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3-3 example10
# 3-3 example10 - Student Grady Silnonton Course Math119...
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Unformatted text preview: Student: Grady Silnonton Course: Math119: Elementary Statistics - Spring 2010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, 11e Time: 11:54 ANI Listed below are the nicotine amounts (in mg per cigarette) for samples of filtered and nonfiltered cigarettes. Find the coefficient of variation for each of the two sets of data, then compare the variation. Nonfiltered 1.1 1.? 1.5 1.3 0.9 1.7 1.0 1.5 1.0 1.4 1.2 1.0 1.1 1.3 0.9 2.1 1.5 1.0 1.4 1.6 1.3 1.1 1.2 1.0 1.3 Filtered 0.2 1.0 0.2 1.0 0.7 0.5 1.2 0.8 1.0 1.0 0.8 1.1 1.0 0.8 0.1 1.4 1.0 0.8 1.2 0.9 0.? 1.0 0.5 1.2 1.2 S The coefficient of variation is given as CV = = - 100%, where s is the sample standard deviation, and E is the sample x mean . First, calculate the sample means. The mean, i, of a sample is found with the formula below, where x is the variable used to represent the individual data values, and n is the number of data values in the sample. 2 x <— sum of all data values x = n <— number of data values Let x1 be the nicotine amounts (in mg per cigarette) for samples of nonfiltered cigarettes. Calculate £1. £1 = 1.284 Let X; be the nicotine amounts (in mg per cigarette) for samples of filtered cigarettes. Calculate £2. £2 = 0.852 Next, calculate the sample standard deviations. — 2 x — x — The standard deviation, s, of a sample is found by using the formula 5 = fl—l) , where x is the mean of the n _ sample, and n is the size of the sample. For x1, calculate the first value of (x — E). (1.1— 1.284) —0.184 (x—;) Calculate (x - ;) for the remaining data values. Page 1 Student: Grady Silnonton Course: Mathll9: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, lle Time: 11:54 AMI Calculate (x — i) 2 for the first data value. (X—;)2 = (1.1—1.284)2 = ( — 0.184)2 = 0.0339 Calculate (x — i) 2 for the remaining data values. Now find the sum of the values of (x — E) 2. 2(x — i) 2 = 2.0947 Finally, divide the sum of the values of (x — E) 2 by (11— l) to find the variance, where n is the sample size. Then take the square root to find the standard deviation, rounding to three decimal places. .1: [Zea—EV n-l 2.0947 25 — l = 0.295 Repeat the calculations for the other sample. Page 2 Student: Grady Silnonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, 11e Time: 11:54 ANI Recall that the mean of X; is 0.852. Calculate (x — E) for the data values. Now find the sum of the values of (x — ;) 2. 2(x — E) 2 = 2.?223 Finally, divide the sum of the values of (x — i) 2 by (n— l) to find the variance, where n is the sample size. Then take the square root to find the standard deviation, rounding to three decimal places. 51: 2(X_;)2 n-l = f 27223 25 — 1 = 0.33? Now that s and K have been calculated for both data sets, the coefficient of variation can be found and the variations can be compared. 5 _ Recall that the coefficient of variation is given as CV = = - 100%, where s is the sample standard deviation, and x is the x Page 3 Student: Grady Silnonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, 11e Time: 11:54 ANI sample mean. Calculate the coefficient of variation for the two samples, rounding to three decimal places. 51 s2 CV1 = ——- 100% CV2 = ——- 100% X1 X2 0 295 0.337 = —- 100% = — 100% 1.284 0.852 = 2.975% = 39.554% When comparing variation in two different sets of data, the standard deviation should be compared only if the two sets of data use the same scale and units and they have the same mean. If the means are substantially different, or if the samples use different scales or measurement units, use the coefficient of variation to compare the variation in the data. The coefficient of variation for the nicotine amounts (in mg per cigarette) for samples of nonfiltcred cigarettes is approximately 2.975%, and the coefficient of variation for the nicotine amounts (in mg per cigarette) for samples of filtered cigarettes is approximately 39.554%. By comparing the coefficients of variation, it can be concluded that the nicotine amounts of nonfiltered cigarettes have considerably less variation than the nicotine amounts of filtered cigarettes. Page 4 ...
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# Slope-Finding Functions
By the end of this post, you will know what a derivative is. You won’t know how to find one yet, but you will know what you have if you are given one!
Let’s say that we were given a function such as f(x) = x3 -4x.
We know it isn’t a linear function so when we graph it, we are not surprised to see the curves. And if you can imagine a tangent line “surfing” along the curve, the slope of the tangent line would clearly be changing as you went along. Here’s what I mean:
If you wanted to know the slope at a given point, you could estimate it by drawing the tangent line and then finding the slope of that tangent line (as we have seen in the last post).
But there is an alternative to that method that is useful and quick. When you have the rule for a function, there is a way to derive a formula for a new function that will give you the slope of the original function’s tangent line at any x-value you choose. Since it is derived from the original function, this new function is called “the derivative” of the original function.
For example, if the original function is:
f(x) = x3 – 4x
The derivative is written:
f’(x) = 3x2 – 4
The symbol f’(x) is pronounced “f-prime of x” and it is the most commonly used symbol to represent the derivative of the function f.
TWO THINGS YOU SHOULD NOT WORRY ABOUT (YET)
1. How did we get from the function f(x) = x3 – 4x to its derivative f’(x) = 3x2 – 4?
I will show you how to “take a derivative” in the next post. Once you learn a handful of rules, the process is not difficult. It’s easier than factoring. It’s easier than long division.
2. Why would anyone want a slope-finding formula ?
Once you know how to take derivatives, I can show you a variety of ways that they are useful. In fact, you’ll even get to use them forward and backwards! (That was supposed to be some fore-shadowing…)
For now, my goal is simpler. I just want to be sure that we understand what it is that the derivative formula tells us. And once again…it’s better with diagrams.
I made the graphs and videos in this post using a web-based applet written by Paul Seeburger, a professor of mathematics at Monroe Community College in New York. You can find the applet at:
http://web.monroecc.edu/manila/webfiles/pseeburger/JavaCode/DerivativeDemo2.htm
The blue line is the original function, f(x) = x3 – 4x
The purple line is the derivative, f’(x) = 3x2 – 4.
The red line is the tangent line. You can calculate the slope of that tangent using rise over run if you want to, but you don’t have to. The applet does the calculation for you – look in the pink box.
Now let’s pick a point on the graph and examine all of the lines on this diagram.
At x = -2, f(-2) = (-2)3 – 4×(-2) = -8 + 8 = 0. This means that (-2,0) is a point on the graph of f.
At that same point, the tangent line has a slope of 8 (as you can see in the pink box).
If we plug x = -2 into the derivative formula, we get f’(-2) = 3(-2)2 – 4 = 12 – 4 = 8. And then, when you look at the graph of the derivative (in purple), you see that (2,8) is a point on that graph. So the derivative formula tells us the slope – at least at that one point. But does it always work?
Well, let’s look at another point: x = -1.
(If you are playing along at home, you can enter any x-value you want into the box on the lower left hand side of the applet where it says “Trace x =”)
At x = -1, f(x) = (-1)3 – 4(-1) = 3. So (-1, 3) is a point on the graph.
Then, f’(x) = 3(-1)2 – 4 = -1. So we expect the slope of the tangent line to be -1, and it is. You can see that (-1,-1) is a point on the graph of the derivative (again, the purple one). So once again, the derivative formula has given the slope of the tangent line.
Here’s one more point that is worth noticing:
Look at the value of the derivative. And then look at the shape of the graph. Where the derivative is zero, the graph has a horizontal tangent. That little fact turns out to be a big part of one of the reasons why we care about derivatives. They help us to find places where the graph has a maximum or (as in this case) a minimum. We’ll be hearing more about that later.
So we have confirmed, at least for the points we checked, that for this particular function, the derivative formula does in fact give values that match the slopes of the original function’s tangent lines. If you want to, you can check many more points:
As you saw in the video above, Professor Seeburger’s applet lets you trace along the graph. As you do, the tangent line surfs along the curve and the applet keeps track of all of the values. Here is that video again, this time with the derivative function and the values all visible:
So if you are skeptical, run the video and then pause it wherever you want. You’ll see that the slope of the tangent line always matches the value given by the derivative. Derivatives really are slope-finding formulas.
## One thought on “Slope-Finding Functions”
1. Wow! Fantastic explanation! I learned about derivatives in my college Calculus class, but I don’t think it was ever explained to me so neatly. I definitely felt the emphasis on how to “perform” derivatives and less on “why.”
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## Question:
The elevator starts from rest at the first floor of the building. It can accelerate at $5 ft/{ s }^{ 2 }$ and then decelerate at $2 ft/{ s }^{ 2 }$. Determine the shortest time it takes to reach a floor 40 ft above the ground. The elevator starts from rest and then stops. Draw the at, vt, and st graphs for the motion.
## Step-by-step
\begin{aligned} + \uparrow \quad &{ v }_{ 2 } = { v }_{ 1 } + { a }_{ c } { t }_{ 1 } \\ &{ v }_{ max } = 0 + 5 { t }_{ t1 } \end{aligned} \\ \begin{aligned} + \uparrow \quad &{ v }_{ 3 } = { v }_{ 2 } + { a }_{ c } { t } \\ & 0 = { v }_{ max } – 2 { t }_{ 2 } \end{aligned}
Thus
\quad\quad\space\space\space { t }_{ 1 } = 0.4 { t }_{ 2 } \\ \begin{aligned} + \uparrow \quad & { s }_{ 2 } = { s }_{ 1 } + { v }_{ 1 } { t }_{ 1 } \frac { 1 } { 2 } { a }_{ c } { t }_{ 1 }^{ 2 } \\ & h = 0 + 0 + \frac { 1 } { 2 } (5)({ t }_{ 1 }^{ 2 }) = 2.5 \space { t }_{ 1 }^{ 2 } \end{aligned} \\ + \uparrow 40 – h = 0 + { { v }_{ max } }^{ t_2 } – \frac{ 1 } { 2 }(2) { t }_{ 2 }^{ 2 } \\ + \uparrow { v }^{ 2 } = { v }_{ 1 }^{ 2 } + 2 { a }_{ c } ( { s } – { s }_{ 1 } ) \\ { v }_{ max }^{ 2 } = 0 + 2(5)(h – 0) \\ { v }_{ max }^{ 2 } = 10h \\ 0 = { v }_{ max }^{ 2 } + 2(-2)(40 – h) \\ { v }_{ max }^{ 2 } = 160 – 4h
Thus,
$10 h = 160 – 4h \\ h = 11.429 ft \\ { v }_{ max } = 10.69 ft/s \\ { t }_{ 1 } = 2.138s \\ { t }_{ 2 } = 5.345s \\ t = { t }_{ 1 } + { t }_{ 2 } = 7.48s$
When t = 2.145, $v = { v }_{ max } = 10.7 ft/s$ and h = 11.4 ft.
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# When Does A Limit Not Exist? (4 Key Cases To Know)
When working with functions in calculus, we will sometimes find that a limit does not exist. There are lots of reasons this can happen, and it helps to be familiar with each one so you know what to look for.
So, when does a limit not exist? The limit of a function at a point does not exist in 4 cases: 1. when the left hand limit does not exist, 2. when the right hand limit does not exist, 3. when the left and right hand limits exist, but have different values, and 4. when the function value is undefined, due to a domain restriction.
Of course, the existence of limits for a function can also tell us about continuity and differentiability of the function at certain points.
In this article, we’ll talk about when a limit does not exist. We’ll look at some examples of each case to make the concept clear.
Let’s get started.
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## When Does A Limit Not Exist?
A limit does not exist in the following cases:
• Left Hand Limit Does Not Exist
• Right Hand Limit Does Not Exist
• Left & Right Hand Limits Both Exist, But They Have Different Values
• Function Is Not Defined Due To Domain Restriction
Note that there are a few ways for a left or right hand limit to not exist, including:
• Oscillating values of the limit
• Unbounded increase or decrease (a vertical asymptote, or a limit of infinity)
Let’s start by looking at cases and examples where the left hand limit does not exist.
### Left Hand Limit Does Not Exist
In order for a limit to exist, both the left and right hand limits must exist, and they must have the same value. Here are some examples where the left hand limit does not exist.
#### Example 1: Left Hand Limit Does Not Exist (Oscillating Values)
Consider the function f(x) = sin(1 / x2). If we take a left hand limit as x approaches zero:
• Limx->0-f(x)
we will find that the limit does not exist.
This is because the sine function oscillates between all values from -1 to 1 as x increases.
As x approaches zero, 1 / x2 increases without bound. So, the limit will never “settle down” to a single value, and will continue oscillating forever.
Thus, the left hand limit does not exist in this case.
#### Example 2: Left Hand Limit Does Not Exist (Vertical Asymptote)
Consider the function f(x) = 1 / x2. If we take a left hand limit as x approaches zero:
• Limx->0-f(x)
we will find that the limit does not exist.
This is because the function 1 / x2 grows without bound as x approaches zero.
As x approaches zero, x2 remains positive but becomes smaller and smaller. Since x2 is in the denominator, we will see 1 / x2 increase without bound.
Thus, the left hand limit does not exist in this case (or we could say that the limit is infinity).
### Right Hand Limit Does Not Exist
Just as a left hand limit can fail to exist, a right hand limit can also fail to exist. Here are some examples where the right hand limit does not exist.
#### Example 1: Right Hand Limit Does Not Exist (Oscillating Values)
Consider the function f(x) = cos(1 / x2). If we take a right hand limit as x approaches zero:
• Limx->0+f(x)
we will find that the limit does not exist.
This is because the cosine function oscillates between all values from -1 to 1 as x increases.
As x approaches zero, 1 / x2 increases without bound. So, the limit will never “settle down” to a single value, and will continue oscillating forever.
Thus, the right hand limit does not exist in this case.
#### Example 2: Right Hand Limit Does Not Exist (Vertical Asymptote)
Consider the function f(x) = 1 / (x – 2). If we take a left hand limit as x approaches 2:
• Limx->2+f(x)
we will find that the limit does not exist.
This is because the function 1 / (x – 2) grows without bound as x approaches 2.
As x approaches 2, (x– 2) approaches zero (and remains positive). Since (x – 2) is in the denominator, we will see 1 / (x – 2) increase without bound.
Thus, the right hand limit does not exist in this case (or we could say that the limit is infinity).
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### Left & Right Hand Limits Both Exist, But They Have Different Values
In some cases, both the left and right hand limits will exist for a function, but they will have different values. This is sometimes called a “jump” discontinuity.
Here are some cases where the left and right hand limits do not agree.
#### Example 1: Left & Right Hand Limits Both Exist, But They Have Different Values (Absolute Value)
Consider the function f(x) = |x| / x.
If we take a left hand limit as x approaches 0:
• Limx->0-f(x)
we will find that the limit is -1.
This is because the numerator and denominator will have opposite signs, but the same absolute value (as you can see in the table below).
Similarly, if we take a right hand limit as x approaches 0:
• Limx->0+f(x)
we will find that the limit is +1. This is because the numerator and denominator will have the same sign and the same absolute value (as you can see in the table below).
So, the left hand limit (-1) and right hand limit (+1) both exist. However, they disagree, since they do not have the same value.
Thus, the limit as x approaches zero does not exist for this function.
• Limx->0f(x): DNE (does not exist)
#### Example 2: Left & Right Hand Limits Both Exist, But They Have Different Values (Piecewise)
Consider the following piecewise function:
If we take a left hand limit as x approaches 1:
• Limx->1-f(x)
we will find that the limit is 2.
Similarly, if we take a right hand limit as x approaches 1:
• Limx->1+f(x)
we will find that the limit is 3.
So, the left hand limit (2) and right hand limit (3) both exist. However, they disagree, since they do not have the same value.
Thus, the limit as x approaches 1 does not exist for this function.
### Function Is Not Defined Due To Domain Restriction
A limit can also fail to exist if a function is not defined due to a domain restriction.
#### Example: Function Is Not Defined Due To Domain Restriction (Square Root)
Consider the function f(x) = √x.
If we take a right hand limit as x approaches 0:
• Limx->0+f(x)
the limit is defined, and the value is 0 (since the square root of zero approaches zero).
However, if we take a left hand limit as x approaches 0:
• Limx->0-f(x)
the limit is undefined in the real numbers.
The reason is that when we approach zero from the left, we are looking at negative values of x. No matter how small these values become, they are still negative.
The square root of any negative number is an imaginary number, so the limit does not exist. Thus, the domain restriction for the square root function (x >= 0) means that we cannot take a left hand limit.
Since the left hand limit at zero does not exist, then the limit at zero does not exist for this function.
## What Does It Mean For A Limit To Not Exist?
When the limit of a function does not exist at a point x = c, it means that:
• the function is not continuous at x = c (since there is an oscillating value, vertical asymptote, jump discontinuity, etc.)
• the function is not differentiable at x = c (since a function cannot be differentiable if it is not continuous).
In other words, the existence of a limit for a function at a point is a prerequisite for continuity of the function at that point, which is in turn a prerequisite for differentiability of the function at that point.
## Conclusion
Now you know about the cases when a limit does not exist and what to look for (oscillating values, vertical asymptotes, and jump discontinuities).
If you have the limit of a quotient that results in an indeterminate form (such as 0/0 or infinity/infinity), you can use L’Hopital’s Rule.
You can learn about the floor function (a type of piecewise function) here.
I hope you found this article helpful. If so, please share it with someone who can use the information.
~Jonathon
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# 2007 Alabama ARML TST Problems/Problem 11
## Problem
In how many distinct ways can a rectangular $3\times 17$ grid be tiled with $17$ non-overlapping $1\cdot 3$ rectangular tiles?
## Solution
In both solutions we consider the grid to have $3$ rows and $17$ columns.
### Solution 1
There are either $17$ vertical tiles, $14$ vertical and $3$ horizontal, $11$ vertical and $6$ horizontal, etc. The horizontal tiles always have to form blocks $3\times 3$.
If we now consider tilings that consist of $a$ vertical tiles and $b$ blocks of three horizontal tiles, it is clear that there are $a+b \choose b$ such tilings. Hence we get that the total number of tilings in our case is:
$$\dfrac{17!}{17!}+\dfrac{15!}{14!}+\dfrac{13!}{11!\cdot 2!}+\dfrac{11!}{8!\cdot 3!}+\dfrac{9!}{5!\cdot 4!}+\dfrac{7!}{2!\cdot 5!}$$
It isn't that such a pain to compute, so we do:
$$1+15+78+165+126+21=\boxed{406}$$
### Solution 2
Let $T_n$ be the number of ways to tile a $3\times n$ rectangle.
Obviously $T_0=T_1=T_2=1$.
Consider a $3\times n$ rectangle with $n\geq 3$. Take a look at the tile that contains the upper right corner. It can be either vertical or horizontal. If it is vertical, we still need to tile the remaining $n-1$ columns. If it is horizontal, the other two cells in the rightmost column must be covered by horizontal tiles as well. We are then left with $n-3$ columns to tile.
This gives us the recurrence $T_n = T_{n-1} + T_{n-3}$ for $n\geq 3$.
Using it we can now compute:
n | T(n)
------------
3 | 2
4 | 3
5 | 4
6 | 6
7 | 9
8 | 13
9 | 19
10 | 28
11 | 41
12 | 60
13 | 88
14 | 129
15 | 189
16 | 277
17 | 406
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# In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?
Given:
In an A.P., the 5th and 12th terms are 30 and 65 respectively.
To do:
We have to find the sum of the first 20 terms.
Solution:
Let the first term be $a$ and the common differnce be $d$.
Fifth term $a_5=a+(5-1)d$
$30=a+4d$
$a=30-4d$......(i)
12th term $a_{12}=a+(12-1)d$
$65=a+11d$
$65=30-4d+11d$ (From (i))
$7d=65-30$
$d=\frac{35}{7}$
$d=5$.....(ii)
This implies,
$a=30-4(5)$
$=30-20$
$=10$
We know that,
Sum of $n$ terms$S_{n} =\frac{n}{2}(2a+(n-1)d)$
$S_{20}=\frac{20}{2}[2(10)+(20-1)5]$
$=10(20+95)$
$=10(115)$
$=1150$
Hence, the sum of the first 20 terms is $1150$.
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Updated on: 10-Oct-2022
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# Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$
All the angles in a triangle $A,B,$ and $C$ are less than $120^{o}$
Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$
-
Consider triangle with angles $\small{A_1=120-A, B_1=120-B, C_1=120-C}$ Applying the triangle inequality in this triangle with angles $A_1, B_1,$ and $C_1$,
$$\small{B_1 C_1+C_1 A_1 > A_1 B_1}$$
$$\small{\sin A_1 +\sin B_1 > \sin C_1}$$ $$\small{\sin (120-A) +\sin (120-B) > \sin (120-C)}$$
which by applying $\sin(x-y)$ identities
$$\small{\frac{\sqrt{3}}{2}(\cos A+\cos B-\cos C) + \frac{1}{2}(\sin A+ \sin B-\sin C) > 0} \tag{1}$$
And since $\small{a+b > c , \sin A+\sin B - \sin C > 0}$, and therefore dividing by this is perfectly legitimate
Using this observation and re-writing $(1)$, we obtain
$$\small{\frac{\sqrt{3}}{2} \frac{\cos A + \cos B -\cos C}{\sin A + \sin B - \sin C} + \frac{1}{2} > 0}$$ $$\small{\Rightarrow \frac{\cos A + \cos B -\cos C}{\sin A + \sin B - \sin C} > \frac{-\sqrt{3}}{3}}$$
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thanks I knew there was a reason why they gave $A,B$ and $C$ are less than $120$. – Kirthi Raman Mar 26 '12 at 12:18
Some brute force...
Let $f$ be your quotient, depending on $a$ and $b$ (I have eliminated $c$, because $c=\pi-a-b$):
In[1]:= f = (Cos[a] + Cos[b] - Cos[Pi - a - b]) / (Sin[a] + Sin[b] - Sin[Pi - a - b])
Cos[a] + Cos[b] + Cos[a + b]
Out[1]= ----------------------------
Sin[a] + Sin[b] - Sin[a + b]
Notice that Mma is being smart and has midly rewritten our expression...
Find the points where the gradient of $f$ vanishes:
In[2]:= Reduce[{D[f, a] == 0, D[f, b] == 0}, {a, b}]
Out[2]= (C[1] | C[2]) \[Element] Integers &&
-2 Pi -2 Pi
> ((a == ----- + 2 Pi C[1] && b == ----- + 2 Pi C[2]) ||
3 3
2 Pi 2 Pi
> (a == ---- + 2 Pi C[1] && b == ---- + 2 Pi C[2]))
3 3
I used Reduce and not Solve, because the latter complains a little, harmlessly in this case; Reduce gives the answer in the annoying form above, though. In any case, this tells us that the critical points of the function occur at the points of the form $(-2\pi/3+2\pi n,-2\pi/3+2\pi m)$ and $(2\pi/3+2\pi n,2\pi/3+2\pi m)$ wit $n$, $m\in\mathbb Z$. To find the value of $f$ at these points it sufficies to take $n=m=0$, because $f$ is periodic in $a$ and in $b$.
In[3]:= f /. {a -> -2 Pi/3, b -> -2 Pi/3}
1
Out[3]= -------
Sqrt[3]
In[4]:= f /. {a -> 2 Pi/3, b -> 2 Pi/3}
1
Out[4]= -(-------)
Sqrt[3]
This tells us that the minimal value of $f$ is $-1/\sqrt{3}$ and proves what you want.
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Notice I have found the minimum of my $f$ with no restriction on $a$ and on $b$, so in principle this is a stronger result than what you wanted (you only wanted to find the minimum of $f$ for pairs $(a,b)$ which come from a triangle). – Mariano Suárez-Alvarez Mar 26 '12 at 2:44
What if the minimum is at a boundary point, and not in the interior of the regions? – marty cohen Mar 26 '12 at 3:09
I showed that the unrestricted minimum is $-1/\sqrt{3}$, so the restricted minimum (to whatever restriction you may want to consider) is at least that. This is enough to prove the inequality in the question. – Mariano Suárez-Alvarez Mar 26 '12 at 3:11
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## Decoding Fractions: Finding Decimal Equivalent Without a Chart
In the meticulous world of metalworking, precision isn't just a goal; it's a necessity. This precision often involves converting fractions to their decimal equivalents, a fundamental skill that's as essential as it is straightforward. Let's jump into the process of how to find an equivalent decimal, and learn an easy way to turn a fraction into a decimal.
## Understanding Fractions in Metalworking
First off, let’s remind ourselves what a fraction represents. In the simplest terms, a fraction is a division problem. It's a way to express parts of a whole. In our line of work, fractions are everywhere – from measurements on blueprints to adjustments on machinery, etc.
## The Forward Slash: More Than Just a Symbol
In fractions, the forward slash (/) is more than just a separator; it's a division symbol. This is where many of the calculations in metalworking hinge. Recognizing this slash as a sign of division is the first step in translating fractions into decimals.
## Divide to Conquer: The Core Principle
To find the decimal equivalent of a fraction, you divide the numerator (the number to the left of the slash) by the denominator (the number on the right of the slash). For example, let's take a common fraction in metalworking: 1/2. This fraction is essentially telling us to divide 1 by 2 (1 ÷ 2). Doing this division gives us 0.5, which is the decimal equivalent of 1/2.
Fractions, like 5/8 or 7/16 work the same way. The same principle applies – divide the top number by the bottom number. For example, for 5/8, 5 divided by 8 is .625, and for 7/16, 7 divided by 16 gives us 0.4375.
## Why This Conversion Matters
In metalworking, precision is a top priority. Working with decimals can often be more practical and accurate, especially when dealing with digital measuring tools, computers, or programable machinery. Understanding how to quickly convert fractions to decimals allows for seamless transitions between different measurement systems, ensuring accuracy in every cut. In addition, utilizing these conversions can help you confidently confirm you've selected the appropriate size equivalent of the drill bits, reamers, or taps utilized for your holemaking and threading applications.
## Save time & convert sizes AT a glance!
We've compiled a decimal equivalent chart that includes the most popular fractional, metric, wire gage and letter sizes used in metalworking and holemaking applicatiions. Download the chart and quickly determine the equivalent size of your cutting tools, allowing you to avoid time-consuming math and potential errors.
Our comprehensive decimal equivalent chart lists:
• Fractional Sizes
• Metric Sizes
• Wire Gage Sizes
• Letter Sizes
• Tap & Drill Sizes
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# Elementary Number Theory Problems 3.3 Solution (David M. Burton's 7th Edition)
My solutions for Burton's Elementary Number Theory Problems 3.3 (7th Edition)
## Background
These are my solutions while I was studying number theory using this book. I will use the theorems and definitions mentioned in the book.
If you don't have the book or are unsure which theorem or definition I am using here, you can check out this article, where I listed all theorems, corollaries, and definitions by following the book's order.
If you haven't heard of this book and are interested in learning number theory, I strongly recommend it since it is quite friendly for beginners. You can check out the book by this link:
I will only use theorems or facts that are proved before this chapter. So, you will not see that I quote theorems or facts from the later chapters.
## All Problems in 3.2 (p. 49 - p. 50)
### Q1
Verify that the integers $1949$ and $1951$ are twin primes.
Solution
### Q2
(a) If $1$ is added to a product of twin primes, prove that a perfect square is always obtained.
(b) Show that the sum of twin primes $p$ and $p + 2$ is divisible by $12$, provided that $p > 3$.
Solution
### Q3
Find all pairs of primes $p$ and $q$ satisfying $p - q = 3$.
Solution
### Q4
Sylvester ($1896$) rephrased the Goldbach conjecture: Every even integer $2n$ greater than $4$ is the sum of two primes, one larger than $n/2$ and the other less than $3n/2$. Verify this version of the conjecture for all even integers between $6$ and $76$.
Solution
### Q5
In 1752, Goldbach submitted the following conjecture to Euler: Every odd integer can be written in the form $p + 2a^2$, where $p$ is either a prime or $1$ and $a \geq 0$. Show that the integer $5777$ refutes this conjecture.
Solution
### Q6
Prove that the Goldbach conjecture that every even integer greater than $2$ is the sum of two primes is equivalent to the statement that every integer greater than $5$ is the sum of three primes.
[Hint: If $2n - 2 =p_{1}+ p_{2}$, then $2n =p_{1}+ p_{2} + 2$ and $2n + 1 =p_{1}+ p_{2} + 3$.]
Solution
### Q7
A conjecture of Lagrange ($1775$) asserts that every odd integer greater than $5$ can be written as a sum $p_{1} + 2p_{2}$, where $p_{1}$, $p_{2}$ are both primes. Confirm this for all odd integers through $75$.
Solution
### Q8
Given a positive integer $n$, it can be shown that there exists an even integer $a$ that is representable as the sum of two odd primes in $n$ different ways. Confirm that the integers $60$, $78$, and $84$ can be written as the sum of two primes in six, seven, and eight ways, respectively.
Solution
### Q9
(a) For $n > 3$, show that the integers $n$, $n + 2$, $n + 4$ cannot all be prime.
(b) Three integers $p$, $p + 2$, $p + 6$, which are all prime, are called a $\textit{prime-triplet}$. Find five sets of prime-triplets.
Solution
### Q10
Establish that the sequence $$(n + 1)! - 2, (n + 1)! - 3, ... , (n + 1)! - (n + 1)$$ produces $n$ consecutive composite integers for $n > 2$.
Solution
### Q11
Find the smallest positive integer $n$ for which the function $f(n) = n^2 + n + 17$ is composite. Do the same for the functions $g(n) = n^2 + 21n + 1$ and $h(n) = 3n^2 + 3n + 23$.
Solution
### Q12
Let $p_{n}$ denote the $n$th prime number. For $n \geq 3$, prove that $p_{n+3}^{2} < p_{n}p_{n+1}p_{n+2}$.
[Hint: Note that $p_{n+3}^{2} < 4p_{n+2}^2 <8p_{n+1}p_{n+2}$.]
Solution
### Q13
Apply the same method of proof as in Theorem 3.6 to show that there are infinitely many primes of the form $6n + 5$.
Solution
### Q14
Find a prime divisor of the integer $N = 4(3 \cdot 7 \cdot 11) - 1$ of the form $4n + 3$. Do the same for $N = 4(3 \cdot 7 \cdot 11 \cdot 15) - 1$.
Solution
### Q15
Another unanswered question is whether there exists an infinite number of sets of five consecutive odd integers of which four are primes. Find five such sets of integers.
Solution
### Q16
Let the sequence of primes, with $1$ adjoined, be denoted by $p_{0} = 1, p_{1} = 2, p_{2} = 3, p_{3} = 5, ....$ For each $n \geq 1$, it is known that there exists a suitable choice of coefficients $\epsilon_{k} = \pm 1$ such that
$$$$\begin{split} p_{2n} = p_{2n - 1} + \sum_{k = 0}^{2n - 2} \epsilon_{k}p_{k} \qquad p_{2n + 1} = 2p_{2n} + \sum_{k = 0}^{2n - 1} \epsilon_{k}p_{k} \end{split} \nonumber$$$$
To illustrate:
$$13 = 1 + 2 - 3 - 5 + 7 + 11$$
and
$$17 = 1 + 2 - 3 - 5 + 7 - 11 + 2 \cdot 13$$
Determine similar representations for the primes $23, 29, 31,$ and $37$.
Solution
### Q17
In $1848$, de Polignac claimed that every odd integer is the sum of a prime and a power of $2$. For example, $55 = 47 + 2^{3} = 23 + 2^{5}$. Show that the integers $509$ and $877$ discredit this claim.
Solution
### Q18
(a) If $p$ is a prime and $p \not \mid b$, prove that in the arithmetic progression $$a, a + b, a + 2b, a + 3b, ...$$ every $pth$ term is divisible by $p$.
[Hint: Because $gcd(p, b) = 1$, there exist integers $r$ and $s$ satisfying $pr+ bs = 1$.
Put $n_{k} = kp - as$ for $k= 1, 2, ...$ and show that $p \mid (a+ n_{k}b)$.]
(b) From part (a), conclude that if $b$ is an odd integer, then every other term in the indicated progression is even.
Solution
### Q19
In $1950$, it was proved that any integer $n > 9$ can be written as a sum of distinct odd primes. Express the integers $25$, $69$, $81$, and $125$ in this fashion.
Solution
### Q20
If $p$ and $p^{2} + 8$ are both prime numbers, prove that $p^{3} + 4$ is also prime.
Solution
### Q21
(a) For any integer $k > 0$, establish that the arithmetic progression $$a + b, a + 2b, a + 3b, ...$$ where $gcd(a, b) = 1$, contains $k$ consecutive terms that are composite.
[Hint: Put $n =(a+ b)(a + 2b) \cdots (a+ kb)$ and consider the $k$ terms $a+ (n + 1)b, a+ (n + 2)b, ... , a+ (n + k)b$.]
(b) Find five consecutive composite terms in the arithmetic progression $$6, 11, 16,21,26,31,36, ...$$
Solution
### Q22
Show that $13$ is the largest prime that can divide two successive integers of the form $n^{2} + 3$.
Solution
### Q23
(a) The arithmetic mean of the twin primes $5$ and $7$ is the triangular number $6$. Are there any other twin primes with a triangular mean?
(b) The arithmetic mean of the twin primes $3$ and $5$ is the perfect square $4$. Are there any other twin primes with a square mean?
Solution
### Q24
Determine all twin primes $p$ and $q = p + 2$ for which $pq - 2$ is also prime.
Solution
### Q25
Let $p_{n}$ denote the $n$th prime. For $n > 3$, show that $$p_{n} < p_{1} + p_{2} + \cdots + p_{n-1}$$
[Hint: Use induction and the Bertrand conjecture.]
Solution
### Q26
Verify the following:
(a) There exist infinitely many primes ending in $33$, such as $233$, $433$, $733$, $1033, ....$
[Hint: Apply Dirichlet's theorem.]
(b) There exist infinitely many primes that do not belong to any pair of twin primes.
[Hint: Consider the arithmetic progression $21k + 5$ for $k= 1, 2, ....$ ]
(c) There exists a prime ending in as many consecutive $1$'s as desired.
[Hint: To obtain a prime ending in $n$ consecutive $1$'s, consider the arithmetic progression $10^{n}k +R_{n}$ for $k= 1, 2, ....$ ]
(d) There exist infinitely many primes that contain but do not end in the block of digits $123456789$.
[Hint: Consider the arithmetic progression $10^{n}k + 1234567891$ for $k= 1, 2, ....$]
Solution
### Q27
Prove that for every $n \geq 2$ there exists a prime $p$ with $p \leq n < 2p$.
[Hint: In the case where $n = 2k + 1$, then by the Bertrand conjecture there exists a prime $p$ such that $k < p < 2k$.]
Solution
### Q28
(a) If $n > 1$, show that $n!$ is never a perfect square.
(b) Find the values of $n \geq 1$ for which $$n! + (n + 1)! + (n + 2)!$$ is a perfect square.
[Hint: Note that $n! + (n + 1)! + (n + 2)! = n!(n + 2)^{2}$ .]
Solution
MathNumber TheorySolution
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# NCERT Solutions Of Chapter 6 Exercise 6.3
NCERT solutions of chapter 6 exercise 6.3 is going to help all the maths students of class 12,in this exercise 6.3 of the maths chapter class 12 -Application of Derivatives you will learn solutions of the questions related to the slope of the curve. All questions are explained by an expert of the subject as per the CBSE norms. NCERT Solutions of the exercise 6.3 Application of Derivatives will give an idea to all students about the concept and the way of solving the questions of exercise 6.3 of chapter 6 which is required to solve all the questions of Application of Derivatives in the Class 12 CBSE Board exam.
You can also study
Exercise 6.1 – Application of Derivatives
Exercise 6.2 – Application of Derivatives
Exercise 6.3 – Application of Derivatives
Exercise 6.4- Application of Derivatives
## NCERT Solutions Of Chapter 6 Exercise 6.3
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Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
Q1.Find the slope of the tangent to the curve $y=3x^{^{4}}-4x$ at x = 4.
Ans. The slope of the tangent to the curve $y=3x^{^{4}}-4x$ is
$\frac{dy}{dx}=12x^{3}-4$
If x = 4, the slope =
$12(4)^{3}-4= 768-4= 764$
Q2.Find the slope of the tangent to the curve $y=\frac{x-1}{x-2}, x\neq 2\, at\, x = 10.$
Ans. We are given the equation of the curve
$y =\frac{x-1}{x-2}$
The slope of the given curve is the differentiation of y with respect to x as follows
$\frac{dy}{dx}= \frac{d}{dx}\left ( \frac{x-1}{x-2} \right )$
$\frac{dy}{dx}= \frac{\left ( x-2 \right )\frac{d}{dx}\left ( x-1 \right )-\left ( x-1 \right )\frac{d}{dx}\left ( x-2 \right )}{\left ( x-2 \right )^{2}}$
$=\frac{x-2-x+1}{\left ( x-2 \right )^{2}}$
$=-\frac{1}{\left ( x-2 \right )^{2}}$
If x = 10, the slope =
$=\frac{-1}{(10-2)^{2}}=-\frac{1}{(64)}$
Q3.Find the slope of the tangent to the curve y = x³ – x + 1 at the point whose x-coordinate is 2.
Ans. The slope of the tangent to the curve
$y=x^{3}-x + 1$ is
y = x³ – x + 1
$\frac{dy}{dx}=\frac{d}{dx}\left ( x^{3}-x+1 \right )$
= 3x²- 1
Slope of the curve at x = 2
3 × 2²- 1
12 -1 = 11
Q4.Find the slope of the tangent to the curve y = x3 – 3x + 2 at the point whose x-coordinate is 3
Ans. We are given the curve y = x3 – 3x + 2
y = x3 – 3x + 2
$\frac{dy}{dx}=\frac{d}{dx}\left ( x^{3}-3x +2\right )$
= 3x² – 3
The slope of the curve at x = 3
3 × 3² – 3 = 27 – 3 = 24
$\left [ \frac{dy}{dx} \right ]_{x=3}=24$
Q5.Find slope of the normal to the curve x = a cos³θ, y = a sin³θ at θ = π/4.
Ans. We are given
x = a cos³θ, y = a sin³θ
$\frac{dx}{d\Theta }=-3a\: cos^{2}\Theta.sin\Theta ,\frac{dy}{d\Theta } =3a\: sin^{2}\Theta .cos\Theta$
$\frac{dy/d\Theta }{dx/d\Theta }=\frac{3a\: sin^{2}\Theta .cos\Theta }{-3a\: cos^{2}\Theta .sin\Theta }$
$\frac{dy}{dx}=\frac{-sin\Theta }{cos\Theta }$
$\frac{dy}{dx}=-tan\Theta$
$\left [ \frac{dy}{dx} \right ]_{\Theta =\frac{\pi }{4}}=-tan\frac{\pi }{4}=-1$
The slope of the curve is = -1
Let the slope of the normal to the curve = m
We know the relation
the slope of the curve × m = -1
-1 × m = -1
m = 1
Q6. Find the slope of the normal to the curve x = 1 – a sinθ, y = b cos”θ at θ = π/2.
Ans. We are given the curve
x = 1 – a sinθ, y = b cos”θ
$\frac{dx}{d\Theta }=\frac{d}{d\Theta }\left ( 1-a sin\Theta \right ),\: \: \frac{dy}{d\Theta }=\frac{d}{d\Theta }\left ( b\: cos^{2}\Theta \right )$
$\frac{dx}{d\Theta }= a\: cos\Theta ,\: \: \frac{dy}{d\Theta }=-2b\: cos\Theta .sin\Theta$
$\frac{dy/d\Theta }{dx/d\Theta }=\frac{-2b\: cos\Theta .sin\Theta }{-a\: cos\Theta }$
$\frac{dy}{dx}=\frac{2b}{a}sin\Theta$
The slope of the given curve is
$=\frac{2b}{a}sin\Theta$
The slope of the curve at θ = π/2
$\left [ \frac{dy}{dx} \right ]_{\Theta =\frac{\pi }{2}}=\frac{2b}{a}sin\frac{\pi }{2}=\frac{2b}{a}$
So, slope of the curve, m = 2b/a
Let slope of the normal to the curve is = M
We have
$m\times M= -1$
$\frac{2b}{a}\times M= -1$
$M= -\frac{a}{2b}$
Hence normal to the given curve is = -a/2b
### NCERT Solutions Of Chapter 6 Exercise 6.3
Q7. Find points at which the tangent to the curve y = x3 – 3x2 – 9x + 7 is parallel to the x-axis.
Ans. The given curve is
y = x3 – 3x2 – 9x + 7
Slope of the tangent dy/dx is evaluated as follows
$\frac{dy}{dx}=\frac{d}{dx}\left ( x^{3}-3x^{2} -9x+7\right )$
$\frac{dy}{dx}=3x^{2}-6x-9$
Since the tangent to the curve is parallel x-axis
The equation of x-axis is
y = 0
$\therefore \frac{dy}{dx}=0$
It is given to us that the tangent to the curve is parallel to each other
So, 3x² – 6x – 9 = 0
x² – 2x – 3 = 0
x² – 3x + x – 3 = 0
x( x – 3) + ( x – 3) = 0
( x – 3)(x + 1) = 0
x = 3, -1
Hence at x = 3, -1 the tangent to the curve is parallel to x-axis
Q8.Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2,0) and (4,4).
Ans.
Calculating the slope of tagent to the given curve, y = (x – 2)²
y = (x – 2)²
$\frac{dy}{dx}=\frac{d}{dx}\left ( x -2 \right )^{2}$
$\frac{dy}{dx}=2\left ( x-2 \right )$
Slope of the chord joining the points (2,0) and (4,4) = m
$m=\frac{4-0}{4-2}=2$
We are given that the tangent is parallel to the given chord of the curve, so their slopes must be equal
2(x – 2) = 2
x = 3
Putting this value of x in the equation of the curve
y = (3 – 2)² = 1
Therefore the point on the given curve is (3,1) in which tangent is parallel to the given chord.
Q9. Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x -11.
Ans. The given curve is
y = x³ – 11x + 5
Finding its slope as following
$\frac{dy}{dx}=\frac{d}{dx}\left ( x^{3}-11x +5 \right )$
= 3x² – 11
The slope of the given tangent y = x -11 is calculated as follows
$\frac{dy}{dx}=1$
The slope of the curve = Slope of the tangent
3x² – 11 = 1
x = ±2
If x = 2, then y = 2 – 11 = -9, if x = -2, then y= -2 -11 = -13
(-2,-13) does’t satisfy the equation of the curve so it does’t lie on the curve
The point (2, -9) satisfy the equation of the curve so the given tangent on the curve is at (2, -9)
## NCERT Solutions of Science and Maths for Class 9,10,11 and 12
### NCERT Solutions for class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
### NCERT Solutions for class 11 maths
Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
### NCERT Solutions for Class 11 Physics
Chapter 1- Physical World
chapter 3-Motion in a Straight Line
### NCERT Solutions for Class 11 Chemistry
Chapter 1-Some basic concepts of chemistry
Chapter 2- Structure of Atom
### NCERT Solutions for Class 11 Biology
Chapter 1 -Living World
### NCERT solutions for class 12 maths
Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Class 12 Maths Important Questions-Application of Integrals
Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution
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Students can go through AP Board 7th Class Maths Notes Chapter 15 Symmetry to understand and remember the concepts easily.
## AP State Board Syllabus 7th Class Maths Notes Chapter 15 Symmetry
→ Line of symmetry: The line which divides a figure into two identical parts is called the line of symmetry or axis of symmetry.
Ex: In the adjacent figure the dotted lines are the line of symmetry.
→ An object can have one or more than one lines of symmetry or axes of symmetry.
Ex: In the above figure there are two lines of symmetry.
→ If we rotate a figure, about a fixed point by a certain angle and the figure looks exactly the same as before, we say that the figure has rotational symmetry.
Ex: An equilateral triangle; a square etc.
→ The angle of turning during rotation is called the angle of rotation (or) the minimum angle rotation of a figure to get exactly the same figure as original is called the angle of rotation.
Ex: i) Angle of rotation of an equilateral triangle = 120°.
ii) Angle of rotation of a square = 90°.
→ All figures having rotational symmetry of order 1, can be rotated completely through 360° to come back to their original position. So we say that an object has rotational symmetry only when the order of symmetry is more than 1.
Eg: The order of rotational symmetry for an equilateral triangle is 3.
ii) For a square is 4.
→ Some shapes only have line symmetry and some have only rotational symmetry and some have both. Squares, equilateral triangles and circles have both line symmetry and rotational symmetry.
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1. ## intersections
Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point
2. Originally Posted by chibiusagi
Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point
we can find how many zeros the function has by simply solving for them.
set $\displaystyle (x - 2) \left( x^2 + 2x + 6 \right) = 0$
$\displaystyle \Rightarrow x - 2 = 0$ or $\displaystyle x^2 + 2x + 6 = 0$
$\displaystyle \Rightarrow x = 2$ or $\displaystyle x = \frac {-2 \pm \sqrt{-20}}2$
clearly the quadratic has no real roots, thus the only root is $\displaystyle x = 2$
to find the tangent line, use the point-slope form.
$\displaystyle y - y_1 = m(x - x_1)$
here, $\displaystyle m = f'(2)$ and $\displaystyle (x_1,y_1) = (2,0)$
3. Originally Posted by chibiusagi
Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point
$\displaystyle y = (x - 2)(x^2 + 2x + 6)$
When this crosses the x-axis we have y = 0. So to find these x values:
$\displaystyle (x - 2)(x^2 + 2x + 6) = 0$
So either
$\displaystyle x - 2 = 0 \implies x = 2$
or
$\displaystyle x^2 + 2x + 6 = 0$
Note, however, that the quadratic factor here has no real zeros. So no real zeros are obtained by solving this equation.
Thus the curve only crosses the x-axis once.
Now to find the equation of the tangent.
$\displaystyle y^{\prime}(x) = 3x^2 + 2$
So at x = 2 the slope of the tangent to the curve is $\displaystyle y^{\prime}(2) = 14$.
So we need the equation of a line with a slope of 14 that passes through the point (2, 0).
$\displaystyle y = 14x + b$
$\displaystyle 0 = 14 \cdot 2 + b \implies b = -28$
So the tangent line at (2, 0) is $\displaystyle y = 14x - 28$.
-Dan
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# Find the equation of the line which passes through the point
Question:
Find the equation of the line which passes through the point (22, -6) and whose intercept on the x-axis exceeds the intercept on the y-axis by 5.
Solution:
To Find:The equation of the line that passes through the point (22, -6) and intercepts on the x-axis exceeds the intercept on the y-axis by 5.
Given : let x-intercept be a and $y$-intercept be $b$.
According to the question : $a=b+5$
Formula used: And the given point satisfies the equation of the line, so
$\frac{x}{a}+\frac{y}{b}=1$
$\frac{22}{b+5}+\frac{-6}{b}=1$
$22 b-6 b-30=b^{2}+5 b$
$11 b-30=b^{2}$
$b^{2}-11 b+30=0$
$b^{2}-6 b-5 b+30=0$
$b(b-6)-5(b-6)=0$
$(b-5)(b-6)=0$
The values are $\mathrm{b}=5, \mathrm{~b}=6$
When $\mathrm{b}=5$ then $\mathrm{a}=10$
and $b=6$ then $a=11$
case 1 : when $b=5$ and $a=10$
Equation of the line: $\frac{x}{a}+\frac{y}{b}=1$
$\frac{x}{10}+\frac{y}{5}=1$
$\frac{x+2 y}{10}=1$
Hence, x + 2y = 10 is the required equation of the line.
case 2 : when b=6 and a=11
Equation of the line : $\frac{x}{a}+\frac{y}{b}=1$
$\frac{x}{11}+\frac{y}{6}=1$
$\frac{6 x+11 y}{66}=1$
Hence, 6x + 11y = 66 is the required equation of the line.
Therefore, $x+2 y=10$ is the required equation of the line when $b=5$ and $a=10 .$ And $6 x+$ $11 \mathrm{y}=66$ is the required equation of the line when $\mathrm{b}=6$ and $\mathrm{a}=11$.
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# Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expressions And Identities – Exercise 6.7
### (i) (x + 4) (x + 7)
Solution:
By simplifying the given expression, we get
x (x + 7) + 4 (x + 7)
x2 + 7x + 4x + 28
x2+ 11x + 28
### (ii) (x – 11) (x + 4)
Solution:
By simplifying the given expression, we get
x (x + 4) – 11 (x + 4)
x2 + 4x – 11x – 44
x2 – 7x – 44
### (iii) (x + 7) (x – 5)
Solution:
By simplifying the given expression, we get
x (x – 5) + 7 (x – 5)
x2 – 5x + 7x – 35
x2 + 2x – 35
### (iv) (x – 3) (x – 2)
Solution:
By simplifying the given expression, we get
x (x – 2) – 3 (x – 2)
x2 – 2x – 3x + 6
x2 – 5x + 6
### (v) (y2 – 4) (y2 – 3)
Solution:
By simplifying the given expression, we get
y2 (y2– 3) – 4 (y2 – 3)
y4 – 3y2 – 4y2 + 12
y4 – 7y2 + 12
### (vi) (x + 4/3) (x + 3/4)
Solution:
By simplifying the given expression, we get
x (x + 3/4) + 4/3 (x + 3/4)
x2 + 3x/4 + 4x/3 + 12/12
x2 + 3x/4 + 4x/3 + 1
x2 + 25x/12 + 1
### (vii) (3x + 5) (3x + 11)
Solution:
By simplifying the given expression, we get
3x (3x + 11) + 5 (3x + 11)
9x2 + 33x + 15x + 55
9x2 + 48x + 55
### (viii) (2x2 – 3) (2x2+ 5)
Solution:
By simplifying the given expression, we get
2x2(2x2 + 5) – 3 (2x2 + 5)
4x4 + 10x2 – 6x2 – 15
4x4 + 4x2 – 15
### (ix) (z2 + 2) (z2– 3)
Solution:
By simplifying the given expression, we get
z2 (z2 – 3) + 2 (z2 – 3)
z4 – 3z2 + 2z2 – 6
z4– z2 – 6
### (x) (3x – 4y) (2x – 4y)
Solution:
By simplifying the given expression, we get
3x (2x – 4y) – 4y (2x – 4y)
6x2 – 12xy – 8xy + 16y2
6x2 – 20xy + 16y2
### (xi) (3x2 – 4xy) (3x2 – 3xy)
Solution:
By simplifying the given expression, we get
3x2 (3x2 – 3xy) – 4xy (3x2 – 3xy)
9x4 – 9x3y – 12x3y + 12x2y2
9x4 – 21x3y + 12x2y2
### (xii) (x + 1/5) (x + 5)
Solution:
By simplifying the given expression, we get
x (x + 1/5) + 5 (x + 1/5)
x2 + x/5 + 5x + 1
x2 + 26/5x + 1
### (xiii) (z + 3/4) (z + 4/3)
Solution:
By simplifying the given expression, we get
z (z + 4/3) + 3/4 (z + 4/3)
z2 + 4/3z + 3/4z + 12/12
z2+ 4/3z + 3/4z + 1
z2 + 25/12z + 1
### (xiv) (x2+ 4) (x2 + 9)
Solution:
By simplifying the given expression, we get
x2 (x2 + 9) + 4 (x2+ 9)
x4 + 9x2 + 4x2 + 36
x4 + 13x2+ 36
### (xv) (y2 + 12) (y2+ 6)
Solution:
By simplifying the given expression, we get
y2 (y2+ 6) + 12 (y2 + 6)
y4+ 6y2 + 12y2 + 72
y4 + 18y2 + 72
### (xvi) (y2 + 5/7) (y2 – 14/5)
Solution:
By simplifying the given expression, we get
y2 (y2 – 14/5) + 5/7 (y2 – 14/5)
y4 – 14/5y2 + 5/7y2 – 2
y4 – 73/35y2 – 2
### (xvii) (p2 + 16) (p2 – 1/4)
Solution:
By simplifying the given expression, we get
p2 (p2 – 1/4) + 16 (p2 – 1/4)
p4 – 1/4p2 + 16p2 – 4
p4 + 63/4p2 – 4
### (i) 102 × 106
Solution:
By simplifying the given expression, we get
102 × 106 = (100 + 2) (100 + 6)
= 100 (100 + 6) + 2 (100 + 6)
= 10000 + 600 + 200 + 12
= 10812
### (ii) 109 × 107
Solution:
By simplifying the given expression, we get
109 × 107 = (100 + 9) (100 + 7)
= 100 (100 + 7) + 9 (100 + 7)
= 10000 + 700 + 900 + 63
= 11663
### (iii) 35 × 37
Solution:
By simplifying the given expression, we get
35 × 37 = (30 + 5) (30 + 7)
= 30 (30 + 7) + 5 (30 + 7)
= 900 + 210 + 150 + 35
= 1295
### (iv) 53 × 55
Solution:
By simplifying the given expression, we get
53 × 55 = (50 + 3) (50 + 5)
= 50 (50 + 5) + 3 (50 + 5)
= 2500 + 250 + 150 + 15
= 2915
### (v) 103 × 96
Solution:
By simplifying the given expression, we get
103 × 96 = (100 + 3) (100 – 4)
= 100 (100 – 4) + 3 (100 – 4)
= 10000 – 400 + 300 – 12
= 10000 – 112
= 9888
### (vi) 34 × 36
Solution:
By simplifying the given expression, we get
34 × 36 = (30 + 4) (30 + 6)
= 30 (30 + 6) + 4 (30 + 6)
= 900 + 180 + 120 + 24
= 1224
### (vii) 994 × 1006
Solution:
By simplifying the given expression, we get
994 × 1006 = (1000 – 6) (1000 + 6)
= 1000 (1000 + 6) – 6 (1000 + 6)
= 1000000 + 6000 – 6000 – 36
= 999964
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# What Is the Square Root of 72?
A perfect square is a number with an integer as its square root. This means that it’s a product of an integer with itself. In decimal representation, the square root of 72 is 8.485 when rounded to four significant figures. Because it’s not a square number or perfect square, you can solve it quickly with a scientific calculator. If this option is unavailable, there are a couple of ways you can use to arrive at a solution. If this alternative can’t be used, there are a couple of tricks you can use to arrive at a solution.
Multiplying Using Two Reference Numbers
To find the square root of 72, you’ll need to know the two closest perfect squares surrounding the number 72. In this case, 8 and 9 are our numbers. The number 72 falls between the squares of 8 and 9, which are 64 and 81, respectively.
Thereafter, divide 72 by 8 or 9. The result will be 9 or 8, depending on the number chosen to divide it with. Next, find the mean of that result and the original square used, which is (9+8)/2=8.5. Finally, repeat the previous two steps until you attain the desired accuracy.
Expressing in simplified radical form entails simplifying a radical till no more square roots, cube roots, fourth roots, and so on can be found. It also involves the removal of radicals in the denominator of a fraction. The square root of 72 can be simplified by breaking the radicand up into a product of known factors. Begin by finding the highest square that divides into 72 evenly. In this case, 36 is the number. Thus, 72 can be expressed as 36 x 2, then proceed as follows:
√ 72 = √2x√36 = √2x√62 = 6√ 2
Cross Multiplication
The method is used to calculate the exact answer for a square root. For even numbers of digits, you need to multiply the first digit by the last digit, the second by the second last, and so on, until all the digits have been multiplied. Find the sum and double the total. Use the same process for an odd number of digits till you reach the middle digit. Find the sum of the answers and double the total. Next, square the middle digit and add it to the total.
Long Division Method
Start by grouping the digits into pairs, beginning with the units digit. The pair formed and remaining digit (if any) is called a period. After establishing the largest square number equal to or slightly less than the first period, use it as both the divisor and the quotient. Multiply the divisor and quotient then subtract it from the first period. Then, write the next period to the right of the remainder. As a result, a new dividend is formed. This becomes the new dividend.
Now, the new divisor is found by using two times the quotient and adding with it an ideal digit that acts as the next digit of the quotient. The number is chosen, such as the product of the digit, and the new divisor’s value equal to or slightly less than the new dividend. Finally, repeat the second, third, and fourth steps until all the periods have been incorporated or taken up. The quotient obtained is the required square root of the given number.
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# How to calculate 36 divided by 9 using long division?
36 ÷ 9
=
4
Division is a fundamental arithmetic operation where we calculate how many times a number (divisor or denominator) can fit into another number (dividend or numerator). In this case, we are dividing 36 (the dividend) by 9 (the divisor).
There are three distinct methods to convey the same information: in decimal, fractional, and percentage formats:
• 36 divided by 9 in decimal = 4
• 36 divided by 9 in fraction = 36/9
• 36 divided by 9 in percentage = 400%
## What is the Quotient and Remainder of 36 divided by 9?
The quotient is calculated by dividing the dividend by the divisor, and the remainder is what's left over if the division doesn't result in a whole number. In this case, however, since 36 is a multiple of 9, there should be no remainder.
The quotient of 36 divided by 9 is 4, and the remainder is 0. Thus,
### 36 ÷ 9 = 4 R 0
When you divide Thirty Six by Nine, the quotient is Four, and the remainder is Zero.
## Verdict
The division of 36 by 9 results in a quotient of 4 and a remainder of 0, meaning 9 goes into 36 Four times with 0 left over. Understanding this division process is crucial in both basic arithmetic and real-life applications where division is used, such as in financial calculations, data analysis, and everyday problem-solving.
## Random Division Problems?
No worries, we got your back! Tell us what are you brainstorming with and we will bring correct answers to you.
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### How do we differentiate between divisor and dividend?
A dividend is a number we divide, while a divisor is a number by which we divide. Divisor comes on second, followed by the dividend that we write first.
For instance, if you have 12 candies and want to distribute them among 3 children, the equation will be 12 ÷ 3. You will put 12 first because this is the number being divided. So here, 12 is a dividend. On the other hand, 3 is written after 12, and it is the number with which we are dividing 12. Hence, 3 is a divisor.
### Which formula is used to find a divisor?
There are two formulas used to find a divisor.
The first one is: Divisor = Dividend ÷ Quotient. This formula is used to find a divisor when the remainder is 0.
Second is: Divisor = (Dividend – Remainder) /Quotient. This formula is used when the remainder is not 0.
### Is there a possibility of a number having the same divisor?
Yes, there is. Every number can be divided by itself, leaving 1 as the quotient. So, it would not be wrong to say that all the numbers can have the same divisors.
Let’s take the example of 5. If we divide 5 by 5 (5 ÷ 5), then 5 will be the divisor of 5. And ultimately, 1 will be the quotient.
### What is the difference between a divisor and a factor?
A divisor is a number with which we can divide any number. However, a factor is different from a divisor. It is the number that can be divided with another number leaving no remainder. All factors are divisors, but not all divisors are factors.
### Is it possible to do division by repeated subtraction?
Fortunately yes. You can do division by repeated subtraction. In repeated subtraction, we continuously subtract a number from a bigger number. It continues until we get the 0 or any other number less than the actual number as a remainder.
However, it can be a lengthy process, so we can use division as a shortcut.
### Can I check the remainder and the quotient in a division problem?
Yes, you can quickly check the remainder and quotient in a division problem by using this relationship:
Dividend = Divisor x Quotient + Remainder
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# Transitive Property of Congruence & Substitution Property of Equality, Vertical Angles, Geometry
substitution property
This is a topic that many people are looking for. thevoltreport.com is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, thevoltreport.com would like to introduce to you Transitive Property of Congruence & Substitution Property of Equality, Vertical Angles, Geometry. Following along are instructions in the video below:
in this video were going to talk about the transitive property the substitution property and also vertical angles so heres the general idea of the transitive property if angles are congruent to the same angle then theyre congruent to each other so for instance lets say if angle 1 is congruent to angle 2 and if angle 3 is congruent to angle 2 then we can make the statement that angle 1 is congruent to angle 3 so now lets look at the sentence carefully if angles that is angle 1 and angle 3 so if lets say two angles are congruent to the same angle that is angle 1 and angle 3 are both congruent to angle 2 then the original two angles are congruent to each other the original two angles being 1 and 3 so 1 in 3 are congruent to each other and thats the basic idea of the transitive property the next property that you need to be familiar with is the substitution property so in a substitution property it just basically replaces something youre substituting a variable with another variable or an angle from that angle so lets say if a is equal to B and if B is equal to C well since a is equal to B I can replace B with a so Im substituting beat with a so I could say that a is equal to C so thats the substitution property but now you could use a transitive property to get the same result but the way in
which you go about doing it is a little bit different but the result is very similar so lets say that a is equal to B and C is equal to B then I can make the statement that a is equal to C so notice that I have lets say that these letters represent angles so angle a and angle C we have two angles that are congruent to the same angle the same angle is angle B a and C are both congruent to angle B so therefore I could say that the two angles angle a and angle C are congruent to each other so its really the way in which you use it the result is the same just remember if you want to use the substitution property youre replacing something youre substituting a variable with another variable if you want to use a transitive property youre basically linked in two equations to another equation now lets move on our discussion into vertical angles whenever two lines intersect they form two pairs of vertical angles lets call this angle one two three and four so the first pair of vertical angles are angle one and three I need to know that vertical angles are congruent so angle 1 is equal to angle 3 angles 2 and 4 are opposite to each other so those are vertical angles so angle 2 is congruent to angle 4 now it turns out that 2 & 3 are supplementary they form a linear pair so angle 2 plus angle
3 adds up to 180 and the same is true for 1 & 4 1 in 4 forms Alinea pair 1 in 2 forms Eleni pair and also 3 & 4 forms alumina pair but our focus is on vertical angles so what exactly are vertical angles there are two angles two angles are vertical angles if the race form in the sides of the other are opposite race so what does that mean well first need to know one whats an opposite rain to clean your rays to have a common endpoint and extend in different directions are opposite race so to illustrate that with draw a line and lets lets put some points on this line lets call this a B and C so ray ba which starts from B and points towards a is congruent or not congruent but its opposite to Ray BC BC starts with the common endpoint b and extends in the other direction so ba and BC are collinear rays because they exist on the same line but they share a common endpoint and they extend in opposite directions BC extends towards the right and the a extends towards the left so those are opposite race now in the case of vertical angles lets redraw so lets call this point a b c d and e so angle 1 and angle 3 are vertical angles angle 1 is formed from Ray CA which is opposite to Ray C II so Im just gonna put angle 1 and angle 3 now angle 1
is also formed two by Ray CB which is opposite to Ray CD so this is Ray CA thats opposite to see and Ray CD is opposite to C V so therefore the vertical angles are formed whenever you have raised forming the sides of those angles and it has to be opposite race so angle 1 and angle 3 are vertical angles now lets work on an example problem so consider the diagram lets call this angle 2 and angle 3 angle 2 is equal to x squared plus 6 angle 3 is equal to 7x plus 14 so given this information what is the measure of angle 2 now we know that vertical angles are congruent so therefore we could say that angle 2 is equal to angle 3 and value and that we could substitute angle 2 with x squared plus 6 unless you place angle 3 with 7x plus 14 our goal is to find the value of x unless we do that we could find the measure of angle 2 now what we have is in a quadratic equation so everything on the right side lets move it to the left so its going to be x squared minus 7x and we got to subtract 14 from both sides so 6 minus 14 is negative 8 and that we could factor this expression so what two numbers multiply it to negative 8 but add to negative 7 so this has to be negative 8 and 1 negative 8 plus 1 adds up to negative
its a factor its going to be X minus 8 times X plus 1 and thats equal to 0 so at this point we need to set each factor equal to zero so if we set X minus 8 equal to 0 X is equal to 8 and if we set X plus 1 equal to 0 X will be negative 1 now both answers may work out so lets try each one so lets start with X equals eight lets find the measure of angle two by plugging it into this expression so its eight squared plus 6 + 8 squared is 64 64 plus 6 is 70 so angle two is equal to 70 and it should be the same as angle three so if we plug it into seven X plus 14 we should get the same answer so seven times eight is fifty-six and 56 plus 14 is 70 so one possible answer is 70 degrees now lets find the other possible answer when X is equal to negative one lets make sure we get the same value so negative one squared plus six negative one squared thats negative one times negative one thats positive one and one plus 6 is 7 now angle three if we replace X with negative 1 7 times negative 1 is negative 7 and negative 7 plus 14 is positive 7 so the second possible answer that fits the equation is 7 degrees so angle 2 could be 70 degrees or 7 degrees both answers are acceptable in this example
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# SOLVING ABSOLUTE VALUE INEQUALITIES
The general form of an absolute value inequality is
|ax + b| k
or
|ax + b| ≥ k
Method 1 : (Less Than or Equal to)
Solve the absolute value inequality given below
|x + 2| ≤ 3
Solution :
We can solve the absolute value inequality |x + 2| ≤ 3 as shown below.
Let us graph the solution of the first branch x ≤ 1.
Let us graph the solution of the second branch x ≥ -5.
If we combine the above two graphs, we will get a graph as given below.
From the above graph, the solution for |x + 2| ≤ 3 is
-5 ≤ x ≤ 1
Method 2 : (Greater Than or Equal to)
Solve the absolute value inequality given below
|x - 3| ≥ 1
Solution :
We can solve the absolute value inequality |x - 3| ≥ 1 as shown below.
Let us graph the solution of the first branch x ≥ 4
Let us graph the solution of the second branch x ≤ 2
If we combine the above two graphs, we will get a graph as given below.
From the above graph, the solution for |x - 3| ≥ 1 is
(-∞, 2] U [3, +∞)
## Solved Examples
Solve the following absolute value inequalities :
Example 1 :
|2x + 1| 5
Solution :
Solve :
2x + 1 ≤ 5 or 2x + 1 ≥ -5
2x ≤ 4 or 2x ≥ -6
x ≤ 2 or x ≥ -3
Hence, the solution is
-3 ≤ x ≤ 2
Example 2 :
|3x + 5| 7
Solution :
Solve :
3x + 5 ≥ 7 or 3x + 5 ≤ -7
3x ≥ 2 or 3x ≤ -12
≥ 2/3 or x ≤ -4
Hence the solution is
(-∞, -4] U [2/3, +∞)
Example 3 :
|x - 1| + 2 5
Solution :
Solve :
|x - 1| + 2 5
Subtract 2 from each side.
|x - 1| ≤ 3
x - 1 ≤ 3 or x - 1 ≥ -3
≤ 4 or x ≥ -2
Hence the solution is
-2 ≤ x ≤ 4
Example 4 :
|2x - 3| - 5 ≥ 7
Solution :
Solve :
|2x - 3| - 5 ≥ 7
|2x - 3| ≥ 12
2x - 3 ≥ 12 or 2x - 3 ≤ -12
2x ≥ 15 or 2x ≤ -9
≥ 15/2 or x ≤ -9/2
Hence the solution is
(-∞, -9/2] U [15/2, +∞)
Example 5 :
2|x + 1| 6
Solution :
Solve :
2|x + 1| ≤ 6
Divide each side by 2.
|x + 1| ≤ 3
x + 1 ≤ 3 or x + 1 ≥ -3
≤ 2 or x ≥ -4
Hence the solution is
-4 ≤ x ≤ 2
Example 6 :
5|x - 3| 15
Solution :
Solve :
5|x - 3| 15
Divide each side by 5.
|x - 3| ≥ 3
x - 3 ≥ 3 or x - 3 ≤ -3
≥ 6 or x ≤ 0
Hence the solution is
(-∞, 0] U [6, +∞)
Example 7 :
2|x + 3| + 5 ≤ 13
Solution :
Solve :
2|x + 3| + 5 13
Subtract 5 from each side.
2|x + 3| ≤ 8
Divide each side by 2.
|x + 3| ≤ 4
x + 3 ≤ 4 or x + 3 ≥ -4
≤ 1 or x ≥ -7
Hence the solution is
-7 ≤ x ≤ 1
Example 8 :
5|x +7| - 2 ≥ 18
Solution :
Solve :
5|x +7| - 2 ≥ 18
5|x +7| ≥ 20
Divide each side by 5.
|x +7| ≥ 4
x + 7 ≥ 4 or x + 7 ≤ -4
≥ -3 or x ≤ -11
Hence the solution is
(-∞, -11] U [-3, +∞)
Example 9 :
|x + 3| < 13
Solution :
Solve :
|x + 3| < 13
x + 3 < 13 or x + 3 > -13
x < 10 or x > -16
Hence the solution is
-16 < x < 10
Example 10 :
|x +7| > 18
Solution :
Solve :
|x +7| > 18
x + 7 > 18 or x + 7 < -18
x > 11 or x < -25
Hence the solution is
(-∞, -25) U (11, +∞)
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Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period – Domain \u0026 Range
Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period – Domain \u0026 Range
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Cosecant
The cosecant function is the reciprocal of the trigonometric function sine. Cosecant is one of the main six trigonometric functions and is abbreviated as csc x or cosec x, where x is the angle. In a right-angled triangle, cosecant is equal to the ratio of the hypotenuse and perpendicular. Since it is the reciprocal of sine, we write it as csc x = 1 / sin x.
In this article, we will explore the concept of cosecant function and understand its formula. We will plot the cosecant graph using its domain and range, explore the trigonometric identities of cosec x, its values, and properties. We will solve a few examples based on the concept of csc x to understand its applications better.
1 What Is a Cosecant Function? 2 Cosecant Function Formula 3 Domain and Range of Cosec x 4 Cosecant Graph 5 Cosecant Identities 6 Properties of Cosecant Function 7 Cosecant Values 8 FAQs on Cosecant Function
## What Is a Cosecant Function?
Cosecant is the reciprocal of sine. We have six important trigonometric functions:
• Sine
• Cosine
• Tangent
• Cotangent
• Secant
• Cosecant
Since it is the reciprocal of sin x, it is defined as the ratio of the length of the hypotenuse and the length of the perpendicular of a right-angled triangle.
Consider a unit circle with points O as the center, P on the circumference, and Q inside the circle and join them as shown above. Since it is a unit circle, the length of OP is equal to the 1 unit. Consider the measure of angle POQ equal to x degrees. Then, using the cosecant definition, we have
csc x = OP/PQ
= 1/PQ
## Cosecant Function Formula
Since the cosecant function is the reciprocal of the sine function, we can write its formula as
Cosec x = 1 / sin x
Also, since the formula for sin x is written as
Sin x = Perpendicular / Hypotenuse and csc x is the reciprocal of sin x, we can write the formula for the cosecant function as
Cosec x = Hypotenuse / Perpendicular
## Domain and Range of Cosec x
As we discussed before, cosecant is the reciprocal of the sine function, that is, csc x = 1 / sin x, cosec x is defined for all real numbers except for values where sin x is equal to zero. We know that sin x is equal to for all integral multiples of pi, that is, sin x = 0 implies that that x = nπ, where n is an integer. So, cosec x is defined for all real numbers except nπ. Now, we know that the range of sin x is [-1, 1] and csc x is the reciprocal of sin x, so the range of csc x is all real numbers except (-1, 1). So the domain and range of cosecant are given by,
• Domain = R – nπ
• Range = (-∞, -1] U [+1, +∞)
## Cosecant Graph
Now that we know the domain and range of cosecant, let us now plot its graph. As we know cosec x is defined for all real numbers except for values where sin x is equal to zero. So, we have vertical asymptotes at points where csc x is not defined. Also, using the values of sin x, we have y = csc x as
• When x = 0, sin x = 0 and hence, csc x = not defined
• When x = π/6, sin x = ½, csc x = 2
• When x = π/4, sin x = 1/√2, csc x = √2
• When x = π/3, sin x = √3/2, csc x = 2/√3
• When x = π/2, sin x = 1, csc x = 1
So, by plotting the above points on a graph and joining them, we have the cosecant graph as follows:
## Cosecant Identities
Let us now go through some of the important trigonometric identities of the cosecant function. We use these identities to simplify and solve various trigonometric problems.
• 1 + cot²x = csc²x
• csc (π – x) = csc x
• csc (π/2 – x) = sec x
• csc (-x) = csc x
• csc x = 1 / sin x
• csc x = sec (π/2 – x)
## Properties of Cosecant Function
We have understood that the cosecant function is the reciprocal of the sine function and its formula. Let us now explore some of the important properties of the cosecant function to understand it better.
• The graph of cosec x is symmetrical about the x-axis.
• Cosecant Function is an odd function, that is, csc (-x) = -csc x
• The cosecant graph has no x-intercepts, that is, the graph of cosecant does not intersect the x-axis at any point.
• The value of csc x is positive when sin x is positive and it is negative when sin x is negative.
• The period of csc x is 2π radians (360 degrees).
• Cosec x is not defined at the integral multiples of π.
## Cosecant Values
To solve various trigonometric problems, we use the trigonometry table to memorize the values of the trigonometric functions which are most commonly used. The table given below shows the values of the cosecant function which help to simplify the problems and are easy to understand and remember.
X (radians) Csc x 0 Not defined π/6 2 π/4 √2 π/3 2/√3 π/2 1 3π/2 -1 2π Not defined
Important Notes on Cosecant Function
• Cosecant is the reciprocal of the sine function.
• It is equal to the ratio of hypotenuse and perpendicular of the right angles triangle.
• The cosecant graph has vertical asymptotes and has no x-intercepts.
• Cosecant Function is defined at integer multiples of π.
☛ Related Topics:
## Cosecant Function Examples
1. Example 1: Find the values of the cosecant of angles A and C of triangle right angled at B, if AB = 12, AC = 13.
Solution: We know that csc x = Hypotenuse / Opposite Side.
Let us evaluate the value of BC first.
AC² = AB² + BC²
BC = √(AC² – AB²)
= √(13² – 12²)
= √(169 – 144)
= √25
= 5 units
So, the values of cosecant of angles A and C are given by,
csc A = AC / BC
= 13/5
csc C = AC / AB
= 13/12
Answer: csc A = 13/5, csc C = 13/12
2. Example 2: Find the value of csc x if cot x = ¾ using cosecant identity.
Solution: To find the value of csc x, we will use the identity 1 + cot²x = csc²x
We have cot x = ¾
So,
1 + cot²x = csc²x
1 + (3/4)² = csc²x
1 + 9/16 = csc²x
csc²x = 25/16
csc x = √(25/16)
= 5/4
3. Example 3: Find the value of cosecant of x if sin x = 4/13.
Solution: As we know that cosec x is the reciprocal of sin x, so we have
csc x = 1 / sin x
= 1 / (4/13)
= 13/4
## FAQs on Cosecant
### What is Cosecant Function in Trigonometry?
The cosecant function is one of the important six trigonometric functions. It is the reciprocal of the sine function and hence, is equal to the ratio of Hypotenuse and Perpendicular of a right-angled triangle.
### What is Cosecant Function Formula?
The cosecant function formula can be written in two different ways:
• csc x = 1/sin x
• csc x = Hypotenuse/Perpendicular OR Hypotenuse/Opposite Side
### What is the Cosecant of an Angle?
The cosecant of an angle is equal to the ratio of the hypotenuse and opposite side of the angle in a right-angled triangle. We can also find the cosecant of angle using trigonometric identities.
### What is the Difference between Secant and Cosecant?
Secant function is the reciprocal of the cosine function and the Cosecant function is the reciprocal of the sine function. Secant is the ratio of hypotenuse and adjacent side whereas cosecant is the ratio of the Hypotenuse and Opposite Side.
### Is Csc the Inverse of Sin?
No, csc x is not the inverse of sin. It is the reciprocal of the sine function. The inverse of sin is called inverse sine or arcsin.
### What is the Reciprocal of Cosecant?
The reciprocal of the cosecant function is the sine function. It is written as sin x = 1/csc x
### What is the Period of Cosecant?
The values of the cosecant function repeat after every 2π radians, so the period of cosec x is equal to 2π radians (360 degrees).
### Why is Cosecant the Reciprocal of Sine?
We know that sin x is the ratio of perpendicular and Hypotenuse of a right-angled triangle and Cosecant is the ratio of perpendicular and Hypotenuse, so cosecant is the reciprocal of sine. Also, the product of these two functions at an angle is always equal to one. Hence, cosecant is the reciprocal of the sine function.
### Is Cosecant Function Graph Continuous?
Cosecant Graph is not continuous as it has vertical asymptotes at points where cosecant function is not defined. We know that cosec x is not defined at integer multiples of pi, so the cosecant function graph has a discontinuity at points nπ, where n is an integer.
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# 2.8: Graphing
Difficulty Level: At Grade Created by: CK-12
## Lesson Objectives
The student will:
• correctly graph data with the proper scale, units, and best fit curve.
• recognize patterns in data from a graph.
• solve for the slope of given line graphs.
## Vocabulary
• extrapolation
• graph
• interpolation
• slope
## Introduction
Scientists search for regularities and trends in data. To make it easier to find these regularities and trends, scientists often present data in either a table or a graph. The table below presents data about the pressure and volume of a sample of gas. You should note that all tables have a title and include the units of the measurements. The unit of pressure used here is atm (atmosphere).
You may note a regularity that appears in this table: as the pressure of the gas increases, its volume decreases. This regularity or trend becomes even more apparent in a graph of this data. A graph is a pictorial representation of the relationship between variables on a coordinate system.
When the data from Data Table A is plotted as a graph, the trend in the relationship between the pressure and volume of a gas sample becomes more apparent. The graph aids the scientist in the search for any regularity that may exist in the data.
## Drawing Line Graphs
Reading information from a line graph is easier and more accurate as the size of the graph increases. In the example below, the graph on the left uses only a small fraction of the space available on the graph paper. The graph on the right shows the same data but uses all the space available. If you were attempting to determine the pressure at a temperature of 110 K\begin{align*}110 \ \text{K}\end{align*}, using the graph on the left would give a less accurate result than using the graph on the right.
When you draw a line graph, you should arrange the numbers on the axes to use as much of the graph paper as you can. If the lowest temperature in your data is 100 K\begin{align*}100 \ \text{K}\end{align*} and the highest temperature in your data is 160 K\begin{align*}160 \ \text{K}\end{align*}, you should arrange for 100 K\begin{align*}100 \ \text{K}\end{align*} to be on the extreme left of your graph and 160 K\begin{align*}160 \ \text{K}\end{align*} to be on the extreme right of your graph. The creator of the graph on the left did not take this advice and did not produce a very good graph. You should also make sure that the axes on your graph are labeled and that your graph has a title.
## Reading Information from a Graph
When we draw a line graph from a set of data points, we are inferring a trend and constructing new data points between known data points. This process is called interpolation. Even though we may only have a few data points, we are estimating the values between measured points, assuming that the line connecting these data points is a good model of what we're studying.
Consider the following set of data for the solubility of KClO3 in water. Data Table B shows that there are exactly six measured data points. When the data is graphed, however, the graph maker assumes that the relationship between the temperature and the solubility exists for all points within the data range. The graph maker draws a line by interpolating the data points between the actual data points. Note that the line is not drawn by just connecting the data points in a connect-the-dot manner. Instead, the line is a smooth curve that reasonably connects the known data points.
We can now read Graph B1, shown below, for points that were not actually measured. If we wish to determine the solubility of KClO3 at 70C\begin{align*}70^\circ\text{C}\end{align*}, we follow the vertical grid line for 70C\begin{align*}70^\circ\text{C}\end{align*} up to where it touches the graphed line and then follow the horizontal grid line to the axis to read the solubility. In this case, we would read the solubility to be 30.0 g/100 mL\begin{align*}30.0 \ \text{g/100 mL}\end{align*} of H2O at 70C\begin{align*}70^\circ\text{C}\end{align*}.
There are also occasions when scientists wish to know more about points that are outside the range of measured data points. Extending the line graph beyond the ends of the original line, using the basic shape of the curve as a guide, is called extrapolation.
Suppose the graph for the solubility of potassium chlorate has been made from just three measured data points. If the actual data points for the curve were the solubility at 60C\begin{align*}60^\circ\text{C}\end{align*}, 80C\begin{align*}80^\circ\text{C}\end{align*}, and 100C\begin{align*}100^\circ\text{C}\end{align*}, the graph would be the solid line shown in Graph B2 above. If the solubility at 30C\begin{align*}30^\circ\text{C}\end{align*} was desired, we could extrapolate the curve (the dotted line) and obtain a solubility of 5.0 g/100 mL of H2O. If we check the more complete graph above (Graph B1), you can see that the solubility at 30\begin{align*}30^\circ\end{align*}C is closer to 10. g/100 mL of H2O. The reason the second graph produces such a different answer is because the real behavior of potassium chlorate in water is more complicated than the behavior suggested by the extrapolated line. For this reason, extrapolation is only acceptable for graphs where there is evidence that the relationship shown in the graph will hold true beyond the ends of the graph. Extrapolation is more dangerous that interpolation in terms of producing possibly incorrect data.
In situations where it is unreasonable to interpolate or extrapolate data points from the actual measured data points, a line graph should not be used. If it is desirable to present data in a graphic form but a line graph is not useful, a bar graph can often be used instead. Consider the data in the following table.
For this set of data, you would not plot the data on a line graph because interpolating between years does not make sense; the concept of the average yearly rainfall halfway between the years 1980 and 1981 would not make sense. Looking at the general trend exhibited by Data Table C also does not provide the slightest amount of evidence about the rainfall in 1979 or 1990. Therefore, the interpolation and extrapolation of the data in this table is not reasonable. If we wish to present this information in a graphic form, a bar graph like the one seen in Graph C would be best.
From this bar graph, you could very quickly answer questions like, “Which year was most likely a drought year for Trout Creek?” and “Which year was Trout Creek most likely to have suffered from a flood?”
## Finding the Slope of a Graph
As you may recall from algebra, the slope of the line may be determined from the graph. The slope represents the rate at which one variable is changing with respect to the other variable. For a straight-line graph, the slope is constant for the entire line, but for a non-linear graph, the slope varies at different points along the line. For a straight-line graph, the slope for all points along the line can be determined from any section of the graph. Consider the following data table and the linear graph that follows.
The relationship in this set of data is linear; in other words, the data produces a straight-line graph. The slope of this line is constant at all points on the line. The slope of a line is defined as the rise (change in vertical position) divided by the run (change in horizontal position). For a pair of data points, the coordinates of the points are identified as (x1, y1)\begin{align*}(x_1, \ y_1)\end{align*} and (x2, y2)\begin{align*}(x_2, \ y_2)\end{align*}. In this case, the data points selected are (40C, 65 mL)\begin{align*}(40^\circ\text{C}, \ 65 \ \text{mL})\end{align*} and (100C, 80 mL)\begin{align*}(100^\circ\text{C}, \ 80 \ \text{mL})\end{align*}. The slope can then be calculated in the following manner:
slope m=riserun=(y2y1)(x2x1)=(80 mL65 mL)(100C40C)=0.25 mL/C\begin{align*}\text{slope} \ m = \frac {rise} {run} = \frac {(y_2 - y_1)} {(x_2 - x_1)} = \frac {(80 \ \text{mL} - 65 \ \text{mL})} {(100^\circ\text{C} - 40^\circ\text{C})} = 0.25 \ \text{mL}/^\circ\text{C}\end{align*}
Therefore, the slope of the line is 0.25 mL/C\begin{align*}0.25 \ \text{mL}/^\circ\text{C}\end{align*}. The fact that the slope is positive indicates that the line is rising as it moves from left to right and that the volume increases by 0.25 mL\begin{align*}0.25 \ \text{mL}\end{align*} for each 1C\begin{align*}1^\circ\text{C}\end{align*} increase in temperature. A negative slope would indicate that the line was falling as it moves from left to right.
For a non-linear graph, the slope must be calculated for each point independently. Since the line will be a curve, the slope is calculated from the tangent to the curve at the point in question. Data Table E and Graph E are for a reaction in which the concentration of one of the reactants, bromine, was measured against time. The concentration is expressed in moles/liter, which is symbolized by M.
In order to determine the slope at some point on a curved line, a tangent (approximate) is drawn in as a line that just touches the point in question. Once the tangent has been drawn, the slope of the tangent is determined, which is also the slope of the curve at that point. In the graph above, the tangent has been drawn at the point where t = 2 seconds. We determine the x\begin{align*}x\end{align*}- and y\begin{align*}y\end{align*}-coordinates for two points along the tangent line (as best we can) and use the coordinates of those two points to calculate the slope of the tangent. The coordinates of the point at the left end of the tangent line is determined to be (1.00 s, 0.056 M). The coordinates of the point at the right end of the line is harder to determine, and we are guessing that the coordinates are (3.25 s, 0.031 M).
slope m=riserun=(y2y1)(x2x1)=(0.031 M0.056 M)(3.25 s1.00 s)=0.011 M/s\begin{align*}\text{slope} \ m = \frac {rise} {run} = \frac {(y_2 - y_1)} {(x_2 - x_1)} = \frac {(0.031 \ \text{M} - 0.056 \ \text{M})} {(3.25 \ \text{s} - 1 .00 \ \text{s})} = -0.011 \ \text{M/s}\end{align*}
Since the slope is a negative number, we know the line is decreasing in height. At t = 2 seconds, the concentration of bromine is decreasing at a rate of 0.011 moles/liter per second. At other points along this curve, the slope would be different. From the appearance of the curve, it is apparent that the slope is negative (the concentration of bromine is decreasing) all along the line, but it appears to be decreasing more quickly at the beginning of the reaction and less quickly as time increases.
## Lesson Summary
• Tables and graphs are two common methods of presenting data that aid in the search for regularities and trends within the data.
• When we draw a line graph from a set of data points, we are inferring a trend and constructing new data points between known data points. This process is called interpolation.
• Constructing data points beyond the end of a line graph, using the basic shape of the curve as a guide, is called extrapolation.
• The slope of a graph represents the rate at which one variable is changing with respect to the other variable.
• For a straight-line graph, the slope for all points along the line can be determined from any section of the graph.
• For a non-linear graph, the slope must be determined for each point by drawing a tangent line to the curve at the point in question.
## Review Questions
1. What would you do to find the slope of a curved line?
2. Andrew was completing his density lab for his chemistry lab exam. He collected the following data in his data table (shown in Table below).
1. Draw a graph to represent the data.
2. Calculate the slope.
3. What does the slope of the line represent?
Data Table for Problem 2
Mass of Solid (g) Volume of Solution (mL)
3.4\begin{align*}3.4\end{align*} 0.3\begin{align*}0.3\end{align*}
6.8\begin{align*}6.8\end{align*} 0.6\begin{align*}0.6\end{align*}
10.2\begin{align*}10.2\end{align*} 0.9\begin{align*}0.9\end{align*}
21.55\begin{align*}21.55\end{align*} 1.9\begin{align*}1.9\end{align*}
32.89\begin{align*}32.89\end{align*} 2.9\begin{align*}2.9\end{align*}
44.23\begin{align*}44.23\end{align*} 3.9\begin{align*}3.9\end{align*}
55.57\begin{align*}55.57\end{align*} 4.9\begin{align*}4.9\end{align*}
1. Donna is completing the last step in her experiment to find the effect of the concentration of ammonia on the reaction. She has collected the following data from her time trials and is ready for the analysis. Her data table is Table below. Help Donna by graphing the data, describing the relationship, finding the slope, and then discussing the meaning of the slope.
Data Table for Problem 3
Time (s) Concentration (mol/L)
0.20\begin{align*}0.20\end{align*} 49.92\begin{align*}49.92\end{align*}
0.40\begin{align*}0.40\end{align*} 39.80\begin{align*}39.80\end{align*}
0.60\begin{align*}0.60\end{align*} 29.67\begin{align*}29.67\end{align*}
0.81\begin{align*}0.81\end{align*} 20.43\begin{align*}20.43\end{align*}
1.08\begin{align*}1.08\end{align*} 14.39\begin{align*}14.39\end{align*}
1.30\begin{align*}1.30\end{align*} \begin{align*}10.84\end{align*}
\begin{align*}1.53\end{align*} \begin{align*}5.86\end{align*}
\begin{align*}2.00\end{align*} \begin{align*}1.95\end{align*}
\begin{align*}2.21\end{align*} \begin{align*}1.07\end{align*}
\begin{align*}2.40\end{align*} \begin{align*}0.71\end{align*}
\begin{align*}2.60\end{align*} \begin{align*}0.71\end{align*}
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# If the last tem in the binomial expansion of (213−1√2)nis(1353)log38 , then 5th term from the beginning is 210 b. 420 c. 105 d. none of these
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## To solve the problem step by step, we need to find the 5th term from the beginning of the binomial expansion of (21/3−1√2)n given that the last term equals (135/3)log38.Step 1: Identify the last term conditionThe last term in the binomial expansion occurs when r=n. The last term can be expressed as:Tn+1=(nn)(21/3)n−n(−1√2)n=(−1√2)nThis must equal (135/3)log38.Step 2: Simplify the right-hand sideWe simplify the right-hand side:(135/3)log38=3−5/3⋅log38=3−53⋅3=3−5=1243Step 3: Set up the equationNow we have:(−1√2)n=1243Taking the absolute values, we get:1(√2)n=1243This implies:(√2)n=243Step 4: Express 243 in terms of powers of 2We know that 243=35. Therefore:(√2)n=35Taking logarithms:n⋅12log2=5log3Thus:n=10log3log2Step 5: Find the 5th term from the beginningThe general term in the binomial expansion is given by:Tr+1=(nr)(21/3)n−r(−1√2)rFor the 5th term, r=4:T5=(n4)(21/3)n−4(−1√2)4Calculating:T5=(n4)(21/3)n−4⋅122=(n4)(21/3)n−4⋅14Step 6: Substitute n and calculateUsing n=10 (as derived from the previous steps):T5=(104)(21/3)10−4⋅14Calculating (104):(104)=10×9×8×74×3×2×1=210Now substituting:T5=210⋅(21/3)6⋅14=210⋅22⋅14=210⋅1=210ConclusionThe 5th term from the beginning is 210.
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Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## In the binomial expansion of (3√3+√2)5, the term which does not contain irrational number is :
A1st
B3rd
C4th
D5th
• Question 2 - Select One
## In the Binomial expansion of (3√2+√3)5 , which term does not contain irrational expression
A2nd term
B3^(rd) term
C4^(th) term
DNone of these
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# Use two ordered pairs to write a prediction equation. Then using your prediction equation to predict the number of comic books sold when the price is \$4.50.
The table is as follows:
Price-\$2.00 # sold was 16
-\$2.50 13
-\$2.75 12
-\$3.00 10
-\$3.50 7
I used (2.00,16) & (3.00, 10) to get the prediction equation y=-6x + 4 but do not know how to solve for the # of books sold when the price is \$4.50.
## The relationship between price and number sold is not quite linear. If you choose prices of = = 2.50 and 3.50 to get a prediction equation, the straight-line fit would be
y = 28 - 6x
Then if x = 4.50, the number sold is
y = 28 - 27 = 1
## To use the prediction equation y = -6x + 4 to predict the number of comic books sold when the price is \$4.50, you can substitute x = 4.50 into the equation and solve for y.
Let's plug in the value x = 4.50 into the prediction equation:
y = -6(4.50) + 4
y = -27 + 4
y = -23
Therefore, when the price is \$4.50, the predicted number of comic books sold is -23. However, it is important to note that predicting a negative number of comic books sold does not make sense in this context. It is likely that the prediction equation may not accurately reflect the relationship between price and the number of books sold for prices beyond the given data points.
## To solve for the number of books sold when the price is \$4.50 using the prediction equation y = -6x + 4, you need to substitute the given price into the equation and calculate the corresponding number of books sold.
Let's start by substituting x = 4.50 into the equation:
y = -6(4.50) + 4
Now, multiply -6 by 4.50:
y = -27 + 4
Next, perform the subtraction:
y = -23
So, when the price is \$4.50, the predicted number of books sold would be -23. However, it is not possible to have a negative number of books sold. Therefore, the prediction equation may not be accurate for prices above \$3.50.
It is important to note that prediction equations use the pattern in the given data to make predictions, but they are not always perfect. In this case, the prediction equation assumes a linear relationship between price and the number of books sold. However, the actual relationship may not be perfectly linear, resulting in a less accurate prediction when the price is outside the range of the given data.
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# NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines
In this chapter, we provide NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 10 Straight Lines for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 10 Straight Lines pdf, free NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 10 Straight Lines book pdf download. Now you will get step by step solution to each question.
## NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines
Q1. Find the equation of the straight line which passes through the point (1, -2)
Q2. Find the equation of the line passing through the point (5,2) and perpendicular to the line joining the points (2, 3) and (3, -1)
Sol:
We have points A(5, 2), B(2, 3) and C(3, -1).
Q3. Find the angle between the lines y = (2 -√3) (x + 5) and y = (2 + -√3) {x – 7)
Sol: Slope of the line = (2 -√3)(x + 5) is: ml = (2 -√3 )
Q4. Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.
Q5. Find the points on the line x+y = 4 which lie at a unit distance from the line 4x + 3y= 10
Sol.
Let the required point be (h, k) lies on the line x + y = 4
Q7. Find the equation of lines passing through (1,2) and making angle 30° with y-axis.
Q8. Find the equation of the line passing through the point of intersection of lx + y = 5 and x + 3 y +8 = 0 and parallel to the line 3x + 4y = 1.
Q9. For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.
Sol:
Given line is:
Q10. If the intercept of a line between the coordinate axes is divided by the point (-5,4) in the ratio 1 : 2, then find the equation of the line.
Sol:
Let the line through the point P(-5, 4) meets axis at A(h, 0) and B(0, k)
According to the question, we have AP: BP =1:2
Q11. Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x-axis.
Sol:
Given that the line makes and angle 120° with positive direction of x-axis.
Q12. Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + Ay = 4 and the opposite vertex of the hypotenuse is (2, 2).
Q13. If the equation of the base of an equilateral triangle is x + y – 2 and the vertex is (2, -1), then find the length of the side of the triangle.
Q14. A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P
Q15. In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x+y = 4 is at a distance √6/3 from the given point.
Q16. A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.
Q17. Find the equation of the line which passes through the point (-4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.
Sol:
Let the line through the point P(-A, 3) meets axis at A(h, 0) and 0(0, k)
Now according to the question AP : BP =5:3
Q18. Find the equations of the lines through the point of intersection of the lines x-y+ 1=0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5
Q19. If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.
Q20. Pl, P2 are points on either of the two lines y — √3 |x| = 2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P1, P2 on the bisector of the angle between the given lines.
Q21. If p is the length of perpendicular from the origin on the line (frac { x }{ a } +quad frac { y }{ b } ) and a2,p2 and are in the A.P , then show that a4+b4 = 0
Objective Type Questions
Q22. A line cutting off intercept -3 from the y-axis and the tangent of angle to the x-axis is 3/5, its equation is
Q23. Slope of a line which cuts off intercepts of equal lengths on the axes is
(a) -1
(b) 0
(c) 2
(d) √3
Sol: (a) Equation of the according to the question is (frac { x }{ a } +quad frac { y }{ a } )
=> x+y = a
Required slope = -1
Q24. The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is
(a) x-y = 5
(b) x+y = 5
(c)x+y=l
(d)x-y=1
Sol:(b) Slope of the given line y = x is 1.
Thus, slope of line perpendicular to y = x is -1.
Line passes through the point (3, 2).
So, equation of the required line is:y-2=-l (x – 3) => x + y = 5
Q25. The equation of the line passing through the point (1,2) and perpendicular to the line x +y + 1 = 0 is
(a) y-x +1=0
(b) y — x—1=0
(c) y-x + 2 = 0
(d) y — x — 2=0
Sol: (b) Slope of the given line +1=0 is-1.
So, slope of line perpendicular to above line is 1.
Line passes through the point (1,2).
Therefore, equation of the required linens:
y-2 = 1(x- 1) => y-x-1=0.
Q26. The tangent of angle between the lines whose intercepts on the axes are a, -b and b, -a, respectively, is
Q27. If the line (frac { x }{ a } +frac { y }{ b } ) passes through the points (2, -3) and (4, -5), then (a, b) a b is
(a) (1,1)
(b) (-1,1)
(c) (1,-1)
(d) (-1,-1)
Q28. The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is
Q29. The equations of the lines which pass through the point (3, -2) and are inclined at 60° to the line √3 x + y = 1 is
Q30. The equations of the lines passing through the point (1,0) and at a distance √3/2 from the origin, are
Q33. If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be
(a) 2x + 3y = 12
(b) 3x + 2y=l2
(c) 4x-3y = 6
(d) 5x- 2y=10
Sol: (a) Since, the middle point is P(3, 2), then line meets axes at A(6, 0) and B (0, 4).
Q34. Equation of the line passing through (1,2) and parallel to the line y = 3x – 1 is
(a)y + 2=x+l
(b) y + 2 = 3(x + 1)
(c) y -2 = 3(x — 1)
(d) y-2=x-l
Sol: (c) Line is parallel to the line y = 3x – 1.
So, slope of the line is‘3’.
Also, line passes through the point (1,2).
So, equation of the line is: y – 2 = 3(x – 1)
Q35. Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are
Q37. The point (4, 1) undergoes the following two successive transformations:
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the positive x-axis Then the final coordinates of the point are
(a) (4,3)
(b) (3,4)
(c) (1,4)
(d) (7/2,7/2)
Sol: (b) Reflection of A (4, 1) in y = x is 5(1,4).
Now translation of point B through a distance ‘2’ units along the positive x-axis shifts B to C( 1 + 2,4) or C(3,4).
Q38. A point equidistant from the lines 4x + 3y+ 10 = 0, 5x – 12y + 26 = 0 and lx + 24y – 50 = 0 is
(a) (1,-1)
(b) (1, 1)
(c) (0,0)
(d) (0, 1)
Sol:
(c) Clearly distance of each of three lines from (0, 0) is 2 units.
Q39. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is
Q40. The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is
(a) 1:2 (b) 3:7 (c) 2:3 (d) 2:5
Sol: (b) Given lines are:
Q41. One vertex of the equilateral triangle with centroid at the origin and one side asx + y- 2 = 0is
(a) (-1,-1) (b) (2,2) (c) (-2,-2) (d) (2,-2)
Fill in the Blanks Type Questions
Q42. If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through __________
Q43.The line which cuts off equal intercept from the axes and pass through the point (1, -2) is
Q44. Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are _____
Q45. The points (3,4) and (2, -6) are situated on the_____ of the line 3x – 4y – 8= 0.
Sol: Given line is 3x – 4y – 8 = 0
For point (3, 4), 3(3) – 4(4) – 8 = -15 < 0
For point (2, -6), 3(2) – 4(—6) – 8 = 22 > 0
Hence, the points (3,4) and (2, -6) lies on opposite side of the line.
Q46. A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is______ .
Sol: Let the moving point be P(h, k).
Given point is A(3, -2).
Q47. Locus of the mid-points of the portion of the line x sin θ + y cos θ = p intercepted between the axes is ______
True/False Type Questions
Q48. If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.
Sol: True
This is a contradiction to the fact that the area is a rational number. Hence, the triangle cannot be equilateral.
Q49. The points A(-2, 1), B(0, 5), C(-l, 2) are collinear.
Sol: False
Q50. Equation of the line passing through the point (a cos3 , a sin3 ) and perpendicular to the line
x sec + y cosec = a isx cos -y sin = a sin 2
Sol: False
Q51. The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y— 10 = 0 and 2x +y + 5 = 0.
Sol: True
Q53. The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x +y – 1 = 0 and lx – 3v – 35 = 0 is equidistant from the points (0, 0) and (8, 34).
Sol: True
Q55. The line ax + 2y + 1 = 0, bx + 2y + 1 = 0 and cx + 4y + 1 = 0 are concurrent, if a, b and c are in GP.
Sol: False
Q56. Line joining the points (3, -4) and (-2, 6) is perpendicular to the line joining the points (-3, 6) and (9, -18).
All Chapter NCERT Exemplar Problems Solutions For Class 11 Maths
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# 1 Meter equals How many Yards?
TopThere are various measures in maths which are written as the units of length like meter, yards, feet, inches etc. These units are basically used for specifying the dimensions of areas like some garden or track or field etc. For instance, if we have a rectangular field having the length and breadth as 40 meter & 50 meter, then its area can be calculated as 40 x 50 = 2000 meter Square. We can convert this value to other units also like yards; for converting meters to yards we would multiply the quantity by a conversion factor 1.09 and back from yards to meters by multiplying by a factor 0.91. Thus we can find in 1 meter equals how many yards are there as follows:
1 meter = 1.09 yards,
Or 1 yard = 0.91 meters,
Let us take an example of the following:
Example 1: Convert 120 meters into equivalent yards?
Solution: We can do so by using the above formula as follows:
1 meter = 1.09 yards
So 120 meters = 120 x 1.09 yards = 130.8 yards.
Example 2: Find the area of a Rectangle in yards square unit whose length and breadth are given as 20 meters and 30 meters respectively?
Solution: We would be using the usual way of evaluating the area of the rectangle as follows:
Area of rectangle = Length x Breadth,
Substituting the values of the length and breadth in the above formula we get area of rectangle as:
Area of rectangle = 20 x 30 = 600 meters square
To convert it into yards square we would follow the steps as:
1 meter = 1.09 yards,
Squaring both sides we get:
1 meter square = 1.1881 approx.
So, the area of the rectangle in yards square can be calculated as:
600 meters = 600 x 1.1881 yards square = 712.86 yards square.
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Trigonometry Lecture Notes_part2
# 2 analyze the identity and look for opportunities to
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other side. 2. Analyze the identity and look for opportunities to apply the fundamental identities. Rewriting the more complicated side of the equation in terms of sines and cosines is often helpful. 3. If sums or differences of fractions appear on one side, use the least common denominator and combine the fractions. 4. Don't be afraid to stop and start over again if you are not getting anywhere. Creative puzzle solvers know that strategies leading to dead ends often provide good problem-solving ideas. Section 7.3 Sum and Difference Formulas
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The Cosine of the Difference of Two Angles cos( ) cos cos sin sin α β α β α β - = + The cosine of the difference of two angles equals the cosine of the first angle times the cosine of the second angle plus the sine of the first angle times the sine of the second angle. Example 49 Use the difference formula for Cosines to find the Exact Value: Find the exact value of cos 15° Solution We know exact values for trigonometric functions of 60° and 45°. Thus, we write 15° as 60° - 45° and use the difference formula for cosines. cos l5° = cos(60° - 45°) = cos 60° cos 45° + sin 60° sin 45° Example 50 Find the exact value of cos 80° cos 20° + sin 80° sin 20°. Example 51 Find the exact value of cos(180º-30º) Example 52 Verify the following identity: cos( ) cot tan sin cos α β α β α β - = + Example 53 Verify the following identity: 5 2 cos (cos sin ) 4 2 x x x π - = - +
cos( ) cos cos sin sin cos( ) cos cos sin sin sin( ) sin cos cos sin sin( ) sin cos cos sin α β α β α β α β α β α β α β α β α β α β α β α β + = - - = + + = + - = - Example 54 Find the exact value of sin(30º+45º) Example 55 Find the exact value of 7 sin 12 π Example 56 Show that 3 sin cos 2 x x π - =
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Example 57 Find the exact value of tan(105º) Example 58 Verify the identity: tan 1 tan 4 tan 1 x x x π - - = + Example 59 Write the following expression as the sine, cosine, or tangent of an angle. Then find the exact value of the expression. 7 7 sin cos cos sin 12 12 12 12 π π π π - Section 7.4 Double-Angle and Half-Angle Formulas Double – Angle Formulas tan tan tan( ) 1 tan tan tan tan tan( ) 1 tan tan α β α β α β α β α β α β + + = - - - = +
2 2 2 sin 2 2sin cos cos 2 cos sin 2tan tan 2 1 tan θ θ θ θ θ θ θ θ θ = = - = - We can derive these by using the sum formulas we learned in section 6.2. Example 60 If 5 sin 13 θ = and θ lies in quadrant II, find the exact value of: a. sin 2 θ b. cos2 θ c. tan 2 θ Example 61 Find the exact value of 2 2tan15 1 tan 15 ° - ° Three Forms of the Double-Angle Formula for cos2 θ 2 2 2 2 cos2 cos sin cos2 2cos 1 cos2 1 2sin θ
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# 4.3: Sum and Difference Identities
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Focus Questions
The following questions are meant to guide our study of the material in this section. After studying this section, we should understand the concepts motivated by these questions and be able to write precise, coherent answers to these questions.
• What are the Cosine Difference and Sum Identities?
• What are the Sine Difference and Sum Identities?
• What are the Tangent Difference and Sum Identities?
• What are the Cofunction Identities?
• Why are the difference and sum identities useful?
The next identities we will investigate are the sum and difference identities for the cosine and sine. These identities will help us find exact values for the trigonometric functions at many more angles and also provide a means to derive even more identities.
Beginning Activity
1. Is $$\cos(A - B) = \cos(A) - \cos(B)$$ an identity? Explain.
2. Is $$\sin(A - B) = \sin(A) - \sin(B)$$ an identity? Explain.
3. Use a graphing utility to draw the graph of $$y = \sin(\dfrac{\pi}{2} - x)$$ and $$y = \cos(x)$$ over the interval $$[-2\pi, 2\pi]$$ on the same set of axes. Do you think $$y = \sin(\dfrac{\pi}{2} - x) = \cos(x)$$ is an identity? Why or why not?
## The Cosine Difference Identity
To this point we know the exact values of the trigonometric functions at only a few angles. Trigonometric identities can help us extend this list of angles at which we know exact values of the trigonometric functions. Consider, for example, the problem of finding the exact value of $$\cos(\dfrac{\pi}{12})$$. The definitions and identities we have so far do not help us with this problem. However, we could notice that $$\dfrac{\pi}{12} = \dfrac{\pi}{3} - \dfrac{\pi}{4}$$ and if we knew how the cosine behaved with respect to the difference of two angles, then we could find $$\cos(\dfrac{\pi}{12})$$. In our Beginning Activity, however, we saw that the equation $$\cos(A - B) = \cos(A) - \cos(B)$$ is not an identity, so we need to understand how to relate $$\cos(A - B)$$ to cosines and sines of $$A$$ and $$B$$.
We state the Cosine Difference Identity below. This identity is not obvious, and a verification of the identity is given later in this section. For now we focus on using the identity.
Cosine Difference Identity
For any real numbers $$A$$ and $$B$$ we have $\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$
Example $$\PageIndex{1}$$: (Using the Cosine Difference Identity)
Let us return to our problem of finding $$\cos(\dfrac{\pi}{12})$$. Since we know $$\dfrac{\pi}{12} = \dfrac{\pi}{3} - \dfrac{\pi}{4}$$, we can use the Cosine Difference Identity with $$A = \dfrac{\pi}{3}$$ and $$B = \dfrac{\pi}{4}$$ to obtain
$\cos(\dfrac{\pi}{12}) = \cos(\dfrac{\pi}{3} - \dfrac{\pi}{4}) = \cos(\dfrac{\pi}{3})\cos(\dfrac{\pi}{4}) + \sin(\dfrac{\pi}{3}\sin(\dfrac{\pi}{4})) = (\dfrac{1}{2})(\dfrac{\sqrt{2}}{2}) + (\dfrac{\sqrt{3}}{2})(\dfrac{\sqrt{2}}{2}) = \dfrac{\sqrt{2}+\sqrt{6}}{4}.$
So we see that $$\cos(\dfrac{\pi}{12}) = \dfrac{\sqrt{2}+\sqrt{6}}{4}.$$
Exercise $$\PageIndex{1}$$
1. Determine the exact value of $$\cos(\dfrac{7\pi}{12})$$ using the Cosine Difference Identity
2. Given that $$\dfrac{5\pi}{12} = \dfrac{\pi}{6} + \dfrac{\pi}{4} = \dfrac{\pi}{6} - (-\dfrac{\pi}{4})$$, determine the exact value of $$\cos(\dfrac{5\pi}{12})$$ using the Cosine Difference Identity.
1. We first note that $$\dfrac{7\pi}{12} = \dfrac{9\pi}{12} - \dfrac{2\pi}{6} = \dfrac{3\pi}{4} - \dfrac{\pi}{6}$$.
$\cos(\dfrac{7\pi}{12}) = \cos(\dfrac{}{})$
2.
$\cos(\dfrac{5\pi}{12}) = \cos(\dfrac{\pi}{6} - (-\dfrac{\pi}{4})) = \cos(\dfrac{\pi}{6})\cos(\dfrac{\pi}{4}) + \sin(\dfrac{\pi}{6})\sin(-\dfrac{\pi}{4}) = (\dfrac{\sqrt{3}}{2})(\dfrac{\sqrt{2}}{2}) + (\dfrac{1}{2})(-\dfrac{\sqrt{2}}{2}) = \dfrac{\sqrt{6} - \sqrt{2}}{4}$
## The Cosine Sum Identity
Since there is a Cosine Difference Identity, we might expect there to be a Cosine Sum Identity. We can use the Cosine Difference Identity along with the negative identities to find an identity for $$\cos(A + B)$$. The basic idea was contained in our last Progress Check, where we wrote $$A + B$$ as $$A - (-B)$$. To see how this works in general, notice that
$\cos(A + B) = \cos(A - (-B)) = \cos(A)\cos(-B) + \sin(A)\sin(-B) = \cos(A)\cos(B) - \sin(A)\sin(B)$
This is the Cosine Sum Identity.
Cosine Sum Identity
For any real numbers $$A$$ and $$B$$ we have
$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$
Exercise $$\PageIndex{2}$$
1. Find a simpler formula for $$\cos(\pi + x)$$ in terms of $$\cos(x)$$. Illustrate with a graph.
2. Use the Cosine Difference Identity to prove that $$\cos(\dfrac{\pi}{2} - x) = \sin(x)$$ is an identity.
1. $$\cos(\pi + x) = \cos(\pi)\cos(x) - \sin(\pi)\sin(x) = -\cos(x)$$. The graphs of $$y = \cos(\pi + x)$$ and $$y = \cos(x)$$ are indentical.
2. $$\cos(\dfrac{\pi}{2} - x) = \cos(\dfrac{\pi}{2})\cos(x) + \sin(\dfrac{\pi}{2})\sin(x) = 0\cdot \cos(x) + 1\cdot \sin(x)$$. So we see that $$\cos(\dfrac{\pi}{2} - x) = \sin(x)$$
## Cofunction Identities
In Progress Check 4.14 we used the Cosine Difference Identity to see that $$\cos(\dfrac{\pi}{2} - x) = \sin(x)$$ is an identity. Since this is an identity, we can replace $$x$$ with $$\dfrac{\pi}{2} - x$$ to see that
$\sin(\dfrac{\pi}{2} - x) = \cos(\dfrac{\pi}{2} - (\dfrac{\pi}{2} - x)) = \cos(x),$
so $\sin(\dfrac{\pi}{2} - x) = \cos(x),$. The two identities
$\cos(\dfrac{\pi}{2} - x) = \sin(x)$ and $\sin(\dfrac{\pi}{2} - x) = \cos(x)$
are called cofunction identities. These two cofunction identities show that the sine and cosine of the acute angles in a right triangle are related in a particular way.
Since the sum of the measures of the angles in a right triangle is $$\pi$$ radians or $$180^\circ$$, the measures of the two acute angles in a right triangle sum to $$\dfrac{\pi}{2}$$ radians or $$90^\circ$$. Such angles are said to be complementary. Thus, the sine of an acute angle in a right triangle is the same as the cosine of its complementary angle. For this reason we call the sine and cosine cofunctions. The naming of the six trigonometric functions reflects the fact that they come in three sets of cofunction pairs: the sine and cosine, the tangent and cotangent, and the secant and cosecant. The cofunction identities are the same for any cofunction pair.
Cofunction Identities
For any real number x for which the expressions are defined,
• $$\cos(\dfrac{\pi}{2} - x) = \sin(x)$$
• $$\sin(\dfrac{\pi}{2} - x) = \cos(x)$$
• $$\tan(\dfrac{\pi}{2} - x) = \cot(x)$$
• $$\cot(\dfrac{\pi}{2} - x) = \tan(x)$$
• $$\sec(\dfrac{\pi}{2} - x) = \csc(x)$$
• $$\csc(\dfrac{\pi}{2} - x) = \sec(x)$$
For any angle $$x$$ in degrees for which the functions are defined,
• $$\cos(90^\circ - x) = \sin(x)$$
• $$\sin(90^\circ - x) = \cos(x)$$
• $$\tan(90^\circ- x) = \cot(x)$$
• $$\cot(90^\circ - x) = \tan(x)$$
• $$\sec(90^\circ - x) = \csc(x)$$
• $$\csc(90^\circ - x) = \sec(x)$$
Exercise $$\PageIndex{3}$$
Use the cosine and sine cofuntion identities to prove the cofunction identity $\tan(90^\circ- x) = \cot(x)$
We will use the identity $$\tan(y) = \dfrac{\sin(y)}{\cos(y)}$$.
$\tan(\dfrac{\pi}{2} - x) = \dfrac{\sin(\dfrac{\pi}{2} - x)}{\cos(\dfrac{\pi}{2} - x)} = \dfrac{\cos(x)}{\sin(x)} = \cot(x)$
### The Sine Difference and Sum Identities
We can now use the Cosine Difference Identity and the Cofunction Identities to derive a Sine Difference Identity:
$\sin(A - B) = \cos(\dfrac{\pi}{2} - (A - B)) = \cos((\dfrac{\pi}{2} - A) + B)) = \cos(\dfrac{\pi}{2} - A)\cos(B) - \sin(\dfrac{\pi}{2} - A)\sin(B)= \sin(A)\cos(B) - \cos(A)\sin(B).$
We can derive a Sine Sum Identity from the Sine Difference Identity:
$\sin(A + B) = \sin(A - (-B)) = \sin(A)\cos(-B) - \cos(A)\sin(-B) = \sin(A)\cos(B) + \cos(A)\sin(B)$
Sine Difference and Sum Identities
For any real numbers $$A$$ and $$B$$ we have
$\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)$ and $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$
Exercise $$\PageIndex{4}$$
Use the Sine Sum or Difference Identities to find the exact values of the following.
1. $\sin(\dfrac{\pi}{12})$
2. $\sin(\dfrac{5\pi}{12})$
1. We note that $$\dfrac{\pi}{12} = \dfrac{\pi}{3} - \dfrac{\pi}{4}$$.
$\sin(\dfrac{\pi}{12}) = \sin(\dfrac{\pi}{3} - \dfrac{\pi}{4}) = \sin(\dfrac{\pi}{3})\cos(\dfrac{\pi}{4}) - \cos(\dfrac{\pi}{3})\sin(\dfrac{\pi}{4}) = \dfrac{\sqrt{3}}{2}\cdot \dfrac{\sqrt{2}}{2} - \dfrac{1}{2}\cdot \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$
2. We note that $$\dfrac{5\pi}{12} = \dfrac{\pi}{4} + \dfrac{\pi}{6}$$.
$\sin(\dfrac{5\pi}{12}) = \dfrac{\pi}{4} + \dfrac{\pi}{6} = \sin(\dfrac{\pi}{4})\cos(\dfrac{\pi}{6}) + \cos(\dfrac{\pi}{4})\sin(\dfrac{\pi}{6}) = \dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2}\cdot \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$
### Using Sum and Difference Identities to Solve Equations
As we have done before, we can use our new identities to solve other types of trigonometric equations.
Example $$\PageIndex{2}$$: (Using the Cosine Sum Identity to Solve an Equation)
Consider the equation
$\cos(\theta)\cos(\dfrac{\pi}{5}) - \sin(\theta)\sin(\dfrac{\pi}{5}) = \dfrac{\sqrt{3}}{2}$
On the surface this equation looks quite complicated, but we can apply an identity to simplify it to the point where it is straightforward to solve. Notice that left side of this equation has the form $$\cos(A)\cos(B) - \sin(A)\sin(B)$$ with $$A = \theta$$ and $$B = \dfrac{\pi}{5}$$
We can use the Cosine Sum Identity $$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$ to combine the terms on the left into a single term, and we can solve the equation from there:
$\cos(\theta)\cos(\dfrac{\pi}{5}) - \sin(\theta)\sin(\dfrac{\pi}{5}) = \dfrac{\sqrt{3}}{2}$
$\cos(\theta + \dfrac{\pi}{5}) = \dfrac{\sqrt{3}}{2}$
Now $$\cos(x) =\dfrac{\sqrt{3}}{2}$$ when $$x = \dfrac{\pi}{6} + 2k\pi$$ or $$x = -\dfrac{\pi}{6} + 2k\pi$$ for integers $$k$$. Thus, $$\cos(\theta + \dfrac{\pi}{5}) = \dfrac{\sqrt{3}}{2}$$ when $$\theta + \dfrac{\pi}{5} = \dfrac{\pi}{6} + 2k\pi$$ or $$\theta + \dfrac{\pi}{5} = -\dfrac{\pi}{6} + 2k\pi$$. Solving for $$\theta$$ gives us the solutions
$\theta = -\dfrac{\pi}{30} + 2k\pi$ or $\theta = -\dfrac{11\pi}{30} + 2k\pi$
where $$k$$ is any integer. These solutions are illustrated in Figure $$\PageIndex{1}$$.
Note
Up to now, we have been using the phrase “Determine formulas that can be used to generate all the solutions of a given equation.” This is not standard terminology but was used to remind us of what we have to do to solve a trigonometric equation. We will now simply say, “Determine all solutions for the given equation.” When we see this, we should realize that we have to determine formulas that can be used to generate all the solutions of a given equation.
Exercise $$\PageIndex{5}$$
Determine all solutions of the equation
$\sin(x)\cos(1) + \cos(x)\sin(1) = 0.2$
Hint: Use a sum or difference identity and use the inverse sine function.
Figure $$\PageIndex{1}$$: Graphs of $$y = \cos(\theta)\cos(\dfrac{\pi}{5}) - \sin(\theta)\sin(\dfrac{\pi}{5})$$ and $$y = \dfrac{\sqrt{3}}{2}$$
We first use the Sine Sum Identity to rewrite the equation as $$\sin(x + 1) = 0.2$$. If we let $$t = x + 1$$, we see that for $$0 \leq t < 2\pi$$,
$t = \arcsin(0.2)$ or $t = (\pi - \arcsin(0.2))$
So we have $$x + 1 = \arcsin(0.2)$$ or $$x + 1 = \pi - \arcsin(0.2)$$. Since the period of the functions we are working with is $$2\pi$$, we see that
$x = (-1 + \arcsin(0.2)) + k(2\pi)$ or $x = (-1 + \pi - \arcsin(0.2)) + k(2\pi)$
where $$k$$ is an integer.
## Appendix – Proof of the Cosine Difference Identity
To understand how to calculate the cosine of the difference of two angles, let $$A$$ and $$B$$ be arbitrary angles in radians. Figure 4.7 shows these angles with $$A > B$$, but the argument works in general. If we plot the points where the terminal sides of the angles $$A$$, $$B$$ and $$A - B$$ intersect the unit circle, we obtain the picture in Figure 4.7.
Figure $$\PageIndex{2}$$: The cosine difference formula
The arc on the unit circle from the point $$(\cos(B), \sin(B))$$ to the point $$(\cos(A), \sin(A))$$ has length $$A - B$$ and the arc from the point $$(1, 0)$$ to the point $$(\cos(A - B), \sin(A - B))$$ also has length $$A - B$$. So the chord from $$(\cos(B), \sin(B))$$ to $$(\cos(A), \sin(A))$$ has the same length as the chord from (1,0) to $$(\cos(A - B), \sin(A - B))$$. To find the cosine difference formula, we calculate these two chord lengths using the distance formula.
The length of the chord from $$(\cos(B), \sin(B))$$ to $$(\cos(A), \sin(A))$$ is
$\sqrt{(\cos(A) - \cos(B))^{2} + (\sin(A) - \sin(B))^{2}}$
and the length of the chord from $$(1, 0)$$ to the point $$(\cos(A - B), \sin(A - B))$$ is
$\sqrt{(\cos(A - B) - 1)^{2} + (\sin(A - B) - 0)^{2}}$
Since these two chord lengths are the same we obtain the equation
$\sqrt{(\cos(A - B) - 1)^{2} + (\sin(A - B) - 0)^{2}} = \sqrt{(\cos(A) - \cos(B))^{2} + (\sin(A) - \sin(B))^{2}}$
The cosine difference identity is found by simplifying Equation (2) by first squaring both sides:
$(\cos(A - B) - 1)^{2} + (\sin(A - B) - 0)^{2} = (\cos(A) - \cos(B))^{2} + (\sin(A) - \sin(B))^{2}$
Then we expand both sides
$[\cos^{2}(A - B) - 2\cos(A - B) + 1] + \sin^{2}(A - B) = [\cos^{2}(A) - 2\cos(A)\cos(B) + \cos^{2}(B)] + [\sin^{2}(A) - 2\sin(A)\sin(B) + \sin^{2}(B)]$
We can combine some like terms:
$[\cos^{2}(A - B) + \sin^{2}(A - B)] - 2\cos(A - B) + 1 = [\cos^{2}(A) + \sin^{2}(A)] + [\cos^{2}(B) + \sin^{2}(B)] - 2\sin(A)\sin(B) - 2\cos(A)\cos(B)]$
Finally, using the Pythagorean identities yields
$1 - 2\cos(A - B) + 1 = 1 + 1 - 2\cos(A)\cos(B) - 2\sin(A)\sin(B)$
$- 2\cos(A - B) = - 2\cos(A)\cos(B) - 2\sin(A)\sin(B)$
$\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$
Summary
In this section, we studied the following important concepts and ideas:
• Sum and Difference Identities
$\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$
$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$
$\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)$
$\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$
• Cofunction Identities
See page 266 for a list of the cofunction identities.
This page titled 4.3: Sum and Difference Identities is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Ted Sundstrom & Steven Schlicker (ScholarWorks @Grand Valley State University) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# How do you use synthetic substitution to find p(-3) for p(x)=4x^3-5x^2+7x-10?
Sep 20, 2015
-184
#### Explanation:
Well, you need to know what a synthetic substitution is. I am not going to explain this here but just show you the steps.
I didn't know what it was until I read this question so I googled it and found some great Youtube videos that explain how to do the trick. Here is one, for example
So in this question, we want to find p(-3) so write down $- 3$ on the left, and the coefficients of the polynomial $4 , - 5 , 7 , - 10$.
Here are the steps:
- drop down the 4.
- multiply 4 by -3. $4 \cdot - 3 = - 12$
- write -12 below the -5.
- add -5 and -12. $- 5 + \left(- 12\right) = - 17$
- multiply -17 by -3. $- 17 \cdot - 3 = 51$
- write 51 below the 7.
- add 51 and 7. $7 + 51 = 58$
- multiply 58 by -3. $58 \cdot - 3 = - 174$
- write -174 below the -10.
- add -174 and -10. $- 10 + \left(- 174\right) = - 184$
i.e. $p \left(- 3\right) = - 184$
You can check if this answer is correct by directly substituting in the original equation:
$4 \cdot {\left(- 3\right)}^{3} = 4 \cdot 9 \cdot - 3 = 36 \cdot - 3 = - 108$
$- 5 \cdot {\left(- 3\right)}^{2} = - 5 \cdot 9 = - 45$
$7 \cdot \left(- 3\right) = - 21$
$- 10 = - 10$
i.e. $p \left(- 3\right) = - 108 - 45 - 21 - 10 = - 184$
That's it. Hope it helped.
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Is there another simpler method to solve this elementary school math problem?
I am teaching an elementary student. He has a homework as follows.
There are $16$ students who use either bicycles or tricycles. The total
number of wheels is $38$. Find the number of students using bicycles.
I have $3$ solutions as follows.
Using a single variable.
Let $x$ be the number of students in question. The number of students using tricycles is $16-x$. The total number of wheels is the sum of the total number of bicycles times $2$ and the total number of tricycles times $3$.
$$2\times x + 3 \times (16-x) = 38$$
The solution is $x=10$.
Using $2$ variables.
Let $x$ and $y$ be the number of students using bicycles and tricycles, respectively.
It implies that
\begin{align}
x+y&=16\\
2x+3y&=38
\end{align}
The solution is $x=10$ and $y=6$.
Using multiples
The multiples of $2$ are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22,24, 26,28,30,32,\dotsc$
The multiples of $3$ are $3,6,9,12,15,18,21,24,27,30,33,36,\dotsc$
The possible wheel combinations with format $(\#\text{bicycle wheels}, \#\text{tricycle wheels})$:
$\hspace{6cm} (32,6)$ but there are $18$ students
$\hspace{6cm} (26,12)$ but there are $17$ students
$\hspace{6cm} (20,18)$ there are $16$ students
$\hspace{6cm} (14, 24)$ there are $15$ students
$\hspace{6cm} (8, 30)$ there are $14$ students
$\hspace{6cm} (2,36)$ there are $13$ students
Thus the correct combination is $10$ bicycles and $6$ tricycles.
My question
Is there any other simpler method?
If everybody had bicycles, there would be $16 \times 2 = 32$ wheels. There are actually $38$, so that’s $6$ additional wheels. Each one must be on a different tricycle.
So there are $6$ tricycles, and the other $10$ students have bicycles.
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# Before we start: get used to the direction of vectors for rotations. Tangential velocity Centripetal acceleration.
## Presentation on theme: "Before we start: get used to the direction of vectors for rotations. Tangential velocity Centripetal acceleration."— Presentation transcript:
Before we start: get used to the direction of vectors for rotations. Tangential velocity Centripetal acceleration
Trying to solve problems in a non- inertial frame? Inertial frames Non-Inertial frames One example is the “centrifugal force”.
Another way of solving the train problem: given =30, the radius of the train’s trajectory, R, what is the speed of the train? In the frame of reference of the train, there is a fictitious force on any object of mass m that aims opposite the acceleration: From here we get the train speed.
First we solve a simple problem. What is the angle of the liquid in the accelerated pail? In the frame of reference of the pail, there is a fictitious force on any object of mass m that aims opposite the acceleration: The surface of the liquid will be normal to the “net force”
What is the shape of the surface of a spinning bucket of water? The normal to the surface will be along the direction of the “net force” (quotes indicate in this non-inertial frame) :
Coriolis: you are on top of a Marry-go-round and through a ball. On the ground the velocity of the ball is constant On the rotating frame the trajectory looks accelerated For t 0 There is an additional funny acceleration that is proportional to the velocity: Coriolis acceleration
Coriolis: you are on top of a Marry-go-round and through a ball. On the ground the velocity of the ball is constant On the rotating frame the trajectory looks accelerated Coriolis Centrifugal
The Earth equator is hot: Air raises, so currents of air flow from poles to equator.
Which way do you prefer it?
Focault’s pendulum: a pendulum at the North pole would oscillate on a plane that would rotate with 24-hour period. What is the period for us? Focault’s pendulum pivot: It allows for the plane of oscillation to rotate freely.
A blimp has a velocity of 8.2 m/s due west relative to the air. There is a wind blowing at 3.5 m/s due north. The speed of the balloon relative to the ground is: A. 11.7 m/s B. 8.9 m/s C. 7.4 m/s D. 4.7 m/s E. Some other speed (specify)
An elevator moves downward with an acceleration of 6.2 m/s 2. A ball dropped from rest by a passenger will have what downward acceleration relative to the elevator? A. 3.6 m/s 2. B. 6.2 m/s 2. C. 9.8 m/s 2. D. 16.0 m/s 2. E. Some other acceleration (specify)
N9T.4 When you go over the crest of a hill in a roller coaster, a force appears to lift you up out of your seat. This is a fictitious force, true (T) or false (F)?
N9T.5 If your car is hit from behind, you are suddenly pressed back into the seat. The normal force that the seats exerts on you is a fictitious force, true (T) or false (F)?
N9T.6 You are pressed downward towards the floor as an elevator begins to move upward. The force pressing you down is fictitious, true (T) or false (F)?
N9T.7 A beetle (black dot on the top view shown at the right) sits on a rapidly rotating turntable. The table rotates faster and faster, and eventually the beetle loses its grip. What is the subsequent trajectory relative to the ground (assuming it cannot fly)?
N9T.8 When is an elevator a reasonably good inertial reference frame? A. Never under any circumstances. B. Always, since it is attached to the surface of the earth. C. It is except when it is changing speed. D. Other conditions (describe).
N9T.9 Is a freely falling elevator an accurate inertial reference frame? A. No, such a frame is accelerating toward the earth. B. It is not a real inertial frame, but we can treat it as one if we ignore the gravitational forces on objects inside. C. Yes. D. It depends (explain).
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Calculus of Parametric Curves
Now that you've learned how to represent curves as parametric equations, we can apply Calculus to parametric curves!
Suppose you wanted to model the way a football flies through the air after leaving a quarterback's hand. Since the ball moves in both the vertical $$y$$ direction and the horizontal $$x$$ direction, it would be useful to model the football's position with parametric equations. Just as we used Calculus to find a particle in motion's speed or acceleration, we can use Calculus to find the motion of an object that follows a parametric curve.
Types of Parametric Curves
Parametric curves can be used to describe curves as surfaces. They're particularly useful in graphic curves that are not functions, such as the unit circle. So, parametric curves can be used to describe a large variety of curves including parabolas, circles, and ellipses. We can also parametrize more complex curves, like the Lissajous curve (used to describe harmonic motion), kites, rounded triangles, and many other funky shapes!
The Meaning of Parametric Curves in Calculus
We can apply all the Calculus we've learned up to this point to parametric curves. The most important Calculus applications to parametric curves are finding a tangent line, finding the area under a parametric curve, calculating the arc length of a parametric curve, and finding the surface area to a parametric curve volume.
Formulas of Parametric Curves in Calculus
There are a few important formulas for applying Calculus to parametric curves. These formulas cover some of the most important aspects of Parametric Curves.
There is stuff that may first seem far from calculus, but calculus proves to be an indispensable tool in our study of Parametric Curves.
The formulas we will take a look at, are:
• Tangent Lines to Parametric Curves
• Area under a Parametric Curve
• Arc length of a Parametric Curve
• Surface Area of a Parametric solid
In the next sections, you will look at each one, along with an example of each.
Tangent Lines to Parametric Curves
To find the equation for a tangent line, we need the derivative of the parametric equations.
Consider a curve defined by the parametric equations $$x(t)$$ and $$y(t)$$. Assume that $$x'(t)$$ and $$y'(t)$$ exist and $$x'(t) \neq 0$$. Then the derivative $$\frac{dy}{dx}$$ is
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y'(t)}{x'(t)}.$
Remember that the parameter need not be $$t$$, it can be anything.
Find the derivative of $$y$$ w.r.t. $$x$$ for the curve given by the equations $$y=e^{2t}$$ and $$x=t^3-t$$ where $$t \neq 0$$ at the point where $$t=2$$.
Solution:
Using the formula that we saw above:
\begin{aligned} \frac{dy}{dx}&=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ & =\frac{\frac{d(e^{2t})}{dt}}{\frac{d(t^3-t)}{dt}} \\ &=\frac{2e^{2t}}{3t^2-1}. \end{aligned}
Evaluating it at $$t=2$$:
$$\frac{dy}{dx}=\frac{2e^{4}}{11}.$$
This is an example of how we can find the derivative of a dependent variable w.r.t. the independent variable, for a pair of parametric equations, using this we can find the equation of the tangent line to the curve (an example of this is at the end).
Area Under a Parametric Curve Using Calculus
You already know how to calculate the area under a curve, whose equation is given by $$y=f(x)$$. The calculus is quite straightforward for that. But what if we are given parametric equations, you need a different formula in this case.
Consider the following theorem for the area under a parametric curve.
Consider a non-self intersecting parametric curve defined by $$x(t)$$ and $$y(t)$$ for $$a \le t \le b$$. Assume that $$x(t)$$ is differentiable. The area under this specified curve is:
$$A=\int_a^{b} y(t)x'(t)dt.$$
Arc Length of a Parametric Curve
Let's go over the theorem for the arc length of a parametric curve.
Consider the curve defined by $$x(t)$$ and $$y(t)$$ for $$a \le t \le b$$. Assume that $$x(t)$$ and $$y(t)$$ are differentiable. We want to find the Arc Length of the curve between that interval. But what is arc length?
Suppose you have a curve, and you want to find the length of the curve between a given interval, the matter is easy for a straight line or even for a circle. But what if the curve is any random function, then we need the help of calculus to find the length of any random curve.
To visualize what exactly is meant by Arc Length, take a look at the figure below:
Figure 1.- Arc length of a curve between two points - StudySmarter Originals
Then, the arc length of the curve is given by
$$S=\int_a^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \,dt.$$
Find the arc length of the parametric curve given by $$x=3t^2$$ and $$y=2t^3$$ for the interval $$1<t<3$$.
Solution:
Firstly, we will calculate the derivatives $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$:
$$\frac{dx}{dt}=\frac{d(3t^2)}{dt}=6t,$$
And for $$\frac{dy}{dt}$$:
$$\frac{dy}{dt}=\frac{d(2t^3)}{dt}=6t^2.$$
Now substituting these in the formula of Arc length:
\begin{aligned} S &=\int_a^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \,dt \\ &=\int_1^{3} \sqrt{(6t)^2+(6t^2)^2} \,dt \\ &=6 \int_1^{3} t \sqrt{1+t^2} \,dt. \end{aligned}
The problem is now of pure integration, here we shall use integration by substitution.
Substituting for $$1+t^2=u^2$$, we get $$t \,dt=u \,du$$. For the integral limits, $$1<t<3$$ becomes $$\sqrt{2}<u<\sqrt{10}$$.
Our integral now becomes:
\begin{aligned} S&=6 \int_\sqrt{2}^{\sqrt{10}} u^2 \,du \\ &= 2u^3 \Biggr|_{\sqrt{2}}^{\sqrt{10}} \\ &=2(10^{3/2}-2^{3/2}). \end{aligned}
Hence, the arc length of the curve is $$2(10^{3/2}-2^{3/2})$$ units.
Surface Area of Parametric Solid of Revolution
There is one last formula you need to go through in this section. This formula involves finding the surface area of a parametric curve that is revolves around the $$x$$-axis.
Again, let's consider the curve defined by $$x(t)$$ and $$y(t)$$ for $$a \le t \le b$$. Assume that $$x(t)$$ and $$y(t)$$ are differentiable. This curve is revolved around the $$x$$-axis to create a Solid of Revolution. The surface area of the solid is defined by
$$S=2\pi \int_a^{b} y(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}} \,dt$$
If this curve is revolved around the $$y$$-axis to create a Solid of Revolution, the surface area of the solid is defined by
$S=2\pi \int_a^{b} x(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}}dt$
Parametric Curve Examples in Calculus
Find the derivative $$\frac{dy}{dx}$$ of the parametric curve defined by $$x(t)=t^{3}-4$$ and $$y(t)=3t$$ for $$t \neq 0$$. Then, find the equation of the line tangent to the curve at the point $$t=2$$.
Solution:
$$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{3}{3t^{2}}=\frac{1}{t^{2}}.$$
Plugging in $$t=2$$ to $$x(t)$$ and $$y(t)$$.
$x(2)=2^{3}-4=4,$
$y(2)=3(2)=6.$
Plugging in $$t=2$$ to $$\frac{dy}{dx}$$ to find the slope of the tangent line at the point $$(4, 6)$$.
$\frac{dy}{dx}=\frac{1}{2^{2}}=\frac{1}{4}.$
Plugging in these values to point-slope form
$y-6=\frac{1}{4}(x-4).$
Figure 2.- The line tangent to the parametric curve at the point $$t=2$$
Consider the curve defined by $$x(t)=3cost$$ and $$y(t)=3sint$$. Find the area under the curve on the interval $$0 \le t \le \pi$$.
Solution:
Based on the above theorem, you need to differentiate $$x(t)$$ before integrating.
$$x'(t)=-3sint.$$
All you have to do is plug in $$x'(t)$$ and $$y(t)$$ and evaluate the integral over the interval $$0 \le t \le \pi$$: $$\int_a^{b} y(t)x'(t)dt$$
\begin{aligned} \int_0^{\pi} (3sint)(-3sint) \,dt &=\int_0^{\pi} -9\sin^{2} \,dt \\ &=-9\int_0^{\pi} \frac{1-\cos(2t)}{2} \,dt \\ &=\frac{-9}{2} \int_0^{\pi} 1-\cos(2t) \,dt \\ &=\frac{-9}{2}[(\pi - \frac{\sin(2\pi)}{2})-(0 - \frac{\sin(2(0))}{2}) \\ &=\frac{-9\pi}{2}. \end{aligned}
Find the arc length of the parametric curve defined by $$x(t)=cost$$ and $$y(t)=sint$$ on the interval $$0 \le t \le 2\pi$$.
Solution:
First, let's differentiate $$x(t)$$ and $$y(t)$$.
\begin{align} x'(t)=-\sin t, \\ y'(t)=\cos t. \end{align}
Recall that the arc length is given as follows:
$$s=\int_a^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \,dt.$$
Now plugging our data into this formula:
\begin{aligned} s &=\int_0^{2\pi} \sqrt{(\cos t)^{2} + (\sin t)^{2}} \,dt \\ s&=\int_0^{2\pi} \sqrt{1} \,dt \\ s&=2\pi - 0 \\ s &=2\pi. \end{aligned}
Essentially, what this tells us is the circumference of the unit circle is $$2\pi$$.
Find the surface area of the curve described by $$x(t)=t^{2}$$ and $$y(t)=t$$ between $$0 \le t \le 3$$ revolved around the $$x$$-axis.
Solution:
First, let's differentiate $$x(t)$$ and $$y(t)$$.
\begin{align} x'(t)=2t, \\ y'(t)=1. \end{align}
Recall that the surface area of revolution is given by:
$$S=2\pi \int_a^{b} y(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}} \,dt.$$
Substituting the given data in the equation:
\begin{aligned} S &=2\pi \int_0^{3} t \sqrt{(2t)^{2} + (1)^{2}} \,dt \\ &=2\pi \int_0^{3} t \sqrt{4t^{2}+1} \,dt. \end{aligned}
To integrate, we need to use $$u$$-substitution.
Let $$u=4t^{2}+1$$ and $$\,du=8t \,dt$$. For the integral limits, $$0 \le t \le 3$$ becomes $$1 \le u \le 37$$.
\begin{aligned} S &=2\pi \int_0^{3} \frac{8}{8}t \sqrt{4t^{2}+1} \,dt \\ &=\frac{\pi}{4} \int_1^{37} \sqrt{u} \,du \\ S &=\frac{\pi}{4} \left( \frac{2}{3}u^{3/2} \right) \Biggr|_{1}^{37} \end{aligned}
Now substituting for the limits, you get:
$$\begin{gathered} S=\frac{\pi}{6}(37^ \frac {3}{2} -1)\\ S \approx 117.3\, \text{square units}. \end{gathered}$$
Calculus of Parametric Curves - Key takeaways
• We can apply all the Calculus we've learned up to this point to parametric curves
• Then the derivative $$\frac{dy}{dx}$$ of a curve $$x(t)$$ and $$y(t)$$ is
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y'(t)}{x'(t)}.$
• The area under the curve $$x(t)$$ and $$y(t)$$ over the interval $$a \le t \le b$$ is
$\int_a^{b} y(t)x'(t)dt.$
• The arc length of the curve $$x(t)$$ and $$y(t)$$ over the interval $$a \le t \le b$$ is given by
$$S=\int_a^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \,dt.$$
• The surface area of the solid obtained by revolving the curve $$x(t)$$ and $$y(t)$$ around the $$x$$-axis is defined by
$S=2\pi \int_a^{b} y(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}}dt.$
• The surface area of the solid obtained by revolving the curve $$x(t)$$ and $$y(t)$$ around the $$y$$-axis is defined by
$S=2\pi \int_a^{b} x(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}}dt.$
The most important Calculus applications to parametric curves are finding a tangent line, finding the area under a parametric curve, calculating the arc length of a parametric curve, and finding the surface area to a parametric curve volume.
Set x(t) = t and plug in t to f(x) to derive y(t).
An example of applying Calculus to a parametric curve is finding the arc length of a parametrized curve over a certain interval.
To find the area under a parametric curve, integrate y(t)x'(t) over the given interval with respect to t.
Set x(t) = t and plug in t to f(x) to derive y(t).
Final Calculus of Parametric Curves Quiz
Question
What is the meaning of Calculus of a parametric equation?
You can apply all the Calculus you've learned up to this point to parametric curves. The most important Calculus applications to parametric curves are finding a tangent line, finding the area under a parametric curve, calculating the arc length of a parametric curve, and finding the surface area to a parametric curve volume.
Show question
Question
What is the formula for a derivative of a parametric equation?
$$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}$$
Show question
Question
How do we find the equation of a tangent line of a parametric equation?
Use the formula for finding the derivative of a parametric equation. This formula produces the slope of the tangent line at a given t. Then, find the corresponding x and y values of the curve in order to write the tangent line in point-slope form.
Show question
Question
What is the formula for the area under a parametric curve?
The area under the curve $$x(t)$$ and $$y(t)$$ over the interval $$a<t<b$$ is
$$\int_{a}^{b} y(t) \ x'(t) \,dt$$
Show question
Question
What is the formula for the arc length of a parametric curve?
The arc length of the curve $$x(t)$$ and $$y(t)$$ over the interval $$a<t<b$$ is
$$S=\int_a^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \,dt$$
Show question
Question
What is the formula for surface area of a parametric solid of revolution around the x-axis?
The surface area of the solid obtained by revolving the curve $$x(t)$$ and $$y(t)$$ around the x-axis is defined by
$$S=2\pi \int_a^{b} y(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}} \,dt$$
Show question
Question
What is the formula for surface area of a parametric solid of revolution around the y-axis?
The surface area of the solid obtained by revolving the curve $$x(t)$$ and $$y(t)$$ around the y-axis is defined by
$$S=2\pi \int_a^{b} x(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}} \,dt$$
Show question
Question
How do you find the parametric equation of a curve?
Set $$x=f(t)$$ and plug in $$f(t)$$ to get $$y(t)$$.
Show question
Question
For a parametric curve given by $$x(t)=t^3$$ and $$y(t)=t^2$$ on the interval $$0<t<1$$, which of the following is the correct surface area of the solid around a revolution about the x-axis?
$$S=\frac{\pi(494\sqrt{13}+128)}{1215}$$
Show question
Question
For which of the parametric equations, the curve formed by them has the arc length of $$3\pi$$, on the interval $$0 \le t \le \pi$$?
$$x=3 \cos t$$ and $$y=3 \sin t$$
Show question
Question
Is it true that for a given curve, the surface area of revolution is the same when revolved around the x-axis and the y-axis?
No.
Show question
Question
Eliminate the parameter for the equations $$x=\log_e t$$ and $$y=e^t$$ by writing $$y$$ as a function of $$x$$.
Solving for $$t$$:
\begin{aligned} &x=\log_e t \\ \Rightarrow \ &e^x=t \end{aligned}
Now substituting for $$t$$ in $$y=e^t$$:
$$y=e^{e^x}$$
Show question
Question
Is the method of arc length an approximating way of finding the length or a precise and exact way?
The method using calculus determines the arc length of any curve precisely and exactly, with no approximations.
Show question
Question
The area under a curve $$y=f(x)$$ is given using the classical method of integration. How will this formula differ for parametric equations?
For the parametric equation, the area under the curve is given by:
$$A=\int_a^{b} y(t)x'(t)dt$$
Show question
Question
What is the area under the curve whose parametric equations are $$x=t-\sin t$$ and $$y=1-\cos t$$ on the interval $$0 \le t \le 2\pi$$?
$$3\pi$$
Show question
Question
Motion in space refers to motion in ____ dimensions.
three.
Show question
Question
The Cartesian plane can be used to model motion in ____ dimensions.
two.
Show question
Question
A curve that can be drawn on a plane is known as a ____ curve.
plane.
Show question
Question
To plot a trajectory in space, you use three axes, which are known as the $$x-$$axis, the $$y-$$axis, and the ____.
$$z-$$axis.
Show question
Question
The function $$z=f(x,y)$$ drawn on the three-dimensional space represents a ____.
surface.
Show question
Question
In general, the intersection of two planes in space is a(n) ____.
curve.
Show question
Question
To describe motion in space you need ____ parametric equation(s).
three.
Show question
Question
The derivative of position with respect to time is known as ____.
velocity.
Show question
Question
The derivative of velocity with respect to time is known as ____.
acceleration.
Show question
Question
Acceleration is the ____ derivative of position with respect to time.
second.
Show question
Question
True/False: Speed is a synonym of velocity.
False.
Show question
Question
In problems about motion, the word initial usually refers to the instant when $$t$$ equals ____.
$$0$$.
Show question
Question
True/False: A trajectory can be described using parametric equations of one parameter.
True.
Show question
Question
True/False: To find the acceleration of an object, you just need to differentiate $$x(t)$$.
False.
Show question
Question
The position of an object is described by the parametric equations
$x(t)=t^2,$
$y(t) = 4,$
and
$z(t)=e^{-t}.$
Does this movement occur on a plane?
Yes.
Show question
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# Integration by Partial Fractions
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#### Integration by Partial Fractions - Lesson Summary
While evaluating integrals, where the integrand is a fraction of polynomials.
Integral contains P(x)/Q(x)
First, check whether the function is a proper fraction.
If the degree of the numerator is less than that of the denominator, then it is a proper fraction.
i .e Degree P(x) < Degree Q(x)
If the function is a proper fraction, then check whether the denominator can be factorised into linear or quadratic factors.
If the denominator cannot be split, then other integration methods are chosen.
If the denominator can be split, then split the function into partial fractions.
We will shortly discuss the ways to split a function into partial fractions.
Improper fraction P(x)/Q(x) Degree P(x) > Degree Q(x)
If the function is a not a proper fraction, that is, the degree of the numerator is greater than or equal to that of the denominator, then it is an improper fraction.
Thus, the degree of the numerator is greater than the degree of the denominator, and hence, it is an improper fraction.
In such cases, we convert the function into a proper fraction by dividing the numerator by the denominator.
After this, we check whether the denominator of this proper fraction can be factorised into linear or quadratic factors.
If the denominator can be split, then split P1(x)/Q(x) into partial fractions.
P(x)/Q(x) = T(x) + P1(x)/Q(x)
This is how we convert an improper fraction into a proper fraction.
Now we split a proper fraction into partial fractions.
Now px+q / (x-a)(x-b) = A/(x-a) + B/(x-b)
px+q / (x-a)(x-b) = A(x-b)+B(x-a) /(x-a)(x-b)
px+q = A(x-b)+B(x-a)
px+q = (A+B)x - (Ab + Ba)
Comparing the coefficients of x:
p = A + B .....(1)
q = -(Ab + Ba).....(2)
(1) x b + (2)
(1) x b bp = Ab + Bb .........(1)
(1) x b + (2): bp + q = (b - a)A
⇒ B = (bp + q)/(b-a)
(1) x a + (2)
(1) x a: ap = Aa + Ba
(1) x a + (2): ap + q = (a - b)A
A = (ap+q)/(a-b)
Ex:
∫ (7x - 4)/((x-1)2(x+2)) dx
Degree of numerator: 1 < Degree of denominator: 2
(7x - 4)/((x-1)2(x+2)) = A/(x-1) + B/(x-1)2 + C/(x+2)
⇒ (7x - 4)/((x-1)2(x+2)) = (A(x-1)(x+2) + B(x+2) + c(x-1)2)/((x-1)2(x+2))
7x - 4 = A(x-1)(x+2) + B(x+2) + c(x-1)2
⇒ 7x - 4 = A(x2 + x -2) + B(x+2) + C(x2 - 2x +1)
⇒ 7x - 4 = x2(A+c) + x(A+B-2C) + (-2A+2B+C)
Comparing the coefficients of x2:
A+C = 0 ……..(1)
Comparing the coefficients of x:
A+ B -2C = 7……..(2)
Comparing the constants:
-2A + 2B + C = -4………(3)
A = 2, B = 1 and C = -2
∫ (7x - 4)/((x-1)2(x+2)) dx = ∫ [A/(x-1) + B/(x-1)2 + C/(x+2)] dx
= ∫ [2/(x-1) + 1/(x-1)2 - 2/(x+2)] dx
= 2 log|x-1| - 1/(x-1) - 2log|x+2| + C ('.' ∫1/x2 dx = -1/x + C)
= 2 (log|x-1| - log|x+2|) - 1/(x-1) + C
= 2(log|(x-1)/(x+2)|) - 1/(x-1) + C
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# Three sides of a triangle are consecutive even integers and the perimeter of the triangle is 180 centimeters. What is the length of each of the three sides?
Apr 23, 2018
$58 , 60 , 62$
#### Explanation:
Three consecutive even numbers can be expressed in terms of $n$. If $n$ is an even number then
$\left(n - 2\right) , \left(n\right) , \left(n + 2\right)$
are three consecutive even numbers.
Therefore
$\left(n - 2\right) + \left(n\right) + \left(n + 2\right) = \text{180 cm}$
We can simplify so...
$3 n + 2 - 2 = 180$
$3 n + 0 = 180$
Divide both sides by $3$ to solve for $n$
$\frac{3 n}{3} = \frac{180}{3}$
$n = 60$
$60$ is an even integer. Input $60$ into the original equation.
$\left(n - 2\right) + \left(n\right) + \left(n + 2\right) = 180$
$\left(60 - 2\right) + \left(60\right) + \left(60 + 2\right) = 180$
$58 + 60 + 62 = 180 \to$ check!
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Every Question Helps You Learn
# Practice - Addition and Subtraction - 02
Hello, young mathematicians! Are you ready to put your addition and subtraction skills to the test? This quiz contains 10 questions that will challenge your understanding of these two fundamental maths operations. So, brace yourselves, it's time to add and subtract!
Question 1
There are 6 sandwiches on a plate and 2 are eaten. How many are left?
3
5
4
2
6 sandwiches - 2 eaten = 4 sandwiches. So, 4 sandwiches are left.
Question 2
You have a box with 7 pencils and you add 3 more. How many do you have now?
8
9
10
11
7 pencils + 3 pencils = 10 pencils. So, you have 10 pencils now.
Question 3
Sam had £20. He spent £5 on a book. How much does he have left?
£15
£16
£14
£13
£20 - £5 = £15. So, Sam has £15 left.
Question 4
There were 8 birds on a tree. 5 flew away. How many are left?
5
4
3
2
8 birds - 5 birds = 3 birds, so 3 birds are left on the tree.
Question 5
Emma is saving £1 coins. She has 3 and her dad gives her another 4. How many does she have now?
7
6
8
5
3 coins + 4 coins = 7 coins. So, Emma has 7 £1 coins now.
Question 6
There are 10 marbles in a bag. You add 5 more. How many are there now?
14
15
16
17
10 marbles + 5 marbles = 15 marbles. So, there are 15 marbles in the bag now.
Question 7
Olivia has 9 sweets and eats 6. How many are left?
13
7
4
3
9 sweets - 6 eaten = 3 sweets. So, Olivia has 3 sweets left.
Question 8
If you have 3 apples and you eat 1, how many are left?
4
3
2
1
3 apples - 1 apple = 2 apples, you are left with 2 apples.
Question 9
There were 15 balls in a basket. 7 were taken out. How many are left?
6
8
9
7
15 balls - 7 balls = 8 balls. So, there are 8 balls left in the basket.
Question 10
Jamie has £7.00 and his mum gives him another £2.00. How much does he have altogether?
£5.00
£9.00
£8.00
£10.00
£7.00 + £2.00 = £9.00, Jamie has £9.00 altogether.
Author: Graeme Haw
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# Here's How Your Watch Can Prove That 2 + 2 Doesn't Equal 4
2 + 2 = ? appears to be one of the easiest problems in mathematics, and it is probably one of the first you ever encountered. If Kate has 2 apples and Matt gives her 2 more apples, then she has 4 apples. Obviously.
But what if we told you that 2 + 2 = ? has stumped even some of the smartest mathematicians because it doesn’t necessarily have to equal 4?
You’re probably wondering how that’s possible.
## But First, Here's An Example
I get to work at 7 o’clock in the morning. This is what my watch looks like. It's first time that the smaller hand of the clock hits 7 on the clock’s face that day.
Later in the day, I leave work at 5 o’clock in the afternoon. When I look down at my wrist, this is what my watch looks like. This is the second time that the smaller hand of the clock hits 5 on the watch’s face. The first time was at 5 AM.
In other words, the smaller hand of my watch has hit all 12 numbers on the face, and then started again from 1. We can think of 1PM as 13 o’clock; 2 PM as 14 o’clock; and 5 PM as 17 o’clock.
However, most people don’t say: “I’ll be done with work at 17.” They generally say: “I’ll be done at 5.” If you do this as well, you’re actually solving a complicated math problem without even realizing.
## The Watch Operates In A Specific System
What’s happening is that the watch’s hands operate in a system (the watch’s face) that has 12 numbers, but the watch’s hands are attempting to represent a system, which has more than 12 numbers (in this case, the system is a day which has 24 hours).
We’re going to class the watch’s system “Modulo 12”, meaning that 12 is the highest number we can have on the watch which has the numbers 1, 2 , 3, 4 … all the way through 12. (Don’t freak out! Modulo is just the fancy math term for the math we are doing).
As a result, to understand how the 17th hour in the day is represented on the watch, one must do 17 (the number outside of Modulo 12) minus 12 (the maximum number in Modulo 12) which equals 5 (which is a number within Modulo 12 to represent a number outside of Modulo 12).
In other words, in the watch system, we can say that 12 + 5 = 5 because 5 represents 17.
Weirder still, even though you’d think that 13 +4 = 17, in this Modulo 12 system, 14 +4 = 6 because 6 pm represents “18 o’clock.”
## Now, Back To 2+2
Using what we learned here, let’s get back to 2 + 2 = ?.
Believe it or not, you can actually create a Modulo system with any numbers. It does not have to be limited to Modulo 12 like with the clock with the numbers 1 through 12.
Now, our new system is going to be Modulo 3 with the numbers 0, 1, 2. This is a little different from the watch, because a watch doesn’t have 0’s. Let’s quickly refresh what that means.
Modulo 3 with numbers 0, 1, 2 means that after we reach the third number in our set of numbers, we start counting from the first number again. In this case, after we reach 2, we start again with 0. This is just like with the watch, when after we reached 12, we started again with 1.
So now, let’s see what happens when we add 2 + 2 in a Modulo 3, (0,1,2) system.
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# Big Ideas Math Answers Grade 3 Chapter 13 Classify Two-Dimensional Shapes
Big Ideas Math Grade 3 Chapter 13 Classify Two-Dimensional Shapes Answers: Students who are studing grade 3 must know about the geomteric shapes like square, triangle, rectangle and others before learning about two dimensional shapes. You can get the direct link to download Big Ideas Math Book Grade 3 Chapter 13 Classify Two-Dimensional Shapes Answers pdf in the below sections. By learning this chapter, you will be ab;e to define various two dimensional shapes, explain their features and compare one with another and draw shapes.
## Big Ideas Math Book Grade 3 Chapter 13 Classify Two-Dimensional Shapes Solutions
After all these lessons, you can see performance task section where you can test your math skills. Improve your performance in the exam byreferring Big Ideas Math Answers Grade 3 Chapter 13 Classify Two-Dimensional Shapes AnswerKey. Scroll down to getthe quick links of all the topics of BIM Grade 3 Chapter 13 Classify Two-Dimensional Shapes.
Lesson 1: Identify Sides and Angles of Quadrilaterals
Lesson 2: Describe Quadrilaterals
Lesson 3: Classify Quadrilaterals
Lesson 4: Draw Quadrilaterals
### Lesson 13.1 Identify Sides and Angles of Quadrilaterals
Explore and Grow
Sort the Polygon Cards.
Structure
Does the sort change if you sort by the number of vertices? Explain.
Think and Grow: Sides and Angles of Quadrilaterals
A polygon is a closed, two-dimensional shape with three or more sides.
A quadrilateral is a polygon with four sides. Quadrilaterals have four vertices and four angles.
Quadrilaterals can have parallel sides and right angles. Parallel sides are two sides that are always the same distance apart. A right angle is an L-shaped angle.
When two straight lines intersect each other at 90˚ or are perpendicular to each other at the intersection, they form the right angle. A right angle is represented by the symbol ∟.
Parallel lines are lines in a plane that are always the same distance apart. Parallel lines never intersect. Parallel lines are represented as ll.
Example
Identify the number of right angles and pairs of parallel sides.
Right angles: ___
Pairs of parallel sides: ___
Right angles: 2 (The symbol in pink shows right angles)
Pairs of parallel sides: 1 ( The sides in yellow shows parallel lines)
Right angles: ___
Pairs of parallel sides: ___
Right angles: 4 (The symbols in pink are right angles).
Pairs of parallel sides: 2 (yellow and green sides are 2 pairs of parallel sides)
Show and Grow
Identify the number of right angles and pairs of parallel sides.
Question 1.
Right angles: ___
Pairs of parallel sides: ___
Explanation:
Right Angles : 0
Pairs of parallel sides: 2 (sides in yellow and green are 2 pairs of parallel sides)
Question 2.
Right angles: ___
Pairs of parallel sides: ___
Right angles: 2 (In L shape)
Pairs of parallel sides:0
Apply and Grow: Practice
Identify the number of right angles and pairs of parallel sides.
Question 3.
Right angles: ___
Pairs of parallel sides: ___
Right angles: 2
Pairs of parallel sides: 1 (1 II 2 as per figure).
Question 4.
Right angles: ___
Pairs of parallel sides: ___
Right angles: 4 (L shape)
Pairs of parallel sides: 2 (sides 1 ll 2 and 3 ll 4 as per above fig).
Question 5.
Right angles: ___
Pairs of parallel sides: ___
Right angles: 2 (L SHAPE)
Pairs of parallel sides: 2 (sides 1 ll 2 and 3 ll 4).
Question 6.
Right angles: ___
Pairs of parallel sides: ___
Right angles: 0
Pairs of parallel sides: 0
Question 7.
Right angles: ___
Pairs of parallel sides: ___
Right angles: 1
Pairs of parallel sides: 0
Question 8.
Right angles: ___
Pairs of parallel sides: ___
Right angles: 1
Pairs of parallel sides: 0
Question 9.
YOU BE THE TEACHER
Your friend says the yellow sides are parallel. Is your friend correct? Explain.
Think and Grow: Modeling Real Life
Use quadrilateral pattern blocks to complete the puzzle.
Show and Grow
Question 10.
Use quadrilateral pattern blocks to complete the puzzle.
### Identify Sides and Angles of Quadrilaterals Homework & Practice
Identify the number of right angles and pairs of parallel sides.
Question 1.
Right angles: ___
Pairs of parallel side: ___
Right angles: 0
Pairs of parallel side: 2 (side 1 ll 2 and 3 ll 4)
Question 2.
Right angles: ___
Pairs of parallel side: ___
Right angles: 2
Pairs of parallel side: 1 (side 1 ll 2).
Question 3.
Right angles: ___
Pairs of parallel side: ___
Right angles: 0
Pairs of parallel side: 0
Question 4.
Right angles: ___
Pairs of parallel side: ___
Right angles: 1 (angle between side 1 and 2 )
Pairs of parallel side: 0
Question 5.
Right angles: ___
Pairs of parallel side: ___
Right angles: 0
Pairs of parallel side: 3 (side 1 ll 5, 2 ll 6, 3 ll 7 and 4 ll 8 as per above figure)
Question 6.
Right angles: ___
Pairs of parallel side: ___
Right angles: 3 ( angles between side 2&3, side 4&5 and 1&5)
Pairs of parallel side: 1 ( sides 1 ll 4).
Question 7.
Reasoning
Can a quadrilateral have exactly three right angles? Explain.
Explanation:
The types of quadrilaterals that have 3 right angles are known as:
1.Squares
2.Rectangles
3.Other shapes where all angles are 90°.
The reason for this is:
All quadrilaterals interior angles must add up to exactly 360°.
So:
= 360−(90+90+90)
= 90°.
And thus, the fourth angle must be 90°.
Thefore, the only quadrilaterals where all angles are 90° are squares and rectangles.
Question 8.
Modeling Real Life
Use pattern blocks to complete the puzzle.
Review & Refresh
Question 9.
Newton has 28 cards. Descartes has 24 cards. Newton divides his cards into 4 equal stacks and gives Descartes one stack. How many cards does Descartes have now?
Explanation:
Given:
Cards with Descartes = 24 cards
Cards with Netwon = 28 cards
Cards divided into 4 stacks .
Each stack contains 7 cards
Total cards with Descartes= 24+7= 31 Cards.
### Lesson 13.2 Describe Quadrilaterals
Explore and Grow
Sort the Quadrilateral Cards.
Structure
What is another way you can sort the quadrilaterals?
Think and Grow: Idenify Quadrilaterals
You can identify a quadrilateral using its sides and angles. A quadrilateral can have more than one name.
4 sides
4 angles
Trapezoid
exactly 1 pair of parallel sides
Parallelogram
2 pairs of parallel sides
Rectangle
2 pairs of parallel sides
4 right angles
Rhombus
2 pairs of parallel sides
4 equal sides
square
2 pairs of parallel sides
4 equal sides
4 right angles
Example
Circle all of the names for the quadrilateral.
Show and Grow
Circle all of the names for the quadrilateral.
Question 1.
Question 2.
Apply and Grow: Practice
Write all of the names for the quadrilateral.
Question 3.
Rhombus
2 pairs of parallel sides
4 equal sides
Question 4.
Trapezoid – only 1 pair of parallel sides
Name all of the quadrilaterals that can have the given attribute.
Question 5.
2 pairs of parallel sides
Square, Rhombus , Rectangle and Parallelogram.
Question 6.
4 right angles
Square and Rectangle.
Question 7.
Precision
Is the shape a rhombus? Explain.
No.
Explanation:
Rhombus contains 2 pairs of parallel sides and 4 equal sides but the given figure doesnot have equal sides and parallel sides.
Question 8.
YOU BE THE TEACHER
Your friend not says the shape is a rhombus. Is your friend correct? Explain.
Yes .
Explanation:
Rhombus contain 2 pairs of parallel sides and 4 equal sides. but it doesn’t contain 4 right angles. so it is square not rhombus.
Think and Grow: Modeling Real Life
Write all of the names for the red quadrilateral in the painting.
Show and Grow
Use the painting above.
Question 9.
Write all of the names for the blue quadrilateral.
1 Square and 4 triangles.
Question 10.
What color is the rhombus that is not a square?
Purple.
Question 11.
How many trapezoids are in the painting? Circle them.
Question 12.
DIG DEEPER!
There are 4 squares and 8 rectangles in a floortile pattern. Find the total number of right angles in the pattern. Explain.
one square contain 4 right angles
4 squares = 4×4=16 right angles
one rectangle contain 4 right angles
8 rectangles = 8×4=32 right angles.
Therefore Total Right angles = 16 +32 = 48 angles.
### Describe Quadrilaterals Homework & Practice 13.2
Circle all of the names for the quadrilateral.
Question 1.
Question 2.
Write all of the names for the quadrilateral.
Question 3.
Parallelogram.
Explanation:
It contains 2 pairs of parallel sides.
Question 4.
Square .
Explanation:
It contains 2 pairs of parallel sides, 4 equal sides and 4 right angles.
Name all of the quadrilaterals that can have the given attributes.
Question 5.
4 sides and 4 angles
Question 6.
exactly 1 pair of parallel sides
Trapezoid.
Question 7.
Writing
Explain how a trapezoid is different from a parallelogram.
A Trapezoid contains only 1 pair of parallel sides where as, parallelogram contains 2 pairs of parallel sides .
Question 8.
Reasoning
Explain why the rectangle shown is not a square.
As per the figure only opposite sides are equal but where as square contains all 4 sides equal.
Question 9.
DIG DEEPER!
What is Descartes’s shape?
Rectangle.
Question 10.
Which One Doesn’t Belong?
Which not does belong with the other three? Explain.
Trapezoid as it contains one one parallel side.
Modeling Real Life
Use the mosaic
Question 11.
Write all of the names for the purple quadrilateral.
Triangle and Rhombus
Question 12.
How many parallelograms are in the mosaic? Circle them.
Review & Refresh
Question 13.
Find the area of the rectangle.
___6_ × __2_ = ______
Area = _________12 sqft____
Explanation:
Area of rectangle is length x Breadth.
Question 14.
_4___ × 5__ = _____
Area = ____________
Area=20 sqcms.
Explanation:
Area of rectangle is length x Breadth.
### Lesson 13.3 Classify Quadrilaterals
Explore and Grow
Use each description to model a quadrilateral on your geoboard. Draw each quadrilateral.
Two pairs of parallel sides
Exactly one pair of parallel sides
Four right angles that do not form a square
Two pairs of parallel sides and no right angles
Structure
Different Answer for 1 condition because Two pairs of parallel sides are rectangle,square,rhombus and parallelogram.
Same Answers for 2,3 and 4 conditions.
Think and Grow:
How are the parallelograms and rhombuses alike? How are they different?
Ways they are alike:
Each has ___ sides.
Each has ___ angles.
Each has ___ pairs of parallel sides.
A parallelogram is a rhombus if the diagonals are perpendicular.
Ways they are alike:
Each has __4_ sides.
Each has __2_ angles.
Each has __2_ pairs of parallel sides.
Ways they are different:
Rhombus always have __4__ equal sides.
What names can you use to classify all parallelograms and rhombuses?
Squares and Rectangles.
A rectangle has two sets of parallel sides, so it is a parallelogram.
A rhombus has four congruent sides and a square has four congruent sides and angles. Therefore, a rhombus is a square when it has congruent angles.
Show and Grow
Question 1.
How are squares and rectangles alike? How are they different?
Squares and rectangles both are a type of quadrilateral, both have their four interior angles equal to 90 degrees and opposite sides parallel to each other.
The difference between square and rectangle is that a square has all its sides equal whereas a rectangle has its opposite sides equal.
Question 2.
What names can you use to classify all squares and rectangles?
Parallelogram because A Rectangle is a Parallelogram with all right angles. A Square is a Rectangle with adjacent sides equal.
Question 3.
Draw a quadrilateral that is not a square or a rectangle. Explain.
Explanation:
Trapezoids contain only 1 pair of parallel sides.
Apply and Grow: Practice
Question 4.
How are trapezoids and parallelograms alike? How are they different?
A trapezoid has one pair of parallel sides and a parallelogram has two pairs of parallel sides.
A trapezoid can be called a parallelogram when it has more than one pair of parallel sides.
Question 5.
What name can you use to classify all trapezoids and parallelograms?
Question 6.
How are rhombuses and squares alike? How are they different?
Rhombuses and Square are alike as they contain 2 pairs of parallel sides and 4 equal sides. Different as Square have 4 right angles, but rhombus does not.
Question 7.
What names can you use to classify all rhombuses and squares?
Parallelogram.
Question 8.
Draw a quadrilateral that is a rhombus but not a square.
Question 9.
Draw a quadrilateral that is not a rhombus or a square. Explain.
Think and Grow: Modeling Real Life
Sort the road signs into two groups by shape. What is alike and what is different between the two groups? What name can you use to classify all of the road sign shapes?
Explanation:
Rectangles contain 2 pairs of parallel sides but where as , Trapezoid contain only 1 pair of parallel sides .
Show and Grow
Question 10.
Sort the road signs into two groups by shape. What is alike and what is different between the two groups? What names can you use to classify all of the road sign shapes?
Explanation:
Square and Rectangles are alike as they contain 2 pairs of parallel sides and
4 right angles. they are different as square contain all sides equal but rectangle contain only opposite sides equal.
### Classify Quadrilaterals Homework & Practice 13.3
Question 1.
How are rectangles and rhombuses alike? How are they different?
Rectangle and Rhombus are alike because both contain 2 pairs of parallel lines. they are different because rectangle contain 4 right angles and opposite sides equal.Rhombus contain all sides equal with 0 right angles.
Question 2.
What names can you use to classify all rectangles and rhombuses?
Parallelogram.
Question 3.
DIG DEEPER!
Your friend says a shape is a rectangle. Newton says the same shape is a rhombus, and Descartes says it is a square. Can they all be correct? Explain.
In rhombus,Square and Rectangle all the opposite sides are equal.
Question 4.
YOU BE THE TEACHER
Is Newton correct? Explain.
Trapezoids have only one pair of parallel sides; parallelograms have two pairs of parallel sides. A trapezoid can never be a parallelogram.
Question 5.
Modeling Real Life
Sort the earrings into two groups by shape. What is alike and what is different between the two groups? What name can you use to classify all of the earring shapes?
Explanation:
Trapezoid contain only 1 pair of parallel sides .Square contains 2 pairs of parallel sides .
Review & Refresh
Question 6.
Use the number line to find an equivalent fraction.
Question 7.
Write two fractions that name the point shown.
### Lesson 13.4 Draw Quadrilaterals
Explore and Grow
Model each quadrilateral on your geoboard. Move one vertex of each quadrilateral to create a new quadrilateral. Draw each new quadrilateral.
Structure
Think and Grow: Draw Quadrilaterals
Example
Draw a quadrilateral that has four right angles. Name the quadrilateral.
Square.
Example
Below are three rhombuses. Draw a quadrilateral not that is a rhombus. Explain why it is not a rhombus.
The quadrilateral is not a rhombus because _______
______________________________________.
Explanation:
The quadrilateral is not a rhombus because rectangle contains only opposite sides equal .
Show and Grow
Question 1.
Draw a quadrilateral that has exactly one pair of parallel sides. Name the quadrilateral.
Trapezoid.
Question 2.
Draw a quadrilateral that is not a square. Explain why it is not a square.
Explanation:
Square contains 2 pairs of parallel sides , 4 equal sides and 4 right angles.
Rectangle contains 2 pairs of parallel sides and 4 right angles .
Apply and Grow: Practice
Name the group of quadrilaterals. Then draw a quadrilateral that does not belong in the group. Explain why it does not belong.
Question 3.
Explanation:
Trapezoid contains only 1 pair of parallel side but whereas rhombus contain 2 pairs of parallel sides .
Question 4.
Explanation:
Rectangle and Square contains 2 pairs of parallel sides but whereas Trapezoid contains only 1 pair of parallel sides .
Question 5.
Precision
Circle the quadrilaterals that are not rhombuses.
Question 6.
YOU BE THE TEACHER
Your friend draws the shape and says it is a parallelogram because it has two pairs of parallel sides. Is your friend correct? Explain.
It is a pentagon as it contains 5 sides, whereas parallelogram contain only 4 sides.
Question 7.
DIG DEEPER!
Draw the quadrilateral with two pairs of parallel sides. One side is given.
Think and Grow: Modeling Real Life
A helicopter travels to various Colorado cities. Draw to show a route that forms a parallelogram. Write the names of the cities you use.
Sterling, eckley, pritchett and La junta forms parallelogram .
Show and Grow
Question 8.
Use the map above. Drawtoshow a route that forms a trapezoid. Write the names of the cities you use.
Meeker,walden,loveland and cedaredge.
Question 9.
You have four markers of equal length. Name all of the quadrilaterals you can make using the markers as sides.
Rhombus and Square .
Question 10.
DIG DEEPER!
Use a ruler to draw a trapezoid for each description.
• one side length of 1 inch
• one side length of 1 inch
• one side length of 2 inches
• one side length of 2 inches
• two right angles
• no right angles
Trapezoid with 2 right angles
### Draw Quadrilaterals Homework & Practice 13.4
Question 1.
Draw a quadrilateral that has two pairs of parallel sides. Name the quadrilateral.
Question 2.
Draw a quadrilateral that is not a rhombus.
Question 3.
Name the group of quadrilaterals. Then draw a quadrilateral that does not belong in the group. Explain why it does not belong.
Explanation:
Square, Rhombus and Rectangle contains 2 pairs of parallel sides. where Trapezoid contains only 1 pair of parallel side .
Question 4.
Precision
Circle the quadrilaterals that are not trapezoids.
Question 5.
Modeling Real Life
A bus tour wants to travel to various locations. Draw to show a route that forms a rectangle. Write the names of the location drop-off points you use.
The Drop off point is The Discovery of king Tut exhibit .
Review & Refresh
Question 6.
Tell whether the shape shows equal parts or unequal parts. If the shape shows equal parts, then name them
_____ parts _____
11 square parts – equal parts .
8 rectangle parts – unequal parts .
### Classify Two-Dimensional Shapes Performance Task
Question 1.
Your teacher wants to create cards using the fossils below. Use each polygon description to write the fossil name on the correct card.
Question 2.
Your teacher uses the cards and a sandbox to create an archaeological dig site. Your teacher lays a grid with eight grid squares over the top of the sandbox. Each square has 10-inch side lengths. How many square inches is the bottom of the sandbox?
### Classify Two-Dimensional Shapes Activity
Directions:
1. Players take turns rolling a die.
2. On your turn, move your piece the number of spaces shown on the die.
3. Cover a space on the board that describes the shape where you landed.
4. If there are no spaces that match your shape, then you lose your turn.
5. The first player to cover six spaces wins!
### Classify Two-Dimensional Shapes Chapter Practice
13.1 Identify Sides and Angles of Quadrilaterals
Identify the number of right angles and pairs of parallel sides.
Question 1.
Right angles: ____
Pairs of parallel sides: ___
Right angles: __0__
Pairs of parallel sides: __2_( sides 1 ll 2 and 3 ll 4 ) .
Question 2.
Right angles: ____
Pairs of parallel sides: ___
Right angles: 2
Pairs of parallel sides: 0
Question 3.
YOU BE THE TEACHER
Your friend says the yellow sides are parallel. Is your friend correct? Explain.
Yes.
Explanation:
It is a Trapezoid it contains only 1 pair of parallel sides.
Question 4.
Write all of the names for the quadrilateral.
Rectangle
Question 5.
Name all of the quadrilaterals that can have no right angles.
Trapezoid, Rhombus and Parallelogram.
Question 6.
How are squares and trapezoids alike? How are they different?
Both are quadrilaterals, so both of them have four sides. The square has two pairs of parallel sides, while the trapezoid only has one pair of parallel sides. So, we can say that both have one pair of parallel sides in common.
squares have all sides equal . but whereas trapezoid sides are different.
squares have 4 right angles but whereas trapezoid have different angles.
Question 7.
What name can you use to classify all squares and trapezoids?
Question 8.
Draw a quadrilateral that has four right angles. Name the quadrilateral.
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# Geometric Proof - Three different sized circles with tangent line.
Two different sized circles lie on an line, touching each other at a point. A small circle is inscribed in the space between. How does its radius depend on the radii of the two larger circles?
• I solved your problem. If you want to see my solution, show us your attempts. – Michael Rozenberg Feb 25 at 19:32
We have 3 triangles such that each are tangent to each other and each are tangent to some line. Lets say the radius a,b are given, and we must find c.
We can use the Pythagorean theorem to find the lengths $$x, y$$ and $$x+y$$ in terms of $$a,b,c$$
Then $$x + y = x+y$$ should give you enough information to solve for $$c$$ in terms of $$a,b.$$
Let $$k_n$$ be the curvature (i.e. $$=r_n^{-1})$$ of the $$n$$th circle
In virtue of Descartes' theorem, the curvature $$k_4$$ from the radius of the smaller circle is $$k_4=k_1+k_2+k_3\pm2\sqrt{k_1k_2+k_2k_3+k_1k_3}$$
Since one of the 'circles' is a line, its curvature is $$0$$. Thus
$$k_4=k_1+k_2\pm2\sqrt{k_1k_2}$$
The $$\pm$$ symbol depends on how the fourth circle is tangent to the others:
Since you mentioned that the fourth circle would be 'a small circle', I guess your formula turns into
$$k_4=k_1+k_2+2\sqrt{k_1k_2}$$
Once obtained the value of $$k_4$$, simply use the fact that $$k_4=\frac{1}{r_4}\iff r_4=\frac{1}{k_4}$$
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## Miscellaneous Topics for Calculus Applied to the Real World This Topic: Graphing the Derivative
To begin, we recall two basic facts about the derivative f'(x) of a function f(x):
1. The value f'(a) of f'(x) at x = a is the slope of the tangent to the graph of the function f at the point where x = a.
2. f'(x) is a function of x: the slope at a point on the graph depends on the x-coordinate of that point.
The graph of the derivative function f'(x) gives us interesting information about the original function f(x). The following example shows us how to sketch the graph of f'(x) from a knowledge of the graph of f(x).
Example 1 Sketching the Graph of the Derivative
Let f(x) have the graph shown below.
Give a rough sketch of the graph of f'(x).
Solution
Remember that f'(x) is the slope of the tangent at the point (x, f(x)) on the graph of f. To sketch the graph of f', we make a table with several values of x (the corresponding points are shown on the graph) and rough estimates of the slope of the tangent f'(x).
x 0 0.5 1 1.5 2 2.5 3 f'(x) 3 0 4 3 0 1 0
(Note that rough estimates are the best we can do; it is difficult to measure the slope of the tangent accurately without using a grid and a ruler, so we couldn't reasonably expect two people's estimates to agree. However, all that is asked for is a rough sketch of the derivative.) Plotting these points suggests the curve shown below.
Notice that the graph f'(x) intersects the x-axis at points that correspond to the high and low points on the graph of f(x). Why is this so?
Here is a more interactive example.
Example 2 Graph of Derivative
Let f(x) have the graph shown below.
Complete the following table, giving rough estimates of the slope of the tangent f'(x) at the given values of x.
x 3 2 1 0 1 2 3 f'(x) Peek at Answers
Now plot these points, and hence make a rough sketch of the graph of f'(x). Which of the following best approximates your sketch of the graph of f'(x)? (click on one)
Last Updated:November, 1997
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# Factors of 21: Understanding Prime Factors and Pairs
If you’ve ever encountered a math problem that required finding factors, then you know how important this concept is. In this article, we will explore what factors are and delve into the factors of 21, including prime factors and factor pairs.
• What are Factors?
• Factors of 21
• Prime Factors of 21
• Factor Pairs of 21
• How to Find Factors of 21
• Properties of Factors
• Applications of Factors
• Conclusion
• FAQs
## What are Factors?
Factors refer to numbers that can divide another number without leaving a remainder. For example, the factors of 21 are numbers that can divide 21 without leaving a remainder.
In simpler terms, if you multiply two factors of a number, you should get the original number. For instance, the factors of 21 are 1, 3, 7, and 21 since:
• 1 x 21 = 21
• 3 x 7 = 21
## Factors of 21
Now that we understand what factors are, let’s delve into the factors of 21. The factors of 21 are 1, 3, 7, and 21.
## Prime Factors of 21
Prime factors refer to the factors of a number that are also prime numbers. Prime numbers are those that are only divisible by 1 and the number itself. The prime factors of 21 are 3 and 7 since both are prime numbers, and 3 x 7 = 21.
## Factor Pairs of 21
Factor pairs refer to the factors of a number that are multiplied together to give the original number. For instance, the factors of 21 in pairs include:
• (1, 21)
• (3, 7)
It’s important to note that when listing factor pairs, order doesn’t matter. Therefore, (1, 21) and (21, 1) represent the same factor pair.
## How to Find Factors of 21
To find the factors of 21, you can start by dividing the number by 1 and itself. Then, divide by 2 and 3, and so on until you reach a quotient that is equal to or less than the divisor.
Once you get a factor, divide the original number by the factor to get another factor. Repeat the process until you find all the factors. You can also list out all the numbers that multiply together to give the original number to find factor pairs.
## Properties of Factors
Factors come with unique properties that make them crucial in mathematics. For instance:
• Every number has at least two factors, 1 and itself.
• Prime numbers have only two factors: 1 and the number itself.
• Every factor pair has an even number of factors.
• A number is said to be prime if it only has two factors.
## Applications of Factors
Factors are applied in numerous fields, including mathematics, computer science, and cryptography. Some of the common applications of factors include:
• Finding the greatest common factor (GCF) of two numbers
• Simplifying fractions
• Factoring polynomials
• Encryption and decryption in cryptography
## Conclusion
In conclusion, understanding factors is essential in mathematics, and mastering prime factors and factor pairs is crucial. In this article, we have explored the factors of 21, including prime factors and factor pairs. We have also discussed how to find factors, the properties of factors, and their applications. With this knowledge, you can easily solve problems that involve factors and apply them in practical situations.
## FAQs
1. What are the factors of 21 besides 1 and 21?
• The other factors of 21 are 3 and 7.
1. Is 21 a perfect number?
• No, 21 is not a perfect number since the sum of its proper divisors (1, 3, and 7) is not equal to the original number.
1. What is the difference between prime factors and composite factors?
• Prime factors are factors that are also prime numbers, while composite factors are factors that are not prime.
1. How many factor pairs does 21 have?
• 21 has two factor pairs: (1, 21) and (3, 7).
1. Why are factors important in mathematics?
• Factors are vital in mathematics since they help us solve problems involving multiplication, division, simplification of fractions, and factoring polynomials, among others.
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<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
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Proportions Using Cross Products
Have you ever been on a swim team? Do you know how to relate proportions to real - world situations? Take a look at this one.
Tony also works at the supermarket, but at school, he is on the swim team.
Tony swims 10 laps in 30 minutes. How long does it take him to swim 15 laps?
Solving this problem involves proportions and cross- products. You will know how to figure this out by the end of the Concept.
### Guidance
Proportions are everywhere in the world around us. Proportions are comparisons that we make between different things. You will often hear the words “in proportion” meaning that there is a relationship between things.
What is the relationship of a proportion? That is exactly what this Concept is going to work on.
What is a proportion?
A proportion is two equal ratios. Remember that a ratio compares two quantities; well, a proportion compares two equal ratios.
While ratios can be written in three different ways, often you will see proportions written as equal fractions. Let’s look at a proportion.
Here we have two ratios. We have four compared to twelve and one compared to three. These two ratios form a proportion. Simplified, they equal the same thing. You can simplify four-twelfths and it equals one-third.
Sometimes one of the trickiest things is figuring out if two ratios form a proportion. In the example above, we can see the equals sign letting us know that the ratios form a proportion.
How can we tell if two ratios form a proportion?
There are two different ways to figure this out. The first has already been mentioned and that is to simplify the two ratios and see if they are equal.
and
One-fourth is already in simplest form, we leave that one alone. If we simplify five-twentieths, we get one-fourth as an answer. One-fourth is equal to one-fourth, so these two ratios do form a proportion.
and
If we simplify these two fractions we get two different answers. Two-eighths simplifies to one-fourth. Three-sixths simplifies to one-half. The two ratios are not equal. Therefore, they DO NOT form a proportion.
The second way of figuring out if two ratios form a proportion is to cross multiply or to use cross products.
What is a cross product?
A cross product is when you multiply the numerator of one ratio with the denominator of another. Essentially you multiply on the diagonals. If the product is the same, then the two ratios form a proportion.
and
Let’s use cross products here.
The two ratios form a proportion.
We can use cross products to figure out whether or not two ratios form a proportion.
Try a few of these on your own. Use cross products to determine if the two ratios form a proportion. Write yes if they form a proportion and no if they do not.
#### Example A
and
Solution: Not a proportion
and
Solution: Yes
#### Example C
and
Solution: Not a proportion
Now back to Tony and the swim team. Here is the original problem once again.
Tony also works at the supermarket, but at school, he is on the swim team.
Tony swims 10 laps in 30 minutes. How long does it take him to swim 15 laps?
Our first step is to write two ratios.
This is our known information.
This is what we are trying to figure out.
Notice that we put the same unit in the numerator of both ratios and the same unit in the denominator of both units.
Now we can write a proportion.
Our answer is not obvious in this problem. Because of this, we need to use cross - products. We multiply 10 times and get and then we multiply 15 times 30 and get 450.
We can ask ourselves, “what times ten will give me 450?” or we can simplify the zeros and solve.
Here is our answer if we simplify the zeros. 1 times 45 equals 45.
Or we can think “10 times 45 equals 450.”
### Vocabulary
Proportion
two equal ratios.
Ratio
a comparison of two quantities can be written in fraction form, with a colon or with the word “to”.
Cross Products
to multiply the diagonals of each ratio of a proportion.
### Guided Practice
Here is one for you to try on your own.
Do these two ratios form a proportion? Why or why not?
To figure this out, we use cross - products.
6 x 4.5 = 27
9 x 3 = 27
The cross - products are equal.
These two ratios form a proportion.
### Practice
Directions: Use cross products or simplifying to identify whether each pair of ratios form a proportion. If they do, write yes. If not, write no.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
### Vocabulary Language: English
Cross Products
Cross Products
To simplify a proportion using cross products, multiply the diagonals of each ratio.
Proportion
Proportion
A proportion is an equation that shows two equivalent ratios.
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NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3
# NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3
## NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3
NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3 are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT Solutions for Class 11 Maths chapter 6 Linear Inequalities Ex 6.3.
Solve the following system of inequalities graphically:
### Ex 6.3 Class 11 Maths Question 1.
x ≥ 3, y ≥ 2
Solution:
x ≥ 3 … (1)
y ≥ 2 … (2)
The graph of the lines, x = 3 and y = 2, are drawn in the figure below.
Inequality (1) represents the region on the right hand side of the line, x = 3 (including the line x = 3), and inequality (2) represents the region above the line, y = 2 (including the line y = 2).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown below.
### Ex 6.3 Class 11 Maths Question 2.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution:
3x + 2y ≤ 12 … (1)
x ≥ 1 … (2)
y ≥ 2 … (3)
The graphs of the lines, 3x + 2y = 12, x = 1 and y = 2, are drawn in the figure below.
Inequality (1) represents the region below the line, 3x + 2y = 12 (including the line 3x + 2y = 12). Inequality (2) represents the region on the right side of the line, x = 1 (including the line x = 1). Inequality (3) represents the region above the line, y = 2 (including the line y = 2).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown below.
### Ex 6.3 Class 11 Maths Question 3.
2x + y ≥ 6, 3x + 4y ≤ 12
Solution:
2x + y ≥ 6 … (1)
3x + 4y ≤ 12 … (2)
The graph of the lines, 2x + y = 6 and 3x + 4y = 12, are drawn in the figure below.
Inequality (1) represents the region above the line, 2x + y= 6 (including the line 2x + y= 6), and inequality (2) represents the region below the line, 3x + 4y =12 (including the line 3x + 4y =12).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown below.
### Ex 6.3 Class 11 Maths Question 4.
x + y ≥ 4, 2x – y > 0
Solution:
x + y ≥ 4 … (1)
2x – y > 0 … (2)
The graph of the lines, x + y = 4 and 2x – y = 0, are drawn in the figure below.
Inequality (1) represents the region above the line, x + y = 4 (including the line x + y = 4).
It is observed that (1, 0) satisfies the inequality, 2x – y > 0. [2(1) – 0 = 2 > 0]
Therefore, inequality (2) represents the half plane corresponding to the line, 2x – y = 0, containing the point (1, 0) [excluding the line 2x – y > 0].
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on line x + y = 4 and excluding the points on line 2x – y = 0 as shown below.
### Ex 6.3 Class 11 Maths Question 5.
2x – y > 1, x – 2y < -1
Solution:
2x – y > 1 … (1)
x – 2y < –1 … (2)
The graph of the lines, 2x – y = 1 and x – 2y = –1, are drawn in the figure below.
Inequality (1) represents the region below the line, 2x – y = 1 (excluding the line 2x – y = 1), and inequality (2) represents the region above the line, x – 2y = –1 (excluding the line x – 2y = –1).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region excluding the points on the respective lines as shown below.
### Ex 6.3 Class 11 Maths Question 6.
x + y ≤ 6, x + y ≥ 4
Solution:
x + y ≤ 6 … (1)
x + y ≥ 4 … (2)
The graph of the lines, x + y = 6 and x + y = 4, are drawn in the figure below.
Inequality (1) represents the region below the line, x + y = 6 (including the line x + y = 6), and inequality (2) represents the region above the line, x + y = 4 (including the line x + y = 4).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown below.
### Ex 6.3 Class 11 Maths Question 7.
2x + y ≥ 8, x + 2y ≥ 10
Solution:
2x + y = 8 … (1)
x + 2y = 10 … (2)
The graph of the lines, 2x + y = 8 and x + 2y = 10, are drawn in the figure below.
Inequality (1) represents the region above the line, 2x + y = 8, and inequality (2) represents the region above the line, x + 2y = 10.
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown below.
### Ex 6.3 Class 11 Maths Question 8.
x + y ≤ 9, y > x, x ≥ 0
Solution:
x + y ≤ 9 ... (1)
y > x ... (2)
x ≥ 0 ... (3)
The graph of the lines, x + y = 9 and y = x, are drawn in the figure below.
Inequality (1) represents the region below the line, x + y = 9 (including the line x + y = 9).
It is observed that (0, 1) satisfies the inequality, y > x. [1 > 0]
Therefore, inequality (2) represents the half plane corresponding to the line, y = x, containing the point (0, 1) [excluding the line y = x].
Inequality (3) represents the region on the right hand side of the line, x = 0 or y-axis (including y-axis).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the lines, x + y = 9 and x = 0, and excluding the points on line y = x as shown below.
### Ex 6.3 Class 11 Maths Question 9.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution:
5x + 4y ≤ 20 … (1)
x ≥ 1 … (2)
y ≥ 2 … (3)
The graph of the lines, 5x + 4y = 20, x = 1 and y = 2, are drawn in the figure below.
Inequality (1) represents the region below the line, 5x + 4y = 20 (including the line 5x + 4y = 20). Inequality (2) represents the region on the right hand side of the line, x = 1 (including the line x = 1). Inequality (3) represents the region above the line, y = 2 (including the line y = 2).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown below.
### Ex 6.3 Class 11 Maths Question 10.
3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Solution:
3x + 4y ≤ 60 … (1)
x + 3y ≤ 30 … (2)
The graph of the lines, 3x + 4y = 60 and x + 3y = 30, are drawn in the figure below.
Inequality (1) represents the region below the line, 3x + 4y = 60 (including the line 3x + 4y = 60), and inequality (2) represents the region below the line, x + 3y = 30 (including the line x + 3y = 30).
Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective line and the axes represents the solution of the given system of linear inequalities.
### Ex 6.3 Class 11 Maths Question 11.
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Solution:
2x + y ≥ 4 … (1)
x + y ≤ 3 … (2)
2x – 3y ≤ 6 … (3)
The graph of the lines, 2x + y = 4, x + y = 3, and 2x – 3y = 6, are drawn in the figure below.
Inequality (1) represents the region above the line, 2x + y = 4 (including the line 2x + y = 4). Inequality (2) represents the region below the line, x + y = 3 (including the line x + y = 3). Inequality (3) represents the region above the line, 2x – 3y = 6 (including the line 2x – 3y = 6).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown below.
### Ex 6.3 Class 11 Maths Question 12.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution:
x – 2y ≤ 3 … (1)
3x + 4y ≥ 12 … (2)
y ≥ 1 … (3)
The graph of the lines, x – 2y = 3, 3x + 4y = 12 and y = 1, are drawn in the figure below.
Inequality (1) represents the region above the line, x – 2y = 3 (including the line x – 2y = 3). Inequality (2) represents the region above the line, 3x + 4y = 12 (including the line 3x + 4y = 12). Inequality (3) represents the region above the line, y = 1 (including the line y = 1).
The inequality, x ≥ 0, represents the region on the right hand side of y-axis (including y-axis).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines and y- axis as shown below.
### Ex 6.3 Class 11 Maths Question 13.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
Solution:
4x + 3y ≤ 60 … (1)
y ≥ 2x … (2)
x ≥ 3 … (3)
The graph of the lines, 4x + 3y = 60, y = 2x and x = 3, are drawn in the figure below.
Inequality (1) represents the region below the line, 4x + 3y = 60 (including the line 4x + 3y = 60). Inequality (2) represents the region above the line, y = 2x (including the line y = 2x). Inequality (3) represents the region on the right hand side of the line, x = 3 (including the line x = 3).
Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown below.
### Ex 6.3 Class 11 Maths Question 14.
3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0
Solution:
3x + 2y ≤ 150 … (1)
x + 4y ≤ 80 … (2)
x ≤ 15 … (3)
The graph of the lines, 3x + 2y = 150, x + 4y = 80 and x = 15, are drawn in the figure below.
Inequality (1) represents the region below the line, 3x + 2y = 150 (including the line 3x + 2y = 150). Inequality (2) represents the region below the line, x + 4y = 80 (including the line x + 4y = 80). Inequality (3) represents the region on the left hand side of the line, x = 15 (including the line x = 15).
Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities.
### Ex 6.3 Class 11 Maths Question 15.
x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Solution:
x + 2y ≤ 10 … (1)
x + y ≥ 1 … (2)
x – y ≤ 0 … (3)
The graph of the lines, x + 2y = 10, x + y = 1 and x – y = 0, are drawn in the figure below.
Inequality (1) represents the region below the line, x + 2y = 10 (including the line x + 2y = 10). Inequality (2) represents the region above the line, x + y = 1 (including the line x + y = 1). Inequality (3) represents the region above the line, x – y = 0 (including the line x – y = 0).
Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities.
You can also like these:
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 12
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Found in: Page 209
### Essential Calculus: Early Transcendentals
Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280
# Find the critical numbers of the function.28. $$g(t) = |3t - 4|$$
The is critical point for the function occurs at $$\frac{4}{3}$$.
See the step by step solution
## Step 1: Given data
The given function is $$g(t) = |3t - 4|$$.
## Step 2: Concept of Differentiation
Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.
## Step 3: Determine the derivative of the function
The given $$g(t)$$ is an absolute function. Express the given function as follows.
Obtain the first derivative of the given function $$g(t) = |3t - 4|$$.
\begin{aligned}{c}g(t) &= \frac{d}{{dt}}\left\{ {\begin{aligned}{*{20}{l}}{3t - 4}&{{\rm{ if }}t \ge \frac{4}{3}}\\{4 - 3t}&{{\rm{ if }}t < \frac{4}{3}}\end{aligned}} \right.\\{g^\prime }(t) &= \left\{ {\begin{aligned}{*{20}{l}}3&{{\rm{ if }}t \ge \frac{4}{3}}\\{ - 3}&{{\rm{ if }}t < \frac{4}{3}}\end{aligned}} \right.\end{aligned}
Here, $${g^\prime }(t)$$ does not exist at $$t = \frac{4}{3}$$ as it cannot be zero ever. Therefore, the only valid critical number occurs at $$t = \frac{4}{3}$$ as it satisfies the condition of the definition of critical number. Thus, the only critical number of the function $$g(t) = |3t - 4|$$ is $$\frac{4}{3}$$.
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## Explore measure of dispersion: range.
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Remember Tania and the carrots in the Mode Concept? Well, now let's think about range.
Here is Tania’s data about the number of carrots picked each week over nine weeks of harvest.
2, 8, 8, 14, 9, 12, 14, 20, 19, 14
What is the range of carrots picked?
This Concept will teach you all about range. Then we will revisit this problem at the end of the Concept.
### Guidance
The range of a set of data simply tells where the numbers fall, so that we know if they are close together or spread far apart. A set of data with a small range tells us something different than a set of data with a large range. We’ll discuss this more, but first let’s learn how to find the range.
Here are the steps for finding the range of a set of data.
1. What we need to do is put the values in the data set in numerical order. Then we know which is the greatest number in the set (the maximum), and which is the smallest number (the minimum).
2. To find the range, we simply subtract the minimum from the maximum.
Take a minute to copy these steps into your notebook.
Take a look at the data set below.
11, 9, 8, 12, 11, 11, 14, 8, 10
First, we arrange the data in numerical order.
8, 8, 9, 10, 11, 11, 11, 12, 14
Now we can see that the minimum is 8 and the maximum is 14. We subtract to find the range.
14 - 8 = 6
The range of the data is 6. That means that all of the numbers in the data set fall within six places of each other. All of the data results are fairly close together.
How can we use a range to help us answer a question? Suppose we wanted to know the effect of a special soil on plant growth. The numbers in this data set might represent the height in inches of 9 plants grown in the special soil. We know that the range is 6, so all of the plants heights are within 6 inches of each other.
6, 11, 4, 12, 18, 9, 25, 16, 22
Let’s reorder the data and find the range.
4, 6, 9, 11, 12, 16, 18, 22, 25
Now we can see that the minimum is 4 and the maximum is 25. Let’s subtract to find the range.
25 - 4 = 21
The range of this data is 21. That means the numbers in the data set can be much farther apart.
What does this mean about plants grown in special soil?
If the first group of plants had a range of only 6, their heights ended up being fairly close together. So they grew about the same in the special soil. In contrast, the second group of plants had a much greater range of heights. We might not be so quick to assume that the special soil had any effect on the plants, since their heights are so much more varied.
The range has helped us understand the results of the experiment.
Here are a few for you to try on your own. Find the range of the following data sets.
#### Example A
4, 5, 6, 9, 12, 19, 20
Solution: 16
#### Example B
5, 2, 1, 6, 8, 20, 25
Solution: 24
#### Example C
65, 23, 22, 45, 11, 88, 99, 123, 125
Solution: 114
Now let's go back to Tania and the carrots.
Here is Tania’s data about the number of carrots picked each week over nine weeks of harvest.
2, 8, 8, 14, 9, 12, 14, 20, 19, 14
What is the range of carrots picked?
To find this out, we need to figure out the difference between the greatest number of carrots picked and the least number of carrots picked.
The greatest number picked was 20.
The least number picked was 2.
\begin{align*}20 - 2 = 18\end{align*}
The range is 18 carrots.
### Guided Practice
Here is one for you to try on your own.
The following is the number of patrons at a local movie theater.
26, 22, 40, 45, 46, 18, 30, 80, 60, 75
What is the range of the data?
To figure this out, we need to find the difference between the highest number of patrons and the lowest number of patrons.
The highest number of patrons was 80.
The lowest number of patrons was 22.
\begin{align*}80 - 22 = 58\end{align*}
The range for the data set is 58.
### Explore More
Directions: Find the range for each set of data.
1. 4, 5, 4, 5, 3, 3
2. 6, 7, 8, 3, 2, 4
3. 11, 10, 9, 13, 14, 16
4. 21, 23, 25, 22, 22, 27
5. 27, 29, 29, 32, 30, 32, 31
6. 34, 35, 34, 37, 38, 39, 39
7. 43, 44, 43, 46, 39, 50
8. 122, 100, 134, 156, 144, 110
9. 224, 222, 220, 222, 224, 224
10. 540, 542, 544, 550, 548, 547
11. 2, 3, 3, 3, 2, 2, 2, 5, 6, 7
12. 4, 5, 6, 6, 6, 7, 3, 2
13. 23, 22, 22, 24, 25, 25, 25
14. 123, 120, 121, 120, 121, 125, 121
15. 678, 600, 655, 655, 600, 678, 600, 600
### Vocabulary Language: English
Maximum
Maximum
The largest number in a data set.
measure
measure
To measure distance is to determine how far apart two geometric objects are by using a number line or ruler.
Median
Median
The median of a data set is the middle value of an organized data set.
Minimum
Minimum
The minimum is the smallest value in a data set.
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REFLECTING IN THE COORDINATE PLANE
About the topic "Reflecting in the coordinate plane"
Reflecting in the coordinate plane :
A point on a coordinate plane can be reflected across an axis. The reflection is located on the opposite side of the axis, at the same distance from the axis.
Reflecting in the coordinate plane - Examples
Example 1 :
Graph (3, −2). Then fold your coordinate plane along the y-axis and find the reflection of (3, −2). Record the coordinates of the new point in the table.
Solution :
Example 2 :
Graph (3, −2). Then fold your coordinate plane along the x-axis and find the reflection of (3, −2). Record the coordinates of the new point in the table.
Solution :
Based on the above examples, answer the question given below.
Question :
What is the relationship between the coordinates of a point and the coordinates of its reflection across each axis ?
Across y-axis :
Opposite x-coordinate and same y-coordinate as the original point.
Across x-axis :
Same x-coordinate and opposite y-coordinate as the original point
Now, let us look at the rules o reflection.
Rules on finding reflected image
Once students understand the rules which they have to apply for reflection transformation, they can easily make reflection -transformation of a figure.
For example, if we are going to make reflection transformation of the point (2,3) about x-axis, after transformation, the point would be (2,-3). Here the rule we have applied is (x, y) ------> (x, -y).
So we get (2,3) -------> (2,-3).
Let us consider the following example to have better understanding of reflection.
Question :
Let A ( -2, 1), B (2, 4) and (4, 2) be the three vertices of a triangle. If this triangle is reflected about x-axis, what will be the new vertices A' , B' and C' ?
Solution:
Step 1 :
First we have to know the correct rule that we have to apply in this problem.
Step 2 :
Here triangle is reflected about x - axis. So the rule that we have to apply here is (x , y) -------> (x , -y)
Step 3 :
Based on the rule given in step 1, we have to find the vertices of the reflected triangle A'B'C'
Step 4 :
(x , y) ----------> (x , -y)
A ( -2, 1 ) ------------ A' ( -2, -1 )
B ( 2, 4 ) ------------ B' ( 2, -4 )
C ( 4, 2 ) ------------ C' ( 4, -2 )
Step 5 :
Vertices of the reflected triangle are
A' ( -2, -1) , B ( 2, -4 ) and C' ( 4, -2)
Reflecting over any line
When we look at the above figure, it is very clear that each point of a reflected image A'B'C' is at the same distance from the line of reflection as the corresponding point of the original figure.
In other words, the line x = -2 (line of reflection) lies directly in the middle between the original figure and its image.
And also, the line x = -2 (line of reflection) is the perpendicular bisector of the segment joining any point to its image.
Students can keep this idea in mind when they are working with lines of reflections which are neither the x-axis nor the y-axis.
After having gone through the stuff given above, we hope that the students would have understood "Reflecting in the coordinate plane".
If you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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## BEATCALC: Beat the Calculator!
Back to Calculation Tips & Tricks
From B. Lee Clay
Squaring Numbers
Multiplying Numbers
Dividing Numbers
Subtracting Numbers
Percents
Calculation
Practice Exercises
Full List
### Adding even numbers from two through a selected 2-digit even number
1. Divide the even number by 2 (or multiply by 1/2).
2. Multiply this result by the next number.
#### Example:
1. If the 2-digit even number selected is 24:
2. Divide 24 by 2 (24/2 = 12) or multiply by 1/2 (1/2 × 24 = 12).
3. The next number is 13; 12 × 13 = 156.
Ways to multiply 13 by 12:
• Square 12, then add 12: 12 × 12 = 144, 144 + 12 = 156.
• Multiply left to right. 12 × 13 can be done in steps: 12 × (10+3) = (12 × 10) + (12 × 3) = 120 + 36 = 156.
4. So the sum of all the even numbers from two through 24 is 156.
See the pattern?
1. If the 2-digit even number selected is 42:
2. Divide 42 by 2 (42/2 = 21) or multiply by 1/2 (1/2 × 42 = 21).
3. The next number is 22; 21 × 22 = 462.
Ways to multiply 22 by 21:
• Square 21, then add 21: 21 × 21 = 441, 441 + 21 = 462.
• Multiply left to right. 21 × 22 can be done in steps: 21 × 22 = (21 × 20) + (21 x 2) = 420 + 42 = 462.
4. So the sum of all the even numbers from two through 42 is 462.
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# The Science That Draws Necessary Conclusions
To Learn Math
## Shortcuts for Finding Derivatives
Using the derivative formula to find the derivative of a function is a hassle. There are many shortcuts to finding the derivative. The first one is the Power Rule. It can be used with functions which are equal to a polynomial like $x^3 + 5x^2 - 7$. Each term’s coefficient is multiplied by the exponent and the exponent decreases by one. Let’s look at $f(x)=3x^2$. The derivative’s coefficient will be 6 because the coefficient of the term, 3, is multiplied by the exponent, 2. The exponent will be 1 because the exponent is decreased by one. Therefore, the derivative is $f'(x)=6x$. If the function has more than one term, use the same procedure for each term. Now let’s use the power rule to find the derivative of $g(x)=2x^4-5x^2+3x-2$. The derivative is $g'(x)=8x^3-10x+3$. It is important to notice that the constant, -2, is eliminated since x has an exponent of zero. Also, 3x becomes 3 since anything raised to the power of 0 is equal to 1.
Another useful shortcut is the Product Rule. It is useful when the definition is two polynomials multiplied by each other but cannot easily be simplified such as $f(x) = (x^2 - 2)(3x^2 + x - 4)$. If u is assigned to the first polynomial and v is assigned to the second polynomial, the derivative is $u\frac{dv}{dx}+v\frac{du}{dx}$. So in $f(x)$ $u = x^2 - 2$ and $v = 3x^2 + x - 4$. Using the power rule, $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = 6x + 1$. Therefore, by using the product rule $f'(x) = (x^2 - 2)(6x + 1) + (3x^2 + x - 4)(2x) = 12x^3 + 3x^2 - 20x - 2$.
The last rule I am going to discuss for now is the Quotient Rule. The Quotient Rule is used when a polynomial is being divided by another polynomial. Where $y=\frac{u}{v}$, $\frac{dy}{dx} = \frac{ v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. For example, let $f(x)=\frac{x^2+2}{x-1}$. The u is $x^2 + 2$ and v is $x - 1$. Once again utilizing the power rule, we find that $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = 1$. Plugging these values into the quotient rule yeilds $f'(x) = \frac{(x-1)(2x)-(x^2+2)(1)}{(x-1)^2} = \frac{x^2-2x-2}{x^2-2x+1}$.
Proper use of these rules save time. On some occasions, multiple rules will need to be applied. There is another rule I have not yet mentioned called the chain rule. It makes taking the derivative of a function within a function possible. That post will be saved for after the explanation of the usefulness of derivatives.
Written by todizzle91
August 13th, 2009 at 10:35 am
Posted in Calculus
### 3 Responses to 'Shortcuts for Finding Derivatives'
1. plz update me with limit and derivatives
prem
21 Mar 10 at 2:10 am
2. You are a cool young man. Thank you for explaining this so clearly.
Juan Wiltocruz
20 May 11 at 9:58 am
3. This was really helpful, especially when my math teacher did properly explain each rule. Thank you. And now I know where to come, if he does it again.
Kaitlin
28 Nov 12 at 3:21 am
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# sinh function
Martin McBride
2020-09-10
The sinh function is a hyperbolic function. It is also known as the hyperbolic sine function.
Here is a video that explains sinh, cosh and tanh:
## Equation and graph
The sinh function is defined as:
$$\sinh{x} = \frac{e^{x}-e^{-x}}{2}$$
Here is a graph of the function:
## sinh as average of two exponentials
The sinh function can be interpreted as the average of two functions, $e^{x}$ and $-e^{-x}$. This animation illustrates this:
## Other forms of the equation
If we multiply the top and bottom of the original equation for the sinh function by $e^{x}$ (see below) we get:
$$\sinh{x} = \frac{e^{2x}-1}{2e^{x}}$$
Alternatively, if we multiply the top and bottom of the original equation for the sinh function by $e^{-x}$ we get:
$$\sinh{x} = \frac{1-e^{-2x}}{2e^{-x}}$$
## Derivation of other forms
To see how the two formulae above were derived, we start with the original definition of sinh:
$$\sinh{x} = \frac{e^{x}-e^{-x}}{2}$$
Multiplying top and bottom by $e^{x}$ gives:
$$\sinh{x} = \frac{e^{x}(e^{x}-e^{-x})}{2e^{x}} = \frac{e^{x}e^{x}-e^{x}e^{-x})}{2e^{x}}$$
Remember that $e^{x}e^{x}$ is $e^{2x}$. Also $e^{x}e^{-x}$ is 1. This gives:
$$\sinh{x} = \frac{e^{2x}-1}{2e^{x}}$$
which is the second form of the sinh equation. The other alternative form is derived in a similar way.
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### Home > INT3 > Chapter 8 > Lesson 8.3.2 > Problem8-112
8-112.
Carlo is trying to factor the polynomial $p\left(x\right) = x^{4} – 4x^{3} – 4x^{2} + 24x – 9$ to determine all of its zeros. He discovers one factor by making a guess and dividing the polynomial, so he has $p\left(x\right) = \left(x – 3\right)\left(x^{3} – x^{2} – 7x + 3\right)$. Now he is trying to factor $x^{3} – x^{2} – 7x + 3$, so he tries dividing it by $\left(x + 3\right),$ then by $\left(x + 1\right)$, and finally by $\left(x – 1\right)$, but none works without a remainder. Then Teo comes by and says, “You should look at the graph.”
1. How does the graph help?
It shows that $x = 3$ is a double root, so $\left(x − 3\right)$ is a repeated factor.
2. Complete the problem.
Divide $x^{3} − x^{2} − 7x + 3$ by $\left(x − 3\right).$2 by 3 rectangle, labeled as follows: left edge, x, minus 3, top edge, left, x, squared, interior top, left, x squared.
Labels added: Interior bottom, left, negative 3, x, squared, interior top, middle, 2, x squared.
Labels added: top edge, middle, 2, x, interior bottom, middle, negative 6, x, interior top, right, negative x.
Labels added: top edge, right, negative 1, interior bottom, right, 3.
Now use the Quadratic Formula to solve for $x$.
$x = 3, -1 \pm \sqrt{2}$
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## How do we use area in everyday life?
What real-life situations require us to use area? ▫ Floor covering, like carpets and tiles, require area measurements. Wallpaper and paint also call for area measurements. Fabric used for clothing and other items also demand that length and width be considered.
## What is area of square?
In other words, the area of a square is the product of the length of each side with itself. That is, Area A = s x s where s is the length of each side of the square. For example, the area of a square of each side of length 8 feet is 8 times 8 or 64 square feet.
## What is perimeter and area with examples?
For Example, to fence the garden at your house, the length required of the material for fencing is the perimeter of the garden. If it’s a square garden with each side as a cm then perimeter would be 4a cm. The area is the space contained in the shape or the given figure. It is calculated in square units.
## What is the definition for area?
In geometry, the area can be defined as the space occupied by a flat shape or the surface of an object. The area of a figure is the number of unit squares that cover the surface of a closed figure. Area is measured in square units such as square centimteres, square feet, square inches, etc.
## What is the formula of area in pressure?
Pressure and force are related, and so you can calculate one if you know the other by using the physics equation, P = F/A. Because pressure is force divided by area, its meter-kilogram-second (MKS) units are newtons per square meter, or N/m2.
## How is area and perimeter used in everyday life?
In everyday life area and perimeter are used constantly – for example, for describing the size of a house by talking about its floor area, or for working out how much wire is needed to fence off a field.
## How do you find area?
To work out the area of a square or rectangle, multiply its height by its width. If the height and width are in cm, the area is shown in cm². If the height and width are in m, the area is shown in m². A square with sides of 5 m has an area of 25 m², because 5 × 5 = 25.
## How do you find the area and perimeter?
The perimeter of a two-dimensional shape is the distance around the shape. It is found by adding up all the sides (as long as they are all the same unit). The area of a two-dimensional shape is found by counting the number of squares that cover the shape.
## Where is area used?
Area is a mathematical term defined as the two-dimensional space taken up by an object, notes Study.com, adding that the use of area has many practical applications in building, farming, architecture, science, and even how much carpet you’ll need to cover the rooms in your house.
## What do area and perimeter have in common?
Perimeter. In geometry, area is the 2-dimensional space or region occupied by a closed figure, while perimeter is the distance around a closed figure i.e. the length of the boundary. For example, the area can be used to calculate the size of the carpet to cover the whole floor of a room.
## What is perimeter in math definition?
Perimeter is the distance around a two-dimensional shape. Add the side lengths to find the perimeter of regular polygons.
## What is perimeter area?
About Transcript. Perimeter is the distance around the outside of a shape. Area measures the space inside a shape.
## What is the perimeter of a shape?
The perimeter is the distance around the edge of a 2D shape.
## What is area and perimeter of square?
The area of a square with sides of length 1cm is 1cm². The area of other squares can be found by counting squares or by multiplying the length of the sides. The perimeter of a square is the total length of the four sides.
## How do you find the perimeter?
To find the perimeter of a rectangle, add the lengths of the rectangle’s four sides. If you have only the width and the height, then you can easily find all four sides (two sides are each equal to the height and the other two sides are equal to the width). Multiply both the height and width by two and add the results.
## How do you explain a perimeter to a child?
Perimeter is the distance around the outside of a shape. Perimeter is found by adding together the length of all a shape’s sides.
## How do you find area in physics?
The area of a rectangle is determined by multiplying the base by the height. A = b • h where b = 4 s and h = 30 m/s. That is, the object was displaced 120 m during the first 4 seconds of motion. The area of a rectangle is determined by multiplying the base by the height.
## How do you find the perimeter and area of a square?
Well, the formula for perimeter is simply 2 times length + 2 times width. In the case of our square, this would simply be 2x+2x (since length = width), or 4x . Therefore, if given an area, all you’d do is take the square root (to get from x2 to x , and then multiply by 4 to find perimeter.
## Who uses perimeter in their job?
A lot of jobs use atea and perimeter such as; Surveying, flooring estimates architecture, mechanical engineering, the list goes on and on.
## What is a real life example of density?
Everyday Density Examples In an oil spill in the ocean, the oil rises to the top because it is less dense than water, creating an oil slick on the surface of the ocean. A Styrofoam cup is less dense than a ceramic cup, so the Styrofoam cup will float in water and the ceramic cup will sink.
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# If d is the HCF of 45 and 27 , then x,y satisfying d=27x+45y are:A) $x=2,y=1$ B) $x=2,y=-1$ C) $x=-1,y=2$ D) $x=-1,y=-2$
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Hint: The given question is related to the highest common factor of two numbers and linear equations in two variables. Find the highest common factor of $45$ and $27$, then draw the line represented by the equation $d=27x+45y$ on a graph and check the points that lie on the line.
To solve the question, first we have to find the highest common factor of $45$ and $27$. We will use the factorization method to find the value of the highest common factor of $45$ and $27$. In factorization method, we write the numbers as a product of prime numbers and then find the highest number that is common in both.
$45$ can be written as $45=3\times 3\times 5$ and $27$ can be written as $27=3\times 3\times 3$. We can see that the highest number common in both is $3\times 3=9$. So, the highest common factor of $45$ and $27$ is $9$. We are given that the highest common factor of $45$ and $27$ is $d$ . So, $d=9$.
Now, we are given the equation $d=27x+45y$. Substituting $d=9$ in the equation, we get $9=27x+45y$. On rearranging the equation to make it of the form $y=mx+c$, we get $y=\dfrac{-27}{45}x+\dfrac{1}{5}$. So, the equation represents a line with slope $m=\dfrac{-27}{45}$ and $y$ intercept $c=\dfrac{1}{5}$.
Now, we will plot the points corresponding to the given options on the line. The points lying on the line will satisfy the equation $d=27x+45y$. The points corresponding to the options are:
Option A. $(2,1)$ ; Option B. $(2,-1)$ ; Option C. $(-1,2)$; Option D. $(-1,-2)$.
From the graph we can see that $B(2,-1)$ is the only point lying on the line. So, the values of $x$ and $y$ satisfying $d=27x+45y$ are $x=2,y=-1$, where $d$ is the highest common factor of $45$ and $27$.
Note: The correct answer can also be found by substituting the values of $x$ and $y$ from each option in the equation. But it will be time taking. So, it is better to plot the line and points on the graph to find the values of $x$ and $y$ satisfying the equation.
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# Number Bonds to Ten
Do you want to learn faster and more easily?
Then why not use our learning videos, and practice for school with learning games.
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## Number Bonds that Add to 10
The local bees are trying to figure out how many more bees can fit into each hive to make ten, but they need help to figure out how many more can fit! You can save the day by using number pairs to 10 to solve the problem!
There are nine different pairs of numbers that add together to make ten. They are one plus nine, two plus eight, three plus seven, four plus six, five plus five, six plus four, seven plus three, eight plus two, and nine plus one!
0 + 10
1 + 9
2 + 8
3 + 7
4 + 6
5 + 5
## Number Bonds – Explanation
What are number bonds to ten? Number bonds to ten are pairs of numbers that, when added together, equal ten. When teaching number bonds to ten we say there are eleven different pairs of numbers that add together to make ten. They are zero plus ten, one plus nine, two plus eight, three plus seven, four plus six, five plus five, six plus four, seven plus three, eight plus two, nine plus one and ten plus zero! We can see them all here in this number bonds to ten display.
## Number Bonds to 10 – Tens Frames
Let’s imagine that one of the hives already has four bees inside! We can use a tens frame to set this up with four counters.
To find the missing number, we can count the empty spaces on the tens frame to find the missing number pair for four that makes ten! There are six empty spaces, so the missing number pair to make ten here is six! 4 + 6 = 10!
## Number Bonds to 10 – Summary
Remember, there are eleven pairs of numbers that make ten. However, you only have to memorise six of them because some are the same reversed. For example, one plus nine is the same as nine plus one!
0 10
1 9
2 8
3 7
4 6
5 5
To practise further, have a look at our number bonds to ten interactive exercises as well as our number bonds to ten worksheets and other activities after watching the video.
### TranscriptNumber Bonds to Ten
"Hey! What's all the buzzing about here?" It appears the bees LOVE to be in groups of ten in their hives, but they need help working out how many more bees are needed for each hive! Let's help Mr. Squeaks sort the bees into groups of ten by learning about "Number Bonds to Ten". Number bonds to ten are pairs of numbers that can be added together to make ten. A tens frame like THIS can help us to make ten. There are eleven different pairs of numbers that add together to make ten. They are: zero plus ten, one plus nine, two plus eight, three plus seven, four plus six, five plus five, six plus four, seven plus three, eight plus two, nine plus one, and ten plus zero. However, you only need to remember six pairs of numbers because they can all be reversed apart from five plus five which stays the same. Zero plus ten is the same as ten plus zero. One plus nine is the same as nine plus one. Two plus eight is the same as eight plus two. What is the same as three plus seven? Three plus seven is the same as seven plus three. What is the same as four plus six? Four plus six is the same as six plus four. Now we are ready to help sort the bees into hives of ten. The first hive already has FOUR bees inside it. The tens frame has four counters to show this. We can count the empty spaces to find the missing number bond! Let's count. One, two, three, four, five, SIX! SIX is the missing number. FOUR plus SIX makes TEN. The next hive already has FIVE bees inside it. The tens frame has five counters to show this. Again, count the empty spaces. How many empty spaces are there? One, two, three, four, FIVE! FIVE is the missing number. FIVE plus FIVE makes TEN. The final hive already has THREE bees inside it. What is the missing number bond to make ten here? SEVEN is the missing number. THREE plus SEVEN makes TEN. While the bees head to their hives, let's review! Remember, today we learnt there are eleven pairs of numbers that make ten. They are: zero plus ten, one plus nine, two plus eight, three plus seven, four plus six, five plus five, six plus four, seven plus three, eight plus two, nine plus one, and ten plus zero.
"Erm...Imani, what are you doing?"
## Number Bonds to Ten exercise
Would you like to apply the knowledge you’ve learnt? You can review and practice it with the tasks for the video Number Bonds to Ten.
• ### Can you match the groups of bees?
Hints
How many bees are in the tens frame? What do you need to add to that to get 10?
Look at the number of spaces in the tens frame to help you.
For example, if there were 9 bees in a tens frame, we could see that there is 1 space left, so we need to find a tens frame with 1 bee in to make 10.
Solution
Here we can see that 6 + 4 = 10.
The tens frame with 6 bees in has 4 spaces, meaning we need to add 4 more to it to get 10.
______________________________________________________
• 5 + 5 = 10
• 3 + 7 = 10
• 2 + 8 = 10
• ### How many more bees are needed for each hive?
Hints
Use the tens frames to help you. How many spaces are there in each one?
For example, this tens frame has 5 spaces so we would need to add another 5 to make 10.
Remember, it doesn't matter which way round a part whole model goes.
Solution
The hive on the left needed 2 more bees.
• There were 8 bees already in it and there were 2 spaces left in the tens frame, meaning we needed to add 2 more.
• The number bond is 2 + 8 = 10.
The hive on the right needed 7 more bees.
• There were 3 bees already in it and there were 7 spaces left in the tens frame, meaning we needed to add 7 more.
• The number bond is 3 + 7 = 10.
• ### Which of these hives have 10 bees?
Hints
Solve the number sentences. If the answer is 10, highlight it in green.
For example, if you were solving 6 + 4:
• draw 6 circles
• draw 4 circles
• count the number of circles - this is the answer!
There are 5 hives with 10 bees in.
Solution
These bee hives have 10 bees in them.
• 6 + 4 = 10
• 7 + 3 = 10
• 0 + 10 = 10
• 5 + 5 = 10
• 9 + 1 = 10
These bee hives do not have 10 bees in them.
• 3 + 6 = 9
• 2 + 9 = 11
• ### How many more bees do the hives need?
Hints
Pick out the key information.
How many bees do we need in the hive?
How many bees are already in the hive?
We know that each hive needs 10 bees.
We can draw out part whole models to help us like this one.
What do we add to 4 to make 10?
We could then use or draw a tens frame to help us.
We can draw 4 bees as we know there are 4 already in the hive and then count the spaces left to work out what we need to add to 4 to make 10.
How many spaces are left in this tens frame?
Solution
• The first hive needed 6 more bees. 4 + 6 = 10
• The second hive needed 9 more bees. 1 + 9 = 10
• The third hive needed 7 more bees. 3 + 7 = 10
• The fourth hive needed 10 more bees. 0 + 10 = 10
• ### How many more bees does this hive need?
Hints
The hive already has 7 bees, how many more are needed to get to 10? Use the tens frame to help you.
Count the number of spaces in the tens frame.
Here the blue spots represent the bees already in the hive. How many red spots are there? This is the number of bees needed to make 10.
Solution
The hive needed 3 more bees.
7 + 3 = 10
We could see in the tens frame that there were 7 bees and 3 spaces, so that means we needed 3 more bees to make 10.
• ### Can you solve the bee dilemmas?
Hints
Look for the key information. How many bees are in the hive at the beginning? Did more join or did some fly away?
• If more bees joined then we need to start with an addition e.g. 3 + 2
• If some bees flew away we need to start with a subtraction e.g. 7 - 5
Once you have worked out how many bees are in the hive now, you need to find out what we add to that number to make 10.
• 3 + 2 = 5; 5 + ? = 10
• 7 - 5 = 2; 2 + ? = 10
For the final question, you will need to subtract and then add.
Solution
For the first hive
• We started with 3 bees and then 2 more joined, so we need to add 3 + 2 which equals 5.
• We then need to solve 5 + ? = 10.
• 5 + 5 = 10
For the second hive
• We started with 7 bees and then 5 flew away, so we need to subtract 5 from 7 which equals 2.
• We then need to solve 2 + ? = 10.
• 2 + 8 = 10.
For the third hive
• We started with 10 bees and then 6 flew away, so we need to subtract 6 from 10 which equals 4.
• 2 bees then flew back in, so we need to add 4 + 2 which equals 6.
• We then need to solve 6 + ? = 10.
• 6 + 4 = 10
Phew! All of the bees are in their hives!
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# Mobile Algebra
Following the introductory use of structure and emoji math to introduce systems, my teaching partner and I continued with mobiles as suggested by the authors of “An Emoji is Worth a Thousand Variables.” EDC has this great website, SolveMe Mobiles, that has 200 mobile puzzles like this:
Each shape in this mobile has a value (or weight) and the total value (or weight) in this mobile is 60 (units). Go ahead and solve the mobile.
This mobile represents a system of four unknowns. Using traditional algebra symbols it might look like this:
A couple of those equations have just one variable, so it may not be quite as intimidating to look at the traditional symbols. On the other hand, the mobile shapes are just so accessible to everyone!
We needed to move our students away from systems that had one variable defined for them, though, and the SolveMe site, as great as it is, always includes some kind of hint. So we started to make up our own mobiles.
As first, students used a lot of educated guessing to solve the mobiles. Then there was a breakthrough.
Take a closer look at the left-hand mobile.
Students realized that they could “cross off” the same shapes on equal branches and the mobile would stay balanced. In the example above, you can “cross off” two triangles and one square. Whatever remains is equivalent, though it no longer totals 36. Therefore, two triangles equals one square. Using that relationship, some students then substituted two triangles for the one square in the left branch. Then they had a branch of 6 triangles with a total of 18. So, each triangle is worth 3. Other students used the same relationship to substitute one square for the two triangles in the right branch, resulting in a branch of 3 squares with a total of 18. So, each square is worth 3.
We were floored. We had never discussed the idea of substitution, but here it was, naturally arising from students reasoning about the structure in the mobile.
Looking closer at the center mobile, students used the same “cross out” method to find the relationship that 2 triangles equals 3 squares. If we’d been teaching the substitution method in a more traditional way, kids would have been pushed to figure out how much 1 triangle (or 1 square) was worth before making the substitution step. We knew substitution was happening here, but we didn’t invent this approach so we just followed closely to see where our students took us. Since 2 triangles equals 3 squares, some kids substituted 3 squares for the two triangles on the right branch of the mobile. Others made two substitutions of 6 squares for the 4 triangles on the left branch. Either way the result was a branch of 7 squares that totaled 14. It seemed quite natural to them.
What would you do with this one?
Next up: Moving to traditional symbols. The final (?) post of this saga.
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# Derivative of square root of sin x using first principle
In this section, we will learn how to find
1. the derivative of the square root of sin x by definition or
2. the derivative of the square root of sin x from first principle.
To answer the question, let us first know the definition of the derivative.
Definition of derivative: Let $f(x)$ be a differentiable function of $x$. From first principle of by definition, the derivative of $f(x)$ is given as follows:
$f'(x)=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$
## Derivative of the Square Root of Sin x from first principle:
Question: Find the Derivative of $\sqrt{\sin x}$ from first principle.
Solution:
Let $f(x)=\sqrt{\sin x}$
So by the definition or from the first principle, we get that
$f'(x)=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0} \dfrac{\sqrt{\sin(x+h)}-\sqrt{\sin x}}{h}$
Rationalizing the numerator, $f'(x)$ is equal to
$=\lim\limits_{h \to 0}[\dfrac{\sqrt{\sin(x+h)}-\sqrt{\sin x}}{h} \times$ $\dfrac{\sqrt{\sin(x+h)}+\sqrt{\sin x}}{\sqrt{\sin(x+h)}+\sqrt{\sin x}}]$
$=\lim\limits_{h \to 0}\dfrac{\sin(x+h)-\sin x}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$
$[\because (a-b)(a+b)=a^2-b^2]$
$=\lim\limits_{h \to 0}\dfrac{\sin x \cos h+\cos x \sin h-\sin x}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$
$[\because \sin(a+b)=\sin a \cos b+\cos a \sin b]$
$=\lim\limits_{h \to 0}\dfrac{\sin x (\cos h-1)+\cos x \sin h}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$
$=\lim\limits_{h \to 0}\dfrac{\sin x (\cos h-1)}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$ $+\lim\limits_{h \to 0}\dfrac{\cos x \sin h}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$
$=0+\lim\limits_{h \to 0}\dfrac{\cos x \sin h}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$ $[\because \lim\limits_{h \to 0} \dfrac{\cos h-1}{h}=0]$
$=\lim\limits_{h \to 0}\dfrac{\cos x \sin h}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$
$=\lim\limits_{h \to 0}\dfrac{\sin h}{h} \times$ $\lim\limits_{h \to 0}\frac{\cos x}{\sqrt{\sin(x+h)}+\sqrt{\sin x}}$
$=1 \times \dfrac{\cos x}{\sqrt{\sin x}+\sqrt{\sin x}}$ $[\because \lim\limits_{h \to 0} \dfrac{\sin h}{h}=1]$
$=\dfrac{\cos x}{2\sqrt{\sin x}}$ ans.
So the derivative of square root of sinx is equal to (cos x)/(2 root sin x), obtained by the first principle of derivatives, that is, the limit definition of derivatives.
RELATED TOPICS:
Derivative of cos(ex)
Derivative of log(3x)
Derivative of e1/x
## FAQs
Q1: What is the Derivative root sinx?
Answer: The derivative of the square root of sinx is (cos x)/(2 √sin x)
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0
# What are the factors of 340?
Updated: 4/28/2022
Wiki User
15y ago
The factors of 340 are 1, 2, 4, 5, 10, 17, 20, 34, 68, 85, 170, and 340.
The factor pairs of 340 are 1 x 340, 2 x 170, 4 x 85, 5 x 68, 10 x 34, and 17 x 20.
The proper factors of 340 are 1, 2, 4, 5, 10, 17, 20, 34, 68, 85, are 170 or, if the definition you are using excludes 1, they are 2, 4, 5, 10, 17, 20, 34, 68, 85, are 170.
The prime factors of 340 are 2, 2, 5, and 17. Note: There is repetition of these factors, so if the prime factors are being listed instead of the prime factorization, usually only the distinct prime factors are listed.
The distinct prime factors of 340 are 2, 5, and 17.
The prime factorization of 340 is 2 x 2 x 5 x 17 or, in exponential form, 22 x 5 x 17.
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15y ago
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Q: What are the factors of 340?
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Related questions
### What are the prime factors for 340?
The prime factors of 340 are 2 5 17.
### Factors of 340?
The factors of 340 are: 1 2 4 5 10 17 20 34 68 85 170 340
### What are the factors 340?
The factors of 340 are: 1, 2, 4, 5, 10, 17, 20, 34, 68, 85, 170, 340
### What are the factors of the number 340?
The factors of 340 are: 1, 2, 4, 5, 10, 17, 20, 34, 68, 85, 170, 340
2 and 17
340
### What is 340 as the product of prime factors?
2 x 2 x 5 x 17 = 340
### What prime numbers go into 340?
To find what prime numbers are in 340, start with any two factors of 340 and keep factoring the composite factors until all factors are prime. 2 x 170 2 x 2 x 85 2 x 2 x 5 x 17 Prime numbers that go into 340 are 2, 5, and 17.
### What are factors of 340?
1, 2, 4, 5, 10, 17, 20, 34, 68, 85, 170, 340
### What are the factors and prime factors of 340?
340 is a composite number because it has factors other than 1 and itself. It is not a prime number.The 12 factors of 340 are 1, 2, 4, 5, 10, 17, 20, 34, 68, 85, 170, and 340.The proper factors of 340 are 1, 2, 4, 5, 10, 17, 20, 34, 68, 85, and 170 or,if the definition you are using excludes 1, they are 2, 4, 5, 10, 17, 20, 34, 68, 85, and 170.The prime factors of 340 are 2, 2, 5, and 17. Note: There is repetition of these factors, so if the prime factors are being listed instead of the prime factorization, usually only the distinct prime factors are listed.The 3 distinct prime factors (listing each prime factor only once) of 340 are 2, 5, and 17.The prime factorization of 340 is 2 x 2 x 5 x 17 or, in index form (in other words, using exponents), 22 x 5 x 17.NOTE: There cannot be common factors, a greatest common factor, or a least common multiple because "common" refers to factors or multiples that two or more numbers have in common.
### What prime factors equal 340?
The Prime Factors are: 2 x 2 x 5 x 17
### What is the LCM of 17 20 85?
Factors of 17= 17x1 Factors of 20=2x2x5 Factors of 85=17x5 Lcm=17x5x2x2 LCM=340
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# NCERT Solution Class 10 Maths Chapter 6 – Triangles
## Exercise 6.1 Class 10 Maths NCERT Solution Page: 122
1. Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
(b) Proportional
2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures
Solution:
3. State whether the following quadrilaterals are similar or not:
Solution:
From the given two figures, we can see their corresponding angles are different or unequal. Therefore they are not similar.
## Exercise 6.2 Class 10 Maths NCERT Solution Page: 128
1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution:
(i) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒1.5/3 = 1/EC
⇒EC = 3/1.5
EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.
(ii) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒ AD/7.2 = 1.8 / 5.4
⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10
2. E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm
Solution:
Given, in ΔPQR, E and F are two points on side PQ and PR respectively. See the figure below;
(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm
Therefore, by using Basic proportionality theorem, we get,
PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3
And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, we get, PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.
(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm
Therefore, by using Basic proportionality theorem, we get,
PE/QE = 4/4.5 = 40/45 = 8/9
And, PF/RF = 8/9
So, we get here,
PE/QE = PF/RF
Hence, EF is parallel to QR.
(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
From the figure,
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55…………. (i)
And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii)
So, we get here,
PE/EQ = PF/FR
Hence, EF is parallel to QR.
3. In the figure, if LM || CB and LN || CD, prove that AM/MB = AN/AD
Solution:
In the given figure, we can see, LM || CB,
By using basic proportionality theorem, we get,
AM/MB = AL/LC……………………..(i)
Similarly, given, LN || CD and using basic proportionality theorem,
From equation (i) and (ii), we get,
Hence, proved.
4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC
Solution:
In ΔABC, given as, DE || AC
Thus, by using Basic Proportionality Theorem, we get,
∴BD/DA = BE/EC ………………………………………………(i)
In ΔABC, given as, DF || AE
Thus, by using Basic Proportionality Theorem, we get,
∴BD/DA = BF/FE ………………………………………………(ii)
From equation (i) and (ii), we get
BE/EC = BF/FE
Hence, proved.
5. In the figure, DE||OQ and DF||OR, show that EF||QR.
Solution:
Given,
In ΔPQO, DE || OQ
So by using Basic Proportionality Theorem,
PD/DO = PE/EQ……………… ..(i)
Again given, in ΔPQO, DE || OQ ,
So by using Basic Proportionality Theorem,
PD/DO = PF/FR………………… (ii)
From equation (i) and (ii), we get,
PE/EQ = PF/FR
Therefore, by converse of Basic Proportionality Theorem,
EF || QR, in ΔPQR.
6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Solution:
Given here,
In ΔOPQ, AB || PQ
By using Basic Proportionality Theorem,
OA/AP = OB/BQ…………….(i)
Also given,
In ΔOPR, AC || PR
By using Basic Proportionality Theorem
∴ OA/AP = OC/CR……………(ii)
From equation (i) and (ii), we get,
OB/BQ = OC/CR
Therefore, by converse of Basic Proportionality Theorem,
In ΔOQR, BC || QR.
7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Solution:
Given, in ΔABC, D is the midpoint of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.
We have to prove that E is the mid point of AC.
Since, D is the mid-point of AB.
In ΔABC, DE || BC,
By using Basic Proportionality Theorem,
From equation (i), we can write,
⇒ 1 = AE/EC
∴ AE = EC
Hence, proved, E is the midpoint of AC.
8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Solution:
Given, in ΔABC, D and E are the mid points of AB and AC respectively, such that,
We have to prove that: DE || BC.
Since, D is the midpoint of AB
Also given, E is the mid-point of AC.
∴ AE=EC
⇒ AE/EC = 1
From equation (i) and (ii), we get,
By converse of Basic Proportionality Theorem,
DE || BC
Hence, proved.
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Solution:
Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.
We have to prove, AO/BO = CO/DO
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔADC, we have OE || DC
Therefore, By using Basic Proportionality Theorem
AE/ED = AO/CO ……………..(i)
Now, In ΔABD, OE || AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/BO…………….(ii)
From equation (i) and (ii), we get,
AO/CO = BO/DO
⇒AO/BO = CO/DO
Hence, proved.
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.
Solution:
Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,
AO/BO = CO/DO.
We have to prove here, ABCD is a trapezium
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔDAB, EO || AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/OB ……………………(i)
Also, given,
AO/BO = CO/DO
⇒ AO/CO = BO/DO
⇒ CO/AO = DO/BO
⇒DO/OB = CO/AO …………………………..(ii)
From equation (i) and (ii), we get
DE/EA = CO/AO
Therefore, By using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
⇒ AB || DC.
Hence, quadrilateral ABCD is a trapezium with AB || CD.
## Exercise 6.3 Class 10 Maths NCERT Solution Page: 138
1. State which pairs of triangles in Figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Solution:
(i) Given, in ΔABC and ΔPQR,
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
Therefore by AAA similarity criterion,
∴ ΔABC ~ ΔPQR
(ii) Given, in ΔABC and ΔPQR,
AB/QR = BC/RP = CA/PQ
By SSS similarity criterion,
ΔABC ~ ΔQRP
(iii) Given, in ΔLMP and ΔDEF,
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF = 2.7/5 = 27/50
Here , MP/DE = PL/DF ≠ LM/EF
Therefore, ΔLMP and ΔDEF are not similar.
(iv) In ΔMNL and ΔQPR, it is given,
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
Therefore, by SAS similarity criterion
∴ ΔMNL ~ ΔQPR
(v) In ΔABC and ΔDEF, given that,
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here , AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.
(vi) In ΔDEF, by sum of angles of triangles, we know that,
∠D + ∠E + ∠F = 180°
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
⇒ ∠F = 30°
Similarly, In ΔPQR,
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
⇒ ∠P = 70°
Now, comparing both the triangles, ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Therefore, by AAA similarity criterion,
Hence, ΔDEF ~ ΔPQR
2. In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
Solution:
As we can see from the figure, DOB is a straight line.
Therefore, ∠DOC + ∠ COB = 180°
⇒ ∠DOC = 180° – 125° (Given, ∠ BOC = 125°)
= 55°
In ΔDOC, sum of the measures of the angles of a triangle is 180º
Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180°
⇒ ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°)
⇒ ∠DCO = 55°
It is given that, ΔODC ∝ ¼ ΔOBA,
Therefore, ΔODC ~ ΔOBA.
Hence, Corresponding angles are equal in similar triangles
∠OAB = ∠OCD
⇒ ∠ OAB = 55°
∠OAB = ∠OCD
⇒ ∠OAB = 55°
3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD
Solution:
In ΔDOC and ΔBOA,
AB || CD, thus alternate interior angles will be equal,
∴∠CDO = ∠ABO
Similarly,
∠DCO = ∠BAO
Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;
∴∠DOC = ∠BOA
Hence, by AAA similarity criterion,
ΔDOC ~ ΔBOA
Thus, the corresponding sides are proportional.
DO/BO = OC/OA
⇒OA/OC = OB/OD
Hence, proved.
4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Solution:
In ΔPQR,
∠PQR = ∠PRQ
∴ PQ = PR ………………………(i)
Given,
QR/QS = QT/PRUsing equation (i), we get
QR/QS = QT/QP……………….(ii)
In ΔPQS and ΔTQR, by equation (ii),
QR/QS = QT/QP
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]
5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
Solution:
Given, S and T are point on sides PR and QR of ΔPQR
And ∠P = ∠RTS.
In ΔRPQ and ΔRTS,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ΔRPQ ~ ΔRTS (AA similarity criterion)
6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Solution:
Given, ΔABE ≅ ΔACD.
∴ AB = AC [By CPCT] ……………………………….(i)
And, AD = AE [By CPCT] ……………………………(ii)
In ΔADE and ΔABC, dividing eq.(ii) by eq(i),
∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [SAS similarity criterion]
7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iv) ΔPDC ~ ΔBEC
Solution:
Given, altitudes AD and CE of ΔABC intersect each other at the point P.
(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (90° each)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by AA similarity criterion,
ΔAEP ~ ΔCDP
(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB ( 90° each)
∠ABD = ∠CBE (Common Angles)
Hence, by AA similarity criterion,
ΔABD ~ ΔCBE
∠PAE = ∠DAB (Common Angles)
Hence, by AA similarity criterion,
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (90° each)
∠PCD = ∠BCE (Common angles)
Hence, by AA similarity criterion,
ΔPDC ~ ΔBEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Solution:
Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,
In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (AA similarity criterion)
9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
Solution:
Given, ABC and AMP are two right triangles, right angled at B and M respectively.
(i) In ΔABC and ΔAMP, we have,
∠CAB = ∠MAP (common angles)
∠ABC = ∠AMP = 90° (each 90°)
∴ ΔABC ~ ΔAMP (AA similarity criterion)
(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)
If two triangles are similar then the corresponding sides are always equal,
Hence, CA/PA = BC/MP
10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:
(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Solution:
Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.
(i) From the given condition,
ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
Since, ∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠A = ∠F
∠ACD = ∠FGH
∴ ΔACD ~ ΔFGH (AA similarity criterion)
⇒CD/GH = AC/FG
(ii) In ΔDCB and ΔHGE,
∴ ΔDCB ~ ΔHGE (AA similarity criterion)
(iii) In ΔDCA and ΔHGF,
∴ ΔDCA ~ ΔHGF (AA similarity criterion)
11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Solution:
Given, ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∴ ΔABD ~ ΔECF (using AA similarity criterion)
12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.
Solution:
Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
i.e. AB/PQ = BC/QR = AD/PM
We have to prove: ΔABC ~ ΔPQR
As we know here,
⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)
⇒ ΔABD ~ ΔPQM [SSS similarity criterion]
∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
⇒ ∠ABC = ∠PQR
In ΔABC and ΔPQR
AB/PQ = BC/QR ………………………….(i)
∠ABC = ∠PQR ……………………………(ii)
From equation (i) and (ii), we get,
ΔABC ~ ΔPQR [SAS similarity criterion]
13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD
Solution:
Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.
∠ACD = ∠BCA (Common angles)
∴ ΔADC ~ ΔBAC (AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
∴ CA/CB = CD/CA
⇒ CA2 = CB.CD.
Hence, proved.
14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
Solution:
Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that;
We have to prove, ΔABC ~ ΔPQR
Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.
In ΔABD and ΔCDE, we have
BD = DC [Since, AP is the median]
and, ∠ADB = ∠CDE [Vertically opposite angles]
∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]
⇒ AB = CE [By CPCT] …………………………..(i)
Also, in ΔPQM and ΔMNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [SAS criterion of congruence]
⇒ PQ = RN [CPCT] ………………………………(ii)
Now, AB/PQ = AC/PR = AD/PM
From equation (i) and (ii),
⇒ CE/RN = AC/PR = 2AD/2PM
⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
∴ ΔACE ~ ΔPRN [SSS similarity criterion]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P …………………………………………….(iii)
Now, in ΔABC and ΔPQR, we have
From equation (iii),
∠A = ∠P
∴ ΔABC ~ ΔPQR [ SAS similarity criterion]
15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Given, Length of the vertical pole = 6m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m
In ΔABC and ΔDEF,
∠C = ∠E (angular elevation of sum)
∠B = ∠F = 90°
∴ ΔABC ~ ΔDEF (AA similarity criterion)
∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)
∴ 6/h = 4/28
⇒h = (6×28)/4
⇒ h = 6 × 7
⇒ = 42 m
Hence, the height of the tower is 42 m.
16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.
Solution:
Given, ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.
∴AB/PQ = AC/PR = BC/QR……………………………(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..(ii)
Since AD and PM are medians, they will divide their opposite sides.
∴ BD = BC/2 and QM = QR/2 ……………..………….(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM ……………………….(iv)
In ΔABD and ΔPQM,
From equation (ii), we have
∠B = ∠Q
From equation (iv), we have,
AB/PQ = BD/QM
∴ ΔABD ~ ΔPQM (SAS similarity criterion)
## Exercise 6.4 Class 10 Maths NCERT Solution Page: 143
1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
Solution: Given, ΔABC ~ ΔDEF,
Area of ΔABC = 64 cm2
Area of ΔDEF = 121 cm2
EF = 15.4 cm
As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
= AC2/DF2 = BC2/EF2
∴ 64/121 = BC2/EF2
⇒ (8/11)2 = (BC/15.4)2
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm
2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Solution:
Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.
In ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [AAA similarity criterion]
As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,
Area of (ΔAOB)/Area of (ΔCOD) = AB2/CD2
= (2CD)2/CD2 [∴ AB = 2CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD2/CD2 = 4/1
Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1
3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.
Solution:
Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.
We have to prove: Area (ΔABC)/Area (ΔDBC) = AO/DO
Let us draw two perpendiculars AP and DM on line BC.
We know that area of a triangle = 1/2 × Base × Height
In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ~ ΔDMO (AA similarity criterion)
∴ AP/DM = AO/DO
⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO.
4. If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Say ΔABC and ΔPQR are two similar triangles and equal in area
Now let us prove ΔABC ≅ ΔPQR.
Since, ΔABC ~ ΔPQR
∴ Area of (ΔABC)/Area of (ΔPQR) = BC2/QR2
⇒ BC2/QR2 =1 [Since, Area(ΔABC) = (ΔPQR)
⇒ BC2/QR2
⇒ BC = QR
Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]
5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.
Solution:
Given, D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC.
In ΔABC,
F is the mid-point of AB (Already given)
E is the mid-point of AC (Already given)
So, by the mid-point theorem, we have,
FE || BC and FE = 1/2BC
⇒ FE || BC and FE || BD [BD = 1/2BC]
Since, opposite sides of parallelogram are equal and parallel
∴ BDEF is parallelogram.
Similarly, in ΔFBD and ΔDEF, we have
FB = DE (Opposite sides of parallelogram BDEF)
FD = FD (Common sides)
BD = FE (Opposite sides of parallelogram BDEF)
∴ ΔFBD ≅ ΔDEF
Similarly, we can prove that
ΔAFE ≅ ΔDEF
ΔEDC ≅ ΔDEF
As we know, if triangles are congruent, then they are equal in area.
So,
Area(ΔFBD) = Area(ΔDEF) ……………………………(i)
Area(ΔAFE) = Area(ΔDEF) ……………………………….(ii)
and,
Area(ΔEDC) = Area(ΔDEF) ………………………….(iii)
Now,
Area(ΔABC) = Area(ΔFBD) + Area(ΔDEF) + Area(ΔAFE) + Area(ΔEDC) ………(iv)
Area(ΔABC) = Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF)
From equation (i)(ii) and (iii),
⇒ Area(ΔDEF) = (1/4)Area(ΔABC)
⇒ Area(ΔDEF)/Area(ΔABC) = 1/4
Hence, Area(ΔDEF): Area(ΔABC) = 1:4
6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.
We have to prove: Area(ΔABC)/Area(ΔDEF) = AM2/DN2
Since, ΔABC ~ ΔDEF (Given)
∴ Area(ΔABC)/Area(ΔDEF) = (AB2/DE2) ……………………………(i)
and, AB/DE = BC/EF = CA/FD ………………………………………(ii)
In ΔABM and ΔDEN,
Since ΔABC ~ ΔDEF
∴ ∠B = ∠E
AB/DE = BM/EN [Already Proved in equation (i)]
∴ ΔABC ~ ΔDEF [SAS similarity criterion]
⇒ AB/DE = AM/DN …………………………………………………..(iii)
∴ ΔABM ~ ΔDEN
As the areas of two similar triangles are proportional to the squares of the corresponding sides.
∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2
Hence, proved.
7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
Given, ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.
Area(ΔBQC) = ½ Area(ΔAPC)
Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,
∴ ΔAPC ~ ΔBQC [AAA similarity criterion]
∴ area(ΔAPC)/area(ΔBQC) = (AC2/BC2) = AC2/BC2
Since, Diagonal = √2 side = √2 BC = AC
⇒ area(ΔAPC) = 2 × area(ΔBQC)
⇒ area(ΔBQC) = 1/2area(ΔAPC)
Hence, proved.
Tick the correct answer and justify:
8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Solution:
GivenΔABC and ΔBDE are two equilateral triangle. D is the midpoint of BC.
∴ BD = DC = 1/2BC
Let each side of triangle is 2a.
As, ΔABC ~ ΔBDE
∴ Area(ΔABC)/Area(ΔBDE) = AB2/BD2 = (2a)2/(a)2 = 4a2/a2 = 4/1 = 4:1
Hence, the correct answer is (C).
9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
Solution:
Given, Sides of two similar triangles are in the ratio 4 : 9.
Let ABC and DEF are two similar triangles, such that,
ΔABC ~ ΔDEF
And AB/DE = AC/DF = BC/EF = 4/9
As, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,
∴ Area(ΔABC)/Area(ΔDEF) = AB2/DE
∴ Area(ΔABC)/Area(ΔDEF) = (4/9)= 16/81 = 16:81
Hence, the correct answer is (D).
## Exercise 6.5 Class 10 Maths NCERT Solution Page: 150
1. Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of the sides of the, we will get 49, 576, and 625.
49 + 576 = 625
(7)2 + (24)2 = (25)2
Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm
(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
Clearly, 9 + 36 ≠ 64
Or, 32 + 62 ≠ 82
Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.
Hence, the given triangle does not satisfies Pythagoras theorem.
(iii) Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 502 + 802 ≠ 1002
As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle does not satisfies Pythagoras theorem.
Hence, it is not a right triangle.
(iv) Given, sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Thus, 144 +25 = 169
Or, 122 + 52 = 132
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Hence, length of the hypotenuse of this triangle is 13 cm.
2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.
Solution:
Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR
We have to prove, PM2 = QM × MR
In ΔPQM, by Pythagoras theorem
PQ2 = PM2 + QM2
Or, PM2 = PQ2 – QM2 ……………………………..(i)
In ΔPMR, by Pythagoras theorem
PR2 = PM2 + MR2
Or, PM2 = PR2 – MR2 ………………………………………..(ii)
Adding equation, (i) and (ii), we get,
2PM2 = (PQ2 + PM2) – (QM2 + MR2)
= QR2 – QM2 – MR2 [∴ QR2 = PQ2 + PR2]
= (QM + MR)2 – QM2 – MR2
= 2QM × MR
∴ PM2 = QM × MR
3. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB2 = BC × BD
(ii) AC2 = BC × DC
(iii) AD2 = BD × CD
Solution:
∠DAB = ∠ACB (Each 90°)
∠ABD = ∠CBA (Common angles)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB2 = CB × BD
(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° – 90° – x
∠CBA = 90° – x
= 90° – x
∠CDA = 180° – 90° – (90° – x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CAB = ∠CDA
∠ACB = ∠DCA (Each 90°)
∴ ΔCBA ~ ΔCAD [AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC2 = DC × BC
(iii) In ΔDCA and ΔDAB,
∠DCA = ∠DAB (Each 90°)
∴ ΔDCA ~ ΔDAB [AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD2 = BD × CD
4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 .
Solution:
Given, ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°
AC = BC (By isosceles triangle property)
AB2 = AC2 + BC2 [By Pythagoras theorem]
= AC2 + AC2 [Since, AC = BC]
AB2 = 2AC2
5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Solution:
Given, ΔABC is an isosceles triangle having AC = BC and AB2 = 2AC2
In ΔACB,
AC = BC
AB2 = 2AC2
AB2 = AC+ AC2
= AC2 + BC[Since, AC = BC]
Hence, by Pythagoras theorem ΔABC is right angle triangle.
6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
Given, ABC is an equilateral triangle of side 2a.
AB = AC
Hence, BD = DC [by CPCT]
⇒ AD2 = 4a2 – a2
7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
Solution:
Given, ABCD is a rhombus whose diagonals AC and BD intersect at O.
We have to prove, as per the question,
AB+ BC+ CD2 + AD= AC+ BD2
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB2 = AO+ BO…………………….. (i) [By Pythagoras theorem]
Similarly,
DC2 = DO+ CO…………………….. (iii)
BC2 = CO+ BO…………………….. (iv)
Adding equations (i) + (ii) + (iii) + (iv), we get,
AB+ AD+ DC+ BC2 = 2(AO+ BO+ DO+ CO2)
= 4AO+ 4BO[Since, AO = CO and BO =DO]
= (2AO)+ (2BO)2 = AC+ BD2
AB+ AD+ DC+ BC2 = AC+ BD2
Hence, proved.
8. In Fig. 6.54, O is a point in the interior of a triangle.
ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Solution:
Given, in ΔABC, O is a point in the interior of a triangle.
And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
Join OA, OB and OC
(i) By Pythagoras theorem in ΔAOF, we have
OA2 = OF2 + AF2
Similarly, in ΔBOD
OB2 = OD2 + BD2
Similarly, in ΔCOE
OC2 = OE2 + EC2
OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE+ EC2
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.
(ii) AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)
∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Given, a ladder 10 m long reaches a window 8 m above the ground.
Let BA be the wall and AC be the ladder,
Therefore, by Pythagoras theorem,
AC2 = AB2 + BC2
102 = 82 + BC2
BC= 100 – 64
BC= 36
BC = 6m
Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.
10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Given, a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.
Let AB be the pole and AC be the wire.
By Pythagoras theorem,
AC2 = AB2 + BC2
242 = 182 + BC2
BC= 576 – 324
BC= 252
BC = 6√7m
Therefore, the distance from the base is 6√7m.
11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after
hours?
Solution:
Given,
Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane flying due north in
hours (OA) = 100 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance covered by second aeroplane flying due west in
hours (OB) = 1200 × 3/2 km = 1800 km
In right angle ΔAOB, by Pythagoras Theorem,
AB2 = AO2 + OB2
⇒ AB2 = (1500)2 + (1800)2
⇒ AB = √(2250000 + 3240000)
= √5490000
⇒ AB = 300√61 km
Hence, the distance between two aeroplanes will be 300√61 km.
12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:
Given, Two poles of heights 6 m and 11 m stand on a plane ground.
And distance between the feet of the poles is 12 m.
Let AB and CD be the poles of height 6m and 11m.
Therefore, CP = 11 – 6 = 5m
From the figure, it can be observed that AP = 12m
By Pythagoras theorem for ΔAPC, we get,
AP2 = PC2 + AC2
(12m)2 + (5m)2 = (AC)2
AC2 = (144+25) m2 = 169 m2
AC = 13m
Therefore, the distance between their tops is 13 m.
13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Solution:
Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
By Pythagoras theorem in ΔACE, we get
AC2 + CE2 = AE2 ………………………………………….(i)
In ΔBCD, by Pythagoras theorem, we get
BC2 + CD2 = BD2 ………………………………..(ii)
From equations (i) and (ii), we get,
AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii)
In ΔCDE, by Pythagoras theorem, we get
DE2 = CD2 + CE2
In ΔABC, by Pythagoras theorem, we get
AB2 = AC2 + CB2
Putting the above two values in equation (iii), we get
DE2 + AB2 = AE2 + BD2.
14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2.
Solution:
Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;
DB = 3CD.
In Δ ABC,
AD ⊥BC and BD = 3CD
AB2 = AD2 + BD2 ……………………….(i)
AC2 = AD2 + DC2 ……………………………..(ii)
Subtracting equation (ii) from equation (i), we get
AB2 – AC2 = BD2 – DC2
= 9CD2 – CD2 [Since, BD = 3CD]
= 8CD2
= 8(BC/4)[Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.
15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.
Solution:
Given, ABC is an equilateral triangle.
And D is a point on side BC such that BD = 1/3BC
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given, BD = 1/3BC
∴ BD = a/3
DE = BE – BD = a/2 – a/3 = a/6
⇒ 9 AD2 = 7 AB2
16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
Given, an equilateral triangle say ABC,
Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
In ΔABE, by Pythagoras Theorem, we get
AB2 = AE2 + BE2
4AE2 = 3a2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)
Hence, proved.
17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution:
Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
We can observe that,
AB2 = 108
AC2 = 144
And, BC2 = 36
AB2 + BC2 = AC2
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct answer is (C).
## Exercise 6.6 Class 10 Maths NCERT Solution Page: 152
1. In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/PQ = SR/PR
Solution:
Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
Given, PS is the angle bisector of ∠QPR. Therefore,
∠QPS = ∠SPR………………………………..(i)
As per the constructed figure,
∠SPR=∠PRT(Since, PS||TR)……………(ii)
∠QPS = ∠QRT(Since, PS||TR) …………..(iii)
From the above equations, we get,
∠PRT=∠QTR
Therefore,
PT=PR
In △QTR, by basic proportionality theorem,
QS/SR = QP/PT
Since, PT=TR
Therefore,
QS/SR = PQ/PR
Hence, proved.
2. In Fig. 6.57, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove that: (i) DM2 = DN . MC (ii) DN2 = DM . AN.
Solution:
1. Let us join Point D and B.
Given,
BD ⊥AC, DM ⊥ BC and DN ⊥ AB
Now from the figure we have,
DN || CB, DM || AB and ∠B = 90 °
Therefore, DMBN is a rectangle.
So, DN = MB and DM = NB
The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.
∴ ∠CDB = 90° ⇒ ∠2 + ∠3 = 90° ……………………. (i)
In ∆CDM, ∠1 + ∠2 + ∠DMC = 180°
⇒ ∠1 + ∠2 = 90° …………………………………….. (ii)
In ∆DMB, ∠3 + ∠DMB + ∠4 = 180°
⇒ ∠3 + ∠4 = 90° …………………………………….. (iii)
From equation (i) and (ii), we get
∠1 = ∠3
From equation (i) and (iii), we get
∠2 = ∠4
In ∆DCM and ∆BDM,
∴ ∆DCM ∼ ∆BDM (AA similarity criterion)
BM/DM = DM/MC
DN/DM = DM/MC (BM = DN)
⇒ DM2 = DN × MC
Hence, proved.
(ii) In right triangle DBN,
∠5 + ∠7 = 90° ……………….. (iv)
In right triangle DAN,
∠6 + ∠8 = 90° ………………… (v)
D is the point in triangle, which is foot of the perpendicular drawn from B to AC.
∴ ∠ADB = 90° ⇒ ∠5 + ∠6 = 90° ………….. (vi)
From equation (iv) and (vi), we get,
∠6 = ∠7
From equation (v) and (vi), we get,
∠8 = ∠5
In ∆DNA and ∆BND,
∴ ∆DNA ∼ ∆BND (AA similarity criterion)
AN/DN = DN/NB
⇒ DN2 = AN × NB
⇒ DN2 = AN × DM (Since, NB = DM)
Hence, proved.
3. In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that
AC2= AB2+ BC2+ 2 BC.BD.
Solution:
By applying Pythagoras Theorem in ∆ADB, we get,
AB2 = AD2 + DB2 ……………………… (i)
Again, by applying Pythagoras Theorem in ∆ACD, we get,
AC2 = AD2 + (DB + BC) 2
AC2 = AD2 + DB2 + BC2 + 2DB × BC
From equation (i), we can write,
AC2 = AB2 + BC2 + 2DB × BC
Hence, proved.
4. In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that
AC2= AB2+ BC2 – 2 BC.BD.
Solution:
By applying Pythagoras Theorem in ∆ADB, we get,
We can write it as;
⇒ AD2 = AB2 − DB2 ……………….. (i)
By applying Pythagoras Theorem in ∆ADC, we get,
From equation (i),
AB2 − BD2 + DC2 = AC2
AB2 − BD2 + (BC − BD) 2 = AC2
AC2 = AB2 − BD2 + BC2 + BD2 −2BC × BD
AC= AB2 + BC2 − 2BC × BD
Hence, proved.
5. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) AC2 = AD2 + BC.DM + 2 (BC/2) 2
(ii) AB2 = AD2 – BC.DM + 2 (BC/2) 2
(iii) AC2 + AB2 = 2 AD2 + ½ BC2
Solution:
(i) By applying Pythagoras Theorem in ∆AMD, we get,
AM2 + MD2 = AD2 ………………. (i)
Again, by applying Pythagoras Theorem in ∆AMC, we get,
AM2 + MC2 = AC2
AM2 + (MD + DC) 2 = AC2
(AM2 + MD2 ) + DC2 + 2MD.DC = AC2
From equation(i), we get,
AD2 + DC2 + 2MD.DC = AC2
Since, DC=BC/2, thus, we get,
AD+ (BC/2) 2 + 2MD.(BC/2) 2 = AC2
AD+ (BC/2) 2 + 2MD × BC = AC2
Hence, proved.
(ii) By applying Pythagoras Theorem in ∆ABM, we get;
AB2 = AM2 + MB2
= (AD2 − DM2) + MB2
= (AD2 − DM2) + (BD − MD) 2
= AD2 − DM2 + BD2 + MD2 − 2BD × MD
= AD2 + BD2 − 2BD × MD
Hence, proved.
(iii) By applying Pythagoras Theorem in ∆ABM, we get,
AM2 + MB2 = AB2 ………………….… (i)
By applying Pythagoras Theorem in ∆AMC, we get,
AM2 + MC2 = AC2 …………………..… (ii)
Adding both the equations (i) and (ii), we get,
2AM2 + MB2 + MC2 = AB2 + AC2
2AM2 + (BD − DM) 2 + (MD + DC) 2 = AB2 + AC2
2AM2+BD2 + DM2 − 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2
2AM2 + 2MD2 + BD2 + DC2 + 2MD (− BD + DC) = AB2 + AC2
2(AM2+ MD2) + (BC/2) 2 + (BC/2) 2 + 2MD (-BC/2 + BC/2) 2 = AB2 + AC2
2AD+ BC2/2 = AB2 + AC2
6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:
Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.
By applying Pythagoras Theorem in ∆DEA, we get,
DE2 + EA2 = DA2 ……………….… (i)
By applying Pythagoras Theorem in ∆DEB, we get,
DE2 + EB2 = DB2
DE2 + (EA + AB) 2 = DB2
(DE2 + EA2) + AB2 + 2EA × AB = DB2
DA2 + AB2 + 2EA × AB = DB2 ……………. (ii)
By applying Pythagoras Theorem in ∆ADF, we get,
Again, applying Pythagoras theorem in ∆AFC, we get,
AC2 = AF2 + FC2 = AF2 + (DC − FD) 2
= AF2 + DC2 + FD2 − 2DC × FD
= (AF2 + FD2) + DC2 − 2DC × FD AC2
AC2= AD2 + DC2 − 2DC × FD ………………… (iii)
Since ABCD is a parallelogram,
AB = CD ………………….…(iv)
And BC = AD ………………. (v)
∠DEA = ∠AFD (Each 90°)
∴ ∆EAD ≅ ∆FDA (AAS congruence criterion)
⇒ EA = DF ……………… (vi)
Adding equations (i) and (iii), we get,
DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2
DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2
From equation (iv) and (vi),
BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2
AB2 + BC2 + CD2 + DA2 = AC2 + BD2
7. In Figure, two chords AB and CD intersect each other at the point P. Prove that :
(i) ∆APC ~ ∆ DPB
(ii) AP . PB = CP . DP
Solution:
Firstly, let us join CB, in the given figure.
(i) In ∆APC and ∆DPB,
∠APC = ∠DPB (Vertically opposite angles)
∠CAP = ∠BDP (Angles in the same segment for chord CB)
Therefore,
∆APC ∼ ∆DPB (AA similarity criterion)
(ii) In the above, we have proved that ∆APC ∼ ∆DPB
We know that the corresponding sides of similar triangles are proportional.
∴ AP/DP = PC/PB = CA/BD
⇒AP/DP = PC/PB
∴AP. PB = PC. DP
Hence, proved.
8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
(i) ∆ PAC ~ ∆ PDB
(ii) PA . PB = PC . PD.
Solution:
(i) In ∆PAC and ∆PDB,
∠P = ∠P (Common Angles)
As we know, exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is opposite interior angle, which are both equal.
∠PAC = ∠PDB
Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)
(ii) We have already proved above,
∆APC ∼ ∆DPB
We know that the corresponding sides of similar triangles are proportional.
Therefore,
AP/DP = PC/PB = CA/BD
AP/DP = PC/PB
∴ AP. PB = PC. DP
9. In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC.
Solution:
In the given figure, let us extend BA to P such that;
AP = AC.
Now join PC.
Given, BD/CD = AB/AC
⇒ BD/CD = AP/AC
By using the converse of basic proportionality theorem, we get,
∠BAD = ∠APC (Corresponding angles) ……………….. (i)
And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)
By the new figure, we have;
AP = AC
⇒ ∠APC = ∠ACP ……………………. (iii)
On comparing equations (i), (ii), and (iii), we get,
Therefore, AD is the bisector of the angle BAC.
Hence, proved.
10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Solution:
Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the
horizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.
To find AC, we have to use Pythagoras theorem in ∆ABC, is such way;
AC2 = AB2+ BC2
AB= (1.8 m) 2 + (2.4 m) 2
AB= (3.24 + 5.76) m2
AB2 = 9.00 m2
⟹ AB = √9 m = 3m
Thus, the length of the string out is 3 m.
As its given, she pulls the string at the rate of 5 cm per second.
Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m
Let us say now, the fly is at point D after 12 seconds.
Length of string out after 12 seconds is AD.
AD = AC − String pulled by Nazima in 12 seconds
= (3.00 − 0.6) m
= 2.4 m
(1.8 m) 2 + BD2 = (2.4 m) 2
BD2 = (5.76 − 3.24) m2 = 2.52 m2
BD = 1.587 m
Horizontal distance of fly = BD + 1.2 m
= (1.587 + 1.2) m = 2.787 m
= 2.79 m
|
SUBSTITUTION METHOD SOLVING SYSTEMS OF EQUATIONS
Problem 1 :
Solve the following system of equations by substitution method.
2x - 3y = -1 and y = x - 1
Solution :
2x - 3y = -1 -----(1)
y = x - 1 -----(2)
Step 1 :
From (2), substitute (x - 1) for y into (1).
(1)-----> 2x - 3(x - 1) = -1
Simplify.
2x - 3x + 3 = -1
-x - 3 = -1
-x = 2
Multiply each side (-1).
x = -2
Step 2 :
Substitute -2 for x into (2).
(2)-----> y = 2 - 1
y = 1
Therefore, the solution is
(x, y) = (-2, 1)
Problem 2 :
Solve the following system of equations by substitution method.
y = -3x + 5 and 5x - 4y = -3
Solution :
y = -3x + 5 -----(1)
5x - 4y = -3 -----(2)
Step 1 :
From (1), substitute (-3x + 5) for y into (2).
(2)-----> 5x - 4(-3x + 5) = -3
Simplify.
5x + 12x - 20 = -3
17x - 20 = -3
17x = 17
Divide each side by 17.
x = 1
Step 2 :
Substitute 1 for x into (1).
(1)-----> y = -3(1) + 5
y = -3 + 5
y = 2
Therefore, the solution is
(x, y) = (1, 2)
Problem 3 :
Solve the following system of equations by substitution method.
y = 5x - 7 and -3x - 2y = -12
Solution :
y = 5x - 7 -----(1)
-3x - 2y = -12 -----(2)
Step 1 :
From (1), substitute (5x - 7) for y into (2).
(2)-----> -3x - 2(5x - 7) = -12
Simplify.
-3x - 10x + 14 = -12
-13x + 14 = -12
Subtract 14 from each side.
-13x = -26
Divide each side by -13.
x = 2
Step 2 :
Substitute 2 for x into (1).
(1)-----> y = 5(2) - 7
y = 10 - 7
y = 3
Therefore, the solution is
(x, y) = (2, 3)
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# How do you use the first and second derivatives to sketch y = x^3 - 12x - 12?
Feb 1, 2016
#### Explanation:
$y = {x}^{3} - 12 x - 12$
First derivative: $y ' = 3 {x}^{2} - 12$
Second derivative: $y ' ' = 6 x$
Using $y ' = 3 {x}^{2} - 12$
$3 \left({x}^{2} - 4\right)$
$3 \left(x - 2\right) \left(x + 2\right)$
$x = - 2 , 2$
Sub x values into $y = {x}^{3} - 12 x - 12$, to find y values
$y = {\left(- 2\right)}^{3} - 12 \left(- 2\right) - 12$
$y = 4$, $\left(- 2 , 4\right)$
$y = {\left(2\right)}^{3} - 12 \left(2\right) - 12$
$y = - 28$, $\left(2 , - 28\right)$
Using $y ' ' = 6 x$
Sub x values into $y ' ' = 6 x$
$6 \left(- 2\right) = - 12$
$x < 0$
$\therefore$ $\left(2 , - 28\right)$ is a maximum point
$6 \left(2\right) = 12$
$x > 0$
$\therefore$ $\left(- 2 , 4\right)$ is a minimum point
When $x = 0$
$y = {0}^{3} - 12 \left(0\right) - 12$
$y = - 12$
$\left(0 , - 12\right)$
Use the information from the first and second derivative to sketch the graph as well as the y intercept.
graph{y=x^3-12x-12 [-10, 10, -5, 5]}
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# How do you find the critical numbers of f(x)=(x^2+6x-7)^2?
##### 1 Answer
Dec 8, 2017
$x = - 7 , x = 1$ and $x = - 3$
#### Explanation:
The critical points of a function are where the function's derivative is either undefined or $0$.
Let's first start by computing $f ' \left(x\right)$. I will not expand the parenthesis (because I'll have to do less factoring later) and instead use the chain rule. If we let $u = {x}^{2} + 6 x - 8$, we get:
$\frac{d}{\mathrm{dx}} \left({\left({x}^{2} + 6 x - 7\right)}^{2}\right) = \frac{d}{\mathrm{du}} \left({u}^{2}\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 6 x - 7\right)$
$2 u \left(2 x + 6\right) = 2 \left({x}^{2} + 6 x - 7\right) \left(2 x + 6\right)$
Now we set this expression equal to $0$.
$2 \left({x}^{2} + 6 x - 7\right) \left(2 x + 6\right) = 0$
Factoring ${x}^{2} + 6 x - 7$ gives:
$2 \left(x + 7\right) \left(x - 1\right) \left(2 x + 6\right) = 0$
This tells us that $x = - 7 , x = 1$ and $x = - 3$ all are solutions and therefor critical points. This function is never undefined, so these are also the only critical points.
|
# 2021 AMC 10B Problems/Problem 1
Redirect page
## Problem
How many integer values of $x$ satisfy $|x|<3\pi$?
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$
## Solution 1
Since $3\pi$ is about $9.42$, we multiply 9 by 2 for the numbers from $1$ to $9$ and the numbers from $-1$ to $-9$ and add 1 to account for the zero to get $\boxed{\textbf{(D)}\ ~19}$~smarty101 and edited by Tony_Li2007
## Solution 2
$3\pi \approx 9.4.$ There are two cases here.
When $x>0, |x|>0,$ and $x = |x|.$ So then $x<9.4$
When $x<0, |x|>0,$ and $x = -|x|.$ So then $-x<9.4$. Dividing by $-1$ and flipping the sign, we get $x>-9.4.$
From case 1 and 2, we know that $-9.4 < x < 9.4$. Since $x$ is an integer, we must have $x$ between $-9$ and $9$. There are a total of $$9-(-9) + 1 = \boxed{\textbf{(D)}\ ~19} \text{ integers}.$$
-PureSwag
## Solution 3
$|x|<3\pi$ $\iff$ $-3\pi. Since $\pi$ is approximately $3.14$, $3\pi$ is approximately $9.42$. We are trying to solve for $-9.42, where $x\in\mathbb{Z}$. Hence, $-9.42 $\implies$ $-9\leq x\leq9$, for $x\in\mathbb{Z}$. The number of integer values of $x$ is $9-(-9)+1=19$. Therefore, the answer is $\boxed{\textbf{(D)}19}$.
~ {TSun} ~
## Solution 4
Looking at the problem, we see that instead of directly saying $x$, we see that it is $|x|.$ That means all the possible values of $x$ in this case are positive and negative. Rounding $\pi$ to $3$ we get $3(3)=9.$ There are $9$ positive solutions and $9$ negative solutions. $9+9=18.$ But what about zero? Even though zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is $9+9+1=18+1=\boxed{\textbf{(D)}19}$.
~DuoDuoling0
~savannahsolver
~IceMatrix
-Interstigation
## See Also
2021 AMC 10B (Problems • Answer Key • Resources) Preceded byFirst Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
2021 AMC 12B (Problems • Answer Key • Resources) Preceded byFirst Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Computer Algorithms: Adding Large Integers
## Introduction
We know how to add two integers using a perfectly simple and useful algorithm learned from school or even earlier. This is perhaps one of the very first techniques we learn in mathematics. However we need to answer few questions. First of all do computers use the same technique, since they use binary representation of numbers? Is there a faster algorithm used by computers? What about boundaries and large integers?
## Overview
Let’s start by explaining how we humans add two numbers. An important fact is that by adding two single-digit numbers we get at most two digit number. This can be proven by simply realizing that 9+9 = 18. This fact lays down in the way we add integers. Here’s how.
We just line-up the integers on their right-most digit and we start adding them in a column. In case we got a sum greater than 9 (let’s say 14) we keep only the right-most digit (the 4) and the 1 is added to the next sum.
Thus we get to the simple fact that by adding two n-digit integers we can have either an n-digit integer or a n+1 digit integer. As an example we see that by adding 53 + 35 (two 2-digit integers) we get 88, which is again 2-digit integer, but 53 + 54 result in 107, which is 3 digit integer.
That fact is practically true, as I mentioned above, for each pair of n-digit integers.
In fact binaries can be added by using the exact same algorithm. At the example below we add two integers represented as binary numbers.
As a matter of fact this algorithm is absolutely wonderful, because it works not only on decimals and binaries but in any base B.
Of course computers tend to perform better when adding integers that “fit” the machine word. However as we can see later this isn’t always the case and sometimes we need to add larger numbers that exceed the type boundaries.
Since we know how to add “small” integers, it couldn’t be so hard to apply the same algorithm on big integers. The only problem is that the addition will be slower and sometimes (done by humans) can be error prone.
So practically the algorithm is the same, but we can’t just put a 1 billion integer into a standard computer type INT, right? That means that the tricky part here is the way we represent integers in our application. A common solution is to store the “big” integer into an array, thus each digit will be a separate array item. Then the operation of addition will be simple enough to be applied.
## Complexity
When we talk about an algorithm that is so well known by every human being (or almost every) a common question is “is there anything faster” or “do computers use a different algorithm”. The answer may be surprising to someone, but unfortunately that is the fastest (optimal) algorithm for number addition.
Practically there’s nothing to optimize here. We just read the two n-digit numbers (O(n)), we apply “simple” addition to each digit and we carry over the 1 from the sums greater than 9 to the next “simple” addition. We don’t have loops or any complex operation in order to search for an optimization niche.
## Application
It’s strange how often this algorithm is asked on coding interviews. Perhaps the catch is whether the interviewed person will start to look for a faster approach?! Thus is cool to know that this algorithm is optimal.
Sometimes we may ask ourselves why we humans use decimals. It’s considered because we have 10 fingers on our hands and this is perhaps true.
An interesting fact though, is that the Mayas (who barely predicted the end of the world a couple of weeks ago) used a system of a base 20. That is logical, since we have not 10, but total of 20 fingers considering our legs.
Finally, this algorithm may seem to easy to be explained but it lays down in more complex algorithms.
# How to Setup Different Error Messages for Each Zend Form Element Validator
Anyone who has worked with that has come across this problem. I’d like to show different error message on each validator attached to a Zend_Form_Element. Let’s say we validate an text input field. We want it to contain only digits, but also we’d like to display different messages when the field is empty and when the user has entered something that is different from digits.
It can be done by attaching to the form element two validators: Zend_Validate_Digits and Zend_Validate_NotEmpty, but first let’s see how to change the default “Value is required and can’t be empty” error message of a form field.
```\$element = \$form->createElement('text', 'phone'); \$element->setLabel('Please, enter your phone number:') ->setRequired(true) ->addValidator('Digits'); \$form->addElement(\$element);```
Here we validate the field with Zend_Validate_Digits and we have set it to be required. Thus everything containing characters, i.e. “my123name” or “007bond”, will be false, while “1234” will be true.
Continue reading How to Setup Different Error Messages for Each Zend Form Element Validator
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# Important Questions for CBSE Class 7 Maths Chapter 7 - Comparing Quantities
Last updated date: 20th Sep 2024
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## CBSE Class 7 Maths Chapter - 7 Important Questions - Free PDF Download
If you're a Year 7 student studying with CBSE, in your NCERT Maths book, there's a chapter called "Comparing Quantities." It's like a special part of maths, and it's important to understand it because the things you learn here can help you in everyday life. This chapter is all about different ways to compare numbers. Knowing these ideas can help you solve more difficult maths problems later on. You can find important questions and answers for Chapter 7 on the Vedantu website. We explain everything in an easy way, so it's not hard to understand. If you need help with Class 7 Science, you can also sign up online for lessons on Vedantu.com.
## Study Important Questions for class 7 Mathematics Chapter 7 – Comparing Quantities
Very Short Answer Questions (1 Mark)
1. Find the ratio of $600\;{\text{g}}$ to $5\;{\text{kg}}$.
Ans: First we will convert masses into the same units.
$5\;{\text{kg}}\; = \;5\; \times \;1000\;{\text{g}}\; = \;5000\;{\text{g}}$
Now we will find ratio of $600\;{\text{g}}$ and $5000\;{\text{g}}$
$600\;{\text{g}}\;:\;5000\;{\text{g}}\;{\text{ = }}\;\dfrac{{600}}{{5000}}\; \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; = \;\dfrac{6}{{50}}\; = \;\dfrac{3}{{25}} \\$
Required ratio is $3:25$
2. Write $\dfrac{2}{5}$ as percent.
Ans: We’ll multiply numerator and denominator by $20$ to make denominator $100$
$\dfrac{2}{5} \times \dfrac{{20}}{{20}} = \dfrac{{40}}{{100}} = 40\%$
3. Convert $0.25$ to percent.
Ans: To convert $0.25$ to percent, we have to multiply it by $100\%$`
$= 0.25 \times 100\% \\ = \dfrac{{25}}{{100}} \times 100\% \\ = 25\% \\$
4. Find $100\% - 55\% =$____
Ans: we have
$100\% = \dfrac{{100}}{{100}} \\ 55\% = \dfrac{{50}}{{100}} \\ 100\% - 55\% = \dfrac{{100}}{{100}} - \dfrac{{55}}{{100}} \\ = \dfrac{{100 - 55}}{{100}} \\ = \dfrac{{45}}{{100}} = 45\% \\$
5. Find $25\%$ of $150$.
Ans: By using formula, we have
$\dfrac{{25}}{{100}} \times 150 = 37.5$
6. In a class of $45$ students $40\%$ are girls. Find the no. of boys.
Ans: Given: Girls percentage in class= $40\%$
Total no. of students =$45$
No. of girls $= 45 \times \dfrac{{40}}{{100}} = 45 \times \dfrac{2}{5} = 18$
No. of boys = Total no. of students – No. of girls
$= 45 - 18 \\ = 27 \\$
7. Find the ratio of
a. $18\,{\text{m }}$ to $45\;{\text{cm}}$
Ans: Let’s convert both lengths into the same unit.
$1\;m$ = $100\;cm$
$\therefore 18\;m$ = $18\times 100\; cm$
$18\;m\:\text{to}\: 45\;cm$ = $\dfrac{1800}{45}$ = $\dfrac{40}{1}$ = $40:1$
b. $20$ days to $48$ hours
Ans: First convert both time into same units
\begin{align} & \text{1 day = 24 hours} \\ & \text{20 days = 20 }\!\!\times\!\!\text{ 24 = 480 hours }\!\!~\!\!\text{ } \\ & \text{20 days : 48 hours = 480 : 48} \\ & \text{ =}\frac{480}{48}\,=\ 10 \\ \end{align}
Required Ratio is $\text{10 : 1}$
8. In a city, $35\%$ of population are males and $25\%$ are females and the remaining are children. What is the percentage of children in the city?
Ans: The total percentage = $100$
$\%$ of males = $35\%$
$\%$ of females = $25\%$
Also, Total $\%$ = $\%$ of (Females $+$ males $+$ children)
$\therefore$ $\%$ of children = Total $\%$ $-$ $\%$ of females $-$ $\%$ of males
$= (100\; - \;35\; - \;25)\;\% \\ = \;\left( {100 - \;60} \right)\;\% \\ = \;40\;\% \\$
9. Find:
a. $80\% \;{\text{of}}\;2\;{\text{kg's}}$
b. $25\% \;{\text{of}}\;3000$
Ans:
a. $80\% \;{\text{of}}\;2\;{\text{kg's}}$
We know, $2\;{\text{kg}}\; = \;2000\;{\text{g}}$
$80\% \;{\text{of}}\;2000\;{\text{g}}\;{\text{ = }}\;80\% \; \times \;2000\;{\text{g}} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \dfrac{{80}}{{100}} \times \;2000\;{\text{g}} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \;80\; \times \;20\;{\text{g}} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \;1600\;{\text{g}}\;{\text{ = }}\;{\text{1}}{\text{.6}}\;{\text{kg}} \\$
b. $25\% \;{\text{of}}\;3000$
$\dfrac{{25}}{{100}} \times 3000 \\ = 25 \times \;30 \\ = 750 \\$
10. Convert given percent to decimal fractions and also to fractions in simplest form
a. $10\%$
b. $40\%$
Ans: To convert percent into decimal fraction we will divide the number by $100$
a. $10\% \; = \;\dfrac{{10}}{{100}}\; = \;\dfrac{1}{{10}}\; = \;0.1$
b. $40\% \; = \;\dfrac{{40}}{{100}}\; = \;\dfrac{4}{{10}}\; = \;0.4$
11. Mithali buys a TV at $12,000$ and sells it at a profit of $20\%$. How much money does she get?
Ans: Mithali buys TV = Cost price of TV = $12,000$
Mithali sells TV at profit = $20\% \;{\text{of}}\;12000$
$= \dfrac{{20}}{{100}} \times 12,000\; \\ = \;20\, \times 120 \\ = {\text{Rs}}{\text{.}}\;2400 \\$
Now, Selling price = cost price $+$ profit
$= 12,000 + 2,400 \\ = {\text{ Rs}}{\text{. }}14,400 \\$
Therefore, Mithali get ${\text{Rs}}{\text{. }}14,400$
12. Find
a. $12\dfrac{1}{2}\% {\text{ of}}\;75$
b. $30\% {\text{ of }}150$
Ans:
a. $12\dfrac{1}{2}\% {\text{ of}}\;75$
We know, $12\dfrac{1}{2}\; = \;\dfrac{{25}}{2}$
$\therefore \dfrac{{25}}{2}\% {\text{ of}}\;75 = \dfrac{{25}}{{2 \times 100}} \times 75 \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \dfrac{1}{{2 \times 4}} \times 75 \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \dfrac{{75}}{8} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 9\dfrac{3}{8} \\$
b. $30\% {\text{ of }}150$
$= \dfrac{{30}}{{100}} \times 150\\ = \;3\; \times 15 \\ = \;45 \\$
13. Find the whole quantity if
a. $10\,\%$ of it is $800$
b. $15\%$ of it is $1050$
Ans:
a. $10\,\%$ of it is $800$
Let quantity is $x$
$\therefore \;10\;\% \;{\text{of}}\;x\; = \;800 \\ \Rightarrow \;10\;\% \; \times \;x\; = \;800 \\ \Rightarrow \;\;\dfrac{{10}}{{100}}\; \times \;x\; = \;800 \\ \Rightarrow \;\;\dfrac{1}{{10}}\; \times \;x\; = \;800 \\ \Rightarrow \;\;\;\;\;\;\;\;\;x\; = \;800\; \times \;10 \\ \;\;\;\;\;\therefore \;\;\;\;\;x = \;8000 \\$
So, whole quantity = $8000$
b. $15\%$ of it is $1050$
Let the quantity is $x$
$\therefore \;15\;\% \;{\text{of}}\;x\; = \;1050 \\ \Rightarrow \;15\;\% \; \times \;x\; = \;1050 \\ \Rightarrow \;\;\dfrac{{15}}{{100}}\; \times \;x\; = \;1050 \\ \Rightarrow \;\;\;\;\dfrac{3}{{20}}\; \times \;x\; = \;1050 \\ \Rightarrow \;\;\;\;\;\;\;\;\;\;x\; = \;1050\; \times \;\dfrac{{20}}{3}\\ \;\;\;\;\;\therefore \;\;\;\;\;\;x = \;350\; \times \;20 \\ \;\;\;\;\;\;\; \Rightarrow \;\;x\; = \;7000 \\$
14. Manvita saves ${\mathbf{Rs}}.{\text{ }}{\mathbf{9000}}$ from her salary every month. If it is ${\mathbf{5}}\%$ of her salary. Find her salary.
Ans: Let Manvita’s salary = $x\;{\text{Rs}}{\text{.}}$
${\mathbf{5}}\%$ of Manvita’s Salary = ${\mathbf{Rs}}.{\text{ }}{\mathbf{9000}}$
$\therefore 5\% {\text{ of x}} = 9000 \\ \Rightarrow \dfrac{5}{{100}} \times x = 9000 \\ \;\;\;\;\;\; \Rightarrow x = \dfrac{{9000 \times 100}}{5}\\ \;\;\;\;\; \Rightarrow \;x\; = \;1800\; \times 100 \\ \;\;\;\;\;\therefore \;\;x = 1,80,000\\$
Therefore, Manvita’s salary is $1,80,000$
15. convert the following fractions to percent.
a. $\dfrac{3}{4}$
b. $\dfrac{2}{5}$
Ans:
a. We’ll multiply the numerator and denominator by $25$to make the denominator $100$.
$\dfrac{3}{4} \times \dfrac{{25}}{{25}}\; = \;\dfrac{{75}}{{100}}\; = \;75\;\%$
b. We’ll multiply the numerator and denominator by $20$to make the denominator $100$.
$\dfrac{2}{5} \times \dfrac{{20}}{{20}}\; = \;\dfrac{{40}}{{100}}\; = \;40\;\%$
16. Convert the following into percent
a. $\dfrac{{34}}{{50}}$
b. $\dfrac{1}{4}$
c. $0.03$
Ans: First, we will make denominator $100$, and then we will convert that number into percent
a. $\dfrac{{34}}{{50}} \times \dfrac{2}{2} = \dfrac{{68}}{{100}} = 68\%$
b. $\dfrac{1}{4} \times \dfrac{{25}}{{25}} = \dfrac{{25}}{{100}} = 25\%$
c. $0.03\;\; = \;\dfrac{3}{{100}} = 3\%$
17. Out of $24,000$voters in a constituency $48\%$ voted. Find the no. of voters who did not vote.
Ans: Total voters = $24,000$
Vote percentage = $48\%$
No. of Voters voted = $48\,\% \;{\text{of}}\;12,000$
$= \dfrac{{48}}{{100}} \times 24,000 \\ = \;48 \times 240 \\ = 11,520 \\$
No. of voters did not vote = Total voters – No. of voters voted
$= 24,000 - 11,520 \\ = 12,480 \\$
Therefore, $12,480$ people did not vote.
18. If ${\text{Rs}}{\text{. }}500$has to be divided among Mishala, Manvita and Meera in the ratio $1{\text{ }}:{\text{ }}3{\text{ }}:{\text{ }}6$. Then how much money will each get and what will be the percentages.
Ans: Total Amount =${\text{Rs}}{\text{. }}500$
Given, ${\text{Rs}}{\text{. }}500$has to be divided among Mishala, Manvita and Meera in the ratio $1{\text{ }}:{\text{ }}3{\text{ }}:{\text{ }}6$
Let common ratio be $x$
Mishala’s share = $x$
Manvita’s share = $3x$
Meera’s share = $6x$
According to question
$x + 3x + 6x\; = \;500 \\ \;\;\;\; \Rightarrow \;10\;x\; = \;500 \\ \;\;\;\;\;\;\;\;\;\;\therefore x = \dfrac{{500}}{{10}}\; = \;50 \\$
Mishala’s share = $50\;{\text{Rs}}.$
$\% = \dfrac{1}{{10}} \times 100 = 10\%$
Manvita’s share = $3x\; = \;3 \times \;50\; = \;150\;{\text{Rs}}.$
$\% = \dfrac{3}{{10}} \times 100 = 30\%$
Meera’s share = $6x\, = \;6\; \times \;\;50\; = \;300\;{\text{Rs}}.$
$\% = \dfrac{6}{{10}} \times 100 = 60\%$
19.Find the amount to be paid at the end of $4$ years at each case
a. Principal = ${\text{Rs}}{\text{. }}1500{\text{ at }}8\%$ PA
b. Principal = ${\text{Rs}}{\text{. }}7500{\text{ at }}5\%$ PA
Ans:
a. Given,
${\text{P}} = 1500,\;{\text{R}} = 8\% \;\,{\text{PA}}\;,\;{\text{T}} = 4{\text{ years }}$
We know, ${\text{Amount}}\;{\text{ = }}\;{\text{principal}}\;{\text{ + }}\;{\text{interest}}$
And, $\text{Interest}\;$ = $\dfrac{\text{Principal}\; \times \text{Rate}\;\times\;\text{Time}}{{100}}$
${\text{I}} = \dfrac{{1500 \times 8 \times 4}}{{100}} = 15 \times 8 \times 4\; = 480\;{\text{Rs}}.$
Amount = ${\text{P}}\;{\text{ + }}\;{\text{I}}\;{\text{ = }}\;{\text{1500}}\;{\text{ + }}\;{\text{480}}\;{\text{ = }}\;{\text{1980}}\;{\text{Rs}}{\text{.}}$
b. ${\text{P}} = 7500,\;{\text{R}} = 5\% \;{\text{PA}},\;\;{\text{T}} = 4{\text{ years }}$
Now, interest ${\text{I = }}\dfrac{{{\text{PRT}}}}{{{\text{100}}}} = \dfrac{7500 \times 5 \times 4}{100} = \;75 \times 5 \times 4\; = \;1500\;{\text{Rs}}.$
Therefore, Amount = ${\text{P}}\;{\text{ + }}\;{\text{I}}\; = \,{\text{7500}}\;{\text{ + }}\;{\text{1500}}\;{\text{ = }}\;{\text{9000}}\;{\text{Rs}}{\text{.}}$
20. What rate gives an interest of ${\text{Rs}}{\text{. }}540$on a sum of ${\text{Rs}}{\text{. }}18000$in $3{\text{ years}}$?
Ans: Given, Interest, ${\text{I}}\;{\text{ = }}\;{\text{Rs}}{\text{. }}540$
Principal,${\text{P}}\;{\text{ = }}\;{\text{Rs}}{\text{. }}18000$
Time, ${\text{T}}\; = \;3{\text{ years}}$
We know, $\text{Interest}\;$ = $\dfrac{\text{Principal}\; \times \text{Rate}\;\times\;\text{Time}}{{100}}$
$\therefore {\text{Rate}}\;{\text{ = }}\;\dfrac{\text{Interest}\;\times\;\text{100}}{\text{Principal}\,\times\;\text{Time}}$
$= \dfrac{{540 \times 100}}{{18000 \times 3}} = \dfrac{{180}}{{180}}\;\; = \;1\%$
## CBSE Important Questions Class 7 Maths Chapter 7
### Solved Examples
1. There are 45 students in a class, and 40% are girls. What is the number of boys?
Solution:
Total number of students in a class: 45
Percentage of girls
So, the number of girls in the class are:
40/100 x 45
= 2 x 9
= 18
So, the total number of boys is 45 - 18 = 27
2. What is 2/5 as a percent?
Solution:
⅖ x 20/20 = 40/100
= 40%
## Why you should Use Vedantu’s Important Questions for Class 7 Math?
Benefits of Important Questions from Vedantu for Class 7 Chapter 7 - Comparing Quantities”:
• Focus on key topics for efficient studying.
• Prepares students for exams and reduces anxiety.
• Reinforces understanding of fundamental concepts.
• Teaches effective time management.
• Enables self-assessment and progress tracking.
• Strategic approach for higher scores.
• Covers a wide range of topics for comprehensive understanding.
• Supports exam preparation and boosts confidence.
## Conclusion
Engaging with Important Questions for CBSE Class 7 Maths Chapter 7 - Comparing Quantities serves as a pivotal step toward mastering the fundamental concepts of quantitative comparison. These questions offer students a comprehensive review of various scenarios, honing their skills in analyzing and contrasting quantities. By tackling these exercises, students develop a solid foundation in mathematical comparisons, fostering their ability to apply these skills in practical scenarios. This practice ensures a thorough understanding of the chapter, empowering students to confidently tackle quantitative challenges and excel in their mathematical journey.
## FAQs on Important Questions for CBSE Class 7 Maths Chapter 7 - Comparing Quantities
1. How to Quickly Grasp the Sums Based on the Concept of Interest?
There are mainly two subtypes of interest and sums are based on various operations on them. These two variants are simple interest and compound interest. Although compound interest is more complicated than simple interest and it is taught in higher standards. You need to learn all the formulas related to simple interest and solve the sums in different procedures. The main formula of simple interest is SI= (PxRxT)/100. Here, P is the principal amount, R denotes the urate of interest and T denotes time. Now, you have to derive either of these three while doing a sum on interest belonging to the exercises of Chapter 7.
2. What is the Right Way to Learn the Sums on Decimals Easily?
The right way to learn the sums on decimal is simple as you need to study the basic concepts of decimal units at first. You can differentiate between integers and decimal numbers and learn how to denote decimal numbers verbally. In the next step, you will come to know about adding two decimal numbers, subtracting them, or multiplying and dividing. Furthermore, decimals’ advanced problems relate to conversations where you have to learn to change the decimal numbers to fractions and work with both decimals and percentages. As per the sums on percentage are concerned, you can learn the basic mathematical works with them like the decimals.
3. What do you mean by comparing quantities?
Two values must have the same units to be compared, and two ratios can be compared by turning them into like fractions. We say the two provided ratios are comparable if the two fractions are equal. Comparing Quantities refers to the quantitative relationship between two quantities that indicates their respective sizes. It is merely a tool for comparing amounts. It has multiple benefits and uses in our day to day life, and hence it is very crucial to learn about the concepts related to it.
4. What is the use of comparing quantities Class 7?
To compare is to perform an analysis between the differences in numbers, amounts, or values in order to validate if one quantity is larger than, lower than, or equal to another. We can define or determine how much a number is bigger or less by comparing. We may also compare decimals and fractions in this way. If you want to know how much something weighs, you may do that by measuring it against another standard or reference unit.
5. What is the use of percentage Class 7?
To make computations easier, we utilise percentages. Working with parts of 100 is considerably easier than working with thirds, twelfths, and so on, especially since many fractions lack a precise (non-recurring) decimal counterpart. When the number of individuals in the groups are not the same, it is critical to compare using percentages. Comparing counts and percentages are similar when the number of persons in the categories is the same. Hence, using percentages can come in handy in multiple operations and can help address different confusions and problems.
6. What does it mean by interest in maths?
The cost of borrowing money is interest, which is paid to the lender as a charge for the loan. Simple interest is a fixed proportion of the principal amount borrowed or lent that is paid or received over a period of time. The amount of money paid for the use of someone else's money is known as interest. It is a very crucial agent in transactions carried out in today’s market. Hence it is important to have a good knowledge of its concepts, processes and uses. To know more, solve the important questions by visiting the page Important questions for Class 7 Maths and download a free PDF of the same.
7. Where can I find the Solutions of Class 7 Maths Chapter 7?
The solutions are easily available on the Vedantu site.
• Click on the link- NCERT Solutions for Class 7 maths chapter 7.
• The webpage with Vedantu’s solutions for Class 7 Maths Chapter 7 will open.
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What is a Polynomial?
An expression that represents a quantity in mathematical form is called a polynomial.
Introduction
The meaning of the polynomial is derived from two words “Poly” and “Nomial”.
1. The prefix “Poly” means many.
2. The word “Nomial” means a term.
The meaning of a Polynomial is many terms as per the meanings of both words.
A quantity can be expressed by only a single expression in mathematics, but it is not possible to write every quantity in a single mathematical expression. In that case, two or more expressions are connected to form another mathematical expression by the operations like addition, subtraction and combination of both. An expression can be formed by either one or more connected expressions but every expression that represents an indeterminate quantity is called a polynomial.
d
Single expression
In mathematics, a quantity is expressed in mathematical form by multiplying a coefficient with either one or product of more variables having exponents. The quantity in mathematical form is called a mathematical expression.
For example, the quantity $10$ is expressed mathematically by multiplying a numerical coefficient $5$ with the product of square of a variable $x$ and an indeterminate $y$.
$10 \,=\, 5x^2y$
The quantity $10$ is written mathematically as $5x^2y$ and the quantity in mathematical form $5x^2y$ is called a mathematical expression.
Connecting many expressions
The quantities can be expressed in a single mathematical expression in some cases but it is not possible in the remaining cases. Hence, either two or more mathematical expressions are connected by the mathematical operations like addition or subtraction or combination of both.
For example, the quantity $13$ cannot be expressed in mathematical form by a single expression. Hence, it can be expressed by connecting two expressions $x^2$ and $3y$ by a plus sign.
$13 \,=\, x^2+3y$
The quantity $-3$ cannot be written in mathematical form by a single expression but it can be obtained by connecting the expression $x^2$ with another $3xy$ by a minus sign and another expression $7$ by a plus sign.
$-3 \,=\, x^3-3xy+7$
In this case, the quantities $13$ and $-3$ are expressed in mathematical form as $x^2+3y$ and $x^3-3xy+7$ respectively.
In this case, $x = 2$ and $y = 3$.
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## Section5.2The Division Algorithm
An application of the Principle of Well-Ordering that we will use often is the division algorithm.
This is a perfect example of the existence-and-uniqueness type of proof. We must first prove that the numbers $q$ and $r$ actually exist. Then we must show that if $q'$ and $r'$ are two other such numbers, then $q = q'$ and $r = r'\text{.}$
Existence of $q$ and $r\text{.}$ Let
\begin{equation*} S = \{ a - bk : k \in {\mathbb Z} \text{ and } a - bk \geq 0 \}\text{.} \end{equation*}
If $0 \in S\text{,}$ then $b$ divides $a\text{,}$ and we can let $q = a/b$ and $r = 0\text{.}$ If $0 \notin S\text{,}$ we can use the Well-Ordering Principle. We must first show that $S$ is nonempty. If $a \gt 0\text{,}$ then $a - b \cdot 0 \in S\text{.}$ If $a \lt 0\text{,}$ then $a - b(2a) = a(1 - 2b) \in S\text{.}$ In either case $S \neq \emptyset\text{.}$ By the Well-Ordering Principle, $S$ must have a smallest member, say $r = a - bq\text{.}$ Therefore, $a = bq + r\text{,}$ $r \geq 0\text{.}$ We now show that $r \lt b\text{.}$ Suppose that $r \gt b\text{.}$ Then
\begin{equation*} a - b(q + 1)= a - bq - b = r - b \gt 0\text{.} \end{equation*}
In this case we would have $a - b(q + 1)$ in the set $S\text{.}$ But then $a - b(q + 1) \lt a - bq\text{,}$ which would contradict the fact that $r = a - bq$ is the smallest member of $S\text{.}$ So $r \leq b\text{.}$ Since $0 \notin S\text{,}$ $r \neq b$ and so $r \lt b\text{.}$
Uniqueness of $q$ and $r\text{.}$ Suppose there exist integers $r\text{,}$ $r'\text{,}$ $q\text{,}$ and $q'$ such that
\begin{equation*} a = bq + r, 0 \leq r \lt b \quad \text{and}\quad a = bq' + r', 0 \leq r' \lt b\text{.} \end{equation*}
Then $bq + r = bq' + r'\text{.}$ Assume that $r' \geq r\text{.}$ From the last equation we have $b(q - q') = r' - r\text{;}$ therefore, $b$ must divide $r' - r$ and $0 \leq r'- r \leq r' \lt b\text{.}$ This is possible only if $r' - r = 0\text{.}$ Hence, $r = r'$ and $q = q'\text{.}$
Let $a$ and $b$ be integers. If $b = ak$ for some integer $k\text{,}$ we write $a \mid b\text{.}$ An integer $d$ is called a common divisor of $a$ and $b$ if $d \mid a$ and $d \mid b\text{.}$ The greatest common divisor of integers $a$ and $b$ is a positive integer $d$ such that $d$ is a common divisor of $a$ and $b$ and if $d'$ is any other common divisor of $a$ and $b\text{,}$ then $d' \mid d\text{.}$ We write $d = \gcd(a, b)\text{;}$ for example, $\gcd( 24, 36) = 12$ and $\gcd(120, 102) = 6\text{.}$ We say that two integers $a$ and $b$ are relatively prime if $\gcd( a, b ) = 1\text{.}$
Let
\begin{equation*} S = \{ am + bn : m, n \in {\mathbb Z} \text{ and } am + bn \gt 0 \}\text{.} \end{equation*}
Clearly, the set $S$ is nonempty; hence, by the Well-Ordering Principle $S$ must have a smallest member, say $d = ar + bs\text{.}$ We claim that $d = \gcd( a, b)\text{.}$ Write $a = dq + r'$ where $0 \leq r' \lt d\text{.}$ If $r' \gt 0\text{,}$ then
\begin{align*} r'& = a - dq\\ & = a - (ar + bs)q\\ & = a - arq - bsq\\ & = a( 1 - rq ) + b( -sq )\text{,} \end{align*}
which is in $S\text{.}$ But this would contradict the fact that $d$ is the smallest member of $S\text{.}$ Hence, $r' = 0$ and $d$ divides $a\text{.}$ A similar argument shows that $d$ divides $b\text{.}$ Therefore, $d$ is a common divisor of $a$ and $b\text{.}$
Suppose that $d'$ is another common divisor of $a$ and $b\text{,}$ and we want to show that $d' \mid d\text{.}$ If we let $a = d'h$ and $b = d'k\text{,}$ then
\begin{equation*} d = ar + bs = d'hr + d'ks = d'(hr + ks)\text{.} \end{equation*}
So $d'$ must divide $d\text{.}$ Hence, $d$ must be the unique greatest common divisor of $a$ and $b\text{.}$
Among other things, Theorem 5.2.2 allows us to compute the greatest common divisor of two integers.
Let us compute the greatest common divisor of $945$ and $2415\text{.}$ First observe that
\begin{align*} 2415 & = 945 \cdot 2 + 525\\ 945 & = 525 \cdot 1 + 420\\ 525 & = 420 \cdot 1 + 105\\ 420 & = 105 \cdot 4 + 0\text{.} \end{align*}
Reversing our steps, $105$ divides $420\text{,}$ $105$ divides $525\text{,}$ $105$ divides $945\text{,}$ and $105$ divides $2415\text{.}$ Hence, $105$ divides both $945$ and $2415\text{.}$ If $d$ were another common divisor of $945$ and $2415\text{,}$ then $d$ would also have to divide $105\text{.}$ Therefore, $\gcd( 945, 2415 ) = 105\text{.}$
If we work backward through the above sequence of equations, we can also obtain numbers $r$ and $s$ such that $945 r + 2415 s = 105\text{.}$ Observe that
\begin{align*} 105 & = 525 + (-1) \cdot 420\\ & = 525 + (-1) \cdot [945 + (-1) \cdot 525]\\ & = 2 \cdot 525 + (-1) \cdot 945\\ & = 2 \cdot [2415 + (-2) \cdot 945] + (-1) \cdot 945\\ & = 2 \cdot 2415 + (-5) \cdot 945\text{.} \end{align*}
So $r = -5$ and $s= 2\text{.}$ Notice that $r$ and $s$ are not unique, since $r = 41$ and $s = -16$ would also work.
To compute $\gcd(a,b) = d\text{,}$ we are using repeated divisions to obtain a decreasing sequence of positive integers $r_1 \gt r_2 \gt \cdots \gt r_n = d\text{;}$ that is,
\begin{align*} b & = a q_1 + r_1\\ a & = r_1 q_2 + r_2\\ r_1 & = r_2 q_3 + r_3\\ & \vdots\\ r_{n - 2} & = r_{n - 1} q_{n} + r_{n}\\ r_{n - 1} & = r_n q_{n + 1}\text{.} \end{align*}
To find $r$ and $s$ such that $ar + bs = d\text{,}$ we begin with this last equation and substitute results obtained from the previous equations:
\begin{align*} d & = r_n\\ & = r_{n - 2} - r_{n - 1} q_n\\ & = r_{n - 2} - q_n( r_{n - 3} - q_{n - 1} r_{n - 2} )\\ & = -q_n r_{n - 3} + ( 1+ q_n q_{n-1} ) r_{n - 2}\\ & \vdots\\ & = ra + sb\text{.} \end{align*}
The algorithm that we have just used to find the greatest common divisor $d$ of two integers $a$ and $b$ and to write $d$ as the linear combination of $a$ and $b$ is known as the Euclidean algorithm.
### Exercises5.2.1Exercises
###### 1.
For each of the following pairs of numbers $a$ and $b\text{,}$ calculate $\gcd(a,b)$ and find integers $r$ and $s$ such that $\gcd(a,b) = ra + sb\text{.}$
1. $14$ and $39$
2. $234$ and $165$
3. $1739$ and $9923$
4. $471$ and $562$
5. $23771$ and $19945$
6. $-4357$ and $3754$
###### 2.
Let $a$ and $b$ be nonzero integers. If there exist integers $r$ and $s$ such that $ar + bs =1\text{,}$ show that $a$ and $b$ are relatively prime.
###### 3.
Let $a$ and $b$ be integers such that $\gcd(a,b) = 1\text{.}$ Let $r$ and $s$ be integers such that $ar + bs = 1\text{.}$ Prove that
\begin{equation*} \gcd(a,s) = \gcd(r,b) = \gcd(r,s) = 1\text{.} \end{equation*}
###### 4.
Let $x, y \in {\mathbb N}$ be relatively prime. If $xy$ is a perfect square, prove that $x$ and $y$ must both be perfect squares.
###### 5.
Using the division algorithm, show that every perfect square is of the form $4k$ or $4k + 1$ for some nonnegative integer $k\text{.}$
###### 6.
Suppose that $a, b, r, s$ are pairwise relatively prime and that
\begin{align*} a^2 + b^2 & = r^2\\ a^2 - b^2 & = s^2\text{.} \end{align*}
Prove that $a\text{,}$ $r\text{,}$ and $s$ are odd and $b$ is even.
###### 7.
Let $n \in {\mathbb N}\text{.}$ Use the division algorithm to prove that every integer is congruent mod $n$ to precisely one of the integers $0, 1, \ldots, n-1\text{.}$ Conclude that if $r$ is an integer, then there is exactly one $s$ in ${\mathbb Z}$ such that $0 \leq s \lt n$ and $[r] = [s]\text{.}$ Hence, the integers are indeed partitioned by congruence mod $n\text{.}$
###### 8.
Define the least common multiple of two nonzero integers $a$ and $b\text{,}$ denoted by $\lcm(a,b)\text{,}$ to be the nonnegative integer $m$ such that both $a$ and $b$ divide $m\text{,}$ and if $a$ and $b$ divide any other integer $n\text{,}$ then $m$ also divides $n\text{.}$ Prove there exists a unique least common multiple for any two integers $a$ and $b\text{.}$
###### 9.
If $d= \gcd(a, b)$ and $m = \lcm(a, b)\text{,}$ prove that $dm = |ab|\text{.}$
###### 10.
Show that $\lcm(a,b) = ab$ if and only if $\gcd(a,b) = 1\text{.}$
###### 11.
Prove that $\gcd(a,c) = \gcd(b,c) =1$ if and only if $\gcd(ab,c) = 1$ for integers $a\text{,}$ $b\text{,}$ and $c\text{.}$
###### 12.
Let $a, b, c \in {\mathbb Z}\text{.}$ Prove that if $\gcd(a,b) = 1$ and $a \mid bc\text{,}$ then $a \mid c\text{.}$
###### 13.Fibonacci Numbers.
The Fibonacci numbers are
\begin{equation*} 1, 1, 2, 3, 5, 8, 13, 21, \ldots\text{.} \end{equation*}
We can define them inductively by $f_1 = 1\text{,}$ $f_2 = 1\text{,}$ and $f_{n + 2} = f_{n + 1} + f_n$ for $n \in {\mathbb N}\text{.}$
1. Prove that $f_n \lt 2^n\text{.}$
2. Prove that $f_{n + 1} f_{n - 1} = f^2_n + (-1)^n\text{,}$ $n \geq 2\text{.}$
3. Prove that $f_n = [(1 + \sqrt{5}\, )^n - (1 - \sqrt{5}\, )^n]/ 2^n \sqrt{5}\text{.}$
4. Show that $\lim_{n \rightarrow \infty} f_n / f_{n + 1} = (\sqrt{5} - 1)/2\text{.}$
5. Prove that $f_n$ and $f_{n + 1}$ are relatively prime.
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# What is the vertex of f(x)=7-x^2?
Aug 28, 2014
The vertex is $\left(0 , 7\right)$.
Sometimes we have problems with the easier questions because it's not exactly in the form we're used to. Normally for a quadratic, you would complete the square to find the vertex. But this quadratic is already in vertex or standard form:
$f \left(x\right) = - {\left(x - 0\right)}^{2} + 7$
Oct 22, 2014
We can also find the vertex by using the expressions:
$\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$
Standard form:
$a {x}^{2} + b x + c = 0$
In this example, a=−1 and $b = 0$
x=−0/(2(−1))=0/2=0
y=f(0)=7−0^2=7
Same result of $\left(0 , 7\right)$
Oct 22, 2014
Yes, the vertex is at (0,7), but I would like to address this problem graphically.
The graph of a function $f \left(x\right) = {x}^{2}$ is a parabola with branches directed upward and a vertex at the point $\left(0 , 0\right)$, as everybody knows.
When you consider a graph of a function $f \left(x\right) = - {x}^{2}$, you just turn the graph of $f \left(x\right) = {x}^{2}$ upside down. The vertex will still be at $\left(0 , 0\right)$, but the branches of a parabola will be directed downwards.
Next we transform our function into $f \left(x\right) = 7 - {x}^{2}$, which adds $7$ to all values of a function $f \left(x\right) = - {x}^{2}$. That shifts an entire graph by $7$ upwards. Vertex also gets shifted by $7$, so its position is shifted from $\left(0 , 0\right)$ to $\left(0 , 7\right)$.
Oct 23, 2014
We could also use Calculus to solve this question.
We have to recognize that this is a quadratic equation which is just a parabola.
We know that a parabola will have either a maximum or minimum at the vertex.
The derivative of a function is the slope of the tangent line at a specific point on the function.
The derivative or tangent line at the vertex will have a slope of 0.
$f \left(x\right) = 7 - {x}^{2}$
$f ' \left(x\right) = 0 - 2 x$
$f ' \left(x\right) = - 2 x$
$- 2 x$ is the derivative, the slope of the tangent line.
Set the derivative equal to zero to find the $x$ value at the vertex.
$- 2 x = 0$
$x = 0$
The $x$ value of the vertex is 0.
Now substitute in $x = 0$ in the original function, $f \left(x\right) = 7 - {x}^{2}$
$f \left(0\right) = 7 - {\left(0\right)}^{2}$
$f \left(0\right) = 7$
The vertex is at point $\left(0 , 7\right)$.
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Fundamental Theorem of Calculus ( Read ) | Analysis | CK-12 Foundation
# Fundamental Theorem of Calculus
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Fundamental Theorem of Calculus
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Velocity due to gravity can be easily calculated by the formula: v = gt , where g is the acceleration due to gravity (9.8m/s 2 ) and t is time in seconds. In fact, a decent approximation can be calculated in your head easily by rounding 9.8 to 10 so you can just add a decimal place to the time.
Using this function for velocity, how could you find a function that represented the position of the object after a given time? What about a function that represented the instantaneous acceleration of the object at a given time?
Embedded Video:
### Guidance
If you think that evaluating areas under curves is a tedious process you are probably right. Fortunately, there is an easier method. In this section, we shall give a general method of evaluating definite integrals (area under the curve) by using antiderivatives.
Definition: The Antiderivative
If F ' ( x ) = f ( x ), then F '( x ) is said to be the antiderivative of f ( x ).
There are rules for finding the antiderivatives of simple power functions such as f ( x ) = x 2 . As you read through them, try to think about why they make sense, keeping in mind that differentiation reverses integration.
Rules of Finding the Antiderivatives of Power Functions
• The Power Rule
$\int x^n dx = \frac{1} {n + 1} x^{n + 1} + C$
where C is constant of integration and n is a rational number not equal to -1.
• A Constant Multiple of a Function Rule
$\int k x^n dx = k \int x^n dx = k \cdot \frac{1} {n + 1} x^{n + 1} + C$
where k is a constant.
• Sum and Difference Rule
$\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx$
• The Constant rule
$\int k \cdot dx = kx + C$
where k is a constant. (Notice that this rule comes as a result of the power rule above.)
The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus makes the relationship between derivatives and integrals clear. Integration performed on a function can be reversed by differentiation.
The Fundamental Theorem of Calculus
If a function f ( x ) is defined over the interval [ a , b ] and if F ( x ) is the antidervative of f on [ a, b ], then
$\int_{a}^{b} f(x) dx = F(x)|^b_a$
$= F(b) - F(a)$
We can use the relationship between differentiation and integration outlined in the Fundamental Theorem of Calculus to compute definite integrals more quickly.
#### Example A
Evaluate $\int_{1}^{2} x^2 dx.$
Solution
This integral tells us to evaluate the area under the curve f ( x ) = x 2 , which is a parabola over the interval [1, 2], as shown in the figure below.
To compute the integral according to the Fundamental Theorem of Calculus, we need to find the antiderivative of f ( x ) = x 2 . It turns out to be F ( x ) = (1/3) x 3 + C , where C is a constant of integration. How can we get this? Think about the functions that will have derivatives of x 2 . Take the derivative of F ( x ) to check that we have found such a function. (For more specific rules, see the box after this example). Substituting into the fundamental theorem,
$\int_{a}^{b} f(x) dx$ $= F(x)|^b_a$
$\int_{1}^{2} x^2 dx$ $= \left [\frac{1} {3} x^3 + C\right ]^2_1$
$= \left [\frac{1} {3} (2)^3 + C\right ] - \left [\frac{1} {3} (1)^3 + C\right ]$
$= \left [\frac{8} {3} + C\right ] - \left [\frac{1} {3} + C\right ]$
$= \frac{7} {3} + C - C$
$= \frac{7} {3}$
So the area under the curve is (7/3) units 2 .
#### Example B
Evaluate $\int x^3dx.$
Solution
Since $\int x^n dx = \frac{1} {n + 1}x^{n + 1} + C$ , we have
$\int x^3 dx$ $= \frac{1} {3 + 1}x^{3 + 1} + C$
$= \frac{1} {4}x^4 + C$
To check our answer we can take the derivative of $\frac{1} {4}x^4 + C$ and verify that it is $\,\! x^3$ , the original function in our integral.
#### Example C
Evaluate $\int 5x^2 dx.$
Solution
Using the constant multiple of a power rule, the coefficient 5 can be removed outside the integral:
$\int 5x^2 dx = 5 \int x^2 dx$
Then we can integrate:
$= 5 \cdot \frac{1} {2 + 1} x^{2 + 1} + C$
$= \frac{5} {3} x^3 + C$
Again, if we wanted to check our work we could take the derivative of $\frac{5} {3} x^3 + C$ and verify that we get $\,\! 5x^2$
### Vocabulary
The Fundamental Theorem of Calculus demonstrates that integration performed on a function can be reversed by differentiation.
Integrals allow for the calculation of the area between a line (such as the x-axis) and a curve, or of the area between two curves. Since the area is generally given in square units, it is technically only an approximation, but can be an effectively infinitely close one!
The antiderivative has much the same relationship to a function that a square root has to a constant. The antiderivative of a function is the function whose derivative is the function you want the antiderivative of.
### Guided Practice
Questions
1) Evaluate $\int (3x^3 - 4x^2 + 2)dx.$
2) Evaluate $\int_{2}^{5} \sqrt{x} dx.$
3)Use the fundamental theorem of calculus to solve: $\int_{4}^{6} \frac{dx}{x}$
4)Use the fundamental theorem of calculus to solve: $\int_{-2p}^{2p} 3cos(x) dx =$
5)Use the 2nd fundamental theorem of calculus to solve: $A(x) = \int_{3}^{x} cot^3 (t) dt$
Solutions
1) Using the sum and difference rule we can separate our integral into three integrals:
$\int (3x^3 - 4x^2 + 2)dx =$
$3 \left( \int x^3 dx \right) - 4 \left( \int x^2 dx \right) + \left( \int 2dx \right)$
$\to 3 \cdot \frac{1} {4} x^4 - 4 \cdot \frac{1} {3}x^3 + 2x + C$ $\to \frac{3} {4} x^4 - \frac{4} {3} x^3 + 2x + C$
2) The evaluation of this integral represents calculating the area under the curve $y = \sqrt{x}$ from x = -2 to x = 3, shown in the figure below.
$\int_{2}^{5} \sqrt{x} dx$ $= \int_{2}^{5} x^{1/2} dx$
$= \left [\frac{1} {\frac{1} {2} + 1} x^{1/2 + 1}\right ]^5_2$
$= \left [\frac{1} {3/2} x^{3/2}\right ]^5_2$
$= \frac{2} {3} \left [x^{3/2}\right ]^5_2$
$= \frac{2} {3}\left [5^{3/2} - 2^{3/2}\right ]$
$= 5.57$
So the area under the curve is 5.57.
3) Given what we know, that if F(x) = ln x, then F'(x) = $\frac {1}{x}$
Thus, we apply the fundamental theorem of calculus:
$\int-{4}^{6} \frac{dx}{x} = ln x |-{4}{6}$
= F(6) - F(4) = [ln(6)] - [ln(4)] = 0.4055
4) Given what we know, that if F(x) = 3sin(x), then F'(x) = 3cos(x)
So we apply the fundamental theorem of calculus:
$\int_{-2p}^{2p} 3cos dx = 3sin(x) | _{-2p}^{2p}$
= F(8) - F(0) = [3sin(2p)] - [3sin(-2p)] = 1 - 0 = 0
5) Find $A'(\frac{3p}{4})$
The second theorem states: $\int_{3}^{x}cot^3 (t) dt$ is the anti-derivative of $cot^3(x)$
So, $A'(x) = cot^3(x)$
Substituting in $x = \frac{3p}{4}$ we get an answer of $A'{4} = -1$
### Practice
Evaluate the Integral:
1. Evaluate the integral $\int_{0}^{3} 5xdx$
2. Evaluate the integral $\int_{0}^{1} x^4dx$
3. Evaluate the integral $\int_{1}^{4} (x - 3)dx$
Find the Integral:
1. Find the integral of ( x + 1)(2 x - 3) from -1 to 2.
2. Find the integral of $\sqrt{x}$ from 0 to 9.
3. Find the integral of $\int_{-1}^{0} - 3 dx =$
4. Find the integral of $\int_{-1}^{3} dx =$
5. Find the integral of $\int_{-p}^{\frac{p}{2}} - 4cos(x) dx =$
6. Find the integral of $\int_{0}^{2} -dx =$
7. Find the integral of $\int_{2}^{7} \frac{dx}{x}$
8. Find the integral of $\int_{-2}^{0}x + 5 dx =$
9. Find the integral of $\int_{-p}^{\frac{3p}{2}}6sin(x) dx =$
10. Find the integral of $\int_{6}^{7} \frac{dx}{x}$
Challenge yourself:
1. Sketch y = x 3 and y = x on the same coordinate system and then find the area of the region enclosed between them (a) in the first quadrant and (b) in the first and third quadrants.
2. Evaluate the integral $\int_{-R}^{R} (\pi R^2 - \pi x^2) dx$ where R is a constant.
Apply the second theorem of calculus:
1. $A(x) = \int_{2}^{x} tan^3(t) dt$
2. Find: $\frac{d}{dx} \int_{1}^{x} csc^2(t) dt$
3. Find: $\frac{d}{dx} \int_{-2}^{x^2} sec(t) dt$
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CHAPTER 1. Mathematics Aviation Applications Ratios have widespread application in the field of aviation. For example: Compression ratio on a reciprocating engine is the ratio of the volume of a cylinder with the piston at the bottom of its stroke to the volume of the cylinder with the piston at the top of its stroke. For example, a typical compression ratio might be 10:1 (or 10 to 1). Aspect ratio is the ratio of the length (or span) of an airfoil to its width (or chord). A typical aspect ratio for a commercial airliner might be 7:1 (or 7 to 1). Air-fuel ratio is the ratio of the weight of the air to the weight of fuel in the mixture being fed into the cylinders of a reciprocating engine. For example, a typical air-fuel ratio might be 14.3:1 (or 14.3 to 1). Glide ratio is the ratio of the forward distance traveled to the vertical distance descended when an aircraft is operating without power. For example, if an aircraft descends 1,000 feet while it travels through the air for a distance of two linear miles (10,560 feet), it has a glide ratio of 10,560:1,000 which can be reduced to 10.56: 1 (or 10.56 to 1). Gear Ratio is the number of teeth each gear represents when two gears are used in an aircraft component. In Figure 1-7, the pinion gear has 8 teeth and a spur gear has 28 teeth. The gear ratio is 8:28 or 2:7. Speed Ratio. When two gears are used in an aircraft component, the rotational speed of each gear is represented as a speed ratio. As the number of teeth in a gear decreases, the rotational speed of that gear increases, and vice-versa. Therefore, the speed ratio of two gears is the inverse (or opposite) of the gear ratio. If two gears have a gear ratio of 2:9, then their speed ratio is 9:2. Example: A pinion gear with 10 teeth is driving a spur gear with 40 teeth. The spur gear is rotating at 160 rpm. Determine the speed of the pinion gear. To solve for SP, multiply 40 x 160, then divide by 10. The speed of the pinion gear is 640 rpm. Example: If the cruising speed of an airplane is 200 knots and its maximum speed is 250 knots, what is the ratio of cruising speed to maximum speed? First, express the cruising speed as the numerator of a fraction whose denominator is the maximum speed. Next, reduce the resulting fraction to its lowest terms. Therefore, the ratio of cruising speed to maximum speed is 4:5. Another common use of ratios is to convert any given ratio to an equivalent ratio with a denominator of 1. Example: Express the ratio 9:5 as a ratio with a denominator of 1. Therefore, 9:5 is the same ratio as 1.8:1. In other words, 9 to 5 is the same ratio as 1.8 to 1. Proportion A proportion is a statement of equality between two or more ratios. For example, This proportion is read as, “3 is to 4 as 6 is to 8.” Extremes and Means The first and last terms of the proportion (the 3 and 8 in this example) are called the extremes. The second and third terms (the 4 and 6 in this example) are called the means. In any proportion, the product of the extremes is equal to the product of the means. In the proportion 2:3 = 4:6, the product of the extremes, 2 x 6, is 12; the product of the means, 3 x 4, is also 12. An inspection of any proportion will show this to be true.
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# مثالی از قانون AND
سرفصل: بخش ریاضی / سرفصل: احتمالات / درس 7
## بخش ریاضی
14 سرفصل | 192 درس
### توضیح مختصر
• زمان مطالعه 4 دقیقه
• سطح خیلی سخت
### دانلود اپلیکیشن «زوم»
این درس را میتوانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید
## Examples of the AND Rule
In the last video we’ve learned about the simplified AND rule. AND just means multiply when we’re dealing with independent events. So, in this video we’re going to look at some examples of this. So, first of all, what is the probability of tossing three coins and getting heads, heads, heads? Well, first of all, notice that the three coins are independent, whatever happens to one coin is gonna have no influence on another coin.
So, we have three independent events, probability is one half for getting heads in one case, one half, one half, and so the three one halves just get multiplied for a probability of one eighth. One eighth is the probability of tossing three heads. Another question.
What is the probability of rolling two six-sided dice and getting snake eyes? That’s one on each die. So snake eyes would look something like this. Well, again, the two die don’t influence each other. So, these are independent events also. The probability of getting a one on one six sided die is one sixth, so one sixth on one, one six on the other, they’re independent so that means and means multiply.
Multiply them we get 1 over 36, the probability of rolling snake eyes when you roll 2 dice. Three cards are selected from a full deck, each time with replacement, what is the probability of selecting three spades in a row? So again. Let’s make sure that we are perfectly clear on this idea of with replacement.
What that means is I shuffle the deck, I pull a card out, I look at it. Then I put that card back in the deck, shuffle it again. So that the second card I pick is being picked from a full shuffled deck. There are 52 cards what I make the second pick. Now I’ll look at that second card, I record it, I put it back into the deck, I shuffle again, so now when I pick the third choice I’m also picking from a full deck of 52 cards, a full shuffled deck, so all three cards are picked from exactly the same conditions.
A full shuffled 52 card deck. That’s what’s going on with replacement. Well, of course, in a modern deck there are four suits. There are hearts, diamonds, clubs and spades and those are evenly divided. So the chance of picking one spade is one quarter. And then, on the second pick it’s also one quarter.
Then on the third pick it’s also one quarter, because we’re selecting with replacement because each time the choice is made from the same initial conditions the events are independent, the choices are independent. And that’s always true with replacement. That means if we get to multiply, we get 1 over 64.
The probability of selecting three spades in a row. Finally, we get this algebraic equation. We don’t even know what the events are. But we’re just told that they’re independent. Sometimes problem will just tell you A and B are independent. Probability of A is 0.6, probability of B is 0.8, what does the probability of A or B mean?
Okay, well this is a bit of a curve-ball. We are talking about an And rule and here we have an Or rule, what’s going on here? Well let’s think about this. This is the general Or rule. And incidentally, if the events are independent that does, that means they can not be mutually exclusive.
If the events are mutually exclusive they are definitely having an effect on one another. Because if one happens it prevents the other one from happening. So that is definitely a very different case from independent. If they’re independent, they’re not mutually exclusive, we need to use the generalized and rule.
So, let’s think about this. We know the probability of A, we know the probability of B. The probability of A and B, well we can get that, because the events are independent, we can get this by multiplying. That would be 0.6 times 0.8, the probability of A and B would be the probability of A times the probability of B, 0.6 times 0.8 is 0.48.
So now we know all 3 terms and now we can say we have 0.6 plus 0.8 minus 0.48. So this is 1.4 minus 0.48 and this winds up being 0.92. Which is the probability of A or B.
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# How to solve 1/2 cup in a gallon ?
Welcome to my article How to solve 1/2 cup in a gallon ?. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve 1/2 cup in a gallon ?, read and understand it carefully till the end.
Let us know how to solve this questionHow to solve 1/2 cup in a gallon ?
First write the question on the page of the notebook.
## How to solve 1/2 cup in a gallon ?
Suppose this question is asked how many 1/2 cups will be equal to the value of one gallon.
Then we need to know how many cups are in 1 gallon –
here
1 gallon = 16 cup
16 cup = 1 gallon
1 cup = 1/16 gallon
Thus, we can use this formula to find the value of 1 gallon in 1/2 cup.
Let’s find the value of 1 gallon in 1/2 cup,
We know that,
\displaystyle 1cup=\frac{1}{{16}}gallon ,
\displaystyle \frac{1}{2}cup=\frac{1}{{16}}\times \frac{1}{2}gallon ,
\displaystyle \frac{1}{2}cup=\frac{1}{{16\times 2}}gallon ,
\displaystyle \frac{1}{2}cup=\frac{1}{{32}}gallon ,
After understanding the whole solution in this way, we know that,
\displaystyle 1cup=\frac{1}{{16}}gallon ,
\displaystyle \frac{1}{2}cup=\frac{1}{{32}}gallon
This article How to solve 1/2 cup in a gallon ? has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need.
See also simplify: a ^ 2 * (a - b + c) + b ^ 2 * (a + b - c) - c ^ 2 * (a - b - c)
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# Ideal (mathematics)
In ring theory, an ideal is defined as a subset of a ring with the following properties:
1. Zero is an element of the ideal
2. The ideal is closed under addition
3. The product of an element of the ideal and an element of the initial ring is an element of the ideal
In general, one must distinguish between left and right ideals, because many rings do not have commutative multiplication. All discussion here applies to commutative rings.
## Examples
An example is the set of even integers, as a subset of the ring of integers. This is an ideal because:
1. 0 is even.
2. The sum of two even integers is even.
3. The product of an even integer with and any other integer is even.
This example is itself an example of a principal ideal. Given any ring R and element $x \in R$, we may define an ideal (r), which consists of all elements of R which may be written as the product of r with some other element a of the ring. The ideal (r) is called a principal ideal. It is an ideal, because:
1. $0 = 0 \cdot r \in (r)$
2. If $x \in (r)$ and $y \in (r)$, then we can write x = ra and y = rb. Then $x+y = ra + rb = r(a+b) \in R$.
3. If $x \in (r)$ and $b \in R$ is an arbitrary element of the ring R, then $xb = (ar)b = (ab)r \in R$.
When $R=\mathbb Z$ is the ring of integers, then every ideal is a principal ideal: that is, any ideal I of the ring of integers is the set of elements "multiples of k" for some integer k. Many familiar rings have the property that every ideal is a principal ideal: such rings are known as principal ideal domains.
## In Polynomial Rings
One familiar ring is $R=\mathbb R[x]$, the ring of polynomials over the integers. One ideal in R is the polynomials with constant term 0, such as x + 5x3. This can be thought of as the set of polynomials p satisfying p(0) = 0. This is an ideal, because:
1. The zero polynomial has constant term 0.
2. If f,g are two polynomials with constant term 0, then (f + g)(0) = f(0) + g(0) = 0 + 0 = 0.
3. If f has constant term 0 and p is arbitrary, then $(f \cdot p)(0) = f(0) \, p(0) = 0 \, p(0) = 0$.
A brief reflection shows that this ideal is actual just the principal ideal (x): every polynomial vanishing at 0 is just the polynomial x times some other polynomial. In fact, it turns out that $\mathbb R[x]$ is a principal ideal domain.
However, consider the ring $\mathbb R[x,y]$ of polynomials in two variables. There is again an ideal I of polynomials vanishing at 0. However, this is not a principal ideal. If it were principal and generated by p(x,y), then this p would divide both the polynomials x and y, since both are in the ideal. Clearly there is no such polynomial. This I is an exactly of an ideal that is not principal, and $\mathbb R[x,y]$ is a ring which is not a principal ideal domain.
## Prime Ideals
Prime ideals provide a way to generalize the prime numbers. An ideal I is called a prime ideal if $ab \in I$ implies that either $a \in I$ or $b \in I$. The basic examples of prime ideas are the principal ideals $(p) \subset \mathbb Z$, where p is a prime number. For example, consider the ring (5), which consists of all integer multiples of 5. If the product of two numbers is a multiple of 5, it must be that one of the numbers is a multiple of 5. This is in contrast with ideals that are not generated by a prime number: for example, (6) is the ideal of all multiples of 6. However, it is possible for the product of two numbers to be a multiple of 6 without either of the numbers being one: for example, $2 \cdot 3 = 6$.
Prime ideals play an important role in many areas of mathematics related to algebra, especially in algebraic number theory and algebraic geometry.
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# McGraw Hill My Math Grade 3 Chapter 9 Lesson 1 Answer Key Take Apart to Multiply
All the solutions provided in McGraw Hill My Math Grade 3 Answer Key PDF Chapter 9 Lesson 1 Take Apart to Multiply will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 3 Answer Key Chapter 9 Lesson 1 Take Apart to Multiply
When you take apart, or decompose, a factor, you 1 have smaller numbers that are easier to multiply.
Build It
Find 4 Ă— 7.
1. Model 4 Ă— 7.
Use color tiles to make a 4 Ă— 7 array. Draw the array.
2. Decompose one factor.
• Take apart the 7.
• Separate 7 columns into 5 columns + 2 columns.
3. Find the products of each part. Then add.
So, 4 Ă— 7 = ______________.
Answer: 4 Ă— 7 = 28
Explanation:
Given, 4 Ă— 7 = ( 4 Ă— 5 ) + ( 4 Ă— 2 )
28 = 20 + 8
28 = 28.
So, 4 Ă— 7 = 28
Try It
Gretchen cut 6 oranges into 9 slices each. How many orange slices are there?
Find 6 Ă— 9.
1. Outline a 6 Ă— 9 array on the grid paper.
2. Decompose one factor.
Draw a vertical line through the array to decompose the factor 9 into 5 + 4. Write the addends above.
3. Find the product of each part.
So, 6 Ă— 9 = _____________. There are _____________ orange slices.
Answer: There are 54 orange slices.
Explanation:
Given, 6 Ă— 9 = ?
decompose the factor 9 into 5 + 4.
6 Ă— 9 = ( 6 Ă— 5) + ( 6 Ă— 4)
54 = 30 + 24
54 = 54
So, There are 54 orange slices.
Question 1.
Mathematical PRACTICE Justify Conclusions In the example above, could the 6 have been decomposed instead of the 9? Explain.
Answer: Yes , 6 could have been decomposed instead of the 9
Explanation:
In the above example we have seen,
Given, 6 Ă— 9 = ?
decompose the factor 9 into 5 + 4.
6 Ă— 9 = ( 6 Ă— 5) + ( 6 Ă— 4)
54 = 30 + 24
54 = 54
What if 6 is decomposed,
decompose the factor 6 into 3 + 3.
6 Ă— 9 = ( 3 Ă— 9) + ( 3 Ă— 9)
54 = 27 + 27
54 = 54
So, What ever the number is decomposed in the given product the result will be same .
Question 2.
How is decomposing a factor helpful in finding products?
Answer: To decompose is to break down numbers into parts.
Explanation:
As we said earlier, To decompose is to break down numbers into parts. which makes the calculation is easier to solve , by making the smaller products of the numbers,
So, it will be easy to decomposing a factor helpful in finding products.
Question 3.
Explain how using a known fact strategy is similar to decomposing a factor.
Answer: Decomposing a Factor (number) means to break apart the number to uncover the numbers within the factor.
Explanation:
known fact is a fact that you know by memory. expression. A combination of numbers and operations that represent a quantity. while decomposing is to break apart the number.
In both the cases we are using the same strategy as to break the numbers down for easy calculations,
So, a known fact strategy is similar to decomposing a factor.
Practice It
Use color tiles to model the array. Decompose one factor. Then find the product for each part and add.
Question 4.
Explanation:
Given, 7 Ă— 6
decompose the factor 6 into 3 + 3.
7 Ă— 6 = ( 7 Ă— 3 ) + ( 7 Ă— 3)
42 = 21 + 21
So, 7 Ă— 6 = 42.
Question 5.
Explanation:
Given,
decompose the factor 7 into 4 + 3.
8 Ă— 7 = ( 8 Ă— 4 ) + ( 8 Ă— 3 )
56 = 32 + 24
56 = 56
So, 8 Ă— 7 = 56.
Question 6.
Decompose one factor. Color the array two colors to represent your numbers. Then find the product for each part and add.
Explanation:
Given, 7 × 9 = ?
decompose the factor 9 into 5 + 4.
7 Ă— 9 = ( 7 Ă— 5 ) + ( 7 Ă— 4 )
63 = 35 + 28
63 = 63
So, 7 Ă— 9 = 63.
Question 7.
Decompose the fact a different way.
Explanation:
Given, Decompose the fact a different way.
decompose the factor 9 into 6 + 3.
7 Ă— 9 = ( 7 Ă— 6 ) + ( 7 Ă— 3 )
63 = 42 + 21
63 = 63
So, 7 Ă— 9 = 63.
Apply It
Mathematical PRACTICE Identify Structure Decompose one factor. Find each product. Then add.
Question 8.
Mr. Daniels bought 9 packages of metal brackets to build some bookshelves. There are 8 brackets in each package. How many brackets did Mr. Daniels buy altogether?
Answer: Mr. Daniels bought 72 brackets altogether.
Explanation:
Given, Mr. Daniels bought 9 packages of metal brackets to build some bookshelves.
There are 8 brackets in each package.
Then, 9 Ă— 8 =?
decompose the factor 8 into 4 + 4.
9 Ă— 8 = ( 9 Ă— 4 ) + ( 9 Ă— 4 )
72 = 36 + 36
72 = 72
So, Mr. Daniels bought 72 brackets altogether.
Question 9.
Jenna is making 6 costumes for the dance recital. Each costume uses 9 feet of fabric. How much fabric does Jenna need altogether?
Answer: Jenna need 54 feet of fabric altogether
Explanation:
Given, Jenna is making 6 costumes for the dance recital. Each costume uses 9 feet of fabric.
Then 6 Ă— 9 = ?
decompose the factor 9 into 5 + 4.
6 Ă— 9 = ( 6 Ă— 5 ) + ( 6 Ă— 4 )
54 = 30 + 24
54 = 54
So, 6 Ă— 9 = 54.
Question 10.
Eight horses each ate the number of apples shown. How many apples did they eat altogether?
Answer: 8 Ă— 4 = 32 , Horses ate 32 apples altogether
Explanation:
Given, Eight horses each ate the number of apples shown.
There are 4 apples
then, 8 Ă— 4 = ?
decompose the factor 8 into 4 + 4.
4 Ă— 8 = ( 4 Ă— 4 ) + ( 4 Ă— 4 )
32 = 16 + 16
32 = 32
So, Horses ate 32 apples altogether
Question 11.
Mathematical PRACTICE Reason How could you change Exercise 8 so that Mr. Daniels buys a total of 81 brackets?
Answer:Â There should be 9 brackets in each package.
Explanation:
Given, Mr. Daniels bought 9 packages of metal brackets to build some bookshelves.
There are 8 brackets in each package.
To have a total of 81 brackets , we should change the number of brackets in each package
Then 8 brackets should be replaced by 9 brackets
thus, 9 Ă— 9 = 81
So, There should be 9 brackets in each package.
Question 12.
How does decomposing a factor allow you to group numbers differently?
Answer: Decomposition of a composite number means converting it into a product of smaller integers. this allows us to find factors of the numbers.so we can group number differently by this method using factors.
To decompose a number or set is to break it apart or separate it into two or more parts. To decompose numbers can mean to determine a sum positive rational numbers that give a specific value. To decompose numbers can also mean to factor a composite number.
### McGraw Hill My Math Grade 3 Chapter 9 Lesson 1 My Homework Answer Key
Practice
Decompose one factor. Color the array two colors to represent your numbers. Then find the product for each part and add.
Question 1.
Explanation;
7 Ă— 7 = ( 7 Ă— 4 ) + ( 7 Ă— 3 )
49 = 28 + 21
49 = 49
So, 7 Ă— 7 = 49.
Question 2.
Explanation:
6 Ă— 8 = ( 6 Ă— 4 ) + ( 6 Ă— 4 )
48 = 24 + 24
48 = 48
So, 6 Ă— 8 = 48
Decompose one factor. Find each product. Then add. Decompose the fact a different way below.
Question 3.
Explanation:
Given, 8 Ă— 8
decompose the factor 8 into 4 + 4.
8 Ă— 8 = ( 8 Ă— 4 ) + ( 8 Ă— 4 )
64 = 32 + 32
64 = 64
So, 8 Ă— 8 = 64
Another way is
Given, 8 Ă— 8
decompose the factor 8 into 6 + 2.
8 Ă— 8 = ( 8 Ă— 6 ) + ( 8 Ă— 2 )
64 = 48 + 16
64 = 64
So, 8 Ă— 8 = 64
Question 4.
Explanation:
Given, 5 Ă— 7
decompose the factor 7 into 5 + 2.
5 Ă— 7 = ( 5 Ă— 5 ) + ( 5 Ă— 2 )
35 = 25 + 10
35 = 35
So, 5 Ă— 7 = 35
Another way is
Given, 5 Ă— 7
decompose the factor 7 into 4 + 3.
5 Ă— 7 = ( 5 Ă— 4 ) + ( 5 Ă— 3 )
35 = 20 + 15
35 = 35
So, 5 Ă— 7 = 35
Problem Solving
Decompose one factor. Find each product. Then add.
Question 5.
Mathematical PRACTICE Identify Structure Orlando’s baby sister takes 3 naps a day. How many naps does she take in 9 days?
Answer: 3 Ă— 9Â = 27 , She takes 27 naps in 9 days
Explanation:
Given, Orlando’s baby sister takes 3 naps a day.
How many naps does she take in 9 days?
for each day 3 naps
for 9 days 3 Ă— 9 = 27 naps,
So, She takes 27 naps in 9 days
Question 6.
Carli gets to the bus stop 5 minutes early each morning. How many minutes does she wait at the bus stop in 5 days?
Answer: 5 Ă— 5 = 25, Totally she waits for 25 minutes in 5 days
Explanation:
Given, Carli gets to the bus stop 5 minutes early each morning.
She waits for 5 minutes for 5 days,
Then 5 Ă— 5 = 25
So, Totally she waits for 25 minutes in 5 days
Question 7.
Every Monday, Wednesday, and Friday, Mr. Brennan walks 2 miles and jogs 4 miles. What is the total number of miles Mr. Brennan walks and jogs in two weeks?
Answer: Mr. Brennan walks and jogs 36 miles in two weeks
Explanation:
Given, Every Monday, Wednesday, and Friday,
Mr. Brennan walks 2 miles and jogs 4 miles.
total miles 2 + 4 = 6 miles
Number of days, are 3 days per week
Then for 2 weeks 3 Ă— 2 = 6 days
thus 6 Ă— 6 = 36.
So, Mr. Brennan walks and jogs 36 miles in two weeks.
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Question of Solved Question
# Question The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the ratio between the height and radius of the cylinder.
Represent Root 9 point 3 on the number line
Solution:
Explanation:
Step 1: Draw a line segment AB of length 9.3 units.
Step 2: Now, Extend the line by 1 unit more such that BC=1 unit .
Step 3: Find the midpoint of AC.
Step 4: Draw a line BD perpendicular to AB and let it intersect the semicircle at point D.
Step 5: Draw an arc DE such that BE=BD.
Hence, Number line of √ 9.3 is attached below.
Which one of the following statement is true
A: Only one line can pass through a single point.
B: There are an infinite number of lines which pass through two distinct points.
C: Two distinct lines cannot have more than one point in common.
D: If two circles are equal, then their radii are not equal.
Solution:
Explanation:
From one point there is an uncountable number of lines that can pass through.
Hence, the statement “ Only one line can pass through a single point” is false.
We can draw only one unique line passing through two distinct points.
Hence, the statement “There are an infinite number of lines which pass through two distinct points” is false.
Given two distinct points, there is a unique line that passes through them.
Hence, the statement “Two distinct lines cannot have more than one point in common” is true.
If circles are equal, which means the circles are congruent. This means that circumferences are equal and so the radii of two circles are also equal.
Hence, the statement “If two circles are equal, then their radii are not equal” is false.
The correct option is (C) Two distinct lines cannot have more than one point in common.
The class mark of the class 90-120 is
A: 90
B: 105
C: 115
D: 120
Solution:
Explanation:
To find the class mark of a class interval, we find the sum of the upper limit and lower limit of a class and divide it by 2 Thus,
Class -mark=Upper limit + Lower limit/2
Here, the lower limit of 90-120=90
And the upper limit of 90-120=120
So,
Class -mark=120+90/2
=210/2
=105
Hence, the class mark of the class 90-120 is 105
That is, option (B) is correct.
Option (B) 105 is correct.
ABCD is a parallelogram and AP
and CQ are perpendiculars from vertices A and C on diagonal BD Show that i)ΔAPBΔCQD ii) AP = CQ
Solution:
Find the roots of the following equation
A: - 1, - 2
B: - 1, - 3
C: 1,3
D: 1,2
Solution:
Explanation:-
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# The Distributive Property for Seventh Graders
🏆Practice distributive property for seventh grade
Solving algebraic equations is made easier by understanding some basic rules and properties. A few examples of properties that we will learn to use in the seventh grade are: the distributive, associative and commutative properties. These properties get learned and relearned throughout our time in school, each time adding new layers to or understanding. Today we will focus on the distributive property. We will go into depth on what it is and how to use it, and we will briefly get to know the associative and commutative properties as well.
## What is the distributive property?
The distributive property is a method to simplify multiplication and division exercises. Essentially, it breaks down expressions into smaller, easier to manage terms.
Let's see some examples:
• $6 \times 26 = 6 \times (20 + 6) = 120 + 36 = 156$
• $7 \times 32 = 7 \times (30 + 2) = 210 + 14 = 224$
• $104:4 = (100+4):4 = 100:4 + 4:4 = 25+1 = 26$
If we look at the following examples, we can see that we have broken down the larger number into several smaller numbers that are more manageable. The value is the same as before, but now we can distribute a complex operation into several easy operations.
The distributive property can be described as:
$Z \times (X + Y) = ZX + ZY$
$Z \times (X - Y) = ZX - ZY$
## Test yourself on distributive property for seventh grade!
$$140-70=$$
## The distributive property
Sometimes, an expression will require us to perform both addition and subtraction within our parentheses. Not to worry! The distributive property can simplify these expressions too.
Let's see some examples:
• $(X + 2) \times (X + 3) =$
$X² + 3X + 2X + 6 = X² + 5X + 6$
• $(X - 4) \times (X - 3) =$
$X² - 3X - 4X + 12 = X² - 7X + 12$
How would you go about using the distributive property in an equation with two sets of parentheses?
First, we multiply the first term of the first parenthesis by the first and second terms of the second parenthesis..
Next, we multiply the second term of the first parenthesis and multiply it by the first and second terms of the second parenthesis.
Remember to place the addition and subtraction signs in the correct places.
Another way to describe the distributive property:
$(Z + T) \times (X + Y) = ZX + ZY + TX + TY$
$(Z - T) \times (X - Y) = ZX - ZY - TX + TY$
## The distributive property in elementary school
When first learning about the distributive property, we practice only with known whole numbers (without variables or fractions) in order to understand the idea of breaking down a larger number into smaller numbers. In this first stage, we use the distributive property mainly to simplify calculation, especially with large numbers.
For example:
$3\times 102=3\times(100+2)=300+6=306$
$7\times 96=7\times(100-4)=700-28=672$
By this point, most students have already mastered long addition and subtraction, but they might have less experience multiplying large numbers. The distributive property helps them to solve these problems by reducing into simpler multiplications equations.
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## The distributive property in middle school
In middle school, the distributive property gets more interesting. Now, we will start to use not just whole numbers, but variables and exponents too!
For example:
• $(X + 5) \times (X + 6) =$
$X² + 6X + 5X+ 30 = X² + 11X + 30$
• $(X- 7) \times (X- 8) =$
$X² - 8X - 7X + 56 = X² - 15X + 56$
## Other properties
As we mentioned earlier, there are other rules and properties out there that help us to simplify algebraic expressions. In this section we will briefly look at two of them: the associative property and the commutative property.
Do you know what the answer is?
### The associative property
The associative property allows us to group several terms of an equation together without changing the final results, by moving the parentheses. However, we can only use this property to solve addition or multiplication exercises.
For example:
• $(10 + 2) + 8 =$
$10 + (2 + 8) = 10 + 2 + 8 = 20$
• $2 \times (3 \times 6) =$
$(2 \times 3) \times 6 = 2 \times 3 \times 6 = 36$
### The commutative property
The commutative property allows us to change the order of the terms in an equation without changing the outcome of the equation. Like the associative property, the commutative property can only be used for addition and multiplication.
Let us look at some examples:
• $2 + 6 = 6 + 2 = 8$
• $3 \times 4 = 4 \times 3 = 12$
## Example exercises for seventh graders
### Exercise 1
Using the distributive property, solve the following:
• $294:3=$
• $105\times 4=$
• $505:5=$
• $207\times 5=$
• $168:8=$
Solutions:
• $294:3=(300-6):3=300:3-6:3=100-2=98$
• $105\times 4=(100+5)\times 4=100\times 4+5\times 4=400+20=420$
• $505:5=(500+5):5=500:5+5:5=100+1=101$
• $207\times 5=(200+7)\times 5=200\times 5+7\times 5=1000+35=1035$
• $168:8=(160+8):8=160:8+8:8=20+1=21$
### Exercise 2
Three hundred fifty-one students from a high school were divided into nine equal groups.
How many students are in each group?
Solve the problem using the distributive property.
Solution:
We begin by expressing the problem numerically:
$351:9=(360-9):9=360:9-9:9=40-1=39$
There are 39 students in each group.
Do you think you will be able to solve it?
### Exercise 3
Dani bought 15 boxes. In each box there were 9 pieces of candy.
How many pieces of candy did Dani buy in total?
Use the distributive property to solve the problem.
Solution:
We begin by expressing the problem numerically:
$15\times9=(10+5)\times9=10\times9+5\times9=90+45=135$
Dani bought 135 pieces of candy.
### Exercise 4
Isabel has packed 246 notebooks into 6 equal packages.
How many notebooks has Isabel put in each package?
Use the distributive property to solve the problem.
Solution:
We begin by expressing the problem numerically:
$246:6=(240+6):6=240:6+6:6=40+1=41$
Isabel has put 41 notebooks in each packet.
### Exercise 5
A mother had 894 euros. She divided the money equally among her three children.
How much money did each child receive?
Use the distributive property to solve the problem.
Solution:
We begin by expressing the problem numerically:
$894:3=(900-6):3=900:3-6:3=300-2=298$
### Exercise 6
Solve the following:
$93:3=\text{?}$
Solution:
We simplify the number 93 into 4 smaller numers to make it easier for us to divide it by 3.
For example $30:3=10$
$93:3=\text{?}$
$(30+30+30+3):3=$
$+30:3=10$
$+30:3=10$
$+30:3=10$
$+3:3=1$
Then we will add the results and we will get:
$93:3=31$
$31$
Do you know what the answer is?
### Exercise 7
Solve the following:
$=90:5$
Solution:
We simplify the number 90 into 3 smaller numbers:
(50,20,20)
$90=5-+20+20$
Then we divide each of them by 5 and add the three results.
$+50:5=10$
$+20:5=4$
$+20:5=4$
Which gives us:
$90:5=10+4+4=18$
$18$
### Exercise 8
Solve the following:
$=72:18$
Solution:
We simplify the number 72 into two smaller numbers $\left(36+36\right)$ and then divide each of them by 18.
$72:18=$
$(36+36):18=$
$+36:18=2$
$+36:18=2$
$72:18=4$
$4$
$4$
### Exercise 9
Solve the following:
$\left(40+70+35−7\right)×9=$
Solution:
$\left(40+70+35−7\right)×9=$
$40\times9+70\times9+35\times9-7\times9=$
$=360+630+(30+5)9-63$
$=360+630+270+45-63=1242$
$1242$
### Exercise 10
Solve the following:
$\left(35+4\right)×\left(10+5\right)=$
Solution:
$(35+4)(10+5)=35\times10+35\times5+4\times10+4\times5$
$=350+175+40+20=585$
$585$
Do you think you will be able to solve it?
### Exercise 11
Solve the following:
$\left(7x+3\right)×\left(10+4\right)=$
Solution:
$(7X+3)(10+4)=7X\times10+7X\times4+3\times10+3\times4$
$=70X+28X+30+12=98X+42=238$,$-42$
$98X=196$,$:98$
$X=\frac{196}{98}=2$
$2$
## More practice
• $187\times (8-5)=$
• ${2\over3}\times (12+0-5)=$
• $5\times (2{1\over2}+1{1\over6}+{3\over4})=$
• $(10+5+18)\times 4=$
• $(5.5-0.8)\times 5=$
• $340:(12-7)=$
• $(29-4):5=$
• ${5\over1}:(6+1-5)=$
• $18:(5+7+4)=$
• $({1\over7}-{1\over3}):4=$
• $97\times 12=$
• $3\times 36=$
• $120:97=$
• $8:{1\over2}=$
• $151\times 23=$
## FAQs about the distributive property
What is the distributive property?
The distributive property is a method used to simplify expressions into smaller, more manageable pieces.
How is the distributive property used?
In an equation, we use the distributive property to break down a large number into two or more smaller numbers (using addition and subtraction), and then by distributing the multiplication.
Example
• $20 \times 8\times 7=20+8\times 7=20\times 7+8\times 7=140+56=196$
Can we use the distributive property in division?
Of course we can! In an expression with division, we break down the numerator into smaller numbers (using addition and subtraction), and then the division is distributed.
Example
• $150:6=120+30:6=120:6+30:6=20+5=25$
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Do you know what the answer is?
## examples with solutions for distributive property for seventh grade
### Exercise #1
Solve the exercise:
84:4=
### Step-by-Step Solution
There are several ways to solve the exercise,
We will present two of them.
In both ways, in the first step we divide the number 84 into 80 and 4.
$\frac{4}{4}=1$
And thus we are left with only the 80.
From the first method, we will decompose 80 into$10\times8$
We know that:$\frac{8}{4}=2$
And therefore, we reduce the exercise $\frac{10}{4}\times8$
In fact, we will be left with$2\times10$
which is equal to 20
In the second method, we decompose 80 into$40+40$
We know that: $\frac{40}{4}=10$
And therefore: $\frac{40+40}{4}=\frac{80}{4}=20=10+10$
which is also equal to 20
Now, let's remember the 1 from the first step and add them:
$20+1=21$
And thus we manage to decompose that:$\frac{84}{4}=21$
21
### Exercise #2
$94+72=$
### Step-by-Step Solution
To facilitate the resolution process, we break down 94 and 72 into smaller numbers. Preferably round numbers
We obtain:
$90+4+70+2=$
Using the associative property, we arrange the exercise in a more comfortable way:
$90+70+4+2=$
We solve the exercise in the following way, first the round numbers and then the small numbers.
$90+70=160$
$4+2=6$
Now we obtain the exercise:
$160+6=166$
166
### Exercise #3
$140-70=$
### Step-by-Step Solution
To facilitate the resolution process, we use the distributive property for 140:
$100+40-70=$
Now we arrange the exercise using the substitution property in a more convenient way:
$100-70+40=$
We solve the exercise from left to right:
$100-70=30$
$30+40=70$
70
### Exercise #4
$63-36=$
### Step-by-Step Solution
To solve the problem, first we will use the distributive property on the two numbers:
(60+3)-(30+6)
Now, we will use the substitution property to arrange the exercise in the way that is most convenient for us to solve:
60-30+3-6
It is important to pay attention that when we open the second parentheses, the minus sign moved to the two numbers inside.
30-3 =
27
27
### Exercise #5
$133+30=$
### Step-by-Step Solution
To solve the question, we first use the distributive property for 133:
$(100+33)+30=$
Now we use the distributive property for 33:
$100+30+3+30=$
We arrange the exercise in a more comfortable way:
$100+30+30+3=$
We solve the middle exercise:
$30+30=60$
Now we obtain the exercise:
$100+60+3=163$
|
# Area and perimeter of parallelograms
### Area and perimeter of parallelograms
#### Lessons
In this lesson, we will learn:
• The basic properties of the shape of the parallelogram
• How to understand and calculate the perimeter of a parallelogram
• How to understand and calculate the area of a parallelogram
Notes:
• A parallelogram is a 2D shape with 4 straight sides
• Each pair of sides (across from each other) are parallel and the same length
• The internal angles are not right angles (90°); otherwise it would be a rectangle
• Opposite angles are the same size
• A special parallelogram with all sides of equal length is called a rhombus
• The perimeter is the exact distance around the shape.
• Perimeter is a 1D (one-dimensional) quantity
• It uses units such as meters (m, cm, mm, km), miles (mi), yards (yd), or inches and feet (in and ft)
• For parallelograms, the formula is written as:
• Pparallelogram = $( 2 \, \times \,a) \, + \, (2\, \times \, b)$
• Where $a$ is the length of one of the sides, and $b$ is the length of the other side
• The area is the space that is covered by the shape.
• Area is a 2D (two-dimensional) quantity
• It uses squared units such as square meters (m2, cm2, mm2, km2), square miles (mi2), square yards (yd2), or square inches and feet (in2 and ft2)
• For parallelograms, the formula is written as:
• Aparallelogram = $b \, \times \, h$
• Where $b$ is base, and $h$ is height
• A parallelogram’s height is not the same as the slant (side) length.
• Introduction
Introduction to Area and Perimeter of Parallelograms:
a)
What are the basic properties of a parallelogram?
b)
How do you calculate the perimeter of a parallelogram?
c)
How do you calculate the area of a parallelogram?
• 1.
Perimeter of Parallelograms
Calculate the perimeter of each parallelogram.
a)
b)
c)
• 2.
Area of Parallelograms
Calculate the area of each parallelogram.
a)
b)
c)
• 3.
Perimeter and Area of Compound Parallelograms
Calculate the perimeter and area of each compound shape. Break down the complex shapes into components (parallelograms and other polygons) first!
a)
b)
c)
• 4.
Parallelograms Word Problem - 1
Sophia wants to make a special picture frame using parallelograms. It will use four of the same parallelograms with the measurements: 8cm along the base and 3cm in height.
a)
What is the area of the picture frame?
b)
If the slanted part of the parallelogram is 5cm, what is the perimeter of the picture frame? Do not count the edges of the parallelograms that are glued together.
|
# Problem
Given K variables, $a_1, a_2, a_3 \dots a_K$ and a value $N$, how many ways can we write $a_1 + a_2 + a_3 \dots a_K = N$, where $a_1, a_2, a_3 \dots a_K$ are non-negative integers?
For example, if $K = 3$ and $N=2$, there are 6 solutions.
a b c
-----
2 0 0
0 2 0
0 0 2
1 1 0
1 0 1
0 1 1
# Solution
The solution is very simple. In order to understand it, let us visualize the problem from a different angle.
Another way to represent the solutions visually is to use “Stars and Bars”. In the visualization below, we imagine we have $K$ partitions and we need to distribute $N$ stars in them. Like before, if we have $K = 3$ and $N=2$, then:
a b c
--------
**| |
|**|
| |**
* |* |
* | |*
|* |*
This visualization gives us the foundation of the solution. If we have $K$ partitions and $N$ stars to distribute in them, we can imagine that we have $N+K-1$ blanks. For example, when $K=3$ and $N=2$, we have $N+K-1=2+3-1=4$ blanks.
- - - - 4 Blanks
So, if we have $N+K-1$ blanks, how many of them will be stars? $N$ of the blanks need to be stars. And how many of the blanks need to be bars? Remember, if we have $K$ partition, then we just need $K-1$ bars to create $K$ partitions. So there will be $K-1$ bars.
So, the problem boils down to which blanks get to become stars and which blanks become bars. If we have $N+K-1$ blanks, how many ways are there to select $N$ of them to become stars? $\binom{N+K-1}{N}$ or $\binom{N+K-1}{K-1}$
For our example this equals to $\binom{4}{2} = 6$.
## Stars and Bars Theorem
Theorem: For any pair of positive integers $N$ and $K$, the number of $K$-tuples of non-negative integers whose sum is $N$ is equal to the binomial coefficient $\binom{N+K-1}{N}$ or $\binom{N+K-1}{K-1}$.
Or we can rephrase “Stars and Bars Theorem” as “Ball and Urn Theorem”:
Theorem: If we have K distinguishable containers and N indistinguishable balls, then we can distribute them in $\binom{N+K-1}{N}$ ways.
The “Stars and Bars” theorem is also known as “Ball and Urn” theorem. Don’t get confused. They are the same thing.
# Extensions
## Positive Number of Stars in Each Partition
What if every partition needs to have at least one star? Simple. Place a star in each partition and subtract those placed stars from total stars. Then apply the Stars and Bars theorem normally.
For example, find the number of solution to $a + b + c = 10$, where $a,b,c > 0$.
Let $a = 1 + a’$, $b = 1 + b’$ and $c = 1 + c’$, where $a’,b’,c’ \geq 0$
$a+b+c = 10$
$(1+a’)+(1+b’)+(1+c’) = 10$
$a’+b’+c’ = 10 – 3$
$a’+b’+c’ = 7$, where $a’,b’,c’ \geq 0$
This can now be solved using normal “Stars and Barsh” theorem, which equals to $\binom{9}{2}$.
Theorem: For any pair of positive integers $N$ and $K$, the number of $K$-tuples of positive integers whose sum is $N$ is equal to the binomial coefficient $\binom{N-1}{K-1}$.
## Lower Bound with Each Partition
The extension above can be further generalized. For each partition, if we are given a lower bound $x$, then we simply place $x$ stars in that partition and subtract with from total stars $K$. Finally, we apply “Stars and Bars” theorem normally.
So, for $a_1 + a_2 + a_3 \dots a_K = N$, where $a_1 \geq L_1, a_2 \geq L_2, a_3 \geq L_3 \dots a_K \geq L_n$ is same as $a’_1 + a’_2 + a’_3 \dots a’_K = N – L_1 – L_2 – L_3 \dots L_K$, where $a’_1, a’_2, a’_3 \dots a’_K \geq 0$.
## Equation with Inequality
What if we are asked to find:
$$a_1 + a_2 + a_3 \dots a_K \leq N$$
This is a very interesting extension. There are two ways to think about the solution to this problem.
### Garbage Partition
Imagine that you have an extra partition which acts like a garbage bin. Then if you try to apply normal “Stars and Bars Theorem”, what happens?
So the equation now becomes: $a_1 + a_2 + a_3 \dots a_K + a_{K+1} = N$. What is the solution to this problem? Ans: $\binom{N+K}{K}$.
Now, simply imagine, anything that gets assigned to the extra garbage partition gets thrown out. So the sum of remaining partitions will be: $a_1 + a_2 + a_3 \dots a_K \leq N$
Theorem: For any pair of positive integers $N$ and $K$, the number of $K$-tuples of non-negative integers whose sum is less than or equal to $N$ is $\binom{N+K}{K}$.
### Vertical Sum of Pascal Triangle
Another way to solve the problem is to simply loop over the solution from 0 to $N$.
$$a_1 + a_2 + a_3 \dots a_K = 0 \\ a_1 + a_2 + a_3 \dots a_K = 1 \\ a_1 + a_2 + a_3 \dots a_K = 2 \\ a_1 + a_2 + a_3 \dots a_K = 3 \\ \dots \\ a_1 + a_2 + a_3 \dots a_K = N \\$$
If we find the sum of all the results above, then we will get result for $a_1 + a_2 + a_3 \dots a_K \leq N$. So what is the sum?
$$\binom{N+K-1}{K-1} + \binom{N+K-2}{K-1} + \binom{N+K-3}{K-1} \dots + \binom{K-1}{K-1}$$
Do you know what this is? This is the vertical sum of a column in pascal triangle. I will discuss the properties of Pascal Triangle someday else. For today, understanding the “Garbage Partition” logic will be enough.
So vertical sum of pascal triangle, or $\sum_{i=k}^{n}\binom{i}{k}$, is equal to $\binom{n+1}{k+1}$.
## Upper Bound with Each Partition
Now, this is tricky. This time we are given the upper limit for each partition. So the number of stars we place must be smaller or equal to the upper limit for that partition.
One possible solution is Dynamic Programming. We iterate over each partition and keep track of remaining stars in our hand. When at each partition, we run a loop from 0 to upper limit and try placing the different amount of stars in each container.
It will be something similar to this ( not exactly this ):
int dp ( int partition, int stars ) {
int res = 0;
for ( int i = 0; i < upperLimit[pos]; i++ ) {
res += dp ( partition + 1, stars - i );
}
}
Now, the above DP solution will work only if we are lucky to get constraints that fit the complexity of the DP. When we are unlucky, we will need to be more creative.
|
# What is Associative Property, Definition, Types, and Examples
0
47
The associative property is a numeric law of mathematics. It states that adding and multiplying three or more numbers produce the same results irrespective of how they are grouped.
You might be thinking about how to group the numbers, right?
Grouping of numbers is done with parenthesis (brackets). Let’s take an example.
Assume a + b + c are three different numbers. It is a simple expression.
a + (b + c) This is the same expression, but the only difference is that numbers are grouped with brackets.
You must be clear on how to group numbers. Now you can group numbers for different expressions easily.
In this article, we will learn about different terms, like the definition of associative properties, associative law, associative property of addition, associative property of multiplication, and some examples.
## Definition
It is defined as the addition and multiplication of more than two numbers that result in the same no matter how they are grouped.
Example
If you want to add three numbers, like 4, 5, and 6 altogether. You have to group these numbers in the addition procedure; it will be something like this
4+ (5 + 6) = (4 + 5) + 6
4+11 = 9+6
15=15
The same rule applies to multiplication;
If you want to multiply three numbers, like 4, 5, and 6 altogether. You have to group these numbers in a multiplication procedure; it will be something like this
4 x (5 x 6) = (4 x 5) x 6
4×30 = 20×6
120 =120
## Associative Law
When the exact numbers are grouped differently for addition and multiplication, they give the same resuls.
Associative law applies to addition and multiplication only.
A general expression for associative properties
A general expression for associative properties is given right below;
For the associative property of addition:
a + (b + c) = (a + b) + c
For the associative property of multiplication:
a x (b x c) = (a x b) x c
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According to the associative property of addition, when three numbers are added, the result remains the same regardless of how they are grouped.
Suppose we add three numbers: a, b, and c.
### Associative Property of Addition Formula
(A + B) + C = A + (B + C)
To understand this, we take an example.
Example
Suppose we want to add 2, 4, and 5.
First of all, group the numbers according to the general formula
(A + B) + C = A + (B + C)
(2+ 4) + 5 = 2 + (4 + 5)
6+5 = 2+9
11=11
L.H.S = R.H.S
You can see that the result in both cases is the same. However, the numbers are grouped in different ways.
Let’s take another example
Suppose we want to add 5, 6, and 8.
First of all, group the numbers according to the general formula
(A + B) + C = A + (B + C)
(5+ 6) + 8 = 5 + (6 + 8)
11+8 = 5+14
19=19
L.H.S = R.H.S
Thus, the result in both cases is the same.
## Associative Property of Multiplication
According to the associative property of multiplication,
When three numbers are multiplied, the result remains the same regardless of how they are grouped.
Suppose we have to multiply three numbers a, b, and c.
### Associative Property of multiplication Formula:
(A x B) x C = A x (B x C)
To understand this, we take an example.
Example
Suppose we want to multiply 2, 4, and 5.
First of all, group the numbers according to the general formula
(A x B) x C = A x (B x C)
(2x 4) x 5 = 2 x (4 x 5)
8×5 = 2×20
40=40
L.H.S = R.H.S
You can see that the result in both cases is the same. However, the numbers are grouped in different ways.
Let’s take another example
Suppose we want to add 5, 6, and 8.
First of all, group the numbers according to the general formula
(A x B) x C = A x (B x C)
(5x 6) x 8 = 5 x (6 x 8)
30 x 8 = 5 x 48
240=240
L.H.S = R.H.S
Thus, the result in both cases is the same.
## FAQ’s
### Which two operations satisfy the condition of the associative properties?
The two operations which fulfill the condition of the associative property are;
Multiplication
It means that the associative property is only applicable to addition and multiplication.
### Define the associative property in math with an example.
The associative property is the addition and multiplication of three or more numbers that result in the same irrespective of how brackets group them.
Example
If you want to add three numbers, like 2, 3, and 6, together, you have to group these numbers in the addition procedure; it will be something like this
2+ (3 + 6) = (2 + 3) + 6
2+9 = 5+6
11=11
The same rule applies to multiplication.
### Does the associative property apply to fractional numbers?
Yes, the associative property applies to fractional numbers. When we add more than two fractions in any order, the sum of the fractions remains the same.
### How can you differentiate Associative Property and Commutative Property?
Associative and commutative properties are applicable only for addition and multiplication operations.
The associative property states that adding and multiplying three or more numbers produces the same results irrespective of how they are grouped by brackets.
A + (b + c) = (a + b) + c
Associative property of multiplication
A x (b x c) = (A x b) x c
While Commutative property states;
The addition and multiplication of numbers result the same regardless of how the numbers are ordered.
a + b = b + a
a x b = b x a
### Why does associative law not apply to subtraction and division?
Associative law does not apply to subtraction and division. Let’s take an example to explain this;
In the case of subtraction
10 – (5 – 2) = (10 – 5) – 2
10- 3 = 5 – 2
7 ≠ 3
L.H.S ≠ R.H.S
Thus, the associative property does not apply to subtraction.
In the case of division:
(18 ÷ 3) ÷ 2 = 18 ÷ (4 ÷ 2)
6 ÷ 2 = 18 ÷ 2
3 ≠ 9
L.H.S ≠ R.H.S
Thus, the associative property does not apply to the division.
I hope all your confusion related to associative properties has been cleared. You can contact us at scientific Sarkar.com if you need help with any problem.
Kishwar is a former content writer at scientific Sarkar. She believes that research makes sense of it and put it in simple language. Her specialties include creative writing, Seo content, blog post, and product description. She helps businesses to increase their google visibility with optimized and engaging content.
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# Math 189 Exam 1 Solutions
40 %
60 %
Information about Math 189 Exam 1 Solutions
Education
Published on February 19, 2014
Author: tylerisaacmurphy
Source: slideshare.net
## Description
Jason Smith Exam 1 Solutions for Math 189, Boise State University, Spring 2014
1 True or False. Justify your answer. 1.1 A. The Contrapositive of A → B is ¬B → ¬A. True. A B T T T F F T F F 1.2 ¬A F F T T ¬B F T F T A→B T F T T ¬B → ¬A. T F T T B. If you want to prove a statement is true, it is enough to find 867 examples where it is true. False The case may not be true for the 868th example.You must show it is true for all cases. However, to disprove it, you only need to show one counter example. 1.3 C. If P ∧ Q is true, then P ∨ Q is true. True P ∧ Q is true means that P is true and Q is true. In order for P ∨ Q to be true, we only need only one of P, Q to be true. Since we have both, then P ∨ Q is true. 1.4 D. P → Q and Q → P are logically equivalent. False P Q P → Q Q → P. T T T T T F F T F T T F F F T T Consider the bolded cases that have different truth values. 1
1.5 E. If p → q is false, then the truth value of (¬p ∨ ¬q) → (p ↔ q) is also false. True. Realize that the only case where p → q is false is when p is true and q is false. So ¬p is false and ¬q is true. So (¬p ∨ ¬q) is true since ¬q is true. Also, (p ↔ q) means (p → q) ∧ (q → p). However, since p → q is false, then (p ↔ q) is false. Therefore, since the first part of the implication(¬p∨¬q) is true and the second (p ↔ q) is false, we have that (¬p ∨ ¬q) → (p ↔ q) is false when p → q is false. 2 2.1 Prove that if a and b are odd integers, then a ∗ b is an odd integer. Proof. Assume that a, b are odd integers. Then a = 2k + 1 for some integer k. Also, b = 2m + 1 for some integer m. We have that a ∗ b = (2k + 1)(2m + 1). = 4km + 2k + 2m + 1. = 2(2km + m + k) + 1 . Since k, m are integers, so is 2km + k + m. So a ∗ b is odd. 3 3.1 For any mathematical statement, say C, fn (C) denotes the mathematical statement: ¬¬ . . . ¬C, where there are n¬ symbols in front of C. Prove that if A is a True mathematical statement and B is a False mathematical statement, then ¬(f3 (A) ∨ f2 (B)) is a True mathematical statement. 2
Proof. Assume that A is a True mathematical statement and B is a False mathematical statement. We have that f3 (A) = ¬¬¬A = ¬(¬¬)A = ¬A since every pair of ¬ symbols cancels out by the Double Negative Law. We also have f2 (B) = ¬¬B = (¬¬)B = B. So we have ¬(¬A ∨ B), which is (¬¬A ∧ ¬B). So we have A ∧ ¬B. Since A is true and B is false (that is, ¬B is true), we have two true statements. The joining of two true statements with ”and” is true, so ¬(f3 (A) ∨ f2 (B)) is a True mathematical statement 4 4.1 4.1.1 p T T T T F F F F Find the Disjunctive Normal Form of ((p → q) ∧ (q → r)) → (p → r) Method: Truth Table q T T F F T T F F r T F T F T F T F ¬p F F F F T T T T ¬q F F T T F F T T ¬r F T F T F T F T p→q T T F F T T T T q→r T F T T T F T T p→r T F T F T T T T ((p → q) ∧ (q → r)) T F F F T F T T ((p → q) ∧ (q → r)) → (p → r) T T T T T T T T So we have (p ∧ q ∧ r) ∨ (p ∧ q ∧ ¬r) ∨ (p ∧ ¬q ∧ r) ∨ (p ∧ ¬q ∧ ¬r) ∨ (¬p ∧ q ∧ r) ∨ (¬p ∧ q¬r) ∨ (¬p ∧ ¬q ∧ r) ∨ (¬p ∧ ¬q ∧ ¬r). Note that this the original statement is true no matter what the truth values of p, q, r. So Disjunctive Normal Form addresses each of the possible cases represented in the truth table. 3
5 5.1 Let a, b, c be real numbers. Prove that if a + b ≥ c, then a ≥ b ≥ c2. c 2 or Note: This can be proven directly but then you must address each of the following cases: c c (i). a = 2 , b = 2 . c c (ii) a = 2 , b > 2 . c c (iii) a > 2 , b = 2 (similar to ii) c c (iiii) a > 2 , b > 2 . This is why contradiction proofs are easier to use in situations where you have OR statements to deal with. In this situation, a contradiction gives us an AND statement, which is a single case. When writing proofs it is often helpful to ask yourself ”When can this be false?” Then show that that case cannot happen. So, c c Proof. Suppose for contradiction that a + b = c and a < 2 and b < 2 . c c Then a + b < 2 + 2 . So a + b < c, which contradicts our original assumption that a + b = c. 4
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## Application of Integrals Class 12 Important Questions with Solutions Previous Year Questions
Question 1.
Using integration, find the area of ∆ ABC, the coordinates of whose vertices are A (2, 5), B(4, 7) and C(6, 2). (Delhi 2019, 2011: All India 2010C)
Given, the vertices of ∆ABC are A(2, 5), B(4, 7) and C(6, 2).
By plotting these points on the graph, we find the required region.
Equation of the line AB is given by
Now, required area = (Area under line segment AB) + (Area under line segment BC) – (Area under line segment AC)
Question 2.
Using integration, find the area of triangle whose vertices are (2, 3), (3, 5) and (4, 4). (Delhi 2019)
32 sq units.
Question 3.
Find the area of the region lying above X-axis and included between the circle x2 + y2 = 8x and inside the parabola y2 = 4x (Delhi 2019)
The equation of circle is
x2 + y2 = 8x …… (i)
and the equation of parabola is
y2 = 4x …… (ii)
Eq. (i) can be written as
(x2 – 8x) + y2 = 0
⇒ (x2 – 8x + 16) + y2 = 16
⇒ (x – 4)2 + y2 = (4)2 …… (iii)
which is a circle with centre C(4, 0) and radius = 4.
From Eqs. (i) and (ii), we get
x2 + 4x = 8x
⇒ x2 – 4x = 0
⇒ x(x – 4) = 0
⇒ x = 0, 4
Now, from Eq. (ii), we get
y = 0, 4
∴ Points of intersection of circle (i) and parabola (ii), above the A-axis, are 0(0, 0) and P(4, 4).
Now, required area = area of region OPQCO = (area of region OCPQ + (area of region PCQP)
Question 4.
Using integration, prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts. (Delhi 2019, 2009; All India 2015)
First, we draw a square formed by the lines x = 0, x = 4, y = 4, and y = 0 and after that, we draw given parabolas which intersect each other on the square such that the whole region divided into three parts. Now, we find separately area of each part and show that area of each part is equal.
Let OABC be the square whose sides are represented by following equations
Equation of OA is y = 0
Equation of AB is x = 4
Equation of BC is y = 4
Equation of CO is x = 0
On solving equations y2 = 4x and x2 = 4y, we get A(0, 0) and B(4, 4) as their points of intersection. Now, area bounded by these curves is given by.
Hence, area bounded by curves y2 = 4x and x2 = 4y is \frac{16}{3} sq units ……. (i)
Now, area bounded by curve x2 = 4y and the lines x = 0, x = 4 and X-axis
Similarly, the area bounded by curve y2 = 4x, the lines y = 0, y = 4 and Y-axis
From Eqs. (i), (ii) and (iii), it is clear that area bounded by the parabolas y2 = 4x and x2 = 4y divides the area of square into three equal parts.
Hence proved.
Question 5.
Using method of integration, find the area of the triangle whose vertices are (1, 0), (2, 2) and (3, 1). (All India 2019)
32 sq units.
Question 6.
Using integration, find the area of the region enclosed between the two circles
x2 + y2 = 4 and (x – 2)2 + y2 = 4. (All India 2019, 2010C: Delhi 2013, 2008)
First, find the intersecting points of two circles and then draw a rough sketch of these two circles. The common shaded region is symmetrical about X-axis. So, we find area of one part only, i.e. upper part of X-axis.
After that required area is twice of that area.
Given circles are
x2 + y2 = 4 ……… (i)
and (x – 2)2 + y2 = 4 …… (ii)
Eq. (i) is a circle with centre origin and radius 2, Eq. (ii) is a circle with centre C (2, 0) and radius 2.
On solving Eqs. (i) and (ii), we get
(x – 2)2 + y2 = x2 + y2
⇒ x2 – 4x + 4 + y2 = x2 + y2
⇒ x = 1
On putting x = 1 in Eq. (i), we get
y = ± √3
Thus, the points of intersection of the given circles are A (1, √3) and A'(1, – √3) as shown in the figure given below:
Clearly, required area = Area of the enclosed
region OACA’O between circles
= 2 [Area of the region ODCAO]
= 2 [Area of the region ODAO + Area of the region DCAD]
Question 7.
Using integration, find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4. (All India 2019, 2011C; Delhi 2011)
Given, equation of sides are
y = 2x + 1, y = 3x + 1 and x = 4
On drawing the graph of these equations, we get the following triangular region
By solving these equations we get the vertices of triangle as A(0, 1), B(4, 13) and C(4, 9).
∴ Required area = Area (OABDO) – area (OACDO)
Question 8.
Find the area of the region in the first quadrant enclosed by the X-axis, the line y = x and the circle x2 + y2 = 32. (CBSE 2018; Delhi 2014)
Given the circle x2 + y2 = 32 ….. (i)
having centre (0, 0) and radius 4√2 and the line y = x ….. (ii)
Let us find the point of intersection of Eqs. (i) and (ii).
On substituting y = x in Eq. (i), we get
x2 + x2 = 32
⇒ 2x2 = 32
⇒ x2 = 16
⇒ x = ± 4
Thus, the points of intersection are (4, 4) and (-4, -4). [∵ y = x]
Clearly, the required area
= Area of shaded region OABO
Question 9.
Using integration, find the area of the region : {(x, y): 0 ≤ 2y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x ≤ 3}. (CBSE 2018 C)
Given region is
{(x, y): 0 ≤ 2y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x ≤ 3}
which can be represent graphically as shown below.
Now, let us find the point of intersection of y = x and y = x22.
For this consider,
x = x22
⇒ x2 – 2x = 0
⇒ x(x – 2) = 0
⇒ x = 0 or 2
Clearly, when x = 0, then y = 0 and when x = 2, then y = 2
Thus, the points of intersection are (0, 0) and (2, 2).
Question 10.
Using integration, find the area of region bounded by the triangle whose vertices are (- 2, 1), (0, 4) and (2, 3). (Delhi 2017)
4 sq units
Question 11.
Find the area bounded by the circle x2 + y2 = 16 and the line √3y = x in the first quadrant, using integration. (Delhi 2017)
Given equations of circle is x2 + y2 = 16 and x = √3y.
⇒ y = 13x represents a line through the origin.
The line y = 13x intersects the circle
∴ x2 + x23 = 16
⇒ 3x2+x23 = 16
⇒ 4x2 = 48 ⇒ x2 = 12 ⇒ x = ±2√3
when x = 2√3, then y = 233 = 2
= (Area under the line y = 13x from x = 0 to 2√3) + (Area under the circle from x = 2√3 to x = 4)
Question 12.
Using the method of integration, find the area of the ∆ABC, coordinates of whose vertices are A (4, 1), B(6, 6) and C (8, 4). (All India 2017)
7 sq units
Question 13.
Find the area enclosed between the parabola 4y = 3x2 and the straight line 3x – 2y + 12 = 0. (All India 2017)
Given parabola is 4y = 3x2 …… (i)
represent an upward parabola with vertex (0, 0) and equation of line is 2y = 3x + 12 …… (ii)
From Eqs. (j) and (ii), we get
2(3x + 12) = 3x2
⇒ 3x2 – 6x – 24 = 0
⇒ x2 – 2x – 8 = 0
⇒ (x – 4) (x + 2) = 0
⇒ x = 4, – 2
when x = 4, then y = 3×4+122 = 12 [from Eq. (ii)]
when x = – 2, then y = 3×(2)+122 = 3
Thus, intersection points are (- 2, 3) and (4, 12).
Question 14.
Using integration, find the area of the region
{(x, y): x2 + y2 ≤ 2ax, y2 ≥ ax;x, y ≥ 0}. (Delhi 2016)
Given region is
{(x, y): x2 + y2 ≤ 2ax, y2 ≥ ox; x, y ≥ 0}
Now, we have x2 + y2 ≤ 2ax
⇒ x2 + y2 – 2ax ≤ 0
⇒ x2 – 2ax + a2 + y2 < a2
[adding a2 both sides of inequality]
⇒ (x – a)2 + y2 < a2
which is the interior of the circle having centre (a, 0) and radius a.
Also, we have y2
which is the exterior of the parabola having vertex (0, 0) and axis is X-axis.
Now, let us find the intersection point of circle
(x – a)2 + y2 = a2 and parabola y2 = ax.
On substituting y2 = ax in the given circle, we get
(x – a)2 + ax = a2
⇒ x2 + a2 – 2ax + ax = a2
⇒ x2 – ax = 0
⇒ x(x – a) = 0
⇒ x = 0, a
When x = 0, then
y2 = 0 ⇒ y = 0
When x = a, then
y2 = a2 ⇒ y = ± a
So, the points of intersection are 0(0, 0), A (a, a) and B(a, -a). (1)
Now, draw the graph of given curve as shown below:
Clearly, the required area of region will lie in first
quadrant as x, y ≥ 0.
∴ Required area = Area of shaded region
Question 15.
Using integration, find the area of the triangular region whose vertices are (2, – 2), (4, 3) and (1, 2). (All India 2016)
132 sq units
Question 16.
Using integration, find the area of the region bounded by the curves y = 4x2−−−−−√, x2 + y2 – 4x = 0 and the x-axis. (Foreign 2016)
Given equations of curves are
y = 4x2−−−−−√ ……. (i)
and x2 + y2 – 4x = 0
Consider the curve
y = 4x2−−−−−√ = y2 = 4 – x2
⇒ x2 + y2 = 4 which represents a circle with centre (0, 0) and radius 2 units.
Now, consider the curve x2 + y2 – 4x = 0
⇒ (x – 2)2 + y2 = 4, which also represents a circle with centre (2, 0) and radius 2 units.
Now, let us sketch the graph of given curves and find their points of intersection.
On substituting the value of y from Eq. (i) in Eq. (ii), we get
x2 + (4 – x2) – 4x = 0
⇒ 4 – 4x = 0
⇒ x = 1
On substituting x = 1 in Eq. (1). we get y = √3
Thus, the point of intersection is (1, √3).
Clearly, required area = Area of shaded region OABO
Question 17.
Using integration, find the area of the region in the first quadrant enclosed by the Y-axis, the line y = x and the circle x2 + y2 = 32. (Delhi 2015C)
Given, equation of circle is x2 + y2 = 32
and equation of line is
y = x
Consider x2 + y2 = 32
⇒ x2 + y2 = (4√2)2
So, given circle has centre (0, 0) and radius 4√2 units.
Now, let us sketch the graph of given curves and find their points of intersection
On substituting y = x in Eq. (i), we get
2x2 = 32 ⇒ x2 = 16 ⇒ x = ± 4
Thus, the points of intersection are (4, 4) and (-4, -4).
So, given line and the circle intersect in the first quadrant at point A(4, 4) and the circle cut the y-axis at point B (0, 4√2).
Let us draw AM perpendicular to y-axis.
Clearly, required area
= Area of shaded region OABO
Question 18.
Using integration, find the area of the triangle formed by positive X-axis and tangent and normal to the circle x2 + y2 = 4 at (1, √3). (Delhi 2015)
First, differentiate the given curve w.r.t. x and determine the value of dydx at (1, √3).
Find the equations of tangent and normal at
point(1, √3) by using formula y – y1 = dydx(x – x1) and y – y1 = – 1dy/dx(x – x1)
Further, plot the above lines on a graph paper and find the area by using integration.
Given equation of circle is
x2 + y2 = 4
On differentiating both sides of Eq. (i) w.r.t. x,
we get
2x + 2ydydx = 0
On putting y = 0 in Eq. (ii), we get
x + 0 = 4 ⇒ x = 4
∴ the tangent line x + √3y = 4 cuts the X-axis at A(4, 0).
∴ Required area = Area of shaded region OAB
Question 19.
Using integration, find the area of the region bounded by the line x – y + 2 = 0, the curve x = √y and Y-axis. (Foreign 2015)
Given curves are
x – y + 2 = 0 ….. (i)
and x = √y ….. (ii)
Consider x = √y ⇒ x2 = y, which represents the parabola whose vertex is (0, 0) and axis is Y-axis.
Now, the point of intersection of Eqs.(i) and (ii)
is given by x = x+2−−−−√
⇒ x2 = x + 2
⇒ x2 – x – 2 = 0
⇒ (x – 2) (x + 1) = 0
⇒ x = – 1, 2
But x = – 1 does not satisfy the Eq. (ii).
∴ x = 2
Now, putting x = 2 in Eq. (ii), we get
2 = √y ⇒ y = 4
Hence, the point of intersection is (2, 4).
But actual equation of given parabola is x = √y, it means a semi-parabola which is on right side of Y – axis.
The graph of given curves are shown below:
Question 20.
Find the area of the region {(x, y): y2 ≤ 4x, 4x2 + 4y2 ≤ 9}, using method of integration. (All India 2015C, 2013, 2008C: Delhi 2008C)
First, find the intersection points of given curves and then draw a rough diagram to represent the required area. If it is symmetrical about X-axis or Y-axis, then we first find area of only one portion , from them and then required area is twice of that area.
Given curves are
y2 = 4x …… (i)
and 4x2 + 4y2 = 9
⇒ x2 + y2 = 94 …… (ii)
Eq. (i) represents a parabola having vertex (0, 0) and axis is X-axis and Eq. (ii) represents a circle having centre (0, 0) and radius 32.
On substituting y2 = 4x in Eq. (ii), we get
x2 + 4x = 94
⇒ 4x2 + 18x – 2x – 9 = 0
⇒ 2x(2x + 9) – 1 (2x + 9) = 0
⇒ (2x + 9) (2x – 1) = 0
⇒ x = 12,92
On putting x = 12 in Eq. (i), we get
y = ±√2
At x = –92, y have imaginary values.
So, intersection points are p(12,2–√) and p(12,2–√).
Now, the shaded region represents the required region as shown below:
∴ Required area = 2[Area of the region ORPO + Area of the region RAPR]
Question 21.
Using integration, find the area of the region in the first quadrant enclosed by the X-axis, the line y = x and the circle x2 + y2 = 18. (All India 2014C)
9π4 sq units
Question 22.
Using integration, find the area of the region bounded by the curves y = |x + 1| + 1, x = – 3, x = 3 and y = 0. (Delhi 2014C)
Given curves are
Eq. (ii) represents the line parallel to Y-axis and passes through the point (- 3, 0).
Eq. (iii) represents the line parallel to Y-axis and passes through the point (3, 0).
Eq. (iv) represent X-axis.
Now, Eqs. (i), (ii) (iii) and (iv) can be represented in graph as shown below:
Clearly, required area
Hence, the required area is 16 sq units.
Question 23.
Using integration, find the area of ∆ PQR, coordinates of whose vertices are P(2, 0), Q(4, 5) and R (6, 3). (All India 2014C)
7 sq units
Question 24.
Using integration, find the area of the region bounded by the triangle whose vertices are (- 1, 2), (1, 5) and (3, 4). (All India 2014)
4 sq units
Question 25.
Find the area of the smaller region bounded by the ellipse x29+y24 = 1 and the line x3+y2 = 1. (Foreign 2014)
Or
Using integration, find the area of the following region. (Delhi 2010)
{(x,y):x29+y241x3+y2}
Given equation of ellipse is
x29+y24 = 1
and equation of line is x3+y2 = 1
For the points of intersection of ellipse and line, put the value of x from Eq. (ii) in Eq. (i), we get
(1y2)2+y24 = 1
⇒ 1 + y24 – y + y24 = 1
⇒ y2 – 2y = 0
⇒ y = 0, 2
When y – 0, then x = 3 and point is A(3, 0).
When y = 2, then x = 0 and point is B{0, 2).
Question 26.
Using integration, find the area of the region bounded by the lines 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0. (All India 2014C; Foreign 2011; Delhi 2009)
Given lines are
2x + y = 4 ……… (i)
3x – 2y = 6 ……… (ii)
and x – 3y = – 5 ……… (iii)
Clearly, the line 2x + y = 4 passes through the points (2, 0) and (0, 4), the line 3x – 2y = 6 passes through the points (2, 0) and (0, – 3) and the line x – 3y = – 5 passes through the points (-5, 0) and (0, 53).
Now, the region bounded by these lines is shown below:
On solving Eqs. (i) and (ii),
we get x = 2 and y = 0
So, lines 2x + y = 4 and 3x – 2y = 6 meet at the point C(2,0).
Again, solving Eqs. (ii) and (iii), we get
x = 4 and y = 3
So, lines 3x – 2y = 6 and x – 3y = – 5 meet at the point B(4, 3).
On solving Eqs. (iii) and (i), we get
x = 1 and y = 2
So, lines 2x + y = 4 and x – 3y = – 5 meet at the point A (1, 2).
Now, required area of ∆ABC
= Area of region ABNMA – (Area of ∆AMC + Area of ∆BCN)
Question 27.
Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y – 2. (Delhi 2014C, 2013, 2010)
Given curves are
x2 = 4y ……… (i)
and x = 4y – 2 …….. (ii)
Eq. (i) represents a parabola with vertex at origin and axis along positive direction of Y-axis. Eq. (ii) represents a straight line which meets the coordinate axes at (- 2, 0) and (0, 12) respectively. To find the points of intersection of the given parabola and the line, we solve Eqs. (i) and (ii), simultaneously.
On substituting x = 4y – 2 in Eq. (i), we get
(4y – 2)2 = 4y
⇒ 16y2 + 4 – 16y = 4y
⇒ 16y2 – 20y + 4 = o
⇒ 4y2 – 5y + 1 = 0
⇒ (4y – 1) (y – 1) = 0
⇒ y = 1, 14
On putting the values of y in Eq. (ii), we get x = 2, – 1
So, the points of intersection of the given parabola and the line are (2, 1) and (- 1, 1/4).
The region whose area is to be found out is shaded in figure.
∴ Required area, A is given by
Question 28.
Using integration, find the area of the region bounded by the curves y = x2 and y = x. (Delhi 2013C)
16 sq units
Question 29.
Find the area of the region {(x, y):y2 ≤ 6ax and x2 + y2 ≤ 16a2}, using method of integration. (All India 2013)
(433a2 + 16a2π3) sq units.
Question 30.
Find the area of the region bounded by the parabola y = x2 and the line y = |x|. (All India 2013)
Or
Find the area of the region given by {(x, y): x2 ≤ y ≤ |x|}. (Delhi 2011C; All India 2009, 2008C)
Given curves x2 = y ……. (i)
and y = |x| …… (ii)
From Eqs. (i) and (ii), we get
x2 = |x|
Case I When x ≤ 0
Then, x2 = – x
⇒ x(x + 1) = 0
∴ x = 0, – 1
On putting the values of x in Eq. (i), we get
y = 0, 1
Case II When x ≥ 0
Then, x2 = x ⇒ x(x – 1) = 0
∴ x = 0, 1
On putting the values of x in Eq. (j), we get
y = 0, 1
So, both curves cut each other at points
A (-1, 1), 0(0, 0) and B(1, 1).
The graphs of given curves is shown below,clearly the shaded region is symmetrical about the Y-axis.
Now, area of region OPBO
Hence, required area = 2 × Area of region OPBO
[∵ region is symmetrical about Y-axis]
= 2 ×16 = 13 sq unit
Question 31.
Using integration, find the area of the circle x2 + y2 = 16, which is exterior to the parabola y2 = 6x. (All India 2012C)
Given, equation of circle is x2 + y2 = 16 ….. (i)
and equation of parabola is y2 = 6x ……. (ii)
Clearly, the given circle has centre (0, 0) and radius 4 units and the given parabola has vertex (0, 0) and axis parallel to X-axis.
Now, let us sketch the graph of given curves and find their points of intersection.
On substituting y2 = 6x in Eq. (i), we get
x2 + 6x -16 = 0
⇒ (x + 8)(x – 2) = 0
⇒ x = – 8 or x = 2
Clearly, from Eq. (ii), when x = – 8, then y2 = – 48, which is not possible. So, x ≠ – 8
∴ x = 2
Now, on substituting x = 2in Eq. (ii), we get
y2 = 12 ⇒ y = ± 2√3
Thus, the points of intersection are (2, – 2√3) and (2, 2√3).
Clearly, required area = Area of shaded region
= Area of circle – Area of region OABCO
= π(4)2 – 2(Area of region OBCO)
Question 32.
Using method of integration, find the area of region bounded by lines 3x – 2y + 1 = 0, 2x + 3y – 21 = 0 and x – 5y + 9 = 0. (Delhi 2012)
132 sq units
Question 33.
Using integration, find the area of the region bounded by the lines 3x – y – 3 = 0, 2x + y – 12 = 0, x – 2y – 1 = 0. (Delhi 2012)
10 sq units
Question 34.
Using integration, find the area of the region bounded by the lines 5x – 2y – 10 = 0, x + y – 9 = 0 and 2x – 5y – 4 = 0. (Delhi 2012)
212 sq units
Question 35.
Find the area of the region
{(x, y): x2 + y2 ≤ 4, x + y ≥ 2}. (All India 2012).
Given region is {(x, y): x2 + y2 ≤ 4, x + y ≥ 2}.
The above region has a circle with equation
x2 + y2 = 4 … (i)
whose centre is (0, 0) and radius is 2, and line with equation
x + y = 2 …(ii)
Point of intersection is calculated as follows
x2 + y2 = 4
⇒ x2 + (2 – x)2 = 4 [from Eq. (ii)]
⇒ x2 + 4 + x2 – 4x = 4
⇒ 2x2 – 4x = 0
⇒ 2x (x – 2) = 0
⇒ x = 0 or 2
When x = 0, then y = 2 – 0 = 2 and when x = 2, then y = 2 – 2 = 0
So, points of intersection are (0, 2) and (2, 0).
On drawing the graph, we get the shaded region as shown below:
Question 36.
Find the area of the region
{(x, y):(x2 + y2) ≤ 1 ≤ x + y}. (All India 2011; Delhi 2010C)
(π412) sq units
Question 37.
Sketch the graph of y = | x + 3 | and evaluate the area under the curve y = |x + 3| above X-axis and between x = – 6 to x = 0. (All India 2011 )
First, we sketch the graph of
y = |x + 3|
So, we have y = x + 3 for x ≥ – 3 and y = – x – 3 for x < – 3
A sketch of y = |x + 3| is shown below:
Here, y = x + 3 is the straight line which cuts X and Y-axes at (- 3, 0) and (0, 3) respectively.
Thus, y = x + 3 for x ≥ – 3 represents the part of line which lies on the right side of x = – 3.
Similarly, y = – x – 3, x < – 3 represents the part of line y = – x – 3, which lies on left side of x = – 3.
Clearly, required area
= Area of region ABPA + Area of region PCOP
Question 38.
Using integration, find the area of the region {(x, y): x2 + y2 ≤ 16, x2 ≤ 6y} (Delhi 2010C)
Given region is {(x, y): x2 + y2 ≤ 16, x2 ≤ 6y}
Above region has a circle x2 + y2 = 16 whose centre is (0, 0) and radius 4 and a parabola whose vertex is (0, 0) and axis along y-axis.
First, let us sketch the region, as shown below:
For finding the points of intersection of two curves, we have
x2 + y2 = 16
and x2 = 6y
On putting x2 = 6y from Eq. (ii) in Eq. (i), we get
y2 + 6y -16 = 0
⇒ y2 + 8y – 2y -16 = 0
⇒ y(y + 8) – 2(y + 8) = 0
⇒ (y – 2) (y + 8) = 0
⇒ y = 2 or – 8
When y = 2, then from Eq. (ii), we get
x = ± √12 = ± 2√3
and when y = – 8, then from Eq. (ii), we get
x2 = – 48 which is not possible.
So, y = – 8 is rejected.
Thus, the two curves meet at points C(2√3, 2) and D(- 2√3, 2).
Now, required area
= Area of shaded region OCBDO
= 2 [ Area of region OACO + Area of region ABCA]
Question 39.
Find the area of circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y. (All India 2010)
26+94 sin-1(223)sq units
Hint:
Question 40.
Using integration, find the area of the following region.
{(x, y):|x – 1| ≤ y ≤ 5x2−−−−−√} (Delhi 2010)
First, write the given curves separately, i.e. y = |x – 1| and y = 5x2−−−−−√.
Then, sketch all the above defined functions and find the required area.
Given region is {(x, y): |x – 1| ≤ y ≤ 5x2−−−−−√}
Above region has two equations
y = |x – 1| and y = 5x2−−−−−√
Also, other curve is y = 5x2−−−−−√
On squaring both sides, we get
y2 = 5 – x2
⇒ x2 + y2 = 5
which represents equation of circle with centre (0, 0) and radius, r = √5.
But the actual equation of curve is y = V5- x2 which represents a semi-circle whose centre is (0, 0) and radius r = √5.
On drawing the rough sketch, we get the following graph:
For finding the points of intersection of the curves, we have
y = 1 – x … (i)
y = x – 1 …… (ii)
and y = 5x2−−−−−√ ….. (iii)
On putting y = 1 – x from Eq. (i) in Eq. (iii), we get
(1 – x) = 5x2−−−−−√
⇒ x2 + (1 – x)2 = 5
⇒ x2 + 1 + x2 – 2x = 5
⇒ 2x2 – 2x – 4 = 0
⇒ x2 – x – 2 = 0
⇒ x2 – 2x + x – 2 = 0
⇒ x(x – 2) + 1(x – 2) = 0
⇒ (x + 1) (x – 2) = 0
∴ x = – 1 or 2
Now, when x = -1, then
y = 5x2−−−−−√ = 51−−−−√ = √4
⇒ y = 2
and when x = 2, then y = 5x2−−−−−√ = 54−−−−√ = 1
⇒ y = 1
So, points of intersection of Eqs. (i) and (iii) are (-1, 2) and (2, 1).
Similarly, on solving Eq. (ii) and Eq. (iii), we get
x = – 1 or 2
From Eq. (iii), at x = – 1, y = 2and at x = 2, y = 1.
Hence, the two curves intersect at (-1, 2) and (2, 1).
Now, required area
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Home | | Maths 7th Std | Exercise 5.5 (Construction of special angles without using protractor)
# Exercise 5.5 (Construction of special angles without using protractor)
7th Maths : Term 1 Unit 5 : Geometry : Construction of special angles without using protractor
Exercise 5.5
1. Construct the following angles using ruler and compass only.
(i) 60°
(ii) 120°
(iii) 30°
(iv) 90°
(v) 45°
(vi) 150°
(vii) 135
(i) 60o
Step 1 : Draw a line. Mark a point A on it.
Step 2 : With A as center draw an arc of convenient radius, to the line at a point B.
Step 3 : With the same radius and B as center draw an arc to cut the previous arc at C.
Step 4 : Join AC. Now BAC is the required angle with the measure 60°.
(ii) 120°
Step 1 : Draw a line. Mark a point A on it.
Step 2 : With A as center, draw an arc of convenient radius to the line at a point B.
Step 3 : With the same radius and B as center draw an arc to cut the previous arc at C.
Step 4 : With the same radius and C as center draw an arc to cut previous arc at D.
Step 5 : Join AD. Now BAD is the required angle with the measure 120°.
(iii) 30°
Step 1 : Draw a line. Mark a point A on it.
Step 2 : With A as center, draw an arc of convenient radius to the line at a point B.
Step 3 : With the same radius and B as center draw an arc to cut the previous arc at C.
Step 4 : Join AC. BAC = 60°
Step 5 : With B as center, draw an arc of convenient radius in the interior of BAC.
Step 6 : With the same radius and C as center draw an arc to cut the previous arc at D.
Step 7 : Join AD. Now BAD is the required angle with measure 30°.
(iv) 90o
Step 1 : Draw a line. Mark a point A on it.
Step 2 : With A as center, draw an arc of convenient radius to the line at a point B.
Step 3 : With the same radius and B as center draw an arc to cut the previous arc at C.
Step 4 : With the same radius and C as center draw an arc to cut the previous arc at D.
Step 6 : With C as center, draw an arc of convenient radius in the interior of CAD
Step 7 : With the same radius and D as center draw an arc to cut the previous arc at E.
Step 8 : Join AE. Now BAE = 90° is the required angle.
(v) 45o
Step 1 : Draw a line. Mark a point A on it.
Step 2 : With A as center, draw an arc of convenient radius to the line at a point B.
Step 3 : With the same radius and B as center draw an arc to cut the previous arc at C.
Step 4 : With the same radius and C as center draw an arc to cut the previous arc at D.
Step 6 : With C as center, draw an arc of convenient radius in the interior of CAD.
Step 7 : With the same radius and D as center draw an arc to cut the previous arc at E.
Step 8 : Join AE. It cuts the previous BCD arc at F B AF = 90°.
Step 9 : With B and F as centers draw arcs of convenient radius in the interior of BAF. They cut G.
Step 10 : Join AG Now BAG the required angle with measure 45°.
(vi) 150°
Step 1 : Draw a line. Mark a point A on it.
Step 2 : With A as center, draw an arc of convenient radius to the line at a point B.
Step 3 : With the same radius and B as center draw an arc to cut the previous arc at C.
Step 4 : With the same radius and C as center draw an arc to cut the previous arc at D.
Step 5 : With the same radius and D as center draw an arc to cut the previous arc at E.
Step 6 : With D and E as centers and with the convenient radius draw arcs in the interior of DAE to cut at F.
Step 7 : JoinAF. Now BAF = 150° is the required angle.
(vii) 135o
Step 1 : Draw a line. Mark a point A on it.
Step 2 : With A as center, draw an arc of convenient radius to the line at a point B.
Step 3 : With the same radius and B as center draw an arc to cut the previous arc at C.
Step 4 : With the same radius and C as center draw an arc to cut the previous arc at D.
Step 5 : With the same radius and D as center draw an arc to cut the previous arc at E.
Step 6 : With the convenient radius and D and E as centers draw arcs in the interior of DAE to cut at F.
Step 7 : Join AF. It cuts the arc at G. Now GAB = 150°
Step 8 : With the convenient radius and G and D as centers draw arcs in the interior of DAG to cut at H.
Step 9 : Join AH. Now BAH = 135° is the required angle.
Tags : Questions with Answers, Solution | Geometry | Term 1 Chapter 5 | 7th Maths , 7th Maths : Term 1 Unit 5 : Geometry
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
7th Maths : Term 1 Unit 5 : Geometry : Exercise 5.5 (Construction of special angles without using protractor) | Questions with Answers, Solution | Geometry | Term 1 Chapter 5 | 7th Maths
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# How do you simplify 5/6 + 4/12?
Aug 2, 2016
$\frac{5}{6} + \frac{4}{12} = \frac{7}{6} = 1 \frac{1}{6}$
#### Explanation:
To add the two fractions, we first need to make denominators equal. As first denominator is $6$ and second one is $12$, we can do this by multiplying numerator and denominator of first fraction by $2$.
Hence $\frac{5}{6} + \frac{4}{12}$
= $\frac{5 \times 2}{6 \times 2} + \frac{4}{12}$
= $\frac{10}{12} + \frac{4}{12}$
= $\frac{10 + 4}{12}$
= $\frac{14}{12}$, but as both numerator and denominator are divisible by $2$, this is equal to
$\frac{2 \times 7}{2 \times 6} = \frac{7}{6} = 1 \frac{1}{6}$
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## Estimation
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#### Study Guide Estimation Mathematics, Grade 3
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ESTIMATION What Is Estimation? When you make an estimate, you are making a guess that is approximate. This is often done by rounding. How to Estimate: When estimating, you will often need to round off the numbers with which you are working. For example: There are 148 gumballs in three jars. If the jars have about the same amount, estimate how many gumballs are in each jar? To solve this problem: o First round 148 to 150. 150 is an easier number to work with when estimating. o Now take 150 and divide it into three parts. Since 50 + 50 + 50 = 150, then there are approximately 50 gumballs in each jar. When estimating to add or subtract, round each number before adding or subtracting. For example: To estimate the sum of 23 and 17: o First round 23 and 17 to the nearest ten. 23 rounded to the nearest ten is 20 17 rounded to the nearest ten is 20 o Then add 20 and 20 together. 20 + 20 = 40 To estimate the difference between 32 and 14: o First round 32 and 14 to the nearest ten. 32 to the nearest ten is 30 14 to the nearest ten is 10 o Then subtract 10 from 30. 30 10 = 20 © Copyright NewPath Learning. All Rights Reserved. Permission is granted for the purchaser to print copies for non-commercial educational purposes only. Visit us at www.NewPathLearning.com.
Try This! Estimate the sum of the following. 24 + 37 = ______ 62 + 19 =______ 31 + 46 = ______ 18 + 48 =______ Estimate the difference of the following. 48 31 = ______ 52 29 = ______ 88 53 = ______ 76 20 = ______ © Copyright NewPath Learning. All Rights Reserved. Permission is granted for the purchaser to print copies for non-commercial educational purposes only. Visit us at www.NewPathLearning.com.
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# 29. Height and Distance Solutions Part 2#
1. The diagram is given below:
Let $AB$ be the lamp post having height $h$ m, and $BD$ be the girl having height $1.6$ m. The distance of the grl from the lamp post is $AC = 3.2$ m. $CE$ is the langeth of the shadow given as $4.8$ m. In the $\triangle ABE$ and $\triangle CDE, \angle E$ is common, $\angle A = \angle C = 90^\circ$ so third angle will be also equal. This makes the triangles similar.
$\therefore \frac{AB}{CD} = \frac{AE}{CE} \Rightarrow h = \frac{8}{3}$ m.
2. The diagram is given below:
Let $AC$ be the building having a height of $30$ m. Let $E$ and $G$ point of observations where angles of elevation are $60^\circ$ and $30^\circ$ respectively. Let $AEF$ be the line of foot of the building and foot of the observer which is a horizontal line. Let $DE$ and $FE$ are the heights of the observer. Draw $BEG\parallel ADF$ so that $AB = DE = FG = 1.5$ m. Thus, $BC = 28.5$ m. We have to find $DF = EG$.
In $\triangle BCE, \tan60^\circ = \sqrt{3} = \frac{BC}{CE} \Rightarrow CE = \frac{28.5}{\sqrt{3}}$ m.
In $\triangle BCG, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BC}{CG} \Rightarrow CG = 28.5\sqrt{3}$ m.
Thus, $DF = EG = CG - CF = \frac{57}{\sqrt{3}}$ m, which is the distance walked by the observer.
3. The diagram is given below:
Let the height of the tower $AB$ is $h$ m. When the altitude of the sun is $60^\circ$ let the length of the shadown be $AC = x$ m. Then according to question length of shadow when the sun’s altitude i $30^\circ$ the length of shadow will be $AD, 40$ m longer i.e. $AD = x + 40$.
In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} = \frac{h}{x} \Rightarrow x = \sqrt{3}h$ m.
In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AC} = \frac{h}{x + 40} \Rightarrow x = 20$ m and $h = 20\sqrt{3}$ m.
4. The diagram is given below:
Let $AB$ be the building with $20$ m height. Let the height of tower be $h$ m represented by $BC$ in the figure. Let $D$ be the point of observation at a distance $x$ from the foot of the building $AB$.
In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{20}{x} \Rightarrow x = 20\sqrt{3}$ m.
In $\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{AC}{AD} = \frac{h + 20}{x}\Rightarrow h = 40$ m.
5. The diagram is given below:
Let $DE$ be the building having a height of $8$ m. Let $AC$ be the multistoried building having height $h + 8$ m. Foot of both the buildings are joined on horizontal plane i.e. $AD$. Draw a line parallel to $AD$ which is $BE$. So $BE$ is equal to $AD$ which we have let as $x$ m. Clearly, $AB = 8$ m. Let height of $BC$ to be $h$ m.
In $\triangle CBE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BC}{BE} = \frac{h}{x} \Rightarrow \sqrt{3}h = x$.
In $\triangle ACD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AC}{AD} = \frac{h + 8}{x}\Rightarrow h = \frac{8}{\sqrt{3} - 1} \Rightarrow h + 8 = \frac{8\sqrt{3}}{\sqrt{3} - 1}$ m.
6. The diagram is given below:
Let $AB$ be the pedestal having height $h$ m and $BC$ be the statue having height $1.6$ m on top of pedestal. Let $D$ be the point of observation from where the angles of elevation as given in the question are $45^\circ$ and $60^\circ$.
In $\triangle ABD, \tan45^\circ = 1 = \frac{AB}{BD} = \frac{h}{x} \Rightarrow h = x$.
In $\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{AC}{CD} = \frac{h + 1.6}{x} \Rightarrow h = \frac{1.6}{\sqrt{3} - 1}$ m.
7. This problem is similar to 55 and has been left as an exercise.
8. The diagram is given below:
Let $AB$ be the tower having height $75$ m. Let $C$ and $D$ be the position of two ships and angles of elevation are as given in the question. Let foor of the tower be in line with ships such that $AC = x$ m and distance between the ships as $d$ m.
In $\triangle ABC, \tan45^\circ = 1 = \frac{AB}{AC} = \frac{75}{x} \Rightarrow x = 75$ m.
In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{75}{x + d} \Rightarrow d = 75(\sqrt{3} - 1)$ m.
9. The diagram is given below:
Let $AB$ be the building and $CD$ be thw tower having height $50$ m. The angles of elevation are shown as given in the question. Let distance between the foot of the tower and the building be $d$ m and height of the building be $h$ m.
In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{d} \Rightarrow d = \sqrt{3}h$.
In $\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{50}{d} \Rightarrow 3h = 50 \Rightarrow h = \frac{50}{3}$ m.
10. The diagram is given below:
Let $DE$ represent the banks of river and $BC$ the bridge. Given that height of the bridge is $30$ m. $\therefore BD = CE = 30$ m. The angles of depression from point $A$ is shown as given in the question. We have to find $DE = BC$ i.e. width of the river.
In $\triangle ACE, \tan45^\circ = 1 = \frac{CE}{AC} \Rightarrow AC = 30$ m.
In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BD}{AB} \Rightarrow AB = 30\sqrt{3}$ m.
Thus, width of river $= 30 + 30\sqrt{3} = 30(\sqrt{3} + 1)$ m
11. The diagram is given below:
Let $BC$ and $DE$ be the two poles. Let $A$ be the point between them such that $AB = x$ m and, thus $AD = 80 - x$ m. Let the elevation from $A$ to $C$ is $60^\circ$ and to $E$ is $30^\circ$. Let the height of poles be $h$ m.
In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{BC}{AB} = \frac{h}{x} \Rightarrow h = \sqrt{3}x$ m.
In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{DE}{AD} = \frac{h}{80 - x} \Rightarrow 3x = 80 -x \Rightarrow x = 20$ m. $\Rightarrow h = 20\sqrt{3}$ m.
12. The diagram is given below:
Let $BD$ and $CE$ be the poles and $AJ$ be the tree. Given, $AJ = 20$ m and angles of depression to base of poles are $60^\circ$ and $30^\circ$. Let $\angle DAB = 6-00^\circ$ and $\angle EAC = 30^\circ$.
Clearly, $AB = DJ = y$ m(say) and $AC = EJ = x$ m(say).
In $\triangle AEJ, \tan60^\circ = \sqrt{3} = \frac{AJ}{EJ} \Rightarrow x = \frac{20}{\sqrt{3}}$ m.
Similarly, $y = 20\sqrt{3}$ m.
Thus, width of river $x + y = \frac{80}{\sqrt{3}}$ m.
13. This problem is similar to $56$ and has been left as an exercise.
14. This problem is similar to $58$ and has been left as an exercise.
15. This problem is similar to $49$ and has been left as an exercise.
16. The diagram is given below:
Let $A$ be the point on the ground, $AC$ be the string and $BC$ the height of balloon. Then given, angle of elevation $\angle BAC = 60^\circ$.
In $\triangle ABC, \sin60^\circ = \frac{\sqrt{3}}{2} = \frac{BC}{AC} = \frac{BC}{215} = 107.5\sqrt{3}$ m.
17. The diagram is given below:
Let $AB$ be the cliff having a height of $80$ m. Let $C$ and $D$ be two points on eihter side of the cliff from where angle of elevations are $60^\circ$ and $30^\circ$ respectively.
In $\triangle ABC, \tan60^\circ = \frac{AB}{AC} \Rightarrow AC = \frac{80}{\sqrt{3}}$ m.
In $\triangle ABD, \tan30^\circ = \frac{AB}{AD}\Rightarrow AD = 80\sqrt{3}$ m.
Distance bettwen points of observation $CD = AC + AD = \frac{320}{\sqrt{3}}$ m.
18. Since the length of shadow is equal to height of pole the angle of elevation would be $45^\circ$ as $\tan45^\circ = 1$.
19. This problem is similar to $62$ and has been left as an exercise.
20. This problem is similar to $25$ and has been left as an exercise.
21. The diagram is given below:
Let $AB$ be the lighthouse having a height of $200$ m. Let $C$ and $D$ be the ships. The angles of depression are converted to angles of elevation.
In $\triangle ABC, \tan45^\circ = 1 = \frac{AB}{AC}\Rightarrow AC = 200$ m.
In $\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{AB}{AD} \Rightarrow AD = \frac{200}{\sqrt{3}}$ m.
Thus distance between ships $CD = AC + AD = \frac{200(\sqrt{3} + 1)}{\sqrt{3}}$ m.
22. The diagram is given below:
Let $AB$ be the first pole and $CD$ be the second pole. Given, $CD = 24$ m and $AC = 15$ m. Draw $BE || AC \Rightarrow BE = 15$ m. Angle of depression is converted to angle of elevation.
In $\triangle BDE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{ED}{BE} \Rightarrow ED = \frac{15}{\sqrt{3}} = 5\sqrt{3}$ m.
$\Rightarrow CE = BD - ED = 24 - 5\sqrt{3} = AB$ which is height of the first pole.
23. This problem is similar to $71$ and has been left as an exercise.
24. The diagram is given below:
xLet $AB$ be the tower and $C$ and $D$ are two points at a distance of $4$ m and $9$ m respectively. Because it is given that angles of elevations are complementary we have chosen and angle of $\theta$ for $C$ and $90^\circ - \theta$ for $D$.
In $\triangle ABC, \tan\theta = \frac{AB}{AC} = \frac{h}{4}$
In $\triangle ABD, \tan(90^\circ - \theta) = \cot\theta = \frac{AB}{AD} = \frac{h}{9}$
Substituting for $\cot\theta$, we get
$\frac{4}{h} = \frac{h}{9}\Rightarrow h^2 = 36 \Rightarrow h = 6$ m.
25. This problem is similar to $72$ and has been left as an exercise.
26. This problem is similar to $56$ and has been left as an exercise.
27. This problem is similar to $55$ and has been left as an exercise.
28. This problem is similar to $71$ and has been left as an exercise
29. This problem is similar to $55$ and has been left as an exercise.
30. This problem is similar to $58$ and has been left as an exercise.
31. This problem is similar to $26$ annd has been left as an exercise.
32. This problem is similar to $71$ and has been left as an exercise.
33. This problem is similar to $26$ annd has been left as an exercise.
34. This problem is similar to $28$ annd has been left as an exercise.
35. This problem is similar to $71$ and has been left as an exercise
36. This problem is similar to $23$ and has been left as an exercise
37. The diagram is given below:
Let $AB$ be the tower and $BC$ be the flag-staff having a height of $h$ m. Let $D$ be the point of observation having angle of elevations $\alpha$ and $\beta$ as given in the question.
In $\triangle ABC, \tan\alpha = \frac{AB}{AD} \Rightarrow AB = AD\tan\alpha$
In $\triangle ABD, \tan\beta = \frac{AC}{AD} = \frac{AB + BC}{AD}$
$\Rightarrow \frac{AB\tan\beta}{\tan\alpha} = AB + h \Rightarrow AB = \frac{h\tan\alpha}{\tan\beta - \tan\alpha}.$
38. This proble is similar to $74$ and has been left as an exercise.
39. The diagram is given below:
Let $BE$ be the tower leaning northwards and $AB$ be the vertical height of tower taken as $h$. Let $C$ and $D$ be the points of observation. Given that angle of leaning is $\theta$ and angles of elevation are $\alpha$ at $C$ and $\beta$ at $D$. Let $AB = x$. Given $BC = a$ and $BD = b$.
In $\triangle ABE, \cot\theta = \frac{x}{h}$, in $\triangle ACE, \cot\alpha = \frac{x + a}{h}$ and in $\triangle ADE, \cot\beta = \frac{x + b}{h}$.
$\Rightarrow b\cot\alpha = \frac{bx + ab}{h}, a\cot\beta = \frac{ax + ab}{h}$
$\Rightarrow b\cot\alpha - a\cot\beta = \frac{bx - ax}{h}\Rightarrow \frac{x}{h} = \cot\theta = \frac{b\cot\alpha - a\cot\beta}{b - a}$.
40. The diagram is given below:
Let $AE$ be the plane of lake and $AC$ be the height of the cloud. $F$ is the point of observation at a height $h$ from lake. $AD$ is the reflection of cloud in the lake. Clearly, $AC = AD$. Draw $AE || BF$ and let $BF = x$. $\alpha$ and $\beta$ are angles of elevation and depression as given.
In $\triangle BCF, \tan\alpha = \frac{BC}{BF} = \frac{BC}{x}\Rightarrow BC = x\tan\alpha$
$AC = AD = AB + BC = h + x\tan\alpha$
In $\triangle BDF, \tan\beta = \frac{AB + AD}{BF} = \frac{h + h + x\tan\alpha}{x} \Rightarrow x = \frac{2h}{\tan\beta - \tan\alpha}$
$AC = AB + BC = h + x\tan\alpha = \frac{h(\tan\alpha + \tan\beta)}{\tan\beta - \tan\alpha}$.
41. The diagram is given below:
Let the cicle represent round balloon centered at $O$ having radius $r$. $B$ is the point of observation from where angle of elevation to the center of the balloon is given as $\beta$. $BL$ and $BM$ are tangents to the balloon and $OL$ and $OM$ are perpendiculars. Clearly $OL = OM = r$. GIven $\angle LBM = \alpha$ and $\angle OBL = \angle OBM = \alpha/2$.
In $\triangle OBL, \sin\alpha/2 = \frac{OL}{OB} \Rightarrow OB = r\cosec\alpha/2$.
In $\triangle ABO, \sin\beta = \frac{AO}{OB}\Rightarrow AO = r\sin\beta\cosec\alpha/2$.
42. The diagram is given below:
Let $AB$ be the cliff having a height $h$ and $F$ be the initial point of observation from where the angle of elevation is $\theta$. Let $D$ be the point reached after walking a distance $k$ towards the top at an angle $\phi$. The angle of elevation at $D$ is $\alpha$.
In $\triangle DEF, \sin\phi = \frac{DE}{DF} \Rightarrow DE = k\sin\phi, \cos\phi = \frac{EF}{DF} \Rightarrow EF = k\cos\phi$.
In $\triangle ABF, \tan\theta = \frac{AB}{BF} \Rightarrow \frac{x}{k\cos\phi + (x - k\sin\phi)\cot\alpha}$
$\Rightarrow x\cot\theta = k\cos\phi + x\cot\alpha - k\sin\phi\cot\alpha \Rightarrow x(\cot\theta - \cot\alpha) = k(\cos\phi - \sin\phi\cot\alpha)$
$\Rightarrow x = \frac{k(\cos\phi - \sin\phi\cot\alpha)}{\cot\theta - \cot\alpha}$.
43. The diagram is given below:
Let $CD$ be the tower having a height $h$. Point $A$ is due south of $A$ making an angle of elevation $\alpha$ and $B$ is due east of tower making an angle of elevation $\beta$. Clearly, $\angle ACB = 90^\circ$. Given that $AB = d$.
In $\triangle ACD, \tan\alpha = \frac{CD}{AC} \Rightarrow AC = h\cot\alpha$ and in $\triangle BCD, \tan\beta = \frac{CD}{BC} \Rightarrow BC = h\cot\beta$.
In $\triangle ABC, AB^2 = AC^2 + AD^2 \Rightarrow d^2 = h^2\cot^2\alpha + h^2\cot^2\beta \Rightarrow h = \frac{d}{\sqrt{\cot^2\alpha + \cot^2\beta}}$.
44. This problem is similar to $93$ and has been left as an exercise.
45. The diagram is given below:
Let $AB$ be the girl having a height of $1.2$ m, $C$ and $F$ be the two places of balloon for which angle of elevations are $60^\circ$ and $30^\circ$ respectively. Height of ballon above ground level is given as $88.2$ m and thus height of balloon above the girl’s eye-level is $88.2 - 1.2 = 87$ m.
In $\triangle ACD, \tan60^\circ = \frac{CD}{AD} \Rightarrow AD = 87/\sqrt{3}$ m.
In $\triangle AFG, \tan30^\circ = \frac{FG}{AG} \Rightarrow AG = 87\sqrt{3}$ m.
Thus distance trarvelled by the ballon $= 87\sqrt{3} - 87/\sqrt{3} = 174/\sqrt{3}$
46. The diagram is given below:
Let $AB$ represent the tower with a height $h$. Let $C$ and $D$ be the points to which angles of depression are given as $60^\circ$ and $30^\circ$ which are shown as angles of elevation at these points.
In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} \Rightarrow AC = h/\sqrt{3}$
In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} \Rightarrow AD = h\sqrt{3}$
$CD = AD - AC = 2h/\sqrt{3}$
The car covers the distance $CD$ in six seconds. Thus speed of the car if $2h/(6\sqrt{3}) = h/3\sqrt{3}$
Time taken to cover $AC$ to reach the foot of the tower is $\frac{h}{\sqrt{3}}\times\frac{3\sqrt{3}}{h} = 3$ seconds.
47. Proceeding like previous problem the answer would be three minutes.
48. This problem is similar to $96$ and has been left as an exercise.
49. The diagram is given below:
Let $AB$ be the building having height $h$ m. Let $C$ and $D$ be the fire stations from which the angles of elevation are $60^\circ$ and $45^\circ$ separated by $20,000$ m.
In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC}\Rightarrow AC = h/\sqrt{3}$ m.
In $\triangle ABD, \tan45^\circ = h = \frac{AB}{AD}\Rightarrow AD = h$ m.
Since $AD < AD$ so the fire station at $C$ will reach the building faster.
$AD = AC + CD \Rightarrow h = h/\sqrt{3} + 20000 \Rightarrow h = \frac{20000\sqrt{3}}{\sqrt{3} - 1}$
$\therefore AC = \frac{2000}{\sqrt{3} - 1}$ m.
50. The diagram is given below:
Let $AB$ be the deck of the ship with given height of $10$ m. Let $CE$ be the cliff with base at $C$. Let the height of portion $DE$ be $x$ m. The angles of elevation of the top and of the bottom of the cliff are shown as given in the question.
In $\triangle BDE, \tan45^\circ = DE/BD \Rightarrow BD = x$ m.
In $\triangle, \tan30^circ = CD/BD \Rightarrow BD = 10\sqrt{3} = x$
Thus, $CE = 10 + 10\sqrt{3} = 27.32$ m.
So height of the cliff is $27.32$ m and distance of cliff from the ship is $10$ m.
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Analytical geometry examples
Analytical geometry solved examples
18 Circle given by two points and tangent to the y axis ▲
Given two points on a circle (6 , 2) and (3 , -1), the circle is tangent to the y axis. Find the radius and the center coordinate of the circle.
Draw the circle and mark the point of the circle center that is at a = R and b (because the circle is tangent to the y axis, b is unknown)
We can write the equation of the circle related to the points and the center of the circle:
( x1 − R )2 + ( y1 − b )2 = R2 (1) ( x2 − R )2 + ( y2 − b )2 = R2 (2)
From first eq. (1) x12 − 2x1R + R2 + y12 − 2y1b + b2 = R2 x12 − 2x1R + y12 − 2y1b + b2 = 0 From first eq. (2) x22 − 2x2R + R2 + y22 − 2y2b + b2 = R2 x22 − 2x2R + y22 − 2y2b + b2 = 0
Substitute R from eq. (3) into the last equation and arrange terms:
b2(2x1 − 2x2) + b(4x2y1 − 4x1y2) + 2x1x22 − 2x12x2 + 2x1y22 − 2x2y12 = 0
b2(x1 − x2) + b(2x2y1 − 2x1y2) + x1x22 − x12x2 + x1y22 − x2y12 = 0
We get a quadratic equation with the unknown b (all the points are given) the solution is:
Now insert the value of b into eq. (3) to get the radius R of the circle.
Insert the given values of the points (6 , 2) and (3 , − 1) to get b:
The radius from eq. (3) for b = 2 is: And the second radius for b = − 10 is:
We can see that there are two circles that fulfills the requirements of the given data.
The two circles centers are at the points (3 , 2) and (15 , − 10)
19 Circle given by two points and the radius ▲
Given two points on a circle (2 , -1) and (-2 , 7) and the radius which is equal to 5. Find the center coordinate of the circle.
We can write the equation of the circle with two points which are located on the circle as:
(x1 - a)2 + (y1 - b)2 = R2 (x2 - a)2 + (y2 - b)2 = R2
We get two equations with two unknowns a and b but to find their values from this equations is not an easy job, you can try it.
So we will try another method to solve the problem.
Connect the two points so it make a lineThe mid point coordinate of this two points is: xm = (x1 + x2) / 2 ym = (y1 + y2) / 2
The slope of the line connecting the two points is: Line AB is perpendicular to this line and the slope is:
The equation of line AB based on point (xm , ym) is: y − ym = mAB (x − xm)
Line AB is passing through the center of the circle (point a,b) so now we search a point on this line which is at a distance of R fron the points (x1 , y1), so we have the equations
Distance from point 1 to (a , b) is R: (1) Point (a , b) is on the line AB so: (2)
Substituting the value of b from eq. (2) into eq. (1) we get:
(x1 − a)2 + [y1 + mAB(xm − a) − ym]2 = R2
In order to simplify the equation we will mark z as: z = y1 + mABxm − ym
x12 − 2x1a + a2 + (z − mABa)2 = R2
And we get the quadratic equation: a2(1 + mAB2) − a(2x1 + 2zmAB) + x12 + z2 − R2 = 0
The value of b is found by substituting the values of a into eq. (2).
According to given data:
z = − 1 + 0.5 * 0 − 3 = − 4
And b is according to eq. (2): b1 = 3 + 0.5(2 − 0) = 4 b2 = 3 + 0.5(− 2 − 0) = 2
So the two possible circles centers are at: (2 , 4) and (− 2 , 2)
Point 1: ( , ) Point 2: ( , ) Radius:
Circles centers:
Two circles tangency - example 20 ▲
Determine the tangency point of two circles of the form
(x − x1)2 + (y − y1)2 = r12 (x − x2)2 + (y − y2)2 = r22
Then solve numerically with the values: y1 = y2 = 2 x1 = − 10
and x2 = 4 r1 = 5 r2 = 9
The condition for tangency are:
Outer tangency: d = r1 + r2 Inner tangency: d = | r1 − r2 |
The slope of the line that connectes circles centers m is: (1)
Because the tangency point is on this line we can use simple trigonometry to find the point location:
From the similarity of two trangles we have:
For the x axis we have: ⟶ For y axis we have: ⟶
Equations of the circles are: (x + 10)2 + (y − 2)2 = 52 (x − 4)2 + (y − 2)2 = 92
The tangency point is:
Determine the tangency point of two circles of the form
(x − 1)2 + (y + 4)2 = 92 (x − 1)2 + (y − 1)2 = 42
The distance between circles centers is:
Because |r1 − r2| = |9 − 4| = 5 = d
then the circles are inner tangent.
Because the tangency point is on this line we can use simple trigonometry to find the point location:
From the similarity of two trangles we have:
For the x axis we have: ⟶ For y axis we have: ⟶
The tangency point is:
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Upper and Lower Fences: Definition & Example
In statistics, the upper and lower fences represent the cut-off values for upper and lower outliers in a dataset. They are calculated as:
• Lower fence = Q1 – (1.5*IQR)
• Upper fence = Q3 + (1.5*IQR)
where IQR stands for “interquartile range” which represents the difference between the 75th percentile (Q3) and 25th percentile (Q1) in a dataset.
An observation that lies above the upper fence or below the lower fence is often considered to be an outlier.
Example: Calculating the Upper and Lower Fence
Suppose we have the following dataset:
```Dataset: 11, 13, 14, 14, 15, 16, 18, 22, 24, 27, 34, 36, 38, 41, 45
```
We can use the following steps to calculate the upper and lower fence of the dataset:
Step 1: Find Q1 and Q3.
Q1 represents the 25th percentile of the dataset and Q3 represents the 75th percentile. According to the Interquartile Range Calculator, Q1 and Q3 for this dataset are as follows:
• Q1: 14
• Q3: 36
Step 2: Find the IQR (Interquartile Range).
The interquartile range represents the difference between Q3 and Q1, which is calculated as:
• Interquartile Range: Q3 – Q1 = 36 – 14 = 22
Step 3: Find the Upper and Lower Fence
We can use the following formulas to calculate the upper and lower fences:
• Lower fence: Q1 – (1.5*IQR) = 14 – (1.5*22) = -19
• Upper fence: Q3 + (1.5*IQR) = 36 + (1.5*22) = 69
Since none of the observations in our dataset lie below the lower fence or above the upper fence, none of the observations would be considered outliers.
We can also create a boxplot to visualize our distribution of data values along with the upper and lower fences:
Bonus: Upper and Lower Fence Calculator
Instead of calculating the upper and lower fence of a dataset by hand, feel free to use the Upper and Lower Fence Calculator:
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# Is roulette really random? – How To Win Roulette Every Spin
What percentage chance did you have that a roulette wheel would return you one of the possible outcomes of six different numbers? This is a very easy question, but the answer is, the probability that 6 numbers will appear is . The probability of seeing one of the numbers is . And I mean real probability: a roulette wheel can return any number between 1 and 6… not just 6 or 6 plus 1. So the probability that it’ll return me a number between 1 and 6 multiplied by each of the six possible outcomes is .
So when you have a probability of 1/6 that a random number that appears on a roulette wheel would return a number between 1 and 6 (in other words, the probability that a random number that appears has a greater than 5% chance of appearing), you can use it to predict the distribution of numbers that appeared on the wheel and the probabilities of those numbers that appeared on the wheel appearing.
So we’ll apply this method to another example of the same probability problem. Instead, we are going to solve for a random event that we want to predict, and determine our final probability. In this case, it is our outcome of one of our players picking up the last card of the deck: the last card of the deck is either . The last card is either played or . Now, we expect that there is 1 in 6 chance that one of the players will be playing . For this reason, we want to determine the probability that 1 of the next two cards that appear will be . We don’t know which of those two will be played as yet but we can use the distribution shown above to infer the probabilities. The probability of either of the outcomes is … 1/6
Therefore our final probability is … 6/6 which is 1/6 times the probability that each card of the deck is .
The probability that the remaining 3 cards will be is 1/6. Hence, it also remains a 1/6 possibility that the next card will be .
So we have derived a probability for each of the other outcomes for each random card. What is the probability that the outcome of that last card that the player picked up is of our outcome? It is … 1/6!
The last card that the player picked up is the last card in the deck, which means that our last card is of our outcome if the outcome was the last card in the deck. Note that this outcome is our outcome just as well, so now we also know that our outcome
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Is roulette really random? – How To Win Roulette Every Spin
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Question Video: Identifying the Graph of a Quadratic Function and Determining the Relationship between Two Quadratic Graphs | Nagwa Question Video: Identifying the Graph of a Quadratic Function and Determining the Relationship between Two Quadratic Graphs | Nagwa
Question Video: Identifying the Graph of a Quadratic Function and Determining the Relationship between Two Quadratic Graphs Mathematics • Third Year of Preparatory School
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Answer the following questions. Which graph represents the quadratic function π(π₯) = π₯Β² + 3? Which graph represents the quadratic function π(π₯) = π₯Β² + 4? Which of the following is true about the two graphs? [A] The two curves are identical. [B] The first curve is just a stretched form of the second curve. [C] The two curves have the same shape but the second is a horizontal shift of the first. [D] The two curves have the same shape but the second is a vertical shift of the first. [E] One curve is obtained by rotating the other by 90Β° about the origin.
07:17
Video Transcript
Answer the following questions. Which graph represents the quadratic function π of π₯ equals π₯ squared plus three? Which graph represents the quadratic function π of π₯ equals π₯ squared plus four?
There is also a third part to this question, which we will look at when weβve completed the first two parts. To answer the first part of the question then, letβs think about the properties of the graph of the function π of π₯ equals π₯ squared plus three. First, this is a quadratic function. So we know that the shape of its graph will be a parabola. The parabola will open upwards if the coefficient of π₯ squared is positive, and it will open downwards if the coefficient of π₯ squared is negative. In this question, the coefficient of π₯ squared is one, which is positive. So we know that the shape of the graph will be a parabola that opens upwards, or we might say a U-shaped parabola.
Now, all parabolas are symmetric with a vertical line of symmetry. But furthermore, we know that any quadratic function of the form π of π₯ equals ππ₯ squared plus π, which is what we have here, is a parabola with its line of symmetry on the π¦-axis.
Next, letβs consider the π¦-intercept of the graph of this function. We know that the π¦-intercept occurs when π₯ is equal to zero because π₯ is equal to zero everywhere on the π¦-axis. We can therefore find the π¦-value of the π¦-intercept by substituting π₯ equals zero into the function, or in other words evaluating π of zero. We have zero squared plus three, which is equal to three. We know then that the coordinates of the π¦-intercept for the given quadratic function are zero, three.
Now, in fact, as the line of symmetry is the π¦-axis and the line of symmetry passes through the vertex of a quadratic function, we know that the vertex will also have the coordinates zero, three. We now look at the five graphs we were given. Graph (B) is a U-shaped parabola with its line of symmetry along the π¦-axis and the π¦-intercept and vertex with coordinates zero, three. So graph (B) is the correct graph.
If we were to look at the others, we could rule graph (E) out because it is a negative parabola, that is, a parabola that opens downwards. And we could rule out options (A), (C), and (D) because they all have π¦-intercepts that are not the point zero, three. So weβve answered the first part of the question and found the graph which represents the quadratic function π of π₯ equals π₯ squared plus three.
Letβs now look at the second part, in which weβre finding the graph which represents the quadratic function π of π₯ equals π₯ squared plus four. Well, for the same reasons as in the first part of the question, we know that this will be a positive parabola, or a parabola which opens upwards, with its line of symmetry along the π¦-axis. To find the π¦-intercept and the coordinates of the vertex, we evaluate π of zero, giving zero squared plus four, which is four. So the coordinates of the π¦-intercept and the coordinates of the vertex are zero, four.
We now know that weβre looking for a parabola that opens upwards with a line of symmetry along the π¦-axis and a π¦-intercept and vertex with coordinates of zero, four. Looking at the five graphs, we can see that the one which has these correct properties is graph (D).
Weβll now clear some space to answer the final part of the question.
Which of the following is true about the two graphs? (A) The two curves are identical. (B) The first curve is just a stretched form of the second curve. (C) The two curves have the same shape, but the second is a horizontal shift of the first. (D) The two curves have the same shape, but the second is a vertical shift of the first. Or (E) one curve is obtained by rotating the other by 90 degrees about the origin.
We have here the two graphs that we identified earlier in the question: the graph representing the function π of π₯ equals π₯ squared plus three and the graph representing the function π of π₯ equals π₯ squared plus four. We can see by looking at the two graphs side by side that they do have the same shape. But the graph of π of π₯ equals π₯ squared plus four is above the graph of π of π₯ equals π₯ squared plus three. If we label the vertex of each graph, we have zero, three for the first function and zero, four for the second. So the vertex of the graph of the function π of π₯ equals π₯ squared plus four is one unit vertically above the vertex of the graph of the first function.
We can also see this if we label other points on the two curves. The point two, seven is on the graph of π of π₯ equals π₯ squared plus three. And the point two, eight is on the graph of π of π₯ equals π₯ squared plus four. So, for the same π₯-coordinate, the π¦-coordinate, or the value of the function, is one more for the second function than for the first. If we write the second function as π₯ squared plus three plus one, then this confirms that we are indeed just adding one to the first function. And from our knowledge of transformations of graphs, we know that adding a constant to a function is a vertical shift or vertical translation by that constant. So the relationship between the two graphs is that the two curves have the same shape, but the second is a vertical shift of the first. And in fact it is a vertical shift by one unit.
Looking quickly at the other options, we can see that none of them are also true. Firstly, option (a), the two curves are not identical; they donβt have the same vertex or indeed any of the same points. The two curves have the exact same shape. So the first curve is not a stretched form of the second. Considering option (c), the two curves do have the same shape as weβve already said. But as the vertex of each graph has the same π₯-coordinate, we can see that the second is not a horizontal shift of the first. And finally, in (e), if one curve was obtained by rotating the other by 90 degrees about the origin, then the second curve would be a parabola in a different orientation. And as we can see, the two parabolas are in exactly the same orientation.
So we have identified the graph that represents the function π of π₯ equals π₯ squared plus three and the graph that represents the function π of π₯ equals π₯ squared plus four. And weβve found that the relationship between the two graphs is that the two curves have the same shape, but the second is a vertical shift of the first.
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# 1.10: Function Notation
Difficulty Level: Basic Created by: CK-12
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Practice Function Notation
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Suppose that you want to set up a function that allows you to input a dog's age in human years and which outputs the dog's age in dog years. How would you go about setting up such a function, and what notation would you use? Would the notation be the same as it was for the equations that you've looked at in previous Concepts? In this Concept, you'll learn what is needed to write a function such as this.
### Guidance
#### Example C
Write functions to represent the total each friend spent at the park.
Solution:
J(r)=2r\begin{align*}J(r)= 2r\end{align*} represents Joseph’s total,
L(r)=2r\begin{align*}L(r)= 2r\end{align*} represents Lacy's total,
K(r)=2r\begin{align*}K(r)= 2r\end{align*} represents Kevin's total, and
A(r)=2r\begin{align*}A(r)= 2r\end{align*} represents Alfred’s total.
### Guided Practice
Recall the example from a previous Concept where a student organization sells shirts to raise money. The cost of printing the shirts was expressed as 100+7x\begin{align*} 100+7x\end{align*} and for the revenue, we had the expression 15x\begin{align*} 15x\end{align*}, where x\begin{align*}x\end{align*} is the number of shirts.
a. Write two functions, one for the cost and one for revenue.
b. Express that the cost must be less than or equal to $800. c. Express that the revenue must be equal to$1500.
d. How many shirts must the students sell in order to make 1500? Solution: a. The cost function we will write as C(x)=100+7x\begin{align*}C(x)=100+7x\end{align*} and the revenue function we will write as R(x)=15x\begin{align*}R(x)=15x\end{align*}. b. Since C(x)\begin{align*}C(x)\end{align*} represents the costs, we substitute in800 for C(x)\begin{align*}C(x)\end{align*} and replace the equation with the appropriate inequality symbol
100+7x800\begin{align*}100+7x \le 800\end{align*}
This reads that 100+7x\begin{align*}100+7x\end{align*} is less than or equal to $800, so we have written the inequality correctly. c. We substitute in$1500 for R(x)\begin{align*}R(x)\end{align*}, getting
1500=15x.\begin{align*}1500=15x.\end{align*}
d. We want to find the value of x\begin{align*}x\end{align*} that will make this equation true. It looks like 100 is the answer. Checking this we see that 100 does satisfy the equation. The students must sell 100 shirts in order to have a revenue of 1500. 1500=15(100)\begin{align*}1500=15(100)\end{align*} 1500=1500\begin{align*}1500=1500\end{align*} ### Practice 1. Rewrite using function notation: y=56x2\begin{align*}y= \frac{5}{6} x-2\end{align*}. 2. Rewrite using function notation: m=n2+2n3\begin{align*}m=n^2+2n-3\end{align*}. 3. What is one benefit of using function notation? 4. Write a function that expresses the money earned after working some number of hours for10 an hour.
5. Write a function that represents the number of cuts you need to cut a ribbon in x\begin{align*}x\end{align*} number of pieces.
6. Jackie and Mayra each will collect a \$2 pledge for every basket they make during a game. Write two functions, one for each girl, expressing how much money she will collect.
Mixed Review
1. Compare the following numbers 23 21.999\begin{align*}23 \ \underline{\;\;\;\;\;} \ 21.999\end{align*}.
2. Write an equation to represent the following: the quotient of 96 and 4 is g\begin{align*}g\end{align*}.
3. Write an inequality to represent the following: 11 minus b\begin{align*}b\end{align*} is at least 77.
4. Find the value of the variable k:13(k)=169\begin{align*}k:13(k)=169\end{align*}.
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### Vocabulary Language: English Spanish
dependent variable
A dependent variable is one whose values depend upon what is substituted for the other variable.
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
independent variable
The independent variable is the variable which is not dependent on another variable. The dependent variable is dependent on the independent variable.
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# wikiHow to Simplify Math Expressions
Math students are often asked to give their answer in "simplest terms"—in other words, to write answers as elegantly as possible. Though a long, ungainly expression and a short, elegant one may technically equal the same thing, often, a math problem isn't considered "done" until the answer has been reduced to simplest terms. In addition, answers in simplest terms are almost always the easiest expressions to work with. For these reasons, learning how to simplify expressions is a crucial skill for aspiring mathematicians.
### Method 1 Using the Order of Operations
1. 1
Know the order of operations. When simplifying math expressions, you can't simply proceed from left to right, multiplying, adding, subtracting, and so on as you go. Some math operations take precedence over others and must be done first. In fact, doing operations out of order can give you the wrong answer. The order of operations is: terms in parentheses, exponents, multiplication, division, addition, and, finally, subtraction. A handy acronym you can use to remember this is "Please excuse my Dear Aunt Sally," or "PEMDAS".
• Note that, while basic knowledge of the order of operations makes it possible to simplify most basic expressions, specialized techniques are needed to simplify many variable expressions, including nearly all polynomials. See Method Two below for more information.
2. 2
Start by solving all of the terms in parentheses. In math, parentheses indicate that the terms inside should be calculated separately from the surrounding expression. Regardless of the operations being performed within them, be sure to tackle the terms in parentheses as your first act when you attempt to simplify an expression. Note that, however, within each pair of parentheses, the order of operations still applies. For instance, within parentheses, you should multiply before you add, subtract, etc.
• As an example, let's try to simplify the expression 2x + 4(5 + 2) + 32 - (3 + 4/2). In this expression, we would solve the terms in parentheses, 5 + 2 and 3 + 4/2, first. 5 + 2 = 7. 3 + 4/2 = 3 + 2 = 5.
• The second parenthetical term simplifies to 5 because, owing to the order of operations, we divide 4/2 as our first act inside the parentheses. If we simply went from left to right, we might instead add 3 and 4 first, then divide by 2, giving the incorrect answer of 7/2.
• Note - if there are multiple parentheses nested inside one another, solve the innermost terms first, than the second-innermost, and so on.
3. 3
Solve the exponents. After tackling parentheses, next, solve your expression's exponents. This is easy to remember because, in exponents, the base number and the power are positioned right next to each other. Find the answer to each exponent problem, then substitute the answers back into your equation in place of the exponents themselves.
• After dealing with the parentheses, our example expression is now 2x + 4(7) + 32 - 5. The only exponent in our example is 32, which equals 9. Add this back into the equation in the place of 32 to get 2x + 4(7) + 9 - 5.
4. 4
Solve the multiplication problems in your expression. Next, perform any necessary multiplication in your expression. Remember that multiplication can be written several ways. A × symbol, a dot, or an asterisk are all ways to show multiplication. However, a number hugging a parentheses or a variable (like 4(x)) also denotes multiplication.
• There are two instances of multiplication in our problem: 2x (2x is 2 × x) and 4(7). We don't know the value of x, so let's leave 2x as it is.. 4(7) = 4 × 7 = 28. We can rewrite our equation as 2x + 28 + 9 - 5.
5. 5
Move on to division. As you search for division problems in your expression, keep in mind that, like multiplication, division can be written multiple ways. The simple ÷ symbol is one, but also remember that slashes and bars in a fraction (like 3/4, for instance) signify division.
• Because we already solved a division problem (4/2) when we tackled the terms in parentheses, our example no longer has any division in it, so we will skip this step. This brings up an important point - you don't have to perform every operation in the PEMDAS acronym when simplifying an expression, just the ones that are present in your problem.
6. 6
Add. Next, perform any addition problems in your expression. You can simply proceed from left to right through your expression, but you may find it easiest to add numbers that combine in simple, manageable ways first. For instance, in the expression 49 + 29 + 51 +71, it's easier to add 49 + 51 = 100, 29 + 71 = 100, and 100 + 100 = 200, rather than 49 + 29 = 78, 78 + 51 = 129, and 129 + 71 = 200.
• Our example expression has been partially simplified to "2x + 28 + 9 - 5". Now, we must add what we can - let's look at each addition problem from left to right. We can't add 2x and 28 because we don't know the value of x, so let's skip it. 28 + 9 = 37, so let's rewrite or expression as "2x + 37 - 5".
7. 7
Subtract. The very last step in PEMDAS is subtraction. Proceed through your problem, solving any remaining subtraction problems. You may address the addition of negative numbers in this step, or in the same step as the normal addition problems - it won't effect your answer..
• In our expression, "2x + 37 - 5", there is only one subtraction problem. 37 - 5 = 32
8. 8
Review your expression. After proceeding through the order of operations, you should be left with your expression in simplest terms. However, if your expression contains one or more variables, understand that the variable terms will remain largely untouched. Simplifying variable expressions requires you to find the values of your variables or to use specialized techniques to simplify the expression (see below).
• Our final answer is "2x + 32". We can't address this final addition problem until we know the value of x, but when we do, this expression will be much easier to solve than our initial lengthy expression.
### Method 2 Simplifying Complex Expressions
1. 1
Add like variable terms. When dealing with variable expressions, it's important to remember that terms with the same variable and exponent (or "like terms") can be added and subtracted like normal numbers. The terms must not only have the same variable, but also the same exponent. For example, 7x and 5x can be added to each other, but 7x and 5x2 can not.
• This rule also extends to terms with multiple variables. For instance, 2xy2 can be added to -3xy2, but not -3x2y or -3y2.
• Let's look at the expression x2 + 3x + 6 - 8x. In this expression, we can add the 3x and -8x terms because they are like terms. Simplified, our expression is x2 - 5x + 6.
2. 2
Simplify numerical fractions by dividing or "canceling out" factors. Fractions that have only numbers (and no variables) in both the numerator and denominator can be simplified in several ways. First, and perhaps easiest, is to simply treat the fraction as a division problem and divide the denominator by the numerator. In addition, any multiplicative factors that appear both in the numerator and denominator can be "canceled" because they divide to give the number 1. In other words, if both the numerator and denominator share a factor, this factor can be removed from the fraction, leaving a simplified answer.
• For example, let's consider the fraction 36/60. If we have a calculator handy, we can divide to get an answer of .6. If we don't, however, we can still simplify by removing common factors. Another way to think of 36/60 is (6 × 6)/(6 × 10). This can be rewritten as 6/6 × 6/10. 6/6 = 1, so our expression is actually 1 × 6/10 = 6/10. However, we're not done yet - both 6 and 10 share the factor 2. Repeating the above procedure, we are left with 3/5.
3. 3
In variable fractions, cancel out variable factors. Variable expressions in the form of fractions offer unique opportunities for simplification. Like normal fractions, variable fractions allow you to remove factors that are shared by both the numerator and denominator. However, in variable fractions, these factors can be both numbers and actual variable expressions.
• Let's consider the expression (3x2 + 3x)/(-3x2 + 15x).This fraction can be rewritten as (x + 1)(3x)/(3x)(5 - x), 3x appears both in the numerator and in the denominator. Removing these factors from the equation leaves (x + 1)/(5 - x). Similarly, in the expression (2x2 + 4x + 6)/2, since every term is divisible by 2, we can write the expression as (2(x2 + 2x + 3))/2 and thus simplify to x2 + 2x + 3.
• Note that you can't cancel just any term - you can only cancel multiplicative factors that appear both in the numerator and denominator. For instance, in the expression (x(x + 2))/x, the "x" cancels from both the numerator and denominator, leaving (x + 2)/1 = (x + 2). However, (x + 2)/x does not cancel to 2/1 = 2.
4. 4
Multiply parenthetical terms by their constants. When dealing with variable terms in parentheses with an adjacent constant, sometimes, multiplying each term in the parentheses by the constant can result in a simpler expression. This holds true for purely numeric constants and for constants that include variables.
• For instance, the expression 3(x2 + 8) can be simplified to 3x2 + 24, while 3x(x2 + 8) can be simplified to 3x3 + 24x.
• Note that, in some cases, such as in variable fractions, the constant adjacent to the parentheses gives an opportunity for cancellation and thus shouldn't be multiplied through the parentheses. In the fraction (3(x2 + 8))/3x, for instance, the factor 3 appears both in the numerator and the denominator, so we can cancel it and simplify the expression to (x2 + 8)/x. This is simpler and easier to work with than (3x3 + 24x)/3x, which would be the answer we would get if we had multiplied through.
5. 5
Simplify by factoring. Factoring is a technique by which some variable expressions, including polynomials, can be simplified. Think of factoring as the opposite of the "multiplying through parentheses" step above - sometimes, an expression can be rendered more simply as two terms multiplied by each other, rather than as one unified expression. This is especially true if factoring an expression allows you to cancel part of it (as you would in a fraction). In special cases (often with quadratic equations), factoring even allows you to find answers to the equation.
• Let's consider the expression x2 - 5x + 6 once more. This expression can factor to (x - 3)(x - 2). So, if x2 - 5x + 6 is the numerator of a certain expression with one of these factor terms in the denominator, like is the case with the expression (x2 - 5x + 6)/(2(x - 2)), we may want to write it in factored form so that we can cancel it with the denominator. In other words, with (x - 3)(x - 2)/(2(x - 2)), the (x - 2) terms cancel, leaving us with (x - 3)/2.
• As hinted at above, another reason you may want to factor your expression has to do with the fact that factoring can reveal answers to certain equations, especially when those equations are written as expressions equal to 0. For instance, let's consider the equation x2 - 5x + 6 = 0. Factoring gets us (x - 3)(x - 2) = 0. Since any number times zero equals zero, we know that if we can get either of the terms of parentheses to equal zero, the whole expression on the left side of the equals sign will equal zero as well. Thus, 3 and 2 are two answers to the equation.
## Community Q&A
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• How do I solve one without parentheses?
To simplify a math expression without parentheses, you follow the order of operations, which is PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction). Because the expression has no parentheses, you can start checking the expression for exponents. If it does, simplify that first. Then, you can move on to multiplication and division, then finally, addition and subtraction.
• What is x + (-y) + 1/2 z when x= -2, y=3 and z=-2?
Replace each letter in the expression with its given value: (-2) + [-(3)] + (½)(-2) = (-2) - (3) +(-1) = -2 -3 -1 = -6.
• What is the hcf of 1/2, 2/3, 3/4, and 4/5?
wikiHow Contributor
First, write all those fractions with a common denominator: 30/60, 40/60, 45/60, and 48/60. The highest common (integer) factor of those fractions is the hcf of their numerators over that common denominator. The hcf of 30, 40, 45, and 48 is 1 (as it always is when you use the least common denominator), so the hcf of 1/2, 2/3, 3/4, and 4/5 is 1/60.
• What is the coefficient of 24x + 32 + 4x + 3?
Naples opti sailor
32 and 3
• How do I simplify a math expression?
200 characters left
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Categories: Mathematics
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# 3-3
3-3. Solving Systems of Linear Inequalities. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 2. Holt Algebra 2. 2 x + y = 2. 2 y – x = –6. x – y = – 1. x + 2 y = 2. Warm Up. 1. Graph 2 x – y > 4. Determine if the given ordered pair is a
## 3-3
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### Presentation Transcript
1. 3-3 Solving Systems of Linear Inequalities Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2
2. 2x + y = 2 2y– x = –6 x– y = –1 x+ 2y = 2 Warm Up 1. Graph 2x– y > 4. Determine if the given ordered pair is a solution of the system of equations. 2. (2, –2) yes 3. (–4, 3) no
3. Objective Solve systems of linear inequalities.
4. Vocabulary system of linear inequalities
5. When a problem uses phrases like “greater than” or “no more than,” you can model the situation using a system of linear inequalities. A system of linear inequalities is a set of two or more linear inequalities with the same variables. The solution to a system of inequalities is often an infinite set of points that can be represented graphically by shading. When you graph multiple inequalities on the same graph, the region where the shadings overlap is the solution region.
6. y< – 3 y ≥–x + 2 For y < – 3, graph the dashed boundary line y =– 3, and shade below it. Example 1A: Graphing Systems of Inequalities Graph the system of inequalities. For y ≥ –x + 2, graph the solid boundary line y = –x + 2, and shade above it. The overlapping region is the solution region.
7. 0< (0)–3 y< x – 3 1 1 1 1 1 1 0 < –3 2 2 2 2 2 2 –2< (5) –3 –2 <– y ≥–x + 2 Check Test a point from each region on the graph. 0 ≥ –(0) + 2 x 0 ≥ 2 x –2≥–(5) + 2 –2≥ –3 3≥–(0) + 2 3< (0)–3 x 3≥2 2 < –3 –4≥–(0) + 2 –4< (0)–3 –4 < –3 x –4≥2 Only the point from the overlapping (right) region satisfies both inequalities.
8. Helpful Hint If you are unsure which direction to shade, use the origin as a test point.
9. Example 1B: Graphing Systems of Inequalities Graph each system of inequalities. y< –3x + 2 y ≥ –1 For y < –3x + 2, graph the dashed boundary line y =–3x + 2, and shade below it. For y ≥–1, graph the solid boundary line y =–1, and shade above it.
10. Example 1B Continued CheckChoose a point in the solution region, such as (0, 0), and test it in both inequalities. y < –3x + 2 y ≥ –1 0 < –3(0)+ 2 0≥–1 0 < 2 0≥–1 The test point satisfies both inequalities, so the solution region is correct.
11. For x – 3y < 6, graph the dashed boundary line y = – 2, and shade above it. Check It Out! Example 1a Graph the system of inequalities. x – 3y< 6 2x + y > 1.5 For 2x + y > 1.5, graph the dashed boundary line y = –2x + 1.5, and shade above it. The overlapping region is the solution region.
12. x – 3y< 6 2x + y > 1.5 Check Test a point from each region on the graph. 0– 3(0)< 6 2(0) + 0 >1.5 x 0 > 1.5 x 0 < 6 4– 3(–2)< 6 2(4)– 2 >1.5 6 > 1.5 10 < 6 x 0– 3(3)< 6 2(0) + 3 >1.5 –9 < 6 3 > 1.5 0– 3(–4)< 6 2(0)– 4 >1.5 –12 < 6 –4 > 1.5 x Only the point from the overlapping (top) region satisfies both inequalities.
13. Check It Out! Example 1b Graph each system of inequalities. y ≤4 2x + y< 1 For y ≤4, graph the solid boundary line y =4, and shade below it. For 2x + y < 1, graph the dashed boundary line y =–3x +2, and shade below it. The overlapping region is the solution region.
14. Check It Out! Example 1b Continued CheckChoose a point in the solution region, such as (0, 0), and test it in both directions. 2x + y< 1 y ≤ 4 2(0) + 0< 1 0≤ 4 0 < 1 0 ≤ 4 The test point satisfies both inequalities, so the solution region is correct.
15. Example 2: Art Application Lauren wants to paint no more than 70 plates for the art show. It costs her at least \$50 plus \$2 per item to produce red plates and \$3 per item to produce gold plates. She wants to spend no more than \$215. Write and graph a system of inequalities that can be used to determine the number of each plate that Lauren can make.
16. Example 2 Continued Let x represent the number of red plates, and let y represent the number of gold plates. The total number of plates Lauren is willing to paint can be modeled by the inequality x + y ≤ 70. The amount of money that Lauren is willing to spend can be modeled by 50 + 2x + 3y ≤ 215. x 0 y 0 The system of inequalities is . x + y ≤ 70 50 + 2x + 3y ≤ 215
17. Example 2 Continued Graph the solid boundary line x + y = 70, and shade below it. Graph the solid boundary line 50 + 2x + 3y ≤ 215, and shade below it. The overlapping region is the solution region.
18. Example 2 Continued Check Test the point (20, 20) in both inequalities. This point represents painting 20 red and 20 gold plates. x + y ≤ 70 50 + 2x + 3y ≤ 215 20 + 20 ≤ 70 50 + 2(20) + 3(20) ≤ 215 40 ≤ 70 150 ≤ 215
19. Check It Out! Example 2 Leyla is selling hot dogs and spicy sausages at the fair. She has only 40 buns, so she can sell no more than a total of 40 hot dogs and spicy sausages. Each hot dog sells for \$2, and each sausage sells for \$2.50. Leyla needs at least \$90 in sales to meet her goal. Write and graph a system of inequalities that models this situation.
20. Check It Out! Example 2 Continued Let d represent the number of hot dogs, and let s represent the number of sausages. The total number of buns Leyla has can be modeled by the inequality d + s ≤ 40. The amount of money that Leyla needs to meet her goal can be modeled by 2d + 2.5s ≥ 90. d 0 s 0 The system of inequalities is . d + s ≤ 40 2d + 2.5s ≥ 90
21. Check It Out! Example 2 Continued Graph the solid boundary line d + s = 40, and shade below it. Graph the solid boundary line 2d + 2.5s ≥ 90, and shade above it. The overlapping region is the solution region.
22. Check It Out! Example 2 Continued Check Test the point (5, 32) in both inequalities. This point represents selling 5 hot dogs and 32 sausages. 2d + 2.5s ≥ 90 d + s ≤ 40 2(5) + 2.5(32) ≥ 90 5 + 32 ≤ 40 37 ≤ 40 90 ≥ 90
23. Systems of inequalities may contain more than two inequalities.
24. Example 3: Geometry Application Graph the system of inequalities, and classify the figure created by the solution region. x ≥ –2 x ≤ 3 y ≥ –x + 1 y ≤ 4
25. Example 3 Continued Graph the solid boundary line x = –2 and shade to the right of it. Graph the solid boundary line x = 3, and shade to the left of it. Graph the solid boundary line y = –x + 1, and shade above it. Graph the solid boundary line y = 4, and shade below it. The overlapping region is the solution region.
26. Check It Out! Example 3a Graph the system of inequalities, and classify the figure created by the solution region. x ≤ 6 y ≤ x + 1 y ≥ –2x + 4
27. Graph the solid boundary line x = 6 and shade to the left of it. Graph the solid boundary line, y ≤x + 1 and shade below it. Graph the solid boundary line y ≥ –2x + 4, and shade below it. The overlapping region is the solution region. The solution is a triangle. Check It Out! Example 3a Continued
28. Check It Out! Example 3b Graph the system of inequalities, and classify the figure created by the solution region. y ≤ 4 y ≥–1 y ≤ –x + 8 y ≤ 2x + 2
29. Check It Out! Example 3b Continued Graph the solid boundary line y = 4 and shade to the below it. Graph the solid boundary line y = –1, and shade to the above it. Graph the solid boundary line y = –x + 8, and shade below it. Graph the solid boundary line y = 2x + 2, and shade below it. The overlapping region is the solution region.
30. Check It Out! Example 3b Continued The solution region is a four-sided figure, or quadrilateral. Notice that the boundary lines y = 4 and y = –1 are parallel, horizontal lines. The boundary lines y = –x + 8 and y = 2x + 2are not parallel since the slope of the first is –1 and the slope of the second is 2. A quadrilateral with one set of parallel sides is called a trapezoid. The solution region is a trapezoid.
31. Lesson Quiz: Part I 1. Graph the system of inequalities and classify the figure created by the solution region. y ≤ x – 2 y ≥ –2x – 2 x ≤ 4 x ≥ 1 trapezoid
32. Lesson Quiz: Part II 2. The cross-country team is selling water bottles to raise money for the team. The price of the water bottle is \$3 for students and \$5 for everyone else. The team needs to raise at least \$400 and has 100 water bottles. Write and graph a system of inequalities that can be used to determine when the team will meet its goal.
33. Lesson Quiz: Part II Continued x + y ≤ 100 3x + 5y ≥ 400
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# 4.5 Integration of Rational Functions by Partial Fractions Brian E.
Veitch
## 4.5 Integration of Rational Functions by Partial Fractions
From algebra, we learned how to find common denominators so we can do something like this,
2 3 2(x 3) 3(x + 1) 5x 3
+ = + = 2
x+1 x3 (x + 1)(x 3) (x + 1)(x 3) x 2x 3
So why do we need this? We need to know how to do this in the reverse order. If were
5x 3 2 3
given 2 , we need to know that it can be rewritten as + .
x 2x 3 x+1 x3
5x 3
Z
Lets consider the integral dx.
x2 2x 3
We dont have a method that can do this. We cant use u-substitution, trig substitution,
integration by parts, and there are no powers of trig functions. But what if we wrote the
integral as,
5x 3
Z Z
2 3
2
dx = + dx
x 2x 3 x+1 x3
## Now we can evaluate this,
Z
2 3
+ dx = 2 ln |x + 1| + 3 ln |x 3| + C
x+1 x3
148
4.5 Integration of Rational Functions by Partial Fractions Brian E. Veitch
5x 3
The next objective is given a rational function like , how do we break it up?
x2 2x 3
## 1. Factor the denominator
x2 2x 3 = (x + 1)(x 3)
## 2. Next, rewrite the rational function as
5x 3 A B
= +
x2 2x 3 x+1 x3
The numerators are A and B because the denominators are linear factors.
## 3. Clear denominators by multiplying by (x + 1)(x 3)
5x 3 = A(x 3) + B(x + 1)
5x 3 = Ax 3A + Bx + B
5x 3 = (A + B)x + (3A + B)
## 4. Match the coefficients
5 = A+B
3 = 3A + B
149
4.5 Integration of Rational Functions by Partial Fractions Brian E. Veitch
## (c) Using row reducing with matrices
Solution: A = 2 and B = 3.
5x 3 2 3
Therefore, = +
x2 2x 3 x+1 x3
The next methods require the degree of the numerator to be less than the degree of
the denominator. If the degree of the numerator is the same or higher, you must do long
division before proceeding to the following methods.
## 4.5.2 Case 1: Denominator is a product of distinct linear factors
P (x)
Suppose you have a rational function, , where the degree of P (x) is smaller than Q(x).
Q(x)
If the degree of P (x) is greater than or equal to Q(x), you must use long division.
## Find the factors of Q(x).
x2 + 4x + 1
Z
Example 4.11. Find dx
(x 1)(x + 1)(x + 3)
## 1. Set up the fraction as,
x2 + 4x + 1 A B C
= + +
(x 1)(x + 1)(x + 3) x1 x+1 x+3
150
4.5 Integration of Rational Functions by Partial Fractions Brian E. Veitch
## x2 + 4x + 1 = (A + B + C)x2 + (4A + 2B)x + (3A 3B C)
4. Match coefficients
A+B+C = 1
4A + 2B = 4
3A 3B C = 1
(a) Add equations (1) and (3) together and you get
4A 2B = 2
4A 2B = 2
4A + 2B = 4
## (c) Add the equations together
3
8A = 6 A =
4
151
4.5 Integration of Rational Functions by Partial Fractions Brian E. Veitch
3
5. Pluggin A = into the other equations, we get
4
3 1 1
A = ,B = ,C =
4 2 4
6. Rewrite the integral
x2 + 4x + 1
Z Z
3/4 1/2 1/4
dx = + dx
(x 1)(x + 1)(x + 3) x1 x+1 x+3
## 7. Now we can integrate
x2 + 4x + 1
Z
3 1 1
dx = ln |x 1| + ln |x + 1| ln |x + 3| + C
(x 1)(x + 1)(x + 3) 4 2 4
## 4.5.3 Case 2: Denominator is a product of repeated linear factors
If Q(x) has a factor of (x r)n , then you have the following partial fraction breakdown,
A B C
+ + + ... +
x r (x r)2 (x r)3 (x r)n
x4 2x2 + 4x + 1
Z
Example 4.12. Find dx
x3 x2 x + 1
1. Note the numerator has a larger degree than the numerator. You need to do long
division. After you finish, you should have
152
4.5 Integration of Rational Functions by Partial Fractions Brian E. Veitch
Z
4x
x+1+ dx
x3 x2
x+1
2. We know how to integrate x + 1, so well just focus on the fraction. Set up the fraction
as,
4x A B C
= + +
x3 x2
x+1 x 1 (x 1)2 (x + 1)
## 4x = Ax2 A + Bx + B + Cx2 2Cx + C
4x = (A + C)x2 + (B 2C)x + (A + B + C)
4. Match coefficients
A+C = 0
B 2C = 4
A + B + C = 0
## 6. Rewrite the integral
x4 2x2 + 4x + 1 1
Z Z
1 2
dx = x+1+ + + dx
x3 x2 x + 1 x 1 (x 1) 2 x+1
x4 2x2 + 4x + 1
Z
1 1
3 2
dx = x2 + x + ln |x 1| ln |x + 1| + C
x x x+1 2 x1
153
4.5 Integration of Rational Functions by Partial Fractions Brian E. Veitch
## 4.5.4 Case 3: Denominator has irreducible quadratic factors
P (x)
If Q(x) in has irreducible factors, the partial fraction breakdown will have the term
Q(x)
Ax + B
ax2 + bx + c
where ax2 + bx + c is the irredicuble quadratic.
## You can tell if a quadratic is irreducible if b2 4ac < 0.
2x2 x + 4
Z
Example 4.13. Find dx
x3 + 4x
## 1. Since the degree of the denominator is bigger, we can start by writing
2x2 x + 4 A Bx + C
2
= + 2
x(x + 4) x x +4
## 2x2 x + 4 = A(x2 + 4) + (Bx + C)x
2x2 x + 4 = (A + B)x2 + Cx + 4A
## 3. Match the coefficients
A+B = 2
C = 1
4A = 1
154
4.5 Integration of Rational Functions by Partial Fractions Brian E. Veitch
## 4. Solving the system of equations we get
A = 1, B = 1, C = 1
## 5. Rewrite the integral
2x2 x + 4 1 1x 1
Z Z
dx = + dx
x3 + 4x x x2 + 4
x1
We really cant do much with except to break it up into two separate integrals.
x2 + 4
2x2 x + 4
Z Z
1 x 1
dx = + 2 2 dx
x3 + 4x x x +4 x +4
A quick note:
Z
1 1 1 x
dx = tan
x 2 + a2 a a
and
Z
x
dx requires u-substitution
x2 +4
2x2 x + 4
Z
1 1 2
+ ln x + 4 1 tan1 x + C
dx = ln
x3 + 4x x 2 2 2
155
4.5 Integration of Rational Functions by Partial Fractions Brian E. Veitch
## 4.5.5 Case 4: Denonminator has repeated irreducible quadratic factors
P (x) n
If Q(x) in has a repeated irreducible factor (ax2 + bx + c) , the partial fraction break-
Q(x)
down will have the following terms
Ax + B Cx + D Ex + F
2
+ 2 2
+ + ... +
ax + bx + c (ax + bx + c) (ax + bx + c)3
2 (ax2 + bx + c)n
## where ax2 + bx + c is the irredicuble quadratic.
x2 + x + 1
Z
Example 4.14. Find dx
(x2 + 1)2
## 1. Write your fraction as
x2 + x + 1 Ax + B Cx + D
= +
(x2 + 1)2 x2 + 1 (x2 + 1)2
## 2. Multiply through by (x2 + 1)2
x2 + x + 1 = (Ax + B)(x2 + 1) + Cx + D
x2 + x + 1 = Ax3 + Bx2 + Ax + B + Cx + D
## x2 + x + 1 = Ax3 + Bx2 + (A + C)x + (B + D)
156
4.5 Integration of Rational Functions by Partial Fractions Brian E. Veitch
A = 0
B = 1
A+C = 1
B+D = 1
## 4. Solving this system, we get
A = 0, B = 1, C = 1, D = 0
## 5. Rewrite the integral
x2 + x + 1
Z Z
1 x
dx = + 2 dx
(x2 + 1)2 x2 + 1 (x + 1)2
Z
x
6. dx requires u-substitution.
(x2 + 1)2
Z
x 1
dx = +C
(x2 + 1) 2 2
2(x + 1)
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# Presentation
on Concept Delivery
Concept :
PRIOR KNOWLEDGE
## Quadratic equation in one variable
THE AIM
To teach the students how to solve a quadratic
equation.
INTRODUCTION
Some questions will be asked to check if the students know
Q.
Q.
Q.
x2 3 5
mx 2 4 0
x 23
x 8 2
2x
4
3
## Next the following equations will be given to write in standard form
y2
3 x 4 5Z
4 16 y
3
3 2 5
2
16 x 4 4 x 2
x x 0
7
14 the
7 following two equations:
Now the class will be told to consider
2
z ( z 1) 0
(i )
## Q. What do( ythe
2)equations
( y 3) represent?
(ii )
A. These represent product of linear factors.
## Q. For what values of z is equation (i) true?
A. z = 0 or z = 1
Q. For what values of y is the equation (ii) true?
A. y = 2 or y = -3
Q. We thus got two values of variables satisfying the given
equation. So these are the solutions of the given equation. So
solution set of the given equations are {0,1} and {2,-3}
The class will now be told that they will now begin the study
of the solution of quadratic equation.
## The topic SOLUTION OF QUADRATIC EQUATION will
be written on the board.
DEVELOPMENT
CONCEPT:
DLO:
The students will understand the technique to solve
## TREATMENT AND PROCEDURE
The class will be asked to consider the equation
x 2 7 x 12 0
(iii )
## Let us find the solution of the quadratic equation
Q. How should we proceed?
A. Let us factorize the given equation
Q. How will you factorize x2 + 7x + 12 = 0
A. It can be factorized by splitting the middle term into two terms
x2 + 4x + 3x + 12 = 0
Q. What will be the next step?
A. x (x + 4) + 3 (x + 4) = 0
(x + 4) (x + 3) = 0
or
## Q. Now how will you proceed?
x + 3 = 0 or x + 4 = 0
Q. Which two values of x satisfy these two equations?
A. x = -3 , x = -4
## Tell the class now that -3 and -4 is the solution of given
equation and it is written as {-3, -4}
Tell the class now we look at the principles that govern the
Q. What must be our first step
A. The equation must be in standard form
Q. Suppose the following equation
z2 + 13 = 4z
Q.
A.
Q.
A.
Q.
A.
Q.
A.
Q.
A.
## Is it in the standard form?
No
Why
Because 4z must be on the left side of the equation
Now look at the following equation
5x2 + x -3 = 0
Can you factorize this equation?
No
Why
Because the middle term cannot be splitted into two terms
What are the essential rules to find the solution of quadratic equation?
(a) The equation must be in standard form
(b) If the coefficients of an equation are fractions then convert them in standard
form by multiplying the equation with the L.C.M of the denominators.
BOARD SUMMARY
1.
2.
form?
RECAPITULATION
1.
2.
3.
## Teacher will go through the main points and students
will be asked if they have understood everything we
discussed.
Anything that they like to be revised, will be revised.
On green signal from the class next step of the lesson
will be taken up.
CONSOLIDATION
Q.1
Solve
2y2 + 5y -3 = 0
Q.2
1 2 3
x x 20
Solve 2
2
HOMEWORK
## Ex 1.10 (Page 36)
Q.1 (ii), Q.2 (iii), Q.3 (viii)
CONCLUSION
Today we have discussed the method to solve quadratic
equation by factorization.
Next time we will try to learn the other methods to solve
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Title: Sec Math II
1
Similarity of Triangles
• Sec Math II
• Unit 8 Lesson 3
• Class Notes
Click one of the buttons below or press the
enter key
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2
In geometry, two polygons are similar when one is
a replica (scale model) of the other.
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3
Consider Dr. Evil and Mini Me from Mike Meyers
hit movie Austin Powers. Mini Me is supposed to
be an exact replica of Dr. Evil.
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4
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5
The following are similar figures.
I
II
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6
The following are non-similar figures.
I
II
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7
Feefee the mother cat, lost her daughters, would
daughters have the similar footprint with their
mother.
Feefees footprint
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8
1.
Which of the following is similar to the above
triangle?
B
A
C
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9
Similar triangles are triangles that have the
same shape but not necessarily the same size.
?ABC ? ?DEF
When we say that triangles are similar there are
several repercussions that come from it.
?A ? ?D
?B ? ?E
?C ? ?F
10
Six of those statements are true as a result of
the similarity of the two triangles. However, if
we need to prove that a pair of triangles are
similar how many of those statements do we need?
Because we are working with triangles and the
measure of the angles and sides are dependent on
each other. We do not need all six. There are
three special combinations that we can use to
prove similarity of triangles.
1. PPP Similarity Theorem ? 3 pairs of
proportional sides
2. PAP Similarity Theorem ? 2 pairs of
proportional sides and congruent angles between
them
3. AA Similarity Theorem ? 2 pairs of
congruent angles
11
1. PPP Similarity Theorem ? 3 pairs of
proportional sides
?ABC ? ?DFE
12
2. PAP Similarity Theorem ? 2 pairs of
proportional sides and congruent angles between
them
m?H m?K
?GHI ? ?LKJ
13
The PAP Similarity Theorem does not work
unless the congruent angles fall between the
proportional sides. For example, if we have the
situation that is shown in the diagram below, we
cannot state that the triangles are similar. We
do not have the information that we need.
Angles I and J do not fall in between sides GH
and HI and sides LK and KJ respectively.
14
3. AA Similarity Theorem ? 2 pairs of
congruent angles
m?N m?R
?MNO ? ?QRP
m?O m?P
15
It is possible for two triangles to be similar
when they have 2 pairs of angles given but only
one of those given pairs are congruent.
m?T m?X
m?S m?Z
m?S 180?- (34? 87?)
?TSU ? ?XZY
m?S 180?- 121?
m?S 59?
16
Note One triangle is a scale model of the other
triangle.
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17
How do we know if two triangles are similar or
proportional?
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18
Triangles are similar () if corresponding angles
are equal and the ratios of the lengths of
corresponding sides are equal.
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19
Interior Angles of Triangles
The sum of the measure of the angles of a
triangle is 1800. Ð A Ð B ÐC 1800
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20
Example 6-1b
Determine whether the pair of triangles is
Answer Since the corresponding angles have
equal measures, the triangles are similar.
21
If the product of the extremes equals the product
of the means then a proportion exists.
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22
This tells us that ? ABC and ? XYZ are similar
and proportional.
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23
Q Can these triangles be similar?
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24
AnswerYes, right triangles can also be similar
but use the criteria.
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25
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26
Do we have equality?
This tells us our triangles are not similar. You
cant have two different scaling factors!
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27
If we are given that two triangles are similar or
proportional what can we determine about the
triangles?
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28
The two triangles below are known to be similar,
determine the missing value X.
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29
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30
In the figure, the two triangles are similar.
What are c and d ?
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31
In the figure, the two triangles are similar.
What are c and d ?
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32
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33
Two triangles are called similar if their
corresponding angles have the same measure.
?
?
?
?
?
?
34
Two triangles are called similar if their
corresponding angles have the same measure.
?
Ratios of corresponding sides are equal.
C
A
?
a
c
?
?
?
?
b
B
a
A
b
B
c
C
35
Mary is 5 ft 6 inches tall. She casts a 2 foot
is the tree?
36
Mary is 5 ft 6 inches tall. She casts a 2 foot
is the tree?
Marys height
Trees height
x
5.5
2
7
37
Mary is 5 ft 6 inches tall. She casts a 2 foot
is the tree?
5.5
x
2
7
Marys height
Trees height
x
5.5
2
7
38
5.5
x
2
7
7 ( 5.5 ) 2 x 38.5 2 x
x 19.25
The height of the tree is 19.25 feet
39
Congruent Figures
• In order to be congruent, two figures must be the
same size and same shape.
40
Similar Figures
• Similar figures must be the same shape, but their
sizes may be different.
41
Similar Figures
• This is the symbol that means similar.
• These figures are the same shape but different
sizes.
42
SIZES
• Although the size of the two shapes can be
different, the sizes of the two shapes must
differ by a factor.
4
2
6
6
3
3
2
1
43
SIZES
• In this case, the factor is x 2.
4
2
6
6
3
3
2
1
44
SIZES
• Or you can think of the factor as 2.
4
2
6
6
3
3
2
1
45
Enlargements
• When you have a photograph enlarged, you make a
similar photograph.
X 3
46
Reductions
• A photograph can also be shrunk to produce a
slide.
4
47
Determine the length of the unknown side.
15
12
?
4
3
9
48
These triangles differ by a factor of 3.
15 3 5
15
12
?
4
3
9
49
Determine the length of the unknown side.
?
2
24
4
50
These dodecagons differ by a factor of 6.
?
2 x 6 12
2
24
4
51
Sometimes the factor between 2 figures is not
obvious and some calculations are necessary.
15
12
10
8
18
12
?
52
To find this missing factor, divide 18 by 12.
15
12
10
8
18
12
?
53
18 divided by 12 1.5
54
The value of the missing factor is 1.5.
15
12
10
8
18
12
1.5
55
When changing the size of a figure, will the
angles of the figure also change?
?
40
?
?
70
70
56
Nope! Remember, the sum of all 3 angles in a
triangle MUST add to 180 degrees. If the size of
the angles were increased, the sum would
exceed 180 degrees.
40
40
70
70
70
70
57
We can verify this fact by placing the smaller
triangle inside the larger triangle.
40
40
70
70
70
70
58
The 40 degree angles are congruent.
40
70
70
70
70
59
The 70 degree angles are congruent.
40
40
70
70
70
70
70
60
The other 70 degree angles are congruent.
4
40
70
70
70
70
70
61
Find the length of the missing side.
50
?
30
6
40
8
62
This looks messy. Lets translate the two
triangles.
50
?
30
6
40
8
63
Now things are easier to see.
50
30
?
6
40
8
64
The common factor between these triangles is 5.
50
30
?
6
40
8
65
So the length of the missing side is?
66
Thats right! Its ten!
50
30
10
6
40
8
67
Similarity is used to answer real life questions.
• Suppose that you wanted to find the height of
this tree.
68
Unfortunately all that you have is a tape
measure, and you are too short to reach the top
of the tree.
69
You can measure the length of the trees shadow.
10 feet
70
10 feet
2 feet
71
If you know how tall you are, then you can
determine how tall the tree is.
6 ft
10 feet
2 feet
72
The tree must be 30 ft tall. Boy, thats a tall
tree!
6 ft
10 feet
2 feet
73
Similar figures work just like equivalent
fractions.
30
5
11
66
74
These numerators and denominators differ by a
factor of 6.
30
5
6
11
6
66
75
Two equivalent fractions are called a proportion.
30
5
11
66
76
Similar Figures
• So, similar figures are two figures that are the
same shape and whose sides are proportional.
77
Practice Time!
78
1) Determine the missing side of the triangle.
?
9
5
?
3
4
12
79
1) Determine the missing side of the triangle.
15
9
5
?
3
4
12
80
2) Determine the missing side of the triangle.
36
36
6
6
?
4
?
81
2) Determine the missing side of the triangle.
36
36
6
6
?
4
24
82
3) Determine the missing sides of the triangle.
39
?
?
33
?
8
24
83
3) Determine the missing sides of the triangle.
39
13
?
33
11
8
24
84
4) Determine the height of the lighthouse.
?
?
8
2.5
10
85
4) Determine the height of the lighthouse.
?
32
8
2.5
10
86
5) Determine the height of the car.
?
?
3
5
12
87
5) Determine the height of the car.
7.2
?
3
5
12
|
# Calculus 2 : Area Under a Curve
## Example Questions
← Previous 1 3 4 5 6
### Example Question #1 : Area Under A Curve
Find the area under the curve from to .
Explanation:
The area under a curve f(x) between two x-values corresponds to the integral
.
In this case, , and we can find by the fundamental theorem of calculus that...
evaluated from to
### Example Question #2 : Area Under A Curve
Find the area between and , such that and
Explanation:
The first step to find the area between two curves is to set them equal to each other and find their intersection points.
The next step is to find which curve is "on the top" and which is "on the bottom" inside the interval . We find that is on the top and is on the bottom.
Now, to find the area between the two curves, we take the integral from to of the top curve minus the bottom curve.
evaluated from to
### Example Question #2 : Area Under A Curve
Find the total area under the curve from
Explanation:
The definite integral of f(x) over the interval [0, 5] is the area under the curve.
A substitution makes this integral easier to evaluate. Let . Then . We can also change the limits of integration so that they are in terms of u. When x=5, . When x=0, . Making these substitutions, the integral becomes
### Example Question #4 : Area Under A Curve
What is the area under the curve over the interval ?
Explanation:
The area under a curve over the interval is
In this example, this leads to the definite integral
.
A substitution makes the antiderivative of this function more obvious.
Let
.
We can also convert the limits of integration to be in terms of to simplify evaluation. When , and when .
Making these substitutions results in
.
Recall that in general , so evaluating the integral leads to
### Example Question #5 : Area Under A Curve
Find the following indefinite integral:
Explanation:
In order to find the anti-derivative, lets first factor the denominator.
Now we use partial-fraction decomposition to get this into a form we know how to take the anti-derivative.
Now we need to find the values of A and B. First lets write out the equation.
Now we can get our system of equations in order to find A and B.
From equation 1, we get
Now we can substitute this into equation 2.
Now we substitue B back into equation 1.
Now we can put the values of A, and B into our problem.
Now we can simply take the anti-derivative.
### Example Question #6 : Area Under A Curve
Evaluate:
Explanation:
To evaluate the integral, we use inverse power law.
Remember that the inverse power law is
.
Lets apply this to our problem.
From here we plug in the bounds and subtract the lower bound function value from the upper bound function value.
### Example Question #7 : Area Under A Curve
Find the area under the curve of the function between and .
Explanation:
Asking for the area under the curve means you are going to be doing an integral of the function provided. The bounds are given for you. The integral thus looks like this: .
When integrated, you get
evaluated at and .
At the expression evaluates to and at it evalutes to .
### Example Question #8 : Area Under A Curve
Determine the area under the curve of
.
Explanation:
For this particular function we will need to preform a "u-substitution".
In our case let
which will make .
Now we will substitute these into our integral to get the following.
if
Then we integrate using the power rule which states,
Now plug back in the original variable and then subtract the function values at the bounds.
### Example Question #3 : Area Under A Curve
Find the area under the curve of the following function from to :
Explanation:
To find the area under the curve, we must integrate over the given bounds:
To integrate, we must do the following substitution:
The derivative was found using the following rule:
Now, rewrite the integral, and change the bounds in terms of u:
Note that during the rewriting, the bounds changed to the upper bound being -1 and the lower bound being 1, but the negative sign that came from the derivative of u allowed us to make the bounds as they are seen above.
Now perform the definite integration:
The integral was found using the following rule:
and the definite integration was finished by plugging in the upper bound into the resulting function, plugging the lower bound into the resulting function, and subtracting the two, as seen above.
### Example Question #9 : Area Under A Curve
Find the area underneath the curve of the function
on the interval
square units
square units
square units
square units
square units
Explanation:
To find the area underneath the curve of the function on the interval , we find the definite integral
Because the function in this problem is always positive on the interval, or
on the interval the area underneath the curve can be found using the definite integral
and rewriting the function the definite integral is
Using the inverse power rule which states
we find the definite integral to be
And by the corollary of the Fundamental Theorem of Calculus, the definite integral equals
As such, the area is square units
← Previous 1 3 4 5 6
|
# What percentage is 20 out of 69?
Now we can see that our fraction is 28.985507246377/100, which means that 20/69 as a percentage is 28.9855%.
## What is 88 out of 90 as a percentage?
What is this? Now we can see that our fraction is 97.777777777778/100, which means that 88/90 as a percentage is 97.7778%.
## How do I deduct 20% from a figure?
To subtract any percentage from a number, simply multiply that number by the percentage you want to remain. In other words, multiply by 100 percent minus the percentage you want to subtract, in decimal form. To subtract 20 percent, multiply by 80 percent (0.8).
## How do you take 20% off in Excel?
If you want to calculate a percentage of a number in Excel, simply multiply the percentage value by the number that you want the percentage of. For example, if you want to calculate 20% of 500, multiply 20% by 500.
## What number is 75 of 52?
So, how did we get to the solution that 75 percent of 52 = 39? Step 1: As we know, a whole of something is equal to 100%. In this case, we want to find what 75% of 52 is. We know that 100% of 52 is, well, just 52.
## What is 1/8 as a decimal?
To convert 1/8 to a decimal, divide the denominator into the numerator. 1 divided by 8 = . 125. To convert the decimal .
## When one finds what percent one number is of another he looks for?
To find what percent one number is of another: Multiply or divide both terms so that the Base becomes 100.
## What percentage is 30 out of 120?
Solution and how to convert 30 / 120 into a percentage 0.25 times 100 = 25. That’s all there is to it!
## What percentage is 60 out of 80?
60/80 x 100 = 6/8 = 3/4 =75% of 100! that’s how you work out the percentage. you have 20 left out of 80 when you take 60.
## What is a 16 out of 20 on a quiz?
Now we can see that our fraction is 80/100, which means that 16/20 as a percentage is 80%. And there you have it!
## What is 40 as a percentage of 8?
Now we can see that our fraction is 20/100, which means that 8/40 as a percentage is 20%. And there you have it! Two different ways to convert 8/40 to a percentage.
## What percent is 48 out of 120?
Solution and how to convert 48 / 120 into a percentage 0.4 times 100 = 40. That’s all there is to it!
## What percent is 125 out of 200?
Solution and how to convert 125 / 200 into a percentage 0.62 times 100 = 62.5. That’s all there is to it!
## What percent is 89 out of 90?
What is this? Now we can see that our fraction is 98.888888888889/100, which means that 89/90 as a percentage is 98.8889%.
## What percentage is 18 out of 120?
What is this? Now we can see that our fraction is 15/100, which means that 18/120 as a percentage is 15%.
## How do you break down percentages?
To calculate a percentage of a percentage, convert both percentages to fractions of 100, or to decimals, and multiply them. For example, 50% of 40% is calculated; (50/100) x (40/100) = 0.50 x 0.40 = 0.20 = 20/100 = 20%.
## How do you calculate 20 percent?
First, convert the percentage discount to a decimal. A 20 percent discount is 0.20 in decimal format. Secondly, multiply the decimal discount by the price of the item to determine the savings in dollars. For example, if the original price of the item equals \$24, you would multiply 0.2 by \$24 to get \$4.80.
|
# Average – Definition – How to Calculate Average
## What is an Average?
The word “average” is used in everyday life to describe where the middle number of a data set is. It’s the typical number you would expect to find in a series of numbers. In statistics, the average is called the “arithmetic mean,” usually just shortened to the mean. Both the average and the mean use the same formula:
avg = total sum of all the numbers / number of items in the set.
In other words, to find the average, add up all of the numbers in the set, and then divide by however many items you have. Let’s say you have 5,10, and 15. Add them all up to get 5 + 10 + 15 = 30, then divide by 3(the number of items). The answer is 30 / 3 = 10.
## How to Calculate Average: Examples
Example 1: You earned \$129, \$139, \$155 and \$176 over the last 4 weeks. What is your average pay?
Step 1: Add up all of the numbers in the set. \$129 + \$139 + \$155 + \$176 = \$599.
Step 2: Divide Step 1 by the total number of items in the set. There are 4 items in the set, so \$599 / 4 = \$149.75.
Example 2: You have semester grades of B, C, D, A, B and B. What is your average grade?
Step 1: Add up all of the numbers in the set. You have grades here, so you need to convert them on a 4.0 scale:
B = 3.0
C = 2.0
D = 1.0
A = 4.0
B = 3.0
B = 3.0
So we have: 3.0 + 2.0 + 1.0 + 4.0 + 3.0 + 3.0 = 16.0.
Step 2: Divide Step 1 by the total number of items in the set. There are 6 items in the set, so 16.0/6 = 2.66.
## A More Formal Definition
In probability and statistics, you’ll see the following formula used:
Don’t let the formula scare you! The summation sign (Σ) just means to “add up” and the letter n stands for the number of items in the set. Taking example 1 from above, the formula could be used to find the same answer:
AM = 1/4 * (\$129 + \$139 + \$155 + \$176 = \$599.
------------------------------------------------------------------------------
Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free!
|
## Sunday, 22 July 2012
### Trigonometry Activity 1
Group Activity 1
1. Farrell, Dylan, Bryan, Sherwin, Harindrar, Edwin7/24/2012 10:03 am
Solution 1: Take a picture of a ruler beside the flag pole. Then use a photo editing software to copy, paste and stack up the ruler until it reaches the height of the flagpole. Read the length from the ruler.
Solution 2: Put a ruler a certain distance away from the flagpole. Imagine a line from the top of the flagpole to the top of the ruler, to the floor. Mark the point where the line hits the floor. Measure to ratio of the distance of the flagpole to the point to the distance of the ruler to the point. The ratio of the height of the ruler to the height of the flag pole is the same ratio.
2. Yan Feng,Bowen,Praveen,Ryan,Darshan
Solution 1:The triangle of interest is FAB. We have measured two angles of this
triangle: angle A (measured elevation from A), angle B (equal to 180
deg - elevation from B), and angle F (180 deg - A - B).
Using the Law of Sines, we can write
dist(BA) dist(FA)
------- = ------- this allows us to calculate the
sin(F) sin(B) distance FA.
And then we apply trigonometry to the right triangle FOA:
Solution 2:
Find a stick the length of your arm. Hold your arm out straight with the stick pointing straight up (90 degree angle to your outstretched arm). Walk backwards until you see the tip of the stick line up with the top of the tree. Your feet are now at approximately the same distance from the tree as it is high. (Provided the tree is significantly taller than you are, and the ground is relatively level.)
3. This comment has been removed by the author.
4. Solution: Ask a person to to stand beside the flagpole, take a picture. Then measure the the height of the person and the flagpole in the picture and then you can find out the scale ratio by the the real height of the person and the height o the person int he image. Using that ratio you can find out the height of the flagpole.
Image (object) Image (pole) Real (object) Real (pole)
1.4cm 4.6cm 1.75m 5.75m
1.8cm 6.15cm 1.565m 5.35m
2.1cm 6.4cm 1.8m 5.48m
Done by : Chinni,Wei qin,wai kit,jereme,lindsay,shiying
|
# 2015 AMC 12A Problems/Problem 13
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A league with 12 teams holds a round-robin tournament, with each team playing every other team exactly once. Games either end with one team victorious or else end in a draw. A team scores 2 points for every game it wins and 1 point for every game it draws. Which of the following is NOT a true statement about the list of 12 scores?
$\textbf{(A)}\ \text{There must be an even number of odd scores.}\\ \qquad\textbf{(B)}\ \text{There must be an even number of even scores.}\\ \qquad\textbf{(C)}\ \text{There cannot be two scores of }0\text{.}\\ \qquad\textbf{(D)}\ \text{The sum of the scores must be at least }100\text{.}\\ \qquad\textbf{(E)}\ \text{The highest score must be at least }12\text{.}$
## Solution
We can eliminate answer choices $\textbf{(A)}$ and $\textbf{(B)}$ because there are an even number of scores, so if one is false, the other must be false too. Answer choice $\textbf{(C)}$ must be true since every team plays every other team, so it is impossible for two teams to lose every game. Answer choice $\textbf{(D)}$ must be true since each game gives out a total of two points, and there are $\binom{12}{2} = 66$ games, for a total of $132$ points. Answer choice $\boxed{\textbf{(E)}}$ is false. If everyone draws each of their 11 games, then every team will tie for first place with 11 points each.
|
# How do you simplify 8^(2/3)?
Jul 24, 2016
$= 4$
#### Explanation:
${8}^{\frac{2}{3}}$ can be written as
root3(8^2
$= \sqrt[3]{64}$
=root3((4)(4)(4)
$= 4$
Jun 30, 2017
$4$
#### Explanation:
One of the laws of indices states: ${x}^{\frac{m}{n}} = \sqrt[n]{\left({x}^{m}\right)} = {\left(\sqrt[n]{x}\right)}^{m}$
Therefore ${8}^{\frac{2}{3}}$ can be written as $\sqrt[3]{\left({8}^{2}\right)} \mathmr{and} {\left(\sqrt[3]{8}\right)}^{2}$
I prefer to use the second form because it uses smaller numbers - the root is found first and then that is squared.
$\sqrt[3]{8} = 2 \mathmr{and} {2}^{2} = 4$
So: ${\left(\textcolor{b l u e}{\sqrt[3]{8}}\right)}^{2} = {\textcolor{b l u e}{2}}^{2} = 4$
Consider a question such as ${32}^{\frac{3}{5}}$
$\sqrt[5]{\textcolor{b l u e}{\left({32}^{3}\right)}}$ would mean finding ${32}^{3}$ first ... ouch!
${\left(\textcolor{b l u e}{\sqrt[5]{32}}\right)}^{3}$ would mean finding $\sqrt[5]{32}$ first. That is $\textcolor{b l u e}{2}$
${\left(\textcolor{b l u e}{\sqrt[5]{32}}\right)}^{3} = {\textcolor{b l u e}{2}}^{3} = 8$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 4.17: Points in the Coordinate Plane
Difficulty Level: Basic Created by: CK-12
Estimated5 minsto complete
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Progress
Practice Points in the Coordinate Plane
Progress
Estimated5 minsto complete
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Megan is on vacation in Rome and is trying to find the restaurant where she will meet her friends for lunch. She got a map from her hotel that is formatted like a coordinate grid. The hotel is at the origin of the map and she was told that the restaurant is at the point (4,5)\begin{align*}(4, 5)\end{align*} on the map. How can Megan find this location on the map so she can make it to the restaurant?
In this concept, you will learn how to name and graph ordered pairs of integer coordinates in a coordinate plane.
### Naming and Graphing Points in the Coordinate Plane
The coordinate plane is a grid created by a horizontal number line, the x\begin{align*}x\end{align*}-axis, intersecting a vertical number line, the y\begin{align*}y\end{align*}-axis. The point of intersection is called the origin.
The coordinate plane allows you to describe locations in two-dimensional space.
Each point on the coordinate plane can be named by a pair of numbers called an ordered pair in the form (x,y)\begin{align*}(x,y)\end{align*}.
• The first number in an ordered pair identifies the x\begin{align*}x\end{align*}-coordinate of the point. This coordinate describes how far to the right or left of the y\begin{align*}y\end{align*}-axis a point is.
• The second number in an ordered pair identifies the y\begin{align*}y\end{align*}-coordinate of the point. This coordinate describes how far above or below the x\begin{align*}x\end{align*}-axis a point is.
Here is an example.
Plot the point (3,4)\begin{align*}(3,-4)\end{align*} on the coordinate plane.
This point has an x\begin{align*}x\end{align*}-coordinate of 3 and a y\begin{align*}y\end{align*}-coordinate of -4.
To plot this point, first start at the origin.
Then, find the location of the x\begin{align*}x\end{align*}-coordinate. Because the x\begin{align*}x\end{align*}-coordinate is positive, you will be moving to the right. Move to the right along the x\begin{align*}x\end{align*}-axis 3 units until you find 3.
Next, look at your y\begin{align*}y\end{align*}-coordinate. Because the y\begin{align*}y\end{align*}-coordinate is negative, you will be moving down. Move down from the 3 on the x\begin{align*}x\end{align*}-axis 4 units until you are lined up with the -4 on the y\begin{align*}y\end{align*}-axis.
The answer is shown plotted on the coordinate plane below.
Here is another example.
Give the ordered pair for the point plotted below.
To write the ordered pair you need both the x\begin{align*}x\end{align*}-coordinate and the y\begin{align*}y\end{align*}-coordinate.
Start at the origin. You need to figure out how far to the right/left you need to move and then how far up/down you need to move to reach your point.
First notice that you need to move 5 units to the right from the origin to reach the point. This means the x\begin{align*}x\end{align*}-coordinate is 5.
Next notice that you do not need to move up or down from the x\begin{align*}x\end{align*}-axis at all to reach the point. This means the y\begin{align*}y\end{align*}-coordinate is 0.
The answer is that the ordered pair is (5,0)\begin{align*}(5, 0)\end{align*}.
### Examples
#### Example 1
Earlier, you were given a problem about Megan, who is on vacation in Rome.
She's at her hotel and trying to use a map to find the location of the restaurant where she will be meeting her friends for lunch. The restaurant is at the point (4,5)\begin{align*}(4, 5)\end{align*} on the map.
This point has an x\begin{align*}x\end{align*}-coordinate of 4 and a y\begin{align*}y\end{align*}-coordinate of 5.
To find this point, Megan should start at the origin on the map (her hotel).
Then, she should find the location of the x\begin{align*}x\end{align*}-coordinate. Because the x\begin{align*}x\end{align*}-coordinate is positive, she will be moving to the right. She needs to move to the right 4 units until she reaches the 4 on the x\begin{align*}x\end{align*}-axis.
Next, Megan should look at her y\begin{align*}y\end{align*}-coordinate. Because the y\begin{align*}y\end{align*}-coordinate is positive, she will be moving up. She should move up 5 units from the 4 on the x\begin{align*}x\end{align*}-axis.
The answer is shown plotted on the coordinate plane below.
#### Example 2
This coordinate grid shows locations in Jimmy’s city. Name the ordered pair that represents the location of the city park.
To write the ordered pair you need both the x\begin{align*}x\end{align*}-coordinate and the y\begin{align*}y\end{align*}-coordinate.
Start at the origin. You need to figure out how far to the right/left you need to move and then how far up/down you need to move to reach your point.
First notice that you need to move 2 units to the left from the origin in order to be exactly above the point for the city park. This means the x\begin{align*}x\end{align*}-coordinate is -2.
Next notice that you need to move down 6 units from the -2 on the x\begin{align*}x\end{align*}-axis to reach the point. This means that the y\begin{align*}y\end{align*}-coordinate is -6.
The arrows below show how you should have moved your finger to find the coordinates.
The answer is that the ordered pair for the city park is (2,6)\begin{align*}(-2, 6)\end{align*}.
#### Example 3
Plot the point (0,3)\begin{align*}(0, 3)\end{align*} on the coordinate plane.
This point has an x\begin{align*}x\end{align*}-coordinate of 0 and a y\begin{align*}y\end{align*}-coordinate of 3.
To plot this point, first start at the origin.
Then, find the location of the x\begin{align*}x\end{align*}-coordinate. Because the x\begin{align*}x\end{align*}-coordinate is 0, you do not need to move to the right or to the left from the origin. Stay at the origin.
Next, look at your y\begin{align*}y\end{align*}-coordinate. Because the y\begin{align*}y\end{align*}-coordinate is positive you will be moving up. Move up from the origin 3 units until you are lined up with the 3 on the y\begin{align*}y\end{align*}-axis.
The answer is shown plotted on the coordinate plane below.
#### Example 4
Give the ordered pair for the point plotted below.
To write the ordered pair you need both the x\begin{align*}x\end{align*}-coordinate and the y\begin{align*}y\end{align*}-coordinate.
Start at the origin. You need to figure out how far to the right/left you need to move and then how far up/down you need to move to reach your point.
First notice that you need to move 6 units to the right from the origin to be exactly above the point. This means the x\begin{align*}x\end{align*}-coordinate is 6.
Next notice that you need to move down 3 units from the 6 on the \begin{align*}x\end{align*}-axis to reach your point. This means your \begin{align*}y\end{align*}-coordinate is -3.
The answer is that the ordered pair is \begin{align*}(6, -3)\end{align*}.
#### Example 5
Plot the point \begin{align*}(-2, -5)\end{align*} on the coordinate plane.
This point has an \begin{align*}x\end{align*}-coordinate of -2 and a \begin{align*}y\end{align*}-coordinate of -5.
To plot this point, first start at the origin.
Then, find the location of the \begin{align*}x\end{align*}-coordinate. Because the \begin{align*}x\end{align*}-coordinate is negative, you will be moving to the left. Move to the left 2 units until you reach the -2 on the \begin{align*}x\end{align*}-axis.
Next, look at your \begin{align*}y\end{align*}-coordinate. Because the \begin{align*}y\end{align*}-coordinate is negative, you will be moving down. Move down 5 units from the -2 on the \begin{align*}x\end{align*}-axis.
The answer is shown plotted on the coordinate plane below.
### Review
1. Name the ordered pair that represents each of these points on the coordinate plane.
Below is a map of an amusement park. Name the ordered pair that represents the location of each of these rides.
1. roller coaster
2. Ferris wheel
3. carousel
4. log flume
Name the ordered pairs that represent the vertices of triangle FGH.
1. F
2. G
3. H
Name the ordered pairs that represent the vertices of pentagon ABCDE.
1. A
2. B
3. C
4. D
5. E
6. On the grid below, plot point \begin{align*}V\end{align*} at \begin{align*}(-6, 4)\end{align*}.
1. On the grid below, plot point a triangle with vertices \begin{align*}R(4, -1), S(4, -4)\end{align*} and \begin{align*}T (-3, -4)\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
$x-$axis
The $x-$axis is the horizontal axis in the coordinate plane, commonly representing the value of the input or independent variable.
$x-$coordinate
The $x-$coordinate is the first term in a coordinate pair, commonly representing the value of the input or independent variable.
$y-$axis
The $y-$axis is the vertical number line of the Cartesian plane.
Abscissa
The abscissa is the $x-$coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 3 is the abscissa.
Cartesian Plane
The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.
Coordinate Plane
The coordinate plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin. The coordinate plane is also called a Cartesian Plane.
Ordinate
The ordinate is the $y$-coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 7 is the ordinate.
Origin
The origin is the point of intersection of the $x$ and $y$ axes on the Cartesian plane. The coordinates of the origin are (0, 0).
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# Test yourself now
High marks in science are the key to your success and future plans. Test yourself and learn more on Siyavula Practice.
## 10.6 Constructing triangles
In this section, you will learn how to construct different triangles using given information. A triangle has three sides and three interior angles. However, we do not need to know the lengths of all three sides and the sizes of all three angles in order to construct a triangle.
To construct a triangle, we only need to know three properties of the triangle. Any of these three combinations provides enough information to construct a triangle:
• the lengths of all three sides (side, side, side)
• the sizes of two interior angles and the length of the side between them (angle, side, angle)
• the lengths of two sides and the size of the angle between them (side, angle, side).
It is helpful to remember the following important properties of triangles:
• The longest side of a triangle lies opposite the biggest angle of the triangle. (It is impossible to construct a triangle where the longest side is not opposite the biggest angle.)
• The shortest side of a triangle lies opposite the smallest angle of the triangle.
• The sum of the lengths of the two shorter sides of the triangle must be greater than the length of the longest side of the triangle.
### Constructing a triangle given two sides and an included angle
In this section, you will construct triangles for which you are given two sides as well as the included angle.
included angle
the angle between two sides of a triangle
The diagram below illustrates an included angle.
Before starting a construction, it is a good idea to draw a rough sketch of the triangle. Label the sketch correctly and write down all the given information.
## Worked Example 10.11: Constructing a triangle given two sides and the included angle
Construct $$\triangle ABC$$ with $$AB = 4 \text{ cm}$$, $$A\hat{B}C = 65^{\circ}$$ and $$BC = 8 \text{ cm}$$. You may use a protractor for this construction.
### Draw a rough sketch of the triangle.
Read the given information carefully and fill everything in on the rough sketch. Label the vertices of the triangle.
### Draw a line segment $$BC$$.
Draw a line segment (longer than $$8 \text{ cm}$$) and mark point $$B$$. Use the compass to measure a length of $$8 \text{ cm}$$ on your ruler.
Keep the compass width the same and place the compass on point $$B$$. Then draw a small arc to mark off a length of $$8\text{ cm}$$. Label the point $$C$$.
### Use a protractor to measure and draw the included angle.
Put the origin of the protractor on point $$B$$. Use the inner scale of the protractor to find $$65^{\circ}$$. Make a small dot at the edge of the protractor. Then draw a line through the dot and point $$B$$.
### Measure the length of the second side of the triangle.
Use the compass to measure a length of $$4 \text{ cm}$$ on your ruler. Keeping the compass width the same, place the compass on point $$B$$ and draw a small arc to mark off a length of $$4 \text{ cm}$$. Label the point $$A$$.
### Draw the third side of the triangle.
Use a ruler to join points $$A$$ and $$C$$.
This completes the construction of $$\triangle ABC$$.
• Never erase the construction lines after you have completed a construction. These may be important for giving marks for your work.
• The constructions provided here may not display at the exact size on all devices. Check your own constructions to make sure they are correct.
### Construct triangles given two sides and the included angle
Construct $$\triangle DEF$$ with $$DE = 6 \text{ cm}$$, $$\hat{E} = 50^{\circ}$$ and $$EF = 7 \text{ cm}$$.
Construct $$\triangle PQR$$ with $$PQ = 7 \text{ cm}$$, $$\hat{Q} = 120^{\circ}$$ and $$QR = 5 \text{ cm}$$.
Construct $$\triangle XYZ$$ with $$XY = 2 \text{ cm}$$, $$\hat{Q} = 90^{\circ}$$ and $$YZ = 8 \text{ cm}$$.
Make an accurate construction of $$\triangle KLM$$.
Make an accurate construction of $$\triangle STU$$.
Make an accurate construction of $$\triangle VWX$$.
### Construct a triangle given two angles and the side between them
In this section, you will construct triangles for which you are given two angles and the length of the side between the two angles.
## Worked Example 10.12: Constructing a triangle given two angles and the side between them
Construct $$\triangle ABC$$ with $$\hat{B} = 35^{\circ}$$, $$\hat{C} = 70^{\circ}$$ and $$BC = 7 \text{ cm}$$. You may use a protractor for this construction.
### Draw a rough sketch of the triangle.
Read the given information carefully and fill everything in on the rough sketch. Label the vertices of the triangle.
### Construct the line segment $$BC$$.
Draw a line segment (longer than $$7 \text{ cm}$$) and mark point $$B$$. Use the compass to measure a length of $$7 \text{ cm}$$ on your ruler.
Keeping the compass width the same, place the compass on point $$B$$ and make an arc on the line. Label point $$C$$.
### Use a protractor to measure and draw $$\hat{B}$$.
Put the origin of the protractor on point $$B$$. Use the inner scale of the protractor to find $$35^{\circ}$$. Make a small dot at the edge of the protractor and then draw a line through the dot and point $$B$$.
### Use a protractor to measure and draw $$\hat{C}$$.
Put the origin of the protractor on point $$C$$. On the protractor, use the outer scale from the left and find $$70^{\circ}$$. Make a small dot at the edge of the protractor.
### Draw the line segment $$CA$$.
Use a ruler to draw a line through the dot and point $$C$$. Label the point where the two lines intersect, point $$A$$.
This completes the construction of $$\triangle ABC$$.
### Construct a triangle given two angles and the side between them
Construct $$\triangle DEF$$ with $$\hat{E} = 50^{\circ}$$, $$EF = 8 \text{ cm}$$ and $$\hat{F} = 60^{\circ}$$. You may use a protractor.
Construct $$\triangle PQR$$ with $$\hat{R} = 130^{\circ}$$, $$\hat{Q} = 30^{\circ}$$ and $$RQ = 4 \text{ cm}$$. You may use a protractor.
Use a compass and a ruler to construct $$\triangle XYZ$$ with $$YZ = 6 \text{ cm}$$, $$\hat{Y} = 60^{\circ}$$ and $$\hat{Z}$$, a right angle.
Make an accurate construction of $$\triangle DTM$$ as shown below. You may use a protractor. (Note that the diagram is not to scale.)
Make an accurate construction of $$\triangle JME$$ as shown below. You may use a protractor. (The given diagram is not to scale.)
Use a compass and a ruler to construct $$\triangle CDE$$ as shown below. (The given diagram is not to scale.)
Why is it impossible to construct $$\triangle KLM$$ if $$\hat{L} = 95^{\circ}$$, $$LM = \text{7,5} \text{ cm}$$ and $$\hat{M} = 110^{\circ}$$?
## Worked Example 10.13: Constructing a triangle given all three sides
Construct $$\triangle ABC$$ with $$AB = 6 \text{ cm}$$, $$BC = 5 \text{ cm}$$ and $$CA = 7 \text{ cm}$$.
### Draw a rough sketch of the triangle.
Read the given information carefully and draw a rough sketch. Label the vertices of the triangle and the lengths of the sides.
### Draw a line segment $$BC$$.
Draw a line segment (longer than $$5 \text{ cm}$$) and mark point $$B$$. Use a compass to measure a length of $$5 \text{ cm}$$ on your ruler. Keeping the compass width the same, place the compass on point $$B$$ and draw a small arc to mark off a length of $$5 \text{ cm}$$. Label the point $$C$$.
### Draw an arc to mark the length of side $$AB$$.
Use the compass to measure a length of $$6 \text{ cm}$$ on your ruler. Keeping the compass width the same, place the compass on point $$B$$ and draw an arc above line $$BC$$ and between $$B$$ and $$C$$.
### Construct an arc to mark the length of side $$CA$$.
Use your compass to measure a length of $$7 \text{ cm}$$ on your ruler. Keeping the width of the compass the same, place the compass on point $$C$$ and make a third arc that intersects the second arc. Label the point of intersection $$A$$.
### Draw $$AB$$ and $$CA$$ to complete $$\triangle ABC$$.
Use a ruler to draw $$AB$$ and $$CA$$.
### Construct a triangle given all three sides
Construct these triangles. Note that diagrams are not to scale.
1. $$\triangle ABC$$ with sides $$AB = 6 \text{ cm}$$, $$BC = 7 \text{ cm}$$ and $$CA = 4 \text{ cm}$$.
2. $$\triangle KLM$$ with sides $$KL = 10 \text{ cm}$$, $$LM = 5 \text{ cm}$$ and $$MK = 8 \text{ cm}$$.
3. $$\triangle PQR$$ with sides $$PQ = 5 \text{ cm}$$, $$QR = 9 \text{ cm}$$ and $$RP = 11 \text{ cm}$$.
4. Use a ruler and a compass to construct $$\triangle PQR$$.
5. Use a ruler and a compass to construct $$\triangle KMN$$.
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Question
# Use the family in Problem 1 to find a solution of y+y''=0 that satisfies the boundary conditions y(0)=0,y(1)=1.
Second order linear equations
Use the family in Problem 1 to find a solution of $$y+y''=0$$ that satisfies the boundary conditions $$y(0)=0,y(1)=1.$$
2021-05-19
We have to find the solution of the given initial value problem, $$y+y''=0, \ y(0)=0,\ y(1)=1$$
The differential equation can be written as, $$\displaystyle{\left({D}^{{2}}+{1}\right)}{y}={0}\ldots.{\left({1}\right)}$$
where $$\displaystyle{D}==\frac{{d}}{{\left.{d}{x}\right.}},{D}^{{2}}==\frac{{{d}^{{2}}}}{{{\left.{d}{x}\right.}^{{2}}}}$$
So the auxiliari equation of (1) is, $$D^2+1=0 \rightarrow D^2=-1 \rightarrow D=+-i$$
The required general solution is, $$\displaystyle{y}={c}{1}{\cos{{x}}}+{c}{2}{\sin{{x}}}\ldots.{\left({2}\right)}$$
Putting $$y=0$$ for $$x=0$$ in equation (2), we get, $$0=c1\cos0+c2sin0 \rightarrow c1=0$$
And putting $$y=1$$ for $$x=1$$ in equation (2), we get, $$1=c1\cos1+c2\sin1 \rightarrow c2=\frac{1}{\sin1} [as\ c1=0]$$
Putting the values of c1 and c2 in equation (2), we get, $$\displaystyle{y}={\left(\frac{{1}}{{\sin{{1}}}}\right)}{\sin{{x}}}$$
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