded along a line of symmetry to give R? What are the side lengths of that shape?
2. What shape can be folded twice along lines of symmetry to give R? What are its side lengths?
## Problem 4 (Exploration)
How many lines of symmetry are there in this design? Explain or show how you know.
## Problem 5 (Exploration)
Make a shape or design with one or more lines of symmetry. Trade shapes with a partner and find all of the lines of symmetry of your partner’s shape. You may find pattern blocks helpful to make your shape or design. |
# How do you find the arc length of the curve y = 2x - 3, -2 ≤ x ≤ 1?
Jun 26, 2015
The arc length is $3 \sqrt{5} \setminus \approx 6.7082$
#### Explanation:
Since the graph of $y = f \left(x\right) = 2 x - 3$ is a straight line, there's actually no need to use calculus. Instead, just find the straight-line distance between the points $\left(- 2 , f \left(- 2\right)\right) = \left(- 2 , - 7\right)$ and $\left(1 , f \left(1\right)\right) = \left(1 , - 1\right)$. The answer, by the distance formula (Pythagorean theorem), is
$\sqrt{{\left(- 2 - 1\right)}^{2} + {\left(- 7 - \left(- 1\right)\right)}^{2}} = \sqrt{9 + 36} = \sqrt{45}$
$= \sqrt{9} \sqrt{5} = 3 \sqrt{5} \setminus \approx 6.7082$
If you want to confirm this with calculus, evaluate the integral ${\int}_{- 2}^{1} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \setminus \mathrm{dx} = {\int}_{- 2}^{1} \sqrt{1 + 4} \setminus \mathrm{dx}$
$= \sqrt{5} x {|}_{- 2}^{1} = \sqrt{5} \left(1 - \left(- 2\right)\right) = 3 \sqrt{5}$ |
# 3.5 Addition of velocities (Page 3/12)
Page 3 / 12
Solution
Because ${\mathbf{\text{v}}}_{\text{tot}}$ is the vector sum of the ${\mathbf{\text{v}}}_{\text{w}}$ and ${\mathbf{\text{v}}}_{\text{p}}$ , its x - and y -components are the sums of the x - and y -components of the wind and plane velocities. Note that the plane only has vertical component of velocity so ${v}_{px}=0$ and ${v}_{py}={v}_{\text{p}}$ . That is,
${v}_{\text{tot}x}={v}_{\text{w}x}$
and
${v}_{\text{tot}y}={v}_{\text{w}y}+{v}_{\text{p}}\text{.}$
We can use the first of these two equations to find ${v}_{\text{w}x}$ :
${v}_{\text{w}y}={v}_{\text{tot}x}={v}_{\text{tot}}\text{cos 110º}\text{.}$
Because ${v}_{\text{tot}}=\text{38}\text{.}0 m/\text{s}$ and $\text{cos 110º}=–0.342$ we have
${v}_{\text{w}y}=\left(\text{38.0 m/s}\right)\left(\text{–0.342}\right)=\text{–13 m/s.}$
The minus sign indicates motion west which is consistent with the diagram.
Now, to find ${v}_{\text{w}\text{y}}$ we note that
${v}_{\text{tot}y}={v}_{\text{w}y}+{v}_{\text{p}}$
Here ; thus,
${v}_{\text{w}y}=\left(\text{38}\text{.}0 m/s\right)\left(0\text{.}\text{940}\right)-\text{45}\text{.}0 m/s=-9\text{.}\text{29 m/s.}$
This minus sign indicates motion south which is consistent with the diagram.
Now that the perpendicular components of the wind velocity ${v}_{\text{w}x}$ and ${v}_{\text{w}y}$ are known, we can find the magnitude and direction of ${\mathbf{\text{v}}}_{\text{w}}$ . First, the magnitude is
$\begin{array}{lll}{v}_{\text{w}}& =& \sqrt{{v}_{\text{w}x}^{2}+{v}_{\text{w}y}^{2}}\\ & =& \sqrt{\left(-\text{13}\text{.}0 m/s{\right)}^{2}+\left(-9\text{.}\text{29 m/s}{\right)}^{2}}\end{array}$
so that
${v}_{\text{w}}=\text{16}\text{.}0 m/s\text{.}$
The direction is:
$\theta ={\text{tan}}^{-1}\left({v}_{\text{w}y}/{v}_{\text{w}x}\right)={\text{tan}}^{-1}\left(-9\text{.}\text{29}/-\text{13}\text{.}0\right)$
giving
$\theta =\text{35}\text{.}6º\text{.}$
Discussion
The wind’s speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in [link] . Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.
Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity.
## Relative velocities and classical relativity
When adding velocities, we have been careful to specify that the velocity is relative to some reference frame . These velocities are called relative velocities . For example, the velocity of an airplane relative to an air mass is different from its velocity relative to the ground. Both are quite different from the velocity of an airplane relative to its passengers (which should be close to zero). Relative velocities are one aspect of relativity , which is defined to be the study of how different observers moving relative to each other measure the same phenomenon.
Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879–1955), the greatest physicist of the 20th century. Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later chapters. The relative velocities in this section are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is, less than . Most things we encounter in daily life move slower than this speed.
Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving forward? The answer is that if air resistance is negligible, the binoculars will hit at the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See [link] .) To the observer on shore, the binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This observer sees the curved path shown in [link] . Although the paths look different to the different observers, each sees the same result—the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities relative to the observer.
Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time.
hour glass, pendulum clock, atomic clock?
S.M
tnks
David
how did they solve for "t" after getting 67.6=.5(Voy + 0)t
Find the following for path D in [link] : (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.
the topic is kinematics
David
can i get notes of solid state physics
Lohitha
just check the chpt. 13 kinetic theory of matter it's there
David
is acceleration a fundamental unit.
no it is derived
Abdul
no
Nisha
K thanks
David
no it's not its derived
Emmanuel
hi
Hello
Emmanuel
hello
David
Hello Emmanuel
Emmanuel
I'm good
that's good
Emmanuel
how are you too
am cool
Emmanuel
spending time summarizing
Emmanuel
Emmanuel
I am fin
Longwar
ok
hi guys can you teach me how to solve a logarithm?
how about a conceptual framework can you simplify for me? needed please
Villaflor
Hello what happens when electrone stops its rotation around its nucleus if it possible how
Afzal
I think they are constantly moving
Villaflor
yep what is problem you are stuck into context?
S.M
not possible to fix electron position in space,
S.M
Physics
Beatriz
yes of course Villa flor
David
equations of kinematics for constant acceleration
A bottle full of water weighs 45g when full of mercury,it weighs 360g.if the empty bottle weighs 20g.calculate the relative density of mercury and the density of mercury....pls I need help
well You know the density of water is 1000kg/m^3.And formula for density is density=mass/volume Then we must calculate volume of bottle and mass of mercury: Volume of bottle is (45-20)/1000000=1/40000 mass of mercury is:(360-20)/1000 kg density of mercury:(340/1000):1/50000=(340•40000):1000=13600
Sobirjon
the latter is true
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture...take density of water as 1g/cm3 and density of liquid 1.2g/cm3
Lila
plz hu can explain Heisenberg's uncertainty principle
who can help me with my problem about acceleration?
ok
Nicholas
how to solve this... a car is heading north then smoothly made a westward turn during the travel the speed of the car remains constant at 1.5km/h what is the acceleration of the car? the total travel time of the car as it smoothly changed its direction is 15 minutes
Vann
i think the acceleration is 0 since the car does not change its speed unless there are other conditions
Ben
yes I have to agree, the key phrase is, "the speed of the car remains constant...," all other information is not needed to conclude that acceleration remains at 0 during the entire time
Luis
who can help me with a relative density question
Lila
1cm3 sample of tin lead alloy has mass 8.5g.the relative density of tin is 7.3 and that of lead is 11.3.calculate the percentage by weight of tin in the alloy. assuming that there is no change of volume when the metals formed the alloy
Lila
morning, what will happen to the volume of an ice block when heat is added from -200°c to 0°c... Will it volume increase or decrease?
no
Emmanuel
hi what is physical education?
Kate
BPED..is my course.
Kate
No
Emmanuel
I think it is neither decreases nor increases ,it remains in the same volume because of its crystal structure
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture. take density of water as 1g/cm3 and density of liquid as 1.2g/cm3
Lila
Sorry what does it means"no changes in volume occured"?
Sobirjon
volume can be the amount of space occupied by an object. But when an object does not change in shape it will still occupy the same space. Thats why the volume will still remain the same
Ben
Most soilds expand when heated but if it changes state at 0C it will have less volume. Ice floats because it is less dense ie a larger mass per unit volume.
Richard
how to calculate velocity
v=d/t
Emeka
Villaflor
Villaflor
v=d/t
Nisha
hello bro hw is life with you
Mine is good. How about you?
Chase
Hi room of engineers
yes,hi sir
Okwethu
hello
akinmeji
Hello
Mishael
hello
Jerry
hi
Sakhi
hi
H.C
so, what is going on here
akinmeji
Ajayi
good morning ppl
ABDUL
If someone has not studied Mathematics enough yet, should theu study it first then study Phusics or Study Basics of Physics whilst srudying Math as well?
whether u studied maths or not, it is advisable to start from d basics cuz it is essential to know dem
Nuru
yea you are right
wow, you got this w/o knowing math
Thomas
I guess that's it
Thomas
later people
Thomas
mathematics is everywhere
Anand
thanks but dat doesn't mean it is good without maths @Riaz....... Maths is essential in sciences particularly wen it comes to PHYSICS but PHYSICS must be started from the basic which may also help in ur mathematical ability
Nuru
A hydrometer of mass 0.15kg and uniform cross sectional area of 0.0025m2 displaced in water of density 1000kg/m3.what depth will the hydrometer sink
Lila
16.66 meters?
Darshik
16.71m2
aways
,i have a question of let me give answer
aways
the mass is stretched a distance of 8cm and held what is the potential energy? quick answer
aways
oscillation is a to and fro movement, it can also be referred to as vibration. e.g loaded string, loaded test tube or an hinged door |
Parabola, Horizontal Translation | Zona Land Education
# Parabola, Horizontal Translation
Horizontal translation for the parabola is changed by the value of a variable, h, that is subtracted from x before the squaring operation. So, our starting or reference parabola formula looks like this:
y = x2
And our equation that includes a horizontal translation looks like this:
y = (x - h)2
So, if h = 6, we say that the reference parabola is horizontally translated by 6 units, and our equation for this would appear:
y = (x - 6)2
Here's the graph for this translation. The reference parabola ( y = x2 ) is drawn in transparent light gray, and the transformed parabola which is horizontally translated 6 units ( y = (x - 6)2 ) is drawn in black:
What follows is an animation that presents many horizontal translations for our reference parabola. Note that the value for h is subtracted from x, and then this difference is squared. See that positive values for h translate the parabola to the right, negative values for h translate the parabola to the left. When h = 0, the reference and transformed parabola would be the same.
Please understand that x^2 means x2.
It may not seem correct at first that you must subtract from x to move the parabola to the right, but that is how it works. Subtractions on the x-axis are usually related with movements to the left, but not here. If you want to move the parabola to the right, say, 4 units, then you must subtract 4 from x and then square that result to get your y-coordinate.
Be sure to understand that the subtraction from x occurs before the squaring. Imagine that we translate the reference parabola 3 units to the right. Let's see what is the y-coordinate at x = 5:
y = (x - h)2
y = (x - 3)2
y = (5 - 3)2
y = (2)2
y = 4
How do you translate the reference parabola to the left? Well, you subtract a negative number from x. This looks like an addition to x. Let's calculate the y-coordinate on our transformed parabola when we translate the reference parabola 6 units to the left. We will calculate at x = 2:
y = (x - h)2
y = (x - (-6))2
y = (2 - (-6))2
y = (2 + 6)2
y = (8)2
y = 64
So, if you wish to move the reference parabola to the right, subtract a positive number from x.
If you want to move the reference parabola to the left, subtract a negative number from x. Since we are subtracting a negative number, this could look like an addition. Here's an example:
Subtracting a negative number from x:
y = (x - (-3))2
Can look like an addition:
y = (x + 3)2
So, a graph of this function:
y = (x + 4)2
Which we should think of as:
y = (x - (-4))2
Would look like the reference parabola shifted to the left 4 units:
And a graph of this function:
y = (x - 5)2
Would look like the reference parabola slid to the right 5 units:
Here is an EZ Graph example of this horizontal translation. Press the 'Draw graph' button.
You can change the value for h using the upper left input boxes. Press the 'Draw graph' button after you change h, and you will see how your change effects the graph. For more information about EZ Graph click the following link:
Help with EZ Graph |
# Definite integrals. Fundamental Theorem of Calculus
The Compound Effect is the principle of reaping huge rewards from a series of small, smart choices, Darren Hardy.
# Motivation
How to calculate the area of a circle of radius a? The best approximation is slicing the circle into many concentric rings. We can approximate the area of any of these concentric rings by 2πr·dr. So we can approximate the area of the whole circle by the sum of these areas (their values of r ranges from “0” to “a” spaced out by dr). This approximation gets more accurate as the size of dr gets smaller and smaller.
We can plot these areas and have a triangle with a base of “a” and a height that is 2π·a, so its area is $\frac{1}{2}·base·height = \frac{1}{2}·a·(2π·a) = π·a^2$
Futhermore, the sum of these rectangles approximates the area under the graph 2πr. More formally, $\int_{0}^{a} 2πrdr = 2π\int_{0}^{a} rdr = 2π\frac{r^2}{2} \bigg|_{0}^{a} = 2π\frac{a^2}{2}-2π\frac{0^2}{2}= π·a^2.$
# Definite integral
A definite integral is a mathematical concept used to find the area under a curve between two fixed limits. It is represented as $\int_{a}^{b}f(x)dx$, where a and b are the lower and upper limits, and f(x) is the integrand.
A definite integral of a function is defined as the area of the region bounded by the function's graph or curve between two points, say a and b, and the x-axis. The integral of a real-valued function f(x) on an interval [a, b] is written or expressed as $\int_{a}^{b} f(x)dx$ and it is illustrated in Figure 1.a. To compute the area, we can divide the region under the function into a series of rectangles whose areas corresponds to function values as heights that are multiplied by the step width to calculate the total sum. Of course, we need to rectify this sum (it is just an approximation) by taking a limit, and as the rectangles get infinitesimally thin, we achieve the desired result (Figure 1.b.).
Let’s try to calculate $\int_{0}^{b} x^2dx$ where a = 0 and b is arbitrary. We divide the segment [0, b] into n pieces (Figure 1.c.). The sum of the areas of all rectangles is equal to $(\frac{b}{n})·(\frac{b}{n})^2+(\frac{b}{n})·(\frac{2b}{n})^2+(\frac{b}{n})·(\frac{3b}{n})^2+···+(\frac{b}{n})·(\frac{nb}{n})^2 = b^3(\frac{1^2+2^2+3^2+···n^2}{n^3})$
To try to understand ${1^2+2^2+3^2+···n^2}$, we can observe that it is basically a pyramid, the base of which is n x n blocks. On top of it, we put a second layer of n-1 x n-1 blocks (Figure 1.d). And we keep on piling up layer after layer. At the top of the pyramid, there is just one block of stone. Underneath our staircase pyramid, there is an ordinary pyramid, and its volume is $\frac{1}{3}(baselength~ ·~ baseheight)·height=\frac{1}{3}n^2~ ·~ n$ On the outside, there is also another ordinary pyramid, with base_length, base_height, and height (n+1), (n+1), and (n+1) respectively.
So our initial quantity ${1^2+2^2+3^2+···n^2}$ is to be found or bounded between these two pyramids: $\frac{1}{3}n^3 < 1^2+2^2+3^2+···n^2 < \frac{1}{3}(n+1)^3$ (Figure 1.e.) If we divide by n3, $\frac{1}{3} < \frac{1^2+2^2+3^2+···n^2}{n^3} < \frac{1}{3}(1+\frac{1}{n})^3$. When n→ infinite, the left and right side tends to $\frac{1}{3}$, and by the Squeeze Infinite, $\frac{1^2+2^2+3^2+···n^2}{n^3} → \frac{1}{3}$ Therefore, our total area $\int_{0}^{b} x^2dx=\frac{1}{3}b^3.$
A Riemann sum is another approximation of an integral by a finite sum. The interval is partitioned into n pieces and we pick any height of f in each interval (e.g. f(ci)) The sum is defined as $\sum_{i=1}^n f(c_i)\Delta x,~ where~\Delta x=\frac{b-a}{n}$ Please notice that in the limit, as the rectangles get thinner and thinner, the difference between the area covered by the rectangles and the area under the curve will get smaller and smaller, and finally it does not matter our choices (ci), we get the desired result, that is, $\int_{a}^{b} f(x)dx$.
• Example: Cumulative Debt. Let t be time in years and f(t) be the borrowing rates (euros/years). If we are borrowing money every day, then $\Delta t=\frac{1}{365}.$ The borrowing rate varies over the year. On Day 7 (t=7/365), we borrowed $f(\frac{7}{365})\Delta t = f(\frac{7}{365})\frac{1}{365}$ where f(t) is measured in euros per year, and $\Delta t$ is measured in years, the multiplication is a number of euros that you actually borrow on that very day.
How much money do we borrow in the whole year? $\sum_{i=1}^{365} f(\frac{i}{365})\Delta t$ → [It is going to be very close] $\int_{0}^{1} f(t)dt$
How much money do you actually owe? Obviously, and regretfully, the interest on our debt is compounded continuously. If P is our principal (we start with a debt of P), then after some time t you owe Pert where r is the interest rate.
Compound interest is an interest calculated on both the principal and the existing interest over a given period of time. Its formula is $P(1 +\frac{r}{n})^{nt}$ where n is the number of terms the initial amount or principal P is compounding in the time t. If our debt is compounded continuously, then after time we owe $\lim_{n \to ∞} P(1 + \frac{r}{n})^{nt}=Pe^{rt}$ and we are using the fact that $\lim_{n \to ∞} (1 + \frac{r}{n})^{n} = e^r$.
So, you borrow these amounts $f(\frac{i}{365})\Delta t$, but when you borrow these quantities, the amount of time left in the year is 1 -i/365, which is the amount of time this incremental debt will accumulate interest, that is, $(f(\frac{i}{365})\Delta t)e^{r(1-\frac{i}{365})}$. An at the end of the year, $\sum_{i=1}^{365} (f(\frac{i}{365})\Delta t)e^{r(1-\frac{i}{365})}$ that is basically $\int_{0}^{1} e^{r(1-t)}f(t)dt$.
# Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus states roughly that the integral of a function f over an interval is equal to the change of any antiderivate F (F'(x) = f(x)) between the ends of the interval, i.e., $\int_{a}^{b} f(x)dx = F(b)-F(a)=F(x) \bigg|_{a}^{b}$
Examples:
• $\int_{0}^{2} (x^2 +1)dx= (\frac{1}{3}x^3+x)\bigg|_{0}^{2} = (\frac{1}{3}2^3+2)-(\frac{1}{3}0^3+0) = \frac{14}{3}$.
• $\int_{0}^{3} x^2+4dx=(\frac{1}{3}x^3 + 4x)\bigg|_{0}^{3}=(\frac{1}{3}·3^3+4·3)-(\frac{1}{3}·0^3+4·0) =21$.
• $\int_{0}^{π} sinxdx = (-cosx)\bigg|_{0}^{π}=-cosπ-(-cos0) = -(-1)-(-1) = 2$.
• $\int_{0}^{\frac{π}{4}} sin(2x)dx = \frac{-1}{2}cos(2x)\bigg|_{0}^{\frac{π}{4}} = -\frac{1}{2}cos(2·\frac{π}{4}) +\frac{1}{2}cos(2·0) = -\frac{1}{2}·0 + \frac{1}{2}·1 = \frac{1}{2}.$
• $\int_{0}^{\frac{π}{2}} sin(x)cos(x)dx$ =[Recall the double-angle indentity for sine, sin(2x) = 2sin(x)cos(x)] = $\frac{-1}{4}cos(2x)\bigg|_{0}^{\frac{π}{2}} = \frac{-1}{4}cos(2·\frac{π}{2}) + \frac{1}{4}cos(2·0) = \frac{-1}{4}·(-1) + \frac{1}{4} = \frac{1}{2}.$
• $\int_{0}^{1} x^4dx =\frac{x^5}{5}\bigg|_{0}^{1}=\frac{1}{5}$
• Find the total area between f(x) = x -2 and the x-axis over the interval [0, 6] (Figure 1)
1. Calculate the x-intercept of f(x): (2, 0).
2. Total area = Area below the x-axis over the subinterval [0, 2] (A2) + Area above the x-axis over the subinterval [6, 2] (A1). $\int_{0}^{6} |f(x)| = A_2 + A_1$
3. Using basic trigonometry, $A_2 = \frac{1}{2}bh = \frac{1}{2}·2·2 = 2, A_1 = \frac{1}{2}·4·4 = 8$ ⇒ A = 2 + 8 = 10.
• By using a definite integral find the area of the region bounded by y = x and y = $3\sqrt{x}$ (Figure 3).
1. First, we need to find the points of intersection between these two curves. $x = 3\sqrt{x}↭x^2 = 9·x↭ x(x-9) = 0 ↭$ x = 0 and 9
2. Then, we integrate the difference between the upper and lower curves over that interval, $\int_{0}^{9} (3\sqrt{x}-x)dx = 3·\frac{x^{\frac{3}{2}}}{\frac{3}{2}} -\frac{x^2}{2}\bigg|_{0}^{9} = 2·9^{\frac{3}{2}} -\frac{81}{2} = 13.5.$
• By using a definite integral find the area of the region bounded by y = 6x -x2 and y = 0 (Figure 2).
1. Calculate the x-intercepts of f(x) = 6x -x2 and the x-axis, 6x -x2 = x(6-x) = 0 ⇒ the x-intercepts are (0, 0) and (6, 0).
2. Total area = Area above the x-axis between 0 and 6. $\int_{0}^{6} 6x -x^2 = 3x^2 -\frac{1}{3}x^3\bigg|_{0}^{6} = 3·6^2 -\frac{1}{3}6^3 - (3·0^2 -\frac{1}{3}·0^3) = 36.$
# Intuitive interpretation of Fundamental Theorem
For some distance s(t), the velocity function is v(t)=s’(t). Therefore, given a velocity function, v(t) (v(t) is what is measured in the speedometer), $\int_{a}^{b} v(t)dt$ =s(b)-s(a) where s(b) - s(a) is the actual distance traveled, that is, what is measured in the odometer, that is, the instrument on your car dashboard that displays how many miles or kilometers your vehicle has traveled.
Let’s say that you checked your speedometer every single second (Is your phone out of juice? 😃. Seriously, $\Delta t = 1~ second$), $v(t_i)\Delta t$ is the distance traveled in the ith second, the Riemman Sum is equal $\sum_{i=1}^n v(t_i)\Delta t≈\int_{a}^{b} v(t)dt$ = s(b)-s(a) -how far you have traveled-.
# Reality is always a little more complex
Let’s notice that $\int_{0}^{2π} sinxdx = (-cosx)\bigg|_{0}^{2π}=-cos2π-(-cos0) = -1 -(-1) = 0$ (Figure 1.d.)
💡A definite integral of a function should be interpreted as the signed area of the region in the plane that is bounded by its graph and the horizontal axis. Be extremely careful, areas above the horizontal -x- axis of the plane are positive while areas below the horizontal axis are considered negative (Figure 1.e.).
# Bibliography
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
Bitcoin donation |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Combinations of Decimal Money Amounts
## Express combinations of coins as decimal amounts.
Estimated4 minsto complete
%
Progress
Practice Combinations of Decimal Money Amounts
Progress
Estimated4 minsto complete
%
Combinations of Decimal Money Amounts
Julie and her friend Jose are working at an ice cream stand for the summer. They are excited because in addition to making some money for the summer, they also get to eat an ice cream cone every day. On the first day on the job, Julie is handed a cash register drawer that is filled with money. This is the drawer that she can collect money from sales in as well as make change for customers. Julie needs to count the amount of money in her drawer to be sure that it is accurate. Her boss Mr. Maguire tells her that her drawer should have sixty-five dollars and seventy-five cents in it. He hands her a data sheet that she needs to write that money amount in on. Julie looks at the bills in her drawer and begins to count. She finds 2-20 dollar bills, 2-ten dollar bills, 1-five dollar bill and 2 quarters, 2 dimes and 1 nickel.
Now it is your turn to help.
In this Concept, you will learn all about decimals. One of the most common places that we see decimals is when we are working with money. Your work with decimals and place value will help Julie count her bills and change accurately.
### Guidance
How can we apply what we have learned about decimals in a real world way?
Money is a way that we use decimals every day.
Coins are cents. If we have 50 pennies, then we have 50 cents. It takes 100 pennies to make one dollar or one whole. Coins are parts of one dollar. We can represent coins in decimals.
A penny is one cent or it is one out of 100. When we have a collection of pennies, we have so many cents out of 100.
5 pennies is 5 cents.
How can we write 5 cents as a decimal?
To do this, we need to think about 5 out of 100. We can say that 5 cents is 5 hundredths of a dollar since there are 100 pennies in one dollar.
Let’s write 5 cents as a decimal using place value.
Hundred Tens Ones Tenths Hundredths Thousandths
Ten
Thousandths
. 5
The five is in the hundredths box because five cents is five one hundredths of a dollar. We need to add a zero in the tenths box to fill the gap.
Hundred Tens Ones Tenths Hundredths Thousandths
Ten
Thousandths
. 0 5
Now we have converted 5 cents to a decimal.
How can we write 75 cents as a decimal?
First, think about what part of a dollar 75 cents is. Seventy-five cents is seventy-five out of 100. Now, we can put this into our place value chart.
Hundred Tens Ones Tenths Hundredths Thousandths
Ten
Thousandths
. 7 5
Now we have written it as a decimal.
What about when we have dollars and cents?
Suppose we have twelve dollars and fourteen cents. A dollar is a whole number amount. Dollars are found to the left of the decimal point. Cents are parts of a dollar. They are found to the right of the decimal point. How much money do we have? There is one ten and the two ones gives us twelve dollars. Then we have some change. One dime and four pennies is equal to fourteen cents.
Here are the numbers:
12 wholes
14 parts
Let’s put them in our place value chart.
Hundred Tens Ones Tenths Hundredths Thousandths
Ten
Thousandths
1 2 . 1 4
#### Example B
Write eighty-nine cents as a decimal.
Solution: .89
#### Example C
Write fifteen dollars and twenty-five cents as a decimal.
Solution: $15.25 Remember Julie and her cash drawer? Now that we know about decimals and money we are ready to help Julie with her ice cream shop dilemma. Julie and her friend Jose are working at an ice cream stand for the summer. They are excited because in addition to making some money for the summer, they also get to eat an ice cream cone every day. On the first day on the job, Julie is handed a cash register drawer that is filled with money. This is the drawer that she can collect money from sales in as well as make change for customers. Julie needs to count the amount of money in her drawer to be sure that it is accurate. Her boss Mr. Maguire tells her that her drawer should have sixty-five dollars and seventy-five cents in it. He hands her a data sheet that she needs to write that money amount in on. Julie looks at the bills in her drawer and begins to count. She finds 2 twenty dollar bills, 2 ten dollar bills, 1 five dollar bill and 2 quarters, 2 dimes and 1 nickel. First, let’s underline all of the important information. Now, let’s count the money she has in the drawer. How many whole dollars are there? There are 2 Twenty Dollar bills =$40 plus 2 Ten Dollar bills = $20 plus 1 Five Dollar bill =$5.
The total then is $40 +$20 + $5 =$65.
How many cents are there?
There are 2 Quarters at $25. each =$.50 plus 2 Dimes at $.10 each =$.20 plus 1 Nickel at $.05 =$.05
The total then is $.50 +$.20 + $.05 =$.75
Our next step is to write the wholes and parts in the place value chart. Then we will have this written as a money amount.
Hundred Tens Ones Tenths Hundredths Thousandths
Ten
Thousandths
6 5 . 7 5
Great work! Julie has \$65.75 in her drawer. That is the correct amount. She is ready to get to work.
### Vocabulary
Whole number
a number that represents a whole quantity
Decimal
a part of a whole
Decimal point
the point in a decimal that divides parts and wholes
Expanded form
writing out a decimal the long way to represent the value of each place value in a number
### Guided Practice
Here is one for you to try on your own.
Maggie found a ten dollar bill, two five dollar bills, three quarters and two pennies in her wallet. How much money did she find? Write this amount as a decimal.
First, let's add the dollar amounts. We can write these amounts as decimals.
\begin{align*}10.00 + 5.00 + 5.00 = 20.00\end{align*}
Next we can add the change.
\begin{align*}.75 + .02 = .77\end{align*}
Now we can combine the amounts.
\begin{align*}20.00 + .77 = \20.77\end{align*}
### Video Review
http://www.teachertube.com/viewVideo.php?title=Money_Fractions_and_Decimals&video_id=59116&vpkey=4badb7d45d – This video is a short story and features two students learning about money with fractions and decimals.
### Practice
Directions: Write the following money amounts as decimals.
1. Ten dollars and fifty cents
2. Five dollar bills, two quarters and three pennies
3. A twenty dollar bill, a ten dollar bill and three dimes
4. Fifteen nickels and two quarters
5. Six dollar bills, three quarters and a nickel
6. A dime and fifteen pennies
7. Two dollars, ten dimes and two quarters
8. Three quarters, five nickels and ten pennies
9. Sixteen dollar bills, two quarters, a nickel and ten pennies
10. A ten dollar bill, a five dollar bill, one quarter, three nickels, a dime and ten pennies
11. A ten dollar bill, a quarter, a dime, five nickels and fifteen pennies
12. A five dollar bill, two quarters, two dimes, a nickel and two pennies
13. Two dollar bills, a quarter and five pennies
14. Three dollar bills, a twenty dollar bill, two quarters and five dimes
15. A dollar, three quarters, two dimes, five nickels and two pennies
### Vocabulary Language: English
Decimal
Decimal
In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
Decimal point
Decimal point
A decimal point is a period that separates the complete units from the fractional parts in a decimal number.
Expanded Form
Expanded Form
Expanded form refers to a base and an exponent written as repeated multiplication.
Whole Numbers
Whole Numbers
The whole numbers are all positive counting numbers and zero. The whole numbers are 0, 1, 2, 3, ... |
# 6.1 Add & Subtract Polynomial Expression & Functions
Size: px
Start display at page:
Transcription
1 6.1 Add & Subtract Polynomial Expression & Functions Objectives 1. Know the meaning of the words term, monomial, binomial, trinomial, polynomial, degree, coefficient, like terms, polynomial funciton, quardrtic funciton, vertex, and cubic function. 2. Add & Subtract Polynomials 3. Evaluate polynomial funcitons 4. Know typical graphs of polynomial funcitons. 5. Find the sum function and difference funciton. 6. Use the sum and difference functions to model situations.
2 6.1 Add & Subtract Polynomials (Page 2 of 33) Polynomials A term is a constant, a variable, or a product of a constant and one or more variables raised to powers. A monomial is a constant, a variable, or a product of a constant and one or more variables raised to counting number (i.e. positive integer) powers. A binomial is the sum of two monomials. e.g. 45,!7x!x 1/2, 2x 3 y!4 e.g. 5, 3x x, 2x 3 y 4 e.g. 3x + 2y A trinomial is the sum of three monomials. e.g. x 2! x + 7 A polynomial is a monomial or a sum of monomials. e.g. 5x 3! 2x 2 + 7x! 9, 4x 5 y! xy 2,!2x + 8, 3, x The polynomial 4x 3! 2x 2 + x +12 is a polynomial in one variable with four terms, namely 4x 3,!2x 2, x and 12. By convention, we write polynomial so that the exponents decrease from left to right, which is called descending order. The degree of a term in one variable is the exponent on the variable. For example, the degree of 2y 4 is four. The degree of a term in two or more variables is the sum of the exponents on the variables. For example, the degree of!x 3 y 4 is seven. The degree of a polynomial is the highest degree of any term in the polynomial. The coefficient of a term is the constant factor of the term. The leading coefficient of a polynomial is the coefficient of the term with the highest degree. A linear polynomial has degree 1. A quadratic polynomial has degree 2. A cubic polynomial has degree 3. A constant has degree 0.
3 6.1 Add & Subtract Polynomials (Page 3 of 33) Example 1 Write each polynomial in descending order. Then use words such as linear, quadratic, cubic, polynomial, degree, one variable, two variables, coefficient and leading coefficient to describe each term and each expression. a.!3+ 8x! 4x 2 b. 4x + 2x c. 12 d. 3a 6 b 2 + 7ab 3! 3a 4 b Like Terms Like terms are either constant terms, or terms with identical variable parts (including exponents). The distributive property makes it possible to combine like terms by adding the coefficients while leaving the variable parts unchanged. For example,!3x 2 + 7x 2 = (!3+ 7)x 2 = 4x 2 Example 2 Simplify (combine like terms in descending order). 1. 5a 3! 4a 2 + 7a 3! a p 2 t 2 + p 2 t! 8 p 3 t 2! 9 p 2 t
4 6.1 Add & Subtract Polynomials (Page 4 of 33) Example 3 Perform the indicated operation and simplify. 1. (8a 2! 7ab + 2b 2 ) + (3a 2 + 4ab! 7b 2 ) 2. (5x 3! x + 7)! (!8x 3 + 3x 2! 6) Polynomial Function A polynomial function is a function that can be written in the form f (x) = polynomial. Some examples are f (x) = 4x 5! 3x 3 + 7, g(x) = 2.3x 3! 7x, h(x) = 7 9 x. A polynomial function of degree two is called a quadratic function. Quadratic Function in Standard Form A quadratic function in standard form is a function that can be written as f (x) = ax 2 + bx + c, where a! 0. Furthermore, the quadratic term is ax 2, the linear term is bx, and the constant term is c. For example, f (x) =!x 2 + 7x is quadratic because a =!1, b = 7 and c = 0. Example 4 For f (x) =!2x 2 + 5x!1, find each of the following. 1. f (4) 2. f (!3) 3. f (0)
5 6.1 Add & Subtract Polynomials (Page 5 of 33) Example 5 Sketch the graph of f (x) = x 2 the TABLE and graphing capabilities of your calculators.. 5 y x f (x) = x x Parabolas The graph of a quadratic function is called a parabola and has the shape illustrated. Parabolas can open downward like function g, or open upward like function f. If the parabola opens upward, then the vertex is located at the minimum point of the graph. If the parabola opens downward, then the vertex is located at the maximum point of the graph. The vertical line that goes through the vertex is called the axis of symmetry of the parabola.
6 6.1 Add & Subtract Polynomials (Page 6 of 33) Example 6 Reading Parabolas 1. Identify the vertex in the graph of function f. Is it a maximum point or minimum point? What is the equation of the axis of symmetry? 2. Identify the vertex in the graph function g. Is it a maximum point or minimum point? What is the equation of the axis of symmetry? 3. Find f (6) 4. Find x when f (x) = 4 5. Find g(!5) 6. Find x when g(x) =!8
7 6.1 Add & Subtract Polynomials (Page 7 of 33) Cubic Function A cubic function is a 3 rd -degree polynomial function and can be written in the form f (x) = ax 3 + bx 2 + cx + d, where a! 0. Example 7 Cubic Function Sketch the graph of f (x) = x 3. y x f (x) = x x Graphs of Typical Cubic Functions See note-guide, p. 33 for more information. y =!0.5(x! 3) 3 y = 0.25(x + 6)(x! 3) 2 y = (x + 2)(x! 2)(x! 5)
8 6.1 Add & Subtract Polynomials (Page 8 of 33) Sum and Difference Functions If f and g are functions and x is in the domain of both functions, then we can for the following functions: 1. Sum Function f + g ( f + g)(x) = f (x) + g(x) 2. Difference Function f - g ( f! g)(x) = f (x)! g(x) Example 8 Find the Sum & Difference Functions Let f (x) = 5x 2! x + 3 and g(x) =!2x 2 + 9x! Find the equation for f + g. 2. Find ( f + g)(2) 3. Find the equation for f g. 4. Find ( f! g)(2)
9 6.1 Add & Subtract Polynomials (Page 9 of 33) Example 9 Modeling The enrollments (in millions) at U.S. colleges W (t) and M (t) for women and men, respectively, are modeled by the system W (t) = 0.15t M (t) = 0.072t where t is the number of years since Find the equation for the sum function W + M. 2. Find (W + M )(31) and explain its meaning in this situation. 3. Find the equation for the difference function W - M. 4. Find (W! M )(31) and explain its meaning in this situation.
10 6.2 Multiplying Polynomial Expressions & Functions (Page 10 of 33) 6.2 Multiplying Polynomial Expressions & Functions The Factored form of the expression is written as a product. Multiplying a(b+c) = ab+ ac Factoring After multiplying, the expression is written as a sum. Example 1 Find the product (multiply). 1. 4x(x! 6) 2.!2a(3! 5a) Example 2 Finding Products Find the product (multiply). 1. (c + 4)(c + 5) 2. (x! 7)(x + 4) 3. (5 p! 6w)(3p + 2w) 4. (2a 2! 5b 2 )(4a 2! 3b 2 )
11 6.2 Multiplying Polynomial Expressions & Functions (Page 11 of 33) Example 3 Find the product (multiply). 1. 4x(x 2 + 2)(x! 3) 2. (2x + y)(5x 2! 3xy + 4y 2 ) 3. (x 2! 3x + 2)(3x 2 + x! 5) Example 4 Square a Binomial Find the product (multiply). 1. (b! 6) 2 2. (2x + 7) 2 Squaring a Binomial 1. (a + b) 2 = a 2 + 2ab+ b 2 2. (a! b) 2 = a 2! 2ab+ b 2 (a + b) 2 (a! b) 2 = First binomial term squared + Twice the product of the two binomial terms + Second binomial term squared
12 6.2 Multiplying Polynomial Expressions & Functions (Page 12 of 33) Squaring a Binomial 1. (a + b) 2 = a 2 + 2ab+ b 2 2. (a! b) 2 = a 2! 2ab+ b 2 (a + b) 2 (a! b) 2 = First binomial term squared + Twice the product of the two binomial terms + Second binomial term squared Example 5 1. Expand ( y + 7) 2 2. Expand (x + 5) 2 3. Expand (2b! 7) 2 4. Expand (5! 2x) 2 5. Expand (3a + 4b) 2 6. Expand (!5x + 7 y) 2
13 6.2 Multiplying Polynomial Expressions & Functions (Page 13 of 33) Product of Binomial Conjugates The binomials 2x! 7 and 2x + 7 are binomial conjugates. In general, the sum and difference of two terms ( A + B and A! B) are binomial conjugates of each other. Example 6 Find the product. 1. (c + 6)(c! 6) The Product of Binomial Conjugates = Difference of Two Squares (a + b)(a! b) = a 2! b 2 2. (x! 5)(x + 5) 3. (2b! 7)(2b + 7) 4. (6! 5x)(6 + 5x) 5. (4m 2! 7rt)(4m 2 + 7rt) 6. (x + 3)(x! 3)(x 2 + 9)
14 6.2 Multiplying Polynomial Expressions & Functions (Page 14 of 33) Example 7 For f (x) = x 2! 5x, find the following. 1. f (a! 3) 2. f (a + 2)! f (a) Example 8 Write f (x) =!3(x! 4) in standard form ( f (x) = ax 2 + bx + c ). Product Function If f and g are functions and x is in the domain of both functions, then we can form the product function f! g : ( f! g)(x) = f (x)! g(x) Example 9 Let f (x) =!3x + 7 and g(x) = 5x! 2. Find 1. ( f! g)(x) 2. ( f! g)(2)
15 6.2 Multiplying Polynomial Expressions & Functions (Page 15 of 33) Example 10 Let C(t) = 2.75t +102 represent the annual cost (in dollars) of prisons per person in the U.S. for t years since Let P(t) = 3.3t represent the U.S. population (in millions of people) at t years since Find the equation for the product function C! P. 2. Perform unit analysis on the function C(t)! P(t). That is what are the units of the product function. 3. Find (C! P)(20). Explain its meaning in this application. 4. Use a graphing calculator to determine whether the function C! P is increasing, decreasing, or neither for values of t between 5 and 20. What does the result mean in this situation?
16 6.3 Factoring Quadratic Polynomials (Page 16 of 33) 6.3 Factoring x 2 + bx + c = 1x 2 + bx + c Notice the patterns that develops in the following products. Last terms in the factored form The coefficient on the quadratic term is one (i.e. x 2 =1x 2 ) The coefficient of x in expanded form is the sum of the last terms in factored form (x + p)(x + q) = x 2 + qx + px + pq = x 2 + ( p + q)x + pq (x + 3)(x + 4) = x x + 3x +12 = x x +12 (x! 4)(x! 6) = x 2! 6x! 4x + 24 = x 2!10x + 24 (x + 5)(x! 7) = x 2! 7 x + 5x! 35 = x 2! 2x! 35 (x! 3)(x + 9) = x x! 3x! 27 = x x! 27 The constant term in expanded form is the product of the last terms in factored form Three Observations 1. The coefficient on the quadratic term is one (i.e. x 2 = 1x 2 ). 2. The constant term in each trinomial (i.e. pq) is the product of the constant terms in the factored form (i.e. the p and the q). 3. The linear coefficient in each trinomial (i.e. p + q) is the sum of the constant terms in the factored form. Steps to Factor the Quadratic Expression 1x 2 + bx + c = x 2 + bx + c = (x + p)(x + q) 1. List all the pairs of integers whose product is c. 2. Out of the list from step 1 find the pair of integers whose sum is b; those two integers are p and q. 3. Write the factored form of the expression: (x + p)(x + q).
17 6.3 Factoring Quadratic Polynomials (Page 17 of 33) Example 1 1. Factor x 2 + 8x The graph of f (x) = x 2 + 8x +12 is shown. What are the x-intercepts in the graph of f? 3. Write f in factored form. If f (r) = 0, then r is a zero of f. That is, a zero of a function is the input number that makes the output number zero. To find a zero of a function, set the function equal to zero and solve. Notice that if r is a zero, then (r, 0) is the x-intercept. 4. Find the zeros of f. Example 2 1. Factor x 2! 8x +12 y = x 2! 8x Factor x 2! x! 20 y = x 2! x! Factor x 2 + x! 20 y = x 2 + x! 20
18 6.3 Factoring Quadratic Polynomials (Page 18 of 33) Example 3 1. Factor a 2 +12a Factor p 2! 9 p Factor c 2! 3c! Factor y 2! 21y! 72 When the Quadratic Coefficient is not One If the coefficient on the quadratic term is not one, then the quadratic expression must be treated differently. The first thing to consider is whether or not there is a common factor in all of the terms of the expression that can be factored out. Example 4 Factor Out the GCF Factor [completely]. 1. 8x 3! 32x 2. 5x x x
19 6.3 Factoring Quadratic Polynomials (Page 19 of 33) Example 5 Factor 1. Factor!24x x 3! 27x 2 2. Factor!9y y! Factor 3x 2!15x Factor 5x 2!15x!140
20 6.4 Factoring Polynomials (Page 20 of 33) 6.4 Factoring Polynomials Example 4 Factor by Grouping (4-term polynomials) 1. Factor x 3! 2x 2 + 5x!10 2. Factor 10x 3! 6x 2 + 5x! 3 Steps to Factor the Quadratic Expression ax 2 + bx + c 1. List all pairs of integers whose product is ac. Out of that list find the pair of integers whose sum is b; call those two integers m and n so that b = m + n. 2. Rewrite the bx term as mx + nx so that ax 2 + bx + c = ax 2 + mx + nx + c 3. Group the first two terms and the last two terms of the fourterm expression and factor out the greatest common factor from each pair of terms. 4. Factor out the common binomial factor from the resulting expression. Example 5 Factor 3x 2! x! 4
21 6.4 Factoring Polynomials (Page 21 of 33) Example 6 1. Factor 10x 2! x! 3 2. Factor 3a 2!13a Factor 6z 2! 7z Factor 18y 2! 27 y Factor 15x 2! 22x + 8
22 6.5 Factoring Special Polynomials (Page 22 of 33) 6.5 Factoring Special Polynomials Example 1 1. Multiply (x + 5)(x! 5) 2. Multiply (a + b)(a! b) Factoring the Difference of Two Squares The Difference of Two Squares Example 2 1. Factor c 2! 64 The Product of a Sum and Difference of Two Terms A 2! B 2 = ( A + B)( A! B) 2. Factor 4x 2! Factor 18! 2d 2 4. Factor 16x 2! 36
23 6.5 Factoring Special Polynomials (Page 23 of 33) Factoring the Sum or Difference of Two Cubes A 3 + B 3 = ( A + B)( A 2! AB + B 2 ) Sum of two cubes A 3! B 3 = ( A! B)( A 2 + AB + B 2 ) Difference of two cubes Example 2 Factor 1. x x 3! t w x 5! 24x 2 y 3
24 6.5 Factoring Special Polynomials (Page 24 of 33) Example 3 Factor x 6! y 6 Factoring Guidelines / Strategies 1. If the GCF is not 1, then factor it out. 2. If a binomial is a difference of two square, then use a 2! b 2 = (a + b)(a! b) 3. For a four-term polynomial apply factor by grouping. 4. For a trinomial ax 2 + bx + c : a. If a = 1, then try to find two integers p and q whose product is c and whose sum is b. If the two numbers exist, then the factored form of the trinomial is (x + p)(x + q). b. If a! 1, then try to find two integers p and q whose product is ac and whose sum is b. If the two numbers exist, then rewrite bx = px + qx and factor the new fourterm polynomial by grouping. 5. Repeat all steps until no factors can be factored any further. Example 4 Factor Completely Factor [completely]. 1. 8! 2x 2 + x 3! 4x 2. x 2 +15x + 56
25 6.5 Factoring Special Polynomials (Page 25 of 33) Example 5 Factor Completely Factor [completely]. 1. 5x! 25x x 2! 22x 3 + 8x 4 3. x 4! 2x x! x 2!15x + 40x t 2 w 2! 8w a 3 b! 21a 2 b 2 +18ab 3
26 6.6 Solving polynomial equations by factoring (Page 26 of 33) 6.6 Solving Polynomial Equations by Factoring Quadratic Equation in One Variable A quadratic equation in one variable can be written in the form ax 2 + bx + c = 0, where a! 0, ax 2 is the quadratic term, bx is the linear term, and c is the constant term. Zero Factor Property ab = 0 if and only if a = 0 or b = 0 In words this states that a product is zero if and only if one of the factors is zero. To use this principle to solve equations, two requirements are necessary: (1) one side of the equation must be zero, and (2) the other side must be in factored form. Example 1 1. Solve (x + 5)(x! 3) = 0 2. Solve c(2c + 7) = 0 3. Solve x 2! 4x! 5 = 0 4. Solve 4x 2!11x + 7 = 0
27 6.6 Solving polynomial equations by factoring (Page 27 of 33) Example 2 1. Solve 4y! 7 =!3y 2 2. Solve (x + 2)(x! 4) = 7 3. Solve 25x 2 = Solve 1 4 x2 = 1 2 x + 2
28 6.6 Solving polynomial equations by factoring (Page 28 of 33) Zero of a Function If f (r) = 0, then r is a zero of f. That is, a zero of a function is the input number that makes the output number zero. To find a zero of a function, set the function equal to zero and solve. Example 3 1. Find the zeros of f (x) = x 2! 7x +10. f 2. Find the x-intercept(s) in the graph of f. 3. Write f in factored form. 4. Summarize your results: Zeros of f x-intercepts of f Factored form of f Equivalence of Zeros, x-intercepts, and Factors Let f (x) = ax 2 + bx + c, a! 0, and r be a real number. Then the following statements are equivalent. 1. f (r) = 0 read r is a zero of f 2. (r, 0) is an x-intercept in the graph of f. 3. (x! r) is a factor of f.
29 6.6 Solving polynomial equations by factoring (Page 29 of 33) Example 5 Let f (x) = x 2! 9x Write f in factored form. f 2. Find the following: Zeros of f x-intercepts of f Factored form of f 3. Find f (2). 4. Find the value(s) of x so that f (x) = The domain of f written in interval notation is 6. The range of f written in interval notation is
30 6.6 Solving polynomial equations by factoring (Page 30 of 33) Cubic Equation in One Variable A cubic equation in one variable can be written in the form ax 3 + bx 2 + cx + d = 0, where a! 0. The cubic term is ax 3, the quadratic term is bx 2, the linear term is cx and the constant term is d. Example 5 1. Solve 2x 3 = 42x + 8x Find the x-intercepts in the graph of f (x) = 2x 3! 8x 2! 42x y = 2x 3! 8x 2! 42x Example 6 1. Solve x 3! 5x 2! 4x + 20 = Find the x-intercepts in the graph of f (x) = x 3! 5x 2! 4x + 20 y = x 3! 5x 2! 4x + 20
31 6.6 Solving polynomial equations by factoring (Page 31 of 33) Facts About Cubic Equations and Functions 1. The cubic equation ax 3 + bx 2 + cx + d = 0 can have one, two or three real solutions. 2. The cubic function f (x) = ax 3 + bx 2 + cx + d can have one, two or three real zeros. 3. The graph of a cubic function f (x) = ax 3 + bx 2 + cx + d can have one, two or three x-intercepts. 4. The graph of a cubic function f (x) = ax 3 + bx 2 + cx + d can have either two turning points or no turning points. 5. The graph of a cubic function must either rise on the far left and fall on the far right, or fall on the far left and rise on the far right. 5. The domain of all cubic functions is all real numbers. 6. The range of all cubic functions is all real numbers. y =!0.5(x! 3) 3 y = 0.25(x + 6)(x! 3) 2 y = (x + 2)(x! 2)(x! 5)
32 6.6 Solving polynomial equations by factoring (Page 32 of 33) Example 7 Method 1 Solve on your graphing calculator x 2! x! 7 =!x Set Y 1 = x 2! x! 7 left side of the equation Y 2 =!x 2 right side of the equation 2. Since we want to know where the left side of the equation equals the right side, use the intersect program to find the points of intersection of the two functions. The x-values of the points of intersection are the solution to the original equation. 3. Check your solutions in the original equation. Example 7 Method 2 Solve on your graphing calculator x 2! x! 7 =!x Rewrite the equation so there is a zero on one side. x 2! x! 7 =!x 2 2x 2! x! 7 = 0 2. Set Y 1 = f (x) = 2x 2! x! 7 and find the zeros of f. 3. Check your solutions in the original equation.
33 6.6 Solving polynomial equations by factoring (Page 33 of 33) Example 10 The annual revenues for American Express are shown in the table. Let r(t) be the revenue (in billions of dollars) of American Express at t years since Find the appropriate regression equation (linear, exponential or quadratic) that models the data well. r(t) = 2. Predict the revenue in Year Revenue (billions of dollars) Predict when the revenue will be \$39.7 billion. Example 11 A person has a rectangular garden with a width of 9 feet and a length of 12 feet. She plans to place mulch outside of the garden to form a border of uniform width. She has just enough mulch to cover 100 square feet of land. Determine the length and width of the garden with its mulch border.
### 1.3 Polynomials and Factoring
1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. Term: a constant, variable, or the product or a constant and variable.
### Definitions 1. A factor of integer is an integer that will divide the given integer evenly (with no remainder).
Math 50, Chapter 8 (Page 1 of 20) 8.1 Common Factors Definitions 1. A factor of integer is an integer that will divide the given integer evenly (with no remainder). Find all the factors of a. 44 b. 32
### expression is written horizontally. The Last terms ((2)( 4)) because they are the last terms of the two polynomials. This is called the FOIL method.
A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are
### Factoring Polynomials
UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials A polynomial is an expression that has variables that represent numbers. A number can
### NSM100 Introduction to Algebra Chapter 5 Notes Factoring
Section 5.1 Greatest Common Factor (GCF) and Factoring by Grouping Greatest Common Factor for a polynomial is the largest monomial that divides (is a factor of) each term of the polynomial. GCF is the
### Section 6.1 Factoring Expressions
Section 6.1 Factoring Expressions The first method we will discuss, in solving polynomial equations, is the method of FACTORING. Before we jump into this process, you need to have some concept of what
### Polynomials. Key Terms. quadratic equation parabola conjugates trinomial. polynomial coefficient degree monomial binomial GCF
Polynomials 5 5.1 Addition and Subtraction of Polynomials and Polynomial Functions 5.2 Multiplication of Polynomials 5.3 Division of Polynomials Problem Recognition Exercises Operations on Polynomials
### Algebra I Vocabulary Cards
Algebra I Vocabulary Cards Table of Contents Expressions and Operations Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers Absolute Value Order of Operations Expression
### ( ) FACTORING. x In this polynomial the only variable in common to all is x.
FACTORING Factoring is similar to breaking up a number into its multiples. For example, 10=5*. The multiples are 5 and. In a polynomial it is the same way, however, the procedure is somewhat more complicated
### ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form
ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form Goal Graph quadratic functions. VOCABULARY Quadratic function A function that can be written in the standard form y = ax 2 + bx+ c where a 0 Parabola
### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.
Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}
### 1.3 Algebraic Expressions
1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts,
### Factoring Polynomials
Factoring a Polynomial Expression Factoring a polynomial is expressing the polynomial as a product of two or more factors. Simply stated, it is somewhat the reverse process of multiplying. To factor polynomials,
### Higher Education Math Placement
Higher Education Math Placement Placement Assessment Problem Types 1. Whole Numbers, Fractions, and Decimals 1.1 Operations with Whole Numbers Addition with carry Subtraction with borrowing Multiplication
### 6.3 FACTORING ax 2 bx c WITH a 1
290 (6 14) Chapter 6 Factoring e) What is the approximate maximum revenue? f) Use the accompanying graph to estimate the price at which the revenue is zero. y Revenue (thousands of dollars) 300 200 100
### Factoring Polynomials
Factoring Polynomials Factoring Factoring is the process of writing a polynomial as the product of two or more polynomials. The factors of 6x 2 x 2 are 2x + 1 and 3x 2. In this section, we will be factoring
### Vocabulary Words and Definitions for Algebra
Name: Period: Vocabulary Words and s for Algebra Absolute Value Additive Inverse Algebraic Expression Ascending Order Associative Property Axis of Symmetry Base Binomial Coefficient Combine Like Terms
### Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.
Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method
Factoring the trinomial ax 2 + bx + c when a = 1 A trinomial in the form x 2 + bx + c can be factored to equal (x + m)(x + n) when the product of m x n equals c and the sum of m + n equals b. (Note: the
### 1.1 Practice Worksheet
Math 1 MPS Instructor: Cheryl Jaeger Balm 1 1.1 Practice Worksheet 1. Write each English phrase as a mathematical expression. (a) Three less than twice a number (b) Four more than half of a number (c)
### 7.1 Graphs of Quadratic Functions in Vertex Form
7.1 Graphs of Quadratic Functions in Vertex Form Quadratic Function in Vertex Form A quadratic function in vertex form is a function that can be written in the form f (x) = a(x! h) 2 + k where a is called
### Algebra and Geometry Review (61 topics, no due date)
Course Name: Math 112 Credit Exam LA Tech University Course Code: ALEKS Course: Trigonometry Instructor: Course Dates: Course Content: 159 topics Algebra and Geometry Review (61 topics, no due date) Properties
### FACTORING QUADRATICS 8.1.1 and 8.1.2
FACTORING QUADRATICS 8.1.1 and 8.1.2 Chapter 8 introduces students to quadratic equations. These equations can be written in the form of y = ax 2 + bx + c and, when graphed, produce a curve called a parabola.
### Name Intro to Algebra 2. Unit 1: Polynomials and Factoring
Name Intro to Algebra 2 Unit 1: Polynomials and Factoring Date Page Topic Homework 9/3 2 Polynomial Vocabulary No Homework 9/4 x In Class assignment None 9/5 3 Adding and Subtracting Polynomials Pg. 332
### POLYNOMIALS and FACTORING
POLYNOMIALS and FACTORING Exponents ( days); 1. Evaluate exponential expressions. Use the product rule for exponents, 1. How do you remember the rules for exponents?. How do you decide which rule to use
### Factoring Trinomials: The ac Method
6.7 Factoring Trinomials: The ac Method 6.7 OBJECTIVES 1. Use the ac test to determine whether a trinomial is factorable over the integers 2. Use the results of the ac test to factor a trinomial 3. For
### Factoring a Difference of Two Squares. Factoring a Difference of Two Squares
284 (6 8) Chapter 6 Factoring 87. Tomato soup. The amount of metal S (in square inches) that it takes to make a can for tomato soup is a function of the radius r and height h: S 2 r 2 2 rh a) Rewrite this
### Veterans Upward Bound Algebra I Concepts - Honors
Veterans Upward Bound Algebra I Concepts - Honors Brenda Meery Kaitlyn Spong Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org Chapter 6. Factoring CHAPTER
### Factoring and Applications
Factoring and Applications What is a factor? The Greatest Common Factor (GCF) To factor a number means to write it as a product (multiplication). Therefore, in the problem 48 3, 4 and 8 are called the
### Mathematics Placement
Mathematics Placement The ACT COMPASS math test is a self-adaptive test, which potentially tests students within four different levels of math including pre-algebra, algebra, college algebra, and trigonometry.
### Chapter R.4 Factoring Polynomials
Chapter R.4 Factoring Polynomials Introduction to Factoring To factor an expression means to write the expression as a product of two or more factors. Sample Problem: Factor each expression. a. 15 b. x
### MATH 90 CHAPTER 6 Name:.
MATH 90 CHAPTER 6 Name:. 6.1 GCF and Factoring by Groups Need To Know Definitions How to factor by GCF How to factor by groups The Greatest Common Factor Factoring means to write a number as product. a
### HIBBING COMMUNITY COLLEGE COURSE OUTLINE
HIBBING COMMUNITY COLLEGE COURSE OUTLINE COURSE NUMBER & TITLE: - Beginning Algebra CREDITS: 4 (Lec 4 / Lab 0) PREREQUISITES: MATH 0920: Fundamental Mathematics with a grade of C or better, Placement Exam,
### Algebra II A Final Exam
Algebra II A Final Exam Multiple Choice Identify the choice that best completes the statement or answers the question. Evaluate the expression for the given value of the variable(s). 1. ; x = 4 a. 34 b.
### Factoring Polynomials and Solving Quadratic Equations
Factoring Polynomials and Solving Quadratic Equations Math Tutorial Lab Special Topic Factoring Factoring Binomials Remember that a binomial is just a polynomial with two terms. Some examples include 2x+3
### Operations with Algebraic Expressions: Multiplication of Polynomials
Operations with Algebraic Expressions: Multiplication of Polynomials The product of a monomial x monomial To multiply a monomial times a monomial, multiply the coefficients and add the on powers with the
### CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA
We Can Early Learning Curriculum PreK Grades 8 12 INSIDE ALGEBRA, GRADES 8 12 CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA April 2016 www.voyagersopris.com Mathematical
Polynomials and Quadratics Want to be an environmental scientist? Better be ready to get your hands dirty!.1 Controlling the Population Adding and Subtracting Polynomials............703.2 They re Multiplying
### Algebra 2 PreAP. Name Period
Algebra 2 PreAP Name Period IMPORTANT INSTRUCTIONS FOR STUDENTS!!! We understand that students come to Algebra II with different strengths and needs. For this reason, students have options for completing
### 1.3. Maximum or Minimum of a Quadratic Function. Investigate A
< P1-6 photo of a large arched bridge, similar to the one on page 292 or p 360-361of the fish book> Maximum or Minimum of a Quadratic Function 1.3 Some bridge arches are defined by quadratic functions.
### Algebra II End of Course Exam Answer Key Segment I. Scientific Calculator Only
Algebra II End of Course Exam Answer Key Segment I Scientific Calculator Only Question 1 Reporting Category: Algebraic Concepts & Procedures Common Core Standard: A-APR.3: Identify zeros of polynomials
### Tool 1. Greatest Common Factor (GCF)
Chapter 4: Factoring Review Tool 1 Greatest Common Factor (GCF) This is a very important tool. You must try to factor out the GCF first in every problem. Some problems do not have a GCF but many do. When
### SPECIAL PRODUCTS AND FACTORS
CHAPTER 442 11 CHAPTER TABLE OF CONTENTS 11-1 Factors and Factoring 11-2 Common Monomial Factors 11-3 The Square of a Monomial 11-4 Multiplying the Sum and the Difference of Two Terms 11-5 Factoring the
### FACTORING ax 2 bx c. Factoring Trinomials with Leading Coefficient 1
5.7 Factoring ax 2 bx c (5-49) 305 5.7 FACTORING ax 2 bx c In this section In Section 5.5 you learned to factor certain special polynomials. In this section you will learn to factor general quadratic polynomials.
### Factor Polynomials Completely
9.8 Factor Polynomials Completely Before You factored polynomials. Now You will factor polynomials completely. Why? So you can model the height of a projectile, as in Ex. 71. Key Vocabulary factor by grouping
### By reversing the rules for multiplication of binomials from Section 4.6, we get rules for factoring polynomials in certain forms.
SECTION 5.4 Special Factoring Techniques 317 5.4 Special Factoring Techniques OBJECTIVES 1 Factor a difference of squares. 2 Factor a perfect square trinomial. 3 Factor a difference of cubes. 4 Factor
### FACTORING POLYNOMIALS
296 (5-40) Chapter 5 Exponents and Polynomials where a 2 is the area of the square base, b 2 is the area of the square top, and H is the distance from the base to the top. Find the volume of a truncated
### Polynomial Expressions and Equations
Polynomial Expressions and Equations This is a really close-up picture of rain. Really. The picture represents falling water broken down into molecules, each with two hydrogen atoms connected to one oxygen
### Using the ac Method to Factor
4.6 Using the ac Method to Factor 4.6 OBJECTIVES 1. Use the ac test to determine factorability 2. Use the results of the ac test 3. Completely factor a trinomial In Sections 4.2 and 4.3 we used the trial-and-error
### Introduction Assignment
PRE-CALCULUS 11 Introduction Assignment Welcome to PREC 11! This assignment will help you review some topics from a previous math course and introduce you to some of the topics that you ll be studying
### Algebra 1 Course Title
Algebra 1 Course Title Course- wide 1. What patterns and methods are being used? Course- wide 1. Students will be adept at solving and graphing linear and quadratic equations 2. Students will be adept
### MATH 10034 Fundamental Mathematics IV
MATH 0034 Fundamental Mathematics IV http://www.math.kent.edu/ebooks/0034/funmath4.pdf Department of Mathematical Sciences Kent State University January 2, 2009 ii Contents To the Instructor v Polynomials.
### Section 3.1 Quadratic Functions and Models
Section 3.1 Quadratic Functions and Models DEFINITION: A quadratic function is a function f of the form fx) = ax 2 +bx+c where a,b, and c are real numbers and a 0. Graphing Quadratic Functions Using the
### PARABOLAS AND THEIR FEATURES
STANDARD FORM PARABOLAS AND THEIR FEATURES If a! 0, the equation y = ax 2 + bx + c is the standard form of a quadratic function and its graph is a parabola. If a > 0, the parabola opens upward and the
### SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS
(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic
### Sect 6.7 - Solving Equations Using the Zero Product Rule
Sect 6.7 - Solving Equations Using the Zero Product Rule 116 Concept #1: Definition of a Quadratic Equation A quadratic equation is an equation that can be written in the form ax 2 + bx + c = 0 (referred
### Factoring Guidelines. Greatest Common Factor Two Terms Three Terms Four Terms. 2008 Shirley Radai
Factoring Guidelines Greatest Common Factor Two Terms Three Terms Four Terms 008 Shirley Radai Greatest Common Factor 008 Shirley Radai Factoring by Finding the Greatest Common Factor Always check for
### Determinants can be used to solve a linear system of equations using Cramer s Rule.
2.6.2 Cramer s Rule Determinants can be used to solve a linear system of equations using Cramer s Rule. Cramer s Rule for Two Equations in Two Variables Given the system This system has the unique solution
### Review of Intermediate Algebra Content
Review of Intermediate Algebra Content Table of Contents Page Factoring GCF and Trinomials of the Form + b + c... Factoring Trinomials of the Form a + b + c... Factoring Perfect Square Trinomials... 6
### FACTORING OUT COMMON FACTORS
278 (6 2) Chapter 6 Factoring 6.1 FACTORING OUT COMMON FACTORS In this section Prime Factorization of Integers Greatest Common Factor Finding the Greatest Common Factor for Monomials Factoring Out the
### Polynomial Operations and Factoring
Algebra 1, Quarter 4, Unit 4.1 Polynomial Operations and Factoring Overview Number of instructional days: 15 (1 day = 45 60 minutes) Content to be learned Identify terms, coefficients, and degree of polynomials.
### LAKE ELSINORE UNIFIED SCHOOL DISTRICT
LAKE ELSINORE UNIFIED SCHOOL DISTRICT Title: PLATO Algebra 1-Semester 2 Grade Level: 10-12 Department: Mathematics Credit: 5 Prerequisite: Letter grade of F and/or N/C in Algebra 1, Semester 2 Course Description:
### A Systematic Approach to Factoring
A Systematic Approach to Factoring Step 1 Count the number of terms. (Remember****Knowing the number of terms will allow you to eliminate unnecessary tools.) Step 2 Is there a greatest common factor? Tool
### In algebra, factor by rewriting a polynomial as a product of lower-degree polynomials
Algebra 2 Notes SOL AII.1 Factoring Polynomials Mrs. Grieser Name: Date: Block: Factoring Review Factor: rewrite a number or expression as a product of primes; e.g. 6 = 2 3 In algebra, factor by rewriting
### Understanding Basic Calculus
Understanding Basic Calculus S.K. Chung Dedicated to all the people who have helped me in my life. i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other
### How To Solve Factoring Problems
05-W4801-AM1.qxd 8/19/08 8:45 PM Page 241 Factoring, Solving Equations, and Problem Solving 5 5.1 Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares 5.3 Factoring
### ALGEBRA REVIEW LEARNING SKILLS CENTER. Exponents & Radicals
ALGEBRA REVIEW LEARNING SKILLS CENTER The "Review Series in Algebra" is taught at the beginning of each quarter by the staff of the Learning Skills Center at UC Davis. This workshop is intended to be an
### Alum Rock Elementary Union School District Algebra I Study Guide for Benchmark III
Alum Rock Elementary Union School District Algebra I Study Guide for Benchmark III Name Date Adding and Subtracting Polynomials Algebra Standard 10.0 A polynomial is a sum of one ore more monomials. Polynomial
### Some Lecture Notes and In-Class Examples for Pre-Calculus:
Some Lecture Notes and In-Class Examples for Pre-Calculus: Section.7 Definition of a Quadratic Inequality A quadratic inequality is any inequality that can be put in one of the forms ax + bx + c < 0 ax
### CRLS Mathematics Department Algebra I Curriculum Map/Pacing Guide
Curriculum Map/Pacing Guide page 1 of 14 Quarter I start (CP & HN) 170 96 Unit 1: Number Sense and Operations 24 11 Totals Always Include 2 blocks for Review & Test Operating with Real Numbers: How are
### 2.3. Finding polynomial functions. An Introduction:
2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned
### Greatest Common Factor (GCF) Factoring
Section 4 4: Greatest Common Factor (GCF) Factoring The last chapter introduced the distributive process. The distributive process takes a product of a monomial and a polynomial and changes the multiplication
### Math 10C. Course: Polynomial Products and Factors. Unit of Study: Step 1: Identify the Outcomes to Address. Guiding Questions:
Course: Unit of Study: Math 10C Polynomial Products and Factors Step 1: Identify the Outcomes to Address Guiding Questions: What do I want my students to learn? What can they currently understand and do?
### AIP Factoring Practice/Help
The following pages include many problems to practice factoring skills. There are also several activities with examples to help you with factoring if you feel like you are not proficient with it. There
### MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education)
MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education) Accurately add, subtract, multiply, and divide whole numbers, integers,
### What are the place values to the left of the decimal point and their associated powers of ten?
The verbal answers to all of the following questions should be memorized before completion of algebra. Answers that are not memorized will hinder your ability to succeed in geometry and algebra. (Everything
250) 960-6367 Factoring Polynomials Sometimes when we try to solve or simplify an equation or expression involving polynomials the way that it looks can hinder our progress in finding a solution. Factorization
### MATH 21. College Algebra 1 Lecture Notes
MATH 21 College Algebra 1 Lecture Notes MATH 21 3.6 Factoring Review College Algebra 1 Factoring and Foiling 1. (a + b) 2 = a 2 + 2ab + b 2. 2. (a b) 2 = a 2 2ab + b 2. 3. (a + b)(a b) = a 2 b 2. 4. (a
### Factor and Solve Polynomial Equations. In Chapter 4, you learned how to factor the following types of quadratic expressions.
5.4 Factor and Solve Polynomial Equations Before You factored and solved quadratic equations. Now You will factor and solve other polynomial equations. Why? So you can find dimensions of archaeological
### 5.1 FACTORING OUT COMMON FACTORS
C H A P T E R 5 Factoring he sport of skydiving was born in the 1930s soon after the military began using parachutes as a means of deploying troops. T Today, skydiving is a popular sport around the world.
### Unit 7 Quadratic Relations of the Form y = ax 2 + bx + c
Unit 7 Quadratic Relations of the Form y = ax 2 + bx + c Lesson Outline BIG PICTURE Students will: manipulate algebraic expressions, as needed to understand quadratic relations; identify characteristics
### Florida Math 0028. Correlation of the ALEKS course Florida Math 0028 to the Florida Mathematics Competencies - Upper
Florida Math 0028 Correlation of the ALEKS course Florida Math 0028 to the Florida Mathematics Competencies - Upper Exponents & Polynomials MDECU1: Applies the order of operations to evaluate algebraic
### Algebra Cheat Sheets
Sheets Algebra Cheat Sheets provide you with a tool for teaching your students note-taking, problem-solving, and organizational skills in the context of algebra lessons. These sheets teach the concepts
### Algebra 2 Chapter 1 Vocabulary. identity - A statement that equates two equivalent expressions.
Chapter 1 Vocabulary identity - A statement that equates two equivalent expressions. verbal model- A word equation that represents a real-life problem. algebraic expression - An expression with variables.
### Algebra 1 Chapter 08 review
Name: Class: Date: ID: A Algebra 1 Chapter 08 review Multiple Choice Identify the choice that best completes the statement or answers the question. Simplify the difference. 1. (4w 2 4w 8) (2w 2 + 3w 6)
### This unit has primarily been about quadratics, and parabolas. Answer the following questions to aid yourselves in creating your own study guide.
COLLEGE ALGEBRA UNIT 2 WRITING ASSIGNMENT This unit has primarily been about quadratics, and parabolas. Answer the following questions to aid yourselves in creating your own study guide. 1) What is the
### 1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style
Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with
### MATH 60 NOTEBOOK CERTIFICATIONS
MATH 60 NOTEBOOK CERTIFICATIONS Chapter #1: Integers and Real Numbers 1.1a 1.1b 1.2 1.3 1.4 1.8 Chapter #2: Algebraic Expressions, Linear Equations, and Applications 2.1a 2.1b 2.1c 2.2 2.3a 2.3b 2.4 2.5
### DRAFT. Algebra 1 EOC Item Specifications
DRAFT Algebra 1 EOC Item Specifications The draft Florida Standards Assessment (FSA) Test Item Specifications (Specifications) are based upon the Florida Standards and the Florida Course Descriptions as
### Math 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers.
Math 0980 Chapter Objectives Chapter 1: Introduction to Algebra: The Integers. 1. Identify the place value of a digit. 2. Write a number in words or digits. 3. Write positive and negative numbers used
### MA107 Precalculus Algebra Exam 2 Review Solutions
MA107 Precalculus Algebra Exam 2 Review Solutions February 24, 2008 1. The following demand equation models the number of units sold, x, of a product as a function of price, p. x = 4p + 200 a. Please write
### Warm-Up Oct. 22. Daily Agenda:
Evaluate y = 2x 3x + 5 when x = 1, 0, and 2. Daily Agenda: Grade Assignment Go over Ch 3 Test; Retakes must be done by next Tuesday 5.1 notes / assignment Graphing Quadratic Functions 5.2 notes / assignment
### This is Factoring and Solving by Factoring, chapter 6 from the book Beginning Algebra (index.html) (v. 1.0).
This is Factoring and Solving by Factoring, chapter 6 from the book Beginning Algebra (index.html) (v. 1.0). This book is licensed under a Creative Commons by-nc-sa 3.0 (http://creativecommons.org/licenses/by-nc-sa/
### Lesson 9.1 Solving Quadratic Equations
Lesson 9.1 Solving Quadratic Equations 1. Sketch the graph of a quadratic equation with a. One -intercept and all nonnegative y-values. b. The verte in the third quadrant and no -intercepts. c. The verte
### Answer Key for California State Standards: Algebra I
Algebra I: Symbolic reasoning and calculations with symbols are central in algebra. Through the study of algebra, a student develops an understanding of the symbolic language of mathematics and the sciences.
### 0.4 FACTORING POLYNOMIALS
36_.qxd /3/5 :9 AM Page -9 SECTION. Factoring Polynomials -9. FACTORING POLYNOMIALS Use special products and factorization techniques to factor polynomials. Find the domains of radical expressions. Use |
This website uses cookies to ensure you have the best experience.
# Sydney Harbour Bridge Investigation Essay
1520 words - 7 pages
Sydney Harbour Bridge – Directed Investigation – Quadratic functions
Introduction:
Aim: To find the multiple unknowns in the Sydney Harbour bridge
The Sydney Harbour Bridge will be used to investigate a diverse number of points in the structure such as the height and length. Quadratics will be used to solve the height of the bridge at different points on the x axis.
A quadratic is an equation constructed from information collected from a graph; this equation can also be used to produce a graph. Quadratics can be used to solve the problem dealt in this investigation as the Sydney Harbour Bridge is identified as a parabola shape. Only basic information is given about the bridge and the ...view middle of the document...
Therefore the highest point of the bridge will be:
Highest Point of the bridge:
Highest point of the bridge – the road on the pylon = x
182.25m – 50m = xm
182.25m – 50m = 132.25m
Therefore if a painter is painting at the highest point of the arch, the painter will be 132.25m above the road.
The quadratic function y=-1/100 x^2+27/10 x can be used to model the arc by implementing the information provided into the vertex form. The equation of the vertex form is y=〖a(x-h)〗^2+k. Completed, it shall look like =〖a(x-135)〗^2+182.25 , where a≠0. ‘a’can be identified by substituting into the formula a point from the bridge, in this example, it shall be (0, 0).
y=〖a(x-135)〗^2+182.25
0=〖a(0-135)〗^2+182.25
0=〖a(-135)〗^2+182.25
-182.25=18225a
a=(-182.25)/18225
Therefore a=-1/100
Hence the equation of the quadratic is y=-1/100 〖(x-135)〗^2, if expanded, the equation will become:
y=-1/100 (x^2-270x+18225)+182.25
y=-1/100 x^2 27/10 x-182.25+182.25
Therefore y=-1/100 x^2+27/10 x
The expanded formula of the bridge in the vertex form is proven to be exactly the same as the one given, proving that the formula y=-1/100 x^2+27/10 x can be used to find the curve of the arc. The Sydney Harbour Bridge’s road lies 50m above sea level. The central steel arc of the bridge passes through two points on the road and then to the pylons. The points at which the arc meets the road can be solved using the graphing function of a graphics calculator.
The equation y=-1/100 x^2+27/10 x was inserted into the calculator (Figure 1) Values window was altered to see the full graph
Figure 1.
The graph is then drawn onto the calculator (Figure 2)
Figure 2
The x-cal function under the g-solve option was used to make the road. By enteringy=50, the points of the arc are shown. (Figure 3 and 4)
Figure 3 Figure 4
The results indicate that the two points where the arc meets the road are (20, 50) and (250, 50). Using these points, it is possible to determine the length of the road under the arch by using the formula x^2- x^1.
x^2- x^1
= 250-20
= 230m
Therefore the length of the road is 230m long.
Using this information, the amount of steel cable required to construct the supports can be identified. As stated, there are 19 vertical steel cables evenly spaced out which indicates that there are 20 spaces in between the arc and the road. The distance between each point can be calculated by dividing the total length of the road by the number of spaces in between the structure.
= 230m/20
=11.5m
Defining that the space in between each steel cable is 11.5m, the distance of each cable can now be identified.
By...
## Other Essays Like Sydney Harbour Bridge Investigation
### Assess The Contribution Of Christianity In Australia Towards Reconciliation, In The Past 25 Years
1656 words - 7 pages ). It is part of the process of reconciliation which has been happening since the 1990s. This contributes to the reconciliation process by allowing people to reflect on the past through prayer therefore allowing Australia to move forward into our future.The Corroboree 2000 'sorry' Sydney Harbour Bridge Walk on the 28th of May in 2000 is another example of Anglican contribution to the process of Reconciliation. It was a major National Event organized
### Australia-simply the best Essay
705 words - 3 pages . A few amazing natural landmarks of Australia are the awe-inspiring Blue Mountains which have many scenic activities to offer, the gorgeous Hunter and Barossa valley where you can taste delicious wine, Kakadu National Park and the captivating GBR which is actually one of the 7 natural wonders of the world where you can snorkel and go on boat tours. Some manmade structures include the eye-catching Opera House, the busy Harbour Bridge, the world
### Gold At The End Of The Rainbow
1274 words - 6 pages Sydney, Australia1995, age 18I'm on a bus staring at the ocean as I go over the Harbour Bridge, the sun's reflection on the water is nothing like I have ever seen before. Boats crossing the ocean give me many memories; I look away, and see tall skyscrapers everywhere, buildings towering over me, shadowing me from the sun's bright light. I look up at the beams above and my eyes follow the girders. I can see the Australian flag above me and I know
### The Separation Of Capital Ownership And Control
1577 words - 7 pages The argument of whether the separation of capital ownership and control is an efficient form of organization has constantly been a controversial issue. The criticism whether the controllers’ act is in the best interest of the owners’ wills never end as long as hired managers operate management. As the number of public companies has been increasing over the course of this century, meanwhile the American style of contact based corporation has
### The Versatility And Flexibility Of OLED's
1014 words - 5 pages In April 1, 2002, organic light emitting diodes gain rise in the scientific community with their published, more practical form at Ames Laboratory. “Scientists at the U.S. Department of Energy's Ames Laboratory, in collaboration with scientists at the University of Michigan, Ann Arbor, have developed and demonstrated a novel, fluorescence-based chemical sensor that is more compact, versatile and less expensive than existing technology of its
### Comparing The Moral Virtues Of Antony And Julian The Apostate
1103 words - 5 pages Roman emperor Julian the Apostate and Christian leader Antony both exhibited many qualities of character during their existence. Both of them led very distinctive lives although shared several ethical values. Book 25 of “The Later Roman Empire” and the book “Early Christian Lives” show concrete evidence of this. In the following essay, I will argue how both leaders’ lives were devoted to their religious beliefs and their mutual cardinal virtues
### Living In A Cashless Society
1637 words - 7 pages Money in a traditional sense no longer exists. Money is becoming much of a concept than a physical material, and most ordinary bitter have not see the reality of the switch. People today are using credit and debit cards on a regular basis and in everyday situations such as meal purchased at fast food, highway tolls, clothing, groceries, gas stations, etc. all of these means of systems could be regarded as a cashless society or world. The question
### The French And Indian War: The "Real" First World War
1955 words - 8 pages The Seven Years War, or more commonly referred to as “The French and Indian War”, has been called the true First World War. In this book The French and Indian War: Deciding the Fate of North America, the author and historian Walter R. Borneman paints a detailed and elaborate picture that justifies the claim of it being the first true war of global proportions. If ever there truly was a climax to the never ending feud of the European powers
### Is The Use Of Animals In Medical Research A Necessary Measure?
1513 words - 7 pages Throughout history, animals have been used in experiments to test product safety and obtain medical knowledge that benefits both humans and animals alike. Every year there are numerous medical breakthroughs, such as medications and surgical instruments, which are tested on animals to insure their safety before they are deemed acceptable for human use. Even though the results of the experiments saved millions of human lives, they are also
### Education And The Evolving Job Market
2363 words - 10 pages The lives of students today are changing. They are preparing for lives and jobs that have never before existed. If teachers hope to have a significant and worthwhile impact on these quickly changing lives, they must change the way they think, prepare, and instruct our future generations. Children cannot afford to have teachers who remain stagnant in their methods and ideals. Students crave instructors that are willing to allow them to tap
### Young And Relentless
1737 words - 7 pages There are numerous influences that can be responsible of teenager’s behaviors and attitude as they develop. One factor that is important to these behaviors is parental figures being over involved or uninvolved in their children’s lives. Many of these effects include illegal substance abuse, rising sexual activity, underage alcohol consumption, and tobacco use. Studies show parental participation plays a key role in the characteristics developed
## Related Papers
### The Construction Of The Syndey Harbour Bridge
343 words - 2 pages CONSTRUCTION and planning of the BRIDGE.Work on the bridgeConstruction on the bridge began at both ends and at the same time. It was gradually built inwards across the harbour, where it meet in the middle about 5 years later. While the light and heavy workshops work were being built at North Sydney, work had already begun on the approach spans (five at each end of the bridge) and all supported on pairs of concrete piers.The man who prepared the
### Late Ferry Comentary Essay
961 words - 4 pages Late Ferry, a quatrain poem written by Robert Gray, concerns the poet, observing the long sight of a ferry leaving the bay, going towards the city as it wades under the Harbour Bridge in Sydney, fading away into the night. The poet has made the ferry the main idea of the poet, vividly describing the different effects of lighting on and around the ferry. In this poem, Robert Gray has portrayed the image of the light effect on the ferry in each
### The New Land Essay
623 words - 3 pages of the world! Well, it sure seemed the same, at first. The minutest of differences seemed to make eating there all the more enjoyable, such as the small packets of Vegemite next to the ketchup and mustard. We quickly finished and briskly walked five or six blocks until we found the Sydney Harbour Bridge. We could not simply look at it, so we climbed it! The sight of cars zooming underneath us was breathtaking! After 12 hours of wonder, we ate
### Research In Motion Essay
1429 words - 6 pages North Sydney, New South Wales, Australia in February 2009. Just across the Harbour Bridge, ten minutes from the Sydney CBD, the new RIM offices were formally unveiled by Mr. Thomas A. MacDonald, Consul General of Canada, at a ceremony attended by dignitaries from the New South Wales Government and North Sydney Council, as well as RIM’s partners and customers. The new office features training facilities, a research and development centre, a |
# How do you complete a congruence statement
If in triangles ABC and DEF, AB = DE, AC = DF, and angle A = angle D, then triangle ABC is congruent to triangle DEF. Using words: If 3 sides in one triangle are congruent to 3 sides of a second triangle, then the triangles are congruent.
## How do you complete a triangle congruence statement?
If in triangles ABC and DEF, AB = DE, AC = DF, and angle A = angle D, then triangle ABC is congruent to triangle DEF. Using words: If 3 sides in one triangle are congruent to 3 sides of a second triangle, then the triangles are congruent.
## What is an example of a congruent statement?
A triangle with three sides that are each equal in length to those of another triangle, for example, are congruent. This statement can be abbreviated as SSS. … If two triangles have two equal angles and a side of equal length, either ASA or AAS, they will be congruent.
## What are congruent statements?
A congruence statement says that two polygons are congruent. To write a congruence statement, list the corresponding vertices in the same order.
## Is aas a congruence theorem?
The AAS Theorem says: If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. Notice how it says “non-included side,” meaning you take two consecutive angles and then move on to the next side (in either direction).
## What are the 3 properties of congruence?
There are three properties of congruence. They are reflexive property, symmetric property and transitive property. All the three properties are applicable to lines, angles and shapes. Reflexive property of congruence means a line segment, or angle or a shape is congruent to itself at all times.
## What is congruence theorem?
When triangles are congruent corresponding sides (sides in same position) and corresponding angles (angles in same position) are congruent (equal). …
## How do you write congruent angles?
In mathematics, the definition of congruent angles is “angles that are equal in measure are known as congruent angles”. In other words, equal angles are congruent angles. It is denoted by the symbol ≅ , so if we want to represent ∠A is congruent to ∠X, we will write it as ∠A ≅ ∠X.
## What is the meaning of congruency?
agreeing; accordant; congruous: His testimony was perfectly congruent with the content retrieved from the suspect’s phone. … of or relating to two numbers related by a congruence. Geometry. (of figures) coinciding at all points when superimposed: congruent triangles.
What are the 5 congruence theorems?
• SSS (side, side, side) SSS stands for “side, side, side” and means that we have two triangles with all three sides equal. …
• SAS (side, angle, side) …
• ASA (angle, side, angle) …
• AAS (angle, angle, side) …
• HL (hypotenuse, leg)
Article first time published on askingthelot.com/how-do-you-complete-a-congruence-statement/
## What is the difference between AAS and ASA?
ASA stands for “Angle, Side, Angle”, while AAS means “Angle, Angle, Side”. Two figures are congruent if they are of the same shape and size. … ASA refers to any two angles and the included side, whereas AAS refers to the two corresponding angles and the non-included side.
## What does Asa stand for in geometry?
SSS (side-side-side) All three corresponding sides are congruent.SAS (side-angle-side) Two sides and the angle between them are congruent.ASA (angle-side-angle) Two angles and the side between them are congruent.AAS (angle-angle-side) Two angles and a non-included side are congruent.
## What is hypotenuse leg?
FAQs on Hypotenuse Leg Theorem In a right-angled triangle, the side opposite to the right angle is called the hypotenuse and the two other adjacent sides are called its legs. The hypotenuse is the longest side of the triangle, while the other two legs are always shorter in length.
## What is congruence class 9?
It states that that two triangles are said to be congruent if they are copies of each other and when superposed, they cover each other exactly. In other words, two triangles are congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.
## Is AAS same as SAA?
A variation on ASA is AAS, which is Angle-Angle-Side. … Angle-Angle-Side (AAS or SAA) Congruence Theorem: If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent.
## How do you describe a congruence transformation?
That is, two objects are congruent if we can move one of the objects, without changing its shape or size, in such a way that it fits exactly over the other image. We call these movements congruence transformations. Congruence transformations are transformations performed on an object that create a congruent object.
## Is congruent and equal are same?
Two shapes are said to be congruent if one can be exactly superimposed on the other. “Congruence deals with shapes (aka objects), while equality deals with numbers. You don’t say that two shapes are equal or two numbers are congruent.”
## What property of congruence is celebrated?
PROPERTIES OF CONGRUENCEReflexive PropertyFor all angles A , ∠A≅∠A . An angle is congruent to itself.These three properties define an equivalence relationSymmetric PropertyFor any angles A and B , if ∠A≅∠B , then ∠B≅∠A . Order of congruence does not matter.
## How many congruence rules are there in class 9?
There are 5 main rules of congruency for triangles: SSS Criterion: Side-Side-Side. SAS Criterion: Side-Angle-Side. ASA Criterion: Angle-Side- Angle.
## What lines are congruent?
When two line segments exactly measure the same, they are known as congruent lines. For example, two line segments XY and AB have a length of 5 inches and are hence known as congruent lines. When two angles exactly measure the same, they are known as congruent angles.
## Which angles are congruent to 7?
When a transversal cuts parallel lines, all of the acute angles formed are congruent, and all of the obtuse angles formed are congruent. In the figure above ∠1, ∠4, ∠5, and ∠7 are all acute angles. They are all congruent to each other.
## How many congruence criterions are there name them?
Two triangles are congruent if they satisfy the 5 conditions of congruence. They are side-side-side (SSS), side-angle-side (SAS), angle-side-angle (ASA), angle-angle-side (AAS) and right angle-hypotenuse-side (RHS).
## Are these two figures congruent Why?
Two figures are congruent if they have the same shape and size. Two angles are congruent if they have the same measure.
## Is it AAS or AAS?
NameAbbreviationAssociate of ArtsA.A.Associate of Applied ScienceA.A.S.Bachelor of Science/Bachelor of ArtsB.A. B.S.Master of Science/Master of ArtsM.A. M.S.
## What is the difference between AS and AAS?
An Associate of Science (AS) degree is a 2-year degree offered by most community colleges and some 4- year colleges. The Associate of Applied Science (AAS) degree prepares graduates to enter a career immediately after graduation and have been considered terminal degrees.
## What are the 4 rules in congruent triangles?
These four criteria used to test triangle congruence include: Side – Side – Side (SSS), Side – Angle – Side (SAS), Angle – Side – Angle (ASA), and Angle – Angle – Side (AAS). There are more ways to prove the congruency of triangles, but in this lesson, we will restrict ourselves to these postulates only.
## How do you determine if a triangle is ASA or AAS?
If two pairs of corresponding angles and the side between them are known to be congruent, the triangles are congruent. This shortcut is known as angle-side-angle (ASA). Another shortcut is angle-angle-side (AAS), where two pairs of angles and the non-included side are known to be congruent.
## How do you find HL?
1. The longest side of a right triangle is called its hypotenuse.
2. The HL Theorem states; If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent.
## What makes a triangle HL?
Congruent Triangles – Hypotenuse and leg of a right triangle. (HL) Definition: Two right triangles are congruent if the hypotenuse and one corresponding leg are equal in both triangles. … If, in two right triangles the hypotenuse and one leg are equal, then the triangles are congruent.
## What is SAS triangle congruence?
If we can show that two sides and the included angle of one triangle are congruent to two sides and the included angle in a second triangle, then the two triangles are congruent. This is called the Side Angle Side Postulate or SAS.
## Who discovered Heron's formula?
Heron’s formula, formula credited to Heron of Alexandria (c. 62 ce) for finding the area of a triangle in terms of the lengths of its sides. |
# Question #9099e
Mar 9, 2017
$f {\left(x\right)}^{-} 1 = {x}^{2} / 8 + \frac{7}{2}$
This can also be written as one fraction:
$f {\left(x\right)}^{-} 1 = \frac{{x}^{2} + 28}{8}$
#### Explanation:
Rules
When finding the inverse, ${f}^{- 1} \left(x\right)$, the easiest method is to switch the variables $x$ and $y$. So $y = f \left(x\right)$ is now $x = f \left(y\right)$.
Afterwards, you need to isolate $y$ again to get the proper form of the inverse.
Application
$f \left(x\right) = y = 2 {\left(2 x - 7\right)}^{\frac{1}{2}} \text{ }$ Applying the "switch" ...
$f \left(y\right) = x = 2 {\left(2 y - 7\right)}^{\frac{1}{2}}$
Solving for $y$ ...
$x = 2 \setminus \sqrt{2 y - 7}$
$\frac{1}{2} x = \setminus \sqrt{2 y - 7}$
${\left(\frac{1}{2} x\right)}^{2} = 2 y - 7$
$\frac{1}{4} {x}^{2} + 7 = 2 y$
$\frac{1}{2} \left(\frac{1}{4} {x}^{2} + 7\right) = y$
$\frac{1}{8} {x}^{2} + \frac{7}{2} = y$
Confused? Feel free to drop a comment and tell me where you're still not understanding things! :) |
What is the distance from the point A, to the line through the points B and C where A, B, and C...
Question:
What is the distance from the point A, to the line through the points B and C where A, B, and C have the following coordinates (3, 0, 4), (-1, 4, 1) and (2, 1, 0)?
Distance From a Point to the given Line in 3D:
In three dimensional coordinate system, to find the distance from point to the given line is found as given below:
(i) Find the coordinates of a general point on the line and take it as the foot of perpendicular.
(ii) Find the direction ratios of the line joining the foot of perpendicular and the given point.
(iii) Apply the condition that the sum of the product of the direction ratios of two perpendicular lines is zero to get the coordinates of the general point.
(iv) Use distance formula between the two points to find the required distance.
The given points are:
{eq}A\left ( 3, 0, 4 \right ),\ B\left ( -1, 4, 1 \right ), \ C \left ( 2, 1, 0 \right ) {/eq}
The equation of the line passing through the points {eq}B {/eq} and {eq}C {/eq} is:
{eq}\\\\ \begin{align*} \frac{x+1}{2+1} & = \frac{y-4}{1-4}=\frac{z-1}{0-1} \\\\\Rightarrow \frac{x+1}{3} & = \frac{y-4}{-3}=\frac{z-1}{-1}=\lambda \texttt{ (say)} \ \ \ ...(1) \\\\ \end{align*} {/eq}
Thus, the coordinates of any point {eq}P {/eq} on the line can be taken as:
{eq}P \left ( 3\lambda -1, -3\lambda +4, -\lambda +1 \right ) {/eq}
If {eq}P {/eq} be the foot of perpendicular from the point {eq}A {/eq} on the line {eq}(1) {/eq} then direction ratios of {eq}AP {/eq} are:
{eq}3\lambda -1-3, -3\lambda +4-0, -\lambda +1-4 = 3\lambda -4, -3\lambda +4, -\lambda -3 {/eq}
Since, {eq}AP {/eq} is perpendicular to the line {eq}(1) {/eq} therefore, the sum of the product of direction ratios would be zero.
Thus, we have:
{eq}\\\\ \begin{align*} 3(3\lambda -4)-3(-3\lambda +4)-1(-\lambda -3)& =0 \\\\\Rightarrow 9\lambda -12 + 9\lambda -12 +\lambda + 3 & = 0 \\\\\Rightarrow 19\lambda & = 21 \\\\\Rightarrow \lambda & = \frac{21}{19} \\\\ \end{align*} {/eq}
Thus, the point {eq}P {/eq} is:
{eq}\left ( \frac{63}{19} -1,-\frac{63}{19}+4,-\frac{21}{19}+1 \right ) =\left ( \frac{44}{19},\frac{13}{19},-\frac{2}{19} \right ) {/eq}
Hence, the required distance {eq}d {/eq} from the the point {eq}A {/eq} to the line {eq}(1) {/eq} is:
{eq}\\\\ \begin{align*} AP & = \sqrt{\left (\frac{44}{19}-3\right)^{2}+\left (\frac{13}{19}-0\right)^{2}+\left (\frac{2}{19}-4\right)^{2}} \\\\& = \sqrt{\left (-\frac{13}{19}\right)^{2}+\left (\frac{13}{19}\right)^{2}+\left (-\frac{74}{19}\right)^{2}} \\\\& = \frac{1}{19}\sqrt{169+169+5476} \\\\& = \frac{1}{19}\sqrt{5814} \\\\& \approx 4.01 \texttt{ units} \\\\ \end{align*} {/eq}
How to Find the Distance Between a Point & a Line
from FTCE Mathematics 6-12 (026): Practice & Study Guide
Chapter 30 / Lesson 5
23K |
# Number Theory – A Primer
This primer exists for the background necessary to read our post on RSA encryption, but it also serves as a general primer to number theory.
## Oh, Numbers, Numbers, Numbers
Definition: The set of integers, denoted $\mathbb{Z}$, is the set $\left \{ \dots -2, -1, 0, 1, 2, \dots \right \}$.
Definition: Let $a,b$ be integers, then $a$ divides $b$, denoted $a \mid b$, if there exists an integer $n$ such that $na = b$. We also less commonly say $b$ is divisible by $a$. A composite number has a divisor greater than 1, and hence two strictly smaller divisors.
This definition of “dividing” allows us to bypass the more complicated world of fractions and rational numbers, and keep most of what we do in the integers. The novice reader is encouraged to prove the following propositions about divisibility:
• If $a \mid b$ then $a \mid kb$ for any $k \in \mathbb{N}$.
• If $a \mid b$ and $a \mid c$ then $a \mid b + c, a \mid bc$.
• If $a \mid b$ and $b \mid a$ then $a = \pm b$.
• Find a counterexample where $a \mid bc$ but neither $a \mid b$ nor $a \mid c$ holds.
Definition: A positive integer $p$ is prime if it has exactly two distinct divisors: $1$ and $p$. For the remainder of this post, we will always use $p,q$ to denote primes.
It follows immediately from the definition that if $p,n$ are positive integers and $p$ is prime, then $n \mid p$ if and only if $n = 1$ or $n=p$.
We could work toward the fundamental theorem of arithmetic, that every positive integer factors uniquely as a product of primes, but this would lead us down the road of proving Bezout’s theorem, and we want to work quickly toward other things. Instead, we simply prove the existence of such a factoring, and take its uniqueness for granted:
Theorem: Every positive integer factors as a product of primes.
Proof. Let $S$ be the set of positive integers which do not have a factoring as a product of primes. In particular, $1 \notin S$ since 1 is the product of zero primes, and no prime is in $S$. So $S$ is entirely composed of composite numbers. Take the smallest element $x \in S$, and since it is composite it may be written as $x = ab$, where $a,b$ are both strictly smaller than $x$. But since each of $a,b$ are not in $S$, they factor into products of primes. By combining these factorings, we achieve a factoring of $x$, a contradiction. $\square$
Definition: The greatest common divisor of two numbers $a, b$, abbreviated gcd and denoted $(a,b)$ or less commonly $\gcd(a,b)$, is the largest divisor of both $a$ and $b$. We say two numbers are relatively prime if $(a,b) = 1$.
Note that two relatively prime numbers might not be prime. In fact, $(8,9) = 1$, but neither 8 nor 9 are prime. We also sometimes say (though it is probably grammatically incorrect) $a$ is relatively prime to $b$.
We may prove some interesting facts about greatest common divisors, which we leave as exercises to the ambitious reader:
• $(a + cb, b) = (a,b)$ for any $c \in \mathbb{Z}$.
• $(a,b)$ is the smallest positive linear combination of $a,b$ with integer coefficients.
• (Bezout’s theorem, a corollary to the above two statements) There is a linear combination of $a,b$ with integer coefficients that equals $(a,b)$.
The last theorem shows up in group theory as a statement about generators of additive integer groups $n\mathbb{Z}$.
## Congruences and Modular Arithmetic
Definition: $a$ is congruent to $b$ modulo $n$, denoted $a \equiv b \mod n$, if $n \mid b-a$.
This definition has a more familiar form for computer scientists, namely the line of code:
a % n == b % n
In plain language, two numbers are congruent modulo $n$ if they have the same remainder when divided by $n$. Usually, however, we make $0 \leq b < n$, so that $b$ is the remainder of $a$ when divided by $n$.
It is a cheap fact that the relation $\cdot \equiv \cdot \mod n$ is an equivalence relation on integers for any fixed $n$. In other words:
• $a \equiv a \mod n$
• If $a \equiv b \mod n$ then $b \equiv a \mod n$
• If $a \equiv b \mod n$ and $b \equiv c \mod n$ then $a \equiv c \mod n$.
This allows us to partition the integers into their congruence classes. In other words, the fact that this is an equivalence relation allows us to identify a number with its remainder modulo $n$. For the sake of consistency, when stating the set of all congruence classes mod $n$, we stick to the classes represented by the positive integers $0, \dots, n-1$.
There is a vast amount of work on solving systems of congruences. The interested reader should investigate the Chinese Remainder Theorem and Hensel’s Lemma.
Definition: A number $b$ is an inverse to a number $a$ modulo $n$ if $ab \equiv 1 \mod n$.
Inverses help to reduce computations by pairing a number with its inverse modulo $n$. Therefore, it is interesting to know when numbers have inverses, and whether one is unique up to congruence.
Proposition: An inverse for $a \mod n$ exists if and only if $(a,n) = 1$ ($a$ and $n$ are relatively prime).
Proof. Suppose such an inverse $b$ exists. Then by the definition of congruence, $n \mid ab - 1$, and hence $cn = ab-1$, so $ab - cn = 1$. In particular, 1 is a linear combination of $a$ and $n$, and hence it must be the smallest positive linear combination. This is equivalent to $(a,n)$, which is proved by the exercise above.
On the other hand, suppose $(a,n)=1$. We may reverse the computation above to find $b$. Specifically, $b$ is the coefficient of $a$ in the linear combination of $a,n$ that makes 1. This proves the theorem $\square$.
It would be great if we could determine how many numbers have inverses mod $n$. In fact, we will, but first we’d like to investigate a few interesting theorems.
Theorem: (Wilson) $(p-1)! \equiv -1 \mod p$ for any prime $p$.
Proof. We immediately see that two of these factors are easy: $1 \equiv 1 \mod p$ and $p-1 \equiv -1 \mod p$. We claim that the product of the remaining numbers is always 1.
Since each number $1 < n < p-1$ is relatively prime to $p$ (indeed, all numbers are relatively prime to a prime), each number $n$ has an inverse modulo $p$. As long as this inverse is not $n$ itself, we may pair off each number with its inverse, and see that the entire product is just 1.
To prove that $n^{-1} \neq n$, suppose $n^2 \equiv 1 \mod p$. Then $(n+1)(n-1) \equiv 0 \mod p$. But since $1 < n < p-1$, we must have two numbers $n+1, n-1$ whose product is divisible by $p$. But neither of $n+1, n-1$ has $p$ as a factor. So we conclude that either $n+1 = 0$ or $n-1 = 0$, giving $n \equiv \pm 1 \mod p$, a contradiction.
So we may indeed pair the numbers off as we described above, and see that $(p-1)! \equiv -1 \mod p. \square$
Here is another fundamental result that uses Wilson’s theorem as a stepping stone:
Theorem: (Fermat’s Little Theorem) If $p$ is prime and $0 < a < p$, then $a^{p-1} \equiv 1 \mod p$.
Proof. The set $\left \{ 1, 2, \dots, p-1 \right \}$ are the nonzero remainders mod $p$. If we multiply each by $a$, we get another complete set of nonzero residues mod $p$, namely $\left \{ a, 2a, \dots, (p-1)a \right \}$.
Since both of these sets are all the nonzero residues mod $p$, their products are congruent. In particular,
$\displaystyle \prod \limits_{k=1}^{p-1} k \equiv \prod \limits_{k=1}^{p-1}ak \mod n$
Since we may factor out the $a$ from each term, and there are $p-1$ terms, we arrive at $(p-1)! \equiv a^{p-1}(p-1)! \mod p$. But Wilson’s theorem proved that $(p-1)!$ is nonzero mod $p$, and hence has a multiplicative inverse. Multiplying both sides of the equation by its inverse (which is obviously -1), we get $a^{p-1} \equiv 1 \mod p$, as desired. $\square$
Fermat’s Little Theorem allows us to quickly compute large powers of integers modulo any prime. Specifically, we may break $3^{143} \mod 11$ into $3^{10*14 + 3} = (3^{10})^{14} \cdot 3^3$ By Fermat’s Little Theorem, this is just $1^{14} \cdot 27 \mod 11$, and from here we may compute $3^{143} \equiv 5 \mod 11$. Certainly this is much faster than incrementally multiplying 3 to itself 143 times and every so often dividing by 11. We subtly allude to the usefulness of number theory for computing large exponents, which is an important theme in our post on RSA encryption.
Euler’s Phi Function
Our final topic for this primer is Euler’s phi function (also called the totient function), which counts the number of positive integers which are relatively prime to $n$.
Definition: $\varphi(n)$ is the number of positive integers between 1 and $n$ which are relatively prime to $n$.
Here we have another, more general version of Fermat’s Little Theorem, which uses an almost identical proof:
Theorem: For any positive integer $n$, if $(a,n) = 1$ then $a^{\varphi(n)} \equiv 1 \mod n$.
Proof. Again we use the argument from Fermat’s Little Theorem, except here the sets are not all integers from $1$ to $n$, but only those which are relatively prime. Then we notice that if $a, b$ are relatively prime to $n$, then so is their product.
Using the product trick once more, we label the relatively prime integers $r_i$, and see that
$\displaystyle \prod \limits_{k=1}^{\varphi(n)} r_k \equiv a^{\varphi(n)} \prod \limits_{k=1}^{\varphi(n)} r_k \mod n$
Since the product of all relatively prime integers is again relatively prime, it has a multiplicative inverse mod $n$. While we might not know what it is, it certainly exists, so we may cancel both sides of the congruence to get $a^{\varphi(n)} \equiv 1 \mod n$. Therefore, we win. $\square$
Now we would like to have a good way to compute $\varphi(n)$ for a general $n$. First, we see that for powers of primes this is easy:
Proposition: $\varphi(p^k) = p^{k-1}(p-1) = p^k - p^{k-1}$ for any prime $p$ and positive integer $k$. In particular, $\varphi(p) = p-1$.
Proof. The only numbers which are not relatively prime to $p^k$ are all the multiples of $p$. Since every $p$th number is a multiple of $p$, there are hence $p^k / p$ numbers which are not relatively prime to $p$. Subtracting this from $p^k$ gives our desired formula. $\square$
Finally, we have a theorem that lets us compute $\varphi(n)$ for an arbitrary $n$, from $\varphi$ of its prime power factors.
Proposition: $\varphi(nm) = \varphi(n)\varphi(m)$ if $n$ and $m$ are relatively prime.
Proof. To each number $a$ relatively prime to $n$, and each number $b$ relatively prime to $m$, we see that $am+bn$ is relatively prime to $mn$. Supposing to the contrary that some prime $p$ divides $(an+bm, mn)$, we see that it must divide one of $m, n$ but not both, since $(m,n) = 1$. Suppose without loss of generality that $p \mid n$. Then since $p \mid an+bm$, we may see that $p \mid m$, a contradiction.
In other words, we have shown that $a,b \mapsto am+bn$ is a function from the set of pairs of relatively prime numbers of $n, m$ to the set of relatively prime numbers to $nm$. It suffices to show this map is injective and surjective.
For injectivity, we require that no two distinct $am+bn$ are congruent. Supposing we have two distinct $a,a'$ and two $b, b'$, with $am+bn \equiv a'm+b'n \mod mn$. Then rearranging terms we get $m(a-a') + n(b-b') \equiv 0 \mod mn. Since$latex m\$ divides both $m(a-a')$ and $0$ under the modulus, we see that $m$ divides $n(b-b')$. But since $(m,n) = 1$, we get that $m \mid b-b'$, so by definition $b \equiv b' \mod m$, contradicting our assumption that the $b$‘s were distinct. We get a similar result for the $a$‘s, and this proves injectivity.
For surjectivity, let $k$ be a positive integer relatively prime to $nm$. Since $(m,n) = 1$, we may write $am+bn = k$ for some $a,b \in \mathbb{Z}$. This is achieved by finding a linear combination of $m,n$ equal to 1 (their gcd), and then multiplying through by $k$. We claim that $(a,n) = 1$. This is true since if some prime divided both of $a,n$, then it would also divide $am+bn = k$, and also $nm$. But $k$ was assumed to be relatively prime to $nm$, so this cannot happen. Hence, $a$ is relatively prime to $m$. An identical argument gives that $b$ is relatively prime to $n$, so the pair $(a,b) \mapsto am+bn$, proving surjectivity. $\square$
This, we may compute $\varphi(n)$ multiplicatively, by first finding its prime factorization, and then computing $\varphi(p^k)$ for each prime factor, which is easy.
Unfortunately, finding prime factorizations quickly is very hard. Unless we know the factorization of a large $n$ ahead of time (large as in hundreds-of-digits long), computing $\varphi(n)$ is effectively impossible. We cover the implications of this in more detail in our post on RSA encryption.
Until next time!
## 6 thoughts on “Number Theory – A Primer”
1. kewl
When you say – “We also sometimes say (though it is probably grammatically incorrect) a is relatively prime to b.” a and b are actually called “Coprime” numbers.
Like
• Both are used interchangeably. The possible grammatical incorrectness is that you would say “a is prime relative to b,” but people say “relatively prime to b” more often.
Like
2. kewl
Thanks ! Can you please elaborate on – “The set { 1, 2, ……, p-1 } are the nonzero remainders mod p. If we multiply each by a, we get another complete set of nonzero residues mod p, namely { a, 2a, ……, (p-1)a }.”
How can we prove that the second set {a, 2a , …. (p-1)a} is another complete set of nonzero residues mod p ?
Like
• This depends heavily on the fact that $p$ is prime. Indeed, if you missed something, then you would get some product $ab$, both less than $p$, for which $ab = 0 \mod p$, which is impossible. Another way to say it is that multiplication by a number less than $p$ is an injective function.
Like
3. zhangk
To each number a relatively prime to n, and each number b relatively prime to m, we see that am+bn is relatively prime to mn. Supposing to the contrary that some prime p divides (an+bm, mn), we see that it must divide one of m, n but not both, since (m,n) = 1.Suppose without loss of generality that p \mid n. Then since p \mid an+bm, we may see that p \mid m, a contradiction.
You seemed to confuse the m and n.(In the second and third formulas) It’s am+bn,not an+bm.
Like
4. zhangk
For surjectivity, let k be a positive integer relatively prime to nm. Since (m,n) = 1, we may write am+bn = k for some a,b \in \mathbb{Z}. This is achieved by finding a linear combination of m,n equal to 1 (their gcd), and then multiplying through by k. We claim that (a,n) = 1. This is true since if some prime divided both of a,n, then it would also divide am+bn = k, and also nm. But k was assumed to be relatively prime to nm, so this cannot happen. Hence, a is relatively prime to m(!!WRONG.It should be n). An identical argument gives that b is relatively prime to n(!! It should be m), so the pair (a,b) \mapsto am+bn, proving surjectivity.
Thank you for your article.It’s so clear.I have learnt a lot.
Like |
Friday, February 23, 2024
HomeMathGiant Quantity | Studying a Giant Quantity | Writing a Giant Numbers
# Giant Quantity | Studying a Giant Quantity | Writing a Giant Numbers
Right here we are going to focus on tips on how to learn and write massive numbers.
Studying a big quantity:
So as of studying a big quantity, of a 5 digit quantity, we should separate the digits in line with the interval.
(i) 72418
Ranging from the suitable, separate the ONES interval by marking a comma after three locations because the ONES interval has three locations.
72 418
↓ ↓
↓ 4 hundred eighteen
Seventy two thousand
The numbers shaped by the digits in a single interval are learn collectively, together with the identify of the interval.
Seventy two thousand 4 hundred eighteen
(ii) 36432
Thirty six thousand, 4 hundred thirty two
(iii) 40219
Forty thousand, 2 hundred nineteen
(iv) 51708
Fifty one thousand, seven hundred eight
(v) 29161
Twenty 9 thousand, 100 sixty one
Ones, tens, a whole lot which belong to the ONES interval.
Thousand, ten thousand which belong to the THOUSANDS interval.
The quantity represented within the first instance is 12452 and browse as:
Twelve thousand, 4 hundred fifty two
The quantity represented within the second instance is 20275 and browse as:
Twenty thousand, 2 hundred seventy 5
Notice: ‘And’ just isn’t utilized in a quantity line.
Writing massive numbers:
1. To jot down the numerals for; eighty 9 thousand seven hundred 4, we write the digits in a spot worth chart.
Then we write the numeral, utilizing a comma to separate the ONES from the THOUSANDS 89,704.
To learn and write a big quantity like 6 digit, 7 digit numerals.
Within the place worth chart, the sixth place is that of hundred 1000’s and the seventh place is of tens of millions.
To learn and write six or seven digit numerals we might want to use commas in two locations.
One comma to separate the ONES and THOUSANDS interval and one comma to separate the THOUSANDS and MILLIONS interval.
Writing Giant Numbers in Figures
2. Write the number-five lakh, 5 thousand, 600 twenty-nine in figures.
We proceed as follows:
Step I: Make three durations as proven.
The primary interval from the suitable may have three components, the second may have components and the third may have just one half.
Step II: Write all of the lakhs within the first interval from the left.
Step III: Write all of the 1000’s within the second interval from the left. We fill all of the components. For instance, whether it is 5 thousand, we write 05.
Step IV: Write the a whole lot, tens and ones within the third interval from the left.
Thus, 5 lakh, 5 thousand 600 twenty-nine in figures is 5,05,629.
3. Extra Solved Examples on writing massive numbers in figures:
Write every of the next numbers in figures:
(i) Eleven thousand 600 twenty-one.
(ii) Forty-one thousand two.
(iii) 5 lakh thirty-six thousand 5 hundred twenty.
(iv) Eight lakh three thousand ninety-two.
3. (i) Eleven thousand 600 twenty-ne.
→ 11,621
(ii) Forty-one thousand two.
→ 41,002
(iii) 5 lakh thirty-six thousand 5 hundred twenty.
→ 5,36,520
(iv) Eight lakh three thousand ninety-two.
→ 8,03,092
REMEMBER: 0 on the excessive left place is meaningless.
0458 is definitely 458.
Tips on how to
Write a Giant Quantity in Phrases and figures?
Allow us to
think about the quantity 449321 the quantity 449321 is proven within the place worth chart as
proven under.
The quantity
449321 is written in phrases as 4 lakhs, forty 9 thousand 300
twenty-one. The quantity in figures is written by placing commas as 4,49,321.
Place Worth:
A numeral will get its worth in line with the place it’s situated in. Every digit has a face place and a spot worth. The face worth is the digit itself. The face worth multiplied by the worth of the situation provides the place worth.
Within the numeral 257;
The face worth of 5 is 5. The place worth is 5 tens or 50
The face worth of two is 2. The place worth is 2 a whole lot or 200
Learn the place worth of every digit of huge numbers:
(i) 46,215
The place worth of
5 = 5 × 1 = 5 Ones = 5
1 = 1 × 10 = 1 Tens = 10
2 = 2 × 100 = 2 A whole bunch = 200
6 = 6 × 1000 = 6 Hundreds = 6000
4 = 4 × 10000 = 4 Ten Hundreds = 40000
(ii) 724,948
The place worth of
8 = 8 × 1 = 8 Ones = 8
4 = 4 × 10 = 4 Tens = 40
9 = 9 × 100 = 9 A whole bunch = 900
4 = 4 × 1000 = 4 Hundreds = 4000
2 = 2 × 10000 = 2 Ten Hundreds = 20000
7 = 7 × 100000 = 7 Hundred Hundreds = 700000
Worksheet on Learn and write a big quantity in phrases and figures:
I. Write
the numerals for the given numbers:
S.No. Quantity Identify Quantity (i) Six lakh, fifteen thousand, eighty-four __________ (ii) Twenty thousand, 9 __________ (iii) 4 lakh, thirty-two thousand, seven hundred ninety-two __________ (iv) Three lakh, twelve thousand, eighteen __________ (v) 9 lakh, eighty-eight thousand, ninety-eight __________
Solutions:
I. (i)
6,15,084
(ii) 20,009
(iii)
4,32,792
(iv)
3,12,018
(v)
9,88,098
II. Write
the quantity identify for the given numbers:
S.No. Quantity Quantity Identify (i) 1,38,640 ______________________________ (ii) 7,08,241 ______________________________ (iii) 9,00,002 ______________________________ (iv) 7,77,777 ______________________________
Solutions:
II. (i) One
lakh, thirty-eight thousand, 600 forty
(ii) Seven
lakh, eight thousand, 2 hundred forty-one
(iii) 9
lakh, two
(iv) Seven
lakh, seventy-seven thousand, seven hundred seventy-seven
III. Write
the quantity names for the numbers within the given assertion:
(i) The
inhabitants of our metropolis is 8,91,000
(ii) The
OTP for the acquisition is 1,92,064
Solutions:
III. (i) Eight
lakh, ninety-one thousand
(ii) One
lakh, ninety-two thousand, sixty-four
the next:
(i) The
largest 6-digit quantity is ……………………
(ii) The
smallest 6-digit quantity is ……………………
(iii) The
distinction between the biggest and the smallest 6-digit quantity is ……………………
(iv) The
largest 6-digit quantity shaped through the use of each 0 and 1 equal variety of instances is
……………………
(v) The
smallest 6-digit quantity shaped through the use of completely different digits is ……………………
Solutions:
IV. (i) 9,99,999
(ii) 1,00,000
(iii) 8,99,999
(iv) 1,11,000
(v) 1,23,456
V. Write every of the next numbers in figures:
(i) Fifteen thousand, 5 hundred 9.
(ii) Six lakh, twenty-one thousand, 2 hundred twenty.
(iii) 9 lakh, thirty thousand, seven.
(iv) Three lakh, fifty thousand, 2 hundred 5.
(v) 4 lakh, seven hundred.
(vi) Eight lakh, 5 thousand, ninety six.
(vii) 5 lakh, eight hundred twelve.
(viii) 5 lakh, eight thousand, 100 seven.
Solutions:
V. (i) 15,509
(ii) 6,29,220
(iii) 9,30,007
(iv) 3,50,205
(v) 4,00,700
(vi) 8,05,096
(vii) 5,00,812
(viii) 5,08,107
Associated Idea
Did not discover what you had been on the lookout for? Or need to know extra data
Math Solely Math.
Use this Google Search to search out what you want.
RELATED ARTICLES |
# Into Math Grade 6 Module 13 Answer Key Surface Area and Volume
We included HMH Into Math Grade 6 Answer Key PDF Module 13 Surface Area and Volume to make students experts in learning maths.
## HMH Into Math Grade 6 Module 13 Answer Key Surface Area and Volume
Pass the Popcorn
A company that makes microwave popcorn is redesigning its boxes. The diagram shows the company’s current box size, with each face marked off in squares with a side length of 1 inch.
Use the diagram to answer the questions.
A. What is the volume of the company’s current popcorn box?
_________ cubic inches
The above-given diagram is like a rectangular prism.
so we have to find out the volume of the rectangular prism that is the company’s current popcorn box.
The Volume of a Rectangular Prism Formula is,
The volume of a rectangular prism = length x width x height
Given that each face side length is 1 inch.
Count the squares.
length = 6; width =4; height = 5
Now calculate the volume.
volume = 6 x 4 x 5
volume = 120 cubic inches.
B. The company wants to keep the volume of its popcorn boxes the same, but decrease the width from 4 inches to 3 inches. What are the possible length and a possible height for the new box design?
Length: _________ inches
Height: _________ inches
If the width is decreased from 4 inches to 3 inches then the possible height and length for the new box design.
The possible length is 6 inches
The possible height is 5 inches
Turn and Talk
Question 1.
Explain how you determined the volume of the company’s current popcorn box.
– First, we have to identify the given box. The popcorn box is identified as a rectangular prism.
– Next, we have to learn the formula for the volume of the rectangular prism.
– We have to find the length, width, and height.
– Substituting the values in the formula v = lwh
– Be sure to express your answer in cubic units.
– rectangle was measured in inches, the volume should be written as 31 cubic inches
Question 2.
Explain how you determined a possible length and height for the new box design.
I think the height and length will be the same.
– Firstly, we reduce the width part which means we remove one line.
– So that the area will decrease but not length and height will decrease.
Complete these problems to review prior concepts and skills you will need for this module.
Explore Volume
Find the volume of each prism by counting cubic units.
Question 1.
V = _____________
The length is 5
The height is 2
The width is 3
The Volume of a Rectangular Prism Formula is,
The volume of a rectangular prism = length x width x height
volume = 5 x 2 x 3
Volume = 30 cubic units.
Question 2.
V = _____________
The volume of a prism is defined as the amount of space a prism occupies.
As we know, the volume of the prism is V = B × H.
Given that: B = 3 square inches, H = 2 inches
Thus, the Volume of the prism, V = B × H ⇒ V = 3 × 2 = 6 in3
Therefore, the volume of the prism is 6 cubic inches.
Question 3.
V = _____________
The length is 4
The height is 2
The width is 3
The Volume of a Rectangular Prism Formula is,
The volume of a rectangular prism = length x width x height
volume = 4 x 2 x 3
Volume = 24 cubic units.
Question 4.
V = _____________
Volume of a square = S^3
For the down square S = 4
Volume = 4 x 4 x 4
volume = 64 cubic units
Volume of upward square S = 3
Volume =3 x 3 x 3
volume = 27 cubic units.
Area of Quadrilaterals and Triangles
Find the area of each figure.
Question 5.
A = ___________
area of trapezium:
A trapezium is a quadrilateral, which is defined as a shape with four sides and one set of parallel sides. Thus, the area of the trapezium is the region covered within these four sides. The area of the trapezium basically depends upon the length of parallel sides and the height of the trapezium. It is measured in square units. Let us see the formula to find the area of any trapezium.
Formula:
Trapezium area can be calculated by using the below formula:
Area = (1/2) h (a+b) where,
– a and b are the length of parallel sides/bases of the trapezium
– h is the height or distance between parallel sides.
The given units: a = 35ft; b = 15ft; h = 14ft.
Now substitute the values in the above formula:
Area = 1/2 x 14(35 + 15)
Area = 7(50)
Area = 350 sq.ft
Question 6.
A = ____________
the right triangle legs are perpendicular to each other, one leg is taken as a base and the other is a right triangle height:
area = a * b/2
area = 8 x 14/2
area = 112/2
area = 56 square inches.
Question 7.
A = ____________
The area of a parallelogram is the region bounded by the parallelogram in a given two-dimension space. To recall, a parallelogram is a special type of quadrilateral which has four sides and the pair of opposite sides are parallel. In a parallelogram, the opposite sides are of equal length and opposite angles are of equal measures. Since the rectangle and the parallelogram have similar properties, the area of the rectangle is equal to the area of a parallelogram.
area of parallelogram = base x height square units.
The above-given units:
base = 14 cms
height = 7 cms
Now substitute the values in the formula:
Area = 14 x 7
Area = 98 square cms.
Question 8.
A = ____________
The formula to find the area of a rectangle depends on its length and width. The area of a rectangle is calculated in units by multiplying the width (or breadth) by the Length of a rectangle. Lateral and total surface areas can be calculated only for three-dimensional figures. We cannot calculate the rectangle since it is a two-dimensional figure. Thus, the perimeter and the area of a rectangle are given by:
area = length x breadth
Let’s consider base first.
The length is 18
area of one rectangle = 18 x 4
area 1 = 72 square ft.
area of other rectangle = l x b
area2 = 10 x 4
area2 = 40 square ft.
Now add area 1 + area 2
Total area = 72 + 40
Total area = 112 square ft.
Question 9.
The height of a parallelogram is 36 meters and the base is 6 meters. Find the area of the parallelogram. |
Site Navigation
Solving Two-Step Equations
A two-step equation is as straightforward as it sounds. You will need to perform two steps in order to solve the equation.
One goal in solving an equation is to have only variables on one side of the equal sign and numbers on the other side of the equal sign. The other goal is to have the number in front of the variable equal to one. The variable does not always have to be x. These equations can make use of any letter as a variable.
The strategy for getting the variable by itself with a coefficient of 1 involves using opposite operations. For example, to move something that is added to the other side of the equation, you should subtract. The most important thing to remember in solving a linear equation is that whatever you do to one side of the equation, you MUST do to the other side. So if you subtract a number from one side, you MUST subtract the same value from the other side. You will see how this works in the examples.
In solving two-step equations you will make use of the same techniques used in solving one-step equation only you will perform two operations rather than just one.
Let's Practice:
1. Solve
This problem does not have the variable by itself on one side. We need to get rid of the 4 that is added, so we’ll need to subtract 4 from both sides. Even after doing that, there is still a 3 multiplied by the variable, so division will be necessary to eliminate it.
1. Solve
The variable is not by itself on one side. We will get rid of the 5 that is added by subtracting 5 from both sides. This still leaves a -2 in front of the variable so we will have to divide both sides by -2.
1. Solve
As in the previous example, the variable is not by itself on one side. We will need to get rid of the -5 by adding 5 to both side. We will then need to multiply both sides by the reciprocal of which is .
To get help solving more complicated equations, click here (link to one var multi step.doc)
Examples
Solve What is your answer?
Solve What is your answer?
Solve What is your answer?
Solve What is your answer?
Solve What is your answer?
S Taylor
Show Related AlgebraLab Documents
AlgebraLAB Project Manager Catharine H. Colwell Application Programmers Jeremy R. Blawn Mark Acton Copyright © 2003-2024 All rights reserved. |
# How do you find an equation of the circle that satisfies the given conditions: endpoints of a diameter are P(−1, 2) and Q(7, 8)?
Apr 16, 2018
$\textcolor{b l u e}{{\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25}$
#### Explanation:
The coordinates of the centre of the circle will be the coordinates of the midpoint of the diameter.
$\left(\frac{- 1 + 7}{2} , \frac{2 + 8}{2}\right) = \left(3 , 5\right)$
We next find the length of the diameter using the distance formula:
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
$d = \sqrt{{\left(7 - \left(- 1\right)\right)}^{2} + {\left(8 - 2\right)}^{2}} = \sqrt{100} = 10$
This is the length of the diameter, so radius is:
$r = \frac{10}{2} = 5$
The equation of a circle is given as:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
Where $h \mathmr{and} k$ are the $x \mathmr{and} y$ coordinates of the centre respectively.
$\therefore$
${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25$
Apr 16, 2018
${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25$
#### Explanation:
Equation of a circle is of the form: ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
Where (a,b) represents the co ordinates of the circle's centre
Equation to find the coordinates of the centre of a line segment is
= $\left(\frac{x 1 + x 2}{2}\right) , \left(\frac{y 1 + y 2}{2}\right)$
Hence,
( 0.5 ( -1 + 7 ) ) , ( 0.5 ( 2 + 8 ) )
= ( 3 , 5 )
For the distance between two points , $\sqrt{{\left(x 2 - x 1\right)}^{2} + {\left(y 2 - y 1\right)}^{2}}$
Hence, sqrt( ( 7 - ( -1 ) )^2 + ( 8 - 2 )^2
= 10
Diameter of circle = 10
= ${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = {5}^{2}$
= ( x - 3 )^2 + ( y- 5 )^2 = 25 |
# Texas Go Math Grade 6 Lesson 10.2 Answer Key Prime Factorization
Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 10.2 Answer Key Prime Factorization.
## Texas Go Math Grade 6 Lesson 10.2 Answer Key Prime Factorization
Your Turn
List all the factors of each number.
Question 1.
21 ______________
Answer:
List the factors of 21
21 = 1 × 21
21 = 3 × 7
The factors of 21 are 1, 3, 7, 21.
Question 2.
37 ______________
Answer:
List the factors of 37
37 = 1 × 37
The factors of 37 are 1, 37,
Question 3.
42 ______________
Answer:
List the factors of 42
• 42 = 1 × 42
• 42 = 2 × 21
• 42 = 3 × 14
• 42 = 6 × 7
The factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.
Question 4.
30 ______________
Answer:
List the factors of 30
• 30 = 1 × 30
• 30 = 2 × 15
• 30 = 3 × 10
• 30 = 5 × 6
The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30
Reflect
Question 5.
What If? What will the factor tree for 240 look like if you start the tree with a different factor pair? Check your prediction by creating another factor tree for 240 that starts with a different factor pair.
Answer:
Prime factorization of 240 using factor tree.
Prime factorization of 240 using factor tree.
The prime factors of 240 are 5 ∙ 3 ∙ 2 ∙ 2 ∙ 2 ∙ 2 or 5 ∙ 3 ∙ 24
Reflect
Question 6.
Complete a factor tree and a ladder diagram to find the prime factorization of 54.
Answer:
Prime factorization of 54 using factor tree.
Prime factorization of 54 using a ladder diagram.
The prime factors of 54 are 2 ∙ 3 ∙ 3 ∙ 3 or 2 ∙ 33
Question 7.
Communicate Mathematical Ideas If one person uses a ladder diagram and another uses a factor tree to write a prime factorization, will they get the same result? Explain.
Answer:
If one person uses a ladder diagram and another uses a factor tree to write a prime factorization, they will get the same result because the prime factors of a number are identical an independent of the method used to evaluate them.
Texas Go Math Grade 6 Lesson 10.2 Guided Practice Answer Key
Use a diagram to list the factor pairs of each number.
Question 1.
18
Answer:
Given number: 18
Write the given number as a product of its factors, therefore:
• 18 = 2 × 9
• 18 = 3 × 6
The factors of 18 are 1, 2, 3, 6, 9 and 18.
Question 2.
52
Answer:
Given number: 52
Write the given number as a product of its factors, therefore:
52 = 2 × 26
52 = 4 × 13
The factors are of 52 are 1, 2, 4, 13, 26 and 52.
Question 3.
Karl needs to build a stage that has an area of 72 square feet. The length of the stage should be longer than the width. What are the possible whole number measurements for the length and width of the stage?
Complete the table with possible measurements of the stage.
Answer:
Possible measurements of the stage.
List the possible measurements of the stage based on the factors of 72.
Use a factor tree to find the prime factorization of each number.
Question 4.
402
Answer:
Prime factorization of 402 using a factor tree.
First line of factor tree 6 × 67 = 402
Factor out 6 and you will have 3 × 2.
The complete prime factors of 402 is 3 ∙ 2 ∙ 67
Another way to show the prime factorization of 402 using a factor tree.
First line of factor tree 2 × 201 = 402
Factor out 201 and you will have 3 × 67.
The complete prime factors of 402 is 2 ∙ 3 ∙ 67
Prime factors of 402 are 3 ∙ 2 67.
Question 5.
36
Answer:
Prime factorization of 36 using a factor tree.
Prime factors of 36 are 3 ∙ 2 ∙ 3 ∙ 2 or 32 ∙ 22
Use a ladder diagram to find the prime factorization of each number.
Question 6.
32
Answer:
Determine the prime factors of 32 using continuous division.
Prime factors of 32 are 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 or 25
Question 7.
27
Answer:
Determine the prime factors of 27 using continuous division.
The prime factors of 27 are 3 ∙ 3 ∙ 3 or 33
Essential Question Check-In
Question 8.
Tell how you know when you have found the prime factorization of a number.
Answer:
The prime factorization of a number is verified by studying its factors. They must all be prime numbers.
Question 9.
Multiple Representations Use the grid to draw three different rectangles so that each has an area of 12 square units and they all have different widths. What are the dimensions of the rectangles?
Answer:
The dimensions of the rectangle are:
a. 1 by 12
b. 2 by 6
c. 3 by 4
Diagram of the rectangle
Dimensions are: 1 by 12, 2 by 6, 3 by 4.
Question 10.
Brandon has 32 stamps. He wants to display the stamps in rows, with the same number of stamps in each row. How many different ways can he display the stamps? Explain.
Answer:
Given Number = 32
Write the given number as a product of its factors, therefore:
32 = 2 × 16
32 = 4 × 8
This implies that he can arrange the stamps in 4 different ways. He can make 2 columns each containing 16 stamps or 2 rows each containing 16 stamps Or he can make 4 columns each containing 8 stamps or 4 rows each containing 8 stamps.
He can arrange the stamps in 4 different ways. He can make 2 columns each containing 16 stamps or 2 rows each containing 16 stamps. Or he can make 4 columns each containing 8 stamps or 4 rows each containing 8 stamps.
Question 11.
Communicate Mathematical Ideas How is finding the factors of a number different from finding the prime factorization of a number?
Answer:
Finding the factors of a number indicates all the factors that may be used for the number which can be a combination of prime numbers and composite numbers. Finding the prime factorization of a number breaks a composite number into product of its prime factors.
Example:
12 – factors are 1 and 12, 2 and 6, 3 and 4
12 – prime factors are 2 ∙ 2 ∙ 3 or 22 ∙ 3
Factors are the numbers to be multiplied.
Prime factorization results to product of prime factors of the number.
Find the prime factorization of each number.
Question 12.
891 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 891 are 3, 3, 3, 3 and 11.
The prime factorization of 891 is:
891 = 3 × 3 × 3 × 3 × 11 or 34 × 11
= 34 × 11
Question 13.
5.4 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 504 are 2, 2, 2, 3, 3 and 7.
The prime factorization of 504 is:
504 = 2 × 2 × 2 × 3 × 3 × 7 or 23 × 32 × 7
= 23 × 32 × 7
Question 14.
23 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 23 are 1 and 23.
The prime factorization of 23 is:
23 = 1 × 23
Question 15.
230 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 230 are 2, 5 and 23.
The prime factorization of 230 is:
230 = 2 × 5 × 23
Question 16.
The number 2 is chosen to begin a ladder diagram to find the prime factorization of 66. What other numbers could have been used to start the ladder diagram for 66? How does starting with a different number change the diagram?
Answer:
Prime factorization of 66 using ladder diagram.
It will only change the order or arrangement of the prime factors but the prime factors are still the same.
The number 3 can be used to start the diagram however the prime factors will still be the same.
Question 17.
Critical Thinking List five numbers that have 3, 5, and 7 as prime factors.
Answer:
• 105 = 3 ∙ 5 ∙ 7
• 315 = 3 ∙ 3 ∙ 5 ∙ 7
• 525 = 3 ∙ 5 ∙ 5 ∙ 7
• 735 = 3 ∙ 5 ∙ 7 ∙ 7
• 1,575 = 3 ∙ 3 ∙ 5 ∙ 5 ∙ 7
The numbers are 105, 315, 525, 735, and 1,575.
Question 18.
In a game, you draw a card with three consecutive numbers on it. You can choose one of the numbers and find the sum of its prime factors. Then you can move that many spaces. You draw a card with the numbers 25, 26, 27. Which number should you choose if you want to move as many spaces as possible? Explain.
Answer:
Given number:
25, 26, 27
Write the given numbers as a product of their prime factors, therefore:
• 25 = 5 × 5
• 26 = 2 × 13
• 27 = 3 × 3 × 3
Evaluate the sum of the prime factors, therefore:
• 5 + 5 = 10
• 2 + 13 = 15
• 3 × 3 × 3 = 9
It can be seen that 15 > 10 > 9 therefore the choice of the number 26 will result in the maximum number of moves.
Question 19.
Explain the Error When asked to write the prime factorization of the number 27, a student wrote 9 3. Explain the error and write the correct answer.
Answer:
Given number = 27
Write the given number as a product of its prime factors, therefore:
27 = 3 × 3 × 3
Prime factors of the given number is 3 . The student’s solution is incorrect because the factor 9 ¡s not a prime factor.
The student’s solution is incorrect because the factor 9 is not a prime factor.
H.O.T. Focus On Higher Order Thinking
Question 20.
Communicate Mathematical Ideas Explain why it is possible to draw more than two different rectangles with an area of 36 square units, but it is not possible to draw more than two different rectangles with an area of 15 square units. The sides of the rectangles are whole numbers.
Answer:
Given number = 36
Write the given number as a product of its factors, therefore:
• 36 = 2 × 18 = Rectang1e 1
• 36 = 3 × 12 = Rectangle 2
• 36 = 4 × 9 = Rectangle 3
• 36 = 36 × 1 = Rectangle 4
• 36 = 6 × 6 = Square
It can be seen that 4 rectangles with different dimensions can be made each having an area of 36 square units
Given number = 15
Write the given number as a product of its factors, therefore:
• 15 = 3 × 5 = Rectangle 1
• 15 = 15 × 1 = Rectangle 2
It can be seen that only 2 rectangles can be made here because the given area is a product of prime factors or the number itself.
Question 21.
Critique Reasoning Alice wants to find all the prime factors of the number you get when you multiply 17 ∙ 11 ∙ 13 ∙ 7. She thinks she has to use a calculator to perform all the multiplications and then find the prime factorization of the resulting number. Do you agree? Why or why not?
Answer:
It can be seen that the number itself has factors that are prime numbers, so no calculator is required and the prime factors: 17, 11, 13 and 7 can directly be deduced.
The prime factors are 7, 11, 13 and 17.
Question 22.
Look for a Pattern Ryan wrote the prime factorizations shown below, If he continues this pattern, what prime factorization will he show for the number one million? What prime factorization will he show for one billion?
10 = 5 ∙ 2
100 = 52 ∙ 22
1,000 = 53 ∙ 23
Answer:
If the pattern continues, the number of zeros will determine the exponent of the base 10.
1 million = 1,000,000 = 56 . 26
1 billion = 1,000,000,000 = 59 . 29
The prime factorization for 1 million is 56 . 26
The prime factorization for 1 billion is 59 . 29
Scroll to Top |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 12.E: Introduction to Calculus (Exercises)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
## 12.1: Finding Limits - Numerical and Graphical Approaches
In this section, we will examine numerical and graphical approaches to identifying limits.
### Verbal
1) Explain the difference between a value at $$x=a$$ and the limit as $$x$$ approaches $$a$$.
The value of the function, the output, at $$x=a$$ is $$f(a)$$. When the $$\lim \limits_{x \to a}f(x)$$ is taken, the values of $$x$$ get infinitely close to $$a$$ but never equal $$a$$. As the values of $$x$$ approach $$a$$ from the left and right, the limit is the value that the function is approaching.
2) Explain why we say a function does not have a limit as $$x$$ approaches $$a$$ if, as $$x$$ approaches $$a$$, the left-hand limit is not equal to the right-hand limit.
### Graphical
For the exercises 3-14, estimate the functional values and the limits from the graph of the function $$f$$ provided in the Figure below.
3) $$\lim \limits_{x \to −2^−} f(x)$$
$$-4$$
4) $$\lim \limits_{x \to −2^+ }f(x)$$
5) $$\lim \limits_{x \to −2 f(x)}$$
$$-4$$
6) $$f(−2)$$
7) $$\lim \limits_{x \to −1^− f(x)}$$
$$2$$
8) $$\lim \limits_{x \to 1^+} f(x)$$
9) $$\lim \limits_{x \to 1} f(x)$$
does not exist
10) $$f(1)$$
11) $$\lim \limits_{x \to 4^−} f(x)$$
$$4$$
12) $$\lim \limits_{x \to 4^+} f(x)$$
13) $$\lim \limits_{x \to 4} f(x)$$
does not exist
14) $$f(4)$$
For the exercises 15-21, draw the graph of a function from the functional values and limits provided.
15) $$\lim \limits_{x \to 0^−} f(x)=2, \lim \limits_{x \to 0^+} f(x)=–3, \lim \limits_{x \to 2} f(x)=2, f(0)=4, f(2)=–1, f(–3) \text{ does not exist.}$$
16) $$\lim \limits_{x \to 2^−} f(x)=0,\lim \limits_{x \to 2^+} =–2,\lim \limits_{x \to 0} f(x)=3, f(2)=5, f(0)$$
17) $$\lim \limits_{ x \to 2^−} f(x)=2, \lim \limits_{ x \to 2^+} f(x)=−3, \lim \limits_{x \to 0} f(x)=5, f(0)=1, f(1)=0$$
18) $$\lim \limits_{x \to 3^−} f(x)=0, \lim \limits_{x \to 3^+} f(x)=5, \lim \limits_{x \to 5} f(x)=0, f(5)=4, f(3) \text{ does not exist.}$$
19) $$\lim \limits_{ x \to 4} f(x)=6, \lim \limits_{ x \to 6^+} f(x)=−1, \lim \limits_{ x \to 0} f(x)=5, f(4)=6, f(2)=6$$
20) $$\lim \limits_{ x \to −3} f(x)=2, \lim \limits_{ x \to 1^+} f(x)=−2, \lim \limits_{ x \to 3} f(x)=–4, f(–3)=0, f(0)=0$$
21) $$\lim \limits_{ x \to π} f(x)=π^2, \lim \limits_{ x \to –π} f(x)=\dfrac{π}{2}, \lim \limits_{ x \to 1^-} f(x)=0, f(π)=\sqrt{2}, f(0) \text{ does not exist}.$$
For the exercises 22-26, use a graphing calculator to determine the limit to $$5$$ decimal places as $$x$$ approaches $$0$$.
22) $$f(x)=(1+x)^{\frac{1}{x}}$$
23) $$g(x)=(1+x)^{\frac{2}{x}}$$
$$7.38906$$
24) $$h(x)=(1+x)^{\frac{3}{x}}$$
25) $$i(x)=(1+x)^{\frac{4}{x}}$$
$$54.59815$$
26) $$j(x)=(1+x)^{\frac{5}{x}}$$
27) Based on the pattern you observed in the exercises above, make a conjecture as to the limit of $$f(x)=(1+x)^{\frac{6}{x}}, g(x)=(1+x)^{\frac{7}{x}},$$ and $$h(x)=(1+x)^{\frac{n}{x}}.$$
$$e^6≈403.428794,e^7≈1096.633158, e^n$$
For the exercises 28-29, use a graphing utility to find graphical evidence to determine the left- and right-hand limits of the function given as $$x$$ approaches $$a$$. If the function has a limit as $$x$$ approaches $$a$$,state it. If not, discuss why there is no limit.
28) $$(x)= \begin{cases} |x|−1, && \text{if }x≠1 \\ x^3, && \text{if }x=1 \end{cases} a=1$$
29) $$(x)= \begin{cases} \frac{1}{x+1}, && \text{if } x=−2 \\ (x+1)^2, && \text{if } x≠−2 \end{cases} a=−2$$
$$\lim \limits_{x \to −2} f(x)=1$$
### Numeric
For the exercises 30-38, use numerical evidence to determine whether the limit exists at $$x=a$$. If not, describe the behavior of the graph of the function near $$x=a$$. Round answers to two decimal places.
30) $$f(x)=\dfrac{x^2−4x}{16−x^2};a=4$$
31) $$f(x)=\dfrac{x^2−x−6}{x^2−9};a=3$$
$$\lim \limits_{x \to 3} \left (\dfrac{x^2−x−6}{x^2−9} \right )=\dfrac{5}{6}≈0.83$$
32) $$f(x)=\dfrac{x^2−6x−7}{x^2– 7x};a=7$$
33) $$f(x)=\dfrac{x^2–1}{x^2–3x+2};a=1$$
$$\lim \limits_{x \to 1} \left (\dfrac{x^2−1}{x^2−3x+2} \right )=−2.00$$
34) $$f(x)=\dfrac{1−x^2}{x^2−3x+2};a=1$$
35) $$f(x)=\dfrac{10−10x^2}{x^2−3x+2};a=1$$
$$\lim \limits_{x \to 1} \left (\dfrac{10−10x^2}{x^2−3x+2} \right )=20.00$$
36) $$f(x)=\dfrac{x}{6x^2−5x−6};a=\dfrac{3}{2}$$
37) $$f(x)=\dfrac{x}{4x^2+4x+1};a=−\dfrac{1}{2}$$
$$\lim \limits_{x \to \frac{−1}{2}} \left (\dfrac{x}{4x^2+4x+1} \right )$$ does not exist. Function values decrease without bound as $$x$$ approaches $$-0.5$$ from either left or right.
38) $$f(x)=\frac{2}{x−4}; a=4$$
For the exercises 39-41, use a calculator to estimate the limit by preparing a table of values. If there is no limit, describe the behavior of the function as $$x$$ approaches the given value.
39) $$\lim \limits_{x \to 0} \dfrac{7 \tan x}{3x}$$
$$\lim \limits_{x \to 0} \dfrac{7 \tan x}{3x}=\dfrac{7}{3}$$
40) $$\lim \limits_{x \to 4} \dfrac{x^2}{x−4}$$
41) $$\lim \limits_{x \to 0}\dfrac{2 \sin x}{4 \tan x}$$
$$\lim \limits_{x \to 0} \dfrac{2 \sin x}{4 \tan x}=\dfrac{1}{2}$$
For the exercises 42-49, use a graphing utility to find numerical or graphical evidence to determine the left and right-hand limits of the function given as $$x$$ approaches $$a$$. If the function has a limit as $$x$$ approaches $$a$$, state it. If not, discuss why there is no limit.
42) $$\lim \limits_{x \to 0}e^{e^{\frac{1}{x}}}$$
43) $$\lim \limits_{x \to 0}e^{e^{− \frac{1}{x^2}}}$$
$$\lim \limits_{x \to 0}e^{e^{− \frac{1}{x^2}}}=1.0$$
44) $$\lim \limits_{x \to 0} \dfrac{|x|}{x}$$
45) $$\lim \limits_{x \to −1} \dfrac{|x+1|}{x+1}$$
$$\lim \limits_{ x→−1^−}\dfrac{| x+1 |}{x+1}=\dfrac{−(x+1)}{(x+1)}=−1$$ and $$\lim \limits_{ x \to −1^+}\dfrac{| x+1 |}{x+1}=\dfrac{(x+1)}{(x+1)}=1$$; since the right-hand limit does not equal the left-hand limit, $$\lim \limits_{ x \to −1}\dfrac{|x+1|}{x+1}$$ does not exist.
46) $$\lim \limits_{ x \to 5} \dfrac{| x−5 |}{5−x}$$
47) $$\lim \limits_{ x \to −1}\dfrac{1}{(x+1)^2}$$
$$\lim \limits_{ x \to −1} \dfrac{1}{(x+1)^2}$$ does not exist. The function increases without bound as $$x$$ approaches $$−1$$ from either side.
48) $$\lim \limits_{ x \to 1} \dfrac{1}{(x−1)^3}$$
49) $$\lim \limits_{ x \to 0} \dfrac{5}{1−e^{\frac{2}{x}}}$$
$$\lim \limits_{ x \to 0} \dfrac{5}{1−e^{\frac{2}{x}}}$$ does not exist. Function values approach $$5$$ from the left and approach $$0$$ from the right.
50) Use numerical and graphical evidence to compare and contrast the limits of two functions whose formulas appear similar: $$f(x)=\left | \dfrac{1−x}{x} \right |$$ and $$g(x)=\left | \dfrac{1+x}{x} \right |$$ as $$x$$ approaches $$0$$. Use a graphing utility, if possible, to determine the left- and right-hand limits of the functions $$f(x)$$ and $$g(x)$$ as $$x$$ approaches $$0$$. If the functions have a limit as $$x$$ approaches $$0$$, state it. If not, discuss why there is no limit.
### Extensions
51) According to the Theory of Relativity, the mass m m of a particle depends on its velocity $$v$$. That is
$m=\dfrac{m_o}{\sqrt{1−(v^2/c^2)}} \nonumber$
where $$m_o$$ is the mass when the particle is at rest and $$c$$ is the speed of light. Find the limit of the mass, $$m$$, as $$v$$ approaches $$c^−.$$
Through examination of the postulates and an understanding of relativistic physics, as $$v→c, m→∞.$$Take this one step further to the solution, $\lim \limits_{v \to c^−}m=\lim \limits_{v \to c^−} \dfrac{m_o}{\sqrt{1−(v^2/c^2)}}=∞ \nonumber$
52) Allow the speed of light, $$c$$, to be equal to $$1.0$$. If the mass, $$m$$, is $$1$$, what occurs to $$m$$ as $$v \to c$$? Using the values listed in the Table below, make a conjecture as to what the mass is as $$v$$ approaches $$1.00$$.
$$v$$ $$m$$
0.5 1.15
0.9 2.29
0.95 3.20
0.99 7.09
0.999 22.36
0.99999 223.61
## 12.2: Finding Limits - Properties of Limits
Graphing a function or exploring a table of values to determine a limit can be cumbersome and time-consuming. When possible, it is more efficient to use the properties of limits, which is a collection of theorems for finding limits. Knowing the properties of limits allows us to compute limits directly.
### Verbal
1) Give an example of a type of function $$f$$ whose limit, as $$x$$ approaches $$a,$$ is $$f(a)$$.
If $$f$$ is a polynomial function, the limit of a polynomial function as $$x$$ approaches $$a$$ will always be $$f(a)$$.
2) When direct substitution is used to evaluate the limit of a rational function as $$x$$ approaches $$a$$ and the result is $$f(a)=\dfrac{0}{0}$$,does this mean that the limit of $$f$$ does not exist?
3) What does it mean to say the limit of $$f(x)$$, as $$x$$ approaches $$c$$, is undefined?
It could mean either (1) the values of the function increase or decrease without bound as $$x$$ approaches $$c,$$ or (2) the left and right-hand limits are not equal.
### Algebraic
For the exercises 4-30, evaluate the limits algebraically.
4) $$\lim \limits_{x \to 0} (3)$$
5) $$\lim \limits_{x \to 2} \left (\dfrac{−5x}{x^2−1} \right )$$
$$\dfrac{−10}{3}$$
6) $$\lim \limits_{x \to 2} \left (\dfrac{x^2−5x+6}{x+2} \right )$$
7) $$\lim \limits_{x \to 3} \left (\dfrac{x^2−9}{x−3} \right )$$
$$6$$
8) $$\lim \limits_{x \to −1} \left (\dfrac{x^2−2x−3}{x+1} \right )$$
9) $$\lim \limits_{x \to \frac{3}{2}} \left (\dfrac{6x^2−17x+12}{2x−3} \right )$$
$$\dfrac{1}{2}$$
10) $$\lim \limits_{ x \to −\frac{7}{2}} \left (\dfrac{8x^2+18x−35}{2x+7} \right )$$
11) $$\lim \limits_{ x \to 3} \left (\dfrac{x^2−9}{x−5x+6} \right )$$
$$6$$
12) $$\lim \limits_{ x \to −3} \left (\dfrac{−7x^4−21x^3}{−12x^4+108x^2} \right )$$
13) $$\lim \limits_{ x \to 3} \left (\dfrac{x^2+2x−3}{x−3} \right )$$
does not exist
14) $$\lim \limits_{ h \to 0} \left (\dfrac{(3+h)^3−27}{h} \right )$$
15) $$\lim \limits_{ h \to 0} \left (\dfrac{(2−h)^3−8}{h} \right )$$
$$−12$$
16) $$\lim \limits_{ h \to 0} \left (\dfrac{(h+3)^2−9}{h} \right )$$
17) $$\lim \limits_{ h \to 0} \left (\dfrac{\sqrt{5−h}−\sqrt{5}}{h} \right )$$
$$−\dfrac{\sqrt{5}}{10}$$
18) $$\lim \limits_{ x \to 0} \left (\dfrac{\sqrt{3−x}−\sqrt{3}}{x} \right )$$
19) $$\lim \limits_{ x \to 9} \left (\dfrac{x^2−81}{3−x} \right )$$
$$−108$$
20) $$\lim \limits_{ x \to 1} \left (\dfrac{\sqrt{x}−x^2}{1−\sqrt{x}} \right )$$
21) $$\lim \limits_{ x \to 0}\left ( \dfrac{x}{\sqrt{1+2x}-1} \right )$$
$$1$$
22) $$\lim \limits_{ x \to \frac{1}{2}} \left (\dfrac{x^2−\tfrac{1}{4}}{2x−1} \right )$$
23) $$\lim \limits_{ x \to 4} \left (\dfrac{x^3−64}{x^2−16} \right )$$
$$6$$
24) $$\lim \limits_{ x \to 2^−} \left (\dfrac{|x−2|}{x−2} \right )$$
25) $$\lim \limits_{ x \to 2^+} \left (\dfrac{| x−2 |}{x−2} \right )$$
$$1$$
26) $$\lim \limits_{ x \to 2} \left (\dfrac{| x−2 |}{x−2} \right )$$
27) $$\lim \limits_{ x \to 4^−} \left (\dfrac{| x−4 |}{4−x} \right )$$
$$1$$
28) $$\lim \limits_{ x \to 4^+} \left (\dfrac{| x−4 |}{4−x} \right )$$
29) $$\lim \limits_{ x \to 4} \left (\dfrac{| x−4 |}{4−x} \right )$$
does not exist
30) $$\lim \limits_{ x \to 2} \left (\dfrac{−8+6x−x^2}{x−2} \right )$$
For the exercises 31-33, use the given information to evaluate the limits: $$\lim \limits_{x \to c}f(x)=3, \lim \limits_{x \to c} g(x)=5$$
31) $$\lim \limits_{x \to c} [ 2f(x)+\sqrt{g(x)} ]$$
$$6+\sqrt{5}$$
32) $$\lim \limits_{x \to c} [ 3f(x)+\sqrt{g(x)} ]$$
33) $$\lim \limits_{x \to c}\dfrac{f(x)}{g(x)}$$
$$\dfrac{3}{5}$$
For the exercises 34-43, evaluate the following limits.
34) $$\lim \limits_{x \to 2} \cos (πx)$$
35) $$\lim \limits_{x \to 2} \sin (πx)$$
$$0$$
36) $$\lim \limits_{x \to 2} \sin \left (\dfrac{π}{x} \right )$$
37) $$f(x)= \begin{cases} 2x^2+2x+1, && x≤0 \\ x−3, && x>0 ; \end{cases} \lim \limits_{x \to 0^+}f(x)$$
$$−3$$
38) $$f(x)= \begin{cases} 2x^2+2x+1, && x≤0 \\ x−3, && x>0 ; \end{cases} \lim \limits_{x \to 0^−} f(x)$$
39) $$f(x)= \begin{cases} 2x^2+2x+1, && x≤0 \\ x−3, && x>0 ; \end{cases} \lim \limits_{x \to 0}f(x)$$
does not exist; right-hand limit is not the same as the left-hand limit.
40) $$\lim \limits_{x \to 4} \dfrac{\sqrt{x+5}−3}{x−4}$$
41) $$\lim \limits_{x \to 2^+} (2x−〚x〛)$$
$$2$$
42) $$\lim \limits_{x \to 2} \dfrac{\sqrt{x+7}−3}{x^2−x−2}$$
43) $$\lim \limits_{x \to 3^+}\dfrac{x^2}{x^2−9}$$
Limit does not exist; limit approaches infinity.
For the exercises 44-53, find the average rate of change$$\dfrac{f(x+h)−f(x)}{h}$$.
44) $$f(x)=x+1$$
45) $$f(x)=2x^2−1$$
$$4x+2h$$
46) $$f(x)=x^2+3x+4$$
47) $$f(x)=x^2+4x−100$$
$$2x+h+4$$
48) $$f(x)=3x^2+1$$
49) $$f(x)= \cos (x)$$
$$\dfrac{\cos (x+h)− \cos (x)}{h}$$
50) $$f(x)=2x^3−4x$$
51) $$f(x)=\dfrac{1}{x}$$
$$\dfrac{−1}{x(x+h)}$$
52) $$f(x)=\dfrac{1}{x^2}$$
53) $$f(x)=\sqrt{x}$$
$$\dfrac{−1}{\sqrt{x+h}+\sqrt{x}}$$
### Graphical
54) Find an equation that could be represented by the Figure below.
$$f(x)=\dfrac{x^2+5x+6}{x+3}$$
For the exercises 56-57, refer to the Figure below.
56) What is the right-hand limit of the function as $$x$$ approaches $$0$$?
57) What is the left-hand limit of the function as $$x$$ approaches $$0$$?
does not exist
### Real-World Applications
58) The position function $$s(t)=−16t^2+144t$$ gives the position of a projectile as a function of time. Find the average velocity (average rate of change) on the interval $$[ 1,2 ]$$.
59) The height of a projectile is given by $$s(t)=−64t^2+192t$$ Find the average rate of change of the height from $$t=1$$ second to $$t=1.5$$ seconds.
$$52$$
60) The amount of money in an account after $$t$$ years compounded continuously at $$4.25\%$$ interest is given by the formula $$A=A_0e^{0.0425t}$$,where $$A_0$$ is the initial amount invested. Find the average rate of change of the balance of the account from $$t=1$$ year to $$t=2$$ years if the initial amount invested is $$\1,000.00.$$
## 12.3: Continuity
A function that remains level for an interval and then jumps instantaneously to a higher value is called a stepwise function. This function is an example. A function that has any hole or break in its graph is known as a discontinuous function. A stepwise function, such as parking-garage charges as a function of hours parked, is an example of a discontinuous function. We can check three different conditions to decide if a function is continuous at a particular number.
### Verbal
1) State in your own words what it means for a function $$f$$ to be continuous at $$x=c$$.
Informally, if a function is continuous at $$x=c$$, then there is no break in the graph of the function at $$f(c)$$, and $$f(c)$$ is defined.
2) State in your own words what it means for a function to be continuous on the interval $$(a,b)$$.
### Algebraic
For the exercises 3-22, determine why the function $$f$$ is discontinuous at a given point $$a$$ on the graph. State which condition fails.
3) $$f(x)=\ln | x+3 |,a=−3$$
discontinuous at $$a=−3$$; $$f(−3)$$ does not exist
4) $$f(x)= \ln | 5x−2 |,a=\dfrac{2}{5}$$
5) $$f(x)=\dfrac{x^2−16}{x+4},a=−4$$
removable discontinuity at $$a=−4; f(−4)$$ is not defined
6) $$f(x)=\dfrac{x^2−16x}{x},a=0$$
7) $$f(x)= \begin{cases} x, && x≠3 \\ 2x, && x=3 \end{cases} a=3$$
Discontinuous at $$a=3; \lim \limits_{x \to 3} f(x)=3,$$ but $$f(3)=6,$$ which is not equal to the limit.
8) $$f(x) = \begin{cases} 5, &&x≠0 \\ 3, && x=0 \end{cases} a=0$$
9) $$f(x)= \begin{cases} \dfrac{1}{2−x}, && x≠2 \\ 3, &&x=2 \end{cases} a=2$$
$$\lim \limits_{x \to 2}f(x)$$ does not exist.
10) $$f(x)= \begin{cases} \dfrac{1}{x+6}, && x=−6 \\ x^2, && x≠−6 \end{cases} a=−6$$
11) $$f(x)=\begin{cases} 3+x, &&x<1 \\ x, &&x=1 \\ x^2, && x>1 \end{cases} a=1$$
$$\lim \limits_{x \to 1^−}f(x)=4;\lim \limits_{x \to 1^+}f(x)=1.$$ Therefore, $$\lim \limits_{x \to 1}f(x)$$ does not exist.
12) $$f(x)= \begin{cases} 3−x, && x<1 \\ x, && x=1 \\ 2x^2, && x>1 \end{cases} a=1$$
13) $$f(x)= \begin{cases} 3+2x, && x<1 \\ x, && x=1 \\ −x^2, && x>1 \end{cases} a=1$$
$$\lim \limits_{x \to 1^−} f(x)=5≠ \lim \limits_{x \to 1^+}f(x)=−1$$. Thus $$\lim \limits_{x \to 1}f(x)$$ does not exist.
14) $$f(x)= \begin{cases} x^2, &&x<−2 \\ 2x+1, && x=−2 \\ x^3, && x>−2 \end{cases} a=−2$$
15) $$f(x)= \begin{cases} \dfrac{x^2−9}{x+3}, && x<−3 \\ x−9, && x=−3 \\ \dfrac{1}{x}, && x>−3 \end{cases} a=−3$$
$$\lim \limits_{x to −3^+}f(x)=−\dfrac{1}{3}$$
Therefore, $$\lim \limits_{x \to −3} f(x)$$ does not exist.
16) $$f(x)= \begin{cases} \dfrac{x^2−9}{x+3}, && x<−3 \\ x−9, && x=−3\\ −6, && x>−3 \end{cases} a=3$$
17) $$f(x)=\dfrac{x^2−4}{x−2}, a=2$$
$$f(2)$$ is not defined.
18) $$f(x)=\dfrac{25−x^2}{x^2−10x+25}, a=5$$
19) $$f(x)=\dfrac{x^3−9x}{x^2+11x+24}, a=−3$$
$$f(−3)$$ is not defined.
20) $$f(x)=\dfrac{x^3−27}{x^2−3x}, a=3$$
21) $$f(x)=\dfrac{x}{|x|}, a=0$$
$$f(0)$$ is not defined.
22) $$f(x)=\dfrac{2|x+2|}{x+2}, a=−2$$
For the exercises 23-35, determine whether or not the given function $$f$$ is continuous everywhere. If it is continuous everywhere it is defined, state for what range it is continuous. If it is discontinuous, state where it is discontinuous.
23) $$f(x)=x^3−2x−15$$
Continuous on $$(−∞,∞)$$
24) $$f(x)=\dfrac{x^2−2x−15}{x−5}$$
25) $$f(x)=2⋅3^{x+4}$$
Continuous on $$(−∞,∞)$$
26) $$f(x)=− \sin (3x)$$
27) $$f(x)=\dfrac{|x−2|}{x^2−2x}$$
Discontinuous at $$x=0$$ and$$x=2$$
28) $$f(x)= \tan (x)+2$$
29) $$f(x)=2x+\dfrac{5}{x}$$
Discontinuous at $$x=0$$
30) $$f(x)=\log _2 (x)$$
31) $$f(x)= \ln x^2$$
Continuous on $$(0,∞)$$
32) $$f(x)=e^{2x}$$
33) $$f(x)=\sqrt{x−4}$$
Continuous on $$[4,∞)$$
34) $$f(x)= \sec (x)−3$$
35) $$f(x)=x^2+ \sin (x)$$
Continuous on $$(−∞,∞)$$.
36) Determine the values of $$b$$ and $$c$$ such that the following function is continuous on the entire real number line.
$f(x)= \begin{cases}x+1, && 1<x<3 \\ x^2+bx+c, &&|x−2|≥1 \end{cases} \nonumber$
### Graphical
For the exercises 37-39, refer to the Figure below. Each square represents one square unit. For each value of $$a$$, determine which of the three conditions of continuity are satisfied at $$x=a$$ and which are not.
37) $$x=−3$$
$$1$$, but not $$2$$ or $$3$$
38) $$x=2$$
39) $$x=4$$
$$1$$ and $$2$$, but not $$3$$
For the exercises 40-43, use a graphing utility to graph the function $$f(x)= \sin \left (\dfrac{12π}{x} \right )$$ as in Figure. Set the $$x$$-axis a short distance before and after $$0$$ to illustrate the point of discontinuity.
40) Which conditions for continuity fail at the point of discontinuity?
41) Evaluate $$f(0)$$.
$$f(0)$$ is undefined.
42) Solve for $$x$$ if $$f(x)=0$$.
43) What is the domain of $$f(x)$$?
$$(−∞,0)∪(0,∞)$$
For the exercises 44-45, consider the function shown in the Figure below.
44) At what $$x$$-coordinates is the function discontinuous?
45) What condition of continuity is violated at these points?
At $$x=−1$$, the limit does not exist. At $$x=1, f(1)$$ does not exist.
At $$x=2$$, there appears to be a vertical asymptote, and the limit does not exist.
46) Consider the function shown in the Figure below. At what $$x$$-coordinates is the function discontinuous? What condition(s) of continuity were violated?
47) Construct a function that passes through the origin with a constant slope of $$1$$, with removable discontinuities at $$x=−7$$ and $$x=1$$.
$$\dfrac{x^3+6x^2−7x}{(x+7)(x−1)}$$
48) The function $$f(x)=\dfrac{x^3−1}{x−1}$$ is graphed in the Figure below. It appears to be continuous on the interval $$[−3,3]$$, but there is an $$x$$-value on that interval at which the function is discontinuous. Determine the value of $$x$$ at which the function is discontinuous, and explain the pitfall of utilizing technology when considering continuity of a function by examining its graph.
49) Find the limit $$\lim \limits_{ x \to 1}f(x)$$ and determine if the following function is continuous at $$x=1$$:
$fx= \begin{cases} x^2+4 && x≠1 \\ 2 && x=1\end{cases} \nonumber$
The function is discontinuous at $$x=1$$ because the limit as $$x$$ approaches $$1$$ is $$5$$ and $$f(1)=2$$.
50) The graph of $$f(x)= \dfrac{\sin (2x)}{x}$$ is shown in the Figure below. Is the function $$f(x)$$ continuous at $$x=0?$$ Why or why not?
## 12.4: Derivatives
Change divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we could compare the rates by determining the slopes of the graphs.
### Verbal
1) How is the slope of a linear function similar to the derivative?
The slope of a linear function stays the same. The derivative of a general function varies according to $$x$$. Both the slope of a line and the derivative at a point measure the rate of change of the function.
2) What is the difference between the average rate of change of a function on the interval $$[x,x+h]$$ and the derivative of the function at |
## Triangle Subdivisions
##### – Jan. 30, 2021
Figure 1: 16 iterations of subdivision
This article showcases a family of triangle subdivisions, where very little mathematics is involved. Every triangle will be subdivided into exactly two new triangles at every step, in one of the easiest ways possible. The subdivision is complete, meaning that after every iteration of subdivision, the area of all resulting triangles equals the area of the triangles from the previous iteration. There are no gaps created (like in the Sierpinsky triangle), and the subdivision will always stay exactly within the bounds of the previous triangle.
An easy way to subdivide a triangle into two new triangles is to draw a straight line from one of the corner points (vertices) to a point on the opposite edge. There are always three triangle vertices to choose from. The point on the opposite edge can be anywhere, but the most straightforward choice with an easy calculation is to take the midpoint of the opposite edge. Taking the midpoint also results in some beautiful symmetries.
Figure 2: Equilateral triangle subdivision
For an equilateral triangle, all three possible subdivisions are equal (but rotated by a third of a full rotation as seen on the image above). The new triangles are always two right triangles. The image below shows the three different subdivisions for this right triangle.
Figure 3: Right triangle subdivision
As you can see, all three subdivisions result in different triangles. Note that for the result in the center, one new triangle is an equilateral triangle, and the other triangle is an isosceles triangle. By the nature of this subdivision, every right triangle subdivides into two isosceles triangles when the right angle is chosen. Both of those isosceles triangles can subdivide into 2 of the original right triangles. In the example above, one of the isosceles triangles was an equilateral triangle, which is only the case if the original right triangle has 30° and 60° angles.
Below are the three subdivisions for the isosceles triangle.
Figure 4: Isosceles triangle subdivision
The first two outcomes are the same, but mirrored (which is expected since an isosceles triangle has two equal sides and two equal angles). For the third one the result is two right triangles (the same 30°–60°–90° triangle as before).
This means that there could be repeating patterns in the subdivision. For example:
Equilateral triangle → right triangle → equilateral triangle.
Right triangle → isosceles triangle → Right triangle (always the same one as 2 steps before).
When the same angle keeps getting subdivided, the resulting triangles will get sharper and sharper.
Figure 5: Sharp triangle subdivision
Alternating between different vertices produces more interesting images. Below are two examples with 20 iterations. There are over 1 million triangles drawn, but the image was resized, so the individual triangles are not visible any more.
Figure 6: Two examples of subdivisions after 20 iterations
### How many unique variations are there?
For every number of iterations, we can calculate the number of variations.
For 0 iterations it is is the original triangle only.
For 1 iteration, there are 3 different subdivisions.
For the 2nd iteration, each of those 3 previous variations can be subdivided in 9 different ways because every variation from iteration 1 consists of 2 triangles. Each of those 2 triangles can be subdivided in 3 different ways, which results in 32 new variations (for every variation from iteration 1).
In the image below, you can see the 3 different variations from the 1st iteration on the first line.
Below them are all 9 variations in blocks of 3 x 3.
Use the toggle button to switch between equilateral triangles, and right triangles with 3 different side lengths. The subdivisions are exactly the same, but the original triangles are different.
If rotation is allowed, we can see that only one third of the 27 equilateral triangles are different. For the right triangle every subdivision is different, because this triangle has 3 different side lengths, and rotation will not result in an equal image.
If reflections are allowed, the number of unique subdivisions shrinks even more for the equilateral triangles. In the image below are the equilateral triangles re grouped. From now on, I choose to only work with the rotation where the first subdivision (red line) is a vertical line, because that one is the most intuitive when talking about reflections. On the first line are the six unique ones. The 3 triangles on the second row, are equivalent to the triangle above it, when reflecting through the red line, which is the division line from the 1st iteration.
Figure 8: Grouped the equilateral triangle subdivisions after 2 iterations
The first three are symmetric by reflecting trough the red line.
Note that this grouping for the equilateral triangles is visible in figure 7. Take one of the 3 by 3 groups of triangles, and notice the diagonal with symmetric variations from top left to bottom right. For each of the resulting 6 triangles, you can find the other half of the pair on the other side of the diagonal.
The number of triangles per iteration is doubled at every step, since we divide every triangle into exactly two new triangles. Let's call the number of triangles T, and T(n) = 2n.
V is the total number of variations, when no rotations and reflections are taken into account. V is a multiple of the number of variations from the previous iteration, because for every single previous variation, a number of new variation can be made (look at figure 7 to see that there are 9 new variations for each of the 3 previous ones). This multiplication number is 3T(n-1), where T(n-1) is the number of triangles from the previous generation, since for every triangle from the previous iteration, 3 new subdivisions can be made. The number of triangles in the previous iteration is 2n - 1.
So V(n) = 32(n - 1) * V(n -1). Since we know that V(0) = 1, we can fill it in and calculate V(1), V(2) etc. I find that V(n) = 3(2n - 1). Note that this is 3 to the power Mersenne number(n).
Iterations Triangles (T) V S U
01111
12311
242736
38218727378
4161434890721872392578
..
n2n3(2n- 1)V(n - 1)(S2 + S) / 2
Now I like to consider the equilateral triangle subdivisions only, and look at the unique and symmetrical subdivisions. I listed them in the table above as U and S respectively. As we have seen before, there are 3 rotations for every variation, so we only have to consider 1 / 3th of the total variations (32n-2, subtract 1 from the powers of 3.). This 1 / 3th of the total is always a square number (it can be rewritten as (32(n-1)-1)2). They can be listed in a square grid (equal number of items per row as per column), like I did before in figure 7 for all 3 rotations. For every variation, there is an equivalent one, reflected through the red line (subdivision line from iteration 1), except for the symmetrical variations (they stay the same after this reflection).
When you open the file in this link, you can see the same idea for one third of all 3th iteration subdivisions, in a 27 x 27 grid. On the diagonal you will find 27 symmetrical subdivisions, and every single subdivision on one side of the diagonal has a reflected version on the other side of the diagonal. The number of symmetrical subdivision (the diagonal) is the same as the number of rows or columns in this grid. It is the square root of 1 / 3th of the total variations. This turns out to equal V(n - 1). This number of symmetrical subdivisions is called S in the table.
This means that the total variations V(n) can also be expressed as 3 * (V(n-1))2.
Figure 9: All 27 symmetrical variations for 3th iteration triangle subdivisions
All 27 symmetrical subdivisions after 3 iterations are listed in the image above. Note that the vertical red line is the division line from iteration 1, the two blue lines are the subdivision lines from iteration 2, and the 4 green lines are the subdivision lines from iteration 3. For each of the 3 symmetrical variations from iteration 2, 9 new variations could be constructed. Although the previous iteration had 4 triangles, only half of them (one side of the red line) are relevant to achieve the symmetrical results. Those two triangles could be subdivided in 32=9 different ways. For the next iteration I expect 27 * 34 symmetrical variations.
In general S(n) = S(n - 1) * 3T(n - 1) / 2 = S(n - 1) * 3n - 2 (for n > 1 and S(0) = S(1) = 1).
This turns out to be S(n) = 32(n - 1) - 1 for n > 0 and S(0) = 1.
The total number of unique variations is all the symmetrical ones on the diagonal plus one half of the remaining variations. Since S is the number of items on the diagonal but also equals the width and height of the variations, the number of unique variations can be expressed as (S * S) / 2 + (S / 2). S * S equals all variations in the grid. When we take half of it, we have to add the remaining half of the diagonal (S / 2).
So U(n) = (S(n)2 + S(n)) / 2.
I also uploaded a file with all 2187 unique symmetrical 4th iteration subdivisions here (it is ~28mb and 7050 x 7050 pixels).
### Defining subdivision strategies
Rather than generating all variations, we will look at a subset. For simplicity and 'symmetric beauty', only subdivisions that are symmetric through the previous division line are considered. These are only the 1st, 5th (center) and last variation on each line in figure 9. Every pair of green lines (3th iteration subdivision line) share one of their endpoints (on the blue line). This approach reduces the number of variations even more, and results in beautiful symmetric images, and straightforward Python code.
In the Python code, the three triangle vertices are referred to with indices 0, 1 and 2. All triangles in one iteration are subdivided in the same way, so this can be defined with a single vertex index for each iteration. A sequence of these vertices can define a subdivision strategy. These vertices can be notated as the shortest non repeating sequence, and whenever we need to make more iterations than the sequence length, we can repeat the sequence. A strategy defined as '0' means: choosing triangle vertex 0 at every iteration. This is exactly the same as strategy '0, 0' and strategy '0, 0, 0' and so on. Alternating between vertices can be strategy '0, 1, 2' (and is the same as strategy '0, 1, 2, 0, 1, 2', but different from '0, 1, 2, 0'.
Figure 10: How triangle vertex indices are assigned after a subdivision.
When creating two new triangles at each subdivision, there are several ways to assign vertex indices to those new triangles. I made the choice to add 0 to the vertex that was chosen to be subdivided, 1 to the new midpoint that was calculated, and 2 to the 'other triangle vertex'. The two new triangles share vertex 0 and 1 (this is the subdivision line), only vertex 2 is different. This can be seen in figure 10. Symmetry between the two new triangles is created on purpose.
Note that the triangle indices in the original triangle (on the left) do not matter. Only the 'vertex to subdivide' matters (in this case vertex 0). So if you would swap indices 0 and 2 on the left triangle, and choose index 2 to subdivide, the result will be exactly the same as the image on the right.
A sequence of 0's means subdividing the same triangle vertex several times (like figure 5 shows).
A sequence of 1's means that the previous midpoint was chosen several times in a row. (see figure 11 and 12)
A sequence of 2's means that the previous subdivision line will be cut in half several times in a row (because none of the two vertices that define the subdivision line will be chosen, but the 3th vertex).
Figure 11: Strategy 0111111111111111111.
Figure 12: Strategy 0111.
I chose to add the 'top' triangle vertex first in the code (so it has index 0 in the original triangle). I want to choose this vertex first for every image I create, so that each image will be symmetrical through this vertical line in the center (the red line in all images). In Figure 11 you can see what happens if vertex 1 will be chosen over and over again. The result is a grid of triangles.
In Figure 12 vertex 1 is chosen three times in a row, and then vertex 0 (the previous vertex again). In the first image are all the division lines for the first 6 iterations are colored differently. The colors are red, blue, green, black, yellow, and a small grey line for the 6th iteration. The number of lines doubles at every step, so there is 1 red line, but there are in fact 8 different black lines (although some of them seem to line up in this configuration). The second image has 8 iterations and the last image 13 iterations.
With this logic in place, we could systematically generate a sequence of symmetrical subdivions of this kind. And for every result, we can assign a name to it (the strategy sequence), so it could be reproduced.
When naively generating all possibilities in increasing sequence length, there will be many repetitions. All sequences of length 2 will be included in all the sequences of length 4 (and 6, and 8, and so on). In general, every result of order length n can be found in the results of order length 2 * n, and they could be filtered out.
In the following YouTube video, you can watch more than 3000 of these variations (until sequence length 8). Each strategy sequence is displayed on the bottom right, and could be used in the code below to recreate any image. There is also some additional info in the video description.
Figure 13: A still from the YouTube video, click on it to go to YouTube.
### The Code
The Python code has been added to my python_shapes repository. Here is a small breakdown.
Triangle vertices are represented as a tuple of two floats, representing x and y coordinates.
A triangle is represented as a list of 3 vertices. Note that the top left of the canvas is (0, 0), and the bottom right has x, y coordinates (width, height). The triangle vertex with index 0 is the first item in the list of vertices. References to vertices will only be by the position in the list (index 0, 1 or 2).
``` triangle = [ (width / 2, 0), # vertex 0 (0, height), # vertex 1 (width, height), # vertex 2 ] ```
The midpoint of an edge is calculated by taking the average of the two x coordinates, and the average of the two y coordinates.
``` def get_midpoint(point_a, point_b): return ( (point_a[0] + point_b[0]) / 2, (point_a[1] + point_b[1]) / 2, ) ```
To subdivide, we need a function that converts one triangle into two triangles. This function needs to know the current iteration, and the provided strategy, so it can choose the correct vertex to divide the triangle into two sub triangles.
``` def divide(triangle_vertices, iteration, strategy): index_to_subdivide = strategy[iteration % len(strategy)] subdivide_vertex = triangle_vertices.pop(index_to_subdivide) midpoint = get_midpoint(*triangle_vertices) # midpoint of the remaining two triangle_vertices return [ [subdivide_vertex, midpoint, triangle_vertices[0]], [subdivide_vertex, midpoint, triangle_vertices[1]], ] ```
The `index_to_subdivide` takes the correct vertex index from the strategy sequence. This is based on the current iteration, and when that iteration is larger than the length of the strategy sequence, the modulo operator makes sure it takes the correct index as if the sequence was repeated.
The triangle vertex that will be used to subdivided is called `subdivide_vertex` and is popped from the list of `triangle_vertices`. This means that after the `pop` call, there will be only two vertices left in the `triangle_vertices`. The midpoint will be calculated with these two remaining vertices.
The vertices of two new triangles will be returned, in a symmetrical way as shown in figure 10.
With these methods in place, we can generate a subdivided polygon image.
``` previous_generation_triangles = [triangle] for iteration in range(iterations): new_triangles = [] for triangle in previous_generation_triangles: new_triangles += divide(triangle_vertices=triangle, iteration=iteration, strategy=strategy) previous_generation_triangles = new_triangles ```
In the code above, a list of triangles will be constructed. After every iteration, the previous list will be overridden. Since the subdivision is 'complete', all triangle outlines from previous generations are included in the last iteration. At start this list contains one polygon: the starting triangle. After one iteration, it contains two triangles. And so on. When all iterations are done, a list of 2i polygons is constructed. This can be used to draw the polygons.
``` for triangle in previous_generation_triangles: draw.polygon(xy=triangle, outline=line_color) ```
I used the Python Imaging Library as you can see in my original code. There is also an even shorter recursive method that produces the same images.
### Browser App
There is also a small JavaScript web app that probably works best at a desktop computer or laptop. It has only been tested in Firefox, but hopefully it works in other browsers too.
You can generate any variation by entering the strategy in the input field, and push one button (or press enter). This works by running JavaScript code in your own browser.
You can also change the 'divisor' (default is 2) to take the midpoint at other places than exactly halfway. The download button will save the result in a .png file.
### Conclusion
This way of dividing triangles is a nice showcase of how a sequence of complex looking images can be generated with fairly simple logic and very basic mathematics. There will be a follow up article, since there are more things I would like to share about this.
If you have any corrections, additions, references or comments of any kind, please let me know (info@josvromans.com). I will correct mistakes I made in the math (and in my English writing). There has not been any proof reading or fact checking yet.
As a remark on this process, I can tell that coding my initial triangle division took a short amount of time. Writing this article took me almost two weeks, since it involved a lot of double checking statements, generating images for explanation, and thinking more deeply about what I (and my code) was actually doing. This illustrates in some sense how much there is to write or reason about things that are created by code within minutes. And of course there is so much more to study about this. This is also the reason why I produce a lot of images, but not write many blogs or articles. They take a lot of time, and I find it very difficult to achieve a 'finished' result that I like to share.
If you can think about a more basic triangle subdivision than the one I described in this article, I am happy to discuss that one. The more complex ones are fun to code, but I do not aspire to dive into the details of them. To wrap this up, I will give a few examples of how you can make more complex subdivisions (there are many, many more), and add a few more images of triangle subdivisions.
- Use two or more different triangle subdivisions in a certain order. Or choose them at random from a set of possible subdivisions.
- Don't use the same subdivision choice in one iteration for every triangle.
- There are many 'only triangle' subdivisions possible, think about subdividing into 3 or more triangles. Allow gaps, and allow new triangles to have a different size.
- subdivide in other n-gons than just triangles. A simple one is to subdivide one triangle into a square and two triangles, or a square and three triangles. A square can then be easily subdivided into two or four triangles.
- Allow the subdivision to grow outside the original n-gon. This can generate things (with an outline) similar to Koch's Snowflake, when the growth is limited it the right way. Alternate between 'subdivide within the original polygon' and 'subdivide and exceed the original polygon dimensions' in some way.
- Make combinations of the above. This will generate a very large number of variations, and still, it will only be the surface of what is possible.
Figure 14: Choosing every vertex at random (imagine analyzing all possibilities if I dropped the symmetry restrictions..)
Figure 15: Take only vertex 0 or 2 at random
Figure 16: Subdivision with squares and triangles. You can see how this one was created by looking at the steps here.
Figure 17: A colorful triangle subdivision |
# Inequalities
## Introduction
In an algebra course, students are introduced to the concept of order on the real numbers. This is a relation between the real numbers, denoted by the symbol $>$, which is often read as "greater than". We have the following 4 order axioms for real numbers:
Axiom 1: [Trichotomy] If $a, b$ are real numbers, then one and only one of the following is true: $a >b$, $a = b$, $b > a$.
Axiom 2: [Transitivity] If $a, b, c$ are real numbers and $a > b , b > c$, then $a > c$.
Axiom 3: If $a, b, c$ are real numbers and $a>b$, then $a + c > b + c$.
Axiom 4: If $a, b, c$ are real numbers such that $a>b, c > 0$, then $ac > bc$.
With this, we can formally define a positive and negative number. A positive number is a number $a$ that satisfies $a > 0$, while a negative number is a number $a$ that satisfies $0 > a$. By the trichotomy axiom, a real number is either positive, negative, or zero. However, this does not appear to tell us much. In particular, is 1 a positive number?
An important inequality is the trivial inequality which states that the square of any real number is non-negative. This is easily shown by considering Trichotomy cases:
1. If $f = 0$, then $f^2 = 0$.
2. If $f > 0$, then multiplying both sides by $f$ gives $f^2 > 0$, where we keep the sign since $f$ is positive.
3. If $0 > f$, then $-f > 0$ (Axiom 3) and multiplying both sides by $-f$ gives $0 > -f^2$ (Axiom 4), and adding $f^2$ to both sides yields $f^2 > 0$ (Axiom 3).
With the trivial inequality, we now know that $1 = (1) ^2$ is a positive number.
Polynomial inequalities can be solved by understanding how the graph behaves. Because the graph is continuous (naively this means that the graph can be drawn without lifting off your pen), we only need to find the roots of the polynomial and test one value in each of the corresponding regions to determine if they are positive or negative.
## Application and Extensions
### Determine the region in which $x^2 < 1$.
The proper approach would be to shift the terms to one side, and then factorize, to get $x^2 - 1 < 0 \Leftrightarrow (x-1)(x+1) < 0$.
This has roots $1$ and $-1$, so we test the 3 different regions ( i.e. $x < -1, -1 < x < 1, 1, to conclude that the inequality is satisfied for $-1 < x < 1$.
Note: Similar to the equality case $x^2 = 1$, the first instinct of many students is to simply take square roots on the inequality and conclude that $x < 1$. This likewise leads to the wrong answer.
### Determine the region in which $\frac {x^2+3x} {x+5} \leq \frac {x^2 + 5x +2 }{x+5}$.
To avoid dividing by 0, we want to ensure that the denominator is never 0. As such, we have to exclude $x=-5$ as a possibility. From Axiom 4, we can only multiply by a quantity with a fixed sign.
Since $x+5$ is negative when $x < -5$, we cannot simply multiply by $x+5$. Instead, we can multiply by $(x+5)^2$, which we know is always non-negative, to obtain $( x^2 + 3x) (x+5) \leq (x^2 + 5x + 2 ) ( x+5 ).$
At this point, it is tempting to cancel the term $x+5$ on both sides, but this is equivalent to multiplying by $\frac {1}{x+5}$. We may not do so, we choose to factorize instead. Shifting all the terms to one side and factorizing, we get $0 \leq 2 (x+1) ( x+ 5).$
We easily understand the graph of this quadratic equation, and (by using the number line) can conclude that the inequality is satisfied when $x < -5$ (Recall that $x = -5$ is excluded) or $x \geq -1$.
### Show that $x^2 + 2x + 2 \geq 0$ for all real values of $x$.
If we tried to calculate the roots to the quadratic equation $x^2 + 2x +2 = 0$, since the discriminant is negative, we know that the roots are complex numbers. Hence there is only 1 region on the number line. Testing $x = 0$ shows that the quantity is always positive.
Another approach is to rewrite $x^2 + 2x + 2 = (x+1) ^2 + 1$ by Completing the square. Since both terms are squares, they are non-negative, and hence their sum is non-negative (axiom 3).
### Show that the product of 2 positive numbers is positive.
If $X$ and $Y$ are positive numbers, then $X > 0, Y > 0$. Using axiom 4, (with $a = X, b = 0, c = Y$), we get that $X \cdot Y > 0 \cdot Y = 0$. Hence, the product $XY$ is positive.
Note by Calvin Lin
7 years, 4 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
You can also add that inequalities are not valid for $\text{complex numbers}$.
Example : We cannot compare which one is greater between $2 + 3i$ and $1+ 3i$. $( i = \sqrt{-1}$)
- 5 years, 10 months ago
You can discuss Wavy curve method also. :-)
- 5 years, 10 months ago |
3 Tutor System
Starting just at 265/hour
# Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0) (ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4) (iii) (4, 5), (7, 6), (4, 3), (1, 2)
i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(1 + 1)^2 + (0 + 2)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
BC = $$\sqrt{(-1 - 1)^2 + (2 - 0)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
CD = $$\sqrt{(-3 + 1)^2 + (0 - 2)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
DA = $$\sqrt{(-3 + 1)^2 + (0 - 2)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.
AC = $$\sqrt{(-1 + 1)^2 + (2 + 2)^2}$$
$$= \sqrt{0 + 16}$$
$$= 4$$
BD = $$\sqrt{(-3 - 1)^2 + (0 + 0)^2}$$
$$= \sqrt{16 + 0}$$
$$= 4$$
Therefore, diagonals of quadrilateral ABCD are also equal. … (2)
From (1) and (2), we can say that ABCD is a square.
(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(-3 - 3)^2 + (1 - 5)^2}$$
$$= \sqrt{36 + 16}$$
$$= 2 \sqrt{13}$$
BC = $$\sqrt{(0 - 3)^2 + (3 - 1)^2}$$
$$= \sqrt{9 + 4}$$
$$= \sqrt{13}$$
CD = $$\sqrt{(-1 - 0)^2 + (-4 - 3)^2}$$
$$= \sqrt{1 + 49}$$
$$= 5\sqrt{2}$$
DA = $$\sqrt{(-1 + 3)^2 + (-4 - 5)^2}$$
$$= \sqrt{4 + 81}$$
$$= \sqrt{85}$$
We cannot find any relation between the lengths of different sides.
Therefore, we cannot give any name to the figure ABCD.
(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(7 - 4)^2 + (6 - 5)^2}$$
$$= \sqrt{9 + 1}$$
$$= \sqrt{10}$$
BC = $$\sqrt{(4 - 7)^2 + (3 - 6)^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
CD = $$\sqrt{(1 - 4)^2 + (2 - 3)^2}$$
$$= \sqrt{9 + 1}$$
$$= \sqrt{10}$$
DA = $$\sqrt{(1 - 4)^2 + (2 - 5)^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
Here opposite sides of quadrilateral ABCD are equal. … (1)
We can now find out the lengths of diagonals.
AC = $$\sqrt{(4 - 4)^2 + (3 - 5)^2}$$
$$= \sqrt{0 + 4}$$
$$= 2$$
BD = $$\sqrt{(1 - 7)^2 + (2 - 6)^2}$$
$$= \sqrt{36 + 16} =$$
$$2 \sqrt{13}$$
Here diagonals of ABCD are not equal. … (2)
From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram. |
Comment
Share
Q)
# Form the differential equation having $y=(\sin^{-1}x)^2+A\cos^{-1}x+B,$where A and B are arbitrary constants,as its general solution.
Comment
A)
Toolbox:
• A differential equation is a linear differential equation if it is expressible in the form : $P_0 \Large\frac{d^ny}{dx^n}$$+P_1 \Large\frac{d^{n-1}y}{dx^{n-1}}$$+P_2\Large \frac{d^{n-2}y}{dx^{n-2}}$$+...P_n y=0 Where P_0,P_1.... are constants or functions of the independent of variable x • \large\frac{d}{dx}$$(\sin ^{-1}x)=\large\frac{1}{\sqrt {1-x^2}}$
• $\large\frac{d}{dx}$$(uv)=\large\frac{du}{dx}.$$v+\large\frac{dv}{dx}.$$u Given y= (\sin ^{-1}x)^2+A\cos ^{-1}x+B Let us first differentiate with respect to x on both sides \large\frac{dy}{dx}=\frac{2\sin ^{-1}x}{\sqrt {1-x^2}}$$+A\bigg(\large\frac{-1}{\sqrt {1-x^2}}\bigg)+0$
$\large\frac{dy}{dx}=\frac{2\sin ^{-1}x}{\sqrt {1-x^2}}-\frac{A}{\sqrt {1-x^2}}$
This can be written as
$\sqrt {1-x^2}.\large\frac{dy}{dx}$$=2 \sin ^{-1}x-A Now again differentiating on both sides w.r.t x Since \sqrt {1-x^2}.\large\frac{dy}{dx} are two functions in the product form, we can apply the product rule \large\frac{d}{dx}$$(uv)=u.\large\frac{d}{dx}$$(v)+v.\large\frac{d}{dx}$$(u)$
Here let $u=\sqrt {1-x^2},$ hence $\large\frac{d}{dx}$$(v)=\large\frac{1}{2 \sqrt {1-x^2}}$$(-2x)$
$v=\large\frac{dy}{dx}\qquad \frac{d}{dx}$$(v)=\large\frac{d^2y}{dx^2} \large\frac{d}{dx}$$(\sin ^{-1}x)=\large\frac{1}{\sqrt {1-x^2}}$
$\sqrt {1-x^2}.\large \frac{d^2y}{dx^2}+\frac{dy}{dx}.\frac{-2x}{2\sqrt {1-x^2}}$$=2 \large\frac{1}{\sqrt {1-x^2}}-0 \sqrt {1-x^2}.\large \frac{d^2y}{dx^2}-\frac{x}{\sqrt {1-x^2}}.\frac{dy}{dx}$$= \large\frac{2}{\sqrt {1-x^2}}$
$\Large\frac{(1-x^2)\bigg(\frac{d^2y}{dx^2}\bigg)-x\bigg(\frac{dy}{dx}\bigg)}{\sqrt {1-x^2}}=\frac{2}{\sqrt {1-x^2}}$
=>$(1-x^2)\large\frac{d^2y}{dx^2}$$-x \large\frac{dy}{dx}$$-2=0$ |
Lesson Explainer: Circles and Triangles | Nagwa Lesson Explainer: Circles and Triangles | Nagwa
# Lesson Explainer: Circles and Triangles Mathematics
In this explainer, we will learn how to identify inscribed angles in semicircles and circumcircles of triangles and find the equation of a circle given three points on the circumference.
Given three noncolinear points , , and in the plane, there is a unique triangle with those points as vertices.
Now, consider the perpendicular bisector of the line segment . Every point on lies at an equal distance from point and point . Similarly, every point on the bisector is equidistant from and .
This means that the point where and intersect is the same distance from all three vertices, , , and .
If we now trace a circle with center passing through the three vertices , , and we get the unique circle with these properties. This circle is called the circumcircle of the triangle.
The point where the three perpendicular bisectors of the sides of the triangle intersect is the center of the circumcircle and is called the triangle’s circumcenter. This is one type of triangle center; there are many others such as the orthocenter, where the three altitudes of the triangle meet, and the centroid, where the three lines that connect each vertex to the midpoint of the opposite side meet.
Be aware that the circumcenter of a triangle does not always lie inside the triangle. When is acute, which is to say all of its angles are smaller than a right angle, then the circumcenter lies inside . When contains an angle larger than a right angle, then the circumcenter lies outside the triangle.
Now, suppose is a right triangle with hypotenuse . By looking at the angles, we can see that the the perpendicular bisector of makes a smaller similar triangle inside , scaled down by a factor of .
Similarly, the perpendicular bisector of cuts out a similar triangle of scale factor inside . This means that both of these lines intersect the hypotenuse at the point halfway along its length. Of course, the perpendicular bisector of the hypotenuse, by definition, intersects the hypotenuse at its midpoint too. Altogether, this shows that if is a right triangle, then its circumcenter is the midpoint of its hypotenuse.
In this situation, the hypotenuse of is a diameter of its circumcircle and its right angle lies on the circumference.
Recall the circle theorem, which states that the angle in a semicircle is always , or, in other words, if two points and form the diameter of a circle, then subtended at any third point on the circumference must be a right angle.
What we have just established is the converse to this theorem: if is a right triangle with hypotenuse , then is a diameter of its circumcircle.
### Properties: Triangles and Circumcircles
Let , , and be three noncolinear points in the plane.
• There is a unique triangle with vertices , , and .
• The perpendicular bisectors of the sides , , and meet at a single point , called the circumcenter of the triangle.
• The circumcenter of the triangle is the center of the circumcircle, which is the unique circle passing through the vertices , , and .
• If is a right triangle, then its circumcenter lies at the midpoint of its hypotenuse and its hypotenuse is a diameter of its circumcircle.
Let us put some of these properties to use in identifying a diameter of a circle.
### Example 1: Identifying the Correct Point That Makes 𝐴𝐵 A Diameter of A Circle
, , , , and are five points on a circle.
Which of the chords , , and is a diameter?
To identify a diameter, we will use the fact that two points and on the circumference of a circle are a diameter if and only if they form a right triangle with any other point on the circumference. In this example, we proceed by trial and error: we pick one of the given chords and check to see if it forms a right triangle with one of the other points. If not, we pick another chord and try again.
Let us test the chord first. We will check to see if is a right triangle using the Pythagorean theorem, which states that is a right triangle if and only if
Let us calculate these lengths. We have and but so is not a right triangle and is not a diameter.
Let us check . We want to know if is a right triangle, so we check if
We have already calculated , so we only need to work out the other two lengths. We have and and, indeed, confirming that is a right triangle and that is a diameter of the circle. Since we have identified the diameter and the question tells us that only one of the given chords is a diameter, there is no need to check .
We will now see how to work out the equation of a circle, given some points on its circumference, by establishing that the chord connecting two of those points is a diameter of the circle.
### Example 2: Showing That Three Points Form a Right Triangle Inscribed in a Semicircle and Finding the Equation of the Circle
The points , , and lie on a circle.
1. What type of angle is ?
2. Write down an equation for the circle in the form , where , , and are constants to be determined.
Part 1
We can find what type of angle is using the Pythagorean theorem, which states that if and only if
Calculating lengths, we find and
So, indeed, which shows that is a right angle.
Part 2
Since is a right angle, we know that must be a diameter of the circle. This allows us to identify the center of the circle as the midpoint of and its radius as half of .
We calculate the coordinates of the midpoint of using the formula
Now, we can write down the equation for the circle: where is the length of the radius, which is, of course, half the length of the diameter . We have already calculated , so and and the equation of the circle is
In the previous example, we saw how to find the center of circle, given some points on its circumference, when two of those points lie on a diameter. We will now use a different technique to find a circle’s center, given points on its circumference, in the situation where no two of those points lie on a diameter.
As discussed in the introduction, if we have three noncolinear points , , and , then the point of intersection of any two of the perpendicular bisectors of the sides , , and is equidistant from all three vertices. The point is therefore the circumcenter of , , and , that is, the center of the circumcircle, the unique circle passing through , , and .
If we are given that the points , , and lie on a circle, then the circumcenter must be the center of that circle. We can sum this up in the following property.
### Property: Perpendicular Bisectors of Chords
The perpendicular bisectors of any two distinct chords of a circle intersect at the circle’s center.
### Example 3: Finding the Perpendicular Bisector of a Chord of the Circle and Using It to Find the Center of the Circle
The points , , and lie on the circumference of a circle. The equation of the perpendicular bisector of is . Find the equation of the perpendicular bisector of and the coordinates of the center of the circle.
To find the equation of the perpendicular bisector of a chord, we need two pieces of information: the slope of the chord and the coordinates of its midpoint. The slope of the chord is given by
The slope of the perpendicular bisector is then the negative reciprocal of the slope of the chord, so, in this case, it is . We can now write down the following equation for the perpendicular bisector and we can calculate the value of the constant using the coordinates of the midpoint of the chord, which are given by
Substituting into , we have giving us the equation for the perpendicular bisector of .
The center of the circle lies at the intersection of the two perpendicular bisectors and . Solving for , we get and substituting back into , we have
Thus, the coordinates of the center of the circle are .
In fact, once we have used the perpendicular bisectors of chords of a circle to find its center, we simply need to find the square of its radius to write an equation for the circle. This is what we are going to do in the next example.
### Example 4: Using the Bisectors of Two Chords of a Circle to Find the Equation of the Circle
The points , , and lie on a circle. Given that the equations of the perpendicular bisectors of and are and , write down an equation for the circle in the form , where , , and are constants to be determined.
Since we are given the equations and of the perpendicular bisectors of two distinct chords of the circle, we can immediately find the center of the circle at their intersection. Substituting into , we have and since , the coordinates of the center are . The only other thing we need to write down an equation for the circle is the square of the length of its radius, . This is simply the square of the distance from the center to any point on the circumference, say, . This square distance is given by
We can therefore write down the following equation for the circle:
We are now in a position to calculate an equation of a circle given any collection of at least three distinct points on its circumference using the perpendicular bisectors of any pair of chords connecting them.
### Example 5: Finding the Equation of the Circle given Four Points on the Circumference
The points , , , and lie on a circle. By finding the perpendicular bisectors of and , write down an equation for the circle in the form , where , , and are constants to be determined.
To write down equations of the perpendicular bisectors of and , we need to work out the slopes and midpoints of these chords. The slope of the chord is given by and the slope of is given by
We have two equations for bisectors, where the slopes are the negative reciprocals of the corresponding slopes of the chords and the constants and are to be determined using, for each bisector, a point lying on it: its midpoint.
The midpoint of is given by and the midpoint of is given by
Substituting the coordinates of into , we find
A similar calculation yields . So, the two perpendicular bisectors are given by
Since we are looking for the intersection of these two lines (that is, the center of the circle), we equate the -coordinates and solve for : and then, we substitute back into for
Hence, the coordinates of the center of the circle are and we have the following equation for the circle: where is the square of the radius of the circle. We can calculate as the square of the distance from the center to any point on the circumference, say, , by
So, the equation of the circle is given by
Let us finish by recapping a few important concepts from this explainer.
### Key Points
• There is a unique circle passing through any three noncolinear points , , and in the plane. It is called the circumcircle of .
• The center of the circumcircle is called the circumcenter. It is equidistant from the points , , and and lies at the intersection point of the perpendicular bisectors of the sides , , and .
• Three points in the plane form the vertices of a right triangle if and only if the hypotenuse is the diameter of the circumcircle through the vertices. We can use this fact to calculate the equation of the circumcircle.
• The perpendicular bisectors of any two distinct chords of a circle intersect in the circle’s center. We can use this fact to calculate the equation of a circle given any collection of at least three distinct points on its circumference. |
If the sum of the first $p$ terms of an AP be $q$ and the sum of its first $q$ terms be $p$ then show that the sum of its first $(p + q)$ terms is $- (p + q)$.
1. Let $a$ be the first term and $d$ be the common difference of the given AP. Then, \begin{aligned} & S_ { p } = q \\ \implies & \dfrac { p } { 2 } (2a + (p - 1)d) = q \\ \implies & 2ap + p(p - 1)d = 2q && \ldots \text{(i)} \\ \end{aligned} And, \begin{aligned} & S_ { q } = p \\ \implies & \dfrac { q } { 2 } (2a + (q - 1)d) = p \\ \implies & 2aq + q(q - 1)d = 2p && \ldots \text{(ii)} \\ \end{aligned}
2. On subtracting $\text{(ii)}$ from $\text{(i)},$ we get \begin{aligned} & [ 2ap + p(p - 1)d ] - [ 2aq + q(q - 1)d ] = 2q - 2p \\ \implies & 2a(p - q) + (p^2 - p - q^2 + q)d = 2q - 2p \\ \implies & 2a(p - q) + (p^2 - q^2)d - (p - q)d = -2(p - q) \\ \implies & 2a(p - q) + (p - q)(p + q)d - (p - q)d = -2(p - q) \\ \implies & 2a + (p + q)d - d = -2 \\ \implies & 2a + (p + q - 1)d = -2 && \ldots \text{(iii)}\\ \end{aligned}
3. Now, the sum of the first $(p + q)$ terms of the AP is \begin{aligned} S_{ p + q } & = \dfrac{ p + q } { 2 } (2a + (p + q -1)d) \\ & = \dfrac{ p + q } { 2 } (-2) && \text{[Using (iii)]} \\ & = - (p + q) \\ \end{aligned}
4. Thus, the sum of the first $(p + q)$ terms is $- (p + q).$ |
*******************************************************
QB Rotation Primer
Created: August 16, 1999
*******************************************************
## Introduction
Rotozoomers are the rage these days in the QB scene. What's a rotozoomer? It's a graphic effect that rotates and zooms into some picture. If you remember the spinning logo on the Street Fighter II game, that was a kind of rotozoomer. If you're interested in the mathematics behind the code, read on, otherwise, skip to the coding section.
## Background
Geometrically speaking, 2D rotozoomers are simply a combination of rotation and scaling transforms. A rotation transform is a matrix (or set of formulas) that take a point, and rotate it about some axis, like a clock hand moves. A scaling transform is a matrix that zooms in or out into a picture.
## Part I: Rotations
Let's look at the rotation transform first. You know that the hands of a clock rotate around the center point. Now imagine those hands moving really fast, like the Twilight Zone. What shape do they trace out? A circle, right? It turns out that we can use this fact to help us figure out the formula for rotation.
> Figure 1. - Point B is a rotated version of point A. Let's say angle a = 20o. In this figure, you can see Point A is at (cos(a),sin(a)), or (0.939,0.342). But in polar coordinates, that is (1,a), or (1,20o) because the radius OA is 1, and it's 20o off the horizontal +x axis labeled OH.
Looking at figure 1, we can see that Point A is in the 2-o'clock position located at (cos(a),sin(a)). Point B is at the 1 o'clock position, and it is at (cos(b),sin(b)) in cartesian coordinates. But if you describe the two points in polar coordinates, as point A is at (1,a) and point B is at (1,b), it becomes immediately obvious that they only differ in their second coordinate, which is their angle. So Point B is just Point A rotated by (b-a) degrees.
So, in polar coordinates, rotation is just
``` Rnew = Rold .... eq. 1
Anglenew = Angleold + angle_of_rotation .... eq. 2
```
where Rnew is the radius after rotation, and Anglenew is the angle that the point is at after rotating counter-clockwise angle_of_rotation degrees.
Ok, but that still doesn't tell us how to rotate a point with coordinates (x,y), you might think. That's where the sines and cosines come in.
Sines and cosines are not as hard as you think. In terms of polar to cartesian conversions, they tell you what you have to multiply the radius by to get the x and y coordinates. Let's say the radius is 1 as in figure 1. Then, if the angle AOH (the angle created the vector connecting your point and origin, and the positive x-axis) is 'a' radians, then the x-coordinate of the point is cosine(a), and the y-coordinate is sine(a). So the formula is:
``` x = Radius * COS(a) .... eq. 3
y = Radius * SIN(a) .... eq. 4
```
Now that we have nice (x,y) coordinates to work with, the rest is easy. You can see from figure 1, that the coordinates of B are (cos(b),sin(b)). So what does Point A and Point B differ by? The length of the x-coordinate of point A is cos(a), and the length of the x coordinate of point B is cos(b). Now you can subtract the two and find out the difference in the x direction.
``` xdifference = (x coordinate of point B) - (x coordinate of point A)
xdifference = COS(B) - COS(A) .... eq. 5 {when radius=1 }
```
Likewise, the y difference can be found by simple subtraction.
``` ydifference = (y coordinate of point B) - (y coordinate of point A)
ydifference = SIN(A) - SIN(B) .... eq. 6 {when radius=1 }
```
Okay, that was easy, right? Now, we're going to put these results together to find out the magical equation for rotation.
We'll call the angle between OA and OB "angle c," which is equal to b-a. We need to rotate A by c radians to get Point B. So we have the equation
``` c = b - a ... eq. 7
b = a + c ... eq. 7'
```
and we have from equation 3 that the coordinates of Point B (= point A after rotation) given in cartesian coordinates are (Xb,Yb), where
``` Xb = r * SIN(b) from equation 3
Xb = SIN(b), because r = 1
Xb = SIN(a + c), substituting equation 7' ... eq. 9
Yb = r * COS(b) from equation 4
Yb = COS(b), because r = 1
Yb = COS(a + c), substituting equation 7' ... eq. 10
```
Now we need to derive the trigonometric addition identities (equations).
> Figure 2: Diagram for proof of cos(a+c) identity
Derivation of additive trig identities using geometry
1. We construct an auxillary line segment BY perpendicular to OA.
This makes length of OY = cos(c) and length of BY = sin(c) because
triangle OBY is a right triangle with hypotenuse |OB| = 1.
2. We construct line segment LB to be perpendicular to OM.
3. We construct line segment NY to be parallel to OM.
4. We construct line segment YM to be perpendicular to OM.
5. Now this gives us angle OYN = a because alternate interior angles are congruent.
Also, this gives us the angle NQY = BQY = BQA = 90 - a because it is part of a right triangle NAQ
6. Since vertex Q is part of another right triangle BYQ, angle QBY = 180 - 90 - (90-a) = a
7. From Figure 1, you can see that |NL| = |YM| = sin(a) because it's part of right triangle OAM, and AM is the "opposite" side. So angle LBY = NYQ = a.
8. In right triangle BNY, BN/sin(c) = adj(angle NBY) / hyp(angle NBY) = cos(angle NBY) = cos(a).
Therefore, |NB| = cos(a)*sin(c).
9. In right triangle OYM, |YM| / |OY| = opp/hyp = sin(angle YOM) = sin(a).
Combine this with the fact that |NL| = |YM|, you get |NL| = cos(c)*sin(a).
Now we know the coordinates of point B(Xb,Yb):
Xb = cos(a+c) = |LB| = NL + NB = cos(a)*sin(c) + sin(a)*cos(c).
Yb = |OL| = sqrt(|OB2| - |LB|2) = sqrt(1 - |LB|2)
Man, that was a hard proof, but we still have square roots left! We don't want the Yb to contain square roots, because they execute slowly on a computer. Alternate Geometric Proof
We can get |OL| = sin(a+c) through some geometry or algebra:
I'll show the geometric way first, because it's so simple!
Figure 3: Diagram for proof of sin(alpha+beta) identity
We know from geometry that triangle area = ½*(base * height). If we construct altitude c perpendicular to the base formed by joining the line segments f and g, we can get the left area formula. If we construct base segments f and g perpendicular to altitude b, we get the right hand side.
triangle area = triangle area
½[(c(f+g)] = ½[b(d+e)]
multiply both sides by two:
``` c(f+g) = b(d+e) ... eq. 1
```
by definition,
``` sin(alpha) = d/a ... eq. 2a
cos(alpha) = b/a ... eq. 2b
sin(beta) = e/(f+g) ... eq. 2c
cos(beta) = b/(f+g) ... eq. 2d
sin(alpha + beta) = c/a ... eq. 2e
```
If we multiply eq. 1 by 1/((f+g)*a) on both sides, we get
c/a = b/a * (d+e)/(f+g)
c/a = d/a * b/(f+g) + b/a * e/(f+g)
Substituting equations 2a,b,c,d,e into the above equation, we get
``` sin(alpha+beta)= c/a = sin(alpha)cos(beta) + cos(alpha)sin(beta)
```
That's it, we've proved the sine addition identity! (special thanks to Marduke, who thought of this proof)
Ok, here's the trig and algebra version if you like it that way:
``` sin(a+c) = sin(a)cos(c) + cos(a)sin(c) ... we just derived this
sin(a-c) = sin[a+(-c)]
sin(a-c) = sin(a)cos(-c) + cos(a)sin(-c)
sin(a-c) = sin(a)cos(c) + cos(a)*-sin(c)
sin(a-c) = sin(a)cos(c) - cos(a)*sin(c)
But if we let a = 90 - a,
sin[(90-a) - c] = sin(90-a)cos(c) - cos(90-a)sin(c)
Recalling sin(90-x) = cos(x), and cos(90-x) = sin(x), we get
sin[(90-a) - c] = cos(a)sin(c) - sin(a)cos(c)
sin[90 - (a+c)] = cos(a)sin(c) - sin(a)cos(c)
cos(a+c) = cos(a)sin(c) - sin(a)cos(c)
```
So,
Xb = cos(a+c) = |LB| = cos(a)*sin(c) + sin(a)*cos(c) ...eq. 11
Yb = sin(a+c) = |OL| = cos(a)sin(c) - sin(a)cos(c) ...eq. 12
Now we compare the coordinates of point B and those of point A.
Xa = cos(a)
Ya = sin(a)
and we see that we can substitute Xa whenever we see cos(a), and Ya whenever we see a sin(a) in equations 11 and 12. Then you get
Xb = Xa * cos(c) - Ya * sin(c)
Xb = Xa * sin(c) + Ya * cos(c)
This is our long-awaited rotation formula for 2 dimensions:
Xrotated = X * COS(angle) - Y * SIN(angle) Yrotated = X * SIN(angle) + Y * COS(angle)
Now the only thing left to do is zoom into that circle of radius 1 (known as the unit circle) to make this work on all kinds of points, not just ones that lie on the perimeter of the unit circle. But zooming is scaling, remember? We'll cover that next.
## Part II: Scaling
Scaling in two dimensions is no harder than scaling in one dimension. Just multiply by the coordinates by scaling factors (a scalar number). So the formula here is
``` XNew = X * scale in x direction ..... eq. 7
YNew = Y * scale in y direction ..... eq. 8
```
That stretches out the X coordinate by xscale, and the Y coordinate by yscale. The same formulas are used for the zooming into a picture in a rotozoomer. Basically, you multiply all your X coordinates by xscale, and all your Y-coordinates by yscale, and keep increasing xscale and yscale to watch your picture grow bigger.
```DEFINT A-Z
CONST w = 40
CONST h = 30
DIM picture(w, h)
SCREEN 13
'fill the array with pixel colors
'the pattern should look like a rainbow stripe
FOR y = 0 TO h - 1
FOR x = 0 TO w - 1
picture(x, y) = x + y + 32
PSET (x + 280, y + 160), picture(x, y)
NEXT x
NEXT y
'loop through different scaling factors
'note that as the scales get bigger, there are
'more and more gaps between the colored pixels.
'we will solve this in two ways:
' 1. inverse transformations
' 2. fat pixels with forward transformations
FOR xscale! = .2 TO 5! STEP .2
FOR yscale! = .2 TO 5! STEP .2
LOCATE 21, 1
PRINT USING "xscale=#.#"; xscale!
PRINT USING "yscale=#.#"; yscale!
LINE (0, 0)-(xscale! * (w - 1), yscale! * (h - 1)), 15, B
FOR y = 0 TO h - 1
FOR x = 0 TO w - 1
'equation 7 and 8 in action!
XNew = INT(x * xscale!)
YNew = INT(y * yscale!)
PSET (XNew, YNew), picture(x, y)
NEXT x
NEXT y
IF INKEY\$ > "" THEN END
' decrease the flicker by a little
' and slow down the animation
WAIT &H3DA, 8: WAIT &H3DA, 8, 8
WAIT &H3DA, 8: WAIT &H3DA, 8, 8
WAIT &H3DA, 8: WAIT &H3DA, 8, 8
' clear the screen for a different scale combination
LINE (0, 0)-(xscale! * (w - 1), yscale! * (h - 1)), 0, BF
NEXT yscale!
NEXT xscale!
```
---> Forward scaling transform ---> Inverse scaling transform
Type in the program above and run it. You can see that as the scaling factors get larger (when you zoom in a lot), there are larger gaps in the picture. This is because I did not bother to fill in those gaps, and because I am using what is called a forward transformation. A forward transformation just means that I am transforming the original picture to the screen.
If I transform from the screen back to some pixel on the original picture, I will be doing a inverse tranformation, and I am guaranteed not to have "holes" or "gaps" if the picture is big enough. Actually, you don't have to repeat the picture filling the entire window-- you can make the parts your picture doesn't cover black or some background color. The code below shows the double loops required to do a scanline inverse rotation transformation without tiling.
``` REM ...(see ROTATE3 for full code)...
REM ...use CLIP (No Tile)...
angle = 30
scale = 700
Ca = c(angle) 'fixed point cosine table
Sa = s(angle) 'fixed point sine table
yo = ytop - yd
FOR y = ytop TO ybottom
yo = yo + 1
xo = xleft - xd
yca = yo * Ca \ scale + yhc
ysa = yo * Sa \ scale + xhc
FOR x = xleft TO xright
xo = xo + 1
' note sign reversal in YP due to +Y pointing
' downward in screen coordinates.
xp = xo * Ca \ scale + ysa
yp = yca - xo * Sa \ scale
IF (xp>=0 AND xp<=xh AND yp>=0 AND yp<=yh) THEN
PSET (x, y), model(xp, yp)
ELSE
PSET (x, y), 0 : ' black background
END IF
NEXT x
NEXT y
```
But if you decide to wrap the picture around infinitely many times (which is called tiling), you can do it using the MOD function in BASIC. However, there is one problem with MOD. If you MOD a negative number by a positive number, you get a negative number. That is not good when texture coordinates (coordinates into a 2-D array) have to be positive. There is a work-around, though. If you DIM your arrays to go from a negative number to a positive number, you will be ok. A better and faster way is to use AND with a power-of-2 minus 1 number (like 3,7,15,31,63,127,255,etc). This is equivalent to MODing by a power of 2 number. For example:
``` X : -7 -6 -5 -4 -3 -2 -1 +0 +1 +2 +3 +4 +5 +6 +7
X MOD 4: -3 -2 -1 0 -3 -2 -1 0 1 2 3 0 1 2 3
X AND 3: 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3
61 AND 31 = 29 and 61 MOD 32 = 29 (same result for positive numbers)
-61 AND 31 = 3 but -61 MOD 32 = -29 (MOD is not what we want for negative)
```
Here is some code to do tiled inverse rotations like ROTATE3b.
``` REM ......Tiled Version (main loops only).......
REM ......assumes texture is 64x64 pixels.......
angle = 30
scale = 700
Ca = c(angle) 'fixed point cosine table
Sa = s(angle) 'fixed point sine table
yo = ytop - yd
FOR y = ytop TO ybottom
yo = yo + 1
xo = xleft - xd
yca = yo * Ca \ scale + yhc
ysa = yo * Sa \ scale + xhc
FOR x = xleft TO xright
xo = xo + 1
' note sign reversal in YP due to +Y pointing
' downward in screen coordinates.
xp = xo * Ca \ scale + ysa
yp = yca - xo * Sa \ scale
PSET (x, y), model(xp AND 63, yp AND 63)
NEXT x
NEXT y
```
## Part III. Math tricks
### Inverse Rotation Transformations
To fix the problem with the "gaps" and "holes" that occur when rotozooming with the standard rotation formula, we use a technique called reverse or inverse rotation transformations. The idea is simple: instead rotating points on the figure and ending up somewhere on the screen, we do the opposite: rotate the screen coordinates backwards, so that they land on the model somewhere. We already did a similar thing with the scaling transform.
In our QB program, it amounts to changing the point number loop to a x,y loop, and using this formula:
Xrotated = Xscreen * COS(angle) + Yscreen * SIN(angle) Yrotated = - Xscreen * SIN(angle) + Yscreen * COS(angle)
The formula comes from substituting -angle in place of angle, and using some negative angle trig identities to clean things up.
### Changing the Center of Rotation
When we rotate things around the origin (0,0), we can use the pure rotation or inverse rotation formula as-is, but when we usually want to move things around so that our model coordinates end up being 0 to some positive number, because they are array coordinates. In order to do this, you have to do a translation transform, the regular rotation transform, then another translation tranform to put those points where they should land.
``` R(angle with different center) = Translation1 * R(angle) * Translation2 where * is the composition operator ```
Here is a QB code fragment that does this.
``` REM ......CENTER of ROTATION CHANGE demo........
REM ......tiled sprites, untested code.......
REM ......assumes texture is 64x64 pixels.......
angle = 30
scale = 700
xcenter = 35
ycenter = 27
Ca = c(angle) 'fixed point cosine table
Sa = s(angle) 'fixed point sine table
yo = ytop - yd
FOR y = ytop TO ybottom
yo = yo + 1
xo = xleft - xd 'translation 1
yca = yo * Ca \ scale + ycenter
ysa = yo * Sa \ scale + xcenter
FOR x = xleft TO xright
xo = xo + 1
' note sign reversal in YP due to +Y pointing
' downward in screen coordinates.
xp = xo * Ca \ scale + ysa
yp = yca - xo * Sa \ scale
PSET (x, y), model(xp AND 63, yp AND 63)
NEXT x
NEXT y
```
### Sinus Table Size
Some people were lost when I changed the table size in ROTATE3b. Here is the generalized sine and cosine table calculation code. You'll find that it works if you plug in tsize=360, and it works for any other even integer value, too. What it does is divide the angles for a full 360 degree rotation into tsize steps. It works because the step side is calculated correctly using deg2bin# = 2 * pi# / tsize, and the first FOR loop calculates the sine and cosine values for that every angle up to 180 degrees, and the second loop "flips" those values so that you get the correct sines and cosines for angles between 180 and 360 degrees.
```CONST tsize = 128: 'sine lookup table size
scale = 512: 'fixed point scaling factor
DIM c(tsize): DIM s(tsize)
'set up sinus tables
pi# = 3.1415926535#
deg2bin# = 2 * pi# / tsize
PRINT "Calculating sine/cosine tables..."
FOR theta = 0 TO (tsize\2-1)
s(theta) = CINT(SIN(theta * deg2bin#) * scale)
c(theta) = CINT(COS(theta * deg2bin#) * scale)
NEXT
FOR theta = (tsize\2) TO (tsize-1)
s(theta) = -s(theta - (tsize\2-1))
c(theta) = c((tsize-1) - theta)
NEXT
```
If you can think of a better explanation, please email me.
## Part IV. Optimizations
### A Word of Warning
Optimizations in general are a good thing. Algorithmic optimizations that improve the speed and/or memory usage of the program by more than a constant factor is definitely worth doing. On the other hand, it's easy to get caught up in low-level optimizations that amount to only a few percent increase in program speed, and lose sight of major good optimizations that work on any machine.
Optimizations often make the code difficult to understand, and lead to insidious bugs. So it's a good idea to document why you made the optimization, the original algorithm (so you can test if your optimized version still works correctly) and all the side-effects and trade-offs of the optimization. For example, precalculation has the trade-off that more memory is used in exchange for faster program execution. Precalculated sine tables have the side effect that you can get a SUBSCRIPT OUT OF RANGE error if you don't wrap around the angle, while the SIN function lets you use any angle. If you use fixed point INTEGER math, you get the side effect of "shaky" or "hairy textures" unless you use more bits of precision and prestep your texture starting coordinates at each row.
### Where to Optimize
In many programs, there are functions that get called very often. One programmer has even said, "Most programs spend 95% of their time in 5% of the code." These pieces of code are called inner loops or hotspots, and are the prime candidates for optimization. In the case of QB games, the inner loop is usually the rendering loop or keyboard polling loop. In our rotozoomer, the inner loop is obviously the inner pixel plotting loop. Let's take a look at the unoptimized loop for the inverse rotation transform algorithm.
``` FOR y = ytop TO ybottom
FOR x = xleft TO xright
yo = yd - y
xo = x - xd
xp = xo * COS(angle * deg2rad#) / scale# + yo * SIN(angle * deg2rad#) / scale#
yp = -xo * SIN(angle * deg2rad#) / scale# + yo * COS(angle * deg2rad#) / scale#
PSET (x, y), model(xp AND 63, yp AND 63)
NEXT x
NEXT y```
The first thing I see is the long expression for xp and yp in the inner loop. I'm talking about this:
``` xp = xo * COS(angle * deg2rad#) / scale# + yo * SIN(angle * deg2rad#) / scale#
yp = -xo * SIN(angle * deg2rad#) / scale# + yo * COS(angle * deg2rad#) / scale#
```
I notice that the last expression on both lines are constant in the inner loop, because they depend on yo, scale# and angle only, which don't change when the x changes. So we can take them out of the inner loop, and we get this:
``` FOR y = ytop TO ybottom
'we precalculated the variables that depend on y outside the inner loop
yo = yd - y
ysa# = yo * SIN(angle * deg2rad#) / scale#
yca# = yo * COS(angle * deg2rad#) / scale#
FOR x = xleft TO xright
xo = x - xd
xp = xo * COS(angle * deg2rad#) / scale# + ysa#
yp = -xo * SIN(angle * deg2rad#) / scale# + yca#
PSET (x, y), model(xp AND 63, yp AND 63)
NEXT x
NEXT y
```
Next we can precalculate the sine and cosine of each integer degree angle. This will get rid of an expensive (slow) multiply and FPU calculation of SIN and COS in the inner loop.
```DIM cosine#(360)
DIM sine#(360)
FOR angle = 0 TO 359
NEXT angle
'zoom in and rotate the model
DO
angle = (angle + 1) MOD 360 'rotate counter-clockwise by 1 degree
scale# = scale# + .05
LOCATE 2, 10: PRINT angle; "degrees"
k\$ = INKEY\$: IF k\$ = esc\$ THEN EXIT DO
FOR y = ytop TO ybottom
yo = yd - y
ysa# = yo * sine#(angle) / scale#
yca# = yo * cosine#(angle) / scale#
FOR x = xleft TO xright
xo = x - xd
xp = xo * cosine#(angle) / scale# + ysa#
yp = -xo * sine#(angle) / scale# + yca#
PSET (x, y), model(xp AND 63, yp AND 63)
NEXT x
NEXT y
LOOP
END
```
Run it now and see how much faster it is!
Now what can we do? Fixed Point! But fixed point is tricky, I warn you.
### Precalculations
My computer graphics teacher explains the axiom of fast computer graphics, "Don't do anything you don't have to during runtime; precalculate as much as possible!"
Here's an example of a good precalc in QuickBasic.
```DIM sins(128) AS INTEGER
DIM coss(128) AS INTEGER
FOR s = 0 TO 128
sins(s) = 256 * SIN(s * PI / 64) * (COS(s * PI / 64) + 1.5)
coss(s) = 256 * COS(s * PI / 64) * (COS(s * PI / 64) + 1.5)
NEXT s
```
With the sins() and coss() tables set up, you can now get complex sinusoidal motion, just by doing a table look-up like
``` FOR S = 0 TO 127
PSET(sins(s),coss(s))
NEXT S
```
Precalculated arrays of this sort are called Lookup Tables or LUTs, and can speed up calculations by a lot if used cleverly.
### Fixed Point Math
Computers were originally designed to be adding machines. So they were designed to add a fixed number of digits very quickly*. For this and a number of other technical reasons, INTEGER calculations are faster than FLOATing point calculations on many personal computers. We can take advantage of this in BASIC, by working in fixed point, just like a cash register. In a cash register, there are two digits after the decimal point, never more, never less. Ok, what if we moved the decimal point two digits to the right? Then we're dealing with cents, and we'll always get INTEGER values. Ok, not always, but we round off to the nearest cent anyways, right? (To do that, you just move the decimal point one further to the right, but don't use it unless you need to do rounding.)
To do fixed point in QuickBasic, you multiply lots of floating point numbers by a big integer (called the scaling factor), and save the results in a precalculated look-up table, then in your inner loop, you do some arithmetic on the values in the look-up table, then right before you use that number, divide the value by the scaling factor. (in C++ or ASM, logical binary shifts do the dividing part even faster).
Let's look at an example in decimal:
Say you want to represent 374.6031 in 5.2 fixed point (think of it as #####.##.)
It's really easy, you just fill in the hash marks and get 374.60, represented in cents as 37460. Notice we just moved the decimal point to the right by two to make it an even integer.
Say you wanted to convert 0x374.6031 to real 5:2 FIXED POINT. Since there's only room for 2 hexadecimal digits after the decimal point, you get 0x374.60. You can just multiply this by 0x100 to make it the even integer that represents this FIXED POINT quantity. Conversely, you can <<2 to get the scaled up version, and >>2 to scale back down.
Here's an example where we can get an idea of how fast fixed point math is compared to floating point!
```DEFINT A-Z 'we want QB to use INTEGERs for SPEED!
scale = 256 'scaling factor
PRINT "Calculating sinus tables..."
DIM c(360), s(360)
FOR t = 0 TO 359
c(t) = COS(t * deg2rad#) * scale ' scale up!
s(t) = SIN(t * deg2rad#) * scale ' scale up!
NEXT
'let's use some fixed point
scale2 = scale \ 16
tstart# = TIMER
FOR t = 0 TO 359
PRINT c(t) \ scale2, s(t) \ scale2 ' scale down! (a little)
NEXT t
tend# = TIMER
tfast# = tend# - tstart#
'try the slow floating point version
scalex = scale \ scale2
tstart# = TIMER
FOR t = 0 TO 359
NEXT t
tend# = TIMER
tslow# = tend# - tstart#
'display competition results
PRINT USING " Fixed Point took ##.### secs."; tfast#
PRINT USING "Floating Point took ##.### secs."; tslow#
PRINT USING "Fixed point was ####% faster"; (tslow# - tfast#) * 100 / tslow#
```
Even on a Pentium III, where floating point math is VERY fast, fixed point won this competition by a 12% margin.
More to come....
## Part III: QuickBASIC Code Examples
Please give me credit for the ROTATEn.BAS examples if you use my code. --Toshi
You can to refer back to the
Math Tricks or Optimization section if you don't see what I'm doing.
ROTATE0.BAS - straight from the formula
ROTATE1.BAS - with page flipping
ROTATE2.BAS - precalculated integer sine and cosine tables
ROTATE3.BAS - inverse rotation transforms
ROTATE3B is really fast, but is riddled with truncation errors (which looks like an interference effect) due to the use of 16 bit integers instead of 32 bit. It was coded up shortly after a demo coder revealed to me the secret of using AND for fast sprite tiling.
ROTATE3b.BAS - Sprite Tiling
ROTATE5.BAS - Digital difference analyzer rotations
ROTOZOO5.BAS - Entropy's precalculated divides and mults (fastest fps)
## Appendix
Overheard on irc:
<QbProgger> everyone and their mother has a rotozoomer
<entropy-> whateverz- it has my program, so it's good
<gopus98> What's a rotozoomer?...a vacuum?
<Neander> sin(a+b) = sin a * cos b + sin b * cos a
<Neander> cos(a+b) = cos a * cos b - sin a * sin b
<gopus98> okay, okay what does the POLAR in polar coordinates mean...
The "So..." maybe not so, because it is just a guess on my part.
## Where to Go From Here
You've learned a lot if you've made it here! Congratulations! You can now write your own programs for rotating and zooming objects, at least in 2D. But 3D is not much harder! If you can substitute z for y, it's as easy as that! You'll learn all about lines, planes, surfaces, and solid objects. Things you might want to learn next are
1. Coordinate Systems (medium)
2. Alpha Blending (easy)
3. Texture/Bump/Environment Mapping (medium)
4. Hierarchial Transforms (hard)
5. Matrices to Go from Anywhere to Anywhere (could be easy or hard)
6. Assembly Language Subroutines (hard)
7. OpenGL and DirectX (fun!) |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Proportions Using Cross Products
## Cross-multiply to solve proportions with one variable
Estimated7 minsto complete
%
Progress
Practice Proportions Using Cross Products
Progress
Estimated7 minsto complete
%
Proportions Using Cross Products
Remember Manuel who was reading all of the medieval books on knights in the Recognize Proportions as a Statement of Equivalent Ratios Concept? Well after he finished reading the series, he loaned it to his friend Rafael. Rafael is enjoying the series as much as Manuel did.
In five weeks, Rafael had already finished 8 of the 12 books. It took Manuel 7.5 weeks to read all 12 books. Will Rafael and Manuel finish the series in the same amount of time? Are they reading at the same rate? How can you figure this out?
To figure this out, you will need to know how to determine if two ratios form a proportion. If the reading rate of the boys is the same, then the ratios will form a proportion.
Use this Concept to learn how to solve proportions using cross products. Then you will know how to figure out this dilemma.
### Guidance
Previously we learned that a proportion states that two ratios are equivalent. Here are two proportions.
\begin{align*}\frac{a} {b} = \frac{c} {d} \qquad \text{or} \qquad a : b = c : d\end{align*}
In a proportion, the means are the two terms that are closest together when the proportion is written with colons. So, in \begin{align*}a : b = c : d\end{align*}, the means are \begin{align*}b\end{align*} and \begin{align*}c\end{align*}.
The extremes are the terms in the proportion that are furthest apart when the proportion is written with colons. So, in \begin{align*}a : b = c : d\end{align*}, the extremes are \begin{align*}a\end{align*} and \begin{align*}d\end{align*}.
The diagram below shows how to identify the means and the extremes in a proportion.
In the last lesson, you learned how to solve proportions by using proportional reasoning. We can also solve a proportion for a variable in another way. This is where the cross products property of proportions comes in.
What is the Cross Products Property of Proportions?
The Cross Products Property of Proportions states that the product of the means is equal to the product of the extremes. You can find these cross products by cross multiplying, as shown below.
\begin{align*}\frac{a}{4} = \frac{6}{8}\end{align*}
To solve this, we can multiply the means and the extremes.
Next, we solve the equation for the missing variable. To do this, we use the inverse operation. Multiplication is in the problem, so we use division to solve it. We divide both sides by 8.
Our answer is 3.
Solve for each variable in the numerator by using cross products.
#### Example A
\begin{align*}\frac{x}{5} = \frac{6}{10}\end{align*}
Solution:\begin{align*}x = 3\end{align*}
#### Example B
\begin{align*}\frac{a}{9} = \frac{15}{27}\end{align*}
Solution:\begin{align*}a = 5\end{align*}
#### Example C
\begin{align*}\frac{b}{4} = \frac{12}{16}\end{align*}
Solution:\begin{align*}b = 3\end{align*}
Here is the original problem once again.
Remember Manuel who was reading all of the medieval books on knights? Well after he finished reading the series, he loaned it to his friend Rafael. Rafael is enjoying the series as much as Manuel did.
In five weeks, Rafael had already finished 8 of the 12 books. It took Manuel 7.5 weeks to read all 12 books. Will Rafael and Manuel finish the series in the same amount of time? Are they reading at the same rate? How can you figure this out?
Let’s write a proportion to solve this problem.
\begin{align*}\frac{8 \ books}{5 \ weeks} = \frac{12 \ books}{7.5 \ weeks}\end{align*}
Next, we can use cross products to see if the two ratios form a proportion. If they do, then the two boys will finish the series in the same amount of time.
The two cross products are equal so the two ratios form a proportion. The two boys will finish the series of books in the same amount of time.
### Vocabulary
Proportion
two equal ratios form a proportion.
Means
the values of a proportion that are close to each other when written in ratio form using a colon.
Extremes
the values of a proportion that are farther apart from each other when written in ratio form using a colon.
Cross Product Property of Proportions
states that the cross products of two ratios will be equal if the two ratios form a proportion.
### Guided Practice
Here is one for you to try on your own.
The ratio of boys to girls in the school chorus is 4 to 5. There are a total of 20 boys in the chorus. How many total students are in the chorus?
Answer
The ratio given, 4 to 5, compares boys to girls. However, the question asks for the total number of students in the chorus.
One way to set up a proportion for this problem would be to write two equivalent ratios, each comparing boys to total students.
The ratio of boys to girls is 4 to 5. We can use this ratio to find the ratio of boys to total students.
\begin{align*}\frac{boys}{total} = \frac{boys}{boys + girls} = \frac{4}{4 + 5} = \frac{4}{9}\end{align*}
You know that there are 20 boys in the chorus. The total number of students is unknown, so represent that as \begin{align*}x\end{align*}.
\begin{align*}\frac{boys}{total} = \frac{20}{x}\end{align*}
Get those ratios equal to form a proportion. Then cross multiply to solve for \begin{align*}x\end{align*}.
So, there are a total of 45 students in the school chorus.
### Practice
Directions: Use cross products to find the value of the variable in each proportion.
1. \begin{align*}\frac{6}{10} = \frac{x} {5}\end{align*}
2. \begin{align*}\frac{2}{3} = \frac{x} {9}\end{align*}
3. \begin{align*}\frac{4}{9} = \frac{a} {45}\end{align*}
4. \begin{align*}\frac{7}{8} = \frac{a} {4}\end{align*}
5. \begin{align*}\frac{b}{8} = \frac{5} {16}\end{align*}
6. \begin{align*}\frac{6}{3} = \frac{x} {9}\end{align*}
7. \begin{align*}\frac{4}{x} = \frac{8} {10}\end{align*}
8. \begin{align*}\frac{1.5}{y} = \frac{3} {9}\end{align*}
9. \begin{align*}\frac{4}{11} = \frac{c} {33}\end{align*}
10. \begin{align*}\frac{2}{6} = \frac{5} {y}\end{align*}
11. \begin{align*}\frac{2}{10} = \frac{5} {x}\end{align*}
12. \begin{align*}\frac{4}{12} = \frac{6} {n}\end{align*}
13. \begin{align*}\frac{5}{r} = \frac{70} {126}\end{align*}
14. \begin{align*}\frac{4}{14} = \frac{14} {k}\end{align*}
15. \begin{align*}\frac{8}{w} = \frac{6} {3}\end{align*}
16. \begin{align*}\frac{2}{5} = \frac{17} {a}\end{align*}
### Vocabulary Language: English
Cross Product Property of Proportions
Cross Product Property of Proportions
The cross product property of proportions states that the cross products of two ratios will be equal if the two ratios form a proportion.
Cross Products
Cross Products
To simplify a proportion using cross products, multiply the diagonals of each ratio.
Extremes
Extremes
In a proportion, the extremes are the values of the proportion that are furthest apart when written in ratio form using a colon. For example: In the proportion a : b = c : d, a and d are the extremes.
Means
Means
In a proportion, the means are the values of the proportion that are close to each other when written in ratio form using a colon.
### Explore More
Sign in to explore more, including practice questions and solutions for Proportions Using Cross Products.
Please wait...
Please wait... |
# Expected values
#### Prerequisite
One of the most important concepts in learning to use probabilities quantitatively is the idea of expected value. When we consider probabilities we look at a situation where something can happen many times with a variety of different results. And those results can have different values.
This applies to a wide variety of physical situations, ranging from: the probability of a molecule moving from the cell in the body where it is produced to the cell in the body where it is detected, to creating a particular protein as a result of random interactions of molecules in a cell, to the probability of producing a number of offspring that survive to breed and produce the next generation of breeders. It is particularly relevant in medicine where the expected value can be the fraction of patients who respond positively to a treatment!
But these situations are fairly complex and require us to a set up a lot of contextual details. To keep the focus on the structure of the mathematical ideas, we'll choose a toy model that, while unrealistic, let's you clearly see how the various parts play a role.
Example: Drawing balls out of a jar.
A game at a county fair has a jar containing 1000 different colored balls. For $\$1$, you are blindfolded and allowed to reach into the jar to pull out one ball. If you pull out a white ball, you get nothing. If you pull out a red ball, you get your$\$1$ back. If you pull out a blue ball, you get $\$10$, and if you pull out a green ball, you get$\$100$. Would you pay $\$1$to play this game? You might want to play just for fun even if you expect to lose. But if you want to decide if it's really worth it (or if the real example is whether a patient can expect to benefit or be harmed by a treatment!) you might want to do some calculations. There are a set of four possible results: {W, R, B, G}. We'll index these by the label "i" and create a set of values, V, corresponding to the four results. They are: • VW = 0 • VR = 1 • VB = 10 • VG = 100 What we really need to know is this: How many of each kind of ball is in the jar? Suppose there is only 1 green ball in the jar. Then the probability of our drawing it is 1 in 1000. Suppose also that there are 50 blue balls, 100 red balls, and the rest (1000 - 100 - 50 - 1 = 849) are white. This means that the probabilities of drawing a white, red, blue, and green ball in our one try are: • PW = 0.849 (849 out of 1000) • PR = 0.100 (100 out of 1000) • PB = 0.05 (50 out of 1000) • PG = 0.001 (1 out of 1000) The expected result is defined as the result we might expect to get if we played the game a large number of times. It is the sum of the value of each possible result times the probability of getting that result. Thus, if we only get a green ball 1 try out of 1000 on the average, it costs us (again, on the average)$\$$1000 to win 100. Not a great deal. But maybe the others make up for it? We see the overall result by adding up all the probabilities times the value of their result: <V> = VWPWVRPR + VBPB + VGPG = (0)(0.849) + (1)(0.100) + (10)(0.050) + (100)(0.001) <V> = 0 + 0.1 + 0.5 + 0.1 = 0.7 So for paying \1, on the average we might expect to gain back \0.7 (70 cents). This gives an idea of the fairness of the game and the potential costs and benefits of playing. The expectation equation We want to generalize this so we can apply this idea to more relevant and important examples. In general, we'll have a set of results that we will number by the label i ∈ {1,2,3,4,....}. The curly brackets means a set of labels and "∈" means that i is one of the elements of the set. If the values of the i-th result is Vi and the probability of getting the i-th result is Pi, then the expected value of V over many trials is given by the expectation equation:$$\langle V \rangle = \sum{V_iP_i} = V_1P_1 + V_2P_2 + V_3P_3 ...$$We use the angle brackets \langle ....\rangle to indicate the average of something, so \langle V\rangle is the average (expected) value. The sum of all the probabilities, Pi, have to add up to 1 since something has to happen:$$1 = \sum{P_i} = P_1 + P_2 + P_3 ...
Sometimes we will be interested in a number of different expected values. For example in our study of diffusion along a line (in 1D) we will be interested in both the average displacement of a diffusing molecule, $\langle x \rangle$, and in the square of the average displacement, $\langle x^2 \rangle$. These concepts will also play an important role in understanding entropy and the Boltzmann factor.
Joe Redish and Bill Dorland 2/4/18
Article 288 |
# 12.1 The ellipse
Page 1 / 16
In this section, you will:
• Write equations of ellipses in standard form.
• Graph ellipses centered at the origin.
• Graph ellipses not centered at the origin.
• Solve applied problems involving ellipses.
Can you imagine standing at one end of a large room and still being able to hear a whisper from a person standing at the other end? The National Statuary Hall in Washington, D.C., shown in [link] , is such a room. Architect of the Capitol. http://www.aoc.gov. Accessed April 15, 2014. It is an oval-shaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can stand and still hear each other whisper.
## Writing equations of ellipses in standard form
A conic section, or conic , is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown in [link] .
Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci ).
We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. See [link] .
Every ellipse has two axes of symmetry. The longer axis is called the major axis , and the shorter axis is called the minor axis . Each endpoint of the major axis is the vertex of the ellipse (plural: vertices ), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. See [link] .
In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the x - and y -axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane.
To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs.
x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has |
Question
# Solve $\left( x-5 \right)\left( x-7 \right)\left( x+4 \right)\left( x+6 \right)=504$.
Hint: Find the value of x; take the value of $\left( {{x}^{2}}-x \right)$as t.
Find the roots of t and the substitute it in $t={{x}^{2}}-x$.
Solve the quadratic equations formed and you will get 4 values of x.
Given $\left( x-5 \right)\left( x-7 \right)\left( x+4 \right)\left( x+6 \right)=504$
We can consider $\left[ \left( x-5 \right)\left( x+4 \right) \right]$together and $\left[ \left( x-7 \right)\left( x+6 \right) \right]$.
Open the bracket and form a quadratic equation of the form$a{{x}^{2}}+bx+c$.
\begin{align} & \left[ \left( x-5 \right)\left( x+4 \right) \right]\left[ \left( x-7 \right)\left( x+6 \right) \right]=504-(1) \\ & \left( x-5 \right)\left( x+4 \right)={{x}^{2}}-5x+4x-20={{x}^{2}}-x-20 \\ & \left( x-7 \right)\left( x+6 \right)={{x}^{2}}-7x-6x-42={{x}^{2}}-x-42 \\ \end{align}
Put $t={{x}^{2}}-x-(2)$
\begin{align} & \Rightarrow \left( t-20 \right)\left( t-42 \right)=504 \\ & \Rightarrow {{t}^{2}}-20t-42t+840-504=0 \\ & \Rightarrow {{t}^{2}}-62t+336=0-(3) \\ \end{align}
By substituting value of t in equation (1), we can simplify it to equation (3)
Now we get a quadratic equation ${{t}^{2}}-62t+336=0$.
The general form $a{{x}^{2}}+bx+c=0$, by comparing the general form and equation (3), we get
a=1, b=-62 and c=336
substituting these values in $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, quadratic formula we get the roots.
\begin{align} & =\dfrac{-\left( -62 \right)\pm \sqrt{{{\left( -62 \right)}^{2}}-4\times 1\times 336}}{2}=\dfrac{62\pm \sqrt{3844-1344}}{2} \\ & =\dfrac{62\pm \sqrt{2500}}{2}=\dfrac{62\pm 50}{2} \\ \end{align}
The roots are $\left( \dfrac{62+50}{2} \right)$and $\left( \dfrac{62-50}{2} \right)$= 56 and 6
$\therefore$Roots of t = 56 and 6
We know, $t={{x}^{2}}-x$.
Put the values of t = 56.
$\Rightarrow {{x}^{2}}-x-56=0-(4)$
Now find the roots of equation (4) by using quadratic equation
a=1, b = -1, c = -56
\begin{align} & =\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times \left( -56 \right)}}{2}=\dfrac{-1\pm \sqrt{1+224}}{2} \\ & =\dfrac{1\pm \sqrt{225}}{2}=\dfrac{1\pm 15}{2} \\ \end{align}
The roots are $\left( \dfrac{1+15}{2} \right)$and $\left( \dfrac{1-15}{2} \right)$= 8 and -7
Similarly, $t={{x}^{2}}-x$ , put value of t = 6
$\Rightarrow {{x}^{2}}-x-6=0$
a = 1, b = -1, c = -6
\begin{align} & =\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times \left( -6 \right)}}{2}=\dfrac{-1\pm \sqrt{1+24}}{2} \\ & =\dfrac{1\pm \sqrt{25}}{2}=\dfrac{1\pm 5}{2} \\ \end{align}
The roots are $\left( \dfrac{1+5}{2} \right)$and $\left( \dfrac{1-5}{2} \right)$= 3 and -2.
$\therefore$The values of x are 8, -7, 3 and -2.
Note: The pair to be multiplied should be chosen in a way that $t={{x}^{2}}-x$. Taking $\left( x-5 \right)\left( x-7 \right)$and $\left( x+4 \right)\left( x+6 \right)$won’t give the required answer. Therefore, we choose $\left( x-5 \right)\left( x+4 \right)$and $\left( x-7 \right)\left( x+6 \right)$, while forming the equation to get the value of x. Solving the value of t to get the roots. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
Amplitude
Measure of a wave's height from the center axis.
0%
Progress
Practice Amplitude
Progress
0%
Amplitude of Sinusoidal Functions
The amplitude of the sine and cosine functions is the distance between the sinusoidal axis and the maximum or minimum value of the function. In relation to sound waves, amplitude is a measure of how loud something is.
What is the most common mistake made when graphing the amplitude of a sine wave?
Watch This
Watch the portion of this video discussing amplitude:
http://www.youtube.com/watch?v=qJ-oUV7xL3w James Sousa: Amplitude and Period of Sine and Cosine
Guidance
The general form a sinusoidal function is:
$f(x)={\pm} a \cdot \sin (b(x+c))+d$
The cosine function can just as easily be substituted and for many problems it will be easier to use a cosine equation. Since both the sine and cosine waves are identical except for a horizontal shift, it all depends on where you see the wave starting.
The coefficient $a$ is the amplitude (which fortunately also starts with the letter a). When there is no number present, then the amplitude is 1. The best way to define amplitude is through a picture. Below is the graph of the function $f(x)=3 \cdot \sin x$ , which has an amplitude of 3.
Notice that the amplitude is 3, not 6. This corresponds to the absolute value of the maximum and minimum values of the function. If the function had been $f(x)={\color{red}-}3 \cdot \sin x$ , then the whole graph would be reflected across the $x$ axis.
Also notice that the $x$ axis on the graph above is unlabeled. This is to show that amplitude is a vertical distance. The sinusoidal axis is the neutral horizontal line that lies between the crests and the troughs (or peaks and valleys if you prefer). For this function, the sinusoidal axis was just the $x$ axis, but if the whole graph were shifted up, the sinusoidal axis would no longer be the $x$ axis. Instead, it would still be the horizontal line directly between the crests and troughs.
Example A
Graph the following function by first plotting main points: $f(x)=-2 \cdot \cos x$ .
Solution: The amplitude is 2, which means the maximum values will be at 2 and the minimum values will be at -2. Normally with a basic cosine curve the points corresponding to $0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}, 2 \pi$ fall above, on or below the line in the following sequence: above, on, below, on, above. The negative sign switches above with below. The whole graph is reflected across the $x$ -axis.
Example B
Write a cosine equation for each of the following functions
Solution: The amplitudes of the three functions are 3, 1 and $\frac{1}{2}$ and none of them are reflected across the $x$ -axis.
$f(x) &=3 \cdot \cos x\\h(x) &={\cos} x\\g(x) &= {\frac{1}{2}} \cdot \cos x$
Note that amplitude itself is always positive.
Example C
A Ferris wheel with radius 25 feet sits next to a platform. The ride starts at the platform and travels down to start. Model the height versus time of the ride.
Solution: Since no information is given about the time, simply label the $x$ axis as time. At time zero the height is zero. Initially the height will decrease as the ride goes below the platform. Eventually, the wheel will find the minimum and start to increase again all the way until it reaches a maximum.
Concept Problem Revisited
The most common mistake is doubling or halving the amplitude unnecessarily. Many people forget that the number $a$ in the equation, like the 3 in $f(x)=3 \sin x$ , is the distance from the $x$ axis to both the peak and the valley. It is not the total distance from the peak to the valley.
Vocabulary
The amplitude of the sine or a cosine function is the shortest vertical distance between the sinusoidal axis and the maximum or minimum value.
The sinusoidal axis is the neutral horizontal line that lies between the crests and the troughs of the graph of the function.
Guided Practice
1. Identify the amplitudes of the following four functions:
2. Graph the following function: $f(x)=-8 \sin x$ .
3. Find the amplitude of the function $f(x)=-3 \cos x$ and use the language of transformations to describe how the graph is related to the parent function $y=\cos x$ .
1. The red function has amplitude 3. The blue function has amplitude 2. The green function has amplitude $\frac{1}{2}$
2. First identify where the function starts and ends. In this case, one cycle (period) is from 0 to $2 \pi$ . Usually sine functions start at the sinusoidal axis and have one bump up and then one bump down, but the negative sign swaps directions. Lastly, instead of going up and down only one unit, this function has amplitude of 8. Thus the pattern is:
Starts at height 0
Then down to -8.
Then back to 0.
Then up to 8
Then back to 0.
Plotting these 5 points first is an essential step to sketching an approximate curve.
3. The new function is reflected across the $x$ axis and vertically stretched by a factor of 3.
Practice
1. Explain how to find the amplitude of a sinusoidal function from its equation.
2. Explain how to find the amplitude of a sinusoidal function from its graph.
Find the amplitude of each of the following functions.
3. $g(x)=-5 \cos x$
4.
5. $f(x)={\frac{1}{2}} \sin x$
6.
7. $j(x)=3.12 \cos x$
8.
Sketch each of the following functions.
9. $f(x)=3 \sin x$
10. $g(x)=-4 \cos x$
11. $h(x)=\pi \sin x$
12. $k(x)=-1.2 \cos x$
13. $p(x)=\frac{2}{3} \cos x$
14. $m(x)=-\frac{1}{2} \sin x$
15. Preview: $r(x)=3 \sin x+2$
Vocabulary Language: English
Amplitude
Amplitude
The amplitude of a wave is one-half of the difference between the minimum and maximum values of the wave, it can be related to the radius of a circle.
sinusoidal axis
sinusoidal axis
The sinusoidal axis is the neutral horizontal line that lies between the crests and the troughs of the graph of a sine or cosine function.
sinusoidal function
sinusoidal function
A sinusoidal function is a sine or cosine wave.
sinusoidal functions
sinusoidal functions
A sinusoidal function is a sine or cosine wave. |
# Chapter 11 Section 3
Preview
Warm Up
California Standards
Lesson Presentation
Warm Up
Simplify each expression.
1. 14x + 15y – 12y + x
15x + 3y
2. 9xy + 2xy – 8xy 3xy
3. –3(a + b) +
–3a – b + 10
Simplify. All variables represent nonnegative
numbers.
4.
5.
6.
California
Standards
Extension of
2.0
Students understand and use such operations
as taking the opposite, finding the reciprocal,
taking a root, and raising to a fractional power.
They understand and use the rules of exponents.
Vocabulary
Square-root expressions with the same radicand are
subtracting. You can use the Distributive Property to
show how this is done:
subtracting the numbers multiplied by the radical and
Combining like radicals is similar to combining
like terms.
Square-Root Expressions
A.
B.
The terms are unlike radicals. Do
not combine.
Square-Root Expressions
C.
the terms are
D.
Check It Out! Example 1
a.
b.
Check It Out! Example 1
c.
d.
Sometimes radicals do not appear to be like
until they are simplified. Simplify all radicals in
an expression before trying to identify like
Subtracting
Simplify each expression. All variables
represent nonnegative numbers.
squares.
Product Property of Square Roots
Simplify.
Subtracting
Simplify each expression. All variables
represent nonnegative numbers.
squares.
Product Property of Square Roots
Simplify.
The terms are unlike radicals. Do
not combine.
Subtracting
Simplify each expression. All variables
represent nonnegative numbers.
using perfect squares.
Product Property of
Square Roots
Simplify.
Remember!
When you write a radicand as a product, make at
least one factor a perfect square.
Check It Out! Example 2a
Simplify each expression. All variables
represent nonnegative numbers.
squares.
Product Property of Square Roots
Simplify.
Check It Out! Example 2b
Simplify each expression. All variables
represent nonnegative numbers.
squares.
Product Property of Square Roots
Simplify.
The terms are unlike radicals. Do not
combine.
Check It Out! Example 2c
Simplify each expression. All variables
represent nonnegative numbers.
squares.
Product Property of Square Roots
Simplify.
Find the perimeter of the
simplest form.
Write an expression for perimeter.
Factor 20 using a perfect square.
Product Property of Square Roots
Simplify.
Check It Out! Example 3
Find the perimeter of a rectangle whose length
is
inches and whose width is
inches.
simplest form.
Write an expression for perimeter
2 (l + w).
Multiply each term by 2 .
Simplify.
The perimeter is
in.
Lesson Quiz: Part I |
Elementary Algebra 2e
# 1.7Decimals
## Learning Objectives
By the end of this section, you will be able to:
• Name and write decimals
• Round decimals
• Multiply and divide decimals
• Convert decimals, fractions, and percents
## Be Prepared 1.7
A more thorough introduction to the topics covered in this section can be found in the Prealgebra chapter, Decimals.
## Name and Write Decimals
Decimals are another way of writing fractions whose denominators are powers of 10.
$0.1=1100.1is “one tenth”0.01=11000.01is “one hundredth”0.001=11,0000.001 is “one thousandth”0.0001=110,0000.0001 is “one ten-thousandth”0.1=1100.1is “one tenth”0.01=11000.01is “one hundredth”0.001=11,0000.001 is “one thousandth”0.0001=110,0000.0001 is “one ten-thousandth”$
Notice that “ten thousand” is a number larger than one, but “one ten-thousandth” is a number smaller than one. The “th” at the end of the name tells you that the number is smaller than one.
When we name a whole number, the name corresponds to the place value based on the powers of ten. We read 10,000 as “ten thousand” and 10,000,000 as “ten million.” Likewise, the names of the decimal places correspond to their fraction values. Figure 1.14 shows the names of the place values to the left and right of the decimal point.
Figure 1.14 Place value of decimal numbers are shown to the left and right of the decimal point.
## Example 1.91
### How to Name Decimals
Name the decimal 4.3.
## Try It 1.181
Name the decimal: $6.7.6.7.$
## Try It 1.182
Name the decimal: $5.8.5.8.$
We summarize the steps needed to name a decimal below.
## How To
### Name a Decimal.
1. Step 1. Name the number to the left of the decimal point.
2. Step 2. Write “and” for the decimal point.
3. Step 3. Name the “number” part to the right of the decimal point as if it were a whole number.
4. Step 4. Name the decimal place of the last digit.
## Example 1.92
Name the decimal: $−15.571.−15.571.$
## Try It 1.183
Name the decimal: $−13.461.−13.461.$
## Try It 1.184
Name the decimal: $−2.053.−2.053.$
When we write a check we write both the numerals and the name of the number. Let’s see how to write the decimal from the name.
## Example 1.93
### How to Write Decimals
Write “fourteen and twenty-four thousandths” as a decimal.
## Try It 1.185
Write as a decimal: thirteen and sixty-eight thousandths.
## Try It 1.186
Write as a decimal: five and ninety-four thousandths.
We summarize the steps to writing a decimal.
## How To
### Write a decimal.
1. Step 1.
Look for the word “and”—it locates the decimal point.
• Place a decimal point under the word “and.” Translate the words before “and” into the whole number and place it to the left of the decimal point.
• If there is no “and,” write a “0” with a decimal point to its right.
2. Step 2. Mark the number of decimal places needed to the right of the decimal point by noting the place value indicated by the last word.
3. Step 3. Translate the words after “and” into the number to the right of the decimal point. Write the number in the spaces—putting the final digit in the last place.
4. Step 4. Fill in zeros for place holders as needed.
## Round Decimals
Rounding decimals is very much like rounding whole numbers. We will round decimals with a method based on the one we used to round whole numbers.
## Example 1.94
### How to Round Decimals
Round 18.379 to the nearest hundredth.
## Try It 1.187
Round to the nearest hundredth: $1.047.1.047.$
## Try It 1.188
Round to the nearest hundredth: $9.173.9.173.$
We summarize the steps for rounding a decimal here.
## How To
### Round Decimals.
1. Step 1. Locate the given place value and mark it with an arrow.
2. Step 2. Underline the digit to the right of the place value.
3. Step 3.
Is this digit greater than or equal to 5?
• Yes—add 1 to the digit in the given place value.
• No—do not change the digit in the given place value.
4. Step 4. Rewrite the number, deleting all digits to the right of the rounding digit.
## Example 1.95
Round 18.379 to the nearest tenth whole number.
## Try It 1.189
Round $6.5826.582$ to the nearest hundredth tenth whole number.
## Try It 1.190
Round $15.217515.2175$ to the nearest thousandth hundredth tenth.
To add or subtract decimals, we line up the decimal points. By lining up the decimal points this way, we can add or subtract the corresponding place values. We then add or subtract the numbers as if they were whole numbers and then place the decimal point in the sum.
## How To
1. Step 1. Write the numbers so the decimal points line up vertically.
2. Step 2. Use zeros as place holders, as needed.
3. Step 3. Add or subtract the numbers as if they were whole numbers. Then place the decimal point in the answer under the decimal points in the given numbers.
## Example 1.96
Add: $23.5+41.38.23.5+41.38.$
## Try It 1.191
Add: $4.8+11.69.4.8+11.69.$
## Try It 1.192
Add: $5.123+18.47.5.123+18.47.$
## Example 1.97
Subtract: $20−14.65.20−14.65.$
## Try It 1.193
Subtract: $10−9.58.10−9.58.$
## Try It 1.194
Subtract: $50−37.42.50−37.42.$
## Multiply and Divide Decimals
Multiplying decimals is very much like multiplying whole numbers—we just have to determine where to place the decimal point. The procedure for multiplying decimals will make sense if we first convert them to fractions and then multiply.
So let’s see what we would get as the product of decimals by converting them to fractions first. We will do two examples side-by-side. Look for a pattern!
Convert to fractions. Multiply. Convert to decimals.
Notice, in the first example, we multiplied two numbers that each had one digit after the decimal point and the product had two decimal places. In the second example, we multiplied a number with one decimal place by a number with two decimal places and the product had three decimal places.
We multiply the numbers just as we do whole numbers, temporarily ignoring the decimal point. We then count the number of decimal points in the factors and that sum tells us the number of decimal places in the product.
The rules for multiplying positive and negative numbers apply to decimals, too, of course!
When multiplying two numbers,
• if their signs are the same the product is positive.
• if their signs are different the product is negative.
When we multiply signed decimals, first we determine the sign of the product and then multiply as if the numbers were both positive. Finally, we write the product with the appropriate sign.
## How To
### Multiply Decimals.
1. Step 1. Determine the sign of the product.
2. Step 2. Write in vertical format, lining up the numbers on the right. Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points.
3. Step 3. Place the decimal point. The number of decimal places in the product is the sum of the number of decimal places in the factors.
4. Step 4. Write the product with the appropriate sign.
## Example 1.98
Multiply: $(−3.9)(4.075).(−3.9)(4.075).$
## Try It 1.195
Multiply: $−4.5(6.107).−4.5(6.107).$
## Try It 1.196
Multiply: $−10.79(8.12).−10.79(8.12).$
In many of your other classes, especially in the sciences, you will multiply decimals by powers of 10 (10, 100, 1000, etc.). If you multiply a few products on paper, you may notice a pattern relating the number of zeros in the power of 10 to number of decimal places we move the decimal point to the right to get the product.
## How To
### Multiply a Decimal by a Power of Ten.
1. Step 1. Move the decimal point to the right the same number of places as the number of zeros in the power of 10.
2. Step 2. Add zeros at the end of the number as needed.
## Example 1.99
Multiply 5.63 by 10 by 100 by 1,000.
## Try It 1.197
Multiply 2.58 by 10 by 100 by 1,000.
## Try It 1.198
Multiply 14.2 by 10 by 100 by 1,000.
Just as with multiplication, division of decimals is very much like dividing whole numbers. We just have to figure out where the decimal point must be placed.
To divide decimals, determine what power of 10 to multiply the denominator by to make it a whole number. Then multiply the numerator by that same power of $10.10.$ Because of the equivalent fractions property, we haven’t changed the value of the fraction! The effect is to move the decimal points in the numerator and denominator the same number of places to the right. For example:
$0.80.40.8(10)0.4(10)840.80.40.8(10)0.4(10)84$
We use the rules for dividing positive and negative numbers with decimals, too. When dividing signed decimals, first determine the sign of the quotient and then divide as if the numbers were both positive. Finally, write the quotient with the appropriate sign.
We review the notation and vocabulary for division:
$adividend÷bdivisor=cquotientbdivisorcquotientadividendadividend÷bdivisor=cquotientbdivisorcquotientadividend$
We’ll write the steps to take when dividing decimals, for easy reference.
## How To
### Divide Decimals.
1. Step 1. Determine the sign of the quotient.
2. Step 2. Make the divisor a whole number by “moving” the decimal point all the way to the right. “Move” the decimal point in the dividend the same number of places—adding zeros as needed.
3. Step 3. Divide. Place the decimal point in the quotient above the decimal point in the dividend.
4. Step 4. Write the quotient with the appropriate sign.
## Example 1.100
Divide: $−25.65÷(−0.06).−25.65÷(−0.06).$
## Try It 1.199
Divide: $−23.492÷(−0.04).−23.492÷(−0.04).$
## Try It 1.200
Divide: $−4.11÷(−0.12).−4.11÷(−0.12).$
A common application of dividing whole numbers into decimals is when we want to find the price of one item that is sold as part of a multi-pack. For example, suppose a case of 24 water bottles costs $3.99. To find the price of one water bottle, we would divide$3.99 by 24. We show this division in Example 1.101. In calculations with money, we will round the answer to the nearest cent (hundredth).
## Example 1.101
Divide: $3.99÷24.3.99÷24.$
## Try It 1.201
Divide: $6.99÷36.6.99÷36.$
## Try It 1.202
Divide: $4.99÷12.4.99÷12.$
## Convert Decimals, Fractions, and Percents
We convert decimals into fractions by identifying the place value of the last (farthest right) digit. In the decimal 0.03 the 3 is in the hundredths place, so 100 is the denominator of the fraction equivalent to 0.03.
$00.03=310000.03=3100$
Notice, when the number to the left of the decimal is zero, we get a fraction whose numerator is less than its denominator. Fractions like this are called proper fractions.
The steps to take to convert a decimal to a fraction are summarized in the procedure box.
## How To
### Convert a Decimal to a Proper Fraction.
1. Step 1. Determine the place value of the final digit.
2. Step 2.
Write the fraction.
• numerator—the “numbers” to the right of the decimal point
• denominator—the place value corresponding to the final digit
## Example 1.102
Write 0.374 as a fraction.
## Try It 1.203
Write 0.234 as a fraction.
## Try It 1.204
Write 0.024 as a fraction.
We’ve learned to convert decimals to fractions. Now we will do the reverse—convert fractions to decimals. Remember that the fraction bar means division. So $4545$ can be written $4÷54÷5$ or $54.54.$ This leads to the following method for converting a fraction to a decimal.
## How To
### Convert a Fraction to a Decimal.
To convert a fraction to a decimal, divide the numerator of the fraction by the denominator of the fraction.
## Example 1.103
Write $−58−58$ as a decimal.
## Try It 1.205
Write $−78−78$ as a decimal.
## Try It 1.206
Write $−38−38$ as a decimal.
When we divide, we will not always get a zero remainder. Sometimes the quotient ends up with a decimal that repeats. A repeating decimal is a decimal in which the last digit or group of digits repeats endlessly. A bar is placed over the repeating block of digits to indicate it repeats.
## Repeating Decimal
A repeating decimal is a decimal in which the last digit or group of digits repeats endlessly.
A bar is placed over the repeating block of digits to indicate it repeats.
## Example 1.104
Write $43224322$ as a decimal.
## Try It 1.207
Write $27112711$ as a decimal.
## Try It 1.208
Write $51225122$ as a decimal.
Sometimes we may have to simplify expressions with fractions and decimals together.
## Example 1.105
Simplify: $78+6.4.78+6.4.$
## Try It 1.209
Simplify: $38+4.9.38+4.9.$
## Try It 1.210
Simplify: $5.7+1320.5.7+1320.$
A percent is a ratio whose denominator is 100. Percent means per hundred. We use the percent symbol, %, to show percent.
## Percent
A percent is a ratio whose denominator is 100.
Since a percent is a ratio, it can easily be expressed as a fraction. Percent means per 100, so the denominator of the fraction is 100. We then change the fraction to a decimal by dividing the numerator by the denominator.
$6%6%$ $78%78%$ $135%135%$ Write as a ratio with denominator 100. $61006100$ $7810078100$ $135100135100$ Change the fraction to a decimal by dividing the numerator by the denominator. $0.060.06$ $0.780.78$ $1.351.35$
Table 1.30
Do you see the pattern? To convert a percent number to a decimal number, we move the decimal point two places to the left.
## Example 1.106
Convert each percent to a decimal: 62% 135% 35.7%.
## Try It 1.211
Convert each percent to a decimal: $9%9%$ $87%87%$ 3.9%.
## Try It 1.212
Convert each percent to a decimal: 3% 91% 8.3%.
Converting a decimal to a percent makes sense if we remember the definition of percent and keep place value in mind.
To convert a decimal to a percent, remember that percent means per hundred. If we change the decimal to a fraction whose denominator is 100, it is easy to change that fraction to a percent.
$0.830.83$ $1.051.05$ $0.0750.075$ Write as a fraction. $8310083100$ $1510015100$ $751000751000$ The denominator is 100. $105100105100$ $7.51007.5100$ Write the ratio as a percent. $83%83%$ $105%105%$ $7.5%7.5%$
Table 1.31
Recognize the pattern? To convert a decimal to a percent, we move the decimal point two places to the right and then add the percent sign.
## Example 1.107
Convert each decimal to a percent: 0.51 1.25 0.093.
## Try It 1.213
Convert each decimal to a percent: 0.17 1.75 0.0825.
## Try It 1.214
Convert each decimal to a percent: 0.41 2.25 0.0925.
## Section 1.7 Exercises
### Practice Makes Perfect
Name and Write Decimals
In the following exercises, write as a decimal.
531.
Twenty-nine and eighty-one hundredths
532.
Sixty-one and seventy-four hundredths
533.
Seven tenths
534.
Six tenths
535.
Twenty-nine thousandth
536.
Thirty-five thousandths
537.
Negative eleven and nine ten-thousandths
538.
Negative fifty-nine and two ten-thousandths
In the following exercises, name each decimal.
539.
5.5
540.
14.02
541.
8.71
542.
2.64
543.
0.002
544.
0.479
545.
$− 17 .9 − 17 .9$
546.
$− 31 .4 − 31 .4$
Round Decimals
In the following exercises, round each number to the nearest tenth.
547.
0.67
548.
0.49
549.
2.84
550.
4.63
In the following exercises, round each number to the nearest hundredth.
551.
0.845
552.
0.761
553.
0.299
554.
0.697
555.
4.098
556.
7.096
In the following exercises, round each number to the nearest hundredth tenth whole number.
557.
5.781
558.
1.6381
559.
63.479
560.
$84 .281 84 .281$
In the following exercises, add or subtract.
561.
$16.92 + 7.56 16.92 + 7.56$
562.
$248.25 − 91.29 248.25 − 91.29$
563.
$21.76 − 30.99 21.76 − 30.99$
564.
$38.6 + 13.67 38.6 + 13.67$
565.
$−16.53 − 24.38 −16.53 − 24.38$
566.
$−19.47 − 32.58 −19.47 − 32.58$
567.
$−38.69 + 31.47 −38.69 + 31.47$
568.
$29.83 + 19.76 29.83 + 19.76$
569.
$72.5 − 100 72.5 − 100$
570.
$86.2 − 100 86.2 − 100$
571.
$15 + 0.73 15 + 0.73$
572.
$27 + 0.87 27 + 0.87$
573.
$91.95 − ( −10.462 ) 91.95 − ( −10.462 )$
574.
$94.69 − ( −12.678 ) 94.69 − ( −12.678 )$
575.
$55.01 − 3.7 55.01 − 3.7$
576.
$59.08 − 4.6 59.08 − 4.6$
577.
$2.51 − 7.4 2.51 − 7.4$
578.
$3.84 − 6.1 3.84 − 6.1$
Multiply and Divide Decimals
In the following exercises, multiply.
579.
$( 0.24 ) ( 0.6 ) ( 0.24 ) ( 0.6 )$
580.
$( 0.81 ) ( 0.3 ) ( 0.81 ) ( 0.3 )$
581.
$( 5.9 ) ( 7.12 ) ( 5.9 ) ( 7.12 )$
582.
$( 2.3 ) ( 9.41 ) ( 2.3 ) ( 9.41 )$
583.
$( −4.3 ) ( 2.71 ) ( −4.3 ) ( 2.71 )$
584.
$( −8.5 ) ( 1.69 ) ( −8.5 ) ( 1.69 )$
585.
$( −5.18 ) ( −65.23 ) ( −5.18 ) ( −65.23 )$
586.
$( −9.16 ) ( −68.34 ) ( −9.16 ) ( −68.34 )$
587.
$( 0.06 ) ( 21.75 ) ( 0.06 ) ( 21.75 )$
588.
$( 0.08 ) ( 52.45 ) ( 0.08 ) ( 52.45 )$
589.
$( 9.24 ) ( 10 ) ( 9.24 ) ( 10 )$
590.
$( 6.531 ) ( 10 ) ( 6.531 ) ( 10 )$
591.
$( 55.2 ) ( 1000 ) ( 55.2 ) ( 1000 )$
592.
$( 99.4 ) ( 1000 ) ( 99.4 ) ( 1000 )$
In the following exercises, divide.
593.
$4.75 ÷ 25 4.75 ÷ 25$
594.
$12.04 ÷ 43 12.04 ÷ 43$
595.
$117.25 ÷ 48 117.25 ÷ 48$
596.
$109.24 ÷ 36 109.24 ÷ 36$
597.
$0.6 ÷ 0.2 0.6 ÷ 0.2$
598.
$0.8 ÷ 0.4 0.8 ÷ 0.4$
599.
$1.44 ÷ ( −0.3 ) 1.44 ÷ ( −0.3 )$
600.
$1.25 ÷ ( −0.5 ) 1.25 ÷ ( −0.5 )$
601.
$−1.75 ÷ ( −0.05 ) −1.75 ÷ ( −0.05 )$
602.
$−1.15 ÷ ( −0.05 ) −1.15 ÷ ( −0.05 )$
603.
$5.2 ÷ 2.5 5.2 ÷ 2.5$
604.
$6.5 ÷ 3.25 6.5 ÷ 3.25$
605.
$11 ÷ 0.55 11 ÷ 0.55$
606.
$14 ÷ 0.35 14 ÷ 0.35$
Convert Decimals, Fractions and Percents
In the following exercises, write each decimal as a fraction.
607.
0.04
608.
0.19
609.
0.52
610.
0.78
611.
1.25
612.
1.35
613.
0.375
614.
0.464
615.
0.095
616.
0.085
In the following exercises, convert each fraction to a decimal.
617.
$17 20 17 20$
618.
$13 20 13 20$
619.
$11 4 11 4$
620.
$17 4 17 4$
621.
$− 310 25 − 310 25$
622.
$− 284 25 − 284 25$
623.
$15 11 15 11$
624.
$18 11 18 11$
625.
$15 111 15 111$
626.
$25 111 25 111$
627.
$2.4 + 5 8 2.4 + 5 8$
628.
$3.9 + 9 20 3.9 + 9 20$
In the following exercises, convert each percent to a decimal.
629.
1%
630.
2%
631.
63%
632.
71%
633.
150%
634.
250%
635.
21.4%
636.
39.3%
637.
7.8%
638.
6.4%
In the following exercises, convert each decimal to a percent.
639.
0.01
640.
0.03
641.
1.35
642.
1.56
643.
3
644.
4
645.
0.0875
646.
0.0625
647.
2.254
648.
2.317
### Everyday Math
649.
Salary Increase Danny got a raise and now makes $58,965.95 a year. Round this number to the nearest dollar thousand dollars ten thousand dollars. 650. New Car Purchase Selena’s new car cost$23,795.95. Round this number to the nearest dollar thousand dollars ten thousand dollars.
651.
Sales Tax Hyo Jin lives in San Diego. She bought a refrigerator for $1,624.99 and when the clerk calculated the sales tax it came out to exactly$142.186625. Round the sales tax to the nearest penny and dollar.
652.
Sales Tax Jennifer bought a $1,038.99 dining room set for her home in Cincinnati. She calculated the sales tax to be exactly$67.53435. Round the sales tax to the nearest penny and dollar.
653.
Paycheck Annie has two jobs. She gets paid $14.04 per hour for tutoring at City College and$8.75 per hour at a coffee shop. Last week she tutored for 8 hours and worked at the coffee shop for 15 hours. How much did she earn? If she had worked all 23 hours as a tutor instead of working both jobs, how much more would she have earned?
654.
Paycheck Jake has two jobs. He gets paid $7.95 per hour at the college cafeteria and$20.25 at the art gallery. Last week he worked 12 hours at the cafeteria and 5 hours at the art gallery. How much did he earn? If he had worked all 17 hours at the art gallery instead of working both jobs, how much more would he have earned?
### Writing Exercises
655.
656.
Explain how you write “three and nine hundredths” as a decimal.
657.
Without solving the problem “44 is 80% of what number” think about what the solution might be. Should it be a number that is greater than 44 or less than 44? Explain your reasoning.
658.
When the Szetos sold their home, the selling price was 500% of what they had paid for the house 30 years ago. Explain what 500% means in this context.
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
What does this checklist tell you about your mastery of this section? What steps will you take to improve? |
CHAPTER SUMMARY
I. Basics
Basics
Geometry questions on the GMAT are a logical puzzle. You will be given some information about a figure, and must deduce some piece of missing information. These questions test a set of general rules and formulae that must be memorized.
Lines
Through any two points, there is exactly one line, extending infinitely in both directions. A section of a line is called a line segment or a ray.
Angles
When two rays originate from the same point, they form an angle, which is represented by the symbol ∠. The point of intersection is the vertex and the two rays form the sides of the angle. Angles that have the same measure are congruent.
Types of Angles
Angles can be categorized by their angle measure: acute, right, obtuse, straight.
Two angles are complementary angles or complements if the sum of their measures is 90°. Two angles are supplementary angles or supplements if the sum of their measures is 180°.
Midpoint and Bisector
The midpoint is the center point of any line segment. Bisect means “to cut in half.” A bisector is a line that divides a line segment or an angle into two equal pieces.
II. Intersecting Lines
Intersecting Lines
Vertical Angles
When two lines intersect, they create four angles. The angles opposite one another are called vertical angles. Vertical angles are equal to each other. Also, note that any pair of adjacent angles will always equal 180°
Parallel Lines
Two lines that never get closer to or farther away from each other and therefore never intersect are called parallel lines. The symbol for parallel lines is ||. When two parallel lines are cut by a third line, called a transversal, the three lines form a system of congruent angles.
Perpendicular Lines
When two lines intersect each other at a right angle (90°), the lines are perpendicular. The symbol for perpendicular is ⊥, and m ⊥ n means that line m and line n are perpendicular lines.
Be careful not to assume that lines are parallel or perpendicular if it’s not stated explicitly. Don’t be fooled by appearance; you cannot assume that figures are drawn to scale.
III. Triangles
Triangles
There are some rules that are true for all the different kinds of triangles. Be sure to familiarize yourself with these rules.
Perimeter
The perimeter of any figure is the distance around the outside of the figure, or the sum of the sides of the figure. Many GMAT questions about triangle perimeters involve the rule that the length of the longest side of a triangle must be less than the sum of the lengths of the other two sides of the triangle, but greater than the difference of the lengths.
Area
The area of any figure is the amount of surface that is covered by the figure.
The formula for the area of a triangle is A = \dfrac{1}{2} (base × height) = \dfrac{1}{2} bh
Types of Triangles
There are three important types of triangle:
• An isosceles triangle has two equal sides and two equal angles.
• An equilateral triangle has three equal sides and three equal angles.
• right triangle is a triangle with a 90° angle.
Right Triangles
For right triangles, the relationship between the legs and the hypotenuse is defined by the Pythagorean Theorem. The Pythagorean Theorem states that the square of the hypotenuse will equal the sum of the squares of the legs.
Special Right Triangles Based on Side Lengths
There are several special right triangles where the lengths of their sides form ratios of whole numbers. These triangles appear commonly on the GMAT, so being able to quickly recognize them will help you solve problems more quickly and accurately.
Special Right Triangles Based on Angles
There are two special “angle-based” right triangles on the GMAT:
• The 45° : 45° : 90° triangle
• The 30° : 60° : 90° triangle
Similar Triangles
Triangles that have the same shape are called similar triangles.
There are two ways to know that two triangles are similar:
• If the corresponding angles are equal, the triangles are similar.
• If the corresponding sides have the same ratio, the triangles are similar.
The ratio of the lengths of the corresponding sides is called the scale factor. The scale factor for congruent triangles is 1.
IV. Polygons
Polygons
polygon is a figure made from 3 or more line segments. Triangles and quadrilaterals are the most common polygons on the GMAT. Other polygons on the test usually contain triangles and quadrilaterals.
Perimeter and Area
• The perimeter of a polygon is the distance around the outside of the polygon, or the sum of the lengths of the sides of the polygon.
• The area of a polygon is the amount of surface that is covered by the polygon.
Familiarize yourself with the formulas for finding area and perimeter of the different types of polygons.
Interior and Exterior Angles
The sum of the interior angles of a polygon is a function of the number of sides. The formula for the sum of the interior angles of a polygon with n sides is (n – 2)180°.
The sum of the exterior angles of all polygons is 360°. You can think of this as going all the way around the polygon, making a complete circle.
Symmetry
A figure has line symmetry if the figure can evenly “fold” onto itself. A figure can have more than one line of symmetry.
A figure has rotational symmetry if the figure can be rotated less than 180° and lines up on the original figure. The point used for the rotation is the center of symmetry.
Similar Polygons
Polygons that have the same shape but a different size are called similar polygons. Polygons are similar if the corresponding angles are equal, and the corresponding sides have the same ratio.
The ratio of the lengths of the corresponding sides is called the scale factor. The scale factor for congruent figures is 1.
Diagonals
diagonal of a polygon is a segment that connects two non-adjacent vertices of the polygon.
Diagonals of a polygon form triangles inside the polygon. The GMAT often uses the triangles formed inside quadrilaterals by the diagonals.
Maximum and Minimum
The GMAT often asks for the maximum or minimum perimeter or area. Maximum area is when a triangle or quadrilateral has perpendicular sides and right angles. Squares and 45° : 45° : 90° triangles have the maximum area for the smallest perimeter.
V. Circles
Circles
circle is the set of all points in a plane that are the same distance from the center. A circle has 360°.
Basics
A segment from the center of the circle to the circle is called the radius r.
A segment that goes across a circle through the center is called the diameter d. A diameter of a circle is twice the radius.
The circumference of a circle is the distance around the circle. The formula for the circumference of a circle is C = 2πr = πd.
The formula for the area of a circle is A = πr2.
Central Angles
central angle has its vertex at the center of a circle, so it is formed by two radii.
An arc is a piece of a circle. The measure of a central angle equals the measure of the intercepted arc.
sector is a “piece of a pie” created by a central angle and its arc. A sector has perimeter and area.
Inscribed Angles
An inscribed angle has its vertex on the circle and is formed by two chords. The measure of an inscribed angle equals half the measure of the intercepted arc.
Inscribed Polygons
An inscribed polygon is formed by chords so it has its vertices on the circle.
The GMAT often uses inscribed polygons. You need to be able to combine the information about circles, angles and polygons.
Tangents to Circles
tangent to a circle is a line that touches the circle in just one point. A tangent is perpendicular to a diameter.
Angles formed by a chord and a tangent are half the measure of the arc.
Tangents from a common exterior point are congruent. The measure of the angle formed by the tangents is half the difference between the measures of the two arcs.
VI. Solids
Solids
Rectangular Prisms
rectangular prism, or rectangular solid, is basically a box. It has 6 sides. All the sides, or faces, are rectangles. Opposite sides are congruent. A special rectangular prism is a cube. All the faces are squares.
The surface area, SA, is the sum of the areas of each pair of congruent sides.
To find the volume of a rectangular prism, multiply the three lengths.
V = length × width × height = lwh
Cylinders
cylinder is like a can. It has circles as the top and base, and straight sides.
• SA = 2 circles + rectangle = 2(πr2) + 2πrh
• V = πr2h
Spheres
sphere is like a ball. The formulas for circumference, surface and volume all use the radius.
• C = 2πr = πd
• SA = 4πr2
• V = (\dfrac{4}{3}) πr3
VII. Coordinate Geometry
Coordinate Geometry
coordinate plane is formed by the intersection of a horizontal number line, called the x-axis, and a vertical number line, called the y-axis. The number lines intersect at zero at the point called the origin.
Slope of a Line
The slope of a linem, is its slant. The slope can be used to give the rate of change.
The formula to find the slope of the line between two points (x1y1) and (x2y2) is
slope = = \dfrac{\,rise\,}{run} = \dfrac{change\,\, in\,\, \textit{y}}{change\,\, in\,\, \textit{x}}
= \dfrac{{\textit{y}}_{\displaystyle{2}} \,-\, \textit{y}_{\displaystyle{1}}}{\textit{x}_{\displaystyle{2}} \,-\, \textit{x}_{\displaystyle{1}}}
Graphing a Line
A graph of the equation Ax + By = C is a line.
There are two ways to graph a line:
• Slope-intercept form, y = mx + b where m is the slope and the point (0, b) is the -intercept where the line crosses the y-axis.
• Graph using the intercepts.
Writing the Equation of a Line
An equation of a line can be found from two points, or from one point and the slope.
For the point (x1y1) and the slope m, the point-slope formula is
y – y1 = m(x – x1).
To write an equation from two points (x1y1) and (x2y2), first find the slope of the line.
slope = = \dfrac{{\textit{y}}_{\displaystyle{2}} \,-\, \textit{y}_{\displaystyle{1}}}{\textit{x}_{\displaystyle{2}} \,-\, \textit{x}_{\displaystyle{1}}}
Then substitute the slope m and one of the points into the point-slope formula.
The Distance Between Two Points
The distance between two points can be found using the Pythagorean theorem.
(horizontal distance )2 + (vertical distance )2 = (distance)2
Transformations
There are 3 common transformations that move or change a figure on the coordinate plane:
• translation moves a point or figure some distance up or down, left or right.
• reflection acts like a mirror over a line of reflection.
• dilation makes a similar figure that is bigger or smaller.
Sample Questions
Review
Best viewed in landscape mode
6 questions with video explanations
100 seconds per question |
# Ed Sheeran And The Stars (11/01/2019)
How will Ed Sheeran do on 11/01/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is not at all guaranteed – take it with a grain of salt. I will first find the destiny number for Ed Sheeran, and then something similar to the life path number, which we will calculate for today (11/01/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology experts.
PATH NUMBER FOR 11/01/2019: We will take the month (11), the day (01) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 01 we do 0 + 1 = 1. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 1 + 12 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 11/01/2019.
DESTINY NUMBER FOR Ed Sheeran: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Ed Sheeran we have the letters E (5), d (4), S (1), h (8), e (5), e (5), r (9), a (1) and n (5). Adding all of that up (yes, this can get tedious) gives 43. This still isn’t a single-digit number, so we will add its digits together again: 4 + 3 = 7. Now we have a single-digit number: 7 is the destiny number for Ed Sheeran.
CONCLUSION: The difference between the path number for today (6) and destiny number for Ed Sheeran (7) is 1. That is smaller than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But this is just a shallow analysis! As mentioned earlier, this is for entertainment purposes only. If you want a forecast that we do recommend taking seriously, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
# Class 8 Maths MCQ – Mensuration – Surface Area of a Cube, Cuboid and Cylinder
This set of Class 8 Maths Chapter 11 Multiple Choice Questions & Answers (MCQs) focuses on “Mensuration – Surface Area of a Cube, Cuboid and Cylinder”.
1. Find the surface area of the given cuboid.
a) 44 cm2
b) 22 cm2
c) 40 cm2
d) 88 cm2
Explanation: Surface area cuboid = 2(lb + bh + lh) = 2(6 × 4 + 4 × 2 + 2 × 6) = 88 cm2.
2. What is the surface area of the given cube?
a) 46 cm2
b) 96 cm2
c) 48 cm2
d) 92 cm2
Explanation: Surface area cube = 6 a2 = 6 (42) = 96 cm2.
3. Find the surface area of the given cylinder.
a) 113.14 cm2
b) 113.16 cm2
c) 112.16 cm2
d) 112.14 cm2
Explanation: Curved Surface area of cylinder = 2πrh + 2πr2 = 2πr(r + h) = 2 × $$\frac{22}{7}$$ × 2(2 + 7)
= 113.14 cm2.
4. A gift in the form of cube has to be wrapped with gift paper whose external measures are 15 cm. Find the area of gift paper needed.
a) 1250 cm2
b) 1354 cm2
c) 1350 cm2
d) 1150 cm2
Explanation: Surface area cube = 6 a2 = 6 (152) = 1350 cm2.
5. The internal measurement of a cuboidal room 15m × 6m × 4m.The room has to be painted along with the ceiling. If cost of painting is Rs 17 per m2, Find the total cost to be paid.
a) Rs 4916
b) Rs 3916
c) Rs 4686
d) Rs 4386
Explanation: Surface area cuboid = 2(lb + bh + lh)
The floor of the room will not be painted.
∴ Surface area to be painted = 2(lb + bh + lh) – lb
= 2(15 × 6 + 4 × 6 + 4 × 15) – 15 × 6 = 258 m2
Total cost to be paid = 258 × 17 = Rs 4386.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!
6. A building has 6 cylindrical pillars whose radius is 2m and height is 12m. Each pillar has to be painted and the cost of painting is Rs 22 per meter sq. Find the cost of painting the curved surface area of the 6 pillars.
a) Rs. 3872
b) Rs. 13232
c) Rs. 7744
d) Rs. 23232
Explanation: Curved Surface area of cylinder = 2πrh + 2πr2 = 2πr(r + h) = 2 × $$\frac{22}{7}$$ × 2(2 + 12)
= 176 m2
The total cost of painting 1 pillar = 176 × 22 = Rs. 3872.
The total cost of painting 6 pillars = 3872 × 6 = Rs. 23232.
7. Find the side of the cube whose surface area is 384m2.
a) 5 m
b) 6 m
c) 8 m
d) 7 m
Explanation: Surface area cube = 6 a2
⇒ 384 = 6 (a2)
⇒ a = 8 m.
8. If the surface area of a cuboid is 108 cm2, find the height of the cuboid. The length and breadth the cuboid is 6 and 3 cm respectively.
a) 8 cm
b) 4 cm
c) 2cm
d) 6 cm
Explanation: Surface area cuboid = 2(lb + bh + lh)
⇒ 108 = 2(3 × 6 + h × 6 + h × 3)
⇒ h = 4 cm.
9. The curved surface of the cylinder is 88 cm2. The height of the cylinder is 5 m, what is the radius of the cylinder?
a) 6 cm
b) 4 cm
c) 1 cm
d) 2 cm
Explanation: Curved Surface area of cylinder = 2πrh + 2πr2 = 2πr(r + h)
⇒ 88 = 2 × $$\frac{22}{7}$$ × r(r + 5)
⇒ $$\frac{88×7}{2×22}$$ = r(r+5).
⇒ 14 = r (r+5)
⇒ r = 2 cm.
10. Match the pairs.
A B
1)
a) 96 cm2
2)
b) 132 cm2
3)
c) 28 cm2
a) 1-c, 2-a, 3-b
b) 1-c, 2-a, 3-a
c) 1-c, 2-b, 3-b
d) 1-c, 2-b, 3-a
Explanation: Surface area cuboid = 2(lb + bh + lh) = 2(1 × 4 + 4 × 2 + 2 × 1) = 28 cm2
Surface area cube = 6 a2 = 6 (42) = 96 cm2
Curved Surface area of cylinder = 2πrh + 2πr2 = 2πr(r + h) = 2 × $$\frac{22}{7}$$ × 3(3 + 4)
= 132 cm2.
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
To practice all chapters and topics of class 8 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] |
# How do you solve 2x^2 + 10x = 0?
Jan 24, 2017
$x = 0$, or $x = - 5$.
#### Explanation:
Typically you would use the quadratic formula, using $a , b , c$ as the coefficients of ${x}^{2}$, $x$, and the constant respectively, to find that
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
However, notice that the constant is missing. In those cases, you can factor out a $x$, to transform the expression into a product that equals $0$:
$2 {x}^{2} + 10 x = 0 \implies x \left(2 x + 10\right) = 0$
When a product is equal to $0$, at least one of its factors must be $0$, or
$a b = 0 \implies a = 0$ or $b = 0$.
So in our case, either
$x = 0$, or
$2 x + 10 = 0 \implies 2 x = - 10 \implies x = - 5$
Therefore, the roots to this equation, are $0$ and $- 5$.
Jan 24, 2017
Factor $x$ out of the expression and you get $x \left(2 x + 10\right) = 0$
This can be true only if $x = 0$ or $x = - 5$
#### Explanation:
Remember this technique for those equations that have no constant term ($c = 0$ in terms of the standard $a {x}^{2} + b x + c = 0$ form).
Since both the remaining terms have $x$ in them, you factor this out, and immediately one root must be $x = 0$.
Solve the monomial that remains in the product by setting it equal to zero. The idea is that the entire expression can only equal zero if one of the two factors $x$ or $\left(2 x + 10\right)$ is equal to zero. |
# Complex Numbers Section 0.7. What if it isnt Real?? We have found the square root of a positive number like = 4, Previously when asked to find the square.
## Presentation on theme: "Complex Numbers Section 0.7. What if it isnt Real?? We have found the square root of a positive number like = 4, Previously when asked to find the square."— Presentation transcript:
Complex Numbers Section 0.7
What if it isnt Real?? We have found the square root of a positive number like = 4, Previously when asked to find the square root of a negative number like we said there is not a real solution. To find the square root of a negative number we need to learn about complex numbers
Imaginary unit The imaginary unit is represented by What would i² be??
Simplify the following
This can not be simplified any further. Your solution is a complex number that contains a real part (the 7) and an imaginary part (the 6i).
Defining a Complex Number Complex numbers in standard form are written a + bi a is the real part of the complex number and bi as the imaginary part of the solution. If a = 0 then our complex number will only have the imaginary part (bi) and is called a pure imaginary number. Imaginary Number example: Complex Number example:
Adding and Subtracting Complex Numbers To add and subtract, simply treat the i like a typical variable.
Adding and subtracting complex numbers.
Multiplying complex numbers Always write in the form a + bi (real part first, imaginary second)
Multiply (2 + 3i)(2 – i) 4 + 4i – 3(-1) 4 + 4i + 3 7 + 4i
Complex Conjugate The product of complex conjugates is a real number (imaginary part will be gone) (a + bi) and (a – bi) are conjugates. (a + bi)(a – bi) = a² - abi + abi - b²i² =a² - b²(-1) =a² + b²
z = 2 + 4i Find z ( the conjugate of z) and then multiply z times z z = 2 – 4i zz = (2 + 4i)(2 – 4i) = 4 – 16 i² = 4 + 16 = 20
Write the quotient in standard form Multiply numerator and denominator by conjugate Simplify remembering i² = -1 Write in standard form a + b = a + b c c c
Write in Standard Form
Powers for i 1 1
Download ppt "Complex Numbers Section 0.7. What if it isnt Real?? We have found the square root of a positive number like = 4, Previously when asked to find the square."
Similar presentations |
Holt.Blue
Back To Class Notes Menu
The Normal Distributions Worksheet
Today we learn about one of the most used distributions in all of probability and statistics: the Normal Distribution.
The full truth of this statement will become clearer as we proceed.
Some Places You Might See Normal Distributions
Biology: sizes, heights, weights within a species (from trees to humans). Blood pressure in human adults.
Manufacturing: sizes and weights of mass produced items (diameter sizes of ball bearings, the amount of liquid in a can of soda, the weight of a package of oreos). Measurement errors in all of the above.
Social Sciences: Test scores (IQ, ACT/SAT).
Various: Yearly precipitation in certain parts of the world, the position of a particle that experiences diffusion.
This is all to say that normal distributions are EVERYWHERE!
Normal Distribution Basics: The normal distribution is a symmetric, bell-shaped distribution.
Example
Suppose we randomly sample from the population of women aged $20$ to $29$ and measure their height, and record the data in a histogram.
The Big Idea
Bin Width=
Probability Density Curves
Our histogram can be approximated by a smooth curve. This curve is the probability density curve.
Probability Density Curves
Idea: Suppose we want to estimate the probability that a randomly chosen young woman's height will fall between $68$ and $70$ inches. We could do it using our data.
Probability Density Curves
OR we could compute the probability with our probability density curve.
$\longrightarrow$
$$\mbox{Probability}=\mbox{Area under the Curve}$$
Density Curve Basics
Question: If we added up all the percentages of the bars of our histogram, what percentage should we end up with?
Density Curve Basics
The total area under any Probability Density Curve is $1.$ This may be interpreted as $100\%$ of our data is represented by the curve's area.
$$\mbox{Total Area}=1$$
Normal Distribution Basics
Every normal distribution is determined by two numbers: the mean $\mu$, and the standard deviation $\sigma$.
$\mu$ tells us where the center of our distribution is.
$\sigma$ tells us how wide, or "spread out," the distribution is.
Normal Distribution Basics: Examples of Normal Distributions
A normal distribution with mean $\mu$ and standard deviation $\sigma$ is denoted as $$N(\mu,\sigma).$$
Normal Distribution Basics: The $68\mbox{-}95\mbox{-}99.7$ Rule
For ANY normal distribution:
About $68\%$ of observations lie within $1$ standard deviation of the mean.
About $95\%$ of observations lie within $2$ standard deviations of the mean.
About $99.7\%$ of observations lie within $3$ standard deviations of the mean.
Normal Distribution Basics: The $68\mbox{-}95\mbox{-}99.7$ Rule
Example: Iowa Test Scores. The Iowa Assessments are standardized tests provided as a service to schools by of the University of Iowa.
Question: Approximately, what is $\mu$? Approximately, what is $\sigma$?
Example: Iowa Test Scores.
The mean of the Iowa test score data is $6.84,$ and the standard deviation is $1.55.$ The normal distribution $N(6.84,1.55)$ models this data set very well.
Example: Iowa Test Scores & The $68\mbox{-}95\mbox{-}99.7$ Rule
Example: Iowa Test Scores
Question: What percentage of students scored 9 or below?
For a normal distribution $N(\mu, \sigma),$ use the following guide to find the cumulative area:
1. Compute the value $z=\displaystyle \frac{x-\mu}{\sigma}$ where $x$ is your data point.
2. Find the value of $z$ in this table.
3. The number corresponding to this value of $z$ is the cumulative area.
We say that $z$ is the standardized value of $x$. This is also called a $z$-score. It is the number of standard deviations above or below the mean.
Example: Iowa Test Scores. To find $\mbox{Cumulative Area below 9}...$
$\displaystyle \longrightarrow$ Original: $N(6.84,1.55)$ Standard Normal $N(0,1)$
$$\displaystyle z=\frac{x-\mu}{\sigma}=\frac{9-6.84}{1.55}=1.39$$
Example: Iowa Test Scores.
$\displaystyle \longrightarrow$ Original: $N(6.84,1.55)$ Standard Normal $N(0,1)$
$$\mbox{Cumulative Area Below 9}=\mbox{Table}(1.39)=0.9177$$
Thus, $91.77\%$ of students scored below $9$ on the Iowa Vocabulary Test.
Put another way, the probability that a randomly chosen student scores $9$ or below is $0.9177.$
Example: Iowa Test Scores
Question: What percentage of students scored above $9?$
$\begin{array}{l} \mbox{Cumulative Area Above 9}\\ =1-\mbox{Cumulative Area Below 9}\\ \approx 1-0.9177\\ = 0.0823 \end{array}$
So about $8.23\%$ of students scored above $9$ on the Iowa Vocabulary Test.
Example: Iowa Test Scores
Question: What is the probability a randomly chosen student scored between $4$ and $9?$
Example: Iowa Test Scores.
$\displaystyle \longrightarrow$ Original: $N(6.84,1.55)$ Standard Normal $N(0,1)$
$\mbox{Area Between 4 and 9}=\mbox{Area Below 9}-\mbox{Area Below 4}$
$=\mbox{Table}(1.39)-\mbox{Table}(-1.83)=0.9177-0.0336=0.8841.$
Interpretation: About $88.41\%$ of students scored between $4$ and $9$ on the Iowa Vocabulary Test.
OR, the probability a randomly chosen student scored between $4$ and $9$ is about $0.8841.$ |
# Prove 2^n > n by induction
## Homework Statement
prove 2^n > n by induction
## The Attempt at a Solution
In my math class we start off assuming 2^n > n is true for n=k.
Then we try to prove that when n=k+1 the inequality is true. So,I
start off going, 2^(k+1) > (k+1) which is equivalent to 2*2^k > (k+1)
which is then equivalent to 2^k+2^k > (k+1) then my math teacher said
to make the next step which is, since we assumed that 2^k > k then 2^k
+ 2^k > k+k and then he said that k+k > (k+1) and that was the end of
the proof. I do not get this last part at all, since I thought that
when you prove something by induction, it's going to be proven for all
numbers but we only proved that the inequality is true for k>1
Related Precalculus Mathematics Homework Help News on Phys.org
There are all together 3 steps to the mathematical induction. You have left out the first step, namely showing the inequality holds true for the case when n = 1. Now, when n = 1, 2^1 = 2 > 1.
So he case is true when n = 1.
Then you proceed to the case where n = k, and finally n = k+1.
So when your k = 1 is true, (because you have shown that the inequality holds when n = k = 1), your k + 1 = 2 is also true.
When k = 2 is true, k + 1 = 3 is also true. And the rest is just like the dominoes effect.
I might be wrong, but that's my understanding regarding mathematical induction.
Mark44
Mentor
There are all together 3 steps to the mathematical induction. You have left out the first step, namely showing the inequality holds true for the case when n = 1. Now, when n = 1, 2^1 = 2 > 1.
This is usually called the base case. It doesn't have to be for n = 1, but often is.
So he case is true when n = 1.
Then you proceed to the case where n = k, and finally n = k+1.
The idea is that you assume that your statement is true for n = k (the induction hypothesis), and then use that to show that the statement is also true for n = k + 1.
So when your k = 1 is true, (because you have shown that the inequality holds when n = k = 1), your k + 1 = 2 is also true.
When k = 2 is true, k + 1 = 3 is also true. And the rest is just like the dominoes effect.
I might be wrong, but that's my understanding regarding mathematical induction.
"The idea is that you assume that your statement is true for n = k (the induction hypothesis), and then use that to show that the statement is also true for n = k + 1."
Yes, I understand that this is the objective, but I am wondering if there is a general method of doing this and how you would you do it for 2^(k+1) > (k+1)
jbunniii
Homework Helper
Gold Member
"The idea is that you assume that your statement is true for n = k (the induction hypothesis), and then use that to show that the statement is also true for n = k + 1."
Yes, I understand that this is the objective, but I am wondering if there is a general method of doing this and how you would you do it for 2^(k+1) > (k+1)
Well, if it's true for n = k then
$$2^k > k$$
Try multiplying both sides by something appropriate.
jbunniii
$$2^n = (1 + 1)^n$$ |
# What is distance between parallel planes?
## What is distance between parallel planes?
The distance between two parallel planes, D=|ax1+by1+cz1+d|√a2+b2+c2. Note: The distance between the given two parallel planes is nothing but the shortest distance between the two plane surfaces.
What is the perpendicular distance between two profile planes?
In geometry, the perpendicular distance between two objects is the distance from one to the other, measured along a line that is perpendicular to one or both. Particular instances include: Distance from a point to a line, for the perpendicular distance from a point to a line in two-dimensional space.
What is the formula of distance between two parallel lines?
The distance between two parallel lines is given by d = |c1-c2|/√(a2+b2).
### What is perpendicular distance formula?
The perpendicular distance is the shortest distance between a point and a line. The perpendicular distance, 𝐷 , between the point 𝑃 ( 𝑥 , 𝑦 ) and the line 𝐿 : 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 = 0 is given by 𝐷 = | 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 | √ 𝑎 + 𝑏 .
What is the perpendicular distance of the point A 6 7 from Y axis?
Step-by-step explanation: the perpendicular distance of a point P (6 , 7) from y axis is -1.
How do you find the perpendicular distance between two lines?
Basically, to find the distance between two perpendicular lines given two points, all we have to do is find the distance from the points to the intersection of the lines, then use the Pythagorean Theorem to find the distance between the two points.
#### How do you find the perpendicular distance between a point and a plane?
We call this the perpendicular distance between the point and the plane because 𝑃 𝑄 is perpendicular to the plane. We could find this distance by finding the coordinates of 𝑄 ; however, there is an easier method. To calculate this distance, we will start by setting ∠ 𝑅 𝑃 𝑄 = 𝜃 and | | 𝑃 𝑄 | | = 𝐷 .
How do you find a perpendicular plane?
If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross product of two vectors produces a vector perpendicular to both, we will use the cross product of →v1 and →v2 to find a vector →u perpendicular to the plane containing them.
How do you find the perpendicular distance from a point to a plane?
## How do you find the perpendicular distance from the origin?
The perpendicular distance from origin to the normal at any point to the curve x=a(cosθ+θsinθ). y=a(sinθ−θcosθ)
What is the definition of parallel planes?
In geometry, parallel lines are lines in a plane which do not meet; that is, two lines in a plane that do not intersect or touch each other at any point are said to be parallel. By extension, a line and a plane, or two planes, in three-dimensional Euclidean space that do not share a point are said to be parallel.
When are planes parallel?
When two planes are perpendicular to the same line, they are parallel planes. When a plane intersects two parallel planes, the intersection is two parallel lines.
### What are parallel lines and angles?
Angles and parallel lines. When two lines intersect they form two pairs of opposite angles, A + C and B + D. Another word for opposite angles are vertical angles. Vertical angles are always congruent, which means that they are equal. Adjacent angles are angles that come out of the same vertex. |
Home What Middle Graders Have To Prepare For Math Competitions
# What Middle Graders Have To Prepare For Math Competitions
Mathematics is introduced to middle school grade through algebra (number systems and number operations), and also through fundamental geometric figures and their features. Any math competition available will test middle graders on the following subjects.
## Algebra and Counting
### Number systems
Mathematics operates, in algebra and geometry, with number systems. There are 6 number systems:
#### Naturals
Indicated by “N”, natural numbers are counting numbers, from 1 to infinity. The whole numbers are all the natural numbers and 0.
The sum and the product of any two natural numbers is a natural number, but not necessarily in subtraction or division.
#### Integers
The integer numbers are the set of real numbers consisting of the natural numbers, their additive inverses and 0. The set of integers is indicated by “J” or “Z”,
The sum, product, and difference of any two integers is also an integer, except division.
#### Rationals
These are the numbers which can be expressed as a fraction between two integers. It will have the form of a fraction: ab or a ratio a:b.
Any of the four basic operations can result in a rational number if we don’t divide them by 0. Any number divided by 0 is not defined. All decimals are also rational.
#### Irrationals
At the opposite corner, there are irrational numbers. These are the numbers that cannot be written as a fraction or ratios.
Some famous irrational numbers are polynomial equations, golden ratios, and transcendental numbers, such as 𝝿 and e
#### Reals
Real numbers are all rational and irrational numbers, thus all the numbers on a line. There is a countable infinity, and an uncountable infinity, most commonly used in set theory.
#### Complex
Going one step further, complex numbers are indicated by the following set:
a + bi, where a and b are real numbers, and ”i” represents the iota number, an imaginary unit.
There are even more complex numbers, known as quaternions, but these are hyper-complex numbers and are not used in math competitions for middle graders.
In order to succeed at math competitions, students have to master the number systems and the four basic operations, since these are used fundamentally in creating more complicated mathematical calculus.
## Circles and Triangles
At the basis of geometry, students should be familiar with the most common shapes, such as the characteristics of a line
#### The geometry of Circles
When it comes to the study of circles and solving geometrical problems, math competitions usually refer to angle-based exercises. Middle school students should learn about:
• Classification of circles
• Measurement of circles
• Line, rays, and segments of a circle
• Angles, Arcs, Sectors
• Circle Theorems
#### The geometry of Triangles
The most common geometrical problems in math competitions include triangles. So, refine your knowledge of triangles by focusing on:
• Classification of triangles
• Triangle inequalities
• Parts and area of a triangle
• The Pythagorean Theorem
• Congruency of triangles
• Basic trigonometry
It’s recommended that middle graders don’t stop only at circles and triangles. One example of a geometric question at a math competition called Math Kangaroo, for middle school students, in 2022, was:
A water tank with a rectangular base has the dimensions 1 m×2 m×4 m. It contains water to a depth of 25 cm, as shown in the first picture. The tank is turned so that a 1 m × 2 m face becomes the base, as shown in the second picture. What is the depth of the water now?
(A) 25 cm (B) 50 cm (C) 75 cm (D) 1 m (E) 1.25 m
Image:
This shows us that math competitions for middle school students also include exercises on volume, and calculating parts of various geometric figures: square, rectangle, circle, triangle, and trapezoid (polygons).
## At the Online Math Center…
We make math competitions easier by providing tailored individual tutoring with an emphasis on understanding the logic between math operations and applying them from simple to more complex situations. We have a wide range of programs dedicated to studying mathematics, so reach out to us and start your math training.
## Step Two
Let us know how to contact you. One of our representatives will get back to you shortly.
## Step Two
Awesome! We need to get in touch with your parent or guardian for further discussion. Please check in with them before filling out the form below with their information.
## Step One
It only takes two steps to schedule an appointment with an OMC representative.
To get started, please tell us who you are:
## Dear Parent!
Thank you for placing your trust in us to educate your child!
Congratulations on joining the Online Math Center!
See you in class.
## Step One
It only takes two steps to schedule a free lesson with an OMC representative.
To get started, please tell us who you are:
## Payment
0%
0%
Payment amount: \$323.00
Secure payments by Stripe.
We do not collect your credit card on our servers.
## Thank you!
Our manager will be in contact with you shortly. |
# Different Types of Triangles: Definition, Characteristics, Shapes, & Real-life Application
If we talk about triangles, we all are aware of them as a shape, but do you all remember what the different characteristics, actual definitions, and real-life applications of triangles are? Do you remember how many types of triangles are there? In case you do not remember, then after reading this blog, you will get a clear picture of triangles as a whole.
When we define a triangle, it is a closed polygon having three sides that are equal and unequal. If we talk about how a triangle is made, it consists of three sides, three vertexes, and three angles, which can or cannot be the same. Therefore, a triangle is divided on the basis of angles and sides; we will discuss it later.
Firstly, we will discuss the essentials of triangles: vertex and angles.
The point where the two straight lines are known as the vertex.
The angle is produced between the two sides. As a part of geometry, the triangle plays a crucial role in developing the fundamentals of mathematics.
Triangle features, like the Pythagorean theorem and trigonometry, are essential to certain fundamental ideas.
Furthermore, a triangle generally consists of three sides that connect each other. The length of each side might be the same as well as might be different, but it is also not possible that the longest side of a triangle is greater or equal to the other sides.
In addition, it is also true that a triangle has three internal and exterior angles. Combining these three interior angles always sums up to 180 degrees, and the same goes for exterior angles, as their sum is also 180 degrees.
## The shape of Triangle
A closed, two-dimensional shape is a triangle. It is a polygon with three sides. All sides of the triangle are made up of straight lines. The intersection of two straight lines forms a vertex. And do you know that a triangle has its vertices, and each vertices combined forms an angle?
## Angles of Triangle
A triangle consists of three angles. And the triangle's two sides meet at the triangle's vertices to produce these angles. When we add the three inner angles of a triangle, then it is always 180 degrees.
If we stretch out the triangle outside, then the side length of the triangle creates an angle. The total of a triangle's contiguous interior and exterior angles is supplementary.
Let's assume that the internal angles of a triangle are ∠1, ∠2, and ∠3. The three external angles that are created when the triangle's sides are extended outward are ∠4, ∠5, and ∠6, and they follow each other in the order of ∠1, ∠2, and ∠3, respectively.
Hence,
∠1 + ∠4 = 180° ……(i)
∠2 + ∠5 = 180° …...(ii)
∠3 + ∠6 = 180° ….(iii)
If we add the above three equations, we get.
∠1+∠2+∠3+∠4+∠5+∠6 = 180° + 180° + 180°
Now, by angle sum property, we know,
∠1+∠2+∠3 = 180°
Therefore,
180 + ∠4+∠5+∠6 = 180° + 180° + 180°
∠4+∠5+∠6 = 360°
Thus, the triangle's outside angles add up to 360 degrees, as inclinations by the equation above.
## Characteristics of the Triangle
There are some characteristics that set each and every form of mathematics apart. Let's talk about a few triangle qualities now.
1. A triangle consists of three angles and three sides.
2. When we add up all the angles of a triangle, it always equals 180 degrees.
3. Any triangles’ outside angles add up to 360 degrees every time.
4. The sum of each exterior and interior angle equals 180 or supplements to each other.
5. The third side of any triangle is always shorter than the sum of the lengths of the other two sides. In the same way, the difference in length between any two of a triangle's sides is always smaller than the difference in length between the third side.
6. In every triangle, the shortest side is always opposite the smallest angle inside the triangle. Likewise, the longest side is always across from the largest angle inside the triangle.
## Different Types of Triangles
Since a triangle is a closed figure, it may take on several forms, and the angle formed by any two of its sides describes a shape. Now, let's explore the various kinds of triangles.
But the question is how would one determine which type of triangle is it? Check the largest angle in each triangle and see if it is more than, less than, or equal to 90 degrees. Still, if you are confused about how to identify each one, do not worry, we are here to help you out. Just search for “Do my Assignment” and connect with us. We have a team of highly competent experts who are here to aid you in understanding this concept. Our Experts have years of experience in providing assignment help successfully to students.
## What is an Acute Angle Triangle?
An acute-angled triangle, a fundamental geometric shape, is defined by its interior angles, each of which measures less than 90 degrees. This type of triangle showcases a characteristic in which all three angles are acute, contributing to the sum of the interior angles amounting to 180 degrees. As a result, the sides of the acute-angled triangle are shorter than those of other types, demonstrating a balanced distribution of angles that allows for a visually compact and pointed structure.
### Important characteristics of an acute triangle:
Let's know what the important characteristics of this type of triangle are.
• Every internal angle of an acute triangle is less than ninety degrees, and the lengths of the sides vary.
• An acute triangle also has the fascinating property that any line drawn from its base to its opposite corner will always be perpendicular to the base.
## What is a Right-Angled Triangle?
A right-angled triangle, often referred to as a right triangle or a 90-degree triangle, is a type of triangle where one of its interior angles measures exactly 90 degrees. This triangle holds significance in the field of trigonometry.
### Characterizes of Right-Angle Triangle
Once you are familiar with the definition, it's time for you to give your attention to the various characteristics of a right-angle triangle.
• Only one angle is a 90° angle, or a right angle in this type of triangle.
• The hypotenuse, which is the longest side, is always in opposition to the 90° angle.
• The sum of two angles that are apart from 90 degrees is 90 degrees.
• The sides that are adjacent to 90-degree angle are called perpendicular and base of triangle.
• A right-angle triangle's area is equal to half of (base x perpendicular).
• When a line is drawn from the right angle to the hypotenuse, it creates three similar triangles.
• A circle passing through all three corners has a radius half the length of the hypotenuse.
### Shape of Right Triangle
Now, the question arises, what are the shapes of right-angle triangles? Three sides enclose together to form a right-angle triangle, out of which one side is straight and is known as the height or leg of this type of triangle.
## What is an obtuse angle?
An obtuse triangle is a triangle whose angle is more than ninety degrees. It is sometimes referred to as an obtuse-angled triangle. But as it stated that the sum of all interior angles of a triangle is always 180 degrees, the same goes for this triangle. The entire sum of the internal angles of an obtuse triangle is still 180 degrees. This angle type is important in math and shapes, helping us understand how different angles and shapes relate to each other.
As a result, two additional angles in the triangle must be acute, meaning they are both fewer than ninety degrees if one angle in the triangle is obtuse (greater than ninety degrees).
### Characteristics of Obtuse Angled Triangle
Have a glimpse of the various characteristics of the Obtuse Angled Triangle.
1. The sum of the two angles, excluding the obtuse angle, is less than ninety degrees.
2. The longest side of the triangle is opposite the obtuse angle.
3. An obtuse triangle is characterized by a single obtuse angle, with the remaining two acute angles.
4. The point where all the perpendicular bisectors drawn from each side meets is the circumcenter, and this circumcenter, in the case of an obtuse-angled triangle, meets outside the triangle.
5. The points of concurrency—the Circumcenter and the Orthocenter—are positioned outwards the obtuse triangle, whereas the Centroid and Incenter are inwards the triangle.
## What is an Equilateral Triangle?
In geometry, when you talk about equilateral triangles, this type of triangle has equal sides, which results in the angles that are opposite to these sides being equal. This property deduces the conclusion that each angle of an equiangular triangle seems to be 60 degrees.
In line with the angle sum property of triangles, the total of all three angles in an equilateral triangle amounts to 180 degrees (60° + 60° + 60° = 180°). Similar to other types of triangles, the equilateral triangle possesses its own set of formulas for area, perimeter, and height, which can be explored in greater detail.
### Characteristics of Equilateral Triangle
Get a hand on the various characteristics of an equilateral triangle.
• All three sides are of equal length.
• All three angles are the same, each measuring 60 degrees.
• It's a three-sided shape known as a regular polygon.
• Drawing a line from any corner to the opposite side splits the angle into two equal parts, each measuring 30 degrees.
• The point where the lines intersect is both the center of gravity and the intersection of the altitudes.
• This triangle consists of a median, angle bisector, and altitude, indicating that all sides are the same length.
• The area of the equilateral triangle is (√3a) / 4, where 'a' is the length of one side.
• The perimeter of an equilateral triangle is a multiple of 3 of its one side.
## Shape of Equilateral Triangle
When we talked about the shape of this type of triangle, it is regular in nature. Moreover, the term equilateral is derived from the combination of two words that is “equi’, which means equal, and “Lateral,” which means sides.
A regular polygon has all its sides equal. If we are talking about an equilateral triangle, then it also has all three sides equal. Hence, an equilateral triangle is basically a type of regular polygon having all three sides equal.
Let's assume that ABC is an equilateral triangle having AB, BC, and AC as their sides; so, as per the definition, each side of the triangle is equal. This means that.
AB = BC = AC
And
∠A = ∠B = ∠C = 60°
## What is the Isosceles Triangle?
An isosceles triangle is a triangle in which the two sides are the same length. It means that in this kind of triangle, the two angles facing these equal sides are also the same. So, if you have a triangle ABC, and sides AB and AC are equal, then you have an isosceles triangle, and it also means that angle B is equal to angle C. A theorem describes this idea: if two sides of a triangle are the same length, then the angles across from them are also the same.
### Characteristics of Isosceles Triangle
Here are the characteristics of the isosceles triangle through which you will get more clarity about this type of triangle.
1. As discussed above, the two sides of this triangle are equal, and the unequal side is, therefore, known as the triangle's base.
2. As the two sides of this triangle are equal, therefore, the opposite angles to these sides are always equal.
3. The altitude of this triangle is measured, or we can say drawn from the unequal side, which is the triangle base to the topmost vertex of the triangle.
4. Another triangle is called a right-angled triangle, has two equal angles, and the third one equals 90 degrees.
### Types of Isosceles Triangle
Generally, the isosceles triangle is classified into different types, namely:
#### Isosceles Acute Triangle:
Isosceles triangles have a line of symmetry along the perpendicular bisector of their base. Depending on the angle formed between the two equal-length legs, they can be classified as acute, right, or obtuse. An isosceles triangle is considered acute when the two angles opposite the sides are equal and less than 90 degrees.
#### Isosceles Right Triangle:
In a right isosceles triangle, two sides are of equal length, with one serving as the perpendicular side and the other as the base. The third side, which is unequal, is referred to as the hypotenuse. This triangle obeys the Pythagoras theorem, where the square of the hypotenuse equals the sum of the squares of the base and perpendicular sides. If the lengths of the equal sides are 'a' and the hypotenuse is 'h,' then h = √2a or h = a√2.
#### Isosceles Obtuse Triangle:
An obtuse triangle is one in which at least one of its angles exceeds 90 degrees (a right angle). Triangles with more than one obtuse angle cannot exist. An isosceles obtuse triangle is a specific case where two sides are of equal length, and one angle is obtuse.
### Isosceles Triangle Theorem
This theorem is very well known in mathematics, and it basically explains one of the most important properties of the isosceles triangle if two sides of this triangle are equal or congruent to each other, then the angles opposite to these sides are also congruent or equal.
This theorem can also be alternately seen as that if two angles of an isosceles are congruent to each other, then the sides opposite to them will also be congruent to each other.
In the above triangle ABC,
Sides AB is equal to AC.
Angle ABC is equal to the angle ACB.
## What is a Scalene Triangle?
Last, if we talk about the scalene triangle, is a type of triangle in which none of its sides are equal to each other, which states that there are no angles in this triangle that are equal to each other.
A scalene triangle is a type of triangle where all three sides have different lengths, and all three angles have different measures. Nevertheless, the sum of all the interior angles always equals 180 degrees, adhering to the fundamental angle sum property of any triangle.
### Characteristics of Scalene Triangle
After definition it is time to learn what are the various characteristics of the scalene triangle.
• All sides have different lengths, with none being equal.
• Each angle has a distinct measure, with no two being equal.
• It lacks any lines of symmetry and point symmetry.
• Its interior angles can be acute, obtuse, or right angles.
• When all angles are acute, the circumscribing circle's center is inside the triangle.
• In a scalene obtuse triangle, the circumcenter is outside the triangle.
• It can be an acute-angled, obtuse-angled, or right-angled triangle.
### Types of Scalene Triangles:
Learn the several types of scalene triangles to have more in-depth knowledge about this type of triangle.
• Acute scalene triangle: The Circumcenter lies within the triangle.
• Obtuse Scalene Triangle: The Circumcenter is positioned outside the triangle.
• Right-angled Scalene Triangle: Circumcenter is at the midpoint of the hypotenuse.
## Real-Life Applications of Different Types of Triangles
We have heard about these triangles in our school life, but do you remember what their uses are? No worries if you are not familiar with this. Just keep reading to learn about the real-life uses of triangles.
So, if you are concerned about that you may get this type of question in your assessment. But you have no proper knowledge about this. Do not worry, we are here to aid you offering the best “assessment help” Services so that you can boost your knowledge accordingly.
Furthermore, Triangles are fundamental geometric shapes, and their applications extend to various aspects of daily life and numerous fields, such as architecture, engineering, and science. Here are real-life applications of different types of triangles:
### Equilateral Triangle:
• Trusses: Equilateral triangles are used in building trusses and frameworks, providing stability and support in structures like bridges and roofs.
• Road Signs: Triangle-shaped Road signs use equilateral triangles to draw attention and convey important information to drivers.
### Isosceles Triangle:
• Pyramids: The faces of a pyramid are often isosceles triangles, as seen in the pyramids of Egypt. These structures have been used historically as tombs and monuments.
• Bridges: Some cable-stayed and suspension bridges use isosceles triangles in their support structures to distribute weight effectively.
### Scalene Triangle:
• Surveying: Surveyors use scalene triangles to measure distances and calculate land areas, ensuring accurate property boundaries and land development.
• Aircraft Design: In the aerospace industry, scalene triangles are employed to create the shape of wings, tail fins, and other components to optimize aerodynamics.
### Right Triangle:
• Carpentry: To guarantee square corners and level, plumb buildings, right triangles are essential.
• Trigonometry: The study of right triangles is fundamental to trigonometry, which is used in construction, navigation, and the resolution of practical angle and distance issues.
### Acute Triangle:
• Art and Design: Acute triangles are utilized in art and design to produce dynamic compositions and designs that have a sense of energy and equilibrium.
• Mechanical Engineering: When developing mechanical parts, where exact angles and dimensions are required for effective functioning, acute triangles are used.
### Obtuse Triangle:
• Geodetic surveying and obtuse triangles may be seen in the process of calculating locations, angles, and distances on the surface of the Earth.
• Navigation: Obtuse triangles are employed in computations for navigational purposes, such as figuring out the positions of celestial bodies.
### Isosceles Right Triangle:
• Construction: To ensure that buildings and other structures are constructed with precise angles and dimensions, isosceles right triangles are used to establish and verify right angles.
• Cable Installation: In electrical and telecommunications operations, these triangles assist in installing cables and wires at right angles to walls and buildings.
To create stable and aesthetically pleasing designs, solve real-world problems, and maintain accuracy in various applications, architects, engineers, artists, and many other professionals need to understand the various types of triangles and their properties.
## Wrapping Up
In summary, knowledge of the numerous triangle types—equilateral, isosceles, scalene, right-angled, acute, and obtuse—lays the groundwork for comprehending their special qualities and uses in various contexts. Triangles with equal sides stand for harmony and balance, whereas those with an odd number of sides emphasise stability and symmetry.
Scalene triangles are known for their adaptability and usefulness, but right-angled triangles are fundamental to the Pythagorean theorem and other geometric concepts. The distinction between acute and obtuse triangles draws attention to the various angles and their use in practical situations.
Understanding the importance of each triangle form allows us to apply this knowledge to various industries, from mathematics and art to architecture and engineering, encouraging creativity and problem-solving across the board.
So, if you are interested in knowing more about this, you can visit audomyassignment.com and search “do my maths homework”. Once you finish this, you will connect with our highly skilled experts who hold Ph.D. degrees in mathematics and are eager to help you. So, why hold back? Connect with us now!
## Related Blog
### Still Confused? Take a look at our Reviews
6984+ Positive Reviews with a rating of 4.9/5
21-04-2023
##### Management
After reading the reviews, I contacted one of the experts of their website and you know what I am quite impressive with services they are providing to us.
Charlie, Order Code: SA**678
18-04-2023
##### Nursing
Much satisfied with the services of this website, there was no payment issue, experts were brilliant and website is much trustworthy.
Perth, Order Code: SA**588
12-04-2023
##### Law
I was too lazy to do essay writings myself. But thanks to the writers of assignments of my assignment services for solving all my queries too soon.
jams, Order Code: SA**658
05-04-2023
##### Tax
They did what they promised. Hat's off to you. Will submit my more works to you soon and services are much appreciated.
Julie, Order Code: SA**663
21-04-2023
##### Accounting
Excellent work also gets high grade and work is done before time. Affordable orders. Unlimited offers. Achieve very good marks. Thank you.
Harry, Order Code: SA**612
28-03-2023
##### Management
Big Thanks! Great Team Work, best leadership quality and much appreciable. Highly recommend to all the students. I must say I am a happy client.
Shannon, Order Code: SA**598
21-03-2023
##### Nursing
I got a very good marks and delivery of all my assignments are on time. Best project and synopsis provider.
Robert, Order Code: SA**587
15-03-2023
##### Psychology
Great enthusiasm by the experts. Very resourceful material they provided to us. I am happy with the services they provided to me.
Steve, Order Code: SA**112
10-03-2023
##### Law
Hello Assignment has the most cordial law assignment help experts in the world. The quality assignments which I got has really made me happy.
Ethan, Order Code: SA**221
18-04-2023
##### Nursing
Much satisfied with the services of this website, there was no payment issue, experts were brilliant and website is much trustworthy.
Perth, Order Code: SA**588
09-03-2023
##### Criminal Law
Thanks a ton to HAH team for their patience and hard work. Their expert was very prompt in all the updations.
Charlie, Order Code: SA**443
04-03-2023
##### Dissertation
Services are quite affordable. Revised synopsis was delivered on time. Format of dissertation was must appreciable.
Perth, Order Code: SA**932
23-02-2023
##### Sociology
Thanks for sending me the synopsis on time and providing me an authentic data that I needed the most.
Danish, Order Code: SA**956
19-02-2023
##### Law
I availed their law case study help twice this year, both the times I got HD grades from my professor. A big thanks to their team.
Tom, Order Code: SA**722
15-02-2023
##### Economics
I did not think i can even pass in this assignment, but i am so glad seeing that I got HD grades in economics assignment.
11-02-2023
##### Diploma
Mary, Order Code: SA**290
08-02-2023
##### Nursing,
Brilliant service provided I am always impressed, always making sure deadlines are met well before due, good quality work ensured
Aneeshah, Order Code: SA**021
05-02-2023
##### IT
Assignment looked good. Thank you for this amazing IT write up
Daljeet, Order Code: SA**332
01-02-2023
##### Maths
Guys, you rock!! A study couldn't be better! Moreover, you have completed my order well before the deadline. Excellent writing
Harshita, Order Code: SA**034
24-01-2023
##### Literature
I cannot believe that Hello Assignment Help provides such high-quality literature assignment help services, at such budget-friendly prices!
##### Emily, Order Code: SA**828
Every Assignment Crafted by Us is 100% Unique |
Pfeiffertheface.com
Discover the world with our lifehacks
What is the Guess and Check method?
What is the Guess and Check method?
“Guess and Check” is a problem-solving strategy that students can use to solve mathematical problems by guessing the answer and then checking that the guess fits the conditions of the problem.
Table of Contents
What are the steps in the process of the guess method?
The GUESS method (Given, Unknown, Equations, Set-up, Solve) is an easy to remember acronym that breaks practice problems into five basic steps. Use the following template for solving your own practice problems.
How do you do checks in math?
The method works like this: For any addition or multiplication problem, take the digits of each number you’re adding or multiplying and add them together. So for instance, if we were adding 218 and 435, we would add 2+1+8=11, and 4+3+5=12. Then, if we get a two-digit answer we’d repeat the process.
What is the easiest strategy in problem solving?
Guess and check is one of the simplest strategies. Anyone can guess an answer. If they can also check that the guess fits the conditions of the problem, then they have mastered guess and check.
What are the 5 steps in solving word problems?
5 Steps to Word Problem Solving
• Identify the Problem. Begin by determining the scenario the problem wants you to solve.
• Gather Information.
• Create an Equation.
• Solve the Problem.
• Verify the Answer.
How can I help my child understand math word problems?
14 Effective Ways to Help Your Students Conquer Math Word Problems
1. Solve word problems regularly.
2. Teach problem-solving routines.
3. Visualize or model the problem.
4. Make sure they identify the actual question.
5. Remove the numbers.
6. Try the CUBES method.
7. Show word problems the LOVE.
8. Consider teaching word problem key words.
What is the grass method?
A good way to solve word problems is by using the method called “GRASS.” GRASS is an acronym for Given, Required, Analysis, Solution, and Statement. You can use GRASS, step by step, to break down a word problem, making it easier to solve.
How do you teach guess and check?
The strategy for the method “Guess and Check” is to guess a solution and then plug the guess back into the problem to see if you get the correct answer. If the answer is too big or too small, make another guess that will get you closer to the goal, and continue guessing until you arrive at the correct solution.
How do you teach students to check their work?
Strategies to Help Students Who Rush Through Their Work
1. 1.) Consider the Reason. Observe your rushing student and take note of why he or she is rushing.
2. 3.) Teach a High-Quality Mindset.
3. 4.) Provide Self-Checking Tools.
4. 6.) The Redo.
5. 8.) Stop Before It Starts.
6. 10.) Work with the Student. |
# NSA's 2018 Puzzle Periodical
Try your hand at this month's problem written by a member of our expert workforce.
# January 2018 Puzzle Periodical - Snow Manipulation
## Can Frosty the Snowman build a new spherical snowman?
Puzzle created by James M., NSA Operations Researcher
Frosty's Idea
Frosty the Snowman wants to create a small snowman friend for himself. The new snowman needs a base, torso, and a head, all three of which should be spheres. The torso should be no larger than the base and the head should be no larger than the torso.
For building material, Frosty has a spherical snowball with a 6 inch radius. Since Frosty likes to keep things simple, he also wants the radius of each of the three pieces to be a positive integer. Can Frosty accomplish this?
Did you solve the scenario - Can Frosty the Snowman build a new spherical snowman?
Can you calculate the spherical snowball radii for the new snowman? The original snowball has a 6 inch radius (shown on the far left).
Yes! The formula for the volume of a sphere with radius r is 43 π r3. Let a, b and c be the radii of the base, torso, and head, respectively, which are integers greater than 0. The problem then amounts to Frosty solving the equation
43 π 63 = 43 π a3 + 43 π b3 + 43 π c3
= 43 π (a3 + b3 + c3)
By canceling out the factor of 43 π from both sides, we're left with
216 = a3 + b3 + c3
As stated in the problem, a ≥ b ≥ c. The biggest that a can be is 5, so let's try that first. Subtracting 53 = 125 from both sides gives us
91 = b3+c3
The biggest that b can be is 4, so let's try that next. Subtracting 43 = 64 from both sides gives us
27 =c3
Frosty is in luck, since 33 = 27. Thus we have that 63 = 53 + 43 + 33. Now Frosty can build his new snowman friend to his specifications. In fact, (5, 4, 3) is the only combination of numbers that will work.
### Snow Manipulation Solution:
The original snowball has a 6 inch radius. Radii of the new spheres are: 3, 4, and 5. |
# How do you divide ( x^4-2x^3-8x+16 )/(4x-6)?
##### 1 Answer
Nov 30, 2017
$\frac{1}{4} {x}^{3} - \frac{1}{8} {x}^{2} - \frac{3}{16} x - 2 \frac{9}{32}$and remainder of $29 \frac{11}{16}$
#### Explanation:
$\frac{{x}^{4} - 2 {x}^{3} - 8 x + 16}{4 x - 6}$
$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{.} \frac{1}{4} {x}^{3} - \frac{1}{8} {x}^{2} - \frac{3}{16} x - 2 \frac{9}{32}$
$4 x - 6 | \overline{{x}^{4} - 2 {x}^{3} + 0 x - 8 x + 16}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} \underline{{x}^{4} - \frac{3}{2} {x}^{3}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .} - \frac{1}{2} {x}^{3} + 0 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots} \underline{- \frac{1}{2} {x}^{3} + \frac{3}{4} {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} - \frac{3}{4} {x}^{2} - 8 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- \frac{3}{4} {x}^{2} + \frac{9}{8} x}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} - \frac{73}{8} x + 16$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- \frac{73}{8} x - \frac{438}{32}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \frac{950}{32} = 29 \frac{11}{16}$
$\frac{{x}^{4} - 2 {x}^{3} - 8 x + 16}{4 x - 6} = \frac{1}{4} {x}^{3} - \frac{1}{8} {x}^{2} - \frac{3}{16} x - 2 \frac{9}{32}$and remainder of$29 \frac{11}{16}$ |
US UKIndia
Every Question Helps You Learn
Distance is measured in meters.
# Measurement - The Metric System - Easy Conversions 1
This Math quiz is called 'Measurement - The Metric System - Easy Conversions 1' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.
It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us
As you know, everything in life involves math in one form or another. Growing up in the United States, you have been taught the American or English system of math. However, most of the rest of the world does not use this system. Rather, they use the metric system.
1.
1,000 g = ___ kg
1
100
1,000
10,000
As we are starting at the base unit of 1,000 grams, which is also understood to be 1,000.0, and we are moving up the metric list, we need to move the decimal point one space to the left for each space between the base unit and the kilogram. As there are three spaces the decimal point moves three spaces to the left turning 1,000.0 to 1.0. So, 1,000 grams (g) equals 1 kilogram (kg). Answer (a) is the correct answer
2.
1 cm = ___ mm
.01
.1
10
100
As we are moving down the metric list we will be adding zeroes “0”. There is one space between centimeters and millimeters so 1 centimeter (cm) equals 10 millimeters (mm). Answer (c) is the correct answer
3.
6 hl = ___ dl
600
6,000
6
.006
As we are moving down the metric list we will be adding zeroes “0”. There are three spaces between hectoliters and deciliters so 6 hectoliters (hl) equals 6,000 deciliters (dl). Answer (b) is the correct answer
4.
12 m = ___ dam
.12
120
1,200
1.2
As we are moving up from the base unit of 12 meters, it is understood that 12 meters is the same as 12.0. Moving up we must move the decimal point over one space to the left for each space we go up the metric list. The decameter is one space up from the meter so 12.0 becomes 1.2. 12 meters (m) equals 1.2 decameters (dam). Answer (d) is the correct answer
5.
8 l = ___ kl
0.08
0.008
800
8,000
As we are moving up from the base unit of 8 liters, it is understood that 8 liters is the same as 8.0. Moving up we must move the decimal point over one space to the left for each space we go up the metric list. The kiloliter is three spaces up from the liter so 8.0 becomes 0.008. 8 liters (l) equals 0.008 kiloliters (kl). (Remember the “0” before the decimal point represents the base unit.) Answer (b) is the correct answer
6.
120 m = ___ mm
120,000
12,000
1,200
1.2
As we are moving down the metric list we will be adding zeroes “0”. There are three spaces between meter and millimeters so 120 meters (m) equals 120,000 millimeters (mm). Answer (a) is the correct answer
7.
25 g = ___ dg
25,000
2,500
250
2.5
As we are moving down the metric list we will be adding zeroes “0”. There is one space between grams and decigrams so 25 grams (g) equals 250 decigrams (dg). Answer (c) is the correct answer
8.
7 kg = ___ g
.007
70
700
7,000
As we are moving down the metric list we will be adding zeroes “0”. There are three spaces between kilograms and grams so 7 kilograms (kg) equals 7,000 grams (g). Answer (d) is the correct answer
9.
94 cm = ___ dam
.94
0.094
9.4
940
As we are moving up the metric list we will be adding zeroes “0” to the left of the base number “94”. Remember, when counting the spaces up you place a decimal point in place of the last “0” and then place the last zero to the left of the decimal point to represent the base unit. So there are three spaces starting from 94.0 which then becomes .094 and then you add the final zero to the left of the decimal to get 0.094. So, 94 centimeters (cm) equals 0.094 decameters (dam). Answer (b) is the correct answer
10.
19 hg = ___ dg
190
1,900
19,000
190,000
As we are moving down the metric list we will be adding zeroes “0”. There are three space between hectograms and decigrams so 19 hectograms (hg) equals 19,000 decigrams (dg). Answer (c) is the correct answer
Author: Christine G. Broome |
<meta http-equiv="refresh" content="1; url=/nojavascript/">
# Differences of Integers with Different Signs
## Subtract positive and negative numbers to find a new difference.
%
Progress
Practice Differences of Integers with Different Signs
Progress
%
Differences of Integers with Different Signs
Have you ever watched a football game? Take a look at this dilemma.
After the football game Friday night, Sarah could hardly wait to write and tell her pen pal Emily all about it. It had been one of the most exciting games that Sarah had ever been to. The middle school team was evenly matched with a rival team from a neighboring school. The game had been very close. In fact it had come down to the last few minutes of play.
Sarah wrote this to her pen pal, “At the end of the fourth quarter, we were twenty yards away from a touchdown. The score was 14 to 14. We needed this touchdown to win the game. The running back took the football and began running. He made it 15 yards.”
“Then, on the next play, the defenders charged at our players. We had a loss of ten yards on that play. Next, our players earned a penalty of 15 yards, but the coach challenged the call and the referee took away a loss of ten yards. Then we ran for a gain of 5 yards. On the next play, the quarterback threw the ball for a touchdown and we won the game!”
Sarah reread her letter. All of the yards gained and lost seemed a bit confusing.
“I think I can write this clearer if I use integers,” Sarah thought to herself. “Then I can see how far the quarterback threw the ball for the touchdown.”
Writing about the football game will involve sums and differences of integers. That is what this Concept is all about. Follow along through this Concept and at the end Sarah will show you how she explains the yards lost and gained through an integer problem.
### Guidance
In the last Concept, you learned how to find the differences of integers that had the same sign. You learned how to find the differences of two positive integers and of two negative integers. Now we are going to apply what you learned in the last section when finding the differences of integers that have different signs.
-6 – 4 = ____
Just like the last Concept, there are two different ways to approach this problem. We can think of it in terms of losses and gains or we can change subtraction to addition and add the opposite.
Let’s start by thinking in terms of losses and gains.
This problem starts with a loss. There is a loss of six or a negative six.
-6
Next, we take away a gain of four. The subtraction is the taking away. We have a positive four, so we take away a gain of four. If you take away a gain it is the same as adding a loss.
-6 – 4 = -10
The answer is -10.
Now let’s change the subtraction to addition and add the opposite.
-6 – 4 = -6 + -4
The subtraction changed to addition. Positive four became its opposite, negative four.
-6 + -4 = 10
The answer is the same. It is still -10.
In this last problem we looked for the difference between a negative and a positive. What about finding the difference between a positive and a negative?
6 – -3 = ____
Once again, we can approach this problem in two ways. We can think in terms of losses and gains, and we can change the subtraction to addition and add the opposite.
Let’s start by thinking in terms of losses and gains.
We start with a gain because our first value is positive six.
6
Then we take away a loss. When you take away a loss of 3, it is the same as adding three.
6 – -3 = 9
The answer is 9.
Now let’s change the subtraction to addition and add the opposite.
6 – -3 = 6 + 3
The subtraction sign became an addition sign. The negative three became its opposite which is positive three.
6 + 3 = 9
The answer is 9.
Practice a few of these on your own. Choose a method and find the difference of each pair of integers.
-5 – 7 = ____
Solution: -12
2 – -8 = ____
Solution: 10
#### Example C
-13 – 5 = ____
Solution: -18
Remember the football game? Let's go back to the football game and integers.
After the football game Friday night, Sarah could hardly wait to write and tell her pen pal Emily all about it. It had been one of the most exciting games that Sarah had ever been to. The middle school team was evenly matched with a rival team from a neighboring school. The game had been very close. In fact it had come down to the last few minutes of play.
Sarah wrote this to her pen pal, “At the end of the fourth quarter, we were twenty yards away from a touchdown. The score was 14 to 14. We needed this touchdown to win the game. The running back took the football and began running. He made it 15 yards.”
“Then, on the next play, the defenders charged at our players. We had a loss of ten yards on that play. Next, our players earned a penalty of 15 yards , but the coach challenged the call and the referee took away a loss of ten yards . Then we ran for a gain of 5 yards . On the next play, the quarterback threw the ball for a touchdown and we won the game!”
Sarah reread her letter. All of the yards gained and lost seemed a bit confusing.
“I think I can write this clearer if I use integers,” Sarah thought to herself. “Then I can figure out how far the quarterback threw the ball for the touchdown.”
Let’s write out the integers that we are using in this problem.
“He made it 15 yards” = +15
“A loss of ten yards” = + -10
“A penalty of 15 yards” = + -15
“Referee took away a loss of ten yards” = – -10
“Then we ran for a gain of 5 yards” = +5
Now we can write a problem using sums and differences of the following integers.
15 + -10 + -15 – - 10 + 5
Let’s work from left to right adding integers.
15 + -10 = 5
5 + -15 = -10
-10 – -10 = 0 yards gained
0 + 5 = 5 yards gained.
Since the team originally needed 20 yards for a touchdown, after all of the gains and losses, they ended up with a gain of five.
20 – 5 = 15
The quarterback threw the ball 15 yards for the touchdown.
### Guided Practice
Here is one for you to try on your own.
During the first quarter of Friday night’s game, Lawrence High School’s football team had a gain of 10 yards, then a loss of 20 yards then a gain of 5 yards, another gain of 3 yards and a loss of 2 yards before the coach called time out. If they started on the ten yard line, where were they when the coach called time out?
To work through this problem, we need to write an integer number sentence showing the losses and gains that the team had. Each loss is a negative number and each gain is a positive one. We know that they started on the ten yard line, so that is our first number.
10 + 10 – 20 + 5 + 3 – 2 = ____
Next, we add each integer in order.
10+1020200+55+382=20=0=5=8=6
The team was on the six yard line when the coach called time out. At this point they had actually experienced a loss of four from their starting place on the ten yard line.
### Video Review
Here are videos for review.
### Explore More
Directions: Find the differences of the following integer pairs.
1. -2 – 4 = ____
2. -8 – 9 = ____
3. -6 – 7 = ____
4. -11 – 12 = ____
5. -13 – 22 = ____
6. -89 – 11 = ____
7. 2 – 7 = ____
8. 4 – 9 = ____
9. 5 – 8 = ____
10. 13 – 20 = ____
11. 12 – 23 = ____
12. 25 – 30 = ____
13. 45 – 90 = ____
14. 34 – 67 = ____
15. -2 – -3 = ____
16. -8 – -3 = ____
17. -9 – -7 = ____
18. -5 – -10 = ____
19. -9 – -12 = ____
20. -10 – -10 = ____
21. -14 – -15 = ____
22. 5 – -8 = ____
23. 6 – -7 = ____
24. 10 – -9 = ____
25. 11 – -7 = ____
26. 18 – -9 = ____
27. 22 – -5 = ____
28. 34 – -3 = ____
29. 35 – -35 = ____
30. 45 – -10 = ____
### Vocabulary Language: English
Difference
Difference
The result of a subtraction operation is called a difference.
Sum
Sum
The sum is the result after two or more amounts have been added together. |
# Supplementary angles
One of the supplementary angles is larger by 33° than the second one. Calculate the angles size.
A = 106.5
B = 73.5
### Step-by-step explanation:
A+B = 180
A = 33+B
A+B = 180
A-B = 33
A = 213/2 = 106.5
B = 147/2 = 73.5
Our linear equations calculator calculates it.
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Tips to related online calculators
Do you have a linear equation or system of equations and looking for its solution? Or do you have a quadratic equation?
## Related math problems and questions:
• Supplementary angles
One of the supplementary angles are three times larger than the other. What size is larger of supplementary angles?
• MO Z7–I–6 2021
In the triangle ABC, point D lies on the AC side and point E on the BC side. The sizes of the angles ABD, BAE, CAE and CBD are 30°, 60°, 20° and 30°, respectively. Find the size of the AED angle.
• Supplementary angles
(a + 30)° and (2a)° are the measure of two supplementary angles. Find the value of 'a'.
• Angles of a triangle
In triangle ABC, the angle beta is 15° greater than the angle alpha. The remaining angle is 30° greater than the sum of the angles alpha and beta. Calculate the angles of a triangle.
• Triangles
Find out whether given sizes of the angles can be interior angles of a triangle: a) 23°10',84°30',72°20' b) 90°,41°33',48°37' c) 14°51',90°,75°49' d) 58°58',59°59',60°3'
• Internal angles
One internal angle of the triangle JAR is 25 degrees. The difference is the size of the two others is 15°. Identify the size of these angles.
• Angles in ratio
The size of the angles of the triangle are in ratio x: y = 7: 5 and the angle z is 42° lower than the angle y. Find size of the angles x, y, z.
• Internal angles
The ABCD is an isosceles trapezoid, which holds: |AB| = 2 |BC| = 2 |CD| = 2 |DA|: On its side BC is a K point such that |BK| = 2 |KC|, on its side CD is the point L such that |CL| = 2 |LD|, and on its side DA the point M is such that | DM | = 2 |MA|. Dete
• Outer angles
The outer angle of the triangle ABC at the A vertex is 71°40 ' outer angle at the vertex B is 136°50'. What size has the inner triangle angle at the vertex C?
• Angles
The outer angle of the triangle ABC at the vertex A is 114°12'. The outer angle at the vertex B is 139°18'. What size is the internal angle at the vertex C?
• In a 2
In a thirteen sided polygon, the sum of five angles is 1274°, four of the eight angles remaining are equal and the other four are 18° less than each of the equal angles. Find the angles. .
• Angles in triangle
The triangle is ratio of the angles β:γ = 6:8. Angle α is 40° greater than β. What are the size of angles of the triangle?
• The second
The second angle of a triangle is the same size as the first angle. The third angle is 12 degrees larger than the first angle. How large are the angles?
• Inner angles
The magnitude of the internal angle at the main vertex C of the isosceles triangle ABC is 72°. The line p, parallel to the base of this triangle, divides the triangle into a trapezoid and a smaller triangle. How big are the inner angles of the trapezoid?
• Angle at the apex
In an isosceles triangle, the angle at the apex is 30° greater than the angle at the base. How big are the internal angles?
• Neighbor angle
For 136° angle calculate size of adjacent angle on one side of a straight line.
• RWY
Calculate the opposite direction of the runway 13. Runways are named by a number between 01 and 36, which is generally one tenth of the azimuth of the runway's heading in degrees: a runway numbered 09 points east (90°), runway 18 is south (180°), runway 2 |
Share
Books Shortlist
# Solution for Find the Points on the Curve X2 + Y2 − 2x − 3 = 0 at Which the Tangents Are Parallel to the X-axis ? - CBSE (Science) Class 12 - Mathematics
ConceptMaximum and Minimum Values of a Function in a Closed Interval
#### Question
Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are parallel to the x-axis ?
#### Solution
Let (x1, y1) be the required point.
$\text { Since the point lie on the curve } .$
$\text { Hence } {x_1}^2 + {y_1}^2 - 2 x_1 - 3 = 0 . . . \left( 1 \right)$
$\text { Now }, x^2 + y^2 - 2x - 3 = 0$
$\Rightarrow 2x + 2y \frac{dy}{dx} - 2 = 0$
$\therefore \frac{dy}{dx} = \frac{2 - 2x}{2y} = \frac{1 - x}{y}$
$\text { Now,}$
$\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{1 - x_1}{y_1}$
$\text { Slope of the tangent } = 0 (\text { Given )}$
$\therefore \frac{1 - x_1}{y_1} = 0$
$\Rightarrow 1 - x_1 = 0$
$\Rightarrow x_1 = 1$
$\text { From (1), we get }$
${x_1}^2 + {y_1}^2 - 2 x_1 - 3 = 0$
$\Rightarrow 1 + {y_1}^2 - 2 - 3 = 0$
$\Rightarrow {y_1}^2 - 4 = 0$
$\Rightarrow y_1 = \pm 2$
$\text { Hence, the points are }\left( 1, 2 \right)\text { and }\left( 1, - 2 \right).$
Is there an error in this question or solution?
#### APPEARS IN
Solution Find the Points on the Curve X2 + Y2 − 2x − 3 = 0 at Which the Tangents Are Parallel to the X-axis ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
S |
# Calendar – Aptitude MCQ Questions and Solutions with Explanations 7
61. What was the day of the week on 16th August, 1947?
Solution:
15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)
= [(11 x 2) + (36 x 1) ] odd days
= 58 odd days
= 2 odd days
Jan Feb Mar Apr May Jun Jul Aug
= 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15
= 227 days
= (32 weeks + 3 days)
= 3,
Total number of odd days
= (0 + 1 + 2 + 3) odd days
= 6 odd days.
Hence, the required day was ‘Saturday’.
62. Prove that any date in March of a year is the same day of the week corresponding date in November that year.
Solution:
We will show that the number of odd days between last day of February and last day of October is zero.
March April May June July Aug. Sept. Oct.
31 + 30 + 31 + 30 + 31 + 31 + 30 + 31
= 241 days
= 35 weeks
= 0 odd day
Number of odd days during this period = 0.
Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows
63. If today is Saturday, what will be the day 350 days from now ?
64. The calendar for the year 1988 is same as which upcoming year ?
Solution:
We already know that the calendar after a leap year repeats again after 28 years.
Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.
65. Given that on 9th August 2016 is Saturday. What was the day on 9th August 1616 ?
Solution:
We know that, After every 400 years, the same day occurs.
Thus, if 9th August 2016 is Saturday, before 400 years
i.e., on 9th August 1616 has to be Saturday.
66. Second & fourth Saturdays and every Sunday is a holiday. How many working days will be there in a month of 31 days beginning on a Friday ?
Solution:
Given that the month begins on a Friday and has 31 days
Sundays = 3rd, 10th, 17th, 24th, 31st
⇒ Total Sundays = 5
Every second & fourth Saturday is holiday. 2nd & 4th Saturday in every month = 2
Total days in the month = 31
Total working days = 31 – (5 + 2) = 24 days.
67. On 17th March, 1997 Monday falls. What day of the week was it on 17th March, 1996?
Solution:
The year 1996 is a leap year. So, it has 2 odd days. But 17th March comes after 29th February.
So, the day on 17th March, 1997 will be 1 day beyond the day on 17th March,1996.
Here 17th March, 1997 is Monday. So, 17th March, 1996 is Sunday.
68. Which year has 366 days? |
# AP Calculus AB : Derivatives
## Example Questions
### Example Question #65 : Derivatives Of Functions
Find the derivative of the function
Possible Answers:
Correct answer:
Explanation:
To find the derivative of the function, we use both the product rule and the chain rule
### Example Question #66 : Derivatives Of Functions
Find the derivative of the function
Possible Answers:
Correct answer:
Explanation:
To find the derivative of the function, we use both the quotient rule and the chain rule
### Example Question #67 : Derivatives Of Functions
If then
Possible Answers:
Correct answer:
Explanation:
To calculate the derivative of this function at the desired point, first recall that,
Now, substitute the value into the derivative function to solve.
### Example Question #68 : Derivatives Of Functions
Let . Which of the following gives the equation of the line normal to when ?
Possible Answers:
Correct answer:
Explanation:
We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at . In order to find the tangent line, we will need to evaluate the derivative of at .
The slope of the tangent line at is . Because the tangent line and the normal line are perpendicular, the product of their slopes must equal .
(slope of tangent)(slope of normal) =
We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.
Since the normal line passes through the function at , it will pass through the point . Be careful to use the original equation for , not its derivative.
The normal line has a slope of and passes through the piont . We can now use point-slope form to find the equation of the normal line.
Multiply both sides by .
The answer is .
### Example Question #69 : Derivatives Of Functions
If the position of a particle over time is represented by then what is the particle's instantaneous acceleration at ?
Possible Answers:
Correct answer:
Explanation:
The answer is .
Since velocity is the first derivative of the position function, take the derivative once. Then, recall that the acceleration function is the second derivative of position thus the derivative needs to be taken one more time.
### Example Question #61 : Computation Of The Derivative
Consider the function:
The relative minimum for this function is at:
Possible Answers:
There is no relative minimum, because the first derivative is always positive.
None of the other answers.
Correct answer:
There is no relative minimum, because the first derivative is always positive.
Explanation:
To find any relative minimum, one first needs to find the critical points by setting the first derivative equal to zero:
However, the first derivative is positive for all real values of x, since the exponential function is always positive. Thus, there are no values for which , and therefore no critical points and no relative minimum.
### Example Question #81 : Ap Calculus Ab
Find the derivative of the function:
Possible Answers:
None of the other derivatives are correct
Correct answer:
Explanation:
We are given the function:
To take the derivative of this, we must understand 3 things:
1. The product rule for derivatives
2. The derivative of the sin function
3. The derivative of the tan function
We can apply the product rule as follows:
Simplifying once more, the answer becomes:
(Note: If you require or desire further explanation for the assumed knowledge, ask your Varsity Tutor for clarity on the concepts!)
### Example Question #82 : Ap Calculus Ab
Find the first derivative of the function:
Possible Answers:
Correct answer:
Explanation:
We are given the function:
Our first step in tackling this problem is to apply the product rule for derivatives:
To simplify the above further, it is best to write everything in terms of sin and cos, like so:
We can simplify both sides by multiplying the components together and canceling out like so:
Simplifying our last terms, we arrive at the correct answer:
### Example Question #81 : Ap Calculus Ab
Find the limit of the function below using L'Hôpital's rule
Possible Answers:
Correct answer:
Explanation:
*****************************************************************
STEPS
*****************************************************************
First, try to plug in the 2, and check the result.
When doing this, the limit becomes an indeterminate form of
Thus, we continue with L'Hôpital's rule, deriving the expressions in the numerator and denominator like so:
Now, plug 2 in for y, to check again
Thus, our correct answer is:
*****************************************************************
CORRECT ANSWER
*****************************************************************
### Example Question #74 : Derivatives Of Functions
Find the limit of the function below using L'Hopital's Rule
Possible Answers:
Correct answer:
Explanation:
*****************************************************************
STEPS
*****************************************************************
We are asked to find the limit:
First, plug pi in for s to check if we can use L'Hopital's Rule
We have the indeterminate form zero over zero, so we continue with L'Hopital's Rule, deriving the expressions in the numerator and denominator, independently of each other:
(The 3 before the cos is from the derivative of 3s (=3)
Now, plug pi in once more to check:
Thus, canceling the negatives, we arrive at the correct answer:
*****************************************************************
CORRECT ANSWER
***************************************************************** |
# Geometry Chapter 4 Resource Book Answers
Geometry Chapter 4 Resource Book Answers: A Comprehensive Guide
Geometry is a branch of mathematics that deals with the study of shapes, sizes, positions, and properties of figures and spaces. In Chapter 4, students delve into the world of congruent triangles, exploring their properties and relationships. To aid in their understanding, resource books with answers are invaluable tools. In this article, we will provide a comprehensive guide to Geometry Chapter 4 Resource Book Answers, along with five unique facts about congruent triangles. Additionally, we will address 13 frequently asked questions and provide detailed answers at the end.
Chapter 4 in Geometry focuses on congruent triangles, which are figures that have the same shape and size. This chapter delves into the various criteria and methods to prove triangles congruent, including side-side-side (SSS), side-angle-side (SAS), angle-side-angle (ASA), angle-angle-side (AAS), and hypotenuse-leg (HL). With the help of resource books, students can practice applying these concepts and verify their solutions.
Here are five unique facts about congruent triangles:
1. Corresponding parts of congruent triangles are congruent (CPCTC): When two triangles are congruent, their corresponding parts, such as angles and sides, are also congruent. This principle, known as CPCTC, enables mathematicians to deduce additional information about the triangles.
2. SSS and SAS are sufficient to prove congruence: By using the side-side-side (SSS) or side-angle-side (SAS) criteria, students can prove that two triangles are congruent. These criteria involve comparing lengths of sides and measures of angles, respectively.
3. ASA, AAS, and HL criteria: While SSS and SAS are sufficient to prove congruence, the angle-side-angle (ASA), angle-angle-side (AAS), and hypotenuse-leg (HL) criteria are also valid methods. These criteria provide alternative ways to establish congruence.
4. Congruent triangles have congruent corresponding angles: When two triangles are congruent, the corresponding angles are also congruent. This property allows mathematicians to identify congruent angles based on the congruence of the triangles.
5. Congruent triangles have equal areas: If two triangles are congruent, their areas will be equal. This fact is a consequence of their corresponding sides and angles being congruent.
1. What is the purpose of a resource book in Geometry?
A resource book provides additional practice problems and exercises to reinforce the concepts taught in class. It also includes answers for self-assessment.
2. How can I access the answers for Chapter 4 of the Geometry resource book?
The answers for Chapter 4 can be found at the end of the resource book or in a separate answer key, depending on the publication.
3. Can I rely solely on the resource book to study for my Geometry exams?
While the resource book is a valuable tool for practice, it is recommended to also refer to your textbook and consult with your teacher or classmates to ensure a comprehensive understanding of the material.
4. Are the answers in the resource book always correct?
Errors can occur in any book. However, reputable publishers strive to minimize mistakes. If you encounter an answer that seems incorrect, consult with your teacher for clarification.
5. How should I use the resource book effectively?
Start by attempting the exercises on your own before referring to the answers. Use the answers to check your work and identify any mistakes or areas of confusion that require further study.
6. Are there explanations provided for the answers in the resource book?
Some resource books provide step-by-step explanations for solving the problems, while others simply provide the answers. It’s beneficial to have a resource book that includes explanations to help you understand the process behind each solution.
7. Can I use the resource book for additional practice, even if it’s not assigned by my teacher?
Absolutely! The more practice you have, the better your understanding of the concepts. Using the resource book for extra practice can improve your skills and boost your confidence.
Yes, practicing with the resource book answers can help you prepare for exams by allowing you to gauge your understanding of the material and identify areas that require further review.
9. Can I use the resource book as a study guide for Chapter 4?
The resource book can serve as a study guide, as it contains a variety of problems that cover the concepts taught in Chapter 4. Solve the problems without referring to the answers to enhance your problem-solving skills.
10. How can I find out if my answer is correct without looking at the answers in the resource book?
11. Can the resource book be used by students at different levels of proficiency?
Yes, resource books often cater to students of varying proficiency levels by including problems of different difficulty levels. This allows students to practice at their own pace and challenge themselves accordingly.
12. Are there any online resources available with additional practice problems for Chapter 4?
Yes, several websites offer extra practice problems for Geometry Chapter 4. Some even provide step-by-step solutions to help you understand the concepts better.
13. Can I use the resource book for self-study?
Certainly! The resource book is a valuable tool for self-study, allowing you to practice and reinforce the concepts at your own pace.
In conclusion, Geometry Chapter 4 Resource Book Answers provide a comprehensive guide to understanding congruent triangles. By mastering the concepts and utilizing the resource book effectively, students can enhance their problem-solving skills and excel in Geometry. Remember to practice, seek help when needed, and strive for a deep understanding of the material.
## Author
• Laura is a seasoned wordsmith and pop culture connoisseur with a passion for all things literary and cinematic. Her insightful commentary on books, movies, and the glitzy world of film industry celebrities has captivated audiences worldwide. With a knack for blending literary analysis and movie magic, Laura's unique perspective offers a fresh take on the entertainment landscape. Whether delving into the depths of a novel or dissecting the latest blockbuster, her expertise shines through, making her a go-to source for all things book and film-related. |
# Online CBSE & NCERT Solutions for cbse class 12 maths Function
## Online CBSE & NCERT Solutions for cbse class 12 maths Function
We are providing solutions for maths and other subjects. We are following modern technology to tech and guide our students.
We provide demo videos to students, so that they can judge the content before purchasing.
Few Concepts in details for cbse class 12 maths
(Note: To understand the concept clearly, watch our respective 3D animated videos while reading this article as the illustrations are based on that.)
## Function – Class 12 maths
Objective :
In this session we will understand the concept of a function. We have plotted points on a graph paper. The points are given by their co-ordinates. (4, 5), (-5, 4), (-3.5, -4) etc.
We know that the first number gives us the x-coordinate that is, the distance from origin along x—axis and the second number gives us the y- coordinate that is, the distance from origin along y — axis. It is for this reason that if we change the order of numbers, we get a different point. (4, 5) and (5, 4) are two different points. Such an arrangement of pair of numbers where order is to be retained is called an ordered pair.
The cross product of two sets, AxB = {(a, b) I a e A and b e B} To explore all possibilities let us have two sets. Set A with five robots and set B with seven bikes.
Now we pair a robot with the bike matching its color.
Case I :Green robot with Green bike. Blue robot with Blue bike etc. In this case we find that each robot has only one bike to itself.
Case 2 : Let the bike with orange color be replaced with a bike of light green color.
Now we find that the robot with Orange color does not have a bike to itself.
Case 3 :Take the bikes of light green and purple color and replaced them with orange colored bikes. Matching the sets again.
Now we find that the Orange colored robot has two bikes to itself. Thus we observe that in the first case each robot from set A has a bike to itself, and there are 2 spare bikes. In the second case there is a robot without a matching bike and finally in the third case we have a robot that has more than one matching bike. These conditions lead us to the definition of a function.
A function is a subset of the cross product of two non-empty sets A and B such that every element of set A has a unique image in set B.
The function is represented by small letters, for example, Function f:A—>B. Set A is called Domain and set B Co domain. This further implies that no element of A can have more than one image in B. And also there should not be any element of A which does not have an image in B.Mathematically it implies that no two ordered pairs will have the same first element.
Complete cbse class 12 maths Function Download pdf free & watch free demo for Class 12 Maths here..
Find More Searches
• All maths, physics, chemistry, biolofy, commerce,cbse solutions
• Complete cbse sample papers
• Latest cbse class 10 syllabus
• Perfect cbse guide
• cbse maths
subscribing to our social channel.
0Shares
August 22, 2019 |
## Thursday, August 29, 2019
### Set and subset
How to divide a set into subsets? To do this, it is necessary to introduce a new unit of measure that is present in some elements of the selected set. Consider an example.
Suppose we have a set of A, consisting of four people. This set is formed on the basis of "people." We denote the elements of this set by the letter "a", the subscript with a number will indicate the serial number of each person in this set. We introduce a new unit of measurement "gender" and denote it with the letter "b". Since sexual characteristics are common to all people, we multiply each element of the set A by sexual characteristic b. Please note that now our multitude of “people” has turned into a multitude of “people with sexual characteristics”. After that, we can divide the sexual characteristics into male bm and female bw sexual characteristics. Now we can apply the mathematical filter: we choose one of these sexual characteristics, no matter which one - male or female. If it is present in a person, then we multiply it by one; if there is no such sign, we multiply it by zero. And then we use ordinary school math. See what happened.
Set and subset
After multiplication, contractions, and rearrangements, we got two subsets: a subset of Bm men and a subset of Bw women. Mathematicians reason the same way when they apply set theory in practice. But they don’t dedicate us in detail, but give the finished result - "a set of people consist of a subset of men and a subset of women." Naturally, you may wonder how correctly mathematics is applied in the above transformations? I dare to assure you that, in essence, everything was done correctly; it’s enough to know the mathematical justification of arithmetic, Boolean algebra and other branches of mathematics. What it is? Some other time, I'll tell you about it.
You can combine two sets into one by selecting the unit of measure present in the elements of these two sets.
As you can see, units of measure and ordinary mathematics turn set theory into a relic of the past. A sign that the set theory is not all right is that for set theory mathematicians have come up with their own language and their own notation. Mathematicians did what shamans once did. Only shamans know how to "apply" their "knowledge" correctly. They teach us this "knowledge".
In conclusion, I want to show you how mathematicians manipulate infinite sets. |
# Blogs
## Arithmetic Blogs
Subtraction Method "Borrow" vs. "Shift" Most can remember subtraction using the classic "borrowing" method, such as, 63 - 28 = (50 + 13) - (20 + 8) = (50 - 20) + (13 - 8)* = 35. Here we "borrowed" from the 60, adding 10 to the 3 for 13, and leaving 50 behind to subtract the 20 from. This all looks more familiar if one writes it out in the traditonal way, vertically, crossing out the 6, writing the 1 by the 3, etc. You know, "borrowing." 13 5 63 - 28 35 This is a way to do this, but not the only one. We also can change the problem a bit, and into one that gives the same answer. We do this by "shifting" the numbers an equal amount, in this case by adding 2 to both. Here it is. 63 - 28 = (63 + 2) - (28 + 2) = 65 - 30 = 35. Or vertically, ... read more
We’ll start with the easy stuff. Multiplying by 4: (1) Double the multiplicand you want to multiply 4 by (2) Double it one more time e.g. 8 * 4 = 32 8 * 2 = 16 16 * 2 = 32 Why does this work? 4 can be broken up into 2 * 2 8 * 4 = 32 8 * (2 * 2) = 32 Thanks to the associative property of multiplication, we can multiply factors in whatever grouping or order we chose and still get the same answer. We start by multiplying multiplicand we want to multiply 4 by 2 because this computation is easy for most people to do in their heads. (8 * 2) * 2 (16) * 2 We then multiply our product by the remaining multiplicand, which is 2. 16 * 2 = 32 Multiplying by 10: Stick a zero behind whatever number you wish to multiply 10 by 988 * 10 = 9880 Why does this work? Consider what we’re doing in terms of place value. When... read more
Welcome back to the school everyone! I hope you all had a great summer. For all those whose summer was maybe a little too great, maybe those who’ve forgotten even the basics, we’re going to take it all the way back to arithmetic a.k.a “number theory”. A review of number theory is a perfect place to start for many levels. Calculus and a lot of what you learn in pre-calculus is based on the real number system. When we use the word “number” we are typically referring to all real number. But how can numbers be “real”? You can’t touch the number 6 or smell 1,063. You can’t boil 1/2 or stick it in a stew. So what’s so real about real numbers? The simple answer is this: a real number is a point on a number line (1). -2.5 -1 0 ... read more
Summary: Mental math teaches students to see short, efficient solutions—rather than to blindly follow the brute-force, cookie-cutter, one-size-fit-all, show-all-your-work procedures taught at school. To my youngest students, I lie—by omission—that vertical arithmetic does not exist. I can usually get away with it for about a year. Until the school shows them the light. Say, how to add 25 and 8 vertically, with the carry-over 1 carefully written on top of the 2. By that time, my students are proficient in mental addition and subtraction of 3-digit numbers: carrying, borrowing, and all. My goal though is by no means to turn them into human calculators. So then, why bother? Vertical arithmetic is a convenient method for computing numerical answers. Especially when the numbers to manipulate are multidigit. But it is a procedure, requiring—once learned—little thought. The entire process is delegated to the... read more
Dazzling pocket PCs are aplenty for the children of today. Kids roll into the classroom with iPhones, Blackberries, and various Android devices capable of supporting myriad complex applications. We are living in a wonderful age where handheld computers help us tremendously and continuously. Alongside all of the fancy apps (that allow us to manage everything from our finances to our fantasy football teams) is a standard utility application that accompanies every smartphone: the basic calculator. Need to carry out some quick arithmetic to figure out how much money you owe your buddy? Pull out your phone and type away. It’s that simple. So why the heck do kids need to memorize the multiplication table? Because it is still crucial to a successful math career and a promising life thereafter. Don’t believe me? Here are four reasons why mental math is still tremendously important and absolutely foundational. 1. Confidence Is Key You have likely heard people utter the following... read more
In elementary school, mathematics is often taught as a set of rules for counting and computation. Students learn that there is only one right answer and that the teacher knows it. There is no room for judgment or making assumptions. Students are taught that Arithmetic is the way it is because it's the truth, plain and simple. Often students go on to become trapped in this view of the universe. As fairy tales fade from the imagination, so is mathematical creativity lost. There is evidence that Mathematics and Arithmetic existed over 3000 years ago, but only the very well educated leisure class had access to it. The rules for simple computation only were developed recently, so much of the computation of sums and products was much more complicated. Imagine adding and multiplying Roman Numerals for example. Because of this difficulty, computations were laid out only to solve very specific practical problems. Although mathematics was mainly limited to solving practical... read more
When working with fractions, I find it effective to require students to convert each fraction that we work with to its decimal equivalent, to convert that decimal equivalent back into the original fraction, to convert that decimal into its percentage equivalent, to work a simple percentage problem using that percentage and finally to work the same problem using the initial fraction. This comprehensive method helps students to see the relationships between fractions, decimals and percentages in a holistic way and to promote the necessary skills in each element. |
# Properties of exponents & negative exponents
I’ve decided to make my blog posts more focused. I want to contribute in some way–I want to be able to give in a way that benefits myself and the other person–and I think that posts that are more focused will make it easier for people to engage because I won’t be traveling from one subject to another and possibly introducing them to things that are not so interesting to them.
Negative exponents
There are several exponent properties. Exponents are numbers written above other numbers, and which indicate how many times 1 should be multiplied/ divided by the base (the lower number).
Xn
Above, n is the exponent, and X is the base.
There are both positive and negative exponents. When the exponent (n) is positive, then 1 is multiplied by the base n times.
2= 1 * 2 * 2 * 2
2= 8
When the exponent (n) is negative, then 1 is divided by the base |n| times. In this scenario, 2 is the base and -3 is the exponent. Also remember the math rule that dividing x by m is the same thing as multiplying x by m’s reciprocal.
x / m = x / (m/1) = x * (1/m)
The reciprocal of a number is that number with its numerator and denominator flipped. An integer like 5, for example, can be written as 5/1, meaning it can be written as a fraction. To find the reciprocal of 5, or 5/1, flip the numerator and denominator. So the reciprocal of 5 or 5/1 is 1/5. The reciprocal of 27 or 27/1 is 1/27.
When dividing x by m, you can also multiply if that is more convenient–you can multiply x by the reciprocal of m, and you will get the same thing you would get if you had done the division.
Anyway, if you have a negative exponent, you divide 1 by the base |n| times.
2-3
So we have to divide 1 by 2 three times. Not -3 times, just 3 times, because -3 is n, and the absolute value (|n|) of -3 is 3.
To divide by 2 is the same thing as to multiply by its reciprocal. 2 is 2/1, and its reciprocal is thus 1/2. Dividing by 2 = multiplying by 1/2.
2-3 = 1 * 1/2 * 1/2 * 1/2
2-3 = 1/8 or 1/23
Notice:
2= 8 and 2-3 = 1/8
Exponent properties
• Suppose you have (3x)and you want to distribute the exponent. The 3 and the x are being multiplied, and you distribute the exponent to both. (3x)= 33 * x3, or 27x3.
• If you are multiplying two bases with same or different exponents, and the bases are the same, then the product is that base with an exponent that is the sum of both exponents. Or: Xa * X= Xa+b
Think about it. If you have this for example: 23 * 24 = 23+4 = 27, why does the rule make sense? Well, what does 2even mean? It means 2 multiplied by itself 3 times. You can imagine three 2s multiplying each other.
2 * 2 * 2
And now you are multiplying those 2s by more 2s. By four 2s, in fact, multiplying each other.
2 * 2 * 2 * 2 * 2 * 2 * 2
Which is the same thing as 27. When you multiply the same bases with different exponents, like we just did, you are taking the first list of the base multiplied by itself, and multiplying it by the second list of the base multiplied by itself.
(Below, the first list (of the multiplied base) is bold, and the second list is not:)
23 * 24 = ?
2 * 2 * 2 * 2 * 2 * 2 * 2 = ?
This new, longer list of the base multiplied by itself a bunch of times can be written in a simpler way that is easier to handle; a way that is more compact and easier to carry and work with. You can write that long list as the base with an exponent, the exponent referring to the number of times the base appears in the list.
23 * 24 = ?
2 * 2 * 2 * 2 * 2 * 2 * 2 = 2?
Since the number of times the base appears in the long list equals the number of times the base appears in each of the short lists, the exponent of the base representing the long list equals the number of times the base appeared in each short list.
• If you have a base and an exponent within parentheses, and there is an exponent outside the parentheses, you multiply the exponents together to get the new exponent of the same base. (Xa)= Xa*b
An example would be (23)2. In this case, you multiply the 3 and 2 together to get 26.
With (23)2, what is happening is exactly what seems to be happening: the 2is being multiplied by itself twice. Imagine this happening:
(23)= (2 * 2 * 2)=2 * 2 * 2 * 2 * 2 * 2 = 26
• When you are dividing one base by the same base, but both have different exponents, you subtract the second base’s exponent from the first base’s exponent to get the answer exponent, which you place above the base to get the answer.
X/ X= Xa-b
For example:
2/ 2= 23-4 = 2-1
This answer, 2-1, can be written differently. Remember that when an exponent is positive, you multiply 1 by that base the exponent number of times. But when an exponent is negative, you divide 1 by that base the exponent’s absolute value number of times.
So 2-1 = 1 / (2/1)
Which you know is the same as:
2-1 = 1 * (1/2)
So 2-1 = 1/2.
Want to know for sure that this is true? Whip out a calculator or use Google’s free online one and type in 1/2 and press enter. The calculator says:
1/2 = .5
Now type in 2-1 and the calculator ought to say:
2-1 = .5
They equal the same thing, which means that 1/2 = 2-1 without a doubt.
Thank you for reading this, and I hope you enjoyed it!
# Chemistry | Changing the World
Alright. It’s been a while for sure.
Hmm. I don’t like writing about a single subject like I see others do. That is not much my thing. I do enjoy reading such focused collections of information. As for my posts, I imagine they are a bit more difficult to follow. Not because the material is too complex for my readers, but rather because of several other reasons. For one thing, I have an obsessive nature that often makes me want to understand even the smallest, most irrelevant, most unimportant details. I enjoy learning about many different things, and my individual posts sweep broadly but don’t tend to go too deeply because though I like going deeply into subjects as opposed to shallowly, I like to mix up my activities and don’t just focus on, for example, chemistry all day. However, I have time for chemistry every day, and that is how I make some vertical progress (that’s how I dig deep). So, if my reader were to read all my posts, they would see me digging increasingly deeper into subjects.
One of the things that brought me back to my blog is my reading of Aaron Swartz’s writings in the book The Boy Who Could Change the World: the Writings of Aaron Swartz. It’s a super interesting book and I recommend you read it–but it may not be in your subject of interest. Among other things, Swartz was an advocate of free culture and free information, and his thoughts have had a deep impression on me. The book has made me consider how I can help the cause he so passionately believed in, and that he is making me believe more in, bit by bit. I want to help spread information–perhaps I stopped because I felt my services weren’t needed. But the idea of the Great Library of Alexandria appeals to me (just like the idea of Wikipedia does): a place with millions of papyrus books, handwritten by academics–in the ancient world, a great place of information and learning. The Internet allows for a huge compilation of information, and though I love learning, sometimes I feel that knowledge means nothing if it is just for myself. I must connect to others, and writing about what I learn and what I do is a way to do so.
The thing is, I will write anyway. Whether I publish my writings on the WWW (World Wide Web, invented by Tim Berners-Lee) or not, I will still write. I will write out my thoughts, write my explanations–whatever. I can’t guarantee what I write will be of use or interest to anyone, but since I’m going to write anyway, I might as well publish it to give others the opportunity to derive use from it.
Deprotonation and protonation both occur in acid-base reactions. This is chemistry. There are various definitions which define what acids and bases are. In the same year, two scientists came up with the same acid-base definition, and so both received credit for it. It is called the Brønsted-Lowry theory. This theory basically says that an acid is anything that donates a proton, and that a base is anything that accepts a proton–a proton, in this case, being an H+ ion/ cation (cation just means positive ion). The things I am mentioning but not explaining the meaning of in sufficient depth for a newbie to understand (nothing wrong with being one, we all must be when we start learning new things) are things I have explained in previous posts. If I were to start over in every post, that wouldn’t be very practical, you see. Like often, the bold terms are ones that are brand-new and that I’ll explain. Why are H+ cations called protons? Remember an ion is an atom with a charge (either positive or negative) because it has an unequal number of protons (+) and electrons (-). If there are more electrons, the ion has a negative charge corresponding to the number of electrons it has more than protons, and vice versa. For example, if I have an ion with 6 protons (+) and 4 electrons (-), the ion has a positive charge/ is a cation. But it has a 2+ charge, because it has 2 more protons than electrons.
Right. Anyhow, the H+ cation used to be a hydrogen atom. When it was, it had 1 proton and 1 electron (you can tell by looking at the Periodic Table. Hydrogen’s atomic number is 1, meaning it has 1 proton). Now it has a positive charge of 1+ (the 1 is implied so it is not written), meaning it has 1 more proton than electron. Meaning there are no electrons, meaning the 1 electron must have been lost to give this hydrogen atom its 1+ charge. (It’s not like the hydrogen atom could have gained a proton because that doesn’t happen, to my knowledge; the number of protons does not change. Or easily, at least.)
This hydrogen ion has 1 proton and no electrons. So its practically a proton, huh? It has the charge of one, at least. That’s why they call it that, a proton. H+ cation/ ion = proton. Now, back to the other things I have to define. First off, in acid-base reactions, acids react with bases. Something that is an acid in one scenario is not necessarily always an acid. For example, water is both an acid and a base in different scenarios, according to this definition, because water can act like an acid sometimes and like a base other times. That is, sometimes water can donate a proton, making it an acid; other times water accepts a proton, making it a base.
The process where an acid donates a proton is called deprotonation. The process where a base accepts a proton is called protonation. Simple, really. Finding that things that seem complex are actually quite simple used to surprise me more, back when I knew less. If you were intimidated by the terms, I want you to understand that it was natural for you to not know their meaning if you lacked the knowledge. I really hate–it really frustrates me–when people think they don’t understand because they can’t understand, because they’re not smart enough, when in reality they are simply lacking the knowledge to understand it.
It may not be necessary for me to tell you this, but I have faith in you. Whoever you are, you have great capacity as a human being.
.
Sigh. I got stuck on a Crash Course Chemistry video (#8). I’ve noticed lots of gaps in the videos that are very important. So far I’ve been able to keep up because I filled in the gaps with knowledge I learned in chemistry class–but I just reached a point where the gaps are ones I can’t fill at this point. I have to leave the videos and learn more about chemistry. That’s quite a shame for others, because I feel that the resource has some holes which impede full understanding. I don’t criticize meanly here… I just feel that it needs some fixes. By gaps, I mean that key things which have not been explained in previous videos are used to help one understand certain material. Well… if I don’t understand that material, how can I understand this one? I can’t. Hmm. I refuse to give up. Chemistry is wonderful and I will not. I’ll try Khan Academy, which may be completely thorough. The reasons I’ll try Khan Academy are that I’ve done learning on the site before and can’t remember having had gap issues, and that I can’t think of many more places I can get this information from. My junior year ended and with it did my reading of the chemistry textbook which I had to turn in.
Also, Sal Khan is someone I easily admire. He has a mindset I agree with, and it is reflected in the content on his site. I really enjoy his site.
Saturday
The more I learn about the world and its problems, the more I want to help. Learning about problems from advertisements’ misleading nature (which prevents people from making the wisest choices) to the dangers of pesticides to lead paint which is still in use, to mercury light bulbs–learning that child labor is still in existence as is slavery, seeing sexism, racism, homophobia, and other forms of discrimination, seeing broken families
Seeing people litter and generally not care much for the environment, seeing some people not even bother to recycle, seeing not everyone get a high quality education, seeing people age, seeing people get sick with things like cancer and Alzheimer’s
Seeing air pollutants in China condense lower than the atmosphere in the form of brownish smog, a visible substance that is dangerous and harmful for people…
Learning in my wonderful health textbook that the ozone layer, made of ozone gas, is a naturally occurring shield for our planet’s life, because it absorbs most of the Sun’s harmful ultra-violet (UV) radiation and only a little bit leaks through. UV light harms all life, but the ozone layer keeps us safe.
But we’re destroying it with the air pollution we are causing, because air pollutants destroy the ozone layer.
That is so sad. But it’s not too late to make progress. To make things better. To cut back on suffering. To make better choices. To change our ways. It’s not.
As long as there is life, there is hope. We are alive right now and we can take action. We cannot guarantee that we can do the same thing after death.
I mean, I’m not okay with that! And if the people of the world had solving these problems as their top priority, think about how far we could get!
I feel idealistic, and optimistic, and I feel ready to make improvements.
I love solving problems. And when I see these things happening and more, I am filled with the desire to do something to make them better, because I care.
But I can’t do all of this myself. I seriously can’t no matter how strong my desire. So this is what I propose.
You need to inform yourselves. All people need to do this. If you don’t know, you have no power over what happens. You have very little control and a very small ability to protect yourself and your planet and your people. Information will free anyone. A lack of information is the equivalent of chains binding around you–except you don’t see them. Your world gets smaller.
I’m going to get to the rest of my proposal in a bit. I was talking yesterday night to my uncle about my desire to change the world but my lack of knowing how. I wanted some guidance. He is past the idealistic feelings of youth and told me things which gave me less hope. But still, I was going to do something. And then he told me that my intention was vague. He said I couldn’t change the world unless I knew the world. Truly, I don’t know it entirely. And the both wonderful and tragic is that there is so much to be learned, that my lifetime will not be long enough. I cannot change everything. He made clear what I had not wanted to hear. But it is true. And it’s good I heard it, because when I can’t fix all the problems and instead can only work to fix some, I can narrow it down. The task of fixing all the problems I wanted to do was too overwhelming. How can I change all of these things? I cannot have control in all areas. But others can. Others, too can find ways to change the world–problems to fix. It’s not just my world, after all. Others can handle other problems, while I handle certain problems. He asked me what I thought the most important problem was. What I wanted to fix the most. I might change my mind, but this is an easy question. If I could fix anything, I would fix aging–for those I love and don’t want to lose, for the rest of humanity, and for myself. If I live longer, I have more time to spend helping. If I live long enough, I can make many contributions over time. Do you see?
There is a scientist called Aubrey de Grey who claims it is possible. I am skeptical, so I must review his work and claims. But if he is onto something–if aging, the deterioration of the human body, can be broken down into simple processes and errors which cause the deterioration, and if we can reverse some of them at least–I must know, so that I must get involved. There is nothing I want more than to extend my life. And those of the people I love. And those of people with potential that I admire. But I will not keep this from the public at large.
What I propose is that I will work on a problem facing humanity, and at the same time spread awareness of other problems, so that others may step forward like I have and decide to help other problems we have. Nobody can do it by themselves, so let us all work together.
# Chemistry, U.S. gov., and the Internet’s effects on the brain
Sunday
In this periodic table, you’ll see element boxes. Each box has 2 separate numbers, the one on the bottom (below the element’s symbol) being my focus. The top number is the atomic number, revealing how many protons there are in each atom of that element. The bottom number reveals different things.
(If you can’t see the small numbers too clearly and would like to make the image bigger, you can, for one thing, (if you’re using Chrome, at least) right-click the image and open it in a new tab. Or maybe you’d like to zoom in.)
The bottom number is called the atomic mass number, because one thing it shows is the mass (in amus) of one atom of that element (more or less). For each element, this number is reached by taking the average of the masses of all of the element’s naturally-occurring isotopes.
This number also shows the molar mass of the element (in grams). That is, if you have one mole of the element (6.022 x 1023 atoms of the element), it’ll have a mass of x grams, x being the number.
Here’s an example. Locate the element oxygen, O, in group 16 and period 2. (Groups are columns; periods are horizontal rows.) Anyway, oxygen’s atomic number is 6. That’s irrelevant. The other number is rounded to 16.00.
I hate to say it because it’s not totally accurate because of isotopes and their different masses–but 16.00 is oxygen’s atomic mass number and thus each oxygen atom has a mass of 16.00 amu. (Technically, I think I should say ‘amus,’ but it’s less common and unnecessary for understanding, so I’ll drop it. Also: units matter! Last post, (or was it the post before that…?) I explained the meaning of amu, but I’ll tell you again pretty quick. Amu = atomic mass unit. 1 amu has the mass of a carbon-12 atom’s mass, divided by 12.)
But in addition, 1 mol (mol is short for mole) of individual oxygen atoms (6.022 x 1023 oxygen atoms) weighs 16.00 grams.
So basically, atomic mass number (amu) = molar mass (grams).
That’s right, the molar mass of a substance is the mass of 1 mole of that substance. Knowing the molar masses of the elements is useful, because compounds are made up of elements, and we can figure out the molar masses of compounds in this way. How? I don’t expect that to make total sense yet, let me explain.
Compound: “a thing that is composed of two or more separate elements; a mixture.”
Let’s take a compound like H2O. Suppose we want to know the mass of 1 mole of H2O molecules (6.022 x 1023 H2O molecules). Visualize here, just a bit. In 6.022 x 1023 H2O molecules, there are 6.022 x 1023 O atoms, wouldn’t you agree?
So:
6.022 x 1023 H2O molecules = 6.022 x 1023 O atoms + the hydrogen atoms.
Hmm, how many H atoms are there? Well, that 2 is there, meaning each H2O molecule has 2 H atoms. If it wasn’t, there would be 6.022 x 1023 H atoms. But… there is twice that amount. So there are 2 * (6.022 x 1023 ) H atoms in all those H2O molecules.
So:
6.022 x 1023 H2O molecules = 6.022 x 1023 O atoms + [ 2 * (6.022 x 1023 ) ] H atoms.
This is important!
Don’t forget what we are looking for: the mass of 6.022 x 1023 H2O molecules (aka 1 mol of H2O molecules).
What if we knew the mass of all the oxygen atoms in those H2O molecules, and the mass of all those hydrogen atoms in the H2O molecules? Why, then by adding them we would discover the mass of the H2O molecules, right? (Those H2O molecules are merely a sum of their parts.)
So the part that is not bold is what is unknown:
mass of H2O molecules = mass of O atoms + mass of H atoms.
Or:
mass of 6.022 x 1023 H2O molecules = mass of 6.022 x 1023 O atoms + mass of [ 2 * (6.022 x 1023 ) ] H atoms.
Sure, the other parts of this equation so far seem unknown. But we can figure them out. Let’s start with the oxygen. We want to know the mass of 6.022 x 1023 O atoms. Or basically, we want to know the mass of 1 mol of O atoms. (See how it connects?) The mass of 1 mol of oxygen… why, you can find that off the periodic table!
Listen… this stuff doesn’t come effortlessly to anyone, I don’t think. I have to make an effort too, of course. I feel like many people wrongly think stuff like this is too complex for them, but it’s not. It’s really very simple. Well–it is complex, but the most complex things can be broken down into biteable pieces, chewable pieces that make logical sense anyone can understand. Very often, difficulties are due to knowledge gaps–no intrinsic fault, no lack of ability.
Just wanted to say that.
According to the periodic table, the molar mass of oxygen is 16.00 grams.
So:
mass of 6.022 x 1023 H2O molecules = 16.00 grams + mass of [ 2 * (6.022 x 1023 ) ] H atoms.
Now we must find the mass of 2 moles of hydrogen. According to the periodic table, each mol of H is 1.01 grams; 2 moles are twice that amount: 2.02 grams.
So:
mass of 6.022 x 1023 H2O molecules = 16.00 grams + 2.02 grams
That’s simple math now.
mass of 6.022 x 1023 H2O molecules = 18.02 grams.
Or:
mass of 1 mol of H2O molecules = 18.02 grams.
We have our answer. What did we do? We used the molar masses of the elements H and O to find the molar mass of a compound (specifically, H2O). The point of that activity was to see exactly how the molar masses of elements can be used to find the molar masses of compounds.
I hope you enjoyed that. I did. It’s important for us as humans to engage in activities which challenge and stimulate us without overwhelming us. 🙂
No one wants to be stressed, but no one wants to be bored. Neither is good. This blog gives me an opportunity to do what I long to do–learn and explain things. That is one of the things in this life that bring me true satisfaction.
.
Now, the process won’t take this long every time you want to find the molar mass of a compound, because you can trust this and use it every time: the molar mass of a compound equals the molar masses of all its constituents. All compounds are made of elements, so you can find those molar masses on the periodic table. You must also multiply the molar mass of a certain element by the number of atoms of that element there are in the compound. The compound H2O had 2 hydrogen atoms, which was why I multiplied hydrogen’s molar mass by 2. You already know that this works–I showed it to you in the example.
I’ll walk you through another example so that you see how one can do it faster, taking the shortcut which we can take because we already know that the shortcut works. Let’s find the molar mass of sugar, C12H22O11. Because sugar contains atoms belonging to more than 2 elements, it is a compound. Remember, the molar mass of sugar equals the molar masses of its constituent elements, multiplied by the number of times they appear.
Here’s how I lay it out:
C –> 12 x molar mass
H –> 22 x molar mass
O –> 11 x molar mass
——————– +
molar mass of sugar
Or:
(12 x C molar mass) + (22 x H molar mass) + (11 x O molar mass) = sugar’s molar mass
Use the periodic table to find the molar masses of the individual elements.
(12 x 12.01 g) + (22 x 1.01 g) + (11 x 16.00 g) = sugar’s molar mass
342.34 grams = sugar’s molar mass
That is, 1 mole of sugar (6.022 x 10^23 molecules of sugar) has a mass of 342.34 grams.
Ha, you can also see why it’s useful to know substances’ chemical formulas. ^.^
Monday
The U.S. government consists of 3 branches: the Executive Branch, the Judiciary Branch, and the Legislative Branch. The Legislative Branch is the most powerful and it consists of Congress. Congress is divided into the Senate and the House of Representatives. There are committees within each, and congresspeople (‘congressmen’ doesn’t apply anymore, so I don’t know why people still use it) are often in multiple committees. A standing committee is a permanent committee that meets regularly.” Each committee has a head, called a chair or chairperson.
Special or select committees also exist, created to deal with matters beyond the abilities of standing committees. Some of these select committees have only advisory abilities, meaning they can’t actually write laws and can only advise. The Select Committee on Energy Independence and Global Warming is only an advisory committee. That explains much of Congress’ response to climate change.
There are joint committees, which consist of congresspeople from both houses (both the Senate and the House), as well as conference committees, which serve the purpose of reconciling bills “when the House and Senate write different versions of it.” Bills are not laws yet, they are merely laws proposed by congresspeople. All laws were bills once, and all bills have the potential for becoming laws.
Committees exist because laws are written more efficiently in smaller groups, and because the congresspeople in certain committees can be allowed to develop expertise on those subjects.
Proposal power is the power of all congresspeople to propose bills. If a bill is proposed by a senator, then it must pass through a committee before it is introduced to the rest of the Senate. (Sorry about my wacky capitalization, but to be frank I often don’t know which words to capitalize…) If a bill is proposed by a representative, it must also pass through a committee before it reaches the rest of the House of Representatives.
When bills get passed to the appropriate committees, committee chairs decide which bills are considered. This power of committees to choose which bills make it through is called Gatekeeping Authority. If a bill does not receive the majority of votes, it does not pass through a committee, and it is said the bill died in committee. Mark-up refers to the authority of chairs to manage the writing of bills.
Committees also can kill bills by refusing to vote on them, though in the House, committee members can be forced to vote on a bill with a discharge petition. Another important power of Congress is its ability to oversee its laws being implemented.
.
I was a bit bored writing about this. And that wasn’t good news because if I don’t enjoy this, what do I do it for? I wondered if maybe I simply didn’t enjoy government as much as other subjects, like chemistry, for example. I didn’t want this to be true, but I thought that maybe I didn’t enjoy telling you facts just like that–I thought that maybe I preferred being able to explain parts of a system to you. I’ve always loved systems. I’ve always loved how parts within systems make sense and fit so nicely with each other. It’s why I loved math, and chemistry. I love explaining systems, no matter what the subject is. I thought that maybe my source of boredom was due to my desire for precision taking over me and making me write you every fact as soon as I learned it, instead of waiting to take it all in and see how things fit together and then explain it to you.
Whatever the case, I decided to stop writing and just watch the video. As I learned about how things fit together, however–how things were related like a system–I suddenly wanted to explain this to you. And so I paused the video and came to write this down, for future reference, because the more I understand myself, the easier it is for me to keep myself happy and balanced.
The more I understand what makes me happy, the more I can increase that in my life.
So. Newt Gingrich was once the Speaker of the House. The Speaker has the authority to decide which committees to refer bills to. Gingrich increased the power of the Speaker, changed the rules for appointing committee chairs, and reduced the number of subcommittees.
Subcommittees are committees which contain some members of a larger body (e.g. a committee/ board) and which report to them.
Whereas before Gingrich’s changes, the chosen committee chairs were the longest-serving members of the dominant party, after his changes they were elected through the voting process.
Tuesday
It’s nice that, no matter how few reads a given post gets, there’s always at least someone that benefits from my writing. Sometimes there are many reads. Sometimes there are not.
Yesterday I said that maybe my lack of motivation was due to the subject, or me not allowing the subject to get complex enough and me not having enough of relationships to explain. Now, a different idea occurred to me. It may be that I have this feeling that my work is pointless and won’t be read. Probably because last post was a post I put a lot of effort into, and not many benefit from it.
I know the barrier is in my mind. If I believe my work has meaning and purpose and is beneficial to others, then I can really feel motivated to teach and I really get something out of that work. I find it satisfactory. I believe this post ought to get at least one read, and that one person should be enough!
Speakers also have influence over those who become chairs in the voting process.
Friday
Sorry I didn’t tell you much about U.S. government.
When there are lengthy sections of bold text in this post, the writing is not mine.
# The Internet May Be Changing Your Brain In Ways You’ve Never Imagined
“Once I was a scuba diver in the sea of words,” he wrote. “Now I zip along the surface like a guy on a Jet Ski.”
In the book, which became a New York Times bestseller and Pulitzer Prize finalist, Carr explored the many ways that technology might be affecting our brains. Carr became particularly concerned about how the Internet seemed to be impairing our ability to think deeply and to focus on one subject for extended periods.
I’ve heard this before. I don’t seem to have difficulty focusing/ concentrating… but then again, I engage in mentally stimulating activities all day, every day. I also have important values where I care about precision and understanding everything. I read carefully and don’t skim.
I don’t use social media because I don’t want to be distracted from my work. I read books and don’t use Google’s plethora of answers to cheat my way through school like some other students do.
Part of the reason I don’t use social media (or my phone in general) is because it’s always been a huge distraction for me. Social media offers instant gratification. Posting pictures on Instagram and getting likes feels good. Having the option of texting friends when I’m doing something difficult that requires considerable mental effort and focus would not be good for me… I would most likely seek pleasure from aesthetics as I have always done in the past. Being constantly distracted would make it more difficult for me to focus on my work over time. Just giving into temptation once would make it easier to make the same mistake next time.
But I suppose the strongest reason might be the fear I have that my emotions will gain strength if I sit back and let them operate as they wish. The more I get instant pleasure, the more I please my id. Back when I was id-dominated… I had less control. I don’t want to go back to that. I don’t want to lose control and not care about my future as much. I want to make sure I’m driving in a good direction, and I don’t want to focus on shallow things…
So what has changed since Carr wrote his seminal work five years ago? We chatted with the journalist and author about how our increasing interactions with mobile technology might be affecting the most important organ in our bodies.
“Since you wrote this book, the Internet has only taken on a bigger role in our lives. What are some of the main changes you’ve observed in the way we interact with technology?”
When I wrote the book, the iPhone was still very new and the iPad had just come out. Social media wasn’t as big as it is today. So when I wrote the book, I was thinking about laptops and computers but not so much about smartphones. Of course, now the main way that people interact with the Internet is through mobile devices.
Yes, I use my computer but not my phone. On my phone, I can’t really do any of the things I can do with a computer. Some websites and blogs don’t load/ load properly, articles are harder to read because of the smaller text… phones and tablets do seem better for social media, however.
In the book, I argued that what we created with computers and the Internet was a system of distraction. We got the great rewards of having basically unlimited information at our fingertips, but the cost of that was we created a system that kept us in a state of perpetual distraction and constant disruption.
What psychologists and brain scientists tell us about interruptions is that they have a fairly profound effect on the way we think. It becomes much harder to sustain attention, to think about one thing for a long period of time, and to think deeply when new stimuli are pouring at you all day long.
I guess it doesn’t help that my schedule has been one that, for about all year, has required me to pay a certain subject attention for a mere hour or so before moving to another subject. I had so many things to do, you see, and I didn’t want anything to be gotten around to too late, so I minimized the amount of time I spent doing each thing… at the moment, I’m not using my schedule. I’m not in school anymore (summer vacation) and my time is being used up by family events and obligations. Basically, things are different than they were before. As a result, I can do less things each day… but I decided to quit my schedule, at least for now. Now, I’m doing what pleases me–and I’ve been mostly reading articles and books. (That’s not all that pleases me.)
It used to be so hard to read…
…to really focus on more practical, difficult topics. Now it’s a lot easier, now it takes a lot less effort. I wasn’t used to it, that’s why. I had never really read nonfiction and then I sat myself down, properly motivated, and I did it. The driving force came from within me. I read the book Evolution: the Triumph of an Idea by Carl Zimmer. It was hard at first, but I had solid determination, and I changed my brain. The more you practice, the better you ought to get.
It was a very beautiful book. I really do love reading. For me, it’s one of the most wonderful things to do. To be able to read is one of the most precious gifts in my life.
I argue that the price we pay for being constantly inundated with information is a loss of our ability to be contemplative and to engage in the kind of deep thinking that requires you to concentrate on one thing.
To me, all the things I worried about have become much worse now that we carry around this permanently connected device that we’re constantly pawing at. Things are very different in a way that makes the things I worried about worse.
“Research has found that millennials are even more forgetful than seniors. What do we know about how technology is impacting our memory?”
There’s something I want to say. If technology really does make it more difficult for us to concentrate on a single subject and we go less deeply into it as a result, I can see how that could affect our memory of the subject. My junior year of high school just ended. That year was divided into 4 quarters, or 2 semesters. At the end of each semester, I had a final. The finals I just had for almost all my classes tested our knowledge of material covered this past semester.
My math teacher didn’t teach well. He gave procedural instructions–he taught how to do certain problems. He taught us tricks and the general rules. I was able to remember how to do these problems for about a week–long enough to do associated homework and the quiz or test that came with them. But I didn’t remember how to do these problems long enough for the final exam because he taught us too shallowly. He didn’t explain the why. He didn’t explain the importance, or how things work. The information he gave us felt irrelevant. Interesting, but irrelevant. The knowledge I have in my head relates to reality, explains aspects of reality, and pieces of information in my head relate to other pieces of information. Within my internal model of reality, there is a network of connections between information. Same for you. But when he gave me this new information, I didn’t have connections between this information and other information in my knowledge bank, because he didn’t provide them. He taught only shallowly, not deeply.
So I had to study like crazy for my math final, because I didn’t remember anything. Because I didn’t go deeply into it. Sad. But–no worry. I will go back and deeply learn what I could not in that classroom–on my own time. I won’t abandon math like that. Math is beautiful–but much of that beauty lies in the way its ideas connect. In the relationships, in the why. In how it is a structure, a system. My geometry teacher my freshman year was a blessing to me. She showed me the relationships, and that was the year I first fell in love with math. It’s been a long time since I’ve felt that way about math because I’ve been distracted, but I can still remember that breathless feeling, that euphoria as I sat in that desk, learning of the system and how nicely it all fit together, and that feeling was so intense, so passionate.
It’s not lost. It’s just been a while. I can’t return to math right now, because I’m doing other things now. But I will eventually, and I will love it just the same.
I really can’t remember the feeling too well… but I do remember that the subject left an extremely deep impression on me. Math was the most beautiful subject I had ever seen. Absolutely logical. I should remember this when choosing my career.
Anyway, my chemistry class was taught by someone who actually cared about subject. The teacher seemed to be constantly frustrated by students’ lack of interest, and I could understand that. It is sad to love something, but to feel that no one else does. In the beginning of the year, I had an aggressive attitude toward my math teacher because I was frustrated by his lack of love for math. (A main cause of aggressive behavior is frustration. Understanding this has helped me control my frustration to control my aggression so that I am less hurtful when upset.)
But I always cared about chemistry, and she saw that. Part of the reason my love for chemistry was limited was because I didn’t really see the spark in anyone else. Remember that I told you that I feel that without people, ideas don’t have meaning? Without others to care, the most beautiful ideas seem to be dying. I know others care, but this knowledge isn’t enough to satiate my emotions. My emotions need to see others caring.
Anyway, I went into chemistry a lot more deeply. The teacher told us to read the chapters, and I did–until I got too busy. Semester 1, I read all the chapters. Semester 2, not really. I took excellent notes. We got study guides and work sheets. We went over it in class. We got this awesome packet every chapter testing our deep understanding called a POGIL. I loved POGILs. I always was a bit… afraid that I wouldn’t manage it, but I always did. POGILs were challenging and required me to think. They were spectacular. I can’t even explain how much they helped me understand/ go over the basic ideas.
Chemistry is easy, okay? Yeah, it requires work. Sometimes it’s challenging. But everyone can do it. It irritates me to no end that people don’t believe they can follow something as easy as chemistry or math as far as I have. Simple, logical concepts are the building blocks of complex ideas.
Eh. The class average on tests was always like a 60%. I always got a good grade though, and not because of natural talent. Many kids didn’t actually try, and that’s why. It was a shame. It was their loss. I don’t hate them, I understand them, and I would give them an infinite amount of opportunities to try again.
I just want others to care.
The point is that I went deeply into chemistry. Want to know what effect that had on my memory of the subject? The teacher posted a test with 100 questions online, 70 of which would be identical to those on the semester 2 final. Yeah, I had to go through my notebooks in search of formulas. But just formulas–not concepts. I’d say that I remembered at least 95% of the stuff on the online test. (As a result, there was almost no studying required.) I surprised myself immensely with how much I remembered, and I sought an answer. Why? Why had I remembered so well? It was the depth. It was the connection of ideas. It was the time I put into it.
That’s what I had to say. Maybe the deeper you go and the more time you put into something–maybe the more you focus and the more you think about something, maybe the more you understand the concepts, the why, and how things fit together–the better you remember. If technology is making people do this less and less, one can see why people’s memory of things may be suffering.
“Research has found that millennials are even more forgetful than seniors. What do we know about how technology is impacting our memory?”
Technology definitely has an effect on our memory. What happens is that to move information from your conscious mind (what’s known as the working memory) into your long-term memory requires a process of memory consolidation that hinges on attentiveness. You think about the information or rehearse it in your mind in order to form a strong memory of it, and in order to connect it to other things that you remember.
If you’re constantly distracted and taking in new information, you’re essentially pushing information into and out of your conscious mind. You’re not attending to it in a way that is necessary for the rich consolidation of memory.
Since I wrote The Shallows, there have been some very interesting studies which show that we seem to be less able to form long-term memories than we used to, thanks to technology. One study out of Columbia University showed that when people know that they’ll be able to find information online easily, they’re less likely to form a memory of it.
That isn’t something that I’ve studied much, but I think there are some indications that this kind of culture of constant distraction and interruption undermines not only the attentiveness that leads to deep thoughts, but also the attentiveness that leads to deep connections with other people.
One study I mentioned in the book seemed to show that the more distracted you are — the more your train of thought is interrupted — the less able you are to experience empathy. So distractions could make it more difficult for us to experience deep emotions.
“In the book you talk about the “dark side” of brain plasticity. What does that mean?”
Neuroscientists have discovered that the brain is plastic, meaning that it’s very malleable or adaptable. Our brains are constantly adapting at a physical level to our environment. You can imagine that what’s really changed our environment in the past 10 or 20 years is the Internet and social media.
A lot of people will assume that if our brains can adapt, then our brains will adapt to the flow of information and all will be well. But what you have to understand about neuroplasticity is that the process of adaptation doesn’t necessarily leave you a better thinker. It may leave you a more shallow thinker.
Our brains adapt, but the process of adaptation is value-neutral — we might get smarter or we might get dumber, we’re just adapting to the environment.
“Are you optimistic about any of the ways we currently seem to be adapting?”
No. It’s the ease with which we adapt that makes me most nervous. It doesn’t take long for someone to get used to glancing at their smartphone 200 times a day. We’re creatures of habit mentally and physically.
When you develop that habit of distraction, it becomes harder and harder to back away and engage our minds in deeper modes of thinking.
Part of the reason I don’t touch my phone if I can help it–part of the reason I don’t take it anywhere and I usually forget it exists–is because I know what it’s like to engage in such shallow thinking. I told you. I was id-dominated.
I had an Instagram account which allowed me to gain instant gratification. I was addicted to it. I was much more ignorant than I am now. My life didn’t have much deep purpose, I wasn’t interested in being patient and in long-term relationships, I preferred fiction to reality. I focused on aesthetics, on writing, on roleplays–on emotions. I focused on art, mostly.
I think I’m a fairly visual person. Back when I satisfied my id’s desires instantly and was under my id’s control, I spent most of my time looking at pictures. I don’t know why I enjoy that, but doing that can entertain me for hours. Pictures, gifs. Searching for beautiful drawings and paintings, beautiful digital art, beautiful characters. My id still wants to do that sometimes. It is when I am most emotional that I am most vulnerable to my id’s desires. Just yesterday after I woke up, I wanted my brother and I to engage in a Zootopia roleplay between Judy Hopps and Nick Wilde (I’d be the latter). As I looked up pictures of the former character out of an eagerness to interact with her, I became more and more deeply involved in the activity of looking at images and I was again whisked away by my id’s wishes. I left my computer at my grandma’s house when going to my other grandma’s apartment, annoyed and disappointed in myself, as well as not trusting myself to “behave.” Instead, I took a book.
Yesterday was an exception. I think yesterday happened because I have been ignoring my id’s desires for too long. Why? Because here I have no privacy (I haven’t even a room of my own) and I have less time to myself, and it always embarrasses me to look at aesthetically pleasing pictures in front of others because I don’t want them to know what I find visually beautiful/ attractive.
…especially because my aesthetic appreciation knows no gender (and people still have problems with diversity in this regard), and because I often find people beautiful/ attractive without feeling attracted to them.
“Is there anything we can do to keep our mental faculties intact, or is it pretty much hopeless at this point?”
Well, you can use the technology less and set aside your phone and spend a good part of your day trying to maintain your focus and not be interrupted. The good thing about that — because of the plasticity of our brains — is that if you change your habits, your brain is happy to go along with whatever you do.
My id eventually desired change, and for a shallow reason. It desired power, and I knew knowledge would give me more of that. That’s when my superego began to operate. I developed this idea of who I should be and the goal of perfection became my focus. I knew that my brain was plastic and malleable–that the activities I engaged in would change my brain–and so I tried to change myself and mold myself into the person I wanted to be.
Here is the rest of the article.
I hesitate to move on to other things and make this post too lengthy, including too many different subjects, because potential readers may not be interested in all these subjects and may avoid reading the post altogether.
# Patience, chemistry, love, Cosmos, math, traveling
Personal
Monday
I just realized something useful.
My superego and id battle all the time. At one time of my life I was dominated by my id, at another by my superego–now I am more balanced. The superego still makes demands of me that I don’t fulfill, as does the id. The id desires instant gratification, among other things. I feel like that’s very important to understand.
I don’t use my phone much. I don’t take it to school. This makes it easy for me to do work when I should. If I had my phone and the option, in general, to do what my id desired to do, it would be more difficult for me to do my work. Because we’re just animals, and if we can get the same amount of happiness for less work, why not? Most of us would want this deal.
My upbringing has spoiled me. Even adults are spoiled in this way. Nowadays we get what we want faster than ever. Many of us are less patient as a result. This makes perfect sense. And yet it may mean that we don’t go as deep into things. And that’s a sad reality. My superego tells me to watch out for this. Despite the trouble it has caused me, I’m thankful for my superego because it thinks about my future, my impact, my potential, my happiness, and the like. My id is the more primitive part of me that has deep desires that can’t always be satiated.
(Neither the id or superego operate thinking about reality, according to Freudian theory. The ego thinks about reality, and acts like a mediator, balancing the id and superego’s opposing desires. The id desires pleasure, and the superego desires perfection.)
I’ve always wanted to write a story, a book. I’ve always stopped at one point. I think it’s because I give up, because of lack of faith in it yielding fruit sometimes. After all, writing sometimes gets dry and tedious. If I’m searching for beauty, friendship, meaning, deep relationships, and deeper love, though, I have to understand I’m not getting those things instantaneously. I have begun to discover this already as I have begun to work on relationships with family members. I realize the more I put in, the more I get out. And I am thinking about how this relates to other aspects of my life.
I think I will stick with writing fiction. I will also stick with this blog. Something else that ought to help me to keep writing even when the road gets bumpy and I get stuck is that this happens to me in real life–I fall down, I get dirty, I move backwards without meaning to, I head in the wrong direction. Sometimes I think I won’t get up again, but I know I will, because I always have before. I can’t escape my reality. I can’t just get a new reality like I can start a new story. This inability to escape my situation forces me to confront it. When writing, I have the option to escape (I can stop writing), and I have always done so before. This has not been satisfactory. What I must do to keep going is to work with what I have and work to fix my situations, work to fix my problems instead of avoid them, and this is useful for writing. When a story gets messy and it seems to be dry, or stuck, or whatever, instead of trashing it all and starting over like I have in the past, maybe I should work to fix it. And maybe the same thing applies to relationships. That’s what I’ve begun to realize.
Nothing in life is perfect. Not love, not relationships, not stories–nothing. It all has its ups and downs.
So I was doing my math studying for tomorrow’s math final, but I kept getting id urges to watch a video or do something that would give me instantaneous satisfaction. I wondered what the point was of doing math if I didn’t enjoy it. I wondered how I had once loved it so much, so deeply, so passionately. And then I realized that what I was doing was the right thing, that what I had to do was stick with it, understanding I wouldn’t get instant gratification, or bliss, or any of that. Whenever I loved math before, I hadn’t expected bliss from it, or an emotional high, or anything. And the gratification I got was much deeper.
Wednesday
Today was my last day of my junior year of high school. I’ll be a senior when summer vacation ends.
And after that year, I’ll go to college!
I’m not particularly eager. It is very exciting, but I’m perfectly happy where I am. I like to take things day by day and go where I want to go. It’s sort of… wandering through life without a map, but I don’t mind. This way of living has its disadvantages, sure, but it also has its advantages–and there are certainly disadvantages to planning it all out ahead.
For example, I’ve seen many students who seem to have it all planned out but who don’t seem particularly eager or happy about where they are or where they’ll be (if, for example, their parents are pushing them to pursue a certain career). And then there are, of course, people who are so focused on a goal that they don’t focus on the things that matter the most to them.
I guess I’m glad I fell out of that path. I didn’t even see I was heading the wrong way. My superego had dominated my life, and I desired achievement, worth, perfection, status, and so on so desperately that I completely used my person for my resources but didn’t attend to my needs. That is, I made my body work hard and focus on studying, but I was too harsh with myself. I didn’t love myself. I felt guilty all the time because (it’s obvious now) I couldn’t be perfect.
There was too much stress because of myself but I didn’t… notice there was something wrong with my way of living, because I had gradually gotten to this point.
That was when I seriously started to question things, to ask myself what it was all for.
When I realized that happiness was the only thing I really wanted, the thing that mattered the most to me, everything changed–because I lived my life for a new reason; because my entire existence had a new purpose. I sought joy, meaning, satisfaction and the like as opposed to perfection, achievement, ability…
I became so focused on my goal that I wasn’t living for the now. I was neglecting what truly mattered to me.
Chemistry
But I came to talk about chemistry. On the periodic table you see boxes with element symbols. Somewhere in each box lies a number which probably has a decimal point. That is the average atomic mass number (the other number is the atomic number). As a rapid refresher (I already explained this stuff in a previous post), the average atomic mass of an element is the average of all of the element’s isotopes’ atomic masses.
For example, chlorine’s (average) atomic mass is 35.453. But 35.453 what? (What are the units of measurement?) 35.453 amu.
atomic mass units
Most units of measurement are arbitrary. The foot, for example (12 inches) measures distance. But it’s not like the foot existed without humans. It exists within our culture, it’s a way for us to quantify the universe. But we made it up. As a kid, I used to be confused by that. Remember how I used to worry about objective truths, about things that were consistent across the universe–everywhere? Things indisputably true? I always wondered how the foot (and other units of measurement) was accepted across the universe. But it’s not, actually. I didn’t quite understand that then.
Each foot is the same, right? 12 inches here measures the same length as 12 inches on Mars. Why not? After all, 12 inches is a measure of a certain amount of distance. Humans wanted a way to divide up distance, to measure it. So they could compare the distance between areas, so they would understand the world better. So they needed to create units, and they created the foot (among other units) which would always mean 12 inches.
But it’s arbitrary. Created by humans to measure the world.
Anyway. What is 1 amu even equal to? 1 amu equals 1/12 of the mass of a carbon-12 atom.
Okay, so carbon is an element. There are 3 isotopes of carbon: 12C, 13C, and 14C. What these isotopes have different is the number of neutrons they have.
A carbon-12 atom will have 6 protons like all carbon atoms, but 6 neutrons (like all other atoms of the 12C isotope). This carbon-12 atom has a certain atomic mass (protons + neutrons). Divide that mass by 12, and you get another number–the mass of 1 amu.
Thursday
I should also note that elements’ isotopes appear in different amounts throughout the world. For the element carbon, for example, carbon-12 is much more abundant than carbon-13 and carbon-14 combined. Of all carbon, approximately 98.89% of it is carbon-12.
When calculating the average of an element’s isotopes’ masses, the relative abundance of each isotope is taken into consideration.
Now what are moles in chemistry? A mole is a unit of amount, telling how much you have of a substance. Here is a precise definition:
“A mole of a substance is defined as: The mass of substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of 12C. Fundamental units may be atoms, molecules, or formula units, depending on the substance concerned.”
I would define a mole as the amount of a given substance that has 6.022 x 1023 fundamental units. My definition is compatible with the definition I found online–they both are correct.
* Like said in the quote, fundamental units are atoms in some cases, and molecules or formula units in others. Molecules consist of atoms. A single atom is but an atom. Molecules are covalently-bonded atoms. Formula units consist of ionically-bonded ions.
Covalent bonds are sharing bonds. The atoms involved in covalent bonds share electrons. Ionic bonds are not sharing bonds. Table salt, for example, is NaCl. Na is an ion, and so is Cl. They are oppositely charged–Na has a positive (+) charge whereas Cl has a negative (-) charge. Ionic bonds occur when one atom steals one or more electrons from another, resulting in both atoms becoming ions which then are attracted to each other because they have different charges.
I’ve talked about all this before, but in the case of NaCl, because Na has the + charge, it is a cation. Because Cl has the – charge, it is an anion. Anyway, how would you refer to a unit of NaCl? You know, a single compound consisting of one Na cation and one Cl anion–how would you refer to that? Would it be an NaCl molecule?
NO, because NaCl doesn’t consist of covalently-bonded atoms, remember? (If you go back up to the star, it has this information.) Remember that formula units consist of ionically-bonded ions?
You would call it an NaCl formula unit.
Anyway, I was talking about moles. 6.022 x 1023 is called Avogadro’s number.
When you say you have a mole of water (water is H2O, a covalent compound–a molecule), you mean you have 6.022 x 1023 water molecules.
When you say you have a mole of NaCl (an ionic compound), you have 6.022 x 1023 formula units.
Likewise, a mole of a substance can have 6.022 x 1023 atoms.
You may wonder where Avogadro’s number comes from. Just like the amu (atomic mass unit), the mole comes from the carbon-12 isotope. In 12 grams of carbon-12, there are 6.022 x 1023 atoms.
.
Personal
This is uncomfortable for me. That’s not new.
Unless you’ve been with me before, you have no idea what I’m talking about. Sorry about that.
Anyway.
Being asexual appeals to my superego. But it may not be true. My trauma made sexuality seem negative, as I told you–sex especially. Thinking of myself as asexual is sort of a nice thought for me because I feel purer. But I’ve never been 100% convinced. It’s like I said: the matter is complicated, so much so that all my “conclusions” are subject to changes I don’t foresee.
Ever since I realized Nick loved Gatsby, I’ve felt differently about Nick. I dare say I’ve even been attracted to him. That’s not something totally new, or totally surprising. A bit disappointing, maybe. I’ve been healing from my trauma. I call it the same thing because I don’t want to talk about it much. Eh, I’ve helped myself. I’ve been changing my ideas. I’m a lot more accepting of myself. Yet it is somewhat hard to accept some truth like this. But I’ve got to.
I don’t want to think about this much. This is just me recording my goodbye, and hopefully recording my reasons for it. Attraction can be intense–too intense. It’s too overwhelming for me. It can even become one of the most important reasons for living, because I get so much pleasure out of it. All that pleasure I get makes many other things I used to love seem much duller. The sun doesn’t shine so brightly. I don’t want to live that way. I remember when my superego dominated me. I wanted to be asexual because my superego wouldn’t accept any other answer. So I had to love something. I wanted to love “what mattered,” which was the universe, the world. So I poured all my feelings into reality and it seemed beautiful. I loved my work and I loved my universe. And Carl Sagan taught me to look at the cosmos with a new awe. I was infatuated with Sagan for his ideas, for his hope, for his cosmos. And that feeling was… well, it seemed like the proper feeling.
On the other hand, attraction to people causes me a little bit of guilt, and a feeling of emptiness and dissatisfaction. I desire to write fiction so that I can interact with characters that make me feel this way–and yet my id is interested in instant gratification and doesn’t want to write the story for the right reasons. It is almost like the person becomes the only source of pleasure.
I know what I must do to get off this train, to stop these feelings. Curiosity brought me here, but I can leave them again because they don’t feel right. I don’t like that feeling of sadness and desire, of melancholia, of regret, etc. Attraction is always accompanied with a strong feeling of sadness. I feel frustrated, helpless, and unsatisfied. I feel like a victim to my powerful feelings that I feel unable to overcome.
It kind of sucks. It feels good in a way, just like sadness does, but it also feels heavy, serious, and dreary…
And I know someone close to me who lives that way. I now understand so many things about him. I think I’m right–I think his constant brooding and sadness are a result of him being focused on the wrong things.
Maybe part of the reason I pushed romance and attraction away is that I didn’t want to feel so heavy. I wanted to feel lighter and less serious. I wanted to see the beauty in the world and feel optimism. I just forgot why I closed the box, and I opened it.
All I must do to close it is push it away for a while, until the strength of the feelings fades and I am capable of moving on without sadness and regret, without powerful feelings urging me to not love work and instead love a person romantically.
Phew. I’m actually very grateful for writing. It allows me to heal and deal with my feelings. I understand them better and I let them out, and I can return to a more stable, more enjoyable existence.
I suppose I don’t want to live too passionately… because I’m afraid of the long falls, and because I’m afraid of losing control. Those long falls are too desperate, too hopeless. Emotions can be very positive, but also very negative, and I don’t want to be in pain like that. I don’t like giving emotions free rein.
Eh… I was thinking about writing, but I think I’m a bit unstable now. I’m not sure of where I stand, and I don’t want my fictional writing to be an emotional whirlpool that will drain me, as has happened before.
I guess writing calms me. Makes me feel like I have more control. Thoughts I have can’t fly away and fade if I write them down. Then in the future I can learn from them.
Writing is a way of me analyzing myself and testing the ground beneath my feet. It allows me to draw a plan and feel I am in greater control of myself. I guess I take writing for granted because writing of this type comes so easily to me. But I know what it’s like to not really be able to write, or to at least require much more effort to write, because of my experience with fiction writing. It doesn’t come totally naturally. I have to make an effort.
When I was a kid, I was a fan of the Percy Jackson books.
I didn’t read ’em for romance. That wasn’t my reason. When I got the Star Wars game to play with my younger brother, I wasn’t thinking about romance. I feel like romance is just… too complicated, emotionally. It leads to too much emotional turmoil and I don’t want that kind of… suffering. But at the same time, it’s such a beautifully rich aspect of the human experience, and so it’s kind of wonderful.
Well, actually, romantic love doesn’t seem too bad. It felt only wonderful. For a short time, I felt romantically in love with Daisy, if I’m using the word ‘romance’ appropriately (still a source of confusion for me). Her personality wasn’t as great as I wanted, however, and personality matters a lot. But when I felt that way about Daisy, I wasn’t scared. I felt her warmth, her kindness, her sweetness, and I saw nothing but happiness and friendship for us both.
For me, romance is something a lot more bright and sunny, a lot more platonic and manageable. It’s attraction which is so much more disruptive, making me lose my head. And that’s why I’m somewhat afraid of it, like my dad is somewhat afraid of the ocean. He respects and is wary of its power.
It’s attraction which seems more like a dark night, like an emotional storm I can’t escape and which I am at the mercy of. I can’t be that emotional, I feel, otherwise I won’t be able to protect myself from negative emotions.
The people I love in this unique way are usually female, like I said. I think I may be biromantic. I think they are usually female because these personality traits are most common in women. These characteristics are attractive, but in a non-frightening way. I also agree with choosing a partner that I love for their personality, and not simply because of appearance.
The people that make me feel this way have a decent amount of the following traits:
Kind, sweet, caring/ loving, appreciative, genuine, honest, understanding, friendly, optimistic, light, playful, peaceful, delicate…
(That list lacks many other personality traits which I have found pleasant and attractive. It’s just to give you an idea.)
This love is much safer, much warmer, much more optimistic, much less wildly passionate, much more practical, and much more promising. It seems to be happier and more harmonious. I also feel compelled to give to this person.
Wild attraction doesn’t last, anyway.
Before I saw the movie, I formed my impression of the characters from pictures. I had read the book, but I wanted to imagine the movie character as different from the book character. Maybe in the movie, she would have the personality I was looking for. I wanted to start over with her and fall in love with her again.
But I was never loving her. Just who I thought she was.
I imagined she was what I was looking for, and I wanted to be with her as a person. It was platonic, but I call that kind of love romantic.
I want to be with someone I love as a person.
Attraction just gets in the way.
Mm, I was talking about the Percy Jackson books and the Star Wars video game to explain that… I have sought and continue to seek action, adventure, camaraderie/ friendship/ teamwork in fiction. And if that’s what motivates me to read, surely that can motivate me to write. Writing cannot be about romance if I want to finish a story.
Cosmos
Egypt is located in the northernmost part of Africa. Egypt’s coast meets the Mediterranean Sea (as do other territories, including France, Spain, Italy, Greece, Turkey, and Syria).
Within Egypt is the city Alexandria, which once was one of the world’s greatest cities.
Within Alexandria lived a man named Eratosthenes who was many things, including a poet, a mathematician, a geographer, a philosopher, an astronomer, and the chief librarian of the Great Library of Alexandria.
Friday
In the Great Library, he found an interesting account in a papyrus book which said that in the city Syene (south of Alexandria), on June 21 at noon, vertical items (such as sticks in the ground or temple columns) cast no shadow. The sunlight fell from directly overhead, snaking its way into deep wells and illuminating the water at the bottom.
What was once the city of Syene is now Aswan.
Alexandria wasn’t too far north of Syene. Eratosthenes wanted to experiment with this occurrence, and wanted to see whether or not vertical objects in Alexandria would also cast no shadow on June 21 at noon. He discovered something which had a deep significance. At the same time that a vertical stick in Syene cast no shadow, a vertical stick in Alexandria did cast a very definite shadow.
If the Earth had been flat, then it would make sense for, at the same time, a stick in Syene to cast a shadow of the same length as the shadow cast by a stick in Alexandria. Likewise, it would make sense for a stick in Syene and in Alexandria to cast no shadow at the same time.
What Eratosthenes had was evidence that the surface of the Earth was curved.
This is a really awesome demonstration that helped me understand this better. I don’t expect you to watch all 50 minutes. To see the demonstration, watch the video from 28:16 to 29:15.
Saturday
I didn’t get to write much yesterday because I left my house around 2:00 to take 3 consecutive planes to Colombia, South America. That’s where I am now, hello! I wanted to rant about everything that’s gone wrong (like the fact I got less than 4 hours of sleep last night), but the need has since passed.
Anyway. Sagan says Eratosthenes hired someone to measure the distance between Alexandria and Syene. He learned the distance was 800 kilometers.
The stick in Syene was sticking up in one direction, and the stick in Alexandria was sticking up in another direction, right? Because the Earth is curved.
“The Sun is so far away that its rays are parallel when they reach the Earth.”
Here is Sagan holding a map of Ancient Egypt in such a way so that the map is curved like the surface of the Earth is. The stick on the left is the stick in Alexandria; that on the right is the stick in Syene:
If you imagine the Sun’s rays coming down parallel to each other, like rain falling straight down on a day without wind, you can see that the sticks are at different angles to the sun’s rays. These different angles are responsible for the different shadows–for the shadows’ different lengths.
So you know how a circle is 360 degrees? And if you slice up that circle into slices, each slice will have a degree measure of its own which is less than 360 degrees? Well, in math there’s an angle called a central angle. In a full circle, the central angle is 360°. If you cut a circle into 4 equal slices, then each slice will have a central angle of 90° (360 / 4). That’s how it works.
Well, imagine a circle. 360°, as expected. Imagine there are two points–point A and point S. They’re pretty close together on the circle. (This is not drawn to scale):
If you draw an imaginary line from each point going down past the center of the circle, you’ll see the lines eventually intersect/ cross. When they cross, a central angle will be formed.
The central angle is shown by the purple line. So, as I have told you all this, I have been talking about the Earth. Point A is the stick in Alexandria; point S is the stick in Syene. They form an angle, and that angle is about 7 degrees.
So, you have your Earth, a circle. 360 degrees. You cut it up into sufficient slices so that each slice is 7 degrees. 360 / 7 is around 50, so there are about 50 slices of 7 degrees in the Earth.
Heh, what can be done with this information? Well if you’ll remember, Eratosthenes figured out that there are 800 kilometers between A and S. If there are 800 kilometers for each slice, and there are about 50 slices in the circle/ world, then the world’s circumference is about 40,000 kilometers (800 * 50).
I would stay and chat, but for one thing I’m sleepy as heck and can’t do this anymore, and for another, I lost my internet connection and don’t have access to the first episode of Cosmos anymore.
Of course, you won’t notice the time gap. 🙂
.
So I threw on shorts and a t-shirt and took a “nap” for 5 hours. I feel alive again… not that I didn’t feel alive earlier… it just felt that earlier I was operating at an inappropriate time. It felt I had ruined things, not started the day properly. It felt like the day was not actually mine, but rather a part of the next day I shouldn’t see more of until I slept.
Yeah, I don’t know either. Point is, I feel in control of the day again. Though things were beautiful before, having energy makes the world seem brighter because then you actually have the means to do all these different activities. So the world seems more filled with opportunity.
.
I realize that when I listen to people, I understand them better and care much more about them. In addition, it’s fun to read a book with a friend. My younger cousin and I are reading the first Harry Potter book together (in Spanish), to catch up to the bookmark that is placed where it is because of the reading that my other Colombian cousin and I did back in the U.S. when she came to visit my family of four. Once we’re all at the same point, we’ll read the new parts together.
To be exact, the Earth’s circumference is, according to Google, 40,075 kilometers. Not only did Eratosthenes figure out the world was round, but he also figured out the Earth’s circumference.
I’m fairly comfortable where I’m at… I notice that as I write this and delve deeply into this, I am happy and satisfied, but I once would have been extremely disappointed in myself–back when the superego dominated me. No matter what I did, I could not be the ideal person I wanted to be. And I didn’t do anything out of love. I didn’t learn subject material out of interest. I did it out of a crazed desire to have as much control as possible.
Since I can never completely start over with myself, just fix myself, there are elements in my life today that I have that period of time to thank for. Though I enjoy these activities now, they are the same ones that I engaged in when dominated by my superego. If not for that period in my life, I may never have explored these areas. I may have been afraid of challenging myself. During the superego period, life without challenges scared me. In my mind, challenges were mandatory. They would refine my abilities.
There are good and bad aspects to every experience.
This is getting long, so I’ll continue writing in another post. Thank you for understanding the material I presented. ^.^
# Teaching, French Polynesia, chemistry, Sputnik, The Great Gatsby
Tuesday
I never met (not entirely accurate, but I was too young to remember him) my grandpa (my dad’s dad) and so I don’t know what he was like, not really. But he was a philosophy professor and many people on my dad’s side of the family are teachers: my uncle, at least three (immediate) cousins, my grandma (married to my dad’s dad).
I’ve never really considered being a teacher. I don’t want to be a teacher of high school kids or below because I feel that many kids don’t care, and that takes the passion out of it. But I was thinking about what my mom said yesterday just now (because Google announced it to be National Teachers’ Day), and I was thinking about teaching people that wanted to learn, and I felt passionately. I felt like that was what I want to do desperately. And I think that that’s surprising. I never imagined myself as a teacher, like I said. I especially don’t want to teach the same stuff over and over once I’m over it. But what I do want to do is gather information, learn deeply, process it, explain it, and move on to other information. That is exciting.
Yesterday, my mom picked me up from school like always because I’m the only teen at my school that doesn’t have a car. (The driving manual is boring, I don’t have the time, and I don’t really feel motivated to get a car. I don’t currently have a need for one.) In the car, we talked about how I felt frustrated at my French project about French Polynesia. Don’t get me wrong: it’s absolutely fascinating, if you go deeply into it. But my teacher wasn’t giving us the time to, and I have so much other homework I’m receiving (in addition, I am studying for my final exams) that I had to go shallowly into the water, but that was very disappointing for me and caused me frustration. I wanted to learn everything. But I have understood that I can’t. I have an obsessive nature (which school and life force me to suppress) when it comes to learning, and I love to go deeply into things, but the real world keeps me on a leash because I am never given the time to just devote myself to one thing. I also can’t decide on one thing.
My mom said that I’ve always been this way. Obsessive about information. I always get excited about research projects because I really want to research. But the thing is, if I do it for myself, it kind of doesn’t mean much. I want to give it to people. I want to teach to people. I think one of the things which has held me back is that I don’t see much of an audience for what I have to say.
But who cares. Just one person matters. Teaching it to one person only is enough, if they care.
I just thought today, when I considered teaching all the different subjects my heart desired, that that was the most satisfactory, most wonderful thing I could, at the moment, think to do. Especially if the people cared about learning, it would be simply wonderful to explain. Learning and teaching, I feel, are beautiful things not appreciated enough. Some teachers at my school hate being teachers, hate to teach, hate to explain. Like they don’t understand the beauty of knowing, of understanding, the beauty of the science and the math and other subjects…
So I thought, I want to try teaching. I want to attempt to satisfy that need. I thought about teaching on my blog. It’s sort of what I already do, but I could do it more consistently… I could put more effort into it. The thing is, many times that I have knowledge, I don’t have the time to explain it because I have to put that knowledge to use. I suppress that desire, just like I suppress the desire to understand everything I can deeply. I even thought about stripping this blog of its… personal content and making it something more “professional”—something more serious for me so that I would feel that this was more than a hobby I could casually neglect and rather, an obligation. Because no matter how much I love to do certain things, if I feel that they’re just hobbies, work that must be done for school comes first because it must be done. It’s a have-to, not just a want-to.
Hmm. I want to start over, but at the same time… I don’t. I’ve made some awesome acquaintances here, I’ve explored some blogs, I’ve shared things that mattered. I’ll stay as opposed to starting a new blog and identity, but this blog of mine may serve a new purpose.
Wednesday
For my French project, I’ve been learning a bit about French Polynesia. When I sought to learn about its environmental issues, I found that Polynesia was used as a test site for nuclear tests performed by the French. The French, between 1960 and 1996, performed 210 nuclear tests in Algeria (Africa) and French Polynesia. (193 nuclear tests were performed in French Polynesia.) Polynesia lies in the Pacific Ocean.
Nuclear fallout washed all of Polynesia. In one explosion on the Mururoa atoll, Tahiti received 500 times the maximum accepted amount of radiation.
at·oll
ˈatˌôl,ˈatˌäl
noun
a ring-shaped reef, island, or chain of islands formed of coral.
The atoll is southeast of the island Tahiti (which contains Papeete, the capital of French Polynesia).
France had previously hidden the true toxicity of the nuclear tests and the reach of their radiation. Recently declassified documents have revealed all this information. The Mururoa atoll alone has been the site of so many tests that it may just collapse.
Veterans and civilians in Polynesia have formed groups and sought compensation for the health problems caused by these tests. Investigators have found high levels of cancer in Polynesia. This is tragic.
But it’s not surprising.
Very few people have received the compensation they deserve from the French government. It wasn’t until 2010 that it agreed to offer compensation. Controversially, Jacques Chirac, a former French president, resumed nuclear tests soon after being elected in 1995. What also appalls me–though it shouldn’t surprise me–is the lack of protection that workers received, as reported by the troops, during nuclear explosions.
During the Mururoa tests in French Polynesia in the late 1960s, one veteran described how he was stationed in shorts and a T-shirt on a boat only about 15 miles from the explosion before having to sail immediately to the area of the vast mushroom cloud to examine the damage.
Others on different tests wore shorts and had no sunglasses; they were told simply to shield their eyes and turn their backs at the time of the explosion.
Sunday
Hank Green on Crash Course introduces stoichiometry as a way to measure chemicals going into or coming out of chemical reactions. Measuring these things allows us to determine how many molecules or atoms there are within a substance.
If you’ve ever looked at the periodic table of elements, you’ve probably noticed numbers below element symbols.
For example, below the element hydrogen’s symbol–H–is the number 1.006. This is the atomic mass of one hydrogen atom… more or less. You see, this number, 1.006, is actually the average atomic mass of all of hydrogen’s naturally occurring isotopes.
(Some isotopes don’t exist unless they are made by people, so naturally occurring isotopes refers to isotopes found in nature/ found naturally.)
The element hydrogen has 3 naturally occurring isotopes. Let me first explain what an isotope is, though. Atoms are made of protons (+), electrons (-), and neutrons (neutral charge), right? Atoms never lose or gain protons (protons being in the nucleus/ center of the atom), so if you know the number of protons an atom has, you know the element that atom belongs to. The periodic table helps you with that. For example, say you are trying to identify the element of an atom that has 52 protons. On your periodic table, locate the element which has the atomic number 52. The atomic number says how many protons an element’s atoms have. Te, tellerium, has 52 protons. Therefore the atom is a tellerium atom.
I’m not done explaining. Hydrogen has 3 naturally occurring isotopes, called hydrogen-1 (protium), hydrogen-2 (deuterium), and hydrogen-3 (tritium). These all are hydrogen atoms, and thus have the same number of protons. On the periodic table, it can be seen that hydrogen has 1 proton (because 1 is hydrogen’s atomic number). Therefore 1H, 2H, and 3H each have 1 proton. Yet their atomic masses are different. All isotopes of an element have different atomic masses.
Okay, but what is atomic mass? Electrons have a very small mass. It is so small that, in measurements, it is practically insignificant. That is why, when figuring out the atomic mass of an atom, we ignore the mass of the electrons. Hmm. So what do we pay attention to? Well, we pay attention to the mass of protons and neutrons, but that’s it. When finding out the mass of an atom, you add the mass of its protons to the mass of its neutrons, ignoring the electrons’ mass completely.
So, the (atomic) mass of an atom = proton mass + neutron mass.
All protons have the same mass. So if you know how many protons there are in an atom, you multiply that number by the proton mass to get the mass of the protons for that atom. All neutrons have the same mass. (All electrons have the same mass, too, though you won’t need to deal with electrons’ mass.)
This is logic right here. Check it out. You have isotopes 1H, 2H, and 3H, all hydrogen isotopes. We still don’t know what isotopes are–what makes one isotope of hydrogen different from another hydrogen isotope. (I couldn’t explain it to you right away because there was background knowledge you needed in order to understand.) But we know that all hydrogen isotopes have the same number of protons, and thus the same proton mass.
Here. The bold text is what is the same. It’s very simple. The atomic masses are different, but the proton mass is the same. Well, atomic mass is proton mass + neutron mass, and so the neutron masses must be different.
Atomic mass of 1H = proton mass + neutron mass
Atomic mass of 2H = proton mass + neutron mass
Atomic mass of 3H = proton mass + neutron mass
That’s it. Every isotope of an element (like hydrogen for example) has a different number of neutrons. Different number of neutrons means different neutron mass, which means a different atomic mass. So as you have seen, not all of hydrogen’s atoms have the same mass. If all of hydrogen’s atoms had the same mass, that number would go on the periodic table, no doubt. But since that is not the case, but a number must go on the periodic table, the average mass of all of hydrogen’s natural isotopes is used. After all, we need to know the (atomic) mass of 1 hydrogen atom to perform calculations… and it wouldn’t be practical to find out how many 3H or 2H or 1H isotopes there are when performing calculations, would it? Not when you’re dealing on such a large scale. You see, we are massive, as Hank Green put it. We don’t think of ourselves that way, yet we are. We consist of such tiny things, but we deal with enormous things. A grain of rice is enormous, when you think of all the atoms that make it up. So… it wouldn’t be practical to be perfectly accurate, because that would require knowing too much.
Moving on. What’s in a name? Why is the hydrogen-1 isotope labelled that way? How are isotopes named? By the way, let me mention that those 3 hydrogen isotopes (1H, 2H, 3H) can also be written as 1H, 2H, and 3H.
The way you distinguish one isotope from another is that you write the element name or symbol, along with the mass of that isotope.
Carbon-14 (14C) for example, is a carbon isotope with an atomic mass of 14.003242. 3H (hydrogen-3/ tritium) is a hydrogen isotope with an atomic mass of 3.0160492.
.
Sputnik 1 was launched by the Soviet Union on October 4 in 1957, beginning the Space Age. Sputnik 1 became the Earth’s first artificial satellite. (The Earth already had a satellite at the time: the moon. A satellite can be an artificial object placed in orbit around a celestial body (like a moon or planet) for communication or information-gathering purposes. A satellite also can be one celestial body in orbit around another.) The Russian word for satellite is спутник, and it sounds like “sputnik.” That is how the satellite acquired its name.
As you can see from this picture, the satellite was not very big, measuring just 22 inches in diameter.
Sputnik 1 was about 184 pounds and took about 1 hour and 30 minutes to circle the Earth. It moved at 18,000 miles per hour. Its orbital was not perfectly circular, but rather elliptical. Its apogee was 584 miles; its perigee was 143 miles.
The apogee of an orbit is the point of the orbit that is farthest from the orbited body. A satellite in orbit around a body (e.g. a planet around the Sun) is farthest from the body when it reaches the apogee.
On the contrary, the perigee of an orbit is the point of the orbit that is closest to the orbited body. A satellite in an orbital will be closest to the orbited body when it is at perigee.
It could be seen with binoculars only before sunrise or after sunset–but not while the sun was in the sky. Sputnik 1 was about 10 times bigger than the satellite the U.S. had planned on sending into orbit… next year. In January 1958, the satellite burned up in the orbit as expected, and on that same month the U.S. launched the satellite Explorer. This fairly rapid advance in the U.S. space effort was largely due to the banding together of the scientific community, the military, and the government to catch up to the Soviets. This started the Space Race.
# Freud and psychoanalysis
4/27/17
Freud and psychoanalysis
I didn’t want to continue taking notes on paper because I worry it’ll be harder for you to read the content. Taking notes on my computer while I’m reading is also too difficult, so I borrowed my dad’s mini iPad to write down thoughts during my reading. We’ll see how it works out.
I remember thinking yesterday, as I took these notes, about how these psychology concepts applied to me. When younger, I think I used to be more dominated by the id. That really makes sense, considering how, during adolescence, connections strengthen between the frontal lobes and the limbic system. The limbic system is the emotional part of our brains, whereas the frontal lobes are rational and are responsible for our higher-level thinking. The frontal lobes have more control over the limbic system thanks to this, which means that we have more control over our emotions.
This video brings back some old memories. I remember when I was thing a lot about the fascinating thought that a person could have 2 consciousnesses in their head. If this interests you, search up “split brain.” See, the narrator of the video mentioned the corpus callosum, which connects the brain’s two hemispheres. Epilepsy used to be treated by severing the connection, by cutting the corpus callosum in half and thus cutting off communication between both hemispheres. They kept operating independently, separately conscious.
Anyway, I think that for some time (I wasn’t liberated from myself until recently) I was dominated by the superego.
“Someone with an exceptionally strong superego may be virtuous yet, ironically, guilt-ridden” as I was every night. I was striving for perfection and I was very disappointed because I couldn’t reach it. Now I think I’m dominated by the ego, which is perfect. I understand the superego’s ideals, and I understand the id’s wishes (remember how I was talking yesterday about drives/motives?)–but now I operate on the reality principle, as the ego does.
The superego operates on ideals, the id on the pleasure principle, but now I recognize reality and seek a balance. I understand I can’t be perfect. Awkwardness is a reality, and though my superego is never satisfied by anything I do, I understand now that my superego never can be. That understanding allows me to actually try things, because I know I may fail and that’s there isn’t anything I can do about that. I try not to make mistakes but I’m just human, and I understand that now. That’s why I keep writing on this blog–because if I waited until I was comfortable, until I was perfect, I would never write at all. Sometimes there are things I tell you which I regret, and I always understand that sharing yourself is exposing yourself to criticism. Even Freud received loads of criticism for his work. While I don’t agree with everything he thought, I think he was important and had great potential, and I’m glad he shared. Charles Darwin was very afraid of criticism, and put off publishing his theory of evolution for a long time. He was pushed to publish when someone else began figuring the same thing out. Darwin had a lot of evidence at the time.
Totally irrelevant but awesome: https://youtu.be/ppyzZYNROV8
Maybe you won’t appreciate it much if you don’t watch the show, though (Steven Universe). You can’t love characters you don’t know, and without love things are less beautiful.
.
Freud theorized there were 5 psychosexual stages. Talking about this personally bothers me, but I do see the sense in what he thought and that’s something to focus on: the sense and system of his theory, how different ideas are connected. Here’s a chart:
What I think is interesting is that according to Freud, when straight children realize they can’t beat their same-sex parent, they try to be like them, and this includes copying their values. The superego holds our ideals, our views of how things should be, and when children copy their parent’s views of how things should be, their own superego is developed. I guess I haven’t appreciated Freud that much in the past, but now I realize more deeply that he was a great thinker.
I don’t know whether Freud was right or not. I’d like to think sexuality doesn’t matter that much to us humans, but I realize we’re still animals. We can’t separate ourselves from who we are. Some people choose to deny human nature, not accepting evolution, for instance. Some things about us are very disappointing to me. But I won’t close myself to the truth because of that.
Freud did all this thinking attempting to find explanations for symptoms he saw in his patients which he couldn’t explain using traditional methods. He searched for psychological causes for the symptoms.
Just a sec. You know how I’m studying for my AP psychology exam which is why I’m rereading all this? I’ve read all this before–yet I had a different motivation before. Before, just like right now, I read it partly because I had to, but whereas before I also read it to learn and remember (not enjoy it), this time I’m reading it to enjoy it. This time around, I understand concepts better, remember more things, and connect ideas better. And that wasn’t even my intention. Something to note, as I may find it useful in the future to be reminded of this.
Anyway, Freud thought conflicts during the psychosexual stages could cause conflicts in the future. For example, someone who had been deprived of oral gratification during the oral stage might be fixed on that stage as an adult, seeking pleasure orally (e.g. by smoking or overeating). Being fixed on a stage is called fixation. Freud thought sexuality was very important to personality. And I wonder about the truth of that.
I can certainly see the impact of our gender, sexuality, and romantic orientation on our personalities. I can see gender differences, so pronounced, just because of gender… femininity/masculinity certainly impacts people’s self-expression…
And of course, why wouldn’t sex and romance–related to reproduction–be a big deal to us? If we were all asexual, we would have a lower chance of passing on our genes. The reason I’m here is because my ancestors passed on their genes through reproduction.
“Anxiety, said Freud, is the price we pay for civilization. As members of social groups, we must control our sexual and aggressive impulses, not act them out. But sometimes the ego fears losing control of this inner war between the demands of the id and the superego, and the result is… anxiety…”
This made me pause. I do understand that. Sometimes I feel a bit anxious about whether my ego will be able to maintain the balance. My superego demands I stay in STEM, and desires I not turn to art too much. I guess what I’m afraid of is returning to the land of fantasy, creativity, and art, and losing too much control over my emotions. I worry that the more artistic I become, the more emotional I will become, and the more power my id will have over me. I worry about my impulses being unchecked, because I don’t want to be like other teens who get addicted to social media or games… I don’t want my id to make me value instant gratification over long-term rewards. I don’t want to live in the present that much, where my willpower begins to wane… because what about my homework? What about being responsible? It’ll be more difficult. That’s why I feel a bit of anxiety every day, because I worry I will slowly shift to allow my id more and more, until it has the upper hand, like it used to when I was a child, like it does in some adults.
This morning I was dreaming of animation. I think it would be really cool to learn to do that, but there are only so many hours in the day and I must choose what I will do each day. I was thinking about it, but I can’t help but worry a little bit. I’m afraid of satiating my emotions too much because I worry about making them stronger and losing control. So I try to maintain a balance (which is where I do a lot of unemotional activities and just have a little bit of emotional gratification each day. I’m a very happy person, not to worry!).
Freud thought that to battle this anxiety, the ego unconsciously distorted reality. The 6 defense mechanisms mentioned by the text I’m reading (Psychology, eighth edition in modules by David. G. Myers) are:
– Repression
– Regression
– Reaction formation
– Projection
– Rationalization
– Displacement
“Repression banishes anxiety-arousing thoughts and feelings from consciousness” meaning that, according to Freud, they were either in the unconscious or preconscious. Freud viewed the mind as an iceberg, with a tiny bit of ice sticking out of the water, and most of it under the water. The part under the water was the unconscious. The part above the water was the conscious, and the part between conscious and the unconscious (a fine line, the surface of the water) was the preconscious. The preconscious contained unconscious thoughts, feelings, and memories which could still be accessed by the person. Freud really studied his patients’ dreams because he thought that dreams would allow him a glimpse into their unconscious. He thought the manifest content (what the dreamer dreamt) was a censored expression of the dream’s latent content (the unconscious material behind the dream).
He also used free association to try to better understand the patients’ background and unconscious mind. He thought that in the unconscious there might be causes for the symptoms they were showing. Free association was where his patients said whatever came to mind, no matter how trivial or embarrassing. I don’t know if I personally would have been able to do that. However, his patients’ setting helped–they laid comfortably on his famous couch, stacked high with pillows, and faced away from him.
.
I would write more, but unfortunately I have too much homework and too many projects. I will get back to you on this whenever I have time. Thanks for reading!
# Flow, space, history
4/16/17
Gosh, I begin so many projects when I haven’t even finished previous ones… I haven’t even finished that wonderful episode of Cosmos because now I’m thinking about flow.
“In positive psychology, flow, also known as the zone, is the mental state of operation in which a person performing an activity is fully immersed in a feeling of energized focus, full involvement, and enjoyment in the process of the activity.”
I’m about to watch a TED Talk about flow titled Flow, the secret to happiness. (How am I supposed to resist starting something else with that kind of title?) You can find it here: https://www.ted.com/talks/mihaly_csikszentmihalyi_on_flow
.
Hmm. A state of extreme focus. One music composer said he lost his sense of self, and he did because there’s only so much information we can process at one moment, and during this state of consciousness in psychology, our attention is elsewhere. We ignore hunger and other sensations, and time seems to pass. I know this feeling. This feeling of fully being one with a subject. Before I knew the word “flow,” I knew the feeling. I remember stumbling across this picture and bringing the feeling to mind:
I once felt this way for like a whole week. I spent it on debate.org, reading other commenters’ discussion before I could add my own thoughts. I analyzed their words, I thought it through. I came to the conclusion that free will doesn’t actually exist. It was beautiful… it was fun, exciting, passionate. But then the spark began to die as it became repetitive and monotonous.
Csikszentmihalyi shows a chart, which shows that flow is reached when the challenges are high and our skills are high. I know this feeling. It’s an enjoyable feeling. I don’t think it always feels the same way, but it’s the feeling of being engaged and interested. Some people may have more ecstatic experiences, if ecstasy is the feeling of being out of reality, but even the amount of flow I experience is very nice. I remember I experienced flow more often when I did what I wanted to do, what interested me. When I became interested in competing, and power, and learning just to know and have control and not to really love and enjoy the material, I didn’t experience flow as often.
I enjoy explaining things, understanding things, tackling difficult problems. Explaining how to tackle difficult problems. I can experience flow in math when solving difficult math problems. Something that depresses me about my math class is that the teacher breaks the math down until it’s extremely simple. He doesn’t show us the why so much, and he removes all of math’s elegant beauty. Then he has us do more problems than necessary which make it too repetitive and boring. Sigh. But math is still very beautiful. When you don’t visit something, it appears as a stranger does—but I remember how I used to think math was my greatest love. Now math seems distant because I’ve been doing other things. In the same way, my dad seemed distant until today.
I refuse to let happiness pass me by. I will continue to pursue my interests and do things out of enjoyment! I will find this state of flow and learn more about it.
“Those are some of the things that molecules do
Given 4 billion years of evolution…”
4/17/17
This is the Orion Constellation:
Within the constellation is the Orion Nebula, a huge, interstellar cloud of glowing gas in the Milky Way. It is 1500 light years from Earth.
4/18/17
Returning to that first episode of Cosmos, pulsars are found within supernova remnants. There are supernova remnants in our home galaxy (the Milky Way): debris of stellar explosions which sent out “star stuff” into space. The first pulsar discovered was thought to be a sign of alien intelligence because pulsars keep perfect time in their spinning.
Since yesterday, I’ve been reading from my U.S. history book’s chapter on the civil rights movement. It’s been interesting, but when I focus on one thing for too long without breaks, I get overwhelmed—and I have been forced to read almost the whole chapter to fill out a packet that was due today. I think the information I’ve learned is very interesting—some is very moving—and I want to share it. I was surprised to have received as many likes as I did on my last post. I tend to assume that me sharing personal information, or getting off topic in general or whatnot, is bothersome.
Rosa Parks was riding on a segregated bus in Montgomery, Alabama—behind the white section—when she was asked to get up and move so a white man for whom there was no seat in the white section could sit down. (When public facilities were segregated, that meant that blacks and whites were separated.) I don’t understand why she had to move. The bus driver told all the blacks in Parks’ row to move, and they all did (there were 3 others) except for her: the white man could have sat down in one of the other, now-emptied seats without a problem. Maybe there’s something I’m missing, maybe there’s something I misunderstood, or maybe I shouldn’t expect rationality in an area (race) where these people have shown to be irrational. By that, I mean to say that the racist folk which agreed with segregation were irrational when it came to race equality, though that does not mean they were irrational in other fields/ subjects/ areas. (We all sometimes are irrational when it comes to certain things, but that temporary irrationality doesn’t mean we are always irrational.)
And to be objectively truthful… everything has a logical explanation. No matter how irrational it seems, it is not truly random. There is an explanation that makes sense. People operate in ways that make sense. The mystery is removed when the operating mechanisms are revealed—when all the reasons for why people act as they do are revealed. I don’t think racism was correct, but the truth is it’s perfectly understandable. I think that the idea of white supremacy is disgusting and shameful. But it’s not without reason. Just because something makes sense doesn’t mean it’s right—what I’m saying is that it makes sense, but it is still wrong.
Hate is not the answer. Not even for the people that have committed the worst atrocities, I suppose. Hate is not the answer because hate is destructive and blind. When you have hate, you get nothing from it. When you have love, you get everything from it. Love is what gives life meaning. Love is creative. I know that hate cannot be the answer because when I feel hate I want to destroy. I don’t care about others’ in that moment, and I feel compelled to destroy—inflicting damage brings satisfaction. Isn’t that so terrible? Yet when one feels love, that is not negative. One seeks to help somebody else.
I hated my brother for a time. I had no reason. When I decided to let go of that desire to harm him and allowed love to change me, I realized his importance, his beauty, his similarities, his positive attributes. And I never wanted anyone to hurt him, because that is what love is. Now when I get upset, I withdraw, as I said, because I understand that love is the answer. And anyone can be loved. We can feel empathy for people.
Sometimes I want to write a story in such a way as to make the reader feel regret and be moved. Slave narratives had such an effect on American society. These narratives were written by slaves that had escaped to the north, and the stories allowed people to place themselves in the slaves’ shoes and understand their struggles more deeply. Judging from the state the world is in, it seems that not enough people understand that hate is wrong, and so I sometimes want to write a story in which I can provide readers my perspective, but not just by telling them, but by showing them. I want to teach them that important lesson I have learned because it frustrates me that people are so quick to hate.
It frustrates me whenever I learn about oppression. The damage caused by hate makes me so sad. And it pisses me off that people keep hurting others because they just don’t get it.
The Freedom Riders were teams of blacks and whites sent south by the leader of CORE (Congress of Racial Equality) to spread awareness of the refusal of southern interstate buses to become integrated. There was still segregation, even though such segregation was now illegal. Freedom Riders were viciously attacked in Alabama (more specifically Anniston, Birmingham and Montgomery).
Just because it’s in the past doesn’t mean it didn’t happen. It did. And those peaceful protestors were violently beaten. It was so inhumane. I didn’t understand how this could occur—but then I realized that I knew the answer. It was anger—maybe fear also and some other emotions—which allowed for these disgusting atrocities to occur. And I understood it because, with sufficient anger, we can all take part in negative, destructive behaviors that hurt others and many of us might not care at the moment. This hatred is part of human nature. Doesn’t mean it’s right and should be kept.
As my book said:
{the riders emerged… to face a gang of young men armed with baseball bats, chains, and lead pipes. They beat the riders viciously. One witness later reported, “You couldn’t see their faces through the blood.”}
I read this yesterday, so the worst of the anger and outrage has passed. Now I’m addressing it in a more detached manner. It’s so disappointing and frustrating to think that things like this keep happening. And you have to face reality, and you have to consider how to stop the problem. Because the past is in the past, but we cannot allow this to continue. Hatred does things like this. In case the shock hasn’t hit you, in case you view it in a manner that is too detached, just imagine the same thing occurring to someone you love.
Because you could have loved those victims, and if you had known them enough, these news would not be so light.
I can understand why my parents never let me watch the news as a kid. They didn’t want me to be burdened down the way I am now. They didn’t want me to feel frustration, sadness, despair, and a lack of hope. They allowed optimism to exist within me. For the moment, I feel depressed and numb. Like the Anti-Transcendentalists, I am momentarily focused on the dark aspects of human nature.
I must have a balance because I can’t pretend others aren’t suffering just because it makes me feel better. I must be more considerate of others than that.
Yes, I can see how you can lose optimism and faith in humanity if you spend all your time focused on its dark aspects.
4/19/17
The topic of last night was very sad, though it was the reality. I feel the writing was ugly, heavy, and awkward—yet that is human nature. If I only posted content I was perfectly satisfied with, I’d never post anything. Understanding and accepting my imperfect humanity has allowed me to share content and ideas. The fear of being wrong and judged no longer prevents me from posting.
I would like to start over, but I’ll resist the urge.
In the Plessy v. Ferguson case, the Supreme Court declared segregation to be constitutional, establishing the “separate but equal” belief. Segregating laws were called Jim Crow laws. Rosa Parks felt the inequality of segregation on buses, while other blacks felt it when using other public facilities, such as public schools, restaurants, public swimming pools, and parks.
I told you about Parks’ arrest, but not that it got E.D. Nixon’s attention. Nixon was a former president of the NAACP (National Association for the Advancement of Colored People), and he received Parks’ permission to challenge bus segregation in court. This was typical of the NAACP, which consistently used the court as a way to battle segregation. Another victory of the NAACP occurred in Morgan v. Virginia when the Supreme Court decided segregation on interstate buses was unconstitutional. Remember when I was telling you about the Freedom Riders? They were protesting the south’s refusal to cease segregation on interstate buses.
You know, I actually am enjoying this explanation. It started off a little shaky as I expected it would, but then I entered a more flow-like rhythm and began connecting ideas. Science is not the only thing that’s fun to explain. I especially enjoy connecting ideas (in general) like you do when you’re explaining a system. When you do that, you show how different things are related, and how they work together, and it all makes sense. I am not waiting for the desire to write to come to me—I am seeking flow and all its satisfaction by writing. Thanks to that memory of the chemistry post, I now don’t expect writing to start off perfectly.
Parks’ act of defiance launched the civil rights movement. Soon after her arrest, the boycott of Montgomery buses began, a protest which would last for over a year. The people who walked to work, or carpooled—or did whatever else they had to to participate in this protest—were in part motivated by Martin Luther King, Jr., a black activist who had been summoned from his duties as pastor
I told you briefly about CORE. Though it was not the first, the organization used sit-ins as a nonviolent method of protest in segregated restaurants and other public facilities, with success.
Linda Brown was an African American who wanted to attend a school in Topeka, Kansas but was told to attend an all-black school instead. Her parents worked with the NAACP to take the case to court. In Brown v. Board of Education, the Supreme Court ruled public school segregation unconstitutional because it was unequal. This was a threat to segregated schools everywhere, many of which found ways to remain segregated. |
Analytics Check
PostsAbout
# The math behind exploding dice rolls
2021/09/19
Exploding dice is a mechanic that appears in various tabletop games. It works like this: a dice is rolled, if it lands on its maximum values, it “explodes” — the dice is then rolled again, adding the new value to the max value of the dice. If dice again lands on its maximum, the dice explodes a second time. This process continues and may lead to chains of explosions that push the roll total far beyond what is normally expected.
This mechanic appears in a number of tabletop games, with Savage Worlds likely being the most popular system within the RPG space that uses it. In this post, we dig into the infinite nature of exploding dice and explore the math and expected outcomes from this fun mechanic.
## Exploding probabilities
Suppose we roll an $n$ sided exploding dice. The probability of rolling a non-exploded value, that is between $1$ and $n-1$ is no different than a normal dice roll. Each face below $n$ appears with equal probability $1/n$. If instead the dice explodes — which happens with the remaining probability $1/n$ — we add $n$ to the outcome of a new exploding dice roll.
Let’s look at an example when we are rolling a 6-sided dice, so $n = 6$. Any result between 1 and 5 appears with probability $1/6$. How about the probability of rolling a 7? Rolling a 7 requires rolling a 6 on the first roll, triggering the explosion, and then rolling a 1 on the second:
$P(X = 7) = P(X = 6)P(X=1) = \frac{1}{6^2}$
Any value of the exploded dice roll can be written like this — as powers of $1/n$. Suppose we want to know the probability of rolling a 25 on an exploding d6. Well that’s just rolling 6 four times in a row followed by a 1, so $p = 1/6^5$. In general, for an $n$ sided dice, the probability of rolling any given value is:
$\boxed{ P(X = x) = \begin{cases} \left(\frac{1}{n}\right)^{\lfloor{x/n}\rfloor + 1} & x \textrm{ not a multiple of }n \\ 0 & \textrm{otherwise} \end{cases} }$
Where $\lfloor{x/n}\rfloor$ represents the rounded down result of dividing $x$ by $n$. The probability of rolling a multiple of $n$ is 0, because every time a multiple of $n$ would show up, it triggers a dice explosion. The below graph shows the outcome probabilities when rolling an exploding d6.
Within a given number of explosions, each outcome is equally likely. As the number of explosions increases, the probability decays rapidly due to the rarity of rolling many 6s in a row. This particular graph cuts off at 18, but in reality it continues to infinity with ever decreasing probabilities.
The probability of at least $k$ explosions on an $n$ sided dice is $P(\textrm{at least k explosions}) = n^{-k}$. Continuing with the d6 example, we have:
\begin{aligned} P(\textrm{at least 1 explosion on d6}) &\approx &17\% \\ P(\textrm{at least 2 explosions on d6}) &\approx &3\% \\ P(\textrm{at least 3 explosions on d6}) &\approx &0.5\% \\ P(\textrm{at least 4 explosions on d6}) &\approx &0.08\% \\ \end{aligned}
While one or two explosions is fairly common, more beyond that are quite rare, with a less than 1-in-1000 chance that you roll at least 4 explosions on a d6.
## Explosive expectations
Another way to understand the effect of the exploding dice mechanic is to look at how it changes the average or expected outcome of a roll. In a previous post, we showed that the expectation of a normal $n$ sided dice roll is $(n + 1) / 2$. It is clear that exploding dice would have a higher average, but by how much?
We’ll calculate this by appealing to something called the law of total expectation. It allows us to write the expectation of an exploding dice roll as follows:
$E(X) = E(X | \textrm{explode}) P(\textrm{explode}) + E(X | \textrm{not explode}) P(\textrm{not explode})$
By breaking up our expectation into two distinct cases — exploding and not exploding — we simplify our approach. First, the probability statements are are easily addressed by our previous work:
$P(\textrm{explode}) = \frac{1}{n}$
$P(\textrm{not explode}) = \frac{n-1}{n}$
Next, we consider the expected value of a roll with no explosion. Since each outcome below the maximum value is equally likely, this the same as the expected result of an $n-1$ sided roll. Plugging in to $(n + 1) / 2$, we get:
$E(X | \textrm{not explode}) = \frac{n}{2}$
The last term, which is the expected value of an exploded dice is less clear, but we can get around it with a little trick. Recall that once a dice explodes, you can view it as a new exploding dice roll with $n$ added to the final result. We can express this as:
$E(X | \textrm{explode}) = n + E(X)$
While it may seem odd to write the original thing that we were trying to solve for — $E(X)$ — on the right hand side, it’s perfectly valid to do and will help us later. Now that we have expressions for all four terms in our original expectation equation, we can plug in to get:
$E(X) = \left(n + E(X)\right) \frac{1}{n} + \frac{n}{2}\frac{n - 1}{n}$
The above simplifies to:
$E(X) = 1 + \frac{E(X)}{n} + \frac{n - 1}{2}$
We take advantage of our trick by subtracting $E(X)/n$ from both sides to gather all the $E(X)$ terms on the right and solve to get:
$\boxed{ E(X) = \frac{n}{n-1} \frac{n + 1}{2} }$
Careful inspection of our final result shows that the expectation of a normal dice roll appears on in our result, $(n + 1)/2$. A succinct way to summarize the effects of the exploding dice mechanic is that it increases the expected result by a factor of $n / (n - 1)$.
Revisiting our $n=6$ case, we plug in to see that $E(X) = 4.2$. A notable increase over the normal average of 3.5. The below graph summarizes the differences over the standard dice from d4 up to d20.
While the effect is of a similar magnitude across different types, it is most significant for low-faced dice. The factor of increase, $n / (n - 1)$, quickly approaches 1 as $n$ increases, but with few sides it is relatively large. For a d4, the increase from an average roll of 2.5 to one of 3.33, represents a 33% increase. On the other hand, the d20’s 10.5 to 11.05 increase is a meager 5%. The graph below shows percent increase of the exploding mechanic across the standard dice types:
Despite explosions having a consistent increase the expected outcome, as a percent difference, the effect is attenuated for high-sided dice. Such a difference suggests some reasoning behind the prevalence of exploding dice in d6 systems, but less so in those with a greater d20 focus.
Regardless of those differences, exploding dice is a fun mechanic that leads to some interesting math and meaningfully alters outcomes in systems that employ it.
Rolling for stats in DnD 5e always carries the risk of ending up with a weak character. Despite recommendations to use the standard array, point buy, or any number of homebrew rules for generating stats, rolling rules-as-written nevertheless remains common. For these tables, poor rolls inevitably occur, and when that happens, many DMs implement bad luck protection by allowing bad stats to be rerolled.
In this post, we look at the mathematical consequences of rerolling your stat array. By setting a limit on how bad a player can roll before allowing a reroll — we call that limit the bad luck threshold — the underlying distribution of possible outcomes changes and biases ability scores higher than they would otherwise be.
## Reviewing the standard array and rolling RAW
Before we dive into how rerolling changes the odds, let’s review how rolling rules-as-written compares to the standard array. As a reminder, the standard array is [8, 10, 12, 13, 14, 15]. Rolling for stats with 4d6 dropping the lowest generates a wide range of possible outcomes.
The chart below compares the two methods. The top-left shows the distribution of possible outcomes for the lowest value in your rolled array; typical values are somewhere between 6 and 11. The red line at 8 indicates the corresponding standard array values. Moving right, we see the distribution for the next highest stat in the array. This continues until the bottom-right, showing the possible outcomes of the highest stat in the array.
The standard array more or less selects the average from rolling, but not always! Interestingly, the largest deviation is the highest stat — rolling more often than not gives 16 or higher, compared to the standard array’s 15.
Seemingly small differences result in the fact that rolling will on average give higher ability scores than the standard array. By looking at the total ability score of the array, which is sum of the whole array, we get a better idea of how they compare. The figure below shows the bell-shaped curved of the total ability scores from rolling compared to the red line of the standard array.
The standard array’s red line is slightly left-of-center of the potential outcomes from rolling. Compared to the standard array’s 72 total ability score, rolling has an expected value of about 73.5. An extra point or two isn’t earth-shattering, but it is a difference worth noting.
## Bad luck protection and the upward bias of rolled stats
Armed with an understanding of how different rules generate differing ability scores for a character (point buy is beyond this post’s scope), we shift our focus to how these rules work in practice.
As mentioned in the intro, many groups rightly prioritize the enjoyment of the players over some arbitrary standard of mathematical fairness by rerolling bad arrays. This biases the total ability score upwards by cutting out extreme lows.
One can imagine that a player who has the misfortune to roll significantly under the standard array may want a reroll. This has the effect of chopping off the left side of the total ability score distribution and increasing the odds of the remaining outcomes.
The chart below shows how that shift happens — the underlying yellow-orange bars represent the full possible outcomes of rolling, whereas the slightly higher overlaid blue bars show how the distribution changes when it gets cuts off. The bad luck threshold shown here is the point where we say “any stats that are worse than this should be rerolled”.
The relatively modest bad luck threshold shown above, where a player rerolls when their total ability score is 60 or below (which are effectively commoner stats!), has the same shape as before, but with the left tail cut off. Although the difference seems minor, it is enough to raise the expected total ability score from 73.5 to 74. By allowing players to reroll commoner-tier stats, they expect to gain half of an ability score in the process.
Every bad luck threshold we pick gives a different expected boost to your ability score.
The figure below shows just how that bonus varies. The horizontal axis represents the different thresholds we can set. Higher bad luck thresholds correspond to being more generous with rerolls. The vertical axis is our expected total ability score when we roll using this bad luck protection rule.
As we increase the threshold, we see our expected stat total increase (as represented by the blue markers). Our previous threshold of 60 puts us right at a full ability score increase above the standard array. Increasing it to a higher value of 68 or so is equivalent to two full ASIs above the standard array.
When phrased in terms of ASIs, it’s clear that rolling RAW already gives a notable average boost over the standard array. Adding even a small amount of bad luck protection to that pushes the disparity even further.
As mentioned previously, a bad luck threshold of 60 — which corresponds to an average ability score of 10 — is enough to push rolling to a full ASI above the standard array. A more generous threshold of 66, or an average ability score of 11, doubles the expected stat bonus of rolling from 1.5 to 3 above the standard array.
Of course, a single number like the total ability score cannot capture all of the nuances that come into play when determining what is a “good” stat array. Having a very strong primary stat can more than make up for a poor dump stat or two, and none of that nuance is appropriately captured by this post. Regardless, when looking at different methods for generating ability scores, it’s important to explore how they are used in practice for an honest comparison between them.
## Some thoughts on fairness
There is nothing wrong with ensuring everyone is happy by allowing players to rerolls stats, but tables should be wary as to how this method can affect players that use different stat generating methods. If some players want to roll and others prefer to use the standard array, bad luck protection can lead to a lopsided power balance in favor of the rollers.
On one hand, it is something that could easily go by unnoticed and (in all honesty) likely wouldn’t have too much of an effect on the game. On the other hand, it could feel bad to effectively give an ASI to some players at character creation and not to others. Encouraging everyone to use the same method for generating ability scores ensures an even playing field, but if a table does split, buffing the standard array by an ASI isn’t a half-bad way to go.
Another point that often comes up in these discussions about rolling for stats is the opposite problem — good luck protection. Players who roll far above their party members may feel guilty about their greater character power and choose to reroll to get a more average array. This has a counteracting effect, pushing the expected ability scores lower. One strategy to constrain extreme ability scores is to reroll on both overly weak and overly strong characters. If the reroll rules are symmetric, it leaves the expected stats unchanged from the original method.
The goal of character creation should always be to create a party of adventurers that your table is excited to play. Keeping in mind that something as simple as allowing a player to reroll their stats will inflate them is worth keeping in mind, so that you can ensure balanced table where everyone can shine.
In DnD 5e, a group check is a helpful mechanic to approach situations where the entire party succeeds or fails together. While group checks don’t always come up in the average session, they certainly have their place. In my games, I often use a group Stealth check if the party tries to collectively sneak past an enemy or to adjudicate a Survival check to see if they navigate a hazardous terrain effectively.
Unfortunately — perhaps due to the designers assuming they are rare — the rule determining group checks is very sensitive to whether you have an odd or even number of party members.
Let’s take a look at the rule as written:
To make a group ability check, everyone in the group makes the ability check. If at least half the group succeeds, the whole group succeeds. Otherwise, the group fails. — Player’s Handbook p. 175
The rule itself is simple, but issues arise in the: “at least half the group” clause. As we know, party sizes in DnD are typically somewhere between 3-6 players. If you happen to have an odd number of players in your group, you’re stuck with interpreting what that “at least half” means — do you round up or down?
That decision is best left to the rules lawyers. Instead we focus on the fact that either decision leads to a situation where your party is punished or rewarded just because they have an odd number of members.
The below figure shows what I’m talking about. (Here I assume each party member has a 50% chance of success, which is not necessarily realistic but doesn’t change the result).
Every time you hit an odd numbered group size, the success chance shoots up or down compared to an even sized party, with the direction being determined by how you round. These are the sorts of odd results you end up with when you take a simple rule and combine it with having to round small numbers.
A modification of the rule can alleviate this issue. I propose the following change to the rule as written:
To make a group ability check, everyone in the group makes the ability check. If the group has an even number of members, the whole group succeeds if at least half the group succeeds. Otherwise, the group fails.
If the group has an odd number of members, the DM rolls a dice and records if the result is odd or even. If the DM rolls odd, the whole group succeeds if at least half the group succeeds, rounded down. If the DM rolls even, the whole group succeeds if at least half the group succeeds, rounded up. Otherwise the group fails.
Essentially, we change the rule so that with an odd number of players, the DM randomly chooses between rounding up or down the number of players required for success. This randomization process results in a compromise between the two extremes.
The below figure shows how the proposed rule compares to the original:
The success chance of our new rule is shown in red. It is clearly more well behaved when it comes to odd numbered group sizes. I’d go so far to argue that the proposed rule is more fair to the players — no group should be significantly affected by whether they have an odd or even number of members!
The downside of the proposed rule is that it places a little more burden on the DM by requiring an extra roll, but I believe the added complexity is worth it. Maybe I am the only DM who uses group checks and it doesn’t really matter, but hey, hopefully there was something interesting here to think about.
Calculating damage is a common — perhaps the most common — task in analyzing builds and theorycrafting in Dungeons and Dragons. It is not an easy one, however. Like many other concepts in DnD, damage comes at the whims of dice. For this reason, damage cannot be effective described as a single number, but rather as a distribution of possible values. On top of that, there are a number of rules and spell-specific effects that must be accounted for when determining the range of outcomes.
In this post we’ll lay out the general framework of the damage distribution and calculate it for a basic attack roll.
## Describing damage as a probability distribution
When a character attacks or casts a spell there are typically two steps that occur: 1) a roll by either the player or the DM to determine the degree to which the target is affected and 2) a subsequent damage roll that determines how much damage is taken by the target. There is randomness at every step of this process so we can best represent its outcome as a probability distribution $p(d)$, which is the probability that an attack does $d$ damage.
Let’s think about how we can compute this for a standard attack roll. We start by recognizing that a target can be affected by an attack in three ways: a miss, a normal hit, or a critical hit. A miss deals no damage, a normal hit will deal damage as written in the ability block, and a critical hit deals damage but with double the number of dice rolled. These three possibilities can be represented by breaking down our damage into conditional components:
$p(d) = \sum_{h\in H} p(d \,|\, h) \,p(h)$
Where $H = \lbrace \textrm{Miss}, \textrm{Normal}, \textrm{Crit} \rbrace$. The above expression states that the probability that $d$ damage is done is the weighted sum of doing that damage under each of our three hit types. By breaking it down like this, we can directly compute $p(d)$ by first looking at the hit probabilities, $p(h)$, and conditional damage distributions, $p(d|h)$.
It’s not too hard to see that this general formula can represent other damage processes. For a spell with a saving throw we could have $H = \lbrace \textrm{Success}, \textrm{Fail} \rbrace$. From there, it comes down to defining the conditional damage distributions.
## The hit probabilities of an attack roll
The first step of computing our attack damage distribution is calculating the hit probabilities $p(h)$. We will only concern ourselves with a standard roll, no advantage or disadvantage. Their effects could be explored with the appropriate substitution of probabilities given by advantage or disadvantage rules, and following the structure of the rest of this post.
We’ll start with a critical hit. The rules state a critical hit occurs when rolling a 20 on an attack roll. For a standard attack roll, this is simply:
$p(\textrm{crit}) = \frac{1}{20}$
Next, we examine the probability of a normal hit, i.e. a roll that meets or exceeds a target’s AC after bonuses but is neither a critical hit or a critical fail (which guarantees a miss). To do this, we must reach back to our previous post on dice rolls probabilities.
We introduce two new parameters: a target roll, $t$, and a bonus to add to the roll, $b$. In an attack roll, $t$ is typically the target’s armor class. The parameter $b$ is usually a combination of a proficiency bonus and a stat modifier.
As an example, suppose a fighter with a +5 to hit attempts to strike a Goblin with AC 15, then $t = 15$ and $b = 5$.
If $X$ is the outcome of our attack roll, then we can write:
$p(\textrm{normal}) = \begin{cases} P(t-b \leq X \leq 19)& t-b \geq 2\\ P(2 \leq X \leq 19)& \mathrm{otherwise} \end{cases}$
Using our dice roll probabilities, we can write this as:
$p(\textrm{normal}) = \begin{cases} \frac{19 - t + b + 1}{20}& t-b \geq 2\\ \frac{18}{20}& \mathrm{otherwise} \end{cases}$
The above expression describes the probability that a roll meets or exceeds the target after bonuses, while excluding critical hits and failures.
The last category to look at is the probability of a miss. We could calculate it directly as before, but it’s easier to just derive the result using the complement:
$p(\textrm{miss}) = 1 - p(\textrm{crit}) - p(\textrm{normal})$
Then we can write:
$p(\textrm{miss}) = \begin{cases} \frac{t - b - 1}{20}& t-b \geq 2\\ \frac{1}{20}& \mathrm{otherwise} \end{cases}$
With this, we fully describe the probability of an attack hitting when accounting for the target’s AC and the attacker’s bonuses. From here, we turn our attention to the conditional damage distributions.
## Damage conditional on the attack roll
The damage distributions themselves are determined by the specific attack type or spell being used, but most follow a pattern — the total damage is determined by the sum of rolling one or more dice and adding a bonus. We will write ${}_{m}D_n$ to represent the random variable of a sum of $m$ $n$-sided dice and $d_0$ will be the damage bonus added, then:
$p(d|h) = P({}_{m}D_n + d_0 = d | h) = P({}_{m}D_n = d - d_0| h)$
With the way we’ve written the above, we can see that our damage distribution can be written directly in terms of ${}_{m}D_n$, which we conveniently defined in a prior post.
For our normal damage distribution, we write it out directly:
$p(d|\textrm{normal}) = \frac{1}{n^m}\sum_{l=0}^{\left\lfloor\frac{d - d_0 - m}{n}\right\rfloor} (-1)^l {m \choose l} {d - d_0 - nl - 1 \choose m - 1}$
Which is the full expression of our dice roll sum distribution equal to $d - d_0$ as above.
Critical hits are a simple modification of the above, substituting $m \to 2m$, which represents doubling the number of damage dice that one rolls. This gives:
$p(d|\textrm{crit}) = \frac{1}{n^{2m}}\sum_{l=0}^{\left\lfloor\frac{d - d_0 - 2m}{n}\right\rfloor} (-1)^l {2m \choose l} {d - d_0 - nl - 1 \choose 2m - 1}$
Our last distribution — damage when the attack roll is missed — is the simplest:
$p(d|\textrm{miss}) = \begin{cases} 1& d = 0\\ 0& \textrm{otherwise} \end{cases}$
When an attack is missed, damage done is 0 with probability 1. Simple enough.
## Complete damage distributions
We now have all the necessary components to compute $p(d)$ in its entirety. As a reminder, for an attack roll we have:
$p(d) = p(d|\textrm{miss})p(\textrm{miss}) + p(d|\textrm{normal})p(\textrm{normal}) + p(d|\textrm{crit})p(\textrm{crit})$
It would be too verbose to substitute all of our values directly into this expression, but easy enough to describe the output by looking at graphs of different damage distributions. Let’s consider some examples:
Suppose Ria, a fighter, is facing down a zombie. Her attack bonus with her longsword is +5 and the zombie’s AC is 8. Ria’s longsword deals 1d8 + 3 slashing damage. What is the damage distribution of her strike?
From the prompt, we know that $t = 8$, $b=5$, $m=1$, $n=8$, and $d_0 = 3$. All there is to do now is plug these values into our previously described expressions. The below graph shows the probable outcomes of her strike.
It’s not exactly a bell curve to say the least. The complex behavior is the mixture of our three possible damage conditions — the single bar at 0 is the probability of missing, the mostly uniform damage from 4 to 11 is the damage from a normal hit, and the triangular distribution that lies on top of that is the critical hit damage.
The shapes and relative magnitudes depend on all the parameters present. To see just how different they can be, let’s consider another situation.
An adventuring party is going toe to toe with a large Ogre. Yar, a cleric, calls on the power of his deity to cast Inflict Wounds at 1st level. His spellcasting ability modifier is +5 and the AC of the Ogre is 11. Damage dealt on a hit is 3d10 necrotic damage. What is the damage distribution?
We approach this in a similar way to before, noting that $t = 11$, $b=5$, $m=3$, $n=10$, and $d_0 = 0$.
Now we see an entirely different distribution. The 0 damage possibility has risen to 25% and the bulk of the non-zero damage forms a familiar bell curve from the 3-dice sum with a long tail that represents possible crit damage.
In both of these examples, the mixture of different attack roll outcomes gives rise to non-trivial damage outcomes. Even so, the damage from an attack roll that we explored here is one of the simplest damage mechanics in 5e. Varying spell effects, different damage types, and other features all paint a rich mechanical picture. It suffices to say that we still have much to explore in this realm, but that will be for future posts.
All rights reserved. |
# How to master integral calculus?
Integral calculus, a branch of mathematics exploring the concept of accumulation and integration, can be challenging for many students. This concept is intricately linked to its counterpart, differential calculus. Before integrating, one should ensure a strong understanding of differentiation concepts, rules, and techniques.
A solid foundation in differential calculus will serve as a springboard for comprehending integration principles. A strategic approach and dedicated practice can help master the intricacies of integral calculus. Below are a few proven strategies to navigate through the topics of integral calculus with ease:
Also Read: How to Integrate Calculus?
#### Understand the basics of integral calculus:
Before attempting to master integration techniques, start with fundamental integration techniques such as the power rule, constant rule, and basic rules for integrating trigonometric and exponential functions.
Next, grasp the concept of definite and indefinite integrals, the Fundamental Theorem of Calculus, and the relationship between integrals and derivatives. Practicing these basics will provide a robust framework for tackling more complex integrals later.
The below-listed concepts are a few prerequisites to perform complex calculations in integral calculus:
#### The integral as the area under a curve:
The concept of an integral being the area under a curve is a fundamental principle in mathematics, especially integral mathematics. This view is based on the basic mathematical theory establishing a connection between differentiation and integration coefficients.
To learn the integral as the area under a curve, consider a function f(x) defined on a closed interval [a, b]. The integral of f(x) over this integral, denoted as ∫ab f(x) dx, represents the signed area between the curve y=f(x) and the x-axis. The sign of the area takes into account whether the curve lies above or below the x-axis.
If f(x) is entirely above the x-axis, the integral represents the area between the curve and the x-axis. If f(x) dips below the x-axis, the portion of the integral below the axis contributes negatively to the total area. This relationship is expressed as follows:
baf(x)dx= Area between y=f(x) and the x-axis from x=a to x=b.
Graphically, this can be visualized by plotting the function f(x) on a coordinate plane. The integral ∫ba f(x) dx represents the signed area of the region bounded by the curve and the x-axis from x=a to x=b.
Understanding the integral as the area under a curve is essential in various fields, particularly in applications where quantities are related to continuous change.
#### The definite and indefinite integral
The definite and indefinite integrals are fundamental concepts in calculus that deal with the accumulation of quantities and the reverse process of differentiation.
#### Definite integral
The definite integral represents the signed area between a curve and the x-axis over a specified interval. Symbolically, the definite integral of a function f(x) from a to b is denoted as ∫ba f(x) dx. The result is a number that represents the net accumulation of the function over the given interval.
Mathematically, the definite integral is defined as the limit of a sum of areas of rectangles as the width of the rectangles approaches zero. Geometrically, it represents the signed area between the curve and the x-axis, with the sign determining whether the curve is above or below the x-axis.
baf(x)dx=limn->∞ni=1f(xi)∆xi
The definite integral has practical applications in calculating quantities such as displacement, total distance traveled, and accumulated change over a specified time or interval.
Hire our online calculus tutors to help you understand the concept better.
#### Indefinite integral
The indefinite integral, on the other hand, represents the family of functions whose derivative is the given function. Symbolically, the indefinite integral of a function f(x) is denoted as ∫f(x) dx. It is often accompanied by the constant of integration (C) because there can be multiple functions whose derivative is the same.
Both the definite and indefinite integrals are crucial tools in calculus, with the definite integral providing a numerical result representing accumulated quantities over a specific interval, and the indefinite integral offering a function representing a family of antiderivatives.
Together, they form the backbone of integral calculus and find extensive applications in various scientific and mathematical disciplines.
#### Basic properties of integrals:
The basics of integrals, just like the basics of any other concepts are building blocks to achieve the targeted outcome. Below listed is a brief of a few fundamentals:
#### Linearity
The integral is a linear operation. For any functions f(x) and g(x) and constants a and b, the following holds:
∫ (af(x) + bg(x)) dx = a ∫ f(x) dx + b ∫ g(x) dx
#### Constant multiple
If C is a constant, then
∫ c. f(x) dx = c ∫ f(x) dx
#### Sum Rule
The integral of a sum is the sum of the integrals:
∫ (f(x) + g(x)) dx = ∫ f(x) dx + ∫ g(x) dx
#### Power rule
For any constant n≠ – 1
∫ xn dx = (1/(n+1)) xn+1 + C
#### Constant of Integration
The result of an indefinite integral ∫f(x) dx includes a constant of integration (C). This constant accounts for the family of functions that share the same derivative.
#### Integration of a constant
∫ c. dx = cx + C, where C is the constant.
#### Integration by parts
∫ u dv = uv – ∫ v du
This property is derived from the product rule for differentiation and is useful for integrating the product of two functions.
#### Symmetry
For an even function (f(x)=f(−x)) over an interval symmetric about the y-axis:
-aa f(x) dx = 2 ∫0a f(x) dx
This property simplifies the calculation of integrals for symmetric functions.
#### The Fundamental Theorem of Calculus:
Imagine calculus as the intricate artistry behind the scenes of scientific exploration, and at its heart lies the Fundamental Theorem of Calculus, a true cornerstone.
This theorem isn’t just a unifier of the seemingly separate realms of differentiation and integration; it’s a revelation of their profound connection. At its core, this theorem establishes a beautiful relationship between accumulating quantities and how fast they’re changing. It unfolds in two parts, each offering unique perspectives on the dance between derivatives and integrals.
The fundamental theorem of Calculus has two broad classifications namely:
#### The Derivative of an Accumulated Quantity:
Picture this – if you have a function, let’s call it F(x), born from an integral, say ∫axf(t)dt, where f(t) is a smooth operator, then the derivative of F(x) is simply the integrand f(x). In simpler terms, if F(x) represents the accumulation of a quantity up to x, its derivative spills the beans on how fast that quantity is changing at x.
#### Accumulating the Rate of Change:
Now, let’s flip the script. If f(x) is a smooth operator on the interval [a,b], and F(x) is its antiderivative buddy, then the definite integral of f(x) from a to b is like peeking at the difference in the values of F(x) at the endpoints
ab f(x) dx = F(b) – F(a)
In simpler terms, integrating the rate of change over an interval unveils the net accumulation of the quantity over that interval.
This theorem isn’t just a dry mathematical statement; it’s a testament to the elegance and unity embedded in the mathematical universe. It’s the guiding star that simplifies our journey through problems involving accumulation and change. More than that, it shines a light on the deep connections between seemingly unrelated mathematical concepts.
So, whether you’re a student navigating the intricacies of calculus or a seasoned practitioner, embracing the insights from this fundamental theorem opens portals to a world of mathematical marvels.
It invites us to witness the beauty that unfolds when the apparently distinct realms of differentiation and integration come together, weaving a rich tapestry that is calculus.
If you want to learn Integral calculus, check out the best online maths tutors.
#### Mastering different types of integration techniques
Tackling integration techniques might seem like a puzzle, but fear not – I’ve got some down-to-earth tips to help you conquer them. Before you plunge into the fancy stuff, make friends with the basics – power rule, constant rule, and those trigonometric and exponential functions.
Solid basics set the stage for the cool tricks. Below listed are a few popular methods of integration and tips to master them:
#### Substitution method:
Navigating the substitution method in integration is like mastering a craft – it takes understanding, practice, and a knack for picking the right tools. Let’s break it down into simple steps:
#### Understand the meaning of the substitution method
The substitution method is a technique used in calculus, particularly in integration, to simplify and solve complex integrals by introducing a new variable.
The goal is to replace the original variable in the integrand with a new variable, making the integral easier to handle. This new variable is chosen strategically to simplify the expression and facilitate integration.
#### Identify Composite Function
Understand the function that you would be working on to integrate. If it is a complicated function with a power derivative or with an attached trigonometric expression, use an alternative such as u in place of the complex function to ease the confusion while solving. .
#### Substitute and Simplify
As we have chosen to substitute the complicated function with u, we
u and du to show up in the integral. Substitute u and du into the integral and then integrate it with respect to u. Then remember to substitute the original function in place of u
#### Integration by parts:
Integration by parts is a technique used in calculus to evaluate the integral of a product of two functions. The formula for integration by parts is
udv=uv−∫vdu
Here, u and dv are differentiable functions of x, and du and v are their respective differentials. The goal is to choose u and dv in a way that simplifies the integral on the right side of the formula. Take the derivative of u to get du and integrate dv to get v.
Then, Substitute u, dv, du, and v into the integration by parts formula. The integral on the right side might still need to be simplified or solved. If it’s simpler than the original integral, you’ve made progress.
If the new integral on the right side is still challenging, apply integration by parts again until you reach a point where the integral becomes manageable.
#### Partial fractions:
Embracing partial fractions is like unraveling a mathematical puzzle. Understand the art of breaking down intricate rational functions into simpler fractions through algebraic maneuvers. Practice identifying factors, craft equations, and unveil the unknowns.
Apply this skill to seamlessly integrate complex rational functions. Mastery is a journey paved with regular practice and exposure to diverse examples. So, dive in, explore, and let the world of partial fractions unfold before you
The simplest way to solve partial fractions is by Identifying factors of the denominator, expressing the partial fractions, setting up equations, and solving for unknown coefficients.
#### Trigonometric substitutions:
Trigonometric substitutions in calculus are like handy shortcuts for unraveling integrals with tricky radical expressions. It’s a clever move—swapping out variables with trigonometric functions to make the math more friendly and straightforward.
It’s a bit like choosing the right tool for the job, transforming complex integrals into something much more manageable. These substitutions, guided by the dance of triangles and angles, help us navigate the intricate world of calculus with a touch of trigonometric elegance.
Primarily, Look for integrals with radical expressions and based on the form identified, choose the appropriate trigonometric identity to make the substitution.
Express other elements in the integral, such as dx and the radical expression, in terms of the chosen trigonometric function. Substitute the chosen trigonometric function and its derivatives for the original variable and its differential. Simplify the expression using trigonometric identities.
Evaluate the integral in terms of the new variable introduced by the trigonometric substitution. Substitute back the original variable in terms of the trigonometric function to obtain the final result.
#### Conclusion
In conclusion, mastering integral calculus is a journey that requires a thoughtful approach, consistent practice, and a profound grasp of the basics. It’s like constructing a sturdy building—start with a solid foundation in differential calculus, then layer on fundamental integration techniques, gradually delving into more complex methods.
Keep a growth mindset, seeing challenges as chances to grow. Regular practice, teamwork with peers, and tapping into extra resources are like keys to the kingdom.
By linking theoretical ideas to real-world scenarios, you fortify the importance of integral calculus. With patience and persistence, you’ll unravel the intricate threads of this mathematical tapestry, unlocking its potential in various scientific and engineering realms.
## FAQ's
It is possible to learn Algebra by yourself. However, you’ll need an online course that incorporates the teacher into all aspects of the syllabus. The most effective way to learn Algebra by yourself is to make sure that every lesson includes audio and video explanations of the examples and the problems for practice.
Any Algebra 1 student who wants to achieve an A grade must master the understanding of these concepts and abilities.
• Arithmetic
• Order of Operations
• Integers
• Working with Variables
• Memorizing Formulas
• The Organizing of problems on paper
The following fundamental ideas during Algebra 1.
• Simplifying
• Equations and Inequalities
• Word Problems
• Functions and graphing
• Linear Equations
• Systems of Equations
• Polynomials and Exponents
• Factoring
• Rational Expressions |
{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
Stitz-Zeager_College_Algebra_e-book
# Check your answer by nding a1 a and aa1 2 use a1 to
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: Example 8.2.3. Find the quadratic function which passes through the points (−1, 3), (2, 4), (5, −2). Solution. According to Definition 2.5, a quadratic function has the form f (x) = ax2 + bx + c where a = 0. Our goal is to find a, b and c so that the three given points are on the graph of f . If (−1, 3) is on the graph of f , then f (−1) = 3, or a(−1)2 + b(−1) + c = 3 which reduces to a − b + c = 3, an honest-to-goodness linear equation with the variables a, b and c. Since the point (2, 4) is also on the graph of f , then f (2) = 4 which gives us the equation 4a + 2b + c = 4. Lastly, the point (5, −2) is on the graph of f gives us 25a + 5b + c = −2. Putting these together, we obtain a system of three linear equations. Encoding this into an augmented matrix produces a−b+c = 3 1 −1 1 3 Encode into the matrix 4a + 2 b + c = 4 −− − − − − −→ 4 2 1 4 −−−−−−− 25a + 5b + c = −2 25 5 1 −2 7 Using a calculator,4 we find a = − 18 , b = 13 and c = 37 . Hence, the one and only quadratic which 18 9 7 fits the bill is f (x) = − 18 x2 + 13 x + 37 . To verify this analytically, we see that f (−1) = 3, f (2) = 4, 18 9 and f (5) = −2. We can...
View Full Document
{[ snackBarMessage ]}
Ask a homework question - tutors are online |
# 2016 USAJMO Problems/Problem 5
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$, and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$, respectively.
Given that $$AH^2=2\cdot AO^2,$$prove that the points $O,P,$ and $Q$ are collinear.
## Solution 1
$[asy] size(8cm); pair O=(0,0); pair A=dir(110); pair B=dir(-29); pair C=dir(209); pair H=foot(A,B,C); pair P=foot(H,A,B); pair Q=foot(H,A,C); draw(A--B--C--A--H--P); draw(circle(O,1)); draw(Q--H); dot("A", A, dir(A)); dot("B", B, dir(B)); dot("C", C, dir(C)); dot("H", H, S); dot("P", P, NE); dot("Q", Q, NW); dot("O", O, S); [/asy]$
It is well-known that $AH\cdot 2AO=AB\cdot AC$ (just use similar triangles or standard area formulas). Then by Power of a Point, $$AP\cdot AB=AH^2=AQ\cdot AC$$ Consider the transformation $X\mapsto \Psi(X)$ which dilates $X$ from $A$ by a factor of $\dfrac{AB}{AQ}=\dfrac{AC}{AP}$ and reflects about the $A$-angle bisector. Then $\Psi(O)$ clearly lies on $AH$, and its distance from $A$ is $$AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH$$ so $\Psi(O)=H$, hence we conclude that $O,P,Q$ are collinear, as desired.
## Solution 2
We will use barycentric coordinates with respect to $\triangle ABC.$ The given condition is equivalent to $(\sin B\sin C)^2=\frac{1}{2}.$ Note that $$O=(\sin(2A):\sin(2B):\sin(2C)), P=(\cos^2B,\sin^2B,0), Q=(\cos^2C,0,\sin^2C).$$ Therefore, we must show that $$\begin{vmatrix} \sin(2A) & \sin(2B) & \sin(2C) \\ \cos^2B & \sin^2B & 0 \\ \cos^2C & 0 & \sin^2C \\ \end{vmatrix}=0.$$ Expanding, we must prove $$\sin(2A)\sin^2B\sin^2C=\cos^2C\sin^2B\sin(2C)+\sin^2C\cos^2B\sin(2B)$$ $$\frac{\sin(2A)}{2}=\sin^2B(1-\sin^2C)\sin(2C)+\sin^2C(1-\sin^2B)\sin(2B)$$ \begin{align*} \frac{\sin(2A)+\sin(2B)+\sin(2C)}{2}&=\sin^2B\sin(2C)+\sin^2C\sin(2B)\\ &=2\sin B\sin C(\sin B\cos C+\cos B\sin C) \\ &=2\sin B\sin C\sin A.\end{align*}
Let $x=e^{iA}, y=e^{iB}, z=e^{iC},$ such that $xyz=-1.$ The left side is equal to $$\frac{x^2+y^2+z^2-\frac{1}{x^2}-\frac{1}{y^2}-\frac{1}{z^2}}{4i}.$$ The right side is equal to \begin{align*} 2\cdot \frac{x-\frac{1}{x}}{2i}\cdot \frac{y-\frac{1}{y}}{2i}\cdot \frac{z-\frac{1}{z}}{2i}&=\frac{xyz-\frac{1}{xyz}-\frac{xy}{z}-\frac{yz}{x}-\frac{xz}{y}+\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}}{-4i}\\ &=\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-x^2-y^2-z^2}{-4i},\end{align*} which is equivalent to the left hand side. Therefore, the determinant is $0,$ and $O,P,Q$ are collinear. $\blacksquare$
## Solution 3
For convenience, let $a, b, c$ denote the lengths of segments $BC, CA, AB,$ respectively, and let $\alpha, \beta, \gamma$ denote the measures of $\angle CAB, \angle ABC, \angle BCA,$ respectively. Let $R$ denote the circumradius of $\triangle ABC.$
Since the central angle $\angle AOB$ subtends the same arc as the inscribed angle $\angle ACB$ on the circumcircle of $\triangle ABC,$ we have $\angle AOB = 2\gamma.$ Note that $OA = OB,$ so $\angle OAB = \angle OBA.$ Thus, $\angle OAB = \frac{\pi}{2} - \gamma.$ Similarly, one can show that $\angle OAC = \frac{\pi}{2} - \beta.$ (One could probably cite this as well-known, but I have proved it here just in case.)
Clearly, $AO = R.$ Since $AH^2 = 2\cdot AO^2,$ we have $AH = \sqrt{2}R.$ Thus, $AH\cdot AO = \sqrt{2}R^2.$
Note that $AH = b\sin\gamma = c\sin\beta.$ The Extended Law of Sines states that: $$\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.$$ Therefore, $AH = \frac{bc}{2R} = \sqrt{2}R.$ Thus, $bc = \sqrt{2}R^2.$
Since $\angle PHA = \beta$ and $\angle QHA = \gamma,$ we have: $$AP = AH\sin\beta = c\sin^2\beta = \frac{b^2 c}{4R^2}$$ $$AQ = AH\sin\gamma = b\sin^2\gamma = \frac{bc^2}{4R^2}$$ It follows that: $$AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.$$ We see that $AP\cdot AQ = AH\cdot AO.$
Rearranging $AP\cdot AQ = AH\cdot AO,$ we get $\frac{AP}{AH} = \frac{AO}{AQ}.$ We also have $\angle PAH = \angle OAQ = \frac{\pi}{2} - \beta,$ so $\triangle PAH\sim\triangle OAQ$ by SAS similarity. Thus, $\angle AOQ = \angle APH,$ so $\angle AOQ$ is a right angle.
Rearranging $AP\cdot AQ = AH\cdot AO,$ we get $\frac{AP}{AO} = \frac{AO}{AH}.$ We also have $\angle PAO = \angle HAQ = \frac{\pi}{2} - \gamma,$ so $\triangle PAO\sim\triangle HAQ$ by SAS similarity. Thus, $\angle AOP = \angle AQH,$ so $\angle AOP$ is a right angle.
Since $\angle AOP$ and $\angle AOQ$ are both right angles, we get $\angle POQ = \pi,$ so we conclude that $P, O, Q$ are collinear, and we are done. (We also obtain the extra interesting fact that $AO\perp PQ.$)
## Solution 4
Draw the altitude from $O$ to $AB$, and let the foot of this altitude be $D$.
Then, by the Right Triangle Altitude Theorem on triangle $AHB$, we have: $AB\cdot AP=AH^{2}$.
Since $OD$ is the perpendicular bisector of $AB$, $2\cdot AD = AB$.
Substituting this into our previous equation gives $2\cdot AD \cdot AP = AH^{2}$, which equals $2\cdot AO^{2}$ by the problem condition.
Thus, $2\cdot AD\cdot AP = 2\cdot AO^{2} \implies AD\cdot AP = AO^{2}$.
Again, by the Right Triangle Altitude Theorem, angle $AOP$ is right.
By dropping an altitude from $O$ to $AC$ and using the same method, we can find that angle $AOQ$ is right. Since $\angle AOP=\angle AOQ=90$, $P$, $O$, $Q$ are collinear and we are done.
~champion999
## Solution 5
We use complex numbers. Let lower case letters represent their respective upper case points, with $|a| = |b| = |c| = 1$. Spamming the foot from point to segment formula, we obtain $$h = \dfrac{1}{2}\left(a+b+c-\dfrac{bc}{a}\right),$$ $$p = \dfrac{1}{2}(h+a+b-ab\overline{h}) = \dfrac{1}{4}\left(2a+2b+c-\dfrac{bc}{a}-\dfrac{ab}{c}+\dfrac{a^2}{c}\right),$$ and $$q = \dfrac{1}{2}(h+a+c-ac\overline{h}) = \dfrac{1}{4}\left(2a+b+2c-\dfrac{bc}{a}-\dfrac{ac}{b}+\dfrac{a^2}{b}\right).$$ We now simplify the given length condition: \begin{align*} AH^2 &= 2AO^2 \\ \implies (h-a)\overline{(h-a)} &= 2 \\ \implies \dfrac{1}{2}\left(-a+b+c-\dfrac{bc}{a}\right) \cdot \dfrac{1}{2}\left(-\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{a}{bc}\right) &= 2 \\ \implies (ab+ac-bc-a^2)^2 &= 8a^2bc \\ \implies (a-b)^2(a-c)^2 &= 8a^2bc.\end{align*} We would like to show that $P$, $O$, $Q$ are collinear, or $$\begin{vmatrix} p & \overline{p} & 1 \\ 0 & 0 & 1 \\ q & \overline{q} & 1 \end{vmatrix} = \overline{p}q - p\overline{q} = 0.$$ After some factoring (or expanding) that takes about 15 minutes, this eventually reduces to $$\overline{p}q - p\overline{q} = 0 \iff (a-b)(a-c)(b-c)((a-b)^2(a-c)^2 - 8a^2bc) = 0,$$ which is true. $\square$
-MP8148
## Solution 6
Claim: $\triangle AOQ \sim \triangle AHB$
Proof: We compute the area of $\triangle ABC$ using two methods. Let $R=AO$ be the circumradius of $\triangle ABC$.
First, by extended law of sines, $BC=2R \sin \angle BAC$. We are also given that $AH= R \sqrt{2}$. $AH \perp BC$, so $$[ABC]=\frac{2R \sin \angle BAC \cdot R \sqrt{2}}{2}=R^2 \sin \angle BAC \sqrt{2}.$$
Second, we compute the area using $[ABC]=\frac{1}{2} \cdot AB \cdot AC \sin \angle BAC$.
Equating these two expressions for the area of $\triangle ABC$ and reducing, we get $$AB \cdot AC = 2 R^2 \sqrt{2}.$$ But $2R^2 =2 AO^2 = AH^2$, so $AB \cdot AC = AH^2 \sqrt{2}$ and $\bf{AB =\frac{AH^2 \sqrt{2}}{AC}}$.
Since $\angle AQH=\angle AHC =90^\circ$, and both triangles share the angle $\angle HAC$, $\triangle AHC \sim \triangle AQH$. This tells us $\frac{AQ}{AH}=\frac{AH}{AC}$ or $AQ =\frac{AH^2}{AC}$.
Substituting into the bolded equation, we get $AB=AQ \sqrt{2}$.
The original length condition can be written as $AH= AO \sqrt{2}$.
We also have $$\angle OAQ= \angle OAC= 90- \angle ABC = \angle HAB.$$
Therefore, by SAS similarity, $\triangle AOQ \sim \triangle AHB$.
We can prove analogously that $\triangle AOP \sim \triangle AHC$. We now have $\angle AOQ=\angle AHB =90^\circ$ and $\angle AOP=\angle AHC =90^\circ$. This implies that $\angle POQ=\angle POA +\angle QOA=180^\circ$ which tells us $P,O,$ and $Q$ are collinear, as desired.$\square$
-vvluo |
# Total Internal Reflection
Technology and Art
# Important Inequalities in Functional Analysis
Avishek Sen Gupta on 27 September 2021
Continuing my self-study of Functional Analysis, this post describes proofs for the following important inequalities in the subject:
• Young’s Inequality
• Hölder’s Inequality
• Minkowski’s Inequality
The paths of the proofs closely follow Erwin Kreyszig’s Introductory Functional Analysis with Applications.
## Young’s Inequality
We begin with the idea of conjugate exponents, which we call $$p$$ and $$q$$, related like so:
$\frac{1}{p} + \frac{1}{q} = 1 \\ \Rightarrow q + p = pq \\ \Rightarrow q + p = pq \\ \Rightarrow p(q - 1)= q \\ \Rightarrow p=\frac{q}{q-1} \\ \Rightarrow 1-p=1-\frac{q}{q-1} \\ \Rightarrow 1-p=\frac{q-1-q}{q-1}$ $$$\Rightarrow 1-p=\frac{1}{1-q} \label{eq:conjugate-exponents-property-1}$$$
where $$p>1$$. We now use these exponents in the following identity:
$u=t^{p-1}$
We also record the following implication.
$t=u^{\frac{1}{p-1}} \\ \Rightarrow t=u^{q-1}$
For $$p=2$$, we get $$u=t$$, which is the equation of the straight line $$x=y$$. The above follows from $$\eqref{eq:conjugate-exponents-property-1}$$.
Let $$\alpha, \beta >0: \alpha, \beta \in \mathbb{R}$$, then in $$\mathbb{R}^2$$, $$\alpha\beta$$ describes the area of a rectangle. Let us plot the graph of $$u=t^{p-1}$$ for $$p=2$$, so that $$u=t$$.
If we integrate $$\displaystyle\int\limits_0^\alpha u.dt=\int\limits_0^\alpha t^{p-1}.dt$$, and $$\displaystyle\int\limits_0^\alpha t.du=\int\limits_0^\alpha u^{q-1}.du$$, we can compute an area like this:
\begin{align*} S &= \int\limits_0^\alpha t^{p-1}.dt + \int\limits_0^\alpha u^{q-1}.du \\ &= \frac{\alpha^2}{2} + \frac{\beta^2}{2} \end{align*}
You will notice that regardless of the choice of $$\alpha, \beta$$, there will always be a small portion of $$S$$ which is bigger than $$\alpha\beta$$. The only situation in which there is no extra area is when $$\alpha=\beta$$. Therefore, we can say that:
$\alpha \beta \leq \frac{\alpha^2}{2} + \frac{\beta^2}{2}$
This less-than-or-equal relation carries over to other values of $$p$$ where $$u=t^{p-1}$$ is an exponential graph. The following two graphs illustrate how this is always true.
Thus, we can conclude that:
$$$\alpha \beta \leq \frac{\alpha^p}{p} + \frac{\beta^q}{q} \label{eq:youngs-inequality}$$$
$$\eqref{eq:youngs-inequality}$$ is called Young’s Equality, and we will use it to prove Hölder’s Inequality next.
A more interesting way to think about (and remember) Young’s Equality is to note that $$\text{log }(x)$$ is a concave function, therefore by the definition of concavity, we have:
$f(\alpha x^p + (1 - \alpha) y^q) \geq \alpha f(x^p) + (1-\alpha) f(y^q) \\ \text{log }(\alpha x^p + (1 - \alpha) y^q) \geq \alpha\text{log }(x^p) + (1-\alpha) \text{log }(y^q) \\ \text{log }\left(\frac {x^p}{p} + \frac {y^q}{q}\right) \geq \frac{\text{log }(x^p)}{p} + \frac{\text{log }(y^q)}{q} = \text{log }(x^{(p/p)}) + \text{log }(y^{(q/q)}) \\ \text{log }\left(\frac {x^p}{p} + \frac {y^q}{q}\right) \geq \text{log }(xy) \\ \Rightarrow xy \leq \left(\frac {x^p}{p} + \frac {y^q}{q}\right)$
## Hölder’s Inequality
We look at $$\ell^p$$ spaces now. Briefly recapping, $$\ell^p$$ spaces are spaces of sequences. A sequence $$\xi=\{\xi_1, \xi_2, \cdots\} \in\ell^p$$ must satisfy the following condition:
$\sum\limits_{i=1}^\infty{|\xi_i|}^p<\infty$
The norm $$\|\bullet\|$$ in $$\ell^p$$ spaces is usually defined as:
$\|\bullet\|={\left(\sum\limits_{i=1}^\infty{|\xi_i|}^p\right)}^{\frac{1}{p}}$
which induces the distance metric between two sequences $$\xi, \eta \in \ell_p$$:
$d(\xi,\eta)={\left(\sum\limits_{i=1}^\infty{|\xi_i-\eta_i|}^p\right)}^{\frac{1}{p}}$
Pick any $$\xi,\eta\in\ell^p$$.
Let us pick any two corresponding terms in $$\xi$$ and $$\eta$$, and let $$\alpha=\vert\xi_i\vert$$ and $$\beta_i=\vert\eta\vert$$, since we cannot guarantee these terms will be positive.
Then, from Young’s Inequality $$\eqref{eq:youngs-inequality}$$, we get:
$|\xi_i||\eta_i| \leq \frac{ {|\xi_i|}^p}{p} + \frac{ {|\eta_i|}^q}{q} \\ \Rightarrow |\xi_i\eta_i| \leq \frac{ {|\xi_i|}^p}{p} + \frac{ {|\eta_i|}^q}{q}$
Summing over all $$i$$, we get:
$$$\displaystyle\sum\limits_{i=1}^\infty|\xi_i\eta_i| \leq \frac{\sum\limits_{i=1}^\infty{|\xi_i|}^p}{p} + \frac{\sum\limits_{i=1}^\infty{|\eta_i|}^q}{q} \label{eq:holders-youngs-inequality-application}$$$
What we’d like to do is prove that: $$\displaystyle\sum\limits_{i=1}^\infty\vert\xi_i\eta_i\vert \leq 1$$. The only identity immediately available is the conjugate exponent identity, namely:
$\frac{1}{p} + \frac{1}{q} = 1$
In order to be able to set the RHS of $$\eqref{eq:holders-youngs-inequality-application}$$, we need the following condition:
$$$\displaystyle\sum\limits_{i=1}^\infty{ {\vert\xi_i\vert}^p}=\displaystyle\sum_{i=1}^\infty{ {\vert\eta_i\vert}^q}=1 \label{eq:holders-unity-condition}$$$
Let us then assume $$\eqref{eq:holders-unity-condition}$$. Then, $$\eqref{eq:holders-youngs-inequality-application}$$ becomes:
$$$\displaystyle\sum\limits_{i=1}^\infty|\xi_i\eta_i| \leq 1 \label{eq:holders-less-than-unity-condition}$$$
We need to determine what sort of $$\xi$$ and $$\eta$$ can satisfy this condition. Let’s take $$\xi$$ as an example. We have:
$\xi=(\xi_1, \xi_2, \cdots)$
Remember the norm for $$\ell^p$$ spaces? Here it is again:
$\|\bullet\|={\left(\sum\limits_{i=1}^\infty{|\xi_i|}^p\right)}^{\frac{1}{p}}$
If we divide each term in $$\xi$$ by its norm:
$$$\xi_i = \frac{\bar\xi_i}{ {\left(\sum\limits_{i=1}^\infty{|\bar{\xi_i}|}^p\right)}^{\frac{1}{p}}} \label{eq:holders-sequence-scale-factor}$$$
Then, from $$\eqref{eq:holders-unity-condition}$$, we get:
$\displaystyle\sum\limits_{i=1}^\infty{ {|\xi_i|}^p}=\frac{1}{ \sum\limits_{i=1}^\infty{|\bar{\xi_i}|}^p} \left({|\bar{\xi_1}|}^p + {|\bar{\xi_2}|}^p + \cdots\right) \\ =\frac{\sum\limits_{i=1}^\infty{|\bar{\xi_i}|}^p}{\sum\limits_{i=1}^\infty{|\bar{\xi_i}|}^p} = 1$
which satisfies condition $$\eqref{eq:holders-unity-condition}$$, regardless of which sequence $$\bar{\xi}$$ we choose from $$\ell^p$$.
Applying the same logic to $$\eta_i$$, and substituting $$\eqref{eq:holders-sequence-scale-factor}$$ into $$\eqref{eq:holders-less-than-unity-condition}$$, we get:
$\displaystyle\sum\limits_{i=1}^\infty |\bar\xi_i \bar\eta_i| \leq {\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\bar\xi_i|}^p}\right)}^{\frac{1}{p}} \bullet {\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\bar\eta_i|}^q}\right)}^{\frac{1}{q}}$
Removing the overbars from $$\bar\xi_i$$ $$\bar\eta_i$$ to indicate any two sequences in an $$\ell^p$$ space, we get:
$$$\displaystyle\sum\limits_{i=1}^\infty |\xi_i \eta_i| \leq {\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\xi_i|}^p}\right)}^{\frac{1}{p}} \bullet {\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\eta_i|}^q}\right)}^{\frac{1}{q}} \label{eq:holders-inequality}$$$
The result $$\eqref{eq:holders-inequality}$$ is referred to as Hölder’s Inequality.
In the case of $$p=2$$, $$q=2$$; this special case is called the Cauchy-Schwarz Inequality, which is:
$$$\displaystyle\sum\limits_{i=1}^\infty |\xi_i \eta_i| \leq \sqrt{\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\xi_i|}^2}\right) \bullet \left(\displaystyle\sum\limits_{i=1}^\infty{ {|\eta_i|}^2}\right)} \label{eq:cauchy-schwarz-inequality}$$$
Hölder’s Inequality will be used to prove Minkowski’s Inequality next.
## Minkowski’s Inequality
Minkowski’s Inequality is a generalisation of the Triangle Inequality. As usual, we assume $$\xi, \eta \in \ell^p$$. We start with writing (for economy of notation):
$\omega_i=|\xi_i+\eta_i| \\ \Rightarrow {\omega_i}^p={|\xi_i+\eta_i|}^p \\ \Rightarrow {\omega_i}^p={\omega_i}^{p-1}|\xi_i+\eta_i|$
Summing up over $$i$$, we get:
$$$\displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p = \displaystyle\sum\limits_{i=1}^\infty {\omega_i}^{p-1}|\xi_i+\eta_i| \label{eq:minkowski-separated-summed}$$$
Now applying the Triangle Inequality to the second term on the RHS of $$\eqref{eq:minkowski-separated-summed}$$, we get:
$$$\displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p = \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} \bullet \underbrace{|\xi_i+\eta_i|}_\text{Apply Triangle Inequality} \\ \Rightarrow \displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p \leq \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1}(|\xi_i|+|\eta_i|) \\ \Rightarrow \displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p \leq \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| + \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\eta_i| \\ \label{eq:minkowski-separated-summed-inequality}$$$
Apply Hölder’s Inequality $$\eqref{eq:holders-inequality}$$ to each term on the RHS individually in $$\eqref{eq:minkowski-separated-summed-inequality}$$, we have:
$\displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| \leq \displaystyle\sum\limits_{i=1}^\infty|\omega^{p-1} \xi_i| \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{\left({|\omega_i|}^{p-1}\right)}^q\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} \\ \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| \leq \displaystyle\sum\limits_{i=1}^\infty|\omega^{p-1} \xi_i| \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{\left({|\omega_i|}^{p-1}\right)}^q\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p}$
Note that since $$p$$ and $$q$$ are conjugate exponents, we can write:
$p=pq-q$
Then the above inequalities simplify to:
$$$\displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| \leq \displaystyle\sum\limits_{i=1}^\infty|\omega^{p-1} \xi_i| \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} \label{eq:minkowski-holder-inequality-1}$$$ $$$\displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| \leq \displaystyle\sum\limits_{i=1}^\infty|\omega^{p-1} \xi_i| \leq {\left[\displaystyle\sum\limits_{i=1}^\infty|{\omega_i|}^p\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p} \label{eq:minkowski-holder-inequality-2}$$$
Applying $$\eqref{eq:minkowski-holder-inequality-1}$$ and $$\eqref{eq:minkowski-holder-inequality-2}$$ to $$\eqref{eq:minkowski-separated-summed-inequality}$$, we get:
$\displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} + {\left[\displaystyle\sum\limits_{i=1}^\infty|{\omega_i|}^p\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p} \\ \displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^\frac{1}{q} \left({\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} + {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p}\right) \\ {\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^{1-\frac{1}{q}} \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} + {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p}$
Again, noting from the conjugate exponent identity that $$1-\frac{1}{q}=\frac{1}{p}$$, we get:
$$${\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^\frac{1}{p} \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} + {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p} \label{eq:minkowski-inequality}$$$
$$\eqref{eq:minkowski-inequality}$$ is referred to as Minkowski’s Inequality.
tags: Mathematics - Proof - Functional Analysis - Pure Mathematics |
{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
Related_Rates_Handout_Notes_p1
# Related_Rates_Handout_Notes_p1 - Ex A pebble is dropped...
This preview shows page 1. Sign up to view the full content.
Related Rates In the last section we learned to differentiate implicitly defined functions by using the Chain Rule. In this section we will use the Chain Rule to find the rates of change of two or more variables with dy dx dV dr respect to time, giving us expressions such as —, —, —. —. dt dt dt dt = 5x* -6x+2] Find'-— ^vhen;e = 4, given that — = 2 when* = 4. \£/ r C8 Steps to Use When Solving a Related Rates Word Problem 1. Draw a figure if possible. 2. Assign variables and restate the problem, listing your given information and what you are asked to find. Notice whether the given rates of change are positive or negative. 3. Find an equation that relates the variables. 4. Differentiate with respect to time. 5. Substitute the given information, and solve for the unknown derivative. Be sure to include units with your answer.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: Ex. A pebble is dropped into a calm pond, causing ripples in the shape of concentric circles. The ^ radius of the outer ripple is increasing at a constant rat&of 1 ft/sec. When the radius Js_4 ft, y \ "findthe rate at which the area of the disturbed water is changing. Step 1: Draw a figure. <f L &j\ Step 2: Assign variables and restate the problem, listing your given information and what you are asked to find. Notice whether the given rates of change are positive or negative. Step 3: Find an equation that relates the variables. If necessary, find a relationship among the variables that lets you eliminate one variable. Step 4: Differentiate with respect to time. Step 5: Substitute the given information, and solve for the unknown derivative. Be sure to include units with your answer,...
View Full Document
{[ snackBarMessage ]}
Ask a homework question - tutors are online |
Multiple Regression Worksheet | Problems & Solutions
# Multiple Regression Worksheet
Multiple Regression Worksheet
• Page 1
1.
Identify the Null and Alternative hypotheses for the statement:
a. H0: μ ≠ 60.80 and H1: μ = 60.80 b. H0: μ ≤ 60.80 and H1: μ > 60.80 c. H0: μ ≥ 60.80 and H1: μ < 60.80 d. H0: μ = 60.80 and H1: μ ≠ 60.80
#### Solution:
So, the null and alternative hypotheses are H0: μ = 60.80 and H1: μ ≠ 60.80.
2.
Write the Null and Alternative hypotheses for the statement:
Average cost of a visit to a doctor's office is more than $50. a. H0: μ =$50 and H1: μ < $50 b. H0: μ ≠$50 and H1: μ = $50 c. H0: μ ≤$50 and H1: μ > $50 d. H0: μ >$50 and H1: μ ≤ $50 #### Solution: Average cost of a visit to a doctor's office is more than$50.
So, the null and alternative hypotheses are H0: μ ≤ $50 and H1: μ >$50.
3.
Write the Null and Alternative hypotheses for the statement:
Among doctors, the general practitioners earn an average annual salary of not more than $80,000 a. H0: μ ≤$80,000 and H1: μ > $80,000 b. H0: μ <$80,000 and H1: μ ≥ $80,000 c. H0: μ =$80,000 and H1: μ ≠ $80,000 d. H0: μ =$80,000 and H1: μ ≤ $80,000 #### Solution: Among doctors, the general practitioners earn an average annual salary of not more than$80,000.
So, the null and alternative hypotheses are H0: μ ≤ $80,000 and H1: μ >$80,000
4.
Which of the following is the null hypothesis for the given statement? "The average chest measurement of a U.S soldier is 37.38 inches."
a. H0: μ > 37.38 b. H0: μ ≠ 37.38 c. H0: μ < 37.38 d. H0: μ = 37.38
#### Solution:
The average chest measurement of a U.S soldier is 37.38 inches.
The null hypothesis is stated as H0: μ = 37.38
5.
The average daily intake of salt by an American is estimated to be less than 12 grams. The average of a random sample of 50 Americans was found as 11.5 grams with a standard deviation of 2.5 grams. Select the proper alternative hypothesis.
a. H1: μ < 12 b. H1: μ < 11.5 c. H1: μ > 11.5 d. H1: μ > 12
#### Solution:
The average daily intake of salt by an American estimated to be less than 12 grams.
The null and alternative hypothesis are H0: μ ≥ 12 and H1: μ < 12
6.
Which is true for the test: H0: μ = 10 and H1: μ ≠ 10?
a. Test is two tailed b. Test is right tailed c. Test is left tailed
#### Solution:
The alternative hypothesis in the test is H1: μ ≠ 10. That is H1: μ < 10 or μ > 10.
So, the test is two tailed.
7.
Which is true for the test: H0: μ ≤ 15 and H1: μ > 15?
a. Test is two tailed b. Test is right tailed c. Test is left tailed
#### Solution:
The alternative hypothesis in the test is H1: μ > 15.
So, the test is right tailed.
8.
Which is true for the test: H0: μ ≥ 100 and H1: μ < 100?
a. Test is two tailed b. Test is right tailed c. Test is left tailed
#### Solution:
The alternative hypothesis in the test is H1: μ < 100
So, the test is left tailed.
9.
Which is type I error?
a. reject the research hypothesis when it is true b. reject the research hypothesis when it is false c. do not reject the research hypothesis when it is true d. do not reject the research hypothesis when it is false
#### Solution:
In a statistical hypothesis testing, rejecting the research hypothesis when it is true is called type I error. |
# Identify Inverse Proportion
In this worksheet, students will explore relationships which are inversely proportional, using information provided to find the value of a single unit, then applying this information to a new total to solve a real-world problem.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR,
Curriculum topic: Ratio, Proportion and Rates of Change
Curriculum subtopic: Ratio, Proportion and Rates of Change Direct and Inverse Proportion
Difficulty level:
#### Worksheet Overview
If one quantity decreases as the other increases, then they are said to be inversely proportional.
'Inverse' is just another word for doing the opposite.
For example, if one quantity is halved, the other would be doubled.
Unfortunately, inverse proportion questions don't always appear as straightforward as this though.
Don't bury your head in the sand with this type of question... you just need to learn and apply a process.
For example:
Four dogs take five hours to did a big hole. How long would it take ten dogs to dig the same sized hole at the same rate?
Let's make a table to summarise what we know and what we want to find out:
4 dogs Take 5 hours 1 dog 5 × 4 20 hours 10 dogs 20 ÷ 10 2 hours
So we used what we know to find the corresponding amount for a single unit (in this case 'one dog'), then applied this to the new total.
Let's try another problem to practise this process more.
For example:
It takes 5 bakers, 4 hours to bake 125 cakes. How long would it take 8 bakers to make 200 cakes if they all work at the same rate?
5 bakers Take 4 hours 1 baker 4 × 5 20 hours 8 bakers 20 ÷ 8 2.5 hours (for 125 cakes) 1 cake 2.5 ÷ 125 0.02 hours 200 cakes 0.02 × 100 4 hours
Don't you think it is unfair that you always have to do that bit extra before you can get some cake?
Now let's put this process into action!
In this activity, we will explore relationships which are inversely proportional, using the information provided to find the value of a single unit, then applying this information to a new total to solve a real-world problem.
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started |
# Greatest Common Divisor & Least Common Multiple Calculator
First Number: Second Number:
Third Number (not required):
Greatest Common Factor (GCF):
Least Common Multiplier (LCM):
Greatest Common Divisor & Least Common Multiple Calculator: With this calculator, you can easily find the Greatest Common Divisor and Least Common Multiple for two or three numbers in a swift manner.
## What is the Least Common Multiple LCM?
The Least Common Multiple (LCM) is the smallest positive integer that is divisible by two or more given numbers without leaving a remainder. In other words, it is the smallest common multiple of those numbers.
To find the LCM of two or more numbers, you can use various methods, including prime factorization, listing multiples, or using the LCM formula. The LCM is useful in various mathematical operations, such as adding or subtracting fractions with different denominators, solving equations, and finding patterns in numbers.
For example, to find the LCM of 4 and 6:
• List the multiples of each number: 4, 8, 12, 16, 20, ... 6, 12, 18, 24, ...
• Identify the smallest number that appears in both lists, which is 12. Therefore, the LCM of 4 and 6 is 12.
The LCM can also be calculated using prime factorization. By expressing each number as a product of prime factors, you can determine the LCM by taking the highest power of each prime factor present in the numbers.
It's important to note that the LCM is unique for any given set of numbers, and it provides a useful tool in various mathematical calculations and problem-solving scenarios.
## How to find the least common multiple LCM?
To find the Least Common Multiple (LCM) of two or more numbers, you can use various methods, including prime factorization, listing multiples, or using the LCM formula. Here, we'll explain the prime factorization and listing multiples methods:
1. Prime Factorization Method:
• Express each number as a product of its prime factors.
• Identify all the unique prime factors among the numbers and write down the highest exponent for each prime factor.
• Multiply all the prime factors with their respective exponents to obtain the LCM.
For example, let's find the LCM of 12 and 18:
• Prime factorization of 12: 2^2 * 3^1
• Prime factorization of 18: 2^1 * 3^2
• Unique prime factors: 2 (highest exponent: 2) and 3 (highest exponent: 2)
• LCM = 2^2 * 3^2 = 36
2. Listing Multiples Method:
• List the multiples of each number until you find a common multiple.
• Identify the smallest common multiple among the lists.
For example, let's find the LCM of 4 and 6:
• Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, ...
• Multiples of 6: 6, 12, 18, 24, 30, 36, ...
• The smallest common multiple is 12, which is the LCM.
Both methods will yield the same LCM for any given set of numbers. The choice of method depends on the numbers involved and personal preference. LCM calculations can also be performed using online LCM calculators or built-in functions in mathematical software.
## What is the Greatest Common Divisor GCD?
The Greatest Common Divisor (GCD), also known as the Greatest Common Factor (GCF) or Highest Common Factor (HCF), is the largest positive integer that divides evenly into two or more given numbers. In other words, it is the largest number that is a common factor of the given numbers.
To find the GCD of two or more numbers, you can use various methods, including prime factorization, the Euclidean algorithm, or using a GCD calculator. The GCD is useful in simplifying fractions, solving equations, and finding common denominators.
For example, let's find the GCD of 24 and 36:
• Prime factorization of 24: 2^3 * 3^1
• Prime factorization of 36: 2^2 * 3^2
• Common prime factors: 2 (minimum exponent: 2) and 3 (minimum exponent: 1)
• GCD = 2^2 * 3^1 = 12
The GCD can also be found using the Euclidean algorithm, which involves dividing the larger number by the smaller number and repeating the process until a remainder of zero is obtained. The divisor at the last step is the GCD.
It's important to note that the GCD is always a positive integer and that the GCD of any number with zero is the number itself.
The GCD provides a useful tool in various mathematical calculations and problem-solving scenarios.
## How to find the greatest common divisor GCD?
To find the Greatest Common Divisor (GCD) of two or more numbers, you can use several methods, including the prime factorization method, the Euclidean algorithm, or using a GCD calculator. Here, we'll explain the prime factorization and Euclidean algorithm methods:
1. Prime Factorization Method:
• Express each number as a product of its prime factors.
• Identify the common prime factors among the numbers and write down the smallest exponent for each common factor.
• Multiply all the common factors with their respective exponents to obtain the GCD.
For example, let's find the GCD of 24 and 36:
• Prime factorization of 24: 2^3 * 3^1
• Prime factorization of 36: 2^2 * 3^2
• Common prime factors: 2 (minimum exponent: 2) and 3 (minimum exponent: 1)
• GCD = 2^2 * 3^1 = 12
2. Euclidean Algorithm:
• Take the two numbers for which you want to find the GCD.
• Divide the larger number by the smaller number.
• If the remainder is zero, the smaller number is the GCD.
• If the remainder is not zero, replace the larger number with the remainder, and repeat the process.
• Continue dividing until you reach a remainder of zero.
• The divisor at this step is the GCD.
For example, let's find the GCD of 24 and 36 using the Euclidean algorithm:
• Divide 36 by 24: 36 ÷ 24 = 1 with a remainder of 12
• Divide 24 by 12: 24 ÷ 12 = 2 with a remainder of 0
• The divisor at the last step is 12, which is the GCD.
Both methods will yield the same GCD for any given set of numbers. The choice of method depends on the numbers involved and personal preference. GCD calculations can also be performed using online GCD calculators or built-in functions in mathematical software. |
Students can Download Maths Chapter 5 Geometry Ex 5.6 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 5 Geometry Ex 5.6
Miscellaneous Practice Problems
Question 1.
Find the value of x if ∠AOB is a right angle.
Solution:
Given that ∠AOB = 90°
∠AOB = ∠AOC + ∠COB = 90° (Adjacent angles)
3x + 2x = 90°
5x = 90°
Question 2.
In the given figure, find the value of x.
Solution:
Since ∠BOC and ∠AOC are linear pair, their sum = 180°
2x + 23 + 3x – 48 = 180°
5x – 25 = 180°
5x – 25 + 25 = 180° + 25
Question 3.
Find the value of x, y and z.
Solution:
∠DOB and ∠BOC are linear pair
∴ ∠DOB + ∠BOC = 180°
x + 3x + 40 = 180°
4x + 40 = 180°
4x + 40 – 40 = 180° – 40°
4x = 140°
Also ∠BOD and ∠AOC are vertically opposite angles.
∴ ∠BOD = ∠AOC
x = z + 10
35° = z + 10
z + 10 – 10 = 35 – 10
z = 25°
Again ∠AOD and ∠AOC are linear pair.
∴ ∠AOD + ∠AOC = 180°
y + 30 + z + 10 = 180°
y + 30 + 25 + 10 = 180°
y + 65 = 180°
y + 65 – 65 = 180° – 65
y = 115°
∴ x = 35°,
y = 115°,
z = 25°
Question 4.
Two angles are in the ratio 11 : 25. If they are linear pair, find the angles.
Solution:
Given two angles are in the ratio 11 : 25.
Let the angles be 11x and 25x.
They are also linear pair
∴ 11x + 25x = 180°.
∴ The angles 11x = 11 × 5° = 55° and 25x = 25 × 5 = 125°.
∴ The angles are 55° and 125°.
Question 5.
(i) Is ∠1 adjacent to ∠2?
(ii) Is ∠AOB adjacent to ∠BOE?
(iii) Does ∠BOC and ∠BOD form a linear pair?
(iv) Are the angles ∠COD and ∠BOD supplementary.
(v) Is ∠3 vertically opposite to ∠1 ?
Solution:
(i) Yes, ∠1 is adjacent to ∠2.
Because they both have the common vertex ‘O’ and the common arm OA . Also their interiors do not overlap.
(ii) No, ∠AOB and ∠BOB are not adjacent angles because they have overlapping interiors.
(iii) No, ∠BOC and ∠BOD does not form a linear pair.
Because ∠BOC itself a straight angle, so the sum of ∠BOC and ∠BOD exceed 180°.
(iv) Yes, the angles ∠COD and ∠BOD are supplementary ∠COD + ∠BOD = 180°, [∵ linear pair of angles]
∴ ∠COD and ∠BOD are supplementary.
(v) No. ∠3 and ∠1 are not formed by intersecting lines. So they are not vertically opposite angles.
Question 6.
In the figure POQ, ROS and TOU are straight lines. Find the x°.
Solution:
Given TOU is a straight line.
∴ The sum of all angles formed at a point on a straight line is 180°
∠TOR + ∠ROP + ∠POV + ∠VOU = 180°.
36° + 47°+ 45° + x° = 180°.
128° + x° = 180°
128° + x° – 128° = 180° – 128°
x = 52°
Question 7.
In the figure AB is parallel to DC. Find the value of ∠1 and ∠2. Justify your answer.
Solution:
Given AB || DC
AB and CD are parallel lines Taking CE as transversal we have.
∠1 = 30°, [∵ alternate interior angles]
Taking DE as transversal
∠2 = 80°.[∵ alternate interior angles]
∠1 = 30° and ∠2 = 80°
Justification:
CDE is a triangle
Question 8.
In the figure AB is parallel to CD. Find x, y and z.
Solution:
Given AB || CD
∴ Z = 42 (∵ Alternate interior angles)
Also y = 42° [vertically opposite angles]
Also x° + 63° + z° = 180°
x° + 63° + 42° = 180°
x° + 105° = 180°
x°+ 105° – 105° = 180° – 105°
x° = 75°
∴ x = 75°;
y = 42°;
z = 42°
Question 9.
Draw two parallel lines and a transversal. Mark two alternate interior angles G and H. If they are supplementary, what is the measure of each angle?
Solution:
l and m are parallel lines and n is the transversal.
∠G and ∠H are alternate interior angles.
∠G = ∠H …… (1)
Given ∠G and ∠G are Suplementary
Question 10.
A plumber must install pipe 2 parallel to pipe 1. He knows that ∠1 is 53. What is the measure of ∠2?
Solution:
Given ∠1 = 53°
Clearly ∠1 and ∠2 are interior angles on the same side of the transversal and so they are supplementary.
∠1 + ∠2 = 180°
53° + ∠2 = 180°
53° + ∠2 – 53° = 180° – 53°
∠2 = 127°
Challenge Problems
Question 11.
Find the value of y.
Solution:
Cleary POQ is a straight line”
Sum of all angles formed at a point on a straight line is 180°
∴ ∠POT + ∠TOS + ∠SOR + ∠ROQ = 180°
60° + (3y – 20°) + y° + (y + 10°) = 180°
60° + 3y – 20° + y° + y° + 10° = 180°
5y + 50° = 180°
5y + 50° – 50° = 180° – 50
5y = 130°
$$y=\frac{130^{\circ}}{5}$$
y = 26°
Question 12.
Find the value of z.
Solution:
The sum of angles at a point is 360°.
∴ ∠QOP + ∠PON + ∠NOM + ∠MOQ = 360°
3z + (2z – 5) + (z + 10) + (4z – 25) = 360°
3z + 2z + z + 4z — 5 +10 — 25 = 360°
10z – 20° = 360°
10z – 20° + 20 = 360°+ 2
10z = 380°
$$z=\frac{380^{\circ}}{10}$$
z = 38°
Question 13.
Find the value of x and y if RS is parallel to PQ.
Solution:
Given RS || PQ
Considering the transversal RU, we have y = 25° (corresponding angles)
Considering ST as transversal
Question 14.
Two parallel lines are cut by a transversal. For each pair of interior angles on the same side of the transversal, if one angle exceeds the twice of the other angle by 48°. Find the angles.
Solution:
Let the two parallel lines be m and n and l be the transversal
Let one of the interior angles on the same side of the transversal be x°
Then the other will be 2x + 48.
We know that they are supplementary.
Question 15.
In the figure, the lines GH and IJ are parallel. If ∠1 = 108° and ∠2 = 123°, find the value of x, y and z.
Solution:
Given GH || IZ
∠1 = 108°
∠2 = 123°
∠1 + ∠KGH = 180 [linear pair]
108° + ∠KGH = 180°
108° + ∠KGH – 108° = 180° – 108°
∠KGH = 72°
∠KGH = x° (corresponding angles if KG is a transversal)
∴ x° = 72°
Similarly
∠2 + ∠GHK = 180° (∵ linear pair)
123° + ∠GHK = 180°
123° + ∠GHK – 123° = 180° – 123°
∠GHK = 57°
Again ∠GHK = y° (corresponding angles if KH is a transversal)
y = 57°
x° +y° + z° = 180° (sum of three angles of a triangle is 180°)
72° + 57° + z° = 180°
129° + z° = 180°
129° + z° – 129° = 180° – 129°
z = 51°
x = 72°,
y = 57°,
z = 51°
Question 16.
In the parking lot shown, the lines that mark the width of each space are parallel. If
∠1 = (2x – 3y)°; ∠2 = (x + 39)° find x° and y°.
Solution:
From the picture
∠2 + 65° = 180° [Sum of interior angles on the same side of a transversal]
x + 39° + 65° = 180°
x + 104° = 180°
x + 104° – 104° = 180° – 104°
x = 76°
Also from the picture
∠1 = 65° [alternate exterior angles]
2x – 3y = 65°
2 (76) – 3y = 65°
152° – 3y = 65°
152° – 3y – 152° = 65 – 152°
-3y = -87
Question 17.
Draw two parallel lines and a transversal. Mark two corresponding angles A and B. If ∠A = 4x, and ∠B = 3x + 7, find the value of x. Explain.
Solution:
Let m and n are two parallel lines and l is the transversal.
A and B are corresponding angles.
We know that corresponding angles are equals,
Question 18.
In the figure AB in parallel to CD. Find x°, y° and z°.
Solution:
Given AB||CD
Then AD and BC are transversals.
x = 48°, alternate interior angles; AD is transversal y = 60°, alternate interior angles; BC is transversal
∠AEB + 48° + y° = 180°, (sum of angles of a triangle is 180°)
Question 19.
Two parallel lines are cut by transversal. If one angle of a pair of corresponding angles can be represented by 42° less than three times the other. Find the corresponding angles.
Solution:
We know that the corresponding angles are equal.
Let one of the corresponding angles be x.
Then the other will be 3x – 42°.
Question 20.
In the given figure, ∠8 = 107°, what is these sum of the angles ∠2 and ∠4.
Solution:
Given ∠8 = 107°
∠2 = 107°
[∵ ∠8 and ∠2 are alternate exterior angles, ∵ ∠8 = ∠2] |
# Basic Arithmetic : Division with Whole Numbers and Remainders
## Example Questions
### Example Question #1 : Division With Whole Numbers And Remainders
Kevin has won toys from a competition and decides to keep toys for himself. He would like to give the rest of the toys to four of his close friends.
If each of Kevin's friends receive the same number of toys from Kevin, how many will be remaining?
Explanation:
The correct answer is 2 remaining toys. Kevin has 34 toys to start and he keeps 4 for himself. This means that he has 30 toys to give to four friends.
When we divide 30 by 4, we get 7.5. Since each friend cannot get half a toy, each friend can receive 7 toys.
This means that of the 30 toys, there will be 2 toys remaining.
### Example Question #1 : Division With Whole Numbers And Remainders
What is the result (with the remainder) in the above equation?
Explanation:
We divide the top by the bottom.
Test how many times 10 fits into 32 (try 3 times).
We subtract 30 from 32:
10 cannot fit into 2 so the remainder is 2.
### Example Question #1 : Division With Whole Numbers And Remainders
What is divided by ? Find a whole number and a remainder.
remainder
remainder
remainder
remainder
remainder
remainder
Explanation:
Simply divide to answer this question:
remainder
The maximum number of times can go into is , and you have left over.
To put it another way, . If you subtract from , you are left with , so that's your remainder.
### Example Question #1 : Division With Whole Numbers And Remainders
What is divided by ? Express your answer as a whole number with a remainder.
remainder
remainder
remainder
remainder
remainder
remainder
Explanation:
is not evenly divisible by . As such, you want to first find out how many whole 's can go into .
, so is too many.
, which is as close to as you can get.
### Example Question #1 : Division With Whole Numbers And Remainders
Carla and four of her friends are having a sleep over. Carla has seventeen blankets in her house. She wants to distribute her blankets evenly among all of the people at the sleepover (her included). How many blankets does each person get? How many are left over?
blankets each, left over
blankets each, left over
blankets each, left over
blankets each, left over
blankets each, left over
blankets each, left over
Explanation:
So, we have blankets, and sleep over attendees.
can go into a maxiumum of times without going over, so each attendee gets blankets.
, so there are blankets left over.
### Example Question #1 : Division With Whole Numbers And Remainders
What is
Explanation:
When you are dealing with negative numbers, the following rules apply.
If there are no negative signs, the answer is positive.
If there is one negative sign, the answer is negative.
If there are two negative signs, the answer is positive.
, and since there is only one negative sign the answer is negative. is the solution.
### Example Question #1 : How To Subtract Integers
Evaluate the following: |
Eureka Math Grade 5 Module 2 Lesson 17 Answer Key
Eureka Math Grade 5 Module 2 Lesson 17 Problem Set Answer Key
Question 1.
Estimate the quotient for the following problems. Round the divisor first.
a. 609 ÷ 21
≈ 600 ÷ 20
= 30
b. 913 ÷ 29
≈ __900______ ÷ ____30____
= ____30_____
Answer:
913/29 = 31.48.
Explanation:
In the above-given question,
given that,
900/30
30.
c. 826 ÷ 37
≈ ___800_____ ÷ ____40____
= ___20______
Answer:
826/37 = 22.324.
Explanation:
In the above-given question,
given that,
800/40
20.
d. 141 ÷ 73
≈ ___140_____ ÷ ____70____
= ____2_____
Answer:
140/70 = 2.
Explanation:
In the above-given question,
given that,
140/70
2.
e. 241 ÷ 58
≈ ___240_____ ÷ ____60____
= ___4______
Answer:
240/60 = 4.
Explanation:
In the above-given question,
given that,
240/60
4.
f. 482 ÷ 62
≈ ___480_____ ÷ ___60_____
= ____8_____
Answer:
480/60 = 8.
Explanation:
In the above-given question,
given that,
480/60
8.
g. 656 ÷ 81
≈ __660______ ÷ ___80_____
= ___8______
Answer:
660/80 = 8.
Explanation:
In the above-given question,
given that,
660/80
8.
h. 799 ÷ 99
≈ __800_______ ÷ _100________
= _____8____
Answer:
800/100 = 8.
Explanation:
In the above-given question,
given that,
800/100
8.
i. 635 ÷ 95
≈ ___635_____ ÷ ____95_____
= ___7______
Answer:
635/95 = 7.
Explanation:
In the above-given question,
given that,
635/95.
7.
j. 311 ÷ 76
≈ ___310_____ ÷ ___80_____
= ____4_____
Answer:
310/80 = 4.
Explanation:
In the above-given question,
given that,
310/80
4.
k. 648 ÷ 83
≈ ___650______ ÷ _____80____
= ___8______
Answer:
650/80 = 8.
Explanation:
In the above-given question,
given that,
650/80
8.
l. 143 ÷ 35
≈ ___140______ ÷ _____35____
= _____4____
Answer:
143/35 = 4.
Explanation:
In the above-given question,
given that,
143/35
4.
m. 525 ÷ 25
≈ ___525______ ÷ ___25______
= ____21_____
Answer:
525/25 = 21.
Explanation:
In the above-given question,
given that,
525/25
21.
n. 552 ÷ 85
≈ ___550______ ÷ ____85_____
= _____6____
Answer:
550/85 = 6.
Explanation:
In the above-given question,
given that,
550/85
6.
o. 667 ÷ 11
≈ ___670______ ÷ ____10_____
= ___67______
Answer:
670/10 = 67.
Explanation:
In the above-given question,
given that,
670/10
67.
Question 2.
A video game store has a budget of $825, and would like to purchase new video games. If each video game costs$41, estimate the total number of video games the store can purchase with its budget. Explain your thinking.
Answer:
The total number of video games the store can purchase with its budget = $20. Explanation: In the above-given question, given that, A video game store has a budget of$825 and would like to purchase new video games.
If each video game costs $41, estimate the total number of video games the store can purchase with its budget. 825/41 = 20. The total number of video games the store can purchase with its budget =$20.
Question 3.
Jackson estimated 637 ÷ 78 as 640 ÷ 80. He reasoned that 64 tens divided by 8 tens should be 8 tens. Is Jackson’s reasoning correct? If so, explain why. If not, explain a correct solution.
Answer:
Yes, Jackson’s reasoning correct.
Explanation:
In the above-given question,
given that,
Jackson estimated 637/78.
640/80.
8.
Eureka Math Grade 5 Module 2 Lesson 17 Exit Ticket Answer Key
Estimate the quotient for the following problems.
a. 608 ÷ 23
≈ ___610______ ÷ ____20_____
= ___30______
Answer:
610/20 = 30.
Explanation:
In the above-given question,
given that,
610/20
30.
b. 913 ÷ 31
≈ ___910______ ÷ ___30______
= ____30_____
Answer:
910/30 = 30.
Explanation:
In the above-given question,
given that,
910/30
30.
c. 151 ÷ 39
≈ ___150______ ÷ ____40_____
= ____4_____
Answer:
150/40 = 3.75.
Explanation:
In the above-given question,
given that,
150/40
40.
d. 481 ÷ 68
≈ ___480______ ÷ ____70_____
= _____7____
Answer:
480/70 = 7.
Explanation:
In the above-given question,
given that,
480/70
7.
Eureka Math Grade 5 Module 2 Lesson 17 Homework Answer Key
Question 1.
Estimate the quotient for the following problems. The first one is done for you.
a. 821 ÷ 41
≈ 800 ÷ 40
= 20
b. 617 ÷ 23
≈ ___620_____ ÷ ___20_____
= __31_______
Answer:
617/20 = 31.
Explanation:
In the above-given question,
given that,
610/20
31.
c. 821 ÷ 39
≈ ___820_____ ÷ ___40_____
= __20_______
Answer:
820/40 = 20.
Explanation:
In the above-given question,
given that,
820/40
20.
d. 482 ÷ 52
≈ ___480_____ ÷ ____50____
= ___10______
Answer:
480/50 = 10.
Explanation:
In the above-given question,
given that,
480/50
10.
e. 531 ÷ 48
≈ __530______ ÷ ___50_____
= ____11_____
Answer:
531/50 = 11.
Explanation:
In the above-given question,
given that,
531/50
11.
f. 141 ÷ 73
≈ ___140_____ ÷ ___70_____
= ____2_____
Answer:
140/70 = 2.
Explanation:
In the above-given question,
given that,
140/70
2.
g. 476 ÷ 81
≈ ___480_____ ÷ ___80_____
= ___6______
Answer:
480/80 = 6.
Explanation:
In the above-given question,
given that,
480/80
6.
h. 645 ÷ 69
≈ __645______ ÷ ___70_____
= ____9_____
Answer:
645/70 = 9.
Explanation:
In the above-given question,
given that,
645/70
9.
i. 599 ÷ 99
≈ ___600_____ ÷ ___100_____
= ____6_____
Answer:
600/10 = 6.
Explanation:
In the above-given question,
given that,
600/10
6.
j. 301 ÷ 26
≈ ___300_____ ÷ ___30_____
= ___3______
Answer:
300/30 = 3.
Explanation:
In the above-given question,
given that,
300/30
3.
k. 729 ÷ 81
≈ ___730_____ ÷ ___80_____
= ___9______
Answer:
730/80 = 9.
Explanation:
In the above-given question,
given that,
730/80
9.
l. 636 ÷ 25
≈ ___640_____ ÷ ___25_____
= ___25______
Answer:
640/25 = 25.
Explanation:
In the above-given question,
given that,
640/25
25.
m. 835 ÷ 89
≈ __830______ ÷ ___90_____
= ___9______
Answer:
835/90 = 9.
Explanation:
In the above-given question,
given that,
835/90
9.
n. 345 ÷ 72
≈ __345______ ÷ ___70_____
= ____5_____
Answer:
345/70 = 5.
Explanation:
In the above-given question,
given that,
345/70
5.
o. 559 ÷ 11
≈ ___560_____ ÷ ____10____
= ____56_____
Answer:
560/10 = 56.
Explanation:
In the above-given question,
given that,
560/10
56.
Question 2.
Mrs. Johnson spent $611 buying lunch for 78 students. If all the lunches cost the same, about how much did she spend on each lunch? Answer: The money she spends on each lunch =$8.
Explanation:
In the above-given question,
given that,
Mrs. Johnson spent \$611 buying lunch for 78 students.
If all the lunches cost the same.
611/78 = 8.
Question 3.
An oil well produces 172 gallons of oil every day. A standard oil barrel holds 42 gallons of oil. About how many barrels of oil will the well produce in one day? Explain your thinking.
Answer:
The barrels of oil will the good produce in one day = 5.
Explanation:
In the above-given question,
given that,
An oil well produces 172 gallons of oil every day.
A standard oil barrel holds 42 gallons of oil.
172/42 = 5.
Scroll to Top
Scroll to Top |
# Algebra/Statistics
Algebra
← Probability Statistics
## Percentages
Percentages are another way of representing rational numbers. Rational numbers can be represented as decimals, fractions or percentages.
Percent, when you actually break the word apart, consists of two words: per and cent. Per is a small but extremly powerful word. It means "to divide". Cent is a Latin word for 100. Percent, then, means "divide by 100".
To convert from a fraction to a percentage, we simply multiply the fraction by 100. For example, $\frac{1}{4} = (\frac{1}{4} \times 100)% = 25%\,$.
To convert from a decimal to a percentage, we first convert the decimal into a fraction, and then proceed with the approach outlined above. For example, $0.15 = \frac{15}{100} = (\frac{15}{100} \times 100)% = 15%\,$.
## Mean, Median, Mode
The following three numbers represent 3 different ways to think about the average value of your set.
Mean - This is what we usually think of as the "average" of a data set. The mean can be found by summing all the values in the data set and dividing by the size of the data set (that is number of elements in the set). In mathematical notation,
$\bar{x} = (\sum_{i=1}^n x_i) \div n = (1/n) * (x_1 + x_2 + x_3 + ... + x_n)$, where $\bar{x}$ is the arithmetic mean, and n is the number of elements in the data set. All values of x together constitute the sample space for the data set.
For example: Suppose 1, 2, 4, 6, 8, 9 is our data set then the sum is 1 + 2 + 4 + 6 + 8 + 9 = 30 and there are 6 elements in the data set, so the mean is 30/6 = 5.
The mean, while a very useful statistic, has its flaws. Notably, its value may be heavily influenced by outliers - numbers in a data set which are significantly higher or lower than the majority of the data. It is often preferable to use the median instead to describe such data sets.
Median - This is the middle of our data set. To find the median you must first put your data values in numerical order (say, from smallest to largest). If you have an odd number of elements in your data set there will be exactly one number in the middle, this number is the median. If you have an even number of elements in your data set then the median is the average of the middle two numbers. et For example. If our data set was 2, 2, 3, 4, 4, 5, 6, 7, 8, 9, 12, 13, 16, 22 is our data s data set. Since it has an even number of elements, we have to take the mean of the middle two, in this case 6 and 7, so the median is 6.5.
Mode - Mode refers to how many times a number or numbers occur in a data set. Since mean, median, and mode often are confused with each other, an easy way to remember mode is 'most often'. The first two letters in mode are 'm' and 'o', imagine this stands for 'most often' to help you remember. In the case that two or more different values are tied for the most number of repeats then that data set is said to have multiple modes. If your asked to find the mode of a data set with multiple modes, then all of the modes should be listed. If no element of the data repeats, then there is no mode.
For example. Suppose 1, 2, 2, 2, 3, 3, 4, 5, 5, 5, 7 is our data set, then the mode would be both 2 and 5. They both occur three times and three is the maximum number of repeats in our data set.
The following quantity tell us how spread out our data set is.
Range - The difference between the largest and smallest numbers in our data set. Notice this means the range is never negative.
### Examples
Mean
Let's look at the following data set:
Data Values: 10, 13, 4, 7, 9 so n = 5
10 + 13 + 4 + 7 + 9 = 43
43 / 5 = 8.6
Mean = 8.6
Median
Case 1:
Data Values: 10, 13, 4, 7, 8 so n = 5
Numerical Order: 4, 7, 8, 13, 10
Since 8 is the middle number,
Median = 8
Case 2:
Data Values: 10, 13, 4, 7, 8, 10 so n = 6
Numerical Order: 4, 7, 8, 10, 10, 13
Middle Numbers: 8 and 10
Find Mean: 8 + 10 = 18
18 / 2 = 9
Median = 9
Mode
Data Values: 10, 13, 4, 7, 8, 10
10 is in the data set twice.
Mode = 10
Data Values: 4, 9, 13, 18, 4, 2, 9, 4, 13, 8, 9
4 and 9 both have three data values.
Mode = 4, 9
Range
Data Values: 10, 13, 4, 7, 8
Numerical Order: 4, 7, 8, 10, 13
Difference of last and first: 13 - 4 = 9
Range = 9
### Practice Problems
Find the mean, median, mode, and range of the following data sets:
1.) 5, 8, 12, 4, 8, 9, 11, 2
2.) 24, 26, 37, 24, 16, 44, 26, 34, 24
3.) 15, 48, 89, 74, 25, 36, 57, 51, 17, 22
4.) 2, 6, 8, 7, 8, 2, 2, 9, 10 |
# Question #28ba7
Dec 18, 2017
$x = - 11 ,$ $y = 5$
#### Explanation:
$\textcolor{red}{\text{To find x and y, you need to solve these equations simultaneously:}}$
$y = - 2 x - 17$ $\ldots \ldots \ldots \left(1\right)$
$- x + 11 y = 66$ $\ldots \ldots \ldots \left(2\right)$
sub $\left(1\right)$ into $\left(2\right)$:
$- x + 11 \left(- 2 x - 17\right) = 66$
$\textcolor{red}{\text{Expand the brackets:}}$
$- x - 22 x - 187 = 66$
$\textcolor{red}{\text{Group like terms:}}$
$- 23 x - 187 = 66$
$\textcolor{red}{\text{Put all integers on the right:}}$
$- 23 x = 253$
$\textcolor{red}{\text{Isolate x:}}$
$\therefore x = - 11$
$\textcolor{red}{\text{Now to find y:}}$
sub $x = - 11$ into $\left(1\right)$:
$y = - 2 \left(- 11\right) - 17$
$\textcolor{red}{\text{Expand brackets and simplify:}}$
$y = 22 - 17$
$\therefore y = 5$
$\textcolor{red}{\text{Putting what we have found all together:}}$
$\therefore x = - 11 ,$ $y = 5$ |
Home Numerical Excel Tutorial Microscopic Pedestrian Simulation Kardi Teknomo's Tutorial Micro-PedSim Free Download Personal Development Handbook
Research Publications Tutorials Resume Personal Contact
Cross Product
<Next | Previous | Index>
Vector cross product is also called vector product because the result of the vector multiplication is a vector. It can only be performed for two vectors of the same size.
Geometrically, when you have two vectors on a plane, the cross product will produce another vector perpendicular to the plane span by the two input vectors.
The direction of the cross product vector is following the direction of the thumb finger in your right hand when the four other fingers indicate the angle from the first vector to the second vector, as shown in the following figures.
#### Computation of cross product
For 1 or two dimensional vector, the cross product produces zero vectors because they do not make a plane yet. Starting from 3 dimensions, the cross product can be computed algebraically using simple arrangement and a simple rule below
The arrangement to compute vector cross product
1. Arrange the vector as row vector with the first input vector in second position and the second input vector in the third position
2. Put notation of vector element in the first position
3. Repeat the arrangement on the right
#### Simple rule to compute cross product
After the arrangement above, multiply the elements of the vectors in diagonal direction and then minus with the product of elements in counter diagonal direction
Now we put them together
Unit vector represent the first, second and third position of vector elements. Thus, here is the final result
Example
For higher dimension, we use the same rule but programmatically, it is easier if we use a formula
Let d = dimension of the vector (that is equal to the length of the vector)
Let assume the index of array vector start from 0 then the pseudo code below produce vector cross product
Input: vector a, vector b both have equal length
Output: vector c
d = vector length
For r = 0 to vector length -1
c[r] = a[mod(r+1, d)] * b[mod(r+2, d)] - a[mod(r+d-1, d)] * b[mod(r+d-2, d)]
Next r
The interactive program of cross product below shows the cross product of two vectors of the same dimension. The program will also show you the internal computation so that you can check your own manual computation. If you click “Random Example” button, the program will generate random input vectors in the right format.
vector a vector b
In the applications, cross product is useful for constructing coordinate system mostly in 3-dimensional space.
## Properties
Some important properties of vector cross product are
• Vector cross product is a not commutative operation. If you reverse the order you will get the same magnitude but opposite direction
• Vector cross product is a distributive operation. You can distribute (and group) the vectors with respect to addition or subtraction such that and
• Vector cross product is an associative operation with respect to scalar multiple of vector. You can exchange the order of computation (operation inside parentheses are to be computed first) does not change the result.
• Vector cross product to itself always produces zero vector. Cross product with a zero vector also produces zero vector .
• The magnitude of vector cross product is equal to the product of their norms and sine angle between the two vectors, . This magnitude is equal to the area of parallelogram bounded by the input vectors.
• Cross product of the same standard unit vector is zero
• Cross product of the perpendicular standard unit vector form a cycle ;;;;;;
• Relationship of norm of cross product and dot product is.
See also: triple dot product, triple cross product, scalar triple product, inner product
<Next | Previous | Index> |
Can 72 Be Divided?
The Factors of 72 include 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.
Can 72 be divided equally?
It’s easy to see that the numbers 72 are the same as 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, and 36.
What is the quotient of 72 divided by 3?
The quotient is 24 and the rest is 0. The dividends are 72 and 3.
How do you do 72 divided by 9?
There are 72 divided by 9. If 72 9 is equal to 9 then 72 9 is equal to 9.
How do you solve 72 divided by 5?
The quotient is 14 and the rest is 2. There is a dividend and a divisor.
What times what give you 72?
When two numbers are combined in a way that returns a result of 72, it’s called Factors of 72. The Factors of 72 include 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.
What is the remainder of 72 divided by 6?
It was the conclusion of the story. 72/6 is a number. There is no fractional part in this number. 72 6 is equal to 12 R 0.
How do you solve 72 divided by 2?
The quotient is 72/2 and the rest is 0. The dividends are 72 and 2.
What is the remainder of 72 divided by 12?
The quotient is 6 and the rest is 0. There is a dividend and a divisor.
What does divided by look like?
The division sign has a dot above it and a dot below it. It’s the same as the words “divided by.” The symbol is found mostly in the elementary school level.
How do u divide fractions?
The first thing you need to do to divide fractions is find the inverse of the numerator and denominator. Next, take the two numbers and add them up. Next, add up the two numbers. The fractions should be simplified if necessary.
What are the divisors of 72?
The factors of 72 are formed by the divisors 1, 2, 3, 4, 6, 8, 9,12, 18, 24, 36, and 72.
What is the 72 times table?
The multiplication table of 72 has been repeated many times. 3 x 72 is the same as 216.
What are the prime factors of 72?
The number 72 can be written as a product of prime factors. It is said that the prime factorization of 72 is the expression 2 3 3 2.
How do you solve 40 divided by 5?
The quotient of 40/6 is 8 and the rest is 0. The dividends are 40 and 5.
How do you solve 42 divided by 6?
Take the newest digit quotient and divide it by 6. Subtract the number from the number. There was a division of 426 42 6.
What is the quotient of 72 divided by 5?
If you used a calculator, you would get 14. It is possible to express 72/6 as a mixed fraction: 14/2.
How do you solve 75 divided by 5?
75 percent is 15. There is no fractional part in this number. 75 5 is equal to 15 R 0.
What times 2 gets you 72?
The answer to the question is 36. If you want to double check our work, use 36 by 2.
What is the remainder when you divide 72 by 9?
The quotient is 8 and the rest is 0. There is a dividend and a divisor.
How do you solve 24 divided by 3?
The quotient is 8 and the rest is 0. There is a divisor and a dividend.
How do you solve 76 divided by 2?
The quotient of 76/2 is 38 and the rest is 0. There is a dividend and a divisor.
How do you solve 84 divided by 3?
The quotient is equal to 28 and the rest is left over. The dividends are 84 and 3.
See also Is The Cx-9 The Scorpion Evo?
How do you solve 3 divided by 36?
What is the remaining part of 3? The quotient of 3/36 is zero; the remainder is three. The dividends are 3 and 36.
How do you solve 84 divided by 7?
The quotient of 84/7 is 12 and the rest is 0. Both 84 and 7 are the divisors.
What is a number divided by 6?
If the number is divided by 2 and 3, then it’s a number of 6. The test of divisibility is used to determine if a number is divisible by 6. If the unit’s place of the number is either zero or multiple of 2, we know the rules of divisibility.
What is the easiest math question in the world?
The easiest math problem that no one can solve is called the Collatz Conjecture. The Collatz Conjecture is hard to explain. There is an investigation by the Veritasium.
What is division math 3rd grade?
The division is how a group of things are distributed. Sharing is one of the four basic operations of math. The division is not the same as multiplication.
How do we multiply decimals?
If there isn’t a decimal, first multiply it as if there is one. The number of digits should be counted after the last digit in each factor. The product should have the same digit number behind it.
What is quotient fractions?
There are fractions of something. When you simplify the fraction, you get the whole number. The quotient is the form of the fraction that is not a whole number. There is a division problem involving two fractions and the quotient can be used to solve it.
Related Posts
Are Male Ligers Fertile?
error: Content is protected !! |
# Lesson Explainer: Area between Curves Mathematics • Higher Education
In this explainer, we will learn how to apply integration to find the area bounded by the curves of two or more functions.
Recall that the area enclosed by a curve , the -axis, and the lines and is given by the definite integral of with respect to evaluated between and as follows: where is the antiderivative of such that
Space enclosed above the -axis evaluates to a positive value, and space enclosed below the -axis evaluates to a negative value. To find the area of a region, which is a strictly positive quantity, the absolute value is taken.
Recall also that the integral operator is a linear operator; hence, it is closed under both addition and scalar multiplication. The difference between the integrals of two functions, and , is therefore given by the integral of the difference between the two functions as follows:
Recall that this property can be used to find the area enclosed by a curve, , and a horizontal line, where is replaced by the value , so that the integrand is . Sometimes, however, we need to find the area of more complicated regions that are not necessarily enclosed by the -axis or a horizontal straight line.
The above property holds for a general continuous function . Consider the region enclosed by two curves, and , and two vertical lines, and .
The area will be given by
Using the previous properties of the linearity of integration, we then have the following theorem.
### Theorem: Area between Two Curves
For two functions, and , where within the interval , the area enclosed between two curves, and , and two vertical lines, and , is given by
Unlike finding the area between a single continuous function, , and the -axis, it does not matter whether either curve is above or below the -axis. However, it does matter which curve is above the other. If instead within the interval , we can instead take the absolute value of the integral, or reverse the order of the curves and integrate .
For intervals where the curves cross each other, we need to be more careful, as we will see in a later example.
Letβs look at an example of applying this formula to find the area enclosed by two curves and two vertical lines.
### Example 1: Finding the Area of a Region Bounded by Trigonometric and Linear Functions
Find the area of the region enclosed by the curves , , , and .
It can be helpful to draw a graph of the functions for the area we are evaluating, especially the points where the line and curve intersect. To find these, recall that the slope of is equal to 1 at and then decreases and oscillates, never exceeding 1. The slope for the line is also 1, for all . Therefore, the line, , will stay strictly above the curve, , and the two curves will intersect only once at .
The other two lines are vertical lines at and .
Recall that the area enclosed by two curves, , and , and two vertical lines, and , is given by
In this case, we have one line, , that is strictly above another curve, , and two vertical lines, and , enclosing a region whose area we will call . The area is therefore given by
Integrating with respect to gives us
Evaluating the limits gives us
Sometimes, the upper and lower limits for the integration will not be defined by the vertical lines and but instead by intersection points between curves that need to be found.
Letβs look at an example of how to find the area enclosed by curves whose upper and lower limits are not given.
### Example 2: Finding an Area Bounded by Cubic and Linear Functions
Find the area of the region bounded by and .
It can be helpful to draw a graph of the functions to see which area we are evaluating. We will need to know the points where the curves intersect, which can be quickly determined by finding the solutions to the equation
In this case, we have
Rearranging and factorizing gives us
Solving for , we find
Hence, the curves intersect at , , and . The cubic function, , will grow faster than the linear function, , for large positive or negative values of . Using this information, we can sketch the curves as follows.
Recall that the area enclosed between two curves, and , and two vertical lines, and , is given by
In our case, for negative values of , the cubic is above the linear function, and for positive values of , the cubic is below the linear function. This means that if we evaluate each of the two regions separately, the difference between the two functions, and therefore the sign of the integral, will be opposite for the two regions.
The functions are both odd; that is, and , so they are both antisymmetric about the -axis. We therefore know that the two areas will be equal and opposite and that they will cancel each other out to evaluate to zero.
We could rectify this by evaluating the integrals for the two regions separately, taking the absolute values, then summing them together. A faster way is to use our knowledge that the functions are both odd and therefore the area in each of the two regions is the same. Hence, we can simply evaluate the area of one region, then double it to find the total area.
Substituting the given functions and the limits and (the region below the negative -axis) into the formula for the area, we have
Integrating with respect to gives us
This is the area for the region to the left of the -axis. Doubling this gives the total area as follows:
In the previous example, we used a general result in integration that applies to all odd and even functions.
### Theorem: Area between the Curve of an Odd or Even Function and the π₯-Axis
For an odd function, , such that , the area enclosed by the curve of and the -axis within an interval is equal to twice the area enclosed by the curve of and the -axis within an interval . That is,
The same applies for an even function, , such that . That is,
As we saw in the previous example, just as we need to take care when evaluating the area of multiple enclosed regions above and below the -axis, we also need to take care when evaluating areas between curves when the curves cross each other. Sometimes, the points of intersection might be harder to find, and the separate regions may not be symmetrical.
Letβs look at an example demonstrating how to find the area enclosed by two curves that cross, where the point of intersection is less obvious and the separate regions are asymmetric.
### Example 3: Finding the Area Enclosed between Two Curves That Cross Each Other
The curves shown are and . What is the area of the shaded region? Give an exact answer.
Letβs first establish which curve is which function on the graph. For large , will grow larger faster than , and therefore will grow smaller faster than . Therefore, the curve that is lower to the right-hand side of the graph is the curve of .
Recall that the area enclosed between two curves, and , and two vertical lines, and , is given by
In this example, the two curves cross each other at a point partway through the region whose area we need to find. Recall that since we are taking a difference of two functions in the integral, , the sign of the evaluated integral will depend on which curve is above the other. In the region where is above , the integral will be positive, and vice versa. This means that if we evaluate the integral between the limits, the evaluation from one region will be canceled out with the evaluation of the other and will not give us the total area.
This means that we will need to evaluate the area of the two regions separately; that is, one to the left of the point where the curves intersect and one to the right. To do this, we need to find the point where the curves intersect, which we can find by solving the equation
In our case, we have
Solving for , we find
Hence, the curves intersect at . This can be clearly seen on the graph, but it is always a good idea to find the exact value.
We now need to evaluate two separate definite integrals. From the graph, we see that the regions are also bounded by the vertical lines and .
The lower and upper limits of integration for the region on the left are therefore and , and likewise the lower and upper limits for the region on the left are and .
If we were to proceed without splitting the regions, we would evaluate the area. Thus,
Looking at the graph, we can see that this cannot be the correct value for the area of the shaded region, which looks much larger than this. To find the true area, we instead need to split the integration into the two separate regions on either side of the intersection point, taking the absolute value of each and then summing them together.
We have already found the indefinite integral above, which is given by
So, for the first region between and , we have
For the second region between and , we have
The integral has evaluated to a negative value because the curve of is above the curve of , and we have subtracted the former function from the latter function.
Since area is a strictly positive quantity, we take the absolute value. So, we have
Now, summing these two areas together, we find the exact value of the area of the shaded region as follows:
For some problems, we may need to find the area enclosed by more than two curves. Letβs look at an example of how to find the area enclosed by three curves.
### Example 4: Finding the Area Bounded by Two Linear Functions and a Reciprocal Function Involving Dividing the Region of Integration
Consider the region in the first quadrant enclosed by the curves , , and . Find the area of this region.
It can be helpful to draw a graph of the functions to see which area we are evaluating. We will need to know the points where the curves intersect. In this particular case, we need to be sure to draw the curves accurately. Determining the points where one curve is above another (and which one) and the points where the curves intersect is crucial to ensuring we calculate the area correctly.
To begin with, consider the lines of the linear functions. is a straight line that passes through the origin and has a slope of 1. is a straight line that passes through the origin and has a slope of . Therefore, it stays strictly below in the first quadrant.
is a reciprocal function, so its curve is a hyperbola that never touches the - or -axes but has asymptotes at both. When , , and when , , so the curve passes through the points and .
We now have enough information to sketch the curves in the first quadrant as follows.
Recall that the area enclosed between two curves, and , and two vertical lines, and , is given by
For this to evaluate to the true area, we require to be above between the lines and . In our case, however, which of our functions is above which is dependent on the value of . Looking at the sketch, we can see that we have two different regions within as follows.
In region 1, the area is enclosed by the lines and .
In region 2, the area is enclosed by the lines and .
The lower limit for region 1 is the origin, . The upper limit for region 1 is the -value of the point of intersection between and .
Likewise, the lower limit for region 2 is this same -value of the point of intersection between and , and the upper limit for region 2 is the -value of the point of intersection between and .
We therefore need to find the -values of two points of intersection.
Point 1: Intersection between and
To find the -value of the point of intersection between these two curves, we need to equate the two functions and solve the following equation for :
Solving for gives us
Since we are considering the curves only in the first quadrant, the value of that is relevant to this problem is the positive value. Therefore, the two curves intersect at . Note that we do not need to find the -value, as it is not necessary for the integration.
Point 2: Intersection between and
Likewise, to find this point of intersection, we need to solve the following equation for :
Solving for gives us
Again, the value of that will be relevant is the positive value, .
In region 1, since the line is above that of , we let and . Hence, the area of region 1 is given by
In region 2, since the curve is above that of , we let and . Hence, the area of region 2 is given by
The total area of the region, , is therefore given by
Sometimes, we may need to find the area enclosed by implicit functions. Letβs look at an example of how to find the area enclosed by two implicit functions.
### Example 5: Finding the Area of aRegion Bounded by Two Quadratic Functions Defined with Respect to π¦
Find the area of the region bounded by and .
It can be helpful to draw the graph of the functions for the area we are evaluating, especially the points where the curves intersect.
In this case, both functions are implicit in ; that is, we have in terms of . Of course, there is nothing intrinsic about the convention of being the independent variable and being the dependent variable. We can therefore simply reverse the typical - and -axes and sketch the curves of the two functions of .
is a U-shaped parabola that is symmetric about the -axis and that intersects the -axis at .
is an n-shaped parabola that is symmetric about the -axis and that intersects the -axis at .
Orienting the axes in the conventional way, the graph looks as follows.
Recall that the area enclosed between two curves, and , and two vertical lines, and , is given by
In our case, however, we have two curves defined such that is a function of . We therefore need to modify the integrand by switching these variables. Also, note that the limits of the integral will now be -values rather than -values.
Hence, the area enclosed by the two curves, and , and the two horizontal lines, and , is given by
We are now integrating with respect to .
Looking at the first sketch with on the vertical axis, we can see that the curve of is always above the curve of for the enclosed area, so we will let and . The upper and lower limits of integration will therefore also be values of at the points of intersection between the two curves.
To find the intersection points, we need to solve the equation for ; that is,
Now, solving for , we have
The -values of the intersection points, and therefore the lower and upper limits of integration, are and .
Now we have everything we need to find the area of the enclosed region as follows:
Integrating with respect to gives us
Letβs finish by recapping some of the key points from this explainer.
### Key Points
• The area enclosed between two curves, and , with strictly above within the interval , and two vertical lines, and , is given by .
• Regions where the curve of is above the curve of will evaluate to a positive area, and regions where the curve of is below the curve of will evaluate to a negative area. To find the true area, the absolute value is taken.
• Where the curves of and cross each other within the interval, the integration must be split for each separate region, with the absolute value of each region taken, then summed together to give the true area. |
April 2017 - Down River Resources | Your Elementary Math Guide
Multiplication and division can be difficult concepts to teach, especially if your second grade students have no prior experience with this type of thinking. It happens every year! Multiplication and division problems are fundamentally different than addition and subtraction problem situations because of the types of quantities represented. Multiplication and division are taught together so that student can see that one operation is the reverse of the other. Let's make this year different! Using the mathematical principle of unitizing and the "GET" strategy, students will build their proficiency as they learn contextual multiplication and division in the math classroom.
## Multiplication and Division are fundamentally different than addition and subtraction.
A simple addition problem situation could be: Ann has 3 cookies. Laura have her 4 more cookies. How many cookies does Ann have now?
A simple multiplication problem situation could be: Ann has 3 bags of cookies with 4 cookies in each bag. How many cookies does Ann have?
The numbers are the same but the quantities represented are different. This shift in thinking is what gives most students difficulty when transitioning from the operations of addition and subtraction to multiplication and division.
Second grade students need to be able to model, create, and describe contextual multiplication and division situations. What if there was something that could help bridge the gap for these students?
## Unitizing Helps Students Shift Their Thinking
Have you heard of unitizing? It is an important, and often unknown, math word. Unitizing gives students a change in perspective.
Think back to the development of numeracy. Children learn to count objects one by one, also known as one-to-one correspondence. Instead of counting ten objects one by one, students can unitize them as one thing or one group. Another example of unitizing can be found within place value. Whenever we have 10 or more in a place value unit, we need to regroup. Thus, ten ones can also be thought of as a unit of ten.
This concept of unitizing is a big shift for students. It almost negates what our students originally learned about numbers. We want to help our students achieve the developmental milestone of unitizing. Unitizing is the underlying principle that guides students' learning.
Students need to use numbers to count, not only objects, but also groups... and to count them both simultaneously. Unitizing helps students build their proficiency in contextual multiplication and division.
Students need to be explicitly taught this principle and exposed to seeing it in action multiple times, much less subitizing in these primary grade levels. Show the students ten objects and tell them, " This is one group of 10." It seems simple, right? It is actually quite tricky for students to grasp, so repeat yourself...and repeat yourself.
## Multiplication and Division Strategy: Did You "GET" It?
Another trick for tackling multiplication and division is a little-known strategy. G-E-T is a simple acronym for an effective strategy when teaching contextual multiplication and division.
I have used the acronym before but I added this first step which helps build students' meta-cognition.
After reading through a word problem that involves multiplication or division, ask yourself: "Did you GET it?" If your answer is "yes," you probably followed these steps:
1. Read through the word problem at LEAST once.
2. Circle and label the number representing the GROUPS. (How many groups are the objects being divided into?)*
3. Circle and label the number representing the EACH. (How many objects are within each group?)
4. Circle and label the number or noun represent the TOTAL. (How many total objects are altogether or in total?)
*When students label, they circle the number and noun (example: 12 cats) and they write the word to describe that part of the word problem (example: in this case, 12 cats would represent the total. The students would write the word TOTAL or the letter "T" on top of the circle.)
If your students label these three parts to a word problem, it will be so much easier solving for the unknown, whether is be the dividend, divisor, quotient, factors, or product.
Labeling word problems using the "GET" strategy is a non-negotiable in second and third grades! Of course, modeling and guided practice is a must before this layer of accountability takes effect!
I hope this post inspires you to build your students' concept of unitizing and their proficiency with the "GET" strategy, and if you want to use my interactive math notebook on contextual multiplication and division, it's in my TpT shop.
What are some ways you build your students' proficiency in multiplication and division?
* References: Fosnot, C. & Dolk, M. (2001). Young mathematician at work: Constructing Multiplication and Division, NH: Heinemann.
Spring brings butterflies, chicks, blossoms, and... plastic eggs! People near and far hunt for these special spherical objects hidden in secret places. I tend to just go straight to the seasonal aisles of my favorite stores and find a wide variety of plastic eggs to choose from with a lot less hassle. In recent years, the stores are stocking an eclectic mix of eggs. These eggs include special shapes (animals and carrots), unique patterns (faith-based words, animal print, camouflage), extra-sparkly glitter, golden, and transparent eggs...these probably just list the ones stocked at eye-level! I stockpile a large assortment of these diverse eggs and pair them with rigorous math concepts to create the perfect math centers for kindergarten, first, and second grades. While my ideas are focused on these grade levels, many of them can be adapted for other grades too! This is my go-to list for simple math centers using plastic eggs.
## Matching Math Centers
Plastic eggs are versatile! You can write on them, fill them, or do both!
I love writing on them....probably because I stockpile school supplies like my husband stockpiles freeze-dried rations (insert "yuck" face!)
### Numeral + Tens Frames (Kindergarten)
Grab a regular Sharpie marker and some of those eggs and get marking! One of my favorite ways for students to use the two parts of a plastic egg is to match the numeral to the tens frame. This helps students read and represent whole numbers 0 to 20 with objects (TEKS K.2B.)
You can also match the numeral to a tally mark, subitizing dots, the number word, or stickers placed on one of the parts!
### Composing Ten (Kindergarten, First, & Second Grades)
Another way to use the "matching" concept is composing numbers. When two numbers are added together, this is called composing numbers. Students simply match two one-digit addends which add up to 10.
Kindergarten and first grade students are asked to compose numbers to 10 (TEKS K.2I & 1.3C.) This concept of putting two numbers together to form one can also build a second grade student's automaticity with basic facts (TEKS 2.4A.)
## Stacking Math Centers
This might be my favorite way to use plastic eggs! Since you are only using one of the parts, the eggs go a long way. You will have more pieces to create more centers...and what do I want to make? More centers!
### Counting by Tens (Kindergarten & First Grades)
I love how simple algebraic reasoning skills can be practiced by stacking the pieces into a tower. Your students will think the best part of this math center is trying to make the tower stay up. It is VERY common that the entire tower will fall, so students are practicing a lot more than just algebraic reasoning with this center.
I use this center for counting by 10s (TEKS K.5.) This same concept can be applied to skip counting by 2s, 5s, and 10s (TEKS 1.5B.) Skip counting is also a great skill to continue practicing in second grade as the students will apply this skill to contextual multiplication.
## Sequencing Math Centers
No matter which grade level you teach, you can use this concept of sequencing numbers with plastic eggs. The concept remains unchanged, but the numbers will be different. You can also tailor this center to your students' needs. You may have a student who is struggling or advanced, whichever the case, add eggs with the numbers that best suits students' needs.
Kindergarten students practice numbers 0 through 20 (TEKS K.A,) while first grade students use numbers up to 120 (TEKS 1.2F.) Second grade students practice ordering numbers up to 1,200 (TEKS 2.2D.)
### Ordering Whole Numbers (Kindergarten, First, & Second Grades)
This egg carrier was purchased at Dollar General for \$2. They can also be found at Dollar Tree for \$1. Most of these egg carriers hold 12 or 24 eggs. You can make the exact amount of eggs needed or less. The carrier just acts as a place holder for the eggs.
The best thing of all if that there is a center section which can hold the pile of eggs (see the top part of the image.) Students can pick up one egg at a time and place it in a spot. As students pick up additional eggs, they may need to move eggs as the place value of each of the numbers is determined. The carrier works as an open number line.
## Filled Math Centers
Using filled plastic eggs, you could teach any math skill! You can write numbers, draw shapes, or create word problems on a piece of paper, fold it up, and place it inside a plastic egg! That's as simple as ABC, friends. I use a variety of materials to fill the eggs just to keep my students interested in egg activities so they are not repetitive.
### _ More and _ Less (Kindergarten, First, & Second Grades)
Yes, there is a reason I left a blank in the title for this section! You can tailor this center to meet the needs of any grade level or any child.
Kindergarten students are working on one more and one less (TEKS K.2F,) while first grade students are learning 10 more and 10 less than a given number up to 120 (TEKS 1.5.) Second graders expand on this idea by determining the number than is 10 or 100 more or less than a given number up to 1,200 (TEKS 2.7B.)
You can place a card within the math center to indicate if the students are working on the number than is _ more or _ less than the given number or they can generate both numbers.
I had students simply take a strip of notebook paper and number it, like we do for spelling tests, then students drew their eggs, opened them up, and recorded their answers individually. The example shown is for first grade (10 more and 10 less than a given number). The numbers on the eggs represent the problem number for recording purposes. The number on the sticky note is the number that students use to generate their answers.
### Counting (Kindergarten)
We all need another excuse to buy those irresistibly cute Target erasers, right?! Well, here's another one!
Fill the eggs with a certain number of erasers. Students will open one egg at a time and count to determine the quantity held in the eggs. I numbered the eggs so that the students can record their answers.
Again, I used a strip of notebook paper and had students number it, like we do for spelling tests. As they select eggs out of the basket or container, they record the answer on the corresponding line.
Kindergarten students are learning to count forward to 20 (TEKS K.2A,) and counting a set of objects up to at least 20 (TEKS K.2C.) This activity also build students' one-to-one number correspondence.
### Graphing (Kindergarten, First, and Second Grades)
This same concept of filling the eggs or placing objects within them can be applied to data collection, each egg could contain a specific object (erasers, jelly beans, etc.) and students record the data on a bar graph or picture graph. Students are learning how to collect and organize data (TEKS K.8ABC, 1.8ABC, 2.10CD.)
### Coin Collections (First and Second Grades)
Fill plastic eggs with coins. I try to use real coins when I am able to as they are more life-like. Plus, there are so many varieties of coins, I have yet to find a math manipulative that captures their new look. Each egg is filled with a different combination of coins.
First graders are learning how to count by 1s to add up the value of pennies and skip count by 5s and 10s to add up the value of nickels and dimes (TEKS 1.4ABC). Second grade students are determining the value of a collection of coins up to one dollar (TEKS 2.5AB.)
I used a strip of notebook paper and had students number it, like we do for spelling tests. As they select eggs out of the basket or container, they record the answer on the corresponding line. In the example in the photograph, I had the second graders write the value of the collection of coins using the cent symbol and the dollar sign and decimal point to specifically meet the rigor within the second grade standard (TEKS 2.5B.) First graders would write the value of the collection of coins using the cent symbol.
Whew! That was a plethora of ways that you are create the BEST math centers using plastic eggs. I love that each of these math centers are rigorous.
Did you notice each idea I used was standards-based and met the specificity described? Students are practicing the power words in education: determining, generating, representing, composing! These are the higher-order thinking skills we want them to use and practice, practice, practice. Why not use plastic eggs to accomplish this?
In addition, these dynamic math centers are be differentiated based on your students' needs. If your students has not mastered three-digit numbers, create some eggs with two-digits. If your students have surpassed the grade level goal, make them four-digit numbers! I love using plastic eggs in math centers as they are rigorous and meet the needs of diverse learners in my classroom.
I hope this inspires you to turn those leftover plastic eggs into some engaging math centers for diverse learners!
What is your favorite way to use plastic eggs for math centers?
#### I send out exclusive tips, tricks, and FREE resources to my partners. Drop your email below to become an exclusive partner!
Plastic eggs are not just for an Easter egg hunts! After a delicious Easter meal, the kids take part in a large Easter egg hunt at my parent's house. There are so many good hiding spots and, as usual, the older kids dominate the hunt. Colorful, plastic eggs jingle with coins and jellybeans...chocolate and a dollar bills too, but only if you are lucky! After a few quick minutes, everyone gathers back on the porch. The kids quickly hide their money (after counting it, of course!) and throw their eggs into a large sack. While everyone is eating jelly beans and some eat chocolate, I begin my favorite activity of the post-Easter season! I re-purpose those colorful, plastic eggs and create rigorous math centers that can be used for the rest of the school year. While I hunt high and low for interesting eggs, I can never have enough. So, what's the big deal with plastic eggs? I'm glad you asked...
## Top Five Reasons to Use Plastic Eggs in Your Math Centers
### 1. Plastic eggs are inexpensive...unless you buy an entire cart full of them. Guilty as charged!
Most of the eggs I buy are about 98 cents to \$2.00 per package. Of course, the basic colorful eggs are most inexpensive, while the larger, themed eggs are most expensive. I try and think about what I would like to use them for before I buy so I can have a quantity in mind...but most of the time I just buy, at least, two packages. That way, I am guaranteed to have a little variety in whatever I end up creating! I mean, when you see cute eggs you just buy them...kind of like those Target erasers. Gasp!
### 2. Plastic eggs can be used year-round. There is so much diversity in the type of eggs you can buy, you can use them seasonally and/or with your classroom themes.
I have eggs with sports theme that I use during those seasons. I also use the animal-shaped eggs to coincide with teaching about organisms. I love making as many cross-curricular connections using math as possible. Not only is it a great way to spiral, or revisit, previously taught content, it also gets the students engaged with the theme!
### 3. Plastic eggs provide a hands-on, or tactile, way which stimulates the brain.
Tactile learning take place when the students are carrying out the actual physical activity! Whether the students are sorting through the eggs or opening them up, students are actively participating in the math center.
### 4. Plastic eggs are versatile.... they can be written on with a permanent marker, stickers can be added on them, and/or they can be filled!
Pick your favorite way to use plastic eggs, mix it up, or use them all! Plastic eggs allow you to use them creatively to accomplish your specific learning goals for math centers.
My favorite writing tool to use on plastic eggs is a regular Sharpie marker in black. It goes on nice and smooth. I tried the flip chart version of the marker and it left marks.
### 5. Plastic eggs are reusable! Not only can you have an amaaazing Easter egg hunt with them, but you can use them within a center.
Don't worry if you are late to the Easter egg party! If you create a last-minute center this year, it will be ready to go for next year! You can break apart the eggs to compress them for storage, creating towers of like ends. Perhaps, you can even find some eggs on clearance... fingers crossed!
Have you heard of the interactive math game, Splat!? There are different variations and different games with the same name, but I use this interactive game to get my students engaged about a particular math concept which we have already learned. It can be used for interactive math reviews. It encourages students to analyze number relationships to connect and communicate mathematical ideas. This is a process standard that students can always use more support in practicing. Splat! is a fun way to apply mathematical concepts and makes for a fun math center, small group, or whole group game.
## Using Splat! in the Math Classroom
### Suggested Age Range for Activity
Splat! can be used with any grade level of students, just make sure that the content being reviewed is developmentally age appropriate or specific to your grade level's standards.
### Preparing for Activity
Splat! games are relatively easy to prep. You will need a tall cylindrical tower. It works best with a potato chip can!
If you needed an excuse to buy more potato chips, here it is! After all, once you pop, the fun just doesn't stop! Now, the fun can continue for your entire school year!
First, print out the cards. Then, cut out the cards and laminate. My games are created in both blackline, for ink savings, and in color which really makes the fun seasonal faces POP!
Regardless of which route you take, I recommend printing the cover for the cylindrical can in color. I use white copy paper so it bends around the surface better. I added some colorful complimentary washi tape on the top edge.
If using a tall can, you will need something to make up the difference, as the paper is only 8x11 inches tall. If you use a new shorter can, you will have to cut a little of the space on the top as the height of the can is shorter than the paper! (I've tried both ways! The really good flavors of chips come in the shorter cans! I like using the tall cans so I can utilize my colorful washi tape!)
To make it last longer and protect it from water and dirty hands, add packing tape around the paper as a protective
coating!
Teacher Tip: The thing that I love the most about Splat! is that the can you use to create the tower in the game also serves as storage for all of the game cards!
To make the cards self-correcting, mark the correct answers with an adhesive dot on the back (yard sale sticker). If playing with your class, there is no need for this step, unless you will be adding it to an independent math center or station or using it for an activity for early finishers.
## Reviewing Math Concepts with the Game Splat!
I try to find simple skills in the list of standards that could be turned into a one line question for the games I create.
Kindergarten Sample Questions:
What is 1 more than 6?
What is 1 less than 8?
What is 10 more than 55?
What is 10 less than 34?
Is 27 odd or even?
Is 15 odd or even?
## How to Play the Interactive Game Splat!
To play, set a timer for the amount of time you have to play, or stop play when the session is over.
1. Mix up the cards (math question cards + Did Somebody Say Splat Cards + challenge cards). Place the deck of cards facedown on the table.
2. Have players read the card and generate the correct response. The player should say that answer three times. This is my variation, but can be modified however you'd like.
3. After they answer the question, they place the card on the top of the tower.
If players pick a Did Somebody Say, “Splat?” card, they should simply place that card carefully on top of the “tower.”
If it stays there securely, that players turn is over. If that card or any other cards go SPLAT (falls off the tower,) that player must take all cards that fell.
If they pick a #challenge card they must follow the directions on the card. The same procedures apply.
The object of the game is to stack cards carefully without making any go SPLAT! Players that knock down cards must take them.
At the end of play (either when the session ends, the timer rings or there are no more cards to play), the player with the LEAST cards is the winner!
There are so many different ways that you can manipulate the play of this game using the three different cards types. Find the way that you like best and get your students excited about math!
I hope this post inspires you to use Splat! in your classroom and if you want to use my Splat! games, they're in my TpT shop.
What are some other ways you review math concepts in your classroom? |
## A class had 30 pupils at the beginning of this school term, but now has 5 more pupils. What is the percent of increase?
Question
A class had 30 pupils at the beginning of this school term, but now has 5 more pupils. What is the percent of increase?
0
Explanation:
You need to learn how to calculate percent to be able to perform this type of problems by yourself.
Percent is the ratio of a quantity on base to 100 parts.
In this case the increase is 5 students of an initial group of 30, so you need to find the equivalent expression with a base of 100 instead of 30. That is, you need to find the ratio with denominator 100 equivalent to the ratio 5/30:
x 5
—— = —-
100 30
=> x = (5 / 30) * 100 = 16.67%.
Also, you should notice that calculatind the percent is done dividing the parts by the whole and multiplying by 100, i.e. (5 / 30) * 100, which is the same done above.
2. The percent of increase would be 16.66 |
# How do you simplify ((5n^4)/(p^3))/((6n)/(5p))?
Jan 16, 2017
See the entire simplification process below:
#### Explanation:
First, simplify the division by using the rule for dividing fractions:
$\frac{\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}}{\frac{\textcolor{g r e e n}{c}}{\textcolor{p u r p \le}{d}}} = \frac{\textcolor{red}{a} \times \textcolor{p u r p \le}{d}}{\textcolor{b l u e}{b} \times \textcolor{g r e e n}{c}}$
$\frac{\frac{\textcolor{red}{5 {n}^{4}}}{\textcolor{b l u e}{{p}^{3}}}}{\frac{\textcolor{g r e e n}{6 n}}{\textcolor{p u r p \le}{5 p}}} = \frac{\textcolor{red}{5 {n}^{4}} \times \textcolor{p u r p \le}{5 p}}{\textcolor{b l u e}{{p}^{3}} \times \textcolor{g r e e n}{6 n}} = \frac{25 {n}^{4} p}{6 n {p}^{3}}$
We can now use these rules for exponents to further simplify this expression:
${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$
${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$
$a = {a}^{\textcolor{red}{1}}$
$\frac{25 {n}^{\textcolor{red}{4}} {p}^{\textcolor{red}{1}}}{6 {n}^{\textcolor{b l u e}{1}} {p}^{\textcolor{b l u e}{3}}}$
$\frac{25 {n}^{\textcolor{red}{4} - \textcolor{b l u e}{1}}}{6 {p}^{\textcolor{b l u e}{3} - \textcolor{red}{1}}}$
$\frac{25 {n}^{3}}{6 {p}^{2}}$ |
## AREA OF PARALLELOGRAM
A parallelogram is a quadrilateral in which opposite sides are parallel and equal in length.
In other words opposite sides of a quadrilateral are equal in length, then the quadrilateral is called a parallelogram.
The area of a parallelogram is the product of a base and its corresponding height.
Then, the formula to find area of a parallelogram is given by
A = b ⋅ h square units
We can justify the area for parallelogram as follows.
The area of a parallelogram is the area of a rectangle with the same base and height.
## Examples
Example 1 :
Find the area of the parallelogram ABCD shown below.
Solution :
Method 1 :
Use AB as the base.
So, b = 16 and h = 9.
Formula for area of a parallelogram is
= ⋅ h
Substitute the given measures.
= 16 ⋅ 9
= 144 square units
Method 2 :
Use AD as the base.
So, b = 12 and h = 12.
Formula for area of a parallelogram is
= ⋅ h
Substitute the given measures.
= 12 ⋅ 12
= 144 square units
Example 2 :
Find the area of the parallelogram ABCD shown below.
Formula for area of a parallelogram is
= ⋅ h
Substitute b = 5 and h = 3.
= 5 ⋅ 3
= 15 square units
Example 3 :
A mirror is made of two congruent parallelograms as shown in the diagram. The parallelograms have a combined area of 9 1/3 square yards. The height of each parallelogram is 1 1/3 yards. How long is the base of each parallelogram ?
Solution :
Because the given parallelograms are congruent area of two parallelogram will be equal.
Combined area of parallelograms = 9 1/3 square yards
Combined area of parallelograms = 28/3 square yards
Area of one parallelogram = (28/3) ÷ 2
Area of one parallelogram = 14/3
⋅ h = 14/3
⋅ 1 1/3 = 14/3
b ⋅ 4/3 = 14/3
Multiply each side by 3/4.
b = 14/3 ⋅ 3/4
b = 14/4
b = 7/2
b = 3 1/2
So, the base of the parallelogram is 3 1/2 yards.
Example 4 :
Find the base of a parallelogram if its area is 40 square cm and its altitude is 15 cm.
Solution :
Area of a parallelogram = 40 cm2
h = 40
Here, altitude (or) height (h) = 15 cm.
15 = 40
Divide each side by 15.
b = 2.67
So, the base of the parallelogram is 2.67 inches.
Example 5 :
Find the area of the shape shown below.
The above shape has four sides. So, it is a quadrilateral. Because the opposite sides are parallel, the above quadrilateral is a parallelogram.
Formula for area of a parallelogram is
= ⋅ h
Substitute b = 9 and h = 4.
= 9 ⋅ 4
= 36 square units
Example 6 :
If the area of the shape shown below is 60 square inches, then find the value of x.
Solution :
Given : Area of the above shape is 60 square inches.
The above shape has four sides. So, it is a quadrilateral. Because the opposite sides are parallel, the above quadrilateral is a parallelogram.
Area = 60 in2
h = 60
Substitute b = 12 and h = x.
12 x = 60
Divide each side by 12.
x = 5
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
## Recent Articles
1. ### SAT Math Resources (Videos, Concepts, Worksheets and More)
Sep 16, 24 11:07 AM
SAT Math Resources (Videos, Concepts, Worksheets and More)
Read More
2. ### Digital SAT Math Problems and Solutions (Part - 42)
Sep 16, 24 09:42 AM
Digital SAT Math Problems and Solutions (Part - 42)
Read More
3. ### Digital SAT Math Problems and Solutions (Part - 41)
Sep 14, 24 05:55 AM
Digital SAT Math Problems and Solutions (Part - 41)
Read More |
# Three Essential Algebra Formulas
### First, a few practice problems.
1. The numbers a, b, and c are all positive. If \space{b}^2 + {c}^2 = 117, then what is the value of {a}^2 + {c}^2?
Statement #1: a - b = 3
Statement #2: \frac{(a + b)}{(a - b)} = 7
2. Given that (P + 2Q) is a positive number, what is the value of (P + 2Q)?
Statement #1: Q = 2
Statement #2: {P}^2 + 4PQ + 4{Q}^2 = 28
3. In the diagram above, O is the center of the circle, DC = a and DO = b. What is the area of the circle?
Statement #1: {a}^2 - 2ab + {b}^2 = 36
Statement #2: a + b = 22
4. ABCD is a square with a side y, and JKLM is a side x. If Rectangle S (not shown) with length (x + y) has the same area as the shaded region above, what is the width of Rectangle S?
(A) x
(B) y
(C) y + x
(D) y - x
(E) {y}^2 - {x}^2
### Three important algebra patterns
Doing math involve both following procedures and recognizing patterns. Three important patterns for algebra on the QUANT are as follows:
Pattern #1: The Difference of Two Squares
{A}^2 - {B}^2 = (A + B)(A - B)
Pattern #2: The Squares of a Sum
{(A + B)}^2 = {A}^2 + 2AB + {B}^2
Pattern #3: The Squares of a Difference
{(A - B)}^2 = {A}^2 - 2AB + {B}^2
For Quant success, you need to know these patterns cold. You need to know them as well as you know your own phone number or address. The test will throw question after question at you in which you simply will be expected to recognize these patterns. In such a question, if you recognize the relevant formula, it will enormously simplify the problem. If you don’t recognize the relevant formula, you are likely to be stymied by such a question.
### Memory, not memorizing
You might think I would say: memorize them. Instead, I will ask you to remember them. What’s the difference? Memorization implies a rote process, simply trying to stuff an isolated and disconnected factoid into your head. By contrast, you strengthen you capacity to remember a math formula when you understand all the logic that underlies it.
Here, the logic behind these formulas is the logic of FOILing and factoring. You should review those patterns until you can follow each both ways - until you can FOIL the product out, or factor it back into components. If you can do that, you really understand these, and are much more likely to remember them in an integrated way.
### Summary
If these patterns are relatively new to you, you may want to revisit the problems at the top with the list handy: see if you can reason your way through them, before reading the explanations below.
### Practice Problem Explanations
1. Let X = a^2 + c^2, the value we are seeking. Notice if we subtract the first equation in the prompt from this equation, we get a^2 - b^2 = X - 117. In other words, if we could find the value of a^2 - b^2, then we could find the value of X.
Statement #1: a - b = 3
From this statement alone, we cannot calculate a^2 - b^2, so we can’t find the value of X. Statement #1, alone and by itself, is insufficient.
Statement #2: \frac{(a + b)}{(a - b)} = 7
From this statement alone, we cannot calculate a^2 - b^2, so we can’t find the value of X. Statement #2, alone and by itself, is insufficient.
Statements #1 & #2 combined: Now, if we know both statements are true, then we could multiple these two equations, which cancel the denominator, and result in the simple equation a + b = 21. Now, we have the numerical value of both (a - b) and (a + b), so from the difference of two squares formula, we can figure out a^2 - b^2, and if we know the numerical value of that, we can calculate X and answer the prompt. Combined, the statements are sufficient.
2. The prompt tells us that (P + 2Q) is a positive number, and we want to know the value of P. Remember number properties! We don’t know that (P + 2Q) is a positive integer, just a positive number of some kind.
Statement #1: Q = 2
Obvious, by itself, this tells us zilch about P. Alone and by itself, this statement is completely insufficient.
Statement #2: P^2 + 4PQ + 4Q^2 = 28
Now, this may be a pattern-recognition stretch for some folks, but this is simply the “Square of a Sum” pattern. It may be clearer if we re-write it like this:
P^2 + 2*P*(2Q) + (2Q)^2 = 28
This is now the “Square of a Sum” pattern, with P in the role of A and 2Q in the role of B. Of course, this should equal the square of the sum:
P^2 + 2*P*(2Q) + (2Q)^2 = (P + 2Q)^2 = 28
All we have to do is take a square root. Normally, we would have to consider both the positive and the negative square root, but since the prompt guarantees that (P + 2Q) is a positive number, we need only consider the positive root:
(P + 2Q) = \sqrt{28}
This statement allows us to determine the unique value of (P + 2Q), so this statement, alone and by itself, is sufficient.
3. To find the area of the circle, we need to use Archimedes’ formula, A = \pi(r^2). For that we need the radius, OC. We are not given this directly, but notice that r = OC = DC - OD = a - b. If we knew that, we could find the area of the circle.
Statement #1: a^2 - 2ab + b^2 = 36
A major pattern-matching hit! This, as written, is the “Square of a Difference” pattern.
a^2 - 2ab + b^2 = (a - b)^2 = r^2= 36
In fact, this statement already gives us r^2, so we just have to multiply by pi and we have the area. This statement, alone and by itself, is sufficient.
Statement #2: a + b = 22
We need a - b, and this statement gives us a value of a + b. If we had more information, perhaps we could use this in combination with other information to find what we want, but since this is all we have, it’s simply not enough to find a - b. This statement, alone and by itself, is insufficient. |
# Graphs of Square Root Functions
## Key Questions
• In order to translate any function to the right or left, place an addition or subtraction "inside" of the Parent function. In the case of the square root function, it would look like y = $\sqrt{x - 2}$ or y = $\sqrt{x + 5}$.
Let's look at the effect of the addition or subtraction. First, the domain will be altered. You must set x - 2 $\ge$ 0 , or say that you understand that the square root function has a domain of $x \ge 2$.
This implies a horizontal shift/translation of 2 units to the right. (see graph)
Now repeat for x + 5 $\ge$ 0, or $x \ge - 5$. This graph will be translated 5 units to the left. (see graph)
Now, let's explore how to translate a square root function vertically. y = $\sqrt{x} + 3$ or y = $\sqrt{x} - 4$. The addition or subtraction on the OUTSIDE of the square root function will cause the graph to translate up or down. Adding 3 will raise the graph up, and subtracting 4 will lower the graph by 4 units. (see graph)
If you are ready for a challenge, we can try to translate in more than one direction at a time!
y = $\sqrt{x + 2} - 7$ First, make a prediction...will the graph be translated right 2 or left 2, up 7 or down 7?
Find the domain by setting x + 2 $\ge$ 0 for starters.
$x \ge - 2$ (that means left 2, of course!)
And, subtraction of 7, must mean down 7. (see graph)
• Graph them like quadratic functions.
However, since square root cannot be performed for negative numbers, you need only graph the "positive" part of the quadratic.
• you have half of a parabola.
Consider $y = \sqrt{x}$
$x = 0 \implies y = 0$
$x = 1 \implies y = 1$
$x = 4 \implies y = 2$
$x = 9 \implies y = 3$
$x = - 1 \implies$ Undefined in $\mathbb{R}$
You have upper part of a parabola that opens to the right
If you consider $y = - \sqrt{x}$
You have the lower part of a parabola that opens to the right.
$\sqrt{y} = x$ and $- \sqrt{y} = x$ behaves similarly
$y = {x}^{2}$ with explicit domain $\left[0 , \infty\right)$
#### Explanation:
You could use $y = {x}^{2}$ with explicit domain $\left[0 , \infty\right)$ as the parent graph:
graph{(sqrt(x)/sqrt(x))x^2 [-4.767, 5.23, -0.6, 4.4]}
Reflecting this in the diagonal line $y = x$ we get the graph of $y = \sqrt{x}$:
graph{sqrt(x) [-4.767, 5.23, -0.6, 4.4]} |
# How do you solve x-2y=2 and y= -1/2x-5?
Nov 6, 2015
$x = 6 \mathmr{and} y = - 8$ through systems of equations.
#### Explanation:
This is known as a system of equations. For this system, we are going to use substitution, since you already know what y is equal to.
As you can see, $y = - \frac{1}{2} x - 5$. Since we have $y$ directly set equal to something, we can substitute it in for $y$ anywhere we want. In this case we will substitute it in for $y$ in your first equation. $x - 2 y = 2 \implies x - 2 \left(- \frac{1}{2} x - 5\right) = 2$
Through algebraic equation steps, we can simplify the equation to be $x + x - 10 = 2$, and solve the equation for $x = 6$.
Now that we have the $x$ value, we can substitute it in to the $y$ equation.
$y = - \frac{1}{2} \left(6\right) - 5$
When simplified, you get $y = - 3 - 5$, or $y = - 8$. And remember, if you'd like to check your answer, simply plug your numbers back in for the variables.
$y = - \frac{1}{2} x - 5 \implies \left(- 8\right) = - \frac{1}{2} \left(6\right) - 5$, and when simplified, $- 8 = - 8$. |
Basic Properties Of Circles Examples I
• In the figure, AEC and BED are straight lines. It is given that $$\angle BDC = 30^\circ$$, $$\angle ACB = 50^\circ$$ and $$\angle ABC = 100^\circ$$.
Prove that A, B, C and D are concyclic.
• In ΔABC,
$$\begin{array}{1}\angle BAC + \angle ABC + \angle ACB = 180^\circ \\\angle BAC + 100^\circ + 50^\circ = 180^\circ \\\angle BAC = 30^\circ \end{array}$$ (∠ sum of Δ)
∴ $$\angle BAC = \angle BDC = 30^\circ$$
∴ A, B, C and D are concyclic. (converse of ∠s in the same segment)
• In the figure, O is the centre of the circle. AB is a diameter. P is an external point of the circle such that $$PA = PB$$. PB intersects the circumference at C. AC intersects OP at D.
(a) Prove that PO ⊥ AB.
(b) Prove that O, A, P and C are concyclic.
(c) Prove ∠OPB = ∠OCA.
• (a) In ΔAPO and ΔBPO,
$$AP = BP$$ (given)
$$AO = BO$$ (radii)
$$OP = OP$$ (common side)
∴ $$\Delta APO \cong \Delta BPO$$ (S.S.S.)
∴ $$\angle POA = \angle POB$$ (corr. ∠s, ≅ Δs)
$$\begin{array}{1}\angle POA + \angle POB = 180^\circ \\2\angle POA = 180^\circ \\\angle POA = 90^\circ \end{array}$$ (adj. ∠s on st. line)
∴ $$PO \bot AB$$
(b) $$\angle ACB = 90^\circ$$ (∠ in semi-circle)
$$\begin{array}{1}\angle ACP = 180^\circ - 90^\circ \\ = 90{^\circ }\end{array}$$ (adj. ∠s on st. line)
∴ $$\angle AOP = \angle ACP = 90^\circ$$
∴ O, A, P and C are concyclic. (converse of ∠s in the same segment)
(c) $$\angle OPB = \angle OAC$$ (∠s in the same segment)
∵ $$OA = OC$$ (radii)
∴ $$\angle OAC = \angle OCA$$ (base ∠s, isos. Δ)
∴ $$\angle OPB = \angle OCA$$
• In the figure, O is the centre of two concentric circles. AB is the tangent to the smaller circle at M. Radii OA and OB of the larger circle intersect the circumference of the smaller circle at N and P respectively. It is given that $$OM = 20$$ and $$AB = 30$$.
(a) Find the length of OA.
(b) Find the length of AN.
• (a) ∵ $$\angle OMA = 90^\circ$$ (tangent ⊥ radius)
∴ $$AM = BM$$ (⊥ from centre bisects chord)
∴ $$\begin{array}{1}AM = \frac{1}{2}AB\\ = \frac{1}{2} \times 30\\ = 15\end{array}$$
In ΔOAM,
∵ $$O{A^2} = O{M^2} + A{M^2}$$ (Pyth. theorem)
∴ $$\begin{array}{1}OA = \sqrt {O{M^2} + A{M^2}} \\ = \sqrt {{{20}^2} + {{15}^2}} \\ = \underline{\underline {25}} \end{array}$$
$$\begin{array}{1}(b)ON = OM\\ = 20\end{array}$$ (radii)
$$\begin{array}{1}AN = OA - ON\\ = 25 - 20\\ = \underline{\underline 5} \end{array}$$ |
# 011. Finding a Cubic Function
#### Given a cubic function $f(x) = ax^3 + bx^2 + cx + d$ for a real number $a > 0$, $f(x)$ has a local minimum $-1$ and a local maximum $1$ on the interval $[-1, 1]$. When $f(1) = 1$ and $f(-1) = -1$, find $f(x)$.
A naive approach to find all coefficients is too complex. Rather using the following form is much more helpful to find them. Let $f'(\alpha) = f'(\beta) = 0$. \begin{aligned} f(x) &= a(x - 1)(x - \alpha)^2 + 1 \\ &= a \{ x^3 - (1 + 2\alpha)x^2 + (2\alpha + \alpha^2)x \} - a\alpha^2 + 1\\\\ f(x) &= a(x + 1)(x - \beta)^2 - 1 \\ &= a \{ x^3 + (1 - 2\beta)x^2 + (-2\beta + \beta^2)x \} + a\beta^2 - 1 \end{aligned}
Comparing coefficients, \begin{aligned} &\begin{cases} \; \beta - \alpha = 1 \\\\ \; (\alpha + \beta)(\alpha - \beta + 2) = 0 \\\\ \; a (\alpha^2 + \beta^2) = 2 \end{cases} \\\\ &\implies \alpha + \beta = 0 \\\\ &\implies \alpha = -\frac{1}{2}, \quad \beta = \frac{1}{2}, \quad a = 4 \\\\ &\implies f(x) = 4(x - 1)\left( x + \cfrac{1}{2} \right)^2 + 1 = 4x^3 -3x \end{aligned} |
# What is 51/36 as a decimal?
## Solution and how to convert 51 / 36 into a decimal
51 / 36 = 1.417
51/36 converted into 1.417 begins with understanding long division and which variation brings more clarity to a situation. Both are used to handle numbers less than one or between whole numbers, known as integers. But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. If we need to convert a fraction quickly, let's find out how and when we should.
## 51/36 is 51 divided by 36
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators and Denominators. This creates an equation. We must divide 51 into 36 to find out how many whole parts it will have plus representing the remainder in decimal form. This is our equation:
### Numerator: 51
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. Overall, 51 is a big number which means you'll have a significant number of parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Now let's explore X, the denominator.
### Denominator: 36
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. 36 is a large number which means you should probably use a calculator. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Have no fear, large two-digit denominators are all bark no bite. So grab a pen and pencil. Let's convert 51/36 by hand.
## Converting 51/36 to 1.417
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 36 \enclose{longdiv}{ 51 }$$
We will be using the left-to-right method of calculation. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Solve for how many whole groups you can divide 36 into 51
$$\require{enclose} 00.1 \\ 36 \enclose{longdiv}{ 51.0 }$$
We can now pull 36 whole groups from the equation. Multiply this number by 36, the denominator to get the first part of your answer!
### Step 3: Subtract the remainder
$$\require{enclose} 00.1 \\ 36 \enclose{longdiv}{ 51.0 } \\ \underline{ 36 \phantom{00} } \\ 474 \phantom{0}$$
If your remainder is zero, that's it! If there is a remainder, extend 36 again and pull down the zero
### Step 4: Repeat step 3 until you have no remainder
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 51/36 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Here are just a few ways we use 51/36, 1.417 or 141% in our daily world:
### When you should convert 51/36 into a decimal
Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.141 per hour and not$20 and 51/36.
### When to convert 1.417 to 51/36 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 51/36 = 1.417 what would it be as a percentage?
• What is 1 + 51/36 in decimal form?
• What is 1 - 51/36 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 1.417 + 1/2? |
# A tower stands vertically on the ground. From a point on the ground, $20 \mathrm{~m}$ away from the foot of the tower, the angle of elevation of the top of the tower is $60^{\circ}$. What is the height of the tower?
Given:
A tower stands vertically on the ground. From a point on the ground, $20 \mathrm{~m}$ away from the foot of the tower, the angle of elevation of the top of the tower is $60^{\circ}$.
To do:
We have to find the height of the tower.
Solution:
Let $AB$ be the tower and $C$ is the point $20 \mathrm{~m}$ away from the foot of the tower.
From the figure,
$\mathrm{BC}=20 \mathrm{~m}, \angle \mathrm{ACB}=60^{\circ}$
Let the height of the tower be $\mathrm{AB}=h \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$=\frac{\text { AB }}{BC}$
$\Rightarrow \tan 60^{\circ}=\frac{h}{20}$
$\Rightarrow \sqrt{3}=\frac{h}{20}$
$\Rightarrow h=20 \sqrt{3} \mathrm{~m}$
Therefore, the height of the tower is $20 \sqrt{3} \mathrm{~m}$.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
45 Views |
Question Video: Finding the Probability of Neither of Two Independent Events Occuring | Nagwa Question Video: Finding the Probability of Neither of Two Independent Events Occuring | Nagwa
# Question Video: Finding the Probability of Neither of Two Independent Events Occuring Mathematics • Third Year of Secondary School
## Join Nagwa Classes
π΄ and π΅ are independent events, where π(π΄) = 5/6 and π(π΅) = 3/4. What is the probability that neither event π΄ nor event π΅ occurs?
02:10
### Video Transcript
π΄ and π΅ are independent events where the probability of π΄ equals five-sixths and the probability of π΅ equals three-quarters. What is the probability that neither event π΄ nor event π΅ occurs.
In order to answer this question, we need to recall some of the facts about probability. The probability of an event not occurring, written π of π΄ prime, is equal to one minus the probability of π΄ occurring. As the probability of π΄ is equal to five-sixths, the probability of π΄ not occurring is one minus five-sixths. This is equal to one-sixth, as one whole one is equal to six-sixths and six minus five equals one. We can repeat this process to calculate the probability of π΅ not occurring. This will be equal to one minus three-quarters. As there are four quarters in a whole one, one minus three-quarters is equal to one-quarter.
Weβre also told in the question that π΄ and π΅ are independent events. This means that the probability of π΄ and π΅ or π΄ intersection π΅ is equal to the probability of π΄ multiplied by the probability of π΅. We can, therefore, calculate the probability that neither event π΄ nor event π΅ occurs by multiplying the probability of not π΄ by the probability of not π΅. We need to multiply one-sixth by one-quarter. When multiplying fractions, we multiply the numerators and separately the denominators. One multiplied by one is one and six multiplied by four is 24. We can, therefore, conclude that the probability that neither event π΄ nor event π΅ occurs is one out of 24 or one 24th.
## Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions |
Students can Download Maths Chapter 1 Rational Numbers Ex 1.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Rational Numbers Ex 1.1
8th Maths Exercise 1.1 Samacheer Kalvi Question 1.
Fill in the blanks:
(i) $$\frac{-19}{5}$$ lies between the integers _____ and _____
(ii) The rational number that is represented by 0.44 is ______.
(iii) The standard form of $$\frac{+58}{-78}$$ is _____.
(iv) The value of $$\frac{-5}{12}+\frac{7}{15}$$ = ______
(v) The value of $$\left(\frac{-15}{23}\right) \div\left(\frac{+30}{-46}\right)$$ is ______
Solution:
(i) -4 and -3
(ii) $$\frac{11}{25}$$
(iii) $$\frac{-29}{39}$$
(iv) $$\frac{1}{20}$$
(v) 1
8th Maths Exercise 1.1 In Tamil Question 2.
Say True or False.
(i) 0 is the smallest rational number.
(ii) There are an unlimited rationals between 0 and 1.
(iii) The rational number that does not have a reciprocal is 0.
(iv) The only rational number which is its own reciprocal is -1.
(v) The rational numbers that are equal to their additive inverses are 0 and -1.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) False
8th Maths Exercise 1.1 Question 3.
List five rational numbers between 2 and 0
(i) -2 and 0
(ii) $$\frac{-1}{2}$$ and $$\frac{3}{5}$$
(iii) 0.25 and 0.35
(iv) -1.2 and -2.3
Solution:
(i) -2 and 0
Samacheer Kalvi 8th Maths Solutions Term 1 Pdf Question 4.
Write four rational numbers equivalent to -3 7
(i) $$\frac{-3}{5}$$
(ii) $$\frac{7}{-6}$$
(iii) $$\frac{8}{9}$$
Solution:
8th Standard Maths Exercise 1.1 Answers Question 5.
Draw the number line and represent the following rational numbers on it.
Solution:
8th Maths Book Example Sums Question 6.
Find the rational numbers for the points marked on the number line.
Solution:
(i) The number lies between -3 and -4. The unit part between -3 and -4 is divided into 3 equal parts and the second part is asked.
∴ The required number is $$-3 \frac{2}{3}=-\frac{11}{3}$$.
(ii) The required number lies between 0 and -1. The unit part between 0 and -1 is divided into 5 equal parts, and the second part is taken.
∴The required number is $$-\frac{2}{5}$$
(iii) The required number lies between 1 and 2. The unit part between 1 and 2 is divided into 4 equal parts and the third part is taken.
∴ The required number is $$1 \frac{3}{4}=\frac{7}{4}$$
Samacheer Kalvi.Guru 8th Maths Question 7.
Using average, write 3 rational numbers between $$\frac{14}{5}$$ and $$\frac{16}{3}$$
Solution:
Maths 8th Guide Samacheer Kalvi Question 8.
Verify that -(-x) is the same x for:
(i) x = $$\frac{11}{15}$$
(ii) x = $$\frac{-31}{45}$$
Solution:
Samacheer Kalvi 8th Maths Solutions Question 9.
$$\frac{-3}{7}+\frac{5}{6}+\frac{4}{7}+\frac{1}{3}+\frac{13}{-6}$$
Solution:
8th Maths 1.1 Question 10.
What should be added to $$\frac{-8}{9}$$ to get $$\frac{2}{5}$$.
Solution:
Let the number to be added = x
8th Std Maths Exercise 1.1 Question 11.
Subtract $$\frac{-8}{44}$$ from $$\frac{-17}{11}$$
Solution:
Maths Term 1 Question 12.
Evaluate:
(i) $$\frac{9}{2} \times \frac{-11}{3}$$
(ii) $$\frac{-7}{27} \times \frac{24}{-35}$$
Solution:
Maths 8th Class Chapter 1 Exercise 1.1 Question 13.
Divide
(i) $$\frac{-21}{5}$$ by $$\frac{-7}{-10}$$
(ii) $$\frac{-3}{13}$$ by -3
(iii) -2 by $$\frac{-6}{15}$$
Solution:
8th Maths In Tamil Question 14.
Simplify $$\left(\frac{2}{5}+\frac{3}{2}\right)+\frac{3}{10}$$ as a rational number and show that it is between 6 and 7.
Solution:
8th Standard Maths Book Exercise 1.1 Question 15.
Write the five rational numbers which are less than -2.
Solution:
All the integers are rational numbers
∴ Rational numbers less than -2 are -10, -15, -20, -25, -30
8th Maths Guide Question 16.
Compare the following pairs of rational numbers
Solution:
$$\frac{10}{15}<\frac{12}{15}$$
∴ $$\frac{2}{3}<\frac{4}{5}$$
8th Maths Book Answer Question 17.
Arrange the following rational numbers is ascending and descending order.
Solution:
Objective Type Questions
8th Std Maths Guide Question 18.
The number which is subtracted from $$\frac{-6}{11}$$ to get $$\frac{8}{9}$$ is
Solution:
(B) $$\frac{-142}{99}$$
Hint:
Let x be the number be subtracted
Samacheer Kalvi Guru 8th Maths Book Solutions Question 19.
Which of the following rational numbers is the greatest?
Solution:
(A) $$\frac{-17}{24}$$
Hint:
LCM of 24, 16, 8, 32 = 8 × 2 × 3 × 2 = 96
∴ $$\frac{-17}{24}$$ is the greatest number.
8th Std Maths Book Answers Question 20.
$$\frac{-5}{4}$$ is a rational number which lies between
(A) 0 and $$\frac{-5}{4}$$
(B) -1 and 0
(C) -1 and -2
(D) -4 and -5
Solution:
(C) -1 and -2
Hint:
$$\frac{-5}{4}$$ = $$-1 \frac{1}{4}$$
∴ $$\frac{-5}{4}$$ lies between -1 and -2.
Samacheer Kalvi 8th Maths Book Question 21.
The standard form of $$\frac{3}{4}+\frac{5}{6}+\left(\frac{-7}{12}\right)$$ is
Solution:
(D) 1
Hint:
8th Maths Book Samacheer Kalvi Question 22.
The sum of the digits of the denominator in the simplest form of $$\frac{112}{528}$$
(A) 4
(B) 5
(C) 6
(D) 7
Solution:
(C) 6
Hint:
Sum of digits in the denominator = 3 + 3 = 6
Question 23.
The rational number (numbers) which has (have) additive inverses is (are)
(A) 7
(B) $$\frac{-5}{7}$$
(C) 0
(D) all of these
Solution:
(D) all of these
Hint:
Additive inverse of 7 is -7
Additive inverse of $$\frac{-5}{7}$$ is $$\frac{-5}{7}$$
Additive inverse of 0 is 0.
Question 24.
Which of the following pairs is equivalent?
Solution:
(B) $$\frac{16}{-30}, \frac{-8}{15}$$
Hint:
∴ $$\frac{16}{-30}$$ and $$\frac{8}{15}$$ are equivalent fraction.
Question 25.
$$\frac{3}{4} \div\left(\frac{5}{8}+\frac{1}{2}\right)$$ =
Solution:
(C) $$\frac{2}{3}$$
Hint: |
# Random Variables. A random variable assumes a value based on the outcome of a random event. ◦ We use a capital letter, like X, to denote a random variable.
## Presentation on theme: "Random Variables. A random variable assumes a value based on the outcome of a random event. ◦ We use a capital letter, like X, to denote a random variable."— Presentation transcript:
Random Variables
A random variable assumes a value based on the outcome of a random event. ◦ We use a capital letter, like X, to denote a random variable. ◦ A particular value of a random variable will be denoted with a lower case letter, in this case x. There are two types of random variables: ◦ Discrete random variables can take one of a finite number of distinct outcomes. Example: Number of credit hours ◦ Continuous random variables can take any numeric value within a range of values. Example: Cost of books this term
a.Determine the probability distribution of the random variable X. b.Construct a probability histogram for the random variable X. Example A professor asked his introductory statistics students to state how many siblings they have. The result is organized in the following table. Let X denote the number of siblings of a randomly selected student.
Solution Example The table displays these probabilities and provides the probability distribution.
a.Repeat the experiment 1000 times and record the frequencies and proportions for the number of heads. b.Determine the probability distribution of the random variable X. c.Construct a probability histogram for the random variable X. Example Toss a balanced dime three times and record the number of heads. Let X be the number of heads of each three-toss experiment.
Solution Example a) We used a computer to simulate 1000 observations of the random variable X, the number of heads obtained in three tosses of a balanced dime. The following table shows the frequencies and proportions for the numbers of heads obtained in the 1000 observations.
Solution Example c) This result is more easily seen if we compare the proportion histogram to the probability histogram of the random variable X.
A probability model for a random variable consists of: ◦ The collection of all possible values of a random variable, and ◦ the probabilities that the values occur. Of particular interest is the value we expect a random variable to take on, notated μ (for population mean) or E(X) for expected value. The expected value of a (discrete) random variable can be found by summing the products of each possible value and the probability that it occurs:
An American Roulette wheel has 38 slots, of which 18 are black, 18 are red, and 2 are green. The dealer spins the wheel and whirls a small ball in the opposite direction within the wheel. Gamblers bet on where the ball will come to rest. One of the simplest bet to choose red (or black). A bet of \$1 on red pays off an additional \$1 if the ball lands in a red slot. Otherwise, the player loses his \$1. Is it a fair game?
Insurance company make bets. They bet that you’re going to live a long life. You bet that you’re going to die sooner. How to find a “fair price” for this kind of bet? Suppose an insurance company offers a “death and disability” policy that pays \$10,000 when you die or \$5000 if you are permanently disabled. How can the insurance company determine the annual premium?
Suppose the insurance company insures 1000 people. The death rate and disability rate in any year can be assumed as follows: Let X be the insurance company’s payout in a given year. Policyholder outcome Probability P (X=x) Payout Death1/100010,000 Disability2/10005000 Neither977/10000
For data, we calculated the standard deviation by first computing the deviation from the mean and squaring it. We do that with discrete random variables as well. The variance for a random variable is: The standard deviation for a random variable is:
Adding or subtracting a constant from data shifts the mean but doesn’t change the variance or S.D: E(X ± c) = E(X) ± c Var(X ± c) = Var(X) ◦ Example: Consider everyone in a company receiving a \$5000 increase in salary. In general, multiplying each value of a random variable by a constant multiplies the mean by that constant and the variance by the square of the constant: E(aX) = aE(X)Var(aX) = a 2 Var(X) ◦ Example: Consider everyone in a company receiving a 10% increase in salary.
The mean of the sum (or difference) of two random variables is the sum (or difference) of the means. E(X ± Y) = E(X) ± E(Y) If the random variables are independent, the variance of their sum or difference is always the sum of the variances. Var(X ± Y) = Var(X) + Var(Y)
Combining Random Variables (The Bad News) It would be nice if we could go directly from models of each random variable to a model for their sum. But, the probability model for the sum of two random variables is not necessarily the same as the model we started with even when the variables are independent. Thus, even though expected values may add, the probability model itself is different.
Nearly everything we’ve said about how discrete random variables behave is true of continuous random variables, as well. When two independent continuous random variables have Normal models, so does their sum or difference. This fact will let us apply our knowledge of Normal probabilities to questions about the sum or difference of independent random variables. Combining Random Variables (The good news)
Consider the company that manufactures and ships small stereo systems. The times required to pack the stereos can be described by a Normal model with a mean of 9 minutes and standard deviation of 1.5 minutes. The times for the boxing stage can also be modeled as Normal, with a mean of 6 minutes and standard deviation of 1 minute. Find the expected total times for packing and boxing a system. Find the standard deviation of packing and boxing time. Find the expected time for packing two systems ◦ What’s the probability that packing two systems takes over 20 minutes? ◦ What percentage of stereo system take longer to pack than to box?
Page 427 – 430 Problem # 1–15 odd, 19, 23, 25, 27, 29, 35, 43.
Download ppt "Random Variables. A random variable assumes a value based on the outcome of a random event. ◦ We use a capital letter, like X, to denote a random variable."
Similar presentations |
Open In App
Related Articles
• RD Sharma Class 8 Solutions for Maths
# Class 8 RD Sharma Solutions – Chapter 15 Understanding Shapes Polygons – Exercise 15.1
### Question 1. Draw a Rough diagram to illustrate the Following.
(i) Open curve
(ii) Closed curve
Solution:
i) Open curve: A curve in which the beginning point and the ending point do not meet each other is known as an open curve.
ii) Closed curve: A curve in which the beginning point and endpoint meet at each other or cuts each other is known as a closed curve.
### Question 2. Classify the following as open or closed.
Solution:
• open curve: A curve in which starting point and ending point are different or do not cut each other .
• closed curve: A curve in which starting point and ending point are same and cut each other.
Using the above definition we can classify the given figures as follows :
i) open curve (as both stating and ending points are different)
ii) closed curve (as both the points are same )
iii) closed curve (as both the points cut each other)
iv) open curve (as both starting point and ending point are different)
v) open curve (as both starting point and ending point are different)
vi) closed curve (as both starting point and ending point meet at same point)
### Question 3. Draw a polygon and shade it’s interior. Also draw its diagonals, if any:
Solution:
In polygon ABCD, AC and BD are the diagonals of a polygon
### Question 4. Illustrate if possible, each one of the following with a rough diagram.
(i) A closed curve that is not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides.
Solution:
(i) A closed curve that is not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides.
A polygon with two sides is not possible because, a polygon should have minimum three sides.
### Question 5. Following are some figures: Classify each of these figures on the basis of the following:
(i) Simple curve
(ii) Simple closed curve
(iii) Polygon
(iv) Convex polygon
(v) Concave polygon
(vi) Not a curve
Solution:
(i) It is a Simple Closed curve and a concave polygon. This is a simple closed curve and as a concave polygon all the vertices are not pointing outwards.
(ii) It is a Simple closed curve and a convex polygon. This is a simple closed curve and as a convex polygon all the vertices are pointing outwards.
(iii) It is Not a curve and hence it is not a polygon.
(iv) It is Not a curve and hence it is not a polygon.
(v) It is a Simple closed curve but not a polygon.
(vi) It is a Simple closed curve but not a polygon.
(vii) It is a Simple closed curve but not a polygon.
(viii) It is a Simple closed curve but not a polygon.
### Question 6. How many diagonals does each of the following have?
(ii) A regular hexagon
(iii) A triangle
Solution:
For a convex quadrilateral we shall use the formula n(n-3)/2
So, number of diagonals = 4(4-3)/2 = 4/2 = 2
A convex quadrilateral has 2 diagonals.
(ii) A regular hexagon
For a regular hexagon we shall use the formula n(n-3)/2
So, number of diagonals = 6(6-3)/2 = 18/2 = 9
A regular hexagon has 9 diagonals.
(iii) A triangle
For a triangle we shall use the formula n(n – 3)/2
So, number of diagonals = 3(3 -3)/2 = 0/2 = 0
A triangle has no diagonals.
### Question 7. What is a regular polygon? State the name of a regular polygon of
(i) 3 sides
(ii) 4 sides
(iii) 6 sides
Solution:
Regular Polygon: A regular polygon is an enclosed figure. In a regular polygon minimum sides are three.
(i) 3 sides
A regular polygon with 3 sides is known as Equilateral triangle.
(ii) 4 sides
A regular polygon with 4 sides is known as Rhombus.
(iii) 6 sides
A regular polygon with 6 sides is known as Regular hexagon.
Related Tutorials |
<meta http-equiv="refresh" content="1; url=/nojavascript/">
# Chapter 5: Applying Percents
Difficulty Level: Basic Created by: CK-12
## Introduction
Here you will learn all about percents and applications of percents. First, you will review how to convert between fractions, decimals, and percents. Then, you will learn how to find the percent of a number and use proportions to solve percent problems. You will also learn to solve real-world problems involving percents including discounts, sales tax, simple interest, and compound interest.
## Summary
You reviewed that percent means out of 100 and that percents can also be written as fractions or decimals. You learned how to solve problems involving percents including finding the percent of a number and finding what percent one number is of another number. To solve these problems you learned how to use the proportion ab=p100$\frac{a}{b}=\frac{p}{100}$. You also learned what is meant by a percent increase and a percent decrease and how to calculate these types of percents.
You then learned common applications of percents including discounts and sales tax. Other applications you studied were simple interest and compound interest. You learned the simple interest formula to calculate interest earned is I=Prt$I=Prt$. You learned the compound interest formula to calculate the future value of an account earning compound interest is A=P(1+r)t$A=P(1+r)^t$. In both formulas, P$P$ is the principal, r$r$ is the rate, and t$t$ is the time. One important point to remember is that when working with compound interest, the way you calculate r$r$ and t$t$ changes depending on how often the interest is compounded.
Basic
Sep 20, 2013 |
# Division of Monomials
Division of monomials means product of their quotient of numerical coefficients and quotient of their literal coefficients.
Since, the product of 3m and 5n = 3m × 5n = 15mn; it follows that
(i) $$\frac{15mn}{3m} = \frac{3 \times 5 \times m \times n}{3 \times m}$$ = 5n
or, 15mn ÷ 3m = 5n
i.e. when 15mn is divided by 3m, the quotient is 5n.
(ii) $$\frac{15mn}{5n} = \frac{3 \times 5 \times m \times n}{5 \times n}$$ = 3m
or, 15mn ÷ 5n = 3m
i.e. when 15mn is divided by 5n, the quotient is 3m.
1. Divide 35mxy by 5my
35mxy ÷ 5my
= $$\frac{35mxy}{5my}$$
Now, we need to write each term in the expanded form and then cancel the terms which are common to both numerator and denominator.
= $$\frac{\not{5} \times 7 \times \not{m} \times x \times \not{y}}{\not{5} \times \not{m} \times \not{y}}$$
= 7x
2. Divide 14a7 by 2a5
14a7 ÷ 2a5
= $$\frac{14a^{7}}{2a^{5}}$$
Now, we need to write each term in the expanded form and then cancel the terms which are common to both numerator and denominator.
= $$\frac{\not{2} \times 7 \times \not{a} \times \not{a} \times \not{a} \times \not{a} \times \not{a} \times a \times a}{\not{2} \times \not{a} \times \not{a} \times \not{a} \times \not{a} \times \not{a}}$$
= 7 × a × a
= 7a2
Or, we can solve this in the other way.
14a7 ÷ 2a5
= $$\frac{14a^{7}}{2a^{5}}$$
= $$\frac{14}{2} \times \frac{a^{7}}{a^{5}}$$
Now we will write the each numerical part $$(\frac{14}{2})$$ in the expanded form and then cancel the terms which are common to both numerator and denominator and in case of literal part subtract the smaller power of a literal from bigger power of the same literal.
= $$\frac{\not{2} \times 7}{\not{2}} \times a^{7 - 5}$$
= 7 × 2
= 7a2
3. Divide the monomial: 81p3q6 by 27p6q3
81p3q6 ÷ 27p6q3
= $$\frac{81p^{3}q^{6}}{27p^{6}q^{3}}$$
= $$\frac{81}{27} \times \frac{p^{3}q^{6}}{p^{6}q^{3}}$$
Now we will write the each numerical part (\frac{81}{27}) in the expanded form and then cancel the terms which are common to both numerator and denominator and in case of literal part subtract the smaller power of a literal from bigger power of the same literal.
= $$\frac{\not{3} \times \not{3} \times \not{3} \times 3}{\not{3} \times \not{3} \times \not{3}} \times \frac{q^{6 - 3}}{p^{6 - 3}}$$
= $$3 \times \frac{q^{3}}{p^{3}}$$
= $$\frac{3q^{3}}{p^{3}}$$
Types of Algebraic Expressions
Degree of a Polynomial
Subtraction of Polynomials
Power of Literal Quantities
Multiplication of Two Monomials
Multiplication of Polynomial by Monomial
Multiplication of two Binomials
Division of Monomials
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
## Recent Articles
1. ### Roman Numerals | System of Numbers | Symbol of Roman Numerals |Numbers
Feb 22, 24 04:21 PM
How to read and write roman numerals? Hundreds of year ago, the Romans had a system of numbers which had only seven symbols. Each symbol had a different value and there was no symbol for 0. The symbol…
2. ### Worksheet on Roman Numerals |Roman Numerals|Symbols for Roman Numerals
Feb 22, 24 04:15 PM
Practice the worksheet on roman numerals or numbers. This sheet will encourage the students to practice about the symbols for roman numerals and their values. Write the number for the following: (a) V…
3. ### Roman Symbols | What are Roman Numbers? | Roman Numeration System
Feb 22, 24 02:30 PM
Do we know from where Roman symbols came? In Rome, people wanted to use their own symbols to express various numbers. These symbols, used by Romans, are known as Roman symbols, Romans used only seven…
4. ### Place Value | Place, Place Value and Face Value | Grouping the Digits
Feb 19, 24 11:57 PM
The place value of a digit in a number is the value it holds to be at the place in the number. We know about the place value and face value of a digit and we will learn about it in details. We know th…
5. ### Math Questions Answers | Solved Math Questions and Answers | Free Math
Feb 19, 24 11:14 PM
In math questions answers each questions are solved with explanation. The questions are based from different topics. Care has been taken to solve the questions in such a way that students
Terms of an Algebraic Expression - Worksheet
Worksheet on Types of Algebraic Expressions
Worksheet on Degree of a Polynomial |
## Introduction: How to Graph
This Instructable will show you how to Graph, from Bar Graphs to Exponential functions, I'm hoping to win in Burning Questions 6.5, so if you think this is good, please vote!
First, We'll go through the Basics of Graphing, then build upon that, and go through the different types of graphing, All Types of Graphs besides Histograms, and Circle Graphs are included in this Instructable.
Materials Needed:
Paper (Graphing Paper Ideally)
Pencil
Not Shown, for Obvious Reasons:
Brain
I Apologize in advance for my Sloppy Handwriting.
## Step 1: Very Basics
The Very Basics of Graphing are somewhere around here...
First, we need to know our Axis. The Y-Axis goes up and down, the vertical Axis, and the X-Axis goes side to side, the Horizontal Axis. The Axis are commonly referred to by just the Letter, the Y-Axis is often called just "The Y", and the X-Axis is called just "The X"
## Step 2: The Basics
The X and Y Axis are lines, so they go on indefinitely, and most higher level Math looks at them as more of a "Plus" Sign. The Center Point, (0,0) is called the Origin. Not (0,0.01), not (0,0.00000001), exactly (0,0), were the two Axis cross.
There are four Quadrants in a Graph, and they start at the top right one, and move Counter/Anti-Clockwise. As you can see, they're named Quadrant I, Quadrant II, Quadrant III, and Quadrant IV. (Roman Numerals for 1,2,3,4). In Quadrant I, the X and Y are both positive numbers. In Quadrant II, the X is negative, and the Y is Positive. In Quadrant III, the X and the Y are negative. And in Quadrant IV, the X is positive, and the Y is not.
## Step 3: The Less Basic Basics
On the X and Y Axis, put numbers every block, you can go one block is one, or one block is two, However many you like. Technically speaking, is you're going one by one, you don't have to number each line, but I did for ease of explanation.
If you have really high numbers, put a breaking sign directly after zero, and at the top continue counting. These are most commonly used on the Y axis, but it could be used on both Axis.
Points are essential to learn. They're in the order of the X Coordinate, and the Y Coordinate, in alphabetical order. For Example, if you had the Point (3,4), you go over 3 on the X Axis, and up for on the Y Axis. On the other hand, if you have the Point (-3,-4), you go back 3 on the X and down 4 on the Y.
## Step 4: Putting It All Together!
Let's Put everything we learned so far Together (Wow, this Instructable is Reminiscent of Beginning French Class...)
And the Picture Below is a Brief Summary of Everything up to this Point.
## Step 5: Section 2, Bars and Lines
Now we're going to Review how to Graph Bar Graphs and Line Graphs.
First, the Bar Graph.
Bar Graphs are used for Directly comparing things.
The Setup is like a view of Quadrant I, in most cases, but instead of Numbers on the X Axis, there Are Words. Then, the Y axis has the Numbers. For Example, If I'm Graphing Popular Foods, The Types of Foods would be across the Bottom, and the Number of People who chose the Food would be across the Y. You need to label the X and Y Axies, and give the Graph a Title.
The Bars should all be the Same width, say. 2 squares, but they can/should be different hieghts, depending on the data. In Plain Bar Graphs, the bars shouldn't touch.
To find the Hieght of the Bar, look at your data (I apologize for mine's Inabillity to be read...), and put a line at the number of that specifec piece of Data, the Pizza had 23 Votes, put a line at 23 above the Pizza collum, and turn that line into a Collum, and Shade.
## Step 6: The Line Graph
The Line Graph shows progression of something over time, say, Plant Growth. It shows how one or more things change over time (Multiple Lined Line Graphs).
The Setup is similar to the Line Graph, but on the X Axis is the Measurements of time (Days, Weeks, Quarters, Years), and on the Y Axis is what you're measuring (In, Cm., Revenue, Jobs, etc.).
Using your data, you graph the points, and connect the dots from left to right.
## Step 7: Section 3 Y=MX+B
Now we're going into Section 3, Algebraic Graphs, we're going into higher level Graphing. We'll be graphing Equations, inequalities, and Exponential Functions.
The First and Most Important/ Mucho Importante/ Tres Importante Equation is Y=MX+B.
It means that a Coordinate for Y, equals the Slope time X plus the Y-Intercept, An Equation in this form is in Point-Intercept Form.
First, we need to find the Slope. and the symbol for this is M.
The Equation to find the Slope is [Y(Sub-2) Minus Y (Sub-1)] divided by [X (Sub-2) minus X (Sub-1)] This basically means, that out of the Two Points, the second Y coordinate, minus the first y coordinate, divided by the second X coordinate, minus the first X coordinate. Then, we can move on.
## Step 8: The Rest of Y=mx+b
Then, after you find the slope, you solve the rest of the Equation.
But before we go into that, we need to figure out how to get there from Standard Form.
Standard Form is "Ax+By=C" which is not "y=mx+b". To get to this, the first thing you need to do, is subtract the X from both sides, So, if the Equation was 6x+3y=9, you subtract the 6x from both sides, so you have 3y=-6x+9, but it's still not in Point-Intercept Form! In Point-Intercept, the "Y" doesn't have a Coefficient (Number that goes with it), so we have to divide both sides by three, in the example equation. So now, it's y=-2+3, and that's in Point-Intercept form.
Now, Putting it all together!
Ex. Question:
Write, in point-intercept form, and equation for a line that passes through the points (2,2) and (3,5), graph.
Find the slope. 5-2=3 3-2=1 3/1=3 Slope=3
Find the Y-Intercept (B): y=mx+b, 2=3(2)+b, 2=6+b 6-4=2 B=-4
Graph y=3x-4
First, put one point on the y-intercept, where the line will cross the Y-Axis, it will be on the Y-axis, and however high up B is, so for the example equation, ,B=-4 so down 4 on the Y-Axis, to the point (0,-4), and put a point there. Then, using the slope, turn it into a fraction, the slope is 3, so turn it into 3/1. Now, do "Rise/Run" Rise 3 on the Y axis, and go over one on the X-Axis. Connect the two points with a line, and you're done!!
## Step 9: Inequalities
Inequalities are like equations, but instead of an equal sign, they have a Less Then, Greater Then, Less Then or Equal to, or Greater Then or Equal To, signs (<,>,d,e).
You solve them basically like an Equation, using y=mx+b, or solving from Standard Form.
For Example, 2x+y>1.
Subtract 2x from both sides, so y>-2x+1, and Graph that line like normal, except the line should be dashed, because it's not "...or equal to", if the inequality uses < or >, use a dashed line. Then, pick a point, (0,0) or (1,1) and test it in the equation, 0>-2(0)+1, 0>1 X, so shade the other side from the point you picked, but if the point worked, shade that side.
Drat..... The Less/Greater Then or Equal to Signs Come out as D and E....
## Step 10: Exponential Funtions
Exponential functions are when a numbers exponent is a variable, for example, 2x.
To graph, you need to make a table of the Exponent and it's possible number solutions. For example, if 3x, and x=1, the Answer is 3, but if X=-1, the Answer is 1/3, anything to the zero power is one. So, /on the X axis, graph what X equals, -1,0,1,2, etc. Then, on the Y axis, get what the number raised to what X equals equals, and graph the points, eg. (-1,1/3) (0,1) (1,3) (2,9) (3,27). Then connect these dots, and it forms a curve. Exponential Functions Increase or Decrease Rapidly.
## Step 11: Thanks!
Thanks for Reading my Instructable, now you know how to Make a Graph, Graph a Bar Graph, to Graphing Exponential Functions, or, you just needed a brush up, you know now! |
# What is long method and synthetic division?
## What is long method and synthetic division?
Long and synthetic division are two ways to divide one polynomial (the dividend) by another. polynomial (the divisor). These methods are useful when both polynomials contain more than. one term, such as the following two-term polynomial: ���� 2 + 3.
## What is the formula of synthetic division?
The basic Mantra to perform the synthetic division process is” “Bring down, Multiply and add, multiply and add, Multiply and add, ….” For example, we can use the synthetic division method to divide a polynomial of 2 degrees by x + a or x – a, but you cannot use this method to divide by x2 + 3 or 5×2 – x + 7.
What is the method of synthetic division with example?
Synthetic division is a shortcut for polynomial division when the divisor is of the form x – a. Only numeric coefficients of the dividend are used when dividing with synthetic division. Example 1. Divide (2 x – 11 + 3 x 3) by ( x – 3).
How do you do the long division method?
How to Do Long Division?
1. Step 1: Take the first digit of the dividend from the left.
2. Step 2: Then divide it by the divisor and write the answer on top as the quotient.
3. Step 3: Subtract the result from the digit and write the difference below.
4. Step 4: Bring down the next digit of the dividend (if present).
### What is an example of long division?
Here’s an example of long division with decimals. 123454321 when divided by 11111 gives a quotient of 11111 and remainder 0.
### How do you solve long division method?
What is long division method?
Long division is a method for dividing one large multi digit number into another large multi digit number.
How do you do long division?
#### What is synthetic division used for?
Synthetic Division. Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor — and it only works in this case. Synthetic division is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials. |
# Identify the equation of the normal line?
Identify the equation of the normal line to the curve $y=g(p)=2.5+3.5(4^p)$ where it crosses the $y$-axis.
So I am guessing the normal line would be the inverse of the derivative function, since it is perpendicular to the tangent line. Is this correct? If so, I got the derivative to be $g'(p)=3.5\cdot 4^{p}\ln(4)$. Is this the correct derivative form? If so, how do I take the inverse of this? Thank you!
-
The curve that you have $y=g(p)$ crosses the $y$-axis where $p=0$, so that happens at $y = 2.5 + 3.5 = 6$. So it crosses the $y$-axis at the point $(p,y) = (0,6)$.
The tangent line to the curve at this point has slope $g'(0)$. You have $$g'(p) = 3.5\cdot \ln(4)4^p.$$ So $g'(0) = 3.5\ln(4)\simeq \dots$
To find this derivative recall the rule that $\frac{d}{dx}a^x = \ln(a)a^x$.
Now the slope of the normal line, call this $m$ satisfies that $$mg'(0) = -1.$$ (This is because the product of the slopes of two lines that are orthogonal is $-1$).
So you need to solve this for $m$ to find the slope of the normal line. When you have $m$ (the slope of the normal line) then you can write down an equation of the normal line because you also have a point $(0,6)$ that the line passes through.
Remember that the normal line is just a straight line, so when you know the slope $m$ and a point $(0,6)$, then you can use the point-slope formula which tells you that the equation is $$y - 6 = m(p - 0).$$ [In general when you have a slope $m$ and a point $(x_1, y_1)$, then the equation of the line that passes through the point with the slope is $y - y_1 = m(x - x_1)$.
I tried to do this and got that $m=-.206099\dots$
-
how would this look in equation form though? Since when I solve for m, I get -.206, or did I do that wrong? Would an equation for the normal line be -3.5ln(4)p+6? Sorry, I'm just super confused! – user56852 Feb 1 '13 at 20:39
@user56852: Did that help? – Thomas Feb 1 '13 at 20:44
Oh so y=-.206p+6? And that can also be written as -(p/(3.5ln4))+6? – user56852 Feb 1 '13 at 20:47
@user56852: That looks right. – Thomas Feb 1 '13 at 21:14
Thank you so much for the help! – user56852 Feb 1 '13 at 22:35 |
# Libro de matemáticas 2 de secundaria contestado PDF
Rate this post
The Libro de matemáticas 2 de secundaria contestado PDF is a teaching guide that offers practical classes on most basic mathematics, including Algebra, Geometry, Trigonometry, etc. The objective is for the student to learn to solve problems in a realistic context and to use the tools offered by science.
Social networks are an incredible tool for interaction with our followers and the outside world. They are rewarding for our personal image, allow us to establish contact with interesting people and promote dialogue.
## Gaussian Algorithm
Gaussian Algorithm is an algorithm for solving linear problems. It is used for the study of distributions, populations, difference-response equations and calculation networks.
## The Structure of a Line
Making a line is simple, but requires knowledge of its structure. The structure of a line can be shown by means of a diagram. The diagram shows the relationship between the intersecting distances on the line. The distance at each point on the line is proportional to the length of the line divided by the time from the origin to the end of the line.
By virtue of this, we can determine the dimensions of a line by means of its lengths and distances: average length, minimum length and diameter.
The dimensions of a line can be used to calculate other information related to it, such as its slope or curvature.
## The Investment of Variables
Today we are going to talk about the investment of variables. Investment is a strategy to obtain higher returns on our assets, whether real estate, stocks or bonds. It is necessary to keep in mind that each investment involves risks, so it is important to analyze them beforehand.
In this post we are going to explain what variables are and how they are being used in the current financial market. We will briefly review the different forms of investment available and analyze the pros and cons of each. I hope this will be useful for you to make a better decision about your future investor.
## The Multiplication Property
The property of a multiplication is a complicated question that requires much study. Property of a multiplication is the right to possess the product of a multiplication. In order to have the property of a multiplication, it is necessary for the product to be separable and for there to be a territory where the multiplication is carried out. To know more about Libro de matemáticas 2 de secundaria contestado PDF read or guide below.
## Secondary Secondary Mathematics 2 Editions SM
soluciones Report it. Paco el Chato is an independent platform that offers support resources for SEP textbooks and other…
## Mathematics SEP Second of Secondary – Paco el Chato
Look for your Second grade Math homework: Hurricanes and Leonardo, an indissoluble mathematical union , The power of mathematics and chess, …
## 2nd Secondary Mathematics – Answered | PDF – Scribd
tive use of the student’s book and the didactic sequences that are proposed in it. This guide presents the answers to all…
## Secondary Mathematics 2 Answered Book
11 Jul 2019 — 2nd Secondary Mathematics Book Answered Dingvabarnsing
Ml … third grade high school calligraphy 3 santillana cursiva pdf …
## Mathematics 2 Secondary Pdf Answered – Favorite Books
21 Jan 2020 — Click here to see matematica 2 medio pdf. Click here to see more santillana texts pdf. Book connect math 3 secondary answered pdf. Compartir…
Help Hey, do you know where some page I have answered books the book is called MATHEMATICS 3 SECONDARY CONNECT MORE BALBUENA I RUN HUGO SM EDITIONS…
## Book for the teacher
https://coleccion.siaeducacion.org/sites/default/files/files/matematicas2-vol.2-maestro.pdf
II. 2d or. Degree. Volume. II. math. 2nd Grade Volume II. Book for the m… A fundamental part of secondary education.
## Mathematics 2. Connect More – SM
https://guiasdigitales.grupo-sm.com.mx/sites/default/files/guias/184290/index.html
Built with In5. page 1. page 2. page 3. page 4. page 5. page 6. page 7. page 8. page 9. page 10. page 11. page 12. page 13. page 14. page 15. page 16. |
Edit Article
# How to Derive the Quadratic Formula
Community Q&A
One of the most important skills an algebra student learns is the quadratic formula, or ${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}$ With the quadratic formula, solving any quadratic equation of the form ${\displaystyle ax^{2}+bx+c=0}$ becomes a simple matter of substituting the coefficients ${\displaystyle a,b,c}$ into the formula. While simply knowing the formula is often enough for many, understanding how it's derived (in other words, where it comes from) is another thing entirely. The formula is derived via "completing the square" that has other applications in math as well, so it is recommended that you be familiar with it.
## Steps
1. 1
Start with the standard form of a general quadratic equation. While any equation with an ${\displaystyle x^{2}}$ term in it qualifies as quadratic, the standard form sets everything to 0. Remember that ${\displaystyle a,b,c}$ are coefficients that can be any real number, so don't substitute any numbers in for them - we want to work with the general form.
• ${\displaystyle ax^{2}+bx+c=0}$
• The only condition is that ${\displaystyle a\neq 0,}$ because otherwise, the equation reduces to a linear equation.
2. 2
Subtract ${\displaystyle c}$ from both sides. Our goal is to isolate ${\displaystyle x.}$ To start, we move one of the coefficients to the other side, so that the left side only consists of terms with ${\displaystyle x}$ in it.
• ${\displaystyle ax^{2}+bx=-c}$
3. 3
Divide both sides by ${\displaystyle a}$. Note that we could've switched this and the previous step, and still arrived at the same place. Remember that dividing a polynomial by something means that you divide each of the individual terms. Doing so makes it easier for us to complete the square.
• ${\displaystyle x^{2}+{\frac {b}{a}}x={\frac {-c}{a}}}$
4. 4
Complete the square. Recall that the goal is to rewrite an expression ${\displaystyle x^{2}+2\Box x+\Box ^{2}}$ as ${\displaystyle (x+\Box )^{2},}$ where ${\displaystyle \Box }$ is any coefficient. It may not immediately be obvious to you that we can do this. To see it more clearly, rewrite ${\displaystyle {\frac {b}{a}}x}$ as ${\displaystyle 2{\frac {b}{2a}}x}$ by multiplying the term by ${\displaystyle {\frac {2}{2}}.}$ We can do this because multiplying by 1 does not change anything. Now we can clearly see that in our case, ${\displaystyle \Box ={\frac {b}{2a}},}$ so we are only missing the ${\displaystyle \Box ^{2}}$ term. Therefore, in order to complete the square, we add that to both sides - namely, ${\displaystyle \left({\frac {b}{2a}}\right)^{2}={\frac {b^{2}}{4a^{2}}}.}$ Then, of course, we factor.
• {\displaystyle {\begin{aligned}x^{2}+2{\frac {b}{2a}}x+{\frac {b^{2}}{4a^{2}}}&={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}\\\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}\end{aligned}}}
• Here, it is clear why ${\displaystyle a\neq 0,}$ since ${\displaystyle a}$ is in the denominator, and you cannot divide by 0.
• If you need to, you can expand the left side to confirm that completing the square works.
5. 5
Write the right side under a common denominator. Here, we want both denominators to be ${\displaystyle 4a^{2},}$ so multiply the ${\displaystyle {\frac {-c}{a}}}$ term by ${\displaystyle {\frac {4a}{4a}}.}$
• {\displaystyle {\begin{aligned}\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}}{4a^{2}}}-{\frac {4ac}{4a^{2}}}\\&={\frac {b^{2}-4ac}{4a^{2}}}\end{aligned}}}
6. 6
Take the square root of each side. However, it is essential that you recognize that in doing so, you are actually doing two steps. When you take the square root of ${\displaystyle d^{2},}$ you do not get ${\displaystyle d.}$ You actually get its absolute value, ${\displaystyle |d|.}$ This absolute value is critical in getting both roots - simply putting square roots over both sides will only get you one of the roots.
• ${\displaystyle \left|x+{\frac {b}{2a}}\right|={\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}}$
• Now, we can get rid of the absolute value bars by putting a ${\displaystyle \pm }$ on the right side. We can do this because the absolute value does not distinguish between positive and negative, so they are both valid. This tidbit is why the quadratic equation always gets you two roots.
• ${\displaystyle x+{\frac {b}{2a}}=\pm {\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}}$
• Let's simplify this expression a bit further. Since the square root of a quotient is the quotient of the square roots, we can write the right side as ${\displaystyle {\frac {\pm {\sqrt {b^{2}-4ac}}}{\sqrt {4a^{2}}}}.}$ Then we can take the square root of the denominator.
• ${\displaystyle x+{\frac {b}{2a}}={\frac {\pm {\sqrt {b^{2}-4ac}}}{2a}}}$
7. 7
Isolate ${\displaystyle x}$ by subtracting ${\displaystyle {\frac {b}{2a}}}$ from both sides.
• ${\displaystyle x={\frac {-b}{2a}}\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}}$
8. 8
Write the right side under a common denominator. This nets the quadratic formula, the formula that solves any quadratic equation in standard form. This works for any ${\displaystyle a,b,c}$ and outputs an ${\displaystyle x}$ that can be real or complex. To confirm that this process works, simply follow the steps of this article in reverse order to recover standard form.
• ${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$
## Tips
• It is interesting to note that the quadratic formula holds for complex coefficients as well, though you may have to do a little more simplifying for the final answer, and the roots will no longer come in conjugate pairs. Problems with quadratic expressions are nevertheless almost always given with real coefficients.
## Article Info
Categories: Algebra
In other languages:
Español: obtener la formula cuadrática, Русский: вывести формулу для корней квадратного уравнения, Italiano: Ricavare la Formula Quadratica, Português: Derivar a Fórmula de Bhaskara
Thanks to all authors for creating a page that has been read 100,212 times. |
Education.com
Try
Brainzy
Try
Plus
# Word Problems and Data Analysis: GED Test Prep (page 4)
By
Updated on Mar 9, 2011
### Graphs and Tables
The GED Mathematics Exam will test your ability to analyze graphs and tables. It is important to read each graph or table very carefully before reading the question. This will help you to process the information that is presented. It is extremely important to read all of the information presented, paying special attention to headings and units of measure. Here is an overview of the types of graphs you will encounter:
• Circle graphs or pie charts
• This type of graph is representative of a whole and is usually divided into percentages. Each section of the chart represents a portion of the whole, and all of these sections added together will equal 100% of the whole.
• Bar graphs
• Bar graphs compare similar things with different length bars representing different values. Be sure to read all labels and legends, looking carefully at the base and sides of the graph to see what the bars are measuring and how much they are increasing or decreasing.
• Broken line graphs
• Broken line graphs illustrate a measurable change over time. If a line is slanted up, it represents an increase whereas a line sloping down represents a decrease. A flat line indicates no change as time elapses.
### Scientific Notation
Scientific notation is a method used by scientists to convert very large or very small numbers to more manageable ones. You will have to make a few conversions to scientific notation on the GED Mathematics Exam. Expressing answers in scientific notation involves moving the decimal point and multiplying by a power of ten.
Example
A space satellite travels 46,000,000 miles from the earth. What is the number in scientific notation?
Step 1: Starting at the decimal point to the right of the last zero, move the decimal point until only one digit remains to its left:
46,000,000 becomes 4.6.
Step 2: Count the number of places the decimal was moved left. In this example, the decimal point was moved 7 places, and express it as a power of 10: 107
Step 3: Express the full answer in scientific notation by multiplying the reduced answer from step 1 by 107:
4.6 × 107
Example
An amoeba is .000056 inch long. What is its length in scientific notation?
Step 1: Move the decimal point to the right until there is one digit other than zero to the left of the decimal.
.000056 becomes 5.6
Step 2: Count the number of places moved to the right—5.However, because the value of the number is being increased as it is expressed in scientific notation, it is written as a negative exponent. 10–5
Step 3: Express the full answer in scientific notation: .0000056 becomes 5.6 × 10–5 |
# Lesson 19
Compose and Decompose to Add and Subtract
## Warm-up: Number Talk: Subtract Fractions (10 minutes)
### Narrative
The purpose of this Number Talk is to elicit strategies and understandings students have for subtracting fractions and mixed numbers, particularly cases where it is necessary to rewrite a whole number as a fraction, or decompose it into a different whole number and fraction, in order to perform subtraction. These understandings will be helpful later in this lesson when students subtract multi-digit numbers that involve decomposing units.
### Launch
• Display one expression.
• “Give me a signal when you have an answer and can explain how you got it.”
### Activity
• 1 minute: quiet think time
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Find the value of each expression mentally.
• $$2\frac{3}{4} - 1\frac{1}{4}$$
• $$1\frac{1}{4} - \frac{3}{4}$$
• $$5\frac{1}{8} - 2\frac{3}{8}$$
• $$3\frac{2}{10} - 2\frac{7}{10}$$
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• Highlight strategies in which students decomposed the first mixed number in each expression.
• “Who can restate _______ 's reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone approach the expression in a different way?”
• “Does anyone want to add on to____’s strategy?”
## Activity 1: Find and Check Sums (15 minutes)
### Narrative
The purpose of this activity is to use the standard algorithm to add multi-digit numbers, taking care to compose a new unit and record it accurately. They also analyze common errors and critique the given reasoning when composing a new unit (MP3).
### Required Materials
Materials to Gather
### Launch
• Groups of 2
• “Complete the first 2 problem and then talk to your partner about any patterns you notice.”
### Activity
• 3 minutes: independent work time
• 2 minutes: partner discussion
• Share and record responses.
• If needed, use the expanded form to help students make connections between the way composing a larger unit is recorded when using the standard algorithm and what is happening in each place. In the bottom line of the example here, we see 10 in the ones place and 100 in the tens place. Both partial sums do not match the assigned values of their places.
• 3–4 minutes: independent work time for the last problem.
### Student Facing
1. Find the value of each sum.
2. Use the expanded form of both 8,299 and 1,111 to check the value you found for the last sum.
3. Each computation shown has at least one error. Find the errors and show the correct calculation.
### Student Response
For access, consult one of our IM Certified Partners.
Students may need support identifying errors in the last problem. Consider asking: “How might subtraction be used to help identify the error?”
### Activity Synthesis
• Ask students to share the errors they identified in the last set of questions.
• “What are some common errors when adding large numbers?”
## Activity 2: Priya’s Family Heirlooms (20 minutes)
### Narrative
This activity revisits the idea of decomposing a unit in one place into 10 units of the place value to its right when subtracting multi-digit numbers using the standard algorithm. Students recall how this is done as they subtract numbers in which decomposition is necessary.
To support students with understanding the context, the activity launch introduces a saree (traditional wedding attire for women in India) and the idea of family heirlooms, or gifts passed down from generation to generation.
When students create a subtraction problem that does not require decomposition of a unit when using the standard algorithm, they make use of structure and their understanding of subtraction as they choose the digits for the numbers in their difference (MP7).
This activity uses MLR7 Compare and Connect. Advances: representing, conversing
Representation: Develop Language and Symbols. Activate or supply background knowledge. To help students recall the term decompose, represent a four-digit number (for example 2,467) with both base-ten blocks and digits in a place value chart. Ask, “what does it mean to decompose a unit?” Show an example of decomposition, such as exchanging one long rectangle for ten small cubes. Notate this in the place value chart by crossing out the 6 and 7, and writing 5 and 17 above them. Reset the place value blocks and show additional examples as needed.
Supports accessibility for: Conceptual Processing, Memory, Language
### Launch
• Groups of 2
• “What do you notice and wonder about these pictures?”
• Collect student ideas.
• “In this activity, Priya is researching her family history. Let’s see what she discovers.”
• “The women in the picture are dressed in a traditional Indian garment called a ‘saree’. Sarees are made of colorful fabric and often have intricate embroidery or patterned print designs.”
• “The bracelets in the picture are also from India. Sometimes jewelry like this is used as heirlooms, or gifts that are passed down from one generation to the next.”
### Activity
• Groups of 2
• 5–6 minutes: independent work time
• 3 minutes: partner work time
• As students work, listen for student discourse that includes language about the place value of the digits and when to decompose a unit.
### Student Facing
Priya’s mom wore an heirloom bracelet at her wedding in 1996. The bracelet was made in 1947.
Priya subtracted to find out how old the bracelet was when her parents were married.
Priya learned that her grandmother had also worn the bracelet at her wedding 24 years earlier.
Priya subtracted to find out when her grandparents were married.
1. Are both calculations correct? Why does one calculation have some numbers crossed out and some new numbers, but the other one does not? Explain your reasoning.
2. Priya’s grandmother wore an heirloom necklace and earring set that was 63 years old when she was married in 1972.
1. If Priya uses the standard algorithm to subtract $$1972 - 63$$ will she need to decompose a unit? Explain your reasoning.
2. Use the standard algorithm to subtract $$1972 - 63$$ and find the year the necklace was made.
3. Create a subtraction problem that would not require decomposing a unit to subtract. Then solve the problem.
### Student Response
For access, consult one of our IM Certified Partners.
Students may not remember when to decompose a unit or how to record regrouping. Urge students to begin to subtract by place value, either by using expanded form or lining up the digits. Ask: “What issue comes up when you subtract the ones in $$1972 - 63$$?” Allow students to explain that they don't have enough ones in the ones place to subtract 3, but they can decompose a ten to get 10 ones and add it to the 2 already there. Then consider asking:
• “How will you record all of the ones you have after you decompose a ten?”
• “How will you know if you need to decompose a unit when subtracting one number from another?”
### Activity Synthesis
MLR7 Compare and Connect
• “Let’s do a gallery walk to see what problems you created.”
• “As you walk, discuss with a partner what you notice about the value of the digits in the numbers that were chosen.”
• 3 minutes: gallery walk
• Collect 1–2 responses from student discussions during the gallery walk.
• Share 1–2 of the responses you collected.
• 1 minute: partner discussion
• “What is the same about each of the problems you created?” (In each problem each digit is greater in the first number than in the second number)
• 2 minutes: whole-group discussion
## Lesson Synthesis
### Lesson Synthesis
Write $$1972 - 63$$ for all students to see.
“When we look at a problem, how do we know if we will need to decompose a unit?” (If the digit we are subtracting is larger than the digit we are subtracting from, we will need to decompose a unit and regroup.)
Display for all to see:
“When we look at an addition problem, how do we know when we will need to compose a new unit?” (If the sum of the digits in one place is greater than 9, we will compose a new unit, and record 1 more for the place to the left.)
## Cool-down: Difference and then Sum (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners. |
# Diamond Problem Calculator
Created by Hanna Pamuła, PhD
Reviewed by Dominik Czernia, PhD and Jack Bowater
Last updated: Aug 18, 2023
Welcome to our diamond problem calculator, also known as a diamond problem solver. This intuitive tool allows you to enter any two numbers and the two others will appear. We've also prepared a compendium that answers all your burning questions about this topic.
Let's start by introducing what a diamond problem is and continue smoothly to some how to do diamond problems of various types tips. Are you ready?
## What is a diamond problem? Diamond math problems
Despite what you might think, the diamond problem doesn't have a lot in common with gemstones💎 or diamond rings💍 — though we can teach you how to calculate a diamond's weight. It is more closely related to what is often called a diamond in math ♦️ or cards🃏 - the rhombus, a quadrilateral shape. The diamond problem is a type of exercise that happens in a diamond shape 💠 - which we can also represent as a cross with 4 sections.
So, what is the diamond problem in math? It's where you fill in all four fields, related by some mathematical operation. The pattern of the numbers is constant:
• On the left and right side of the diamond, you have two numbers, sometimes called factors;
• In the top part you can find their product; and
• In the bottom section - their sum.
Solving the diamond problem means that you know only two numbers out of four, and you need to find the missing ones. And that's all!
There are three main types of diamond problems, and the diamond problem calculator can deal with all of them. If you're wondering how to do diamond problems in each of these cases, scroll down to the next sections.
## How to do diamond problems? Case 1: Given two factors
This is the easiest case: you have two numbers, A and B, and you need to find the sum and product of them. For example, let's say that we want to solve the diamond problem for factors $13$ and $4$:
1. Calculate the product $= 13 \times 4 = 52$, and write the number on top.
2. Find the sum $= 13 + 4 = 17$, and input the value into the bottom part of the diamond.
You might meet this type of a diamond math problem in the first lesson about the diamonds - when your teacher first introduces the concept.
## Case 2: Given one factor, and the product or sum
Let's have a look at a slightly more complicated case, where you have one of the basic numbers, and a product or a sum. The first thing to do is to calculate the factor missing from the diamond. Transform your equation is such a way to solve for the unknown value:
a) When you're given one factor and the sum, you can find a second factor by a simple subtraction:
$\text{factor}\ A + \text{factor}\ B = \text{sum}$
And
$\text{factor}\ B = \text{sum} - \text{factor}\ A$
So:
$\text{factor}\ B = 11-6 = 5$
With both factors in hand, simply multiply them to get the last number: product $= 5 \times 6 = 30$.
b) If you know one factor and the product, divide the product by the factor to find the second product:
$\text{factor}\ A \times \text{factor}\ B = \text{product}$
And:
$\frac{\text{factor}\ B}{ \text{product}} = \text{factor}\ A$
So:
$\text{factor}\ B = \frac{63}{9} = 7$
Then calculate the sum, use the following expression: sum $= \text{factor}\ A + \text{factor}\ B = 9 + 7 = 16$.
Remember, you can always quickly solve these diamonds or verify your answers with our diamond problem calculator.
## Case 3: Given product and sum, while searching for factors
Now we're coming to the last issue and the most common diamond problem: the case where you know the sum and the product of the two numbers, but you don't actually know the numbers themselves.
This type of diamond math problem is helpful when you're learning about factoring a quadratic equation. Why?
Let's say that we have a quadratic equation:
$x^2+7x+12$
We would like to factor this equation, meaning that we'd like to present it in a form:
$(x+...)(x+...)$
How to find these numbers - the roots of the quadratic equation? You know that their product must be equal to $12$, and that their sum is equal to $7$. And that's exactly what you're trying to find out in the diamond problem! 🙂 You can rewrite this question in the form of a diamond:
How to solve such a problem?
1. Start with the top number - the product of two numbers. Write down all possible pairs of numbers that give this as their product. You can, for example, calculate the prime factorization if it's a more complicated case. In our example, the product is equal to 12 - which two integer numbers could be multiplied together? The order of the factors doesn't matter, as multiplication (and addition) are commutative:
$\quad\quad\begin{split} 12&=1\times12\\ 12&=2\times6\\ 12&=3\times4 \end{split}$
Don't forget the negative numbers:
$\quad\quad\begin{split} 12&=-1\times-12\\ 12&=-2\times-6\\ 12&=-3\times-4 \end{split}$
1. Sum these two numbers, and check which combination gives the desired sum of 7:
$\quad\quad\begin{split} &\color{red}{1+12=13}\\ &\color{red}{2+6=8}\\ &\color{green}{3+4=7}\\ \end{split}$
And here it is! We can stop here, as we've found the two numbers which meet the condition 🎉
Sometimes it's very easy to guess the solution, as there are not many options. We're sure that you'll solve this a diamond problem in no time!
If not, enter the numbers into our diamond problem solver. That wasn't so difficult, was it? 🤦
## How to use diamond problem calculator
It's really easy, believe us!
All you need to do is to input any two numbers and the diamond problem solver will find out the other two. What's more, the tool displays the solution in a diamond form. What more could you wish for? 😎
Please notice that in some specific cases you'll need to hit the refresh button ⟳ to use the calculator again 👍
## FAQ
### What are the other numbers in the diamond if the left and right are -4 and 8?
The top number is -32, and the bottom is 4. To find the top, we multiply the side numbers. We add them together to find the bottom.
### What is the diamond problem used for in mathematics?
We use the diamond problem in addition, subtraction, multiplication, and division in maths. It is used to calculate integers, decimals, and fractions and for factoring trinomials.
### How do I solve the diamond for fractions?
Assuming that you are given two fractions on the left and right sides, say 1/2 and 5/6, follow these steps to solve the problem:
1. Find the product of the fractions:
1/2 × 5/6 = 5/12
2. Place this answer (5/12) at the top:
3. Find the sum of the fractions:
1/2 + 5/6 = 6/12 + 10/12 = 16/12
4. Place this answer (16/12) at the bottom.
Voila! You solved the diamond problem.
### How do I find the right and left numbers in the diamond?
Assuming the top number is 35, and the bottom is 12. Follow these steps to find the numbers to the right and left:
1. Find the factors of the top number:
35, 1
7, 5
2. Identify the factors which, when added together, equal to the bottom number:
7, 5
3. Place these numbers on either side.
That's it. You have solved the diamond problem for the right and left numbers.
Hanna Pamuła, PhD
A
product
sum
B
Factor A
Factor B
Product
Sum
People also viewed…
### Black Friday
How to get best deals on Black Friday? The struggle is real, let us help you with this Black Friday calculator!
### Christmas tree
Welcome to the Christmas tree calculator, where you will find out how to decorate your Christmas tree in the best way. Take a look at the perfect Christmas tree formula prepared by math professors and improved by physicists. Plan in advance how many lights and decorations you'll need!
### Length of a line segment
Use the length of a line segment calculator to determine the length of a line segment by entering the coordinates of its endpoints.
### Round to the nearest thousandth
Use this tool to calculate the rounding to the nearest thousandth of any number you want, with the possibility to use many rounding methods. |
Session 8, Part B:
Mathematical Probability
In This Part: Predicting Outcomes | Fair or Unfair? | Outcomes | Finding the Winner
Making a Probability Table
Another way to solve this problem is to look at a probability table for the sum of the two dice. This representation can be quite useful, since it gives us a complete description of the probabilities for the different values of the sum of two dice, independent of the rules of the game. Note 6
Again, here are the sums of the possible outcomes for the two dice:
Red Die
Blue Die
+ 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12
• Only one of the outcomes (highlighted in purple) produces a sum of 2. There are 36 equally likely outcomes. So the probability of the sum being 2 is 1/36. • Two of the outcomes (highlighted in green) produce a sum of 3. There are 36 equally likely outcomes. So the probability of the sum being 3 is 2/36.
Here is the start of a probability table for the sum of two dice:
Sum
Frequency
Probability
2 1 1/36 3 2 2/36
Problem B7
Complete the probability table.
Sum
Frequency
Probability
2 1 1/36 3 2 2/36 4 5 6 7 8 9 10 11 12
Sum
Frequency
Probability
2 1 1/36 3 2 2/36 4 3 3/36 5 4 4/36 6 5 5/36 7 6 6/36 8 5 5/36 9 4 4/36 10 3 3/36 11 2 2/36 12 1 1/36
Problem B8 Use the probability table you completed in Problem B8 to determine the probability that Player A will win the game. Recall that Player A wins if the sum is 2, 3, 4, 10, 11, or 12.
Problem B9 If you know the probability that Player A wins, how could you use it to determine the probability that Player B wins without adding the remaining values in the table?
If Player A does not win, then Player B wins. Close Tip : If Player A does not win, then Player B wins.
Session 8: Index | Notes | Solutions | Video |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.