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What is 1 4 out of 1 360
Represent fractions in drawings
We had learned how to understand fractions as proportions and how to read the proportion (i.e. the value for the numerator and denominator) from drawings. Below we briefly consider how we create a drawing from a fraction can.
Draw fraction (1/4) on the circle
Our fraction is:
We know that the circle has to be broken into. is to be selected or colored.
So we draw a circle and divide it into pieces of equal size. We divide the full circle, i.e. for each segment of the circle. Then we color any piece in color.
We speak “1 of 4” or “a quarter”.
Draw fraction (3/8) on the circle
Our fraction is:
We know that the circle has to be broken into. are to be colored.
So we draw a circle and divide it into pieces of equal size. We divide the full circle, i.e. for each segment of the circle. Then we color arbitrary pieces blue. They don't have to be next to each other!
We speak “3 of 8” or “three eighths”.
Draw fraction (2/4) on the rectangle
Our fraction is:
We know that the rectangle needs to be broken into. are to be colored.
So we draw a rectangle and divide it into equal pieces. To do this, we halve the length and width and draw the lines accordingly. Pieces of the same size are created. Then we color arbitrary pieces blue.
We speak “2 out of 4” or “two quarters”.
Note: corresponds to "1 of 2" and is therefore an abbreviated fraction.
Draw fraction (2/7) on the strip
Our fraction is:
We know the strip needs to be disassembled into. are to be colored.
So we draw a rectangle (our strip) and divide it into pieces of equal size. To do this, we calculate the side length (e.g.), we now enter this distance step by step and draw the lines vertically accordingly. Pieces of the same size are created. Then we color arbitrary pieces blue.
We speak “2 out of 7” or “two sevenths” (or “two sevenths”).
Draw a fraction (2/5) on the pentagon
Our fraction is:
We know that the pentagon can be broken down into. To do this, the corner points are simply to be connected to the center point. Then we color it.
We speak “2 out of 5” or “two fifths”.
Draw a fraction (3/9) on the cylinder
Our fraction is:
We know the cylinder needs to be disassembled into. To do this, we draw a vertical cylinder, measure the height and divide it by. We take this distance value step by step and draw the horizontal lines. This is how parts are created on the cylinder. Then we color it.
We speak “3 out of 9” or “three ninths”.
Note: corresponds to "1 of 3" and is therefore an abbreviated fraction.
Draw fraction (1/8) on the cuboid
Our fraction is:
We know that the cuboid has to be broken down into. There are several ways to do this. For example, the division makes sense. We draw a cuboid, measure the width and divide it into equal distances. Then we measure the length and divide it into equal distances. This is how pieces are made. Then we color it.
We speak “1 of 8” or “one eighth”.
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# Difference between revisions of "1998 AIME Problems/Problem 14"
## Problem
An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$?
## Solution
$2mnp = (m+2)(n+2)(p+2)$
Let’s solve for $p$:
$(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)$
$[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)$
$p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4}$
For the denominator, we will use a factoring trick (colloquially known as SFFT), which states that $xy + ax + ay + a^2 = (x+a)(y+a)$.
$p = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}$
Clearly, we want to minimize the denominator, so $(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9$. The possible pairs of factors of $9$ are $(1,9)(3,3)$. These give $m = 3, n = 11$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$, while the second pair gives $98$. We can quickly test for the denominator assuming other values to convince ourselves that $1$ is the best possible value for the denominator. Hence, the solution is $130$.
Proof that setting the denominator $(m - 2)(n - 2) - 8$ to $1$ is optimal: Suppose $(m - 2)(n - 2) = 9$, and suppose for the sake of contradiction that there exist $m', n'$ such that $(m' - 2)(n' - 2) = 8 + d$ for some $d > 1$ and such that $$\frac{2(m + 2)(n + 2)}{(m - 2)(n - 2) - 8} < \frac{2(m' + 2)(n' + 2)}{(m' - 2)(n' - 2) - 8}.$$ This implies that $$d(m + 2)(n + 2) < (m' + 2)(n' + 2),$$ and $$d((m - 2)(n - 2) + 4(m + n)) < (m' - 2)(n' - 2) + 4(m' + n').$$ Substituting gives $$d(9 + 4(m + n)) < 8 + d + 4(m' + n'),$$ which we rewrite as $$d(8 + 4(m + n)) < 24 + 4((m' - 2) + (n' - 2)).$$ Next, note that for $p'$ to be positive, we must have $m' - 2$ and $n' - 2$ be positive, so $$(m' - 2) + (n' - 2) \leq (m' - 2)(n' - 2) + 1 = 9 + d.$$ So $$d(8 + 4(m + n)) < 8 + 4(9 + d)$$ //shouldn't this be 24 + 4(9+d)? then the entire proof is fallacious (it results in 2(1 + 3 + 3) < 15 which is true) $$d(4 + 4(m + n)) < 44$$ $$d(1 + m + n) < 11$$ Next, we must have that $m - 2$ and $n - 2$ are positive, so $3 \leq m$ and $3 \leq n$. Also, $2 \leq d$ by how we defined $d$. So $$2(1 + 3 + 3) < 11,$$ a contradiction. We already showed above that there are some values of $m$ and $n$ such that $(m - 2)(n - 2) = 9$ that work, so this proves that one of these pairs of values of $m$ and $n$ must yield the maximal value of $p$.
|
### SIMILARITY OF LENGTH AND AREA #28 Example Solutions
```SIMILARITY OF LENGTH AND AREA
#28
Once you know two figures are similar with a SCALE FACTOR or
a
RATIO OF SIMILARITY b , the following proportions for the SMALL (sm)
and LARGE (lg) figures (which are enlargements or reductions of each other) are
true:
a
sidesm
=
side lg
b
a
Psm
=
P lg
b
a2
Asm
= 2
A lg
b
where P = perimeter, A = area of a face or the total surface area.
a
b
In each proportion above, the data from the smaller figure is written on top
(in the numerator) to help be consistent with correspondences. When working
with area, the scale factor (ratio of similarity) is squared. NEVER square the
actual areas themselves, only the scale factor.
Example
The two rectangular prisms above are similar. Suppose the ratio of their vertical edges is 37 .
a)
Find the ratio of their surface areas.
b)
Find the perimeter of the front face of the small prism if the perimeter of the front face of the
large prism is 20 units.
c)
Find the area of the front face of the small prism if the area of the front face of the large prism is
24 square units.
Solutions
a)
The ratio of the surface areas is
parts.
b)
3
7
=
P
20
7P = 60
P =
60
7
≈ 8.57 units
c)
A sm 32
9
=
=
A l g 7 2 49
2
()=
A
24
9
49
A
24
3
7
=
. Use proportions to solve for the rest of the
49A = 216
A = 216
49 ≈ 4.41 units
Note that the ratio of similarity
must be squared when using it in
area calculations.
Problems
1. Two rectangular prisms are similar. The smaller, P, has a height of
four units while the larger, Q, has a height of five units.
a) What is the scale factor from prism P to prism Q?
b) What is the ratio, small to large, of their surface areas?
2.
P
4
x
Q
5
y
figures are for problems 1 and 2
What would be the ratio of the lengths of the edges labeled x and y in the prisms above?
For problems 3–16, you may want to draw two rectangles, label them A and B, and write in the specific data
from each problem.
3.
If rectangle A and rectangle B have a ratio of similarity of 53 , what is the area of rectangle B if
the area of rectangle A is 30 square units?
4.
If rectangle A and rectangle B have a ratio of similarity of 53 , what is the area of rectangle A if
the area of rectangle B is 30 square units?
5.
If rectangle A and rectangle B have a ratio of similarity of 25 , what is the area of rectangle B if
the area of rectangle A is 46 square units?
6.
If rectangle A and rectangle B have a ratio of similarity of 25 , what is the area of rectangle A if
the area of rectangle B is 46 square units?
7.
If rectangle A and rectangle B have a ratio of similarity of 37 , what is the area of rectangle B if
the area of rectangle A is 82 square units?
8.
If rectangle A and rectangle B have a ratio of similarity of 37 , what is the area of rectangle A if
the area of rectangle B is 82 square units?
9.
If rectangle A and rectangle B have a ratio of similarity of 19 , what is the area of rectangle B if
the area of rectangle A is 24 square units?
10.
If rectangle A and rectangle B have a ratio of similarity of 19 , what is the area of rectangle A if
the area of rectangle B is 24 square units?
11.
Rectangle A is similar to rectangle B. The area of rectangle A is 121 square units while the area
of rectangle B is 49 square units. What is the ratio of similarity between the two rectangles?
12.
Rectangle A is similar to rectangle B. The area of rectangle A is 36 square units while the area
of rectangle B is 64 square units. What is the ratio of similarity between the two rectangles?
13.
Rectangle A is similar to rectangle B. The area of rectangle B is 81 square units while the area
of rectangle A is 25 square units. What is the ratio of similarity between the two rectangles?
14.
Rectangle A is similar to rectangle B. The area of rectangle B is 289 square units while the area
of rectangle A is 121 square units. What is the ratio of similarity between the two rectangles?
15.
Rectangle A is similar to rectangle B. The area of rectangle A is 324 square units while the
area of rectangle B is 121 square units. If the perimeter of rectangle A is 12 units, what is the
perimeter of rectangle B?
16.
Rectangle A is similar to rectangle B. The area of rectangle A is 324 square units while the
area of rectangle B is 121 square units. If the perimeter of rectangle B is 12 units, what is the
perimeter of rectangle A?
17.
The corresponding diagonals of two similar trapezoids are in the ratio of 11
23 . What is the ratio
of their areas?
18.
The corresponding diagonals of two similar trapezoids are in the ratio of 17
19 . What is the ratio
of their areas?
19.
5 . What is the ratio of their areas?
The ratio of the perimeters of two similar parallelograms is 13
20.
11 . What is the ratio of their areas?
The ratio of the perimeters of two similar parallelograms is 15
21.
4 . What is the ratio of their heights?
The ratio of the areas of two similar trapezoids is 81
22.
The ratio of the areas of two similar trapezoids is 59 . What is the ratio of their heights?
23.
The areas of two circles are in the ratio of 36
1 . What is the ratio of their radii?
24.
The areas of two circles are in the ratio of 36
49 . What is the ratio of their radii?
1.
a) 45
5.
9.
b) 16
25
2.
x
y
= 45
287.5 units2
6.
7.36 units2
1944 units2
10.
≈ 0.30 units2
11.
11
7
12.
6
8
13.
5
9
14.
11
17
15.
7.3 units
16.
19.6 3 units
17.
121
529
18.
289
361
19.
25
169
20.
121
225
21.
2
9
22.
5
3
23.
6
1
24.
6
7
3.
10.8 units2
4.
83. 3 units2
7.
446.4 units2
8.
≈ 15.06 units2
or 34
```
|
LCM: Least Common Multiple
# LCM: Least Common Multiple
## LCM: Least Common Multiple
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. LCM: Least Common Multiple
2. Multiples • A multiple is formed by multiplying a given number by the counting numbers. • The counting numbers are 1, 2, 3, 4, 5, 6, etc. • DON’T GET THEM CONFUSED WITH FACTORS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! • Factors are few but multiples are many. • Multiples are Mounting, it’s like skip counting!!!!
3. Example: List the multiples of 4: • 4 x 1 = 4 • 4 x 2 = 8 • 4 x 3 = 12 • 4 x 4 = 16 • 4 x 5 = 20 • 4 x 6 = 24 So, the multiples of 4 are 4, 8, 12, 16, 20, 24, 28, etc. Counting Numbers
4. What are the first five multiples of 13? 13 x 1 =13 13 x 2 = 26 13 x 3 = 39 13 x 4 = 52 13 x 5 = 65 13, 26, 39, 52, 65
5. Find the Missing Multiples 24 30 • 6, 12, 18, ____, ____ • ___, 6, 9, 12, ____, ____, 21 • ___, 24, 36, 48, 60, ____ 3 15 18 72 12
6. Least Common Multiple (LCM) • The least common multiple is the smallest number that is common between two lists of multiples.
7. Method #1:Find the LCM of the numbers by listing the multiples till you find a match • The multiples of 12: • 12 x 1 = 12 • 12 x 2 =24 • 12 x 3 = 36 • 12 x 4 = 48 • 12 x 5 =60 • The multiples of 18: • 18 x 1 = 18 • 18 x 2 = 36 • 18 x 3 = 54 • 18 x 4 = 72 • 18 x 5 = 90
8. 12, 24, 36, 48, 60 18, 36, 54, 72, 90 The first number you see in both lists is 36. The least common multiple of 12 and 18 is 36.
9. Example 2: Find the LCM of 9 and 10 9, 18, 27, 36, 45, 54, 63, 72 81, 90, 99 10, 20, 30, 40, 50, 60, 70, 80 90, 100, 110 If you don’t see a common multiple, make each list go further. The LCM of 9 and 10 is 90
|
# Parallel Vectors
In these lessons, we will learn how to determine if the given vectors are parallel.
A vector is a quantity that has both magnitude and direction.
### What Are Parallel Vectors?
Vectors are parallel if they have the same direction.
Both components of one vector must be in the same ratio to the corresponding components of the parallel vector.
Example:
#### How To Define Parallel Vectors?
Two vectors are parallel if they are scalar multiples of one another.
If u and v are two non-zero vectors and u = cv, then u and v are parallel.
The following diagram shows several vectors that are parallel.
#### How To Determine If The Given 3-Dimensional Vectors Are Parallel?
Example: Determine which vectors are parallel to v = <-3, -2, 5>
#### What Are The Conditions For Two Lines To Be Parallel Given Their Vector Equations?
Lines are parallel if the direction vectors are in the same ratio.
Example: If the lines l1: $$r = \left( {\begin{array}{*{20}{c}}1\\{ - 5}\\7\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}{a - 1}\\{ - a - 1}\\b\end{array}} \right)$$ and l2: $$r = \left( {\begin{array}{*{20}{c}}9\\3\\{ - 8}\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}{2a}\\{3 - 5a}\\{15}\end{array}} \right)$$.
Find the values of a and b.
#### What Is A Vector Is, How To Add And How To Prove Vectors Are Parallel And Collinear?
Examples:
1. A, B, C are midpoints of their respective lines. Find the vector OB.
2. N = midpoint of OB, M = midpoint of OA. Show that MN is parallel to AB.
3. Given the vectors, prove that the three given points are collinear.
#### How To Answer A Question That Involves Vectors - Lines, Parallel, Perpendicular & Intersection?
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
|
### DOTS Division
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Novemberish
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.
# Two and Two
##### Age 11 to 16 Challenge Level:
We received lots of great solutions for this problem, unfortunately too many to mention you all by name! Many of you found that there were up to seven possible solutions, by systematically eliminating possibilities - well done. Here is an answer submitted by Charlie, Constance, Gabriel, two Lucases, Matilda, Rachel, Rama, Ronan, Ruby, Sanjay and Stephanie from Strand on the Green Junior School:
The first thing we noticed was that F has to be 1 because the most T + T can be is 19 (if you have already carried 1 from the previous column). This also means that T ≥ 5. We also noticed that R must be even.
We decided to look at the value of O again.
If O = 0, then R would also be 0 so that doesn’t work and O can’t be 1 because F = 1.
If O = 2,
$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{2}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{2}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{2}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
then R = 4 and T = 6 and we also know that W < 5 because there can’t be anything carried to the hundreds column. The only possible value of W that hasn’t already been used is 3 but this would mean that U is 6 which is the same as T.
If O = 3,
$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{3}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{3}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{3}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1$
then R = 6 and T = 6 which doesn’t work.
If O = 4,
$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{4}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{4}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{4}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
then R = 8 and T = 7 and we also know that W < 5 because there can’t be anything carried to the hundreds column. So W could be 0, 2 or 3.
W can’t be 0 because then U would be 0 and it can’t be 2 because U would be 4.
If W = 3, U = 6 which works: 734 + 734 = 1468.
If O = 5,
$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{5}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{5}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{5}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1 \, \: ^1$
then R = 0 and T = 7 and we also know that W ≥ 5 because there has to be 1 carried to the hundreds column.
W can’t be 5 because O = 5.
If W = 6, U = 3 which works: 765 + 765 = 1530.
If W = 7, U = 5 which doesn’t work because O and U are the same.
If W = 8, U = 7 which doesn’t work because T and U are the same.
If W = 9, U = 9 which doesn’t work because W and U are the same.
If O = 6,
$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{6}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{6}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{6}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1$
then R = 2 and T = 8 and we also know that W < 5 because there can’t be anything carried to the hundreds column. So W could be 0, 3 or 4.
If W = 0, U = 1 which doesn’t work because F and U are the same.
If W = 3, U = 7 which works. 836 + 836 = 1672
If W = 4, U = 9 which works. 846 + 846 = 1692
If O = 7,
$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{7}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{7}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{7}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1 \, \: ^1$
then R = 4 and T = 8 and we also know that W ≥ 5 because there has to be 1 carried to the hundreds column.
If W = 5, U = 1 which doesn’t work because F and U are the same.
If W = 6, U = 3 which works. 867 + 867 = 1734
W can’t be 7 because O = 7.
If W = 8 , U = 7 which doesn’t work because O and U are the same.
If W = 9, U = 9 which doesn’t work because W and U are the same.
If O = 8,
$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{8}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{8}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{8}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1$
then R = 6 and T = 9 and we also know that W < 5 because there can’t be anything carried to the hundreds column. So W could be 0, 2, 3 or 4.
If W = 0, U = 1 which doesn’t work because F and U are the same.
If W = 2, U = 5 which works: 928 + 928 = 1856.
If W = 3, U = 7 which works: 938 + 938 = 1876.
If W = 4, U = 9 which doesn’t work because T and U are the same.
If O = 9,
$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{9}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{9}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{9}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1 \, \: ^1$
then R = 8 and T = 9 which doesn’t work because O and T are the same.
So there are seven possible answers:
938+938=1876
928+928=1856
867+867=1734
846+846=1692
836+836=1672
765+765=1530
734+734=1468
The group from Strand on the Green Junior School also sent us a solution for the problem ONE+ONE=TWO using a similar strategy, and found 16 possible answers. Well done on such a thorough solution!
Some of Mr Parkin's Students from Alleyn's School also tried finding some other words sums. They also investigated ONE+ONE=TWO. Ed noticed that the word sum FOUR+FOUR=EIGHT has the interesting property, as 'FOUR' and 'EIGHT' have no letters in common. He could only find two solutions, 5469 + 5469 = 10938 and 8235 + 8235 = 16470 but is not sure how to find more, except by using a computer program.
Students from Angel Road Junior School in Norwich have managed to come up with three more answers to the sum:
FOUR
+ FOUR
EIGHT
Their solutions are:
8523 + 8523 = 17046
9235 + 9235 = 18470
9327 + 9327 = 18654
8652 + 8652 = 17304
Their teacher explained how they went about the problem:
They tell me their process started by trying to find any rules that were evident, for example, would any of the letters have to be 5 or more and if so, could the carrying of a digit help achieve another number which has not already been used?
They found that because there were no repeated letters between FOUR and EIGHT, then that they would have to use all of the digits between 0-9 apart from 1 which would be remaining. With this they used a process of crossing off any numbers that they used to see what was left over.
Well done!
Can you think of any more interesting word sums?
|
# Finding normalized eigen vectors
I want to find normalized eigen vectors for:
$$\begin{pmatrix} 1 & -2 & 0 \\ -2 & 5 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix}$$
The eigen values I found are 5.828,2,0.171
When finding the eigen vectors for $\lambda=5.828$ by letting $AX=\lambda X$
I get the following equations:
$$X_2=-2.41 X_1\\ X_1=-0.41 X_2$$
How can I solve these two and find normalized eigen vectors?
I get $X_3=0$
• The exact eigenvalues are $\,3-2\sqrt2, 2,3+2\sqrt2$. Once you have eigenvectors, divide them by their norms, that's all. Jun 21, 2015 at 13:40
It is clear that the eigenvector for the eiganvalue $2$ is $(0,0,1)$. The remaining two are more interesting.
Notice that the eigenvalues are roots of the polynomial $x^2-6x+1$. (This is what you get from the characteristic polynomial. I assume this is how you get them.)
In particular, any eigenvalue fulfills $\lambda^2=6\lambda-1$.
Now to find an eigenvector for the eignevalue $\lambda$ you solve the linear system (I have omitted the third variable, since we know it must be zero): $$\begin{pmatrix} 1-\lambda & -2 \\ -2 & 5-\lambda \end{pmatrix}\overset{(1)}\sim \begin{pmatrix} 1-\lambda & -2 \\ 0 & 0 \end{pmatrix}$$
(Since we know that $\lambda$ is an eigenvalue, the above matrix cannot have full rank. That's why we could replace the second row immediately by zeroes in the step (1). But you can also check that the second row is indeed a multiple of the first or. Just use the fact that $(\lambda-1)(\lambda-5)=4$ and multiply the first row by the number $\frac{\lambda-5}2=\frac2{\lambda-1}$.)
Clearly, the vector $\vec x=(2,1-\lambda,0)$ is a solution.
We want a normalized vector, so we calculate the norm of this vector: $|\vec x|^2 = 4+(1-\lambda^2)=\lambda^2-2\lambda+5=6\lambda-1-2\lambda+5=4(\lambda-1)$.
So the normalized eigenvectors have the form $$\frac{\vec x}{|\vec x|}=\left(\frac1{2\sqrt{\lambda-1}},-\frac{\sqrt{\lambda-1}}2,0\right).$$
Remark. I have tried to solve the system in a way in which we get a solution for both eigenvalues at the same time. (To get two eigenvectors for the price of one.) Moreover, in this way I avoided calculations with square roots.
Of course, if you prefer, you can find the solutions for $\lambda_1=3-2\sqrt2$ and for $\lambda_2=3+2\sqrt2$ separately.
In response to the following comment I will add more details:
I don't understand how the last row elements were made $0$? Is it like doing row echelon form? Why is it important to do this? From then how do you say $\vec x =(2,1−\lambda)$ is a solution
Yes, a standard way to solve a system of linear equations is get the matrix to row echelon form. In our case we have a homogeneous system represented by the matrix: $$\begin{pmatrix} 1-\lambda & -2 \\ -2 & 5-\lambda \end{pmatrix}$$ As I have mentioned before, we already know that the matrix does not have full rank. Therefore second row is a multiple of the first row, so by doing one simple row operation we get zeroes in the second row.
However, we might want to go a bit slower and perform the actual operations. (What I wrote above is only true if we calculated the eigenvalues correctly. So in this way we doublecheck that they are indeed eigenvalues.)
So we want add a $c$-multiple of the first row to the second row. We want to choose $c$ in such way that we get zero in the first place of the second row. So we need $c=\frac{2}{1-\lambda}$.
What do we get in the second place of the second row? Well, let's calculate: $$5-\lambda-2c = 5-\lambda - \frac{4}{1-\lambda} = \frac{(5-\lambda)(1-\lambda)-4}{1-\lambda} = \frac{\lambda^2-6\lambda+1}{1-\lambda} = 0$$
So we see that after this row operation we have the new matrix: $$\begin{pmatrix} 1-\lambda & -2 \\ 0 & 0 \end{pmatrix}$$
This matrix corresponds to the linear system which has only one equation $$(1-\lambda)x_1 - 2x_2=0.$$ How can we find a solution? For any value we choose for $x_2$, the above equation uniquely determines $x_1=\frac{2x_2}{1-\lambda}$. We can choose arbitrary non-zero value for $x_2$, So I chose ${1-\lambda}$, as to avoid fractions.
• I don't understand how this is done. I don't understand how the last row elements were made 0? Is it lke doing row echelon form? Why is it important to do this? From then how do you say $\vec x=(2,1-\lambda)$ is a solution Jun 21, 2015 at 16:10
• @sam_rox I have edited my post to add a few more details. (And I have also added the third coordinate, which was missing.) Jun 21, 2015 at 16:26
• Thanks now I understand how you have done it. But how I do is if the matrix is A, then by |A-$\lambda$ I|=0 I find the eigen values and from AX=$\lambda X$ I find the eigen vectors.But to solve for eigen vectors you have used a form similar to A-$\lambda$ I=0 . I set AX=$\lambda X$ to obtain $X_2=-2.41 X_1\\ X_1=-0.41 X_2$ . Is there a way to come up with the solution through these equations? Jun 21, 2015 at 17:25
• The (systems of) equations $AX=\lambda X$ and $(A-\lambda I)X=0$ are equivalent. Jun 21, 2015 at 17:30
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# How I Knew Immediately that a Factor Pair of 1224 is . . .
12 = 3 × 4 and 24 is one less than 25. Those two facts helped me to know right away that 35² = 1225 and 34 × 36 = 1224. Study the patterns in the chart below and you will likely be able to remember all of the multiplication facts listed in it!
a² – b² = (a – b)(a + b)
You may remember how to factor that from algebra class. Here when b = 1, it has a practical application that can allow you to amaze your friends and family with your mental calculating abilities!
I’ve only typed a small part of that infinite pattern chart. For example, if you know that 19 × 20 = 380, then you can also know that 195² = 38025 and 194 × 196 = 38024.
Also because of that chart, I know that 3.5² = 12.25 and 3.4 × 3.6 = 12.24
(Also (3½)² = 12¼, but 2½ × 4½ = 11¼ because 3-1 = 2, 3+1 = 4, 12-1 = 11
thus 2.5 × 4.5 = 11.25 and 2½ × 4½ = 11¼)
You could also let b = 2 so b² = 4. Then 25 – 4 = 21, and you could know facts like
33 × 37 = 1221 or 193 × 197 = 38021
I hope you have a wonderful time being a calculating genius!
Now I’ll share some other facts about the number 1224:
• 1224 is a composite number.
• Prime factorization: 1224 = 2 × 2 × 2 × 3 × 3 × 17, which can be written 1224 = 2³ × 3² × 17
• The exponents in the prime factorization are 2, 3 and 1. Adding one to each and multiplying we get (3 + 1)(2 + 1)(1 + 1) = 4 × 3 × 2 = 24. Therefore 1224 has exactly 24 factors.
• Factors of 1224: 1, 2, 3, 4, 6, 8, 9, 12, 17, 18, 24, 34, 36, 51, 68, 72, 102, 136, 153, 204, 306, 408, 612, 1224
• Factor pairs: 1224 = 1 × 1224, 2 × 612, 3 × 408, 4 × 306, 6 × 204, 8 × 153, 9 × 136, 12 × 102, 17 × 72, 18 × 68, 24 × 51 or 34 × 36
• Taking the factor pair with the largest square number factor, we get √1224 = (√36)(√34) = 6√34 ≈ 34.98571
When a number has so many factors, I often will make a forest of factor trees for that number, but today I just want us to enjoy this one tree for 34 × 36 = 1224.
1224 is also the sum of two squares:
30² + 18² = 1224
1224 is the hypotenuse of a Pythagorean triple:
576-1080-1224 which is (8-15-17) times 72
That triple can also be calculated from 30² – 18², 2(30)(18), 30² + 18²
293 + 307 + 311 + 313 = 1224 making 1224 the sum of four consecutive prime numbers.
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Pages
Category
Similar Triangles Calculator
Check whether two triangles are similar or find the missing length of a triangle with the help of this similar triangles calculator.
△ABC
△DEF
Use the similar triangles calculator to check the similarity of two triangles. With it, you can also find the missing length of a triangle.
What Are Similar Triangles?
These are the triangles that have the same shape but different sizes. Meanwhile, similar triangles superimpose each other when they are magnified or demagnified. These triangles are different from congruent triangles. The similarity of triangles is denoted by the ‘~’ symbol.
Similar Triangles Formula:
Two triangles △ABC and △EFG can be said to be similar triangles (△ABC ∼ △EFG) if:
∠A = ∠E, ∠B = ∠F and ∠C = ∠G
AB/EF = BC/FG = AC/EG
Similar Triangles and Congruent Triangles:
Similar Triangles Congruent Triangles They have the same shape but different size They have the same shape and size The symbol is ‘~’ The symbol is ‘≅’ all the corresponding sides have the same Ratio The ratio of corresponding sides is equal to a constant
What are Three Similarities Theorems For Triangles?
Similarity theorems help to prove whether two triangles are similar or not. These theorems are used when all the sides or angles are not given. Three similarity theorems are:
AA (or AAA) or Angle-Angle:
This theorem states that if the two angles of a triangle are equal to the two angles of another triangle then they are similar triangles.
SAS or Side-Angle-Side:
If the two sides of a given triangle are in proportion to the two sides of another triangle and the angles associated with these sides of both triangles are equal then they are similar triangles.
SSS or Side-Side-Side:
If all the corresponding ratios of all the sides of both triangles are equal then it means that they are similar triangles.
How To Find Similar Triangles?
Go through the following steps to determine whether two triangles are similar or not:
• First of all, note down the given dimensions including the sides or angles of the triangles on paper
• See if these dimensions follow any of the above-mentioned conditions of similar triangles theorems(AA, SSS, SAS)
• If the dimensions of these triangles fulfill any of the conditions then represent them with the ‘~’ symbol
Using an online similar triangles calculator is the most convenient way to determine the similarity of the triangles. Just add the available dimensions and get to
Example:
Suppose you have two triangles if △ABC and △PQR that are similar triangles or not using the given data: ∠A = 65°, ∠B = 60º and ∠P = 70°, ∠R = 45°.
Solution:
Given that:
∠A = 65°
∠B = 70º
And
∠P = 70°
∠R = 45°
Now we have to find the third angle of each triangle to conclude: As we know the sum of all the angles is = 180°
First Triangle = 70° + 65º = 135°
Second Triangle = 70° + 45º = 115°
Now the thrid angle of the first triangle = 180° - 135º = 45º
Now the third angle of the second triangle = 180° - 115º = 65º
Here both of the triangles have two same angles so according to the first theorem of similarity and the similar triangles formula these two triangles are similar. If you are not getting the concept, then use a similar triangles calculator. It will let you find out whether the triangles are similar are not by just requiring the dimensions of the triangles.
How Do You Find The Missing Side of a Similar Triangle?
Follow the below-mentioned steps to find the missing side:
• First, determine whether the triangle that has missing sides is smaller or larger
• Calculate the scale factor “k” of the similar triangle by taking the ratio of any known sides
• If the triangle is small then divide the corresponding side in the larger by “k”
• If the triangle is larger, then multiply the corresponding side in the smaller triangle by the value of k
How Does Similar Triangles Calculator Work?
Provide the available sides or the angles to this similar triangle calculator and get to know whether they are similar or not in seconds. Through it, you can also find the missing sides of a triangle. Let's see how it works!
What Do You Need to Enter?
Inputs For Checking Similarity:
• Similarity Criterion: First choose the type of criteria to check the similarity of the triangles according to the available data.
• Input the Triangle Data: Add the lengths of the sides for the triangles or angles in the specified fields.
This is What You Will Get!
• It provides you with a statement defining whether the triangles are similar or not
Inputs For Finding the Missing Sides:
• Corresponding Sides of The Similar Triangle: Add the values of the sides of a similar triangle
• Scale Factor: Calculate the scale factor of the triangle by taking the ratio of the known sides
This is What You Will Get!
• Missing Sides
• Area of The both Triangles
• The perimeter of both Triangles
With the help of this similar figures calculator, you can quickly and easily assess the similarity of two given triangles.
Applications of Similar Right Triangles Calculator:
Let's take a look at the following applications:
• Students can use the triangle similarity calculator to verify their geometry problems. Teachers can use it to explain the concept of similarities
• Graphic designers use similar polygons calculators to achieve precise scaling in images.
Using a similar triangle calculator, allows you to perform the whole calculation in a matter of seconds.
FAQ’s:
Are Similar Triangles And Congruent Triangles The Same?
Similar triangles have the same shape but different sizes, and on the other hand, congruent triangles have the same shape and same size.
Are All Equilateral Triangles Similar?
Yes, all the equilateral triangles have the same features.
How To Find The Proportion of Similar Triangles?
If two similar triangles have sides X, Y, Z and x, y, z then the pair of corresponding sides are proportion:
X : x = Y : y = Z : z
What Are The Properties of Similar Figures?
The properties of similar triangles are:
• Two triangles have the same shape but different sizes
• One pair of corresponding angles is equal
• The ratio of corresponding sides is the same
How Do You Find Similar Right Triangles?
If you have a right triangle in which the lengths of the hypotenuse and the leg of a right triangle are proportional to the parts of another right triangle then they are called similar. For the precise calculation, you should get the assistance of a similarity in right triangles calculator.
References:
Cuemath.com: Similar Triangles. Wikipedia: Similarity system of triangles, Triangles appended to a rectangle, Gallery
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# 2.4 Constant acceleration (Page 4/4)
Page 4 / 4
$\begin{array}{l}{\mathbf{r}}_{1}=0\\ ⇒\Delta \mathbf{r}={\mathbf{r}}_{2}-{\mathbf{r}}_{1}={\mathbf{r}}_{2}=\mathbf{r}\phantom{\rule{2pt}{0ex}}\left(\mathrm{say}\right)\end{array}$
In this case, both final position vector and displacement are equal. This simplification, therefore, allows us to represent both displacement and position with a single vector variable r .
## Graphical interpretation of equations of motion
The three basic equations of motion with constant acceleration, as derived above, are :
$\begin{array}{l}\mathrm{1:}\phantom{\rule{2pt}{0ex}}\mathbf{v}=\mathbf{u}+\mathbf{a}t\\ \mathrm{2:}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{\mathrm{avg}}=\frac{\left(\mathbf{u}+\mathbf{v}\right)}{2}\\ \mathrm{3:}\phantom{\rule{2pt}{0ex}}\mathbf{s}=\Delta \mathbf{r}={\mathbf{r}}_{2}-{\mathbf{r}}_{1}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{t}^{2}\end{array}$
These three equations completely describe motion of a point like mass, moving with constant acceleration. We need exactly five parameters to describe the motion under constant acceleration : $\mathbf{u},\mathbf{v},{\mathbf{r}}_{1},{\mathbf{r}}_{2}\phantom{\rule{2pt}{0ex}}\mathrm{and}\phantom{\rule{2pt}{0ex}}t$ .
It can be emphasized here that we can not use these equations if the acceleration is not constant. We should use basic differentiation or integration techniques for motion having variable acceleration (non-uniform acceleration). These equations serve to be a ready to use equations that avoids differentiation and integration. Further, it is evident that equations of motion are vector equations, involving vector addition. We can evaluate a motion under constant acceleration, using either graphical or algebraic method based on components.
Here, we interpret these vector equations, using graphical technique. For illustration purpose, we apply these equations to a motion of an object, which is thrown at an angle θ from the horizontal. The magnitude of acceleration is "g", which is directed vertically downward. Let acceleration vector be represented by corresponding bold faced symbol g . Let ${\mathbf{v}}_{1}$ and ${\mathbf{v}}_{2}$ be the velocities at time instants ${t}_{1}$ and ${t}_{2}$ respectively and corresponding position vectors are ${\mathbf{r}}_{1}$ and ${\mathbf{r}}_{2}$ .
The final velocity at time instant ${t}_{2}$ , is given by :
$\begin{array}{l}\mathbf{v}=\mathbf{u}+\mathbf{a}t\\ ⇒{\mathbf{v}}_{2}={\mathbf{v}}_{1}+\mathbf{g}\left({t}_{2}-{t}_{1}\right)\end{array}$
Graphically, the final velocity is obtained by modifying initial vector ${\mathbf{v}}_{1}$ by the vector $\mathbf{g}\left({t}_{2}-{t}_{1}\right)$ .
Now, we discuss graphical representation of second equation of motion. The average velocity between two time instants or two positions is given by :
$\begin{array}{l}{\mathbf{v}}_{\mathrm{avg}}=\frac{\left(\mathbf{u}+\mathbf{v}\right)}{2}\\ ⇒{\mathbf{v}}_{\mathrm{avg}}=\frac{\left({\mathbf{v}}_{1}+{\mathbf{v}}_{2}\right)}{2}\end{array}$
The vector addition involved in the equation is graphically represented as shown in the figure. Note that average velocity is equal to half of the vector sum ${\mathbf{v}}_{1}+{\mathbf{v}}_{2}$ .
Third equation of motion provides for displacement in terms of two vector quantities - initial velocity and acceleration. The displacement, s , is equal to addition of two vector terms :
$\begin{array}{l}\mathbf{s}=\Delta \mathbf{r}={\mathbf{r}}_{2}-{\mathbf{r}}_{1}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{t}^{2}\\ ⇒\mathbf{s}=\Delta \mathbf{r}={\mathbf{r}}_{2}-{\mathbf{r}}_{1}={\mathbf{v}}_{1}\left({t}_{2}-{t}_{1}\right)+\frac{1}{2}\mathbf{g}{\left({t}_{2}-{t}_{1}\right)}^{2}\end{array}$
## Equations of motion in component form
The application of equations of motion graphically is tedious. In general, we use component representation that allows us to apply equations algebraically. We use equations of motion, using component forms of vector quantities involved in the equations of motion. The component form of the various vector quantities are :
$\begin{array}{l}\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\\ \Delta \mathbf{r}=\Delta x\mathbf{i}+\Delta y\mathbf{j}+\Delta z\mathbf{k}\\ \mathbf{u}={u}_{x}\mathbf{i}+{u}_{y}\mathbf{j}+{u}_{z}\mathbf{k}\\ \mathbf{v}={v}_{x}\mathbf{i}+{v}_{y}\mathbf{j}+{v}_{z}\mathbf{k}\\ \mathbf{v}={v}_{\mathrm{avgx}}\mathbf{i}+{v}_{\mathrm{avgy}}\mathbf{j}+{v}_{\mathrm{avgz}}\mathbf{k}\\ \mathbf{a}={a}_{x}\mathbf{i}+{a}_{y}\mathbf{j}+{a}_{z}\mathbf{k}\end{array}$
Using above relations, equations of motion are :
$\begin{array}{ll}\mathrm{1:}& {v}_{x}\mathbf{i}+{v}_{y}\mathbf{j}+{v}_{z}\mathbf{k}=\left({u}_{x}\mathbf{i}+{u}_{y}\mathbf{j}+{u}_{z}\mathbf{k}\right)+\left({a}_{x}\mathbf{i}+{a}_{y}\mathbf{j}+{a}_{z}\mathbf{k}\right)\phantom{\rule{2pt}{0ex}}t\\ \mathrm{2:}& {v}_{\mathrm{avgx}}\mathbf{i}+{v}_{\mathrm{avgy}}\mathbf{j}+{v}_{\mathrm{avgz}}\mathbf{k}=\frac{\left({u}_{x}\mathbf{i}+{u}_{y}\mathbf{j}+{u}_{z}\mathbf{k}\right)+\left({v}_{x}\mathbf{i}+{v}_{y}\mathbf{j}+{v}_{z}\mathbf{k}\right)}{2}\\ \mathrm{3:}& \Delta x\mathbf{i}+\Delta y\mathbf{j}+\Delta z\mathbf{k}=\left({u}_{x}\mathbf{i}+{u}_{y}\mathbf{j}+{u}_{z}\mathbf{k}\right)\phantom{\rule{2pt}{0ex}}t+\frac{1}{2}\left({a}_{x}\mathbf{i}+{a}_{y}\mathbf{j}+{a}_{z}\mathbf{k}\right){t}^{2}\end{array}$
## Acceleration in component form
Problem : A particle is moving with an initial velocity $\left(8\mathbf{i}+2\mathbf{j}\right)\phantom{\rule{2pt}{0ex}}m/s$ , having an acceleration $\left(0.4\mathbf{i}+0.3\mathbf{j}\right)\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ . Calculate its speed after 10 seconds.
Solution : The particle has acceleration of $\left(0.4\mathbf{i}+0.3\mathbf{j}\right)\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ , which is a constant acceleration. Its magnitude is $\sqrt{\left({0.4}^{2}+{0.3}^{2}\right)}=0.5\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ making an angle with x-direction. $\theta ={\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)$ . Time interval is 10 seconds. Thus, applying equation of motion, we have :
$\begin{array}{l}{v}_{x}\mathbf{i}+{v}_{y}\mathbf{j}+{v}_{z}\mathbf{k}=\left({u}_{x}\mathbf{i}+{u}_{y}\mathbf{j}+{u}_{z}\mathbf{k}\right)+\left({a}_{x}\mathbf{i}+{a}_{y}\mathbf{j}+{a}_{z}\mathbf{k}\right)\phantom{\rule{2pt}{0ex}}t\\ ⇒\mathbf{v}=\left(8\mathbf{i}+2\mathbf{j}\right)+\left(0.4\mathbf{i}+0.3\mathbf{j}\right)x10=12\mathbf{i}+5\mathbf{j}\end{array}$
Speed i.e. magnitude of velocity is :
$\begin{array}{l}⇒v=\sqrt{\left({12}^{2}+{5}^{2}\right)}=13\phantom{\rule{2pt}{0ex}}m/s\end{array}$
This example illustrates the basic nature of the equations of motion. If we treat them as scalar equations, we may be led to wrong answers. For example, magnitude of initial velocity i.e. speed is $\sqrt{\left({8}^{2}+{2}^{2}\right)}=8.25\phantom{\rule{2pt}{0ex}}m/s$ , Whereas magnitude of acceleration is $\sqrt{\left({0.4}^{2}+{0.3}^{2}\right)}=0.5\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ . Now, using equation of motion as scalar equation, we have : $\begin{array}{l}v=u+at=8.25+0.5x10=13.25\phantom{\rule{2pt}{0ex}}m/s\end{array}$
## Equivalent scalar system of equations of motion
We have discussed earlier that a vector quantity in one dimension can be conveniently expressed in terms of an equivalent system of scalar representation. The advantage of linear motion is that we can completely do away with vector notation with an appropriate scheme of assigning plus or minus signs to the quantities involved. The equivalent scalar representation takes advantage of the fact that vectors involved in linear motion has only two possible directions. The one in the direction of chosen axis is considered positive and the other against the direction of the chosen axis is considered negative.
At the same time, the concept of component allows us to treat a motion into an equivalent system of the three rectilinear motions in the mutually perpendicular directions along the axes. The two concepts, when combined together, renders it possible to treat equations of motion in scalar terms in mutually three perpendicular directions.
Once we follow the rules of equivalent scalar representation, we can treat equations of motion as scalar equations in the direction of an axis, say x - axis, as :
$\begin{array}{ll}\mathrm{1x:}& {v}_{x}={u}_{x}+{a}_{x}t\\ \mathrm{2x:}& {v}_{\mathrm{avgx}}=\frac{\left({u}_{x}+{v}_{x}\right)}{2}=\frac{\left({x}_{2}-{x}_{1}\right)}{t}\\ \mathrm{3x:}& \Delta x={x}_{2}-{x}_{1}={u}_{x}t+\frac{1}{2}{a}_{x}{t}^{2}\end{array}$
We have similar set of equations in the remaining two directions. We can obtain the composite interpretation of the motion by combing the individual result in each direction. In order to grasp the method, we rework the earlier example.
## Acceleration in scalar form
Problem : A particle is moving with an initial velocity $\left(8\mathbf{i}+2\mathbf{j}\right)\phantom{\rule{2pt}{0ex}}m/s$ , having an acceleration $\left(0.4\mathbf{i}+0.3\mathbf{j}\right)\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ . Calculate its speed after 10 seconds.
Solution : The motion in x – direction :
$\begin{array}{l}{u}_{x}=8\phantom{\rule{2pt}{0ex}}m/s\phantom{\rule{2pt}{0ex}};{a}_{x}=0.4\phantom{\rule{2pt}{0ex}}m/{s}^{2};10\phantom{\rule{2pt}{0ex}}s\phantom{\rule{2pt}{0ex}}\mathrm{and,}\\ {v}_{x}={u}_{x}+{a}_{x}t\\ ⇒{v}_{x}=8+0.4x10=12\phantom{\rule{2pt}{0ex}}m/s\end{array}$
The motion in y – direction :
$\begin{array}{l}{u}_{y}=2\phantom{\rule{2pt}{0ex}}m/s\phantom{\rule{2pt}{0ex}};{a}_{y}=0.3\phantom{\rule{2pt}{0ex}}m/{s}^{2};10\phantom{\rule{2pt}{0ex}}s\phantom{\rule{2pt}{0ex}}\mathrm{and,}\\ {v}_{y}={u}_{y}+{a}_{y}t\\ ⇒{v}_{y}=2+0.3x10=5\phantom{\rule{2pt}{0ex}}m/s\end{array}$
Therefore, the velocity is :
$\begin{array}{l}⇒\mathbf{v}=2\mathbf{i}+5\mathbf{j}\\ ⇒v=\sqrt{\left({12}^{2}+{5}^{2}\right)}=13m/{s}^{2}\end{array}$
What are the system of units
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
58asagravitasnal firce
Amar
water boil at 100 and why
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what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
which colour has the shortest wavelength in the white light spectrum
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
x=5.8-3.22 x=2.58
what is the definition of resolution of forces
what is energy?
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
Abdul
motion
Mustapha
highlights of atomic physics
Benjamin
can anyone tell who founded equations of motion !?
n=a+b/T² find the linear express
أوك
عباس
Quiklyyy
Moment of inertia of a bar in terms of perpendicular axis theorem
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In this video, we will multiply radical expressions. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.
$\sqrt{3}\times\sqrt{3}=\sqrt{9}=3$
$\sqrt{5}\times\sqrt{5}=\sqrt{25}=3$
In other words,
$\sqrt{7}\times\sqrt{7}=7$
$\sqrt{12}\times\sqrt{12}=12$
Let’s try other examples:
$\sqrt{5}\times\sqrt{7}$ would equal to $\sqrt{35}$
$\sqrt{18}\times\sqrt{2}$ would equal to $\sqrt{36}$, which is a perfect square so it would be $6$
Let’s try the same problem by using another method:
$\sqrt{18}\times\sqrt{2}$ would equal to $\sqrt{36}$ can also be written as $\sqrt{9}\times\sqrt{2}\times\sqrt{2}$
For $\sqrt{15}\times\sqrt{45}$, it would equal to $\sqrt{675}$
25 goes into 675 twenty-seven times
Since 5 is the square root of 25, $\sqrt{675}$ can be written as $5\sqrt{27}$
This can be further simplified into $15\sqrt{3}$
## Examples of Multiplying Radical Expressions
### Example 1
$\sqrt{18}\times \sqrt{16}$
Use the rule to multiply the radicands $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$
$\sqrt{18\times 16}$
$\sqrt{288}$
Look for perfect squares in the radicand, and rewrite the radicand as the product of two factors.
$\sqrt{144\times2}$
Simplify,
$12\sqrt{2}$
### Example 2
$\sqrt{4x}\times \sqrt{5x^3}$
Use the rule to multiply the radicands $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$
$\sqrt{4\times 5 \times x \times x^3}$
$\sqrt{4\times 5 \times x^4}$
Simplify,
$2x\sqrt{5}$
## Video-Lesson Transcript
Let’s go over how to multiply radical expressions.
$\sqrt{3} \times \sqrt{3} = \sqrt{9} = 3$
$\sqrt{5} \times \sqrt{5} = \sqrt{25} = 5$
So if you look, closely
$\sqrt{7} \times \sqrt{7} = 7$
$\sqrt{12} \times \sqrt{12} = 12$
Keep this in mind as we move forward in the lesson.
Let’s take a look at this
$\sqrt{5} \times \sqrt{7} = \sqrt{35}$
this cannot be simplified.
$\sqrt{18} \times \sqrt{2} = \sqrt{36} = 6$
But what if we reduce this first?
$\sqrt{9} \times \sqrt{2} \times \sqrt{2}$
$3 \times 2$
$6$
So you can multiply it across and get an answer.
Or reduce it first and multiply it out.
Let’s have this one
$\sqrt{15} \sqrt{45}$
I can multiply this out
$\sqrt{15} \sqrt{45} = \sqrt{675}$
Then we should break $675$ down.
But it’s a pretty large number.
$\sqrt{675}$
$\sqrt{25} \sqrt{27}$
$5 \sqrt{27}$
$5 \sqrt{9} \sqrt{3}$
$5 \times 3 \sqrt{3}$
$15 \sqrt{3}$
Let’s see what happens if we simplify the radical expressions first.
$\sqrt{15} \sqrt{45}$
$\sqrt{15} \sqrt{9} \sqrt{5}$
$\sqrt{15} \times (3) \sqrt{5}$
$3 \sqrt{75}$
$3 \sqrt{25} \sqrt{3}$
$3 \times 5 \sqrt{3}$
$15 \sqrt{3}$
We came up with the same answer.
The number of steps in the two methods is pretty much the same.
But I dealt with smaller numbers using the second method.
Let’s look back at a point here.
$\sqrt{15} \sqrt{45}$
$\sqrt{15} \sqrt{9} \sqrt{5}$
At this point, we know that $\sqrt{15}$ can’t be broken down.
But there’s a $\sqrt{5}$ there.
And we know that $5$ goes into $15$.
So, we might as well break it down. So we’ll have
$\sqrt{3} \sqrt{5} \sqrt{9} \sqrt{5}$
Let’s just reorganize this
$\sqrt{9} \sqrt{5} \sqrt{5} \sqrt{3}$
And we’ll solve
$3 \times 5 \sqrt{3}$
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# 7.1: Ratios and Proportions
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Write and solve ratios and proportions.
• Use ratios and proportions in problem solving.
## Review Queue
1. Are the two triangles congruent? If so, how do you know?
2. If \begin{align*}AC = 5\end{align*}, what is \begin{align*}GI\end{align*}? What is the reason?
3. How many inches are in a:
1. foot?
2. yard?
3. 3 yards?
4. 5 feet?
Know What? You want to make a scale drawing of your room and furniture for a little redecorating. Your room measures 12 feet by 12 feet. In your room are a twin bed (36 in by 75 in) and a desk (4 feet by 2 feet). You scale down your room to 8 in by 8 in, so it fits on a piece of paper. What size are the bed and desk in the drawing?
## Using Ratios
Ratio: A way to compare two numbers. Ratios can be written: \begin{align*}\frac{a}{b}\end{align*}, \begin{align*}a:b\end{align*}, and \begin{align*}a\end{align*} to \begin{align*}b\end{align*}.
Example 1: There are 14 girls and 18 boys in your math class. What is the ratio of girls to boys?
Solution: The ratio would be 14:18. This can be simplified to 7:9.
Example 2: The total bagel sales at a bagel shop for Monday is in the table below. What is the ratio of cinnamon raisin bagels to plain bagels?
Type of Bagel Number Sold
Plain 80
Cinnamon Raisin 30
Sesame 25
Jalapeno Cheddar 20
Everything 45
Honey Wheat 50
Solution: The ratio is 30:80. Reducing the ratio by 10, we get 3:8.
Reduce a ratio just like a fraction. Always reduce ratios.
Example 3: What is the ratio of honey wheat bagels to total bagels sold?
Solution: Order matters. Honey wheat is listed first, so that number comes first in the ratio (or on the top of the fraction). Find the total number of bagels sold, \begin{align*}80 + 30 + 25 + 20 + 45 + 50 = 250\end{align*}.
The ratio is \begin{align*}\frac{50}{250} = \frac{1}{5}\end{align*}.
Equivalent Ratios: When two or more ratios reduce to the same ratio.
50:250 and 2:10 are equivalent because they both reduce to 1:5.
Example 4: What is the ratio of cinnamon raisin bagels to sesame bagels to jalapeno cheddar bagels?
Solution: 30:25:20, which reduces to 6:5:4.
## Converting Measurements
How many feet are in 2 miles? How many inches are in 4 feet? Ratios are used to convert these measurements.
Example 5: Simplify the following ratios.
a) \begin{align*}\frac{7 \ ft}{14 \ in}\end{align*}
b) 9m:900cm
c) \begin{align*}\frac{4 \ gal}{16 \ gal}\end{align*}
Solution: Change these so that they are in the same units. There are 12 inches in a foot.
a) \begin{align*}\frac{7 \ \bcancel{ft}}{14 \ \cancel{in}} \cdot \frac{12 \ \cancel{in}}{1 \ \bcancel{ft}} = \frac{84}{14} = \frac{6}{1}\end{align*}
The inches cancel each other out. Simplified ratios do not have units.
b) It is easier to simplify a ratio when written as a fraction.
\begin{align*}\frac{9 \ \bcancel{m}}{900 \ \cancel{cm}} \cdot \frac{100 \ \cancel{cm}}{1 \ \bcancel{m}} = \frac{900}{900} = \frac{1}{1}\end{align*}
c) \begin{align*}\frac{4 \ \bcancel{gal}}{16 \ \bcancel{gal}} = \frac{1}{4}\end{align*}
Example 6: A talent show has dancers and singers. The ratio of dances to singers is 3:2. There are 30 performers total, how many of each are there?
Solution: 3:2 is a reduced ratio, so there is a number, \begin{align*}n\end{align*}, that we can multiply both by to find the total number in each group.
\begin{align*}\text{dancers} = 3n, \ \text{singers} = 2n \ \longrightarrow \ 3n+2n &= 30\\ 5n &= 30\\ n &= 6\end{align*}
There are \begin{align*}3 \cdot 6 = 18\end{align*} dancers and \begin{align*}2 \cdot 6 = 12\end{align*} singers.
## Solving Proportions
Proportion: Two ratios that are set equal to each other.
Example 7: Solve the proportions.
a) \begin{align*}\frac{4}{5} = \frac{x}{30}\end{align*}
b) \begin{align*}\frac{y+1}{8} = \frac{5}{20}\end{align*}
c) \begin{align*}\frac{6}{5} = \frac{2x+5}{x-2}\end{align*}
Solution: To solve a proportion, you need to cross-multiply.
a)
b)
c)
Cross-Multiplication Theorem: \begin{align*}a, b, c,\end{align*} and \begin{align*}d\end{align*} are real numbers, with \begin{align*}b \neq 0\end{align*} and \begin{align*}d \neq 0\end{align*}. If \begin{align*}\frac{a}{b} = \frac{c}{d}\end{align*}, then \begin{align*}ad=bc\end{align*}.
The proof of the Cross-Multiplication Theorem is an algebraic proof. Recall that multiplying by the same number over itself is 1 \begin{align*}(b \div b=1)\end{align*}.
Proof of the Cross-Multiplication Theorem
\begin{align*}\frac{a}{b} &= \frac{c}{d} && \text{Multiply the left side by} \ \frac{d}{d} \ \text{and the right side by} \ \frac{b}{b}.\\ \frac{a}{b} \cdot \frac{d}{d} &= \frac{c}{d} \cdot \frac{b}{b}\\ \frac{ad}{bd} &= \frac{bc}{bd} && \text{The denominators are the same, so the tops are equal}.\\ ad &= bc\end{align*}
Example 8: Your parents have an architect’s drawing of their home. On the paper, the house’s dimensions are 36 in by 30 in. If the shorter length of the house is actually 50 feet, what is the longer length?
Solution: Set up a proportion. If the shorter length is 50 feet, then it lines up with 30 in, the shorter length of the paper dimensions.
\begin{align*}\frac{30}{36} = \frac{50}{x} \longrightarrow \ 1800 &=30x\\ 60 &=x \quad \quad \text{The longer length is 60 feet.}\end{align*}
## Properties of Proportions
The Cross-Multiplication Theorem has several sub-theorems, called corollaries.
Corollary: A theorem that follows directly from another theorem.
Below are three corollaries that are immediate results of the Cross Multiplication Theorem.
Corollary 7-1: If \begin{align*}a, b, c,\end{align*} and \begin{align*}d\end{align*} are nonzero and \begin{align*}\frac{a}{b} = \frac{c}{d}\end{align*}, then \begin{align*}\frac{a}{c} = \frac{b}{d}\end{align*}. Switch \begin{align*}b\end{align*} and \begin{align*}c\end{align*}.
Corollary 7-2: If \begin{align*}a, b, c,\end{align*} and \begin{align*}d\end{align*} are nonzero and \begin{align*}\frac{a}{b} = \frac{c}{d}\end{align*}, then \begin{align*}\frac{d}{b} = \frac{c}{a}\end{align*}. Switch \begin{align*}a\end{align*} and \begin{align*}d\end{align*}.
Corollary 7-3: If \begin{align*}a, b, c,\end{align*} and \begin{align*}d\end{align*} are nonzero and \begin{align*}\frac{a}{b} = \frac{c}{d}\end{align*}, then \begin{align*}\frac{b}{a} = \frac{c}{d}\end{align*}. Flip each ratio upside down.
In each corollary, you will still end up with \begin{align*}ad=bc\end{align*} after cross-multiplying.
Example 9: Suppose we have the proportion \begin{align*}\frac{2}{5} = \frac{14}{35}\end{align*}. Write three true proportions that follow.
Solution: First of all, we know this is a true proportion because you would multiply \begin{align*}\frac{2}{5}\end{align*} by \begin{align*}\frac{7}{7}\end{align*} to get \begin{align*}\frac{14}{35}\end{align*}. Using the three corollaries:
1. \begin{align*}\frac{2}{14} = \frac{5}{35}\end{align*}
2. \begin{align*}\frac{35}{5} = \frac{14}{2}\end{align*}
3. \begin{align*}\frac{5}{2} = \frac{35}{14}\end{align*}
Corollary 7-4: If \begin{align*}a, b, c,\end{align*} and \begin{align*}d\end{align*} are nonzero and \begin{align*}\frac{a}{b} = \frac{c}{d}\end{align*}, then \begin{align*}\frac{a+b}{b} = \frac{c+d}{d}\end{align*}.
Corollary 7-5: If \begin{align*}a, b, c,\end{align*} and \begin{align*}d\end{align*} are nonzero and \begin{align*}\frac{a}{b} = \frac{c}{d}\end{align*}, then \begin{align*}\frac{a-b}{b} = \frac{c-d}{d}\end{align*}.
Example 10: In the picture, \begin{align*}\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}\end{align*}.
Find the measures of \begin{align*}AC\end{align*} and \begin{align*}XY\end{align*}.
Solution: Plug in the lengths of the sides we know.
Example 11: In the picture, \begin{align*}\frac{ED}{AD} = \frac{BC}{AC}\end{align*}. Find \begin{align*}y\end{align*}.
Solution: Substitute in the lengths of the sides we know.
\begin{align*}\frac{6}{y} = \frac{8}{12+8} \ \longrightarrow \ 8y &= 6(20)\\ y &= 15\end{align*}
Example 12: If \begin{align*}\frac{AB}{BE} = \frac{AC}{CD}\end{align*} in the picture above, find \begin{align*}BE\end{align*}.
Solution:
\begin{align*}\frac{12}{BE} = \frac{20}{25} \ \longrightarrow \ {20(BE)} & = 12(25)\\ BE & = 15\end{align*}
Know What? Revisited Everything needs to be scaled down by a factor of \begin{align*}\frac{1}{18} (144 \ in \div 8 \ in)\end{align*}. Change everything into inches and then multiply by the scale factor.
\begin{align*}\underline{\text{Bed}}: \ 36 \ \text{in by}\ 75 \ \text{in} \ \longrightarrow \ 2 \ \text{in by}\ 4.167 \ \text{in}\!\\ \underline{\text{Desk}}: \ 48 \ \text{in by}\ 24 \ \text{in} \ \longrightarrow 2.67 \ \text{in by}\ 1.33 \ \text{in}\end{align*}
## Review Questions
• Questions 1 and 2 are similar to Examples 1-4.
• Questions 7-13 are similar to Example 5.
• Questions 14-19, 26, and 27 are similar to Example 6 and 8.
• Questions 20-25 are similar to Example 7.
• Questions 28-31 are similar to Example 9
• Questions 32-35 are similar to Examples 10-12.
1. The votes for president in a club election were: \begin{align*}\text{Smith}:24 \qquad \text{Munoz}:32 \qquad \text{Park}:20\end{align*} Find the following ratios and write in simplest form.
1. Votes for Munoz to Smith
2. Votes for Park to Munoz
4. Votes for Smith to Munoz to Park
Use the picture to write the following ratios for questions 2-6.
\begin{align*}AEFD \ \text{is a square} \qquad ABCD \ \text{is a rectangle}\end{align*}
1. \begin{align*}AE:EF\end{align*}
2. \begin{align*}EB:AB\end{align*}
3. \begin{align*}DF:FC\end{align*}
4. \begin{align*}EF:BC\end{align*}
5. Perimeter \begin{align*}ABCD\end{align*}:Perimeter \begin{align*}AEFD\end{align*}:Perimeter \begin{align*}EBCF\end{align*}
Convert the following measurements.
1. 16 cups to gallons
2. 8 yards to feet
3. 6 meters to centimeters
Simplify the following ratios.
1. \begin{align*}\frac{25 \ in}{5 \ ft}\end{align*}
2. \begin{align*}\frac{8 \ pt}{2 \ gal}\end{align*}
3. \begin{align*}\frac{9 \ ft}{3 \ yd}\end{align*}
4. \begin{align*}\frac{95 \ cm}{1.5 \ m}\end{align*}
1. The measures of the angles of a triangle are have the ratio 3:3:4. What are the measures of the angles?
2. The length and width of a rectangle are in a 3:5 ratio. The perimeter of the rectangle is 64. What are the length and width?
3. The length and width of a rectangle are in a 4:7 ratio. The perimeter of the rectangle is 352. What are the length and width?
4. A math class has 36 students. The ratio of boys to girls is 4:5. How many girls are in the class?
5. The senior class has 450 students in it. The ratio of boys to girls is 8:7. How many boys are in the senior class?
6. The varsity football team has 50 players. The ratio of seniors to juniors is 3:2. How many seniors are on the team?
Solve each proportion.
1. \begin{align*}\frac{x}{10} = \frac{42}{35}\end{align*}
2. \begin{align*}\frac{x}{x-2} = \frac{5}{7}\end{align*}
3. \begin{align*}\frac{6}{9} = \frac{y}{24}\end{align*}
4. \begin{align*}\frac{x}{9} = \frac{16}{x}\end{align*}
5. \begin{align*}\frac{y-3}{8} = \frac{y+6}{5}\end{align*}
6. \begin{align*}\frac{20}{z+5} = \frac{16}{7}\end{align*}
7. Shawna drove 245 miles and used 8.2 gallons of gas. At the same rate, if she drove 416 miles, how many gallons of gas will she need? Round to the nearest tenth.
8. The president, vice-president, and financial officer of a company divide the profits is a 4:3:2 ratio. If the company made \$1,800,000 last year, how much did each person receive?
Given the true proportion, \begin{align*}\frac{10}{6}= \frac{15}{d} = \frac{x}{y}\end{align*} and \begin{align*}d, x,\end{align*} and \begin{align*}y\end{align*} are nonzero, determine if the following proportions are also true.
1. \begin{align*}\frac{10}{y} = \frac{x}{6}\end{align*}
2. \begin{align*}\frac{15}{10} = \frac{d}{6}\end{align*}
3. \begin{align*}\frac{6+10}{10} = \frac{y+x}{x}\end{align*}
4. \begin{align*}\frac{15}{x} = \frac{y}{d}\end{align*}
For questions 32-35, \begin{align*}\frac{AE}{ED} = \frac{BC}{CD}\end{align*} and \begin{align*}\frac{ED}{AD} = \frac{CD}{DB} = \frac{EC}{AB}\end{align*}.
1. Find \begin{align*}DB\end{align*}.
2. Find \begin{align*}EC\end{align*}.
3. Find \begin{align*}CB\end{align*}.
4. Find \begin{align*}AD\end{align*}.
1. Yes, they are congruent by SAS.
2. \begin{align*}GI = 5\end{align*} by CPCTC
1. 12 in = 1 ft
2. 36 in = 3 ft
3. 108 in = 3 yd
4. 60 in = 5 ft.
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# What is the value of log 1
## Reciprocal theorems of the logarithm
In addition to the Laws of logarithms the so-called also help you to calculate with logarithms Reciprocal rates. First of all, you should remember what the different parts of the logarithm are called. In the case of the reciprocal of the logarithms, the various variables swap their place and you can quickly lose track of things.
Logarithm with base a and number b.
The variable \$ \ textcolor {blue} {a} \$ becomes Base called the variable \$ \ textcolor {black} {b} \$ number or something out of date too Logarithm. The question behind the logarithm is: With which number do I have to increase the base \$ \ textcolor {blue} {a} \$ to get the number \$ \ textcolor {black} {b} \$?
What is the reciprocal?
\$ x ~ \ rightarrow \ frac {1} {x} \$
### Reciprocal of a logarithm
Logarithms can be calculated by swapping base and number and taking the reciprocal.
\$ \ log _ {\ textcolor {blue} {a}} (\ textcolor {black} {b}) ~ = ~ \ frac {1} {\ log _ {\ textcolor {black} {b}} (\ textcolor {blue} {a})} \$
This reciprocal rate is a special case of the basic exchange rate. We choose the number as the new basis.
Basic exchange rate
In the event that a logarithm for the base \$ \ textcolor {blue} {a} \$ is unknown, it can be converted into a quotient of two logarithms for any base (\$ \ textcolor {green} {c} \$).
\$ \ log _ {\ textcolor {blue} {a}} (\ textcolor {black} {b}) ~ = ~ \ frac {\ log _ {\ textcolor {green} {c}} (\ textcolor {black} {b} )} {\ log _ {\ textcolor {black} {\ textcolor {green} {c}}} (\ textcolor {blue} {a})} \$
So we can use the basic exchange theorem to prove the first reciprocal theorem:
\$ \ log_ {a} (b) = \ frac {\ log_ {b} (b)} {\ log_ {b} (a)} \$
Since \$ \ log_ {b} (b) ~ = ~ 1 \$ we get the proposition:
\$ \ log_ {a} (b) = \ frac {1} {\ log_ {b} (a)} \$
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# Introduction to the Decimal System: Golden Bead Material & Decimal Cards
### What is the Decimal System?
The decimal number system, which we commonly use, employs digits ranging from 0 to 9. In this number system, the base number is 10, signifying that there are a total of 10 different digits available for representing numbers.
In Montessori, the decimal system is a cornerstone of mathematical education. It serves as the foundation for understanding numbers and place value, paving the way for more advanced mathematical concepts. This system is not just about numbers; it’s about nurturing a deep comprehension of how our numerical system works.
### Introduction to the Decimal System
Prior to this, the child should have lots of exploration of numbers one through ten to acquire a solid grasp of various fundamental concepts, including quantity, numerical symbols from 0 to 10, the association between quantity and symbols, and the notion of zero.
This early exposure also indirectly imparts insights into the four fundamental mathematical operations, sequential order, and the distinction between odd and even numbers. At this stage, children comprehend that quantity is composed of individual units.
In our Montessori approach, we typically progress from Numbers 1-10 directly to the decimal system, bypassing the intermediary step of working with numbers in the teens and tens. However, it’s important to recognize that some children may benefit from an alternative route, initially engaging with the more straightforward numbers before moving on to larger quantities.
Many children find the manipulation of large quantities with our bead materials particularly engaging. The decision to transition to the decimal system aligns with the child’s absorbent mind, sensitive periods for order, language development, and refinement of the senses. It allows us to introduce bead materials on a primarily sensory level, where the child can visually appreciate the relative differences in volume, supporting the development of the mathematical mind.
### Principles to Remember
1. Hierarchies and Categories:
Recognize that the decimal system is structured around hierarchies and categories. The three primary categories are units, tens, and hundreds. These categories repeat in each hierarchy, whether you’re working with the simple hierarchy or larger ones like thousands and millions. Understanding this hierarchical structure is fundamental to grasping the decimal system.
2. Comprehensive Understanding:
Resist the urge to move on to mathematical operations prematurely. It’s essential to ensure that the child fully comprehends the categories before introducing operations like addition, subtraction, multiplication, and division. Patience is key, even if it may feel repetitive.
Be aware that children may occasionally confuse the terms “hundreds” and “thousands.” In such cases, it can be helpful to explicitly count the zeroes to reinforce the distinction between these categories.
4. Speaking Whole Numbers:
Teach children that when saying whole numbers out loud, the word “and” is not used. The word “and” is reserved for discussing fractions, ensuring clarity in language usage.
### Introduction to the Golden Beads
The first material introduced is the golden bead presentation tray, which contains one representative of each category: a single unit bead, one ten-bar, one hundred-square, and one thousand-cube. In our teaching, we prefer the use of glass beads for a more profound sensory experience. By presenting this material, the child gains an initial understanding of the varied scales of these components. These materials are designed to visually and tangibly represent the concept of place value.
The wonder of golden beads lies in their ability to demonstrate place value. Each component represents a different place in the decimal system, allowing children to see how numbers are composed of units, tens, hundreds, and thousands. This tangible representation makes the abstract concept of place value concrete and comprehensible.
Armed with new information of hundred and thousand, the child can extend their understanding up to one million. They accomplish this by counting these components just as they did with the numbers 1-10, with a clear progression from 1-9 tens to 10 tens (equivalent to 1 hundred), and so forth.
The beauty of this material lies in the fact that the child is never compelled to count beyond 10, as adding 1 to 9 of one category naturally yields 1 of the next. This approach ensures that the child counts with genuine comprehension and meaning.
Occasionally, children may struggle to recall the names of the newer categories, specifically “hundred” and “thousand.” To reinforce this knowledge, we engage in many fetching games designed to solidify their understanding of these terms.
### Introduction to the Decimal Cards
Decimal cards are another valuable tool in the Montessori classroom. These cards visually represent numbers, reinforcing the concept of place value introduced through golden beads. Decimal cards are typically color-coded to help children distinguish between different place values.
Units: green
Tens: blue
Hundreds: red
Thousands: green
### Combining Golden Beads and Decimal Cards or the Formation of Numbers
Golden beads and decimal cards complement each other seamlessly. Children can pair the tactile experience of manipulating golden beads with the visual representation of numbers on decimal cards. This dual approach enhances their understanding and helps consolidate their knowledge of the decimal system.
### DIY Golden Bead Cards and Decimal Cards
As an alternative approach to introducing the decimal system in Montessori education, one can explore the use of DIY golden beads and DIY decimal cards. These homemade materials offer a cost-effective way to provide children with hands-on experiences in comprehending the intricacies of the decimal system.
Crafting DIY golden beads and DIY decimal cards using paper or cardstock allows for a personalized and creative touch to the learning process. Though using alternative golden beads may not give the same impression in both appearance and weight, this DIY approach of also the decimal cards not only promotes resourcefulness but also encourages educators and parents to actively engage in creating educational materials that facilitate a deep understanding of mathematics for young learners.
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# 4.1: Line Integrals
In single-variable calculus you learned how to integrate a real-valued function $$f (x)$$ over an interval $$[a,b]$$ in $$\mathbb{R}^1$$ . This integral (usually called a Riemann integral) can be thought of as an integral over a path in $$\mathbb{R}^1$$ , since an interval (or collection of intervals) is really the only kind of “path” in $$\mathbb{R}^1$$ . You may also recall that if $$f (x)$$ represented the force applied along the $$x$$-axis to an object at position $$x$$ in $$[a,b]$$, then the work $$W$$ done in moving that object from position $$x = a \text{ to }x = b$$ was defined as the integral:
$\nonumber W=\int_a^b f (x)dx$
In this section, we will see how to define the integral of a function (either real-valued or vector-valued) of two variables over a general path (i.e. a curve) in $$\mathbb{R}^2$$ . This definition will be motivated by the physical notion of work. We will begin with real-valued functions of two variables.
In physics, the intuitive idea of work is that
$\nonumber \text{Work = Force × Distance}$
Suppose that we want to find the total amount $$W$$ of work done in moving an object along a curve $$C$$ in $$\mathbb{R}^2$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$, with a force $$f (x, y)$$ which varies with the position $$(x, y)$$ of the object and is applied in the direction of motion along $$C$$ (see Figure 4.1.1 below).
Figure 4.1.1 Curve $$C : x = x(t),\, y = y(t) \text{ for }t \text{ in }[a,b]$$
We will assume for now that the function $$f (x, y)$$ is continuous and real-valued, so we only consider the magnitude of the force. Partition the interval $$[a,b]$$ as follows:
$a = t_0 < t_1 < t_2 < ··· < t_{n−1} < t_n = b ,\text{ for some integer }n ≥ 2$
As we can see from Figure 4.1.1, over a typical subinterval $$[t_i ,t_{i+1}]$$ the distance $$∆s_i$$ traveled along the curve is approximately $$\sqrt{∆x_i^2 +∆y_i^2}$$ , by the Pythagorean Theorem. Thus, if the subinterval is small enough then the work done in moving the object along that piece of the curve is approximately
$\text{Force × Distance} \approx f (x_{i∗}, y_{i∗}) \sqrt{ ∆x_i^2 +∆y_i^2}\label{Eq4.1}$
where $$(x_{i∗}, y_{i∗}) = (x(t_{i∗}), y(t_{i∗}))$$ for some $$t_{i∗} \text{ in }[t_i ,t_{i+1}]$$, and so
$W \approx \sum_{i=0}^{n-1} f (x_{i∗}, y_{i∗}) \sqrt{ ∆x_i^2 +∆y_i^2}\label{Eq4.2}$
is approximately the total amount of work done over the entire curve. But since
$\nonumber \sqrt{ ∆x_i^2 +∆y_i^2} = \sqrt{\left ( \dfrac{∆x_i}{∆t_i} \right )^2 +\left ( \dfrac{∆y_i}{∆t_i}\right )^2}∆t_i$
where $$∆t_i = t_{i+1} − t_i$$ , then
$W \approx \sum_{i=0}^{n-1}f (x_{i∗}, y_{i∗})\sqrt{\left ( \dfrac{∆x_i}{∆t_i} \right )^2 + \left ( \dfrac{∆y_i}{∆t_i} \right )^2}∆t_i \label{Eq4.3}$
Taking the limit of that sum as the length of the largest subinterval goes to 0, the sum over all subintervals becomes the integral from $$t = a \text{ to }t = b$$, $$∆x_i ∆t_i \text{ and }∆y_i ∆t_i$$ become $$x ′ (t) \text{ and }y ′ (t)$$, respectively, and $$f (x_{i∗}, y_{i∗})$$ becomes $$f (x(t), y(t))$$, so that
$W=\int_a^b f (x(t), y(t)) \sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt \label{Eq4.4}$
The integral on the right side of the above equation gives us our idea of how to define, for any real-valued function $$f (x, y)$$, the integral of $$f (x, y)$$ along the curve $$C$$, called a line integral:
Definition 4.1
For a real-valued function $$f (x, y)$$ and a curve $$C$$ in $$\mathbb{R}^2$$ , parametrized by $$x = x(t), y = y(t), a ≤ t ≤ b$$, the line integral of $$f (x, y)$$ along $$C$$ with respect to arc length $$s$$ is
$\int_C f (x, y)\,ds = \int_a^b f (x(t), y(t))\sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt \label{Eq4.5}$
The symbol $$ds$$ is the differential of the arc length function
$s = s(t) = \int_a^t \sqrt{x ′ (u)^2 + y ′ (u)^2}\,du \label{Eq4.6}$
which you may recognize from Section 1.9 as the length of the curve $$C$$ over the interval $$[a,t]$$, for all $$t$$ in $$[a,b]$$. That is,
$ds = s ′ (t)\,dt = \sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt, \label{Eq4.7}$
by the Fundamental Theorem of Calculus.
For a general real-valued function $$f (x, y)$$, what does the line integral $$\int_C f (x, y)\,ds$$ represent? The preceding discussion of $$ds$$ gives us a clue. You can think of differentials as infinitesimal lengths. So if you think of $$f (x, y)$$ as the height of a picket fence along $$C$$, then $$f (x, y)\,ds$$ can be thought of as approximately the area of a section of that fence over some infinitesimally small section of the curve, and thus the line integral $$\int_C f (x, y)\,ds$$ is the total area of that picket fence (see Figure 4.1.2).
Figure 4.1.2 Area of shaded rectangle = height × width ≈ $$f (x, y)\,ds$$
Example 4.1
Use a line integral to show that the lateral surface area $$A$$ of a right circular cylinder of radius $$r$$ and height $$h$$ is $$2\pi rh$$.
Solution:
We will use the right circular cylinder with base circle $$C$$ given by $$x^2 + y^2 = r^2$$ and with height $$h$$ in the positive $$z$$ direction (see Figure 4.1.3). Parametrize $$C$$ as follows:
$x = x(t) = r \cos t , y = y(t) = r \sin t , 0 ≤ t ≤ 2π$
Figure 4.1.3
\nonumber \begin{align} A&=\int_C f (x, y)\,ds = \int_a^b f (x(t), y(t))\sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt \\ \nonumber &=\int_0^{2\pi} h \sqrt{(−r \sin t)^2 +(r \cos t)^2}\,dt \\ \nonumber &=h\int_0^{2\pi} r \sqrt{\sin^2 t+\cos^2 t}\,dt \\ \nonumber &=rh\int_0^{2\pi} 1\,dt = 2\pi rh \\ \end{align}
Note in Example 4.1 that if we had traversed the circle $$C$$ twice, i.e. let t vary from $$0 \text{ to }4\pi$$ then we would have gotten an area of $$4\pi rh$$, i.e. twice the desired area, even though the curve itself is still the same (namely, a circle of radius $$r$$). Also, notice that we traversed the circle in the counter-clockwise direction. If we had gone in the clockwise direction, using the parametrization
$x = x(t) = r \cos (2π− t) , y = y(t) = r \sin (2π− t) , 0 ≤ t ≤ 2π ,\label{Eq4.8}$
then it is easy to verify (see Exercise 12) that the value of the line integral is unchanged.
In general, it can be shown (see Exercise 15) that reversing the direction in which a curve $$C$$ is traversed leaves $$\int_C f (x, y)\,ds$$ unchanged, for any $$f (x, y)$$. If a curve $$C$$ has a parametrization $$x = x(t), y = y(t), a ≤ t ≤ b,$$ then denote by $$−C$$ the same curve as $$C$$ but traversed in the opposite direction. Then $$−C$$ is parametrized by
$x = x(a+ b − t) , y = y(a+ b − t) , a ≤ t ≤ b ,\label{Eq4.9}$
and we have
$\int_C f (x, y)\,ds =\int_{-C}f (x, y)\,ds .\label{Eq4.10}$
Notice that our definition of the line integral was with respect to the arc length parameter $$s$$. We can also define
$\int_C f (x, y)\,dx=\int_a^b f (x(t), y(t)) x ′ (t)\,dt\label{Eq4.11}$
as the line integral of $$f (x, y)$$ along $$C$$ with respect to $$x$$, and
$\int_C f (x, y)\,d y=\int_a^b f (x(t), y(t)) y ′ (t)\,dt \label{Eq4.12}$
as the line integral of $$f (x, y)$$ along $$C$$ with respect to $$y$$.
In the derivation of the formula for a line integral, we used the idea of work as force multiplied by distance. However, we know that force is actually a vector. So it would be helpful to develop a vector form for a line integral. For this, suppose that we have a function $$f(x, y)$$ defined on $$\mathbb{R}^2$$ by
$\nonumber \textbf{f}(x, y) = P(x, y)\textbf{i} + Q(x, y)\textbf{j}$
for some continuous real-valued functions $$P(x, y)$$ and $$Q(x, y) \text{ on }\mathbb{R}^2$$ . Such a function $$f$$ is called a vector field on $$\mathbb{R}^2$$ . It is defined at points in $$\mathbb{R}^2$$ , and its values are vectors in $$\mathbb{R}^2$$ . For a curve $$C$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$, let
$\nonumber \textbf{r}(t) = x(t)\textbf{i} + y(t)\textbf{j}$
be the position vector for a point $$(x(t), y(t))$$ on $$C$$. Then $$\textbf{r}'(t) = x'(t)\textbf{i} + y'(t)\textbf{j}$$ and so
\nonumber \begin{align} \int_C P(x, y)\,dx+ \int_C Q(x, y)\,d y &=\int_a^b P(x(t), y(t)) x ′ (t)\,dt+\int_a^b Q(x(t), y(t)) y ′ (t)\,dt \\ \nonumber &=\int_a^b (P(x(t), y(t)) x ′ (t)+Q(x(t), y(t)) y ′ (t))\,dt \\ \nonumber &=\int_a^b \textbf{f}(x(t), y(t))\cdot \textbf{r} ′ (t)dt \\ \end{align}
by definition of $$f(x, y)$$. Notice that the function $$f(x(t), y(t))\cdot r ′ (t)$$ is a real-valued function on $$[a,b]$$, so the last integral on the right looks somewhat similar to our earlier definition of a line integral. This leads us to the following definition:
Definition 4.2
For a vector field $$\textbf{f}(x, y) = P(x, y)\textbf{i} +Q(x, y)\textbf{j}$$ and a curve $$C$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$, the line integral of f along $$C$$ is
\begin{align} \int_C \textbf{f}\cdot d\textbf{r} &= \int_C P(x, y)\,dx+\int_C Q(x, y)\,d y \label{Eq4.13} \\ &=\int_a^b \textbf{f}(x(t), y(t))\cdot \textbf{r} ′ (t)\,dt \label{Eq4.14} \\ \end{align}
where $$\textbf{r}(t) = x(t)\textbf{i}+ y(t)\textbf{j}$$ is the position vector for points on $$C$$.
We use the notation $$d\textbf{r} = \textbf{r} ′ (t)\,dt = dx\textbf{i}+ d y\textbf{j}$$ to denote the differential of the vector-valued function r. The line integral in Definition 4.2 is often called a line integral of a vector field to distinguish it from the line integral in Definition 4.1 which is called a line integral of a scalar field. For convenience we will often write
$\nonumber \int_C P(x, y)\,dx +\int_C Q(x, y)\,d y =\int_C P(x, y)\,dx+Q(x, y)\,d y ,$
where it is understood that the line integral along $$C$$ is being applied to both $$P \text{ and }Q$$. The quantity $$P(x, y)\,dx +Q(x, y)\,d y$$ is known as a differential form. For a real-valued function $$F(x, y)$$, the differential of $$F$$ is $$dF = \dfrac{∂F}{∂x}\,dx+ \dfrac{∂F}{∂y}\, d y.$$ A differential form $$P(x, y)\,dx+Q(x, y)\,d y$$ is called exact if it equals $$dF$$ for some function $$F(x, y)$$.
Recall that if the points on a curve $$C$$ have position vector $$\textbf{r}(t) = x(t)\textbf{i}+ y(t)\textbf{j}$$, then $$\textbf{r} ′ (t)$$ is a tangent vector to $$C$$ at the point $$(x(t), y(t))$$ in the direction of increasing $$t$$ (which we call the direction of $$C$$). Since $$C$$ is a smooth curve, then $$\textbf{r} ′ (t) \neq \textbf{0} \text{ on }[a,b]$$ and hence
$\nonumber \textbf{T}(t) = \dfrac{\textbf{r}'(t)}{\left \lVert \textbf{r}'(t) \right \rVert}$
is the unit tangent vector to $$C$$ at $$(x(t), y(t))$$. Putting Definitions 4.1 and 4.2 together we get the following theorem:
Theorem 4.1
For a vector field $$\textbf{f}(x, y) = P(x, y)\textbf{i} + Q(x, y)\textbf{j}$$ and a curve $$C$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$ and position vector $$\textbf{r}(t) = x(t)\textbf{i}+ y(t)\textbf{j}$$,
$\int_C \textbf{f}\cdot d\textbf{r} = \int_C \textbf{f}\cdot \textbf{T}\,ds,\label{Eq4.15}$
where $$\textbf{T}(t) = \dfrac{\textbf{r} ′ (t)}{ \left \lVert \textbf{r} ′ (t)\right \rVert }$$ is the unit tangent vector to $$C$$ at $$(x(t), y(t))$$.
If the vector field $$\textbf{f}(x, y)$$ represents the force moving an object along a curve $$C$$, then the work $$W$$ done by this force is
$W = \int_C \textbf{f}\cdot \textbf{T} \, ds = \int_C \textbf{f}\cdot d\textbf{r} \label{Eq4.16}$
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# Vertical Angles and Linear Pairs
## Presentation on theme: "Vertical Angles and Linear Pairs"— Presentation transcript:
Vertical Angles and Linear Pairs
Previously, you learned that two angles are adjacent if they share a common vertex and side but have no common interior points. In this lesson, you will study other relationships between pairs of angles. Two angles are vertical angles if their sides form two pairs of opposite rays. 1 4 3 2 1 and 3 are vertical angles. 2 and 4 are vertical angles.
Vertical Angles and Linear Pairs
Previously, you learned that two angles are adjacent if they share a common vertex and side but have no common interior points. In this lesson, you will study other relationships between pairs of angles. Two angles are vertical angles if their sides form two pairs of opposite rays. Two adjacent angles are a linear pair if their noncommon sides are opposite rays. 1 4 2 5 6 3 1 and 3 are vertical angles. 5 and 6 are a linear pair. 2 and 4 are vertical angles.
Identifying Vertical Angles and Linear Pairs
Answer the questions using the diagram. Are 2 and 3 a linear pair? 1 2 4 3 SOLUTION No. The angles are adjacent but their noncommon sides are not opposite rays.
Identifying Vertical Angles and Linear Pairs
Answer the questions using the diagram. Are 2 and 3 a linear pair? 1 2 4 3 Are 3 and 4 a linear pair? SOLUTION No. The angles are adjacent but their noncommon sides are not opposite rays. Yes. The angles are adjacent and their noncommon sides are opposite rays.
Identifying Vertical Angles and Linear Pairs
Answer the questions using the diagram. Are 2 and 3 a linear pair? 1 2 4 3 Are 3 and 4 a linear pair? Are 1 and 3 vertical angles? SOLUTION No. The angles are adjacent but their noncommon sides are not opposite rays. Yes. The angles are adjacent and their noncommon sides are opposite rays. No. The sides of the angles do not form two pairs of opposite rays.
Identifying Vertical Angles and Linear Pairs
Answer the questions using the diagram. Are 2 and 3 a linear pair? 1 2 4 3 Are 3 and 4 a linear pair? Are 1 and 3 vertical angles? Are 2 and 4 vertical angles? SOLUTION No. The angles are adjacent but their noncommon sides are not opposite rays. Yes. The angles are adjacent and their noncommon sides are opposite rays. No. The sides of the angles do not form two pairs of opposite rays. No. The sides of the angles do not form two pairs of opposite rays.
Finding Angle Measures
In the stair railing shown, 6 has a measure of 130˚. Find the measures of the other three angles. SOLUTION 5 6 7 8 6 and 7 are a linear pair. So, the sum of their measures is 180˚. m6 + m7 = 180˚ 130˚ + m7 = 180˚ m7 = 50˚
Finding Angle Measures
In the stair railing shown, 6 has a measure of 130˚. Find the measures of the other three angles. SOLUTION 5 6 7 8 6 and 7 are a linear pair. So, the sum of their measures is 180˚. m6 + m7 = 180˚ 130˚ + m7 = 180˚ m7 = 50˚ 6 and 5 are also a linear pair. So it follows that m5 = 50˚.
Finding Angle Measures
In the stair railing shown, 6 has a measure of 130˚. Find the measures of the other three angles. SOLUTION 5 6 7 8 6 and 8 are vertical angles. So, they are congruent and have the same measure. m 8 = m 6 = 130˚
• B Finding Angle Measures
Solve for x and y. Then find the angle measure. ( x + 15)˚ ( 3x + 5)˚ ( y + 20)˚ ( 4y – 15)˚ D C • B A • E Use the fact that the sum of the measures of angles that form a linear pair is 180˚. SOLUTION Use substitution to find the angle measures (x = 40, y = 35). m AED = ( 3 x + 15)˚ = (3 • )˚ = 125˚ m AED + m DEB = 180° m AEC + mCEB = 180° m DEB = ( x + 15)˚ = ( )˚ = 55˚ ( 3x + 5)˚ + ( x + 15)˚ = 180° ( y + 20)˚ + ( 4y – 15)˚ = 180° m AEC = ( y + 20)˚ = ( )˚ = 55˚ 4x + 20 = 180 5y + 5 = 180 m CEB = ( 4 y – 15)˚ = (4 • 35 – 15)˚ 4x = 160 = 125˚ 5y = 175 So, the angle measures are 125˚, 55˚, 55˚, and 125˚. Because the vertical angles are congruent, the result is reasonable. x = 40 y = 35
Complementary and Supplementary Angles
Two angles are complementary angles if the sum of their measurements is 90˚. Each angle is the complement of the other. Complementary angles can be adjacent or nonadjacent. 1 2 3 4 complementary adjacent complementary nonadjacent
Complementary and Supplementary Angles
Two angles are supplementary angles if the sum of their measurements is 180˚. Each angle is the supplement of the other. Supplementary angles can be adjacent or nonadjacent. 7 8 5 6 supplementary adjacent supplementary nonadjacent
Identifying Angles State whether the two angles are complementary, supplementary, or neither. SOLUTION The angle showing 4:00 has a measure of 120˚ and the angle showing 10:00 has a measure of 60˚. Because the sum of these two measures is 180˚, the angles are supplementary.
Finding Measures of Complements and Supplements
Find the angle measure. Given that A is a complement of C and m A = 47˚, find mC. SOLUTION mC = 90˚ – m A = 90˚ – 47˚ = 43˚
Finding Measures of Complements and Supplements
Find the angle measure. Given that A is a complement of C and m A = 47˚, find mC. Given that P is a supplement of R and mR = 36˚, find mP. SOLUTION mC = 90˚ – m A mP = 180˚ – mR = 90˚ – 47˚ = 180 ˚ – 36˚ = 43˚ = 144˚
Finding the Measure of a Complement
W and Z are complementary. The measure of Z is 5 times the measure of W. Find m W SOLUTION Because the angles are complementary, m W + m Z = 90˚. But m Z = 5( m W ), so m W + 5( m W) = 90˚. Because 6(m W) = 90˚, you know that m W = 15˚.
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# What Is 30 Of 5? Explained In Relaxed English
## Introduction
If you’re wondering what 30 of 5 means, you’re not alone. Many people struggle with mathematical concepts, and percentages can be particularly tricky. In this article, we’ll explain what 30 of 5 is, and how you can calculate it. We’ll break it down into simple terms, so even those who struggle with math can understand.
### What is a Percentage?
Before we dive into what 30 of 5 means, we need to understand what a percentage is. A percentage is a way of expressing a number as a fraction of 100. For example, if you have 25%, it means you have 25 out of 100.
### What Does “of” Mean?
When we see the word “of” in a mathematical expression, it means multiplication. So, if we have 30 of 5, it means 30 multiplied by 5.
### Calculating 30 of 5
To calculate 30 of 5, we simply need to multiply 30 by 5. This gives us: 30 x 5 = 150 So, 30 of 5 is 150.
## Why is 30 of 5 Important?
You might be wondering why anyone would care about 30 of 5. Well, percentages and fractions are used in many areas of life. For example, if you’re shopping and something is 30% off, you need to know how much you’ll be saving. Or, if you’re cooking and a recipe calls for 30% less salt, you need to know how much salt to use.
### Real-World Examples
Let’s look at a few examples of how 30 of 5 might be used in the real world. Example 1: Shopping You’re shopping for a new shirt, and you see one that’s 30% off. The shirt originally costs \$50. How much will you pay for the shirt? To find out, we need to calculate 30% of \$50. This means we need to multiply \$50 by 30%, which is 0.30. So, we get: \$50 x 0.30 = \$15 So, the shirt will cost \$35 after the discount. Example 2: Cooking You’re making a recipe that calls for 30% less sugar than the original recipe. The original recipe calls for 1 cup of sugar. How much sugar should you use? To find out, we need to calculate 30% of 1 cup. This means we need to multiply 1 cup by 30%, which is 0.30. So, we get: 1 cup x 0.30 = 0.3 cups So, we should use 0.7 cups of sugar in the recipe.
## Conclusion
In conclusion, 30 of 5 means 30 multiplied by 5, which equals 150. Understanding percentages and fractions is important in many areas of life, from shopping to cooking. We hope this article has helped you understand what 30 of 5 means, and how to calculate it.
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# NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 6 Triangles Class 10 NCERT Solutions Ex 6.5.
Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter Chapter 6 Chapter Name Triangles Exercise Ex 6.5 Number of Questions Solved 17 Category NCERT Solutions
## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5
Page No: 150
Question 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 that sides are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides we get 49, 576, and 625.
Clearly, 49 + 576 = 625 or 72 + 242 = 252 .
Therefore, given triangle is satisfying Pythagoras theorem. So, it is a right triangle. The longest side in a right angled triangle is the hypotenuse.
Therefore length of hypotenuse of this triangle = 25 cm.
(ii).Given that sides are 3 cm, 8 cm, and 6 cm. Squaring the lengths of these sides we may get 9, 64, and 36. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle
(iii).Given that sides are 50 cm, 80 cm, and 100 cm. Squaring the lengths of these sides we may get 2500, 6400, and 10000. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle.
(iv).Given that sides are 13 cm, 12 cm, and 5 cm. Squaring the lengths of these sides we may get 169, 144, and 25. Clearly, 144 +25 = 169 Or, 122 + 52 = 132.
Therefore given triangle is satisfying Pythagoras theorem. So, it is a right triangle.
The longest side in a right angled triangle is the hypotenuse.
Therefore length of hypotenuse of this triangle = 13 cm.
Question 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:
Question 3. 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:
iii. In ∆DCA & ∆DAB
∠DCA = ∠DAB = 90º
∠DAC = ∠DBA (remaining angle)
Question 4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 .
Solution:
Question 5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Solution:
Question 6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
Question 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:
In ∆AOB, ∆BOC, ∆COD, ∆AOD
Applying Pythagoras theorem
Page No: 151
Question 8. 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:
Question 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:
Question 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:
Question 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:
Question 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:
Question 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:
Question 14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig. 6.55). Prove that 2AB2 = 2AC2+ BC2.
Solution:
Question 15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.
Solution:
Question 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:
Question 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:
We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5, drop a comment below and we will get back to you at the earliest.
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# In Media Res, or a running start
Let’s look at a system of two equations in two unknowns. We have encountered this topic many times in earlier years, and we have a number of ways to ‘solve’ the system.
$2x - y = 0 \qquad \cdots$ (1)
$-x + 2y = 3 \qquad \cdots$ (2)
Before we go on, let’s solve this system of equations in two different ways that we already know about.
First Method. If we had seen these two equations in middle school, we would have said “Those are the equations of two lines. Two (non-parallel) lines intersect in a point.” And you would be right. We can indeed graph these two lines. One goes through the origin, while the other does not. Using the graphical method, we can identify the point of intersection as the point $(1,2)$. In other words, the point $(1,2)$ is the one and only point that solves both of the equations we started with.
Second Method. (You may know this as Gaussian Elimination; if not, don’t worry ). Note that by solving this system of equations, we are finding the intersection point of two lines, just like before.
If we multiply equation (2) by 2, we can then add equations (1) and (2) together in order to eliminate the $x$ term. We are left with
$3y = 6$ which we can solve by inspection as $y = 2$.
Knowing that $y = 2$, we can substitute for $y$ in either of the equations we started with:
$2x - 2 = 0$, or $x = 1$, giving us the full solution, or point of intersection as $(1,2)$.
We are now ready to start thinking about this same problem viewed through the lens of linear algebra.
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Q) In figure 1, a right triangle ABC in which ∠B = 90 is shown. Taking AB as diameter, a circle has been drawn intersecting AC at point P. Prove that the tangent drawn at point P bisects BC.
Ans: Here is the step by step solution method:
Step 1:
We can see that, Q is an external point to the circle and PQ and BQ are the tangents drawn on a circle
∴ QB = QP………….(i)
Let’s connect Point P and B:
Now since, in a triangle, if the sides are equal, their opposite angles will also be equal,
∴ ∠QBP = ∠BPQ …………… (ii)
Step 2:
Next, since AB is the diameter of the circle and teh angle made by diameter on a cirle is 90,
∴ ∠APB = 90
Next, On point P we have a pair of angles
∴ ∠APB + ∠CPB = 180
∠CPB = 180 − 90 = 90 ……….(iii)
Next, we can see that Line PQ is dividing ∠CPB
∴ ∠CPQ + ∠BPQ = 90 ………..(iv)
Step 3:
Further in △CPB, by Angle sum property, we get:
∠CPB + ∠CBP + ∠BCP = 180
Substituting value of ∠CPB from equation (iii), we get:
90 + ∠CBP + ∠BCP = 180
∴ ∠CBP + ∠BCP = 180 − 90 = 90 ………..(v)
Step 4:
Now by comparing equations (iv) and (v), we get
∠CBP + ∠BCP = ∠CPQ + ∠BPQ ………… (vi)
Since, ∠CBP can be written as ∠QBP and ∠BCP can be written as ∠QCP, therefore equation (vi) can be written as:
∠QBP + ∠QCP = ∠CPQ + ∠BPQ ………… (vii)
Next, we substitute value of ∠QBP from equation (ii) in equation (vi), we get:
(∠BPQ) + ∠QCP = ∠CPQ + ∠BPQ
∴ ∠QCP = ∠CPQ ………… (viii)
Step 5:
Next in △PQC, if the angles are equal, their opposite sides will also be equal,
here, since ∠QCP = ∠CPQ
∴ QP = QC ………..(ix)
Now by comparing equations From (i) and (ix), we get
QB = QP = QC
∵ QB = QC,
Therefore, tangent to the circle PQ, bisects the side BC.
Hence Proved !
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Rational Expressions: Basic Operations
Multiplication and Division
In algebra we multiple, divide, add, and subtract fractions the same way
as we do in arithmetic.
and then, if possible, you try to reduce
the fraction to lowest terms.
But we can make the problem easier if we notice that we can immediately
cross-cancel common factors appearing in the numerators and in the denominators,
.
We should be alert to do this when we multiply or divide algebraic fractions.
The same rules of operation apply in algebra.
When we divide fractions, we invert the divisor and multiply.
For example,
Now factor everything that we can so that we can see what common factors,
if any, can be canceled out.
The factor appearing in the numerator and in the denominator cancel out,
.
See Examples 1 - 3, pages 34 - 36.
As in arithmetic when the denominators are not the same, we need to find the LCD,
the Lowest Common Denominator, to work out the problem.
Subtract:
First, we factor completely the denominators,
is the first denominator
is the second denominator.
Now we can write down the LCD in factored form. We will write it as a product.
If we understand the process for this relatively simple example, we will be able to
do any problem of adding or subtracting fractions. The basic idea will be the same,
no matter how many fractions are included in the problem or how complicated the
problem looks, the basic process will be the same.
Here is the process:
We must write down in an appropriate way all the factors
that appear in all the denominators:
appears as a factor in the first denominator and in the second denominator,
both times it is raised to the same power, namely, the first power, so we include it, raised
to the first, as a factor in the LCD.
appears as a factor in the first denominator only, it does not appear
as a factor in the second denominator, we include it as a factor in the LCD.
appears as factor in the second denominator only, it does not appear
as a factor in the first denominator, we include as a factor in the LCD.
Now there are no more factors that we must include in the LCD, so we are done.
There is a subtlety here that we must explain:
If a factor appears raised to the same power in any number of denominators,
it must be included as a factor raised to that same power in the LCD.
For example, if the denominators were
and
then
.
If a factor raised to different powers appears in any number of denominators,
it must be included raised to the largest power in the LCD.
If the denominators were
and
then
.
That's it! Now we will be able to write down the LCD of any number of fractions.
Let's get back to our original problem:
As in arithmetic, we must rewrite the first and the second fraction as
equivalent fractions with the LCD as their common denominator.
It is easy to do this, now that we have everything factored.
Look at the factored form of the first denominator and look at the LCD.
What factor do we see in the LCD that we don't see in the first denominator?
So we multiply it into the numerator and the denominator of the first fraction
in order to rewrite it as an equivalent fraction with the LCD as its denominator,
.
Likewise, what factor do we see in the LCD that we don't see in the second denominator?
So,
.
Now we can finish the problem:
When we combine the two fractions into one fraction, using the parentheses
to group the relevant terms in the numerator is extremely important.
Not using the parentheses may easily result in getting the wrong answer !
The next step is to drop the parentheses,
.
We must always try to reduce fractions to lowest terms.
If possible, we factor the numerator to see if we can cancel any factors common
to both the numerator and the denominator,
The factor in the top and the bottom cancel,
. Finished!
See Example 4, pages 37 - 38.
Complex Fractions
Another use of the LCD is to simplify complex, or compound, fractions.
A complex fraction is a fraction made up of fractions, appearing in the
numerator or in the denominator.
For example,
, , are complex fractions.
In general, the easiest way to simplify a complex fraction is to multiply
the top and the bottom of the complex fraction by the LCD of its constituent fractions.
The constituent fractions are the fractions inside the big fraction.
Look at the first, algebraic, fraction above.
Explicitly writing out the denominators, the constituent fractions
on the top of the complex fraction (the big fraction) are
and .
The constituent fractions on the bottom of the complex fraction are
and .
If we were to write down all the denominators we see in the complex fraction -
without the qualification that they are to be the denominators of the constituent
fractions - we would include
.
In this method, we use the LCD of the denominators of the constituent fractions,
.
Using the distributive property,
.
Finished!
See Examples 5 - 6, pages 39 - 40.
top
next Integer Exponents
|
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Introduction:
Probability is a super-fascinating topic that keeps on your toes. Along with mathematical ability, it requires a combination of quick thinking and reasoning skills in order to crack these exams quickly. An important part of CAT quantitative aptitude section, it is important you practice problems for this topic. We bring this five-problem set with the same intent and purpose.
Question 1: Let A and B be two events such that
Where Ā stands for complement of event A. Then, events A and B are
(a) Mutually exclusive
(b) Independent but not equally likely
(c) Equally likely but not independent
(d) Equally likely and mutually exclusive
### Answers and Explanations
Option (b)
Hence, the events A and B are independent events but not equally likely.
Question 2: Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house, is
(a) 7/9
(b) 8/9
(c) 1/9
(d) 2/9
### Answers and Explanations
Option c
All the three persons has three options to apply a house.
Therefore total number of cases = 33
Now, favorable cases = 3 (All can apply for house 1 or 2 or 3)
Therefore required probability = {3/(33)} = 1/9
Question 3: The probability that A speaks truth is 4/5 while this probability for B is 3/4. The probability that they contradict each other when asked to speak on a fact, is
(a) 7/20
(b) 1/5
(c) 3/20
(d) 4/5
### Answers and Explanations
Option (a)
Question 4: Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse, is
(a) 4/5
(b) 3/5
(c) 1/5
(d) 2/5
### Answers and Explanations
Option (d)
The probability that Mr. A selected the loosing horse = 4/5 x ¾ = 3/5
The probability that Mr. A selected the winning horse = {1 – (3/5)} = 2/5
Question 5: Events A, B, C are mutually exclusive events such that
The set of possible values of x are in the interval
(a) 1/3 , ½
(b) 1/3, 2/3
(c) 1/3, 13/3
(d) 0,1
### Answers and Explanations
Option (a)
Tips and Tricks you can use CAT Probability Problems
• Do remember that probability is not a pure mathematics topics. It requires you to combine reasoning skills with mathematical ability.
• Some important areas that you should cover to solve CAT Probability Problems: dices, playing cards, and more. A number of problems are based on these concepts for this topic.
• The best resource for learning the basics of Probability is class 11/12 books. You can refer to books by RD Sharma and NCERT to understand the basics of the topic.
• One of the best books for probability is by KC Sinha. It adopts a comprehensive approach for the topic and you can refer to it mastering the topic.
• Make sure you study Permutation and Combinations. Probability is based on the basics of P and C and therefore, you need to make sure you study this topic.
|
# Determine Where a Function is Increasing, Decreasing, or Constant
## Using a Graph to Determine Where a Function is Increasing, Decreasing, or Constant
As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3 shows examples of increasing and decreasing intervals on a function.
Figure 3 The function $f(x)=x^{3}-12 x$ is increasing on $(-\infty,-2) \cup(2, \infty)$ and is decreasing on $(-2,2)$.
While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is the location of a local maximum. The function value at that point is the local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is the location of a local minimum. The function value at that point is the local minimum. The plural form is "local minima". Together, local maxima and minima are called local extrema, or local extreme values, of the function. (The singular form is "extremum"). Often, the term local is replaced by the term relative. In this text, we will use the term local.
Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function's entire domain.
For the function whose graph is shown in Figure 4, the local maximum is 16, and it occurs at $x=-2$. The local minimum is $-16$ and it occurs at $x=2$.
Figure 4
To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure 5 illustrates these ideas for a local maximum.
Figure 5 Definition of a local maximum
#### LOCAL MINIMA AND LOCAL MAXIMA
A function $f$ is an increasing function on an open interval if $f(b) > f(a)$ for any two input values $a$ and $b$ in the given interval where $b > a$.
A function $f$ is a decreasing function on an open interval if $f(b) < f(a)$ for any two input values $a$ and $b$ in the given interval where $b > a$.
A function $f$ has a local maximum at $x=b$ if there exists an interval $(a, c)$ with $a < b < c$ such that, for any $x$ in the interval $(a, c), f(x) \leq f(b)$. Likewise, $f$ has a local minimum at $x=b$ if there exists an interval $(a, c)$ with $a < b < c$ such that, for any $x$ in the interval $(a, c), f(x) \geq f(b)$.
#### EXAMPLE 7
##### Finding Increasing and Decreasing Intervals on a Graph
Given the function $p(t)$ in Figure 6, identify the intervals on which the function appears to be increasing.
Figure 6
##### Solution
We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from $t=1$ to $t=3$ and from $t=4$ on.
In interval notation, we would say the function appears to be increasing on the interval $(1,3)$ and the interval $(4, \infty)$.
##### Analysis
Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at $t=1, t=3$, and $t=4$. These points are the local extrema (two minima and a maximum).
#### EXAMPLE 8
##### Finding Local Extrema from a Graph
Graph the function $f(x)=\dfrac{2}{x}+\dfrac{x}{3}$. Then use the graph to estimate the local extrema of the function and to determine the intervals on which the function is increasing.
##### Solution
Using technology, we find that the graph of the function looks like that in Figure 7 . It appears there is a low point, or local minimum, between $x=2$ and $x=3$, and a mirror-image high point, or local maximum, somewhere between $x=-3$ and $x=-2$.
Figure 7
##### Analysis
Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 8 provides screen images from two different technologies, showing the estimate for the local maximum and minimum.
Figure 8
Based on these estimates, the function is increasing on the interval $(-\infty,-2.449)$ and $(2.449, \infty)$. Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing approximation algorithms used by each. (The exact location of the extrema is at $\pm \sqrt{6}$, but determining this requires calculus).
#### TRY IT #4
Graph the function $f(x)=x^{3}-6 x^{2}-15 x+20$ to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing.
#### EXAMPLE 9
##### Finding Local Maxima and Minima from a Graph
For the function $f$ whose graph is shown in Figure 9, find all local maxima and minima.
Figure 9
##### Solution
Observe the graph of $f$. The graph attains a local maximum at $x=1$ because it is the highest point in an open interval around $x=1$. The local maximum is the $y$-coordinate at $x=1$, which is 2.
The graph attains a local minimum at $x=-1$ because it is the lowest point in an open interval around $x=-1$. The local minimum is the $y$-coordinate at $x=-1$, which is $-2$.
Source: Rice University, https://openstax.org/books/college-algebra/pages/3-3-rates-of-change-and-behavior-of-graphs
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## Where does maximum bending moment occur in a beam?
Explanation: The maximum bending moment occurs in a beam, when the shear force at that section is zero or changes the sign because at point of contra flexure the bending moment is zero. Explanation: The positive bending moment in a section is considered because it causes convexity downwards.
## How do you find the maximum stress of a beam?
Shear Stresses in Rectangular Sections where Ic = b·h3/12 is the centroidal moment of inertia of the cross section. The maximum shear stress occurs at the neutral axis of the beam and is calculated by: where A = b·h is the area of the cross section.
## What is hogging and sagging of ships?
Hogging is the stress a ship’s hull or keel experiences that causes the center or the keel to bend upward. Sagging is the stress a ship’s hull or keel is placed under when a wave is the same length as the ship and the ship is in the trough of two waves.
## What is positive and negative shear?
A shear stress acting on a positive face is positive if it acts in the positive direction of an axis, and negative if it acts in the negative direction of an axis. A shear stress acting on a negative face is positive if it acts in the negative direction of an axis and negative if it acts in the positive direction.
## How can you tell if shear stress is positive or negative?
A shear stress is positive if it acts on a positive face in a positive direction or if it acts on a negative face in a negative direction.
## What does negative shear mean?
Considering equilibrium in the vertical direction, V plus p over 2 must be equal to 0. So clearly, V must be minus p over 2. That means the shear force is not pointing upward, it’s pointing downward on this right phase and therefore this is a negative shear.
## What is positive shear force?
Positive shear forces always deform right hand face downward with respect to the left hand face. Positive bending moments always elongate the lower section of the beam.
## What is the meaning of shear force?
A shear force is a force applied perpendicular to a surface, in opposition to an offset force acting in the opposite direction. When a structural member experiences failure by shear, two parts of it are pushed in different directions, for example, when a piece of paper is cut by scissors.
## Where is the bending moment is always zero?
At the ends of a simply supported beam the bending moments are zero. At the wall of a cantilever beam, the bending moment equals the moment reaction. At the free end, the bending moment is zero. At the location where the shear force crosses the zero axis the corresponding bending moment has a maximum value.
## What will be the value of bending moment at section of a beam where shear force is equal to zero?
The bending moment at any point along the beam is equal to the area under the shear force diagram up to that point. (Note: For a simply-supported beam, the bending moment at the ends will always be equal to zero.)
|
# Solving Quadratic Equations by Graphing
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Category: Education
## Presentation Description
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By: Liel (75 month(s) ago)
By: rjohn (95 month(s) ago)
can u send me this ppt to use in my class, i would really appreciate that.
By: moecool (99 month(s) ago)
Hi lsn can u send it to me plz i need it to explain it to my students in the school since we dont have internet there
By: rfantster (107 month(s) ago)
I do sophisticated animations for my math classes and a lot of these do not transfer / render as accurately as I would like.
## Presentation Transcript
### Slide1:
§6.2 Solving Quadratic Equations by Graphing 1) quadratic equation 2) root 3) zero Solve quadratic equations by graphing. Estimate solutions of quadratic equations by graphing.
### Slide2:
§6.2 Solving Quadratic Equations by Graphing When a quadratic function is set equal to a value, the result is a quadratic equation. The solution of a quadratic equation are called the roots of the equation. One method for finding the roots of a quadratic equation is to find the zeros of the related quadratic function. The zeros of the function are the x-intercepts of its graph. These are the solutions of the related equation because f(x) = 0 at those points. The zeros of the function graphed at the right are -1 and 2.
### Slide3:
§6.2 Solving Quadratic Equations by Graphing
### Slide4:
§6.2 Solving Quadratic Equations by Graphing Two Real Solutions Axis of symmetry: Note: The parabola opens up and the vertex is below the x-axis. Therefore, the parabola must cross the x-axis at two distinct points. Solutions:
### Slide5:
§6.2 Solving Quadratic Equations by Graphing One Real Solution Axis of symmetry: Note: The parabola opens up and the vertex is on the x-axis. Therefore, the parabola only touches the x-axis at the vertex (1 point). Solution:
### Slide6:
§6.2 Solving Quadratic Equations by Graphing No Real Solution Axis of symmetry: Note: The parabola opens down and the vertex is below the x-axis. Therefore, the parabola never crosses the x-axis. Solution: No REAL solution.
### Slide7:
§6.2 Solving Quadratic Equations by Graphing Estimate Roots (Solutions) Axis of symmetry: Solutions: and
### Slide8:
§6.2 Solving Quadratic Equations by Graphing Number Theory Use a quadratic equation to find two real numbers that satisfy the situation, or show that no such number exists. Axis of symmetry: Solution: No REAL solution.
### Slide9:
§6.2 Solving Quadratic Equations by Graphing Application of Physics A tennis ball is hit upward at a velocity of 48 feet per second. 3 seconds
### Slide10:
§6.2 Solving Quadratic Equations by Graphing
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# 2.1 An isometric drawing of a cube
Page 1 / 1
## [lo 1.3]
1. How many visible (continuous) lines are there in the drawing?
2. How many invisible (broken) lines are in the drawing?
3. How many planes does the figure have?
4. How many planes are visible?
5. How many planes are invisible?
Underline:
1. A cube is made up of six (squares, rectangles) but in the sketch they look like six (parallelograms, rhombuses).
2. This cube has three views, namely a FRONT VIEW (right front), a SIDE VIEW (left front) and a TOP VIEW (from right above).
Colour the views in the following way: Front: Blue
Side: Green
Top: Red
## [lo 1.3]
Background:
Before we try individually to draw a three-dimensional object, we are going to compare shapes of planes in a drawing with shapes of the real object.
## [lo 1.3]
Drawing A and B
• Complete the table:
Number of plane Type of face on sketch Shape on real object 1 2 3 4 5 6 7 8
Something interesting:
• Do your answers to 2 and 8 differ on the sketch?
If we suppose that these are sketches of the same shed, try to explain why.
• Do your answers regarding planes 2 and 8 differ on the real object?
• Explain
Background:
A three-dimensional shape such as a cube also has vertical and horizontal planes.
Circle:
A cube has 2/ 4/ 6 horizontal and 2/ 4/ 6 vertical planes.
[LO 1.3]
ASSIGNMENT 4A:
a) How many horizontal planes does a wooden block have that stands like
this?
And vertical planes?
b) How many horizontal planes does a wooden block have that stands like
this?
And vertical planes?
c) What conclusion can you draw from this? Complete:
A wooden block/rectangular shape always has horizontal and
vertical planes.
## [lo 1.3]
a) Redraw the three visible planes on carbon paper (numbers 4, 5 and 6 above).
b) Cut out the three planes.
c) Use three pieces of different coloured cardboard and cut out two examples of each plane from the colours (six parts with two of each colour) (two of number 6, two of number 5, two of number 4).
d) Build the box on the grid (annexure 1) by gluing the parts as shown in the sequence that was indicated above. Use Prestik to stick the parts onto the grid.
e) Now draw the box in pencil on the grid. First draw the visible lines and then the invisible lines in dotted lines.
A Challenge:
f) Suppose that plane 4 is a flap that can open. Close flap 4. Now draw the box in pencil with its open flap on the grid. First draw all the visible lines and then the invisible lines.
## Assessment
Learning Outcomes(LOs)
LO 1
TECHNOLOGICAL PROCESSES AND SKILLS
The learner will be able to apply technological processes and skills ethically and responsibly using appropriate information and communication technologies
Assessment Standards(ASs)
We know this when the learner:
1.3 performs, where appropriate, scientific investigations about concepts relevant to a problem, need or opportunity using science process skills.
## Memorandum
Assignment 1
Learners could do it individually after a learning discussion by the teacher.
Assignment 2A
The teacher could do one together with the learners in order to explain it to them, and the learners could then do the rest individually. Make a transparency of the page and use coloured transparency pens.
Assignment 2B
The teacher explains, using a cigarette box or matchbox. The different positions are then indicated in a practical way before the learners draw it. For instance, B could be done with the learners, and then they could be told to do C and D by themselves.
Assignment 3
Let groups of learners, for instance, do drawing A. They should then check and the reasons for the answers. Then learners should do drawing B individually, after which it could be checked. Have a class discussion about the relevant questions.
Assignment 4A
Discuss the questions by illustrating practically with a wooden block and let learners fill in the answers.
Assignment 4B
This is a practical exercise. Each learner should therefore have scissors, Prestik, a 30 60 grid (Addendum 1) and three pieces of coloured cardboard to complete the assignment. Do the assignment together with the learners.
Assignment 5
Each learner can evaluate his/her progress individually. Emphasise the fact that each learner should be honest, as the teacher would be able to see from the practical work whether the learner was honest or not.
where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Got questions? Join the online conversation and get instant answers!
|
# How do you find all zeros of f(x)=x^3-4x^2-25x+100?
Jan 7, 2017
$x = 5$, $x = - 5$ or $x = 4$
#### Explanation:
Note that the ratio of the first and second terms is the same as that of the third and fourth terms, so this cubic factors by grouping and we find:
$0 = {x}^{3} - 4 {x}^{2} - 25 x + 100$
$\textcolor{w h i t e}{0} = \left({x}^{3} - 4 {x}^{2}\right) - \left(25 x - 100\right)$
$\textcolor{w h i t e}{0} = {x}^{2} \left(x - 4\right) - 25 \left(x - 4\right)$
$\textcolor{w h i t e}{0} = \left({x}^{2} - 25\right) \left(x - 4\right)$
$\textcolor{w h i t e}{0} = \left({x}^{2} - {5}^{2}\right) \left(x - 4\right)$
$\textcolor{w h i t e}{0} = \left(x - 5\right) \left(x + 5\right) \left(x - 4\right)$
Hence $x = 5$, $x = - 5$ or $x = 4$
|
Work A Divison Problem Easy Problem:
Problem: (3x2 +16x-6) ÷ (1x+6)
Divide: x1 1x+6 ) 3x2 +16x -6
### Instructions
This program lets you practice long division of polynomials.
The problems all have a binomial of degree one for their divisor. The dividends are random polynomials of degree three to five.
#### The Steps
There are four steps in the long division process that repeat until the problem is complete.
1.Divide:We divide the highest remaining term in the dividend by the divisor.
We get this answer by looking at the hightest term in the divisor and divide the highest remaining term in the dividend by it. For ecample if 4x3 is the highest remaining term in the dividend and the divisor is 2x-1, we divide 4x3 by 2x and get 2x2. We put that number above the 4x3 above the division line.
2.Multiply:We multiply the answer we got in the first step by the divisor and get
4x4-2x3. We place this below the last partial dividend we have.
3.Subtract:We substract the partial result we just got from the partial dividend above it.
4.Subtract:. We write the next power of the polynomial next to the subtraction we have just done.
The x4 goes away and we are left with an x3 term as the highest remaining one.
Now we 'bring down' the x2 term from the original problem and repeat the three steps above.
We are done when the highest term of remaining dividend is less than the highest power of the divisor. In this case, the remainder is just a number.
We check by multiplying the answer by the divisor and adding the remainder. We should get the original dividend.
#### What we are doing
In the case of polynomial division we are simply doing the division one term at a time. The
process is a little more complicated than we would like, because we have a remainder at each stage
that needs to be incorporated in the division problem we are working at the next step.
If you want to feel more comfortable about this, think back to how we learned to do long division in grade school.
We would do the division one digit at a time. We would divide the first digit by the divisor, multiply the answer by the divisor,
and then subtract to get a remainder. In the next step, we would bring down the next digit and divide these two
numbers by the divisor. We were done when there was nothing left to bring down. The remainder was
the result of the last subtraction we had done.
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```CS 70
Spring 2022
1
Discrete Mathematics and Probability Theory
Course Notes
Note 1
Propositional Logic
In order to be fluent in working with mathematical statements, you need to understand the basic framework
of the language of mathematics. This first lecture, we will start by learning about what logical forms mathematical theorems may take, and how to manipulate those forms to make them easier to prove. In the next
few lectures, we will learn several different methods of proving things.
Our first building block is the notion of a proposition, which is simply a statement which is either true or
false.
These statements are all propositions:
(1)
√
3 is irrational.
(2) 1 + 1 = 5.
(3) Julius Caesar had 2 eggs for breakfast on his 10th birthday.
These statements are clearly not propositions:
(4) 2 + 2.
(5) x2 + 3x = 5.
[What is x?]
These statements aren’t propositions either (although some books say they are). Propositions should not
include fuzzy terms.
(6) Arnold Schwarzenegger often eats broccoli.
(7) Henry VIII was unpopular.
[What is “often?”]
[What is “unpopular?”]
Propositions may be joined together to form more complex statements. Let P, Q, and R be variables representing propositions (for example, P could stand for “3 is odd”). The simplest way of joining these
propositions together is to use the connectives “and”, “or” and “not.”
(1) Conjunction: P ∧ Q (“P and Q”). True only when both P and Q are true.
(2) Disjunction: P ∨ Q (“P or Q”). True when at least one of P and Q is true.
(3) Negation: ¬P (“not P”). True when P is false.
Statements like these, with variables, are called propositional forms.
A fundamental principle known as the law of the excluded middle says that, for any proposition P, either
P is true or ¬P is true (but not both). Thus P ∨ ¬P is always true, regardless of the truth value of P. A
propositional form that is always true regardless of the truth values of its variables is called a tautology.
Conversely, a statement such as P ∧ ¬P, which is always false, is called a contradiction.
CS 70, Spring 2022, Note 1
1
Concept check! If we let P stand for the proposition “3 is odd”, Q stand for “4 is odd”, and R for “4+5 = 49”,
what are the values of P ∧ R, P ∨ R and ¬Q?
A useful tool for describing the possible values of a propositional form is a truth table. Truth tables are the
same as function tables: you list all possible input values for the variables, and then list the outputs given
those inputs. (The order does not matter.)
Here are truth tables for conjunction and negation:
P
T
T
F
F
P∧Q
T
F
F
F
Q
T
F
T
F
P
T
F
¬P
F
T
Concept check! Write down the truth table for disjunction (OR).
The most important and subtle propositional form is an implication:
(4) Implication: P =⇒ Q (“P implies Q”). This is the same as “If P, then Q.”
Here, P is called the hypothesis of the implication, and Q is the conclusion.1
We encounter implications frequently in everyday life; here are a couple of examples:
If you stand in the rain, then you’ll get wet.
If you passed the class, you received a certificate.
An implication P =⇒ Q is false only when P is true and Q is false. For example, the first statement above
would be false only if you stood in the rain but didn’t get wet. The second statement would be false only if
you passed the class but didn’t receive a certificate.
1P
is also sometimes called the antecedent and Q the consequent.
CS 70, Spring 2022, Note 1
2
Here is the truth table for P =⇒ Q (along with an extra column that we’ll explain in a moment):
P
T
T
F
F
Q
T
F
T
F
P =⇒ Q
T
F
T
T
¬P ∨ Q
T
F
T
T
Note that P =⇒ Q is always true when P is false. This means that many statements that sound nonsensical
in English are true, mathematically speaking. Examples are statements like: “If pigs can fly, then horses can
read” or “If 14 is odd then 1 + 2 = 18.” (These are statements that we never make in everyday life, but are
perfectly natural in mathematics.) When an implication is stupidly true because the hypothesis is false, we
say that it is vacuously true.
Note also that P =⇒ Q is logically equivalent to ¬P ∨ Q, as can be seen by comparing the last two columns
of the above truth table: for all possible truth values of P and Q, P =⇒ Q and ¬P ∨ Q take the same values
(i.e., they have the same truth table). We write this as (P =⇒ Q) ≡ (¬P ∨ Q).
P =⇒ Q is the most common form mathematical theorems take. Here are some of the different ways of
saying it:
(1) if P, then Q;
(2) Q if P;
(3) P only if Q;
(4) P is sufficient for Q;
(5) Q is necessary for P;
(6) Q unless not P.
If both P =⇒ Q and Q =⇒ P are true, then we say “P if and only if Q” (abbreviated “P iff Q”). Formally,
we write P ⇐⇒ Q. Note that P ⇐⇒ Q is true only when P and Q have the same truth values (both true or
both false).
For example, if we let P be “3 is odd,” Q be “4 is odd,” and R be “6 is even,” then the following are all true:
P =⇒ R, Q =⇒ P (vacuously), and R =⇒ P. Because P =⇒ R and R =⇒ P, we also see that P ⇐⇒ R
is true.
Given an implication P =⇒ Q, we can also define its
(a) Contrapositive: ¬Q =⇒ ¬P
(b) Converse: Q =⇒ P
The contrapositive of “If you passed the class, you received a certificate” is “If you did not get a certificate,
you did not pass the class.” The converse is “If you got a certificate, you passed the class.” Does the
contrapositive say the same thing as the original statement? Does the converse?
Let’s look at the truth tables:
P
T
T
F
F
Q
T
F
T
F
CS 70, Spring 2022, Note 1
¬P
F
F
T
T
¬Q
F
T
F
T
P =⇒ Q
T
F
T
T
Q =⇒ P
T
T
F
T
¬Q =⇒ ¬P
T
F
T
T
P ⇐⇒ Q
T
F
F
T
3
Note that P =⇒ Q and its contrapositive have the same truth values everywhere in their truth tables, so they
are logically equivalent: (P =⇒ Q) ≡ (¬Q =⇒ ¬P). Many students confuse the contrapositive with the
converse: note that P =⇒ Q and ¬Q =⇒ ¬P are logically equivalent, but P =⇒ Q and Q =⇒ P are not!
When two propositional forms are logically equivalent, we can think of them as “meaning the same thing.”
We will see in the next lecture how useful this can be for proving theorems.
2
Quantifiers
The mathematical statements you’ll see in practice will not be made up of simple propositions like “3 is
odd.” Instead you’ll see statements like:
(1) For all natural numbers n, n2 + n + 41 is prime.
(2) If n is an odd integer, so is n3 .
(3) There is an integer k that is both even and odd.
In essence, these statements assert something about lots of simple propositions (even infinitely many!) all at
once. For instance, the first statement is asserting that 02 + 0 + 41 is prime, 12 + 1 + 41 is prime, and so on.
The last statement is saying that, as k ranges over all possible integers, we will find at least one value of k
for which the statement is satisfied.
Why are the above three examples considered to be propositions, while earlier we claimed that “x2 + 3x = 5”
was not? The reason is that in these examples, there is an underlying “universe” that we are working in, and
the statements are quantified over that universe. To express these statements mathematically we need two
quantifiers: The universal quantifier ∀ (“for all”) and the existential quantifier ∃ (“there exists”). Examples:
(1) “Some mammals lay eggs.” Mathematically, “some” means “at least one,” so the statement is saying
“There exists a mammal x such that x lays eggs.” If we let our universe U be the set of mammals, then
we can write: (∃x ∈ U)(x lays eggs). (Sometimes, when the universe is clear, we omit U and simply
write ∃x(x lays eggs).)
(2) “For all natural numbers n, n2 + n + 41 is prime,” can be expressed by taking our universe to be the
set of natural numbers, denoted as N: (∀n ∈ N)(n2 + n + 41 is prime).
Note that in a finite universe, we can express existentially and universally quantified propositions without
quantifiers, using disjunctions and conjunctions respectively. For example, if our universe U is {1, 2, 3, 4},
then (∃x ∈ U)P(x) is logically equivalent to P(1) ∨ P(2) ∨ P(3) ∨ P(4), and (∀x ∈ U)P(x) is logically equivalent to P(1) ∧ P(2) ∧ P(3) ∧ P(4). However, in an infinite universe, such as the natural numbers, this is not
possible.
Concept check! Use quantifiers to express the following two statements: “For all integers x, 2x + 1 is odd”,
and “There exists an integer between 2 and 4”.
Some statements can have multiple quantifiers. As we will see, however, quantifiers do not commute. You
can see this just by thinking about English statements. Consider the following (rather gory) example:
“Every time I ride the subway in New York, somebody gets stabbed.”
“There is someone, such that every time I ride the subway in New York, that someone gets
stabbed.”
CS 70, Spring 2022, Note 1
4
The first statement is saying that every time I ride the subway someone gets stabbed, but it could be a
different person each time. The second statement is saying something truly horrible: that there is some poor
guy Joe with the misfortune that every time I get on the New York subway, there is Joe, getting stabbed
again. (Poor Joe will run for his life the second he sees me.)
Mathematically, we are quantifying over two universes: T = {times when I ride on the subway} and P =
{people}. The first statement can be written: (∀t ∈ T )(∃p ∈ P)(p gets stabbed at time t). The second
statement says: (∃p ∈ P)(∀t ∈ T )(p gets stabbed at time t).
Let’s look at a more mathematical example:
Consider
1. (∀x ∈ Z)(∃y ∈ Z)(x < y)
2. (∃y ∈ Z)(∀x ∈ Z)(x < y)
The first statement says that, given an integer, I can find a larger one. The second statement says something
very different: that there is a largest integer! The first statement is true, the second is not.
3
What does it mean for a proposition P to be false? It means that its negation ¬P is true. It’s helpful to have
some rules for working with negation, as will become more obvious next lecture when we look at proofs.
First, let’s look at how to negate conjunctions and disjunctions:
¬(P ∧ Q) ≡ (¬P ∨ ¬Q)
¬(P ∨ Q) ≡ (¬P ∧ ¬Q)
These two equivalences are known as De Morgan’s Laws, and they are quite intuitive: for example, if it is
not the case that P ∧ Q is true, then either P or Q must be false (and vice versa).
Concept check! Verify both of De Morgan’s Laws by writing down the appropriate truth tables.
Negating propositions involving quantifiers actually follows analogous laws. Let’s start with a simple example. Assume that the universe is {1, 2, 3, 4} and let P(x) denote the propositional formula “x2 > 10.” Check
that ∃xP(x) is true but ∀xP(x) is false. Observe that both ¬(∀xP(x)) and ∃x¬P(x) are true because P(1) is
false. Also note that both ∀x¬P(x) and ¬(∃xP(x)) are false, since P(4) is true. The fact that each pair of
statements had the same truth value is no accident, as the equivalences
¬(∀xP(x)) ≡ ∃x¬P(x)
¬(∃xP(x)) ≡ ∀x¬P(x)
are laws that hold for any proposition P quantified over any universe (including infinite ones).
It is helpful to think of English sentences to convince yourself (informally) that these laws are true. For
example, assume that we are working within the universe Z (the set of all integers), and that P(x) is the
proposition “x is odd.” We know that the statement (∀xP(x)) is false, since not every integer is odd. Therefore, we expect its negation, ¬(∀xP(x)), to be true. But how would you say the negation in English? Well, if
CS 70, Spring 2022, Note 1
5
it is not true that every integer is odd, then there must exist some integer which is not odd (i.e., even). How
would this be written in propositional form? That’s easy, it’s just: (∃x¬P(x)).
To see a more complex example, fix some universe and propositional formula P(x, y). Assume we have the
proposition ¬(∀x∃yP(x, y)) and we want to push the negation operator inside the quantifiers. By the above
laws, we can do it as follows:
¬(∀x∃yP(x, y)) ≡ ∃x¬(∃yP(x, y)) ≡ ∃x∀y¬P(x, y).
Notice that we broke the complex negation into a smaller, easier problem as the negation propagated itself
through the quantifiers. Note also that the quantifiers “flip” as we go.
Let’s look at a trickier example:
Write the sentence “there are at least three distinct integers x that satisfy P(x)” as a proposition using
quantifiers! One way to do it is
∃x∃y∃z(x 6= y ∧ y 6= z ∧ z 6= x ∧ P(x) ∧ P(y) ∧ P(z)).
(Here all quantifiers are over the universe Z of integers.) Now write the sentence “there are at most three
distinct integers x that satisfy P(x)” as a proposition using quantifiers. One way to do it is
∃x∃y∃z∀d(P(d) =⇒ d = x ∨ d = y ∨ d = z).
Here is an equivalent way to do it:
∀x∀y∀v∀z((x 6= y ∧ y 6= v ∧ v 6= x ∧ x 6= z ∧ y 6= z ∧ v 6= z) =⇒ ¬(P(x) ∧ P(y) ∧ P(v) ∧ P(z))).
[Check that you understand both of the above alternatives.] Finally, what if we want to express the sentence “there are exactly three distinct integers x that satisfy P(x)”? This is now easy: we can just use the
conjunction of the two propositions above.
CS 70, Spring 2022, Note 1
6
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Exterior Angles in Convex Polygons
Angles on the outside of a polygon formed by extending a side.
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Practice Exterior Angles in Convex Polygons
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Exterior Angles in Convex Polygons
What if you were given a twelve-sided regular polygon? How could you determine the measure of each of its exterior angles? After completing this Concept, you'll be able to use the Exterior Angle Sum Theorem to solve problems like this one.
Watch This
Watch the second half of this video.
Guidance
Recall that an exterior angle is an angle on the outside of a polygon and is formed by extending a side of the polygon.
As you can see, there are two sets of exterior angles for any vertex on a polygon. It does not matter which set you use because one set is just the vertical angles of the other, making the measurement equal. In the picture above, the color-matched angles are vertical angles and congruent. The Exterior Angle Sum Theorem stated that the exterior angles of a triangle add up to 360\begin{align*}360^\circ\end{align*}. Let’s extend this theorem to all polygons.
Investigation: Exterior Angle Tear-Up
Tools Needed: pencil, paper, colored pencils, scissors
1. Draw a hexagon like the hexagons above. Color in the exterior angles as well.
2. Cut out each exterior angle and label them 1-6.
3. Fit the six angles together by putting their vertices together. What happens?
The angles all fit around a point, meaning that the exterior angles of a hexagon add up to 360\begin{align*}360^\circ\end{align*}, just like a triangle. We can say this is true for all polygons.
Exterior Angle Sum Theorem: The sum of the exterior angles of any polygon is 360\begin{align*}360^\circ\end{align*}.
Proof of the Exterior Angle Sum Theorem:
Given: Any n\begin{align*}n-\end{align*}gon with n\begin{align*}n\end{align*} sides, n\begin{align*}n\end{align*} interior angles and n\begin{align*}n\end{align*} exterior angles.
Prove: n\begin{align*}n\end{align*} exterior angles add up to 360\begin{align*}360^\circ\end{align*}
NOTE: The interior angles are x1,x2,xn\begin{align*}x_1, x_2, \ldots x_n\end{align*}.
The exterior angles are y1,y2,yn\begin{align*}y_1, y_2, \ldots y_n\end{align*}.
Statement Reason
1. Any n\begin{align*}n-\end{align*}gon with n\begin{align*}n\end{align*} sides, n\begin{align*}n\end{align*} interior angles and n\begin{align*}n\end{align*} exterior angles. Given
2. xn\begin{align*}x_n^\circ\end{align*} and yn\begin{align*}y_n^\circ\end{align*} are a linear pair Definition of a linear pair
3. xn\begin{align*}x_n^\circ\end{align*} and yn\begin{align*}y_n^\circ\end{align*} are supplementary Linear Pair Postulate
4. xn+yn=180\begin{align*}x_n^\circ+ y_n^\circ=180^\circ\end{align*} Definition of supplementary angles
5. (x1+x2++xn)+(y1+y2++yn)=180n\begin{align*}(x_1^\circ+x_2^\circ+\ldots+x_n^\circ)+(y_1^\circ+ y_2^\circ+\ldots+ y_n^\circ)=180^\circ n\end{align*} Sum of all interior and exterior angles in an n\begin{align*}n-\end{align*}gon
6. (n2)180=(x1+x2++xn)\begin{align*}(n-2)180^\circ=(x_1^\circ+ x_2^\circ+\ldots+x_n^\circ)\end{align*} Polygon Sum Formula
7. 180n=(n2)180+(y1+y2++yn)\begin{align*}180^\circ n=(n-2)180^\circ+(y_1^\circ+ y_2^\circ+\ldots+ y_n^\circ)\end{align*} Substitution PoE
8. 180n=180n360+(y1+y2++yn)\begin{align*}180^\circ n=180^\circ n-360^\circ+(y_1^\circ+ y_2^\circ+\ldots+ y_n^\circ)\end{align*} Distributive PoE
9. 360=(y1+y2++yn)\begin{align*}360^\circ=(y_1^\circ+ y_2^\circ+\ldots+ y_n^\circ)\end{align*} Subtraction PoE
Example A
What is y\begin{align*}y\end{align*}?
y\begin{align*}y\end{align*} is an exterior angle, as well as all the other given angle measures. Exterior angles add up to 360\begin{align*}360^\circ\end{align*}, so set up an equation.
70+60+65+40+yy=360=125\begin{align*}70^\circ + 60^\circ + 65^\circ + 40^\circ + y & = 360^\circ\\ y & = 125^\circ\end{align*}
Example B
What is the measure of each exterior angle of a regular heptagon?
Because the polygon is regular, each interior angle is equal. This also means that all the exterior angles are equal. The exterior angles add up to 360\begin{align*}360^\circ\end{align*}, so each angle is 360751.43\begin{align*}\frac{360^\circ}{7} \approx 51.43^\circ\end{align*}.
Example C
What is the sum of the exterior angles in a regular 15-gon?
The sum of the exterior angles in any convex polygon, including a regular 15-gon, is 360\begin{align*}360^\circ\end{align*}.
Watch this video for help with the Examples above.
Concept Problem Revisited
The exterior angles of a regular polygon sum to 360\begin{align*}360^\circ\end{align*}. The measure of each exterior angle in a dodecagon (twelve-sided regular polygon) is 36012=30\begin{align*}\frac{360^\circ}{12} = 30^\circ\end{align*}.
Vocabulary
An exterior angle is an angle that is formed by extending a side of the polygon. A regular polygon is a polygon in which all of its sides and all of its angles are congruent.
Guided Practice
Find the measure of each exterior angle for each regular polygon below:
1. 12-gon
2. 100-gon
3. 36-gon
For each, divide 360\begin{align*}360^\circ\end{align*} by the given number of sides.
1. 30\begin{align*}30^\circ\end{align*}
2. 3.6\begin{align*}3.6^\circ\end{align*}
3. 10\begin{align*}10^\circ\end{align*}
Practice
1. What is the measure of each exterior angle of a regular decagon?
2. What is the measure of each exterior angle of a regular 30-gon?
3. What is the sum of the exterior angles of a regular 27-gon?
Find the measure of the missing variables:
1. The exterior angles of a quadrilateral are x,2x,3x,\begin{align*}x^\circ, 2x^\circ, 3x^\circ,\end{align*} and 4x.\begin{align*}4x^\circ.\end{align*} What is x\begin{align*}x\end{align*}?
Find the measure of each exterior angle for each regular polygon below:
1. octagon
2. nonagon
3. triangle
4. pentagon
5. 50-gon
6. heptagon
7. 34-gon
8. Challenge Each interior angle forms a linear pair with an exterior angle. In a regular polygon you can use two different formulas to find the measure of each exterior angle. One way is 360n\begin{align*}\frac{360^\circ}{n}\end{align*} and the other is 180(n2)180n\begin{align*}180^\circ - \frac{(n-2)180^\circ}{n}\end{align*} (180\begin{align*}180^\circ\end{align*} minus Equiangular Polygon Formula). Use algebra to show these two expressions are equivalent.
9. Angle Puzzle Find the measures of the lettered angles below given that m || n\begin{align*}m \ || \ n\end{align*}.
My Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English
Exterior Angle Sum Theorem
Exterior Angle Sum Theorem states that the exterior angles of any polygon will always add up to 360 degrees.
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BREAKING NEWS
Glossary of calculus
## Summary
Most of the terms listed in Wikipedia glossaries are already defined and explained within Wikipedia itself. However, glossaries like this one are useful for looking up, comparing and reviewing large numbers of terms together. You can help enhance this page by adding new terms or writing definitions for existing ones.
This glossary of calculus is a list of definitions about calculus, its sub-disciplines, and related fields.
## A
Abel's test
A method of testing for the convergence of an infinite series.
Absolute convergence
An infinite series of numbers is said to converge absolutely (or to be absolutely convergent) if the sum of the absolute values of the summands is finite. More precisely, a real or complex series ${\displaystyle \textstyle \sum _{n=0}^{\infty }a_{n}}$ is said to converge absolutely if ${\displaystyle \textstyle \sum _{n=0}^{\infty }\left|a_{n}\right|=L}$ for some real number ${\displaystyle \textstyle L}$. Similarly, an improper integral of a function, ${\displaystyle \textstyle \int _{0}^{\infty }f(x)\,dx}$, is said to converge absolutely if the integral of the absolute value of the integrand is finite—that is, if ${\displaystyle \textstyle \int _{0}^{\infty }\left|f(x)\right|dx=L.}$
Absolute maximum
The highest value a function attains.
Absolute minimum
The lowest value a function attains.
Absolute value
The absolute value or modulus |x| of a real number x is the non-negative value of x without regard to its sign. Namely, |x| = x for a positive x, |x| = −x for a negative x (in which case x is positive), and |0| = 0. For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3. The absolute value of a number may be thought of as its distance from zero.
Alternating series
An infinite series whose terms alternate between positive and negative.
Alternating series test
Is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test, Leibniz's rule, or the Leibniz criterion.
Annulus
A ring-shaped object, a region bounded by two concentric circles.
Antiderivative
An antiderivative, primitive function, primitive integral or indefinite integral[Note 1] of a function f is a differentiable function F whose derivative is equal to the original function f. This can be stated symbolically as ${\displaystyle F'=f}$.[1][2] The process of solving for antiderivatives is called antidifferentiation (or indefinite integration) and its opposite operation is called differentiation, which is the process of finding a derivative.
Arcsin
Area under a curve
Asymptote
In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity. Some sources include the requirement that the curve may not cross the line infinitely often, but this is unusual for modern authors.[3] In projective geometry and related contexts, an asymptote of a curve is a line which is tangent to the curve at a point at infinity.[4][5]
Automatic differentiation
In mathematics and computer algebra, automatic differentiation (AD), also called algorithmic differentiation or computational differentiation,[6][7] is a set of techniques to numerically evaluate the derivative of a function specified by a computer program. AD exploits the fact that every computer program, no matter how complicated, executes a sequence of elementary arithmetic operations (addition, subtraction, multiplication, division, etc.) and elementary functions (exp, log, sin, cos, etc.). By applying the chain rule repeatedly to these operations, derivatives of arbitrary order can be computed automatically, accurately to working precision, and using at most a small constant factor more arithmetic operations than the original program.
Average rate of change
## B
Binomial coefficient
Any of the positive integers that occurs as a coefficient in the binomial theorem is a binomial coefficient. Commonly, a binomial coefficient is indexed by a pair of integers nk ≥ 0 and is written ${\displaystyle {\tbinom {n}{k}}.}$ It is the coefficient of the xk term in the polynomial expansion of the binomial power (1 + x)n, and it is given by the formula
${\displaystyle {\binom {n}{k}}={\frac {n!}{k!(n-k)!}}.}$
Binomial theorem (or binomial expansion)
Describes the algebraic expansion of powers of a binomial.
Bounded function
A function f defined on some set X with real or complex values is called bounded, if the set of its values is bounded. In other words, there exists a real number M such that
${\displaystyle |f(x)|\leq M}$
for all x in X. A function that is not bounded is said to be unbounded. Sometimes, if f(x) ≤ A for all x in X, then the function is said to be bounded above by A. On the other hand, if f(x) ≥ B for all x in X, then the function is said to be bounded below by B.
Bounded sequence
.
## C
Calculus
(From Latin calculus, literally 'small pebble', used for counting and calculations, as on an abacus)[8] is the mathematical study of continuous change, in the same way that geometry is the study of shape and algebra is the study of generalizations of arithmetic operations.
Cavalieri's principle
Cavalieri's principle, a modern implementation of the method of indivisibles, named after Bonaventura Cavalieri, is as follows:[9]
• 2-dimensional case: Suppose two regions in a plane are included between two parallel lines in that plane. If every line parallel to these two lines intersects both regions in line segments of equal length, then the two regions have equal areas.
• 3-dimensional case: Suppose two regions in three-space (solids) are included between two parallel planes. If every plane parallel to these two planes intersects both regions in cross-sections of equal area, then the two regions have equal volumes.
Chain rule
The chain rule is a formula for computing the derivative of the composition of two or more functions. That is, if f and g are functions, then the chain rule expresses the derivative of their composition f g (the function which maps x to f(g(x)) ) in terms of the derivatives of f and g and the product of functions as follows:
${\displaystyle (f\circ g)'=(f'\circ g)\cdot g'.}$
This may equivalently be expressed in terms of the variable. Let F = f g, or equivalently, F(x) = f(g(x)) for all x. Then one can also write
${\displaystyle F'(x)=f'(g(x))g'(x).}$
The chain rule may be written in Leibniz's notation in the following way. If a variable z depends on the variable y, which itself depends on the variable x, so that y and z are therefore dependent variables, then z, via the intermediate variable of y, depends on x as well. The chain rule then states,
${\displaystyle {\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}}.}$
The two versions of the chain rule are related; if ${\displaystyle z=f(y)}$ and ${\displaystyle y=g(x)}$, then
${\displaystyle {\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}}=f'(y)g'(x)=f'(g(x))g'(x).}$
In integration, the counterpart to the chain rule is the substitution rule.
Change of variables
Is a basic technique used to simplify problems in which the original variables are replaced with functions of other variables. The intent is that when expressed in new variables, the problem may become simpler, or equivalent to a better understood problem.
Cofunction
A function f is cofunction of a function g if f(A) = g(B) whenever A and B are complementary angles.[10] This definition typically applies to trigonometric functions.[11][12] The prefix "co-" can be found already in Edmund Gunter's Canon triangulorum (1620).[13][14]
Concave function
Is the negative of a convex function. A concave function is also synonymously called concave downwards, concave down, convex upwards, convex cap or upper convex.
Constant of integration
The indefinite integral of a given function (i.e., the set of all antiderivatives of the function) on a connected domain is only defined up to an additive constant, the constant of integration.[15][16] This constant expresses an ambiguity inherent in the construction of antiderivatives. If a function ${\displaystyle f(x)}$ is defined on an interval and ${\displaystyle F(x)}$ is an antiderivative of ${\displaystyle f(x)}$, then the set of all antiderivatives of ${\displaystyle f(x)}$ is given by the functions ${\displaystyle F(x)+C}$, where C is an arbitrary constant (meaning that any value for C makes ${\displaystyle F(x)+C}$ a valid antiderivative). The constant of integration is sometimes omitted in lists of integrals for simplicity.
Continuous function
Is a function for which sufficiently small changes in the input result in arbitrarily small changes in the output. Otherwise, a function is said to be a discontinuous function. A continuous function with a continuous inverse function is called a homeomorphism.
Continuously differentiable
A function f is said to be continuously differentiable if the derivative f(x) exists and is itself a continuous function.
Contour integration
In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane.[17][18][19]
Convergence tests
Are methods of testing for the convergence, conditional convergence, absolute convergence, interval of convergence or divergence of an infinite series ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$.
Convergent series
In mathematics, a series is the sum of the terms of an infinite sequence of numbers. Given an infinite sequence ${\displaystyle \left(a_{1},\ a_{2},\ a_{3},\dots \right)}$, the nth partial sum ${\displaystyle S_{n}}$ is the sum of the first n terms of the sequence, that is,
${\displaystyle S_{n}=\sum _{k=1}^{n}a_{k}.}$
A series is convergent if the sequence of its partial sums ${\displaystyle \left\{S_{1},\ S_{2},\ S_{3},\dots \right\}}$ tends to a limit; that means that the partial sums become closer and closer to a given number when the number of their terms increases. More precisely, a series converges, if there exists a number ${\displaystyle \ell }$ such that for any arbitrarily small positive number ${\displaystyle \varepsilon }$, there is a (sufficiently large) integer ${\displaystyle N}$ such that for all ${\displaystyle n\geq \ N}$,
${\displaystyle \left|S_{n}-\ell \right\vert \leq \ \varepsilon .}$
If the series is convergent, the number ${\displaystyle \ell }$ (necessarily unique) is called the sum of the series. Any series that is not convergent is said to be divergent.
Convex function
In mathematics, a real-valued function defined on an n-dimensional interval is called convex (or convex downward or concave upward) if the line segment between any two points on the graph of the function lies above or on the graph, in a Euclidean space (or more generally a vector space) of at least two dimensions. Equivalently, a function is convex if its epigraph (the set of points on or above the graph of the function) is a convex set. For a twice differentiable function of a single variable, if the second derivative is always greater than or equal to zero for its entire domain then the function is convex.[20] Well-known examples of convex functions include the quadratic function ${\displaystyle x^{2}}$ and the exponential function ${\displaystyle e^{x}}$.
Cramer's rule
In linear algebra, Cramer's rule is an explicit formula for the solution of a system of linear equations with as many equations as unknowns, valid whenever the system has a unique solution. It expresses the solution in terms of the determinants of the (square) coefficient matrix and of matrices obtained from it by replacing one column by the column vector of right-hand-sides of the equations. It is named after Gabriel Cramer (1704–1752), who published the rule for an arbitrary number of unknowns in 1750,[21][22] although Colin Maclaurin also published special cases of the rule in 1748[23] (and possibly knew of it as early as 1729).[24][25][26]
Critical point
A critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0.[27][28]
Curve
A curve (also called a curved line in older texts) is, generally speaking, an object similar to a line but that need not be straight.
Curve sketching
In geometry, curve sketching (or curve tracing) includes techniques that can be used to produce a rough idea of overall shape of a plane curve given its equation without computing the large numbers of points required for a detailed plot. It is an application of the theory of curves to find their main features. Here input is an equation. In digital geometry it is a method of drawing a curve pixel by pixel. Here input is an array (digital image).
## D
Damped sine wave
Is a sinusoidal function whose amplitude approaches zero as time increases.[29]
Degree of a polynomial
Is the highest degree of its monomials (individual terms) with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.
Derivative
The derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). Derivatives are a fundamental tool of calculus. For example, the derivative of the position of a moving object with respect to time is the object's velocity: this measures how quickly the position of the object changes when time advances.
Derivative test
A derivative test uses the derivatives of a function to locate the critical points of a function and determine whether each point is a local maximum, a local minimum, or a saddle point. Derivative tests can also give information about the concavity of a function.
Differentiable function
A differentiable function of one real variable is a function whose derivative exists at each point in its domain. As a result, the graph of a differentiable function must have a (non-vertical) tangent line at each point in its domain, be relatively smooth, and cannot contain any breaks, bends, or cusps.
Differential (infinitesimal)
The term differential is used in calculus to refer to an infinitesimal (infinitely small) change in some varying quantity. For example, if x is a variable, then a change in the value of x is often denoted Δx (pronounced delta x). The differential dx represents an infinitely small change in the variable x. The idea of an infinitely small or infinitely slow change is extremely useful intuitively, and there are a number of ways to make the notion mathematically precise. Using calculus, it is possible to relate the infinitely small changes of various variables to each other mathematically using derivatives. If y is a function of x, then the differential dy of y is related to dx by the formula
${\displaystyle dy={\frac {dy}{dx}}\,dx,}$
where dy/dx denotes the derivative of y with respect to x. This formula summarizes the intuitive idea that the derivative of y with respect to x is the limit of the ratio of differences Δyx as Δx becomes infinitesimal.
Differential calculus
Is a subfield of calculus[30] concerned with the study of the rates at which quantities change. It is one of the two traditional divisions of calculus, the other being integral calculus, the study of the area beneath a curve.[31]
Differential equation
Is a mathematical equation that relates some function with its derivatives. In applications, the functions usually represent physical quantities, the derivatives represent their rates of change, and the equation defines a relationship between the two.
Differential operator
.
Differential of a function
In calculus, the differential represents the principal part of the change in a function y = f(x) with respect to changes in the independent variable. The differential dy is defined by
${\displaystyle dy=f'(x)\,dx,}$
where ${\displaystyle f'(x)}$ is the derivative of f with respect to x, and dx is an additional real variable (so that dy is a function of x and dx). The notation is such that the equation
${\displaystyle dy={\frac {dy}{dx}}\,dx}$
holds, where the derivative is represented in the Leibniz notation dy/dx, and this is consistent with regarding the derivative as the quotient of the differentials. One also writes
${\displaystyle df(x)=f'(x)\,dx.}$
The precise meaning of the variables dy and dx depends on the context of the application and the required level of mathematical rigor. The domain of these variables may take on a particular geometrical significance if the differential is regarded as a particular differential form, or analytical significance if the differential is regarded as a linear approximation to the increment of a function. Traditionally, the variables dx and dy are considered to be very small (infinitesimal), and this interpretation is made rigorous in non-standard analysis.
Differentiation rules
.
Direct comparison test
A convergence test in which an infinite series or an improper integral is compared to one with known convergence properties.
Dirichlet's test
Is a method of testing for the convergence of a series. It is named after its author Peter Gustav Lejeune Dirichlet, and was published posthumously in the Journal de Mathématiques Pures et Appliquées in 1862.[32] The test states that if ${\displaystyle \{a_{n}\}}$ is a sequence of real numbers and ${\displaystyle \{b_{n}\}}$ a sequence of complex numbers satisfying
• ${\displaystyle a_{n+1}\leq a_{n}}$
• ${\displaystyle \lim _{n\rightarrow \infty }a_{n}=0}$
• ${\displaystyle \left|\sum _{n=1}^{N}b_{n}\right|\leq M}$ for every positive integer N
where M is some constant, then the series
${\displaystyle \sum _{n=1}^{\infty }a_{n}b_{n}}$
converges.
Disc integration
Also known in integral calculus as the disc method, is a means of calculating the volume of a solid of revolution of a solid-state material when integrating along an axis "parallel" to the axis of revolution.
Divergent series
Is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.
Discontinuity
Continuous functions are of utmost importance in mathematics, functions and applications. However, not all functions are continuous. If a function is not continuous at a point in its domain, one says that it has a discontinuity there. The set of all points of discontinuity of a function may be a discrete set, a dense set, or even the entire domain of the function.
Dot product
In mathematics, the dot product or scalar product[note 1] is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. In Euclidean geometry, the dot product of the Cartesian coordinates of two vectors is widely used and often called "the" inner product (or rarely projection product) of Euclidean space even though it is not the only inner product that can be defined on Euclidean space; see also inner product space.
Double integral
The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). Integrals of a function of two variables over a region in R2 are called double integrals, and integrals of a function of three variables over a region of R3 are called triple integrals.[33]
## E
e (mathematical constant)
The number e is a mathematical constant that is the base of the natural logarithm: the unique number whose natural logarithm is equal to one. It is approximately equal to 2.71828,[34] and is the limit of (1 + 1/n)n as n approaches infinity, an expression that arises in the study of compound interest. It can also be calculated as the sum of the infinite series[35]
${\displaystyle e=\displaystyle \sum \limits _{n=0}^{\infty }{\dfrac {1}{n!}}={\frac {1}{1}}+{\frac {1}{1}}+{\frac {1}{1\cdot 2}}+{\frac {1}{1\cdot 2\cdot 3}}+\cdots }$
Elliptic integral
In integral calculus, elliptic integrals originally arose in connection with the problem of giving the arc length of an ellipse. They were first studied by Giulio Fagnano and Leonhard Euler (c. 1750). Modern mathematics defines an "elliptic integral" as any function f which can be expressed in the form
${\displaystyle f(x)=\int _{c}^{x}R\left(t,{\sqrt {P(t)}}\right)\,dt,}$
where R is a rational function of its two arguments, P is a polynomial of degree 3 or 4 with no repeated roots, and c is a constant..
Essential discontinuity
For an essential discontinuity, only one of the two one-sided limits needs not exist or be infinite. Consider the function
${\displaystyle f(x)={\begin{cases}\sin {\frac {5}{x-1}}&{\mbox{ for }}x<1\\0&{\mbox{ for }}x=1\\{\frac {1}{x-1}}&{\mbox{ for }}x>1\end{cases}}}$
Then, the point ${\displaystyle \scriptstyle x_{0}\;=\;1}$ is an essential discontinuity. In this case, ${\displaystyle \scriptstyle L^{-}}$ doesn't exist and ${\displaystyle \scriptstyle L^{+}}$ is infinite – thus satisfying twice the conditions of essential discontinuity. So x0 is an essential discontinuity, infinite discontinuity, or discontinuity of the second kind. (This is distinct from the term essential singularity which is often used when studying functions of complex variables.
Euler method
Euler's method is a numerical method to solve first order first degree differential equation with a given initial value. It is the most basic explicit method for numerical integration of ordinary differential equations and is the simplest Runge–Kutta method. The Euler method is named after Leonhard Euler, who treated it in his book Institutionum calculi integralis (published 1768–1870).[36]
Exponential function
In mathematics, an exponential function is a function of the form
${\displaystyle f(x)=ab^{x},}$
where b is a positive real number, and in which the argument x occurs as an exponent. For real numbers c and d, a function of the form ${\displaystyle f(x)=ab^{cx+d}}$ is also an exponential function, as it can be rewritten as
${\displaystyle ab^{cx+d}=\left(ab^{d}\right)\left(b^{c}\right)^{x}.}$
Extreme value theorem
States that if a real-valued function f is continuous on the closed interval [a,b], then f must attain a maximum and a minimum, each at least once. That is, there exist numbers c and d in [a,b] such that:
${\displaystyle f(c)\geq f(x)\geq f(d)\quad {\text{for all }}x\in [a,b].}$
A related theorem is the boundedness theorem which states that a continuous function f in the closed interval [a,b] is bounded on that interval. That is, there exist real numbers m and M such that:
${\displaystyle m
The extreme value theorem enriches the boundedness theorem by saying that not only is the function bounded, but it also attains its least upper bound as its maximum and its greatest lower bound as its minimum.
Extremum
In mathematical analysis, the maxima and minima (the respective plurals of maximum and minimum) of a function, known collectively as extrema (the plural of extremum), are the largest and smallest value of the function, either within a given range (the local or relative extrema) or on the entire domain of a function (the global or absolute extrema).[37][38][39] Pierre de Fermat was one of the first mathematicians to propose a general technique, adequality, for finding the maxima and minima of functions. As defined in set theory, the maximum and minimum of a set are the greatest and least elements in the set, respectively. Unbounded infinite sets, such as the set of real numbers, have no minimum or maximum.
## F
Faà di Bruno's formula
Is an identity in mathematics generalizing the chain rule to higher derivatives, named after Francesco Faà di Bruno (1855, 1857), though he was not the first to state or prove the formula. In 1800, more than 50 years before Faà di Bruno, the French mathematician Louis François Antoine Arbogast stated the formula in a calculus textbook,[40] considered the first published reference on the subject.[41] Perhaps the most well-known form of Faà di Bruno's formula says that
${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum {\frac {n!}{m_{1}!\,1!^{m_{1}}\,m_{2}!\,2!^{m_{2}}\,\cdots \,m_{n}!\,n!^{m_{n}}}}\cdot f^{(m_{1}+\cdots +m_{n})}(g(x))\cdot \prod _{j=1}^{n}\left(g^{(j)}(x)\right)^{m_{j}},}$
where the sum is over all n-tuples of nonnegative integers (m1, …, mn) satisfying the constraint
${\displaystyle 1\cdot m_{1}+2\cdot m_{2}+3\cdot m_{3}+\cdots +n\cdot m_{n}=n.}$
Sometimes, to give it a memorable pattern, it is written in a way in which the coefficients that have the combinatorial interpretation discussed below are less explicit:
${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum {\frac {n!}{m_{1}!\,m_{2}!\,\cdots \,m_{n}!}}\cdot f^{(m_{1}+\cdots +m_{n})}(g(x))\cdot \prod _{j=1}^{n}\left({\frac {g^{(j)}(x)}{j!}}\right)^{m_{j}}.}$
Combining the terms with the same value of m1 + m2 + ... + mn = k and noticing that m j has to be zero for j > n − k + 1 leads to a somewhat simpler formula expressed in terms of Bell polynomials Bn,k(x1,...,xnk+1):
${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum _{k=1}^{n}f^{(k)}(g(x))\cdot B_{n,k}\left(g'(x),g''(x),\dots ,g^{(n-k+1)}(x)\right).}$
First-degree polynomial
First derivative test
The first derivative test examines a function's monotonic properties (where the function is increasing or decreasing) focusing on a particular point in its domain. If the function "switches" from increasing to decreasing at the point, then the function will achieve a highest value at that point. Similarly, if the function "switches" from decreasing to increasing at the point, then it will achieve a least value at that point. If the function fails to "switch", and remains increasing or remains decreasing, then no highest or least value is achieved.
Fractional calculus
Is a branch of mathematical analysis that studies the several different possibilities of defining real number powers or complex number powers of the differentiation operator D
${\displaystyle Df(x)={\dfrac {d}{dx}}f(x)}$,
and of the integration operator J
${\displaystyle Jf(x)=\int _{0}^{x}\!\!\!\!f(s){ds}}$,[Note 2]
and developing a calculus for such operators generalizing the classical one. In this context, the term powers refers to iterative application of a linear operator to a function, in some analogy to function composition acting on a variable, i.e. f ∘2(x) = f ∘ f (x) = f ( f (x) ).
Frustum
In geometry, a frustum (plural: frusta or frustums) is the portion of a solid (normally a cone or pyramid) that lies between one or two parallel planes cutting it. A right frustum is a parallel truncation of a right pyramid or right cone.[42]
Function
Is a process or a relation that associates each element x of a set X, the domain of the function, to a single element y of another set Y (possibly the same set), the codomain of the function. If the function is called f, this relation is denoted y = f(x) (read f of x), the element x is the argument or input of the function, and y is the value of the function, the output, or the image of x by f.[43] The symbol that is used for representing the input is the variable of the function (one often says that f is a function of the variable x).
Function composition
Is an operation that takes two functions f and g and produces a function h such that h(x) = g(f(x)). In this operation, the function g is applied to the result of applying the function f to x. That is, the functions f : XY and g : YZ are composed to yield a function that maps x in X to g(f(x)) in Z.
Fundamental theorem of calculus
The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function. The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives (also called indefinite integral), say F, of some function f may be obtained as the integral of f with a variable bound of integration. This implies the existence of antiderivatives for continuous functions.[44] Conversely, the second part of the theorem, sometimes called the second fundamental theorem of calculus, states that the integral of a function f over some interval can be computed by using any one, say F, of its infinitely many antiderivatives. This part of the theorem has key practical applications, because explicitly finding the antiderivative of a function by symbolic integration avoids numerical integration to compute integrals. This provides generally a better numerical accuracy.
## G
General Leibniz rule
The general Leibniz rule,[45] named after Gottfried Wilhelm Leibniz, generalizes the product rule (which is also known as "Leibniz's rule"). It states that if ${\displaystyle f}$ and ${\displaystyle g}$ are ${\displaystyle n}$-times differentiable functions, then the product ${\displaystyle fg}$ is also ${\displaystyle n}$-times differentiable and its ${\displaystyle n}$th derivative is given by
${\displaystyle (fg)^{(n)}=\sum _{k=0}^{n}{n \choose k}f^{(n-k)}g^{(k)},}$
where ${\displaystyle {n \choose k}={n! \over k!(n-k)!}}$ is the binomial coefficient and ${\displaystyle f^{(0)}\equiv f.}$ This can be proved by using the product rule and mathematical induction.
Global maximum
In mathematical analysis, the maxima and minima (the respective plurals of maximum and minimum) of a function, known collectively as extrema (the plural of extremum), are the largest and smallest value of the function, either within a given range (the local or relative extrema) or on the entire domain of a function (the global or absolute extrema).[46][47][48] Pierre de Fermat was one of the first mathematicians to propose a general technique, adequality, for finding the maxima and minima of functions. As defined in set theory, the maximum and minimum of a set are the greatest and least elements in the set, respectively. Unbounded infinite sets, such as the set of real numbers, have no minimum or maximum.
Global minimum
In mathematical analysis, the maxima and minima (the respective plurals of maximum and minimum) of a function, known collectively as extrema (the plural of extremum), are the largest and smallest value of the function, either within a given range (the local or relative extrema) or on the entire domain of a function (the global or absolute extrema).[49][50][51] Pierre de Fermat was one of the first mathematicians to propose a general technique, adequality, for finding the maxima and minima of functions. As defined in set theory, the maximum and minimum of a set are the greatest and least elements in the set, respectively. Unbounded infinite sets, such as the set of real numbers, have no minimum or maximum.
Golden spiral
In geometry, a golden spiral is a logarithmic spiral whose growth factor is φ, the golden ratio.[52] That is, a golden spiral gets wider (or further from its origin) by a factor of φ for every quarter turn it makes.
Is a multi-variable generalization of the derivative. While a derivative can be defined on functions of a single variable, for functions of several variables, the gradient takes its place. The gradient is a vector-valued function, as opposed to a derivative, which is scalar-valued.
## H
Harmonic progression
In mathematics, a harmonic progression (or harmonic sequence) is a progression formed by taking the reciprocals of an arithmetic progression. It is a sequence of the form
${\displaystyle {\frac {1}{a}},\ {\frac {1}{a+d}}\ ,{\frac {1}{a+2d}}\ ,{\frac {1}{a+3d}}\ ,\cdots ,{\frac {1}{a+kd}},}$
where −a/d is not a natural number and k is a natural number. Equivalently, a sequence is a harmonic progression when each term is the harmonic mean of the neighboring terms. It is not possible for a harmonic progression (other than the trivial case where a = 1 and k = 0) to sum to an integer. The reason is that, necessarily, at least one denominator of the progression will be divisible by a prime number that does not divide any other denominator.[53]
Higher derivative
Let f be a differentiable function, and let f be its derivative. The derivative of f (if it has one) is written f ′′ and is called the second derivative of f. Similarly, the derivative of the second derivative, if it exists, is written f ′′′ and is called the third derivative of f. Continuing this process, one can define, if it exists, the nth derivative as the derivative of the (n-1)th derivative. These repeated derivatives are called higher-order derivatives. The nth derivative is also called the derivative of order n.
Homogeneous linear differential equation
A differential equation can be homogeneous in either of two respects. A first order differential equation is said to be homogeneous if it may be written
${\displaystyle f(x,y)dy=g(x,y)dx,}$
where f and g are homogeneous functions of the same degree of x and y. In this case, the change of variable y = ux leads to an equation of the form
${\displaystyle {\frac {dx}{x}}=h(u)du,}$
which is easy to solve by integration of the two members. Otherwise, a differential equation is homogeneous if it is a homogeneous function of the unknown function and its derivatives. In the case of linear differential equations, this means that there are no constant terms. The solutions of any linear ordinary differential equation of any order may be deduced by integration from the solution of the homogeneous equation obtained by removing the constant term.
Hyperbolic function
Hyperbolic functions are analogs of the ordinary trigonometric, or circular, functions.
## I
Identity function
Also called an identity relation or identity map or identity transformation, is a function that always returns the same value that was used as its argument. In equations, the function is given by f(x) = x.
Imaginary number
Is a complex number that can be written as a real number multiplied by the imaginary unit i,[note 2] which is defined by its property i2 = −1.[54] The square of an imaginary number bi is b2. For example, 5i is an imaginary number, and its square is −25. Zero is considered to be both real and imaginary.[55]
Implicit function
In mathematics, an implicit equation is a relation of the form ${\displaystyle R(x_{1},\ldots ,x_{n})=0}$, where ${\displaystyle R}$ is a function of several variables (often a polynomial). For example, the implicit equation of the unit circle is ${\displaystyle x^{2}+y^{2}-1=0}$. An implicit function is a function that is defined implicitly by an implicit equation, by associating one of the variables (the value) with the others (the arguments).[56]: 204–206 Thus, an implicit function for ${\displaystyle y}$ in the context of the unit circle is defined implicitly by ${\displaystyle x^{2}+f(x)^{2}-1=0}$. This implicit equation defines ${\displaystyle f}$ as a function of ${\displaystyle x}$ only if ${\displaystyle -1\leq x\leq 1}$ and one considers only non-negative (or non-positive) values for the values of the function. The implicit function theorem provides conditions under which some kinds of relations define an implicit function, namely relations defined as the indicator function of the zero set of some continuously differentiable multivariate function.
Improper fraction
Common fractions can be classified as either proper or improper. When the numerator and the denominator are both positive, the fraction is called proper if the numerator is less than the denominator, and improper otherwise.[57][58] In general, a common fraction is said to be a proper fraction if the absolute value of the fraction is strictly less than one—that is, if the fraction is greater than −1 and less than 1.[59][60] It is said to be an improper fraction, or sometimes top-heavy fraction,[61] if the absolute value of the fraction is greater than or equal to 1. Examples of proper fractions are 2/3, –3/4, and 4/9; examples of improper fractions are 9/4, –4/3, and 3/3.
Improper integral
In mathematical analysis, an improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number, ${\displaystyle \infty }$, ${\displaystyle -\infty }$, or in some instances as both endpoints approach limits. Such an integral is often written symbolically just like a standard definite integral, in some cases with infinity as a limit of integration. Specifically, an improper integral is a limit of the form:
${\displaystyle \lim _{b\to \infty }\int _{a}^{b}f(x)\,dx,\qquad \lim _{a\to -\infty }\int _{a}^{b}f(x)\,dx,}$
or
${\displaystyle \lim _{c\to b^{-}}\int _{a}^{c}f(x)\,dx,\quad \lim _{c\to a^{+}}\int _{c}^{b}f(x)\,dx,}$
in which one takes a limit in one or the other (or sometimes both) endpoints (Apostol 1967, §10.23).
Inflection point
In differential calculus, an inflection point, point of inflection, flex, or inflection (British English: inflexion) is a point on a continuous plane curve at which the curve changes from being concave (concave downward) to convex (concave upward), or vice versa.
Instantaneous rate of change
The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The tangent line is the best linear approximation of the function near that input value. For this reason, the derivative is often described as the "instantaneous rate of change", the ratio of the instantaneous change in the dependent variable to that of the independent variable. .
Instantaneous velocity
If we consider v as velocity and x as the displacement (change in position) vector, then we can express the (instantaneous) velocity of a particle or object, at any particular time t, as the derivative of the position with respect to time:
${\displaystyle {\boldsymbol {v}}=\lim _{{\Delta t}\to 0}{\frac {\Delta {\boldsymbol {x}}}{\Delta t}}={\frac {d{\boldsymbol {x}}}{d{\mathit {t}}}}.}$
From this derivative equation, in the one-dimensional case it can be seen that the area under a velocity vs. time (v vs. t graph) is the displacement, x. In calculus terms, the integral of the velocity function v(t) is the displacement function x(t). In the figure, this corresponds to the yellow area under the curve labeled s (s being an alternative notation for displacement).
${\displaystyle {\boldsymbol {x}}=\int {\boldsymbol {v}}\ d{\mathit {t}}.}$
Since the derivative of the position with respect to time gives the change in position (in metres) divided by the change in time (in seconds), velocity is measured in metres per second (m/s). Although the concept of an instantaneous velocity might at first seem counter-intuitive, it may be thought of as the velocity that the object would continue to travel at if it stopped accelerating at that moment. .
Integral
An integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. Integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. .
Integral symbol
The integral symbol:
(Unicode), ${\displaystyle \displaystyle \int }$ (LaTeX)
is used to denote integrals and antiderivatives in mathematics. .
Integrand
The function to be integrated in an integral.
Integration by parts
In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of their derivative and antiderivative. It is frequently used to transform the antiderivative of a product of functions into an antiderivative for which a solution can be more easily found. The rule can be readily derived by integrating the product rule of differentiation. If u = u(x) and du = u(x) dx, while v = v(x) and dv = v(x) dx, then integration by parts states that:
{\displaystyle {\begin{aligned}\int _{a}^{b}u(x)v'(x)\,dx&={\Big [}u(x)v(x){\Big ]}_{a}^{b}-\int _{a}^{b}u'(x)v(x)\,dx\\&=u(b)v(b)-u(a)v(a)-\int _{a}^{b}u'(x)v(x)\,dx\end{aligned}}}
or more compactly:
${\displaystyle \int u\,dv=uv-\int v\,du.}$
Mathematician Brook Taylor discovered integration by parts, first publishing the idea in 1715.[62][63] More general formulations of integration by parts exist for the Riemann–Stieltjes and Lebesgue–Stieltjes integrals. The discrete analogue for sequences is called summation by parts. .
Integration by substitution
Also known as u-substitution, is a method for solving integrals. Using the fundamental theorem of calculus often requires finding an antiderivative. For this and other reasons, integration by substitution is an important tool in mathematics. It is the counterpart to the chain rule for differentiation. .
Intermediate value theorem
In mathematical analysis, the intermediate value theorem states that if a continuous function, f, with an interval, [a, b], as its domain, takes values f(a) and f(b) at each end of the interval, then it also takes any value between f(a) and f(b) at some point within the interval. This has two important corollaries:
1. If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano's theorem).[64]
2. The image of a continuous function over an interval is itself an interval. .
Inverse trigonometric functions
(Also called arcus functions,[65][66][67][68][69] antitrigonometric functions[70] or cyclometric functions[71][72][73]) are the inverse functions of the trigonometric functions (with suitably restricted domains). Specifically, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle's trigonometric ratios.
## J
Jump discontinuity
Consider the function
${\displaystyle f(x)={\begin{cases}x^{2}&{\mbox{ for }}x<1\\0&{\mbox{ for }}x=1\\2-(x-1)^{2}&{\mbox{ for }}x>1\end{cases}}}$
Then, the point x0 = 1 is a jump discontinuity. In this case, a single limit does not exist because the one-sided limits, L and L+, exist and are finite, but are not equal: since, LL+, the limit L does not exist. Then, x0 is called a jump discontinuity, step discontinuity, or discontinuity of the first kind. For this type of discontinuity, the function f may have any value at x0.
## L
Lebesgue integration
In mathematics, the integral of a non-negative function of a single variable can be regarded, in the simplest case, as the area between the graph of that function and the x-axis. The Lebesgue integral extends the integral to a larger class of functions. It also extends the domains on which these functions can be defined.
L'Hôpital's rule
L'Hôpital's rule or L'Hospital's rule uses derivatives to help evaluate limits involving indeterminate forms. Application (or repeated application) of the rule often converts an indeterminate form to an expression that can be evaluated by substitution, allowing easier evaluation of the limit. The rule is named after the 17th-century French mathematician Guillaume de l'Hôpital. Although the contribution of the rule is often attributed to L'Hôpital, the theorem was first introduced to L'Hôpital in 1694 by the Swiss mathematician Johann Bernoulli. L'Hôpital's rule states that for functions f and g which are differentiable on an open interval I except possibly at a point c contained in I, if ${\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty ,}$ ${\displaystyle g'(x)\neq 0}$ for all x in I with xc, and ${\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$ exists, then
${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}.}$
The differentiation of the numerator and denominator often simplifies the quotient or converts it to a limit that can be evaluated directly.
Limit comparison test
The limit comparison test allows one to determine the convergence of one series based on the convergence of another.
Limit of a function
.
Limits of integration
.
Linear combination
In mathematics, a linear combination is an expression constructed from a set of terms by multiplying each term by a constant and adding the results (e.g. a linear combination of x and y would be any expression of the form ax + by, where a and b are constants).[74][75][76] The concept of linear combinations is central to linear algebra and related fields of mathematics.
Linear equation
A linear equation is an equation relating two or more variables to each other in the form of ${\displaystyle a_{1}x_{1}+\cdots +a_{n}x_{n}+b=0,}$ with the highest power of each variable being 1.
Linear system
.
List of integrals
.
Logarithm
.
Logarithmic differentiation
.
Lower bound
.
## M
Mean value theorem
.
Monotonic function
.
Multiple integral
.
Multiplicative calculus
.
Multivariable calculus
.
## N
Natural logarithm
The natural logarithm of a number is its logarithm to the base of the mathematical constant e, where e is an irrational and transcendental number approximately equal to 2.718281828459. The natural logarithm of x is generally written as ln x, loge x, or sometimes, if the base e is implicit, simply log x.[77] Parentheses are sometimes added for clarity, giving ln(x), loge(x) or log(x). This is done in particular when the argument to the logarithm is not a single symbol, to prevent ambiguity.
Non-Newtonian calculus
.
Nonstandard calculus
.
Notation for differentiation
.
Numerical integration
.
## O
One-sided limit
.
Ordinary differential equation
.
## P
Pappus's centroid theorem
(Also known as the Guldinus theorem, Pappus–Guldinus theorem or Pappus's theorem) is either of two related theorems dealing with the surface areas and volumes of surfaces and solids of revolution.
Parabola
Is a plane curve that is mirror-symmetrical and is approximately U-shaped. It fits several superficially different other mathematical descriptions, which can all be proved to define exactly the same curves.
Paraboloid
.
Partial derivative
.
Partial differential equation
.
Partial fraction decomposition
.
Particular solution
.
Piecewise-defined function
A function defined by multiple sub-functions that apply to certain intervals of the function's domain.
Position vector
.
Power rule
.
Product integral
.
Product rule
.
Proper fraction
.
Proper rational function
.
Pythagorean theorem
.
Pythagorean trigonometric identity
.
## Q
In algebra, a quadratic function, a quadratic polynomial, a polynomial of degree 2, or simply a quadratic, is a polynomial function with one or more variables in which the highest-degree term is of the second degree. For example, a quadratic function in three variables x, y, and z contains exclusively terms x2, y2, z2, xy, xz, yz, x, y, z, and a constant:
${\displaystyle f(x,y,z)=ax^{2}+by^{2}+cz^{2}+dxy+exz+fyz+gx+hy+iz+j,}$
with at least one of the coefficients a, b, c, d, e, or f of the second-degree terms being non-zero. A univariate (single-variable) quadratic function has the form[78]
${\displaystyle f(x)=ax^{2}+bx+c,\quad a\neq 0}$
in the single variable x. The graph of a univariate quadratic function is a parabola whose axis of symmetry is parallel to the y-axis, as shown at right. If the quadratic function is set equal to zero, then the result is a quadratic equation. The solutions to the univariate equation are called the roots of the univariate function. The bivariate case in terms of variables x and y has the form
${\displaystyle f(x,y)=ax^{2}+by^{2}+cxy+dx+ey+f\,\!}$
with at least one of a, b, c not equal to zero, and an equation setting this function equal to zero gives rise to a conic section (a circle or other ellipse, a parabola, or a hyperbola). In general there can be an arbitrarily large number of variables, in which case the resulting surface is called a quadric, but the highest degree term must be of degree 2, such as x2, xy, yz, etc.
.
Quotient rule
A formula for finding the derivative of a function that is the ratio of two functions.
## R
Is the SI unit for measuring angles, and is the standard unit of angular measure used in many areas of mathematics. The length of an arc of a unit circle is numerically equal to the measurement in radians of the angle that it subtends; one radian is just under 57.3 degrees (expansion at ). The unit was formerly an SI supplementary unit, but this category was abolished in 1995 and the radian is now considered an SI derived unit.[79] Separately, the SI unit of solid angle measurement is the steradian .
Ratio test
.
Reciprocal function
.
Reciprocal rule
.
Riemann integral
.
.
Removable discontinuity
.
Rolle's theorem
.
Root test
.
## S
Scalar
.
Secant line
.
Second-degree polynomial
.
Second derivative
.
Second derivative test
.
Second-order differential equation
.
Series
.
Shell integration
.
Simpson's rule
.
Sine
.
Sine wave
.
Slope field
.
Squeeze theorem
.
Sum rule in differentiation
.
Sum rule in integration
.
Summation
.
Supplementary angle
.
Surface area
.
System of linear equations
.
## T
Table of integrals
.
Taylor series
.
Taylor's theorem
.
Tangent
.
Third-degree polynomial
.
Third derivative
.
Toroid
.
Total differential
.
Trigonometric functions
.
Trigonometric identities
.
Trigonometric integral
.
Trigonometric substitution
.
Trigonometry
.
Triple integral
.
Upper bound
.
Variable
.
Vector
.
Vector calculus
.
Washer
.
Washer method
.
[[]]
]]
## References
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2. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 978-0-547-16702-2.
3. ^ "Asymptotes" by Louis A. Talman
4. ^ Williamson, Benjamin (1899), "Asymptotes", An elementary treatise on the differential calculus
5. ^ Nunemacher, Jeffrey (1999), "Asymptotes, Cubic Curves, and the Projective Plane", Mathematics Magazine, 72 (3): 183–192, CiteSeerX 10.1.1.502.72, doi:10.2307/2690881, JSTOR 2690881
6. ^ Neidinger, Richard D. (2010). "Introduction to Automatic Differentiation and MATLAB Object-Oriented Programming" (PDF). SIAM Review. 52 (3): 545–563. doi:10.1137/080743627.
7. ^ Baydin, Atilim Gunes; Pearlmutter, Barak; Radul, Alexey Andreyevich; Siskind, Jeffrey (2018). "Automatic differentiation in machine learning: a survey". Journal of Machine Learning Research. 18: 1–43.
8. ^ "Calculus". OxfordDictionaries. Retrieved 15 September 2017.
9. ^ Howard Eves, "Two Surprising Theorems on Cavalieri Congruence", The College Mathematics Journal, volume 22, number 2, March, 1991), pages 118–124
10. ^ Hall, Arthur Graham; Frink, Fred Goodrich (January 1909). "Chapter II. The Acute Angle [10] Functions of complementary angles". Written at Ann Arbor, Michigan, USA. Trigonometry. Part I: Plane Trigonometry. New York, USA: Henry Holt and Company / Norwood Press / J. S. Cushing Co. - Berwick & Smith Co., Norwood, Massachusetts, USA. pp. 11–12. Retrieved 2017-08-12.
11. ^ Aufmann, Richard; Nation, Richard (2014). Algebra and Trigonometry (8 ed.). Cengage Learning. p. 528. ISBN 978-128596583-3. Retrieved 2017-07-28.
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16. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 978-0-547-16702-2.
17. ^ Stalker, John (1998). Complex Analysis: Fundamentals of the Classical Theory of Functions. Springer. p. 77. ISBN 0-8176-4038-X.
18. ^ Bak, Joseph; Newman, Donald J. (1997). "Chapters 11 & 12". Complex Analysis. Springer. pp. 130–156. ISBN 0-387-94756-6.
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20. ^ "Lecture Notes 2" (PDF). www.stat.cmu.edu. Retrieved 3 March 2017.
21. ^ Cramer, Gabriel (1750). "Introduction à l'Analyse des lignes Courbes algébriques" (in French). Geneva: Europeana. pp. 656–659. Retrieved 2012-05-18.
22. ^ Kosinski, A. A. (2001). "Cramer's Rule is due to Cramer". Mathematics Magazine. 74 (4): 310–312. doi:10.2307/2691101. JSTOR 2691101.
23. ^ MacLaurin, Colin (1748). A Treatise of Algebra, in Three Parts. Printed for A. Millar & J. Nourse.
24. ^ Boyer, Carl B. (1968). A History of Mathematics (2nd ed.). Wiley. p. 431.
25. ^ Katz, Victor (2004). A History of Mathematics (Brief ed.). Pearson Education. pp. 378–379.
26. ^ Hedman, Bruce A. (1999). "An Earlier Date for "Cramer's Rule"" (PDF). Historia Mathematica. 26 (4): 365–368. doi:10.1006/hmat.1999.2247.
27. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 978-0-495-01166-8.
28. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 978-0-547-16702-2.
29. ^ Douglas C. Giancoli (2000). [Physics for Scientists and Engineers with Modern Physics (3rd Edition)]. Prentice Hall. ISBN 0-13-021517-1
30. ^ "Definition of DIFFERENTIAL CALCULUS". www.merriam-webster.com. Retrieved 2018-09-26.
31. ^ "Integral Calculus - Definition of Integral calculus by Merriam-Webster". www.merriam-webster.com. Retrieved 2018-05-01.
32. ^ Démonstration d’un théorème d’Abel. Journal de mathématiques pures et appliquées 2nd series, tome 7 (1862), p. 253-255 Archived 2011-07-21 at the Wayback Machine.
33. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks Cole Cengage Learning. ISBN 978-0-495-01166-8.
34. ^ Oxford English Dictionary, 2nd ed.: natural logarithm
35. ^
36. ^ Butcher 2003, p. 45; Hairer, Nørsett & Wanner 1993, p. 35
37. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 978-0-495-01166-8.
38. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 978-0-547-16702-2.
39. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 978-0-321-58876-0.
40. ^ (Arbogast 1800).
41. ^ According to Craik (2005, pp. 120–122): see also the analysis of Arbogast's work by Johnson (2002, p. 230).
42. ^ William F. Kern, James R. Bland, Solid Mensuration with proofs, 1938, p. 67
43. ^ MacLane, Saunders; Birkhoff, Garrett (1967). Algebra (First ed.). New York: Macmillan. pp. 1–13.
44. ^ Spivak, Michael (1980), Calculus (2nd ed.), Houston, Texas: Publish or Perish Inc.
45. ^ Olver, Peter J. (2000). Applications of Lie Groups to Differential Equations. Springer. pp. 318–319. ISBN 9780387950006.
46. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 978-0-495-01166-8.
47. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 978-0-547-16702-2.
48. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 978-0-321-58876-0.
49. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 978-0-495-01166-8.
50. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 978-0-547-16702-2.
51. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 978-0-321-58876-0.
52. ^ Chang, Yu-sung, "Golden Spiral Archived 2019-07-28 at the Wayback Machine", The Wolfram Demonstrations Project.
53. ^ Erdős, P. (1932), "Egy Kürschák-féle elemi számelméleti tétel általánosítása" [Generalization of an elementary number-theoretic theorem of Kürschák] (PDF), Mat. Fiz. Lapok (in Hungarian), 39: 17–24. As cited by Graham, Ronald L. (2013), "Paul Erdős and Egyptian fractions", Erdős centennial, Bolyai Soc. Math. Stud., 25, János Bolyai Math. Soc., Budapest, pp. 289–309, doi:10.1007/978-3-642-39286-3_9, MR 3203600.
54. ^ Uno Ingard, K. (1988). "Chapter 2". Fundamentals of Waves and Oscillations. Cambridge University Press. p. 38. ISBN 0-521-33957-X.
55. ^ Sinha, K.C. (2008). A Text Book of Mathematics Class XI (Second ed.). Rastogi Publications. p. 11.2. ISBN 978-81-7133-912-9.
56. ^ Chiang, Alpha C. (1984). Fundamental Methods of Mathematical Economics (Third ed.). New York: McGraw-Hill. ISBN 0-07-010813-7.
57. ^ "World Wide Words: Vulgar fractions". World Wide Words. Retrieved 2014-10-30.
58. ^ Weisstein, Eric W. "Improper Fraction". MathWorld.
59. ^ Laurel (31 March 2004). "Math Forum – Ask Dr. Math:Can Negative Fractions Also Be Proper or Improper?". Retrieved 2014-10-30.
60. ^ "New England Compact Math Resources". Archived from the original on 2012-04-15. Retrieved 2019-06-16.
61. ^ Greer, A. (1986). New comprehensive mathematics for 'O' level (2nd ed., reprinted. ed.). Cheltenham: Thornes. p. 5. ISBN 978-0-85950-159-0. Retrieved 2014-07-29.
62. ^ "Brook Taylor". History.MCS.St-Andrews.ac.uk. Retrieved May 25, 2018.
63. ^ "Brook Taylor". Stetson.edu. Retrieved May 25, 2018.
64. ^ Weisstein, Eric W. "Bolzano's Theorem". MathWorld.
65. ^ Taczanowski, Stefan (1978-10-01). "On the optimization of some geometric parameters in 14 MeV neutron activation analysis". Nuclear Instruments and Methods. ScienceDirect. 155(3): 543–546. doi:10.1016/0029-554X(78)90541-4.
66. ^ Hazewinkel, Michiel (1994) [1987]. Encyclopaedia of Mathematics (unabridged reprint ed.). Kluwer Academic Publishers / Springer Science & Business Media. ISBN 978-155608010-4.
67. ^ Ebner, Dieter (2005-07-25). Preparatory Course in Mathematics (PDF) (6 ed.). Department of Physics, University of Konstanz. Archived (PDF) from the original on 2017-07-26. Retrieved 2017-07-26.
68. ^ Mejlbro, Leif (2010-11-11). Stability, Riemann Surfaces, Conformal Mappings - Complex Functions Theory (PDF) (1 ed.). Ventus Publishing ApS / Bookboon. ISBN 978-87-7681-702-2. Archived (PDF) from the original on 2017-07-26. Retrieved 2017-07-26.
69. ^ Durán, Mario (2012). Mathematical methods for wave propagation in science and engineering. 1: Fundamentals (1 ed.). Ediciones UC. p. 88. ISBN 978-956141314-6.
70. ^ Hall, Arthur Graham; Frink, Fred Goodrich (January 1909). "Chapter II. The Acute Angle [14] Inverse trigonometric functions". Written at Ann Arbor, Michigan, USA. Trigonometry. Part I: Plane Trigonometry. New York, USA: Henry Holt and Company / Norwood Press / J. S. Cushing Co. - Berwick & Smith Co., Norwood, Massachusetts, USA. p. 15. Retrieved 2017-08-12. […] α = arcsin m: It is frequently read "arc-sinem" or "anti-sine m," since two mutually inverse functions are said each to be the anti-function of the other. […] A similar symbolic relation holds for the other trigonometric functions. […] This notation is universally used in Europe and is fast gaining ground in this country. A less desirable symbol, α = sin-1m, is still found in English and American texts. The notation α = inv sin m is perhaps better still on account of its general applicability. […]
71. ^ Klein, Christian Felix (1924) [1902]. Elementarmathematik vom höheren Standpunkt aus: Arithmetik, Algebra, Analysis (in German). 1 (3rd ed.). Berlin: J. Springer.
72. ^ Klein, Christian Felix (2004) [1932]. Elementary Mathematics from an Advanced Standpoint: Arithmetic, Algebra, Analysis. Translated by Hedrick, E. R.; Noble, C. A. (Translation of 3rd German ed.). Dover Publications, Inc. / The Macmillan Company. ISBN 978-0-48643480-3. Retrieved 2017-08-13.
73. ^ Dörrie, Heinrich (1965). Triumph der Mathematik. Translated by Antin, David. Dover Publications. p. 69. ISBN 978-0-486-61348-2.
74. ^ Lay, David C. (2006). Linear Algebra and Its Applications (3rd ed.). Addison–Wesley. ISBN 0-321-28713-4.
75. ^ Strang, Gilbert (2006). Linear Algebra and Its Applications (4th ed.). Brooks Cole. ISBN 0-03-010567-6.
76. ^ Axler, Sheldon (2002). Linear Algebra Done Right (2nd ed.). Springer. ISBN 0-387-98258-2.
77. ^ Mortimer, Robert G. (2005). Mathematics for physical chemistry (3rd ed.). Academic Press. p. 9. ISBN 0-12-508347-5. Extract of page 9
78. ^ "Quadratic Equation -- from Wolfram MathWorld". Retrieved January 6, 2013.
79. ^ "Resolution 8 of the CGPM at its 20th Meeting (1995)". Bureau International des Poids et Mesures. Retrieved 2014-09-23.
## Notes
1. ^ The term scalar product is often also used more generally to mean a symmetric bilinear form, for example for a pseudo-Euclidean space.[citation needed]
2. ^ j is usually used in Engineering contexts where i has other meanings (such as electrical current)
1. ^ Antiderivatives are also called general integrals, and sometimes integrals. The latter term is generic, and refers not only to indefinite integrals (antiderivatives), but also to definite integrals. When the word integral is used without additional specification, the reader is supposed to deduce from the context whether it refers to a definite or indefinite integral. Some authors define the indefinite integral of a function as the set of its infinitely many possible antiderivatives. Others define it as an arbitrarily selected element of that set. Wikipedia adopts the latter approach.[citation needed]
2. ^ The symbol J is commonly is used instead of the intuitive I in order to avoid confusion with other concepts identified by similar I–like glyphs, e.g. identities.
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Oh, the days – weeks, even – of my university life I spent working out the determinants of matrices. The 3×3 version was the main culprit, of course, usually needing to be split down into three smaller determinants, and usually requiring a sign change in one or two that I’d almost always miss.
A mess, if ever there was one.
Luckily, there’s another method for 3×3 matrices that I find much easier than the old-fashioned way.
We’re going to be looking at diagonal lines in matrices, and there are six of them. There don’t seem to be six, until you realise that the diagonals wrap around. There are three forward diagonals:
• (top left, middle middle, bottom right) is the obvious one;
• (top middle, middle right, bottom left) and
• (top right, middle left, bottom middle)
The three backward ones are:
• (top left, middle right, bottom middle);
• (top middle, middle left, bottom right) and
• (top right, middle middle, bottom left)
Once you recognise the diagonals, the recipe is simple: you’re going to multiply together the numbers on each diagonal to get six products. Add the three forward ones, subtract the three backward ones and boom! There’s your determinant.
For example, let’s take the matrix $\left( \begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 1\\ 3 & 2 & 1 \end{array}\right)$.
The forward diagonals are (1, 3, 1), (2, 1, 3) and (3, 2, 2), giving products of 3, 6 and 12 – these add up to 21.
The backward diagonals are (1, 1, 2), (2, 2, 1) and (3, 3, 3), giving products of 2, 4 and 27, which add up to 33.
$21 - 33 = -12$, which is indeed the determinant of the matrix!
UPDATE: @DrTrapezio, who is Luciano Rila in real life, has very kindly drawn out the calculation as he would do it:
Thanks for sharing!
* Edited 2015-12-01 to fix LaTex. Thanks, @fenneklyra, for spotting the error!
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# Prime number
Here is another way to think of prime numbers. The number 12 is not prime, because a rectangle can be made, with sides of lengths 4 and 3. This rectangle has an area of 12, because all 12 blocks are used. This cannot be done with 11. No matter how the rectangle is arranged, there will always be blocks left over. 11 must therefore be a prime number
A composite number is a natural number of a particular kind. Any natural number is equal to 1 times itself. If the number is equal to any other numbers multiplied, then the number is a composite number. The smallest composite number is 4, because 2 x 2 = 4. 1 is not a composite number. Every other number is a prime number. The prime numbers are the numbers other than 1 which are not equal to m x n (except 1 x itself). The smallest prime number is 2. There is no largest prime number. The next smallest prime numbers are 3, 5, 7, 11, 13, and so on.
The way that the prime numbers occur is a difficult problem for mathematicians. When a number is larger, it is more difficult to know if it is a prime number. One of the answers is the Prime number theorem. One of the unsolved problems is Goldbach's conjecture.
## How to find small prime numbers
There is a simple method to find a list of prime numbers. Eratosthenes created it. It has the name Sieve of Eratosthenes. It catches numbers that are not prime (like a sieve) and lets the prime numbers pass through.
• On a sheet of paper, write all the whole numbers from 2 up to the number being tested. Do not write down the number 1, because it is not a prime number.
• At the start, all numbers are not crossed out.
The method is always the same:
1. Start with 2. Let p equal 2. Go to the next step.
2. If p is the the end of the list, then go to the last step. If p is not the end of the list, go to the next step.
3. Start with p and count out p numbers then cross out that number. Repeat counting off p numbers and crossing out numbers until the end of the list. (This means that all the numbers p x n are crossed out. p x n is a composite number. The first time, this step will cross out 4, 4, 6, and so on. The second time, this step will cross out 3, 6, 9, 12, and so on. 6 and 12 have already been crossed out. Cross them out again. The next time, this will cross out 10, 15, 20, and so on.) Go to the next step.
4. Add 1 to p. Go to the next step.
5. If the new value of p has been crossed out, go back to the last step. (If the new value of p is crossed out, it is composite. Try another value of p)
6. Go back to the 2nd step. (If the new value of p has not been crossed out, then it is a prime number.)
7. (This is the last step). All of the composite numbers are crossed out and all of the prime numbers are not crossed out.
As an example, if this is done up to the number 10, the numbers 2, 3, 5 and 7 are prime numbers, and 4, 6, 8, 9 and 10 are composite numbers.
This method or algorithm takes too long to find very large prime numbers. But it is less complicated than methods used for very large primes, like Fermat's primality test (a primality test is to test whether a number is prime or not) or the Miller-Rabin primality test.
## What prime numbers are used for
Prime numbers are very important in mathematics and computer science. Some real-world uses are given below. Very long prime numbers are hard to solve. It is difficult to find their prime factors, so most of the time, numbers that are probably prime are used for encryption and secret codes.[1]
• Most people have a bank card, where they can get money from their account, using an ATM. This card is protected by a secret access code. Since the code needs to be kept secret, it cannot be stored in cleartext on the card. Encryption is used to store the code in a secret way. This encryption uses multiplications, divisions, and finding remainders of large prime numbers. An algorithm called RSA is often used in practice. It uses the Chinese remainder theorem.
• If someone has a digital signature for their email, encryption is used. This makes sure that no one can fake an email from them. Before signing, a hash value of the message is created. This is then combined with a digital signature to produce a signed message. Methods used are more or less the same as in the first case above.
• Finding the largest prime known so far has become a sport of sorts. Testing if a number is prime can be difficult if the number is large. The largest primes known at any time are usually Mersenne primes because the fastest known test for primality is the Lucas-Lehmer test, which relies on the special form of Mersenne numbers. A group that searches for Mersenne primes is here[1].
## References
1. Graham Templeton (25 October 2013). "Geek Answers: Why should we care about prime numbers?". Retrieved 10 January 2015.
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## What is the formula for a surface area?
Variables:
Surface Area Formula Surface Area Meaning
SA=B+12sP Find the area of each face. Add up all areas.
SA=2B+2πrh Find the area of the base, times 2, then add the areas to the areas of the rectangle, which is the circumference times the height.
SA=4πr2 Find the area of the great circle and multiply it by 4.
## How do you find the surface area and volume of a cylinder?
A cylinder’s volume is π r² h, and its surface area is 2π r h + 2π r².
What is total surface area of a shape?
It is measured in terms of square units. In other words, the surface area is the sum of all the areas of all the shapes that cover the surface of the object. On the other hand, the lateral surface area refers to the area of the sides of a shape, excluding its base and top area.
### What is the volume and surface area of a cylinder?
A cylinder’s volume is π r² h, and its surface area is 2π r h + 2π r². Learn how to use these formulas to solve an example problem.
### How do you learn all surface area and volume formulas?
Surface Areas And Volumes
1. To find the surface area of a solid, add the areas of all the faces. You can remember the formula as sum of areas of the all the faces.
2. Volume of a prism is area of its base times height.
3. Volume of a pyramid is one third of area of its base times height.
What is flat surface area of cylinder?
Remember the formulas for the lateral surface area of a prism is ph and the total surface area is ph+2B . Since the base of a cylinder is a circle, we substitute 2πr for p and πr2 for B where r is the radius of the base of the cylinder.
## How do you find the surface area of a circle?
The area of a circle is pi times the radius squared (A = π r²).
## What do you mean by frustum of cone?
Frustum of cone is the part of cone when it is cut by a plane into two parts. The upper part of cone remains same in shape but the bottom part makes a frustum. To get this part of the right circular cone we have to slice it horizontally or parallel to the base.
How many surface has a cylinder?
A cylinder has 2 flat surfaces and one curved surface. A cone has one flat surface and a curved surface.
### What is volume of hollow cylinder?
If both the outer and inner radius of a hollow cylinder along with the height is given, then the volume of the hollow cylinder is given by the formula: Volume of Hollow cylinder, V = π (R2 – r2) h cubic units.
### What is solid cylinder?
A solid bounded by a cylindrical surface and two parallel planes is called a (solid) cylinder . The line segments determined by an element of the cylindrical surface between the two parallel planes is called an element of the cylinder. All the elements of a cylinder have equal lengths.
What is the formula of curved surface area of a cylinder?
The radius ‘r’ of a cylinder is the radius of its base. Now, the area of the rectangle = length × breadth. 2πr is the circumference of the circle and h is the height. Area of the curved surface will be = 2πr × h = 2πrh.
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# Solve 3x^2 = 8x - 2 giving your answers to 2 d.p.
• 767 views
First things first, we have to decide what type of equation this is, so we can choose how to deal with it. If it were linear (which means every term involving an 'x' has no power/superscript on it. I.e. x2 or x3 do not appear in your equation), then we could simply rearrange to solve. But our equation has a 3x2 term, so we need to do something else.
First, let's put everything on one side, so we can recognise that this equation is actually a quadratic equation: subtracting 8x and adding 2 to both sides gives 3x2 - 8x + 2 = 0.
Now, there are 3 ways we can deal with a quadratic equation: Factorisation, Completeing the Square, and the Quadratic Formula.
The first two methods won't really help us here, but how did I figure that out? Well, the long way would be to do some trial and error, attempting to factorise. You would quickly see you can't make any combination work. When factorising fails, you normally try the quadratic formula next. Here's a tip: if the question asks you to give your answer to a certain number of decimal places, it means you cannot factorise the quadratic because the answer is a fraction or surd, so always use the quadratic formula in this case!
So now we know we're going to apply the quadratic formula to this equation. Here is the formula in words (because I can't format it clearly on here), with brackets added to show you in what order to do operations:
x = [(-b) plus or minus the (square root of {b2 - 4ac})] all over 2a
(Remember to divide everything by the 2a! It is a common mistake not to do this).
We have 3 variables, a,b and c, which we can find from our quadratic equation we are trying to solve. Note: our quadratic MUST be in the form ax2+bx+c=0 for this to work. We did this earlier, so we can proceed.
From our equation, a (the bit in front of the x2) equals +3. b (the bit in front of the x) equals -8. (The minus is important!!). c (the number without an x) equals +2.
Now we can put a=3, b=-8, c=2 into the quadratic equation.
Note this common mistake: If b is negative, like we have in this question, what happens? -b, here, equals -(-8), and two minuses make a plus, so this becomes +8.
For the -b2, it is important to know where the brackets should be. This is the same thing as (-b)2 and therefore is ALWAYS positive, because all square numbers are positive numbers. So in our question, (-b)2 is (-8)2=64.
4ac = 4*3*2 = 24. Therefore, inside the square root we have 64-24=40. 2a = 2*3 = 6.
Now we can put (8 plus/minus the square root of 40)/6 into our calculator, to give us two answers (one for the plus and one for the minus). These are 2.3874... and 0.2792...
Rounding these to 2 decimal places gives us our final solutions to our quadratic: 2.39 and 0.28.
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# Factoring the Sum & Difference of Two Cubes
## Presentation on theme: "Factoring the Sum & Difference of Two Cubes"— Presentation transcript:
Factoring the Sum & Difference of Two Cubes
This is a piece of cake, if you have perfect cubes.
What are perfect cubes?
This is a piece of cake, if you have perfect cubes.
What are perfect cubes? Something times something times something. Where the something is a factor 3 times. 8 is 2 2 2, so 8 is a perfect cube. x6 is x2 x2 x2 so x6 is a perfect cube. It is easy to see if a variable is a perfect cube, how?
This is a piece of cake, if you have perfect cubes.
What are perfect cubes? Something times something times something. Where the something is a factor 3 times. 8 is 2 2 2, so 8 is a perfect cube. x6 is x2 x2 x2 so x6 is a perfect cube. It is easy to see if a variable is a perfect cube, how? See if the exponent is divisible by 3. It’s harder for integers.
The sum or difference of two cubes will factor into a
binomial trinomial. same sign always + always opposite same sign always + always opposite
Now we know how to get the signs, let’s work on
what goes inside. Square this term to get this term. Cube root of 1st term Cube root of 2nd term Product of cube root of 1st term and cube root of 2nd term.
Try one. Make a binomial and a trinomial with the correct signs.
Try one. Cube root of 1st term Cube root of 2nd term
Try one. Square this term to get this term.
Try one. Multiply 3x an 5 to get this term.
Try one. Square this term to get this term.
Try one. You did it! Don’t forget the first rule of factoring is to look for the greatest common factor. I hope you took notes!
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# Percent ICSE Class-6th Concise Selina Mathematics-(Percentage)
Percent ICSE Class-6th Concise Selina Mathematics Solutions Chapter-16 (Percentage) . We provide step by step Solutions of Exercise / lesson-16 Percent for ICSE Class-6 Concise Selina Mathematics. Our Solutions contain all type Questions of Exe-16 A, Exe-16 B, Exe-16 C and Revision Exercise to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6 .
## Percent ICSE Class-6th Concise Selina Mathematics Solutions Chapter-16 (Percentage)
–: Select Topics :–
Exe-16 A,
Exe-16 B,
Exe-16 C,
### Exercise – 16 A Percent ICSE Class-6th Concise Maths (Percentage) Selina Solutions
#### Question -1.
Express each of the following statements in the percentage form :
(i) 13 out of 20
(ii) 21 eggs out of 30 are good
(i) 13 out of 20
.=…132020
=…13 x 5= 65 %……..
(i) 21 eggs out of 30
.=…310030
=…7 x 10= 70 %……..
#### Question- 2.
Express the following fractions as percent :
#### Question -3.
Express as percent:
(i) 0.10
(ii) 0.02
(iii) 0.7
(iv) 0.15
(v) 0.032
#### Question- 4.
Convert into fractions in their lowest terms:
(i) 8%
(ii) 20%
(iii) 85%
(iv) 250%
(v) 12$\frac { 1 }{ 2 }$ %
#### Question 5.
Express as decimal fractions :
(i) 25%
(ii) 108%
(iii) 95%
(iv) 4.5%
(v) 29.2%
#### Question- 6.
Express each of the following natural numbers as percent :
(i) 7
(ii) 2
(iii) 19.5
(iv) 5.37
### ICSE Class-6th Concise Selina Mathematics Solutions Percent (Percentage) Exercise – 16 B
#### Question -1.
Express :
(i) Rs 5 as a percentage of Rs 25.
(ii) 80 paise as a percent of Rs 4.
(iii) 700 gm as a percentage of 2.8 kg.
(iv) 90 cm as a percent of 4.5 m.
(i) 525 x 100 = 20%
(ii) 80 paise as a percent of 400 paise (as/rupee = 100 paise)
#### Question- 2.
Express the first quantity as a percent of the second :
(i)) 40 P, ₹ 2
(ii) 500 gm, 6 kg
(iii) 42 seconds, 6 minutes
40 p, ₹ 2 = 40 p to 200 p
(1 Rupee = 100 paise)
(iii) 42 second , 6 minute
=42 second , 360 second
#### Question -3.
Find the value of each of the following:
(i) 20% of ₹ 150
(ii) 90% of 130
(iii) 15% of 2 minutes
(iv) 7.5 % of 500 kg.
#### Question- 4.
If a man spends 70% of his income, what percent does he save ?
Total income = Rs 100
70% expenses = 100 x 70100
= Rs 70
His saving 100- 70 =30 %
Question- 5.
A girl gets 65 marks out of 80. What percent marks did she get?
#### Question- 6.
A class contains 25 children, of which 6 are girls. What percentage of the class are the boys.
Total number of students = 25
#### Question -7.
A tin contains 20 litres of petrol. Due to leakage, 3 litres of petrol is lost. What percent is still present in the tin?
Total petrol in tin = 20 litres
last due to leakage = 3 litres
Balance petrol in tin = (20 – 3) = 17 litres
Percentage of petrol in tin = 1720 x 100 = 85%
#### Question -8.
An alloy of copper and zinc contains 45% copper and the rest is zinc. Find the weight of zinc in 20 kg of the alloy.
Total weight of alloy = 20 kg
Weight copper = 20 x 45% = 20 x 45100 = 9 kg
Weight of zinc = (total weight of alloy – weight of copper) = 20 – 9 = 11 kg
#### Question -9.
A boy got 60 out of 80 in Hindi, 75 out of 100 in English and 65 out of 70 in Arithmetic. In which subject his percentage of marks the best ? Also, find his overall percentage.
A boy get in Hindi =60 out of 80
We see that he gets best marks in arithmetic
Now total marks he gets = 60 + 75 + 65 = 200
Total marks = 80 + 100 + 70 = 250
Percent marks obtained = 200250 x 100 = 80%
#### Question -10.
In a camp, there were 500 soldiers. 60 more soldiers joined them. What percent of the earlier (original) number have joined the camp.
Number of soldiers = 500
More joined them = 60
Percentage to join the earlier = 60500 x 100 = 12%
#### Question -11.
In a plot of ground of area 6000 sq. m, only 4500 sq. m is allowed for construction. What percent is to be left without construction?
Total ground area = 6000 sq. m.
Allowed for construction = 4500 sq.m.
Area left without construction = 6,000 sq. m – 4500 sq. m = 1500 sq. m
Percentage of construction left = 15006000 x 100 = 25%
#### Question- 12. Percent ICSE Class-6th Concise
Mr. Sharma has a monthly salary of ₹ 8,000. If he spends ₹ 6,400 every month; find :
(i) his monthly expenditure as percent.
(ii) his monthly savings as percent.
Monthly salary of Mr. Sharma = ₹ 8000
He spends every month = ₹ 6400
His savings = ₹ 8000 – 6400 = ₹ 1600
(i) Percent expenditure = 64008000 x 100% = 80%
(ii) Percent savings = 16008000 x 100% = 20%
#### Question -13.
The monthly salary of Rohit is ₹ 24,000. If his salary increases by 12%, find his new monthly salary
Salary = ₹ 24000
New salary = ₹ 24000 + 12% of 24000
= ₹ 24000 + 12100 x 24000
= ₹ 24000 + 2880 = ₹ 26880
New salary = ₹ 26880
#### Question -14.
In a sale, the price of an article is reduced by 30%. If the original price of the article is ₹ 1,800, find :
(i) the reduction in the price of the article
(ii) reduced price of the article.
(i) Original price of article = ₹ 1800
Reduction = 30%
Reduction in price = 30% of 1800
= 30100 x 1800 = ₹ 540
(ii) Reduced price of the article = Original price – Reduction = ₹ 1800 – ₹ 540 = ₹ 1260
#### Question -15.
Evaluate :
(i) 30% of 200 + 20% of 450 – 25% of 600
(ii) 10% of ₹ 450 – 12% of ₹ 500 + 8% of ₹ 500.
### Selina Solutions of Percent (Percentage) Exercise – 16 C for ICSE Class-6th Concise Mathematics
#### Question -1.
The price of rice rises from Rs. 30 per kg to Rs. 36 per kg. Find the percentage rise in the price of rice.
First price of rice = Rs. 30 per kg
Rised price = Rs. 36 per kg
Rise per kg = 36 – 30 = Rs. 6
Percent rise = 630 x 100 = 20%
#### Question- 2.
The population of a small locality was 4000 in 1979 and 4500 in 1981, By what percent had the population increase ?
Year 1979 population = 4,000
Year 1981 population = 4,500
Increase in population = (4,500 – 4,000) = 500
percentage of increase in population = 5004000 x 100 = 12.5%
#### Question -3.
The price of a scooter was ₹ 8000 in 1975. It came down to ₹ 6000 in 1980. By what percent had the price of the scooter came down?
Original cost of scooter = ₹ 8,000
Reduced cost of scooter = ₹ 6000
Reduction in price of scooter = ₹ 8,000 – ₹ 6,000 = ₹ 2,000
Percentage of reduction = 20008000 x 100 = 25%
#### Question -4.
Find the resulting quantity when :
(i) ₹ 400 is decreased by 8%.
(ii) 25 km is increased by 5%.
(iii) a speed of 600 km/h is increased by 1212 %
(iv) there is 2.5% increase in a salary of ₹ 62, 500.
= ₹ 1562.50
Resulting quantity (salary) = ₹ 62500 + ₹ 1562.50 = ₹ 64062.50
#### Question- 5.
The population of a village decreased by 12%. If the original population was 25,000, find the population after decrease?
Original population = 25,000
Decrease in population = 12% Population after decrease
= 25,000 – 12% of 25,000
= 25,000 – 12100 x 25,000
= 25,000 – 3,000 = 22,000
#### Question -6.
Out of a salary of Rs. 13,500, I keep 1/3 as savings. of the remaining money, I spend 50% on food and 20% on house rent. How much do I spend on food and house rent?
Total salary = Rs. 13,500
spend on food and house rent= 4500 + 1800 = 6300
#### Question -7.
A tank can hold 50 litres of water. At present, it is only 30% full. How many litres of water shall I put into the tank so that it becomes 50% full?
Capacity of tank = 50 litres
30% of capacity = 30% of 50 litres
Water need into the tank so that it becomes 50% full = 25-15 = 10 litres
#### Question- 8.
In an election, there are a total of 80,000 voters and two candidates, A and B. 80% of the voters go to the polls out of which 60% vote for A. How many votes does B get.
Member of voters = 80,000
Total vote polled = 80% of 80,000 = 80100 x 80,000 = 64,000
Vote polled to A = 60% of 64,000 = 60100 x 64000 = 38,400
Vote polled to B = Total vote polled – vote polled to A = 64,000 – 38,400 = 25,600
#### Question -9.
70% of our body weight is made up of water. Find the weight of water in the body of a person whose body weight is 56 kg.
Water in human body = 70%
Weight of a man = 56 kg
Quantity of water in him = 70% of 56
= 70100 = x 56 = 39.2 kg
#### Question- 10. Percent ICSE Class-6th Concise
Only one-fifth of water is available in liquid form. This limited amount of water is replenished and used by man recurrently. Express this information as percent, showing :
(i) water available in liquid form.
(ii) water available in frozen form.
Let total quantity of water = 1
water is available in liquid form = 15
Water available in frozen form
#### Question- 11.
By weight, 90% of tomato and 78% of potato is water. Find :
(i) the weight of water in 25 kg of tomato.
(ii) the total quantity, by weight, of water in 90 kg of potato and 30 kg of tomato
(iii) the weight of potato which contains 39 kg of water.
Water in tomato = 90% and water in potato = 78%
(i) Weight of water in 25 kg of tomato
### Revision Exercise of Percent (Percentage) for ICSE Class-6th Concise Selina Mathematics
#### Question-1
Rohit’s age is 12 years and Geeta’s age is 15 years. Express :
(i) Rohit’s age as a percent of Geeta’s age.
(ii) Geeta’s age as a percent of Rohit’s age.
#### Question -2.
A class has 30 boys and 20 girls. Find:
(i) the percentage of girls in the class
(ii) the percentage of boys in the class
(iii) percentage of number of boys as compared with number of girls.
Total students in class = Number of boys + Number of girls = 30 + 20 = 50
#### Question -3.
Mrs. Sharma went to the market with ₹ 800 in her purse. When she returned to her home, ₹ 240 were still left in her purse. What percent of her money did she spend in the market?
Money in her purse = ₹ 800
Balance in her purse = ₹ 240
Money spent = ₹ 800 – ₹ 240 = ₹ 560
Percentage of money spent = 560800 x 100 = 70%
#### Question -4. Percent ICSE Class-6th Concise
In a mixture of two liquids A and B, 35% is liquid B. If the total quantity of the mixture is 20 kg, find the quantity of A, by weight.
Total quantity of A and B = 20 kg
Quantity of B = 35% of 20 = 35100 x 20 = 7kg
Quantity of A = Total quantity – Quantity of B = 20 kg – 7 kg = 13 kg
Hence, quantity of A = 13 kg
#### Question -5.
A girl got 375 marks out of 500 in the first term examination, 560 marks out of 800 in the second term examination and 840 marks out of 1200 in the third term examination. Find:
(i) her percentage score in the first term examination.
(ii) her percentage score in the second term examination.
(iii) her percentage score in the third term examination.
(iv) the total marks secured in all the three examinations.
(v) the total marks scored in all the three examinations.
(vi) her percentage score on the whole in all the three examinations.
#### Question- 6.
Out of his monthly income of ₹ 2,500; a man spends ₹ 1,750. What percent of his income does he save every month?
Monthly income = ₹ 2,500
Spending = ₹ 1,750
Saving = monthly income – spending = 2,500- 1,750 = ₹ 750
Percentage of income he saves = 7502500 x 100 = 30%
#### Question -7.
Mr. Singh’s monthly salary is ₹ 15,000. This month he was promoted with an increment of ₹ 3,000 in his salary. Express his increment as a percent of his original salary.
Monthly salary = ₹ 15,000
Increment on promotion = ₹ 3,000
Percentage of increment to monthly salary = 300015000 x 100 = 20%
#### Question -8.
(i) The price of an article increased from ₹ 16 to ₹ 20 ; find the percentage increase.
(ii) The price of an article decreased from Rs 20 to Rs 16 ; find the percentage decrease.
(i) Original price = ₹ 16
Increased price = ₹ 20
Amount of increase = 20 – 16 = ₹ 4
Percentage of increase = 416 x 100 = 25%
(ii) Original price = ₹ 20
Decrease price = ₹ 16
Amount of decrease = 20 – 16 = ₹ 4
Percentage of decrease = 420 x 100 = 20%
#### Question- 9.
(i)) The salary of a man is ₹ 7,200 per month, which is now increased by 8%. Find his new salary per month.
(ii) The salary of Mr. Sahni is ₹ 8,400 per month, which is now decreased by 8%. Find his new salary per month.
(i)) The salary of a man is = ₹ 7,200 per month
New Salary =Monthly salary – decrease
= 8400 – 420
= 7980
#### Question- 10.
Find the percentage change from the first quantity to the second :
(i) ₹ 80, ₹ 120
(ii) 75 kg, 60 kg
(iii) 50 cm, 45 cm
#### Question -11.
The original price of an article is ₹ 640. Find its new price when its price is :
(i) increased by 30%
(ii) decreased by 20%
original price of an article is = ₹ 640
#### Question- 12. Percent ICSE Class-6th Concise
Find the number that is :
(i) 50% more than 48
(ii) 30% less than 70
#### Question- 13.
Evaluate :
(i) 8% of 900 – 12% of 750 + 20% of 165.
(ii) 70% of 70 + 90% of 90 – 120% of 120.
(i) 8% of 900 – 12% of 750 + 20% of 165.
= 49 + 81 – 144
=130 – 144 = -14
#### Question- 14.
Approximately 97.3% water on the earth is not fit for drinking. Find :
(i) the percentage of water on the earth that is fit for drinking.
(ii) The total volume of water available in certain part of the earth where there is 21,600 m3 of drinking water.
Approximately water on earth which is not fit for drinking = 97.3%
(i) Water fit for drinking = 100 – 97.3 = 2.7%
(ii) At a certain place, the water which is fit for drinking = 21600 m3
Volume of total water on that place
= 800 x 1000
= 8,00,000 m³
#### Question -15.
Air is an important inexhaustible natural resource. It is essential for the survival of human beings, microbes, plants and animals. The following table shows the percentage of various gases in air.
(i) In 800 m3 of air, calculate the approximate quantities of nitrogen, oxygen and other gases.
(ii) If a certain quantity (by volume) of air contains 4,200 litres of oxygen, find the total quantity of air taken and the amount of nitrogen in it.
Nitrogen = 78 %
Oxygen = 21 %
Others = 1%
Quantity of air = 800 m ³
= 200 x 100 L = 20000 L
amount of Nitrogen = 20000 x 78 %
= 20000 x 78100
=15600 Litres
— End of Percent ICSE Class-6th Concise Solutions :–
Thanks
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# Physics Snap
## 1. A ball (ball A) is dropped from the top of a $80 \mathrm{~m}$ building. At the same time, another ball (ball B) is thrown upwards from the same top of the building. If the time delay between both balls hitting the ground below is 1.2 seconds, with what initial velocity was ball B thrown at?
#### STEP 1
1. Air resistance is assumed to be negligible. 2. Gravity is assumed to be constant and equal to $9.8 \text{m/s}^2$. 3. The initial positions for both balls are the same, at $80 \text{m}$ above the ground. 4. The motion of both balls is only affected by gravity (no horizontal motion or other forces).
#### STEP 2
To solve this problem, we need to: 1. Use the kinematic equation for the displacement of ball A (dropped) and ball B (thrown upwards) to find the time it takes for each to hit the ground. 2. The equation to use is $y = y_0 + v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2$, where $y_0$ is the initial height, $v_0$ is the initial velocity, $a$ is the acceleration due to gravity, and $t$ is the time. 3. Set the displacement $y$ to zero (ground level) and solve for $t$ for both balls. 4. Find the initial velocity of ball B such that the difference in time to hit the ground between ball A and ball B is 1.2 seconds.
#### STEP 3
Start by solving for the time it takes for ball A (dropped) to hit the ground. Since it is dropped, its initial velocity $v_0 = 0 \text{m/s}$. The equation simplifies to: $0 = 80 \text{m} + 0 \cdot t + \frac{1}{2} \cdot -9.8 \text{m/s}^2 \cdot t^2$
#### STEP 4
Simplify the equation for ball A: $0 = 80 - 4.9t^2$
#### STEP 5
Solve for $t^2$: $t^2 = \frac{80}{4.9}$
#### STEP 6
Calculate $t^2$: $t^2 \approx 16.33 \text{s}^2$
#### STEP 7
Take the square root to find $t$ for ball A: $t \approx \sqrt{16.33} \approx 4.04 \text{s}$
#### STEP 8
Now, solve for the time it takes for ball B (thrown upwards) to hit the ground. Let $v_0$ be the initial velocity of ball B. The equation is: $0 = 80 \text{m} + v_0 \cdot t - 4.9 \text{m/s}^2 \cdot t^2$
#### STEP 9
We know the time difference between ball A and ball B hitting the ground is 1.2 seconds. Therefore, the time for ball B to hit the ground is $t_B = t_A - 1.2 \text{s} = 4.04 \text{s} - 1.2 \text{s} = 2.84 \text{s}$.
#### STEP 10
Plug $t_B = 2.84 \text{s}$ into the equation for ball B: $0 = 80 + v_0 \cdot 2.84 - 4.9 \cdot (2.84)^2$
#### STEP 11
Simplify and solve for $v_0$: $0 = 80 + v_0 \cdot 2.84 - 4.9 \cdot 8.07$
#### STEP 12
Calculate $4.9 \cdot 8.07$: $4.9 \cdot 8.07 \approx 39.54$
#### STEP 13
Substitute back and solve for $v_0$: $0 = 80 + v_0 \cdot 2.84 - 39.54$
#### STEP 14
Simplify further: $v_0 \cdot 2.84 = 39.54 - 80$
#### STEP 15
Calculate $39.54 - 80$: $39.54 - 80 = -40.46$
##### SOLUTION
Solve for $v_0$: $v_0 = \frac{-40.46}{2.84} \approx -14.24 \text{m/s}$ Ball B was thrown upwards with an initial velocity of approximately $14.24 \text{m/s}$.
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# Example of an ANOVA Calculation
One factor analysis of variance, also known as ANOVA, gives us a way to make multiple comparisons of several population means. Rather than doing this in a pairwise manner, we can look simultaneously at all of the means under consideration. To perform an ANOVA test, we need to compare two kinds of variation, the variation between the sample means, as well as the variation within each of our samples.
We combine all of this variation into a single statistic, called the F statistic because it uses the F-distribution. We do this by dividing the variation between samples by the variation within each sample. The way to do this is typically handled by software, however, there is some value in seeing one such calculation worked out.
It will be easy to get lost in what follows. Here is the list of steps that we will follow in the example below:
1. Calculate the sample means for each of our samples as well as the mean for all of the sample data.
2. Calculate the sum of squares of error. Here within each sample, we square the deviation of each data value from the sample mean. The sum of all of the squared deviations is the sum of squares of error, abbreviated SSE.
3. Calculate the sum of squares of treatment. We square the deviation of each sample mean from the overall mean. The sum of all of these squared deviations is multiplied by one less than the number of samples we have. This number is the sum of squares of treatment, abbreviated SST.
4. Calculate the degrees of freedom. The overall number of degrees of freedom is one less than the total number of data points in our sample, or n - 1. The number of degrees of freedom of treatment is one less than the number of samples used, or m - 1. The number of degrees of freedom of error is the total number of data points, minus the number of samples, or n - m.
5. Calculate the mean square of error. This is denoted MSE = SSE/(n - m).
6. Calculate the mean square of treatment. This is denoted MST = SST/m - `1.
7. Calculate the F statistic. This is the ratio of the two mean squares that we calculated. So F = MST/MSE.
Software does all of this quite easily, but it is good to know what is happening behind the scenes. In what follows we work out an example of ANOVA following the steps as listed above.
## Data and Sample Means
Suppose we have four independent populations that satisfy the conditions for single factor ANOVA. We wish to test the null hypothesis H0: μ1 = μ2 = μ3 = μ4. For purposes of this example, we will use a sample of size three from each of the populations being studied. The data from our samples is:
• Sample from population #1: 12, 9, 12. This has a sample mean of 11.
• Sample from population #2: 7, 10, 13. This has a sample mean of 10.
• Sample from population #3: 5, 8, 11. This has a sample mean of 8.
• Sample from population #4: 5, 8, 8. This has a sample mean of 7.
The mean of all of the data is 9.
## Sum of Squares of Error
We now calculate the sum of the squared deviations from each sample mean. This is called the sum of squares of error.
• For the sample from population #1: (12 – 11)2 + (9– 11)2 +(12 – 11)2 = 6
• For the sample from population #2: (7 – 10)2 + (10– 10)2 +(13 – 10)2 = 18
• For the sample from population #3: (5 – 8)2 + (8 – 8)2 +(11 – 8)2 = 18
• For the sample from population #4: (5 – 7)2 + (8 – 7)2 +(8 – 7)2 = 6.
We then add all of these sum of squared deviations and obtain 6 + 18 + 18 + 6 = 48.
## Sum of Squares of Treatment
Now we calculate the sum of squares of treatment. Here we look at the squared deviations of each sample mean from the overall mean, and multiply this number by one less than the number of populations:
3[(11 – 9)2 + (10 – 9)2 +(8 – 9)2 + (7 – 9)2] = 3[4 + 1 + 1 + 4] = 30.
## Degrees of Freedom
Before proceeding to the next step, we need the degrees of freedom. There are 12 data values and four samples. Thus the number of degrees of freedom of treatment is 4 – 1 = 3. The number of degrees of freedom of error is 12 – 4 = 8.
## Mean Squares
We now divide our sum of squares by the appropriate number of degrees of freedom in order to obtain the mean squares.
• The mean square for treatment is 30 / 3 = 10.
• The mean square for error is 48 / 8 = 6.
## The F-statistic
The final step of this is to divide the mean square for treatment by the mean square for error. This is the F-statistic from the data. Thus for our example F = 10/6 = 5/3 = 1.667.
Tables of values or software can be used to determine how likely it is to obtain a value of the F-statistic as extreme as this value by chance alone.
Format
mla apa chicago
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# Chapter 11 Motion Section 1 Distance and Displacement
• Slides: 9
Chapter 11 Motion Section 1 Distance and Displacement Vectors and Scalars Video Lecture
Practice Questions • 1. How many m are in 28 Km? • 2. Rearrange the following equation to solve for d ; v = d/t • 3. What are the SI units for distance and time?
Frame of Reference • How fast are you moving right now? – Earth is moving at 1000 mph – If you are sitting in a desk you are not moving at all compared to the person next to you. – In a car you may be moving at 55 km/hr compared to the world around you, but compared to the person sitting across from you, you’re not moving.
Frame of Reference • Frame of Reference- is a system of objects that are not moving with respect to each other.
Relative Motion • How fast you are moving also depends on your relative motion. – Relative motion- is movement in relation to a frame of reference. • Compared to the people in the seat of the car next to you, you are not moving. • Compared to the people on the side of the road you are traveling at 55 km/hr
Distance • Distance- length of a path between two points. – Remember to use appropriate units when measuring distance.
Displacement • Displacement is the length of a straight line from the starting point to the ending point and the direction from starting point to ending point. – Example- distance- I walked 5 blocks – Example- displacement- I walked 5 blocks north
Displacement as a Vector on a straight line • Vector- is a quantity that has magnitude and direction. – You can find displacement using vector addition. • Ex. 5 blocks north then 4 more blocks north = 9 blocks north • Ex. 5 blocks north then 2 blocks back south = 3 blocks north 5 blocks + + 4 blocks 2 blocks
Displacement not along a straight line • If you graph the total distance and draw a straight line from the starting point to the finishing point you can find the total displacement
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# Displacement Formula
Displacement is shortest distance between initial and final point which prefers straight line path over curved paths.
If a body is moving in two different directions x and y then Resultant displacement is
It gives the short cut paths for the given original paths. Generally it is also given by :
Here,
u is the Initial velocity
v is the final velocity
a is acceleration
t is the time taken.
While solving the problems if initial and final velocity both are given we use the first formula if final velocity and time taken are given you can use second formula and If you are interested in finding area under the curve then use the third formula.
Related Calculators Displacement Calculator Acceleration Formula Calculator Area of a Circle Formula Calculator Area of a Cylinder Formula Calculator
## Displacement Problems
Solved problems on displacement which helps you to understand the concept better are given below
### Solved Examples
Question 1: The path from garden to a school is 5m west and then 4m south. A constructor wants to build a short distance path for it. Help him out?
Solution:
Given: Distance to the west x = 5m
Distance to the south y = 4m.
Displacement is given by S = $\sqrt{x^{2} + y^{2}}$
= $\sqrt{5^{2} + 4^{2}}$
= 6.70 m.
The constructor can build a path for displacement length of 6.7 m.
Question 2: A girl walks from the corridor to the gate she moves 3m to the north opposite from her house then takes a left turn and walks for 5m. then she takes right turn and moves for 6m and reaches the gate. What is the displacement, magnitude and distance covered by her?
Solution:
Total distance traveled d = 3m + 5m + 6m = 14m.
Magnitude of the displacement can be obtained by visualizing the walking :The actual path from A to B as 3m then from B to D as 5m and finally from D to E as 6m.
So the magnitude of resultant displacement is |S| = $\sqrt{AC^{2} + CE^{2}}$
From figure AC = AB + BC = 3m + 6m = 9m
BD = CE = 5m
|S| = $\sqrt{9^{2} + 5^{2}}$ = 10.29 m.
The direction of Resultant displacement is South East.
More topics in Displacement Formula Angular Displacement Formula
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# 2000 AIME II Problems/Problem 14
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$, meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$, where each $f_i$ is an integer, $0\le f_i\le i$, and $0. Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$, find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$.
## Solution
### Solution 1
Note that $1+\sum_{k=1}^{n-1} {k\cdot k!} = 1+\sum_{k=1}^{n-1} {((k+1)\cdot k!- k!)} = 1+\sum_{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \cdots + (n! - (n-1)!)) = n!$.
Thus for all $m\in\mathbb{N}$,
$(32m+16)!-(32m)! = \left(1+\sum_{k=1}^{32m+15} {k\cdot k!}\right)-\left(1+\sum_{k=1}^{32m-1} {k\cdot k!}\right) = \sum_{k=32m}^{32m+15}k\cdot k!.$
So now,
\begin{align*} 16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\\ &=16! +\sum_{m=1}^{62}(32m+16)!-(32m)!\\ &=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k! \end{align*}
Therefore we have $f_{16} = 1$, $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$, and $f_k = 0$ for all other $k$.
Therefore we have:
\begin{align*} f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j &= (-1)^{17}\cdot 1 + \sum_{m=1}^{62}\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\ &= -1 + \sum_{m=1}^{62}\left[\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\right]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}1\\ &= -1 + \sum_{m=1}^{62}8\\ &= -1 + 8\cdot 62\\ &= \boxed{495} \end{align*}
### Solution 2 (informal)
This is equivalent to Solution 1. I put up this solution merely for learners to see the intuition.
Let us consider a base $n$ number system. It’s a well known fact that when we take the difference of two integral powers of $n$, (such as $10000_{10} - 100_{10}$) the result will be an integer in base $n$ composed only of the digits $n - 1$ and $0$ (in this example, $9900$). More specifically, the difference $(n^k)_n - (n^j)_n$, $j , is an integer $k$ digits long (note that $(n^k)_n$ has $k + 1$ digits). This integer is made up of $(k-j)$ $(n - 1)$’s followed by $j$ $0$’s.
It should make sense that this fact carries over to the factorial base, albeit with a modification. Whereas in the general base $n$, the largest digit value is $n - 1$, in the factorial base, the largest digit value is the argument of the factorial in that place. (for example, $321_!$ is a valid factorial base number, as is $3210_!$. However, $31_!$ is not, as $3$ is greater than the argument of the second place factorial, $2$. $31_!$ should be represented as $101_!$, and is $7_{10}$.) Therefore, for example, $1000000_! - 10000_!$ is not $990000_!$, but rather is $650000_!$. Thus, we may add or subtract factorials quite easily by converting each factorial to its factorial base expression, with a $1$ in the argument of the factorial’s place and $0$’s everywhere else, and then using a standard carry/borrow system accounting for the place value.
With general intuition about the factorial base system out of the way, we may tackle the problem. We use the associative property of addition to regroup the terms as follows: $(2000! - 1984!) + (1968! - 1952!) + \cdots + (48! - 32!) + 16!$ we now apply our intuition from paragraph 2. $2000!_{10}$ is equivalent to $1$ followed by $1999$ $0$’s in the factorial base, and $1984!$ is $1$ followed by $1983$ $0$’s, and so on. Therefore, $2000! - 1984! = (1999)(1998)(1997)\cdots(1984)$ followed by $1983$ $0$’s in the factorial base. $1968! - 1952! = (1967)(1966)\cdots(1952)$ followed by $1951$ $0$’s, and so on for the rest of the terms, except $16!$, which will merely have a $1$ in the $16!$ place followed by $0$’s. To add these numbers, no carrying will be necessary, because there is only one non-zero value for each place value in the sum. Therefore, the factorial base place value $f_k$ is $k$ for all $32m \leq k \leq 32m+15$ if $1\leq m \in\mathbb{Z} \leq 62$, $f_{16} = 1$, and $f_k = 0$ for all other $k$.
Therefore, to answer, we notice that $1999 - 1998 = 1997 - 1996 = 1$, and this will continue. Therefore, $f_{1999} - f_{1998} + \cdots - f_{1984} = 8$. We have 62 sets that sum like this, and each contains $8$ pairs of elements that sum to $1$, so our answer is almost $8 \cdot 62$. However, we must subtract the $1$ in the $f_{16}$ place, and our answer is $8 \cdot 62 - 1 = \boxed{495}$.
### Solution 3 (less formality)
Let $S = 16!-32!+\cdots-1984!+2000!$. Note that since $|S - 2000!| << 2000!$ (or $|S - 2000!| = 1984! + \cdots$ is significantly smaller than $2000!$), it follows that $1999! < S < 2000!$. Hence $f_{2000} = 0$. Then $2000! = 2000 \cdot 1999! = 1999 \cdot 1999! + 1999!$, and as $S - 2000! << 1999!$, it follows that $1999 \cdot 1999! < S < 2000 \cdot 1999!$. Hence $f_{1999} = 1999$, and we now need to find the factorial base expansion of
$$S_2 = S - 1999 \cdot 1999! = 1999! - 1984! + 1962! - 1946! + \cdots + 16!$$
Since $|S_2 - 1999!| << 1999!$, we can repeat the above argument recursively to yield $f_{1998} = 1998$, and so forth down to $f_{1985} = 1985$. Now $S_{16} = 1985! - 1984! + 1962! + \cdots = 1984 \cdot 1984! + 1962! + \cdots$, so $f_{1984} = 1984$.
The remaining sum is now just $1962! - 1946! + \cdots + 16!$. We can repeatedly apply the argument from the previous two paragraphs to find that $f_{16} = 1$, and $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$, and $f_k = 0$ for all other $k$.
Now for each $m$, we have $-f_{32m} + f_{32m+1} - f_{32m+2} + \cdots + f_{32m + 31}$ $= -32m + (32m + 1) - (32m + 2) + \cdots - (32m - 30) + (32 m + 31)$ $= 1 + 1 + \cdots + 1 + 1$ $= 8$. Thus, our answer is $-f_{16} + 8 \cdot 62 = \boxed{495}$.
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# Factors and Multiples in the Real World
18 teachers like this lesson
Print Lesson
## Objective
SWBAT solve real-world problems in context involving factors and multiples.
#### Big Idea
We can apply our understanding of factors and multiples to solve problems.
## Think About It
7 minutes
Students work in partners on the Think About It problem (part a, above the line). After 2-3 minutes of work time, I ask students to identify what the problem is asking us to find.
I then have students share out the different ways that Jane can make snack packs. As an engagement strategy, I have students call on a peer after they've given their response. This continues until all possibilities are shared.
I frame the lesson by telling students that they're going to have to use what they know about factors and multiples, and decide which of these concepts to apply while solving real-world problems.
## Guided Practice
10 minutes
This lesson doesn't contain new material. Rather, students are applying what they know about GCF and LCM, and will need to decide which concept to use in each situation.
We continue the problem about Jane and her snack packs in the Guided Practice section. Because of their work on the Think About It problem, students are quickly able to share that Jane can make 4 snack packs. I extend student thinking by asking what would be in each snack pack.
I ask everyone to write a response to part c, to summarize how finding factors helped them to answer the questions on this page. Student responses may look like: "We knew that the factors of number divide into a number with no remainder. We also knew that the common factors of two numbers divide evenly into both numbers, which told us how many snack packs we could make."
I use cold call to question students about the second problem in this set.
## Partner Practice
20 minutes
Students work in pairs on the Partner Practice problem set. For each question in this set, I've asked students to decide if they need to use factors and multiples, and I expect them to write a justification for their choices before solving the problems. I want students thinking: To solve problems involving events happening in cycles, it may be helpful to find multiples and common multiples. To solve problems involving dividing groups evenly, it may be helpful to find factors and common factors.
As students are working , I circulate around the room and check in with each pair. I am looking for:
• Are students explaining their thinking to their partner?
• Are students writing their work in the work space?
• Are students using factors and multiples to help them solve?
• Are students annotating the problem?
• Are students answering in full sentences?
I'm asking:
• How did you know to use multiples/factors?
• Explain how you determined this answer.
• How did you use common factors (or common multiples) to help you solve?
• Why did you choose this method (ex - if student used prime factorization or t charts)?
After 15 minutes of partner practice, students work independently on the Check for Understanding problem. I have students vote on whether this problem requires factors or multiples. Seeing this data from kids allows me to make a list of students to support at the start of independent practice. I have one student present his/her work to the class under the document camera.
## Independent Practice
15 minutes
Students work on the Independent Practice problem set.
In this problem set, I decided to not to have students explicitly write whether or not they're using factors or multiples to solve each problem. I want students to go through this thought process on their own, without the scaffold of me asking.
As I circulate, I keep an eye out for student answers to problem 2. Students should be able to quickly determine the LCM of 6 and 15. However, to correctly respond to the question, students need to recontextualize the 30, and write that the two events will both happen again at 12:30. If I see students writing simply 30, I'll ask them if the number 30 makes sense, given what the problem is asking them.
## Closing and Exit Ticket
8 minutes
After independent practice time, I have students share with their partners the strategies they used to solve Problem 5. Students get really excited about chicken nugget problems. As a class, we discuss problem 1. I have students show me on their fingers how many possibilities there are for Part A.
Students then work independently on the Exit Ticket to close the lesson. Prior to teaching, I create my Criteria for Success, so I know exactly what I am looking for in student work.
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# Advanced Signals and Systems - Linear and Cyclic Convolution
### 17. Linear convolution of sequences.
Find the convolution sum $$v(n) = v_1(n) \ast v_2(n)$$ of the following sequences $$v_1(n)$$ and $$v_2(n)$$
\begin{align} v_1(n) =& \rho _1^n \cdot \gamma_{-1}(n) \nonumber \\ v_2(n) =& \rho _2^n \cdot \gamma_{-1}(n)\nonumber \end{align}
where $$0 < \rho_1$$ and $$\rho_2 < 1$$.
## Amount and difficulty
• Working time: approx. xx minutes
• Difficulty: xx
## Solution
\begin{align*} v(n) &= \sum \limits_{k=-\infty}^{\infty} v_1(k) \cdot v_2(n-k) \\ &= \sum \limits_{k=-\infty}^{\infty} \rho_1^k \cdot \gamma_{-1}(k) \cdot \rho_2^{n-k} \cdot \gamma_{-1}(n-k) \\ &= \cdots\\ &= \begin{cases} \frac{\rho_1^{n+1}-\rho_2^{n+1}}{\rho_1-\rho_2}\cdot \gamma_{-1}(n) &, \rho_1\neq \rho_2\\ \rho_2^n\cdot(n+1)\cdot \gamma_{-1}(n) &, \rho_1 = \rho_2 \end{cases}\end{align*}
### 18. Linear and cyclic convolution.
Given two sequences $$v_1(n)$$ and $$v_2(n)$$ of length $$M=5$$:
\begin{align} v_1(n) =& [5,4,3,2,1] \nonumber \\ v_2(n) =& [1,2,3,4,5]\nonumber \end{align}
Determine the linear convolution $$v_3(n)$$ and the cyclic convolution $$v_4(n)$$ of the sequences. Give a method to calculate the linear convolution.
## Amount and difficulty
• Working time: approx. xx minutes
• Difficulty: xx
Linear convolution
Cyclic convolution
Calculating the linear convolution by cyclic convolution:
## Solution
• Linear convolution
\begin{align*} v_3(n) &= v_1(n) * v_2(n) = \sum \limits_{k=-\infty}^{\infty} v_1(k) \cdot v_2(n-k)\\ &\cdots\\ v_3(n) &= [5, 14, 26, 40, 55, 40, 26, 14, 5] \ \ \ \text{ for } \ \ 0\leq n \leq M-1 \end{align*}
• Cyclic convolution
\begin{align*} v_4(n) &= \sum \limits_{k=-\infty}^{\infty} v_1(k) \cdot v_2(\text{mod}(n-k,M))\\ &\cdots\\ v_4(n) &= [45, 40, 40, 45, 55] \end{align*}
• Modified cyclic convolution to get correct result
The cyclic convolution of $$v'_1(n)$$ and $$v'_2(n)$$ is calculated, where $$v'_1(n)$$ and $$v'_2(n)$$ denote the sequences $$v_1(n)$$ and $$v_2(n)$$ padded with $$M-M_2$$ and $$M-M_1$$ zeros, where $$M$$ is the length of the resulting sequence $$v_5(n)$$ and $$M=M_1+M_2+1$$.
\begin{align*} v_5(n) &= \sum \limits_{k=-\infty}^{\infty} v'_1(k) \cdot v'_2(\text{mod}(n-k,M))\\ &\cdots\\ v_5(n) &= [5, 14, 26, 40, 55, 40, 26, 14, 5] \ \ \ \text{ for } \ \ 0\leq n \leq M-1 \end{align*}
### Website News
01.10.2017: Started with a Tips and Tricks section for KiRAT.
01.10.2017: Talks from Jonas Sauter (Nuance) and Vasudev Kandade Rajan (Harman/Samsung) added.
13.08.2017: New Gas e.V. sections (e.g. pictures or prices) added.
05.08.2017: The first "slide carousel" added.
### Recent Publications
J. Reermann, P. Durdaut, S. Salzer, T. Demming,A. Piorra, E. Quandt, N. Frey, M. Höft, and G. Schmidt: Evaluation of Magnetoelectric Sensor Systems for Cardiological Applications, Measurement (Elsevier), ISSN 0263-2241, https://doi.org/10.1016/j.measurement.2017.09.047, 2017
S. Graf, T. Herbig, M. Buck, G. Schmidt: Low-Complexity Pitch Estimation Based on Phase Differences Between Low-Resolution Spectra, Proc. Interspeech, pp. 2316 -2320, 2017
### Contact
Prof. Dr.-Ing. Gerhard Schmidt
E-Mail: gus@tf.uni-kiel.de
Christian-Albrechts-Universität zu Kiel
Faculty of Engineering
Institute for Electrical Engineering and Information Engineering
Digital Signal Processing and System Theory
Kaiserstr. 2
24143 Kiel, Germany
## Recent News
Jens Reermann Defended his Dissertation with Distinction
On Friday, 21st of June, Jens Reermann defended his research on signals processing for magnetoelectric sensor systems very successfully. After 90 minutes of talk and question time he finished his PhD with distinction. Congratulations, Jens, from the entire DSS team.
Jens worked for about three and a half years - as part of the collaborative research center (SFB) 1261 - on all kinds of signal ...
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Alignments to Content Standards: A-REI.B.4.b
The first three steps of three visual patterns are shown below.
Here are functions that define how many tiles are in step $n$, in no particular order: $$f(n)=3n^2$$ $$g(n)=n^2+4$$ $$h(n)=n^2-1$$
1. Decide which function defines which pattern.
2. One of these patterns has a step with $432$ tiles in it, one has a step with $195$ tiles in it, and one of these patterns has a step with $404$ tiles in it. Decide which is which, find the step that contains that number of tiles, and explain how you know.
3. Describe and justify the steps to solve this equation for $x$. $$ax^2+c=d$$
IM Commentary
This task belongs to a series of three tasks that presents students with a sequence of tile figures with the property that the $n$-th figure in the sequence has $f(n)$ tiles, for some quadratic function $f$. Students are asked to analyze the functions in the context of the tile sequence, a process which involves manipulating the quadratic expressions into different forms (identifying square roots at first, then completing the square in the third task.) With solving quadratics, there can be an impulse to put everything in standard form and just use the quadratic formula. However, seeing structure is emphasized in the standards because of how it connects and helps in understanding many foundational concepts, and these tasks develop the ability to see structure when working with quadratic expressions and equations.Â
That the domain of these quadratic functions is the set of positive integers provides an interesting wrinkle which students might not be used to thinking about in the setting of quadratic functions. The functions roughly increase in complexity through the three tasks, with the intent that the techniques learned in each will be used and expanded in the subsequent tasks. The other tasks in the sequence are Quadratic Sequence 2 and Quadratic Sequence 3.
Part (a) has many possible points of entry. For example, students might evaluate each function at $n=1$, or they might analyze the way the patterns grow visually. For example, each step of Pattern A consists of $n^2$ tiles with four tiles tacked on, and each step of Pattern B consists of $n^2$ tiles with one tile removed. (Here is a good primer on what that conversation might look like.)
Ideally, students will have had some experience working with visual patterns prior to this task. If not, students might benefit from additional questions to familiarize themselves with the idea. For example, they could be asked to "draw the next step" for one or more patterns. Part (a) could be scaffolded with more specific instructions; for example, students could be asked to evaluate $g(3)$, find that $g(3) = 13$, and conclude that this means the pattern defined by g(n) must have 13 tiles in step 3.
The purpose of part (b) is to motivate solving a quadratic equation where the variable of interest, $x$, can be isolated by undoing operations.
In part (c), students have an opportunity to express regularity in repeated reasoning (MP.8) by applying these same undoing operations to an equation to isolate a variable of interest in terms of other variables.
This task was inspired by visualpatterns.org.
Solution
1. $f(n)$ defines Pattern C, $g(n)$ defines Pattern A, and $h(n)$ defines Pattern B. There are many ways to determine this. You could evaluate each function for $n=1$. For example, $f(1)=3$. This tells you that the first step of the pattern defined by $f$ has 3 tiles, so $f$ must go with Pattern C. Another method would be to see how each pattern grows visually. For example, step $n$ of Pattern A has a square of $n \times n$ tiles with 4 tiles tacked on, so $g(n)=n^2+4$ makes sense.
2. Pattern A has 404 tiles in step 20, Pattern B has 195 tiles in step 14, and Pattern C has 432 tiles in step 12. We need to see which function has an integer solution for each number of given tiles. For example, to check whether $f(n)$ has a step with $432$ tiles, we can write $$3n^2=432$$ $$n^2=144$$ which has a solution of $n=12$. So, step $12$ of Pattern C has $432$ tiles. However, when we check whether Pattern A can have $195$ tiles and write $$n^2+4=195$$ and then $$n^2=191,$$but this does not have integer solutions. So Pattern A does not have any step with $195$ tiles, so by elimination, Pattern B must be the one with  $195$ tiles. Indeed we find that the equation $$n^2-1=195$$ $$n^2=196$$ has a solution of $n=14$. So Pattern B has $195$ tiles in step $14$. Finally, a solution to $n^2+4=404$ is $n=20$, so step $20$ of Pattern A must have $404$ tiles.
3. In figuring out part (b), we did some simple reasoning about equations to undo operations and find solutions. Now we apply those techniques generally. $$ax^2+c=d$$ $$ax^2=d-c$$ $$x^2=\frac{d-c}{a}$$ $$x=\pm\sqrt{\frac{d-c}{a}}$$ For tiles and steps, solutions can only take positive values, but generally, solutions could be the positive or negative value of the square root.
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The slope of a line is a measure of its steepness. We calculate the slope of a line by comparing the change in y values (vertical change) to the change in x values (horizontal change) while moving along our line from one point to another.
Test Objectives
• Demonstrate an understanding of the slope formula
• Demonstrate the ability to calculate slope using the slope formula
• Demonstrate the ability to calculate slope from the graph of the line
Slope Formula Practice Test:
#1:
Instructions: Find the slope of the line that passes through the given points.
$$a) \hspace{.1em}(-5,14), (5,20)$$
$$b) \hspace{.1em}(7,4), (-3,4)$$
#2:
Instructions: Find the slope of the line that passes through the given points.
$$a) \hspace{.1em}(-6,-10), (0,-18)$$
$$b) \hspace{.1em}(6,11), (9,8)$$
#3:
Instructions: Find the value of y.
$$a) \hspace{.1em}(-3,-2), (-2,y), m=7$$
#4:
Instructions: Find the value of x.
$$a) \hspace{.1em}(x,1), (-1,-3), m=\frac{4}{5}$$
#5:
Instructions: Find the slope of the line from the graph.
a)
b)
Written Solutions:
#1:
Solutions:
$$a) \hspace{.1em}m=\frac{3}{5}$$
$$b) \hspace{.1em}m=0$$
#2:
Solutions:
$$a) \hspace{.1em}m=-\frac{4}{3}$$
$$b) \hspace{.1em}m=-1$$
#3:
Solutions:
$$a) \hspace{.1em}y=5$$
#4:
Solutions:
$$a) \hspace{.1em}x=4$$
#5:
Solutions:
$$a) \hspace{.1em}m=-4$$
$$b) \hspace{.1em}m=\frac{5}{2}$$
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Ch. 2-3 Review of Probability and Statistics
# Ch. 2-3 Review of Probability and Statistics - 2 REVIEW of...
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1 | Reviews 2. REVIEW of PROBABILITY AND STATISTICS (CHAP. 2 - 3) [1] Important Concepts and Formulas (1) Population : The group of interest. EX: Household incomes of Phoenix residents, US GNP. (2) Probability (or frequency): A small island with 12 households Income (per day) # of households Probability \$100 2 2/12 = 1/6 \$200 2 2/12 = 1/6 \$300 4 4/12 = 1/3 \$400 4 4/12 = 1/3 ---------------------------------------------- 1 2 1
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2 | Reviews • Let X = a household's income ( X : random variable ). Describe the probability that X takes a specific value by: f ( x ) = 1/6, if x = 100 or 200; f ( x ) = 1/3 if x = 300 or 400. ( probability density function .) (3) Expected value (Population mean) of X : In the above example, the population mean is (100×2+200×2+300×4+400×4)/12 = 283.3 Expected value of X : () Xx E xx f x E x = 100×(1/6) + 200×(1/6) + 300×(1/3) + 400×(1/3) = 283.3. Lesson: If you know the value of f ( x ) for each possible value of x , can compute the population mean.
3 | Reviews (4) Population Variance of X Wish to know the dispersion of a population of size B : Let x 1 , . .. , x B be all members of population. Use 2 1 1 () ( ) B ii x var x x B . • Alternatively, 22 2 2 var( ) ( ) ( ) ( ) x xx x x f x x f x . In the above example, 12 2 1 var( ) ( ) /12 x = {(100-283.3) 2 + (100-283.3) 2 + (200-283.3) 2 + (200-283.3) 2 + (300-283.3) 2 + (300-283.3) 2 + (300-283.3) 2 + (300-283.3) 2 + (400-283.3) 2 + (400-283.3) 2 + (400-283.3) 2 + (400-283.3) 2 }/12 = 11388.889. var( x ) = Σ x ( x - μ x ) 2 f ( x ) = (100-283.3) 2 ×(1/6) + (200-283.3) 2 ×(1/6) + (300-283.3) 2 ×(1/3) + (400-283.3) 2 ×(1/3) = 11388.889.
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4 | Reviews Finance Question. X = Monthly return rate of Intel stock. Why do you want to know () Ex and var( ) x ? How would you make an investment decision? Suppose: ( ) 0.01(1%) & ( ) var( ) 0.3 (30%) s ex x . monthly interest rate from a saving account = 0.005 (0.5%). What if monthly interest rate = 0%? What if ( ) 1(100%) ? What if ( ) 0 sex ?
5 | Reviews (5) Case of Two Random Variables EX: Income ( X ) and consumption ( Y ) of the 12 households. Y (\$)\ X (\$) 100 200 300 400 30 1 1 2 1 : 5 40 1 0 1 1 : 3 50 0 1 1 2 : 4 ---------------------------------------------------------------------- 2 2 4 4 : 12
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6 | Reviews 1. Joint Probability Density Function Y\X 100 200 300 400 30 1/12 1/12 2/12 1/12 : 5/12 40 1/12 0 1/12 1/12 : 3/12 50 0 1/12 1/12 2/12 : 4/12 ------------------------------------------------------- 2/12 2/12 4/12 4/12 : 1 Joint pdf = f ( x , y ): f (100,50) = 0 1/12 = 8.3%.
7 | Reviews 2. Marginal PDFs of X and Y : Marginal pdf of X = f x ( x ) = Σ y f ( x , y ) = Pr( X = x ) regardless of Y . [Unconditional = marginal] Marginal pdf of Y = f y ( y ) = Σ x f ( x,y ) = Pr( Y = y ) regardless of X . Y\X 100 200 300 400 f y ( y ) 30 1/12 1/12 2/12 1/12 : 5/12 40 1/12 0 1/12 1/12 : 3/12 50 0 1/12 1/12 2/12 : 4/12 ------------------------------------------------------- f x ( x ) 2/12 2/12 4/12 4/12 : 1
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8 | Reviews 3. Conditional pdf : ( | ) fy x = Pr( Y = y , given X = x ) = (, ) () x f x y f x ; ( | ) f x y = Pr( X = x , given Y = y ) = y f x y . (3 0 | 100) x = (100,30) 1/12 1 (100) 2 /12 2 x f f ; (4 0 | 300) x = (300,40) 1 (300) 4 /12 4 x f f . 4. Population means and variances of X and Y xx E f x ; yy E f y ; 2 var( ) ( ) ( ) x f x ; 2 var( ) ( ) ( ) y yyf y
9 | Reviews 5. Conditional means and conditional variances (|) xx E xy x f xy ; y E yx y fyx ; 2 v a r [ ] x x Exy fxy ; 2 v a r [ ] (|) y y Eyx fyx . EX: ( | 200) ( | 200) (200,30) (200,40) (200,50) 30 40 50 (200) (200) (200) 1/12 0 30 40 50 40 2 /12 y xxx Ey x y f y x fff 2 222 var( | 200) ( ( | 200)) ( | 200) 0 (30 40) (40 40) (50 40) 100 2/12 y y Eyx
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# How do you find the distance between points (3,-3), (7,2)?
May 24, 2017
$d = \sqrt{41}$
#### Explanation:
Use the distance formula: $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
If we let $\left(3 , - 3\right) \to \left(\textcolor{b l u e}{{x}_{1}} , \textcolor{red}{{y}_{1}}\right)$ and $\left(7 , 2\right) \to \left(\textcolor{b l u e}{{x}_{2}} , \textcolor{red}{{y}_{2}}\right)$ then...
$d = \sqrt{{\left(\textcolor{b l u e}{7 - 3}\right)}^{2} + {\left(\textcolor{red}{2 - - 3}\right)}^{2}}$
$d = \sqrt{{\left(\textcolor{b l u e}{4}\right)}^{2} + {\left(\textcolor{red}{5}\right)}^{2}}$
d=sqrt((color(blue)(16))+(color(red)(25))
$d = \sqrt{41}$
May 24, 2017
The distance between points (3,-3), (7,2) is $d = \sqrt{41}$
That is $\approx 6.4 u n i t s$
#### Explanation:
Use the distance formula which is derived from the Pythagorean Theorem.
They did it here:
http://www.purplemath.com/modules/distform.htm
${d}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}$
We have points (3,-3), (7,2):
Then: ${d}^{2} = {\left(3 - 7\right)}^{2} + {\left(- 3 - 2\right)}^{2}$
${d}^{2} = {\left(- 4\right)}^{2} + {\left(- 5\right)}^{2}$
See how squaring gets rid of those nasty negatives?
${d}^{2} = \left(16\right) + \left(25\right)$
Too bad we cannot take the roots of the two terms without adding.
${d}^{2} = 41$
$d = \sqrt{41} \approx 6.4$
To check, compare this $r i g h t \angle \triangle 4 , 5 , 6.4$ to a standard $r i g h t \angle \triangle 3 , 4 , 5$, and they appear to be similar.
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# Inverse of a Function
Every function has an inverse relation. If this inverse relation is also a function, then it is called an inverse function. In other words, the inverse of a function is another function such that they undo each other.
Also, if is the input of a function and its corresponding output, then is the input of and its corresponding output.
## Example
Consider a function and its inverse
These functions will be shown to undo each other. To do so, it needs to be proven that and that To start, the first equality will be proven. First, the definition of will be used.
Now, in the above equation, will be substituted for
Simplify left-hand side
A similar procedure can be performed to show that
Definition of First Function Substitute Second Function Simplify
Therefore, and undo each other. The graphs of these functions are each other's reflection across the line This means that the points on the graph of are the reversed points on the graph of
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# 8.1: What's the Value
Difficulty Level: At Grade Created by: CK-12
Look at the equation below. Can you figure out the value of \begin{align*}z\end{align*}? In this section, we will learn how to use the order of operations to help us to solve equations.
\begin{align*}6 \times 3^2 \div 2 + z + 4(7 - 3) + 2^3 \div 4 = 3^2 \times 6 + 1\end{align*}
### What's the Value
The order of operations tells us the correct order of evaluating math expressions. We always do parentheses first and then exponents. Next we do multiplication and division (from left to right) and finally addition and subtraction (from left to right).
In order to evaluate expressions using the order of operations, we can use the problem solving steps to help.
• First, describe what you see in the problem. What operations are there?
• Second, identify what your job is. In these problems, your job will be to solve for the unknown.
• Third, make a plan. In these problems, your plan should be to use the order of operations.
• Fourth, solve.
• Fifth, check. Substitute your answer into the equation and make sure it works.
#### Finding Unknown Values
1. Follow the order of operations and show each step. What is the value of the variable?
\begin{align*} b + 2 \times 3 \times 2^2 \div 3 = 2(5 + 6) - 2\end{align*}
We can use the problem solving steps to help us with the order of operations.
\begin{align*}& \mathbf{Describe:} && \text{The equation has parentheses, exponents, multiplication, division, subtraction and addition.}\\ &&& b \ \text{is the variable.}\\ \\ & \mathbf{My \ Job:} && \text{Apply the order of operations rule to figure out the value of}\ b.\\ \\ & \mathbf{Solve:} && b + 2 \times 3 \times 2^2 \div 3 = 2(5 + 6) - 2\\ &&& \mathbf{Parentheses} \qquad \qquad \quad b + 2 \times 3 \times 2^2 \div 3 = 2 \times 11 - 2\\ &&& \mathbf{Exponents} \qquad \qquad \quad \ b + 2 \times 3 \times 4 \div 3 = 2 \times 11 - 2\\ &&& \mathbf{Multiplication/} \qquad \quad b + 8 = 22 - 2\\ &&& \mathbf{Division}\\ &&& \mathbf{(left \ to \ right)}\\ &&& \mathbf{Addition/} \qquad \qquad \qquad \ b + 8 = 20\\ &&& \mathbf{Subtraction}\\ &&& \mathbf{(left \ to \ right)} \qquad \qquad \ b=20-8\\ &&& \qquad \qquad \qquad \qquad \qquad \quad b=12\\ \\ & \mathbf{Check:} && \text{Replace} \ b \ \text{with 12 in the equation. Check that the two expressions}\\ &&&\text{(to the right and to the left of the = symbol) name the same number.}\\ &&& 12 + 2 \times 3 \times 2^2 \div 3 = 2(5 + 6) - 2\\ &&& 12 + 2 \times 3 \times 2^2 \div 3 = 2\times 11 - 2\\ &&& 12 + 2 \times 3 \times 4 \div 3 = 2\times 11 - 2\\ &&& 12 + 8 = 22 - 2\\ &&& 20 = 20\end{align*}
2. Follow the order of operations and show each step. What is the value of the variable?
\begin{align*}10^2 - 6(4 + 6) - 3^2 - 3 \times 4 - 1 = d + 50 \div 5^2\end{align*}
We can use the problem solving steps to help us with the order of operations.
\begin{align*}& \mathbf{Describe:} && \text{The equation has parentheses, exponents, multiplication, subtraction, and addition.}\\ &&& d \ \text{is the variable.}\\ \\ & \mathbf{My \ Job:} && \text{Apply the order of operations rule to figure out the value of}\ d.\\ \\ & \mathbf{Solve:} && 10^2 - 6(4 + 6) - 3^2 - 3 \times 4 - 1 = d + 50 \div 5^2\\ &&& \mathbf{Parentheses} \qquad \qquad \quad 10^2 - 6\times 10 - 3^2 - 3 \times 4 - 1 = d + 50 \div 5^2\\ &&& \mathbf{Exponents} \qquad \qquad \quad \ 100 - 6\times 10 - 9 - 3 \times 4 - 1 = d + 50 \div 25\\ &&& \mathbf{Multiplication/} \qquad \quad 100 - 60 - 9 - 12 - 1 = d + 2\\ &&& \mathbf{Division}\\ &&& \mathbf{(left \ to \ right)}\\ &&& \mathbf{Addition/} \qquad \qquad \qquad \ 18=d+2\\ &&& \mathbf{Subtraction}\\ &&& \mathbf{(left \ to \ right)} \qquad \qquad \ 18 -2= d\\ &&& \qquad \qquad \qquad \qquad \qquad \quad d = 16\\ \\ & \mathbf{Check:} && \text{Replace} \ d \ \text{with 16 in the equation. Check that the two expressions}\\ &&&\text{(to the right and to the left of the = symbol) name the same number.}\\ &&& 10^2 - 6(4 + 6) - 3^2 - 3 \times 4 - 1 = 16 + 50 \div 5^2\\ &&& 10^2 - 6\times 10 - 3^2 - 3 \times 4 - 1 = 16 + 50 \div 5^2\\ &&& 100 - 6\times 10 - 9 - 3 \times 4 - 1 = 16 + 50 \div 25\\ &&& 100 - 60 - 9 - 12 - 1 = 16 + 2\\ &&& 18 = 18\end{align*}
3. Follow the order of operations and show each step. What is the value of the variable?
\begin{align*}4(9 - 5) + h + 3^2 - 2^3 + 4^1 = 3^2 \times 3 - 2 \times 3\end{align*}
We can use the problem solving steps to help us with the order of operations.
\begin{align*}& \mathbf{Describe:} && \text{The equation has parentheses, exponents, multiplication, subtraction, and addition.}\\ &&& h \ \text{is the variable.}\\ \\ & \mathbf{My \ Job:} && \text{Apply the order of operations rule to figure out the value of}\ h.\\ \\ & \mathbf{Solve:} && 4(9 - 5) + h + 3^2 - 2^3 + 4^1 = 3^2 \times 3 - 2 \times 3\\ &&& \mathbf{Parentheses} \qquad \qquad \quad 4 \times 4 + h + 3^2 - 2^3 + 4^1 = 3^2 \times 3 - 2 \times 3\\ &&& \mathbf{Exponents} \qquad \qquad \quad \ 4 \times 4 + h + 9 - 8 + 4 = 9 \times 3 - 2 \times 3\\ &&& \mathbf{Multiplication/} \qquad \quad 16 + h + 9 - 8 + 4 = 27 - 6\\ &&& \mathbf{Division}\\ &&& \mathbf{(left \ to \ right)}\\ &&& \mathbf{Addition/} \qquad \qquad \qquad \ h + 21 = 21\\ &&& \mathbf{Subtraction}\\ &&& \mathbf{(left \ to \ right)} \qquad \qquad \ h = 21 - 21\\ &&& \qquad \qquad \qquad \qquad \qquad \quad h = 0\\ \\ & \mathbf{Check:} && \text{Replace} \ h \ \text{with 0 in the equation. Check that the two expressions}\\ &&&\text{(to the right and to the left of the = symbol) name the same number.}\\ &&& 4(9 - 5) + 0 + 3^2 - 2^3 + 4^1 = 3^2 \times 3 - 2 \times 3\\ &&& 4\times 4 + 0 + 3^2 - 2^3 + 4^1 = 3^2 \times 3 - 2 \times 3\\ &&& 4\times 4 + 0 + 9 - 8 + 4 = 9 \times 3 - 2 \times 3\\ &&& 16 + 0 + 9 - 8 + 4 = 27 - 6\\ &&& 21 = 21\end{align*}
#### Earlier Problem Revisited
\begin{align*}6 \times 3^2 \div 2 + z + 4(7 - 3) + 2^3 \div 4 = 3^2 \times 6 + 1\end{align*}
We can use the problem solving steps to help us with the order of operations.
\begin{align*}& \mathbf{Describe:} && \text{The equation has parentheses, exponents, multiplication, division, and addition.}\\ &&& z \ \text{is the variable.}\\ \\ & \mathbf{My \ Job:} && \text{Apply the order of operations rule to figure out the value of}\ z.\\ \\ & \mathbf{Solve:} && 6 \times 3^2 \div 2 + z + 4(7 - 3) + 2^3 \div 4 = 3^2 \times 6 + 1\\ &&& \mathbf{Parentheses} \qquad \qquad \quad 6 \times 3^2 \div 2 + z + 4 \times 4 + 2^3 \div 4 = 3^2 \times 6 + 1\\ &&& \mathbf{Exponents} \qquad \qquad \quad \ 6 \times 9 \div 2 + z + 4 \times 4 + 8 \div 4 = 9 \times 6 + 1\\ &&& \mathbf{Multiplication/} \qquad \quad 27 + z + 16 + 2 = 54 + 1\\ &&& \mathbf{Division}\\ &&& \mathbf{(left \ to \ right)}\\ &&& \mathbf{Addition} \qquad \qquad \qquad \ 45 + z = 55\\ &&& \mathbf{(left \ to \ right)} \qquad \qquad \ z = 55 - 45\\ &&& \qquad \qquad \qquad \qquad \qquad \quad z = 10\\ \\ & \mathbf{Check:} && \text{Replace} \ z \ \text{with 10 in the equation. Check that the two expressions}\\ &&&\text{(to the right and to the left of the = symbol) name the same number.}\\ &&& 6 \times 3^2 \div 2 + 10 + 4(7 - 3) + 2^3 \div 4 = 3^2 \times 6 + 1\\ &&& 6 \times 3^2 \div 2 + 10 + 4 \times 4 + 2^3 \div 4 = 3^2 \times 6 + 1\\ &&& 6 \times 9 \div 2 + 10 + 4 \times 4 + 8 \div 4 = 9 \times 6 + 1\\ &&& 27 + 10 + 16 + 2 = 54 + 1\\ &&& 55 = 55\end{align*}
### Vocabulary
The order of operations tells us the correct order of evaluating math expressions. We always do parentheses first and then exponents. Next we do multiplication and division (from left to right) and finally addition and subtraction (from left to right).
### Examples
For each problem, follow the order of operations and show each step. What is the value of the variable?
#### Example 1
\begin{align*}2 + 5^2 \div 5 \times 1 + 0 \times 34 = 2y - 7(4 - 3)\end{align*}
\begin{align*} 2 + 5^2 \div 5 \times 1 + 0 \times 34 &= 2y - 7(4 - 3)\\ 2 + 5^2 \div 5 \times 1 + 0 \times 34 &= 2y - 7 \times 1\\ 2 + 25 \div 5 \times 1 + 0 \times 34 &= 2y - 7 \times 1\\ 2 + 5 + 0 &= 2y - 7\\ 7 &= 2y - 7\\ 7 + 7 &= 2y\\ 14 &= 2y\\ 7 &= y\end{align*}
#### Example 2
\begin{align*}2a + 5(9 - 8) \times 2^2 = 2^2 \times 3^2 + 2\end{align*}
\begin{align*} 2a + 5(9 - 8) \times 2^2 &= 2^2 \times 3^2 + 2\\ 2a + 5 \times 1 \times 2^2 &= 2^2 \times 3^2 + 2\\ 2a + 5 \times 1 \times 4 &= 4 \times 9 + 2\\ 2a + 20 &= 36 + 2\\ 2a + 20 &= 38\\ 2a &= 38 - 20\\ 2a &= 18\\ a &= 9\end{align*}
#### Example 3
\begin{align*}9^2 - 8^2 - 16 + 4^3 + 2^2 = e(5 + 2) - 1\end{align*}
\begin{align*} 9^2 - 8^2 - 16 + 4^3 + 2^2 &= e(5 + 2) - 1\\ 9^2 - 8^2 - 16 + 4^3 + 2^2 &= e \times 7 - 1\\ 81 - 64 - 16 + 64 + 4&= 7e - 1\\ 69 &= 7e - 1\\ 69 + 1 &= 7e\\ 70 &= 7e\\ 10 &= e\end{align*}
### Review
For each problem, follow the order of operations and show each step. What is the value of the variable?
1. \begin{align*}2m + 3 \times 9 \div 3^3 + 4(8 - 3) = 7^2 - 3 \times 6\end{align*}
2. \begin{align*}7^2 \div 7 \times (5^2 - 17) - (2 \times 6) = 4l - 2^2 \times 5\end{align*}
3. \begin{align*}6+3^2 \div 3 \times 2 +1 \times 5 =3y-2(5-3)\end{align*}
4. \begin{align*}5m+2(7-6)\times 2^3 = 2^2 \times 4^2 +7\end{align*}
5. \begin{align*}3^2-2^2-15+3^3+2^3=f(2+1)-2\end{align*}
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# Math Word Problems for Armed Services Vocational Aptitude Battery (ASVAB) Study Guide (page 2)
By
Updated on Jul 5, 2011
Many of the math problems on tests are word problems. A word problem can include any kind of math, including simple arithmetic, fractions, decimals, percentages, even algebra and geometry.
The hardest part of any word problem is translating English into math. When you read a problem, you can frequently translate it word for word from English statements into mathematical statements. At other times, however, a key word in the word problem hints at the mathematical operation to be performed. Here are the translation rules:
### Distance Formula: Distance = Rate x Time
The key words are movement words like: plane, train, boat, car, walk, run, climb, swim.
• How far did the plane travel in 4 hours if it averaged 300 miles per hour?
d = 300 × 4
d = 1,200 miles
• Ben walked 20 miles in four hours. What was his average speed?
20 = r × 4
5 miles per hour = r
### Solving a Word Problem Using the Translation Table
Remember the problem at the beginning of this chapter about the jellybeans?
Example: Juan ate of the jellybeans. Maria then ate of the remaining jellybeans, which left 10 jellybeans. How many jellybeans were there to begin with?
1. 60
2. 90
3. 120
4. 140
We solved it by working backward. Now let's solve it using our translation rules.
Assume Juan started with J jellybeans. Eating of them means eating jellybeans. Maria ate a fraction of the remaining jellybeans, which means we must subtract to find out how many are left: Maria then ate , leaving of the jellybeans, or jellybeans. Multiplying out gives J as the number of jellybeans left. The problem states that there were 10 jellybeans left, meaning that we set equal to 10:
Solving this equation for J gives J = 60. Thus, the right answer is a (the same answer we got when we worked backward). As you can see, both methods—working backward and translating from English to math—work. You should use whichever method is more comfortable for you.
### Practice Word Problems
You will find word problems using fractions, decimals, and percentages in those sections of this chapter. For now, practice using the translation table on problems that just require you to work with basic arithmetic.
1. Amir went shopping with \$250 and returned home with only \$33.56. How much money did he spend?
1. \$208.44
2. \$210.54
3. \$212.44
4. \$216.44
5. \$218.54
2. Mark invited ten friends to a party. Each friend brought three guests. How many people came to the party, excluding Mark?
1. 3
2. 10
3. 30
4. 40
5. 41
3. The office secretary can type 75 words per minute. How many minutes will it take him to type a report containing 1,200 words?
1. 16
2. 17
3. 18
4. Mr. Wallace is writing a budget request to upgrade his personal computer system. He wants to purchase a cable modem, which will cost \$100, two new software programs at \$350 each, a color printer for \$249, and an additional color cartridge for \$25.What is the total amount Mr. Wallace should write on his budget request?
1. \$724
2. \$974
3. \$1,049
4. \$1,064
5. \$1,074
1. d
2. d
3. a
4. e
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# Lesson 12
Piecewise Functions
The practice problem answers are available at one of our IM Certified Partners
### Problem 1
A parking garage charges \$5 for the first hour, \$10 for up to two hours, and \\$12 for the entire day. Let $$G$$ be the dollar cost of parking for $$t$$ hours.
1. Complete the table.
2. Sketch a graph of $$G$$ for $$0 \leq t \leq 12$$.
3. Is $$G$$ a function of $$t$$? Explain your reasoning.
4. Is $$t$$ a function of $$G$$? Explain your reasoning.
$$t$$ (hours) $$G$$ (dollars)
0
$$\frac 12$$
1
$$1\frac 3 4$$
2
5
### Problem 2
Is this a graph of a function? Explain your reasoning.
### Problem 3
Use the graph of function $$g$$ to answer these questions.
1. What are the values of $$g(1)$$, $$g(\text-12)$$, and $$g(15)$$?
2. For what $$x$$-values is $$g(x)=\text{-} 6$$?
3. Complete the rule for $$g(x)$$ so that the graph represents it.
$$\displaystyle g(x) =\ \begin{cases} \text{-}10, & \text{-}15\leq x< \text{-}10 \\ \underline{\hspace {8mm}}, & \text{-}10\leq x<\text{-}8 \\ \text{-}6, & \underline{\hspace {8mm}}\leq x<\text{-}1 \\ \underline{\hspace {8mm}}, & \text{-}1\leq x<1 \\ 4, & \underline{\hspace {8mm}}\leq x<\underline{\hspace {8mm}} \\ 8, & 10\leq x<15 \\ \end{cases}$$
### Problem 4
This graph represents Andre’s distance from his bicycle as he walks in a park.
1. For which intervals of time is the value of the function decreasing?
2. For which intervals is it increasing?
3. Describe what Andre is doing during the time when the value of the function is increasing.
(From Algebra1, Unit 4, Lesson 6.)
### Problem 5
The temperature was recorded at several times during the day. Function $$T$$ gives the temperature in degrees Fahrenheit, $$n$$ hours since midnight.
Here is a graph for this function.
1. Describe the overall trend of temperature throughout the day.
2. Based on the graph, did the temperature change more quickly between 10:00 a.m. and noon, or between 8:00 p.m. and 10:00 p.m.? Explain how you know.
(From Algebra1, Unit 4, Lesson 7.)
### Problem 6
Explain why this graph does not represent a function.
(From Algebra1, Unit 4, Lesson 8.)
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# How much work does it take to lift a 35 kg weight 1/2 m ?
Mar 17, 2018
171.5 J
#### Explanation:
The amount of work required to complete an action can be represented by the expression $F \cdot d$, where F represents the force used and d represents the distance over which that force is exerted.
The amount of force required to lift an object is equal to the amount of force required to counteract gravity. Assuming the acceleration due to gravity is $- 9.8 \frac{m}{s} ^ 2$, we can use Newton’s second law to solve for the force of gravity on the object.
${F}_{g} = - 9.8 \frac{m}{s} ^ 2 \cdot 35 k g = - 343 N$
Because gravity applies a force of -343N, to lift the box one must apply a force of +343N. In order to find the energy required to lift the box half a meter, we must multiply this force by half a meter.
$343 N \cdot 0.5 m = 171.5 J$
Mar 18, 2018
$171.5 \setminus \text{J}$
#### Explanation:
We use the work equation, which states that
$W = F \cdot d$
where $F$ is the force applied in newtons, $d$ is the distance in meters.
The force here is the weight of the box.
Weight is given by
$W = m g$
where $m$ is the mass of the object in kilograms, and $g$ is the gravitational acceleration, which is approximately $9.8 \setminus {\text{m/s}}^{2}$.
So here, the weight of the box is
$35 \setminus \text{kg"*9.8 \ "m/s"^2=343 \ "N}$.
The distance here is $\frac{1}{2} \setminus \text{m"=0.5 \ "m}$.
So, plugging in the given values into the equation, we find that
$W = 343 \setminus \text{N"*0.5 \ "m}$
$= 171.5 \setminus \text{J}$
Note that I used $g = 9.8 \setminus {\text{m/s}}^{2}$ to calculate the weight of the box.
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Double Math
# Set in Mathematics, Examples
A collection of well-defined and distinct objects is called a set in mathematics.
Well-defined collection means we can whether the object belongs collection or not and distinct means no two of which are the same.
We are familiar with the set since the word is frequently used in everyday speech for instance water set, tea set, sofa set. It is a wonder that mathematicians have developed this ordinary word into a mathematical concept. It becomes a language employed in more branches of modern mathematics.
The object in a set in mathematics is called members or elements.
Generally, we denote the set with capital letters A, B, C, D, etc., and elements with small letters a,b,c,d, etc.
## How many ways to describe a set:
There are three ways to describe a set
• Descriptive method
• Tabular method
• Set builder notation
## Descriptive method of set:
In this method, a set may be described in words. For instance,
e.g, D = A set of natural numbers.
## Tabular Method of set:
A set may be described by listing its elements within brackets. If D is the set mentioned above, then we may write: D= {1,2, 3, 4…}
## Set-builder notation:
It is sometimes more suitable. The method of set-builder notation in framing sets. This is done by using a symbol or letter for an unpredictable member of the set and expressing the property common to all the members.
So the above set may be written as
D={x|x∈N }
This is read as D is the set of all x such that x is a member of the natural number.
The symbol used for membership of a set. Thus a ∈ D means a is an element of D or a belongs to D.
c ∉ D means c does not belong to D or c is not a member of D.
Elements of a set can be people, cities, canals, oceans, or anything.
Some sets, along with their names are given below:-
N = The set of all natural numbers = {1,2,3,…}
W = The set of all whole numbers = {0,1,2,…}
Z = The set of all integers = {0,±1,+2…}.
Z ‘ = The set of all negative integers = {-1,-2,-3,…}.
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Chapter 5.3 Notes: Use Angle Bisectors of Triangles
```5.3 Use Angle Bisectors of Triangles
Objectives
Use properties of angle bisectors
Locate the incenter
Vocabulary
Recall, an angle bisector is a ray that
divides an angle into two congruent
b
The distance from a point to a line is the
length of the perpendicular segment from
the point to the line.
Angle Bisector Theorems
Theorem 5.5 - Angle Bisector Theorem:
If a point is on the bisector of an angle, then it is equidistant from
the two sides of the angle.
Since AD is an bisector,
then DE ≅ DF.
E
F
Theorem 5.6 – Converse of the Angle Bisector Theorem:
If a point is in the interior of an angle and is equidistant from the
sides of the angle, then it lies on the bisector of the angle.
Example 1
EXAMPLE 1
Use the Angle Bisector Theorems
Find the measure of GFJ.
SOLUTION
Because JG FG and JH FH and
JG = JH = 7, FJ bisects GFH by the Converse of the
Angle Bisector Theorem. So, mGFJ = mHFJ = 42°.
Example 2
EXAMPLE 2
A soccer goalkeeper’s position relative to the ball and goalposts forms
congruent angles, as shown. Will the goalie have to move farther to
block a shot toward the right goalpost R or the left goalpost L?
SOLUTION
The congruent angles tell you that the goalie is on the bisector of
LBR. By the Angle Bisector Theorem, the goalie is equidistant from
BR and BL .
So, the goalie must move the same distance to block either shot.
Example 3
EXAMPLE 3
Use algebra to solve a problem
For what value of x does P lie on the bisector of A?
SOLUTION
From the Converse of the Angle Bisector Theorem,
you know that if P lies on the bisector of A then P is
equidistant from the sides of A, so BP = CP.
BP = CP
x + 3 = 2x –1
4 =x
Set segment lengths equal.
Substitute expressions for segment lengths.
Solve for x.
Point P lies on the bisector of A when x = 4.
GUIDED PRACTICE
More
Practice Problems
For numbers1–3, find the value of x.
1.
2.
B
P
A
B
P
A
C
C
15
11
P
3.
B
C
A
5
GUIDED PRACTICE
Example
4
Do you have enough information to conclude that
QS bisects PQR? Explain.
No; you need to establish that SR QR and SP QP.
Vocabulary and Another Theorem
Theorem 5.7 Concurrency of Angle Bisectors of a Triangle:
The angle bisectors of a triangle intersect at a point that is
equidistant from the sides of the triangle.
The point of concurrency of the three angle bisectors of a triangle
is called the incenter of the triangle.
The incenter always lies inside the triangle.
Example 5
EXAMPLE 4
Use the concurrency of angle bisectors
In the diagram, N is the incenter of ABC. Find ND.
SOLUTION
By the Concurrency of Angle Bisectors of a
Triangle Theorem, the incenter N is equidistant
from the sides of ABC. So, to find ND, you can
find NF in AF. Use the Pythagorean Theorem,
a2 + b2 = c2.
Example 5 (continued):
EXAMPLE 4
c 2 = a 2 + b2
2
20 2 = NF + 16
2
Pythagorean Theorem
2
Substitute known values.
400 = NF + 256
Multiply.
144 = NF 2
Subtract 256 from each side.
12 = NF
Take the positive square root of each side.
Because NF = ND, ND = 12.
```
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# How do you solve 2x^2 -2x- 2= 0 using completing the square?
Jun 19, 2015
$2 {x}^{2} - 2 x - 2 = 0$
$2 {x}^{2} - 2 x = 2$
${x}^{2} - x = 1$
${x}^{2} - x + {\left(\frac{1}{2}\right)}^{2} = {\left(\frac{1}{2}\right)}^{2} + 1$
${\left(x - \frac{1}{2}\right)}^{2} = \frac{5}{4}$
$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$
$x = \frac{1 \pm \sqrt{5}}{2}$
But what did we do and why?
In order to have a habit for how we do this, let's begin by moving the constant to the other side of the equation. We'll add $2$ to both sides to get:
$2 {x}^{2} - 2 x = 2$
When completed, the square will be on the left, and it will have the form:
${x}^{2} \pm 2 a x + {a}^{2}$,
so the next thing to do is get a $1$ (which we won't write) in front of the ${x}^{2}$. (In fancy terms, we're going to make the coefficient of ${x}^{2}$ equal to $1$.)
Multiply both sides of the equation by $\frac{1}{2}$ (Remember to distribute the multiplication.)
$\frac{1}{2} \left(2 {x}^{2} - 2 x\right) = \frac{1}{2} \left(2\right)$ now simplify:
${x}^{2} - x = 1$
Now that we have just ${x}^{2}$, we can see that the coefficient of $x$ is negative, that tells us that the completed square will look like:
${x}^{2} - 2 a x + {a}^{2}$ which we will be able to factor: ${\left(x - a\right)}^{2}$
We have: ${x}^{2} - x = 1$,
Which we can think of as: ${x}^{2} - 1 x = 1$,
so we must have:
$2 a = 1$.
And that makes $a = \frac{1}{2}$. Square that and add the result to both sides:
${\left(\frac{1}{2}\right)}^{2} = {1}^{2} / {2}^{2} = \frac{1}{4}$
${x}^{2} - x + \frac{1}{4} = \frac{1}{4} + 1$,
Now factor on the left (we already know how to factor that! See above.) And add on the right.
${\left(x - \frac{1}{2}\right)}^{2} = \frac{1}{4} + \frac{4}{4} = \frac{5}{4}$
${\left(x - \frac{1}{2}\right)}^{2} = \frac{5}{4}$
Now use the fact that ${n}^{2} = g$ if and only of $n = \sqrt{g} \mathmr{and} - \sqrt{g}$
(The square of a number equals a given number if and only if the number is either the positive or negative square root of the given.)
$x - \frac{1}{2} = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{\sqrt{4}} = \pm \frac{\sqrt{5}}{2}$
$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$ Add $\frac{1}{2}$ to both sides:
$x = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$ which we often prefer to write as a single fraction:
$x = \frac{1 \pm \sqrt{5}}{2}$
Note: never let yourself think this is some kind pf weird positive and negative number. It is just a convenient way of writing the two solutions:
$x = \frac{1 + \sqrt{5}}{2}$ and $x = \frac{1 - \sqrt{5}}{2}$.
|
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2.1: Substitution
Recall that the chain rule states that
$(f(g(x)))' = f'(g(x))g'(x).$
Integrating both sides we get:
$\int[f(g(x)]'dx = \int[f'(g(x)g'(x)dx]$
or
$\int f'\left( g(x) \right) \, g' (x) \, dx = f\left(g(x)\right) + C$
Example 1
Calculate
$\int \dfrac{2x}{x^2+1}\, dx = \int 2x\left( x^2+1\right)^{-2} \, dx.$
Solution
Let
$u = x^2 +1$
then
$\dfrac{du}{dx} = 2x$
and
$du = 2x \,dx.$
We substitute:
$\int u^{-2} du = -u^{-1} + C = (x^2 +1)^{-1} + C.$
Steps:
1. Find the function derivative pair ($$f$$ and $$f'$$).
2. Let $$u = f(x)$$.
3. Find $$du/dx$$ and adjust for constants.
4. Substitute.
5. Integrate.
6. Resubstitute.
We will try many more examples including those such as
$\int x\, \sin(x^2)\, dx,$
$\int x\, \sqrt{x - 2}\, dx.$
Contributors
• Integrated by Justin Marshall.
|
# 5.5: Inverse Trigonometric Functions
Now that we know the common trigonometric functions, let's look at their inverses.
Recall that if $f$ must be one-to-one for it to have an inverse.
## Inverse sine function
The inverse sine function, denoted $\sin^{-1}$, has domain $[-1, 1]$ and range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ where $\sin^{-1}x = y \text{ when } \sin y = x$ The following cancellation properties hold: \begin{align} \sin(\sin^{-1}x) = x, &\qquad -1 \leq x \leq 1 \\ \sin^{-1}(\sin(x)) = x, &\qquad -\frac{\pi}{2}\leq x \leq \frac{\pi}{2} \end{align}
Suppose $a \in [-1, 1]$. Someone says $\sin^{-1}(a) = \frac{2\pi}{3}$ is possible. Is this true?
Find each value: $\sin^{-1}\frac{1}{2} \qquad\qquad \sin^{-1}\left(-\frac{1}{2}\right) \qquad\qquad \sin^{-1}\frac{3}{2}$
Find each value: $\sin^{-1}\left(\sin\frac{\pi}{3}\right) \qquad\qquad \sin^{-1}\left(\sin\left(-\frac{2\pi}{3}\right)\right)$
Important: remember to always check if the input value for $\sin^{-1}$ is actually in the domain. Keep this in mind for all of the inverse functions.
## Inverse cosine function
The inverse cosine function, denoted $\cos^{-1}$, has domain $[-1, 1]$ and range $\left[0, \pi\right]$ where $\cos^{-1}x = y \text{ when } \cos y = x$ The following cancellation properties hold: \begin{align} \cos(\cos^{-1}x) = x, &\qquad -1 \leq x \leq 1 \\ \cos^{-1}(\cos(x)) = x, &\qquad 0\leq x \leq \pi \end{align}
Find each value: $\cos^{-1}\frac{\sqrt{3}}{2} \qquad\qquad \cos^{-1}0 \qquad\qquad \cos^{-1}\left(-\frac{1}{2}\right)$
Find each value: $\cos^{-1}\left(\cos\left(\frac{2\pi}{3}\right)\right) \qquad\qquad \cos^{-1}\left(\cos\left(-\frac{5\pi}{3}\right)\right)$
## Inverse tangent function
The inverse tangent function, denoted $\tan^{-1}$, has domain $\mathbb{R}$ and range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ where $\tan^{-1}x = y \text{ when } \tan y = x$ The following cancellation properties hold: \begin{align} \tan(\tan^{-1}x) = x, &\qquad x \in \mathbb{R} \\ \tan^{-1}(\tan(x)) = x, &\qquad -\frac{\pi}{2} < x < \frac{\pi}{2} \end{align}
Find each value: $\tan^{-1} 1 \qquad\qquad \tan^{-1} \sqrt{3} \qquad\qquad \tan^{-1}0$
Find each value: $\sin^{-1} \left(\tan\frac{\pi}{4}\right) \qquad\qquad \sin^{-1}\left(\cos\frac{\pi}{6}\right)$
|
Math Calculators, Lessons and Formulas
It is time to solve your math problem
mathportal.org
Progressions: (lesson 1 of 2)
## Arithmetic Progressions
Definition:
By an arithmetic progression of $m$ terms, we mean a finite sequence of the form
$$a, a + d, a + 2d, a + 3d, . . . , a + ( m - 1)d.$$
The real number $a$ is called the first term of the arithmetic progression, and the real number $d$ is called the difference of the arithmetic progression.
Example 1:
Consider the sequence of numbers
$$1,\; 3, \; 5, \; 7, \; 9, \; 11, \; 13, \; 15, \; 17, \; 19, \; 21, \; 23$$
The property of this sequence is that the difference between successive terms is constant and equal to 2.
Here we have: $a = 1$; $d = 2$.
Example 2:
Consider the sequence of numbers:
$$2,\; 5, \; 8, \; 11, \; 14, \; 17, \; 20, \; 23, \; 26, \; 29, \; 32$$
The property of this sequence is that the difference between successive terms is constant and equal to 3.
Here we have: $a = 2$; $d = 3$.
### General term of arithmetic progression:
The general term of an arithmetic progression with first term $a_1$ and common difference $d$ is:
$$\color{blue}{a_k = a_1 + (k - 1)d}$$
Example 3: Find the general term for the arithmetic sequence $-1, \; 3, \; 7, \; 11, \; . . .$ and then find $a_{12}$.
Solution:
Here $a_1 = 1$. To find $d$ subtract any two adjacent terms: $d = 7 - 3 = 4$. The general term is:
\begin{aligned} \color{blue}{a_k} &\color{blue}{=} \color{blue}{a_1 + ( k - 1 ) d} \\ a_k &= -1 + ( k - 1 ) \cdot 4 \\ a_k &= -1 + 4k - 4 \\ \color{blue}{a_k} &\color{blue}{=} \color{blue}{4k - 5} \end{aligned}
To find $a_{12}$, let $k = 12$.
$$a_{12} = 4 \cdot 12 - 5 = 43$$
Example 4: If $a_3 = 8$ and $a_6 = 17$, find $a_{14}$.
Solution:
Use the formula for $a_k$ with the given terms
\begin{aligned} a_3 &= a_1 + (3 - 1) \cdot d \\ 8 &= a_1 + 2d \\ \\ a_6 &= a_1 + (6 - 1) \cdot d \\ 17 &= a_1 + 5d \end{aligned}
This gives us a system of two equations with two variables. By solving them, we can find that $a_1 = 2$ and $d=3$.
Use the formula for $a_k$ to find $a_{14}$
\begin{aligned} a_k &= a_1 + (k - 1) \cdot d \\ a_{14} &= 2 + (14 - 1) \cdot 3 \\ \color{blue}{a_{14}} &\color{blue}{=} \color{blue}{41} \end{aligned}
Exercise:
Level 1
$$\color{blue}{ a_1 = 3, \ d = 4 \\ a_6 = ?? }$$ $a_6 = 20$ $a_6 = 21$ $a_6 = 22$ $a_6 = 23$
Level 2
$$\color{blue}{a_4 = 7, \ a_8 = 15 \\ a_{10} = ?? }$$ $a_{10} = 18$ $a_{10} = 19$ $a_{10} = 20$ $a_{10} = 21$
### Sum of an arithmetic progression:
The sum of the $n$ terms of an arithmetic progression with first term $a_1$ and common difference $d$ is:
$$\color{blue}{a_n = \frac{n}{2} [ 2{a_1} + (n - 1)d ]}$$
Also, the sum of an arithmetic progression is equal to
$$\color{blue}{S_n = \frac{n}{2} (a_1 + a_n)}$$
Example 5: Find the sum of the $10$ terms of the arithmetic progression if $a_1 = 5$ and $d = 4$.
Solution:
\begin{aligned} S_n &= \frac{n}{2} [ 2 a_1 + (n - 1) d ] \\ S_{10} &= \frac {10}{2} [ 2 \cdot 5 + ( 10 - 1 ) \cdot 4 ] \\ S_{10} &= 5 \cdot 46 \\ \color{blue}{S_{10}} &\color{blue}{=} \color{blue}{230} \end{aligned}
Example 6: Find $1 + 2 + 3 + . . . + 100$
Solution:
In this example, we have: $a_1 = 1$, $d = 1$, $n = 100$, $a_{100} = 100$. The sum is:
\begin{aligned} S_n &= \frac{n}{2} ( a_1 + a_n ) \\ S_{100} &= \frac {100}{2} ( 1 + 100 ) \\ \color{blue}{S_{100}} &\color{blue}{=} \color{blue}{5050} \end{aligned}
Exercise:
Level 1
$$\color{blue}{ a_1 = -6, \ d = 2 \\ S_8 = ?? }$$ $S_8 = 0$ $S_8 = 10$ $S_8 = 20$ $S_8 = 30$
Level 2
$$\color{blue}{a_4 = 11, \ a_7 = 20 \\ S_5 = ?? }$$ $S_5 = 20$ $S_5 = 30$ $S_5 = 40$ $S_5 = 50$
|
Algebra proofs
Many algebra proofs are done using proof by mathematical induction. To demonstrate the power of mathematical induction, we shall prove an algebraic equation and a geometric formula with induction.
If you are not familiar with with proofs using induction, carefully study proof by mathematical induction given as a reference above. Otherwise, you could struggle with these algebra proofs below
Algebra equation:
Prove by mathematical induction that 1 + 2 + 4 + 8 + ... + 2n-1 = 2n - 1
Step # 1:
Show that the equation is true for n = 2. n = 2 means adding the first two terms
1 + 2 = 3 and 22 - 1 = 4 - 1 = 3. So, it is true for n =2
Just for fun, let's show it is true also for n = 4. n =4 means adding the first 4 terms
1 + 2 + 4 + 8 = 15 and 24 - 1 = 16 - 1 = 15. So, it is true also for n =4
Step # 2:
Suppose it is true for n = k
Just replace n by k
1 + 2 + 4 + 8 + ... + 2k-1 = 2k - 1
Step # 3:
Prove it is true for n = k + 1
You need to write down what it means for the equation to be true for n = k + 1
Caution: Writing down what it means is not the same as proving the equation is true. In fact, it just shows you what you need to prove
Here is what it means for n = k + 1:
After you replace k by k+1, you get :
1 + 2 + 4 + 8 + ... + 2k + 1 -1 = 2k + 1 - 1
1 + 2 + 4 + 8 + ... + 2k = 2k + 1 - 1
You can now complete the proof by using the hypothesis in step # 2 and then show that
1 + 2 + 4 + 8 + ... + 2k = 2k + 1 - 1
starting with the hypothesis, 1 + 2 + 4 + 8 + ... + 2k-1 = 2k - 1
ask yourself, " What does the next term look like? "
Since the last term now is 2k-1, the next term should be 2k + 1 -1 = 2k after replacing k by k + 1
Add 2k to both sides of the hypothesis
1 + 2 + 4 + 8 + ... + 2k-1 + 2k = 2k - 1 + 2k
The trick here is to see that 2k + 2k = 2 × 2k = 21× 2k = 2k + 1
1 + 2 + 4 + 8 + ... + 2k-1 + 2k = 2k - 1 + 2k
= 2 × 2k -1
= 21× 2k -1
= 2k + 1 -1
Geometric formula:
Show by mathematical induction that
the sum of the angles in an n-gon = ( n - 2 ) × 180°
A couple of good observations before we prove it:
Observation #1:
An n-gon is a closed figure with n sides. For example,
an n-gon with 4 sides is called a quadrilateral
an n-gon with 3 sides is called a triangle
Observation #2:
After a close examination of the figure above, can you see that every time you add a side, you are also adding one more triangle
I am now ready to show the proof.
Step # 1:
Show that the equation is true for n = 3. Notice that n cannot be smaller than 3 since we cannot make a closed figure with just two sides or one side.
When n = 3, we get a triangle and the sum of the angles in a triangle is equal to 180°
When n = 3, ( 3 - 2 ) × 180° = 1 × 180° = 180°
When n = 4, you are adding one more triangle to get two triangles and the sum of the angles of the two triangles is equal to 360°
When n = 4, ( 4 - 2 ) × 180° = 2 × 180° = 360°
Thus, the formula is true for n = 3 and n = 4
Step # 2:
Suppose it is true for n = k
Just replace n by k
The sum of the angles in a k-gon = ( k - 2 ) × 180°
Step # 3:
Prove it is true for n = k + 1
You need to write down what it means for the equation to be true for n = k + 1
Here is what it means for n = k + 1:
After you replace k by k+1, you get :
The sum of the angles in a (k+1)-gon = ( k +1 - 2 ) × 180°
The sum of the angles in a (k+1)-gon = ( k - 1 ) × 180°
You can now complete the proof by using the hypothesis in step # 2 and then show that
The sum of the angles in a (k+1)-gon = ( k - 1 ) × 180°
starting with the hypothesis, the sum of the angles in a k-gon = ( k - 2 ) × 180°
ask yourself, " What does the next term look like? "
Since the last term now is k-gon or a figure with k sides, the next term should be a figure with k + 1 sides after replacing k by k + 1
Now, what are we adding to both sides?
First, recall the meaning of adding one side. It means that you will be adding also one triangle
( k + 1) gon = k-gon + ( 1 side or one more triangle)
( k + 1) gon = k-gon + one more triangle
( k + 1) gon = k-gon + 180°
So add 180° to both sides of the hypothesis
The sum of the angles in a k-gon + 180° = ( k - 2 ) × 180° + 180°
The sum of the angles in a (k+1)-gon = ( k - 2 ) × 180° + 180°
= 180°k + -2 × 180° + 180°
= 180°k + -1 × 180°
= 180° ( k + -1 )
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Other important algebra proofs:
The algebra proofs below don't use mathematical induction.
Proof of the Pythagorean theorem
A proof of the Pythagorean Theorem by president Garfield is clearly explained here
Here we prove the quadratic formula by completing the square.
Law of detachment
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03-203-08hwk-Ch04
# 03-203-08hwk-Ch04 - 03-203-07hwk-Ch04 1 Chapter 4...
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03-203-07hwk-Ch04 1 Chapter 4: Two-Dimensional Kinematics Solutions to Problems 7. Picture the Problem : The arrow falls below the target center as it flies from the bow to the target. Strategy: Treat the vertical and horizontal motions separately. First find the time required for the arrow to drop straight down 52 cm from rest. Then use that time together with the horizontal target distance to find the horizontal speed of the arrow. That must also equal the initial speed because the arrow was launched horizontally. Solution: 1. Find the time to drop 52 cm: ( ) 2 2 0.52 m 2 0.326 s 9.81 m/s y t g Δ Δ = = = 2. Find the speed of the arrow from the horizontal distance and time elapsed: 0 15 m 46 m/s 0.326 s x x v t Δ = = = Δ Insight: We had to bend the significant figures rules a bit to obtain an accurate answer. Another way to solve this problem and avoid the rounding error is to solve equation 4-8 for velocity: ( ) 2 0 2 v g x h y = - . In this case we would set h = 0.52 m , y = 0.0 m, and x = 15 m. ------------------------------------------------------------------------------------------------------------------------------------------------------- 8. Picture the Problem : The water falls down along a parabolic arc, maintaining its horizontal velocity but gaining vertical speed as it falls. Strategy : Find the vertical speed of the water after falling 108 m. The horizontal velocity remains constant throughout the fall. Then find the magnitude of the velocity from the horizontal and vertical components. Solution: 1. Use equations 4-6 to find y v : ( )( ) 2 2 2 2 2 0 2 0 2 9.81 m/s 0 m –108 m 2120 m /s y y v v g y = - Δ = - = 2. Use the components of the velocity to find the speed: ( ) 2 2 2 2 2 3.60 m/s 2120 m /s 46.2 m s x y v v v = + = + = Insight: Projectile problems are often solved by first considering the vertical motion, which determines the time of flight and the vertical speed, and then considering the horizontal motion. ------------------------------------------------------------------------------------------------------------------------------------------------------- 16. Picture the Problem : The pumpkin’s trajectory is depicted in the figure at right. Strategy:
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# Angle Bisector Theorem
As per the Angle Bisector theorem, the angle bisector of a triangle bisects the opposite side in such a way that the ratio of the two line segments is proportional to the ratio of the other two sides. Thus the relative lengths of the opposite side (divided by angle bisector) are equated to the lengths of the other two sides of the triangle. Angle bisector theorem is applicable to all types of triangles.
Class 10 students can read the concept of angle bisector theorem here along with the proof. Apart from the angle bisector theorem, we will also discuss here the external angle theorem, perpendicular bisector theorem, the converse of angle bisector theorem.
## What is Angle Bisector Theorem?
An angle bisector is a straight line drawn from the vertex of a triangle to its opposite side in such a way, that it divides the angle into two equal or congruent angles.
Angle Bisector Theorems of Triangles
The table below shows the statements related to internal and external angle bisector theorems as well as their converse.
Theorem Statement Internal angle bisector theorem The angle bisector of a triangle divides the opposite side into two parts proportional to the other two sides of the triangle. Converse of Internal angle bisector theorem In a triangle, if the interior point is equidistant from the two sides of a triangle, then that point lies on the angle bisector of the angle formed by the two line segments. Perpendicular bisector theorem The perpendicular bisector bisects the given line segment into two equal parts, to which it is perpendicular. For a triangle, if a perpendicular bisector is drawn from the vertex to the opposite side, it divides the side into two congruent segments. External angle bisector theorem The external angle bisector divides the opposite side externally in the ratio of the sides containing the angle, and this condition usually occurs in non-equilateral triangles.
Now let us see, what is the angle bisector theorem.
According to the angle bisector theorem, the angle bisector of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.
## Interior Angle Bisector Theorem
In the triangle ABC, the angle bisector intersects side BC at point D. See the figure below.
As per the Angle bisector theorem, the ratio of the line segment BD to DC equals the ratio of the length of the side AB to AC.
$$\frac{\left | BD \right |}{\left | DC \right |}=\frac{\left | AB \right |}{\left | AC \right |}$$
Conversely, when a point D on the side BC divides BC in a ratio similar to the sides AC and AB, then the angle bisector of ∠ A is AD. Hence, according to the theorem, if D lies on the side BC, then,
$$\frac{\left | BD \right |}{\left | DC \right |}=\frac{\left | AB \right |Sin\angle DAB}{\left | AC \right |Sin\angle DAC}$$
If D is external to the side BC, directed angles and directed line segments are required to be applied in the calculation.
Angle bisector theorem is applied when side lengths and angle bisectors are known.
### Proof of Angle bisector theorem
We can easily prove the angle bisector theorem, by using trigonometry here. In triangles ABD and ACD (in the above figure) using the law of sines, we can write;
$$\frac{AB}{BD}=\frac{sin\angle BDA}{sin\angle BAD}$$ ….(1)
$$\frac{AC}{DC}=\frac{sin\angle ADC}{sin\angle DAC}$$ ….(2)
The angles ∠ ADC and ∠ BDA make a linear pair and hence called adjacent supplementary angles.
Since the sine of supplementary angles are equal, therefore,
Sin ∠ BDA = Sin ∠ ADC …..(3)
Also,
Thus,
Sin ∠ BDA = Sin ∠ ADC …(4)
Hence, from equations 3 and 4, we can say, the RHS of equations 1 and 2 are equal, therefore, LHS will also be equal.
$$\frac{\left | BD \right |}{\left | DC \right |}=\frac{\left | AB \right |}{\left | AC \right |}$$
Hence, the angle bisector theorem is proved.
Condition:
If the angles ∠ DAC and ∠ BAD are not equal, the equation 1 and equation 2 can be written as:
$$\frac{\left | AB \right |}{\left | BD \right |}$$ sin ∠ BAD = sin∠ BDA
$$\frac{\left | AC \right |}{\left | DC \right |}$$ sin ∠ DAC = sin∠ ADC
Angles ∠ ADC and ∠ BDA are supplementary, hence the RHS of the equations are still equal. Hence, we get
$$\frac{\left | AB \right |}{\left | BD \right |}$$ sin ∠BAD = $$\frac{\left | AC \right |}{\left | DC \right |}$$ sin ∠DAC
This rearranges to generalized view of the theorem.
## Converse of Angle Bisector Theorem
In a triangle, if the interior point is equidistant from the two sides of a triangle then that point lies on the angle bisector of the angle formed by the two line segments.
## Triangle Angle Bisector Theorem
Extend the side CA to meet BE to meet at point E, such that BE//AD.
Now we can write,
CD/DB = CA/AE (since AD//BE) —-(1)
∠4 = ∠1 [corresponding angles]
∠1 = ∠2 [AD bisects angle CAB]
∠2 = ∠3 [Alternate interior angles]
∠3 = ∠4 [By transitive property]
ΔABE is an isosceles triangle with AE=AB
Now if we replace AE by AB in equation 1, we get;
CD/DB = CA/AB
Hence proved.
## Perpendicular Bisector Theorem
According to this theorem, if a point is equidistant from the endpoints of a line segment in a triangle, then it is on the perpendicular bisector of the line segment.
Alternatively, we can say, the perpendicular bisector bisects the given line segment into two equal parts, to which it is perpendicular. In the case of a triangle, if a perpendicular bisector is drawn from the vertex to the opposite side, then it divides the segment into two congruent segments.
In the above figure, the line segment SI is the perpendicular bisector of WM.
## External Angle Bisector Theorem
The external angle bisector of a triangle divides the opposite side externally in the ratio of the sides containing the angle. This condition occurs usually in non-equilateral triangles.
### Proof
Given: In ΔABC, AD is the external bisector of ∠BAC and intersects BC produced at D.
To prove : BD/DC = AB/AC
Construction: Draw CE ∥ DA meeting AB at E
Since, CE ∥ DA and AC is transversal, therefore,
∠ECA = ∠CAD (alternate angles) ……(1)
Again, CE ∥ DA and BP is a transversal, therefore,
∠CEA = ∠DAP (corresponding angles) —–(2)
But AD is the bisector of ∠CAP,
As we know, Sides opposite to equal angles are equal, therefore,
∠CEA = ∠ECA
BD/DC = BA/AE [By Thales Theorem]
AE = AC,
BD/DC = BA/AC
Hence, proved.
## Solved Examples on Angle Bisector Theorem
Go through the following examples to understand the concept of the angle bisector theorem.
Example 1:
Find the value of x for the given triangle using the angle bisector theorem.
Solution:
Given that,
AD = 12, AC = 18, BC=24, DB = x
According to angle bisector theorem,
Now substitute the values, we get
12/18 = x/24
X = (⅔)24
x = 2(8)
x= 16
Hence, the value of x is 16.
Example 2:
ABCD is a quadrilateral in which the bisectors of angle B and angle D intersects on AC at point E. Show that AB/BC = AD/DC
Solution:
From the given figure, the segment DE is the angle bisector of angle D and BE is the internal angle bisector of angle B.
Hence, the using internal angle bisector theorem, we get
Similarly,
AE/EC = AB/BC ….(2)
From equations (1) and (2), we get
Hence, AB/BC = AD/DC is proved.
Example 3.
In a triangle, AE is the bisector of the exterior ∠CAD that meets BC at E. If the value of AB = 10 cm, AC = 6 cm and BC = 12 cm, find the value of CE.
Solution:
Given : AB = 10 cm, AC = 6 cm and BC = 12 cm
Let CE is equal to x.
By exterior angle bisector theorem, we know that,
BE / CE = AB / AC
(12 + x) / x = 10 / 6
6( 12 + x ) = 10 x [ by cross multiplication]
72 + 6x = 10x
72 = 10x – 6x
72 = 4x
x = 72/4
x = 18
CE = 18 cm
To learn more important Maths theorems, visit BYJU’S – The Learning App and learn all the concepts easily.
## Frequently Asked Questions on Angle Bisector Theorem
### What does the angle bisector theorem state?
According to the angle bisector theorem, an angle bisector of an angle of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.
### What is the formula of angle bisector?
In the triangle ABC, the angle bisector intersects side BC at point D. Thus,
BD/DC = AB/AC
### The angle bisector of vertex angle of an isosceles triangle bisects the opposite side. True or False.
True. An isosceles triangle has two pairs of equal sides with a common vertex. If the angle bisector of vertex angle is drawn, then it divides the opposite side into equal parts.
### How to find the angle bisector of an angle?
Draw an angle say ∠ABC, angled at B. Using a compass, and taking B as centre and any radius, draw an arc intersecting BA at P and BC at Q, respectively. Now taking P and Q as the centre and with the same radius, draw two arcs intersecting each other at R. Join the vertex B to point R and draw the ray BR. Thus, BR is the angle bisector of ∠ABC.
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back to Trigonometry lessons
# Graph of the Sine Function and Transformation
### Graph of the sine function y = sin x
The figure above is the graph of y = sin x. For y = sin x, there is a constant T, when x is any value, there is sin (x + T) = sin x. The constant T is called the period of the function y = sin x. Look the figure above, sin 0 = sin (0 + 2pi), sin pi/2 = sin (pi/2 + 2pi), sin pi = sin (pi + 2pi), sin 3pi/2 = sin (3pi/2 + 2pi), sin 2pi = sin (2pi + 2pi), so, the period T of y = sin x is 2pi.
### Graph of the sine function y = A sin (Bx + C)
The general form of the sine function is y = A sin (Bx + C), in which A > 0 and B > 0. The coefficient A is the amplitude of the curve, let T be the period of the sine function, then T = 2pi/B and C is the initial phase.
In the figure above, the blue curve is the graph of y = sin x, its period is T = 2pi/B = 2pi/1 = 2pi. Its amplitude A = 1 and initial phase C = 0. The yellow curve is the graph of y = sin (x + 2pi/3). Its amplitude A = 1, period T = 2pi/B = 2pi/1 = 2pi and Initial phase C = 2pi/3. Therefore, we can left-shift the graph of y = sin x by 2pi/3 units to get the graph of y = sin (x + 2pi/3).
# Two methods to draw the graph of the sine function
The graph of y = A sin (Bx + C) is called the curve of the sine function. There are two methods to draw the graph of the sine function. The method one is the five-point method, which selects five points in a period of the sine function. The five points include the maximum point of the function, the minimum point of the function, the zero points of the function. Draw the graph of the sine function in one period and then extend the graph to both direction for every 2pi/B interval to get the graph of the sine function y = A sin (Bx + C). The method two is the transformation method, which is transform the graph of y = sin x to the graph of y = A sin (Bx + C).
### Draw the graph of y = sin x using the five points method
For the graph of y = sin x, its period is T = 2pi/B = 2pi/1 = 2pi. To draw the graph of y = sin x, we divide the interval 2pi by five. So, the x-coordinate of the first point is x = 0. The x-coordinate of the fifth point is x = 2pi. The x-coordinate of the third point is x = pi. The x-coordinate of the second point is x = pi/2. The x-coordinate of the fourth point is x = 3pi/2. For each x-coordinate, find its corresponding y-coordinate. When x = 0, y = sin 0 = 0, when x = pi/2, y = sin pi/2 = 1, when x = pi, y = sin pi = 0, when x = 3pi/2, y = sin 3pi/2 = -1, and when x = 2pi, y = sin 2pi = 0. Therefore, the five important points in one period are (0, 0), (pi/2, 1), (pi, 0), (3pi/2, -1), and (2pi, 0).
### Difference among the graphs of y = sin x, y = (1/2) sin x and y = 2sin x
In the figure above, the blue curve is the graph of y = sin x, the yellow curve is the graph of y = 2 sin x and the grey curve is the graph of y = (1/2) sin x. The three curves have the same period but different Amplitudes. For example, the amplitude of y = sin x is one, the amplitude of y = 2 sin x is two and the amplitude of y = (1/2) sin x is one-half. Because the three curves have the same coefficient B, so they have the same period, T = 2pi/B = 2pi/1 = 2pi. Because the three curves have the same coefficient C, which is 0, so their initial phase is zero.
# Graph of Sine Function Transformation
### Draw the graph of y = sin 3x using transformation method
Step 1. Draw the graph of y = sin x, its period is T = 2pi/B = 2pi/1 = 2pi. Its five important points in one period are (0, 0), (pi/2, 1), (pi, 0), (3pi/2, -1) and (2pi, 0).
Step 2. Draw the graph of y = sin 3x
For the sine function y = sin 3x. Its coefficient B = 3, so its period is T = 2pi/3.
So, for each point on y = sin x, shrink the x-coordinate to one-third and keep the y-coordinate the same to get the graph of y = sin 3x.
Note: the period of y = sin x is 2pi. The period of y = sin 3x is 2pi/3. So, three period of y = sin 3x is 2pi. That is, from x = 0 to x = 2pi, the graph of y = sin 3x repeat three times. So, the coefficient B is the frequency of repeat times of the curve. Therefore, in the interval of x from 0 to 2pi, the graph y = sin x has only one period but the graph of y = sin 3x has three period.
### Draw the graph of y = sin (x - 3pi/4) using the transformation method
In the figure above, the blue curve is the graph of y = sin x. The yellow curve is the graph of y = sin (x - 3pi/4). For sine function y = sin (x - 3pi/4), its initial phase is C = -3pi/4, we can right-shift the graph of y = sin x by 3pi/4 units to get the graph of y = sin (x - 3pi/4).
### Draw the graph of y = sin (x + 3pi/4) using the transformation method
In the figure above, the blue curve is the graph of y = sin x. The yellow curve is the graph of y = sin (x + 3pi/4). For the sine function y = sin (x + 3pi/4), its initial phase is C = 3pi/4, we can left-shift the graph of y = sin x by 3pi/4 units to get the graph of y = sin (x + 3pi/4).
Example
Draw the graph of y = sin (2x - pi/3)
Solution
Method one - shift first then shrink
Step 1. Draw the graph of y = sin x
Step 2. Right-shift the graph of y = sin x by pi/3 units to get the graph of y = sin (x - pi/3)
Step 3. For each point on y = sin (x - pi/3), shrink the x-coordinate to half and keep the y-coordinate the same to get the graph of y = sin (2x -pi/3)
In the figure above, the grey curve is the graph of y = sin (2x - pi/3) and the yellow curve is the graph of y = sin (x - pi/3). For each point on the grey curve, its x-coordinate is one-half the x-coordinate of the yellow curve.
Method two - shrink first then shift
Step 1. Draw the graph of y = sin x
Step 2. Draw the graph of y = sin 2x by shrink the x-coordinate of the graph y = sin x to one-half and keep the y-coordinate the same to get the graph of y = sin 2x.
Step 3. Draw the graph of y = sin (2x - pi/3) by shift the graph of y = sin 2x right pi/6 units to get the graph of y = sin (2x - pi/3).
Look the figure above, the yellow curve is the graph of y = sin 2x. The grey curve is the graph of y = sin (2x - pi/3). By shift the graph of y = sin 2x right pi/6 units, we get the graph of y = sin (2x - pi/3).
Method 3 - Five points
The five important points for y = sin x in one period are (0, 0), (pi/2, 1), (pi, 0), (3pi/2, -1) and (2pi, 0).
Now we will find the five important points for y = sin (2x - pi/3)
The first point, when 2x - pi/3 = 0, y = 0.
2x - pi/3 = 0
2x = pi/3
x = pi/6
so, the first point is (pi/6, 0)
The second point, when 2x - pi/3 = pi/2, y = 1
2x - pi/3 = pi/2
2x = pi/3 + pi/2
2x = 2pi/6 + 3pi/6 = 5pi/6
x = 5pi/12
so, the second point is (5pi/12, 1)
The third point, when 2x - pi/3 = pi, y = 0
2x - pi/3 = pi
2x = pi/3 + pi
2x = pi/3 + 3pi/3 = 4pi/3
x = 4pi/6 = 2pi/3
so, the third point is (2pi/3, 0)
The fourth point, when 2x - pi/3 = 3pi/2, y = -1
2x - pi/3 = 3pi/2
2x = pi/3 + 3pi/2
2x = 2pi/6 + 9pi/6 = 11pi/6
x = 11pi/12
so, the fourth point is (11pi/12, -1)
The fifth point, when 2x - pi/3 = 2pi, y = 0
2x - pi/3 = 2pi
2x = pi/3 + 2pi
2x = pi/3 + 6pi/3 = 7pi/3
x = 7pi/6
so, the fifth point is (7pi/6, 0)
In the figure above, the five important points of y = sin (2x - pi/3) are (pi/6, 0), (5pi/12, 1), (2pi/3, 0), (11pi/12, -1), (7pi/6, 0). Extended the graph for each pi interval in both direction to get the graph of y = sin (2x -pi/3).
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The Fundamental Theorem of Calculus Part 1
# The Fundamental Theorem of Calculus Part 1
We are now going to look at one of the most important theorems in all of mathematics known as the Fundamental Theorem of Calculus (often abbreviated as the F.T.C). Traditionally, the F.T.C. is broken up into two part. We will look at the first part of the F.T.C., while the second part can be found on The Fundamental Theorem of Calculus Part 2 page.
Theorem 1 (The Fundamental Theorem of Calculus Part 1): If a function $f$ is continuous on the interval $[a, b]$, such that we have a function $g(x) = \int_a^x f(t) \: dt$ where $a ≤ x ≤ b$, and $g$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then $g'(x) = f(x)$.
• Proof: Suppose that $x \in (a, b)$, and $(x + h) \in (a, b)$. Then:
(1)
\begin{align} g(x + h) - g(x) = \int_a^{x + h} f(t) \: dt - \int_a^x f(t) \: dt \end{align}
• We can take the first integral and split it up such that:
(2)
\begin{align} \quad g(x + h) - g(x) = \left ( \int_a^x f(t) \: dt + \int_x^{x + h} f(t) \: dt \right ) - \int_a^x f(t) \: dt \\ \quad g(x + h) - g(x) = \int_x^{x + h} f(t) \: dt \end{align}
• Now let's divide both sides by $h > 0$ to obtain:
(3)
\begin{align} \frac{g(x + h) - g(x)}{h} = \frac{1}{h} \cdot \int_x^{x + h} f(t) \: dt \end{align}
• By The Extreme Value Theorem, there exists values $x = u$, and $x =v$ such that $f(u) = m$, and $f(v) = M$, where $m$ is the absolute minimum on the interval $[x, x+h]$, and $M$ is the absolute maximum on the interval $[x, x+h]$. Hence it follows that
(4)
\begin{align} f(u) \: h ≤ \int_x^{x + h} f(t) \: dt ≤ f(v) \: h \end{align}
• Essentially, $\int_x^{x + h} f(t) \: dt$ has an area less or equal to that area contained within the box formed by $Mh - mh$. Now let's divide each term in our inequality by $h > 0$ (there is no problem here since $h ≠ 0$). We obtain the following inequality:
(5)
\begin{align} f(u) ≤ \frac{1}{h} \int_x^{x + h} f(t) \: dt ≤ f(v) \end{align}
• Substituting back, we obtain:
(6)
\begin{align} f(u) ≤ \frac{g(x+h) - g(x)}{h} ≤ f(v) \end{align}
• As $h \to 0$, $u \to x$ (since $u$ is contained on the interval $[x, x + h]$) and $v \to x$ (since $v$ is also contained on the interval $[x, x + h]$). Hence it follows that $\lim_{h \to 0} \frac{g(x + h) - g(x)}{h} = g'(x) = f(x)$
(7)
\begin{align} \lim_{h \to 0} f(x) ≤ \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} ≤ \lim_{h \to 0} f(x) \\ \lim_{u \to x} f(u) ≤ \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} ≤ \lim_{v \to x} f(v) \\ f(x) ≤ g'(x) ≤ f(x) \\ f(x) = g'(x) \end{align}
• Hence it follows that $\frac{d}{dx} \int_a^x f(t) \: dt = f(x)$. $\blacksquare$
Note: The process works analogously for h < 0.
## Example 1
Differentiate the function $g(x) = \int_{0}^{x} \sqrt{3 + t} \: dt$.
We note that $f(t) = \sqrt{3 + t}$ is a continuous function, and by the fundamental theorem of calculus part 1, it follows that:
(8)
\begin{align} \frac{d}{dx} g(x) = \sqrt{3 + x} \end{align}
## Example 2
Differentiate the function $g(x) = \int_{2}^{x} 4t^2 + 1 \: dt$.
Once again, $f(t) = 4t^2 + 1$ is a continuous function, and by the fundamental theorem of calculus part, it follows that:
(9)
\begin{align} \frac{d}{dx} g(x) = 4x^2 + 1 \end{align}
## Example 3
Differentiate the function $g(x) = \int_{1}^{x^3} 3t + \sin t \: dt$.
We know that $3t + \sin t$ is a continuous function. We should note that we must apply the chain rule however, since our function is a composition of two parts, that is $m(x) = \int_{1}^{x} 3t + \sin t \: dt$ and $n(x) = x^3$, then $g(x) = (m \circ n)(x)$. Thus, applying the chain rule we obtain that:
(10)
\begin{align} \frac{d}{dx} g(x) = [3x^4 + \sin (x^4)] \cdot 4x^3 \end{align}
## Example 4
Differentiate the function $g(x) = \int_{x}^{x^3} 2t^2 + 3 \: dt$.
We will first begin by splitting the integral as follows, and then flipping the first one as shown:
(11)
\begin{align} g(x) = \int_{x}^{0} 2t^2 + 3 \: dt + \int_{0}^{x^3} 2t^2 + 3 \: dt \\ \: g(x) = -\int_{0}^{x} 2t^2 + 3 \: dt + \int_{0}^{x^3} 2t^2 + 3 \: dt \end{align}
Since $2t^2 + 3$ is a continuous function, we can apply the fundamental theorem of calculus while being mindful that we have to apply the chain rule to the second integral, and thus:
(12)
\begin{align} \frac{d}{dx} g(x) = -(2x^2 + 3) + (2(x^3)^2 + 3) \cdot 3x^2 \end{align}
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# Lesson Video: Derivatives of Inverse Trigonometric Functions Mathematics • Higher Education
In this video, we will learn how to find the derivatives of the inverses of trigonometric functions.
15:56
### Video Transcript
In this video, weβll learn how to find the derivatives of the inverses of trigonometric functions. We will learn how to do this by using implicit differentiation. And thus, it is important that you understand how to apply the chain rule, if not have a thorough understanding of how implicit differentiation works, before watching this video. After deriving the derivatives of the inverse trigonometric functions, weβll consider the application of these derivatives to more complicated inverse trigonometric functions.
Before we look at finding the derivative of our inverse trigonometric functions, letβs just quickly consider the function π¦ equal sin of π₯. Remember, π₯ is a real number. And of course, since weβre performing calculus with a trigonometric function, we need to make sure this is measured in radians. For this function, we can say that π₯ is equal to the inverse sin of π¦. This subscript negative one denotes the inverse function. Remember though without restricting the domain of arc sin of π₯ or inverse sin of π₯, the function is going to be many to one. We, therefore, restrict the domain for π of π₯ equals inverse sin of π₯. And we say that π₯ has to be greater than or equal to negative one and less than or equal to one.
If we go back to the function π₯ equals the inverse sin of π¦, we can, therefore, see that π₯ will take values greater than or equal to negative π by two and less than or equal to π by two. Weβll also need to recall the chain rule. This says that if π¦ is some function in π’ and π’ is some differentiable function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. Weβll now use everything weβve seen here to find the derivative of the inverse sine function.
Find the derivative of the inverse sine function with respect to π₯.
Weβre differentiating the inverse sin of π₯ with respect to π₯. So weβre going to begin by letting π¦ be equal to the inverse sin of π₯. Then we can say that π₯ must be equal to sin of π¦. Weβre going to differentiate both sides of this equation with respect to π₯. So we say that d by dπ₯ of π₯ is equal to d by dπ₯ of sin of π¦. Well, the derivative of π₯ with respect to π₯ is quite straight forward; itβs one. But weβre going to need to use implicit differentiation which is a special case of the chain rule to differentiate sin π¦ with respect to π₯.
The derivative of sin π¦ with respect to π¦ is cos π¦. So the derivative of sin π¦ with respect to π₯ is cos π¦ times the derivative of π¦ with respect to π₯ which is just dπ¦ by dπ₯. So we currently see that one is equal to cos of π¦ times dπ¦ by dπ₯. We divide both sides of this equation by cos π¦ to form an equation for the derivative. And we see that dπ¦ by dπ₯ is equal to one over cos π¦. Now, weβve got a bit of a problem. We do want an expression for the derivative in terms of π₯ not π¦.
And remember, we said that π₯ was equal to sin of π¦. So weβll use the identity cos squared π plus sin squared π equals one. And Iβve replaced π with π¦. Weβll subtract sin squared π¦ from both sides of the equation. And then weβll take the square root of both sides. And we see that cos of π¦ is equal to the positive and negative square root of one minus sin squared π¦. Remember, inverse sin is restricted to the closed interval negative π by two to π by two.
By our definition, that means π¦ must be greater than or equal to [negative] π by two and less than or equal to π by two which, in turn, means cos of π¦ must be greater than or equal to zero and less than or equal to one. And thatβs because in the interval π¦ is greater than or equal to negative π by two and less than or equal to π by two. The smallest value cos of π¦ weβll take is zero. And the greatest value is one. And what this means here is weβre going to take the positive square root of one minus sin squared π¦ only.
We can now replace sin of π¦ with π₯. And we see that cos of π¦ is equal to the square root of one minus π₯ squared. And, therefore, dπ¦ by dπ₯ equals one over the square root of one minus π₯ squared. And we found the derivative of the inverse sin of π₯. Itβs one over the square root of one minus π₯ squared for values of π₯ in the range π₯ is greater than negative one and less than one.
In our next example, weβll consider an alternative method that will help us find the derivative of the inverse cosine function.
This time, weβre going to need to know the inverse function theorem. This says that if π is a differentiable function with a continuous inverse π prime and π prime of π is not equal zero, then not only is π invertible but it has a differentiable inverse. Such that the derivative of the inverse of π at some π equals π of π is equal to one over the derivative of π at π. This is sometimes written simply as dπ₯ by dπ¦ is equal to one over dπ¦ by dπ₯. Letβs see how this might help us when differentiating the inverse cosine function.
Find the derivative of the inverse cos of π₯ over π with respect to π₯, where π is not equal to zero.
Weβll begin by letting π¦ be equal to the inverse cos of π₯ over π. This can be alternately written as π₯ over π equals cos of π¦. And we can then multiply both sides by π. And we see that π₯ is equal to π times cos of π¦. Weβre going to differentiate our expression for π₯ with respect to π¦. In other words, weβre going to find dπ₯ by dπ¦. Weβll use the general result that the derivative of cos of π₯ with respect to π₯ is negative sin of π₯. And we see that dπ₯ by dπ¦ must be equal to negative π sin of π¦.
Now, before we perform the next step we need to recall the fact that for the inverse trigonometric functions, we restrict their domains. And we know that the domain of the inverse cos of π₯ or our cos of π₯ is greater than or equal to zero and less than or equal to π. This means that π¦ must be greater than or equal to zero and less than or equal to π. Now, weβre going to use the inverse function theorem. So weβre going to take values of π¦ greater than zero and less than π, such that sin of π¦ is not equal to zero.
Using this criteria, we can use dπ₯ by dπ¦ equals one over dπ¦ by dπ₯ which can be rearranged to say that dπ¦ by dπ₯ equals one over dπ₯ by dπ¦. And we see that, for our case, dπ¦ by dπ₯ equals one over negative π sin of π¦. And we have an expression for the derivative in terms of π¦. Remember, we want this to be in terms of π₯. We said that π₯ over π equals cos of π¦. So weβll use the fact that sin squared π¦ plus cos squared π¦ equals one and rearrange this to say that sin π¦ equals plus or minus the square root of one minus cos squared π¦. When π¦ is between zero and π, sin π¦ is greater than zero. So in fact, weβre only interested in the positive root.
So weβll replace this in our expression for the derivative. And we get negative one over π times the square root of one minus cos squared π¦. We then replace cos π¦ with π₯ over π and change π₯ over π all squared to π₯ squared over π squared. And then we bring π into the square root. And we see that dπ¦ by dπ₯ is equal to negative one over the square root of π squared minus π₯ squared. So d by dπ₯ of the inverse cos of π₯ over π is equal to negative one over the square root of π squared minus π₯ squared for values of π₯ between negative π and π.
In our next example, weβll consider how we might apply the process thatβs used so far to find the derivative of the inverse tangent function.
Find an expression for the derivative of π¦ equals the inverse tan of ππ₯ in terms of π₯.
Since π¦ is equal to the inverse tan of ππ₯, we can write ππ₯ as being equal to tan π¦. Weβre going to use implicit differentiation to find the derivative of both sides of this equation. The derivative of ππ₯ with respect to π₯ is simply π. And the derivative of tan π¦ with respect to π₯ is equal to the derivative of tan π¦ with respect to π¦ times the derivative of π¦ with respect to π₯. The derivative of tan π₯ is sec squared π₯. And the derivative of π¦ with respect to π₯ is dπ¦ by dπ₯.
So we see that π is equal to sec squared π¦ times dπ¦ by dπ₯. Dividing through by sec squared π¦ and we see that dπ¦ by dπ₯ equals π over sec squared π¦. Weβre going to need to represent our equation for the derivative in terms of π₯. So weβll use this trigonometric identity. One plus tan squared π₯ equals sec squared π₯. This means we can write dπ¦ by dπ₯ as π over one plus tan squared π¦. And then we replaced tan π¦ with ππ₯. And we see the expression for the derivative of π¦ equals the inverse tan of ππ₯ is π over one plus ππ₯ squared.
Similar rules can be applied to help us find the derivative of the inverse cotangent function. We find that the derivative of the inverse of cotangent of π of π₯ is equal to negative π over one plus ππ₯ squared. The inverse cosecant and secant functions are a little more unusual. So weβll consider next how to find the derivative of the inverse cosecant function.
Find d by dπ₯ of the inverse cosecant of π₯.
We begin by letting π¦ be equal to the inverse cosecant of π₯. And this means we can rewrite this. And we can say that π₯ is equal to the cosecant of π¦.
Weβre next going to use implicit differentiation to find the derivative of both sides of this equation. The derivative of π₯ with respect to π₯ is simply one. Then the derivative of cosec π¦ with respect to π₯ is equal to the derivative of cosec π¦ with respect to π¦ times dπ¦ by dπ₯. And the derivative of cosec π¦ with respect to π¦ is negative cosec π¦ cot π¦. So we see that one is equal to negative cosec π¦ cot π¦ times dπ¦ by dπ₯.
Now, we know that, for the inverse cosecant function, π¦ must be greater than negative π by two and less than π by two and not equal to zero. Using these restrictions cosec π¦ cot π¦ cannot be equal to zero. So we can divide through by negative cosec π¦ cot π¦. And we see that dπ¦ by dπ₯ is as shown. We want to represent our equation for the derivative in terms of π₯. So weβll use this trigonometric identity cot squared π¦ plus one equals cosec squared π¦. And we can rewrite this to say that cot of π¦ is equal to the positive and negative square root of cosec squared π¦ minus one.
We put this into the equation for the derivative in place of cot of π¦. And we then use the fact that π₯ is equal to cosec π¦. But we are going to need to make a decision on the sine of the derivative. And it can help here to look at the graph of the inverse cosecant function. Notice how, for all values of π₯ in the range of the function, the derivative of the slope of the tangent is negative.
And we, therefore, use the absolute value to ensure that our derivative is always negative. We say that dπ¦ by dπ₯ is equal to the negative of the absolute value of one over π₯ times the square root of π₯ squared minus one. Since one and the square root of π₯ squared minus one are always positive, we can rewrite this as shown. So the derivative of the inverse cosecant function π₯ with respect to π₯ is negative one over the modulus or absolute value of π₯ times the square root of π₯ squared minus one.
A similar process can be applied to help us find the derivative of the inverse secant function. And we have the derivatives of all of the inverse trigonometric functions we require. Itβs useful to commit these results to memory but also be prepared to derive them where necessary. Weβll now have a look at the application of these results.
Evaluate the derivative of the inverse cotangent of one over π₯ with respect to π₯.
Here, we have a function of a function or a composite function. Weβre, therefore, going to need to use the chain rule to find the derivative. This says that if π and π are differentiable functions such that π¦ is π of π’ and π’ is π of π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. Weβll let π’ be equal to one over π₯. Then π¦ is equal to the inverse cot of π’. To apply the chain rule, we need to find the derivative of both of these functions. And with π’ it can be useful to write it as π₯ to the negative one.
Then dπ’ by dπ₯ is negative π₯ to negative two or negative one over π₯ squared. We can then use the general derivative of the inverse cotangent function. And we see that dπ¦ by dπ’ is equal to negative one over one plus π’ squared. dπ¦ by dπ₯ is the product of these. Itβs negative one over π₯ squared times negative one over one plus π’ squared.
We can replace π’ with one over π₯ and then multiply through. And we see that the derivative of the inverse cotangent of one over π₯ with respect to π₯ is one over π₯ squared plus one.
Did you notice that the derivative of the inverse cotangent of one over π₯ is equal to the derivative of the tangent of π₯? This is, in fact, no accident. And we can use the identity the inverse cotangent of one over π₯ equals the inverse tan of π₯. That could have saved us a little bit more time in this previous example.
Evaluate the derivative of the inverse sin of the square root of one minus π₯ squared with respect to π₯.
Here, we have a function of a function or a composite function. So weβll use the chain rule to find its derivative. This says that if π¦ is some function in π’ and π’ is some function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. Weβll let π’ be equal to the square root of one minus π₯ squared. Which can, of course, alternatively be written as one minus π₯ squared to the power of one-half. Then π¦ is equal to the inverse sin of π’. To apply the chain rule, weβre going to need to find the derivative of both of these functions. The derivative of the inverse sin of π’ with respect to π’ is one over the square root of one minus π’ squared.
And we can use the general power rule to find the derivative of one minus π₯ squared to the power of one-half. Itβs a half times one minus π₯ squared to the negative one-half times the derivative of the bit inside the brackets which is negative two π₯. That can be written as negative π₯ times one minus π₯ squared to the power of negative one-half.
dπ¦ by dπ₯ is, therefore, negative π₯ over the square root of one minus π₯ squared times one over the square root of one minus π’ squared. We can replace π’ with one minus π₯ squared to the power of one-half. And the second fraction becomes one over the square root of one minus one minus π₯ squared. This further simplifies to one over π₯. And we divide through by π₯. And we see that the derivative of our function is negative one over the square root of one minus π₯ squared.
Once again, weβve stumbled across an interesting result. That is that the derivative of the inverse sin of the square root of one minus π₯ squared is equal to the derivative of the inverse cos of π₯. This comes from the identity of the inverse sin of the square root of one minus π₯ squared is equal to the inverse cos of π₯, the values of π₯ between zero and one. Being familiar with this result could have reduced the amount of what we needed to do in this example.
In this video, weβve seen that we can use implicit differentiation or the inverse function theorem to derive the formulae for the derivatives of inverse trigonometric functions. We saw that the derivatives of the inverse trigonometric functions are as shown. And we also saw that being familiar with certain trigonometric identities can sometimes significantly reduce the process of finding these derivatives.
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INSTRUCTORS Carleen Eaton Grant Fraser
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• ## Related Books
0 answersPost by Samvel Karapetyan on July 8, 2012Great work you're doing sweet doctor. 1 answerLast reply by: Dr Carleen EatonMon Feb 6, 2012 11:18 PMPost by Edmund Mercado on February 5, 2012Dr. Eaton:At 22:25, I think you meant to write -3.5 instead of -3/2 for -7/2.
### Solving Rational Equations and Inequalities
• To solve a rational equation, multiply each term on both sides by the LCD of all the denominators in the equation. Then solve the resulting equation, which has no fractions.
• Always check for extraneous solutions – values that make one or more of the denominators equal to 0. Exclude such values from the solution set. Better yet, before solving the equation, determine the values that must be excluded by setting each denominator equal to 0 and solving. Then you will recognize an extraneous solution as soon as it appears.
### Solving Rational Equations and Inequalities
Solve:[5/3] − [1/3] = [(n + 1)/(3n + 6)]
• List the prime factors of 3n + 6 and 3 to find the LCD. Factoring may be required.
• 3 =
• 3n + 6 =
• 3 = 3
• 3n + 6 = 3(n + 2)
• Use each prime factor the greatest number of times it appears in each of the factorization
• LCD =
• LCD = 3(n + 2)
• Change each rational expression into an equivalent expression with the LCD.
• [5/3] − [1/3] = [(n + 1)/(3n + 6)] → [4/3] = [(n + 1)/(3n + 6)]
• [4/3]( [(3(n + 2))/(3(n + 2))] ) = [(n + 1)/(3(n + 2))]
• [4/]( [((n + 2))/(3(n + 2))] ) = [(n + 1)/(3(n + 2))]
• [(4(n + 2))/(3(n + 2))] = [((n + 1))/(3(n + 2))]
• Cancel out (n + 2)
• [4/3] = [((n + 1))/(3(n + 2))]
• Cross Multiply
• 3(n + 1) = 12(n + 2)
• 3n + 3 = 12n + 24
• − 21 = 9n
n = [( − 21)/9] = − [7/3]
Solve:[1/(2x2 − 14x + 12)] + [5/(2x − 2)] = [1/(x − 6)]
• List the prime factors of 2x2 − 14x + 12, 2x − 2 and x − 6 to find the LCD. Factoring may be required.
• 2x2 − 14x + 12 =
• 2x − 2 =
• x − 6 =
• 2x2 − 14x + 12 = 2*(x − 6)(x − 1)
• 2x − 2 = 2*(x − 1)
• x − 6 = (x − 6)
• Use each prime factor the greatest number of times it appears in each of the factorization
• LCD =
• LCD = 2*(x − 6)(x − 1)
• Change each rational expression into an equivalent expression with the LCD.
• [1/(2x2 − 14x + 12)] + [5/(2x − 2)] = [1/(x − 6)] → [1/(2(x − 1)(x − 6))] + [5/(2(x − 1))] = [1/(x − 6)]
• [1/(2(x − 1)(x − 6))] + [5/(2(x − 1))]( [((x − 6))/((x − 6))] ) = [1/(x − 6)]( [(2(x − 1))/(2(x − 1))] )
• [1/(2(x − 1)(x − 6))] + [(5(x − 6))/(2(x − 1)(x − 6))] = [(2(x − 1))/(2(x − 1)x − 6)]
• Add the left side of the equation
• [(1 + 5(x − 6))/(2(x − 1)(x − 6))] = [(2(x − 1))/(2(x − 1)(x − 6))]
• Subtract the right side of the eqution
• [(5x − 29)/(2(x − 1)(x − 6))] − [(2(x − 1))/(2(x − 1)(x − 6))] = 0
• [(5x − 29)/(2(x − 1)(x − 6))] − [(2x − 2)/(2(x − 1)(x − 6))] = 0
• Subtract
• [((5x − 29) − (2x − 2))/(2(x − 1)(x − 6))] = 0 → [((5x − 29) − 2x + 2))/(2(x − 1)(x − 6))] = 0 → [(3x − 27)/(2(x − 1)(x − 6))] = 0
• [(3x − 27)/(2(x − 1)(x − 6))] = 0
• Multiply both sides of the equation by the denominator, simplify
• ( 2(x − 1)(x − 6) )[(3x − 27)/(2(x − 1)(x − 6))] = 0(2(x − 1)(x − 6))
• 3x − 27 = 0
• 3x = 27
x = 9
Solve:[x/(x − 6)] − [1/(3x − 18)] = [1/3]
• List the prime factors of x − 6;3x − 18 and 3 to find the LCD. Factoring may be required.
• x − 6 =
• 3x − 18 =
• 3 =
• x − 6 = (x − 6)
• 3x − 18 = 3(x − 6)
• 3 = 3
• Use each prime factor the greatest number of times it appears in each of the factorization
• LCD =
• LCD = 3(x − 6)
• Change each rational expression into an equivalent expression with the LCD.
• [x/(x − 6)] − [1/(3x − 18)] = [1/3] → [x/(x − 6)] − [1/(3(x − 6))] = [1/3]
• [x/(x − 6)]( [3/3] ) − [1/(3(x − 6))] = [1/3]( [(x − 6)/(x − 6)] )
• [3x/(3(x − 6))] − [1/(3(x − 6))] = [((x − 6))/(3(x − 6))]
• Add the left side of the equation
• [(3x − 1)/(3(x − 6))] = [((x − 6))/(3(x − 6))]
• Subtract the right side of the eqution
• [(3x − 1)/(3(x − 6))] − [((x − 6))/(3(x − 6))] = 0
• Subtract
• [((3x − 1) − (x − 6))/(3(x − 6))] = 0 → [((3x − 1) − x + 6)/(3(x − 6))] = 0 → [(2x + 5)/(3(x − 6))] = 0
• [(2x + 5)/(3(x − 6))] = 0
• Multiply both sides of the equation by the denominator, simplify
• ( 3(x − 6) )[(2x + 5)/(3(x − 6))] = 0(3(x − 6))
• 2x + 5 = 0
• 2x = − 5
x = − [5/2]
Solve:[1/(2x − 5)] − [1/(x − 2)] = [(x + 5)/(2x2 − 9x + 10)]
• List the prime factors of 2x − 5;x − 2 and 2x2 − 9x + 10 to find the LCD. Factoring may be required.
• 2x − 5 =
• x − 2 =
• 2x2 − 9x + 10 =
• 2x − 5 = (2x − 5)
• x − 2 = (x − 2)
• 2x2 − 9x + 10 = (2x − 5)(x − 2)
• Use each prime factor the greatest number of times it appears in each of the factorization
• LCD =
• LCD = (2x − 5)(x − 2)
• Change each rational expression into an equivalent expression with the LCD.
• [1/(2x − 5)] − [1/(x − 2)] = [(x + 5)/(2x2 − 9x + 10)] → [1/(2x − 5)] − [1/(x − 2)] = [(x + 5)/((2x − 5)(x − 2))]
• [1/(2x − 5)]( [(x − 2)/(x − 2)] ) − [1/(x − 2)]( [(2x − 5)/(2x − 5)] ) = [(x + 5)/((2x − 5)(x − 2))]
• [((x − 2))/((2x − 5)(x − 2))] − [((2x − 5))/((x − 2)(2x − 5))] = [(x + 5)/((2x − 5)(x − 2))]
• Subtract the left side of the equation
• [((x − 2) − (2x − 5))/((2x − 5)(x − 2))] = [(x + 5)/((2x − 5)(x − 2))]
• [(x − 2 − 2x + 5)/((2x − 5)(x − 2))] = [(x + 5)/((2x − 5)(x − 2))]
• [(3 − x)/((2x − 5)(x − 2))] = [(x + 5)/((2x − 5)(x − 2))]
• Subtract the right side of the eqution
• [(3 − x)/((2x − 5)(x − 2))] − [(x + 5)/((2x − 5)(x − 2))] = 0
• Subtract
• [(3 − x − (x + 5))/((2x − 5)(x − 2))] = 0 → [(3 − x − x − 5)/((2x − 5)(x − 2))] = 0 → [( − 2x − 2)/((2x − 5)(x − 2))] = 0
• [( − 2x − 2)/((2x − 5)(x − 2))] = 0
• Multiply both sides of the equation by the denominator, simplify
• ( (2x − 5)(x − 2) )[( − 2x − 2)/((2x − 5)(x − 2))] = 0((2x − 5)(x − 2))
• − 2x − 2 = 0
• − 2x = 2
x = − 1
Solve:[3/x] + [1/(x2 + x)] = [1/(x + 1)]
• List the prime factors of x, x2 + x and x + 1 to find the LCD. Factoring may be required.
• x =
• x2 + x =
• x + 1 =
• x = x
• x2 + x = x(x + 1)
• x + 1 = (x + 1)
• Use each prime factor the greatest number of times it appears in each of the factorization
• LCD =
• LCD = x(x + 1)
• Change each rational expression into an equivalent expression with the LCD.
• [3/x] + [1/(x2 + x)] = [1/(x + 1)] → [3/x] + [1/(x(x + 1))] = [1/(x + 1)]
• [3/x]( [((x + 1))/((x + 1))] ) + [1/(x(x + 1))] = [1/((x + 1))]( [x/x] )
• [(3(x + 1))/(x(x + 1))] + [1/(x(x + 1))] = [x/(x(x + 1))]
• Add the left side of the equation
• [(3(x + 1) + 1)/(x(x + 1))] = [x/(x(x + 1))]
• [(3x + 4)/(x(x + 1))] = [x/(x(x + 1))]
• Subtract the right side of the eqution
• [(3x + 4)/(x(x + 1))] − [x/(x(x + 1))] = 0
• Subtract
• [(3x + 4 − x)/(x(x + 1))] = 0
• [(2x + 4)/(x(x + 1))] = 0
• Multiply both sides of the equation by the denominator, simplify
• ( x(x + 1) )[(2x + 4)/(x(x + 1))] = 0((x(x + 1))
• 2x + 4 = 0
• 2x = − 4
x = − 2
Solve:[6/x] − [1/(x2 − 6x)] = [1/(x − 6)]
• List the prime factors of x, x2 − 6x and x − 6 to find the LCD. Factoring may be required.
• x =
• x2 − 6x =
• x − 6 =
• x = x
• x2 − 6x = x(x − 6)
• x − 6 = (x − 6)
• Use each prime factor the greatest number of times it appears in each of the factorization
• LCD =
• LCD = x(x − 6)
• Change each rational expression into an equivalent expression with the LCD.
• [6/x] − [1/(x2 − 6x)] = [1/(x − 6)] → [6/x] − [1/(x(x − 6))] = [1/(x − 6)]
• [6/x]( [((x − 6))/((x − 6))] ) − [1/(x(x − 6))] = [1/(x − 6)]( [x/x] )
• [(6(x − 6))/(x(x − 6))] − [1/(x(x − 6))] = [x/(x(x − 6))]
• Add the left side of the equation
• [(6(x − 6) − 1)/(x(x − 6))] = [x/(x(x − 6))]
• [(6x − 37)/(x(x − 6))] = [x/(x(x − 6))]
• Subtract the right side of the eqution
• [(6x − 37)/(x(x − 6))] − [x/(x(x − 6))] = 0
• Subtract
• [(6x − 37 − x)/(x(x − 6))] = 0
• [(5x − 37)/(x(x − 6))] = 0
• Multiply both sides of the equation by the denominator, simplify
• ( x(x − 6) )[(5x − 37)/(x(x − 6))] = 0((x(x − 6))
• 5x − 37 = 0
• 5x = 37
x = [37/5]
Solve:[(x − 6)/(x − 1)] + [4/(x2 − 1)] = 1
• List the prime factors of x − 1, x2 − 1 to find the LCD. Factoring may be required.
• x − 1 =
• x2 − 1 =
• x − 1 = (x − 1)
• x2 − 1 = (x − 1)(x + 1)
• Use each prime factor the greatest number of times it appears in each of the factorization
• LCD =
• LCD = (x + 1)(x − 1)
• Change each rational expression into an equivalent expression with the LCD.
• [(x − 6)/(x − 1)] + [4/(x2 − 1)] = 1
• [(x − 6)/(x − 1)]( [(x + 1)/(x + 1)] ) + [4/((x − 1)(x + 1))] = 1( [((x − 1)(x + 1))/((x − 1)(x + 1))] )
• [((x − 6)(x + 1))/((x − 1)(x + 1))] + [4/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
• Add the left side of the equation
• [((x − 6)(x + 1))/((x − 1)(x + 1))] + [4/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
• [((x − 6)(x + 1) + 4)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
• [(x2 − 5x − 2)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
• Subtract the right side of the eqution
• [(x2 − 5x − 2)/((x − 1)(x + 1))] − [((x − 1)(x + 1))/((x − 1)(x + 1))] = 0
• Subtract
• [(x2 − 5x − 2 − (x2 − 1))/((x − 1)(x + 1))] = 0
• [( − 5x − 1)/((x − 1)(x + 1))] = 0
• Multiply both sides of the equation by the denominator, simplify
• ( (x − 1)(x + 1) )[( − 5x − 1)/((x − 1)(x + 1))] = 0((x − 1)(x + 1))
• − 5x − 1 = 0
• − 5x = 1
x = − [1/5]
Solve: [x/(x − 6)] − [1/(3x − 18)] < [1/3]
• Identify excluded values
• x − 6 = 0
• x = 6
• 3x − 18 = 0
• 3x = 18
• x = 6
• You cannot have 6 as part of the solution set
• Solve the related equation
• [x/(x − 6)] − [1/(3x − 18)] = [1/3]
• [x/(x − 6)] − [1/(3x − 18)] = [1/3] → [x/(x − 6)] − [1/(3(x − 6))] = [1/3]
• [x/(x − 6)]( [3/3] ) − [1/(3(x − 6))] = [1/3]( [(x − 6)/(x − 6)] )
• [3x/(3(x − 6))] − [1/(3(x − 6))] = [((x − 6))/(3(x − 6))]
• Add the left side of the equation
• [(3x − 1)/(3(x − 6))] = [((x − 6))/(3(x − 6))]
• Subtract the right side of the eqution
• [(3x − 1)/(3(x − 6))] − [((x − 6))/(3(x − 6))] = 0
• Subtract
• [((3x − 1) − (x − 6))/(3(x − 6))] = 0 → [((3x − 1) − x + 6)/(3(x − 6))] = 0 → [(2x + 5)/(3(x − 6))] = 0
• [(2x + 5)/(3(x − 6))] = 0
• Multiply both sides of the equation by the denominator, simplify
• ( 3(x − 6) )[(2x + 5)/(3(x − 6))] = 0(3(x − 6))
• 2x + 5 = 0
• 2x = − 5
• x = − [5/2]
• Test different intervals to check for the solution set
• Test: x=−4
[x/(x − 6)] − [1/(3x − 18)] < [1/3]
[( − 4)/( − 4 − 6)] − [1/(3( − 4) − 18)] < [1/3]
[4/10] + [1/20] < [1/3]
Not True
• Test: x=0
[x/(x − 6)] − [1/(3x − 18)] < [1/3]
[0/(0 − 6)] − [1/(3(0) − 18)] < [1/3]
[1/18] < [1/3]
True
• Test: x=10
[x/(x − 6)] − [1/(3x − 18)] < [1/3]
[10/(10 − 6)] − [1/(3(10) − 18)] < [1/3]
[10/4] − [1/12] < [1/3]
Not True
• Therefore, the solution is
− [5/2] < x < 6
Solve: [(x − 6)/(x − 1)] + [4/(x2 − 1)] ≤ 1
• Identify excluded values
• x − 1 = 0
• x = 1
• x2 − 1 = 0
• x = 1;x = − 1
• You cannot have 1 and − 1 as part of the solution set
• Solve the related equation
• [(x − 6)/(x − 1)] + [4/(x2 − 1)] = 1
• [(x − 6)/(x − 1)]( [(x + 1)/(x + 1)] ) + [4/((x − 1)(x + 1))] = 1( [((x − 1)(x + 1))/((x − 1)(x + 1))] )
• [((x − 6)(x + 1))/((x − 1)(x + 1))] + [4/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
• Add the left side of the equation
• [((x − 6)(x + 1))/((x − 1)(x + 1))] + [4/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
• [((x − 6)(x + 1) + 4)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
• [(x2 − 5x − 2)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
• Subtract the right side of the eqution
• [(x2 − 5x − 2)/((x − 1)(x + 1))] − [((x − 1)(x + 1))/((x − 1)(x + 1))] = 0
• Subtract
• [(x2 − 5x − 2 − (x2 − 1))/((x − 1)(x + 1))] = 0
• [( − 5x − 1)/((x − 1)(x + 1))] = 0
• Multiply both sides of the equation by the denominator, simplify
• ( (x − 1)(x + 1) )[( − 5x − 1)/((x − 1)(x + 1))] = 0((x − 1)(x + 1))
• − 5x − 1 = 0
• − 5x = 1
• x = − [1/5]
• Test different intervals to check for the solution set
• Test: x=−2
[(x − 6)/(x − 1)] + [4/(x2 − 1)] ≤ 1
[( − 2 − 6)/( − 2 − 1)] + [4/(( − 2)2 − 1)] ≤ 1
[8/3] + [4/3] ≤ 1
Not True
• Test: x=−[1/2]
[(x − 6)/(x − 1)] + [4/(x2 − 1)] ≤ 1
[( − [1/2] − 6)/( − [1/2] − 1)] + [4/(( − [1/2])2 − 1)] ≤ 1
[13/3] − [16/3] ≤ 1
True
• Test: x=0
[(x − 6)/(x − 1)] + [4/(x2 − 1)] ≤ 1
[(0 − 6)/(0 − 1)] + [4/(02 − 1)] ≤ 1
6 − 4 ≤ 1
Not True
• Test: x=2
[(x − 6)/(x − 1)] + [4/(x2 − 1)] ≤ 1
[(2 − 6)/(2 − 1)] + [4/(22 − 1)] ≤ 1
− 4 + [4/3] ≤ 1
True
• Therefore, the solution is
− 1 < x ≤ − [1/5] and x > 1
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Solving Rational Equations and Inequalities
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Rational Equations 0:15
• Example: Algebraic Fraction
• Least Common Denominator
• Example: Simple Rational Equation
• Example: Solve Rational Equation
• Extraneous Solutions 9:31
• Doublecheck
• No Solution
• Example: Extraneous
• Rational Inequalities 14:01
• Excluded Values
• Solve Related Equation
• Find Intervals
• Use Test Values
• Example: Rational Inequality
• Example: Rational Inequality 2
• Example 1: Rational Equation 28:50
• Example 2: Rational Equation 33:51
• Example 3: Rational Equation 38:19
• Example 4: Rational Inequality 46:49
### Transcription: Solving Rational Equations and Inequalities
Welcome to Educator.com.0000
In today's lesson, we are going to continue on with talking about rational expressions.0003
this time extending our discussion to solving rational equations and rational inequalities.0007
The definition of a rational equation is an equation that contains one or more rational expressions.0016
And remember that rational expressions are algebraic fractions.0022
For example, a rational equation could be something like this: (x2 - 3x + 9)/(x2 - 36) + 3/(x + 5) = 1/2.0027
The technique for solving rational equations is going to be to eliminate the fractions.0046
And that can be done by multiplying each term of the equation by the least common denominator of all the fractions in the equation.0051
We are going to find the least common denominator, and then multiply each term in the equation by that expression.0059
Recall, from previous lectures: we discussed the least common denominator.0070
And, if necessary, go back and review that, because we are going to be applying that idea to solve rational equations.0074
Let's look at another, simpler example, and then go ahead and solve it.0082
2 divided by (x + 3), minus 1/2x, equals 1/x.0087
Here is a rational equation, and according to this technique, what I need to do is find the least common denominator of all the fractions in the equation.0095
Recall that the LCD is the product of all the unique factors in the denominator, to the highest power that they are present in any one polynomial.0106
Therefore, if I look at the denominators, they are already factored out; that makes this very simple.0120
I have (x + 3), 2x, and x; therefore, the LCD would be (x + 3)...that is a unique factor; in this second denominator,0127
I have 2 times x; so 2 and x are both factors; and then I have x--however, x is already represented here.0140
And I only need to represent each unique factor the highest number of times that it is present in any one of the denominators.0152
So, x is present here once, and it is present here once; so I only have to represent it here once.0163
If I had x2, for example, here, then I would have written this as 2x2(x + 3).0167
Once I have my least common denominator, I am going to go back and follow this technique, multiplying each term of the equation by the LCD.0174
And then, we will see what happens.0182
For the LCD, 2x times (x + 3), times 2 divided by (x + 3), minus 2x times (x + 3), times 1/2x, equals 1/x times 2x(x + 3).0184
Let's start canceling out to simplify: the (x + 3)'s cancel here, leaving me with 2x times 2, minus...0212
here, the 2x's cancel out, leaving me with (x + 3) times 1, or just (x + 3), equals...0222
here the x's cancel out, leaving 2 times (x + 3).0232
At this point, you can see that the fractions have been eliminated, which was exactly my goal.0239
So now, I am just going to set about simplifying this.0244
2x times 2 is 4x; minus x, minus 3, equals 2x + 6.0247
All I have here now is a linear equation, and I just need to solve for x.0257
4x - x is 3x; minus 3 equals 2x + 6; I am going to subtract 2x from both sides: that is going to give me x;0261
and at the same time, I am going to add 3 to both sides; so moving the variables to the left,0274
and the constants to the right, gives me x = 9.0279
And I always want to just double-check and make sure that this is not going to cause the denominator to become 0, because that is not allowed.0287
And we will talk more about values that would cause the denominator to be 0 in a few minutes.0300
But for right now, I can see that 92 - 36 would not give me 0, and 9 + 5 would not give me 0.0304
So, this is a valid solution.0311
Again, the technique is just to find the LCD and multiply each term in the rational equation by the LCD, in order to eliminate the fraction.0314
Now, it actually would have been possible to solve this by converting all of these rational expressions,0323
all of these fractions, to a common denominator, and then just adding and subtracting.0330
But that is actually a lot more work: it is much easier to go about it this way and just get rid of the fractions.0334
Let's do another example to illustrate this technique.0341
5/(x2 + x - 2) + 4/(x - 1) = 1/(x + 2).0346
Recall that the first step is to find the LCD.0357
In order to do that, I am going to go up here and factor out each of the denominators.0360
x2 + x - 2: this is going to factor to (x + something) (x - something).0366
And I want factors of 2 that add up to the coefficient of 1 in front of the x.0375
Therefore, I am going to put the 2 here and 1 here; and this will give me x2 - x + 2x, which will give me x;0380
and then, the last term is - 2; so that checks out.0392
x - 1 just stays as x - 1; and x + 2 is x + 2.0397
So, the LCD: I have this factor, x + 2, and the highest number of times it is present in any one denominator,0403
in any one polynomial, is once, so I can leave it as (x + 2).0410
(x - 1) is also present only once in each of these two, so I am going to represent it once.0416
The LCD is (x + 2) (x - 1).0422
I am rewriting this with the denominators in factored form, (x + 2) (x - 1), and leaving some room to multiply by the LCD.0426
4/(x - 1) = 1/(x + 2).0440
I am going to multiply each term by the LCD: (x + 2) (x - 1).0448
Again, here, I am multiplying this by (x + 2) (x - 1); and finally, on the right side, (x + 2) (x - 1).0456
The goal is to get rid of the fractions, so let's check out if that worked.0469
These cancel; both factors cancel, leaving behind only a 5.0474
Right here, the (x - 1)s cancel, leaving behind 4 times (x + 2).0478
On the right side, the (x + 2)s cancel, leaving behind 1 times (x - 1), or simply (x - 1).0486
Now, I am left with a linear equation that I can solve by isolating x.0494
5 + (4x + 8) = x - 1.0498
Combining the 8 and the 5 is going to give me 4x + 13 = x - 1.0505
I am going to subtract x from both sides to combine the variables on the left, so 4x - x is 3x + 13 = -1.0512
Now, I am going to go ahead and subtract 13 from both sides to give me 3x = -14.0526
And then, dividing both sides by 3 gives me x = -14/3, which I can either leave like that or convert to a mixed number, -4 2/3.0539
So again, the technique to solve a rational equation is to find the least common denominator,0553
multiply each term in the equation by the LCD to get rid of the fractions, and then proceed as you would to solve an equation.0559
All right, I mentioned a little bit ago that you have to watch out and make sure that the solution that you have is valid.0572
Now, we are going to talk about that in detail.0578
And we used a technique there where we squared both sides of the equation.0590
But the result could be the introduction of solutions that were not valid--extraneous solutions.0594
Here, the same thing can happen: the technique that we are using will find a solution if one exists.0600
But extra solutions can also pop up that are not valid.0607
So, using this technique, you might find that you have one extraneous solution and one valid solution;0613
all valid solutions; all extraneous solutions; it could be any combination--you just have to check.0620
And if you find that none of the solutions are valid, it means that there is no solution,0626
because if a solution exists, you will find it with this technique; but it might be buried among extraneous solutions.0630
If you find that all of the solutions are extraneous, that means there is no solution to the equation.0638
Let's take an example: (x + 1) divided by (x2 + 2x + 15)/(x - 3) = 1/5x.0645
Let's just focus on possible extraneous solutions without solving out this whole thing.0664
To find the extraneous solutions, you need to think about values0670
that would make the denominator 0, because a denominator of 0 is undefined; it is not allowed.0674
So, if we ended up with a solution that made the denominator 0, that solution would be extraneous; it would be invalid.0679
Factoring out this first denominator, x2 + 2x - 15, is going to give me (x + something) (x - something).0686
And factors of 15 are 1 and 15, 3 and 5; and I am looking for factors that will add up to 2.0700
5 and -3 would add up to 2; so I am going to go with 5 here and -3 there.0710
Recall that the problem would occur if there was a value of x such that this polynomial became 0.0720
So, if I had (x + 5) (x - 3) = 0, a value of x that gave me that would be an extraneous solution.0729
And using the zero product property, you can say that, if either (x + 5) is 0 or (x - 3) is 0, this whole expression becomes 0.0736
So, I am going to set (x + 5) equal to 0 and (x - 3) equal to 0, and solve.0746
So, if x equals -5, or if x equals 3, either way, those would be values that are not allowable; and they would be excluded.0754
OK, so these are some excluded values; and these are based on that first denominator.0769
Looking at the second denominator: if x - 3 = 0, I would also end up with a 0 in the denominator, and that is not allowed.0777
So, solving for x, x = 3; however, I already have that accounted for here, so I don't have to write that again under my list.0789
5x--I cannot let 5x equal 0; and the situation where that would equal 0 is if x itself equaled 0, so another excluded value would be x = 0.0797
My excluded values are if x equals -5, 3, or 0.0810
One way to approach this is to find the excluded values first, when you are working with a rational equation.0818
And then, go through and solve, and when you get to your solution, check and make sure it doesn't match an excluded value.0823
If it does, you have to throw that solution out.0831
If it doesn't match an excluded value, then you are left with a solution that is valid.0834
Rational inequalities are a little bit more complex to solve; so we are going to go through, and then we are going to do an example of how to solve these.0842
Recall that a rational inequality is an inequality that contains one or more rational expressions.0853
We talked about rational equations containing algebraic fractions, or rational expressions.0859
The idea is the same here, except we are dealing with inequalities instead of equalities.0864
There are several steps to solve these: the first step is to find the excluded values.0869
We talked about excluded values when working with rational equations; and it is the same idea here.0880
It is going to be any value of x that is going to make the denominator 0.0884
Then, the second step is to solve the related equation.0889
Your result is going to be that you are going to have some values of x from #1; you are going to have values of x from #2;0898
and then, you are going to take the values from steps 1 and 2 and use them to split the number lines into intervals.0904
Those intervals may contain parts of the solution set; so then, you are going to use a test value for each interval.0926
And that test value is going to help you determine whether or not the interval contains parts of the solution set.0942
An example of a rational inequality would be something like x/(x2 - 9) - 3/(x + 4) < 5.0952
We will do an example in a second; but pretty much, what you are going to do is find the excluded values0966
by finding values that would make the denominator 0, then solve the corresponding equation0971
by changing this into an equal sign, and then solving, taking those values (whatever those values are), and then dividing the number line.0977
Let's say I ended up with two solutions and two excluded values; then I would divide the number line into 1, 2, 3, 4, 5 intervals.0985
And then, I would use test points to determine if these intervals contain parts of the solution set.0994
I would use the test values and insert them into the original inequality, and see if the inequality held out, based on those values.1001
Let's illustrate this right now with an example.1009
Given the rational inequality x/(x2 - 9)...actually, let's do a slightly less complex one.1013
3/x + 1/(x + 4) ≥ 01028
I am going to start out by finding the excluded values--values that would make the denominator 0.1041
For this first one, it is very simple: if x equals 0, then this becomes the denominator 0, and that is not allowable.1053
The second one: I get x + 4 = 0; if that were true, this would become an undefined expression.1061
Solving for x, x equals -4: if x is -4, add that to 4 and you get 0.1070
So, I have two excluded values: x = 0 and x = -4; that is my first step.1077
My second step is to solve the related equation, which is 3/x + 1/(x + 4) = 0.1083
Recall that we talked about solving rational equations by multiplying all terms by the LCD.1095
And these are already factored out; so my LCD is the factor in this denominator, which is unique, times the factor in this denominator,1105
x times (x + 4)...so I am going to use the technique of multiplying each term by this LCD to get rid of the fractions.1118
Right now, all I am doing is trying to solve this, so that I can add it to this set of values, and then split up the number line into intervals.1127
That is my first term, x(x + 4) times 1/(x + 4) = 0 times x(x + 4).1137
I see here that the x's cancel out, leaving me with 3(x + 4) +...here the (x + 4)s cancel out...that is x =...this whole thing becomes 0.1155
Now, I need to solve this linear equation: this gives me 3x + 12 + x = 0.1169
3x + x is 4x, plus 12 equals 0; subtract 12 from both sides to get 4x = -12; and then finally, divide both sides by 4 to yield x = -3.1178
And that is not an excluded value, so that is a valid solution to this equation.1193
So, I have excluded values, and I have a solution to the related equation: x = -3.1197
The next step is to use these to divide up the number line into intervals.1209
Let's do that down here, out of the way: here x = 0, -1, -2, -3, -4; so, -4: -1, -2, -3...1216
Now that I have the number line divided up, I need to determine which of these intervals...1237
I have one interval here; I will call this A; it is from right here...this is also dividing it into an interval...1241
eliminate these to make the intervals clearer; I have interval B, C, and D; so I have 4 intervals.1251
I need to find test points and check those in the original inequality for all 4 of these.1262
For the first interval, I am going to use a test point of -5.1269
I am working right over here; let's let x equal -5.1275
If I go back to this original inequality, let's see if this value over here satisfies it: 3/-5 + 1/(-5 + 4) ≥ 0.1282
This gives me -3/5; and this becomes 1/-1, because -5 + 4 is -1.1295
So, this gives me -1 3/5, which is not greater than or equal to 0.1305
Therefore, this interval does not contain any of the solution set, because my test point did not satisfy this inequality.1309
This was for interval A: for interval B, I am going to pick another point, and that is between -3 and -4.1322
So, I can't pick an integer; I have to go with a fraction to get something in between there.1330
So, I am going to pick -3 1/2; so let's let x equal -3 1/2, which actually equals -7/2.1336
This is going to give me 3, divided by -7/2, plus 1, divided by (x + 4), is greater than or equal to 0.1349
3 divided by -7/2...I could simplify this complex fraction by turning this into 3 times -2/7; so this becomes (I'll write it over here) 3 times -2/7.1364
Actually, this should be -7/2 right there; and -7/2 + 4 (let's work on this right here)...that would give me...1384
I need to convert this, so this would give me 8/2, because 8 divided by 2 would give me 4 back.1398
So, that is a common denominator of 2; -7/2 + 8/2 = 1/2.1404
That is 3 times -2/7, plus 1/(1/2), is greater than or equal to 0.1412
This gives me -6/7; getting rid of this complex fraction, 1 divided by 1/2 is equal to 1 times 2, which is 2.1422
-6/7 is smaller than 2, so I know that this becomes positive; therefore, this is true.1439
When I use x = -3/2, I find that I satisfy the inequality; therefore, B contains part of the solution set.1448
So, this value did not satisfy the inequality; this value did satisfy the inequality.1459
Looking at...we are going to move over here...the third interval: for the third interval, I am going to use a value of -2,1467
because it has to be between 0 and -3; so let's go ahead and use -2.1479
When x equals -2, that gives me 3 divided by -2 plus 1, divided by -2 plus 4, is greater than or equal to 0.1483
And that becomes -3/2; and this is +1/2, is greater than or equal to 0.1496
1/2 and -3/2 gives me -2/2, or -1; is that greater than or equal to 0? No, it is not.1509
So, C does not contain the solution set.1525
Finally, I am going to test section D, using the test point of 1.1532
Let's let x equal 1; therefore, this is going to give me 3 divided by 1, plus 1 divided by (1 + 4), is greater than or equal to 0.1536
And I can see that these are all positive numbers, so even if I don't figure this out, I know that this is valid; so yes.1548
I have determined that the intervals B and D contain the solution set.1556
Now, as far as how to write this solution set: let's go up here and write it out.1561
You need to be careful that you don't include excluded values.1572
The solutions lie between -4 and -3; and even though this says "greater than or equal to,"1581
I don't want to include an "equal to" that will include an excluded value.1589
So, I am going to say that x is greater than -4; I am not going to say greater than or equal to, or I would be including a value that I am not allowed to include.1593
So, x is greater than -4 and less than or equal to -3.1605
In addition, x is greater than 0; that is also part of the solution set, which is1615
any of these values (values for x that are greater than 0, or values of x that are greater than -4 and less than or equal to -3)...1630
see, the -3 can be included in the solution set, because it is not an excluded value.1638
It is dividing this up, because it was a solution to this equation, not because it was an excluded value.1643
As you can see, this is pretty complicated; it takes a lot of steps.1649
Your first step is to find the excluded values; I found those.1652
The second step is to solve the related equation; I did that right here, and I got that x equals -3.1658
This gave me three values that I used to divide up the number line into four intervals.1665
My next step was to take test points: I took a test point for A right here, and I found that that did not satisfy the inequality.1670
So, this interval is not part of the solution set; A was right here; B was right here.1678
x equals -3/2; I solved that out, and this did satisfy the inequality, so this is part of the solution set.1685
For C, I took x = -2 and went through; and this did not satisfy the inequality, so this is not part of the solution set.1694
And finally, in D, I let x equal 1; that satisfied the inequality, and therefore, the interval D contains part of the solution set.1702
So, the solutions are x > -4 and x ≤ -3, and also x > 0 is part of the solution set.1712
Let's get some more practice by doing some examples.1726
Going back to rational equations, recall that we need to find the LCD.1731
And in order to find the LCD, I am going to factor out these denominators; so this one is factored.1741
1 - x is already factored; but if we factor out a -1, it is going to make our work easier.1753
Recall that we talked earlier about the fact that, when you had two factors that were close, but not exact, factoring out a -1 could help.1761
Let's look at this second denominator: I am going to rewrite this, instead of as 1 - x, as -x + 1.1772
Now, I can look and see that these two are the same, except the signs of both terms are opposite.1781
What this tells me is that, if I factor -1 out of either one (not both--I can pick one or the other and factor out a -1), -1 pulled out of here will give me x.1786
-1 pulled out of 1 will give me -1, because, if I multiply this by -1, that is going to give me -x; -1 times -1 will give me a 1 back.1797
Now, I can see that I actually have two factors that are the same, plus this -1.1807
I could include this 4 as part of the LCD, and then this fraction would end up getting eliminated as well.1812
But I am not really worried about this fraction, because it is a constant; I can work with the number 1/4.1820
So, you can either include the 4 or not include it when you are using the method of eliminating fractions to solve a rational equation.1824
I am actually going to not use that 4, and I am just going to deal with the constant later on.1833
But we still need to multiply this term by the LCD that we find here, as well.1837
OK, so the LCD is going to be the product of (x - 1), because that is a factor, and -1.1844
So, I am going to multiply each term in this equation by this LCD.1854
I am also going to rewrite this in its factored form, just to make it simpler to see that what I am working with, this, is the same as this.1870
I just factored this into this by factoring out the -1.1880
OK, let's go ahead and start canceling common factors.1894
The (x - 1)s cancel; this gives me -1 times 3, which is simply -3, minus...here I can cancel out the -1 and the (x - 1)s.1897
So, this gives me -4: so -3 - 4 equals -1, times (x - 1), all divided by 4.1914
I am not worried about this, because the denominator does not have a variable in it, so it is not a rational expression.1929
I have gotten rid of all of my rational expressions: remember, rational expressions are algebraic fractions,1934
but we are talking about fractions where there is a variable in the denominator.1938
OK, -3 and -4 is -7; equals...let's write this as -x - 1, over 4; therefore, I can multiply both sides of the equation by a -4.1943
That will cancel this -4 out, and that gives me -4 times -7, which is 28, equals x - 1.1959
I am going to add a 1 to both sides to get 29 = x.1968
Now, I can't forget about my excluded values: these are going to be values that make the denominator 0.1974
And for this first one, I have (x - 1); if that equals 0, then this will be undefined.1983
So, that would occur in cases where x equals 1; so x = 1 is an excluded value.1991
Looking over here, I factored this into -1 times (x - 1); using the zero product property, I could again say, if x - 1 equals 0, then I have a problem.1997
x = 1 is the excluded value for this one and for this one.2009
So, since this solution I got, x = 29, is not an excluded value, then it is a valid solution.2013
If I came up with the solution x = 1, that would have been an extraneous solution that I would have had to throw out.2021
This example is another rational equation that we are again going to solve by finding the LCD.2032
So, let's go ahead and just factor the denominators right here.2040
This is x plus a factor, times x minus a factor; factors of 10 are 1 and 10, 2 and 5.2044
And I need those to add up to 3x; so 5 - 2 would give me a 3.2054
Therefore, the correct factorization would be (x + 5) (x - 2), plus...I am going to leave some room to multiply these by the LCD.2062
So, 2/(x - 2) = 19/(x + 5): I have this factored out right here.2078
And the LCD is going to be (x + 5); that is present once; and (x - 2)--that is also present once; that is the LCD.2091
So, I am going to multiply each term by that LCD, (x + 5) (x - 2),2107
plus (x + 5) (x - 2) times this next term, times (x + 5) (x - 2) times this third term.2118
Go ahead and cancel out common factors: here, both factors are common, leaving a 3 behind.2135
Here, the (x - 2) cancels out, leaving behind 2(x + 5), and on the right, the (x + 5) cancels out; that leaves me with 19(x - 2).2143
Let's go ahead and solve this equation: 3 +...2 times x is 2x, plus 2 times 5 is 10, equals 19x...19 times -2 is -38.2158
Combining the constants on the left, 2x + 13 = 19x - 38.2176
I am going to subtract the 2x, first, from each side: that is going to give me 13 = 17x - 38.2185
And then, I am going to go ahead and add a 38 to both sides to give me 51 = 17x.2197
Divide both sides by 17; this is going to give me 51/17 = x; therefore, x = 3.2205
Now, before I say that this is the actual solution, I need to look for excluded values.2215
And it is easy to do that, because I have already factored these out.2219
Excluded values: this is something you can do right at the beginning, before you start working, or right when you finish.2223
But you have to make sure that you check the solutions, each time, for these.2230
So, for this first one, the denominator is (x + 5) (x - 2) = 0...that would then result in a denominator that is 0, which is not allowed.2234
So, using the zero product property, if (x + 5) equals 0, this whole thing will equal 0.2245
Or if (x - 2) equals 0, this whole denominator will equal 0, when you multiply 0 times the other factor.2254
So, x = -5 and x = 2 are excluded values.2261
Looking right here, this is the same factor as here; so this also has an excluded value of x = 2; I have that accounted for.2269
And here, if x + 5 = 0, that would be the same as this, x = -5; so I don't have to worry about these--they are already accounted for.2277
Excluded values for all 3 (I have covered all 3) are these two.2284
I look over here, and my solution, x = 3, is not an excluded value; therefore, it is a valid solution.2289
OK, again, we have a rational equation we need to solve.2299
And the first step is to find the LCD by factoring out the denominators.2302
So, this first denominator (we will find the LCD)...I just have x + 1; that is already factored.2308
x2 - 1 is the difference of two squares, so that is (x + 1) (x - 1).2318
Therefore, the LCD...I have this (x + 1) factor here and here, and I have an x - 1.2325
So, I am going to solve this by multiplying each term by (x + 1) (x - 1)...times x, equals (x + 1) (x - 1)2335
times (x2 + x + 2)/(x + 1), minus (x + 1) (x - 1) times (x2 - 5), divided by...2351
I am going to rewrite this in the factored form, so it is more obvious what cancels and what doesn't.2368
All right, so over here, I end up with just x, times (x + 1) (x - 1).2376
Here, the (x + 1)s cancel out; that is going to leave me with (x - 1) times (x2 + x + 2),2386
minus...the (x + 1)s cancel; the (x - 1)s cancel; so minus (x2 - 5).2397
Multiplying this out, recall that (x + 1) (x - 1) is (x2 - 1), so this gives me x times (x2 - 1).2406
equals...x times x2 is x3; x times x is x2; x times 2 is 2x.2416
Multiplying -1 times each of these terms: -x2 - x...-1 times 2 is -2.2425
A negative times a positive gives me -x2; a negative times a negative gives me + 5.2435
Now, to take care of this: this is x3 - x.2441
All right, first let's take care of the x3; I am going to subtract x3 from both sides to move this from the right to the left.2457
So, what happens is: I do x3 - x3; this drops out.2464
I have to do the same thing to the left: x3 - x3--that drops out.2470
So, that took care of the x3; this looked worse than it actually was.2475
It is leaving me with -x =...well, let's see if I can do any simplifying here on the right.2480
I have x2 and -x2, so those cancel out.2485
That leaves me with 2x...so this is gone; this is gone; this is gone...minus x, minus 2, minus x2, plus 5.2501
I still have some work to do; but I am going to see that I can combine some of this still,2524
which gives me -x = 2x - x (that is x), and combining the constants is going to give me -2 + 5 (that is positive 3), minus x2.2532
Now, what I am going to end up with here is a quadratic equation that I need to solve.2553
Let's finish doing the simplification, and then go ahead and solve that.2558
I am going to move this x over to the right by adding an x to both sides; that is going to leave me with 0 = 2x + 3 - x2.2561
I want to write this in a more standard form, where the x2 is going to be positive.2572
So, let's go ahead and flip the sides, which will give me x2 - 2x - 3 = 0.2577
All I did is added an x2 to both sides, subtracted a 3 from both sides, and subtracted a 2x from both sides, in order to flip this around.2586
OK, now that I have just a quadratic equation left, all I need to do is solve it.2598
And I can use factoring: this is negative, so I have (x + a factor), times (x - a factor), equals 0.2602
Factors of 3 are 1 and 3; and I want them to add up to a negative number, so I am going to make the larger factor negative and the smaller factor positive.2614
And this does work; you get x2...the outer term is -3x...plus x; that gives a -2x for the middle term; 1 times -3 is -3.2624
That is factored correctly; now I just need to use the zero product property to solve.2635
x + 1 = 0; therefore, x = -1; the other solution is that x - 3 = 0; so x = 3.2640
The two solutions that I have are possible solutions; and I say "possible," because I have to make sure that these are not excluded values.2650
Possible solutions are x = -1 and x = 3.2658
Now, let's look for excluded values: excluded values are going to be any values that make either of the denominators 0.2665
Here, I have x2 - 1; that factored into (x + 1) (x - 1).2679
So, if (x + 1) (x - 1) equals 0, the denominator becomes 0.2685
So, any values of x that result in this being 0 are excluded.2690
Using the zero product property: x + 1 = 0; x - 1 = 0--if either of those occurs, this entire thing will be 0, and you would have an invalid situation.2696
Therefore, for this first one, if x equals -1, this whole thing will become 0,2711
because -1 times -1 is 1, minus 1--this would give you a 0.2719
The other excluded value is x = 1, because 12 is 1, minus 1 would give you 0.2724
So, I have these two excluded values: looking over here, I already took care of that, because this is a factor, right here.2729
So, I already said that, if x + 1 equals 0, x equals -1, and that is excluded; so that one is already covered.2737
Now, I need to compare my solutions to my possible solutions, to these.2743
And I see that I only have one valid solution, because this is actually not a valid solution.2749
It is an excluded value; I can't include that as part of the solution--it is an extraneous solution.2758
x = 3 is the only valid solution.2764
This took quite a few steps: the first step was to find the LCD and multiply each term by the LCD to get rid of the fractions.2768
And then, the hardest work was actually just multiplying all of this out, keeping track of all the signs,2777
and then getting down to where we had a quadratic equation that we solve by factoring to get two possible solutions.2783
The final step was to find the excluded values, based on setting the denominators equal to 0 and then solving for x,2788
and then comparing my possible solutions against these excluded values, thus eliminating x = -1 from the solution set,2796
and being left with one solution, which is x = 3.2804
In Example 4, we are going to be working with a rational inequality.2809
Recall that, for rational inequalities, we are going to find the excluded values.2814
We are going to solve the related equation; and then we are going to use those values to divide the number line into intervals,2819
and then test each interval, using a test value to determine where the solution set is.2824
Let's first find excluded values: these are going to be values for which the denominator becomes 0.2832
If 2x equals 0, that would occur when x equals 0/2, or x equals 0.2846
The same thing is going to occur here with 4x--that if x equals 0, it will be an excluded value.2853
So, the excluded value is x = 0.2858
Now, I need to solve the related equation, 1/2x + 5/4x - 1 = 0.2862
And the least common denominator of 2x and 4x...I could factor this out to 2 times 2.2874
So, the LCD is going to be...there is a 2 present once here, but twice here; and remember, I take each unique factor2887
to the highest power that it is present, so I am going to say 22.2898
There is an x present once here and once here, so it is 22 times x equals 4x.2902
So, the LCD is 4x; so multiplying here, 4x(1/2x) + 4x(5/4x) - 1(4x) = 0(4x).2907
Canceling out common factors, this becomes a 2; 2 cancels out; the x's cancel out.2928
2 times 1 is 2, plus...the 4x's cancel, leaving behind 5; this is -4x = 0.2935
7 - 4x = 0; 7 = 4x; divide both sides by 4; that gives x = 7/4, or this could be rewritten as x = 1 3/4.2949
So, that is my excluded value; and my solution to the related equation is x = 1 3/4.2968
Let's go down here, and I am going to use this value and this value to divide up my number line.2974
I have 0 here, and I have 1 3/4 here.2984
So, I end up with three intervals: A, B, and C; and I need to test a point within each of these intervals.2992
And if that value satisfies the inequality, I know that this interval is at least part of the solution set.3003
For the first interval (for A), I am going to let x equal -1; I need something less than 0; I will let x equal -1--that is simple to work with.3010
That is going to give me 1 over 2(-1), plus 5 divided by 4(-1), minus 1, is greater than 0.3022
And let's see if this holds up: this gives me -1/2, minus 5/4, minus 1, is greater than 0.3031
This is the same as saying -2/4 (just converting to a common denominator) - 5/4 - 1 > 0.3040
I actually don't even need to go farther; in fact, I even kind of just looked at this and said,3049
"I have a bunch of negatives; those are not going to be greater than 0 when I combine them."3053
So, no: this is not valid; therefore, that interval does not contain the solution set or part of the solution set.3057
For B, I have values between 0 and 1 3/4, so I can use 1 as a test point, x = 1.3069
This is going to give me 1 over 2(1), plus 5 divided by 4(1), minus 1, is greater than 0,3077
which is 1/2 + 5/4 - 1 > 0; I can just convert this to 2/4 + 5/4 - 1 > 0.3086
And this is going to give me 5/4 + 2/4, is going to give me 7/4: 7/4 is greater than 1, so when I subtract 1 from 7/4, I am going to end up with 3/4.3101
So, this is going to give me 3/4 > 0.3116
Even if I didn't figure this whole thing out, as soon as I saw that this is a positive number larger than 1,3119
I know that, when I subtract 1 from it, I will get something positive; so it is going to be greater than 0.3125
So, this one is valid; therefore, yes for B: this interval does contain at least part of the solution set.3129
For C, I am going to go right here, and I am going to use 2 as my test point: x = 2.3138
This is going to give me 1 over 2(2), plus 5 over 4(2), minus 1, is greater than 0.3148
That is 1/4 + 5/8 - 1 > 0.3156
I need to find a common denominator here, so I am going to convert 1/4 to 2/8: plus 5/8, minus 1, is greater than 0.3163
That is 7/8 - 1 > 0; well, 7/8 - 1 is going to leave -1/8 > 0.3172
And that is not true, so C is not part of the solution set.3183
Therefore, the solution for this...the possible solution...I am going to say "possible,"3188
because we have to look back at excluded values...is x is greater than 0, but it is less than 1 3/4.3196
Now, let's look back up at excluded values: I cannot let x equal 0--that is an excluded value.3209
But I am OK here, because x is greater than 0; so I am covered, and this is my actual solution--this is valid.3223
And just looking back to review: find your excluded values (x = 0 is an excluded value), then solve the related rational equation.3234
I did that by multiplying by the LCD, multiplying each term, getting rid of those fractions,3247
then finding that x = 1 3/4 was the solution, and it wasn't an excluded value (so it was a valid solution).3253
I used those two values, 0 and 1 3/4, to divide the number line into intervals.3262
And then, I tested each interval: I tested interval A and found that my test value did not satisfy the inequality, so that is not part of the solution set.3268
I tested B, using x = 1, and found the inequality did hold up, so B contains part of my solution set.3276
C--I tested: that value did not satisfy the inequality, so that is not part of the solution set.3284
Therefore, the solution set is that x is greater than 0 and less than 1 3/4.3289
But remember: if you are working with greater than or equal to in the original, and you are thinking,3293
"OK, I am going to put an 'equal to' here," you need to be careful that you don't encompass an excluded value as part of that.3300
That concludes this lesson of Educator.com on rational equations and inequalities.3307
I will see you next lesson!3314
|
# Triangle Problem Solving
Simplify this equation step by step and solve it: x x 7 x 3 = 40, 3x 10 = 40, 3x = 40 - 10, 3x = 30, x = 10. Hence, the second side is 10 cm 7 cm = 17 cm long, and the third side is 17 cm - 4 cm = 13 cm long in accordance with the problem condition.
Simplify this equation step by step and solve it: x x 7 x 3 = 40, 3x 10 = 40, 3x = 40 - 10, 3x = 30, x = 10. Hence, the second side is 10 cm 7 cm = 17 cm long, and the third side is 17 cm - 4 cm = 13 cm long in accordance with the problem condition.
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A good first step, after reading the entire exercise, is to draw a right triangle and try to figure out how to label it.
Once you've got a helpful diagram, the math is usually pretty straightforward.
Therefore, in our case the perimeter of the triangle is equal to 8 cm 5 cm 9 cm = 22 cm. Solution The second side is 8 cm 4 cm = 12 cm long in accordance with the problem condition.
The third side is (8 cm 12 cm)/2 = 10 cm long in accordance with the problem condition.
Then the second side is of x - 6 cm long in accordance with the problem condition, and the third side is of (x (x-6))/2 cm long.
Since the perimeter of the triangle is equal to 30 cm, you can write the equation x (x - 6) (x (x-6))/2 = 30.
For navigation over the lessons on Properties of Triangles use this file/link Properties of Trianles.
To navigate over all topics/lessons of the Online Geometry Textbook use this file/link GEOMETRY - YOUR ONLINE TEXTBOOK.
Simplify this equation step by step and solve it: 2x 2(x - 6) (x (x-6)) = 60 (after multiplying both sides of the previous equation by 2), 2x 2x x x - 12 - 6 = 60, 6x - 18 = 60, 6x = 60 18, 6x = 78, x = 13. Hence, the second side is of 13 cm - 6 cm = 7 cm long, the third side is (13 7)/2 = 20/2 = 10 cm long in accordance with the problem condition.
You can check that the sum of the side measures is equal to the given value of the perimeter: 13 cm 7 cm 10 cm = 30 cm. The triangle side measures are 13 cm, 7 cm and 10 cm.
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Problem solving about triangle and rectangle. You have 500 feet of fencing which you want to use to enclose a right triangular area. What is.…
• ###### Solving similar triangles video Khan Academy
Sal solves two problems where a missing side length is found by proving that triangles are similar and using this to find the measure.…
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Oct 29, 2017. Since triangle problems only account for a small percent of the SAT math questions, you. It's essentially a domino effect of problem-solving.…
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Solution of triangles Latin solutio triangulorum is the main trigonometric problem of finding the.…
|
# 11
## Constructions
#### 11.1 Introduction
In Class IX, you have done certain constructions using a straight edge (ruler) and a compass, e.g., bisecting an angle, drawing the perpendicular bisector of a line segment, some constructions of triangles etc. and also gave their justifications. In this chapter, we shall study some more constructions by using the knowledge of the earlier constructions. You would also be expected to give the mathematical reasoning behind why such constructions work.
#### 11.2 Division of a Line Segment
Suppose a line segment is given and you have to divide it in a given ratio, say 3 : 2. You may do it by measuring the length and then marking a point on it that divides it in the given ratio. But suppose you do not have any way of measuring it precisely, how would you find the point? We give below two ways for finding such a point.
Construction 11.1 : To divide a line segment in a given ratio.
Given a line segment AB, we want to divide it in the ratio m : n, where both m and n are positive integers. To help you to understand it, we shall take m = 3 and n = 2.
#### Fig. 11.1
1. Draw any ray AX, making an acute angle with AB.
2. Locate 5 (= m + n) points A1, A2, A3, A4 and A5 on AX so that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
3. Join BA5.
4. Through the point A3 (m = 3), draw a line parallel to A5B (by making an angle equal to
AA5B) at A3 intersecting AB at the point C (see Fig. 11.1). Then, AC : CB = 3 : 2.
Let us see how this method gives us the required division.
Since A3C is parallel to A5B, therefore,
= (By the Basic Proportionality Theorem)
By construction, .
This shows that C divides AB in the ratio 3 : 2.
#### Fig. 11.2
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray BY parallel to AX by making ABY equal to BAX.
3. Locate the points A1, A2, A3 (m = 3) on AX and B1, B2 (n = 2) on BY such that AA1 = A1A2 = A2A3 = BB1 = B1B2.
4. Join A3B2. Let it intersect AB at a point C (see Fig. 11.2).
Then AC : CB = 3 : 2.
Why does this method work? Let us see.
Here AA3C is similar to BB2C. (Why ?)
Then .
Since by construction, therefore,
In fact, the methods given above work for dividing the line segment in any ratio.
We now use the idea of the construction above for constructing a triangle similar to a given triangle whose sides are in a given ratio with the corresponding sides of the given triangle.
#### Construction 11.2 :
To construct a triangle similar to a given triangle as per given scale factor.
This construction involves two different situations. In one, the triangle to be constructed is smaller and in the other it is larger than the given triangle. Here, the scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle (see also Chapter 6). Let us take the following examples for understanding the constructions involved. The same methods would apply for the general case also.
#### Example 1 :
Construct a triangle similar to a given triangle ABC with its sides equal to of the corresponding sides of the triangle ABC (i.e., of scale factor ).
#### Solution :
Given a triangle ABC, we are required to construct another triangle whose sides are of the corresponding sides of the triangle ABC.
Steps of Construction :
1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
2. Locate 4 (the greater of 3 and 4 in ) points B1, B2, B3 and B4 on BX so that BB1 = B1B2 = B2B3 = B3B4.
#### Fig. 11.3
3. Join B4C and draw a line through B3 (the 3rd point, 3 being smaller of 3 and 4 in ) parallel to B4C to intersect BC at C.
4. Draw a line through C parallel to the line CA to intersect BA at A′ (see Fig. 11.3).
Then, ABC is the required triangle.
Let us now see how this construction gives the required triangle.
By Construction 11.1,
Therefore, , i.e., = .
Also CA is parallel to CA. Therefore, ABC ~ ABC. (Why ?)
So,
#### Example 2 :
Construct a triangle similar to a given triangle ABC with its sides equal to of the corresponding sides of the triangle ABC (i.e., of scale factor ).
#### Solution :
Given a triangle ABC, we are required to construct a triangle whose sides are of the corresponding sides of ABC.
#### Steps of Construction :
1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
2. Locate 5 points (the greater of 5 and 3 in ) B1, B2, B3, B4 and B5 on BX so that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
3. Join B3(the 3rd point, 3 being smaller of 3 and 5 in ) to C and draw a line through B5 parallel to B3C, intersecting the extended line segment BC at C.
4. Draw a line through C parallel to CA intersecting the extended line segment BA at A (see Fig. 11.4).
Then ABC is the required triangle.
#### Fig. 11.4
For justification of the construction, note that ABC ~ ABC. (Why ?)
Therefore,
But,
So, and, therefore,
#### Remark :
In Examples 1 and 2, you could take a ray making an acute angle with AB or AC and proceed similarly.
#### Exercise 11.1
In each of the following, give the justification of the construction also:
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.
3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.
4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle.
5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.
6. Draw a triangle ABC with side BC = 7 cm, B = 45°, A = 105°. Then, construct a triangle whose sides are times the corresponding sides of ABC.
7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are times the corresponding sides of the given triangle.
#### 11.3 Construction of Tangents to a Circle
You have already studied in the previous chapter that if a point lies inside a circle, there cannot be a tangent to the circle through this point. However, if a point lies on the circle, then there is only one tangent to the circle at this point and it is perpendicular to the radius through this point. Therefore, if you want to draw a tangent at a point of a circle, simply draw the radius through this point and draw a line perpendicular to this radius through this point and this will be the required tangent at the point.
You have also seen that if the point lies outside the circle, there will be two tangents to the circle from this point.
We shall now see how to draw these tangents.
#### Construction 11.3 :
To construct the tangents to a circle from a point outside it.
We are given a circle with centre O and a point P outside it. We have to construct the two tangents from P to the circle.
Steps of Construction:
1. Join PO and bisect it. Let M be the mid-point of PO.
2. Taking M as centre and MO as radius, draw a circle. Let it intersect the given circle at the points Q and R.
3. Join PQ and PR.
Then PQ and PR are the required two tangents (see Fig. 11.5).
Now let us see how this construction works. Join OQ. Then PQO is an angle in the semicircle and, therefore,
#### Fig. 11.5
PQO = 90°
Can we say that PQ OQ?
Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.
### Note :
If centre of the circle is not given, you may locate its centre first by taking any two non-parallel chords and then finding the point of intersection of their perpendicular bisectors. Then you could proceed as above.
#### Exercise 11.2
In each of the following, give also the justification of the construction:
1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
#### 11.4 Summary
In this chapter, you have learnt how to do the following constructions:
1. To divide a line segment in a given ratio.
2. To construct a triangle similar to a given triangle as per a given scale factor which may be less than 1 or greater than 1.
3. To construct the pair of tangents from an external point to a circle.
#### A Note to the Reader
Construction of a quadrilateral (or a polygon) similar to a given quadrilateral (or a polygon) with a given scale factor can also be done following the similar steps as used in Examples 1 and 2 of Construction 11.2.
|
Evaluate $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-\cos{x}}{x^2}$
$x$ is a variable but also represents an angle of the right angled triangle. The exponential function and trigonometric function formed a special algebraic trigonometric function. It is required to find the value of this function as $x$ approaches $0$.
Separate the functions
Exponential function in algebraic form and trigonometric function formed a special function. The combination of them creates a problem in simplification. So, it is good idea to separate both functions. Add one and subtract it by one in numerator before separating them.
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1+1-\cos{x}}{x^2}$
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \Bigg[\dfrac{e^{x^2}-1}{x^2}+\dfrac{1-\cos{x}}{x^2}\Bigg]$
Simplify limit of sum of the functions
Use sum law of limits to apply limit value to both functions.
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1-\cos{x}}{x^2}$
Simplify the trigonometric function part
The numerator which contains trigonometric function can be simplified by angle to half angle transforming trigonometric identity.
$1-\cos{\theta} = 2\sin^2{\Bigg(\dfrac{\theta}{2}\Bigg)}$
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{2\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{x^2}$
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{x^2}$
Transform functions as Limit rules
Now, try to change each function as a rule of limits and then the rules of limits can be applied to find the value of the function as $x$ approaches $0$.
The first limit of the function is almost in the form of $\displaystyle \large \lim_{x \,\to\, 0} \dfrac{e^x-1}{x}$. In this case, the exponential function and denominator contains $x^2$ but limit value is $x$ tends to $0$. So, it should be changed to apply this rule.
If $x \,\to\, 0$, then $x^2 \,\to\, {(0)}^2$. Therefore, $x^2 \,\to\, 0$
The second function is almost same as the $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x}$ rule. So, try to change it mathematically.
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{4 \times \dfrac{x^2}{4}}$
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{4 \times {\Bigg(\dfrac{x}{2}\Bigg)}^2}$
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1}{4} \times \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{{\Bigg(\dfrac{x}{2}\Bigg)}^2}$
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2 \times \dfrac{1}{4} \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{{\Bigg(\dfrac{x}{2}\Bigg)}^2}$
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\dfrac{1}{2} \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{{\Bigg(\dfrac{x}{2}\Bigg)}^2}$
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\dfrac{1}{2} \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{{\Bigg(\dfrac{x}{2}\Bigg)}^2}$
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\dfrac{1}{2} \displaystyle \large \lim_{x \,\to\, 0} \normalsize {\Bigg[\dfrac{\sin{\Bigg(\dfrac{x}{2}\Bigg)}}{\Bigg(\dfrac{x}{2}\Bigg)} \Bigg]}^2$
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\dfrac{1}{2} {\Bigg[\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{\Bigg(\dfrac{x}{2}\Bigg)}}{\Bigg(\dfrac{x}{2}\Bigg)} \Bigg]}^2$
The denominator of the second function is adjusted same as the angle of the sin function. But, the limit value is different. If $x \,\to\, 0$, then $\dfrac{x}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{x}{2} \to 0$
Obtain the Solution of Limit Problem
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\dfrac{1}{2} {\Bigg[\displaystyle \large \lim_{\frac{x}{2} \,\to\, 0} \normalsize \dfrac{\sin{\Bigg(\dfrac{x}{2}\Bigg)}}{\Bigg(\dfrac{x}{2}\Bigg)} \Bigg]}^2$
According to lim x tends 0 (e^x-1)/x formula, the value of the first function is $1$. Similarly, the value of the second function is also $1$ as per limit x approaches 0 sinx/x rule.
$= 1+\dfrac{1}{2} \times {(1)}^2$
$= 1+\dfrac{1}{2} \times 1$
$= 1+\dfrac{1}{2}$
$= \dfrac{2+1}{2}$
$= \dfrac{3}{2}$
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# Frequent question: What is the probability of rolling a 5 on a dice?
Contents
## What is the probability of getting 5 when a die is thrown?
Solution: Total number of outcomes when die is thrown twice = 6 × 6 = 36. The probability that 5 will not come up either time is 25/36 and the probability that 5 will come up is 11/36. Check out more about terms of probability.
## What is the probability of rolling a 5 then a 2?
There is a 16 probability of getting a 2 on the first roll and a 16 probability of getting a 5 on the second roll.
## What is the probability of getting 5?
Two (6-sided) dice roll probability table
Roll a… Probability
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
6 15/36 (41.667%)
## What is the probability of getting 1 and 5 If a dice is thrown once?
So they are mutually exclusive events, therefore their probabilities add to 1. By symmetry we expect that each face is equally likely to appear and so each has probability = 1/6. The outcome of a 5 is one of those events and so has probability = 1/6 of appearing.
## What are the odds of rolling a 6 with 2 dice?
Probabilities for the two dice
IT IS INTERESTING: Can you claim the lottery anonymously in New Mexico?
Total Number of combinations Probability
3 2 5.56%
4 3 8.33%
5 4 11.11%
6 5 13.89%
## What are the odds of rolling 5 1s in a row?
The probability is 1216 chance, which is approximately a 0.46% chance.
## What is the probability of rolling an odd number?
The probability when rolling a regular six-sided dice that the score is an odd number is three-sixths or three out of six.
## What is the probability of rolling a 7?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
## What is the probability of rolling a 1 and then a 2?
The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on.
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# How to divide
## How to divide
What are the different ways to divide? Different types of divisions. For division you can use the multiplication tables. Find 10 2.
Step 1 Think about the related fact of multiplication. 2 × = 10.
Step 2 Find the missing factor. 2 × 5 = 10
Step 3 Use the missing coefficient to find the quotient. Solution: 10 ÷ 2 5 There are two ways to record the facts of the divorce. 10 ÷ 2 = 5 or 5 @Quotient.
## What are different ways to write Division?
The general method of short and long division is the same, but with short division you write less of your work, making multiplication and subtraction easier. To understand short parts, you need to master the basics of subtraction and multiplication.
## What is the answer when you divide?
In multiplication, the numbers you multiply are called factors and the answer is the product. In division, the divided number is divisible, the number dividing it is the divisor and the answer is the quotient.
## What are the different methods of Division?
Summary: There are three different types of cell division, binary division in prokaryotes and mitosis and meiosis in eukaryotes.
## What are the different strategies for Division?
The six distribution strategies include: Peer groups. chapter. de facto family. repeated subtraction. long division. short division.
## What is the easiest way to divide numbers?
• Dividing by 1 When dividing by 1, the result is equal to the dividend.
• Dividing by 2 If the last digit of a number is even, the whole number is divided by 2.
• Dividing by 4 If the last two digits are divisible by 4, the whole number is divisible by 4.
• Divide by 5 If the number ends in 5 or 0, divide by 5.
## What are the parts of dividing in math?
Dividend Dividend is the number dividing Divisor Divisor is the number dividing by the quotient The quotient is the answer.
## What are the steps to divide?
The division steps are as follows: Determine the first digit or group of digits that your divisor will contain at least once. Determine the maximum number of times your divisor will enter the digit(s). Enter your answer at the correct number. Multiply your answer above by your dividend.
## How does Division work in math?
Division is repeated subtraction. This is the act of dividing a number (dividend) by another number of the same or lower value (divisor) to find the quotient and remainder. This remainder is when the divisor is a factor of the dividend, it is not zero when the divisor is not a factor.
## How dividing by decimal is similar from dividing whole number?
Division of decimal numbers by whole numbers: Dividing decimal numbers by whole numbers is the same as normal division. Here the dividend is a decimal number and the divisor is an integer, so the quotient has a number of decimal places based on the decimals of the dividend.
## What numbers can be divided by two?
Almost everyone knows this rule, according to which any even number can be divided by 2. Even numbers are multiples of 2. A number is even if it ends in 0, 2, 4, 6, or 8. Examples of even numbers that have hit them this divisibility test.
## How do you divide a number?
To separate two numbers in Excel, enter an equal sign (=) in a cell, then enter the number you want to divide, then a slash, then the number you want to divide by and press Enter to enter the formula to calculate. For example, to divide 10 by 5, enter the following expression in a cell: = 10/5.
## What are the different ways to divide a square into fourths
Solution 3. Another way to divide it is to draw the diagonals of the square. Solution 4. You can also cut the square along a horizontal or vertical line and then cut each half into two equal triangles. There are other ways to divide a square into 4 equal parts.
## How do you divide a square into four equal parts?
Dividing a square into four equal parts is an exercise learned in elementary school. Each part of a square, divided into quarters, is called a quarter. The word "fourth" can be used to describe anything that is divided into four equal parts. Measure each side of the square. Place a dot in the center of each measurement.
## How to draw four equal parts of a word?
Draw 4 equal parts. Circle a word that describes the details. Halves Quarter Half Whole 2 Equal Parts Quarter Whole 4 Equal Parts Lesson 28 Understanding the halves, thirds, and fourths in the form Half Quarter Half Quarter Quarter Half.
## Which is one half or one fourth of the whole?
If the whole is divided into 2 equal parts, each part is half of the whole. If the whole is divided into 4 equal parts, each part is a quarter of the whole. When you go to high school, the fractions look like this: 1/2, which is 1 half.
## Why is the second circle divided into two equal parts?
The second circle is divided into 2 equal parts. Cho's mistake was drawing circles of different sizes. He had to draw circles of the same size. This is because you are trying to represent the same sliced cake in two different ways. Cho should have drawn these circles.
## What are the different ways to divide 4th grade
4 introductory division strategies for fourth graders: simulation, skip counting, multiplication, G7 partial division.
## What are 4th graders taught me?
Fourth graders will be intrigued to learn how the heart pumps blood throughout the body and important blood components such as red and white blood cells. Help your students understand the importance of donating blood and how it helps others. PBS offers multimedia courses and learning resources. Understand how matter works.
## What is the 4th grade math level?
The main mathematical areas of the fourth year are the theory of numbers and systems, algebraic thinking, geometric shapes and objects, the measurement of length, weight, capacity, time and temperature, and data analysis and probability. While these math areas may surprise you, they cover the basics of what a fourth-grader should learn.
## How do you calculate division?
A person can calculate the division on the bills by separating the dividend on the bills repeatedly and then using the divisor to subtract the offset of each digit from the result and count the number of possible divisions in each offset.
## What are the different ways to divide a year
Make equal groups (the number of partial circles according to your example is 3 circles). Divide your dividend by whole numbers. Divide the integers into equal groups so that all groups contain the same number of numbers. Add the integers in each group.
## How do you do long division in Excel?
Do long division and write down the problem. To perform long division, place the divisor, the number you want to divide, outside the long division bar, and dividends, the number you want to divide, inside the long division bar. Divide the divisor by the first digit of the first number (if possible).
## What's the best way to do short division?
Make a small division. Write down the problem. Divide your divisor by the first number in the dividend. Divide the divisor by the number obtained from the remainder and the second number of the dividend. Divide the divisor by the remaining numbers. Give your last answer.
## What do you need to know about Division for kids?
The divisor is the number you are dividing by. The quotient is the amount each divisor gets, which is the answer in most cases. Learning the correct vocabulary for all parts of a division problem will make it much easier for your child to find many of the parts. What should my child know about the short and long division between KS1 and KS2?
## How do you write a division problem?
The traditional way to write division problems is to use division parentheses. Another way to write division calculations is to use fractions. A fraction divides the top number or numerator by the bottom number or denominator.
## How to write division problems?
Method 2 of 5: Short division Use the division bar to write down the problem. Place the separator, the number you want to divide, outside (and to the left) of the separator bar. Divide your divisor by the first digit of the dividend. Divide the divisor by the remainder and the second digit of the dividend. Repeat the process until you have received all the dividends.
## What are different ways to write division problems
Use a spreader bar to record the problem. Place the separator, the number you want to divide, outside (and to the left) of the separator bar. Place the dividend, the number you want to divide by, inside (right and bottom) of the divide bar. To make a short division, your divisor cannot contain more than one digit.
## What is an example of a division problem?
Breakdown Problems (1-Step Problems) Dan went to the market on Friday. He bought two tomatoes. In July, a construction company constructed 360 miles of the highway. The company built 60 miles of highway in February. Lin ran 21 miles. Lin ran to Sophie three times. The area of \u200b\u200bthe room "Mollis" is 220 square meters. Molly's dining room is five times the size of her bedroom.
## What are different ways to write division equations
There are several symbols that people can use to represent divisions. The most commonly used is ÷, but the backslash / is also used. Sometimes people write one number after another with a dash in the middle. This is also called rest. Each part of the division equation has a name.
## How do you multiply and divide equations?
This means you can multiply the equation by any number, as long as you do it on both sides. To move a number that is multiplied or divided by a variable, you must do the opposite on both sides. That is, if you multiply the number, you have to divide it by both sides.
## What is the equation for Division?
However, the division is distributive in the sense that (a + b) c = (a c) + (b ÷ c) for any number. In particular, division has the distributive property of addition and subtraction.
## What word uses all the letters in the alphabet?
A pangram is a phrase or phrase that uses all the letters of the alphabet. adjective: pangrammatical. It is also called a holo-alphabetic sentence or an alphabetical sentence.
## How many letters are in the alphabet?
The modern English alphabet is a Latin alphabet of 26 letters, each of which is case sensitive.
## What are some creative ways to write a letter?
Follow these guidelines to format your letter to include everything: Greetings: Don't just write love . Instead, write "My Beloved" or "My One True Love" or even use their nicknames like "Dear Buttercup". First paragraph: Start by explaining the reasons why you are writing this letter. Write heart: this is where you describe your love and your feelings.
## How do I learn to write a letter?
Write a formal letter Know when to write a formal letter. Write a formal letter when talking to someone you only know professionally. Write your address and today's date at the top of the page. Write your name and address in the top left corner of the page. Write down the recipient's name and address.
## What are different ways to write division sentences
Definition of division. part of a large group. Share example in a sentence A small police station was sent to corner 5 and 4 to issue an arrest warrant.
## What is a division sentence?
How do you use division in a sentence? This process is only done if the separation is interrupted for at least a few days. In the division, he regained energy and based his skills on performance issues.
## What is a division number sentence?
The division is a mathematical way of showing how a sum is divided. You have a total of 18 chips. Divide the pieces into 6 equal groups. Each group contains 3 tiles. First they write the total i.e. 18. The division sign means dividing the quantity before it by the next quantity.
## What is Division in writing?
In classical rhetoric, separation is the part of speech in which the speaker describes the main points and the general structure of the speech. Also known in Latin as divisio or partitio and in English as partition. The etymology comes from the Latin word share.
## What are different ways to write division sign
Division sign. The division symbol o is written as a horizontal line with a dot at the top and a dot at the bottom (obelus), either as a slash or as a horizontal line: ÷ / - the division symbol indicates the division by 2 numbers or expressions. For example: 6 2 = 3. 6/2 = 3. means 6 divided by 2, which is the same as dividing 6 by 2, which is 3.
## How do you type the division sign on a keyboard?
The simple keyboard shortcut for creating a separator in Windows is 0247. Hold down the Alt key and enter 0247 on your Windows keyboard.
## What are the different division signs?
• Sum (+): This symbol means adding or finding a sum that is greater or greater than the result of the sum, it is the sum.
• Subtract (): This symbol means subtract, less or less result is the difference.
• Multiplication (×, ∙, *): All these symbols stand for multiplication or coloring.
## How do you type a division symbol on a computer?
If your computer keyboard has a numeric keypad, you can "enter" the division character as follows: 1 - Activate NumLock (if you haven't already). 2 - Hold down the Alt key while entering the numbers 0247 on the numeric keypad. The division character should appear after entering the last digit of the sequence.
## How do you make a divide symbol on a keyboard?
On most IBM compatible computers, the division character can be generated by activating the NUM-LOCK function, holding down the ALT key, and typing 246 on the numeric keypad in the far right corner of the keyboard. The division symbol appears when the ALT key is released.
## What are different ways to write division with the words
Write each expression in two ways: a division sign and a fraction. 1. 12 divided by 4. 2. 3 divided by 5. 3. a divided by 4. 4. The quotient 6 and m 5. Seven divided by the sum of x plus y. 6.y divided by x minus 11.
The divided number is called a dividend. The number you divide by is the divisor. The answers to your fission problems are called quotients. Six divided by two gives the quotient of three.
## What is the answer called when you subtract?
Subtraction is an arithmetic operation that removes objects from a collection. The result of the subtraction is called the difference. Subtraction is indicated by a minus sign, -.
## What is the answer when you subtract?
If you subtract one number from another, the result is called the difference. This term makes sense when you think about it: when you subtract, you find the difference between a higher number and a lower number. In subtraction, the first number is called a descending number and the second number is called a subtracted number.
## What is the answer to a division problem called?
The answer you get in the partial equation is called the quotient. The division sign (÷) is placed between the dividend and the divisor. It is a short horizontal line with dots at the top and bottom.
## What is the answer when you divide called
In every division problem, one number is divisible by the other. The divided number is called a dividend. The number you're dividing with is the divisor. The answers to your division problems are called quotients. Six divided by two gives the quotient of three.
## What is the answer called in Division?
The number which is divisible is called a divisor and the number by which it is divisible is called a divisor. The answer to the division problem is the quotient. Dividend Divider = Ratio Once students understand the basic facts about division, they should learn how to divide the largest dividends.
## What are the parts of a division problem?
A simple division math problem consists of three main parts: the dividend, the divisor, and the quotient. The dividend is the base of a division or mathematical formula.
## What is the answer when you divide by 0
In ordinary arithmetic, this expression does not make sense because there is no number multiplied by 0 that equals one (assumed), and therefore division by zero is undefined. Since any number multiplied by zero is zero, the expression is also undefined, if it is in the form of a boundary value, it is an undefined form.
## What divided by what equals 0?
ZERO DIVIDED to zero is rather vague. Because if A is equal to B times C, then A divided by B is equal to C. But zero is equal to zero times any number. So zero divided by zero is a random number. Ivor Etherington, Isdale, Argyll.
## Is any number divided by zero equal to zero?
A positive or negative number divided by zero is a fraction with zero in the denominator. Zero divided by a negative or positive number is zero or expressed as a fraction with zero in the numerator and finite magnitude in the denominator. Zero divided by a negative or positive number is zero or expressed as a fraction with zero in the numerator and a finite magnitude in the denominator. Zero divided by zero is zero.
## Why can't you divide by zero?
Dividing by zero is not allowed because it gives the same answer (infinite number) for every record and is therefore considered undefined. Multiplication by zero is allowed even if it gives the same answer for every input (zero).
## What happens when you divide by zero?
In math, division by zero is a division where the divisor (denominator) is zero. This division can be formally expressed as a0, where a is the dividend (numerator). In ordinary arithmetic, this expression makes no sense because no number multiplied by 0 gives (assuming it is 0), and therefore dividing by zero is undefined.
## What are the different methods of division in algebra
Apart from addition, subtraction and multiplication, division is one of the 4 basic operations in arithmetic. In addition to integers, you can divide decimals, fractions, or exponents. You can do long division or, if one of the numbers is a single digit, a short division.
## What are the different methods of division in excel
There are several ways to divide numbers. Use the arithmetic operator / (slash) to perform this task. For example, if you enter =10/5 in a cell, the cell will display 2. Important: Remember to put an equal sign (=) in the cell before entering numbers and the / operator. Otherwise, Excel interprets it as a date.
## How do you do a division in Excel?
The division formula is done using the slash division operator: / In the following steps, I will show you how to divide numbers in Excel. Here is a table of the values you want to divide by:.
## What happens when you combine multiplication and Division in Excel?
When you combine multiple operators in a formula, Excel performs the operations in the order shown in the following table. If a formula contains operators with the same priority, for example if the formula contains a multiplication and division operator, Excel evaluates the operators from left to right.
## What's the difference between divisor and dividend in Excel?
The number that is divisible is called a dividend, and the same number that makes it divisible is called a divisor. The result is often called private. Division in Excel is used the same way with an equal sign in each cell to find the quotient. Share formula in Excel.
## What is the formula for dividing two numbers in Excel?
Divide the two numbers. Distribution of the formula: = number1 / number2. What it means: = number divided / number you are dividing with. It's easy to divide numbers in Excel!
## What is the Divison and division operator in Java?
The division operator in Java includes the division, modulo, and division and assignment operators. Let's work with them one by one. department employee. The division operator divides the left operand by the right operand. Module operator. The modulo operator divides the left operand by the right operand and returns the remainder.
## What is the symbol for Division in Java?
The Java arithmetic division operator takes two operands as input and returns the quotient of the left operand divided by the right operand. The / symbol is used for the division operator.
## What are the different types of methods in Java?
Public: Methods declared publicly accessible from all classes in the application. Protected: Methods declared protected can be accessed from the class in which they are defined and from all subclasses of this class. Private: Methods declared private can only be accessed from the class in which they are defined.
## Can you use zero for Division in Java?
You can use zero for operand 1, but not for operand 2. You will get a syntax error if you divide by a constant zero, and an exception if operand2 is zero at runtime. However, the Java division and the math division are very different.
## What are the steps for division of algebraic expressions?
Dividing algebraic expressions involves the following steps.
Step 1 : Extract common terms directly or create specific expressions to check common terms.
Step 2 : cancel the generic term. Remark. General terms here correspond to one of the following meanings: constants, variables, terms, or simply coefficients.
## How do you divide a number by divisor?
Divide the first digit of the dividend by the divisor. In other words, calculate how many times the divisor (a number outside the division bar) is included in the first digit of the dividend. Place the entire result on top of the divider, just above the first digit of the divisor.
## How to subtract Product from divisor in algebraic long division?
Multiply the divisor by the first term of the quotient. Subtract the profit from the dividend and then reduce the next installment. The difference and the next term is a new dividend. Note: Keep in mind that the subtraction rule changes the sign of subtraction and then changes to addition.
## What are the different methods of division in python
An interesting kind of arithmetic that Python can do is divide the remainder of one number by another. To do this, you must use the modulo operator (also known as the percent sign:%).
## What is division operator in Python?
Python actually has two types of division operators: regular division (/) and root division (//). The only warnings that apply to them are the same warnings that apply to floating point distribution in general.
## What is int Division in Python?
For all divisions, the remainder after the one digit is simply discarded. If they divide 7 by 3, they get 3 stacks of 2 with a remainder, so: 7/3 = 2. There are two types of numbers commonly used in Python: integers, which can only express integers, and numbers with floating point. expressing a decimal number can be numbered.
## Is there a "not equal" operator in Python?
• A simple example of an unequal operator. In this example, the variable int_x is set to 20 and int_y = 30.
• An example of a string object comparison.
• Proof of the equality operator (==) with a while loop.
• An example of obtaining even numbers by another operator.
## How to divide whole numbers
Use strategies and algorithms based on knowledge of location, equality, and properties of operations to divide multi-digit integers by one- or two-digit numbers. Strategies can include mental strategies, quotient, commutative, associative and distributive properties, and repeated subtractions.
## What is the rule for dividing decimals by whole numbers?
Rules for dividing decimals: Rules for dividing a decimal by an integer: Put the decimal values in the quotient immediately above the decimal number of the dividend, then divide as usual. Alignment is crucial in a division.
## How is dividing decimals similar to dividing whole numbers?
Dividing decimal numbers is similar to dividing integers, except you have to learn the decimal point before you start dividing. How to divide decimal numbers step by step: Move the decimal point to the divisor and the dividend. Convert the divisor (the number you're dividing by) to an integer by moving the decimal point all the way to the right.
## How do you divide a whole number with a fraction?
To divide whole numbers by fractions, they use the same procedure as when dividing a fraction by a fraction: you return (back) the denominator (bottom number) and multiply. To do this with whole numbers, remember that a whole number can be written as a fraction by making the denominator equal to 1. So 3 is the same as 3/1, 4 is 4/1, and so on.
## How to divide exponents
The exponential rule for dividing exponential terms is called the quotient rule. The quotient rule for exponents is that dividing exponential terms with the same base keeps the same base and then subtracts the exponents.
## How do you divide exponents with different bases?
Divide the indicators for several reasons. If the bases are different and the exponents of a and b are equal, you can divide a and b first: an / bn = (a / b) n. Example: 6 3/2 3 = (6/2) 3 = 3 3 = 3⋅3⋅3 = 27. If the bases and sizes are different, you need to calculate each size and then divide: an / b m.
## When to multiply exponents?
The members can only be multiplied by an exponent if the bases are equal. Multiply the terms by adding the exponents. Example: 2^3 * 2^4 = 2^(3 + 4) = 2^7. General rule: x^a * x^b = x^(a + b). Calculate each term separately if the basic concepts of the terms are not the same.
## What are the division properties of exponents?
• Division of numbers in exponential notation. Simplify (4 x 109) ÷ (16 x 106) and write your answer in exponential notation.
• The positive power of a particular trait.
• Find degrees of positive quotient.
• The negative power of private property.
• Find negative degrees of quotients.
## How to divide in excel
Divide the two numbers. Distribution of the formula: = number1 / number2. What it means: = the number to divide by / the number to divide by. It's easy to divide numbers in Excel! The division formula is done using the slash division operator: /.
## How do you split a box in Excel?
To divide a cell diagonally in Excel, follow these steps: 1. Right click the specified cell you want to divide diagonally, and select Format Cells from the context menu. See screenshot: 2. In the Format Cells dialog box, click to activate the Borders tab, click the button in the Borders section to select it, and then click OK.
## How do you multiply two cells together in Excel?
To multiply two cells in Excel, use the multiplication formula as in the example above, but use cell references instead of numbers. For example, to multiply the value in cell A2 by the value in B2, enter this expression: =A2 * B2. To multiply multiple cells, add additional cell references to the formula, separated by the multiplication sign.
## What is the function of Division in Excel?
There is no special division function in Excel. Instead, it's pretty easy to use the "/" operator.
## How to divide decimals
You know that the decimal point is to the right of the square. So if you divide by ten (ten times a certain number of times, such as ten, hundred, thousand, etc.), the decimal point shifts to the left. For every power of ten, you move one square to the left.
## How do you divide a decimal by a whole number?
The rule for dividing a decimal number by an integer: put the decimal values in the quotient directly above the decimal number in the dividend, then divide as usual. Alignment is crucial in a division.
## Where does the decimal go when dividing?
When you multiply a decimal number by a power of ten, the decimal point moves to the right by the number of digits indicated by the exponent. When you divide a decimal number by a power of ten, the decimal point moves the number of digits indicated by the exponent to the left in the opposite direction.
## How to divide hostas
When choosing a time to separate hosts, it is recommended to check the forecast. Most gardeners like to divide their hosts before the seasonal spring or fall rains. If you are sharing samples in the spring, try when the plants are growing, but before the leaves begin to bloom.
## When to split and transplant hostas?
The best time to separate and transplant hosts is August or September, about a month before the first frost.
## How and when to transplant hostas?
Hosts prefer moist, well-drained soil in a shady spot. The best time to transplant hosts is spring or fall. Place a measuring spoon 6 to 8 inches from the clump. Dig about 10 centimeters. Lift the bump in one piece.
## Can you split hostas anytime?
While spring is the best time to share your host, plants can be shared anytime from spring to late summer. Common late summer hosts should be covered with a few inches of straw, pine needles or other material in late fall.
## How to do long division
Subject: Re: In what year do you study long division? Overall 4. However, many children have yet to work on this in fifth grade, especially if they have more assignments, and some children who are more advanced in math in primary school take about a year longer.
## What is the formula for long division?
Here's a tip to help you master long throws. Use the abbreviation DMSB, which means: D = Divide. M = multiply. S = subtract. B = precipitate. This sequence of letters can be difficult to remember, so think of the acronym in the context of your family: father, mother, sister, brother.
## What is the best way to explain long division?
• Start with numbers that are evenly distributed. Long division is easier to understand if you start with a large number that divides without remainder.
• Explain how to divide the divisor by the first number in the dividend.
• Show your student or child how to find others to take responsibility.
## How do you do long division with decimals?
How to make long division with decimals. If the number you're dividing with has a decimal point, move the decimal point all the way to the right by counting the number of shifted digits. Then move the decimal point to the right on the number you are dividing by the same number of digits.
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# Cholesky Decomposition Calculator
Created by Rijk de Wet
Reviewed by Wojciech Sas, PhD and Steven Wooding
Last updated: Jan 18, 2024
Welcome to the Cholesky decomposition calculator. In this accompanying text to the tool, we'll learn all there is to know about the Cholesky factorization, which decomposes a matrix into a product of matrices. We'll specifically cover how to calculate the Cholesky decomposition and an example of Cholesky decomposition for a $3×3$ matrix.
## What is a matrix decomposition?
Before we can learn about the Cholesky decomposition, we must define what a matrix decomposition is. In the world of linear algebra, a matrix decomposition (or a matrix factorization) is the factorization of a matrix into a product of matrices. So, just as you can factorize $16$ into groups of products (like $4\times4$ or $2\times8$) you can factorize a matrix like $A$ below… but how?
$\footnotesize A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \\ \end{bmatrix}$
Factorizing a matrix is much harder than a number. Lucky for us, mathematicians have discovered many different methods of performing matrix decompositions. The most famous of these methods are the LU decomposition, the QR decomposition, the singular value decomposition (SVD), and the Cholesky decomposition. Which method you'd want to use depends on the problem you're trying to solve.
## What is the Cholesky decomposition?
Knowing what matrix decomposition is, we can go on to define the Cholesky decomposition. The Cholesky decomposition (or the Cholesky factorization) is the factorization of a matrix $A$ into the product of a lower triangular matrix $L$ and its transpose. We can rewrite this decomposition in mathematical notation as:
$\footnotesize A = L\cdot L^T$
To be Cholesky-decomposed, matrix $A$ needs to adhere to some criteria:
• $A$ must be symmetric, i.e. $A^T = A$.
• By extension, this means $A$ must be square.
• $A$ must be positive definite (meaning its eigenvalues must all be positive).
If $A$ doesn't tick all the items on our list, no suitable $L$ can exist.
## How to calculate the Cholesky decomposition?
The goal of any matrix decomposition method is to find the factorization's terms, and so we want to find the lower triangular matrix $L$. The Cholesky decomposition has no single mathematical formula, but it is easily obtained by hand for a small matrix. For larger matrices, there's an algorithmic process to follow.
Let's begin by looking at the simple $2\times2$ matrix case symbolically. We first define:
$\footnotesize A = \begin{bmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ \end{bmatrix}$
…and:
$\footnotesize L = \begin{bmatrix} b_{1,1} & 0 \\ b_{2,1} & b_{2,2} \\ \end{bmatrix}$
Note that the elements of $L$ above its diagonal are zero, as $L$ is a lower triangular matrix.
We know from the definition of the Cholesky factorization that $A = L\cdot L^T$, so let's take a look at the right-hand side of this equation.
$\footnotesize \begin{split} &\ L\cdot L^T \\ =&\ \begin{bmatrix} b_{1,1} & 0 \\ b_{2,1} & b_{2,2} \\ \end{bmatrix} \cdot \begin{bmatrix} b_{1,1} & b_{2,1} \\ 0 & b_{2,2} \\ \end{bmatrix} \\ =&\ \begin{bmatrix} (b_{1,1})^2 & b_{1,1}\cdot b_{2,1} \\ b_{1,1}\cdot b_{2,1} & (b_{2,1})^2 + (b_{2,2})^2 \\ \end{bmatrix} \end{split}$
Knowing the above and that $A = L\cdot L^T$, we can just equate corresponding elements of $A$ and $L\cdot L^T$ and solve the equations:
\footnotesize \begin{alignat*}{2} & a_{1,1} = (b_{1,1})^2 \\ \therefore\ & b_{1,1} = \sqrt{a_{1,1}} \\ & a_{2,1} = b_{1,1}\cdot b_{2,1} \\ \therefore\ & b_{2,1} = a_{2,1}\ /\ b_{1,1} \\ & a_{2,2} = (b_{2,1})^2 + (b_{2,2})^2 \\ \therefore\ & b_{2,2} = \sqrt{a_{2,2} - (b_{2,1})^2} \end{alignat*}
Notice that we need earlier elements of $L$ to solve for the later elements: $b_{2,2}$ needs $b_{2,1}$, which needs $b_{1,1}$.
We can use the equations above to solve $L$ for the $3\times3$ matrix $A$ that we defined at the top:
$\footnotesize \begin{split} A &= \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} \\ \therefore L &= \begin{bmatrix} \sqrt{2} & 0 \\ \frac{3}{\sqrt{2}} & \sqrt{\frac{1}{2}}\end{bmatrix} \end{split}$
The principles of the above example apply to the Cholesky factorization of any sized matrix. For larger matrices, we can generalize the process with the following two equations.
For elements on $L$'s diagonal:
$\footnotesize b_{j,j} = \sqrt{a_{j,j} - \sum_{k=1}^{j-1} (b_{j,k})^2}$
For elements off $L$'s diagonal:
$\footnotesize b_{i,j} = \frac{1}{b_{j,j}}\left(a_{j,j} - \sum_{k=1}^{j-1} b_{i,k}\cdot b_{j,k}\right)$
## How to use the Cholesky decomposition calculator?
With the help of our calculator, you can easily calculate $L$ if you know what $A$ is. Remember, $A$ must be symmetric and positive definite. If it's not, the calculator can't give you $L$ as no $L$ that complies with $A = L\cdot L^T$ exists.
1. Select $A$'s shape. Our Cholesky decomposition solver supports $2\times2$, $3\times3$, and $4\times4$ matrices.
2. Give the calculator your matrix $A$. The matrix's elements are separated by row — see the graphical representation at the top of the calculator if you're unsure.
3. Find the result $L$ below. Our Cholesky decomposition solver will calculate $L$ and display it below your matrix. If $A$ is not symmetric and positive definite, the calculator will notify you accordingly.
## Example Cholesky decomposition of a 3×3 matrix
Now that we know how to compute the Cholesky factorization for a matrix of any size, let's do an example. We'll use the general-case algorithmic equations we've just discussed to solve $L$ given the following $3\times3$ matrix $A$:
$\footnotesize A = \begin{bmatrix} 25 & 15 & 5 \\ 15 & 13 & 11 \\ 5 & 11 & 21 \\ \end{bmatrix}$
We can jump straight into the solving process, starting at the top left and moving left-to-right, top-to-bottom.
\footnotesize \begin{alignat*}{2} b_{1,1} &= \sqrt{a_{1,1}} &= 5 \\ b_{2,1} &= a_{2,1}\ /\ b_{1,1} &= 3 \\ b_{2,2} &= \sqrt{a_{2,2} - (b_{2,1})^2} &= 2 \\ b_{3,1} &= a_{3,1}\ /\ b_{1,1} &= 1 \\ b_{3,2} &= (a_{3,2} - b_{3,1}\cdot b_{2,1})\ /\ b_{2,2} &= 4 \\ b_{3,3} &= \sqrt{a_{3,3} - (b_{3,1})^2 - (b_{3,2})^2} &= 2 \\ \end{alignat*}
And so, after all our effort, we've fully obtained $L$,
$\footnotesize L = \begin{bmatrix} 5 & 0 & 0 \\ 3 & 2 & 0 \\ 1 & 4 & 2 \\ \end{bmatrix}$
Finally, to test our answer, we can see if $L\cdot L^T$ really is equal to $A$:
$\footnotesize \begin{split} &\ L\cdot L^T \\ =&\ \begin{bmatrix} 5 & 0 & 0 \\ 3 & 2 & 0 \\ 1 & 4 & 2 \end{bmatrix} \cdot \begin{bmatrix} 5 & 3 & 1 \\ 0 & 2 & 4 \\ 0 & 0 & 2 \end{bmatrix} \\ =&\ \begin{bmatrix} 5^2 & 5\!\!\times\!\!3 & 5\!\!\times\!\!1 \\ 3\!\!\times\!\!5 & 3^2 + 2^2 & 3\!\!\times\!\!1 + 2\!\!\times\!\!4 \\ 1\!\!\times\!\!5 & 1\!\!\times\!\!3 + 4\!\!\times\!\!2 & 1^2 + 4^2 + 2^2 \\ \end{bmatrix} \\ =&\ \begin{bmatrix} 25 & 15 & 5 \\ 15 & 13 & 11 \\ 5 & 11 & 21 \end{bmatrix} \\ =&\ A \end{split}$
Perfect! We've just performed a complete decomposition of $A$. You can also try out this example in our Cholesky decomposition calculator.
## What are the applications of the Cholesky decomposition?
Matrix decomposition methods exploit the structure of the factorization's terms to make solving system of equations easier. Remember how we said that $L$ is lower triangular? Lower triangular matrices are especially easy to work with, and therefore the Cholesky decomposition is frequently the method of choice in solving systems of equations.
## FAQ
### How do you determine whether a matrix has a Cholesky decomposition?
For the matrix A to have a Cholesky decomposition, it must be symmetric, and it must be positive definite (meaning it must have only positive eigenvalues). If A does not adhere to these requirements, it cannot have a Cholesky decomposition, meaning no matrix L that satisfies L · LT = A can exist.
### What does the Cholesky decomposition do?
Like any matrix decomposition method, the Cholesky decomposition takes a matrix A and factorizes it. It produces a lower triangular matrix L, which when multiplied with its transpose LT produces the original matrix A. This is valuable in many matrix operations, as the structure of a lower triangular matrix can be exploited to make the operations compute much faster.
### What is a symmetric matrix?
A symmetric matrix is one that is equal to its transpose. Redefined mathematically, symmetric matrices satisfy the condition A = AT. This has the implication that the matrix must also be square. Graphically, a symmetric matrix is symmetric along its main diagonal.
### What is a positive definite matrix?
A positive definite matrix is a matrix with only positive eigenvalues. The formal mathematical definition is that matrix A is positive definite if z⃗T·A·z⃗ > 0 for every nonzero column vector z⃗. Alternatively, A is only positive definite if it has a Cholesky decomposition.
### What is the Cholesky decomposition of the identity matrix?
The identity matrix's Cholesky decomposition is also the identity matrix. This is because the identity matrix's transpose is itself (ɪT = ɪ) and so ɪ · ɪT = ɪ·ɪ = ɪ. In general, the Cholesky decomposition L of a diagonal matrix D is also diagonal, and its diagonal entries are the square roots of D's. When the diagonal entries are all 1 (like they are in ɪ), we get the identity matrix back.
Rijk de Wet
A=
⌈ a₁ a₂ ⌉ ⌊ b₁ b₂ ⌋
Matrix size
2x2
First row
a₁
a₂
Second row
b₁
b₂
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# What is an expression in math terms?
## What is an expression in math terms?
In mathematics, an expression or mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context.
What is an expression in a formula?
An expression is a number, a variable, or a combination of numbers and variables and operation symbols. An equation is made up of two expressions connected by an equal sign.
### What is an expression used to calculate a desired result is called a?
An expression used to calculate a desired result, such as a formula to find volume or a formula to count combinations. Formulas can also be equations involving numbers and/or variables, such as Euler’s formula.
What is a math expression example?
We write an expression by using numbers or variables and mathematical operators which are addition, subtraction, multiplication, and division. For example, the expression of the mathematical statement “4 added to 2”, will be 2+4.
## What are the types of algebraic expressions?
There are 3 main types of algebraic expressions which include:
• Monomial Expression.
• Binomial Expression.
• Polynomial Expression.
How do you find expressions?
To evaluate an algebraic expression, you have to substitute a number for each variable and perform the arithmetic operations. In the example above, the variable x is equal to 6 since 6 + 6 = 12. If we know the value of our variables, we can replace the variables with their values and then evaluate the expression.
### Is an expression a formula?
In context|mathematics|lang=en terms the difference between expression and formula. is that expression is (mathematics) an arrangement of symbols denoting values, operations performed on them, and grouping symbols while formula is (mathematics) any mathematical rule expressed symbolically.
Is 2x an expression?
This type of expression has only one term, for example, 2x, 5x 2 ,3xy, etc. An algebraic expression having two, unlike terms, for example, 5y + 8, y+5, 6y3 + 4, etc. This is an algebraic expression with more than one term and with non -zero exponents of variables.
## What are the 3 types of equations?
Different Types of Equations
• Linear Equation.
• Exponential Equation.
• Rational Equation.
How do you classify expressions?
You call an expression with a single term a monomial, an expression with two terms is a binomial, and an expression with three terms is a trinomial. An expression with more than three terms is named simply by its number of terms. For example a polynomial with five terms is called a five-term polynomial.
### What are the examples of expression?
The definition of an example of expression is a frequently used word or phrase or it is a way to convey your thoughts, feelings or emotions. An example of an expression is the phrase “a penny saved is a penny earned.” An example of an expression is a smile. A facial aspect or a look that conveys a special feeling.
How do you simplify expressions?
To simplify any algebraic expression, the following are the basic rules and steps:
1. Remove any grouping symbol such as brackets and parentheses by multiplying factors.
2. Use the exponent rule to remove grouping if the terms are containing exponents.
3. Combine the like terms by addition or subtraction.
4. Combine the constants.
## How do I evaluate the expression?
Determine how to evaluate the expression. Evaluate means to simplify an expression. When you are asked to evaluate an algebraic expression, you need to plug a given value for the variable into the expression and solve. For example, you might be asked to evaluate 2x{displaystyle 2x} when x=2{displaystyle x=2}.
How do you evaluate each expression?
Evaluate each expression is very simple concept in math. When you substitute a particular value for each variable and then act the operations called evaluating expression. The mathematical expression is algebraic, it involves a finite sequence of variables and numbers and then algebraic operations.
### How do you evaluate expression in Algebra?
To evaluate an algebraic expression, you have to substitute a number for each variable and perform the arithmetic operations. In the example above, the variable x is equal to 6 since 6 + 6 = 12. If we know the value of our variables, we can replace the variables with their values and then evaluate the expression.
How do you simplify an expression in math?
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.
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# Lesson 18
The Quadratic Formula and Complex Solutions
• Let’s use the quadratic formula to find complex solutions to quadratic equations.
### Problem 1
Clare solves the quadratic equation $$4x^2+12x+58=0$$, but when she checks her answer, she realizes she made a mistake. Explain what Clare's mistake was.
\begin{align} x &= \frac{\text-12 \pm \sqrt{12^2 - 4 \boldcdot 4 \boldcdot 58}}{2 \boldcdot 4} \\ x &= \frac{\text-12 \pm \sqrt{144-928}}{8} \\ x &= \frac{\text-12 \pm \sqrt{\text-784}}{8} \\ x &= \frac{\text-12 \pm 28i}{8} \\ x &= \text-1.5 \pm 28i \\ \end{align}
### Problem 2
Write in the form $$a+bi$$, where $$a$$ and $$b$$ are real numbers:
1. $$\frac{5 \pm \sqrt{\text-4}}{3}$$
2. $$\frac{10 \pm \sqrt{\text-16}}{2}$$
3. $$\frac{\text-3 \pm \sqrt{\text-144}}{6}$$
### Problem 3
Priya is using the quadratic formula to solve two different quadratic equations.
For the first equation, she writes $$x= \frac{4 \pm \sqrt{16-72}}{12}$$
For the second equation, she writes $$x=\frac{8 \pm \sqrt{64-24}}{6}$$
Which equation(s) will have real solutions? Which equation(s) will have non-real solutions? Explain how you know.
### Problem 4
Find the exact solution(s) to each of these equations, or explain why there is no solution.
1. $$x^2=25$$
2. $$x^3 = 27$$
3. $$x^2=12$$
4. $$x^3=12$$
(From Unit 3, Lesson 8.)
### Problem 5
Kiran is solving the equation $$\sqrt{x+2} - 5 = 11$$ and decides to start by squaring both sides. Which equation results if Kiran squares both sides as his first step?
A:
$$x + 2 - 25 = 121$$
B:
$$x + 2 + 25 = 121$$
C:
$$x+2 - 10\sqrt{x+2} + 25 = 121$$
D:
$$x+2 + 10\sqrt{x+2} + 25 = 121$$
(From Unit 3, Lesson 9.)
### Problem 6
Plot each number on the real or imaginary number line.
1. $$\text-\sqrt{4}$$
2. $$\sqrt{\text-1}$$
3. $$3\sqrt{4}$$
4. $$\text-3\sqrt{\text-1}$$
5. $$4\sqrt{\text-1}$$
6. $$2\sqrt{2}$$
(From Unit 3, Lesson 10.)
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Power Reduction Calculator
Exact value of
given
=
u
≤
Value of
, u =
(
Solution:
Power Reduction Identity Tutorial
The Power Reduction formulas are shown below:
\text{sin}^2\text{(u)}=\frac{1-(1-2\cdot \text{sin}^2\text{(u)})}{2}
\text{cos}^2\text{(u)}=\frac{1+(1-2\text{sin}^2(\text{u}))}{2}
\tan ^2\text{(u)}=\frac{1+(1-2\text{sin}^2(\text{u}))}{1-(1-2\text{sin}^2(\text{u}))}
\csc ^2\text{(u)}=\frac{2}{1-(1-2\text{sin}^2(\text{u}))}
\sec ^2\text{(u)}=\frac{2}{1+(1-2\text{sin}^2(\text{u}))}
\text{cot}^2\text{(u)}=\frac{1-(1-2\text{sin}^2(\text{u}))}{1+(1-2\text{sin}^2(\text{u}))}
To find the exact value of any of the power reduction identites given a ratio given a ratio and quadrant can be done. For example, to find the exact value of $sin^2u$ given tanu = $\frac{5}{2}$ in quadrant 3 you first have to find the values required for the identity. According to the formula for the power reduction formula for sin, the exact value of sin is needed. This means that the first step in solving the identity is finding the exact value of sin in quadrant 3. Since the quadrant of the given function is 3, the exact value of sin must be negative. The calculation process for finding sin in quadrant 1 is shown below:
\text{ opposite}=5
\text{hypotenuse}=\sqrt{5^2+2^2}
\text{hypotenuse}=\sqrt{29}
\text{sin}=\frac{\text{opposite}}{\text{hypotenuse}}=-\frac{5\sqrt{29}}{29}
\text{sin}=-\frac{5\sqrt{29}}{29}
\text{(sin must be negative in quadrants 3 and 4)}
\text{(in simplest form)}
Now that the exact value of sin in quadrant three is known, it can be substituted into the formula for $sin^2u$. The step by step process of this is shown below:
\text{sin}^2\text{u}=\frac{1-(1-2\cdot \text{sin}^2\text{u})}{2}
\text{sin}^2\text{u}=\frac{1-(1-2(-\frac{5\sqrt{29}}{29})^2)}{2}
\text{sin}^2\text{u}=\frac{25}{29}
\text{(in simplest form)}
Tutorial two
To solve power reduction identites without given values and getting decimal approximations is an easier process. To find these values you simply square any of the trigonometric function values. The value of $\text{sin}^2\13$ is the same as $\sin(13)\cdot\sin(13)$. If told to find the value of $sin^2(22)$ the calculation process below can be used. The value of $\text{cos}^2\text{(u)}$ given cscu = $\frac{3}{2}$ being that the angle is in quadrant 2, the first step would be to know the formula. The formula for $\text{cos}^2\text{(u)}$ is $\frac{1+(1-2\text{sin}^2(\text{u}))}{2}$. According to this formula the only value needed for the calculation is $\sin$. To find $\sin$ in quadrant 2 you must first find the opposite side length since that is needed for the value of sin. to find the opposite side length the calculation process for finding the opposite side length is shown below:
\text{csc}=\frac{3}{2}=\frac{\text{hypotenuse}}{\text{opposite}}
\text{ hypotenuse}=3
\text{sin}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{2}{3}
\text{sin}=\frac{2}{3}
Now that the opposite side length is known, the value of sin can be calculated. since sin is the ratio of the opposite side length and the hypotenuse, the opposite side length will be the numerator and the hypotenuse will be the denominator. The final part of solving this identity is substituting the value of sin into the equation. The calculation process for this is shown below:
\text{sin}^2\text{u}=\frac{1-(1-2\cdot \text{sin}^2\text{u})}{2}
\text{sin}^2\text{u}=\frac{1-(1-2(\frac{2}{3})^2)}{2}
\text{sin}^2\text{u}=\frac{4}{9}
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# Does row addition change determinant?
## Does row addition change determinant?
If we add a row (column) of A multiplied by a scalar k to another row (column) of A, then the determinant will not change. If we swap two rows (columns) in A, the determinant will change its sign.
## How do row and column operations affect the determinant?
You can do the other row operations that you’re used to, but they change the value of the determinant. The rules are: If you interchange (switch) two rows (or columns) of a matrix A to get B, then det(A) = –det(B).
How do you expand the determinant of a matrix?
If you factor a number from a row, it multiplies the determinant. If you switch rows, the sign changes. And you can add or subtract a multiple of one row from another. When the matrix is upper triangular, multiply the diagonal entries and any terms factored out earlier to compute the determinant.
Do row operations change the determinant of a matrix?
Proof: Key point: row operations don’t change whether or not a determinant is 0; at most they change the determinant by a non-zero factor or change its sign. Use row operations to reduce the matrix to reduced row-echelon form.
### Does swapping columns affect determinant?
Yes. Swapping two rows, or columns, changes the sign of the determinant (i.e., has the effect of multiplying the determinant by -1.) so becomes and there’s a change of sign unless the determinant is 0.
### What is row and column in determinant?
Determinant evaluated across any row or column is same. If all the elements of a row (or column) are zeros, then the value of the determinant is zero. Determinant of a Identity matrix ( ) is 1. If rows and columns are interchanged then value of determinant remains same (value does not change).
What is row operation in matrix?
Row operations are calculations we can do using the rows of a matrix in order to solve a system of equations, or later, simply row reduce the matrix for other purposes. This means that if we are working with an augmented matrix, the solution set to the underlying system of equations will stay the same. …
How do you expand a column determinant?
Expanding to Find the Determinant
1. Pick any row or column in the matrix. It does not matter which row or which column you use, the answer will be the same for any row.
2. Multiply every element in that row or column by its cofactor and add. The result is the determinant.
## What is row reduction?
Row reduction (or Gaussian elimination) is the process of using row operations to reduce a matrix to row reduced echelon form. This procedure is used to solve systems of linear equations, invert matrices, compute determinants, and do many other things.
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# Samantha invests $1000 at 6% per annum compounded quarterly. Mark invests $1200 at 5.5% per annum compounded quarterly. When will the balances in their accounts be equal?
Aug 31, 2016
The balances in the accounts will be equal in $37$ years.
#### Explanation:
For problems such as these, we use the formula $A = P {\left(1 + \frac{r}{n}\right)}^{n t}$, where P is the amount of money you begin with, $r$ is the rate of interest, $n$ is the frequency the interest is compounded and A is the final amount. $t$ is the time in years.
We have to write a system of equations and solve for $A$ and for $t$.
Equation 1:
$A = 1000 {\left(1 + \frac{0.06}{4}\right)}^{4 t}$
Equation 2:
$A = 1200 {\left(1 + \frac{0.055}{4}\right)}^{4 t}$
Substitute:
$1200 {\left(1 + \frac{0.055}{4}\right)}^{4 t} = 1000 {\left(1 + \frac{0.06}{4}\right)}^{4 t}$
$1200 {\left(1.01375\right)}^{4 t} = 1000 {\left(1.015\right)}^{4 t}$
$1200 {\left(1.01375\right)}^{4 t} - 1000 {\left(1.015\right)}^{4 t} = 0$
Solve using a graphing calculator. If you're using a TI-83 or a TI-84, change the t's to x's. Enter ${y}_{1} = 1200 {\left(1.01375\right)}^{4 x} - 1000 {\left(1.015\right)}^{4 x}$ and ${y}_{2} = 0$. Then press CALC$\to$ Intersect.
You will of course want a positive intersection. Once you have moved your cursor sufficiently, press ENTER twice before the calculator will say guess? followed by INTERSECTION.
The result it gives you should say $36.99$, or $37$ years.
Hopefully this helps!
|
Working with quadratic equations can be a tricky endeavor. If you want to solve them and use them in your math problems, it’s important to understand how to convert a quadratic expression into vertex form. It's also worth noting that vertex form has some advantages over standard form when solving certain types of problems; for example, it’s much easier to find the maximum and minimum values using vertex form than standard form because everything related to those values (such as coordinates) are already isolated and easy to read off from a single line.
## What is Vertex Form?
Vertex form of a quadratic equation is an equation that is written in the shape of a parabola when graphed. The vertex is the highest point on the graph and usually represents where the maximum or minimum value of the equation lies. It's important to recognize that all quadratics can be written in vertex form, as long as you know how to manipulate the equation correctly.
To convert a standard quadratic expression into vertex form, we start by Completing the square. This involves taking half of the coefficient from the x-squared term and adding it to both sides of the equation, then completing square by squaring that number to complete one side of the equation. We then subtract this number from both sides, which will move all terms with an x-variable onto one side and all constants onto the other side of your equation. Finally, we divide both sides by our coefficient for x-squared so that we can properly isolate our x-variable and put it in a more recognizable form.
## Conclusion:
In conclusion, converting a quadratic expression into vertex form is an essential skill for anyone working with quadratics in mathematics or any other field that requires advanced problem-solving skills. It's easy enough once you get used to it; just remember that completing the square is key. With some practice, soon you'll be able to quickly identify key components like maximum/minimum values or graph points just by glancing at an equation written in its vertex form.
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## Matrix Problems and Solutions (Olympiad Level)
The following matrix problems are provided along with the solutions. The problems are in high school olympiad level, so you need higher thinking skills to tackle them. It may not easy, but keep learning and you can.
### Quote by Arthur Ashe
Success is a journey, not a destination. The doing is often more important than the outcome.
Multiple Choice Section
Problem Number 1
If given $A = \begin{pmatrix} 6 &-3 & 2 \\ 1 & 0 & 4 \\ 5 & 7 & 6 \end{pmatrix}$ and $B = \begin{pmatrix} 1 &-9 & 4 \\ 2 & 2 &-3 \\ 3 & 13 &-10 \end{pmatrix}$, then what is the value of $\det(B^{-1}(A^{-1}B^{-1})^{-1}A^{-1})$?
A. $-1$ C. $1$ E. $5$
B. $0$ D. $3$
Solution
Use the following matrix inversing theorem.
\boxed{\begin{aligned} (A^{-1}B^{-1})^{-1} & = BA \\ A^{-1} \cdot A & = A \cdot A^{-1} = I \\ I \cdot I & = I \end{aligned}}
and note that the determinant of identity matrix $I$ always equals $1$.
Therefore, we have
\begin{aligned} & \det(B^{-1}(A^{-1}B^{-1})^{-1}A^{-1}) \\ & = \det(B^{-1}(BA)A^{-1}) \\ & = \det((B^{-1}B) (AA^{-1})) \\ & = \det(I \cdot I) \\ & = \det(I) = 1 \end{aligned}
Thus, the determinant of $B^{-1}(A^{-1}B^{-1})^{-1}A^{-1}$ equals $\boxed{1}$
(Answer C)
[collapse]
Problem Number 2
Let $M$ be a matrix such that
$M \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ a- c & b- d \end{pmatrix}$. The determinant of matrix $M$ equals $\cdots \cdot$
A. $-1$ C. $1$ E. $3$
B. $0$ D. $2$
Solution
Given that $M \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ a- c & b- d \end{pmatrix}$
Therefore, we will have
\begin{aligned} |M| \cdot \begin{vmatrix} a & b \\ c & d \end{vmatrix} & = \begin{vmatrix} a & b \\ a- c & b- d \end{vmatrix} \\ |M| \cdot (ad- bc) & = a(b- d)- b(a- c) \\ |M| \cdot (ad- bc) & = \cancel{ab}- ad \cancel{- ab} + bc \\ |M| \cdot (ad- bc) & =-(ad- bc) \\ |M| & = \dfrac{-(ad-bc)}{ad-bc} =-1 \end{aligned}
Thus, the determinant of matrix $M$ equals $\boxed{-1}$
(Answer A)
[collapse]
Problem Number 3
If given $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = 3$, then $\begin{vmatrix} 2a + d& a & 4a+2d+g \\ 2b+e & b & 4b+2e+h \\ 2c+f & c & 4c+2f+i \end{vmatrix} = \cdots \cdot$
A. $-3$ C. $0$ E. $3$
B. $-2$ D. $2$
Solution
It is given that
$\begin{vmatrix} a & b & c\\ d & e & f \\ g & h & i \end{vmatrix} = 3$
By swapping the first and second row entries, causing the determinant to be negative, we have
$\begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix} =-3$
Transpose the matrix, so we have
$\begin{vmatrix} d & a & g \\ e & b & h \\ f & c & i \end{vmatrix} =-3$
(Transposing a matrix doesn’t affect the determinant)
After that, add correspondingly each entry in the first column to two times of each entry in the second column (it also won’t change the determinant).
$\begin{vmatrix} 2a+d & a & g \\ 2b + e & b & h \\ 2c + f & c & i \end{vmatrix} =-3$
Finally, add correspondingly each entry in the third column to two times of each entry in the first column (it also won’t change the determinant).
$\begin{vmatrix} 2a+d & a & 4a+2d+g\\ 2b + e & b & 4b+2e+h \\ 2c + f & c & 4c+2f+i \end{vmatrix} =-3$
Thus, the value of
$\boxed{\begin{vmatrix} 2a+d & a & 4a+2d+g\\ 2b + e & b & 4b+2e+h \\ 2c + f & c & 4c+2f+i \end{vmatrix} =-3}$
(Answer A)
[collapse]
Problem Number 4
If matrix $A = \begin{pmatrix} 3 & 7 \\-1 &-2 \end{pmatrix}$, then $A^{27} + A^{31} + A^{40}$ equals $\cdots \cdot$
A. $\begin{pmatrix} 5 & 1 \\ 3 &-4 \end{pmatrix}$
B. $\begin{pmatrix}-7 & 14 \\-2 & 3 \end{pmatrix}$
C. $\begin{pmatrix} 7 &-14 \\ 3 & 2 \end{pmatrix}$
D. $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
E. $\begin{pmatrix}-1 & 0 \\ 0 &-1 \end{pmatrix}$
Solution
Given $A = \begin{pmatrix} 3 & 7 \\-1 &-2 \end{pmatrix}$
Try to spot the pattern by finding the power of matrix $A$.
\begin{aligned} A^2 & = \begin{pmatrix} 3 & 7 \\-1 &-2 \end{pmatrix} \begin{pmatrix} 3 & 7 \\-1 &-2 \end{pmatrix} = \begin{pmatrix} 2 & 7 \\-1 &-3 \end{pmatrix} \\ A^3 & = A^2 \cdot A = \begin{pmatrix} 2 & 7 \\-1 &-3 \end{pmatrix} \begin{pmatrix} 3 & 7 \\-1 &-2 \end{pmatrix} \\ & = \begin{pmatrix}-1 & 0 \\ 0 &-1 \end{pmatrix} =- \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} =-I\end{aligned}with $I$ denotes the identity matrix.
After then,
\begin{aligned} & A^{27} + A^{31} + A^{40} \\ & = A^{27}(I + A^4 + A^{13}) \\ & = (A^3)^9(I + A^3A + (A^3)^4A) \\ & = (-I)^9(I- IA + (-I)^4A) \\ & =-I(I-A + A) \\ & =-I = \begin{pmatrix}-1 & 0 \\ 0 &-1 \end{pmatrix} \end{aligned}
Thus, $\boxed{A^{27} + A^{31} + A^{40} = \begin{pmatrix}-1 & 0 \\ 0 &-1 \end{pmatrix}}$
Note: Any power of a identity matrix always yields a same identity matrix.
(Answer E)
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Problem Number 5
If a constant $k$ satisfies the matrix equation
$\begin{pmatrix} k & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x-1 \\ y-1 \end{pmatrix} = \begin{pmatrix} 0 \\ k \end{pmatrix}$,
then the value of $x+y=\cdots \cdot$
A. $(2+k) (1+k)$
B. $(2-k) (1-k)$
C. $(2+k) (1-k)$
D. $(1-k) (1+k)$
E. $(2-k) (1+k)$
Solution
By applying matrix multiplication rule, we have
\begin{aligned} \begin{pmatrix} k & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x- 1 \\ y- 1 \end{pmatrix} & = \begin{pmatrix} 0 \\ k \end{pmatrix} \\ \begin{pmatrix} k(x-1) + 1(y-1) \\ 1(x-1) + 0(y-1) \end{pmatrix} & = \begin{pmatrix} 0 \\ k \end{pmatrix} \\ \begin{pmatrix} k(x-1)+(y-1) \\ x-1 \end{pmatrix} & = \begin{pmatrix} 0 \\ k \end{pmatrix} \end{aligned}
Hence, we have the following equation system.
$\begin{cases} k(x-1) + (y-1) = 0 & (\cdots 1) \\ x- 1 = k & (\cdots 2) \end{cases}$
Substitute equation $(2)$ into equation $(1)$.
$k(k) + y- 1 = 0 \Leftrightarrow y = 1- k^2$
Therefore, we have
\begin{aligned}x+y & =(k+1)+(1-k^2) \\ & =(k+1)+(k+1)(1-k) \\ & = (k+1)(1+(1-k)) \\ & = (k+1)(2-k) \end{aligned}
So, the value of $x+y$ is $(k+1)(2-k)$ or can be rewritten into $\boxed{(2-k) (1+k)}$
(Answer E)
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Problem Number 6
If $x : y = 5 : 4$, then the value of $x$ and $y$ that satisfy
$\begin{pmatrix} 2 & 10 & 1 \end{pmatrix} \begin{pmatrix} x & y \\ 4 & 5 \\ 30 & 25 \end{pmatrix}\begin{pmatrix} 5 \\ 10 \end{pmatrix} = 1.360$
is $\cdots \cdot$
A. $x = 1$ and $y = \dfrac45$
B. $x=\dfrac45$ and $y=1$
C. $x=5$ and $y=4$
D. $x=-10$ and $y=-8$
E. $x=10$ and $y=8$
Solution
Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$.
Hence, we have
\begin{aligned} \begin{pmatrix} 2 & 10 & 1 \end{pmatrix} \begin{pmatrix} 5k & 4k \\ 4 & 5 \\ 30 & 25 \end{pmatrix}\begin{pmatrix} 5 \\ 10 \end{pmatrix} & = 1.360 \\ \begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \end{pmatrix} \begin{pmatrix} 5 \\ 10 \end{pmatrix} & = 1.360 \\ \begin{pmatrix} 10k + 70 & 8k + 75 \end{pmatrix} \begin{pmatrix} 5 \\ 10 \end{pmatrix} & =1.360 \\ 5(10k+70) + 10(8k+75) & = 1.360 \\ \text{Divide each side by}~& 5 \\10k+70 + 2(8k+75) & = 272 \\ 26k + 220 & = 272 \\ 26k & = 52 \\ k & = 2 \end{aligned}Thus, we have
\boxed{\begin{aligned} x & = 5k = 5(2) = 10 \\ y & =4k=4(2)=8 \end{aligned}}
(Answer E)
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Problem Number 7
It is given that $A = \begin{pmatrix} 2 & 1 \\ 3 & 5 \end{pmatrix}$ holds a certain relationship to matrix $B =\begin{pmatrix}-5 & 3 \\ 1 &-2 \end{pmatrix}$. If matrix $C = \begin{pmatrix} 3 & 2 \\ 1 &-5 \end{pmatrix}$ and matrix $D$ have the same relationship, then the value of $C + D = \cdots \cdot$
A. $\begin{pmatrix} 8 & 3 \\ 3 & 8 \end{pmatrix}$ D. $\begin{pmatrix} 3 & 8 \\ 8 & 3 \end{pmatrix}$
B. $\begin{pmatrix} 8 & 3 \\ 3 &-8 \end{pmatrix}$ E. $\begin{pmatrix}-8 &-3 \\-3 & 8 \end{pmatrix}$
C. $\begin{pmatrix} 3 & 8 \\-8 & 3 \end{pmatrix}$
Solution
Given that $A = \begin{pmatrix} 2 & 1 \\ 3 & 5 \end{pmatrix}$ and $B =\begin{pmatrix}-5 & 3 \\ 1 &-2 \end{pmatrix}$
Observe the matrices’ entries and you will spot its relationship.
Two entries of the main diagonal in matrix $A$ are swapped and multiplied by $-1$ to yield the entries of the main diagonal in matrix $B$.
Two entries of the antidiagonal in matrix $A$ are swapped to yield the entries of the antidiagonal in matrix $B$.
Mathematically, we write as
$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \iff B = \begin{pmatrix}-d & c \\ b &-a \end{pmatrix}$
Since $C = \begin{pmatrix} 3 & 2 \\ 1 &-5 \end{pmatrix}$, then by following the relationship pattern, we have
$D = \begin{pmatrix} 5 & 1 \\ 2 &-3 \end{pmatrix}$
Hence,
\begin{aligned} C + D & = \begin{pmatrix} 3 & 2 \\ 1 &-5 \end{pmatrix} + \begin{pmatrix} 5 & 1 \\ 2 &-3 \end{pmatrix} \\ & = \begin{pmatrix} 8 & 3 \\ 3 &-8 \end{pmatrix} \end{aligned}
(Answer B)
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Problem Number 8
If $A$ be a matrix with an order of $2 \times 2$ that has an inverse and $A^2-2A-I = 0$, then $A-2I = \cdots \cdot$
A. $(2A)^{-1}$ D. $A^2-2A$
B. $A^2+2A$ E. $A^{-1}$
C. $2I-A$
Solution
Start from $A^2-2A-I=0$ with $I$ denotes identity matrix and $0$ denotes zero matrix, we have
\begin{aligned} A^2-2A & =I \\ A^2-2IA & = I \\ A(A- 2I) & = I \end{aligned}
Based on the definition of an inverse of the matrix, $B$ is an inverse of matrix $A$ if it satisfies $AB=BA=I$
Thus, we have $\boxed{A- 2I= A^{-1}}$
(Answer E)
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Problem Number 9
Suppose there are matrices $A$ and $B$ with an order of $2 \times 2$ and each of them has an inverse. Given that $AB = \begin{pmatrix} 4 & 2 \\ 3 & 2 \end{pmatrix}$.$AB(B^{-1}+A)A^{-1}$ equals to $\cdots \cdot$
A. $\begin{pmatrix} 4 & 2 \\ 3 & 2 \end{pmatrix}$ D. $\begin{pmatrix} 4 & 3 \\ 4 & 2 \end{pmatrix}$
B. $\begin{pmatrix} 5 & 2 \\ 3 & 3 \end{pmatrix}$ E. $\begin{pmatrix} 1 & 9 \\ 6 & 1 \end{pmatrix}$
C. $\begin{pmatrix} 5 & 2 \\ 2 & 3 \end{pmatrix}$
Solution
Notice that $AB(B^{-1}+A)A^{-1}$ can be simplified furthermore by applying some matrix operation rules.
\begin{aligned} & AB(B^{-1}+A)A^{-1} \\ & = AB(B^{-1}A^{-1} + AA^{-1}) && (\text{Sifat Dis}\text{tributif}) \\ & = AB((AB)^{-1} + I) && (\text{Inve}\text{rse rules}) \\ & = (AB)(AB^{-1}) + AB(I) && (\text{Distributi}\text{ve Property}) \\ & = I + AB && (\text{Inve}\text{rse rules}) \\ & = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 4 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 3 & 3 \end{pmatrix} \end{aligned}Thus,
$\boxed{AB(B^{-1}+A)A^{-1} = \begin{pmatrix} 5 & 2 \\ 3 & 3 \end{pmatrix}}$
(Answer B)
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Problem Number 10
If $A = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix}-2 & 5 \\ 1 &-3 \end{pmatrix}$, then the matrix expressing
$\left(\left(\left(\left(\left(\left(\left(\left(AB\right)^T\right)^{-1}\right)^T\right)^{-1}\right)^T\right)^{-1}\right)^T\right)^{-1}$
is $\cdots \cdot$
A. $\begin{pmatrix}-4 &-5 \\ 9 & 11 \end{pmatrix}$ D. $\begin{pmatrix} 11 & 5 \\-9 &-4 \end{pmatrix}$
B. $\begin{pmatrix}-4 & 9 \\-5 & 11 \end{pmatrix}$ E. $\begin{pmatrix}-9 & 11 \\ 5 &-4 \end{pmatrix}$
C. $\begin{pmatrix} 11 &-9 \\ 5 &-4 \end{pmatrix}$
Solution
Suppose $A$ expressing a non-singular matrix.
\begin{aligned} (A^T)^{-1} & = (A^{-1})^T \\ (A^{-1})^{-1} & = A \\ (A^T)^T & = A \end{aligned}
You may see that there are an even number times of the transpose and inverse, and it will yield the original matrix. We have
\begin{aligned} & \left(\left(\left(\left(\left(\left(\left(\left(AB\right)^T\right)^{-1}\right)^T\right)^{-1}\right)^T\right)^{-1}\right)^T\right)^{-1} \\ & = AB = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} \begin{pmatrix}-2 & 5 \\ 1 &-3 \end{pmatrix} \\ & = \begin{pmatrix}-4 & 9 \\-5 & 11 \end{pmatrix} \end{aligned}(Answer B)
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Problem Number 11
Given that $A = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\-2 &-1 &-3 \end{pmatrix}$.
The value of $A^{2017} + 2017A^{2018} + 2I^{2018}$ is $\cdots \cdot$
A. $O$ D. $2017A+2I$
B. $2I$ E. $A+2I$
C. $A$
Solution
Notice the pattern.
\begin{aligned} A^2 & = A \times A \\ & = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\-2 &-1 &-3 \end{pmatrix} \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\-2 &-1 &-3 \end{pmatrix} \\ & = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\-1 &-1 &-3 \end{pmatrix} \\ A^3 & = A^2 \times A \\ & = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\-1 &-1 &-3 \end{pmatrix} \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\-2 &-1 &-3 \end{pmatrix} \\ & = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{aligned}Since the product of two matrices involving zero matrix yields a zero matrix, while any power of the identity matrix yields a same identity matrix, then we have
\begin{aligned} & A^{2017} + 2017A^{2018} + 2I^{2018} \\ & = O + 2017O + 2I = 2I \end{aligned}
Thus, $\boxed{A^{2017} + 2017A^{2018} + 2I^{2018} = 2I}$
(Answer B)
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Problem Number 12
If $A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, then $A^{2019} + B^{2020} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is satisfied by $a + b + c + d = \cdots \cdot$
A. $2020$ D. $4040$
B. $2036$ E. $4043$
C. $4038$
Solution
Given
$A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}~~~~~B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$
We will discover the pattern entries of $A^{2019}$.
\begin{aligned} A^2 & = \begin{pmatrix} 1 & 0 \\ 1&1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1&1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \\ A^3 & = A^2 \cdot A = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \\ A^4 & = A^3 \cdot A =\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix} \end{aligned}See that only the entry in the second row and first column of this matrix changed according to the power of $A$. This implies that
$A^{2019} = \begin{pmatrix} 1 & 0 \\ 2019 & 1 \end{pmatrix}$
By applying the same observation and principle, we have
$B^{2020} = \begin{pmatrix} 1 & 2020 \\ 0 & 1 \end{pmatrix}$
Next,
\begin{aligned} A^{2019} + B^{2020} & = \begin{pmatrix} 1 & 0 \\ 2019 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 2020 \\ 0 & 1 \end{pmatrix} \\ & = \begin{pmatrix} 2 & 2020 \\ 2019 & 2 \end{pmatrix} \end{aligned}Lastly, we have
$a = d = 2; b = 2020; c = 2019$
Thus, the value of $\boxed{a+b+c+d=4043}$
(Answer E)
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Problem Number 13
If $A = \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix}$, then $A^{2009} = \cdots \cdot$
A. $\begin{pmatrix} a^{1004} & 0 \\ 0 & a^{1004} \end{pmatrix}$ D. $\begin{pmatrix} a^{2008} & 0 \\ 0 & a^{1004} \end{pmatrix}$
B. $\begin{pmatrix} 0 & a^{1005} \\ a^{1004} & 0 \end{pmatrix}$ E. $\begin{pmatrix} 0 & a^{1004} \\ a^{1004} & 0 \end{pmatrix}$
C. $\begin{pmatrix} 0 & a^{1005} \\ a^{1005} & 0 \end{pmatrix}$
Solution
Given $A = \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix}$
Spot the pattern by finding the matrix expressing the power of $A$.
\begin{aligned} A^2 & = A \times A = \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \\ A^3 & = A^2 \times A = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & a^2 \\ 0 & a \end{pmatrix} \\ A^4 & = A^3 \times A = \begin{pmatrix} 0 & a^2 \\ 0 & a \end{pmatrix} \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} a^2 & 0 \\ 0 & a^2 \end{pmatrix} \\ A^5 & = A^4 \times A = \begin{pmatrix} a^2 & 0 \\ 0 & a^2 \end{pmatrix} \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & a^3 \\ a^2 & 0 \end{pmatrix} \\ A^6 & = A^5 \times A = \begin{pmatrix} 0 & a^3 \\ a^2 & 0 \end{pmatrix}\begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} a^3 & 0 \\ 0 & a^3 \end{pmatrix} \end{aligned}Noticing the pattern, we will have
$A^n = \begin{pmatrix} a^{\frac{n} {2}} & 0 \\ 0 & a^{\frac{n} {2}} \end{pmatrix}$
for every even number $n$, or
$A^n = \begin{pmatrix} 0 & a^{\frac{n+1} {2}} \\ a^{\frac{n-1} {2}} & 0\end{pmatrix}$
for every odd number $n$.
Since $2009$ is an odd number, then
$$A^{2009} = \begin{pmatrix} 0 & a^{\frac{2009+1} {2}} \\ a^{\frac{2009-1} {2}} & 0\end{pmatrix} = \begin{pmatrix} 0 & a^{1005} \\ a^{1004} & 0\end{pmatrix}$$(Answer B)
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Problem Number 14
If $A = \begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{pmatrix}$, then $\det(A) = \cdots \cdot$
A. $(a-b)(b-c)(c-a)(a+b+c)$
B. $(a-b)(b-c)(c-a)(a+b-c)$
C. $(a-b)(b-c)(c-a)(a-b+c)$
D. $(a-b)(b-c)(c+a)(a-b-c)$
E. $(a-b)(b-c)(c+a)(a-b+c)$
Solution
\begin{aligned} \det A & = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} \\ & = 1(bc^3-b^3c)- 1(ac^3-a^3c) + 1(ab^3-a^3b) \\ & = ab^3-bc^3 + bc^3-a^3b + ac^3-a^3c \\ & = (a-c)b^3 + (c^3-a^3)b + ac(c^2-a^2) \\ & = (a-c)b^3-(a-c)[(a^2+ac+c^2)b + ac(a+c)] \\ & = (a-c)\color{red}{[b^3- (a^2+ac+c^2)b + ac(a+c)]} \end{aligned}Next, we simplify the algebraic expression in red color above.
\begin{aligned} & b^3- (a^2+ac+c^2)b + ac(a+c) \\ & = b^3- a^2b-abc- bc^2 + a^2c + ac^2 \\ & = b^3-bc^2-a^2b + a^2c-abc + ac^2 \\ & = b(b^2-c^2)-(b-c)a^2- ac(b-c) \\ & = b(b-c)(b+c)- (b-c)a^2- ac(b-c) \\ & = (b-c)\color{blue}{[b(b+c)-a^2-ac]} \end{aligned}After that, we simplify the algebraic expression in blue color above.
\begin{aligned} b(b+c)- a^2-ac & = b^2+bc-a^2-ac \\ & = b^2-a^2+bc-ac \\ & = (b-a)(b+a)+c(b-a) \\ & = (b-a)(a+b+c) \end{aligned}Hence, we write as
\begin{aligned} \det A & = (a-c)(b-c)(b-a)(a+b+c) \\ & = (a-b)(b-c)(c-a)(a+b+c) \end{aligned}
Thus, the determinant of matrix $A$ equals $\boxed{(a-b)(b-c)(c-a)(a+b+c)}$
(Answer A)
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Problem Number 15
Given that $A = \begin{pmatrix} b &-\sin x \\ \sin x & b \end{pmatrix}$. The product of all possible values of $b$ that satisfy the equation $A^{-1} = A^T$ is $\cdots \cdot$
A. $-\sin^2 x$ D. $\cos x$
B. $-\cos^2 x$ E. $\cos^2 x$
C. $-\cos x$
Solution
Given
$A = \begin{pmatrix} b &-\sin x \\ \sin x & b \end{pmatrix}$ $A^T = \begin{pmatrix} b & \sin x \\-\sin x & b \end{pmatrix}$
Of the equation $A^{-1} = A^T$, multiply both side by $A$, and we have
\begin{aligned} A^{-1}A & = A^TA \\ I & = A^TA \\ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & = \begin{pmatrix} b & \sin x \\-\sin x & b \end{pmatrix} \begin{pmatrix} b &-\sin x \\ \sin x & b \end{pmatrix} \\ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & = \begin{pmatrix} b^2 + \sin^2 x & 0 \\ 0 & \sin^2 x + b^2 \end{pmatrix} \end{aligned}From the equivalence of two matrices, we have
\begin{aligned} \sin^2 x + b^2 & = 1 \\ b^2 & = 1- \sin^2 x \\ b^2 & = \cos^2 x \\ b & = \pm \cos x \end{aligned}
The product of all possible values of $b$ is
$\boxed{b_1b_2 = \cos x(-\cos x) =-\cos^2 x}$
(Answer B)
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Problem Number 16
Which one of the following are true statements about matrix?
1. If $A$ and $B$ both are square matrices, then $(A+B)(A-B) = A^2-B^2$
2. If $AB = C$ and $C$ has $2$ columns, then $A$ also has $2$ columns.
3. If $BC = D$, then $B^{-1}C^{-1} = D^{-1}$
4. If $AC=O$, then it holds $A=O$ or $C=O$
5. If $A$ and $B$ both are $m \times n$ matrices, then $AB^T$ and $A^TB$ both are defined.
Solution
We will check the truth of the given options.
Check option A: False
Since $A, B$ are two square matrices, then by applying the distributive property, it holds
\begin{aligned} (A+B)(A-B) &= A^2-AB+BA-B^2 \\ & \neq A^2-B^2 \end{aligned}
The inequality occured because $AB \neq BA$ (recall that multiplication of matrices is not commutative).
Check option B: False
Given $AB=C$ and $C$ has two columns. Based on the definition of multiplication of matrices, the number of row in $A$ must be equal to the number of row in $C$, while the number of column in $B$ must be equal to the number of column in $C$. Hence, the said statement isn’t always true that $A$ has two columns.
Check option C: False
Given $BC = D$. If we take the inverse to the both sides of equation, we have
\begin{aligned} (BC)^{-1} & = D^{-1} \\ C^{-1}B^{-1} & = D^{-1} \end{aligned}
Thus, the statement is absolutely false.
Check option D: False
Given $AC = O$. Besides $A = O$ or $C = O$, there are other possibilities that the product of two matrices yields a zero matrix as follows.
$ACC^{-1} = OC^{-1} \Leftrightarrow A = C^{-1}$
Thus, the matrix that satisfies $A=C^{-1}$ is a solution to that matrix equation.
Check option E: True
Given $A$ and $B$ are $m \times n$ matrices. Therefore, both $A^T$ and $B^T$ have an order of $n \times m$. Hence, $AB^T$ is a product of two matrices with an order of $m \times n$ and $n \times m$, so it holds the matrix multiplication rule (it is defined). The same also holds for $A^TB$.
(Answer E)
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Problem Number 17
The determinant of $\begin{pmatrix} 3+2\sqrt5+\sqrt7 & 2\sqrt5 \\ 3-\sqrt5-\sqrt7 & 3-\sqrt7 \end{pmatrix}$ equals to $\cdots \cdot$
A. $-8$ C. $2\sqrt35$ E. $7\sqrt15$
B. $2$ D. $12$
Solution
Notice that the determinant of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ equals to the determinant of $\begin{pmatrix} a-b & b \\ c-d & d \end{pmatrix}$.
By using this fact, we have
\begin{aligned} & \det \begin{pmatrix} 3+2\sqrt5+\sqrt7 & 2\sqrt5 \\ 3-\sqrt5-\sqrt7 & 3-\sqrt7 \end{pmatrix} \\ & = \det \begin{pmatrix} 3+2\sqrt5+\sqrt7-\color{red}{2\sqrt5} & 2\sqrt5 \\ 3-\sqrt5-\sqrt7-(\color{red}{3-\sqrt7}) & 3-\sqrt7 \end{pmatrix} \\ & = \det \begin{pmatrix} 3+\sqrt7 & 2\sqrt5 \\ -\sqrt5 & 3-\sqrt7 \end{pmatrix} \\ & = (3+\sqrt7)(3-\sqrt7)-(-\sqrt5)(2\sqrt5) \\ & = (9-7)-(-2 \cdot 5) \\ & = 2-(-10) = 12 \end{aligned}Thus, the determinant of said matrix equals to $\boxed{12}$
(Answer D)
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Problem Number 18
For every natural number $n$, it is defined a matrix $A_n = \begin{pmatrix} 3n & n \\ 4n & 2n \end{pmatrix}$. If $\det(A_1 + A_2 + \cdots +A_t) = 882$, then the value of $\det(A_{2t}) = \cdots \cdot$
A. $288$ C. $122$ E. $70$
B. $144$ D. $72$
Solution
Given that $A_n = \begin{pmatrix} 3n & n \\ 4n & 2n \end{pmatrix}$ with $n \in \mathbb{N}$. The determinant of this matrix is
\begin{aligned} \det(A_n) & = 3n(2n)-4n(n) \\ & = 6n^2-4n^2=2n^2 \end{aligned}
Notice that $A_1+A_2+\cdots+A_t = A(1+2+\cdots+t)$ with $A = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix}$ and $\det(A) = 3(2)-1(4)=2$ so that if we apply the theorem $\det(n \cdot A) = n^2 \det(A)$ with $A$ is a square matrix with an order of $2$ along with the determination equation, we have
\begin{aligned} \det(A_1 + A_2 + \cdots +A_t) & = 882 \\ \det(A(1+2+\cdots t)) & = 882 \\ \det(A) \cdot (1+2+\cdots + t)^2 & = 882 \\ 2(1+2+\cdots + t)^2 & = 882 \\ (1+2+\cdots+t)^2 & = 441 \\ 1+2+\cdots+t & = 21 \\ \dfrac{t}{2}(t+1) & = 21 \\ t^2+t-42 & = 0 \\ (t+7)(t-6) & = 0 \end{aligned}
We obtain $t = -7$ or $t = 6$ (It satisfies since $t \in \mathbb{N}$).
Thus, the value of $\boxed{\det(A_{2 \cdot 6}) = A_{12} = 2 \cdot 12^2 = 288}$
(Answer A)
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Problem Number 19
If $A = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}$, then $A^{2013} \cdot \begin{pmatrix}2 \\ 1 \end{pmatrix} + B^{2014} \cdot \begin{pmatrix} 1 \\ -3 \end{pmatrix}$ equals $\cdots \cdot$
A. $\begin{pmatrix} 2023 \\ -6044 \end{pmatrix}$ D. $\begin{pmatrix} -4023 \\ -6044 \end{pmatrix}$
B. $\begin{pmatrix} -4023 \\ 6044 \end{pmatrix}$ E. $\begin{pmatrix} -6044 \\ -4023 \end{pmatrix}$
C. $\begin{pmatrix} 4023 \\ 6044 \end{pmatrix}$
Solution
Given that $A = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$.
We will find the matrix’s entry pattern for $A^n$.
You should note that
\begin{aligned} A^2 & = \begin{pmatrix} 1 & \color{red}{-2} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & \color{red}{-2} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & \color{red}{-4} \\ 0 & 1 \end{pmatrix} \\ A^3 & = A^2 \cdot A = \begin{pmatrix} 1 & \color{red}{-4} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & \color{red}{-2} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & \color{red}{-6} \\ 0 & 1 \end{pmatrix} \end{aligned}From here, we found that only an entry of the first row second column changed and it’s a multiple of $2$ for corresponding power of $A$.
We can write $A^n = \begin{pmatrix} 1 & \color{red}{-2n} \\ 0 & 1 \end{pmatrix}$.
Given that $B = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}$.
We will find the matrix’s entry pattern for $B^n$.
You should note that
\begin{aligned} B^2 & = \begin{pmatrix} 1 & 0 \\ \color{blue}{-3} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \color{blue}{-3} & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ \color{blue}{-6} & 1 \end{pmatrix} \\ B^3 & = B^2 \cdot B = \begin{pmatrix} 1 & 0 \\ \color{blue}{-6} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \color{blue}{-3} & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ \color{blue}{-9} & 1 \end{pmatrix} \end{aligned}From here, we found that only an entry of the second row first column changed and it’s a multiple of $3$ for corresponding power of $B$.
We can write $B^n = \begin{pmatrix} 1 & 0 \\ -3n & 1 \end{pmatrix}$.
Therefore,
\begin{aligned} & A^{2013} \cdot \begin{pmatrix}2 \\ 1 \end{pmatrix} + B^{2014} \cdot \begin{pmatrix} 1 \\ -3 \end{pmatrix} \\ & = \begin{pmatrix} 1 & -4026 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ -3 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ -6042 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ -3 \end{pmatrix} \\ & = \begin{pmatrix} -4024 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ -6039 \end{pmatrix} = \begin{pmatrix} -4023 \\ -6044 \end{pmatrix} \end{aligned}Thus, it is evaluated that $\boxed{\begin{pmatrix} -4023 \\ -6044 \end{pmatrix}}$
(Answer D)
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Problem Number 20
Given a matrix $A = \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}$ and $a_{ij}$ expresses the entry of matrix $A^{10}$ lies in the row number $i$ and column number $j$. If $a_{21}=p^2-q$ with $p, q$ be the positive integers, then the value of $p+q=\cdots \cdot$.
A. $32$ C. $35$ E. $39$
B. $33$ D. $38$
Solution
Given $A = \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}$.
Notice that
\begin{aligned} A^2 & = A \cdot A = \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 4 \end{pmatrix} \\ A^3 & = A^2 \cdot A = \begin{pmatrix} 1 & 0 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 7 & 8 \end{pmatrix} \end{aligned}Start from here, we already discover the pattern as follows.
$A^n = \begin{pmatrix} 1 & 0 \\ 2^n-1 & 2^n \end{pmatrix}$
so that
$A^{10} = \begin{pmatrix} 1 & 0 \\ 2^{10}-1 & 2^{10} \end{pmatrix}$
$a_{21}$ denotes the entry lies in the second row and first column, that is $a_{21} = 2^{10}-1$.
Hence, we will have
\begin{aligned} a_{21} & = p^2-q \\ 2^{10}-1 & = p^2-q \\ (2^5)^2-1 & = p^2-q \\ 32^2-1 & = p^2-q \end{aligned}
The last equation implies that $p=32$ and $q=1$.
Thus, the value of $\boxed{p+q=32+1=33}$
(Answer B)
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Problem Number 21
If $A = \begin{pmatrix} 2a & 2 \\ -4 & a \end{pmatrix}$ and $B = \begin{pmatrix} 2b & b \\ -4 & b \end{pmatrix}$ are both invertible, then all values of $b$ that satisfy the inequation $\det(ABA^{-1}B^{-1}) > 0$ are $\cdots \cdot$
A. $b < 0$
B. $b > 0$
C. $b > -2$
D. $b < -2$ or $b > 0$
E. $b \neq -2$ or $b \neq 0$
Solution
Given:
\begin{aligned} A & = \begin{pmatrix} 2a & 2 \\ -4 & a \end{pmatrix} \\ B & = \begin{pmatrix} 2b & b \\ -4 & b \end{pmatrix} \end{aligned}
Let’s examine matrix $B$ first.
Since $B$ is invertible (does have an inverse), then its determinant must not be zero.
\begin{aligned} \det(B) & \neq 0 \\ 2b(b)-b(-4) & \neq 0 \\ 2b^2+4b & \neq 0 \\ 2b(b+2) & \neq 0 \end{aligned}
We have $2b \neq 0$ or $b+2 \neq 0$, which means $b \neq 0$ or $b \neq -2$.
In matrix, the following determinant properties hold.
\boxed{\begin{aligned} \det(A^{-1}) & = \dfrac{1}{\det(A)} \\ \det(AB) & = \det A \cdot \det B \end{aligned}}
Therefore, $\det(ABA^{-1}B^{-1}) > 0$ can be rewritten into
\begin{aligned} \det(A) \cdot \det(B) \cdot \det(A^{-1}) \cdot \det(B^{-1}) & > 0 \\ \Leftrightarrow \cancel{\det(A)} \cdot \bcancel{\det(B)} \cdot \dfrac{1}{\cancel{\det(A)}} \cdot \dfrac{1}{\bcancel{\det(B)}} & > 0 \\ \Leftrightarrow 1 & > 0 \end{aligned}The last proposition is true in value that $1 > 0$.
Thus, real numbers $b$ except $0$ or $-2$ always satisfy the inequation. In other words, all real numbers $b$ that satisfy $\det(ABA^{-1}B^{-1}) > 0$ are $b \neq -2$ or $b \neq 0$.
(Answer E)
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Problem Number 22
If $B = \begin{pmatrix} 1 & 9 & 7 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$, then $B^{12345678765432} = \cdots \cdot$
A. $\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$
B. $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
C. $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$
D. $\begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$
E. $\begin{pmatrix} 1 & 9 & 7 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$
Solution
Try to observe the second power of $B$.
\begin{aligned} B^2 & = B \cdot B \\ & = \begin{pmatrix} 1 & 9 & 7 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 9 & 7 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \\ & = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \end{aligned}
We got an identity matrix $I$.
This makes $A^3 = A^2 \cdot A = I \cdot A = A$.
Hence, we can conclude that $A^{n} = A$ for every odd number $n$ and $A^n = I$ for every even number $n$.
Since $12345678765432$ is an even number, then obviously $B^{12345678765432} = I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.
(Answer B)
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Essay Section
Problem Number 1
Evaluate all the values of $a, b, c$ if it is given that $A$ be a symmetric matrix with
$A = \begin{pmatrix} 2 & a- 2b + 2c & 2a+b+c \\ 3 & 5 & a + c \\ 0 &-2 & 7 \end{pmatrix}$.
Solution
Since $A$ be a symmetric matrix, then $A$ will be equal to its transpose.
\begin{aligned} A & = A^T \\ \begin{pmatrix} 2 & a- 2b + 2c & 2a+b+c \\ 3 & 5 & a + c \\ 0 &-2 & 7 \end{pmatrix} & = \begin{pmatrix} 2 & 3 & 0 \\ a-2b + 2c & 5 &-2 \\ 2a+b+c & a+c & 7\end{pmatrix} \end{aligned}Hence, we obtain a system of linear equations as follows.
$\begin{cases} a- 2b + 2c & = 3 && (1) \\ 2a + b + c &= 0 && (2) \\ a + c & =-2 && (-3) \end{cases}$
Eliminate $b$ from the equation $1$ and $2$:
\begin{aligned} \! \begin{aligned} a-2b+2c & = 3 \\ 2a+b+c & = 0 \end{aligned} \left| \! \begin{aligned} \times 1 \\ \times 2 \end{aligned} \right| & \! \begin{aligned} a-2b+2c & = 3 \\~4a +2b+2c & = 0 \end{aligned} \\ & \rule{3.5 cm}{0.6pt} + \\ & \! \begin{aligned} 5a + 4c & = 3~~~~~~(4) \end{aligned} \end{aligned}
Find the value of $a$ and $c$ by using equations $3$ and $4$.
\begin{aligned} \! \begin{aligned} a + c& =-2 \\ 5a + 4c & = 3\end{aligned} \left| \! \begin{aligned} \times 4 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~4a + 4c & =-8 \\~ 5a + 4c & = 3 \end{aligned} \\ & \rule{2.5 cm}{0.6pt}- \\ & \! \begin{aligned} a & = 11 \end{aligned} \end{aligned}
For $a = 11$, we have
\begin{aligned} a + c & =-2 \\ 11 + c & =-2 \\ c & =-13 \end{aligned}
Substitute the value of $a$ and $c$ to one of three initial equations, for instance, equation $1$.
\begin{aligned} a-2b + 2c & = 3 \\ 11- 2b + 2(-13) & = 3 \\-15-2b & = 3 \\-2b & = 18 \\ b & =-9 \end{aligned}
Thus, the values of $a, b$, and $c$ are $\color{red}{a = 11, b =-9}$, and $\color{red}{c =-13}$, respectively.
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Problem Number 2
Show that the determinant of matrix
$\begin{pmatrix} \sin \theta & \cos \theta & 0 \\-\cos \theta & \sin \theta & 0 \\ \sin \theta \cos \theta & \sin \theta + \cos \theta & 1 \end{pmatrix}$
doesn’t depend on the value of $\theta$.
Solution
Let
$X = \begin{pmatrix} \sin \theta & \cos \theta & 0 \\-\cos \theta & \sin \theta & 0 \\ \sin \theta- \cos \theta & \sin \theta + \cos \theta & 1 \end{pmatrix}$
By applying cofactor expansion along the third column, we have
\begin{aligned} \det(X) & = 0 \begin{vmatrix}-cos \theta & \sin \theta \\ \sin \theta- \cos \theta & \sin \theta + \cos \theta \end{vmatrix} \\ &- 0 \begin{vmatrix} \sin \theta & \cos \theta \\ \sin \theta- \cos \theta & \sin \theta + \cos \theta \end{vmatrix} + 1 \begin{vmatrix} \sin \theta & \cos \theta \\-\cos \theta & \sin \theta \end{vmatrix} \\ & = 0- 0 + (\sin^2 \theta- (-\cos^2 \theta)) \\ & = \sin^2 \theta + \cos^2 \theta = 1 \end{aligned}We found that the determinant is always to $1$. In other words, the determinant doesn’t depend on the value of $\theta$. Any $\theta$ will yield the same value.
Note:
Recall the Pythagorean Identity in trigonometry as follows.
$\boxed{\sin^2 \theta + \cos^2 \theta = 1}$
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Problem Number 3
Show that $(ABCD)^{-1} = D^{-1} \cdot C^{-1} \cdot B^{-1} \cdot A^{-1}$.
Solution
You should use the definition of the inverse of a matrix as follows.
$\boxed{AA^{-1} = I}$
with $I$ denotes an identity matrix.
Following this, we have
$(ABCD)^{-1}ABCD = I$
Multiply both side by $D^{-1}$:
\begin{aligned} (ABCD)^{-1}ABCDD^{-1} & = ID^{-1} \\ (ABCD)^{-1}ABC & = D^{-1} \end{aligned}
Multiply both side by $C^{-1}$:
\begin{aligned} (ABCD)^{-1}ABCC^{-1} & = D^{-1}C^{-1} \\ (ABCD)^{-1}AB & = D^{-1}C^{-1} \end{aligned}
Multiply both side by $B^{-1}$:
\begin{aligned} (ABCD)^{-1}ABB^{-1} & = D^{-1}C^{-1}B^{-1} \\ (ABCD)^{-1}A & = D^{-1}C^{-1}B^{-1} \end{aligned}
Finally, multiply both side by $A^{-1}$:
\begin{aligned} (ABCD)^{-1}AA^{-1} & = D^{-1}C^{-1}B^{-1}A^{-1} \\ (ABCD)^{-1} & = D^{-1}C^{-1}B^{-1}A^{-1} \end{aligned}
Thus, it is showed that $(ABCD)^{-1} = D^{-1} \cdot C^{-1} \cdot B^{-1} \cdot A^{-1}$.
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Problem Number 4
Evaluate the determinant of the following matrix.
$A = \begin{pmatrix} n & n + 1 & n + 2 \\ n + 1 & n + 2 & n + 3 \\ n + 2 & n+3 & n+4 \end{pmatrix}$
Solution
We will apply Elementary Matrix Operations, specifically Row Elementary Operations to track the determinant.
Given $A = \begin{pmatrix} n & n + 1 & n + 2 \\ n + 1 & n + 2 & n + 3 \\ n + 2 & n+3 & n+4 \end{pmatrix}$.
Obeying the elementary matrix operations, the steps are as follows.
$|A| = \begin{vmatrix} n & n + 1 & n + 2 \\ n + 1 & n + 2 & n + 3 \\ n + 2 & n+3 & n+4 \end{vmatrix}$
Multiply the elements in the second row by $-1$ then add that result to the third row. Also multiply the elements in the first row by $-1$ then add that result to the second row.
$|A| = \begin{vmatrix} n & n+1 & n+2 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}$
Multiply the elements in the second row by $-1$ then add that result to the third row.
$|A| = \begin{vmatrix} n & n+1 & n+2 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{vmatrix}$
We found that all entries in one row are being zero. It can be cocluded that the determinant of the matrix is $\boxed{|A| = 0}$
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Problem Number 5
Evaluate the determinant of the following matrix.
$A = \begin{pmatrix} n^2 & (n + 1)^2 & (n + 2)^2 \\ (n + 1)^2 & (n + 2)^2 & (n + 3)^2 \\ (n + 2)^2 & (n+3)^2 & (n+4)^2 \end{pmatrix}$
Solution
We will apply Elementary Matrix Operations to track the determinant.
Given $A = \begin{pmatrix} n^2 & (n + 1)^2 & (n + 2)^2 \\ (n + 1)^2 & (n + 2)^2 & (n + 3)^2 \\ (n + 2)^2 & (n+3)^2 & (n+4)^2 \end{pmatrix}$.
Obeying the elementary matrix operations, the steps are as follows.
$|A| = \begin{vmatrix} n^2 & (n + 1)^2 & (n + 2)^2 \\ (n + 1)^2 & (n + 2)^2 & (n + 3)^2 \\ (n + 2)^2 & (n+3)^2 & (n+4)^2 \end{vmatrix}$.
Multiply the elements in the first column by $-1$ then add that result to the second column and also the third column.
$|A| = \begin{vmatrix} n^2 & 2n+1 & 4n+4 \\ (n + 1)^2 & 2n+3 & 4n + 8 \\ (n + 2)^2 & 2n+5 & 4n+12 \end{vmatrix}$
Multiply the elements in the first row by $-1$ then add that result to the second and third rows.
$|A| = \begin{vmatrix} n^2 & 2n+1 & 4n+4 \\ 2n+1 & 2 & 4 \\ 4n+4 & 4 & 8 \end{vmatrix}$
Factorize out $4$ from the third column.
$|A| = 4\begin{vmatrix} n^2 & 2n+1 & n+1 \\ 2n+1 & 2 & 1 \\ 4n+4 & 4 & 2 \end{vmatrix}$
Factorize out $2$ from the third row.
$|A| = 8\begin{vmatrix} n^2 & 2n+1 & n+1 \\ 2n+1 & 2 & 1 \\ 2n+2 & 2 & 1 \end{vmatrix}$
Multiply the elements in the second row by $-1$ then add them to the third row.
$|A| = 8\begin{vmatrix} n^2 & 2n+1 & n+1 \\ 2n+1 & 2 & 1 \\ 1 & 0 & 0 \end{vmatrix}$
Apply a cofactor expansion along the third row.
\begin{aligned}|A| & = 8 \cdot 1 \cdot \begin{vmatrix} 2n+1 & n+1 \\ 2 & 1 \end{vmatrix} \\ & = 8 \cdot [(2n+1)-2(n+1)] \\ & = 8(-1) = -8 \end{aligned}
Thus, the determinant of matrix $A$ is $\boxed{|A| = -8}$
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# Find equation of line that is parallel to given line and includes given point.Line is `y = 2x-1` Point is `(1,a)` Also, find equation of a line that is perpendicular to given line through `(1, a)`.
txmedteach | Certified Educator
Hey, so, you didn't give us the full point, but I made it general for you, so you just need to plug in your value for "a," and it will work out!
So, let's start by recognizing what it means for 2 lines to be parallel: to never intersect. This means that they are the same distance apart in the x-y axis at each point! What does this mean for the lines, though?
This means the lines must have the same slope. If they have even slightly different slopes, the lines will eventually intersect, and they won't be parallel!
So, let's look at a graph of the line we're supposd to be parallel to:
From this, we can see the basic slope of the line and the slope we'd need to match to solve the problem.
So, let's go back to our basic slope-intercept form for a line:
`y = mx + b`
Where `m` is the slope, and `b` is the y-intercept. Well, we already have one of these from the given equation! `m=2` to make sure our slopes match!
`y = 2x + b`
However, we need to find the y-intercept to complete the equation. To do this, we're going to need to plug in the point we want the line to go through: (1, a)
`a = 2*1 + b`
`a = 2 + b`
So, we just need to solve for b by subtracting 2 from both sides:
`a-2 = b`
So, we have our equation for the parallel line:
`y = 2x + a-2`
So, again, what you'd need to do for your exact problem is replace "a" and you're finished!
Now to find the perpendicular line. The slope here (let's call `m_p`) has an interesting rule associated with the slope of the original line:
`m_p = -1/m`
So, we take the negative reciprocal, and we'll have the slope of the perpendicular! This gives us a new slope of `-1/2`.
So, again, we have our basic slope intercept form:
`y = -1/2 x + b`
Now, we put in the point (again):
`a = -1/2*1 + b`
Solving, we add 1/2 to both sides:
`a+1/2 = b`
And now we have our y-intercept and our complete equation!
`y = -1/2x + a + 1/2`
I hope that helps! Again, just plug in the value you have for "a" and this should work out for the particular problem you have. Pay attention to how it was solved, though, so it makes sense when they ask it later!
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# Finding the fourth root?
0
562
2
+564
Find the four fourth roots of -3 + 4i and express the roots in polar coordinates.
chilledz3non May 29, 2014
#1
+26399
+5
If we have a general point (x, y) in Cartesian coordinates, the polar form is (R, Θ) where R = √(x2 + y2) and
Θ = tan-1(y/x).
And if we are in the complex plane, the point z = x + iy is z = Re in polar coordinates.
Your point is z = -3 + 4i so, in polar form it is z = Re where R = 5 and Θ = tan-1(4/-3) (note that this is an angle in the 2nd quadrant, because the x-value is negative while the y-value is positive). Because you can add any multiple of 2pi (360°) to Θ without changing the value of e, we can also write z = rei(Θ+2pi*k) where k is an integer.
To find the fourth root of a complex number in polar form we simply take the fourth root of r and divide the angle by 4. That is: z1/4 = r1/4ei(θ+2pi*k)/4. So z1/4 = 51/4ei(tan-1(4/-3)+2pi*k).
We can let k = 1, 2, 3 and 4 to get the four roots.
We can write this as z1/4 = re, where r = 51/4 and θ = tan-1(4/-3) + 2pi*k.
The result of doing this is summarised in the image below:
Alan May 29, 2014
Sort:
#1
+26399
+5
If we have a general point (x, y) in Cartesian coordinates, the polar form is (R, Θ) where R = √(x2 + y2) and
Θ = tan-1(y/x).
And if we are in the complex plane, the point z = x + iy is z = Re in polar coordinates.
Your point is z = -3 + 4i so, in polar form it is z = Re where R = 5 and Θ = tan-1(4/-3) (note that this is an angle in the 2nd quadrant, because the x-value is negative while the y-value is positive). Because you can add any multiple of 2pi (360°) to Θ without changing the value of e, we can also write z = rei(Θ+2pi*k) where k is an integer.
To find the fourth root of a complex number in polar form we simply take the fourth root of r and divide the angle by 4. That is: z1/4 = r1/4ei(θ+2pi*k)/4. So z1/4 = 51/4ei(tan-1(4/-3)+2pi*k).
We can let k = 1, 2, 3 and 4 to get the four roots.
We can write this as z1/4 = re, where r = 51/4 and θ = tan-1(4/-3) + 2pi*k.
The result of doing this is summarised in the image below:
Alan May 29, 2014
#2
+564
0
Thank you Alan!
chilledz3non May 29, 2014
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# 6.2 One-to-One Functions; Inverse Functions
## Presentation on theme: "6.2 One-to-One Functions; Inverse Functions"— Presentation transcript:
6.2 One-to-One Functions; Inverse Functions
MAT SPRING 2009 6.2 One-to-One Functions; Inverse Functions In this section, we will study the following topics: One-to-one functions Finding inverse functions algebraically Finding inverse functions graphically Verifying that two functions are inverses, algebraically and graphically
MAT SPRING 2009 Inverse Functions Example*: Solution:
Inverse Functions Notice, for each of the compositions f(g(x)) and g(f(x)), the input was x and the output was x. That is because the functions f and g “undo” each other. In this example, the functions f and g are __________ ___________.
Definition of Inverse Functions
MAT SPRING 2009 Definition of Inverse Functions The notation used to show the inverse of function f is f -1 (read “f-inverse”). Definition of Inverse Function f -1 is the inverse of function f f(f -1(x)) = x for every x in the domain of f -1 AND f -1(f(x))=x for every x in the domain of f The domain of f must be equal to the range of f -1, and the range of f must be equal to the domain of f -1. We will use this very definition to verify that two functions are inverses.
Watch out for confusing notation.
MAT SPRING 2009 Watch out for confusing notation. Always consider how it is used within the context of the problem.
MAT SPRING 2009 Example Use the definition of inverse functions to verify algebraically that are inverse functions.
A function f has an inverse if and only if f is ONE-TO-ONE.
MAT SPRING 2009 One-to-one Functions Not all functions have inverses. Therefore, the first step in any of these problems is to decide whether the function has an inverse. A function f has an inverse if and only if f is ONE-TO-ONE. A function is one-to-one if each y-value is assigned to only one x-value.
One-to-one Functions Using the graph, it is easy to tell if the function is one-to-one. A function is one-to-one if its graph passes the H____________________ L_________ T__________ ; i.e. each horizontal line intersects the graph at most ONCE. one-to-one NOT one-to-one
Example Use the graph to determine which of the functions are one-to-one.
Finding the Inverse Function Graphically
To find the inverse GRAPHICALLY: Make a table of values for function f. Plot these points and sketch the graph of f Make a table of values for f -1 by switching the x and y-coordinates of the ordered pair solutions for f Plot these points and sketch the graph of f -1 The graphs of inverse functions f and f -1 are reflections of one another in the line y = x.
Finding the Inverse Function Graphically (continued)
Example: y=x f(x)=3x-5 x f(x) -5 1 -2 2 3 4 x f-1(x) -5 -2 1 2 4 3
Which of the following is the graph of the function below and its inverse?
Finding the Inverse Function Algebraically
To find the inverse of a function ALGEBRAICALLY: First, use the Horizontal Line Test to decide whether f has an inverse function. If f is not 1-1, do not bother with the next steps. Replace f(x) with y. Switch x and y Solve the equation for y. Replace y with f -1(x).
Finding the Inverse Function Algebraically
Example Find the inverse of each of the following functions, if possible. 1.
Finding the Inverse Function Algebraically
Example Find the inverse of each of the following functions, if possible. 2.
Finding the Inverse Function Algebraically
Example (cont.) 3.
Finding the Inverse Function Algebraically
Example (cont.) 4.
Java Applet Click here to link to an applet demonstrating inverse functions.
MAT SPRING 2009 End of Section 6.2
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# SAS Triangle Congruence
## Two sets of corresponding sides and included angles prove congruent triangles.
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SAS Triangle Congruence
### SASTriangle Congruence
An included angle is when an angle is between two given sides of a triangle (or polygon). In the picture below, the markings indicate that AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} and BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} are the given sides, so B\begin{align*}\angle B\end{align*} would be the included angle.
Consider the question: If I have two sides of length 2 in and 5 in and the angle between them is 45\begin{align*}45^\circ\end{align*}, can I construct only one triangle?
#### Investigation: Constructing a Triangle Given Two Sides and Included Angle
Tools Needed: protractor, pencil, ruler, and paper
1. Draw the longest side (5 in) horizontally, halfway down the page. The drawings in this investigation are to scale.
2. At the left endpoint of your line segment, use the protractor to measure a 45\begin{align*}45^\circ\end{align*} angle. Mark this measurement.
3. Connect your mark from Step 2 with the left endpoint. Make your line 2 in long, the length of the second side.
4. Connect the two endpoints by drawing the third side.
Can you draw another triangle, with these measurements that looks different? The answer is NO. Only one triangle can be created from any given two lengths and the INCLUDED angle.
Side-Angle-Side (SAS) Triangle Congruence Postulate: If two sides and the included angle in one triangle are congruent to two sides and the included angle in another triangle, then the two triangles are congruent.
The markings in the picture are enough to say that ABCXYZ\begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}.
#### Information Necessary to Prove Congruency
What additional piece of information would you need to prove that these two triangles are congruent using the SAS Postulate?
a) ABCLKM\begin{align*}\angle ABC \cong \angle LKM\end{align*}
b) AB¯¯¯¯¯¯¯¯LK¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{LK}\end{align*}
c) BC¯¯¯¯¯¯¯¯KM¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{BC} \cong \overline{KM}\end{align*}
d) BACKLM\begin{align*}\angle BAC \cong \angle KLM\end{align*}
For the SAS Postulate, you need two sides and the included angle in both triangles. So, you need the side on the other side of the angle. In ABC\begin{align*}\triangle ABC\end{align*}, that is BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} and in LKM\begin{align*}\triangle LKM\end{align*} that is KM¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{KM}\end{align*}. The correct answer is c.
#### Writing a Two-Column Proof
Write a two-column proof to show that the two triangles are congruent.
Given: C\begin{align*}C\end{align*} is the midpoint of AE¯¯¯¯¯¯¯¯\begin{align*}\overline{AE}\end{align*} and DB¯¯¯¯¯¯¯¯\begin{align*}\overline{DB}\end{align*}
Prove: ACBECD\begin{align*}\triangle ACB \cong \triangle ECD\end{align*}
Statement Reason
1. C\begin{align*}C\end{align*} is the midpoint of AE¯¯¯¯¯¯¯¯\begin{align*}\overline{AE}\end{align*} and DB¯¯¯¯¯¯¯¯\begin{align*}\overline{DB}\end{align*} Given
2. AC¯¯¯¯¯¯¯¯CE¯¯¯¯¯¯¯¯,BC¯¯¯¯¯¯¯¯CD¯¯¯¯¯¯¯¯\begin{align*}\overline{AC} \cong \overline{CE}, \overline{BC} \cong \overline{CD}\end{align*} Definition of a midpoint
3. ACBDCE\begin{align*}\angle ACB \cong \angle DCE\end{align*} Vertical Angles Postulate
4. ACBECD\begin{align*}\triangle ACB \cong \triangle ECD\end{align*} SAS Postulate
#### Determining if Two Triangles are Congruent
Is the pair of triangles congruent? If so, write the congruence statement and why.
While the triangles have two pairs of sides and one pair of angles that are congruent, the angle is not in the same place in both triangles. The first triangle fits with SAS, but the second triangle is SSA. There is not enough information for us to know whether or not these triangles are congruent.
### Examples
#### Example 1
Is the pair of triangles congruent? If so, write the congruence statement and why.
The pair of triangles is congruent by the SAS postulate. CABQRS\begin{align*} \triangle CAB \cong \triangle QRS\end{align*}.
#### Example 2
State the additional piece of information needed to show that each pair of triangles is congruent.
We know that one pair of sides and one pair of angles are congruent from the diagram. In order to know that the triangles are congruent by SAS we need to know that the pair of sides on the other side of the angle are congruent. So, we need to know that EF¯¯¯¯¯¯¯¯BA¯¯¯¯¯¯¯¯\begin{align*}\overline{EF} \cong \overline{BA}\end{align*}.
#### Example 3
Fill in the blanks in the proof below.
Given:
AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯, BE¯¯¯¯¯¯¯¯CE¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{DC}, \ \overline{BE} \cong \overline{CE}\end{align*}
Prove: ABEACE\begin{align*}\triangle ABE \cong \triangle ACE\end{align*}
Statement Reason
1. 1.
2. AEBDEC\begin{align*}\angle{AEB} \cong \angle{DEC}\end{align*} 2.
3. ABEACE\begin{align*}\triangle ABE \cong \triangle ACE\end{align*} 3.
Statement Reason
1. AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯, BE¯¯¯¯¯¯¯¯CE¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{DC}, \ \overline{BE} \cong \overline{CE}\end{align*} 1. Given
2. AEBDEC\begin{align*}\angle{AEB} \cong \angle{DEC}\end{align*} 2. Vertical Angle Theorem
3. ABEACE\begin{align*}\triangle ABE \cong \triangle ACE\end{align*} 3. SAS postulate
### Review
Are the pairs of triangles congruent? If so, write the congruence statement and why.
State the additional piece of information needed to show that each pair of triangles is congruent.
1. Use SAS
2. Use SAS
3. Use SAS
Complete the proofs below.
1. Given: B\begin{align*}B\end{align*} is a midpoint of DC¯¯¯¯¯¯¯¯\begin{align*}\overline{DC}\end{align*} AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \perp \overline{DC}\end{align*} Prove: ABDABC\begin{align*}\triangle ABD \cong \triangle ABC\end{align*}
2. Given: AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} is an angle bisector of DAC\begin{align*}\angle{DAC}\end{align*} AD¯¯¯¯¯¯¯¯AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AD} \cong \overline{AC}\end{align*} Prove: \begin{align*}\triangle ABD \cong \triangle ABC\end{align*}
3. Given: \begin{align*}B\end{align*} is the midpoint of \begin{align*}\overline{DE}\end{align*} and \begin{align*}\overline{AC}\end{align*} \begin{align*}\angle{ABE}\end{align*} is a right angle Prove: \begin{align*}\triangle ABE \cong \triangle CBD\end{align*}
4. Given: \begin{align*}\overline{DB}\end{align*} is the angle bisector of \begin{align*}\angle{ADC}\end{align*} \begin{align*}\overline{AD} \cong \overline{DC}\end{align*} Prove: \begin{align*}\triangle ABD \cong \triangle CBD\end{align*}
For each pair of triangles, write what needs to be congruent in order for the triangles to be congruent by SAS. Then, write the congruence statement for the triangles.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Base Angles
The base angles of an isosceles triangle are the angles formed by the base and one leg of the triangle.
Congruent
Congruent figures are identical in size, shape and measure.
Equilateral Triangle
An equilateral triangle is a triangle in which all three sides are the same length.
Included Angle
The included angle in a triangle is the angle between two known sides.
SAS
SAS means side, angle, side, and refers to the fact that two sides and the included angle of a triangle are known.
Side Angle Side Triangle
A side angle side triangle is a triangle where two of the sides and the angle between them are known quantities.
Triangle Congruence
Triangle congruence occurs if 3 sides in one triangle are congruent to 3 sides in another triangle.
Rigid Transformation
A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure.
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### Home > CCG > Chapter 11 > Lesson 11.1.2 > Problem11-29
11-29.
Prove that when two lines that are tangent to the same circle intersect, the lengths between the point of intersection and the points of tangency are equal. That is, in the diagram below, if $\overleftrightarrow{ A B }$ is tangent to $⊙P$ at $B$, and $\overleftrightarrow{ A C }$ is tangent to $⊙P$ at $C$, prove that ${AB} = {AC}$. Use either a flowchart or a two-column proof.
Prove that $ΔABP≅ΔACP$.
Recall the triangle congruence shortcuts you know.
First, decide if the triangles are congruent.
Which congruency shortcut did you use?
Write your conclusion in an oval near the bottom.
Below right, write the shortcut used. 2 bubbles, with arrow from top to bottom bubble, labeled as follows: top, triangle, A, b, p, congruent to triangle, a, c, p, bottom, segment, a, b, congruent to segment, a, c, and congruent triangles give congruent parts.
If the triangles are congruent, then the lengths are the same. 4 bubbles added above 2 previous bubbles, #1 with arrow to #2, with arrow to top previous bubble, & bubbles #3 & #4 each with arrow to top previous bubble. Labels as follows: #1: Segment, a, b, perpendicular, to segment, p, b, segment, a, c, perpendicular to segment, p, c. #2: angle, b, and angle, c, are right angles. #3: segment, a, p, congruent to segment, a, p. #4: segment, p, b, congruent to segment, p, c.
Write the reasons you knew the triangle to be congruent. Additional labels added outside the following bubbles: #1: tangent is perpendicular to radius. #2, definition of perpendicular. #3, segment is congruent to itself. #4, All radii are congruent. Original top bubble: h, l, congruency.
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Adding linear expressions means, combining the like terms in the given expressions.
Here we can find some practice questions on adding linear expressions.
(1) Add -6p - 8 and -3p - 6
(2) Add -8r - 2 and 4r - 3
(3) Add -z - 8 and 6z - 5
(4) Add s + t, 2s - t, -s + t
(5) Add 3a - 2b and 2p + 3q
(6) Add 2a + 5b + 7, 8a - 3b + 3, - 5a - 7b - 6
(7) Add 6x + 7y + 3,- 8x - y - 7, 4x - 4y + 2
(8) Add (6a - 3b) and (7a - 5b)
## Adding linear expressions worksheet - Solution
Question 1 :
Add -6p - 8 and -3p - 6 using (i) horizontal (ii) Vertical method
Solution :
Horizontal method :
= (-6p - 8) + (-3p - 6)
= -6p - 8 - 3p - 6
= -6p - 3p - 8 - 6
= -9p - 14
Vertical method :
Hence -9p - 14 is the sum of the above linear expressions.
Question 2 :
Add -8r - 2 and 4r - 3 using (i) horizontal (ii) Vertical method
Solution :
Horizontal method :
= (-8r - 2) + (4r - 3)
= -8r - 2 + 4r - 3
= -8r + 4r - 2 - 3
= -4r - 1
Vertical method :
Hence -4r - 1 is the sum of the above linear expressions.
Question 3 :
Add -z - 8 and 6z - 5 using (i) horizontal (ii) Vertical method
Solution :
Horizontal method :
= (-z - 8) + (6z - 5)
= -z - 8 + 6z - 5
= -z + 6z - 8 - 5
= 5z - 13
Vertical method :
Hence 5z - 13 is the sum of the above linear expressions.
Question 4 :
Add s + t, 2s - t, -s + t using (i) horizontal (ii) Vertical method
Solution :
Horizontal method :
= (s + t) + (2s - t) + (-s + t)
= s + 2s - s + t - t + t
= 2s + t
Vertical method :
Hence 2s + t is the sum of the above linear expressions.
Question 5 :
Add 3a - 2b and 2p + 3q
Solution :
= (3a - 2b) + (2p + 3q)
= 3a - 2b + 2p + 3q
Hence 3a - 2b + 2p + 3q is the sum of the above linear expressions.
Question 6 :
Add 2a + 5b + 7, 8a - 3b + 3, - 5a - 7b - 6
Solution :
= (2a + 5b + 7) + (8a - 3b + 3) + (- 5a - 7b - 6)
= 2a + 8a - 5a + 5b - 3b - 7b + 7 + 3 - 6
= 10a - 5a + 5b - 10b + 10 - 6
= 5a - 5b + 4
Hence 5a - 5b + 4 is the sum of the above linear expressions.
Question 7 :
Add 6x + 7y + 3,- 8x - y - 7, 4x - 4y + 2
Solution :
= (6x + 7y + 3) + (- 8x - y - 7) + (4x - 4y + 2)
= 6x + 7y + 3 - 8x - y - 7 + 4x - 4y + 2
= 6x - 8x + 4x + 7y - y - 4y + 3 - 7 + 2
= 10x - 8x + 7y - 5y + 5 - 7
= 2x + 2y - 2
Hence 2x + 2y - 2 is the sum of the above linear expressions.
Question 8 :
Add (6a - 3b) and (7a - 5b)
Solution :
= (6a - 3b) + (7a - 5b)
= 6a - 3b + 7a - 5b
= 3a + 2b
Hence 3a + 2b is the sum of the above linear expressions.
After having gone through the stuff given above, we hope that the students would have understood "Adding linear expressions worksheet".
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# Recursive Sequence – Pattern, Formula, and Explanation
We can observe patterns in our everyday lives – from the number of sunflower petals to snowflakes, they all exhibit patterns. We can model most of these patterns mathematically through functions and recursive sequences.
Recursive sequences are sequences that have terms relying on the previous term’s value to find the next term’s value.
One of the most famous examples of recursive sequences is the Fibonacci sequence. This article will discuss the Fibonacci sequence and why we consider it a recursive sequence.
We’ll also learn how to identify recursive sequences and the patterns they exhibit. We’ll also apply this to predict the next terms of a recursive sequence and learn how to generalize the patterns algebraically.
Let’s begin by understanding the definition of recursive sequences.
## What is a recursive sequence?
Recursive sequences are not as straightforward as arithmetic and geometric sequences. That’s because it relies on a particular pattern or rule and the next term will depend on the value of the previous term.
Let’s take a look at the Fibonacci sequence shown below. Take some time to observe the terms and make a guess as to how they progress.
We can see that for this sequence, we start with two $1$’s. These two terms are crucial in predicting the third term: to find the third term; we need to add the two values.
Next, to find the fourth term, we add the second and third terms. Yes, for the fifth term, we add the fourth term by the third term.
This means that the next three terms are shown below:
\begin{aligned}8 + 13 &= \color{blue}21\\13 + 21 &= \color{blue}43\\43 + 21 &= \color{blue}64\end{aligned}
In the next sections, we’ll go back to this example and better understand what makes up a recursive sequence. But for now, we’ve shown you how trickier patterns can be modelled as recursive sequences and why it’s important for us to understand how these sequences work.
## How to solve recursive sequences?
Since recursive sequences’ rules vary from each other, it is nearly impossible to create a general pattern that applies to all recursive sequences, unlike arithmetic or geometric sequences.
Instead, it helps to observe the patterns exhibited by a given sequence and use the initial values to create a rule that may apply to the sequence.
As a guide, why don’t we go ahead and review the definition of recursive sequences and try constructing recursive sequence formulas from the patterns we’ve observed?
### Recursive sequence definition
As we have mentioned, recursive sequences depend on the previous term and the rules observed for the particular sequence. Here are some examples of recursive sequence along with the rules that they follow:
Recursive Sequence Recursive Sequence Formula $\{1, 3, 7, …\}$ $\left\{\begin{matrix}a_1 = 1\phantom{xxxxxxx}\\a_n = 2a_{n-1} + 1 \end{matrix}\right.$ $\{2, 9, 30, …\}$ $\left\{\begin{matrix}a_1 = 1\phantom{xxxxxxx}\\a_n = 3a_{n-1} + 3 \end{matrix}\right.$
As can be seen, by the two examples, the rules will vary for each sequence and we may even observe several patterns for one sequence.
We can also execute sequences given the recursive sequence formula. We use the previous term and evaluate the next term’s value using the given rules or patterns.
### Recursive sequence formula
When given a recursive sequence, we can predict and establish their formulas and rules.
• An initial value such as $a_1$.
• A pattern or an equation in terms of $a_{n – 1}$ or even $a_{n -2}$ that applies throughout the sequence.
• We can express the rule as a function of $a_{n -1}$.
Before establishing recursive sequence formula, let’s practice determining the next terms of a recursive sequence given its formula.
Why don’t we try finding the next three terms of $\{1, 3, 7, …\}$ and $\{2, 9, 30, …\}$?
For $\{1, 3, 7, …\}$, let’s use the rule, $a_n = 2a_{n-1} + 1$:
$\boldsymbol{a_4}$ $\boldsymbol{a_5}$ $\boldsymbol{a_6}$ \begin{aligned}a_4 &= 2a_3 + 1\\&= 2(7) + 1\\&= 15 \end{aligned} \begin{aligned}a_5 &= 2a_4 + 1\\&= 2(15) + 1\\&= 31 \end{aligned} \begin{aligned}a_6 &= 2a_5 + 1\\&= 2(31) + 1\\&= 63 \end{aligned}
This shows that as long as we have the previous term and we know the rule for the recursive sequence, we can predict the terms that follow. Let’s go ahead and do this for $\{2, 9, 30, …\}$ given that $a_n = 3a_{n-1} + 3$.
$\boldsymbol{a_4}$ $\boldsymbol{a_5}$ $\boldsymbol{a_6}$ \begin{aligned}a_4 &= 3a_3 + 3\\&= 3(30) + 3\\&= 93 \end{aligned} \begin{aligned}a_5 &= 3a_4 + 3\\&= 3(93) + 3\\&= 282 \end{aligned} \begin{aligned}a_6 &= 3a_5 + 3\\&= 3(282) + 3\\&= 849 \end{aligned}
But what if we’re not given the rules for the recursive sequence? It will be helpful if we begin observing the first few terms for patterns. We then use these terms to find a rule for the entire sequence.
Let’s go back to the Fibonacci sequence we’ve observed: $\{1, 1, 2, 3, 5, 8, …\}$. We can start with the fact that the third term adds the first and second terms.
As we have discussed in the earlier section, this pattern continues for each set of three terms. We can establish a rule in terms of $a_{n-2}$, $a_{n-1}$, and $a_n$:
\begin{aligned}a_n = a_{n-2} + a_{n – 1}\end{aligned}
This pattern will only be true if we have $a_1 = 1$ and $a_2 = 2$. Hence, we can further establish the general by including the initial values as shown below.
$\left\{\begin{matrix}a_1 = 1\phantom{xxxxxxxxx}\\a_2 =1\phantom{xxxxxxxxx}\\a_n = a_{n -2} + a_{n – 1} \end{matrix}\right.$
Are you ready to try out more problems involving recursive sequences? Make sure to review your notes first before trying out these examples below.
Example 1
What is the fifth term of the recursive sequence that is defined by the following rules: $a_1 = 4$ and $a_n = -2a_{n – 1} + 4$?
Solution
We’ll use the value of $a_1$ and the expression for $a_n$ to find the value of $a_2$.
\begin{aligned}a_2 &= -2a_1 +4 \\&=-2(4) + 4\\&= -8 + 4\\&= -4\end{aligned}
We repeat a similar process to find the next three terms: by using the previous term into the formula as shown below.
$\boldsymbol{a_3}$ $\boldsymbol{a_4}$ $\boldsymbol{a_5}$ \begin{aligned}a_3 &= -2a_2 +4 \\&=-2(-4) + 4\\&= 8 + 4\\&= 12\end{aligned} \begin{aligned}a_4 &= -2a_3 +4 \\&=-2(12) + 4\\&= -24 + 4\\&= -20\end{aligned} \begin{aligned}a_5 &= -2a_5 +4 \\&=-2(-20) + 4\\&= 40 + 4\\&= 44\end{aligned}
Hence, the fifth term of the sequence is equal to $44$.
Example 2
Write down the first six terms of the sequence that follows the recursive formula shown below.
$\left\{\begin{matrix}a_1 = 0\\a_2 = 1\\a_3 = 1\\a_n = \dfrac{2a_{n – 1}}{a_{n – 2}}\end{matrix}\right.$
Solution
Let’s first understand what $a_n = \dfrac{2a_{n – 1}}{a_{n – 2}}$ means. From this formula, we can see that the nth term of this particular recursive sequence can be determined by finding the last two terms’ ratio and multiplying the result by $2$.
If we want the first six terms, we already have the first three terms, and we only need the three remaining terms: $a_4$, $a_5$, and $a_6$.
To find the value of $a_4$, we simply divide $a_3$ by $a_2$ then multiply the result by $2$. We apply a similar approach to determine the values of $a_5$ and $a_6$.
$\boldsymbol{a_4}$ $\boldsymbol{a_5}$ $\boldsymbol{a_6}$ \begin{aligned}a_4 &= \dfrac{2a_3}{a_2}\\&= \dfrac{2(1)}{1}\\&= 2\end{aligned} \begin{aligned}a_5 &= \dfrac{2a_4}{a_3}\\&= \dfrac{2(2)}{1}\\&= 4\end{aligned} \begin{aligned}a_6 &= \dfrac{2a_5}{a_4}\\&= \dfrac{2(4)}{2}\\&= 4\end{aligned}
Hence, we have the first six terms of the recursive formula: $\{0, 1, 1, 2, 4, 4\}$.
Example 3
What is the recursive formula that can describe the pattern for the following sequences?
a. $\{2, 6, 14, 30,…\}$
b. $\{1, 2, 6, 24, …\}$
Solution
It helps to observe how the two terms are related to each other, and we can then check if the rules will work for the fourth term.
Let’s begin with the first sequence, $\{2, 6, 14, 30,…\}$, and see how $6$ relates with $2$ and $14$ with $6$.
\begin{aligned}6 &= 4 + 2\\&= 2(2) + 2\\14&= 12 + 2\\ &= 2(6) + 2\end{aligned}
We can see that the next term can be determined by multiplying the previous term by $2$ then adding the result with $2$. To see if the pattern fits the next term, let’s try to multiply $14$ by $2$ then adding $2$ to the product.
\begin{aligned}14(2) + 2 &= 30\end{aligned}
Since we got $30$, this confirms that we have the correct observations and rule for the recursive sequence. This means that to find $a_n$, we simply multiply $a_{n -1}$ by $2$ then add $2$ to the result: $a_n = 2a_{n – 1} + 2$.
Including the first term, we have the recursive formula shown below for the first sequence.
$\left\{\begin{matrix}a_1 = 2 \phantom{xxxxxx}\\a_n = 2a_{n – 1} + 2\end{matrix}\right.$
Let’s go ahead and move on to the second sequence, $\{1, 2, 6, 24, …\}$. We can apply a similar process when trying to find a pattern for the sequence.
\begin{aligned} 1&= 1 \cdot 1\\ 2 &= 1 \cdot 2\\6 & = 2 \cdot 3\end{aligned}
What can you notice about the four terms? The position of the terms is actually a factor for the formula.
This means that $2$ is the result of the previous term being multiplied by the place number – $2$. Similarly, $6$ is the result when we multiply the previous term with the place number – $3$.
We can see that the fourth term, $24$ is equal to $6 \times 4$, which confirms that observation is correct. Hence, $a_n$ will result from the previous term, $a_{n – 1}$, is multiplied by the place number, $n$.
\begin{aligned}24 &= 6 \times 4\\ a_4 &= a_3 \cdot 4\\a_n &= a_{n – 1} \cdot n \\&= na_{n- 1} \end{aligned}
Let’s include the initial value for the formula; hence, we have the rule shown below:
$\left\{\begin{matrix}a_1 = 1 \phantom{xxxxxx}\\a_n = na_{n- 1}\end{matrix}\right.$
### Practice Questions
1. What is the fifth term of the recursive sequence that is defined by the following rules: $a_1 = -3$ and $a_n = 4a_{n – 1} + 2$?
2. Write down the first six terms of the sequence that follows the recursive formula shown below.
$\left\{\begin{matrix}a_1 = 0\\a_2 = 1 \\a_3 = 1 \\a_n =\dfrac{1}{3} \dfrac{a_{n – 1}}{a_{n – 2}} \end{matrix}\right.$
3. What is the recursive formula that can describe the pattern for the following sequences?
a. $\{2, 5, 11, 9, 15, …\}$
b. $\{3, 3, 6, 9, 15, …\}$
c. $\{1, 2, \dfrac{1}{2}, 4, \dfrac{1}{8}, 32, \dfrac{1}{256}…\}$
1. $-598$
2. $0, 1, 1, \dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{9}$
a. $\left\{\begin{matrix}a_1 =2\\a_n =2a_{n -1} + 1\end{matrix}\right.$
b. $\left\{\begin{matrix}a_1 =3 \\a_2 = 3\\a_n =a_{n -2} + a_{n-1}\end{matrix}\right.$
c.$\left\{\begin{matrix}a_1 =1\\a_2 = 2\\a_n =\dfrac{a_{n-2}}{a_{n -1}}\end{matrix}\right.$
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# Second Derivative Calculator
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## Second Derivative Calculator
A second derivative calculator is an online tool that performs differentiation twice on a function. It can find both first and second derivatives.
Moreover, the 2nd derivative calculator gives the complete solving process with a step-by-step solution.
## How does the second derivative calculator work?
The second derivative test calculator is an easy-to-use tool. Follow these steps to find the second derivative.
1. Enter the function.
2. Choose the variable.
3. Confirm the displayed function from the display box.
4. Click calculate.
To understand the differentiation procedure, click on the ‘+’ icon in the results. It will give a step-wise guide. You can also download the ‘PDf’ copy of the detailed result.
## What is the second derivative?
The derivative taken of the same function for the second time is known as the second derivative. It is the same as the first derivative except for the notation.
The second derivative is represented by two dots over the variable or two dashes on f in the notation f(x) e.g f’’(x).
A graphical representation of 2nd derivatives can be seen below.
## How to find the second derivative?
There is no separate process or formula for the second derivative. It is the same as the first. The second derivative is differentiation performed on the derivative of a function.
Let’s see an example of the second derivative.
Example 1:
Calculate the second derivative for function x = Sinx + x2
Solution:
Step 1: Arrange the function.
f(x) = x2 + sinx
Step 2: Find the first derivative.
f’(x) = d/dx [x2 + sinx]
f’(x) = d/dx [x2] + d/dx[sinx]
f’(x) = 2x + cosx
Step 3: Find the 2nd derivative.
f’’(x) = d/dx [2x + cosx]
f’’(x) = d/dx [2x] + d/dx[cosx]
f’’(x) = 2 - sinx
Example 2:
Find the second derivative for a*(x2+b).
Solution:
First derivative.
Step 1: Apply derivative.
f’(x)= d/dx [a*(x2+b)]
Step 2: Take constant out.
f’(x)= a d/dx (x2 + b)
Step 3: Apply constant rule and power rule.
f’(x)= a (2*x2-1 + 0)
f’(x)= a (2x + 0)
f’(x)= 2ax
Second Derivative:
Step 4: Apply the second derivative.
f’’(x) = d/dx (2ax)
Step 5: Take the constant out.
f’’(x) = 2a d/dx (x)
f’’(x) = 2a (x1-1)
f’’(x) = 2a
Example 3:
What is the second derivative of sinx + x/2?
Solution:
First derivative:
Step 1: Apply the derivative.
f’(x) = d/dx (sinx + x/2)
Step 2: Apply the Sum rule.
f’(x) = d/dx(sinx) + d/dx(x/2)
Step 3: Take out the constant.
f’(x) = cosx + (½)d/dx(x)
f’(x) = cosx + ½
Second derivative:
Step 4: Apply the second derivative.
f’’(x) = d/dx (cosx + ½ )
Step 5: Apply the sum rule.
f’’(x) = d/dx (cosx) + d/dx(½)
Step 6: Constant rule.
f’’(x) = -sinx + 0
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### Prompt Cards
These two group activities use mathematical reasoning - one is numerical, one geometric.
### Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
# X Is 5 Squares
##### Age 7 to 11Challenge Level
A solution came from Tom of Lexden Primary School in Colchester, Essex. Tom explains how he set about solving the problem.
"I worked out what possibilities there were for multiplying the top two numbers together to equal the middle number. The two possibilities were 6 and 8 for the middle number.
I then worked out the other outside numbers so that when I added them altogether they would make 6 or 8. 8 didn't work."
So what did work? Tom knew, so did Ben and Thomas both pupils at Montgomery Church in Wales School, Powys.
They explain: "The middle number has to be 6 because that is the lowest number that is the one that four of the other lower numbers can total. That leaves 1, 2, 3 and 0 to be used around the outside. We multiplied pairs of these numbers together until we found one pair that made 6. These numbers were 2 and 3."
As Tom said, "The solution won't mirror image top to bottom because the top two numbers have to multiply to make the middle number."
Ben and Thomas agreed. They completed the task this way: "1 and 0 were then used to make the diagonal lines total the same amount."
Emma and Francine found the same solution. They showed it the same way as Tom.
Well done to each of you. Are there any other series of consecutive numbers that will allow you to make a similar puzzle? Abigail who goes to Histon and Impington Infants School sent us in a very detailed solution to this problem. She wrote:
We named the squares with letters and then we wrote down the three rules in letters:
A B
C
D E
1. AxB=C 2. D+E=C 3. B+C+D=A+C+E I then made a guess, that if A=6 and if B=4, C must be 24, but this is too big, so B must be smaller. I tried B=2 which is still too big, then B=1, but then C=6, but this is not different to A, so A cannot be 6.
I guessed that A might be 1 and B=2, but then C = 2 too. I saw that A can't be 1 because then B and C would be the same.
I guessed that A might be 5, and tried B=2, but then C is too big, but B can't be 1 because A would be the same as C so I tried another value for A.
I guessed that A might be 3, and B =2 then C=6 which is possible. I tried other B values.. Not 1, not 3, maybe 4, but that is too big.
Then I guessed A might be 2. If A = 2 and B=3, C=6 is possible. If A= 2 and B=4, C=8 is possible. B can't be 5 because C would have to be 10. B can't be bigger than 5 when A is 2.
I then wrote down what I had found so far.
A = 6 is not possible
A = 1 is not possible
A = 5 is not possible
A = 3 is possible
A = 2 is possible
We could try A = 4, but we don't need to try bigger than 6, because then C would be too big.
So if A = 4. B could be 2 and C could be 8. But if B is 3, C is too big.
With all of the possible answers, C is either 6 or 8.
Then I looked at D + E = C. If C = 6
I wrote a table:
D E
1 5
2 4
3 3
4 2
5 1
0 6
6 0
D or E can't be 2 because 2 is already used in A or B. They can't be 3 because they need to be different. They can't be 6 because that is C.
If C = 8, D and E could be...
D E
0 8 but C=8
1 7
2 6 but A or B = 2
5 3
4 4 but they can't be the same 3 5 6 2 but A or B = 2
7 1
8 0 but C =8
Then Mummy helped me write a big table with all the possible answers, and I worked out the sums for rule 3:
A B C D E B+C+D A+C+E
4 2 8 1 7 11 19
4 2 8 5 3 15 15 OK
4 2 8 3 5 13 17
4 2 8 7 1 17 13
2 4 8 1 7 13 17
2 4 8 5 3 17 13
2 4 8 3 5 15 15 OK
2 4 8 7 1 19 11
2 3 6 1 5 10 13
2 3 6 5 1 14 9
3 2 6 1 5 9 14
3 2 6 5 1 13 10
So the answers are:
@ 4 2 8 5 3 and
@ 2 4 8 3 5 When I looked at these answers, I noticed that they are mirror images.
This is fantastic reasoning, Abigail. You have really gone through the problem in a systematic way.
Well done! Thank you also to Katie, Kirsty, Sam, Lucija, Marek, Rosie, Sarah, Natasha, Nathaniel and William who told us they had each found one of the solutions.
|
Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length.
Question:
Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.
Solution:
Let :
$a=42 \mathrm{~cm}, b=34 \mathrm{~cm}$ and $c=20 \mathrm{~cm}$
$\therefore s=\frac{a+b+c}{2}=\frac{42+34+20}{2}=48 \mathrm{~cm}$
By Heron’s formula, we have :
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{48(48-42)(48-34)(48-20)}$
$=\sqrt{48 \times 6 \times 14 \times 28}$
$=\sqrt{4 \times 2 \times 6 \times 6 \times 7 \times 2 \times 7 \times 4}$
$=4 \times 2 \times 6 \times 7$
$=336 \mathrm{~cm}^{2}$
We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
Area of triangle $=336 \mathrm{~cm}^{2}$
$\Rightarrow \frac{1}{2} \times$ Base $\times$ Height $=336$
$\Rightarrow$ Height $=\frac{336 \times 2}{42}=16 \mathrm{~cm}$
|
# Find the term independent of x in the expansion of
Question:
Find the term independent of x in the expansion of (91 + x +$\left.2 x^{3}\right)$
$\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$
Solution:
To Find : term independent of $x$, i.e. coefficient of $x^{0}$
Formula: $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$
We have a formula
$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$
Therefore, the expansion of $\left(x-\frac{2}{x}\right)^{10}$ is given by,
$\left(x-\frac{2}{x}\right)^{10}=\sum_{r=0}^{10}\left(\begin{array}{c}10 \\ r\end{array}\right)(x)^{10-r}\left(\frac{-2}{x}\right)^{r}$
$=\left(\begin{array}{c}10 \\ 0\end{array}\right)(\mathrm{x})^{10}\left(\frac{-2}{\mathrm{x}}\right)^{0}+\left(\begin{array}{c}10 \\ 1\end{array}\right)(\mathrm{x})^{9}\left(\frac{-2}{\mathrm{x}}\right)^{1}+\left(\begin{array}{c}10 \\ 2\end{array}\right)(\mathrm{x})^{8}\left(\frac{-2}{\mathrm{x}}\right)^{2}+\cdots \ldots \ldots$
$+\left(\begin{array}{l}10 \\ 10\end{array}\right)(\mathrm{x})^{0}\left(\frac{-2}{\mathrm{x}}\right)^{10}$
$=x^{10}+\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{9}(-2) \frac{1}{x}+\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{8}(-2)^{2} \frac{1}{x^{2}}+\cdots+\left(\begin{array}{c}10 \\ 10\end{array}\right)(x)^{0}(-2)^{10} \frac{1}{x^{10}}$
$=x^{10}-(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots \ldots \ldots+(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}$
Now,
$\left(91+x+2 x^{3}\right)\left(x-\frac{2}{x}\right)^{10}$
$=\left(91+x+2 x^{3}\right)\left(x^{10}-(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots \ldots \ldots\right.$
$\left.+(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}\right)$
Multiplying the second bracket by $91, x$ and $2 x^{3}$
$=\left\{91 x^{10}-91(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+91(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots+91(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}\right\}$
$+\left\{x \cdot x^{10}-x \cdot(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+x \cdot(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots \ldots \ldots\right.$
$\left.+x .(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}\right\}$
$+\left\{2 x^{3} \cdot x^{10}-2 x^{3} \cdot(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+2 x^{3} \cdot(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots \ldots \ldots\right.$
$\left.+2 x^{3} \cdot(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}\right\}$
In the first bracket, there will be a $6^{\text {th }}$ term of $x^{0}$ having coefficient $91(-2)^{5}\left(\begin{array}{c}10 \\ 5\end{array}\right)$
While in the second and third bracket, the constant term is absent.
Therefore, the coefficient of term independent of $x$, i.e. constant term in the above expansion
$=91(-2)^{5}\left(\begin{array}{c}10 \\ 5\end{array}\right)$
$=-91 \cdot(2)^{5} \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$
$=-91(2)^{5}(252)$
$\underline{\text { Conclusion: }}$ coefficient of term independent of $x=-91(2)^{5}(252)$
|
# How many squares are there in a 6×6 magic square?
## How many squares are there in a 6×6 magic square?
Twenty Four Magic Squares from a single “Root” Pattern.
## How do you calculate a magic square?
Magic Square Solution
1. List the numbers in order from least to greatest on a sheet of paper.
2. Add all nine of the numbers on your list up to get the total.
3. Divide the total from Step 2 by 3.
4. Go back to your list of numbers and the number in the very middle of that list will be placed in the center of the magic square.
What is the magic number of the magic square of order 6?
I remember there are no pandiagonal squares of order 6 (nor symmetrical). The number of magic squares of order 6 is still unknown; this number has been estimated (see Walter TRUMP’s site)….
Magic conditions: B-A=D-C=F-E and C-B=E-D Note: the 6 rows and the 2 diagonals are regular.
How to construct a 6 x 6 magic square?
How to construct 6 x 6 Magic square | Maths IS Fun! 1)Draw a 6 x 6 empty square. 2)Draw a bold line after the third square, Horizontally and vertically. 3).Now the 6 x 6 magic square will be divided into four 3 x 3 Magic squares.
### What’s the best way to solve a 3×3 magic square?
The only way to use these numbers to solve a 3×3 magic square is by excluding either your highest or your lowest number. Once you have done so, assign the lowest remaining value to 1, the next lowest to 2, the next to 3, and so on an so forth until you assign the highest remaining value to 9.
### Can you make a magic square out of any number?
You can create 4X4 magic square for any number without using consecutive numbers. For example, the numbers 1, 4, 7, 10, 14, 17, 20, 23, 27, 30, 33, 36, 40, 43, 46 & 49 will produce a magic constant of 100.
What is the magic constant for a 6×6 puzzle?
The worksheets with normal variations of these puzzles (6×6 puzzles that contain 1-36 in their cells) have a magic constant of 111 no matter how the numbers are arranged in each puzzle. Just because you know the magic constant, don’t think these are easy though!
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# What’s 40% of 25? The Ultimate Guide to Calculating Percentages
If you’re like most people, calculating percentages can be quite challenging. Whether you’re trying to calculate a discount at a store or determine how much of a tip to leave at a restaurant, percentages are used in a wide range of everyday situations. One question that many people ask is: What’s 40% of 25? This is a common question that requires simple math skills to solve. In this article, we’ll provide you with the ultimate guide to calculating percentages. You’ll learn why percentages are important, how to calculate them, and how they’re used in the real world. By the end of this article, you’ll have the confidence to calculate any percentage with ease.
## Why Percentages Are Important
Percentages are used to represent parts of a whole. In other words, percentages show how much of something is a particular amount out of a total amount. For example, if your math test had 20 questions and you got 16 correct, you could say that you got 80% of the questions right. Percentages are important because they allow us to compare different amounts and make informed decisions. They’re used in business, finance, and everyday life to analyze data and make predictions.
### How to Calculate Percentages
Calculating percentages is a simple process that involves three steps:
• Step 1: Divide the percentage by 100.
• Step 2: Multiply the result by the total number you’re working with.
• Step 3: Round the answer to the desired number of decimal places.
Let’s use the example of calculating 40% of 25.
Step Calculation Result
Step 1 40 divided by 100 0.40
Step 2 0.40 multiplied by 25 10
Step 3 Rounding to nearest whole number 10
That’s it! The answer is 10. So, 40% of 25 is 10.
### Decimal and Fraction Equivalents of Common Percentages
Here is a list of common percentages and their decimal and fraction equivalents:
• 10% = 0.1 = 1/10
• 25% = 0.25 = 1/4
• 50% = 0.5 = 1/2
• 75% = 0.75 = 3/4
• 100% = 1 = 1/1
## Real World Uses of Percentages
Percentages are used in a variety of fields, including finance, business, and sports. Here are some examples:
### Finance
Percentages are used in finance to calculate interest rates, investment returns, and taxes. Banks use percentages to calculate interest on loans and credit cards. Stockbrokers use percentages to calculate investment returns. Tax accountants use percentages to calculate tax rates and deductions.
Percentages are used in business to analyze data and make informed decisions. Businesses use percentages to calculate profit margins, employee productivity, and customer satisfaction. Marketing departments use percentages to analyze market trends and customer behavior.
### Sports
Percentages are used in sports to analyze player performance and team success. In basketball, players are judged on their field goal percentage and free throw percentage. In baseball, players are judged on their batting average and on-base percentage. Coaches use percentages to analyze team performance and make strategic decisions.
## Tips for Calculating Percentages
Here are some tips to help you calculate percentages more quickly and accurately:
• Write out the problem before solving it.
• Use a calculator if necessary.
• Understand the application of percentages in the real world.
• Practice, practice, practice.
### Common Percentage Questions and Answers
• Q: What is 20% of 50?
• A: 10
• Q: What is 10% of 100?
• A: 10
• Q: What is 25% of 80?
• A: 20
• Q: What is 33.33% of 150?
• A: 50
• Q: What is 75% of 30?
• A: 22.5
These are just a few examples of common percentage questions. By practicing with similar questions, you’ll become more confident in your ability to calculate percentages.
## Conclusion
Calculating percentages doesn’t have to be difficult. By following the steps outlined in this article, you’ll be able to calculate any percentage with ease. Remember that percentages are used in a variety of real world situations, from business and finance to sports and entertainment. By understanding how percentages work and practicing your skills, you’ll be able to make informed decisions and solve problems more quickly and accurately.
References:
• https://www.mathsisfun.com/percentage.html
• https://www.investopedia.com/terms/p/percentage.asp
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# Mock AIME 2 2006-2007 Problems/Problem 7
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A right circular cone of base radius $17$cm and slant height $51$cm is given. $P$ is a point on the circumference of the base and the shortest path from $P$ around the cone and back is drawn (see diagram). If the length of this path is $m\sqrt{n},$ where $n$ is squarefree, find $m+n.$
## Solution
"Unfolding" this cone results in a circular sector with radius $51$ and arc length $17\cdot 2\pi=34\pi$. Let the vertex of this sector be $O$. The problem is then reduced to finding the shortest distance between the two points $A$ and $B$ on the arc that are the farthest away from each other. Since $34\pi$ is $1/3$ of the circumference of a circle with radius $51$, we must have that $\angle AOB=\frac{360^{\circ}}{3}=120^{\circ}$. We know that $AO=OB=51$, so we can use the Law of Cosines to find the length of $AB$: $$AB=\sqrt{AO^2+OB^2-2AO\cdot OB\cdot\cos{120^{\circ}}}=\sqrt{51^2+51^2+51^2}=51\sqrt{3}.$$ Hence $m=51$, $n=3$, $m+n=\boxed{054}$.
Mock AIME 2 2006-2007 (Problems, Source) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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# How do you find the sum of (1+1)+(1/3+1/5)+(1/9+1/25)+...+(3^-n+5^-n)+...?
Feb 13, 2017
${\sum}_{n = 0}^{\infty} \left({3}^{- n} + {5}^{- n}\right) = \frac{11}{4}$
#### Explanation:
We have:
${\sum}_{n = 0}^{\infty} \left({3}^{- n} + {5}^{- n}\right) = {\sum}_{n = 0}^{\infty} {\left(\frac{1}{3}\right)}^{n} + {\sum}_{n = 0}^{\infty} {\left(\frac{1}{5}\right)}^{n}$
Both these series are geometric series so the sum is:
${\sum}_{n = 0}^{\infty} \left({3}^{- n} + {5}^{- n}\right) = \frac{1}{1 - \frac{1}{3}} + \frac{1}{1 - \frac{1}{5}} = \frac{3}{2} + \frac{5}{4} = \frac{11}{4}$
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RS Aggarwal Solutions: Integers
# RS Aggarwal Solutions: Integers | Mathematics (Maths) Class 6 PDF Download
``` Page 1
Points to Remember :
Integers : The numbers .........., – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, .......... are called integers.
The numbers 1, 2, 3, 4, 5 ............ are called positive integers and the numbers – 1, – 2, – 3, – 4,
– 5 ............. are called negative integers. 0 is an integer which is neither positive nor negative.
Representation of Integers On Number Line :
We draw a line and fix a point almost in the middle of it and call it O. We set off equal distances on
right hand side as well as on left hand side of 0. We name the points of division as 1, 2, 3, 4 etc.on
left hand side as shown below :
Some results :
(i) Zero is less than every positive integer.
(ii) Zero is greater than every negative integer.
(iii) Every positive integer is greater then every negative integer.
(iv) The greater is the number, the lesser is its opposite.
Absolute Value of an Integer. The absolute value of an integer is the numerical value of the
integer regardless of its sign.
(iv) 10 km below sea level
(v) 5°C above the freezing point
(vi) A withdrawl of Rs. 100
(vii) Spending Rs. 500
(viii) Going 6 m to the west
(ix) – 24
(x) 34
Q. 2. Indicate the following using ‘+’ or ‘–’
sign :
(i) A gain of Rs. 600
(ii) A loss of Rs. 800
(iii) 7ºC below the freezing point
(iv) Decrease of 9
(v) 2 km above sea level
(vi) 3 km below sea level
(vii) A deposit of Rs. 200
(viii) A withdrawl of Rs. 300
( ) EXERCISE 4 A
Q. 1. Write the opposite of each of the
following :
(i) An increase of 8
(ii) A loss of Rs. 7
(iii) Gaining a weight of 5 kg
(iv) 10 km above sea level
(v) 5°C below the freezing point
(vi) A deposit of Rs. 100
(vii) Earning Rs. 500
(viii) Going 6 m to the east
(ix) 24
(x) – 34
Sol. (i) A decrease of 8
(ii) A gain of Rs. 7
(iii) Loosing a weight of 5 kg
Page 2
Points to Remember :
Integers : The numbers .........., – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, .......... are called integers.
The numbers 1, 2, 3, 4, 5 ............ are called positive integers and the numbers – 1, – 2, – 3, – 4,
– 5 ............. are called negative integers. 0 is an integer which is neither positive nor negative.
Representation of Integers On Number Line :
We draw a line and fix a point almost in the middle of it and call it O. We set off equal distances on
right hand side as well as on left hand side of 0. We name the points of division as 1, 2, 3, 4 etc.on
left hand side as shown below :
Some results :
(i) Zero is less than every positive integer.
(ii) Zero is greater than every negative integer.
(iii) Every positive integer is greater then every negative integer.
(iv) The greater is the number, the lesser is its opposite.
Absolute Value of an Integer. The absolute value of an integer is the numerical value of the
integer regardless of its sign.
(iv) 10 km below sea level
(v) 5°C above the freezing point
(vi) A withdrawl of Rs. 100
(vii) Spending Rs. 500
(viii) Going 6 m to the west
(ix) – 24
(x) 34
Q. 2. Indicate the following using ‘+’ or ‘–’
sign :
(i) A gain of Rs. 600
(ii) A loss of Rs. 800
(iii) 7ºC below the freezing point
(iv) Decrease of 9
(v) 2 km above sea level
(vi) 3 km below sea level
(vii) A deposit of Rs. 200
(viii) A withdrawl of Rs. 300
( ) EXERCISE 4 A
Q. 1. Write the opposite of each of the
following :
(i) An increase of 8
(ii) A loss of Rs. 7
(iii) Gaining a weight of 5 kg
(iv) 10 km above sea level
(v) 5°C below the freezing point
(vi) A deposit of Rs. 100
(vii) Earning Rs. 500
(viii) Going 6 m to the east
(ix) 24
(x) – 34
Sol. (i) A decrease of 8
(ii) A gain of Rs. 7
(iii) Loosing a weight of 5 kg
Sol. (i) + Rs. 600 (ii) – Rs. 800
(iii) – 7ºC (iv) – 9
(v) + 2 km (vi) – 3 km
(vii) + Rs. 200 (viii) – Rs. 300
Q. 3. Mark the following integers on a number
line :
(i) – 5 (ii) – 2
(iii) 0 (iv) 7
(v) –13
Sol.
Q. 4. Which number is larger in each of the
following pairs.
(i) 0, – 2 (ii) – 3, – 5
(iii) – 5, 2 (iv) – 16, 8
(v) – 365, – 913 (vi) – 888, 8
Sol. (i) 0 (ii) – 3
(iii) 2 (iv) 8
(v) – 365 (vi) 8
Q. 5. Which number is smaller in each of the
following pairs ?
(i) 6, –7 (ii) 0, – 1
(iii) – 13, – 27 (iv) – 26, 17
(v) – 317, – 603 (vi) – 777, 7
Sol. (i) – 7 (ii) – 1
(iii) – 27 (iv) – 26
(v) – 603 (vi) – 777
Q. 6. Write all integers between
(i) 0 and 6 (ii) – 5 and 0
(iii) – 3 and 3 (iv) – 7 and – 5
Sol. (i) The integers between 0 and 6 are
1, 2, 3, 4, 5.
(ii) The integers between – 5 and 0 are
– 4, – 3, – 2, – 1.
(iii) The integers between – 3 and 3 are
– 2, – 1, 0, 1, 2.
(iv) The integer between – 7 and – 5 is – 6.
Q. 7. Fill in the blanks by appropriate symbol
> or < :
(i) 0 .......... 7 (ii) 0 ....... – 3
(iii) – 5 ........ –2 (iv) – 15 ...... 13
(v) – 231 ............ – 132 (vi) – 6 ...... 6
Sol. (i) 0 < 7 (ii) 0 > – 3
(iii) – 5 < – 2 (iv) – 15 < 13
(v) – 231 < – 132 (vi) – 6 < 6
Q. 8. Write the following integers in the
increasing order :
(i) 5, –7, – 2, 0, 8
(ii) – 23, 12, 0, – 6, – 100, – 1
(iii) – 17, 15, – 363, – 501, 165
(iv) 21, –106, –16, 16, 0, – 2, – 81
Sol. (i) – 7, – 2, 0, 5, 8
(ii) – 100, – 23, – 6, – 1, 0, 12
(iii) – 501, – 363, – 17, 15, 165
(iv) – 106, – 81, – 16, – 2, 0, 16, 21.
Q. 9. Write the following integers in the
decreasing order :
(i) 0, 7, – 3, –9, – 132, 36
(ii) 51, – 53, – 8, 0, – 2
(iii) – 71, – 81, 36, 0, – 5
(iv) – 365, – 515, 102, 413, – 7
Sol. (i) 36, 7, 0, – 3, – 9, – 132
(ii) 51, 0, – 2, – 8, – 53
(iii) 36, 0, – 5, – 71, – 81
(iv) 413, 102, – 7, – 365, – 515.
Q. 10. Using the number line, write the integer
which is
(i) 4 more than 6 (ii) 5 more than – 6
Page 3
Points to Remember :
Integers : The numbers .........., – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, .......... are called integers.
The numbers 1, 2, 3, 4, 5 ............ are called positive integers and the numbers – 1, – 2, – 3, – 4,
– 5 ............. are called negative integers. 0 is an integer which is neither positive nor negative.
Representation of Integers On Number Line :
We draw a line and fix a point almost in the middle of it and call it O. We set off equal distances on
right hand side as well as on left hand side of 0. We name the points of division as 1, 2, 3, 4 etc.on
left hand side as shown below :
Some results :
(i) Zero is less than every positive integer.
(ii) Zero is greater than every negative integer.
(iii) Every positive integer is greater then every negative integer.
(iv) The greater is the number, the lesser is its opposite.
Absolute Value of an Integer. The absolute value of an integer is the numerical value of the
integer regardless of its sign.
(iv) 10 km below sea level
(v) 5°C above the freezing point
(vi) A withdrawl of Rs. 100
(vii) Spending Rs. 500
(viii) Going 6 m to the west
(ix) – 24
(x) 34
Q. 2. Indicate the following using ‘+’ or ‘–’
sign :
(i) A gain of Rs. 600
(ii) A loss of Rs. 800
(iii) 7ºC below the freezing point
(iv) Decrease of 9
(v) 2 km above sea level
(vi) 3 km below sea level
(vii) A deposit of Rs. 200
(viii) A withdrawl of Rs. 300
( ) EXERCISE 4 A
Q. 1. Write the opposite of each of the
following :
(i) An increase of 8
(ii) A loss of Rs. 7
(iii) Gaining a weight of 5 kg
(iv) 10 km above sea level
(v) 5°C below the freezing point
(vi) A deposit of Rs. 100
(vii) Earning Rs. 500
(viii) Going 6 m to the east
(ix) 24
(x) – 34
Sol. (i) A decrease of 8
(ii) A gain of Rs. 7
(iii) Loosing a weight of 5 kg
Sol. (i) + Rs. 600 (ii) – Rs. 800
(iii) – 7ºC (iv) – 9
(v) + 2 km (vi) – 3 km
(vii) + Rs. 200 (viii) – Rs. 300
Q. 3. Mark the following integers on a number
line :
(i) – 5 (ii) – 2
(iii) 0 (iv) 7
(v) –13
Sol.
Q. 4. Which number is larger in each of the
following pairs.
(i) 0, – 2 (ii) – 3, – 5
(iii) – 5, 2 (iv) – 16, 8
(v) – 365, – 913 (vi) – 888, 8
Sol. (i) 0 (ii) – 3
(iii) 2 (iv) 8
(v) – 365 (vi) 8
Q. 5. Which number is smaller in each of the
following pairs ?
(i) 6, –7 (ii) 0, – 1
(iii) – 13, – 27 (iv) – 26, 17
(v) – 317, – 603 (vi) – 777, 7
Sol. (i) – 7 (ii) – 1
(iii) – 27 (iv) – 26
(v) – 603 (vi) – 777
Q. 6. Write all integers between
(i) 0 and 6 (ii) – 5 and 0
(iii) – 3 and 3 (iv) – 7 and – 5
Sol. (i) The integers between 0 and 6 are
1, 2, 3, 4, 5.
(ii) The integers between – 5 and 0 are
– 4, – 3, – 2, – 1.
(iii) The integers between – 3 and 3 are
– 2, – 1, 0, 1, 2.
(iv) The integer between – 7 and – 5 is – 6.
Q. 7. Fill in the blanks by appropriate symbol
> or < :
(i) 0 .......... 7 (ii) 0 ....... – 3
(iii) – 5 ........ –2 (iv) – 15 ...... 13
(v) – 231 ............ – 132 (vi) – 6 ...... 6
Sol. (i) 0 < 7 (ii) 0 > – 3
(iii) – 5 < – 2 (iv) – 15 < 13
(v) – 231 < – 132 (vi) – 6 < 6
Q. 8. Write the following integers in the
increasing order :
(i) 5, –7, – 2, 0, 8
(ii) – 23, 12, 0, – 6, – 100, – 1
(iii) – 17, 15, – 363, – 501, 165
(iv) 21, –106, –16, 16, 0, – 2, – 81
Sol. (i) – 7, – 2, 0, 5, 8
(ii) – 100, – 23, – 6, – 1, 0, 12
(iii) – 501, – 363, – 17, 15, 165
(iv) – 106, – 81, – 16, – 2, 0, 16, 21.
Q. 9. Write the following integers in the
decreasing order :
(i) 0, 7, – 3, –9, – 132, 36
(ii) 51, – 53, – 8, 0, – 2
(iii) – 71, – 81, 36, 0, – 5
(iv) – 365, – 515, 102, 413, – 7
Sol. (i) 36, 7, 0, – 3, – 9, – 132
(ii) 51, 0, – 2, – 8, – 53
(iii) 36, 0, – 5, – 71, – 81
(iv) 413, 102, – 7, – 365, – 515.
Q. 10. Using the number line, write the integer
which is
(i) 4 more than 6 (ii) 5 more than – 6
(iii) 6 less than 2 (iv) 2 less than – 3
Sol. (i) We want to write an integer 4 more
than 6.
So, we start from 6 and proceed 4 steps
to the right to obtain 10, as shown
below :
4 more than 6 is 10.
(ii) We want to write an integer 5 more than
– 6.
So, we start from – 6 and proceed 5
steps to the right to obtain – 1, as shown
below :
5 more than – 6 is – 1.
(iii) We want to write an integer 6 less than
2. So we start from 2 and come back to
the left by 6 steps to obtain – 4, as shown
below :
6 less than 2 is – 4.
(iv) We want to write an integer 2 less than
– 3. So we start from – 3 and come
back to the left by 2 steps to obtain – 5,
as shown below :
2 less than – 3 is – 5.
Q. 11. For each of the following statements,
write (T) for true and (F) for false.
(i) The smallest integer is zero.
(ii) Zero is not an integer.
(iii) The opposite of zero is zero.
(iv) – 10 is greater than – 6.
(v) The absolute value of an integer is
always greater than the integer.
(vi) 0 is larger than every negative integer.
(vii) Every negative integer is less than every
natural number.
(viii) The successor of –187 is –188.
(ix) The predecessor of –215 is –214.
Sol. (i) False, as zero is greater than every
negative integer.
(ii) False, as zero is an integer.
(iii) True, as zero is neither positive nor
negative.
(iv) False, as – 10 is to the left of – 6 on a
number line.
(v) False, as absolute value of an integer is
always equal to the integer.
(vi) True, as 0 is to right of every negative
integer, on a number line.
(vii) False, as every natural number is positive.
(viii) False, the successor is –186
(ix) False, the predecessor is –216
Q. 12. Find the value of
(i) | – 9 | (ii) | – 36 |
(iii) | 0 | (iv) | 15 |
(v) – | – 3 | (vi) 7 + | – 3 |
(vii) | 7 – 4 | (viii) 8 – | – 7 |
Sol. (i) | – 9 | = 9 (ii) | – 36 | = 36
(iii) | 0 | = 0 (iv) | 15 | = 15
(v) – | – 3 | = – 3
(vi) 7 + | – 3 | = 7 + 3 = 10
(vii) | 7 – 4 | = | 3 | = 3
(viii) 8 – | – 7 | = 8 – 7 = 1
Q. 13. (i) Write five negative integers greater
than – 7.
(ii) Write five negative integers less than
– 20.
Sol. (i) The required integers are – 6, – 5,
– 4, – 3, – 2.
(ii) The required integers are – 21, – 22,
– 23, – 24, – 25.
Rules for Addition of Integers :
1. If two positive integers or two negative
regardless of their signs and give the sum
their common sign.
Page 4
Points to Remember :
Integers : The numbers .........., – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, .......... are called integers.
The numbers 1, 2, 3, 4, 5 ............ are called positive integers and the numbers – 1, – 2, – 3, – 4,
– 5 ............. are called negative integers. 0 is an integer which is neither positive nor negative.
Representation of Integers On Number Line :
We draw a line and fix a point almost in the middle of it and call it O. We set off equal distances on
right hand side as well as on left hand side of 0. We name the points of division as 1, 2, 3, 4 etc.on
left hand side as shown below :
Some results :
(i) Zero is less than every positive integer.
(ii) Zero is greater than every negative integer.
(iii) Every positive integer is greater then every negative integer.
(iv) The greater is the number, the lesser is its opposite.
Absolute Value of an Integer. The absolute value of an integer is the numerical value of the
integer regardless of its sign.
(iv) 10 km below sea level
(v) 5°C above the freezing point
(vi) A withdrawl of Rs. 100
(vii) Spending Rs. 500
(viii) Going 6 m to the west
(ix) – 24
(x) 34
Q. 2. Indicate the following using ‘+’ or ‘–’
sign :
(i) A gain of Rs. 600
(ii) A loss of Rs. 800
(iii) 7ºC below the freezing point
(iv) Decrease of 9
(v) 2 km above sea level
(vi) 3 km below sea level
(vii) A deposit of Rs. 200
(viii) A withdrawl of Rs. 300
( ) EXERCISE 4 A
Q. 1. Write the opposite of each of the
following :
(i) An increase of 8
(ii) A loss of Rs. 7
(iii) Gaining a weight of 5 kg
(iv) 10 km above sea level
(v) 5°C below the freezing point
(vi) A deposit of Rs. 100
(vii) Earning Rs. 500
(viii) Going 6 m to the east
(ix) 24
(x) – 34
Sol. (i) A decrease of 8
(ii) A gain of Rs. 7
(iii) Loosing a weight of 5 kg
Sol. (i) + Rs. 600 (ii) – Rs. 800
(iii) – 7ºC (iv) – 9
(v) + 2 km (vi) – 3 km
(vii) + Rs. 200 (viii) – Rs. 300
Q. 3. Mark the following integers on a number
line :
(i) – 5 (ii) – 2
(iii) 0 (iv) 7
(v) –13
Sol.
Q. 4. Which number is larger in each of the
following pairs.
(i) 0, – 2 (ii) – 3, – 5
(iii) – 5, 2 (iv) – 16, 8
(v) – 365, – 913 (vi) – 888, 8
Sol. (i) 0 (ii) – 3
(iii) 2 (iv) 8
(v) – 365 (vi) 8
Q. 5. Which number is smaller in each of the
following pairs ?
(i) 6, –7 (ii) 0, – 1
(iii) – 13, – 27 (iv) – 26, 17
(v) – 317, – 603 (vi) – 777, 7
Sol. (i) – 7 (ii) – 1
(iii) – 27 (iv) – 26
(v) – 603 (vi) – 777
Q. 6. Write all integers between
(i) 0 and 6 (ii) – 5 and 0
(iii) – 3 and 3 (iv) – 7 and – 5
Sol. (i) The integers between 0 and 6 are
1, 2, 3, 4, 5.
(ii) The integers between – 5 and 0 are
– 4, – 3, – 2, – 1.
(iii) The integers between – 3 and 3 are
– 2, – 1, 0, 1, 2.
(iv) The integer between – 7 and – 5 is – 6.
Q. 7. Fill in the blanks by appropriate symbol
> or < :
(i) 0 .......... 7 (ii) 0 ....... – 3
(iii) – 5 ........ –2 (iv) – 15 ...... 13
(v) – 231 ............ – 132 (vi) – 6 ...... 6
Sol. (i) 0 < 7 (ii) 0 > – 3
(iii) – 5 < – 2 (iv) – 15 < 13
(v) – 231 < – 132 (vi) – 6 < 6
Q. 8. Write the following integers in the
increasing order :
(i) 5, –7, – 2, 0, 8
(ii) – 23, 12, 0, – 6, – 100, – 1
(iii) – 17, 15, – 363, – 501, 165
(iv) 21, –106, –16, 16, 0, – 2, – 81
Sol. (i) – 7, – 2, 0, 5, 8
(ii) – 100, – 23, – 6, – 1, 0, 12
(iii) – 501, – 363, – 17, 15, 165
(iv) – 106, – 81, – 16, – 2, 0, 16, 21.
Q. 9. Write the following integers in the
decreasing order :
(i) 0, 7, – 3, –9, – 132, 36
(ii) 51, – 53, – 8, 0, – 2
(iii) – 71, – 81, 36, 0, – 5
(iv) – 365, – 515, 102, 413, – 7
Sol. (i) 36, 7, 0, – 3, – 9, – 132
(ii) 51, 0, – 2, – 8, – 53
(iii) 36, 0, – 5, – 71, – 81
(iv) 413, 102, – 7, – 365, – 515.
Q. 10. Using the number line, write the integer
which is
(i) 4 more than 6 (ii) 5 more than – 6
(iii) 6 less than 2 (iv) 2 less than – 3
Sol. (i) We want to write an integer 4 more
than 6.
So, we start from 6 and proceed 4 steps
to the right to obtain 10, as shown
below :
4 more than 6 is 10.
(ii) We want to write an integer 5 more than
– 6.
So, we start from – 6 and proceed 5
steps to the right to obtain – 1, as shown
below :
5 more than – 6 is – 1.
(iii) We want to write an integer 6 less than
2. So we start from 2 and come back to
the left by 6 steps to obtain – 4, as shown
below :
6 less than 2 is – 4.
(iv) We want to write an integer 2 less than
– 3. So we start from – 3 and come
back to the left by 2 steps to obtain – 5,
as shown below :
2 less than – 3 is – 5.
Q. 11. For each of the following statements,
write (T) for true and (F) for false.
(i) The smallest integer is zero.
(ii) Zero is not an integer.
(iii) The opposite of zero is zero.
(iv) – 10 is greater than – 6.
(v) The absolute value of an integer is
always greater than the integer.
(vi) 0 is larger than every negative integer.
(vii) Every negative integer is less than every
natural number.
(viii) The successor of –187 is –188.
(ix) The predecessor of –215 is –214.
Sol. (i) False, as zero is greater than every
negative integer.
(ii) False, as zero is an integer.
(iii) True, as zero is neither positive nor
negative.
(iv) False, as – 10 is to the left of – 6 on a
number line.
(v) False, as absolute value of an integer is
always equal to the integer.
(vi) True, as 0 is to right of every negative
integer, on a number line.
(vii) False, as every natural number is positive.
(viii) False, the successor is –186
(ix) False, the predecessor is –216
Q. 12. Find the value of
(i) | – 9 | (ii) | – 36 |
(iii) | 0 | (iv) | 15 |
(v) – | – 3 | (vi) 7 + | – 3 |
(vii) | 7 – 4 | (viii) 8 – | – 7 |
Sol. (i) | – 9 | = 9 (ii) | – 36 | = 36
(iii) | 0 | = 0 (iv) | 15 | = 15
(v) – | – 3 | = – 3
(vi) 7 + | – 3 | = 7 + 3 = 10
(vii) | 7 – 4 | = | 3 | = 3
(viii) 8 – | – 7 | = 8 – 7 = 1
Q. 13. (i) Write five negative integers greater
than – 7.
(ii) Write five negative integers less than
– 20.
Sol. (i) The required integers are – 6, – 5,
– 4, – 3, – 2.
(ii) The required integers are – 21, – 22,
– 23, – 24, – 25.
Rules for Addition of Integers :
1. If two positive integers or two negative
regardless of their signs and give the sum
their common sign.
2. If a positive integer and a negative integer
are added, we find the difference
between their values regardless of their
signs and give the sign of the integer
with more numerical value.
Properties of Addition of Integers :
Property 1 (Closure Property). The
sum of two Integers is always an integer.
Property 2 (Commutative law of
Addition). If a and b are any two
integers then a + b = b + a.
Property 3 (Associative law of
Addition). If a, b, c are any three
integers then
(a + b) + c = a + (b + c)
Property 4. If a is any integer then
a + 0 = 0 + a = a
Property 5. The sum of an integer and
its opposite is 0. Thus, if a is an integer
then a + (– a) = 0.
a and – a are called opposites or
negatives or additive inverse of each
other.
Property 6. If a is any integer than (a +
1) is also an integer, called the successor
of a.
( ) EXERCISE 4 B
Q. 1. Use the number line and add the
following integers :
(i) 9 + (– 6) (ii) (–3) + 7
(iii) 8 + (– 8) (iv) (–1) + (–3)
(v) (– 4) + (– 7) (vi) (– 2) + (– 8)
(vii) 3 + (– 2) + (– 4)
(viii) (– 1) + (– 2) + (– 3)
(ix) 5 + (– 2) + (– 6)
Sol. (i) On the number line we start from 0
and move 9 steps to the right to reach a
point A. Now, starting from A, we move
6 steps to the left to reach a point B, as
shown below :
Now, B represents the integer 3
9 + (– 6) = 3
(ii) On the number line, we start from 0 and
move 3 steps to the left to reach a point
A. Now, starting from A, we move 7
steps to the right to reach a point B, as
shown below :
And B represents the integer 4
(– 3) + 7 = 4
(iii) On the number line, we start from 0 and
move 8 steps to the right to reach a point
A. Now, starting from A, we move 8
steps to the left to reach a point B, as
shown below :
And, B represents the integer 0.
8 + (– 8) = 0
(iv) On the number line, we start from 0 and
move 1 step the left to reach a point A.
Now, starting from point A, we move 3
steps to the left to reach a point B, as
shown below :
And, B represents the integer – 4
(– 1) + (– 3) = – 4.
(v) On the number line, we start from 0 and
move 4 steps to the left to reach a point
A. Now, starting from point A, we move
Page 5
Points to Remember :
Integers : The numbers .........., – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, .......... are called integers.
The numbers 1, 2, 3, 4, 5 ............ are called positive integers and the numbers – 1, – 2, – 3, – 4,
– 5 ............. are called negative integers. 0 is an integer which is neither positive nor negative.
Representation of Integers On Number Line :
We draw a line and fix a point almost in the middle of it and call it O. We set off equal distances on
right hand side as well as on left hand side of 0. We name the points of division as 1, 2, 3, 4 etc.on
left hand side as shown below :
Some results :
(i) Zero is less than every positive integer.
(ii) Zero is greater than every negative integer.
(iii) Every positive integer is greater then every negative integer.
(iv) The greater is the number, the lesser is its opposite.
Absolute Value of an Integer. The absolute value of an integer is the numerical value of the
integer regardless of its sign.
(iv) 10 km below sea level
(v) 5°C above the freezing point
(vi) A withdrawl of Rs. 100
(vii) Spending Rs. 500
(viii) Going 6 m to the west
(ix) – 24
(x) 34
Q. 2. Indicate the following using ‘+’ or ‘–’
sign :
(i) A gain of Rs. 600
(ii) A loss of Rs. 800
(iii) 7ºC below the freezing point
(iv) Decrease of 9
(v) 2 km above sea level
(vi) 3 km below sea level
(vii) A deposit of Rs. 200
(viii) A withdrawl of Rs. 300
( ) EXERCISE 4 A
Q. 1. Write the opposite of each of the
following :
(i) An increase of 8
(ii) A loss of Rs. 7
(iii) Gaining a weight of 5 kg
(iv) 10 km above sea level
(v) 5°C below the freezing point
(vi) A deposit of Rs. 100
(vii) Earning Rs. 500
(viii) Going 6 m to the east
(ix) 24
(x) – 34
Sol. (i) A decrease of 8
(ii) A gain of Rs. 7
(iii) Loosing a weight of 5 kg
Sol. (i) + Rs. 600 (ii) – Rs. 800
(iii) – 7ºC (iv) – 9
(v) + 2 km (vi) – 3 km
(vii) + Rs. 200 (viii) – Rs. 300
Q. 3. Mark the following integers on a number
line :
(i) – 5 (ii) – 2
(iii) 0 (iv) 7
(v) –13
Sol.
Q. 4. Which number is larger in each of the
following pairs.
(i) 0, – 2 (ii) – 3, – 5
(iii) – 5, 2 (iv) – 16, 8
(v) – 365, – 913 (vi) – 888, 8
Sol. (i) 0 (ii) – 3
(iii) 2 (iv) 8
(v) – 365 (vi) 8
Q. 5. Which number is smaller in each of the
following pairs ?
(i) 6, –7 (ii) 0, – 1
(iii) – 13, – 27 (iv) – 26, 17
(v) – 317, – 603 (vi) – 777, 7
Sol. (i) – 7 (ii) – 1
(iii) – 27 (iv) – 26
(v) – 603 (vi) – 777
Q. 6. Write all integers between
(i) 0 and 6 (ii) – 5 and 0
(iii) – 3 and 3 (iv) – 7 and – 5
Sol. (i) The integers between 0 and 6 are
1, 2, 3, 4, 5.
(ii) The integers between – 5 and 0 are
– 4, – 3, – 2, – 1.
(iii) The integers between – 3 and 3 are
– 2, – 1, 0, 1, 2.
(iv) The integer between – 7 and – 5 is – 6.
Q. 7. Fill in the blanks by appropriate symbol
> or < :
(i) 0 .......... 7 (ii) 0 ....... – 3
(iii) – 5 ........ –2 (iv) – 15 ...... 13
(v) – 231 ............ – 132 (vi) – 6 ...... 6
Sol. (i) 0 < 7 (ii) 0 > – 3
(iii) – 5 < – 2 (iv) – 15 < 13
(v) – 231 < – 132 (vi) – 6 < 6
Q. 8. Write the following integers in the
increasing order :
(i) 5, –7, – 2, 0, 8
(ii) – 23, 12, 0, – 6, – 100, – 1
(iii) – 17, 15, – 363, – 501, 165
(iv) 21, –106, –16, 16, 0, – 2, – 81
Sol. (i) – 7, – 2, 0, 5, 8
(ii) – 100, – 23, – 6, – 1, 0, 12
(iii) – 501, – 363, – 17, 15, 165
(iv) – 106, – 81, – 16, – 2, 0, 16, 21.
Q. 9. Write the following integers in the
decreasing order :
(i) 0, 7, – 3, –9, – 132, 36
(ii) 51, – 53, – 8, 0, – 2
(iii) – 71, – 81, 36, 0, – 5
(iv) – 365, – 515, 102, 413, – 7
Sol. (i) 36, 7, 0, – 3, – 9, – 132
(ii) 51, 0, – 2, – 8, – 53
(iii) 36, 0, – 5, – 71, – 81
(iv) 413, 102, – 7, – 365, – 515.
Q. 10. Using the number line, write the integer
which is
(i) 4 more than 6 (ii) 5 more than – 6
(iii) 6 less than 2 (iv) 2 less than – 3
Sol. (i) We want to write an integer 4 more
than 6.
So, we start from 6 and proceed 4 steps
to the right to obtain 10, as shown
below :
4 more than 6 is 10.
(ii) We want to write an integer 5 more than
– 6.
So, we start from – 6 and proceed 5
steps to the right to obtain – 1, as shown
below :
5 more than – 6 is – 1.
(iii) We want to write an integer 6 less than
2. So we start from 2 and come back to
the left by 6 steps to obtain – 4, as shown
below :
6 less than 2 is – 4.
(iv) We want to write an integer 2 less than
– 3. So we start from – 3 and come
back to the left by 2 steps to obtain – 5,
as shown below :
2 less than – 3 is – 5.
Q. 11. For each of the following statements,
write (T) for true and (F) for false.
(i) The smallest integer is zero.
(ii) Zero is not an integer.
(iii) The opposite of zero is zero.
(iv) – 10 is greater than – 6.
(v) The absolute value of an integer is
always greater than the integer.
(vi) 0 is larger than every negative integer.
(vii) Every negative integer is less than every
natural number.
(viii) The successor of –187 is –188.
(ix) The predecessor of –215 is –214.
Sol. (i) False, as zero is greater than every
negative integer.
(ii) False, as zero is an integer.
(iii) True, as zero is neither positive nor
negative.
(iv) False, as – 10 is to the left of – 6 on a
number line.
(v) False, as absolute value of an integer is
always equal to the integer.
(vi) True, as 0 is to right of every negative
integer, on a number line.
(vii) False, as every natural number is positive.
(viii) False, the successor is –186
(ix) False, the predecessor is –216
Q. 12. Find the value of
(i) | – 9 | (ii) | – 36 |
(iii) | 0 | (iv) | 15 |
(v) – | – 3 | (vi) 7 + | – 3 |
(vii) | 7 – 4 | (viii) 8 – | – 7 |
Sol. (i) | – 9 | = 9 (ii) | – 36 | = 36
(iii) | 0 | = 0 (iv) | 15 | = 15
(v) – | – 3 | = – 3
(vi) 7 + | – 3 | = 7 + 3 = 10
(vii) | 7 – 4 | = | 3 | = 3
(viii) 8 – | – 7 | = 8 – 7 = 1
Q. 13. (i) Write five negative integers greater
than – 7.
(ii) Write five negative integers less than
– 20.
Sol. (i) The required integers are – 6, – 5,
– 4, – 3, – 2.
(ii) The required integers are – 21, – 22,
– 23, – 24, – 25.
Rules for Addition of Integers :
1. If two positive integers or two negative
regardless of their signs and give the sum
their common sign.
2. If a positive integer and a negative integer
are added, we find the difference
between their values regardless of their
signs and give the sign of the integer
with more numerical value.
Properties of Addition of Integers :
Property 1 (Closure Property). The
sum of two Integers is always an integer.
Property 2 (Commutative law of
Addition). If a and b are any two
integers then a + b = b + a.
Property 3 (Associative law of
Addition). If a, b, c are any three
integers then
(a + b) + c = a + (b + c)
Property 4. If a is any integer then
a + 0 = 0 + a = a
Property 5. The sum of an integer and
its opposite is 0. Thus, if a is an integer
then a + (– a) = 0.
a and – a are called opposites or
negatives or additive inverse of each
other.
Property 6. If a is any integer than (a +
1) is also an integer, called the successor
of a.
( ) EXERCISE 4 B
Q. 1. Use the number line and add the
following integers :
(i) 9 + (– 6) (ii) (–3) + 7
(iii) 8 + (– 8) (iv) (–1) + (–3)
(v) (– 4) + (– 7) (vi) (– 2) + (– 8)
(vii) 3 + (– 2) + (– 4)
(viii) (– 1) + (– 2) + (– 3)
(ix) 5 + (– 2) + (– 6)
Sol. (i) On the number line we start from 0
and move 9 steps to the right to reach a
point A. Now, starting from A, we move
6 steps to the left to reach a point B, as
shown below :
Now, B represents the integer 3
9 + (– 6) = 3
(ii) On the number line, we start from 0 and
move 3 steps to the left to reach a point
A. Now, starting from A, we move 7
steps to the right to reach a point B, as
shown below :
And B represents the integer 4
(– 3) + 7 = 4
(iii) On the number line, we start from 0 and
move 8 steps to the right to reach a point
A. Now, starting from A, we move 8
steps to the left to reach a point B, as
shown below :
And, B represents the integer 0.
8 + (– 8) = 0
(iv) On the number line, we start from 0 and
move 1 step the left to reach a point A.
Now, starting from point A, we move 3
steps to the left to reach a point B, as
shown below :
And, B represents the integer – 4
(– 1) + (– 3) = – 4.
(v) On the number line, we start from 0 and
move 4 steps to the left to reach a point
A. Now, starting from point A, we move
7 steps to the left to reach a point B, as
shown below :
And, B represents the integer – 11.
(– 4) + (– 7) = – 11
(vi) On the number line we start from 0 and
move 2 steps to the left to reach a point
A. Now, starting from A, we move 8
steps to the left to reach a point B, as
shown below :
And, B represents the integer – 10
(– 2) + (– 8) = – 10
(vii) On the number line we start from 0 and
move 3 steps to the right to reach a point
A. Now, starting from A, we move 2
steps to the left to reach a point B and
again starting from left to reach a point
B and again starting from B, we move 4
steps to the left to reach a point C, as
shown below :
And, C represents the integer – 3
3 + (– 2) + (– 4) = – 3
(viii) On the number line we start from 0 and
move 1 step to the left to reach a point
A. Now, starting from A, we move 2
steps to the left to reach a point B and
again starting from B, we move 3 steps
to the left to reach point C, as shown
below :
And, C represents the integer – 6
(– 1) + (– 2) + (– 3) = – 6.
(ix) On the number line we start from 0 and
move 5 steps to the right to reach a point
A. Now, starting from A, we move 2
steps to the left to reach a point B and
again starting from point B, we move 6
steps to the left to reach a point C, as
shown below :
And, C represents the integer – 3.
5 + (– 2) + (– 6) = – 3
Q. 2. Fill in the blanks :
(i) (– 3) + (– 9) = ..........
(ii) (– 7) + (– 8) = .........
(iii) (– 9) + 16 = ............
(iv) (– 13) + 25 =...........
(v) 8 + (– 17) = ............
(vi) 2 + (– 12) = ............
Sol. (i) (– 3) + (– 9) = – 12
(Using the rule for addition of integers
having like signs)
(ii) (– 7) + (– 8) = – 15
(Using the rule for addition of integers
having like signs)
(iii) (– 9) + 16 = 7
(Using the rule for addition of integers
having unlike signs)
(iv) (– 13) + 25 = 12
(Using the rule for addition of integers
having unlike signs)
(v) 8 + (– 17) = – 9
(Using the rule for addition of integers
having unlike signs)
(vi) 2 + (– 12) = – 10
(Using the rule for addition of integers
having unlike signs)
```
## Mathematics (Maths) Class 6
94 videos|347 docs|54 tests
## FAQs on RS Aggarwal Solutions: Integers - Mathematics (Maths) Class 6
1. What are RS Aggarwal Solutions?
Ans. RS Aggarwal Solutions are a comprehensive set of solutions to the problems provided in the RS Aggarwal textbook. These solutions help students understand and solve mathematical problems effectively and efficiently.
2. What is the significance of Integers in Class 6 mathematics?
Ans. Integers play a crucial role in Class 6 mathematics as they introduce students to the concept of negative numbers and their operations. Understanding integers helps in solving real-life problems involving positive and negative quantities.
3. How can RS Aggarwal Solutions for Integers in Class 6 help students prepare for exams?
Ans. RS Aggarwal Solutions for Integers in Class 6 provide step-by-step solutions to all the problems in the textbook. By practicing these solutions, students can enhance their problem-solving skills and gain a better understanding of the concepts. This will help them prepare effectively for exams.
4. Are RS Aggarwal Solutions for Integers Class 6 available online?
Ans. Yes, RS Aggarwal Solutions for Integers Class 6 are available online. Many educational websites and platforms provide these solutions for free or at a nominal cost. Students can access them easily to practice and improve their mathematical skills.
5. Can RS Aggarwal Solutions for Integers Class 6 be used for self-study?
Ans. Absolutely! RS Aggarwal Solutions for Integers Class 6 are designed to be user-friendly and self-explanatory. Students can use these solutions for self-study and practice solving problems independently. The step-by-step explanations will help them grasp the concepts effectively and build their confidence in mathematics.
## Mathematics (Maths) Class 6
94 videos|347 docs|54 tests
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Vinculum, An Unexplored Tool by N. S. Murty
Vinculum, An Unexplored Tool
by N. S. Murty
Introduction
Though the concept of working with reminders is not altogether new, I feel ‘Vinculum’ is one good tool that is not explained or used in our school or college mathematics, or even in our daily lives. Once we understand and start using it, I am sure we can do some very (apparently) tough calculations. It can also be used effectively in algebra as well.
In fact, one can change a long subtraction into addition; or a multiplication of large numbers to an operation with very small numbers like 1, 2, 3 or 4
Wikipedia gives an account of Vinculum as follows:
“A vinculum is a horizontal line used in mathematical notation for a specific purpose. It may be placed as an overline (or underline) over (or under) a mathematical expression to indicate that the expression is to be considered grouped together. Historically, vincula were extensively used to group items together, especially in written mathematics, but in modern mathematics this function has almost entirely been replaced by the use of parentheses. Today, however, the common usage of a vinculum to indicate the repetend of a repeating decimal is a significant exception and reflects the original usage.”
However, we can use Vinculum to indicate
(a) shortfall of a given number with respect to another number (a Ten-multiple normally), or
(b) Surplus over a given number over a negative number (which is also a Ten-multiple normally).
The advantage of this notation is that you can avoid using brackets for negative quantities.
Let me explain with examples:
(Caution: Depending upon the nearest 10 digit number you take, the appearance of a 3 digit number may look like a 4-digit number but its value will not change, as can be seen from examples 4 and 6)
Operations:
Vinculum works like a minus (-) sign.
Ex. 7: We know, 89 + 98 = 187;
Using vinculum we can represent it as
(Literally, we are only adding the differences of the given numbers with respect to a nearest 10 multiple, which are operationally small numbers, and then subtracting from the sum of the 10 multiples which is always easy.)
To get the difference with the nearest 10 multiple, we can use the following Sutra given by Swamy Bharati Tirtha in his book on Vedic Mathematics: All from 9 and the last from 10)
Ex.8: We know, 8979 + 9879 = 18858
Converting the numbers using vinculum,
(1thousand 1 hundred 42 short of 20 thousand). Using the same sutra to convert it back, we get 18858.
2. Subtraction:
In fact, Vinculum comes in very handy in subtractions. In fact, the whole operation of subtraction reduces to subtraction of only one number, and addition of the rest. How?
Let us see.
Ex. 9: Subtract 7768 from 11125
3. Multiplication:
4. Division:
This is equally applicable to division and makes division easy. However, it is advised not to venture into it now, until a fair level of command of using Vinculum for the other three operations is obtained.
In the next article, we shall see how efficient vinculum operator is in finding out answers for large multiplications, finding squares, cubes etc.
Contiued to Next Page
20-Oct-2018
More by : N. S. Murty
Thank you Mr. Peter for your interest.I did not follow it up since there was not much enthusiastic response to this post.Your interest gives me fillip to continue.I will post the article shortlybest regardsNS Murty
NS Murty
11/10/2020 09:11 AM
Sir,Many thanks for truly enlightening article, it is sad that the follow up article seems nowhere to be found.Peter.9
Peter Williams
11/09/2020 18:36 PM
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#### Solving Systems of Equations - Substitution
y = 5x + 10
5x + 2y = 80
What does it mean to solve a system of linear equations?
What does it mean to solve by substitution?
The Mechanics of the Substitution Method
How do I know my solution is correct?
Another Consideration
What does it mean to solve a system of linear equations?
Linear simply means a straight line, and a system in this case means more than one line. Above, there are equations for two lines. Curious minds want to know if these lines intersect. If so, where do they intersect? The point of intersection is considered the solution and designated by the (x, y) coordinates. In the real world, systems of linear equations are used many times to represent and solve situations concerning consumer economics as well as geometry.
What does it mean to solve by substitution?
In sports, substitution means one player will come out of the game and is replaced by another player. Likewise, in an equation, one of the players / variables, “x or “y” will be replaced. Pause. Remember, the “=” sign stands for the word “is”. Thus, the equation, y = 5x + 10, reads “y is 5x + 10,” or the value of y is “5x +10”. So, wherever you see the variable “y” in the other equations, you know that y comes out and is replaced with its value. After the substitution is made for “y,” you will notice there are no longer two variables but one. Thus, you are able to combine like terms and solve for the only variable left, “x”.
The Mechanics of the Substitution Method
The original problem:
y = 5x + 10
5x + 2y = 80
Part One:
5x + 2(5x + 10) = 80 ……… make substitution
5x + 2(5x) + 2(10) = 80 ……. use the distributive property
5x + 10x + 20 = 80 ………... multiply
15x + 20 = 80 ……………… combine like terms
15x + 20 – 20 = 80 – 20 …… subtract twenty from both sides
15x = 60 ……………………. Results of Subtraction
15x / 15 = 60 /15 …………… Divide both sides by 15
X = 4 ……………………….. Results of division
“X = 4” is the first coordinate of the solution.
Part Two:
Find the value of ”Y”
Choose one of the original equations above and make a substitution for “x.”
Y=5x +4
In part one, we determined x is 4.
So, y = 5(4) + 10
Y = 20 + 10
Y = 30 ……………. Is the second coordinate of the solution.
Solution = (X, Y) = (4, 30)
How do I know my solution is correct?
The solution implies that the values of x and y can be replaced in each equation, and the equation will be a true statement.
Therefore, let’s make substitution for both variables.
y =5x + 10 .........original equation
30
= 5(4) + 10 …. Substitute
30
= 20 + 10
30 = 30 True
5x + 2y = 80 ……….... original equation
5(4) + 2(30) = 80 ......... substitution
20 + 60 = 80
80 = 80 True
Thus, the solution (4, 30) is correct!
……………………
Another consideration -
Sometimes, neither equation in the system has been solved for a variable. For example, our previous problem was
y = 5x + 10
5x + 2y = 80
The first equation was already solved for y. However, the problem could have been presented as
y - 5x = 10
5x + 2y =80
Then, you would have to solve for the “x” or “y” variable in one of the equations. Let’s choose
y – 5x = 10
y – 5x + 5x = 10 + 5x …….. add 5x to both sides
y = 5x + 10 ……….. result of addition
Now, you can continue with the mechanics of the substitution method.
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Krishna
0
(i) Given that x_3 = 1
Find the value of p,
Step 1: Before going to do the problem, Find the knowns and unknowns in and note it down.
Step 2: Explore the given rule
EXAMPLE: x_{n+1} = ax_n - 3
Succeeding term = a (preceding term) - 3
Step 3: By using the given rule find the required term ( x_2, x_3,x_4...etc)
NOTE: According to the given rule substitute the values to get the required term.
EXAMPLE: For an attempt to find the x_3
x_{n+1} = ax_n - 3
x_3 = ax_2 - 3
Substitute the values x_2 (\text{Calculate it or given in the question }x_2 = a - 3)
x_ 3 = a(a - 3) - 3
Step 4: Equate the given term value to the term calculated by the rule (both represents the same value, so we can equate them)
EXAMPLE : a^2 – 3a – 3 = 7
Step 5: Simplify the equation to find the requires value(a)
(ii) Write down the value of x_{2008}
Step 1: Make sure that the given set of numbers arranged in some particular order. Because the question says that the set of numbers in sequence.
Step 2: Explore the given rule
EXAMPLE: x_{n+1} = ax_n - 3
Succeeding term = a (preceding term) - 3
Step 3: According to the given rule substitute the (n)values.
[NOTE: To find the twenty-first term, replace n by 21. based up on the rule it(n) may change to lower value or higher value]
EXAMPLE: For an attempt to find the x_2 substitute n=1 in the given rule
x_{n+1} = ax_n - 3
x_2 = ax_1 - 3
Substitute the values x_1 [ take x_1 = 1 since n >1), Some times it mention
in the question]
x_ 2 = a(1) - 3
x_2 = a - 3
Step 4: Simplify further
(Apply the BODMAS rules)
|
# MA.912.GR.5.3
Construct the inscribed and circumscribed circles of a triangle.
### Clarifications
Clarification 1: Instruction includes using compass and straightedge, string, reflective devices, paper folding or dynamic geometric software.
General Information
Subject Area: Mathematics (B.E.S.T.)
Grade: 912
Strand: Geometric Reasoning
Date Adopted or Revised: 08/20
Status: State Board Approved
## Benchmark Instructional Guide
### Terms from the K-12 Glossary
• Inscribed Circle
• Circumscribed Circle
• Triangle
### Vertical Alignment
Previous Benchmarks
Next Benchmarks
### Purpose and Instructional Strategies
In grade 7, students used a relationship between triangles and circles to understand the formula for the area of a circle. In Geometry, students identify and construct two special circles that are associated with a triangle.
• Instruction includes the use of manipulatives, tools and geometric software. Allowing students to explore constructions with dynamic software reinforces why the constructions work.
• Instruction includes the student understanding that in a geometric construction, one does not use the markings on a ruler or on a protractor to construct inscribed and circumscribed circles of a triangle. Students should realize that there are limitations on precision that are inherent in the markings on rulers or protractors.
• It is important to build the understanding that formal constructions are valid when the lengths of segments or measures of angles are not known, or have values that do not appear on a ruler or protractor, including irrational values.
• Instruction includes the connection to logical reasoning and visual proofs when verifying that a construction works.
• Instruction includes the connection to constructing angle bisectors and perpendicular bisectors. (MTR.2.1)
• Instruction includes using various methods, like the one described below, to construct an inscribed circle.
• For example, given triangle ABC, students can construct two of the three angle bisectors to create their point of intersection, D. Students should realize that the point D is the incenter of the triangle and may predict that point D will be the center of the inscribed circle. To prove this prediction, students will need to prove that point D is equidistant from each of the three sides. In order to prove this, students can construct the perpendicular segments from point D to each of the three sides, and show that all three segments are congruent using triangle congruence criteria and D is the intersection of the angle bisectors. Each of these segments will be a radius of the inscribed circle, with the center of the circle at point D
• When constructing an inscribed circle, students should make the connection to constructing perpendicular bisectors when they need to construct a line through the incenter of the triangle that is perpendicular to a side of the triangle.
• For example, to construct such a line, students can place the compass at the incenter, point D, and draw arcs to determine two points, E and F, on one of the sides. These points are equidistant to D. Then they, using the same compass setting, place the compass at E and at F and draw arcs intersecting on the opposite side of EF from D. The intersection of these arcs, Q, is the same distance to E and to F. Therefore, the line passing thru D and Q is the perpendicular bisector of EF so it is also a line perpendicular to the side of the triangle.
• Students should understand that the shortest segment from a point, D, to a line is the segment from D to the line that is perpendicular to the line. Additionally, students should understand that the circle centered at point D, which has this segment as a radius, is tangent to the line.
• Instruction includes using various methods, like the one described below, to construct a circumscribed circle.
• For example, given triangle ABC, students can construct two of the three perpendicular bisectors of the sides of the triangle to create their point of intersection, D. Students should realize that the point D is the circumcenter of the triangle and may predict that point D will be the center of the circumscribed circle. To prove this prediction, students will need to prove that point D is equidistant from each of the three vertices. In order to prove this, students can use the fact that point D is the intersection of the perpendicular bisectors. Each of these segments will be a radius of the circumscribed circle, with the center of the circle at point D. So, to construct the circumscribed circle, one can set the compass equal to the distance between point D and any one of the vertices and then draw the circle centered at point D
• Instruction includes exploring the construction of circumscribed circles about various triangles. Have students explore acute, right and obtuse triangles, and compare the locations of the circumcenter of each. Students should understand that with a right triangle, the circumcenter is located at the midpoint of the hypotenuse.
• For expectations of this benchmark, constructions should be reasonably accurate and the emphasis is to make connections between the construction steps and the definitions, properties and theorems supporting them.
• While going over the steps of geometric constructions, ensure that students develop vocabulary to describe the steps precisely. (MTR.4.1)
• Problem types include identifying the next step of a construction, a missing step in a construction or the order of the steps in a construction.
### Common Misconceptions or Errors
• Students may think that the when constructing a circumscribed circle, the center of the circle cannot be outside the triangle.
### Instructional Tasks
Instructional Task 1 (MTR.2.1, MTR.4.1
• Part A. Construct angle bisectors for the three interior angles of a triangle using folding paper, a compass and straightedge and geometric software. What do you notice about each method of construction?
• Part B. Repeat Part A with a triangle that is obtuse, isosceles, acute and right. Describe your findings.
• Part C. Using the incenter as the center, a circle can be constructed inscribed in the triangle. How can you determine the radius of that circle, the inscribed circle?
• Part D. Construct the inscribed circle of one of the triangles from Part B.
Instructional Task 2 (MTR.2.1, MTR.4.1
• Part A. Construct perpendicular bisector for the three sides of a triangle using folding paper, a compass and straightedge and geometric software. What do you notice about each method of construction?
• Part B. Repeat Part A with a triangle that is obtuse, isosceles, acute and right. Describe your findings.
• Part C. Using the circumcenter as the center, a circle can be constructed circumscribed about the triangle. How can you determine the radius of that circle, the circumscribed circle?
• Part D. Construct the circumscribed circle of one of the triangles from Part B.
Instructional Task 3 (MTR.5.1)
• Part A. Given the line $l$ and the point P external to the line $l$, construct a perpendicular line, $m$, through point P
• Part B. Use the construction from Part A to construct a line, $n$, that is parallel to the line $l$ and contains the point P.
### Instructional Items
Instructional Item 1
• Construct the circle that is circumscribed about ΔXYZ.
*The strategies, tasks and items included in the B1G-M are examples and should not be considered comprehensive.
## Related Courses
This benchmark is part of these courses.
1200400: Foundational Skills in Mathematics 9-12 (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1206310: Geometry (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1206320: Geometry Honors (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1206315: Geometry for Credit Recovery (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
7912065: Access Geometry (Specifically in versions: 2015 - 2022, 2022 and beyond (current))
## Related Access Points
Alternate version of this benchmark for students with significant cognitive disabilities.
MA.912.GR.5.AP.3: Select the inscribed and circumscribed circles of a triangle.
## Related Resources
Vetted resources educators can use to teach the concepts and skills in this benchmark.
## Formative Assessments
The Sprinters’ Race:
Students are given a grid with three points (vertices of a right triangle) representing the starting locations of three sprinters in a race and are asked to determine the center of the finish circle, which is equidistant from each sprinter.
Type: Formative Assessment
Inscribed Circle Construction:
Students are asked to use a compass and straightedge to construct an inscribed circle of an acute scalene triangle.
Type: Formative Assessment
Circumscribed Circle Construction:
Students are asked to use a compass and straightedge to construct a circumscribed circle of an acute scalene triangle.
Type: Formative Assessment
## Lesson Plans
Construction of Inscribed Regular Hexagon:
A GeoGebra lesson for students to become familiar with computer based construction tools. Students work together to construct a regular hexagon inscribed in a circle using rotations. Directions for both a beginner and advanced approach are provided.
Type: Lesson Plan
Inscribe Those Rims:
This lesson will engage students with an interactive and interesting way to learn how to inscribe polygons in circles.
Type: Lesson Plan
Construction Junction:
Students will learn how to construct an equilateral triangle and a regular hexagon inscribed in a circle using a compass and a straightedge.
Type: Lesson Plan
Inscribe it:
This activity allows students to practice the construction process inscribing a regular hexagon and an equilateral triangle in a circle using GeoGebra software.
Type: Lesson Plan
Construct Regular Polygons Inside Circles:
Students will be able to demonstrate that they can construct, using the central angle method, an equilateral triangle, a square, and a regular hexagon, inscribed inside a circle, using a compass, straightedge, and protractor. They will use worksheets to master the construction of each polygon, one inside each of three different circles. As an extension to this lesson, if computers with GeoGebra are available, the students should be able to perform these constructions on computers as well.
Type: Lesson Plan
Determination of the Optimal Point:
Students will use dynamic geometry software to determine the optimal location for a facility under a variety of scenarios. The experiments will suggest a relation between the optimal point and a common concept in geometry; in some cases, there will be a connection to a statistical concept. Algebra can be used to verify some of the conjectures.
Type: Lesson Plan
Paper Plate Origami:
A hands-on activity where students construct inscribed regular polygons in a circle using models. Through guided questions, students will discover how to divide a model (paper plate) into 3, 4, and 6 parts. Using folding, a straightedge, and a compass, they will construct an equilateral triangle, a square, and a regular hexagon in their circles.
Type: Lesson Plan
What's the Point? Part 2:
In this lesson, students use a paper-folding technique to discover the properties of angle bisectors. At the conclusion of the activity, students will be able to compare/contrast the points of concurrency of perpendicular and angle bisectors.
Type: Lesson Plan
Circumnavigating the Circumcenter:
Students use the concurrent point of perpendicular bisectors of triangle sides to determine the circumcenter of three points. Students will reason that the circumcenter of the vertices of a polygon is the optimal location for placement of a facility to service all of the needs of sites at the vertices forming the polygon.
Type: Lesson Plan
Crafty Circumference Challenge:
Students learn about geometric construction tools and how to use them. Students will partition the circumference of a circle into three, four, and six congruent arcs which determine the vertices of regular polygons inscribed in the circle. An optional project is included where students identify, find, and use recycled, repurposed, or reclaimed objects to create "crafty" construction tools.
Type: Lesson Plan
Concurrent Points Are Optimal:
Students will begin with a review of methods of construction of perpendicular bisectors and angle bisectors for the sides of triangles. Included in the review will be a careful discussion of the proofs that the constructions actually produce the lines that were intended.
Next, students will investigate why the perpendicular bisectors and angle bisector are concurrent, that is, all three meet at a single meet.
A more modern point of currency is the Fermat-Torricelli point (F-T). The students will construct (F-T) in GeoGebra and investigate limitations of its existence for various types of triangles.
Then a set of scenarios will be provided, including some one-dimensional and two-dimensional situations. Students will use GeoGebra to develop conjectures regarding whether a point of concurrency provides the solution for the indicated situation, and which one.
A physical model for the F-T will be indicated. The teacher may demonstrate this model but that requires three strings, three weights, and a base that has holes. A recommended base is a piece of pegboard (perhaps 2 feet by 3 feet), the weights could be fishing weights of about 3 oz., the string could be fishing line; placing flexible pieces of drinking straws in the holes will improve the performance.
The combination of geometry theorems, dynamic geometry software, a variety of contexts, and a physical analog can provide a rich experience for students.
Type: Lesson Plan
What's the Point? Part 1:
This is a patty paper-folding activity where students measure and discover the properties of the point of concurrency of the perpendicular bisectors of the sides of a triangle.
Type: Lesson Plan
## Original Student Tutorials
Pennant Company Challenge: Inscribed Circles of Triangles:
Discover how easy it is for Katie to construct an inscribed circular logo on her company's triangular pennant template. If she completes the task first, she will win a \$1000 bonus! Follow along with this interactive tutorial.
Type: Original Student Tutorial
Good as New:
Learn the steps to circumscribe a circle around a triangle in this interactive tutorial about constructions. Grab a compass, straightedge, pencil and paper to follow along!
Type: Original Student Tutorial
## Problem-Solving Tasks
Placing a Fire Hydrant:
This problem solving task asks students to place a fire hydrant so that it is equal distance from three given points.
Type: Problem-Solving Task
Locating Warehouse:
This problem solving task challenges students to place a warehouse (point) an equal distance from three roads (lines).
Type: Problem-Solving Task
Inscribing a triangle in a circle:
This problem introduces the circumcenter of a triangle and shows how it can be used to inscribe the triangle in a circle.
Type: Problem-Solving Task
Circumcenter of a triangle:
This task shows that the three perpendicular bisectors of the sides of a triangle all meet in a point, using the characterization of the perpendicular bisector of a line segment as the set of points equidistant from the two ends of the segment.
Type: Problem-Solving Task
Inscribing a circle in a triangle II:
This problem solving task focuses on a remarkable fact which comes out of the construction of the inscribed circle in a triangle: the angle bisectors of the three angles of triangle ABC all meet in a point.
Type: Problem-Solving Task
## MFAS Formative Assessments
Circumscribed Circle Construction:
Students are asked to use a compass and straightedge to construct a circumscribed circle of an acute scalene triangle.
Inscribed Circle Construction:
Students are asked to use a compass and straightedge to construct an inscribed circle of an acute scalene triangle.
The Sprinters’ Race:
Students are given a grid with three points (vertices of a right triangle) representing the starting locations of three sprinters in a race and are asked to determine the center of the finish circle, which is equidistant from each sprinter.
## Original Student Tutorials Mathematics - Grades 9-12
Good as New:
Learn the steps to circumscribe a circle around a triangle in this interactive tutorial about constructions. Grab a compass, straightedge, pencil and paper to follow along!
Pennant Company Challenge: Inscribed Circles of Triangles:
Discover how easy it is for Katie to construct an inscribed circular logo on her company's triangular pennant template. If she completes the task first, she will win a \$1000 bonus! Follow along with this interactive tutorial.
## Student Resources
Vetted resources students can use to learn the concepts and skills in this benchmark.
## Original Student Tutorials
Pennant Company Challenge: Inscribed Circles of Triangles:
Discover how easy it is for Katie to construct an inscribed circular logo on her company's triangular pennant template. If she completes the task first, she will win a \$1000 bonus! Follow along with this interactive tutorial.
Type: Original Student Tutorial
Good as New:
Learn the steps to circumscribe a circle around a triangle in this interactive tutorial about constructions. Grab a compass, straightedge, pencil and paper to follow along!
Type: Original Student Tutorial
## Problem-Solving Tasks
Placing a Fire Hydrant:
This problem solving task asks students to place a fire hydrant so that it is equal distance from three given points.
Type: Problem-Solving Task
Locating Warehouse:
This problem solving task challenges students to place a warehouse (point) an equal distance from three roads (lines).
Type: Problem-Solving Task
Inscribing a triangle in a circle:
This problem introduces the circumcenter of a triangle and shows how it can be used to inscribe the triangle in a circle.
Type: Problem-Solving Task
Circumcenter of a triangle:
This task shows that the three perpendicular bisectors of the sides of a triangle all meet in a point, using the characterization of the perpendicular bisector of a line segment as the set of points equidistant from the two ends of the segment.
Type: Problem-Solving Task
Inscribing a circle in a triangle II:
This problem solving task focuses on a remarkable fact which comes out of the construction of the inscribed circle in a triangle: the angle bisectors of the three angles of triangle ABC all meet in a point.
Type: Problem-Solving Task
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this benchmark.
## Problem-Solving Tasks
Placing a Fire Hydrant:
This problem solving task asks students to place a fire hydrant so that it is equal distance from three given points.
Type: Problem-Solving Task
Locating Warehouse:
This problem solving task challenges students to place a warehouse (point) an equal distance from three roads (lines).
Type: Problem-Solving Task
Inscribing a triangle in a circle:
This problem introduces the circumcenter of a triangle and shows how it can be used to inscribe the triangle in a circle.
Type: Problem-Solving Task
Circumcenter of a triangle:
This task shows that the three perpendicular bisectors of the sides of a triangle all meet in a point, using the characterization of the perpendicular bisector of a line segment as the set of points equidistant from the two ends of the segment.
Type: Problem-Solving Task
Inscribing a circle in a triangle II:
This problem solving task focuses on a remarkable fact which comes out of the construction of the inscribed circle in a triangle: the angle bisectors of the three angles of triangle ABC all meet in a point.
Type: Problem-Solving Task
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# Concept of Ratio
A ratio is a mathematical expression of comparing two similar or different quantities by division. It helps one to compare two things of same dimensions or of different dimensions. For example, in a box, there are 9 red balls and 12 black balls. When we have to find the ratio of red balls to black balls, it means how many numbers of red balls are there as compared to black balls. Here, a number of red balls to black balls are written in form 9:12 (i.e. nine is to twelve) which is same as 3:4 (simplified form). It tells how much amount of the first quantity is there as compared to that of second quantity.
Several quantities such as height, weight, length, objects, population, etc can be compared in terms of ratio. Generally, it is expressed using the word “to” i.e. such as red balls to black balls. There are other two notations of expressing it, one by using “:” (9:12) which is called odds notation and another one is fractional notation using “/” (9/12). Another important thing to be followed is the order. When we say the ratio of red balls to black balls, the number representing red ball must come first i.e. 9:12 is not same as 12:9.
It does not compare not only the quantities of similar dimensions but also that of different dimensions. For example, women are to men in a population, boys are to girls in a class etc. are a comparison of same dimensions while speed is to time is a comparison of two different dimensions.
Solve: From the figure below, compare the number of stars to circles in the given rectangle.
Solution: The number stars in the rectangle=6
The number of circles in the rectangle=4
Star is to circle= 6:4 = 3:2
IN DAILY LIFE
Numbers are inevitable and life without numbers is unthinkable. So is the ratio; it has various real-time applications which you find in daily life. The most common example is cooking; the ratio of ingredients has to be maintained for a good dish. Other fields where we find its applications are in business, sports, construction of buildings and so on.
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