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3.9.1 Quadratic Functions, SPM Practice (Long Questions)
3.9.1 Quadratic Functions, SPM Practice (Long Questions)
Question 1:
Without drawing graph or using method of differentiation, find the maximum or minimum value of the function y = 2 + 4x – 3x2. Hence, find the equation of the axis of symmetry of the graph.
Solution:
By completing the square for the function in the form of y = a(x + p)2+ q to find the maximum or minimum value of the function.
y = 2 + 4x – 3x2
y = – 3x2 + 4x + 2 ← (in general form)
$\begin{array}{l}y=-3\left[{x}^{2}-\frac{4}{3}x-\frac{2}{3}\right]\\ y=-3\left[{x}^{2}-\frac{4}{3}x+{\left(-\frac{4}{3}×\frac{1}{2}\right)}^{2}-{\left(-\frac{4}{3}×\frac{1}{2}\right)}^{2}-\frac{2}{3}\right]\\ y=-3\left[{\left(x-\frac{2}{3}\right)}^{2}-{\left(-\frac{2}{3}\right)}^{2}-\frac{2}{3}\right]\end{array}$
Since a = –3 < 0,
Therefore, the function has a maximum value of
$\begin{array}{l}x-\frac{2}{3}=0\\ x=\frac{2}{3}\end{array}$
Equation of the axis of symmetry of the graph is $x=\frac{2}{3}.$
Question 2:
The diagram above shows the graph of a quadratic function y = f(x). The straight line y = –4 is tangen to the curve y = f(x).
(a) Write the equation of the axis of symmetry of the function f(x).
(b) Express f(x) in the form of (x + p)2 + q , where p and q are constant.
(c) Find the range of values of x so that
(i) f(x) < 0, (ii) f(x) ≥ 0.
Solution:
(a)
x-coordinate of the minimum point is the midpoint of (–2, 0) and (6, 0)
= $=\frac{-2+6}{2}=2$
Therefore, equation of the axis of symmetry of the function f(x) is x = 2.
(b)
Substitute x = 2 into x + p = 0,
2 + p = 0
p = –2
and q = –4 (the smallest value of f(x))
Therefore, f(x) = (x + p)2 + q
f(x) = (x – 2)2 – 4
(c)(i) From the graph, for f(x) < 0, range of values of x are –2 < x < 6 ← (below x-axis).
(c)(ii) From the graph, for f(x) ≥ 0, range of values of x are x ≤ –2 atau x ≥ 6 ← (above x-axis). |
## ◂Math Worksheets and Study Guides First Grade. Story Problems
### The resources above correspond to the standards listed below:
#### Massachusetts Curriculum Frameworks
MA.1.OA. Operations and Algebraic Thinking
1.OA.A. Represent and solve problems involving addition and subtraction.
1.OA.A.1. Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations (number sentences) with a symbol for the unknown number to represent the problem.
1.OA.B. Understand and apply properties of operations and the relationship between addition and subtraction.
1.OA.B.4. Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8.
1.OA.C. Add and subtract within 20.
1.OA.C.5. Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.C.6. Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use mental strategies such as counting on; making 10 (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a 10 (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
MA.1.NBT. Number and Operations in Base Ten
1.NBT.C. Use place value understanding and properties of operations to add and subtract.
1.NBT.C.4. Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings, and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten. |
# MNOP is similar to QRST with a scale factor of 5 : 4. MP = 85mm. What is the value of QT?
Answer: MNOP is similar to QRST with ratio of 5:4. That means the MP:QT ratio would also be 5:4
The length of QT would be
MP/QT = 5/4
85mm/QT=5/4
85mm*4= 5*QT
5QT= 340
QT=68mm
## Related Questions
Which of the following is the solution to 4|x+2|>16
x < - 6 or x > 6
Step-by-step explanation:
Inequalities of the type | x | > a, always have solutions in the form
x < - a or x > a
given 4| x + 2 | > 16 ( divide both sides by 4 )
| x + 2 | > 4 ← in the above form, hence
x + 2 < - 4 or x + 2 > 4 ( subtract 2 from both sides of both inequalities )
x < - 6 or x > 6 ← solution
3 similar shirts and 4 similar jackets cost \$360. 1 such shirt and 3 such jackets cost \$220. Find the costof each shirt.
3s + 4j = 360
1s + 3j = 220
This one we can solve using substitution, in that
s + 3j = 220 implies s = 220 - 3j (subtract 3j from each side).
Now substitute into the first equation:
3(220 - 3j) + 4j = 360. Solve for jackets.
660 - 9j + 4j = 360 [Distributive property]
660 - 5j = 360 [Add -9j + 4j]
Subtract 360 from each side, and add 5j to each side.
660 - 360 - 5j + 5j = 360 - 360 + 5j
300 = 5j
Divide each side by 5.
\$60 = price of a jacket.
Three of these plus a shirt runs \$220.
3(60) = 180
180 + s = 220
180 - 180 + s = 220 - 180 [Subtract 180 from each side]
s = 40
Now we double check.
3(40) + 4 (60) = 120 + 240 = 360. This satisfies the first equation.
We could check the second one too, but I'm satisfied.
Which point is an x-intercept Og the quadratic function f(x) = (x-4)(x+2)?
Hi there!
To find the x-intercept of the function we need to calculate f(x) = 0.
Rule AB = 0, gives A = 0 or B = 0.
Add 4 to both sides or subtract 2 from both sides.
Hence, the solutions are:
(4, 0) & (-2, 0)
A circle has a radius of 5 ft, and an arc of length 7 ft is made by the intersection of the circle with a central angle. Which equation gives the measure of the central angle, q?
B q=7/5
Step-by-step explanation:
Well Q=s/r and they said a Radius of 5 Which puts 5 at the bottom.
Then an arc length of 7 Which =S. so q=7/5
To work out the central angle, you just re-arrange the equation for the length of an arc:
Equation for length of an arc:
× × π =
We can arrange this to work out the central angle, q. But first, lets substitute in all of the values that we know:
angle = q
diameter = 5 x 2 = 10 ft
length of arc = 7
[Substitute in]
× π = (Now just rearrange for q)
= (multiply both sides by 360 to get q)
= × (now just simplify)
=
= (rounded to 3 decimal places)
______________________________
Therefore:
The equation that gives you ange q is:
= ×
and q = 80.214 when all of the values are substituted in.
It takes Harrison 1 1/2 hours to finish his history project it takes Miles 3 2/3 hours to finish his science project. How many times as long does it take Miles to finish his science project then it takes Harrison to finish his history project
Step-by-step explanation: Given in the question that
Time taken by Harrison to finish his history project is given by
and the time taken by Kiles to finish his science project is
Let the time taken by Miles to finish his science project be 'p' times that of the time taken by Harrison to finish his history project.
Therefore,
Thus, Miles take times more time than Harrison.
The discount points are 3% which represents \$2,700. The buyer down payment is 20%. What is the purchase price of the property? (Please show all steps)
Answer: The purchase price of the property is \$112,500.
Step-by-step explanation:
Let the original value of property be 'x'.
Discount rate = 3%
Amount of discount = \$2700
According to question, it becomes,
Rate of down payment = 20%
So, Remaining rate of payment = 100-20 = 80%
So, Purchase price of the property would be
Hence, the purchase price of the property is \$112,500.
How do you write 485,830 in word form?
485,830 is written as . . .
four hundred eighty-five thousand, eight hundred thirty
Four friends went on vacation with their family over the summer. Harlin’s family drove 363 miles in 6 hours, Kevin went 435 miles in 7 hours, Shanna drove 500 miles in 8 hours, Hector drove 215 miles in 5 hours. Which family drove the fastest?
Shanna's family
Step-by-step explanation:
d/t = r (distance/time = rate)
Harlin: 363/6 = 60.5 mph
Kevin: 435/7 = 62.14 mph
Shanna: 500/8 = 62.5 mph
Hector: 215/5 = 43 mph
Identify the equation of a line parallel to y=3x-6 and passes through (3,5) in slope intercept form |
# How To Find Increasing And Decreasing Intervals On A Graph Parabola Ideas
How To Find Increasing And Decreasing Intervals On A Graph Parabola Ideas. So if we want to find the intervals where a function increases or decreases, we take its derivative an analyze it to find where it’s positive or negative (which is easier to do!). Decreasing intervals represent the inputs that make the graph fall, or the intervals where the function has a negative slope.decreasing on an interval :divide 75 75 by 3 3.estimate the intervals on which the function is increasing or decreasing and any relative maxima or minima.
How to find increasing and decreasing intervals on a graph parabola. A x 2 + b x + c = a ( x + b 2 a) 2 + c − b 2 4 a. A function is considered increasing on an interval whenever the derivative is positive over that interval.
### Decreasing Intervals Represent The Inputs That Make The Graph Fall, Or The Intervals Where The Function Has A Negative Slope.decreasing On An Interval :Divide 75 75 By 3 3.Estimate The Intervals On Which The Function Is Increasing Or Decreasing And Any Relative Maxima Or Minima.
A x 2 + b x + c = a ( x + b 2 a) 2 + c − b 2 4 a. Given the function $p\left(t\right)$ in the graph below, identify the intervals on which the function appears to be increasing. So if we want to find the intervals where a function increases or decreases, we take its derivative an analyze it to find where it’s positive or negative (which is easier to do!).
### The Intervals Where A Function Is Increasing (Or Decreasing) Correspond To The Intervals Where Its Derivative Is Positive (Or Negative).
How to find increasing and decreasing intervals on a graph parabola. Finding increasing and decreasing intervals on a graph. A function is considered increasing on an interval whenever the derivative is positive over that interval.
### How To Find Increasing And Decreasing Intervals On A Graph Calculus.
To find increasing and decreasing intervals, we need to find where our first derivative is greater than or less than zero. To find intervals on which $$f$$ is increasing and decreasing:we can say this because its only a parabola.well, first off, under german, the interval for which the function is increasing so as we can see from the graph deck beyond point x is. |
Tutorial on the nth Term of A Geometric Progression (G.P.) | The Simplest Method
Some students run away from Geometric Progression problems, simply because they find it difficult to differentiate it from Arithmetic Progression.
They believe the common difference is the same thing as the common ratio so they get tired when solving and they make a lot of mistakes.
In this lesson, I will show you the difference between common difference used in A.P and common ratio used in G.P. and as well give you a breakdown of solution to problems in the nth term of a G.P.
At the end of the lesson, you will come to discover how simple problems involving the nth term of a G.P. are.
A Geometric Progression also known as exponential Sequence) is a Sequence in which any two consecutive terms differ by a constant factor.
The common ration is densted by ‘r’ and the first term ‘a’. but in A.P common difference is used if a set of numbers 6,18,54,162…are given
The common ration = 18/6 = 3, 54/18 = 3, 162/54.
While the common difference is subtraction i.e 18-6, 54-18….
Common ratio is division, since the common ratio ‘r’ is known and the first term ‘a’ is also known, the process of finding the second term is by multiplying the first term by the common ratio.
That is a, ar, ar2, ar3…. arn-1
Therefore, an nth term of a G.P. is given as
Tn = arn-1
N/B: Tn = nth term, r=common ration, a = first term
Put this formula down and also save it in your brain. You will really need it.
Having known some important things about nth term of a G.P. like ho to obtain the common ratio and the formula for nth term of a G.P. let me eliminate your fear of G.P problems by solving some examples.
Hope we are good to go!!
Solutions to Geometric Progression Questions
Example 1: the 6th term of a G.P is 1215. Given that the common ration is 3, find the 9th term.
Solution
In this problem common ratio is r=3, n=6
The 6th term = 1215
Using the formular for the nth term of G.P Tn = arn-1 we substitute what is given in order to obtain the first term ‘a’
T6 = ax36-1 = 1215
ax35 = 1215
rem: 35 = 3x3x3x3x3
Example 2: The third term of a GP is 9 and the fifth term is 16. Find the 4th term.
Solution.
In the above problem given, the third term that is T3 = 9 and T5 = 16
We are required to find two equations and then solve them simultaneously.
Using the formular Tn = arn-1
T9 = ar3-1 = 9
ar2 = 9 — (1)
T5 = ar5-1 = 16
ar4 = 16 — (2)
Divide equation two by 1
ar4/ar2 = 16/9
Examples 3: if 3,A,B, 192 are consecutive terms of a GP find the value of A and B.
Solution
In the problem given,
The first term a=3
The second term is A
The third term is B
The fourth term is 192
That means
a=3, T4 = 192
using the formular for nth term of G.P
Tn = arn-1
T4 = 3xr4-1 = 192
Multiply 3 by r3
3r3 = 192
Divide both sides by 3
3r3/3 = 192/3
r3 = 64
Take cube rook of both sides
N/B: Cube rook can be express as 1/3 (2) Cube root of a number means a number that can multiply itself three times to give that number e.g 8, the cube rook of 8 is 2 means 2x2x2 = 8 (3) Cube root of 64 = 4
How? 4x4x4 = 64
r3×1/3 = 4
r = 4
Since we both have gotten a and 8, let solve A and B
Example 4: the first and third term of a GP are 5 and 80, respectively, what is the 4th term?
Solution
In the problem above
The first term a =5
The third term T3 = 80
Using the formula
Tn = arn-1
T3 = ar3-1 = ar2 = 80 —– (1)
Put a=5 into equation (1)
5xr2 = 80
Multiply 5 by r2
5r2 = 80
Divide both sides by 5
The 4th term, n=4
T4 = ar4-1 = ar3
Put a=5 and r=4
T4 = 5×43 = 5x4x4x4
T4 = 5×64
T4 = 320
Recall: Second Term is A, that is T2 = A
T2 = ar2-1 = A
A = ar
Substitute a=3 and r=4
A = 3×4 = 12
Therefore A=12
Again, third Term is B
T3 = ar3-1 = B
B=ar2
B= 3x(4) = 3x4x4 = 3×16
B = 48
Therefore A=12 and B=48
Example 5:
A G.P has its 3rd and 7th term as 45 and 3645 respectively what is the product of its 2nd and 4th term?
Solution
Form the problem
The 3rd term T3 = 45
The 7th term T7 = 3645
Using the formular
Tn = arn-1
T3 = ar3-1 = 45
ar2 = 45 —– (1)
T7 = ar7-1 = 3645
ar6 = 3645 —– (2)
divide equation 2 by 1
ar6/ar2 = 3645/45
rxrxrxrxrxr/rxr = 81 How?
Its because rxrxrxr = r4
r4 = 81
substitute r = 3 into equation 1
ar2= 45
a x (3)2 = 45
ax3x3 = 45,axa = 45
9a = 45
Divide both sides by 9
9a/9a = 45/9 a =5
Therefore a = 5 and r =3
And term T2
T2 = ar2-1 = ar
Substitute a = 5 and r=3
T2 = 5×3 = 15
T4 = ar3-1 = ar2
T4 = 5×32 = 3×3 = 5 x 9
= 45
Product means multiplication so the product of the 2nd and 4th terms is T2xT4 = 15×45 = 673
No doubt you have learnt a lot. What you have learned today point to the fact that you can solve as many as possible problems involving nth term of G.P. and that problems involving nth term of G.P are not difficult and should not scare you. I believe the fear has been eliminated.
See:
Remember to take note of the formulas for today’s lesson. |
Select Page
# Year 2
### We can write numbers in order and position them on a number line.We can use the ‘greater than’ and ‘less than’ signs.
#### What we are learning:
• Ordering numbers is about understanding place value – the value of each digit in a number, the greater than and less than signs tell us which number is bigger and which is smaller in a pair.
• Your child needs lots of experience in putting numbers in order.
• Encourage your child to explain their thinking when they order numbers – the strategies they use. Make sure you both use the correct mathematical vocabulary.
• When positioning numbers on an empty number line you are looking for appropriate size gaps between the numbers which will show you that your child has an understanding of the numbers that are between each of the numbers in the set.
#### Activities you can do at home:
Write a set of numbers onto post-it notes 67, 92, 51, 62, 21, 15, 73, 29, 74. Order these from largest to smallest or from smallest to largest.
Which number is the largest? How do you know? Which number is the smallest? How do you know? Order the numbers from largest to smallest or from smallest to largest. Lay this out horizontally and vertically.
Ask your child to close their eyes and muddle up two of the numbers. Can you spot the mistake/muddle? Make it fun. You do this too –Ask your child to muddle up two of the numbers for you to identify and correct.
Ask your child to position the numbers as if they were on a number line (it is a good idea to do this on the floor) leaving the relative size of gap between the numbers – so the gap between the 15 and the 21 would be smaller (the size of 5 post it notes) than the gap between 73 and 92 (the size of 19 post it notes). Put a post-it note between two of the numbers and ask, What number might be here? Why do you think that? Put a postit note next to one of the numbers and ask, What number would be here?
Why do you think that?
Draw an empty number line and position the numbers on the line.
Draw a number line marked in tens and position the numbers on the line
Move on to using a set of mixed two-digit and three-digit numbers on
cards or post it notes
Take pairs of numbers and set them out side by side. Ask your child to draw the ‘greater than’ and ‘less than’ symbols onto a post it and then insert the correct one between the numbers.
Can you put these numbers in order from the smallest to the largest?
Which number is the smallest / largest?
Which number fits in this gap? |
## Basic Analysis: Sequence Convergence (4)
In this article, we’ll consider the convergence of an infinite sum: $a_1 + a_2 + a_3 + \ldots$. We call this sum an infinite series. Let $s_n = a_1 + a_2 + \ldots + a_n$ be the partial sums of the series.
Definition. We say that $\sum_{n=1}^\infty a_i$ is L (resp. ∞, -∞) if the partial sums $(s_n)$ converge to L (resp. ∞, -∞). The series is said to be convergent if the sum is a real number L.
Example 1.
The series: $\frac 1{1\times 2} + \frac 1{2\times 3} + \frac 1{3\times 4} + \ldots$ converges to 1 since the partial sums
$\sum_{i=1}^n \frac 1{i(i+1)} = \sum_{i=1}^n \left(\frac 1 i-\frac 1{i+1}\right) = \left(1 - \frac 1 2\right) + \left(\frac 1 2 - \frac 1 3\right) + \ldots +\left(\frac 1 n-\frac 1 {n+1}\right)=1 - \frac 1{n+1}$
form a telescoping series. As n→∞, the RHS clearly converges to 1.
Example 2.
The series: $\frac 1 {1^2} + \frac 1 {2^2} + \frac 1{3^2} + \ldots$ converges. Indeed, we have $0 < \frac 1 {n^2} < \frac 1 {n(n-1)}$, so upon dropping the first term (1), we can apply squeeze theorem on the partial sums to obtain:
$1 < \frac 1 {1^2} + \frac 1 {2^2} + \frac 1 {3^2} + \ldots < 1 + \frac 1 {1\times 2} + \frac 1 {2\times 3} + \ldots = 2.$
Computing the exact value is rather difficult and we won’t go into that now.
Example 3.
Consider the series $\sum_{n=1}^\infty \frac 1{4n^2-1}$. We can write the sum in two different ways:
\begin{aligned} \sum_n \frac 1{4n^2-1} &= \sum_n \left(\frac n{2n+1} -\frac {n-1}{2n-1}\right) =\frac 1 3 +\left(\frac 2 5 - \frac 1 3\right)+\left(\frac 3 7-\frac 2 5\right) + \ldots\\ &=\sum_n \left(\frac {1/2} {2n-1}-\frac{1/2}{2n+1}\right)= \left(\frac 1 2-\frac 1 6\right)+\left(\frac 1 6-\frac 1 {10}\right) + \ldots \end{aligned}
All the terms in the first sum appear to cancel each other out, leaving a sum of zero, which is ridiculous since all the terms in the series are positive. However, a moment of thought reveals that the partial sums are really $\frac n{2n+1}$ which approaches 1/2. Hence, the series converges to 1/2, which is also obvious from the second sum.
Proposition. If the series $\sum a_n$ is convergent, then $(a_n) \to 0$.
Proof.
Let ε>0. Since the partial sums converge to a real number, it is a Cauchy sequence so there exists N such that $|s_m - s_n| < \epsilon$ for all mnN. In particular, $|a_n| = |s_n-s_{n-1}| < \epsilon$ for all nN+1. Hence, $(a_n) \to 0$. ♦
Example 4.
The converse is not true: there’re many examples of sequences $(a_n)$ which converge to 0, but whose sums don’t converge. The most famous example is that of the harmonic series $1 + \frac 1 2 + \frac 1 3 + \ldots$, where the sum of the first n terms is approximately log(n).
## Basic Properties
Properties. Suppose $\sum_n a_n = L_1$ and $\sum_n b_n = L_2$. Then:
• $\sum_n (a_n + b_n) = L_1 + L_2$;
• $\sum_n (ca_n) = cL_1$ for any real c.
Since the partial sums of (an + bn) is the sum of the partial sum of an and that of bn, the first property follows easily. Similarly, the second property follows from the fact that the partial sums of (c·an) are just (partial sums of an).
On the other hand, the sum of products ∑(anbn) is quite a pain. One way of handling it is to use Abel transformation: write $B_n = b_1 + b_2 + \ldots b_n$ for the partial sums of bn, then:
\begin{aligned}\sum_{n=1}^m a_n b_n &= \sum_{n=1}^m a_n(B_n - B_{n-1}) = \sum_{n=1}^m a_n B_n - \sum_{n=0}^{m-1} a_{n+1} B_n= \sum_{n=1}^m B_n(a_n - a_{n+1})+ a_m B_m.\end{aligned}
Example 5.
Prove that if the partial sums $s_n = a_1 + a_2 + \ldots + a_n$ are bounded, then $\frac{a_1}1 + \frac{a_2} 2 + \frac{a_3} 3 + \ldots$ converges.
Proof.
Let bn = 1/n. By Abel transformation with a‘s and b‘s swapped, we get:
\begin{aligned}\sum_{n=1}^m \frac{a_n} n = \sum_{n=1}^m s_n(b_n - b_{n+1}) + b_m s_m = -\sum_{n=1}^m \frac{s_n}{n(n+1)} + \frac{s_m}m.\end{aligned}
Since $(s_n)$ is bounded, the second term $\frac{s_m}m \to 0$. Hence it suffices to show that $\sum_n \frac {s_n}{n(n+1)}$ converges. But since |sn|≤L are bounded, we have:
\begin{aligned}\left|\sum_{n=j}^k \frac{s_n}{n(n+1)}\right| \le \sum_{n=j}^k \frac{|s_n|}{n(n+1)} \le L\sum_{n=j}^k \frac 1{n(n+1)}.\end{aligned}
Since $\sum_n \frac 1 {n(n+1)}$ converges, its partial sums form a Cauchy sequence. This in turn implies that the partial sums of $\sum_n \frac{s_n}{n(n+1)}$ form a Cauchy sequence, and we’re done. ♦
Theorem (Alternating Series Test). If $(a_n)$ is a decreasing sequence of positive values converging to 0, then the series $\sum_n (-1)^n a_n$ is convergent.
Proof.
Let bn = (-1)n. Apply the Abel transformation:
\begin{aligned}\sum_{n=1}^m a_n b_n =\sum_{n=1}^m B_n(a_n - a_{n+1})+ a_m B_m,\end{aligned}
where $B_n = b_1 + \ldots + b_n$ and thus |Bn| ≤ 2. Since $a_m \to 0$, we also have $a_m B_m \to 0$ so it suffices to show that $\sum_n B_n(a_n-a_{n+1})$ converges. Now,
$|\sum_{n=j}^k B_n(a_n - a_{n+1})| \overbrace{\le\sum_{n=j}^k |B_n|(a_n-a_{n+1})}^{\because (a_n) \text{ dec.}} \le 2(a_j-a_{k+1}).$
Since $(a_n)$ is convergent, it is also Cauchy, so the partial sums of $\sum_n B_n(a_n-a_{n+1})$ form a Cauchy sequence, and the series is convergent. ♦
Example 6.
Consider the infinite series $1 -\frac 1 2 + \frac 1 3 - \frac 1 4 + \ldots$. By the alternating series test, the series converges, though the test doesn’t give any hint what the sum might be.
## Permutation of a Series
If $\sum_{n\ge 1} a_n$ is a series, then we can permute the terms to obtain $\sum_{n\ge 1} a_{\pi(n)}$, where π:N → N is a bijective function. For example:
• we can switch the first two terms to obtain $a_2 + a_1 + a_3 + a_4 + \ldots$, i.e. π(1) = 2, π(2) = 1, and π(m) = m for all m > 2;
• we can switch each 2k-1 with 2k to obtain $a_2 + a_1 + a_4 + a_3 + a_6 + a_5 + \ldots$, i.e. π(2k-1) = 2k, π(2k) = 2k-1 for k=1, 2, 3, … .
It may surprise the reader that the sum of an infinite series may depend on the order of summation, despite the fact that addition is commutative.
Example 7.
We already saw that $S =1 - \frac 1 2 + \frac 1 3 - \ldots$ is convergent. On the one hand, we can write:
$S = 1 - \left(\frac 1 2 - \frac 1 3\right) - \left(\frac 1 4 - \frac 1 5\right) -\ldots$
where each bracketed term is positive, so S < 1. On the other hand, let’s write S = 1 + sum of terms of the form:
$a_k = \frac 1 {4k-1} + \frac 1 {4k+1} -\frac 1 {2k} = \frac {8k}{16k^2-1} - \frac 1{2k} >0$k = 1, 2, 3, … .
This would imply that S > 1, so the sum of an infinite series can be changed by permuting the terms.
For a series to be better-behaved, we need the following concept:
Definition. The series $\sum_n a_n$ is said to converge absolutely if the series $\sum_n |a_n|$ converges. If $\sum a_n$ converges but not absolutely, we say it is conditionally convergent.
For example, the alternating series $1 - \frac 1 2 + \frac 1 3 - \ldots$ we saw above is only conditionally convergent. On the other hand, the alternating series $1 - \frac 1{2^2} + \frac 1 {3^2}-\ldots$ is absolutely convergent.
The key property we want to state is:
Theorem. If $\sum a_n$ converges absolutely, then every permutation of the series converges, and to the same sum.
Proof.
Let ε>0. Since the partial sums Σ|an| form a Cauchy sequence, for some N,
$|a_m| + |a_{m+1}| + \ldots + |a_n|< \epsilon/2$ for all n > mN.
This immediately implies that the partial sums Σan form a Cauchy sequence and converges to some L. Now for any permutation π of {1, 2, 3, … }, we need to show that $\sum (a_{\pi(n)} - a_n) = 0$. To do that, let:
$M = \max\{\pi^{-1}(1), \pi^{-1}(2), \ldots, \pi^{-1}(N)\}.$
Basically, the crux is that when nM, π(n) > N. So when n > mM, we have
\begin{aligned}|\sum_{i=m}^n a_{\pi(i)} - a_i| \le \sum_{i=m}^n |a_{\pi(i)} - a_i| \le \sum_{i=m}^n (|a_{\pi(i)}| + |a_i|) <\frac\epsilon 2 + \frac\epsilon 2.\end{aligned}
So $\sum (a_{\pi(n)} - a_n) \to 0$ as required. ♦
Hence, absolute convergence of a series ensures well behaviour. This fact carries on over to the “continuous case” where we can exchange integrating variables dx dy and dy dx if the absolute value of the integrand integrates to a finite real value. But that’s another story for another day.
Exercise.
Prove that if $\sum_n a_n$ is conditionally convergent, then for every real L, there is a permutation π of {1, 2, 3, … } such that $\sum_n a_{\pi(n)} = L$.
[ Hint (highlight to read): let bn and cn be the positive and negative terms in the series (an) respectively. Show that Σbn = ∞ and Σcn = -∞. Now given L, pick just enough terms from bn so that the sum exceeds L. Then pick just enough terms from cn so that the sum drops below L. And so on. Show that bn → 0 and cn → 0 which implies the process eventually tends to L. ]
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Mathematical Proof/Relations
Introduction
Intuitively, relation associates pairs of objects based on some rules and properties. That is, relation suggests some kinds of relationship or connection between two objects. Take marriage as an example. In the marriage registry, there is a record in which the names of husbands are associated with the names of their corresponding wives, to keep track of the marriages. The entries in the record can be interpreted as ordered pairs ${\displaystyle (a,b)}$, where ${\displaystyle a}$ is the husband, while ${\displaystyle b}$ is the wife.
Expressing it mathematically, let ${\displaystyle M}$ and ${\displaystyle W}$ be the set of all men and women respectively. Then, the Cartesian product ${\displaystyle M\times W=\{(a,b):a\in M{\text{ and }}b\in W\}}$ consists of all pairs of people (first and second coordinate is a man and woman respectively). After that, we know that the record in the marriage registry, ${\displaystyle R}$, is a subset of ${\displaystyle M\times W}$. If a man and a woman form an ordered pair in ${\displaystyle R}$, say ${\displaystyle (m,w)}$, then it means they are married. Then, it is natural to say that ${\displaystyle m}$ is related to ${\displaystyle w}$. In other words, if we find that ${\displaystyle (m,w)\in R}$, then it means that they are related by this relation ${\displaystyle R}$. Also, knowing what ${\displaystyle R}$ exactly is (we have the record from the marriage registry) is the same as knowing all the husband and wife relationship.
Thus, it is natural to define the set ${\displaystyle R}$ as a relation. Let us formally define relation below.
Terminologies
Definition. (Relation) A relation ${\displaystyle R}$ from a set ${\displaystyle A}$ to a set ${\displaystyle B}$ is a subset of ${\displaystyle A\times B}$, i.e., ${\displaystyle R\subseteq A\times B}$. In particular, when ${\displaystyle A=B}$, we say that ${\displaystyle R}$ is a relation on ${\displaystyle A}$, and we have ${\displaystyle R\subseteq A\times A}$. If ${\displaystyle (a,b)\in R}$, then we say that ${\displaystyle a}$ is related to ${\displaystyle b}$ by ${\displaystyle R}$, and write ${\displaystyle aRb}$. On the other hand, if ${\displaystyle (a,b)\notin R}$, then we say that ${\displaystyle a}$ is not related to ${\displaystyle b}$ by ${\displaystyle R}$, and write ${\displaystyle a\not Rb}$.
Example. Let ${\displaystyle A=\{a,b,c\}}$, ${\displaystyle B=\{1,2,3,4\}}$, and ${\displaystyle R=\{(a,1),(b,1),(b,2),(c,3)\}}$. Since the set ${\displaystyle R}$ is a subset of ${\displaystyle A\times B}$, ${\displaystyle R}$ defines a relation from set ${\displaystyle A}$ to set ${\displaystyle B}$. Also, in this relation, ${\displaystyle a}$ is related to 1, ${\displaystyle b}$ is related to 1 and 2, ${\displaystyle c}$ is related to 3. Thus, we have ${\displaystyle aR1,bR1,bR2}$ and ${\displaystyle cR3}$. But, we have, say ${\displaystyle a\not R2}$ or ${\displaystyle d\not R4}$, since ${\displaystyle (a,2),(d,4)\notin R}$.
Exercise.
(a) Suggest a set ${\displaystyle S}$ that is not a relation from set ${\displaystyle A}$ to set ${\displaystyle B}$.
(b) Suggest a set ${\displaystyle T}$ that is a relation from set ${\displaystyle A}$ to set ${\displaystyle R}$.
Solution
(a) ${\displaystyle S=\{(a,5)\}}$ (${\displaystyle (a,5)\notin A\times B}$, so ${\displaystyle S}$ is not a subset of ${\displaystyle A\times B}$).
(b) ${\displaystyle T=\left\{{\big (}a,(a,1){\big )}\right\}}$.
Exercise. Let ${\displaystyle A}$ and ${\displaystyle B}$ be two nonempty sets. Are ${\displaystyle \varnothing }$ and ${\displaystyle A\times B}$ relations from ${\displaystyle A}$ to ${\displaystyle B}$? Describe the relationships meant by these two sets.
Solution
Both ${\displaystyle \varnothing }$ and ${\displaystyle A\times B}$ are relations from ${\displaystyle A}$ to ${\displaystyle B}$ since both are subsets of ${\displaystyle A\times B}$.
For ${\displaystyle \varnothing }$, it means nothing is related, i.e., there is no relationship between elements in ${\displaystyle A}$ and elements in ${\displaystyle B}$.
For ${\displaystyle A\times B}$, it means everything is related, i.e., every element in ${\displaystyle A}$ is related to every element in ${\displaystyle B}$.
Example. Let ${\displaystyle H}$ be a set of all human beings. Then, ${\displaystyle R=\{(a,b):a{\text{ is the son of }}b\}}$. Then, ${\displaystyle R}$ is a subset of ${\displaystyle H\times H}$, and defines the relation for son and (father or mother).
Example. The concept of "less than" defines a relation on ${\displaystyle \mathbb {R} }$. To be more precise, the relation corresponding to this concept is ${\displaystyle R=\{(a,b):a. For instance, ${\displaystyle 1R2}$ but ${\displaystyle 2\not R1}$.
Example. The congruence of integers defines a relation on ${\displaystyle \mathbb {Z} }$. To be more precise, the corresponding relation is ${\displaystyle R=\{(a,b):a\equiv b{\pmod {n}}\}\subseteq \mathbb {Z} \times \mathbb {Z} }$ (${\displaystyle n\in \mathbb {N} }$ with ${\displaystyle n\geq 2}$). For example, when ${\displaystyle n=3}$, then ${\displaystyle 2R5}$ but ${\displaystyle 1\not R3}$.
Exercise. Consider the relation on ${\displaystyle \mathbb {R} }$, corresponding the concept of "equal to": ${\displaystyle R=\{(x,y):x=y\}\subseteq \mathbb {R} ^{2}}$.
(a) Sketch ${\displaystyle R}$ in the Cartesian coordinate system.
(b) Consider also the relations ${\displaystyle S,T}$ on ${\displaystyle \mathbb {R} }$ corresponding the concept of "less than" and "greater than" respectively. Sketch also ${\displaystyle S}$ and ${\displaystyle T}$ in the graph in (a).
Solution
(a) and (b):
y
| y=x
########|#####/%%%
########|####/%%%%
########|###/%%%%%
########|##/%%%%%%
########|#/%%%%%%%
########|/%%%%%%%%
--------*--------- x
#######/|%%%%%%%%%
######/%|%%%%%%%%%
#####/%%|%%%%%%%%%
####/%%%|%%%%%%%%%
###/%%%%|%%%%%%%%%
*----*
|####|
|####| : y>x (relation T)
*----*
*----*
|%%%%|
|%%%%| : y<x (relation S)
*----*
Remark.
• The sets ${\displaystyle R,S,T}$ give a partition of ${\displaystyle \mathbb {R} }$. As we will see, there is a close relationship between the concept of relation and the concept of partition.
Example. Let ${\displaystyle A=\{1,2\}}$. Define a relation on ${\displaystyle {\mathcal {P}}(A)}$, ${\displaystyle R=\{(S,T)\in {\mathcal {P}}(A)\times {\mathcal {P}}(A):S\subseteq T\}}$. Express ${\displaystyle R}$ by listing method.
Solution. First, we have ${\displaystyle {\mathcal {P}}(A)=\{\varnothing ,\{1\},\{2\},\{1,2\}\}}$. So,
${\displaystyle R=\{(\varnothing ,\varnothing ),(\varnothing ,\{1\}),(\varnothing ,\{2\}),(\varnothing ,\{1,2\}),(\{1\},\{1\}),(\{1\},\{1,2\}),(\{2\},\{2\}),(\{2\},\{1,2\}),(\{1,2\},\{1,2\})\}.}$
Exercise. Let ${\displaystyle A=\{1,2\}}$ and ${\displaystyle B=\{3,4\}}$. Consider a relation ${\displaystyle R}$ on ${\displaystyle A\times B}$ defined by
${\displaystyle (a,b)R(c,d){\text{ if }}a+d=b+c.}$
Express ${\displaystyle R}$ by listing method.
Solution
First, ${\displaystyle A\times B=\{(1,3),(1,4),(2,3),(2,4)\}}$. So,
${\displaystyle R=\{{\big (}(1,3),(1,3){\big )}{\big (}(1,3),(2,4){\big )},{\big (}(1,4),(1,4){\big )},{\big (}(2,3),(2,3){\big )},{\big (}(2,4),(1,3){\big )},{\big (}(2,4),(2,4){\big )}\}.}$
Definition. (Domain and range) Let ${\displaystyle R}$ be a relation from a set ${\displaystyle A}$ to a set ${\displaystyle B}$. The domain of ${\displaystyle R}$, denoted by ${\displaystyle \operatorname {dom} R}$ is the subset of ${\displaystyle A}$ defined by
${\displaystyle \operatorname {dom} R=\{a\in A:(a,b)\in R{\text{ for some }}b\in B\}.}$
The range of ${\displaystyle R}$, denoted by ${\displaystyle \operatorname {ran} R}$, is the subset of ${\displaystyle B}$ defined by
${\displaystyle \operatorname {ran} R=\{b\in B:(a,b)\in R{\text{ for some }}a\in A\}.}$
Remark.
• You may have learnt about the domain and range for functions. Using the same terminologies for both relations and functions is indeed not a coincidence. This is because a function is actually a relation. That is, a function is indeed just a special case of relation.
Example. Let ${\displaystyle A=\{a,b,c\}}$, ${\displaystyle B=\{1,2,3,4\}}$, and ${\displaystyle R=\{(a,1),(b,1),(b,2),(c,3)\}}$. Then, ${\displaystyle \operatorname {dom} R=\{a,b,c\}=A}$ and ${\displaystyle \operatorname {ran} R=\{1,2,3\}}$.
Example. Consider the relation ${\displaystyle R}$ on ${\displaystyle \mathbb {Z} }$: ${\displaystyle R=\{(a,b):a{\text{ and }}b{\text{ are both even}}\}}$. Then, ${\displaystyle \operatorname {dom} R}$ is the set of all even numbers, and ${\displaystyle \operatorname {ran} R}$ is the set of all even numbers also.
Exercise. Consider the relation ${\displaystyle R}$ on ${\displaystyle \mathbb {Z} }$: ${\displaystyle R=\{(a,b):a{\text{ and }}b{\text{ are of different parity}}\}}$. Find ${\displaystyle \operatorname {dom} R}$ and ${\displaystyle \operatorname {ran} R}$.
Solution
${\displaystyle \operatorname {dom} R=\mathbb {Z} }$ and ${\displaystyle \operatorname {ran} R=\mathbb {Z} }$.
Definition. (Inverse relation) Let ${\displaystyle R}$ be a relation from a set ${\displaystyle A}$ to a set ${\displaystyle B}$. The inverse relation ${\displaystyle R^{-1}}$ from ${\displaystyle B}$ to ${\displaystyle A}$ is
${\displaystyle R^{-1}=\{(b,a):(a,b)\in R\}\subseteq B\times A.}$
Remark.
• Again, you may have learnt a similar concept (and also a similar notation) for functions: inverse functions. Indeed, inverse functions can be defined as the inverse relation of a function, provided that the inverse relation is also a function.
• We can see that the inverse relation is obtained by "reversing" the order of elements in every ordered pair in the original relation.
• When ${\displaystyle R}$ is empty, then the inverse relation ${\displaystyle R^{-1}}$ is also empty since there is no ordered pair in the empty set.
Example. Let ${\displaystyle A=\{a,b,c\}}$, ${\displaystyle B=\{1,2,3,4\}}$, and ${\displaystyle R=\{(a,1),(b,1),(b,2),(c,3)\}}$. Then, the inverse relation ${\displaystyle R^{-1}}$ is ${\displaystyle \{(1,a),(1,b),(2,b),(3,c)\}}$.
Exercise. Construct an example of sets ${\displaystyle A,B}$ and a nonempty relation ${\displaystyle R}$ from set ${\displaystyle A}$ to set ${\displaystyle B}$ such that ${\displaystyle R=R^{-1}}$.
Solution
Take ${\displaystyle A=B=\{1\}}$, and ${\displaystyle R=\{(1,1)\}}$. Then, ${\displaystyle R^{-1}=\{(1,1)\}=R}$.
Reflexive, symmetric and transitive relations
After introducing the terminologies related to relations, we will study three properties for a relation defined on a set.
Definition. (Reflexive, symmetric and transitive) Let ${\displaystyle A}$ be a set and ${\displaystyle R}$ be a relation defined on ${\displaystyle A}$. Then,
• ${\displaystyle R}$ is reflexive if ${\displaystyle xRx}$ for every ${\displaystyle x\in A}$.
• ${\displaystyle R}$ is symmetric if for every ${\displaystyle x,y\in A}$, ${\displaystyle xRy\implies yRx}$.
• ${\displaystyle R}$ is transitive if for every ${\displaystyle x,y,z\in A}$, ${\displaystyle (xR{\color {darkgreen}y}{\text{ and }}{\color {darkgreen}y}Rz)\implies xRz}$.
Exercise. Let ${\displaystyle A}$ be a set and ${\displaystyle R}$ be a relation defined on ${\displaystyle A}$. Write down the meaning of (i) ${\displaystyle R}$ is not reflexive; (ii) ${\displaystyle R}$ is not symmetric; (iii) ${\displaystyle R}$ is not transitive.
Solution
(i) There exists ${\displaystyle x\in A}$ such that ${\displaystyle x\not Rx}$.
(ii) There exist ${\displaystyle x,y\in A}$ such that ${\displaystyle xRy}$ but ${\displaystyle y\not Rx}$.
(iii) There exist ${\displaystyle x,y,z\in A}$ such that ${\displaystyle xRy}$ and ${\displaystyle yRz}$, but ${\displaystyle x\not Rz}$.
Example. Let ${\displaystyle A=\{1,2\}}$, and let a relation defined on ${\displaystyle A}$ be
${\displaystyle R=\{(1,1),(2,2),(1,2)\}.}$
Then, ${\displaystyle R}$ is reflexive since ${\displaystyle (1,1),(2,2)\in R}$. But ${\displaystyle R}$ is not symmetric since ${\displaystyle (1,2)\in R}$ but ${\displaystyle (2,1)\notin R}$.
Exercise. Is ${\displaystyle R}$ transitive? Explain why.
Solution
${\displaystyle R}$ is transitive since
• ${\displaystyle 1R{\color {darkgreen}1}{\text{ and }}{\color {darkgreen}1}R2\implies 1R2}$;
• ${\displaystyle 1R{\color {darkgreen}1}{\text{ and }}{\color {darkgreen}1}R2\implies 1R2}$.
(There are no more ordered pairs ${\displaystyle (x,{\color {darkgreen}y})}$ and ${\displaystyle ({\color {darkgreen}y},z)}$ in the relation ${\displaystyle R}$, satisfying ${\displaystyle (x,y)\in R}$ and ${\displaystyle (y,z)\in R}$.)
Example. The congruence of integers
${\displaystyle R=\{(a,b):a\equiv b{\pmod {n}}\}}$
defines a relation on ${\displaystyle \mathbb {Z} }$ (${\displaystyle n\in \mathbb {N} }$ with ${\displaystyle n\geq 2}$). Prove that ${\displaystyle R}$ is a reflexive, symmetric and transitive.
Proof.
Reflexive: For every ${\displaystyle x\in \mathbb {Z} }$, ${\displaystyle x\equiv x{\pmod {n}}}$. So, ${\displaystyle xRx}$ for every ${\displaystyle x\in \mathbb {Z} }$.
Symmetric: For every ${\displaystyle x,y\in \mathbb {Z} }$, ${\displaystyle xRy\implies x\equiv y{\pmod {n}}\implies x-y=kn\implies y-x=(-k)n\implies y\equiv x{\pmod {n}}\implies yRx}$ where ${\displaystyle k}$ is some integer.
Transitive: For every ${\displaystyle x,y,z\in \mathbb {Z} }$,
{\displaystyle {\begin{aligned}xRy{\text{ and }}yRz&\implies x-y=kn{\text{ and }}y-z=k'n{\text{ for some }}k,k'\in \mathbb {Z} \\&\implies x-z=(x-y)+(y-z)=(k+k')n\\&\implies xRz.&(k+k'\in \mathbb {Z} )\end{aligned}}}
${\displaystyle \Box }$
Exercise. Which of the properties, reflexive, symmetric and transitive, does each of the following relations possesses? Prove your answer.
(a) ${\displaystyle R=\{(x,y)\in \mathbb {R} ^{2}:x\leq y\}}$.
(b) ${\displaystyle R=\{(x,y)\in \mathbb {R} ^{2}:x.
(c) ${\displaystyle R=\{(x,y)\in \mathbb {Z} \times \mathbb {Z} :xy>0\}}$.
Solution
(a) ${\displaystyle R}$ is reflexive, not symmetric, and transitive.
Proof.
Reflexive: For every ${\displaystyle x\in \mathbb {R} }$, ${\displaystyle x\leq x}$. So, ${\displaystyle xRx}$.
Not symmetric: Take ${\displaystyle x=1}$ and ${\displaystyle y=2}$. Then, ${\displaystyle 1\leq 2}$, so ${\displaystyle xRy}$. However, ${\displaystyle 2>1}$. So, ${\displaystyle y\not Rx}$.
Transitive: For every ${\displaystyle x,y,z\in \mathbb {R} }$, ${\displaystyle xRy{\text{ and }}yRz\implies x\leq y{\text{ and }}y\leq z{\text{ and }}x\leq z\implies xRz}$ (this actually follows from the property of "${\displaystyle \leq }$").
${\displaystyle \Box }$
(b) ${\displaystyle R}$ is not reflexive, not symmetric, and transitive.
Proof.
Reflexive: Take ${\displaystyle x=1}$. Then, 1 is not less than 1 itself. Hence, ${\displaystyle 1\not R1}$.
Not symmetric: Take ${\displaystyle x=1}$ and ${\displaystyle y=2}$. Then, ${\displaystyle 1<2}$, so ${\displaystyle xRy}$. However, ${\displaystyle 2>1}$. So, ${\displaystyle y\not Rx}$.
Transitive: For every ${\displaystyle x,y,z\in \mathbb {R} }$, ${\displaystyle xRy{\text{ and }}yRz\implies x (this actually follows from the property of "${\displaystyle <}$").
${\displaystyle \Box }$
(c) ${\displaystyle R}$ is reflexive, symmetric and not transitive.
Proof.
Reflexive: For every ${\displaystyle x\in \mathbb {Z} }$,
Case 1: ${\displaystyle x\geq 0}$. Then, ${\displaystyle x(x)\geq 0}$. So, ${\displaystyle xRx}$.
Case 2: ${\displaystyle x<0}$. Then, ${\displaystyle x(x)>0}$. So, ${\displaystyle xRx}$.
Symmetric: For every ${\displaystyle x,y\in \mathbb {Z} }$, ${\displaystyle xRy\implies xy\geq 0\implies yx=xy\geq 0\implies yRx}$.
Not transitive: Take ${\displaystyle x=1,y=0}$ and ${\displaystyle z=-1}$. Then, ${\displaystyle xy=0}$ and ${\displaystyle yz=0}$. So, ${\displaystyle xRy}$ and ${\displaystyle yRz}$. However, since ${\displaystyle xy=-1<0}$, ${\displaystyle x\not Rz}$.
${\displaystyle \Box }$
Exercise. Let ${\displaystyle R}$ be a relation on a set ${\displaystyle A}$. Prove or disprove that
(a) if ${\displaystyle R}$ is reflexive, then ${\displaystyle R^{-1}}$ is reflexive;
(b) if ${\displaystyle R}$ is symmetric, then ${\displaystyle R^{-1}}$ is symmetric;
(c) if ${\displaystyle R}$ is transitive, then ${\displaystyle R^{-1}}$ is transitive.
Equivalence relations and equivalence classes
After studying the three properties that a relation on a set can possess, let us focus on those relations that possess all three properties.
Definition. (Equivalence relation) Let ${\displaystyle A}$ be a set. A relation ${\displaystyle R}$ defined on ${\displaystyle A}$ is an equivalence relation if it is reflexive, symmetric and transitive.
Remark.
• To show that a relation is not an equivalence relation (i.e., to disprove that the relation is an equivalence relation), it suffices to show any one of the following: (i) it is not reflexive; (ii) it is not symmetric; (iii) it is not transitive, by considering the negation of the above definition.
Example. Let ${\displaystyle A=\{1,2\}}$. Also let a relation defined on ${\displaystyle A}$ be
${\displaystyle R=\{(1,1),(2,2),(1,2),(2,1)\}.}$
Prove or disprove that ${\displaystyle R}$ is an equivalence relation.
Proof.
Reflexive: Since ${\displaystyle (1,1),(2,2)\in R}$, ${\displaystyle R}$ is reflexive.
Symmetric: Since ${\displaystyle (1,1)\in R\implies (1,1)\in R}$, ${\displaystyle (2,2)\in R\implies (2,2)\in R}$, ${\displaystyle (1,2)\in R\implies (2,1)\in R}$, and ${\displaystyle (2,1)\in R\implies (1,2)\in R}$, ${\displaystyle R}$ is symmetric.
Transitive: Since for every ${\displaystyle x,y,z\in A}$, ${\displaystyle xRy{\text{ and }}yRz\implies xRz}$, ${\displaystyle R}$ is transitive.
${\displaystyle \Box }$
Exercise. Give another equivalence relation that is defined on ${\displaystyle A}$.
Solution
${\displaystyle R=\{(1,1),(2,2)\}}$. It can be shown to be reflexive, symmetric and transitive.
Example. Since the congruence of integers defines a reflexive, symmetric and transitive relation, it defines an equivalence relation on ${\displaystyle \mathbb {Z} }$.
Suppose ${\displaystyle R}$ is an equivalence relation on a set ${\displaystyle A}$. Intuitively, for elements that are related by ${\displaystyle R}$, they are quite "closely related". Thus, when we consider the set consisting elements that are related to a given element of set A, the elements inside the set are "closely related", so the set, in some sense, forms a "group" of elements that are "relatives". It then appears that we can classify the elements of set ${\displaystyle A}$ into different such "groups", according to an equivalence relation. As we will see, this is roughly the case. Hence, such "group" is quite important. Now, let us formally define what the "group" is:
Definition. (Equivalence class)
Let ${\displaystyle R}$ be an equivalence relation defined on a set ${\displaystyle A}$. For every ${\displaystyle a\in A}$, the set
${\displaystyle [a]=\{x\in A:xRa\}}$
consisting of all elements in ${\displaystyle A}$ that are related to ${\displaystyle a}$, is called an (equivalence) class of ${\displaystyle a}$ by ${\displaystyle R}$.
Example. Let ${\displaystyle A=\{1,2,3,4\}}$, and let a relation defined on ${\displaystyle A}$ be
${\displaystyle R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,3),(2,1),(3,1),(2,3),(3,2)\}.}$
This relation ${\displaystyle R}$ can be shown to be an equivalence relation. The equivalence classes are given by
${\displaystyle [1]=\{1,2,3\},[2]=\{1,2,3\},[3]=\{1,2,3\},[4]=\{4\}.}$
Since ${\displaystyle [1]=[2]=[3]}$, there are only two distinct equivalence classes. Graphically, the situation looks like:
*----------**
| . . / |
| 2 3 / |
| . /. |
| 1 / 4 |
*-----*-----*
^ ^
| |
[1]=[2] [4]
=[3]
Exercise. Construct an equivalence relation ${\displaystyle R}$ on ${\displaystyle A}$ such that the equivalence classes are given by
${\displaystyle [1]=\{1,2\},[2]=\{1,2\},[3]=\{3,4\},[4]=\{3,4\}.}$
Solution
${\displaystyle R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(3,4),(4,3)}$. Graphically, the situation looks like:
*---*-------*
| .| . |
| 2 | 3 |
| .\ . |
| 1 \ 4 |
*-----*-----*
^ ^
| |
[1]=[2] [3]=[4]
Example. (Integers modulo ${\displaystyle n}$) Recall that the congruence of integers defines an equivalence relation ${\displaystyle R}$ on ${\displaystyle \mathbb {Z} }$. Using this equivalence relation, we can define the equivalence class ${\displaystyle [a]}$ for every ${\displaystyle a\in \mathbb {Z} }$, as follows:
• ${\displaystyle [0]=\{x\in \mathbb {Z} :xR0\}=\{x\in \mathbb {Z} :x\equiv 0{\pmod {n}}\}=\{\dotsc ,-2n,-n,0,n,2n,\dotsc \}}$
• ${\displaystyle [1]=\{x\in \mathbb {Z} :xR1\}=\{x\in \mathbb {Z} :x\equiv 1{\pmod {n}}\}=\{\dotsc ,-2n+1,-n+1,1,n+1,2n+1,\dotsc \}}$
• ...
• ${\displaystyle [n-1]=\{x\in \mathbb {Z} :xR(n-1)\}=\{x\in \mathbb {Z} :x\equiv n-1{\pmod {n}}\}=\{\dotsc ,-2n+(n-1),-n+(n-1),n-1,n+(n-1),2n+(n-1),\dotsc \}=\{\dotsc ,-2n-1,-n-1,-1,2n-1,\dotsc \}}$
• ${\displaystyle [n]=\{x\in \mathbb {Z} :xRn\}=\{x\in \mathbb {Z} :x\equiv n{\pmod {n}}\}=\{x\in \mathbb {Z} :x\equiv 0{\pmod {n}}\}=[0]}$
• ${\displaystyle [n+1]=\{x\in \mathbb {Z} :xR(n+1)\}=\{x\in \mathbb {Z} :x\equiv n+1{\pmod {n}}\}=\{x\in \mathbb {Z} :x\equiv 1{\pmod {n}}\}=[1]}$
We can observe that starting from ${\displaystyle [n]}$, the classes are not distinct from the previous classes. Indeed, if we consider the classes "backward":
• ${\displaystyle [-1]=\{x\in \mathbb {Z} :xR(-1)\}=\{x\in \mathbb {Z} :x\equiv -1{\pmod {n}}\}=\{x\in \mathbb {Z} :x\equiv n-1{\pmod {n}}\}=[n-1]}$
• ${\displaystyle [-2]=\{x\in \mathbb {Z} :xR(-2)\}=\{x\in \mathbb {Z} :x\equiv -2{\pmod {n}}\}=\{x\in \mathbb {Z} :x\equiv n-2{\pmod {n}}\}=[n-2]}$
• ...
They also do not give new classes.
Hence, we conclude that there are only ${\displaystyle n}$ distinct equivalence classes, namely ${\displaystyle [0],[1],\dotsc ,[n-1]}$. Usually, the set of these equivalence classes, ${\displaystyle \{[0],[1],\dotsc ,[n-1]\}}$, is denoted by ${\displaystyle \mathbb {Z} _{n}}$, and is called integers modulo ${\displaystyle n}$. Notice that ${\displaystyle \mathbb {Z} _{n}}$ is a finite set itself, but each of its elements is an infinite set.
Remark.
• We can illustrate the equivalence classes as follows (each column is an equivalence class, and the whole table is ${\displaystyle \mathbb {Z} }$):
*----*----*---...---*-----*
| . | | | . |
| . | | | . |
| . | | | . |
|-n |-n+1| |-2n-1|
| 0 | 1 | .... | n-1 |
| n |n+1 | |2n-1 |
| . | | | . |
| . | | | . |
| . | | | . |
*----*----*---...---*-----*
• When we consider integers modulo ${\displaystyle m}$ and integers modulo ${\displaystyle n}$ (${\displaystyle m\neq n}$) together, there may be some ambiguities for the elements. For example, ${\displaystyle \mathbb {Z} _{2}=\{[0],[1]\}}$ and ${\displaystyle \mathbb {Z} _{3}=\{[0],[1],[2]\}}$. However, for instance, the "${\displaystyle [0]}$" in ${\displaystyle \mathbb {Z} _{2}}$ and the "${\displaystyle [0]}$" in ${\displaystyle \mathbb {Z} _{3}}$ are different. One of them contains all even numbers, another contains all multiples of 3.
• To avoid such ambiguities, we may add a subscript to the classes. For instance, we may write ${\displaystyle \mathbb {Z} _{2}=\{[0]_{2},[1]_{2}\}}$ and ${\displaystyle \mathbb {Z} _{3}=\{[0]_{3},[1]_{3},[2]_{3}\}}$.
Exercise. A relation ${\displaystyle R}$ on ${\displaystyle \mathbb {Z} }$ is defined by
${\displaystyle aRb{\text{ if }}3a+b\equiv 0{\pmod {4}}.}$
(a) Prove that ${\displaystyle R}$ is an equivalence relation.
(b) Express each of the equivalence classes by ${\displaystyle R}$ ${\displaystyle [0]_{R},[1]_{R},[2]_{R},[3]_{R}}$ in terms of ${\displaystyle [0]_{4},[1]_{4},[2]_{4},[3]_{4}}$ (Hint: use the symmetric property of ${\displaystyle R}$).
Solution
(a)
Proof.
Reflexive: For every ${\displaystyle x\in \mathbb {Z} }$, since ${\displaystyle 3x+x=4x\equiv 0{\pmod {4}}}$, ${\displaystyle xRx}$.
Symmetric: For every ${\displaystyle x,y\in \mathbb {Z} }$, since ${\displaystyle xRy\implies 3x+y\equiv 0{\pmod {4}}\implies 9x+3y\equiv 0{\pmod {4}}\implies 9x-8x+3y\equiv \underbrace {-8x} _{\text{multiple of 4}}\equiv 0{\pmod {4}}\implies yRx}$.
Transitive: For every ${\displaystyle x,y,z\in \mathbb {Z} }$,
{\displaystyle {\begin{aligned}xRy{\text{ and }}yRz&\implies 3x+y\equiv 0{\pmod {4}}{\text{ and }}3y+z\equiv 0{\pmod {4}}\\&\implies (3x+y)+(3y+z)\equiv 0{\pmod {4}}\\&\implies 3x+z+4y\equiv 0{\pmod {4}}\\&\implies 3x+z\equiv -4y\equiv 0{\pmod {4}}\\&\implies xRz.\end{aligned}}}
${\displaystyle \Box }$
(b)
${\displaystyle [0]_{R}=\{x\in \mathbb {Z} :xR0\}=\{x\in \mathbb {Z} :0Rx\}=\{x\in \mathbb {Z} :x\equiv 0{\pmod {4}}\}=[0]_{4}.}$
(By the symmetric property of ${\displaystyle R}$, ${\displaystyle xR0\implies 0Rx}$. Also, ${\displaystyle 0Rx\implies xR0}$. So, ${\displaystyle xR0\iff 0Rx}$.)
${\displaystyle [1]_{R}=\{x\in \mathbb {Z} :xR1\}=\{x\in \mathbb {Z} :1Rx\}=\{x\in \mathbb {Z} :3+x\equiv 0{\pmod {4}}\}=\{x\in \mathbb {Z} :4+x\equiv 1{\pmod {4}}\}=\{x\in \mathbb {Z} :x\equiv 1{\pmod {4}}\}=[1]_{4}.}$
${\displaystyle [2]_{R}=\{x\in \mathbb {Z} :xR2\}=\{x\in \mathbb {Z} :2Rx\}=\{x\in \mathbb {Z} :6+x\equiv 0{\pmod {4}}\}=\{x\in \mathbb {Z} :8+x\equiv 2{\pmod {4}}\}=\{x\in \mathbb {Z} :x\equiv 2{\pmod {4}}\}=[2]_{4}.}$
${\displaystyle [3]_{R}=\{x\in \mathbb {Z} :xR3\}=\{x\in \mathbb {Z} :3Rx\}=\{x\in \mathbb {Z} :9+x\equiv 0{\pmod {4}}\}=\{x\in \mathbb {Z} :12+x\equiv 3{\pmod {4}}\}=\{x\in \mathbb {Z} :x\equiv 3{\pmod {4}}\}=[3]_{4}.}$
Properties of equivalence classes
In this section, we will discuss some properties of equivalence classes. In particular, we will address these two questions:
1. When are two equivalence classes equal?
2. Can two different equivalence classes contain a common element?
The answer to question 1 is given by the following theorem.
Theorem. Let ${\displaystyle R}$ be an equivalence relation defined on a set ${\displaystyle A}$. For every ${\displaystyle a,b\in A}$,
${\displaystyle [a]=[b]{\text{ if and only if }}aRb.}$
Proof. "${\displaystyle \Rightarrow }$" direction: Assume ${\displaystyle [a]=[b]}$. Since ${\displaystyle R}$ is an equivalence relation, and in particular, is reflexive, we have ${\displaystyle aRa}$. Thus, we have by definition ${\displaystyle a\in [a]}$. Then, since ${\displaystyle [a]=[b]}$ by assumption, we have ${\displaystyle a\in [b]}$.
"${\displaystyle \Leftarrow }$" direction: Assume ${\displaystyle aRb}$. First, for every ${\displaystyle x\in A}$,
{\displaystyle {\begin{aligned}x\in [a]&\implies xRa&({\text{definition}})\\&\implies xRb&(aRb{\text{ and }}{\text{transitivity of }}R)\\&\implies x\in [b].&({\text{definition}})\\\end{aligned}}}
So, ${\displaystyle [a]\subseteq [b]}$.
On the other hand, first, since ${\displaystyle aRb}$ by assumption, we have ${\displaystyle bRa}$ by the symmetry of ${\displaystyle R}$. Now, For every ${\displaystyle y\in A}$,
{\displaystyle {\begin{aligned}y\in [b]&\implies yRb&({\text{definition}})\\&\implies yRa&(bRa{\text{ and transitivity of }}R)\\&\implies y\in [a].&({\text{definition}})\\\end{aligned}}}
So, ${\displaystyle [b]\subseteq [a]}$.
Hence, ${\displaystyle [a]=[b]}$.
${\displaystyle \Box }$
Remark.
• From this theorem, we know that "${\displaystyle aRb}$" is the necessary and sufficient condition for "${\displaystyle [a]=[b]}$". Also, we have ${\displaystyle [a]\neq [b]}$ if and only if ${\displaystyle a\not Rb}$.
Now, let us consider the question 2. The answer to question 2 is, indeed, "No". The following corollary justifies this answer:
Corollary. Let ${\displaystyle R}$ be an equivalence relation defined on a nonempty set ${\displaystyle A}$. For every ${\displaystyle a,b\in A}$,
${\displaystyle [a]\neq [b]{\text{ if and only if }}[a]\cap [b]=\varnothing .}$
Proof. "${\displaystyle \Leftarrow }$" direction:
{\displaystyle {\begin{aligned}{}[a]\cap [b]=\varnothing &\implies [a]{\text{ and }}[b]{\text{ have distinct elements}}&([a]{\text{ and }}[b]{\text{ are nonempty}})\\&\implies [a]\neq [b].\end{aligned}}}
(${\displaystyle [a]}$ and ${\displaystyle [b]}$ are nonempty since ${\displaystyle R}$ is reflexive. So they must contain, at least, ${\displaystyle a}$ and ${\displaystyle b}$ respectively.)
"${\displaystyle \Rightarrow }$" direction: We use proof by contrapositive.
{\displaystyle {\begin{aligned}{}[a]\cap [b]\neq \varnothing &\implies \exists x\in [a]\cap [b]\\&\implies xRa{\text{ and }}xRb\\&\implies aRx{\text{ and }}xRb&(R{\text{ is symmetric}})\\&\implies aRb&(R{\text{ is transitive}})\\&\implies [a]=[b].&({\text{above theorem}})\\\end{aligned}}}
${\displaystyle \Box }$
Remark.
• From this result, we know that two equivalence classes are either equal or disjoint (since they are either equal or not equal). Hence, it is impossible for two different equivalence classes to contain a common element.
Now we have reached a key point in studying equivalence relations (and it is probably the major reason for studying equivalence relations at all): using equivalence relation on a set to construct a partition of that set, and vice versa. Before discussing it, let us define partition of a set:
Definition. (Partition) A partition of a nonempty set ${\displaystyle A}$ is a set of nonempty subset(s) of ${\displaystyle A}$ with the property that every element of ${\displaystyle A}$ belongs to exactly one of the subset(s). In other words, ${\displaystyle P}$ is a collection of pairwise disjoint and nonempty subset(s) of ${\displaystyle A}$ whose union is ${\displaystyle A}$.
Example. Let ${\displaystyle A=\{1,2,3\}}$. Then, a partition of ${\displaystyle A}$ is ${\displaystyle \{\{1\},\{2\},\{3\}\}}$. Another one is ${\displaystyle \{\{1,2\},\{3\}\}}$. However, ${\displaystyle \{\{1,2\},\{2,3\}\}}$ is not a partition of ${\displaystyle A}$ since ${\displaystyle \{1,2\}}$ and ${\displaystyle \{2,3\}}$ are not disjoint (or the element "2" belongs to two sets). Also, ${\displaystyle \{\{1\},\{3\}\}}$ is not a partition of ${\displaystyle A}$ since the union of ${\displaystyle \{1\}}$ and ${\displaystyle \{3\}}$ is not the entire set ${\displaystyle A}$ (or the element "2" does not belong to any of the sets inside the partition).
Exercise. Is ${\displaystyle \{\{1,2,3\}\}}$ a partition of ${\displaystyle A}$?
Solution
Yes, since every element of ${\displaystyle A}$ belongs to exactly one of the set inside the partition, namely the set ${\displaystyle \{1,2,3\}}$.
Example. Let ${\displaystyle A=\{1,2,3,4\}}$, and let a relation defined on ${\displaystyle A}$ be
${\displaystyle R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,3),(2,1),(3,1),(2,3),(3,2)\}.}$
Recall that the two distinct equivalence classes are given by
${\displaystyle [1]=\{1,2,3\},[4]=\{4\}.}$
We can see that every element of ${\displaystyle A}$ belongs to exactly one of these equivalence classes. Hence, the set ${\displaystyle \{[1],[4]\}=\{\{1,2,3\},\{4\}\}}$ gives a partition of ${\displaystyle A}$.
Also, we can define another equivalence relation ${\displaystyle R}$ on ${\displaystyle A}$ such that the two distinct equivalence classes are given by
${\displaystyle [1]=\{1,2\},[3]=\{3,4\}.}$
We can also see that every element of ${\displaystyle A}$ belongs to exactly one of these equivalence classes. hence, the set ${\displaystyle \{[1],[3]\}=\{\{1,2\},\{3,4\}\}}$.
Example. Recall that the congruence of integers defines an equivalence relation ${\displaystyle R}$ on ${\displaystyle \mathbb {Z} }$. Also, there are ${\displaystyle n}$ distinct equivalence classes: ${\displaystyle [0],[1],\dotsc ,[n-1]}$. By Euclid's division lemma, every integer belongs to exactly one of these ${\displaystyle n}$ equivalence classes. Thus, the integers modulo ${\displaystyle n}$
${\displaystyle \mathbb {Z} _{n}=\{[0],[1],\dotsc ,[n-1]\}}$
gives a partition on ${\displaystyle \mathbb {Z} }$.
We can observe from the previous examples that equivalence relation of a set can be used to give a partition of that set. The following theorem suggests that, in general, an equivalence relation on a set ${\displaystyle A}$ can be used to give a partition of that set.
Theorem. Let ${\displaystyle R}$ be an equivalence relation defined on a nonempty set ${\displaystyle A}$. Then the set of all distinct equivalence classes by ${\displaystyle R}$ is a partition of ${\displaystyle A}$.
Proof. It suffices to show that every element of ${\displaystyle A}$ belongs to exactly one of the distinct equivalence classes.
For every ${\displaystyle x\in A}$, since ${\displaystyle R}$ is reflexive, we have ${\displaystyle x\in [x]}$. From this, we can ensure that every element of ${\displaystyle A}$ belongs to at least one of the distinct classes. It now remains to show that every element of ${\displaystyle A}$ also belongs to at most one of the distinct classes.
Assume that ${\displaystyle x}$ also belongs to the class ${\displaystyle [y]}$. Then, we have ${\displaystyle xRy}$. Since ${\displaystyle R}$ is an equivalence relation, it means ${\displaystyle [x]=[y]}$ by a previous theorem. Thus, any two equivalence classes to which ${\displaystyle x}$ belongs are equal. This means every element of ${\displaystyle A}$ cannot belong to more than one of the distinct classes.
Hence, every ${\displaystyle x}$ in ${\displaystyle A}$ belongs to exactly one of the distinct classes, and thus the set of all distinct equivalence classes by ${\displaystyle R}$ gives a partition of ${\displaystyle A}$.
${\displaystyle \Box }$
The following theorem suggests the converse of the above theorem is also true. To be more precise, we can use a partition of a set to construct an equivalence relation on that set. Before introducing the theorem, let us make some intuitive guesses on how to construct the equivalence relation in this way. First, from the previous theorem, roughly speaking, using an equivalence relation on a set, we can create several "groups" of elements in different classes, in which the elements are "relatives".
Now, given a partition of a set, it means we have several "groups" of elements. Such "grouping" intuitively indicates the elements inside the group are "relatives" in some sense. So, intuitively, a relation that relates the "relatives" seems to make the relation quite "close", and hence an equivalence relation.
The following theorem formalizes this intuition:
Theorem. Let ${\displaystyle P}$ be a partition of a nonempty set ${\displaystyle A}$. Define a relation ${\displaystyle R}$ on ${\displaystyle A}$ by
${\displaystyle xRy{\text{ if there exists }}S\in P{\text{ such that }}x,y\in S.}$
Then, the relation ${\displaystyle R}$ is an equivalence relation on the set ${\displaystyle A}$.
Proof. It suffices to prove that ${\displaystyle R}$ defined in this way is reflexive, symmetric, and transitive.
Reflexive: By the definition of partition, for every ${\displaystyle x\in A}$, ${\displaystyle x}$ belongs to exactly one of ${\displaystyle S\in P}$, so there exists ${\displaystyle S\in P}$ such that ${\displaystyle x\in S}$. Hence, ${\displaystyle xRx}$.
Symmetric: For every ${\displaystyle x,y\in A}$,
{\displaystyle {\begin{aligned}xRy&\implies \exists S\in P,x,y\in S\\&\implies \exists S\in P,y,x\in S\\&\implies yRx.\end{aligned}}}
Transitive: For every ${\displaystyle x,y,z\in A}$, assume ${\displaystyle xRy}$ and ${\displaystyle yRz}$. Then, there exist ${\displaystyle S,T\in P}$ such that ${\displaystyle x,{\color {blue}y\in S}}$ and ${\displaystyle {\color {blue}y},z{\color {blue}\in T}}$. But by the definition of partition, ${\displaystyle y}$ belongs to exactly one of the set in the partition ${\displaystyle P}$. So, we have ${\displaystyle S=T}$. Hence, there exists a set ${\displaystyle S}$ (${\displaystyle =T}$) such that ${\displaystyle x,z\in S}$, and thus ${\displaystyle xRz}$.
${\displaystyle \Box }$
Example. Let ${\displaystyle R}$ be a relation defined on ${\displaystyle \mathbb {Z} }$ by
${\displaystyle aRb{\text{ if }}|a|=|b|.}$
(a) Prove that ${\displaystyle R}$ is an equivalence relation.
(b) Determine all distinct equivalence classes by ${\displaystyle R}$ and hence give a partition on ${\displaystyle \mathbb {Z} }$.
Solution.
(a)
Proof. Reflexive: For every ${\displaystyle x\in \mathbb {Z} }$, ${\displaystyle |x|=|x|}$. So, ${\displaystyle xRx}$.
Symmetric: For every ${\displaystyle x,y\in \mathbb {Z} }$, ${\displaystyle xRy\implies |x|=|y|\implies |y|=|x|\implies yRx}$.
Transitive: For every ${\displaystyle x,y,z\in \mathbb {Z} }$, ${\displaystyle xRy{\text{ and }}yRz\implies |x|=|y|{\text{ and }}|y|=|z|\implies |x|=|z|\implies xRz}$.
${\displaystyle \Box }$
(b) First, some equivalence classes are
{\displaystyle {\begin{aligned}{}[0]&=\{x\in \mathbb {Z} :xR0\}=\{x\in \mathbb {Z} :|x|=0\}=\{0\},\\{}[1]&=\{x\in \mathbb {Z} :xR1\}=\{x\in \mathbb {Z} :|x|=1\}=\{-1,1\},\\{}[2]&=\{x\in \mathbb {Z} :xR2\}=\{x\in \mathbb {Z} :|x|=2\}=\{-2,2\},\dotsc \\\end{aligned}}}
So, all distinct equivalence classes are ${\displaystyle [0],[1],[2],\dotsc }$ (${\displaystyle -1\in [1]}$, so ${\displaystyle [-1]}$ is not distinct from them, etc.). Hence, a partition on ${\displaystyle A}$ is
${\displaystyle \{[0],[1],[2],\dotsc \}=\{[n]:n{\text{ is a nonnegative integer}}\}}$
(that is, every integer belongs to exactly one of ${\displaystyle [0],[1],[2],\dotsc }$.)
Exercise. Let ${\displaystyle R}$ be a relation defined on ${\displaystyle \mathbb {Z} }$ by
${\displaystyle aRb{\text{ if }}a+b{\text{ is even}}.}$
(a) Prove that ${\displaystyle R}$ is an equivalence relation.
(b) Determine all distinct equivalence classes by ${\displaystyle R}$, and hence give a partition on ${\displaystyle \mathbb {Z} }$.
Solution
(a)
Proof. Reflexive: For every ${\displaystyle x\in \mathbb {Z} }$, since ${\displaystyle x+x=2x}$ is even, ${\displaystyle xRx}$.
Symmetric: For every ${\displaystyle x,y\in \mathbb {Z} }$, ${\displaystyle xRy\implies x+y{\text{ is even}}\implies y+x=x+y{\text{ is even}}\implies yRx}$.
Transitive: For every ${\displaystyle x,y,z\in \mathbb {Z} }$,
{\displaystyle {\begin{aligned}xRy{\text{ and }}yRz&\implies x+y{\text{ is even}}{\text{ and }}y+z{\text{ is even}}\\&\implies (x+y)+(y+z)=x+2y+z{\text{ is even}}\\&\implies x+2y+z-2y{\text{ is even}}&(-2y{\text{ is even}})\\&\implies x+z{\text{ is even}}\\&\implies xRz.\\\end{aligned}}}
${\displaystyle \Box }$
(b) First, some equivalence classes are
{\displaystyle {\begin{aligned}{}[0]_{R}&=\{x\in \mathbb {Z} :xR0\}=\{x\in \mathbb {Z} :x{\text{ is even}}\}=[0]_{2},\\{}[1]_{R}&=\{x\in \mathbb {Z} :xR1\}=\{x\in \mathbb {Z} :x+1{\text{ is even}}\}=\{x\in \mathbb {Z} :x{\text{ is odd}}\}=[1]_{2}.\\\end{aligned}}}
Since every integer belongs to exactly one of ${\displaystyle [0]_{2}}$ and ${\displaystyle [1]_{2}}$ (i.e., all other equivalence classes are not distinct from them), it follows that ${\displaystyle [0]_{R}}$ and ${\displaystyle [1]_{R}}$ are all distinct equivalence classes.
Remark.
• Recall that the sum of two integers is even if and only if they have the same parity. So, this relation relates every pair of integers that have the same parity. So, intuitively, the relation ${\displaystyle R}$ can partition the integers into two "pieces": set of all odd integers and set of all even integers. |
# How are sine and cosine graphs used in real life?
## How are sine and cosine graphs used in real life?
In real life, sine and cosine functions can be used in space flight and polar coordinates, music, ballistic trajectories, and GPS and cell phones.
## What are sine and cosine graphs?
The basic sine and cosine functions have a period of 2π. The function sin x is odd, so its graph is symmetric about the origin. The function cos x is even, so its graph is symmetric about the y-axis. The graph of a sinusoidal function has the same general shape as a sine or cosine function.
What are the applications of sine and cosine graphs?
Sine and cosine functions can be used to model many real-life scenarios – radio waves, tides, musical tones, electrical currents.
### Why is cosine useful?
The law of cosines is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known.
### Where is cosine used?
To solve a triangle is to find the lengths of each of its sides and all its angles. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle. The cosine rule is used when we are given either a) three sides or b) two sides and the included angle.
Where do cos graphs start?
A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection.
#### What is a cosine graph?
The cosine graph or the cos graph is an up-down graph just like the sine graph. The only difference between the sine graph and the cos graph is that sine graph starts from 0 while the cos graph starts from 90 (or π/2). The cos graph given below starts from 1 and falls till -1 and then starts rising again.
#### How do you graph a cosine graph?
Here are the steps:
1. Find the values for domain and range. Like with sine graphs, the domain of cosine is all real numbers, and its range is.
2. Calculate the graph’s x-intercepts.
3. Calculate the graph’s maximum and minimum points.
4. Sketch the graph of the function.
What are the top 3 careers that use trig functions?
Trigonometry spreads its applications into various fields such as architects, surveyors, astronauts, physicists, engineers and even crime scene investigators.
## How to graph the sine and cosine functions?
You can graph sine and cosine functions by understanding their period and amplitude. Sine and cosine graphs are related to the graph of the tangent function, though the graphs look very different. How to graph the sine and cosine function on the coordinate plane using the unit circle?
## What is the difference between standard cosine and sine?
Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection.
What is the domain of Sine and cosine?
Even symmetry of the cosine function The sine and cosine functions have several distinct characteristics: They are periodic functions with a period of 2π. The domain of each function is (−∞,∞) ( − ∞, ∞) and the range is [−1,1] [ − 1, 1]. x is symmetric about the origin, because it is an odd function.
### What does the gray portion of the sine curve indicate?
The gray portion of the graph indicates that the basic sine curve repeats indefinitely in the positive and negative directions. The graph of the cosine function is shown in Figure 4.48. Recall from Section 4.2 that the domain of the sine and cosine functions is the set of all real numbers.
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# How do you express the complex number in trigonometric form: -6i?
Apr 16, 2016
$- 6 i = 6 \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$
#### Explanation:
Trigonometric form is where
$a + b i = \left\mid z \right\mid \left(\cos x + i \sin x\right)$.
$\left\mid z \right\mid$ is given by Pythagoras', $\sqrt{{a}^{2} + {b}^{2}}$, using the values from $a + b i$.
$a + b i = 0 - 6 i$
$\sqrt{{a}^{2} + {b}^{2}} = \sqrt{{0}^{2} + {\left(- 6\right)}^{2}} = \sqrt{36} = 6$
Now you can divide both sides by $6$, so
$\frac{- 6 i}{6} = \frac{6 \left(\cos x + i \sin x\right)}{6}$
$- i = \cos x + i \sin x$
This gives us, obviously, that
$\cos x = 0$
$\sin x = - 1$
and so, superimposing the graphs and finding points that satisfy the simultaneous equations,
At the point $x = 270$, the $\cos x$ graph is $0$ and the $\sin x$ graph is at $- 1$, which satisfies the equations. Being repeating patterns, obviously there will be other results, such as $x = - 450 , - 90 , 630 , 990$ etc.
Putting all of this back together,
$- 6 i = 6 \left(\cos 270 + i \sin 270\right)$
or
$- 6 i = 6 \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$ |
# Prime Numbers
The numbers $\,2, 3, 4,\, \ldots\,$ can be expressed as products in a very natural way! Just keep ‘breaking them down’ into smaller and smaller factors until you can't get the ‘pieces’ any smaller.
For example:
$360$ $=$ $36$ $\cdot$ $10$ $=$ $6$ $\cdot$ $6$ $\cdot$ $2$ $\cdot$ $5$ $=$ $2$ $\cdot$ $3$ $\cdot$ $2$ $\cdot$ $3$ $\cdot$ $2$ $\cdot$ $5$
OR
$360$ $=$ $4$ $\cdot$ $90$ $=$ $2$ $\cdot$ $2$ $\cdot$ $9$ $\cdot$ $10$ $=$ $2$ $\cdot$ $2$ $\cdot$ $3$ $\cdot$ $3$ $\cdot$ $2$ $\cdot$ $5$
OR
$360$ $=$ $6$ $\cdot$ $60$ $=$ $2$ $\cdot$ $3$ $\cdot$ $6$ $\cdot$ $10$ $=$ $2$ $\cdot$ $3$ $\cdot$ $2$ $\cdot$ $3$ $\cdot$ $2$ $\cdot$ $5$
No matter how the number is ‘broken down’, you'll always get to the same place, except for possibly different orderings of the factors.
In the example above, you always get: $$360 = 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5$$ Three factors of $\,2\,,$ two factors of $\,3\,,$ and one factor of $\,5\,.$
These smallest ‘pieces’ (like $\,2\,,$ $\,3\,$ and $\,5\,$ above) are, in a very real way, basic ‘building blocks’ for numbers being represented as products. They're very, very, very important! So, you shouldn't be surprised that these ‘multiplicative building blocks’ are given a special name:
DEFINITION prime numbers
A counting number greater than $\,1\,$ is called prime if the only numbers that go into it evenly are itself and $\,1\,.$
## Notes on the Definition
• The first few prime numbers are $\,2\,,$ $\,3\,,$ $\,5\,,$ $\,7\,,$ $\,11\,,$ $\,13\,,$ $\,17\,,$ $\,19\,,$ $\,23\,,$ $\,29\,,$ $\,31\,$ and $\,37\,.$ Want more? Hop up to WolframAlpha and type in (say):
prime numbers less than 100
• The number $\,1\,$ is not prime. A reason for this is discussed in the next section.
• Recall that there are several different ways to say ‘goes into evenly’. In particular, ‘$\,n\,$ goes into $\,p\,$ evenly’ is equivalent to ‘$\,p\,$ is divisible by $\,n\,$’.
• Every number is divisible by $\,1\,,$ and every number is divisible by itself. If a number is prime, then—that's all there is!
• The divisibility tests are useful for determining if a number is not prime. That is, by looking at just the last digit of the number we can quickly conclude that certain numbers aren't prime.
For example:
• If a number (greater than $\,2\,$) ends in the digits $\,0\,,$ $\,2\,,$ $\,4\,,$ $\,6\,,$ or $\,8\,,$ then it is divisible by $\,2\,,$ and hence not prime.
• If a number (greater than $\,5\,$) ends in the digit $\,5\,,$ then it is divisible by $\,5\,,$ and hence not prime.
Thus, we've eliminated ending digits of $\,0\,,$ $\,2\,,$ $\,4\,,$ $\,5\,,$ $\,6\,,$ and $\,8\,.$
• It follows that all prime numbers greater than $\,5\,$ must end in one of these digits: $\,1\,,$ $\,3\,,$ $\,7\,,$ $\,9\,$
That is: If a number greater than $\,5\,$ is prime, then its base ten representation must end in one of these digits: $\,1\,,$ $\,3\,,$ $\,7\,,$ $\,9\,$
The other direction is NOT true! If a number ends in (say) $\,7\,,$ then it may or may not be prime! For example, $\,17\,$ is prime, but $\,27\,$ is not prime.
• Here's a ‘pile interpretation’ of prime numbers: Suppose you have a bunch of candy (that can't be broken into smaller pieces without being ruined). You want to break the candy into equal piles. If the number of pieces is prime, then there are only two ways to make the piles:
• one big pile
• individual piles (piles of size $\,1\,$)
For example, five pieces of candy can only be piled in two configurations:
one big pile of five five piles of one XXXXX X X X X X
Compare this with (say) six pieces of candy (which is not prime):
one big pile of six two piles of three XXXXXX XXX XXX three piles of two six piles of one XX XX XX X X X X X X
• Definitions are conventionally given as ‘ if ’ statements, when they're actually ‘ if and only if ’ statements. (Recall that ‘ if and only if ’ is another way to say ‘ is equivalent to ’.) This is only true for definitions!!
That is, the definition of prime numbers actually tells us two things for counting numbers greater than $\,1\,$:
• If the number is prime, then it is divisible only by itself and $\,1\,.$
• If the number is divisible only by itself and $\,1\,,$ then it is prime.
• Here is a more correct (but less conventional) version of the definition:
A counting number greater than $\,1\,$ is called prime if and only if the only numbers that go into it evenly are itself and $\,1\,.$
• Here is another more correct (but less conventional) version of the definition:
Let $\,n\,$ be a counting number greater than $\,1\,.$ Then, $\,n\,$ is prime if and only if the only numbers that go into $\,n\,$ evenly are itself and $\,1\,.$
YES
NO |
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# Exercise Worksheets Using Foil
## Multiply the Binomials
How to Use FOIL
D. Russell
Early algebra requires working with polynomials and the four operations. One acronym to help multiply binomials is FOIL. FOIL stand for First Outer Inside Last. Let's put one towork.
(4x + 6) (x + 3)
We look at the first binomials which are 4x and x which gives us 4x2
Now we look at the two outside binomials which are 4x and 3 which gives us 12x
Now we look at the two inside binomials which are 6 and x whichh gives us 6x
Now we look at the last two binomials which are 6 and 3 which gives us 18
Finally, you add all of them together: 4x2 +18x + 18
All you need to remember is what FOIL stands for, whether you have fractions involved or not, just repeat the steps in FOIL and you will be able to mulitply to binomials. Practice with the worksheets and in no time it will come to you with ease. You are really just distributing both terms of one binomial by both terms of the other binomial. When I was taking algebra, I loved it, for me it was gamey!
Here are 2 PDF worksheets with answers for you to work on to practice multiplying binomials using the FOIL method. There are also many calculators that will do these calculations for you but it is important you understand how to multiply binomials correctly before using calculators.
Here are 10 sample questions, you will need to printe the PDFs in order to see the answers or practice with the worksheets. 1.) (4x - 5) (x - 3)
2.) (4x - 4 (x - 4)
3.) (2x +2) (3x + 5)
4.) (4x - 2) (3x + 3)
5.) (x - 1) (2x + 5)
6.) (5x + 2) (4x + 4)
7.) (3x - 3) (x - 2)
8.) (4x + 1) 3x + 2)
9.) (5x + 3) 3x + 4)
10.) (3x - 3) (3x + 2)
It should be noted that FOIL can only be used for binomial multiplication. FOIL is not the only method that can be used. There are other methods, although FOIL tends to be the most popular. If using the FOIL method is confusing for you, you may wish to try the distributive method, the vertical method or the grid method. Regardless of the strategy you find to work for you, all the methods will lead you to the correct answer. After all, mathematics is about finding and using the most efficient method that works for you.
Working with binomials usually occurs in the ninth or tenth grades in high school. An understanding of variables, multiplication, binomials are required before multiplying binomials. For a visual understanding of using FOIL, be sure to check out the video.After all, if a picture is worth a thousand words, a video must be worth ten times that!
Deb Russell
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# How do you find the relative extrema for f(x)=x^3-4x^2+x+6?
Jul 21, 2018
relative min at $\left(\frac{4}{3} + \frac{\sqrt{13}}{3} , \frac{70}{27} - \frac{26 \sqrt{13}}{27}\right)$
relative max at $\left(\frac{4}{3} - \frac{\sqrt{13}}{3} , \frac{70}{27} + \frac{26 \sqrt{13}}{27}\right)$
#### Explanation:
Given: $f \left(x\right) = {x}^{3} - 4 {x}^{2} + x + 6$
To find relative extrema first find the first derivative :
$f ' \left(x\right) = 3 {x}^{2} - 8 x + 1$
Find the critical value(s) by setting $f ' \left(x\right) = 0$
$f ' \left(x\right) = 3 {x}^{2} - 8 x + 1 = 0$
Use the quadratic formula to find the critical value(s):
$x = \frac{8 \pm \sqrt{{8}^{2} - 4 \left(3\right) \left(1\right)}}{2 \cdot 3} = \frac{4}{3} \pm \frac{\sqrt{52}}{6} = \frac{4}{3} \pm \frac{\sqrt{13}}{3}$
Find critical points :
$f \left(\frac{4}{3} + \frac{\sqrt{13}}{3}\right) = \frac{70}{27} - \frac{26 \sqrt{13}}{27} \approx - .8794$
$f \left(\frac{4}{3} - \frac{\sqrt{13}}{3}\right) = \frac{70}{27} + \frac{26 \sqrt{13}}{27} \approx 6.0646$
Use 2nd derivative test since it is easy to find the 2nd derivative of this function:
$f ' ' \left(x\right) = 6 x - 8$
If $f ' ' \left(c\right) > 0$ we have a relative minimum
If $f ' ' \left(c\right) < 0$ we have a relative maximum
$f ' ' \left(\frac{4}{3} + \frac{\sqrt{13}}{3}\right) > 0$ relative min at $x = \frac{4}{3} + \frac{\sqrt{13}}{3}$
$f ' ' \left(\frac{4}{3} - \frac{\sqrt{13}}{3}\right) < 0$ relative max at $x = \frac{4}{3} - \frac{\sqrt{13}}{3}$ |
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# Convert (a) 16 hours 600 seconds into minutes (b) 9 hours 360 seconds into minutes(c) 2 days 180 minutes into hours (d) 9 days 420 minutes into hours
Last updated date: 13th Aug 2024
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Hint: We convert each of the given values of time into required time using the conversions of hours into minutes, minutes into seconds and days into hours. Use a unitary method to find the value of the number of given hours, days and seconds into required time value.
* 1 day has 24 hours in it
* 1 hour has 60 minutes
* 1 minute has 60 seconds
Complete step-by-step solution:
We convert each part one by one
(a) 16 hours 600 seconds into minutes:
We first convert 16 hours into minutes
Since 1 hour $= 60$ minutes
Use unitary method to convert 16 hours into minutes
$\Rightarrow 16$ hours$= 16 \times 60$ minutes
$\Rightarrow 16$ hours$= 960$ minutes………………..… (1)
Now we convert 600 seconds into minutes
Since $60$ seconds $= 1$ minute
Multiply both sides by 10
$\Rightarrow 60 \times 10$ seconds $= 1 \times 10$ minutes
$\Rightarrow 600$ seconds $= 10$ minutes……………...… (2)
Now we add the minutes from both equations (1) and (2)
Total time in minutes $= 960 + 10$
Total time in minutes $= 970$
$\therefore$16 hours and 600 seconds is equal to 970 minutes
(b) 9 hours 360 seconds into minutes:
We first convert 9 hours into minutes
Since 1 hour $= 60$ minutes
Use unitary method to convert 9 hours into minutes
$\Rightarrow 9$ hours$= 9 \times 60$ minutes
$\Rightarrow 9$ hours$= 540$ minutes……………….… (3)
Now we convert 360 seconds into minutes
Since $60$ seconds $= 1$ minute
Multiply both sides by 6
$\Rightarrow 60 \times 6$ seconds $= 1 \times 6$ minutes
$\Rightarrow 360$seconds $= 6$minutes……………..… (4)
Now we add the minutes from both equations (3) and (4)
Total time in minutes $= 540 + 6$
Total time in minutes $= 546$
$\therefore$9 hours and 360 seconds is equal to 546 minutes
(c) 2 days 180 minutes into hours:
We first convert 2 days into hours
Since 1 day $= 24$ hours
Use unitary method to convert 2 days into hours
$\Rightarrow 2$ days$= 2 \times 24$ hours
$\Rightarrow 2$ days$= 48$ hours……………...… (5)
Now we convert 180 minutes into hours
Since $60$ minutes $= 1$ hour
Multiply both sides by 3
$\Rightarrow 60 \times 3$ minutes $= 1 \times 3$ hours
$\Rightarrow 180$ minutes $= 3$ hours……………….… (6)
Now we add the hours from both equations (5) and (6)
Total time in hours $= 48 + 3$
Total time in hours $= 51$
$\therefore$2 days and 180 minutes is equal to 51 hours
(d) 9 days 420 minutes into hours:
We first convert 9 days into hours
Since 1 day $= 24$ hours
Use unitary method to convert 9 days into hours
$\Rightarrow 9$ days$= 9 \times 24$ hours
$\Rightarrow 9$ days$= 216$ hours…………..… (7)
Now we convert 420 minutes into hours
Since $60$minutes $= 1$ hour
Multiply both sides by 7
$\Rightarrow 60 \times 7$minutes $= 1 \times 7$ hours
$\Rightarrow 420$ minutes $= 7$ hours……………...… (8)
Now we add the hours from both equations (7) and (8)
Total time in hours $= 216 + 7$
Total time in hours $= 223$
$\therefore$9 days and 420 minutes is equal to 223 hours
Note: Many students get confused while converting the given values into required values as they have to calculate the value of single unit first and then multiply by complete quantity that is required, keep in mind we can clearly see from the numbers given to us that they are multiples of 60, so we directly multiply with the factors on both sides. |
# What is 55 percent of 60 + Solution With Free Steps
55 percent of 60 is equal to 33. To get 33 as your answer, simply multiply the fraction 0.55 by the number 60.
The simple calculation of 55 percent of 60 can sum up as a real-life example, suppose, you wish to rent a room for 60 dollars a month. But the dealer says that you have to pay a security of 55% of that amount right now, now if you already know that 55 percent of 60 is equal to 33, you can easily pay the amount without waiting to calculate it. Based on this assumption we can easily cater to such scenarios which require proportion and ratios.
## What Is 55 percent of 60?
The 55 percent of 60 is equal to the number 33. The multiplication result of the fraction 0.55 and the number 60 will produce the numeral 33.
To produce the answer quickly, take the fraction 55/100 and multiply it by 60, the result will produce 33 upon solving the equation.
## How To Calculate 55% of 60?
The below steps are carried out to calculate the percent value of the given amount:
## Step 1
Firstly, translating 55 percent of 60 in terms of multiplication as:
55 percent of 60 = 55% x 60
## Step 2
Inserting 1/100 in place of percentage sign %:
Thus, Utilize the above equation.
55 percent of 60 = (55/100) x 60
## Step 3
Rearranging the above equation gives:
55 percent of 60 = (55 x 60) / 100
## Step 4
Eliminating 10 from the above equation and simplifying:
55 percent of 60 = (55 x 6) / 10
## Step 5
Multiplying 55 by 6 and then dividing it by 10 gives:
55 percent of 60 = 33
Therefore 55 percent of 60 is equal to 33.
The following pie chart displays 55 percent of 60.
In the above Pie chart, the orange portion displays 55% of 60. The whole circle comprises 60 parts, of which 55 percent has to be determined. The calculation shows that it is composed of 33 portions. The green portion in the figure portrays the remaining part of 60 which is 45% equal to 27 as 60 – 33 = 27.
All the Mathematical drawings/images are created using GeoGebra. |
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# Area of a Segment
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The area of a segment in a circle is found by first calculating the area of the sector formed by the two radii and then subtracting the area of the triangle formed by the two radii and chord (or secant). In segment problems, the most challenging aspect is often calculating the area of the triangle. Related topics include area of a sector, area of a circle and area of an annulus.
If you want to find the area of a segment you're going to have to do a little bit of problem solving first. The segment is the shaded region right up here, so let's start off by what do we know, well I know how to calculate the area of a sector so I'm going to write area of a sector down below and I'm going to sketch the area of a sector here. So I'm going to draw my circle and I'm going to draw in that sector, so the sector will calculate this whole region right here and then I can take out the part of this triangle. So if I can calculate the area of a triangle which I know how to do, so I'm going to say subtract in the same circle the area of that triangle. So I'll say area of the triangle, so if I start with this whole sector and I take out the part that's the triangle what's left is the segment. So we can say that the area of the segment is equal to the area of the sector minus the area of the triangle. So now let's write this in terms of what we know, well r is our radius, x is our central angle or the intercepted arch since those are always congruent. So the area of the segment is equal to I'll write it down here the area of the sector which we said was the area of the whole triangle pi r squared times x out of 360 and then we're going to subtract our area of our triangle which is going to be the base times the height divide by 2. So in this triangle if I drew in my height which is h and I said that we had some base of that triangle then we'll have the formula for the area of our segment. So when you think about the segment calculating its area you're going to calculate the sector and then subtract the area of that triangle.
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Difference between revisions of "1994 AHSME Problems/Problem 26"
Problem
A regular polygon of $m$ sides is exactly enclosed (no overlaps, no gaps) by $m$ regular polygons of $n$ sides each. (Shown here for $m=4, n=8$.) If $m=10$, what is the value of $n$? $[asy] size(200); defaultpen(linewidth(0.8)); draw(unitsquare); path p=(0,1)--(1,1)--(1+sqrt(2)/2,1+sqrt(2)/2)--(1+sqrt(2)/2,2+sqrt(2)/2)--(1,2+sqrt(2))--(0,2+sqrt(2))--(-sqrt(2)/2,2+sqrt(2)/2)--(-sqrt(2)/2,1+sqrt(2)/2)--cycle; draw(p); draw(shift((1+sqrt(2)/2,-sqrt(2)/2-1))*p); draw(shift((0,-2-sqrt(2)))*p); draw(shift((-1-sqrt(2)/2,-sqrt(2)/2-1))*p);[/asy]$ $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 26$
Solution
Note: might be a horrible solution but:
To find the number of sides on the regular polygons that surround the decagon, we can find the interior angles and work from there. Knowing that the measure of the interior angle of any regular polygon is $\frac{(n-2)*180}{n}$, the measure of the decagon's interior angle is $\frac{8*180}{10} = 144$ degrees.
The regular polygons meet at every vertex such that the angle outside of the decagon is divided evenly in two. With this, we know that the angle of the regular polygon is $216/2=108$ degrees. Using the previous formula, $n=5$ $\boxed{\textbf{(A) }5}$
- proto-solution by Drakodin - |
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Tangent-Secant Theorem
An Introduction to Tangent-Secant Theorem Formula
Last updated date: 21st Mar 2023
Total views: 45.9k
Views today: 1.24k
The tangent-Secant Theorem formula is a fundamental tool of Geometry found in Euclid’s Elements book. There are many methods to prove the theorem. Most important is using similar triangles. So, in this article, we will be discussing the proof of the Theorem in detail and its applications.
Terms Associated with Circle
History of Euclid
Euclid
Name: Euclid
Born: Mid-4th century BC
Died: Mid-3rd century BC
Field: Mathematics
Nationality: Greek
Statement of Tangent-Secant Theorem
Circle with PT tangent and PQR secant
For a circle with centre O, consider PT to be the tangent to the circle at the point from external point P and PQR to be the secant to the circle with points Q and R on the circle, then Tangent-Secant Theorem is defined as:
$P Q \times P R=P T^{2}$
Proof of Secant Tangent Theorem
Proof of the Tangent Secant Theorem
Given: PAB is secant to the circle with centre 0 and radius r. PT is tangent to the circle.
To prove: $P A \times P B=P T^{2}$
Construction: Draw $O D \perp A B$. Join $O P, O T$, and $O A$.
Proof:
Since $O D \perp A B$
$\therefore A D=D B \ldots(1)$ (Perpendicular from the centre to the chord bisects the chord)
$P A \times P B=(P D-A D)(P D \div B D)$
$\Rightarrow P A \times P B=(P D-A D)(P D+A D) \quad$ (Using 1)
$\Rightarrow P A \times P B=P D^{2}-A D^{2}$
In right $\triangle O P D$,
$O P^{2}=O D^{2}+P D^{2}$
$\Rightarrow P D^{2}=O P^{2}-O D^{2}$
$\therefore P A \times P B=\left(O P^{2}-O D^{2}\right)-A D^{2}$
$\Rightarrow P A \times P B=O P^{2}-\left(O D^{2}+A D^{2}\right)$
In right $\triangle O A D$,
$O A^{2}=O D^{2}+A D^{2}$
$\therefore P A \times P B=O P^{2}-O A^{2}$
$\Rightarrow P A \times P B=O P^{2}-O T^{2} \quad(\because O A=O T)$
In $\triangle O P T$,
$O P^{2}=P T^{2}+O T^{2}$
$\Rightarrow O P^{2}-O T^{2}=P T^{2}$
$\therefore P A \times P B=P T^{2}$
Hence proved.
Limitations of Tangent-Secant Theorem
1. The tangent-Secant theorem doesn’t give any idea if the secant and tangent are not drawn from common points.
2. Tangent-Secant Theorem is only applicable in the case of 2-dimensional circles and not in 3-dimensional figure spheres.
Applications of Tangent-Secant Theorem
• Tangent-Secant Theorem is of great significance. It is used in our day-to-day life such as school buildings, bridges, monuments etc.
• Monuments such as the Statue of Liberty and Pyramids are also based on concepts of secants and tangents.
Solved Examples
1. In the given figure, $P Q=12 \mathrm{~cm}, P T=24 \mathrm{~cm}$, then find $R Q$.
PQ and Tangent PT is given
Ans:
Using Tangent Secant Theorem, we have
$P Q \times P R=P T^{2}$
Putting values, we get
$(12+x) \times 12=24^{2} \\$
$\Rightarrow(12+x) \times 12=576 \\$
$\Rightarrow(12+x)=48 \\$
$\Rightarrow x=48-12 \\$
$\Rightarrow x=36$
So, we get
$R Q=36 \mathrm{~cm}$
2. In a circle, the tangent is drawn from outside point $P$ to the circle at point $T$ and from the same point secant is drawn to the circle intersecting the circle at $Q$ and $R$, respectively, such that $P Q=25 \mathrm{~cm}, R Q=9 \mathrm{~cm}$, then find $P T$.
Ans:
According to the question, we get the following figure:
Secant PQR is given
Using Tangent Secant Theorem, we have
$P Q \times P R=P T^{2}$
Putting values, we get
$(25+9) \times 25=P T^{2} \\$
$\Rightarrow P T^{2}=34 \times 25$
$\Rightarrow P T^{2}=850 \\$
$\Rightarrow P T=\sqrt{850} \\$
$\Rightarrow P T=29.154$
So, we get
$P T=12.154 \mathrm{~cm}$
3. In the given figure, $P R=84 \mathrm{~cm}, P T=42 \mathrm{~cm}$, then find $R Q$.
Secant and tangent from a point are given
Ans:
Using Tangent Secant Theorem, we have
$P Q \times P R=P T^{2}$
Putting values, we get
$P Q \times 84=42^{2} \\$
$\Rightarrow P Q \times 84=1764 \\$
$\Rightarrow P Q=21$
Now,
$R Q=P R-P Q$
And,
$P Q=21 \\$
$\Rightarrow R Q=84-21 \\$
$\Rightarrow R Q=63 \mathrm{~cm}$
Important Formulas to remember
• $P Q \times P R=P T^{2}$, where $\mathrm{PT}$ is tangent and $\mathrm{PQR}$ is secant to the circle with centre $O$ and radius $r$.
Important Points to Remember
• A Tangent line touches the circle at only one point.
• A Secant line touches the circle at exactly two points.
Conclusion
In the article, we have discussed the proof of the Tangent-Secant Theorem and its applications in detail. The Tangent Secant Theorem helps us in solving mathematical problems. In the world of art, architecture, and the growing demands of infrastructure, the tangent secant theorem places itself at the centre of its applications. In all, we can say that the theorem is of great importance.
Competitive Exams after 12th Science
FAQs on Tangent-Secant Theorem
1. What do you mean by tangent at a point on a circle?
A tangent line to any plane curve is defined as the line that just touches the plane curve at a point. Here, the plane curve is a circle. So, the tangent at any point on a circle is the line which touches the circle at most and at only one point. Tangent is also sometimes referred to as slope in Geometry. We find the tangent at a point by finding the first derivative of the plane curve at the point at which we have to find the tangent on the curve.
2. What do you mean by the Secant line?
The word Secant is derived from the Latin word ‘Secure’ which means ‘to cut. Secant to a plane curve is the line segment which cuts the curve at a minimum of two points. In the case of a circle, the Secant Line is defined as the line which cuts the circle at exactly two points. Or we can also say that. The Secant Line to the circle is a chord of a circle which is extended outside the circle to become the secant of the circle.
3. What is the relationship between Secant and Tangent?
Secant and Tangents are different terms and have different meanings. We know that secant intersects the circle at exactly two points while the tangent interests the circle at most and at only one point while there is also a similarity between both terms. The value of the angle formed by two tangents or two secants or a secant and a tangent which intersect at a point outside the circle is equal to one-half the positive difference between the angle formed by intercepted arcs. |
# Partial Fractions - PowerPoint PPT Presentation
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## Partial Fractions
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### Example 1. Decompose into partial fractions: ... If we choose x = 3/2, then 2x 3 = 0 and A will be eliminated when we make the substitution. ... – PowerPoint PPT presentation
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Title: Partial Fractions
1
Section 5.8
• Partial Fractions
2
Partial Fractions
3
Example 1
• Decompose into partial fractions
• Solution The degree of the numerator is less
than the degree of the denominator.
• Begin by factoring the denominator (x 2)(2x ?
3). We know that there are constants A and B such
that
• .
• To determine A and B, we add the expressions
on the right
4
Example 1 continued
• Equate the numerators 4x ? 13 A(2x ? 3) B(x
2)
• Since the above equation containing A and B is
true for all x, we can substitute any value of x
and still have a true equation.
• If we choose x 3/2, then 2x ? 3 0 and A will
be eliminated when we make the substitution.
• This gives us
• 4(3/2) ? 13 A2(3/2) ? 3 B(3/2 2)
• ?7 0 (7/2)B.
• B ?2.
5
Example 1 continued
• If we choose x ?2, then x 2 0 and B will be
eliminated when we make the substitution. So,
4(?2) ? 13 A2(?2) ? 3 B(?2 2)
• ?21 ?7A.
• A 3.
• The decomposition is as follows
6
Example 2
• Decompose into partial fractions
• Solution The degree of the numerator is 2
and the degree of the denominator is 3, so the
degree of the numerator is less than the degree
of the denominator. The denominator is given in
factored form. The decomposition has the
following form
7
Example 2 continued
• Next, we add the expression on the right
• Then, we equate the numerators. This gives us
• Since the equation containing A, B, and C is
true for all of x, we can substitute any value of
x and still have a true equation.
• In order to have 2x 1 0, we let x ½ .
This gives us
Solving, we obtain A 5.
8
Example 2 continued
• In order to have x ? 2 0, we let x 2.
• Substituting gives us
• To find B, we choose any value for x except ½ or
2
• and replace A with 5 and C with ?2 . We let x
1
9
Example 2 continued
• The decomposition is as follows
10
Example 3
• Decompose into partial fractions
11
Example 3 continued
• Solution
• The decomposition has
• the following form.
• Adding and equating numerators, we get
• or
12
Example 3 continued
• We then equate corresponding coefficients
• 11 A 2C, The coefficients of the
x2-terms
• ?8 ?3A B, The coefficients of the
x-terms
• ?7 ?3B ? C. The constant terms
• We solve this system of three equations and
obtain
• A 3, B 1, and C 4.
• The decomposition is as follows |
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# Cartesian Products of Sets
In mathematics, you may come across several relations such as number p is greater than number q, line m parallel to line n, set A subset of set B, etc. In all these, we can notice a relationship that involves pairs of objects in a specific order. Also, you might have learned different set operations in maths. Here, you will learn how to link pairs of elements from two sets and then introduce relations between the two elements in pairs.
## Cartesian Products of Sets Definition
In this section, you will learn the definition for the Cartesian products of sets with the help of an illustrative example. Let A and B be the two sets such that A is a set of three colours of tables and B is a set of three colours of chairs objects, i.e.,
A = {brown, green, yellow}
B = {red, blue, purple},
Let’s find the number of pairs of coloured objects that we can make from a set of tables and chairs in different combinations. They can be paired as given below:
(brown, red), (brown, blue), (brown, purple), (green, red), (green, blue), (green, purple), (yellow, red), (yellow, blue), (yellow, purple)
There are nine such pairs in the Cartesian product since three elements are there in each of the defined sets A and B. The above-ordered pairs represent the definition for the Cartesian product of sets given. This product is denoted by “A × B”.
### Cartesian Product of Sets Formula
Given two non-empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, i.e.,
P × Q = {(p,q) : p ∈ P, q ∈ Q}
If either P or Q is the null set, then P × Q will also be an empty set, i.e., P × Q = φ
### What is the Cardinality of Cartesian Product?
The cardinality of Cartesian products of sets A and B will be the total number of ordered pairs in the A × B.
Let p be the number of elements of A and q be the number of elements in B.
So, the number of elements in the Cartesian product of A and B is pq.
i.e. if n(A) = p, n(B) = q, then n(A × B) = pq.
## How to Find the Cartesian Products?
As defined above, the Cartesian product A × B between two sets A and B is the set of all possible ordered pairs with the first element from A and the second element from B. In this section, you will learn how to find the Cartesian products for two and three sets, along with examples.
Let’s have a look at the example given below. Here, set A contains three triangles of different colours and set B contains five colours of stars.
The Cartesian product of given sets A and B is given as a combination of distinct colours of triangles and stars. Thus, a total of 15 pairs are formed in A × B from the given sets.
## Cartesian Products of Two Sets
For example, A = {a1, a2, a3} and B = {b1, b2, b3, b4} are two sets. The Cartesian product of A and B can be shown as:
Set B Set A a1 a2 a3 b1 (a1, b1) (a2, b1) (a3, b1) b2 (a1, b2) (a2, b2) (a3, b2) b3 (a1, b3) (a2, b3) (a3, b3) b4 (a1, b4) (a2, b4) (a3, b4)
### Cartesian Products of Three Sets
Suppose A be a non-empty set and the Cartesian product A × A × A represents the set A × A × A ={(x, y, z): x, y, z ∈ A} which means the coordinates of all the points in three-dimensional space. This forms the basis for the Cartesian product of three sets. The below example helps in understanding how to find the Cartesian product of 3 sets.
Example:
Find the Cartesian product of three sets A = {a, b}, B = {1, 2} and C = {x, y}.
Solution:
Given,
A = {a, b}
B = {1, 2}
C = {x, y}
The ordered pairs of A × B × C can be formed as given below:
1st pair ⇒ {a, b} × {1, 2} × {x, y} ⇒ (a, 1, x)
2nd pair ⇒ {a, b} × {1, 2} × {x, y} ⇒ (a, 1, y)
3rd pair ⇒ {a, b} × {1, 2} × {x, y} ⇒ (a, 2, x)
4th pair ⇒ {a, b} × {1, 2} × {x, y} ⇒ (a, 2, y)
5th pair ⇒ {a, b} × {1, 2} × {x, y} ⇒ (b, 1, x)
6th pair ⇒ {a, b} × {1, 2} × {x, y} ⇒ (b, 1, y)
7th pair ⇒ {a, b} × {1, 2} × {x, y} ⇒ (b, 2, x)
8th pair ⇒ {a, b} × {1, 2} × {x, y} ⇒ (b, 2, y)
Thus, the ordered pairs of A × B × C can be written as:
A × B × C = {(a, 1, x), (a, 1, y), (a, 2, x), (a, 2, y), (b, 1, x), (b, 1, y), (b, 2, x), (b, 2, y)}
### Cartesian Products of Sets Properties
Some of the important properties of Cartesian products of sets are given below.
(i) Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal.
(ii) If there are m elements in A and n elements in B, then there will be mn elements in A × B. That means if n(A) = m and n(B) = n, then n(A × B) = mn.
(iii) If A and B are non-empty sets and either A or B is an infinite set, then A × B is also an infinite set.
(iv) A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered triplet.
(v) The Cartesian product of sets is not commutative, i.e. A × B ≠ B × A
(vi) The Cartesian product of sets is not associative, i.e. A × (B × C) ≠ (A × B) × C
(vii) If A is a set, then A × ∅ = ∅ and ∅ × A = ∅.
(viii) If A and B are two sets, A × B = B × A if and only if A = B, or A = ∅, or B = ∅.
(ix) Let A, B and C be three non-empty sets, then,
• A × (B ∩ C) = (A × B) ∩ (A × C)
• A × (B ∪ C) = (A × B) ∪ (A × C)
• (A ∩ B) × C = (A × C) ∩ (B × C)
• (A ∪ B) × C = (A × C) ∪ (B × C)
## Solved Examples
Example 1:
If A = {1, 2, 3} and B = {3, 4}, find the Cartesian product of A and B.
Solution:
Given,
A = {1, 2, 3}
B = {3, 4}
The Cartesian product of A and B = A × B
= {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
Example 2:
If P = {5, 6}, form the set P × P × P.
Solution:
P × P × P = {5, 6} × {5, 6} × {5, 6}
= {(5, 5, 5), (5, 5, 6), (5, 6, 5), (5, 6, 6), (6, 5, 5), (6, 5, 6), (6, 6, 5), (6, 6, 6)}
Example 3:
The Cartesian product A × A has 9 elements, among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.
Solution:
As we know, if n(A) = p and n(B) = q, then n(A x B) = pq
Thus, n(A x A) = n(A) x n(A)
Given,
n(A x A) = 9
n(A) x n(A) = 9
⇒ n(A) = 3….(i)
Also, given that (- 1, 0) and (0, 1) are two of the nine ordered pairs of A x A.
We know that A x A = {(a, a) : a ∈ A} .
Therefore, – 1, 0, and 1 are the elements of A…..(ii)
From (i) and (ii),
A = {-1, 0, 1}
Hence, the remaining elements of set A x A are (- 1, – 1), (- 1, 1), (0, – 1), (0, 0), (1, – 1), (1, 0), and (1, 1).
## Video Lesson on What are Sets
### Cartesian Products of Sets Questions
Go through the below sets questions based on the Cartesian product.
1. If A = {3, 4, 5}, B = {5, 6} and C = {6, 7, 8}, then find the following.
(i) A × (B ∩ C) (ii) (A × B) ∩ (A × C) (iii) A × (B ∪ C) (iv) (A × B) ∪ (A × C)
1. If X = {2, 3}, then form the set X × X × X.
2. If A × B = {(a, x),(a , y), (b, x), (b, y)}, then find set A and set B.
3. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.
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## Frequently Asked Questions on Cartesian Products of Sets
### What is meant by Cartesian product?
The Cartesian product is a set formed from two or more given sets and contains all ordered pairs of elements such that the first element of the pair is from the first set and the second is from the second set, and so on.
### What is the Cartesian product of two sets?
The Cartesian product A × B of sets A and B is the set of all possible ordered pairs with the first element from A and the second element from B. This can be represented as:
A × B = {(a, b) : a ∈ A and b ∈ B}
### What is the Cartesian product of three sets?
The Cartesian product A × B × C of sets A, B and C is the set of all possible ordered pairs with the first element from A, the second element from B, and the third element from C. This can be represented as:
A × B × C = {(1, a, x) : 1 ∈ A, a ∈ B, and x ∈ C}
### What is the Cartesian product of A(1, 2) and B(a, b)?
Given sets are:
A(1, 2)
B(a, b)
A × B = {1, 2} × {a, b}
= {(1, a), (1, b), (2, a), (2, b)}
Therefore, the Cartesian product of A(1, 2) and B(a, b) is the set A × B, i.e., {(1, a), (1, b), (2, a), (2, b)}.
### Is the Cartesian product a set?
Yes, the Cartesian product of sets is again a set with ordered pairs. |
# Discussion—Population Growth
To study the growth of a population mathematically, we use the concept of exponential models. Generally speaking, if we want to predict the increase in the population at a certain period in time, we start by considering the current population and apply an assumed annual growth rate. For example, if the U.S. population in 2008 was 301 million and the annual growth rate was 0.9%, what would be the population in the year 2050? To solve this problem, we would use the following formula:
P(1 + r)n
In this formula, P represents the initial population we are considering, r represents the annual growth rate expressed as a decimal and n is the number of years of growth. In this example, P = 301,000,000, r = 0.9% = 0.009 (remember that you must divide by 100 to convert from a percentage to a decimal), and n = 42 (the year 2050 minus the year 2008). Plugging these into the formula, we find:
P(1 + r)n = 301,000,000(1 + 0.009)42
= 301,000,000(1.009)42
= 301,000,000(1.457)
= 438,557,000
Therefore, the U.S. population is predicted to be 438,557,000 in the year 2050.
Let’s consider the situation where we want to find out when the population will double. Let’s use this same example, but this time we want to find out when the doubling in population will occur assuming the same annual growth rate. We’ll set up the problem like the following:
Double P = P(1 + r)n
P will be 301 million, Double P will be 602 million, r = 0.009, and we will be looking for n.
Double P = P(1 + r)n
602,000,000 = 301,000,000(1 + 0.009)n
Now, we will divide both sides by 301,000,000. This will give us the following:
2 = (1.009)n
To solve for n, we need to invoke a special exponent property of logarithms. If we take the log of both sides of this equation, we can move exponent as shown below:
log 2 = log (1.009)n
log 2 = n log (1.009)
Now, divide both sides of the equation by log (1.009) to get:
n = log 2 / log (1.009)
Using the logarithm function of a calculator, this becomes:
n = log 2/log (1.009) = 77.4
Therefore, the U.S. population should double from 301 million to 602 million in 77.4 years assuming annual growth rate of 0.9 %.
Now it is your turn:
• Search the Internet and determine the most recent population of your home state. A good place to start is the U.S. Census Bureau (www.census.gov) which maintains all demographic information for the country. If possible, locate the annual growth rate for your state. If you can not locate this value, feel free to use the same value (0.9%) that we used in our example above.
• Determine the population of your state 10 years from now.
• Determine how long and in what year the population in your state may double assuming a steady annual growth rate.
• Look up the population of the city in which you live. If possible, find the annual percentage growth rate of your home city (use 0.9% if you can not locate this value).
• Determine the population of your city in 10 years.
• Determine how long until the population of your city doubles assuming a steady growth rate.
• Discuss factors that could possibly influence the growth rate of your city and state.
• Do you live in a city or state that is experiencing growth?
• Is it possible that you live in a city or state where the population is on the decline or hasn’t changed?
• How would you solve this problem if the case involved a steady decline in the population (say -0.9% annually)? Show an example.
• Think of other real world applications (besides monitoring and modeling populations) where exponential equations can be utilized.
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# Symbols and Notations
Symbols and Notations is a form of reasoning that uses symbols to represent relationships between things. This notation can be used to represent mathematical and logical relationships between the numbers and letters.
A notations is a system of characters, symbols, or abbreviated expressions used in an art or science or in mathematics or logic to express technical facts or quantities.
Example :
If ‘+’ stands for multiplication, ‘x’ stands for division, ‘-‘ stands for addition and ‘+’ stands for subtraction, what is the answer for the following equation?
20 – 5 + 18 x (3 + 2) =?
Solution : 22
Given expression
20-5÷ 18 x (3 + 2) = ?
After changing the signs
? = 20 + 5- 18 ÷ (3 x 2)
or, ? = 20 + 5- 18 ÷ 6
or, ? = 20 + 5- 3
or, ? = 25 -3 = 22
Example :
If * means /, - means *, / means + and + means -, then (3 - 15 / 19) * 8 + 6 = ?
Solution: 2
Using the correct symbols, we have: Given expression = (3 * 15 + 19) / 8 - 6 = (45 + 19) / 8 - 6 = 64 / 8 - 6 = 8 - 6 = 2
Example :
Which alternative Cleary indicates the rule followed in the following set of numbers?
7 4 8 2 = 24
Solution : x, -, ÷
7 x 4 – 8 + 2 =>28-4 = 24
Example :
In a certain code language, '+' means '*', '*' means '-', '-' means '/' and '/' means '+'.
16 - 2 + 4 / 16 - 8 * 2
Solution : 32
As per the given expression,
16 - 2 + 4 / 16 - 8 * 2
becomes
16 / 2 * 4 + 16 / 8 - 2
Using BODMAS rule,
16 / 2 * 4 + 16 / 8 - 2
= 8 * 4 + 2 - 2
= 32 + 0
= 32
Example :
If ‘+’ stands for multiplication, ‘x’ stands for division, ‘-‘ stands for addition and ‘+’ stands for subtraction, what is the answer for the following equation? If L = +, M = – , N = x, P = ÷, then
5 N 5 P 5 L 5 M 5 =?
Solution : 5
Given expression
5 N 5 P 5 L 5 M 5=?
After changing the signs
?=5 x 5 ÷ 5 + 5 – 5
or, ? = 5 + 5- 5 = 5 |
## Conics: Co-ordinate Geometry for IITJEE Mains: Problem solving: Basics 4
Question 1:
If the points of the intersection of the ellipses $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1$ and $\frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{\beta^{2}}=1$ are the end points of the conjugate diameters of the former, prove that :$\frac{a^{2}}{\alpha^{2}} + \frac{b^{2}}{\beta^{2}} = 2$
Solution 1:
The locus of the middle points of a system of parallel chords of an ellipse is a line passing through the centre of the ellipse. This is called the diameter of the ellipse and two diameters of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$ are said to be conjugate if each bisects the chords, parallel to the other. The condition for this is that the product of their slopes should be equal to $\frac{-b^{2}}{a^{2}}$.
Now, equation of the lines joining the centre $(0,0)$ to the points of intersection of the given ellipses is
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{\beta^{2}}$
$\Longrightarrow (\frac{1}{a^{2}}-\frac{1}{\alpha^{2}})x^{2}+(\frac{1}{b^{2}} - \frac{1}{\beta^{2}})y^{2}=0$…call this equation I;
If $m_{1}$, $m_{2}$ are the slopes of the lines represented by Equation I, then $m_{1}m_{2}=\frac{(\frac{1}{a^{2}}-\frac{1}{\alpha^{2}})}{(\frac{1}{b^{2}}-\frac{1}{\beta^{2}})}$
Since I represents a pair of conjugate diameters, $m_{1}m_{2}=-\frac{b^{2}}{a^{2}}$
Thus, $a^{2}(\frac{1}{a^{2}}-\frac{1}{\alpha^{2}})+b^{2}(\frac{1}{b^{2}}-\frac{1}{\beta^{2}})=0$
$\Longrightarrow \frac{a^{2}}{\alpha^{2}} + \frac{b^{2}}{\beta^{2}} =2$
Question 2:
Find the locus of the mid-points of the chords of the circle $x^{2}+y^{2}=16$ which are tangents to the hyperbola, $9x^{2}-16y^{2}=144$.
Solution 2:
Let $(h,k)$ be the middle point of a chord of the circle $x^{2}+y^{2}=16$.
Then, its equation is $hx + ky-16=h^{2}+k^{2}-16$, that is, $hx + ky=h^{2}+k^{2}$ ….call this equation I.
Let I touch the hyperbola: $9x^{2}-16y^{2}=144$
That is, $\frac{x^{2}}{16} - \frac{y^{2}}{9}=1$ …call this equation II.
at the point $(\alpha, \beta)$ say, then I is identical with
$\frac{x\alpha}{16} - \frac{y\beta}{9}=1$….call this equation III.
Thus, $\frac{\alpha}{16h} = \frac{-\beta}{9k} = \frac{1}{h^{2}+k^{2}}$
Since $(\alpha, \beta)$ lies on the hyperbola II,
$\frac{1}{16}(\frac{16h}{(h^{2}+k^{2})})^{2}-\frac{1}{9}(\frac{9k}{h^{2}+k^{2}})^{2}=1$
$\Longrightarrow 16h^{2}-9k^{2}=(h^{2}+k^{2})^{2}$.
Hence, the required locus of $(h,k)$ is $(x^{2}+y^{2})^{2}=16x^{2}-9y^{2}$.
Question 3:
If P be a point on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1$ whose ordinate is $y^{'}$, prove that the angle between the tangent at P and the focal chord through P is $\arctan{(\frac{b^{2}}{aey^{'}})}$.
Solution 3:
Let the coordinates of P be $(a\cos{\theta}, b\sin{\theta})$ so that $b\sin{\theta}=y^{'}$. Equation of the tangent at P is $\frac{x}{a}\cos{\theta}+\frac{y}{b}\sin{\theta}=1$.
Slope of the tangent is equal to $-\frac{b\cos{\theta}}{a\sin{\theta}}$
Slope of the focal chord SP is $\frac{b\sin{\theta}-0}{a\cos{\theta}-ae}=\frac{b\sin{\theta}}{a(\cos{\theta})-e}$.
If $\alpha$ is the required angle, then $\tan{\alpha}=\frac{-\frac{b\cos{\theta}}{a\sin{\theta}}-\frac{b\sin{\theta}}{a(\cos{\theta})-e}}{1-\frac{b\cos{\theta}b\sin{\theta}}{a\sin{\theta}a(\cos{\theta})-e}}$
which in turn equals $\frac{-b(\cos^{2}{\theta}-e\cos{\theta}+\sin^{2}{\theta})}{a\sin{\theta}(\cos{\theta}-e)} \times \frac{a^{2}(\cos{\theta}-e)}{(a^{2}-b^{2})\cos{\theta}-a^{2}e}$,
$=\frac{-ab(1-e\cos{\theta})}{\sin{\theta}(a^{2}e^{2}\cos{\theta}-a^{2}e)}=\frac{ab(e\cos{\theta}-1)}{a^{2}e\sin{\theta}(e\cos{\theta}-1)}$
$=\frac{ab}{a^{2}e\sin{\theta}}=\frac{ab.b}{a^{2}e.y^{'}}=\frac{b^{2}}{aey^{'}}$
$\Longrightarrow \alpha=\arctan{(\frac{b^{2}}{aey^{'}})}$
More later,
I hope you like it…my students should be inspired to try Math on their own…initially, it is slow, gradual, painstaking, but the initial “roots” pay very very “rich dividends” later…
-Nalin Pithwa.
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Lesson 3: Visualize percents
# Finding the whole with a tape diagram
Keisha can run 170 meters in one minute today, which is 125% of her distance from three years ago. To find her previous distance, we convert 125% to 5/4 and divide 170 by 5, obtaining 34 meters per fourth. Multiplying 34 by 4 reveals that Keisha could run 136 meters in one minute three years ago. Created by Sal Khan.
## Want to join the conversation?
• What job is this used in? In other words why do we need to know this? Also u are really smart
• Any job really. In basic conversations it's common to hear percentages. So it's crucial to understand percentages.
• so the first step you did is point out that 100% of 170m of distance was completed in both cases.
then for the today’s section you combined the 125% with the the first “whole” into a fraction which is 125/100 and then simplified it which gave you 5/4.
This 5/4 im assuming gave us details as to how many “wholes” todays and the 3 years ago “wholes” distance % has. Which is indicated by the denominator.
And since we know in both cases that 170m was completely ran. they will share the same denominator/“wholes” which is 4. However the numerator will be different because well im not really sure why. Lol this logic is escaping me.
After that the logic you used to create the diagram was off to me in the sense that i am unfamiliar with too many concepts :(
• i need so much help this did nothing to help
• just divide 170 by 125 and you get 1.36. it works for every % just divide the number that you're trying to find the % of by the % like if youre trying to find 34% of 100 or 100% just divide 34 by 100 and you get 0.34. try it for yourself
• what is a tape diagram
• It was just the thing he drew with sections of equal amounts. I dont think you had to know it beforehand.
• Could you please slow down. I would really appreciated it 😁
• If you divide 170 by (125/100) you also get 136. Why does it also work this way?
• The way how the video explains it isn't enough when doing the exercise because this video only gives you one example to do it. Also, the exercises are a little different. He should have done more problems like 3 instead of 1 and solved some ones from the exercises.
Step 1. But for anyone, who needs some help a simpler way is to first make a line chart.
Step 2. 170 = 125%. 170 is just a number that equals 125%
Step 3. So to find out the number that equals 100% we first need to find out how many squares or lines make 100%
Step 4. So we divide 125% by 100% which equals = 1.25 (Which is just 5/4 in decimal form or 5 over 4)
Step 5. So 5/4 just means that there are 5 total squares or lines. And the number 5 equals 125% and 170
Step 6. If the problem is about finding who amount of numbers for each square then lastly you find out the amount that goes into each square which would equal 170.
So you divide again. We divide 170 by 5 which gives us 34. So now we know that 34 goes into each of the squares 5x time which all equal to 170 (34 x 5 = 170)
Now if you are dealing with number lines then continue.
Step 7. So since we now know there are 5 total lines we need to find out the number percentage that will equal 125%
Step 8. To know what percentage makes 125% we now must divide again. So we 125 divided by 5 = 25. So now we know 25% would be the first percentage out of 5. (25 x 5 = 125%)
Step 9. Since we now know each number line or square will be 25% to find out any other percentage like 50% or 75% you'll just need to multiply by the same amount of lines or squares you have.
For example 25 x 3 = 75% and since we know that 34 is the total amount for each square we also multiply 34 x 3 = 102.
So I hope that I have given you an idea of how or what to do when doing the exercise problem. |
Consecutive odd primed integers
A few years ago, for some reason (I don’t remember exactly what it was) I thought of the following problem and tackled around a solution for it:
Prove that the triplet $(3, 5, 7)$ is the only triplet of consecutive odd integers such that all integers in it are primes.
In other words, prove that $\exists! n \in N (2n + 1, 2n + 3, 2n + 5)$ are primes.
We already know if we set $n = 1$ that we get the triplet $(3, 5, 7)$, but as for the part of proving that this is unique, we need to show that at least one member of the triplet is divisible by 3. So, for example, for $n = 3$ we have $(7, 9, 11)$, and 9 is not a prime because it’s divisible by 3.
We can use induction on n. The base case is already there for $n = 1$, so for the inductive step we can assume that either of the members is divisible by 3 (but not the remaining). Now assuming one of the members in $(2n + 1, 2n + 3, 2n + 5)$ is divisible by 3, we need to show that one of the members in $(2n + 3, 2n + 5, 2n + 7)$ is divisible by 3.
From here, we have 3 cases:
1. $2n + 1$ is divisible by 3. Then, so is $2n + 4$, and $2n + 7$, thus one of $(2n + 3, 2n + 5, 2n + 7)$ is divisible by 3.
2. $2n + 3$ is divisible by 3. Thus one of $(2n + 3, 2n + 5, 2n + 7)$ is divisible by 3.
3. $2n + 5$ is divisible by 3. Thus one of $(2n + 3, 2n + 5, 2n + 7)$ is divisible by 3.
So, since at least one member of $(2n + 1, 2n + 3, 2n + 5)$ is divisible by 3, but only 3 is a prime (and not multiples of 3 of course), we get that $(3, 5, 7)$ is the only triplet of consecutive odd integers such that all members are prime. |
## What is the cross-product?
The cross product, we also know as the vector product (or directed area product to underline its geometric relevance), is a binary operation on two vectors in a three-dimensional oriented Euclidean vector space (denoted by the symbol x in this case). The cross product, a b (read “a cross b”), of two linearly independent vectors a and b, is a vector perpendicular to both a and b and so normal to the plane containing both. Mathematics, physics, engineering, and computer programming are just a few fields where we may use them. It is not confusing with the dot item (projection product).
Take a look other related calculators, such as:
### Cross product formula
We can calculate the area between any two vectors using the cross-product formula. The magnitude of the resulting vector, which is the area of the parallelogram spanned by the two vectors, is determined by the cross-product formula.
c = a × b = |a| * |b| * sinθ * n
## The cross product of two vectors
Cross product is a type of vector multiplication in which two vectors of different natures or sorts are multiplied. The vector magnitude and direction of a vector are both presents. By using the cross product and dot product, we may multiply two or more vectors. The resultant vector is termed the cross product of two vectors or the vector product when two vectors are multiplied with each other, and the product of the vectors is likewise a vector quantity. The resulting vector is perpendicular to the plane in which the two provided vectors are located.
## Dot product vs. cross product
The product of the magnitude of the vectors and the cos of the angle between two vectors we call a dot product. The magnitude of the vectors and the sine of the angle they subtend on each other form a cross product.
## Right-hand rule cross product
The third vector that is perpendicular to the two initial vectors is the cross-product of two vectors. The size of the parallelogram between them determines its magnitude, and the right-hand thumb rule determines its direction.
Physicists have devised several strategies to assist you in navigating these choppy seas. The “Right-hand rule,” which aids in the computation of the cross vector product, is probably the most well-known. This rule allows you to anticipate the direction of the cross product’s resultant vector using only your hand.
In physics, there are two variants of the right-hand rule: one that involves going from an open hand to a closed fist with the fingers extended and the hand still, and the other that involves going from an open hand to a closed fist with the fingers extended and the hand still.
## Example
We’ll take the vectors a = (2, 3, 7) and b = (2, 3, 7) to calculate the cross-product of two vectors (1, 2, 4).
1. The components of the vector must first introduce. x = 2, y = 3, and ‘z = 7’ are the values.
2. The components of vector b should then be introduced. That is, x equals 1, y equals 2, and z equals 4.
3. Now the cross product calculator analyzes the data, uses the formula we learned about before.
4. c = a b = c = a b = c = a b = c = a b = c = (-2, -1, 1).
5. Repeat until you’ve computed all of the necessary cross products.
## FAQ
### Cross product equation
The cross product formula calculates the cross product of any two vectors by calculating the area between them.
c = a × b = |a| * |b| * sinθ * n
### How to calculate cross-product?
The length is calculated by multiplying the length of a by the length of b by the size of the angle between a and b. Finally, we multiply by the vector n to ensure that it points in the right direction.
### What does the cross-product represent?
The process of multiplying two vectors is called the cross product. This is because the multiplication sign(x) between two vectors denotes a cross product.
### Is cross-product commutative?
The direction of the two vectors is related to the direction of their product using the right-hand rule for cross-multiplication. Because cross multiplication is not commutative, the sequence in which the operations are performed is crucial.
### How to do the cross-product of vectors?
We may express the formula for the cross product in terms of components using these qualities and the cross product of the standard unit vectors.
### Is cross-product associative?
The cross-product does not have any associative properties. By brute force, one may demonstrate this by selecting three vectors and seeing that the two expressions are not equivalent. |
In mathematics, the sign function or signum function (from signum, Latin for "sign") is a function that has the value −1, +1 or 0 according to whether the sign of a given real number is positive or negative, or the given number is itself zero. In mathematical notation the sign function is often represented as ${\displaystyle \operatorname {sgn} x}$ or ${\displaystyle \operatorname {sgn}(x)}$.[1]
## Definition
The signum function of a real number ${\displaystyle x}$ is a piecewise function which is defined as follows:[1] ${\displaystyle \operatorname {sgn} x:={\begin{cases}-1&{\text{if ))x<0,\\0&{\text{if ))x=0,\\1&{\text{if ))x>0.\end{cases))}$
The law of trichotomy states that every real number must be positive, negative or zero. The signum function denotes which unique category a number falls into by mapping it to one of the values −1, +1 or 0, which can then be used in mathematical expressions or further calculations.
For example: ${\displaystyle {\begin{array}{lcr}\operatorname {sgn}(2)&=&+1\,,\\\operatorname {sgn}(\pi )&=&+1\,,\\\operatorname {sgn}(-8)&=&-1\,,\\\operatorname {sgn}(-{\frac {1}{2)))&=&-1\,,\\\operatorname {sgn}(0)&=&0\,.\end{array))}$
## Basic properties
Any real number can be expressed as the product of its absolute value and its sign function: ${\displaystyle x=|x|\operatorname {sgn} x\,.}$
It follows that whenever ${\displaystyle x}$ is not equal to 0 we have ${\displaystyle \operatorname {sgn} x={\frac {x}{|x|))={\frac {|x|}{x))\,.}$
Similarly, for any real number ${\displaystyle x}$, ${\displaystyle |x|=x\operatorname {sgn} x\,.}$ We can also be certain that: ${\displaystyle \operatorname {sgn}(xy)=(\operatorname {sgn} x)(\operatorname {sgn} y)\,,}$ and so ${\displaystyle \operatorname {sgn}(x^{n})=(\operatorname {sgn} x)^{n}\,.}$
## Some algebraic identities
The signum can also be written using the Iverson bracket notation: ${\displaystyle \operatorname {sgn} x=-[x<0]+[x>0]\,.}$
The signum can also be written using the floor and the absolute value functions: ${\displaystyle \operatorname {sgn} x={\Biggl \lfloor }{\frac {x}{|x|+1)){\Biggr \rfloor }-{\Biggl \lfloor }{\frac {-x}{|x|+1)){\Biggr \rfloor }\,.}$ If ${\displaystyle 0^{0))$ is accepted to be equal to 1, the signum can also be written for all real numbers as ${\displaystyle \operatorname {sgn} x=0^{\left(-x+\left\vert x\right\vert \right)}-0^{\left(x+\left\vert x\right\vert \right)}\,.}$
## Properties in mathematical analysis
### Discontinuity at zero
Although the sign function takes the value −1 when ${\displaystyle x}$ is negative, the ringed point (0, −1) in the plot of ${\displaystyle \operatorname {sgn} x}$ indicates that this is not the case when ${\displaystyle x=0}$. Instead, the value jumps abruptly to the solid point at (0, 0) where ${\displaystyle \operatorname {sgn}(0)=0}$. There is then a similar jump to ${\displaystyle \operatorname {sgn}(x)=+1}$ when ${\displaystyle x}$ is positive. Either jump demonstrates visually that the sign function ${\displaystyle \operatorname {sgn} x}$ is discontinuous at zero, even though it is continuous at any point where ${\displaystyle x}$ is either positive or negative.
These observations are confirmed by any of the various equivalent formal definitions of continuity in mathematical analysis. A function ${\displaystyle f(x)}$, such as ${\displaystyle \operatorname {sgn}(x),}$ is continuous at a point ${\displaystyle x=a}$ if the value ${\displaystyle f(a)}$ can be approximated arbitrarily closely by the sequence of values ${\displaystyle f(a_{1}),f(a_{2}),f(a_{3}),\dots ,}$ where the ${\displaystyle a_{n))$ make up any infinite sequence which becomes arbitrarily close to ${\displaystyle a}$ as ${\displaystyle n}$ becomes sufficiently large. In the notation of mathematical limits, continuity of ${\displaystyle f}$ at ${\displaystyle a}$ requires that ${\displaystyle f(a_{n})\to f(a)}$ as ${\displaystyle n\to \infty }$ for any sequence ${\displaystyle \left(a_{n}\right)_{n=1}^{\infty ))$ for which ${\displaystyle a_{n}\to a.}$ The arrow symbol can be read to mean approaches, or tends to, and it applies to the sequence as a whole.
This criterion fails for the sign function at ${\displaystyle a=0}$. For example, we can choose ${\displaystyle a_{n))$ to be the sequence ${\displaystyle 1,{\tfrac {1}{2)),{\tfrac {1}{3)),{\tfrac {1}{4)),\dots ,}$ which tends towards zero as ${\displaystyle n}$ increases towards infinity. In this case, ${\displaystyle a_{n}\to a}$ as required, but ${\displaystyle \operatorname {sgn}(a)=0}$ and ${\displaystyle \operatorname {sgn}(a_{n})=+1}$ for each ${\displaystyle n,}$ so that ${\displaystyle \operatorname {sgn}(a_{n})\to 1\neq \operatorname {sgn}(a)}$. This counterexample confirms more formally the discontinuity of ${\displaystyle \operatorname {sgn} x}$ at zero that is visible in the plot.
Despite the sign function having a very simple form, the step change at zero causes difficulties for traditional calculus techniques, which are quite stringent in their requirements. Continuity is a frequent constraint. One solution can be to approximate the sign function by a smooth continuous function; others might involve less stringent approaches that build on classical methods to accommodate larger classes of function.
### Smooth approximations and limits
The signum function coincides with the limits ${\displaystyle \operatorname {sgn} x=\lim _{n\to \infty }{\frac {1-2^{-nx)){1+2^{-nx))}\,.}$ and ${\displaystyle \operatorname {sgn} x=\lim _{n\to \infty }{\frac {2}{\pi )){\rm {arctan))(nx)\,=\lim _{n\to \infty }{\frac {2}{\pi ))\tan ^{-1}(nx)\,.}$as well as,
${\displaystyle \operatorname {sgn} x=\lim _{n\to \infty }\tanh(nx)\,.}$Here, ${\displaystyle \tanh(x)}$ is the Hyperbolic tangent and the superscript of -1, above it, is shorthand notation for the inverse function of the Trigonometric function, tangent.
For ${\displaystyle k>1}$, a smooth approximation of the sign function is ${\displaystyle \operatorname {sgn} x\approx \tanh kx\,.}$ Another approximation is ${\displaystyle \operatorname {sgn} x\approx {\frac {x}{\sqrt {x^{2}+\varepsilon ^{2))))\,.}$ which gets sharper as ${\displaystyle \varepsilon \to 0}$; note that this is the derivative of ${\displaystyle {\sqrt {x^{2}+\varepsilon ^{2))))$. This is inspired from the fact that the above is exactly equal for all nonzero ${\displaystyle x}$ if ${\displaystyle \varepsilon =0}$, and has the advantage of simple generalization to higher-dimensional analogues of the sign function (for example, the partial derivatives of ${\displaystyle {\sqrt {x^{2}+y^{2))))$).
### Differentiation and integration
The signum function ${\displaystyle \operatorname {sgn} x}$ is differentiable everywhere except when ${\displaystyle x=0.}$ Its derivative is zero when ${\displaystyle x}$ is non-zero: ${\displaystyle {\frac ((\text{d))\,(\operatorname {sgn} x)}((\text{d))x))=0\qquad {\text{for ))x\neq 0\,.}$
This follows from the differentiability of any constant function, for which the derivative is always zero on its domain of definition. The signum ${\displaystyle \operatorname {sgn} x}$ acts as a constant function when it is restricted to the negative open region ${\displaystyle x<0,}$ where it equals -1. It can similarly be regarded as a constant function within the positive open region ${\displaystyle x>0,}$ where the corresponding constant is +1. Although these are two different constant functions, their derivative is equal to zero in each case.
It is not possible to define a classical derivative at ${\displaystyle x=0}$, because there is a discontinuity there. Nevertheless, the signum function has a definite integral between any pair of finite values a and b, even when the interval of integration includes zero. The resulting integral for a and b is then equal to the difference between their absolute values: ${\displaystyle \int _{a}^{b}(\operatorname {sgn} x)\,{\text{d))x=|b|-|a|\,.}$
Conversely, the signum function is the derivative of the absolute value function, except where there is an abrupt change in gradient before and after zero: ${\displaystyle {\frac ((\text{d))|x|}((\text{d))x))=\operatorname {sgn} x\qquad {\text{for ))x\neq 0\,.}$
We can understand this as before by considering the definition of the absolute value ${\displaystyle |x|}$ on the separate regions ${\displaystyle x<0}$ and ${\displaystyle x<0.}$ For example, the absolute value function is identical to ${\displaystyle x}$ in the region ${\displaystyle x>0,}$ whose derivative is the constant value +1, which equals the value of ${\displaystyle \operatorname {sgn} x}$ there.
Because the absolute value is a convex function, there is at least one subderivative at every point, including at the origin. Everywhere except zero, the resulting subdifferential consists of a single value, equal to the value of the sign function. In contrast, there are many subderivatives at zero, with just one of them taking the value ${\displaystyle \operatorname {sgn}(0)=0}$. A subderivative value 0 occurs here because the absolute value function is at a minimum. The full family of valid subderivatives at zero constitutes the subdifferential interval ${\displaystyle [-1,1]}$, which might be thought of informally as "filling in" the graph of the sign function with a vertical line through the origin, making it continuous as a two dimensional curve.
In integration theory, the signum function is a weak derivative of the absolute value function. Weak derivatives are equivalent if they are equal almost everywhere, making them impervious to isolated anomalies at a single point. This includes the change in gradient of the absolute value function at zero, which prohibits there being a classical derivative.
Although it is not differentiable at ${\displaystyle x=0}$ in the ordinary sense, under the generalized notion of differentiation in distribution theory, the derivative of the signum function is two times the Dirac delta function. This can be demonstrated using the identity [2] ${\displaystyle \operatorname {sgn} x=2H(x)-1\,,}$ where ${\displaystyle H(x)}$ is the Heaviside step function using the standard ${\displaystyle H(0)={\frac {1}{2))}$ formalism. Using this identity, it is easy to derive the distributional derivative:[3] ${\displaystyle {\frac ((\text{d))\operatorname {sgn} x}((\text{d))x))=2{\frac ((\text{d))H(x)}((\text{d))x))=2\delta (x)\,.}$
### Fourier transform
The Fourier transform of the signum function is[4] ${\displaystyle \int _{-\infty }^{\infty }(\operatorname {sgn} x)e^{-ikx}{\text{d))x=PV{\frac {2}{ik)),}$ where ${\displaystyle PV}$ means taking the Cauchy principal value.
## Generalizations
### Complex signum
The signum function can be generalized to complex numbers as: ${\displaystyle \operatorname {sgn} z={\frac {z}{|z|))}$ for any complex number ${\displaystyle z}$ except ${\displaystyle z=0}$. The signum of a given complex number ${\displaystyle z}$ is the point on the unit circle of the complex plane that is nearest to ${\displaystyle z}$. Then, for ${\displaystyle z\neq 0}$, ${\displaystyle \operatorname {sgn} z=e^{i\arg z}\,,}$ where ${\displaystyle \arg }$ is the complex argument function.
For reasons of symmetry, and to keep this a proper generalization of the signum function on the reals, also in the complex domain one usually defines, for ${\displaystyle z=0}$: ${\displaystyle \operatorname {sgn}(0+0i)=0}$
Another generalization of the sign function for real and complex expressions is ${\displaystyle {\text{csgn))}$,[5] which is defined as: ${\displaystyle \operatorname {csgn} z={\begin{cases}1&{\text{if ))\mathrm {Re} (z)>0,\\-1&{\text{if ))\mathrm {Re} (z)<0,\\\operatorname {sgn} \mathrm {Im} (z)&{\text{if ))\mathrm {Re} (z)=0\end{cases))}$ where ${\displaystyle {\text{Re))(z)}$ is the real part of ${\displaystyle z}$ and ${\displaystyle {\text{Im))(z)}$ is the imaginary part of ${\displaystyle z}$.
We then have (for ${\displaystyle z\neq 0}$): ${\displaystyle \operatorname {csgn} z={\frac {z}{\sqrt {z^{2))))={\frac {\sqrt {z^{2))}{z)).}$
### Polar decomposition of matrices
Thanks to the Polar decomposition theorem, a matrix ${\displaystyle {\boldsymbol {A))\in \mathbb {K} ^{n\times n))$ (${\displaystyle n\in \mathbb {N} }$ and ${\displaystyle \mathbb {K} \in \{\mathbb {R} ,\mathbb {C} \))$) can be decomposed as a product ${\displaystyle {\boldsymbol {Q)){\boldsymbol {P))}$ where ${\displaystyle {\boldsymbol {Q))}$ is a unitary matrix and ${\displaystyle {\boldsymbol {P))}$ is a self-adjoint, or Hermitian, positive definite matrix, both in ${\displaystyle \mathbb {K} ^{n\times n))$. If ${\displaystyle {\boldsymbol {A))}$ is invertible then such a decomposition is unique and ${\displaystyle {\boldsymbol {Q))}$ plays the role of ${\displaystyle {\boldsymbol {A))}$'s signum. A dual construction is given by the decomposition ${\displaystyle {\boldsymbol {A))={\boldsymbol {S)){\boldsymbol {R))}$ where ${\displaystyle {\boldsymbol {R))}$ is unitary, but generally different than ${\displaystyle {\boldsymbol {Q))}$. This leads to each invertible matrix having a unique left-signum ${\displaystyle {\boldsymbol {Q))}$ and right-signum ${\displaystyle {\boldsymbol {R))}$.
In the special case where ${\displaystyle \mathbb {K} =\mathbb {R} ,\ n=2,}$ and the (invertible) matrix ${\displaystyle {\boldsymbol {A))=\left[{\begin{array}{rr}a&-b\\b&a\end{array))\right]}$, which identifies with the (nonzero) complex number ${\displaystyle a+\mathrm {i} b=c}$, then the signum matrices satisfy ${\displaystyle {\boldsymbol {Q))={\boldsymbol {P))=\left[{\begin{array}{rr}a&-b\\b&a\end{array))\right]/|c|}$ and identify with the complex signum of ${\displaystyle c}$, ${\displaystyle \operatorname {sgn} c=c/|c|}$. In this sense, polar decomposition generalizes to matrices the signum-modulus decomposition of complex numbers.
### Signum as a generalized function
At real values of ${\displaystyle x}$, it is possible to define a generalized function–version of the signum function, ${\displaystyle \varepsilon (x)}$ such that ${\displaystyle \varepsilon (x)^{2}=1}$ everywhere, including at the point ${\displaystyle x=0}$, unlike ${\displaystyle \operatorname {sgn} }$, for which ${\displaystyle (\operatorname {sgn} 0)^{2}=0}$. This generalized signum allows construction of the algebra of generalized functions, but the price of such generalization is the loss of commutativity. In particular, the generalized signum anticommutes with the Dirac delta function[6] ${\displaystyle \varepsilon (x)\delta (x)+\delta (x)\varepsilon (x)=0\,;}$ in addition, ${\displaystyle \varepsilon (x)}$ cannot be evaluated at ${\displaystyle x=0}$; and the special name, ${\displaystyle \varepsilon }$ is necessary to distinguish it from the function ${\displaystyle \operatorname {sgn} }$. (${\displaystyle \varepsilon (0)}$ is not defined, but ${\displaystyle \operatorname {sgn} 0=0}$.) |
Inverse of a Matrix using Gauss-Jordan Elimination
by M. Bourne
In this section we see how Gauss-Jordan Elimination works using examples.
You can re-load this page as many times as you like and get a new set of numbers each time. You can also choose a different size matrix (at the bottom of the page).
(If you need some background first, go back to the Introduction to Matrices).
Choose the matrix size you are interested in and then click the button.
Matrix A:
2×2 matrix
3×3 matrix
4×4 matrix
5×5 matrix
The randomly-generated example appears below.
Continues below
Example
Find the inverse of the matrix A using Gauss-Jordan elimination.
A = 11 10 3 6 5 7 12 8 9
Our Procedure
We write matrix A on the left and the Identity matrix I on its right separated with a dotted line, as follows. The result is called an augmented matrix.
We include row numbers to make it clearer.
11 10 3 6 5 7 12 8 9
1 0 0 Row[1] 0 1 0 Row[2] 0 0 1 Row[3]
Next we do several row operations on the 2 matrices and our aim is to end up with the identity matrix on the left, like this:
1 0 0 0 1 0 0 0 1
? ? ? Row[1] ? ? ? Row[2] ? ? ? Row[3]
(Technically, we are reducing matrix A to reduced row echelon form, also called row canonical form).
The resulting matrix on the right will be the inverse matrix of A.
Our row operations procedure is as follows:
1. We get a "1" in the top left corner by dividing the first row
2. Then we get "0" in the rest of the first column
3. Then we need to get "1" in the second row, second column
4. Then we make all the other entries in the second column "0".
We keep going like this until we are left with the identity matrix on the left.
Let's now go ahead and find the inverse.
Solution
11 10 3 6 5 7 12 8 9
1 0 0 Row[1] 0 1 0 Row[2] 0 0 1 Row[3]
New Row [1]
Divide Row [1] by 11 (to give us a "1" in the desired position):
This gives us:
1 0.9091 0.2727 6 5 7 12 8 9
0.0909 0 0 Row[1] 0 1 0 Row[2] 0 0 1 Row[3]
New Row [2]
Row[2] − 6 × Row[1] (to give us 0 in the desired position):
6 − 6 × 1 = 0
5 − 6 × 0.9091 = -0.4545
7 − 6 × 0.2727 = 5.3636
0 − 6 × 0.0909 = -0.5455
1 − 6 × 0 = 1
0 − 6 × 0 = 0
This gives us our new Row [2]:
1 0.9091 0.2727 0 -0.4545 5.3636 12 8 9
0.0909 0 0 Row[1] -0.5455 1 0 Row[2] 0 0 1 Row[3]
New Row [3]
Row[3] − 12 × Row[1] (to give us 0 in the desired position):
12 − 12 × 1 = 0
8 − 12 × 0.9091 = -2.9091
9 − 12 × 0.2727 = 5.7273
0 − 12 × 0.0909 = -1.0909
0 − 12 × 0 = 0
1 − 12 × 0 = 1
This gives us our new Row [3]:
1 0.9091 0.2727 0 -0.4545 5.3636 0 -2.9091 5.7273
0.0909 0 0 Row[1] -0.5455 1 0 Row[2] -1.0909 0 1 Row[3]
New Row [2]
Divide Row [2] by -0.4545 (to give us a "1" in the desired position):
This gives us:
1 0.9091 0.2727 0 1 -11.8 0 -2.9091 5.7273
0.0909 0 0 Row[1] 1.2 -2.2 0 Row[2] -1.0909 0 1 Row[3]
New Row [1]
Row[1] − 0.9091 × Row[2] (to give us 0 in the desired position):
1 − 0.9091 × 0 = 1
0.9091 − 0.9091 × 1 = 0
0.2727 − 0.9091 × -11.8 = 11
0.0909 − 0.9091 × 1.2 = -1
0 − 0.9091 × -2.2 = 2
0 − 0.9091 × 0 = 0
This gives us our new Row [1]:
1 0 11 0 1 -11.8 0 -2.9091 5.7273
-1 2 0 Row[1] 1.2 -2.2 0 Row[2] -1.0909 0 1 Row[3]
New Row [3]
Row[3] − -2.9091 × Row[2] (to give us 0 in the desired position):
0 − -2.9091 × 0 = 0
-2.9091 − -2.9091 × 1 = 0
5.7273 − -2.9091 × -11.8 = -28.6
-1.0909 − -2.9091 × 1.2 = 2.4
0 − -2.9091 × -2.2 = -6.4
1 − -2.9091 × 0 = 1
This gives us our new Row [3]:
1 0 11 0 1 -11.8 0 0 -28.6
-1 2 0 Row[1] 1.2 -2.2 0 Row[2] 2.4 -6.4 1 Row[3]
New Row [3]
Divide Row [3] by -28.6 (to give us a "1" in the desired position):
This gives us:
1 0 11 0 1 -11.8 0 0 1
-1 2 0 Row[1] 1.2 -2.2 0 Row[2] -0.0839 0.2238 -0.035 Row[3]
New Row [1]
Row[1] − 11 × Row[3] (to give us 0 in the desired position):
1 − 11 × 0 = 1
0 − 11 × 0 = 0
11 − 11 × 1 = 0
-1 − 11 × -0.0839 = -0.0769
2 − 11 × 0.2238 = -0.4615
0 − 11 × -0.035 = 0.3846
This gives us our new Row [1]:
1 0 0 0 1 -11.8 0 0 1
-0.0769 -0.4615 0.3846 Row[1] 1.2 -2.2 0 Row[2] -0.0839 0.2238 -0.035 Row[3]
New Row [2]
Row[2] − -11.8 × Row[3] (to give us 0 in the desired position):
0 − -11.8 × 0 = 0
1 − -11.8 × 0 = 1
-11.8 − -11.8 × 1 = 0
1.2 − -11.8 × -0.0839 = 0.2098
-2.2 − -11.8 × 0.2238 = 0.4406
0 − -11.8 × -0.035 = -0.4126
This gives us our new Row [2]:
1 0 0 0 1 0 0 0 1
-0.0769 -0.4615 0.3846 Row[1] 0.2098 0.4406 -0.4126 Row[2] -0.0839 0.2238 -0.035 Row[3]
We have achieved our goal of producing the Identity matrix on the left. So we can conclude the inverse of the matrix A is the right hand portion of the augmented matrix:
A−1 = -0.0769 -0.4615 0.3846 0.2098 0.4406 -0.4126 -0.0839 0.2238 -0.035
Things to Note
1. The above explanation shows all steps. A human can usually take a few shortcuts. Also, sometimes there is already a "1" or a "0" in the correct position, and in those cases, we would not need to do anything for that step.
2. Always write down what you are doing in each step - it is very easy to get lost!
3. I have shown results correct to 4 decimal place, but best possible accuracy was used throughout. Be aware that small errors from rounding will accumulate throughout the problem. Always use full calculator accuracy! (Make full use of your calculator's memory.)
4. Very occasionally there are strange results because of the computer's internal representation of numbers. That is, it may store "1" as 0.999999999872.
See another?
You can go back up to the top of the page and choose another example.
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# Probability of 5 of them having the same birthday while the other 10 having different birthday.
There are 15 students. Assume that there are 365 days in a year, what is the probability that 5 of them have the same birthday on any day of a year (the other 10 students have different birthday)?
$$P\left(1^{st\ }in\ 5\ to\ have\ same\ birthday\right)=\frac{365}{365}=1$$ The rest of the 4 students then need to have same birthday as the first student, making probability for each of them, $$\frac{1}{365}$$. $$P=\frac{365}{365}\times\frac{1}{365}\times\frac{1}{365}\times \frac{1}{365}\times\frac{1}{365}\times\frac{364}{365}\times \frac{363}{365}\times\frac{362}{365}\times\frac{361}{365}\times\frac{360}{365}\times\frac{359}{365}\times\frac{358}{365}\times\frac{357}{365}\times\frac{356}{365}\times\frac{355}{365} = 4.839 \times 10^{-11}$$
Initially, I am using this way to figure out the probability and I am exploring another way.
$$356^{15} = 2.7189 \times 10^{38}$$
This shows the 365 possibilities for their birthday. There are 11 ways for the birthdays to be selected which decreases after each birthday being selected.
$$N_{1} = 365 \times 364 \times363\times362\times361\times360\times359\times358\times357\times356\times355 = 1.3157 \times10^{28}$$ $$N_2 = {{15}\choose{11}} = 1365$$ $$N_2$$ is that there are 11 possible combinations in the 15 students.
$$P = \frac{{N_1}{N_2}}{D} = 6.605 \times 10^{-8}$$
Can someone please explain why is the right way of doing it and why one of them is wrong?
Your first solution is almost correct; you just have to include the choice of which 5 of the 15 students would share the same birthday; i.e. the answer should be
$$\left[ 4.389 \times 10^{-11} \right] \cdot \binom{15}{5} \approx 1.453 \times 10^{-7}$$
The second solution doesn't work because in your computation $$365 \times 364 \times \cdots \times 355$$, you're assigning an unnecessary ordering to the 10 dates chosen for the people with different birthdays. Also, the second step of calculating $$\binom{15}{11}$$ doesn't work - which 11 people are being chosen?
Note a simple method to count the number of ways is to break the process into smaller steps:
Step (1): Choose the birth date that five people will share - this can be done in $$365$$ ways.
Step (2): Choose which five people will share the above date - this can be done in $$\binom{15}{5}$$ ways.
Step (3): Choose the birth dates for the 10 remaining people - this can be done in $$\binom{364}{10}$$ ways.
Step (4): Assign the 10 birth dates to the 10 remaining people - this can be done in $$10!$$ ways.
Thus the required probability is:
$$\frac{365 \binom{15}{5} \binom{364}{10} 10!}{365^{15}} \approx 1.453 \times 10^{-7}$$
• Thank you for clarifying and pointing out my mistake ^^ Commented Nov 6, 2020 at 6:51 |
## Median
Median means the middle number.
For the given set of numbers, first we have to arrange them in order (either in ascending or descending). The middle number denotes the this.
Example 1:
Find the middle of {3,9,5,12,7}
Solution:
I Step: Arrange them in ascending order 3,5,7,9,12
II Step: The middle value of the given 5(odd) numbers is the value of the 3rd number, that is, 7 is the answer.
Example 2:
Find the middle of {2, 13, 15, 6, 24, 9, 11, 14, 8, 21, 3, 7,17}
Solution:
I step: Arrange them in ascending order 2,3, 6, 7, 8, 9, 11, 13, 14, 15, 17,21,24.
II step: Since there are 13 (odd) numbers, the middle value is the value of the 7th number . So 11 is the answer.
Note:
In the above two examples the given Set has only odd number of values. But if we have to find the middle value of a set which has even number of values, we have to follow the step I as usual and after that in step II we have to find the average of the middle two values.
Example 3:
Find the median value of {1, 5, 7,10}
Solution:
Step I: Already the numbers are arranged in ascending order.
Step II:
• The value of middle numbers are 5 and 7.
• We have to find the average of 5 and 7.
• which is (5+7)/2 = 12/2 =6
• So the middle value is 6.
Example 4:
Find the middle value of {2, 22, 21, 12, 15, 20, 5, 14, 7, 3,1, 10, 4,11}
Solution:
Step I: Arrange them in ascending order.
1,2,3,4,5,7,10,11,12,14,15,20,21,22.
Step II:
• The value of the middle numbers are 10 and 11
• Find the average of 10 and 11
• (10+11)/2 = 21/2 = 10.5
• So the middle value is 10.5
Note: In the above two example we have the middle values which are not given in the original set. But half the set is below the middle value and half the set is above the middle value. So our answer is correct.
Practice problems:
1. Find the middle value of {200, 250, 150, 100, 350.}
2. Find the middle value of { 2,3,4,7,8,1,5,19, 11, 10}
3. Find the middle value of {11,12,13,14,15,16,19,20,21,22,23}
Solutions
1. 200
2. 6
3. 16
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# 1.10: Pythagorean Theorem for Solving Right Triangles
Difficulty Level: At Grade Created by: CK-12
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Practice Pythagorean Theorem for Solving Right Triangles
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You are out on the playground with friends playing a game of tetherball. In this game, a ball is attached by a rope to the top of a pole. Each person is trying to hit the ball in a different direction until it wraps the rope completely around the pole. The first person to get the ball wrapped around the pole in their direction is the winner. You can see an example of a tetherball game on the right hand side of the picture shown here:
You notice that the rope attached to the tetherball is 1 meter long, and that the angle between the rope and the pole is 35\begin{align*}35^\circ\end{align*} degrees. Can you use that information to find out how far the ball is from the pole?
At the end of this Concept, you'll know how to solve this problem.
### Guidance
You can use your knowledge of the Pythagorean Theorem and the six trigonometric functions to solve a right triangle. Because a right triangle is a triangle with a 90 degree angle, solving a right triangle requires that you find the measures of one or both of the other angles. How you solve for these other angles, as well as the lengths of the triangle's sides, will depend on how much information is given.
#### Example A
Solve the triangle shown below.
Solution:
We need to find the lengths of all sides and the measures of all angles. In this triangle, two of the three sides are given. We can find the length of the third side using the Pythagorean Theorem:
82+b264+b2b2b=102=100=36=±6b=6\begin{align*}8^2 + b^2 & = 10^2\\ 64 + b^2 & = 100\\ b^2 & = 36\\ b & = \pm 6 \Rightarrow b = 6\end{align*}
(You may have also recognized that this is a “Pythagorean Triple,” 6, 8, 10, instead of using the Pythagorean Theorem.)
You can also find the third side using a trigonometric ratio. Notice that the missing side, b\begin{align*}b\end{align*}, is adjacent to A\begin{align*}\angle{A}\end{align*}, and the hypotenuse is given. Therefore we can use the cosine function to find the length of b\begin{align*}b\end{align*}:
cos53.130.6b=adjacent sidehypotenuse=b10=b10=0.6(10)=6\begin{align*}\cos 53.13^\circ & = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{b}{10}\\ 0.6 & = \frac{b}{10}\\ b & = 0.6(10) = 6\end{align*}
#### Example B
Solve the triangle shown below.
Solution:
In this triangle, we need to find the lengths of two sides. We can find the length of one side using a trig ratio. Then we can find the length of the third side by using a trig function with the information given originally and a different trig function. Because the side we found is an approximation, using the Pythagorean Theorem would not yield the most accurate answer for the other missing side. Therefore, we should use a trig function with the original information to find the length of the third side instead. Only use the given information when solving right triangles.
We are given the measure of A\begin{align*}\angle A\end{align*}, and the length of the side adjacent to A\begin{align*}\angle A\end{align*}. If we want to find the length of the hypotenuse, c\begin{align*}c\end{align*}, we can use the cosine ratio:
cos40cos40ccos40c=adjacenthypotenuse=6c=6c=6=6cos407.83\begin{align*}\cos 40^\circ & = \frac{adjacent}{hypotenuse} = \frac{6}{c}\\ \cos 40^\circ & = \frac{6}{c}\\ c \cos 40^\circ & = 6\\ c & = \frac{6}{\cos 40^\circ} \approx 7.83\end{align*}
If we want to find the length of the other leg of the triangle, we can use the tangent ratio. This will give us the most accurate answer because we are not using approximations.
tan40a=oppositeadjacent=a6=6tan405.03\begin{align*}\tan 40^\circ & = \frac{opposite}{adjacent} = \frac{a}{6}\\ a & = 6 \tan 40^\circ \approx 5.03\end{align*}
#### Example C
Solve the triangle shown below.
Solution:
In this triangle, we have the length of one side and one angle. Therefore, we need to find the length of the other two sides. We can start with a trig function:
tan30tan307tan30b=oppositeadjacent=b7=b7=b=7tan304.04\begin{align*}\tan 30^\circ & = \frac{opposite}{adjacent} = \frac{b}{7}\\ \tan 30^\circ & = \frac{b}{7}\\ 7 \tan 30^\circ & = b\\ b & = 7 \tan 30^\circ \approx 4.04\end{align*}
We can then use another trig relationship to find the length of the hypotenuse:
sin30sin30csin30c=oppositehypotenuse=4.04c=4.04c=4.04=4.04sin308.08\begin{align*}\sin 30^\circ & = \frac{opposite}{hypotenuse} = \frac{4.04}{c}\\ \sin 30^\circ & = \frac{4.04}{c}\\ c \sin 30^\circ & = 4.04\\ c & = \frac{4.04}{\sin 30^\circ} \approx 8.08\end{align*}
### Vocabulary
Sine: The sine of an angle in a right triangle is a relationship found by dividing the length of the side opposite the given angle by the length of the hypotenuse.
Cosine: The cosine of an angle in a right triangle is a relationship found by dividing the length of the side adjacent the given angle by the length of the hypotenuse.
Tangent: The tangent of an angle in a right triangle is a relationship found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.
### Guided Practice
1. Solve the triangle shown below:
2. Solve the triangle shown below:
3. Solve the triangle shown below:
Solutions:
1. Since the angle given is 40\begin{align*}40^\circ\end{align*}, and the length of the side opposite the angle is 9, we can use the tangent function to determine the length of the side adjacent to the angle:
tan40=9aa=9tan40a=9.839a=10.73\begin{align*} \tan 40^\circ = \frac{9}{a}\\ a = \frac{9}{\tan 40^\circ}\\ a = \frac{9}{.839}\\ a = 10.73\\ \end{align*}
We can then use another trig function to find the length of the hypotenuse:
sin40=9cc=9sin40c=9.643c=13.997\begin{align*} \sin 40^\circ = \frac{9}{c}\\ c = \frac{9}{\sin 40^\circ}\\ c = \frac{9}{.643}\\ c = 13.997\\ \end{align*}
Finally, the other angle in the triangle can be found either by a trigonometric relationship, or by recognizing that the sum of the internal angles of the triangle have to equal 180\begin{align*}180^\circ\end{align*}:
90+40+θ=180θ=1809040θ=50\begin{align*} 90^\circ + 40^\circ + \theta = 180^\circ\\ \theta = 180^\circ - 90^\circ - 40^\circ\\ \theta = 50^\circ\\ \end{align*}
2. Since this triangle has two sides given, we can start with the Pythagorean Theorem to find the length of the third side:
a2+b2=c282+b2=172b2=17282b2=28964=225b=15\begin{align*} a^2 + b^2 = c^2\\ 8^2 + b^2 = 17^2\\ b^2 = 17^2 - 8^2\\ b^2 = 289- 64 = 225\\ b = 15\\ \end{align*}
With this knowledge, we can work to find the other two angles:
tanB=158tanB=1.875B=tan11.87561.93\begin{align*} \tan \angle{B} = \frac{15}{8}\\ \tan \angle{B} = 1.875\\ \angle{B} = \tan^{-1} 1.875 \approx 61.93^\circ\\ \end{align*}
And the final angle is: 1809061.93=28.07\begin{align*} 180^\circ - 90^\circ - 61.93^\circ = 28.07^\circ \end{align*}
3. There are a number of things known about this triangle. Since we know all of the internal angles, there are a few different ways to solve for the unknown sides. Here let's use the 60\begin{align*}60^\circ\end{align*} angle to find the unknown sides:
tan60=a4a=4tan60a=(4)(1.73)=6.92\begin{align*} \tan 60^\circ = \frac{a}{4}\\ a = 4 \tan 60^\circ\\ a = (4)(1.73) = 6.92\\ \end{align*}
and
cos60=4hh=4cos60h=4.5h=8\begin{align*} \cos 60^\circ = \frac{4}{h}\\ h = \frac{4}{\cos 60^\circ}\\ h = \frac{4}{.5}\\ h = 8\\ \end{align*}
So we have found that the lengths of the sides are 4, 8, and 6.92.
### Concept Problem Solution
From our knowledge of how to solve right triangles, we can set up a triangle with the rope and the pole, like this:
From this, it is straightforward to set up a trig relationship for sine that can help:
sin35=opposite1(1)sin35=oppositeopposite.5736\begin{align*} \sin 35^\circ = \frac{opposite}{1}\\ (1)\sin 35^\circ = opposite\\ opposite \approx .5736 \end{align*}
### Practice
Use the picture below for questions 1-3.
1. Find mA\begin{align*}m\angle A\end{align*}.
2. Find mB\begin{align*}m\angle B\end{align*}.
3. Find the length of AC.
Use the picture below for questions 4-6.
1. Find mA\begin{align*}m\angle A\end{align*}.
2. Find mC\begin{align*}m\angle C\end{align*}.
3. Find the length of AC.
Use the picture below for questions 7-9.
1. Find mA\begin{align*}m\angle A\end{align*}.
2. Find mB\begin{align*}m\angle B\end{align*}.
3. Find the length of BC.
Use the picture below for questions 10-12.
1. Find mA\begin{align*}m\angle A\end{align*}.
2. Find mB\begin{align*}m\angle B\end{align*}.
3. Find the length of AB.
Use the picture below for questions 13-15.
1. Find mA\begin{align*}m\angle A\end{align*}.
2. Find mC\begin{align*}m\angle C\end{align*}.
3. Find the length of BC.
4. Explain when to use a trigonometric ratio to find missing information about a triangle and when to use the Pythagorean Theorem.
5. Is it possible to have a triangle that you must use cosecant, secant, or cotangent to solve?
6. What is the minimum information you need about a triangle in order to solve it?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
cosine
The cosine of an angle in a right triangle is a value found by dividing the length of the side adjacent the given angle by the length of the hypotenuse.
sine
The sine of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the hypotenuse.
Tangent
The tangent of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.
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## Introduction to Trigonometry
Short Q and A
Q1: If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.
Answer: sin 3A = cos (A – 26°)
∵ sin 3A = cos (90° – 3A)
∴ cos (90° – 3A) = cos (A – 26°)
Since
90° – 3A and A – 26° are both acute angles, therefore,
90° – 3A = A – 26°
A = 29°
Q2: Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Answer: cot 85° + cos 75° = cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
Q3: Evaluate cos 48° − sin 42°
Answer: cos 48° − sin 42°
= cos (90°− 42°) − sin 42°
= sin 42° − sin 42°
= 0
Q4: Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° − sin 38° sin 52° = 0
(i) tan 48° tan 23° tan 42° tan 67°
= tan (90° − 42°) tan (90° − 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°)
= (1) (1)
= 1
(ii) cos 38° cos 52° − sin 38° sin 52°
= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52°
= sin 52° sin 38° − sin 38° sin 52°
= 0
Q5: If tan 2A = cot (A− 18°), where 2A is an acute angle, find the value of A.
Answer: Given that, tan 2A = cot (A− 18°)
cot (90° − 2A) = cot (A −18°)
90° − 2A = A− 18°
108° = 3A
A = 36°
Q6: If tan A = cot B, prove that A + B = 90°
Answer: Given that, tan A = cot B
tan A = tan (90° − B)
A = 90° − B
A + B = 90°
Q7: If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.
Answer: Given that, sec 4A = cosec (A − 20°)
cosec (90° − 4A) = cosec (A − 20°)
90° − 4A= A− 20°
110° = 5A
A = 22°
Q8: Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Answer: sin 67° + cos 75°
= sin (90° − 23°) + cos (90° − 15°)
= cos 23° + sin 15°
Q9: Show that (1 - cos²θ)cosec²θ = 1
Answer: LHS = (1 - cos²θ)cosec²θ
∵ sin²θ + cos²θ = 1
∴ = sin²θ cosec²θ
= sin²θ × 1/ sin²θ = 1
Q10: sec θ (1 - sin θ)(sec θ + tan θ) = 1
Answer: LHS = sec θ (1 - sin θ)(sec θ + tan θ)
= (sec θ − sec θ sin θ) (sec θ + tan θ)
= (sec θ − sin θ/cos θ) (sec θ + tan θ)
= (sec θ − tan θ) (sec θ + tan θ)
= sec²θ − tan²θ
= 1 = RHS
Q11: Show that sin⁶θ + cos⁶θ = 1 - 3sin²θcos²θ
Answer: LHS = sin⁶θ + cos⁶θ
= (sin²θ)³ + (cos²θ)³
= (sin²θ + cos²θ)(sin⁴θ + cos⁴θ - sin²θcos²θ)
= (1)[(sin²θ)² + (cos²θ)² + 2sin²θcos²θ - 3sin²θcos²θ]
= [(sin²θ + cos²θ)² - 3sin²θcos²θ]
= 1 - 3sin²θcos²θ = RHS
Q12: Prove that cosec(65° + θ) − sec(25° − θ) − tan(55° − θ) + cot(35° + θ) = 0
Answer: cosec(65° + θ) = cosec { 90° − (25° − θ)} = sec(25° − θ)
cot( 35° + θ ) = cot{ 90° − (55° − θ)} = tan(55° − θ)
∴ LHS = cosec(65° + θ) − sec(25° − θ) − tan(55° − θ) + cot(35° + θ)
= sec(25°− θ ) − sec( 25°− θ) − tan(55° − θ) + tan(55° − θ)
= 0 = RHS |
Power of a Product Property of Exponents
To find a power of a product, find the power of each factor and then multiply. In general, ${\left(ab\right)}^{m}={a}^{m}\cdot {b}^{m}$.
Example 1:
Simplify ${\left(3t\right)}^{4}$
${\left(3t\right)}^{4}={3}^{4}\cdot {t}^{4}=81{t}^{4}$
Suppose you want to multiply two powers with the same exponent but different bases. By using the commutative property of multiplication, you can rewrite the rule as
${a}^{m}\cdot {b}^{m}={\left(ab\right)}^{m}$.
Example 2:
Simplify ${3}^{2}\cdot {4}^{2}$
$\begin{array}{l}{3}^{2}\cdot {4}^{2}=\left(3\cdot 3\right)\cdot \left(4\cdot 4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(3\cdot 4\right)\left(3\cdot 4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={12}^{2}\end{array}$
In other words, you can keep the exponent the same and multiply the bases. |
Math Concepts
# Integers: Concepts, Properties and Examples
2.4k views
1 Introduction 2 What are Integers? 3 Operations of Integers 4 Properties of Integers 5 Application of Integers 6 Summary 7 Frequently Asked Questions (FAQs)
## Introduction
• Closure property under different operations in integers
• Commutative property under different operations
• Associative property of integers under different operations
• Distributive property of multiplication over addition and subtraction of integers
### Integers: Concepts, Properties and Examples-PDF
This article will help you learn what integers are, and its use. You will also learn the Closure property, Commutative property, Associative property, and Distributive property. Here is a downloadable PDF to explore more.
## What are Integers?
A quick look at the numbers before getting to learn what integers are. All digits are grouped together as Numbers. Numbers are classified into various types and a few are listed below,
• Natural Numbers
• Whole Numbers
• Integers, etc.,
Natural Numbers starts from 1 and the series goes endless
Whole Numbers starts with 0 and the series goes endless
Integers comprise the Natural and the Whole numbers. The difference it makes is in holding both positive and negative values.
If Natural Numbers are 1 2 3 4 5 . . .and Whole Numbers are 0 1 2 3 4 5 . . .
Then Integers are. . . -5 -4 -3 -2 -1 0 1 2 3 4 5 . . .
Where 0 being the center of the series and the movement of numbers takes place in both the directions. Moving left from 0 adds a negative sign to numbers while moving right from 0 will add a positive sign to the numbers. In short, a positive number is not denoted with symbol +. The absence of symbol + represents that the number is positive, treated as default.
Example 7, means this number is +7 a positive integer
However, a negative number (-) sign should be placed in front of the number.
Example -7, means the number 7 in a negative value a negative integer, so finds its place towards the left-hand side of 0 in the series.
. . . -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 . . .
0 is neither positive nor negative integer
The series is consecutive in nature be it towards the right or towards the left.
Towards left means, a series of negative consecutive integers, while towards the right means series of positive consecutive integers.
## Operations of Integers
This session discusses how operations determine the outcome of integer values.
Addition and subtraction of integers using number line and Rules
Follow the below rules to operate on integers
• RULE 1 - Move towards the right, while adding
• RULE 2 - Move towards left, while subtracting
### Illustration 1
-4 + 5 = 1
The above sum is represented using the number line below
(-4) + 5 = ?
The sum represents -4 being ADDED with 5
Rule 1, states that for addition move towards the right
So,
Starting from -4 moving 5 points towards the right
(-4) + 5 = 1 which is a positive integer
### Illustration 2
3 – 5 = -2
The above sum is represented using the number line below
3 – 5 = ?
The sum represents (-5) being SUBTRACTED from 3
Rule 2, states that for subtraction, move towards left
So,
Starting from 3 moving 5 points towards left
3 – 5 = -2 which is a negative integer
Moving on, the idea of a number line for larger numbers becomes impossible to use.
So memorize the below rules,
• RULE 3 - If two numbers have the same sign, add them and keep the sign
• RULE 4 - If two numbers have different signs, subtract and use the sign of the larger number
### Illustration 3
24 + 56 = ?
24 is a positive integer
56 is also a positive integer
RULE 3, states that if 2 integers have the same sign, then add and keep the sign
(+24) + (+56) = +80
### Illustration 4
-24 + -56 = ?
24 is a negative integer
56 is a negative integer
RULE 3, states that if 2 integers have the same sign, then add and keep the sign
(-24) + (-56) = -80
### Illustration 5
24 - 56 = ?
24 is a positive integer
56 is also a negative integer
RULE 4, states that If two numbers have different signs, subtract and use the sign of the larger number
24 - 56 = -32
Subtraction value is 32
- sign, because the larger value from 24 and 56 is 56. Hence the result should carry - sign as per RULE 4
### Illustration 6
-24 + 56 = ?
24 is a negative integer
56 is also a positive integer
RULE 4, states that If two numbers have different signs, subtract and use the sign of the larger number
(-24) + 56 = 32
Subtraction value is 32
+ sign, because the larger value from 24 and 56 is 56. Hence the result should carry + sign as per RULE 4
The answer is +32, in other forms written as 32.
## Multiplication and division of integers using Rules
• RULE 5 - When you multiply or divide 2 numbers having the same sign, the answer is positive
• RULE 6 - When you multiply or divide 2 numbers having different signs, the answer is negative
NOTE - Do not confuse with rules of addition and subtraction with multiplication and division
30 x 2 = ?
Apply Rule 5
30 x 2 = 60
- 30 x 2 = ?
Apply Rule 6
30 x 2 = -60
30 ÷ 2 =?
Apply Rule 5
30 ÷ 2 = 15
30 ÷ (-2) = ?
Apply Rule 6
30 ÷ (-2) = -15
## Exponents on negative integers
• Rule 7 - Even exponent on a negative number gives a positive answer
• Rule 8 - Odd exponent on a negative number gives a negative answer
(-5)2 = ?
Apply Rule 7
-5 x -5 = 25
### Illustration 12
(-5)3 = ?
Apply Rule 8
(-5) x (-5) x (-5)
Step 1
(-5) x (-5) x (-5)
25 x (-5)
Step 2
25 x (-5)
-125
## Properties of Integers
This session will discuss the change in properties of the integers w.r.t operators used
### Closure property
Closure property means, when an operator is used in between integers and the resulting answer is also an integer, the property is said to be closed when that particular operator is used.
But the closure property holds good only for certain operators. We will look into detail about this property below:
Where a and b are 2 integers
### Illustration 13
a + b = an integer
If a is 3 and b is 5
Then, a + b = 3 + 5 = 8
8 is an integer
### Illustration 14
a + b = an integer
If a is -3 and b is 5
Then, a + b = -3 + 5 = 2
2 is an integer
### Illustration 15
a - b = an integer
If a is 3 and b is 5
Then, a - b = 3 - 5 = -2
-2 is an integer
### Illustration 16
a - b = an integer
If a is -3 and b is - 5
Then, a - b = (-3) -(- 5) = 2 (RULE 3)
2 is an integer
### Illustration 17
a x b = an integer
If a is 3 and b is 5
Then, a x b = 3 x 5 = 15
15 is an integer
### Illustration 18
a x b = an integer
If a is 3 and b is -5
Then, a x b = 3 x (-5) = -15 (RULE 6)
-15 is an integer
### Illustration 19
a ÷ b ≠ an integer
If a is 4 and b is 2
Then, a ÷ b = 4 ÷ 2 = 2
However, if a is 2 and b is 4
Then, a ÷ b = 2 ÷ 4 = 1/2
1/2 is not an integer
So, not all the integers when divided will result as an integer.
Hence, Division is not a closed property of Integer.
### Learning outcome
• Integers are closed under Addition, Subtraction, and Multiplication.
• Integers are not closed under Division.
### Commutative property
The commutative property, as the name suggests (commute ~ commutative), we could commute in any direction. Commutative property means when an operator is used between two numbers or integers, the final answer should be the same when worked either way, from left to right or right to left.
But, the commutative property holds good for only certain operators and the details are given below,
Where a and b are 2 integers
### Illustration 20
a + b = b + a
If a is 4 and b is 2
Then,
a + b = b + a
4 + 2 = 2 + 4
6 = 6
### Illustration 21
a - b ≠ b - a
If a is 4 and b is 2
Then,
a - b ≠ b - a
4 - 2 ≠ 2 - 4
2 ≠ -2
Subtraction is not commutative for integers
### Illustration 22
a x b = b x a
If a is 4 and b is 2
Then,
a x b = b x a
4 x 2 = 2 x 4
8 = 8
Multiplication is commutative for integers
### Illustration 23
a ÷ b ≠ b ÷ a
If a is 4 and b is 2
Then,
a ÷ b ≠ b ÷ a
4 ÷ 2 ≠ 2 ÷ 4
2 ≠ 1/2
The division is not commutative for integers
### Learning outcome
• Addition and Multiplication are commutative for Integers
• Subtraction and Division are not commutative for Integers
### Associative property
The associative property, as the name suggests (Associative ~ Association) means that the grouping of terms will not affect the final answer when grouped in different ways.
But the Associative property holds good only for certain operators. The details are given below:
Where a b c are integers
The distributive property is applicable
1. When a group involving the addition of 2 integers is multiplied by another integer. Here it involves 2 operations. Addition and Multiplication. It indicates that we can either add first and then multiply or Multiply first and then add next.
2. When a group involving the subtraction of 2 integers is multiplied by another integer. So it involves 2 operations. Subtraction and Multiplication. It indicates that we can either subtract first and then multiply or Multiply first and then subtract next.
### Illustration 24
a + [ b + c ] = [ a + b ] + c
If a is 6, b is 4 and c is 2
Then,
a + [ b + c ] = [ a + b ] + c
6 + [ 4 + 2 ] = [ 6 + 4 ] + 2
6 + 6 = 10 + 2
12 = 12
Addition property is associative for integers
### Illustration 25
a - [ b - c ] ≠ [ a - b ] - c
If a is 6, b is 4 and c is 2
Then,
a - [ b - c ] ≠ [ a - b ] - c
6 - [ 4 - 2 ] ≠ [ 6 - 4 ] - 2
6 - 2 ≠ 2 - 2
4 ≠ 0
Subtraction property is not associative for integers
### Illustration 26
a x [ b x c ] = [ a x b ] x c
If a is 6, b is 4 and c is 2
Then,
a x [ b x c ] = [ a x b ] x c
6 x [ 4 x 2 ] = [ 6 x 4 ] x 2
6 x 8 = 24 x 2
48 = 48
Multiplication property is associative for integers
### Illustration 27
a ÷ [ b ÷ c ] ≠ [ a ÷ b ] ÷ c
If a is -8, b is 4 and c is -2
Then,
[(-8) / 4] / -2 = -8 / [(4) / (-2)]
a ÷ [ b ÷ c ] ≠ [ a ÷ b ] ÷ c
-8 ÷ [(4) ÷ (-2)] ≠ [ (-8) ÷ 4 ] ÷ (-2)
-8 ÷ (-2) ≠ (-2) ÷ (-2)
4 ≠ 1
Division property is not associative for integers
### Learning outcome
• Addition and Multiplication are Associative for Integers
• Subtraction and Division are Not Associative for Integers
### Distributive property
As the name (distributive ~ distribution) indicates, a factor or a number or an integer along with the operation multiplication (‘x’), is getting distributed to the numbers separated by either addition or subtraction inside the parenthesis.
Where a b and c are integers
The distributive property is applicable when a group involving the addition of 2 integers is multiplied by another integer. So it involves 2 operations. Addition and multiplication.
It indicates that we can either add first and then multiply or Multiply first and then add next.
### Illustration 28
a x ( b + c ) = ( a x b ) + ( a x c )
If a is -2, b is 4 and c is 3
Then,
a x ( b + c ) = ( a x b ) + ( a x c ) is
-2 x (4 + 3) = [(-2) x 4] + [(-2) x 3]
(-2) x 7 = (-8) + (-6)
-14 = -14
### Illustration 29
a x ( b - c ) = ( a x b ) - ( a x c )
If a is 4, b is 3 and c is 6
Then,
a x ( b - c ) = ( a x b ) - ( a x c ) is
4 x (3 – 6) = 4 x 3 – 4 x 6
4 x -3 = 12 - 24
-12 = -12
Multiplication is a distributive property of integers.
### Learning outcome
• Multiplication is distributive over addition for Integers
• Multiplication is distributive over subtraction for Integers
## Application of Integers
### Illustration 30
Elevator or Lifts we use in daily life
### Illustration 31
Atmospheric Temperature
Temperatures of countries differ based on geographic locations.
For instance, Greenland recorded the coldest temperature of -66 °C and the hottest temperature of 30 °C depending upon the season.
### Illustration 32
Usage of positive and negative integers in coding
Int main ()
{
int a;
printf (“Enter the integer”);
scanf (“%d”, &a );
If (a > 0)
printf (“%d is a positive integer”, a);
elseif (a < 0)
printf (“%d is a negative integer”, a);
elseif (a == 0)
printf (“%d is zero”, a);
return 0;
}
Output
Run 1
Enter the integer = 10
10 is a positive integer
Run 2
Enter the integer = -10
-10 is a negative integer
Run 3
Enter the integer = 0
0 is zero
## Summary
Usage of symbols in the problems with Integers can be easily remembered with the below table:
The properties of integers explained in this document are summarised in the below table. This can be used as a reference while practising integers.
## What is an integer?
An integer is a series of numbers that range from -∞ (infinity) to +∞ (infinity) without decimal values.
## Are Even numbers Integers?
All even numbers are integers.
## Are Odd numbers Integers?
All odd numbers are integers.
## How do you identify, if a number is an Integer or not?
An integer is any number that has no decimals, though it may still be either positive or negative in nature.
## Is Zero an Integer?
Yes. Zero falls between -∞ (infinity) to +∞ (infinity) and hence it is an integer, though it neither carries a positive (+) or a negative (-) sign.
## Is Zero positive or a negative Integer?
Though Zero is an integer, it neither carries a positive (+) or a negative (-) sign. Any number added or subtracted with Zero gives the same number as result, and hence the sign for Zero is ignored or not considered.
## External References
-Written by Vishali B, Cuemath Teacher
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# Golden rectangle
A golden rectangle with longer side a and shorter side b, when placed adjacent to a square with sides of length a, will produce a similar golden rectangle with longer side a + b and shorter side a. This illustrates the relationship $\frac{a+b}{a} = \frac{a}{b} \equiv \varphi\,.$
In geometry, a golden rectangle is a rectangle whose side lengths are in the golden ratio, $1 : \tfrac{1 + \sqrt{5}}{2}$, which is $1:\varphi$ (the Greek letter phi), where $\varphi$ is approximately 1.618.
## Construction
A method to construct a golden rectangle. The square is outlined in red. The resulting dimensions are in the golden ratio.
A golden rectangle can be constructed with only straightedge and compass by four simple steps:
1. Construct a simple square.
2. Draw a line from the midpoint of one side of the square to an opposite corner.
3. Use that line as the radius to draw an arc that defines the height of the rectangle.
4. Complete the golden rectangle.
## Relation to regular polygons and polyhedra
A distinctive feature of this shape is that when a square section is removed, the remainder is another golden rectangle; that is, with the same aspect ratio as the first. Square removal can be repeated infinitely, in which case corresponding corners of the squares form an infinite sequence of points on the golden spiral, the unique logarithmic spiral with this property.
Three golden rectangles in an icosahedron
An alternative construction of the golden rectangle uses three polygons circumscribed by congruent circles: a regular decagon, hexagon, and pentagon. The respective lengths a, b, and c of the sides of these three polygons satisfy the equation a2 + b2 = c2, so line segments with these lengths form a right triangle (by the converse of the Pythagorean theorem). The ratio of the side length of the hexagon to the decagon is the golden ratio, so this triangle forms half of a golden rectangle.[1]
The convex hull of two opposite edges of a regular icosahedron forms a golden rectangle. The twelve vertices of the icosahedron can be decomposed in this way into three mutually-perpendicular golden rectangles, whose boundaries are linked in the pattern of the Borromean rings.[2]
## Applications
According to astrophysicist and mathematics popularizer Mario Livio, since the publication of Luca Pacioli's Divina Proportione in 1509,[3] when "with Pacioli's book, the Golden Ratio started to become available to artists in theoretical treatises that were not overly mathematical, that they could actually use,"[4] many artists and architects have been fascinated by the presumption that the golden rectangle is considered aesthetically pleasing. The proportions of the golden rectangle have been observed in works predating Pacioli's publication.[5] Architechural icons such as the Parthenon in Athens and the Alhambra in Granada have been shown to be based on the use of Golden rectangles. |
# Three Equivalent Conditions for a Ring to be a Field
## Problem 436
Let $R$ be a ring with $1$. Prove that the following three statements are equivalent.
1. The ring $R$ is a field.
2. The only ideals of $R$ are $(0)$ and $R$.
3. Let $S$ be any ring with $1$. Then any ring homomorphism $f:R \to S$ is injective.
## Proof.
We prove the equivalences $(1) \Leftrightarrow (2)$ and $(2) \Leftrightarrow (3)$.
### $(1) \implies (2)$
Suppose that $R$ is a field. Let $I$ be an ideal of $R$.
If $I=(0)$, then there is nothing to prove.
So assume that $I\neq (0)$.
Then there is a nonzero element $x$ in $I$.
Since $R$ is a field, we have $x^{-1}\in R$.
Since $I$ is an ideal, we have
$1=x^{-1}\cdot x\in I.$ This yields that $I=R$.
### $(2) \implies (1)$
Suppose now that the only ideals of $R$ are $(0)$ and $R$.
Let $x$ be a nonzero element of $R$. We show the existence of the inverse of $x$.
Consider the ideal $(x)=xR$ generated by $x$.
Since $x$ is nonzero, the ideal $(x)\neq 0$, and thus we have $(x)=R$ by assumption.
Thus, there exists $y\in R$ such that
$xy=1.$
So $y$ is the inverse element of $x$.
Hence $R$ is a field.
### $(2)\implies (3)$
Suppose that the only ideals of $R$ are $(0)$ and $R$.
Let $S$ be any ring with $1$ and $f:R\to S$ be any ring homomorphism.
Consider the kernel $\ker(f)$. The kernel $\ker(f)$ is an ideal of $R$, and thus $\ker(f)$ is either $(0)$ or $R$ by assumption.
If $\ker(f)=R$, then the homomorphism $f$ sends $1\in R$ to $0\in S$, which is a contradiction since any ring homomorphism between rings with $1$ sends $1$ to $1$.
Thus, we must have $\ker(f)=0$, and this yields that the homomorphism $f$ is injective.
### $(3) \implies (2)$
Suppose that statement 3 is true. That is, any ring homomorphism $f:R\to S$, where $S$ is any ring with $1$, is injective.
Let $I$ be a proper ideal of $R$: an ideal $I\neq R$.
Then the quotient $R/I$ is a ring with $1$ and the natural projection
$f:R\to R/I$ is a ring homomorphism.
By assumption, the ring homomorphism $f$ is injective, and hence we have
$(0)=\ker(f)=I.$ This proves that the only ideals of $R$ are $(0)$ and $R$.
##### If $R$ is a Noetherian Ring and $f:R\to R’$ is a Surjective Homomorphism, then $R’$ is Noetherian
Suppose that $f:R\to R'$ is a surjective ring homomorphism. Prove that if $R$ is a Noetherian ring, then so is... |
# CLASS-5SIMPLE INTEREST BY FORMULAE METHOD
SIMPLE INTEREST BY FORMULAE METHOD -
To calculate simple interest using the formula method, you can use the following formula:
Simple Interest (SI) = (Principal × Rate × Time) / 100
Where :-
• Principal refers to the initial amount of money borrowed or invested.
• Rate represents the interest rate per time period.
• Time denotes the duration for which the money is borrowed or invested.
To calculate the simple interest, follow these steps :-
1. Identify the values of Principal, Rate, and Time.
2. Substitute the values into the formula.
3. Multiply Principal, Rate, and Time.
4. Divide the product obtained in step 3 by 100.
5. The result will be the simple interest.
Here's an example to illustrate the calculation :-
Suppose you borrow \$5,000 at an interest rate of 8% per year for a period of 3 years.
Principal (P) = \$5,000,
Rate (R) = 8%,
Time (T) = 3 years,
Using the formula, we can calculate the simple interest (SI) :-
SI = (P × R × T) / 100 = (5000 × 8 × 3) / 100 = 1200
Therefore, the simple interest on the loan would be \$1,200.
The formula for calculating simple interest is:
I = P ⋅ R ⋅ T
Where :-
• I represents the interest amount
• P is the principal amount (the initial amount of money)
• R is the rate of interest (expressed as a decimal)
• T is the time period (in years)
To find the simple interest using this formula, follow these steps:
1. Determine the principal amount (P), which is the initial sum of money.
2. Identify the rate of interest (R). This can be given as a percentage, so make sure to convert it to a decimal by dividing by 100. For example, if the interest rate is 5%, R would be 0.05.
3. Determine the time period (T), usually in years.
Once you have these values, you can plug them into the formula to calculate the simple interest (I).
For example, let's say you have a principal amount of \$1,000, an interest rate of 5% (0.05 as a decimal), and a time period of 3 years. Using the formula, the calculation would be:
= 1000⋅0 X 05⋅3 = 150 I = 1000⋅0 X 05⋅3 = 150
Therefore, the simple interest on a \$1,000 principal amount at an interest rate of 5% over a period of 3 years would be \$150.
1) Find the S.I. by formula method :-
a) \$ 600 for 1 year at 4 % per annum
Ans.)
P = \$ 600,
R = 4 %,
T = 1 year
P × R × T 600 × 4 × 1
S.I. = ------------- = \$ ---------------
100 100
[600 and 100 have the common factor 100]
6 × 4 × 1
= \$ ------------ = \$ 24 (Ans.)
1 × 1
1
b) Rs. 1800 for 2 years at 4 ------ % per annum
2
Ans.)
1
R = 4 ------ % [Convert the mixed number into improper fraction]
2
(4 X 2) + 1 9
= -------------- % = ------ %
2 2
P = \$ 1800, T = 2 years
1800 × 9 × 2
S.I. = \$ ---------------
100 × 2
18 × 9 × 2
= \$ -------------
1 × 2 × 1
[1800 and 100 have the common factor 100 & equal factors in numerator and denominator is 2]
18 × 9 × 1
= \$ -------------- = \$ 162 (Ans.)
1 × 1 × 1
c) Rs. 9000 for 146 days at 5 % per annum (1 year = 365 days)
Ans.)
146
T = 146 days = -------- year [Since, 1 year = 365 days]
365
P = \$ 9000, R = 5 %
9000 × 5 × 146
S.I. = \$ ------------------
100 × 365
[365 and 5 have the common factor 5, 146 and 73 have the common factor 73 & 9000 and 100 have the common factor 100]
2 × 90 × 1
= \$ -------------- = \$ 180 (Ans.)
1 × 1 × 1
d) Rs. 18,000 for 6 months at 10 % per annum
6
T = 6 months = ------ year [Since, 1 year = 12 months.]
12
[6 and 12 have the common factor 6]
P = \$ 18,000, R = 10 %
18000 × 10 × 1
S.I. = \$ -----------------
100 × 2
180 × 5 × 1
= \$ ---------------
1 × 1 × 1
[18000 and 100 have the common factor 100 & 10 and 2 have the common factor 2]
= \$ 900 (Ans.) |
Laws of Exponents
The laws of exponents refer to a set of rules for multiplying, dividing and even raising exponents to powers to make these operations easier.
The product rule is used to multiply exponents with the same base. It states that to multiply two numbers with the same base all you do is add their exponents.
Example: 22 · 23 = 25 25=32
2 · 2 · 2 · 2 · 2 = 32
am · an = am+n
The quotient rule is used to divide exponents with the same base. It states that to divide two numbers with the same base all you do is subtract their exponents.
Example: 105 ¸ 103 = 102 102=100
105=100,000
103=1,000
100,000 ¸ 1,000 = 100
am ¸ an = am-n
3·2=6
The power rule is used to raise a number already written as an exponent to another power. It states that to raise an exponent to a power you multiply the exponents.
Example: (33)2 = 33 · 33 33 · 33 = 36
If you wrote this out you would get 3 · 3 · 3 · 3 · 3 · 3 or 36
(am)n = am·n
The power of a product rule is as follows
To find the power of a product, find the power of each factor and multiply
(ab)m= ambm
And the power of a quotient rule is as follows
To find the value of a quotient, find the power of each number and divide. |
# Learning Objectives
### Learning Objectives
By the end of this section, you will be able to do the following:
• Describe how motors and meters work in terms of torque on a current loop
• Calculate the torque on a current-carrying loop in a magnetic field
Motors are the most common application of magnetic force on current-carrying wires. Motors have loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft. Electrical energy is converted to mechanical work in the process. (See Figure 5.26.)
Figure 5.26 Torque on a current loop. A current-carrying loop of wire attached to a vertically rotating shaft feels magnetic forces that produce a clockwise torque as viewed from above.
Let us examine the force on each segment of the loop in Figure 5.26 to find the torques produced about the axis of the vertical shaft. (This will lead to a useful equation for the torque on the loop.) We take the magnetic field to be uniform over the rectangular loop, which has width $ww$ and height $l.l.$ First, we note that the forces on the top and bottom segments are vertical and, therefore, parallel to the shaft, producing no torque. Those vertical forces are equal in magnitude and opposite in direction, so that they also produce no net force on the loop. Figure 5.27 shows views of the loop from above. Torque is defined as $τ=rFsinθ,τ=rFsinθ,size 12{τ= ital "rF""sin"θ} {}$ where $FF size 12{F} {}$ is the force, $rr$ is the distance from the pivot that the force is applied, and $θθ$ is the angle between $rr$ and $F.F.$ As seen in Figure 5.27(a), RHR-1 gives the forces on the sides to be equal in magnitude and opposite in direction, so that the net force is again zero. However, each force produces a clockwise torque. Since $r=w/2,r=w/2,$ the torque on each vertical segment is $(w/2)Fsinθ,(w/2)Fsinθ,$ and the two add to give a total torque.
5.19 $τ=w2Fsinθ+w2Fsinθ=wFsinθτ=w2Fsinθ+w2Fsinθ=wFsinθ size 12{τ= { {w} over {2} } F"sin"θ+ { {w} over {2} } F"sin"θ= ital "wF""sin"θ} {}$
Figure 5.27 Top views of a current-carrying loop in a magnetic field. (a) The equation for torque is derived using this view. Note that the perpendicular to the loop makes an angle $θθ size 12{θ} {}$ with the field that is the same as the angle between $w/2w/2 size 12{w/2} {}$ and $F.F.size 12{F} {}$ (b) The maximum torque occurs when $θθ size 12{θ} {}$ is a right angle and $sinθ=1.sinθ=1.size 12{"sin"θ=1} {}$ (c) Zero (minimum) torque occurs when $θθ size 12{θ} {}$ is zero and $sinθ=0.sinθ=0.$ (d) The torque reverses once the loop rotates past $θ=0.θ=0.$
Now, each vertical segment has a length $ll size 12{l} {}$ that is perpendicular to $B,B,size 12{B} {}$ so that the force on each is $F=IlB.F=IlB.size 12{F= ital "IlB"} {}$ Entering $FF size 12{F} {}$ into the expression for torque yields
5.20 $τ=wIlBsinθ.τ=wIlBsinθ. size 12{τ= ital "wIlB""sin"θ} {}$
If we have a multiple loop of $NN size 12{N} {}$ turns, we get $NN size 12{N} {}$ times the torque of one loop. Finally, note that the area of the loop is $A=wl;A=wl; size 12{A= ital "wl"} {}$ the expression for the torque becomes
5.21 $τ=NIABsinθ.τ=NIABsinθ. size 12{τ= ital "NIAB""sin"θ} {}$
This is the torque on a current-carrying loop in a uniform magnetic field. This equation can be shown to be valid for a loop of any shape. The loop carries a current $I,I,size 12{I} {}$ has $NN size 12{N} {}$ turns, each of area $A,A,size 12{A} {}$ and the perpendicular to the loop makes an angle $θθ size 12{θ} {}$ with the field $B.B.size 12{B} {}$ The net force on the loop is zero.
### Example 5.5Calculating Torque on a Current-Carrying Loop in a Strong Magnetic Field
Find the maximum torque on a 100-turn square loop of a wire of 10.0 cm on a side that carries 15.0 A of current in a 2.00-T field.
Strategy
Torque on the loop can be found using $τ=NIABsinθ.τ=NIABsinθ.size 12{τ= ital "NIAB""sin"θ} {}$ Maximum torque occurs when $θ=90ºθ=90º$ and $sinθ=1.sinθ=1.size 12{"sin"θ=1} {}$
Solution
For $sinθ=1,sinθ=1,size 12{"sin"θ=1} {}$ the maximum torque is
5.22 $τmax=NIAB.τmax=NIAB. size 12{τ rSub { size 8{"max"} } = ital "NIAB"} {}$
Entering known values yields
5.23 τmax=10015.0 A0.100 m22.00 T=30.0 N⋅m.τmax=10015.0 A0.100 m22.00 T=30.0 N⋅m.alignl { stack { size 12{τ rSub { size 8{"max"} } = left ("100" right ) left ("15" "." 0" A" right ) left (0 "." "100"" m" rSup { size 8{2} } right ) left (2 "." "00"" T" right )} {} # " "="30" "." "0 N" cdot m "." {} } } {}
Discussion
This torque is large enough to be useful in a motor.
The torque found in the preceding example is the maximum. As the coil rotates, the torque decreases to zero at $θ=0.θ=0.size 12{θ=0} {}$ The torque then reverses its direction once the coil rotates past $θ=0.θ=0.size 12{θ=0} {}$ (See Figure 5.27(d).) This means that, unless we do something, the coil will oscillate back and forth about equilibrium at $θ=0.θ=0.size 12{θ=0} {}$ To get the coil to continue rotating in the same direction, we can reverse the current as it passes through $θ=0θ=0 size 12{θ=0} {}$ with automatic switches called brushes. (See Figure 5.28.)
Figure 5.28 (a) As the angular momentum of the coil carries it through $θ=0,θ=0,size 12{θ=0} {}$ the brushes reverse the current to keep the torque clockwise. (b) The coil will rotate continuously in the clockwise direction, with the current reversing each half revolution to maintain the clockwise torque.
Meters, such as those in analog fuel gauges on a car, are another common application of magnetic torque on a current-carrying loop. Figure 5.29 shows that a meter is very similar in construction to a motor. The meter in the figure has its magnets shaped to limit the effect of $θθ size 12{θ} {}$ by making $BB size 12{B} {}$ perpendicular to the loop over a large angular range. Thus the torque is proportional to $II size 12{I} {}$ and not $θ.θ.size 12{θ} {}$ A linear spring exerts a counter-torque that balances the current-produced torque. This makes the needle deflection proportional to $I.I.size 12{I} {}$ If an exact proportionality cannot be achieved, the gauge reading can be calibrated. To produce a galvanometer for use in analog voltmeters and ammeters that have a low resistance and respond to small currents, we use a large loop area $A,A,size 12{A} {}$ high magnetic field $B,B,size 12{B} {}$ and low-resistance coils.
Figure 5.29 Meters are very similar to motors but only rotate through a part of a revolution. The magnetic poles of this meter are shaped to keep the component of $BB size 12{B} {}$ perpendicular to the loop constant, so that the torque does not depend on $θθ size 12{θ} {}$ and the deflection against the return spring is proportional only to the current $I.I.size 12{I} {}$ |
# How do you prove 2tan(x)sec(x) = (1/(1-sin(x))) - (1/(1+sin(x)))?
Dec 14, 2015
See explanation...
#### Explanation:
Recall these identities involving trig functions:
Quotient Identity: $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$
Reciprocal Identity: $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$
Pythagorean Identity: $1 - {\sin}^{2} \left(x\right) = {\cos}^{2} \left(x\right)$
Using these, we can rewrite the identity:
$2 \tan \left(x\right) \sec \left(x\right) = \frac{1}{1 - \sin \left(x\right)} - \frac{1}{1 + \sin \left(x\right)}$
$2 \sin \frac{x}{\cos} ^ 2 \left(x\right) = \frac{1}{1 - \sin \left(x\right)} - \frac{1}{1 + \sin \left(x\right)}$
We then make common denominators:
$2 \sin \frac{x}{\cos} ^ 2 \left(x\right) = \frac{1 + \sin \left(x\right)}{1 - {\sin}^{2} \left(x\right)} - \frac{1 - \sin \left(x\right)}{1 - {\sin}^{2} \left(x\right)}$
$2 \sin \frac{x}{\cos} ^ 2 \left(x\right) = \frac{1 + \sin \left(x\right) - \left(1 - \sin \left(x\right)\right)}{1 - {\sin}^{2} \left(x\right)}$
$2 \sin \frac{x}{\cos} ^ 2 \left(x\right) = 2 \sin \frac{x}{1 - {\sin}^{2} \left(x\right)}$
And simplify using the Pythagorean Identity from above:
$2 \sin \frac{x}{\cos} ^ 2 \left(x\right) = 2 \sin \frac{x}{\cos} ^ 2 \left(x\right)$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 3.3: The First Derivative Test
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
A student will be able to:
• Find intervals where a function is increasing and decreasing.
• Apply the First Derivative Test to find extrema and sketch graphs.
## Introduction
In this lesson we will discuss increasing and decreasing properties of functions, and introduce a method with which to study these phenomena, the First Derivative Test. This method will enable us to identify precisely the intervals where a function is either increasing or decreasing, and also help us to sketch the graph. Note on notation: The symbol ϵ\begin{align*}\epsilon\end{align*} and \begin{align*} \in \end{align*} are equivalent and denote that a particular element is contained within a particular set.
Definition
A function f\begin{align*}f\end{align*} is said to be increasing on [a,b]\begin{align*}[a,b]\end{align*} contained in the domain of f\begin{align*}f\end{align*} if f(x1)f(x2)\begin{align*}f(x_1) \le f(x_2)\end{align*} whenever x1x2\begin{align*}x_{1} \le x_{2}\end{align*} for all x1,x2[a,b].\begin{align*}x_1, x_2 \in [a,b].\end{align*} A function f\begin{align*}f\end{align*} is said to be decreasing on [a,b]\begin{align*}[a,b]\end{align*} contained in the domain of f\begin{align*}f\end{align*} if f(x1)f(x2)\begin{align*}f(x_1) \ge f(x_2)\end{align*} whenever x1x2\begin{align*}x_1 \ge x_2\end{align*} for all x1,x2[a,b].\begin{align*}x_1, x_2 \in [a,b].\end{align*}
If f(x1)<f(x2)\begin{align*}f(x_1) < f(x_2)\end{align*} whenever x1<x2\begin{align*}x_1 < x_2\end{align*} for all x1,x2[a,b],\begin{align*}x_1, x_2 \in [a,b],\end{align*} then we say that f\begin{align*}f\end{align*} is strictly increasing on [a,b].\begin{align*}[a,b].\end{align*} If f(x1)>f(x2)\begin{align*}f(x_1) > f(x_2)\end{align*} whenever x1>x2\begin{align*}x_1 > x_2\end{align*} for all x1,x2[a,b],\begin{align*}x_1, x_2 \in [a,b],\end{align*} then we say that f\begin{align*}f\end{align*} is strictly decreasing on [a,b].\begin{align*}[a,b].\end{align*}
We saw several examples in the Lesson on Extreme and the Mean Value Theorem of functions that had these properties.
Example 1:
The function f(x)=x3\begin{align*}f(x) = x^3\end{align*} is strictly increasing on (,+)\begin{align*}(-\infty, +\infty)\end{align*}:
Example 2:
The function indicated here is strictly increasing on (0,a)\begin{align*}(0,a)\end{align*} and (b,c)\begin{align*}(b,c)\end{align*} and strictly decreasing on (a,b)\begin{align*}(a,b)\end{align*} and (c,d).\begin{align*}(c,d).\end{align*}
We can now state the theorems that relate derivatives of functions to the increasing/decreasing properties of functions.
Theorem: If f\begin{align*}f\end{align*} is continuous on interval [a,b],\begin{align*}[a,b],\end{align*} then:
1. If f(x)>0\begin{align*}f'(x) > 0\end{align*} for every x[a,b],\begin{align*}x \in [a,b],\end{align*} then f\begin{align*}f\end{align*} is strictly increasing in [a,b].\begin{align*}[a,b].\end{align*}
2. If f(x)<0\begin{align*}f'(x) < 0\end{align*} for every x[a,b],\begin{align*}x \in [a,b],\end{align*} then f\begin{align*}f\end{align*} is strictly decreasing in [a,b].\begin{align*}[a,b].\end{align*}
Proof: We will prove the first statement. A similar method can be used to prove the second statement and is left as an exercise to the student.
Consider x1,x2[a,b]\begin{align*}x_1, x_2 \in [a,b]\end{align*} with x1<x2.\begin{align*}x_1 < x_2.\end{align*} By the Mean Value Theorem, there exists c(x1,x2)\begin{align*}c \in (x_1,x_2)\end{align*} such that
f(x2)f(x1)=(x2x1)f(c).
By assumption, f(x)>0\begin{align*}f'(x) > 0\end{align*} for every x[a,b]\begin{align*}x \in [a,b]\end{align*}; hence f(c)>0.\begin{align*}f'(c) > 0.\end{align*} Also, note that x2x1>0.\begin{align*}x_2 - x_1 > 0.\end{align*}
Hence f(x2)f(x1)>0\begin{align*}f(x_2) - f(x_1) > 0\end{align*} and f(x2)>f(x1).\begin{align*}f(x_2) > f(x_1).\end{align*}
We can observe the consequences of this theorem by observing the tangent lines of the following graph. Note the tangent lines to the graph, one in each of the intervals \begin{align*}(0,a), (a,b),\end{align*} \begin{align*}(b,+\infty).\end{align*}
Note first that we have a relative maximum at \begin{align*}x = a\end{align*} and a relative minimum at \begin{align*}x = b.\end{align*} The slopes of the tangent lines change from positive for \begin{align*}x \in (0,a)\end{align*} to negative for \begin{align*}x \in (a,b)\end{align*} and then back to positive for \begin{align*}x \in (b, +\infty)\end{align*}. From this we example infer the following theorem:
First Derivative Test
Suppose that \begin{align*}f\end{align*} is a continuous function and that \begin{align*}x = c\end{align*} is a critical value of \begin{align*}f.\end{align*} Then:
1. If \begin{align*}f'\end{align*} changes from positive to negative at \begin{align*}x = c,\end{align*} then \begin{align*}f\end{align*} has a local maximum at \begin{align*}x = c.\end{align*}
2. If \begin{align*}f'\end{align*} changes from negative to positive at \begin{align*}x = c,\end{align*} then \begin{align*}f\end{align*} has a local minimum at \begin{align*}x = c.\end{align*}
3. If \begin{align*}f'\end{align*} does not change sign at \begin{align*}x = c,\end{align*} then \begin{align*}f\end{align*} has neither a local maximum nor minimum at \begin{align*}x = c.\end{align*}
Proof of these three conclusions is left to the reader.
Example 3:
Our previous example showed a graph that had both a local maximum and minimum. Let’s reconsider \begin{align*}f(x) = x^3\end{align*} and observe the graph around \begin{align*}x = 0.\end{align*} What happens to the first derivative near this value?
We observe that the tangent lines to the graph are positive on both sides of \begin{align*}x=0\end{align*}. The first derivative test (\begin{align*}f'(x)=3x^2\end{align*}) verifies this fact, and that the slopes of the tangent line are positive for all nonzero \begin{align*}x\end{align*}. Although \begin{align*}f'(0)=0\end{align*}, and so \begin{align*}f\end{align*} has a critical value at \begin{align*}x=0\end{align*}, the third part of the First Derivative Test tells us that the failure of \begin{align*}f'\end{align*} to change sign at \begin{align*}x=0\end{align*} means that \begin{align*}f\end{align*} has neither a local minimum nor a local maximum at \begin{align*}x=0\end{align*}.
Example 4:
Let's consider the function \begin{align*}f (x) = x^2 + 6 x - 9\end{align*} and observe the graph around \begin{align*}x = -3.\end{align*} What happens to the first derivative near this value?
We observe that the slopes of the tangent lines to the graph change from negative to positive at \begin{align*}x = -3.\end{align*} The first derivative test verifies this fact. Note that the slopes of the tangent lines to the graph are negative for \begin{align*}x \in (-\infty, -3)\end{align*} and positive for \begin{align*}x \in (-3, -\infty).\end{align*}
## Lesson Summary
1. We found intervals where a function is increasing and decreasing.
2. We applied the First Derivative Test to find extrema and sketch graphs.
For more examples on determining whether a function is increasing or decreasing (9.0), see Math Video Tutorials by James Sousa, Determining where a function is increasing and decreasing using the first derivative (10:05).
For a video presentation of increasing and decreasing trigonometric functions and relative extrema (9.0), see Math Video Tutorials by James Sousa, Increasing and decreasing trig functions, relative extrema (6:03).
For more information on finding relative extrema using the first derivative (9.0), see Math Video Tutorials by James Sousa, Finding relative extrema using the first derivative (6:19).
## Review Questions
In problems #1–2, identify the intervals where the function is increasing, decreasing, or is constant. (Units on the axes indicate single units).
1. Give the sign of the following quantities for the graph in #2.
1. \begin{align*}f'(-3)\end{align*}
2. \begin{align*}f'(1)\end{align*}
3. \begin{align*}f'(3)\end{align*}
4. \begin{align*}f'(4)\end{align*}
For problems #4–6, determine the intervals in which the function is increasing and those in which it is decreasing. Sketch the graph.
1. \begin{align*}f(x) = x^2 - \frac{1} {x}\end{align*}
2. \begin{align*}f(x) = (x^2 - 1)^5\end{align*}
3. \begin{align*}f(x) = (x^2 - 1)^4\end{align*}
For problems #7–10:
a. Use the First Derivative Test to find the intervals where the function increases and/or decreases
b. Identify all max, mins, or relative max and mins
c. Sketch the graph
1. \begin{align*}f(x) = -x^2 -4x - 1\end{align*}
2. \begin{align*}f(x) = x^3 + 3x^2 - 9x + 1\end{align*}
3. \begin{align*} f(x) = x^\frac{2} {3} (x - 5)\end{align*}
4. \begin{align*} f(x) = 2x{\sqrt{x^2 + 1}}\end{align*}
## Date Created:
Feb 23, 2012
Oct 30, 2015
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CBSE Class 9 Maths Notes Chapter 9 Quadrilaterals Pdf free download is part of Class 9 Maths Notes for Quick Revision. Here we have given NCERT Class 9 Maths Notes Chapter 9 Quadrilaterals. https://www.cbselabs.com/quadrilaterals-class-9-notes/
## CBSE Class 9 Maths Notes Chapter 9 Quadrilaterals
Quadrilateral Class 9 Notes Chapter 9
1. Quadrilateral: A plane figure bounded by four lines segment AB, BC, CD and DA is called a quadrilateral.
The sum of the angles of a quadrilateral is 360°.
2. Various Types of Quadrilaterals
(i) Parallelogram: A quadrilateral in which opposite sides are parallel is called parallelogram and it is written as Parallelogram.
(ii) Rectangle: A parallelogram each of whose angle is 90°, is called a rectangle.
(iii) Square: A rectangle having all sides equal, is called a square.
(iv) Rhombus: A parallelogram having all sides equal is called a rhombus.
(v) Trapezium: A quadrilateral in which two opposite sides are parallel and two opposite sides are non-parallel is called a trapezium.
If two non-parallel sides of a trapezium are equal, then it is called an isosceles trapezium.
(vi) Kite: A quadrilateral in which two pairs of adjacent are equal is known as the kite.
Quadrilaterals Class 9 Notes Chapter 9
3. Important Theorems
• The sum of all the four angles of a quadrilateral is 360°.
• A diagonal of a parallelogram divides it into two congruent triangles.
• A quadrilateral is a parallelogram if
• opposite sides are equal.
• opposite angles are equal,
• diagonals bisect each other.
• A quadrilateral is a parallelogram if
• its opposite angles are equal.
• its opposite sides are equal.
• its diagonals bisect each other.
• a pair of opposite sides is equal and parallel.
• Diagonals of a rectangle bisect each other and they are equal and vice-versa.
• Diagonals of a rhombus bisect each other at right angles and they are not equal and vice-versa.
• Diagonals of a square bisect each other at right angles and they are equal and vice-versa.
• The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. (Mid-point theorem)
• The line drawn through the mid-point of one side of a triangle, parallel to another side, intersects the third side at its mid-point. (By converse of mid-point theorem)
• The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, taken in order, is a parallelogram.
• In a quadrilateral, if diagonals bisect each other, then it forms a parallelogram.
We hope the given CBSE Class 9 Maths Notes Chapter 9 Quadrilaterals Pdf free download will help you. If you have any query regarding NCERT Class 9 Maths Notes Chapter 9 Quadrilaterals, drop a comment below and we will get back to you at the earliest. |
ANOVA PART I: The Introductory Guide to ANOVA
# ANOVA PART I: The Introductory Guide to ANOVA
In this blog, we are going to be discussing a statistical technique, ANOVA, which is used for comparison.
The basic principal of ANOVA is to test for differences among the mean of different samples. It examines the amount of variation within each of these samples and the amount of variation between the samples. ANOVA is important in the context of all those situations where we want to compare more than two samples as in comparing the yield of crop from several variety of seeds etc.
The essence of ANOVA is that the total amount of variation in a set of data is broken in two types:-
1. The amount that can be attributed to chance.
2. The amount which can be attributed to specified cause.
#### One-way ANOVA
Under the one-way ANOVA we compare the samples based on a single factor. For example productivity of different variety of seeds.
Stepwise process involved in calculation of one-way ANOVA is as follows:-
1. Calculate the mean of each sample X ̅
2. Calculate the super mean
3. Calculate the sum of squares between (SSB) samples
1. Divide the result by the degree of freedom between the samples to obtain mean square between (MSW) samples.
2. Now calculate variation within the samples i.e. sum of square within (SSW)
1. Calculate mean square within (MSW)
2. Calculate the F-ratio
3. Last but not the least calculate the total variation in the given samples i.e. sum of square for total variance.
Lets now solve a one-way ANOVA problem.
A,B and C are three different variety of seeds and now we need to check if there is any variation in their productivity or not. We will be using one-way ANOVA as there is a single factor comparison involved i.e. variety of seeds.
The f-ratio is 1.53 which lies within the critical value of 4.26 (calculated from the f-distribution table).
Conclusion:- Since the f-ratio lies within the acceptance region we can say that there is no difference in the productivity of the seeds and the little bit of variation that we see is caused by chance.
Two-way ANOVA will be discussed in my next blog so do comeback for the update.
Hopefully, you have found this blog informative, for more clarification watch the video attached down the blog. You can find more such posts on Data Science course topics, just keep on following the DexLab Analytics blog.
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## Wednesday, March 18, 2015
### Algebra: Multiplying Polynomials, Method 3
If you missed the introductory post to this series, you should go read it.
This is the last method I'll be sharing for how to multiply polynomials. I call this one the "Numerical" method because it is set-up like a standard multiplication problem with numbers. I do not find this method particularly quick (like "The Table" method), but it is organized and easy to understand because it builds upon a skill with which the majority of students are already comfortable.
Our first example is an easy one, as usual. :-)
Step 1: Write the polynomials in a column, aligning them at the right, with like terms stack (introduce missing terms with a zero coefficient), like you would a standard multiplication problem. Write the longer polynomial on top.
In this example, there are no missing terms. But, if there were, I'd introduce them (as in previous methods), with zero coefficients. Like terms should be aligned. Notice in the image above that the term (-2x) in the first polynomial is aligned in the same "column" as the (+x) in the second polynomial.
Step 2: Multiply the last term of the bottom polynomial by each term of the top polynomial, from right to left. Write these products below the solid line, aligned with the term from the top polynomial. This is just like a standard multiplication problem with numbers.
In this step, we are working with the last term of the bottom polynomial (circled in red in the image above). Make sure you include the sign along with the term when multiplying! First, this term is multiplied by the last term of the top polynomial (circled in green). The product is written below the term used from the top polynomial (underlined in green).
Continue multiplying the last term (red circle) of the bottom polynomial by each of the terms in the top polynomial, from right to left. These are shown in purple and blue, respectively.
Step 3: Multiply the next to last term of the bottom polynomial by each term of the top polynomial, from right to left. Write these products below the products from Step 2, aligning them one term from the right side (leave a space below the far-right term of the first row of products). This is just like a standard multiplication problem with numbers.
Continue in this same manner until every term of the bottom polynomial has been used. The number of rows in the "products" section (below the solid line) should be equal to the number of terms in the bottom polynomial.
Now, we are multiplying across the top polynomial using the second to last term of the bottom polynomial (circled in red in the image above). These products are aligned below the products from Step 2, but moved one "space" to the left. This ensures that like terms are aligned in the products! This should look very familiar to you as a standard multiplication problem.
In this example, the multiplying is complete. Every term of the bottom polynomial has been multiplied through the top polynomial. Notice, there are two rows of product polynomials below the solid line, which is equal to the number of terms in the bottom polynomial (2).
Step 4: Draw another solid line below the rows of product polynomials. Add these polynomials together. Like terms are added together. The final product polynomial is yielded. (If there are any terms with a zero coefficient, rewrite the product polynomial and remove them.)
This step is where like terms are combined to form the final product polynomial. Simply add the terms in each column (example, boxed in green, above). The final product polynomial is boxed in red.
This method can be difficult to explain, but it really does work exactly like multiplying numbers with multiple digits that is taught in elementary/primary school.
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And now for an example with missing terms and longer polynomials. :-)
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I hope you have found this series helpful. As always, if you have any questions, comments, or requests for future topics, leave a comment below. :-) |
# Computing: The Science of Nearly Everything
Computer Science…Research, Education and Policy
## Proof that the square root of 2 is irrational
with one comment
(N.B. this blog post is mainly directed at my undergraduate students for their first year Mathematics for Computing module!)
The square root of 2 ($\sqrt{2}$, often known as root 2) is the positive algebraic number that, when multiplied by itself, gives the number 2. Geometrically, the square root of 2 is the length of a diagonal across a square with sides of one unit of length; this follows from the Pythagorean theorem. A quick approximation for the square root of two is $\tfrac{99}{70}$ (despite having a denominator of only 70, it differs from the correct value by less than $\tfrac{1}{10000}$).
In the first few lectures we have covered the basics of logic and the propositional calculus, including the idea of reductio ad absurdum, or more specifically, proof by contradiction (indirect proof). One of the examples given in the lecture was proving that the square root of 2 is irrational using infinite descent:
Assume that $\sqrt{2}$ is rational; this means that we can represent it as the ratio of two integers:
$\dfrac{a}{b} = \sqrt{2}$
Then $\sqrt{2}$ can be written as an irreducible fraction $\tfrac{a}{b}$ such that $a$ and $b$ are coprime integers:
$\left(\dfrac{a}{b}\right)^2 = 2$
It follows that:
$\dfrac{a^2}{b^2} = 2$
$a^2 = 2b^2$
Therefore $a^2$ is even because it is equal to $2b^2$ ($2b^2$ is necessarily even because it is 2 times another whole number and even numbers are multiples of 2); it follows that $a$ must be even (as squares of odd integers are never even). Because a is even, there exists an integer $k$ that fulfils:
$a = 2k$
Substituting in $2k$ for $a$ in the earlier equation:
$2b^2 = (2k)^2$
$2b^2 = 4k^2$
$b^2 = 2k^2$
Because $2k^2$ is divisible by 2 and therefore even, and because $2k^2 = b^2$, it follows that $b^2$ is also even which means that $b$ is even. As we have shown that $a$ and $b$ are both even, this contradicts that $\tfrac{a}{b}$ is irreducible. $\square$
Because there is a contradiction, the original assumption that $\sqrt{2}$ is a rational number must be false. By the law of excluded middle, the opposite is proven: $\sqrt{2}$ is irrational.
This proof was hinted at by Aristotle, in his Analytica Priora, but it appeared first as a full proof in Euclid‘s Elements (as proposition 117 of Book X). There are a number of other methods of proving that the square root of 2 is irrational, including a simple geometric proof and proof by unique factorisation (using that fact that every integer greater than 1 has a unique factorisation into powers of primes). Check them out!
Written by Tom
30 October 2011 at 10:26 am
Posted in Mathematics
Tagged with , |
WRITING EXPRESSIONS INVOLVING PERCENT INCREASE AND DECREASE
Recall that whenever you see the percent symbol, $\,\%\,$, you can trade it in for a multiplier of $\frac{1}{100}$.
(Indeed, per-cent means per-one-hundred.)
For example, $\,20\%\,$ goes by all these names: $$\,20\% = 20\cdot\frac{1}{100}= \frac{20}{100} = \frac{2}{10} = \frac{1}{5} = 0.2$$
In particular, note that $\,100\% = 100\cdot\frac{1}{100} = 1\,$,
so $\,100\%\,$ is just another name for the number $\,1\,$.
Also recall that it's easy to go from percents to decimals:
just move the decimal point two places to the left.
For example: $\,20\% = 20.\% = 0.20$
It's good style to put a zero in the ones place (i.e., write $\ 0.20\$, not $\ .20\$).
To change from decimals to percents,
just move the decimal point two places to the right.
For example: $0.2 = 0.20 = 20.\% = 20\%$
The ‘Puddle Dipper’ memory device may be useful to you:
PuDdLe: to change from Percents to Decimals, move the decimal point two places to the Left.
DiPpeR: to change from Decimals to Percents, move the decimal point two places to the Right.
EXAMPLES:
Here, you will practice writing expressions involving percent increase and decrease, and related concepts.
Another name for the expression ‘$\,20\%\text{ of } x\,$’ is: $0.2x$
Why? The mathematical word ‘of ’ indicates multiplication, so:
$\,(20\%\text{ of } x) = (20\%)(x) = (0.2)(x) = 0.2x\,$.
Another name for the expression ‘$\,100\%\text{ of } x\,$’ is: $x$
Another name for the expression ‘$\,300\%\text{ of } x\,$’ is: $3x$
If $\,x\,$ increases by $\,20\%\,$, then the new amount is: $x + 0.2x = 1x + 0.2x = 1.2x$
If $\,x\,$ has a $\,20\%\,$ increase, then the new amount is: $1.2x$
If $\,x\,$ increases by $\,47\%\,$, then the new amount is: $x + 0.47x = 1.47x$
If $\,x\,$ decreases by $\,30\%\,$, then the new amount is: $x - 0.3x = 1x - 0.3x = 0.7x$
If $\,x\,$ has a $\,30\%\,$ decrease, then the new amount is: $0.7x$
If $\,x\,$ increases by $\,100\%\,$, then the new amount is: $x + x = 1x + 1x = 2x$
If $\,x\,$ increases by $\,182\%\,$, then the new amount is: $x + 1.82x = 2.82x$
If $\,x\,$ increases by $\,200\%\,$, then the new amount is: $x + 2x = 3x$
If $\,x\,$ doubles, then the new amount is: $2x$
If $\,x\,$ triples, then the new amount is: $3x$
If $\,x\,$ quadruples, then the new amount is: $4x$
If $\,x\,$ is halved, then the new amount is: $\displaystyle\frac{1}{2}x = 0.5x$
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Calculating Percent Increase and Decrease
Answers must be input in decimal form to be recognized as correct.
Also, you must exhibit good style by putting a zero in the ones place, as needed.
For example, input 0.5x , not (say) .5x or 1/2x .
(MAX is 11; there are 11 different problem types.) |
# What is the axis of symmetry and vertex for the graph y = x^2 + 6x + 13?
May 3, 2016
Axis of symmetry -> x = -3
Vertex -> (x,y)-> (-3, 4 )
#### Explanation:
Consider the general form $y = a {x}^{2} + b x + c$
Write the general form as $y = a \left({x}^{2} + \frac{b}{a} x\right) + c$
In your case $a = 1$
$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a} \to \left(- \frac{1}{2}\right) \times 6 = - 3}$
$\textcolor{b l u e}{\text{axis of symmetry } \to x = - 3}$
To find ${y}_{\text{vertex}}$ substitute $x = - 3$ in the original equation.
$\implies {y}_{\text{vertex}} = {\left(- 3\right)}^{2} + 6 \left(- 3\right) + 13$
$\textcolor{b l u e}{\implies {y}_{\text{vertex}} = + 4}$
$\textcolor{b r o w n}{\text{Vertex} \to \left(x , y\right) \to \left(- 3 , 4\right)}$ |
Grade Level: 4th Skill: Graphs and Charts Topic: X-Coordinates Goal: Understand that the length of a horizontal line segment equals the difference of the x-coordinates. Skill Description: Use subtraction to find the horizontal length of the line on a coordinate grid.
## Building Blocks/Prerequisites
### Sample Problems
(1) Graph the ordered pairs and connect the points. Find the length of the line segment. (2,6) and (15,6) (15 -2 = 13) (2) What is the length of the line segment with points (9, 8) and (18, 8) (18 – 9 = 9) (3) Use the coordinate grid and find the length of the line segment. [Graph the coordinates below on a coordinate grid] (4,5) and (7,5) (4) Which ordered pairs has the horizontal distance of 19 units: (21,6) (2,6) (6,21) (6,3) (6,3) (21,6) (a) (5) If Jay plots a point at (9, 3) where should he plot his second point to have a horizontal length of 4 units? (13,3 or 5,3)
### Learning Tips
(1) Math Vocabulary: Ordered pair- (x, y) a pair of numbers used to locate a point on a coordinate grid. x-axis- the horizontal line on a coordinate grid (line that goes from left to right) y-axis- the vertical line on a coordinate grid (line that goes from top to bottom) x-coordinate- the first number of an ordered pair, which tells how far to move horizontally along the x-axis. y-coordinate- the second number in an ordered pair, which tells how far to move vertically along the y-axis Horizontal line- a line that goes across from left to right or right to left i.e.) ---------------- Line segment- a line that has two endpoints i.e.) •---------------------• Coordinate grid- a map-like tool that is used to locate specific points. (2) Use a number line to explain length of the x-axis: ‹--׀--------•--------׀--------׀--------׀--------•--------׀--------׀--------׀--------׀---› 1 2 3 4 5 6 7 8 9 10 → → → → The length is 4 units (count the arrows from one number to the next). There are 4 units between 2 and 6. To find the horizontal distance between two points on a coordinate grid, count units or subtract the x-coordinates (6 – 2 =4 units). (3) Color-code the x-axis. Most children get confused on finding the length of the horizontal and vertical distance because they subtract the wrong coordinates. In the beginning color-code the x-coordinates, so they understand that those are the two numbers that are to be subtracted (Hint: the y-coordinate should be the same number) Use a red marker to outline the x-coordinate. For example, What is the horizontal length of (2, 3) and (4, 3)? If they see the x-coordinates color coded it will help them grasp the concept of subtracting the x-coordinates to find the length. (4) Relate coordinate grids to Maps: Use a real map and teach your child how to use the coordinate grid on a map to find cities in the US. On most maps, the coordinates are letters and numbers. To find a location they will be given a coordinate of C-2 and the city will be labeled at that point. Once they understand how to use coordinates to locate locations, have them find the horizontal distance between two cities. (5) Explain the parts of a coordinate grid: Draw a coordinate grid on large chart paper and identify the parts of the grid y-axis x-axis how to plot points (x-coordinate go across on the x-axis; y-coordinate go straight up from the x-coordinate to number on the y-axis) horizontal lines vertical lines Practice making horizontal lines. Parent will give the child two ordered pairs have the child graph and connect the points. Discuss why it is a horizontal line and not vertical. (6) Practice simple subtractions facts (timed drills, flash cards)
### Extra Help Problems
(1) Graph the ordered pairs and connect the points. Find the length of the line segment. (5, 7) and (18, 7) (13) (2) Graph the ordered pairs and connect the points. Find the length of the line segment. (16,3) and (19,3) (3) (3) Graph the ordered pairs and connect the points. Find the length of the line segment. (4,12) and (16,12) (12) (4) Graph the ordered pairs and connect the points. Find the length of the line segment. (9,9) and (18,9) (9) (5) Graph the ordered pairs and connect the points. Find the length of the line segment. (0,1) and (10,1) (10) (6) Graph the ordered pairs and connect the points. Find the length of the line segment. (15,0) and (8, 0) (7) (7) What is the length of the line segment with points (117, 10) and (93, 10)? (24) (8) What is the length of the line segment with points (57, 12) and (29, 12)? (28) (9) What is the length of the line segment with points (15, 15) and (9, 15)? (6) (10) What is the length of the line segment with points (66, 7) and (7, 7)? (59) (11) What is the length of the line segment with points (34, 66) and (33, 66)? (1) (12) What is the length of the line segment with points (12, 15) and (12, 15)? (0) (13) Use the coordinate grid and find the length of the line segment. (Do not show ordered pairs, only the graph) [Graph (4,6) and (8,6)] (14) Use the coordinate grid and find the length of the line segment. . (Do not show ordered pairs, only the graph) [Graph (11,1) and (3, 1)] (15) Use the coordinate grid and find the length of the line segment. . (Do not show ordered pairs, only the graph) [Graph (3,2) and (15, 2)] (16) Use the coordinate grid and find the length of the line segment. . (Do not show ordered pairs, only the graph) [Graph (1, 10) and (6, 10)] (17) Use the coordinate grid and find the length of the line segment. . (Do not show ordered pairs, only the graph) [Graph (7,8) and (9,8)] (18) Which ordered pairs has the horizontal distance of 37 units (13,8) and (45, 13) (8,13) and (45, 13) (21, 8) and (37, 8) (b) (19) Which ordered pairs has the horizontal distance of 2 units (6, 6) and (6, 4) (12,6) and (9, 6) (6, 6) and (4, 6) (c) (20) Which ordered pairs has the horizontal distance of 72 units (111,40) and (39, 40) (8, 40) and (9, 40) (40, 111) and (40, 39) (a) (21) Which ordered pairs has the horizontal distance of 23 units (46, 18) and (2, 18) (23, 18) and (46, 18) (23, 18) and (18, 46) (b) (22) Which ordered pairs has the horizontal distance of 14 units (22, 20) and (14, 20) (20, 22) and (14, 22) (8, 20) and (22, 20) (c) (23) If Allison plots a point at (17, 20) where should she plot her second point to have a horizontal length of 12 units? (29,20 or 5,20) (24) If Craig plots a point at (3, 9) where should he plot his second point to have a horizontal length of 21 units? (24,9) (25) If Sarah needed a horizontal length of 12 units and she plotted her points at (8, 24) and (8, 12), what mistake did she make? (She made a vertical length of 12 units not, horizontal) |
If x-y=2 is the equation of a chord of the circle x^2+y^2+2y=0.Find the equation of the circle of which this chord is a diameter.?
Nov 9, 2017
Given that
• the equation of the chord $\text{ } \textcolor{m a \ge n t a}{x - y = 2. \ldots . \left[1\right]}$
• the equation of the circle $\text{ } \textcolor{b l u e}{{x}^{2} + {y}^{2} + 2 y = 0. \ldots . . \left[2\right]}$
From these two equations we get
${\left(y + 2\right)}^{2} + {y}^{2} + 2 y = 0$
$\implies 2 {y}^{2} + 6 y + 4 = 0$
$\implies 2 {y}^{2} + 6 y + 4 = 0$
$\implies {y}^{2} + 3 y + 2 = 0$
$\implies {y}^{2} + 2 y + y + 2 = 0$
$\implies y \left(y + 2\right) + 1 \left(y + 2\right) = 0$
$\implies \left(y + 2\right) \left(y + 1\right) = 0$
So $y = - 1 \mathmr{and} - 2$
Inserting in [1] we get
$x = 1 \mathmr{and} 0$ respectively
So the coordinates of points intersections of the chord with the circles are $\left(1 , - 1\right) \mathmr{and} \left(0 , - 2\right)$
The line segment of the chord is also the diameter of a circle.
So equation of this circle having diameter $x - y = 2$ will be
$\frac{y + 1}{x - 1} \times \frac{y + 2}{x - 0} = - 1$
$\implies \left(y + 1\right) \times \left(y + 2\right) = - x \left(x - 1\right)$
$\textcolor{g r e e n}{\implies {x}^{2} + {y}^{2} - x + 3 y + 2 = 0. \ldots . \left[3\right]}$
This is the required equation of the circle
Nov 9, 2017
${x}^{2} + {y}^{2} - x + 3 y + 2 = 0 ,$
Explanation:
We will solve this Problem using the following Result R :
R : The Equation (eqn.) of a Circle that passes through
the Points (pt.) of Intersection of a Circle S & Line L
$S : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0 \mathmr{and} L : l x + m y + n = 0 ,$ is
$S + \lambda L : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c + \lambda \left(l x + m y + n\right) = 0 ,$
where, $\lambda \in \mathbb{R} .$
Let, $S : {x}^{2} + {y}^{2} + 2 y = 0 , \mathmr{and} L : x - y - 2 = 0.$
Note that, the Reqd. Circle, say $S ' ,$ passes through the pts. of
$S \cap L .$
Applying R, we may suppose that, $S ' : S + \lambda L = 0 , i . e . ,$
$S ' : {x}^{2} + {y}^{2} + 2 y + \lambda \left(x - y - 2\right) = 0.$
$\therefore S ' : {x}^{2} + \lambda x + {y}^{2} + \left(2 - \lambda\right) y = 2 \lambda .$
$\therefore S ' : {x}^{2} + \lambda x + {\lambda}^{2} / 4 + {y}^{2} + \left(2 - \lambda\right) y + {\left(2 - \lambda\right)}^{2} / 4 = 2 l a m \mathrm{da} + {\lambda}^{2} / 4 + {\left(2 - \lambda\right)}^{2} / 4 , \mathmr{and} ,$
$S ' : {\left(x + \frac{\lambda}{2}\right)}^{2} + {\left(y + \frac{2 - \lambda}{2}\right)}^{2} = 2 l a m \mathrm{da} + {\lambda}^{2} / 4 + {\left(2 - \lambda\right)}^{2} / 4.$
This shows that the Centre $C$ of $S '$ is $C \left(- \frac{\lambda}{2} , - \frac{2 - \lambda}{2}\right) .$
Now, given that $L$ is a diameter of $S ' \Rightarrow C \in L .$
$\Rightarrow - \frac{\lambda}{2} - \left(- \frac{2 - \lambda}{2}\right) = 2.$
$\therefore - \frac{\lambda}{2} + \frac{2 - \lambda}{2} = 2 , \mathmr{and} , - \lambda = 2 - 1 = 1 , i . e . , \lambda = - 1.$
Therefore, $S ' : {x}^{2} + {y}^{2} + 2 y - 1 \left(x - y - 2\right) = 0 , i . e . ,$
$S ' : {x}^{2} + {y}^{2} - x + 3 y + 2 = 0 ,$ is the reqd. eqn. of the Circle, as
already derived by Respected dk_ch Sir!
Enjoy Maths.! |
# 0.12 Probability
Page 1 / 8
This chapter covers principles of probability. After completing this chapter students should be able to: write sample spaces; determine whether two events are mutually exclusive; use the addition rule; calculate probabilities using tree diagrams and combinations; solve problems involving conditional probability; determine whether two events are independent.
## Chapter overview
In this chapter, you will learn to:
1. Write sample spaces.
2. Determine whether two events are mutually exclusive.
4. Calculate probabilities using both tree diagrams and combinations.
5. Do problems involving conditional probability.
6. Determine whether two events are independent.
## Sample spaces and probability
If two coins are tossed, what is the probability that both coins will fall heads? The problem seems simple enough, but it is not uncommon to hear the incorrect answer $1/3$ . A student may incorrectly reason that if two coins are tossed there are three possibilities, one head, two heads, or no heads. Therefore, the probability of two heads is one out of three. The answer is wrong because if we toss two coins there are four possibilities and not three. For clarity, assume that one coin is a penny and the other a nickel. Then we have the following four possibilities.
HH HT TH TT
The possibility HT, for example, indicates a head on the penny and a tail on the nickel, while TH represents a tail on the penny and a head on the nickel.
It is for this reason, we emphasize the need for understanding sample spaces.
An act of flipping coins, rolling dice, drawing cards, or surveying people are referred to as an experiment .
Sample Spaces
A sample space of an experiment is the set of all possible outcomes.
If a die is rolled, write a sample space.
A die has six faces each having an equally likely chance of appearing. Therefore, the set of all possible outcomes $S$ is
$\left\{1,2,3,4,5,6\right\}$ .
A family has three children. Write a sample space.
The sample space consists of eight possibilities.
$\left\{\text{BBB},\text{BBG},\text{BGB},\text{BGG},\text{GBB},\text{GBG},\text{GGB},\text{GGG}\right\}$
The possibility $\text{BGB}$ , for example, indicates that the first born is a boy, the second born a girl, and the third a boy.
We illustrate these possibilities with a tree diagram.
Two dice are rolled. Write the sample space.
We assume one of the dice is red, and the other green. We have the following 36 possibilities.
Green Red 1 2 3 4 5 6 1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
The entry (2, 5), for example, indicates that the red die shows a two, and the green a 5.
Now that we understand the concept of a sample space, we will define probability.
## Probability
For a sample space $S$ , and an outcome $A$ of $S$ , the following two properties are satisfied.
1. If $A$ is an outcome of a sample space, then the probability of $A$ , denoted by $P\left(A\right)$ , is between 0 and 1, inclusive.
$0\le P\left(A\right)\le 1$
2. The sum of the probabilities of all the outcomes in $S$ equals 1.
If two dice, one red and one green, are rolled, find the probability that the red die shows a 3 and the green shows a six.
Since two dice are rolled, there are 36 possibilities. The probability of each outcome, listed in [link] , is equally likely.
Since (3, 6) is one such outcome, the probability of obtaining (3, 6) is $1/\text{36}$ .
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
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
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
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
If March sales will be up from February by 10%, 15%, and 20% at Place I, Place II, and Place III, respectively, find the expected number of hot dogs, and corn dogs to be sold
8. It is known that 80% of the people wear seat belts, and 5% of the people quit smoking last year. If 4% of the people who wear seat belts quit smoking, are the events, wearing a seat belt and quitting smoking, independent?
Mr. Shamir employs two part-time typists, Inna and Jim for his typing needs. Inna charges $10 an hour and can type 6 pages an hour, while Jim charges$12 an hour and can type 8 pages per hour. Each typist must be employed at least 8 hours per week to keep them on the payroll. If Mr. Shamir has at least 208 pages to be typed, how many hours per week should he employ each student to minimize his typing costs, and what will be the total cost?
At De Anza College, 20% of the students take Finite Mathematics, 30% take Statistics and 10% take both. What percentage of the students take Finite Mathematics or Statistics? |
# CIRCLES.
## Presentation on theme: "CIRCLES."— Presentation transcript:
CIRCLES
HINT: Pi =3.14
Geometry Lessons 8.5 and 8.6 Area of Circles, Sectors and Segments
HW: 8.5/1-8, 12-14 8.6/1-12
Circumference 3in C ≈ 18.8in
Finding a formulae for the area of a circle
πr r
Area of Rectangle= Base x Height
Finding the area given the diameter
The radius of a circle is half of its diameter, or r = d 2 We can substitute this into the formula A = πr2
How to find the area of a circle given the radius
Find the area of the circle below: Substitute the values for p and r.
How to find the area of a circle
given the diameter Given the diameter: Radius = diameter divided by 2 Substitute the values for p and r.
How would you find the shaded area?
Find the area of the square and subtract the area of the circle.
The area of a circle Use π = 3.14 to find the area of the following circles: 2 cm 10 m A = πr2 A = πr2 = 3.14 × 22 = 3.14 × 52 Explain that rather than use the formula on the previous slide, it is usually easier to halve the diameter mentally to give the radius, before substituting it into the formula. The most common error is to neglect to half the diameter to find the radius and to substitute this value into the formula. Ensure that pupils do not make this mistake. = cm2 = 78.5 m2 23 mm 78 cm A = πr2 A = πr2 = 3.14 × 232 = 3.14 × 392 = mm2 = cm2
Find the area of this shape
Use π = 3.14 to find area of this shape. The area of this shape is made up of the area of a circle of diameter 13 cm and the area of a rectangle of width 6 cm and length 13 cm. Compare this with slide 74, which finds the perimeter of the same shape. Area of circle = 3.14 × 6.52 13 cm 6 cm = cm2 Area of rectangle = 6 × 13 = 78 cm2 Total area = = cm2
The circumference of a circle is 18
The circumference of a circle is 18. Find the area of the circle, leave answer in terms of . C = 2 r A = r2 A = 92 A = 81 18 = 2r 9 = r
Arc Length Reminder The length of part of the circumference.
The length of the arc depends on what two things? 1) The measure of the arc. 2) The size of the circle. An arc length measures distance while the measure of an arc is in degrees.
m 2πr 360 . Arc Length Formula measure of the central angle or arc
The circumference of the entire circle! 2πr Arc Length = 360 The fraction of the circle! .
REVIEW: Finding Arc Length
Find the arc length. Give answers in terms of and rounded to the nearest hundredth. FG Use formula for area of sector. Substitute 8 for r and 134 for m. Simplify. 5.96 cm cm
Sectors Let’s talk pizza
Sector of a circle A region bounded by 2 radii and their intercepted arc. .
Area of a Sector Formula
measure of the central angle or arc πr2 Area of a sector = 360 The area of the entire circle! The fraction of the circle! .
Finding the Area of a Sector
Find the area of the sector. Give answers in terms of and rounded to the nearest hundredth. sector HGJ Use formula for area of sector. Substitute 12 for r and 131 for m. Simplify. = 52.4 m2 m2
Finding the Area of a Sector
Find the area of the sector. Give answers in terms of and rounded to the nearest hundredth. sector ABC Use formula for area of sector. Substitute 5 for r and 25 for m. Simplify. 1.74 ft2 5.45 ft2
The area of sector AOB is
and Find the radius of ○O. m Area of a sector = πr2 360 9 40 π = πr2 4 360 9 9 1 9 = r2 1 4 9 1 81 = r2 4 9 r = 2
The area of sector AOB is 48π and
Find the radius of ○O. m πr2 Area of a sector = 360 270 48π = πr2 360 A O 4 16 3 4 r2 48 = 3 4 3 64 = r2 r = 8 B
Finding the Area of a Sector
Find the area of each sector. Give your answer in terms of and rounded to the nearest hundredth. sector JKL Use formula for area of sector. Substitute 16 for r and 36 for m. Simplify. = 25.6 in2 in2
Automobile Application
A windshield wiper blade is 18 inches long. To the nearest square inch, what is the area covered by the blade as it rotates through an angle of 122°? Use formula for area of sector. r = 18 in. m˚ = 122˚ Simplify. 345 in2
AREA OF SEGMENT A segment of a circle is a region bounded by an arc and its chord.
= ¼100π= 25π 10 =½∙10∙10=50 Area of Segment = 25π - 50
Check It Out! Find the area of segment RST to the nearest hundredth. Step 1 Find the area of sector RST. Use formula for area of sector. Substitute 4 for r and 90 for m. = 4 m2 Simplify.
Check It Out! Continued Find the area of segment RST to the nearest hundredth. Step 2 Find the area of ∆RST. ST = 4 m, and RS = 4m. Simplify. = 8 m2
Check It Out! Continued Find the area of segment RST to the nearest hundredth. Step 3 area of segment = area of sector RST – area of ∆RST = 4 – 8 4.57 m2
Lesson Quiz: Part I Find each measure. Give answers in terms of and rounded to the nearest hundredth. 1. area of sector LQM 7.5 in2 in2 2. length of NP 2.5 in. 7.85 in.
Lesson Quiz: Part II 3. The gear of a grandfather clock has a radius of 3 in. To the nearest tenth of an inch, what distance does the gear cover when it rotates through an angle of 88°? 4.6 in. 4. Find the area of segment GHJ to the nearest hundredth. m2
Find the area of the shaded region
Find the area of the shaded region. Point O marks the center of the circle. 8. 9. 10. 11. 60˚ 8 12 60 O 6 4 30 O O 160 π units2 9π - 18 units2 24π - 36√3 units2 8π - 8√3 units2 3
Some common fractions and measures!
Arc or Central Angle Measure Fraction of the Circle 36o 108o 1/6 5/6 120o 2/3 30o 11/12 1/8 5/8 3/10 1/10 60o 300o 1/3 240o 1/12 330o 45o 225o |
ISEE Middle Level Math : Fractions
Example Questions
1 2 34 35 36 37 38 39 40 42 Next →
Example Question #552 : Numbers And Operations
Divide the following:
Explanation:
To divide fractions, we will take the first fraction and multiply by the reciprocal of the second fraction. So, we get
We will simplify before we multiply to make things easier. The 8 and the 12 can both be divided by 4. We get
Example Question #561 : Numbers And Operations
Divide the following:
Explanation:
To divide, we will multiply the first fraction by the reciprocal of the second fraction. To find the reciprocal, the numerator becomes the denominator, and the denominator becomes the numerator (in other words, we will flip the fraction).
We get
Example Question #412 : Fractions
Divide the following:
Explanation:
To divide fractions, we will take the first fraction and multiply it by the reciprocal of the second fraction.
To find the reciprocal, the numerator will become the denominator, and the denominator will become the numerator. In other words, we will flip the fraction.
So, we get
Example Question #413 : Fractions
Divide the following:
Explanation:
To divide fractions, we will take the first fraction and multiply it by the reciprocal of the second fraction. To find the reciprocal, the numerator will become the denominator, and the denominator will become the numerator. In other words, we will flip the fraction. So, we get
Example Question #414 : Fractions
Divide the following:
Explanation:
To divide fractions, we will take the first fraction and multiply it by the reciprocal of the second fraction. To find the reciprocal of a fraction, the numerator will become the denominator, and the denominator will become the numerator. In other words, we will flip the fraction.
So, we get
Example Question #411 : Fractions
Divide the following:
Explanation:
To divide fractions, we will take the first fraction and multiply it by the reciprocal of the second fraction. To find the reciprocal of a fraction, the numerator will become the denominator, and the denominator will become the numerator. In other words, we will flip the fraction.
So, we get
1 2 34 35 36 37 38 39 40 42 Next → |
# How do you solve x^2 + 5x – 6 = 0?
Aug 20, 2015
The solutions are
color(green)(x=-6
color(green)(x=1
#### Explanation:
x^2+5x–6=0
We can Split the Middle Term of this expression to factorise it and thereby find solutions.
In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:
${N}_{1} \cdot {N}_{2} = a \cdot c = 1 \cdot - 6 = - 6$
and
${N}_{1} + {N}_{2} = b = 5$
After trying out a few numbers we get :
${N}_{1} = - 1$ and ${N}_{2} = 6$
$6 \cdot - 1 = - 6$, and $6 + \left(- 1\right) = 5$
x^2+5x–6= x^2+6x-1x–6
$x \left(x + 6\right) - 1 \left(x + 6\right) = 0$
$\left(x + 6\right)$ is a common factor to each of the terms
color(green)((x+6)(x-1)=0
We now equate factors to zero:
x+6=0, color(green)(x=-6
x-1=0, color(green)(x=1 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Sum to Product Formulas for Sine and Cosine
## Relation of the sum or difference of two trigonometric functions to a product.
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Sum to Product Formulas for Sine and Cosine
Can you solve problems that involve the sum of sines or cosines? For example, consider the equation:
\begin{align*}\cos 10t + \cos 3t\end{align*}
You could just compute each expression separately and add their values at the end. However, there is an easier way to do this. You can simplify the equation first, and then solve.
### Sine and Cosine Sum to Product Formulas
In some problems, the product of two trigonometric functions is more conveniently found by the sum of two trigonometric functions by use of identities.
Here is an example:
\begin{align*}\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\end{align*}
This can be verified by using the sum and difference formulas:
\begin{align*}& 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}\\ &= 2 \begin{bmatrix} \sin \left( \frac{\alpha}{2} + \frac{\beta}{2} \right) \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) \end{bmatrix} \\ &= 2 \begin{bmatrix} \left( \sin \frac{\alpha}{2} \cos \frac{\beta}{2} + \cos \frac{\alpha}{2} \sin \frac{\beta}{2} \left) \right( \cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \right ) \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cos ^2 \frac{\beta}{2} + \sin ^2 \frac{\alpha}{2} \sin \frac{\beta}{2} \cos \frac{\beta}{2} + \sin \frac{\beta}{2} \cos ^2 \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin ^2 \frac{\beta}{2} \cos \frac{\alpha}{2} \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \left( \sin^2 \frac{\beta}{2} + \cos^2 \frac{\beta}{2} \right) + \sin \frac{\beta}{2} \cos \frac{\beta}{2} \left( \sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2} \right) \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + \sin \frac{\beta}{2} \cos \frac{\beta}{2} \end{bmatrix}\\ &= 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + 2 \sin \frac{\beta}{2} \cos \frac{\beta}{2}\\ &= \sin \left( 2 \cdot \frac{\alpha}{2} \right) + \sin \left( 2 \cdot \frac{\beta}{2} \right)\\ &= \sin \alpha + \sin \beta\end{align*}
The following variations can be derived similarly:
\begin{align*} \sin \alpha - \sin \beta &= 2 \sin \frac{\alpha - \beta}{2} \times \cos \frac{\alpha + \beta}{2}\\ \cos \alpha + \cos \beta &= 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\\ \cos \alpha - \cos \beta &= -2 \sin \frac{\alpha + \beta}{2} \times \sin \frac{\alpha - \beta}{2}\\\end{align*}
Here are some problems using this type of transformation from a sum of terms to a product of terms.
1. Change \begin{align*}\sin 5x - \sin 9x\end{align*} into a product.
Use the formula \begin{align*}\sin \alpha - \sin \beta = 2 \sin \frac{\alpha - \beta}{2} \times \cos \frac{\alpha + \beta}{2}.\end{align*}
\begin{align*} \sin 5x - \sin 9x &= 2 \sin \frac{5x - 9x}{2} \cos \frac{5x + 9x}{2}\\ &= 2 \sin (-2x) \cos 7x\\ &= -2 \sin 2x \cos 7x\end{align*}
2. Change \begin{align*}\cos(-3x) + \cos 8x\end{align*} into a product.
Use the formula \begin{align*} \cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\end{align*}
\begin{align*}\cos (-3x) + \cos (8x) &= 2 \cos \frac{-3x + 8x}{2} \cos \frac {-3x - 8x}{2}\\ &= 2 \cos (2.5x) \cos (-5.5x)\\ &= 2 \cos (2.5x) \cos (5.5x)\end{align*}
3. Change \begin{align*}2 \sin 7x \cos 4x\end{align*} to a sum.
This is the reverse of what was done in the previous two examples. Looking at the four formulas above, take the one that has sine and cosine as a product, \begin{align*}\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}.\end{align*} Therefore, \begin{align*}7x = \frac{\alpha + \beta}{2}\end{align*} and \begin{align*}4x = \frac{\alpha - \beta}{2}\end{align*}.
\begin{align*}7x & = \frac{\alpha + \beta}{2} &&&& 4x = \frac{\alpha - \beta}{2} \\ &&& \text{and} \\ 14x & = \alpha+\beta &&&& 8x= \alpha - \beta \\ &\alpha = 14x - \beta &&&& 8x=[14x-\beta]-\beta \\ &&& \text{so} \\ &&&&&-6x = -2\beta\\ &&&&&3x=\beta\\ \alpha=14x-3x\\ \alpha=11x\end{align*}
So, this translates to \begin{align*}\sin(11x) + \sin(3x)\end{align*}. A shortcut for this problem, would be to notice that the sum of \begin{align*}7x\end{align*} and \begin{align*}4x\end{align*} is \begin{align*}11x\end{align*} and the difference is \begin{align*}3x\end{align*}.
### Examples
#### Example 1
Earlier, you were asked to solve
\begin{align*}\cos 10t + \cos 3t\end{align*}
You can easily transform this equation into a product of two trig functions using:
\begin{align*}\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\end{align*}
Substituting the known quantities:
\begin{align*}\cos 10t + \cos 3t = 2 \cos \frac{13t}{2} \times \cos \frac{7t}{2} = 2\cos(6.5t) \cos(3.5t)\end{align*}
#### Example 2
Express the sum as a product: \begin{align*}\sin 9x + \sin 5x\end{align*}
Using the sum-to-product formula:
\begin{align*}& \sin 9x + \sin 5x\\ & 2 \left(\sin \left(\frac{9x + 5x}{2} \right) \cos \left(\frac{9x - 5x}{2} \right) \right)\\ & 2 \sin 7x \cos 2x\end{align*}
#### Example 3
Express the difference as a product: \begin{align*}\cos 4y - \cos 3y \end{align*}
Using the difference-to-product formula:
\begin{align*}& \cos 4y - \cos 3y\\ & -2 \sin \left(\frac{4y + 3y}{2} \right) \sin \left(\frac{4y - 3y}{2} \right)\\ & -2 \sin \frac{7y}{2} \sin \frac{y}{2}\end{align*}
#### Example 4
Verify the identity (using sum-to-product formula): \begin{align*}\frac{\cos 3a - \cos 5a}{\sin 3a - \sin 5a} = - \tan 4a\end{align*}
Using the difference-to-product formulas:
\begin{align*}& \qquad \qquad \frac{\cos 3a - \cos 5a}{\sin 3a - \sin 5a} = - \tan 4a\\ & \frac{-2 \sin \left (\frac{3a + 5a}{2} \right ) \sin \left (\frac {3a - 5a}{2} \right )}{2 \sin \left (\frac{3a - 5a}{2} \right ) \cos \left(\frac{3a + 5a}{2} \right)}\\ & \qquad \qquad \qquad \ \ - \frac{\sin 4a}{\cos 4a}\\ & \qquad \qquad \qquad \ \ - \tan 4a\end{align*}
### Review
Change each sum or difference into a product.
1. \begin{align*}\sin 3x + \sin 2x\end{align*}
2. \begin{align*}\cos 2x + \cos 5x\end{align*}
3. \begin{align*}\sin (-x) - \sin 4x\end{align*}
4. \begin{align*}\cos 12x + \cos 3x\end{align*}
5. \begin{align*}\sin 8x - \sin 4x\end{align*}
6. \begin{align*}\sin x + \sin \frac{1}{2}x\end{align*}
7. \begin{align*}\cos 3x - \cos (-3x)\end{align*}
Change each product into a sum or difference.
1. \begin{align*}-2\sin 3.5x \sin 2.5x\end{align*}
2. \begin{align*}2\cos 3.5x \sin 0.5x\end{align*}
3. \begin{align*}2\cos 3.5x \cos 5.5x\end{align*}
4. \begin{align*}2\sin 6x \cos 2x\end{align*}
5. \begin{align*}-2\sin 3x \sin x\end{align*}
6. \begin{align*}2\sin 4x \cos x\end{align*}
7. Show that \begin{align*}\cos\frac{A+B}{2}\cos\frac{A-B}{2}=\frac{1}{2}(\cos A + \cos B)\end{align*}.
8. Let \begin{align*}u=\frac{A+B}{2}\end{align*} and \begin{align*}v=\frac{A-B}{2}\end{align*}. Show that \begin{align*}\cos u\cos v =\frac{1}{2}(\cos (u+v)+\cos(u-v)).\end{align*}
To see the Review answers, open this PDF file and look for section 3.13.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Sum to Product Formula
A sum to product formula relates the sum or difference of two trigonometric functions to the product of two trigonometric functions. |
## Question
Gauthmathier7111
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Two forces of $$30$$ and $$70$$ pounds act on a point in a plane. If the angle between the force vectors is $$40^{\circ }$$. what are the magnitude and direction (relative to the $$70$$-pound force) of the resultant force? The magnitudes of the forces are to two significant digits and the angles to the nearest degree.
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## Gauth Tutor Solution
Emma
Yale University
Tutor for 2 years
equivalent to a single force of $$95$$ pounds in the direction of $$12^{\circ }$$
Explanation
We start with a diagram (Fig. 12), letting vectors represent the various forces. Because adjacent angles in a parallelogram are supplementary, the measure of angle $$OCB=180^{\circ }-40^{\circ }=140^{\circ }$$. We can now find the magnitude of the resultant vector $$R$$ using the law of cosines (Fig. 13).
$$|R|^{2}=30^{2}+70^{2}-2(30)(70)\cos 140^{\circ }$$
$$|R|=\sqrt {30^{2}+70^{2}-2(30)(70)\cos 140^{\circ }}=95$$ pounds
To find $$θ$$, the direction of $$R$$, we use the law of sines (Fig. 14).
$$\dfrac {\sin \theta }{30}=\dfrac {\sin 140^{\circ }}{95}$$
$$\sin \theta =\dfrac {30\sin 140^{\circ }}{95}$$
$$\theta =\sin ^{-1}\left(\dfrac {30\sin 140^{\circ }}{95}\right)=12^{\circ }$$
The two given forces are equivalent to a single force of $$95$$ pounds in the direction of $$12^{\circ }$$ (relative to the $$70$$-pound force).
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# Lesson 11: Representing Ratios with Tables
Let’s use tables to represent equivalent ratios.
## 11.1: How Is It Growing?
Look for a pattern in the figures.
1. How many total tiles will be in:
1. the 4th figure?
2. the 5th figure?
3. the 10th figure?
2. How do you see it growing?
## 11.2: A Huge Amount of Sparkling Orange Juice
Noah’s recipe for one batch of sparkling orange juice uses 4 liters of orange juice and 5 liters of soda water.
1. Use the double number line to show how many liters of each ingredient to use for different-sized batches of sparkling orange juice.
GeoGebra Applet ycJSQpXT
2. If someone mixes 36 liters of orange juice and 45 liters of soda water, how many batches would they make?
3. If someone uses 400 liters of orange juice, how much soda water would they need?
4. If someone uses 455 liters of soda water, how much orange juice would they need?
5. Explain the trouble with using a double number line diagram to answer the last two questions.
## 11.3: Batches of Trail Mix
A recipe for trail mix says: “Mix 7 ounces of almonds with 5 ounces of raisins.” Here is a table that has been started to show how many ounces of almonds and raisins would be in different-sized batches of this trail mix.
almonds (oz) raisins (oz)
row 1 7 5
row 2 28
row 3 10
row 4 3.5
row 5 250
row 6 56
1. Complete the table so that ratios represented by each row are equivalent.
2. What methods did you use to fill in the table?
3. How do you know that each row shows a ratio that is equivalent
to $7:5$? Explain your reasoning.
## Summary
A table is a way to organize information. Each horizontal set of entries is called a row, and each vertical set of entries is called a column. (The table shown has 2 columns and 5 rows.) A table can be used to represent a collection of equivalent ratios.
Here is a double number line diagram and a table that both represent the situation: “The price is \\$2 for every 3 mangos.”
## Glossary
table
#### table
A table is a way to organize information. Each rectangle in the table is called a cell. Each horizontal set of entries is called a row, and each vertical set of entries is called a column. The first row in a table often contains headers to explain what information is in each column.
This table shows the tail-lengths of three different pets. It has four rows and two columns.
The first cell in each column tells you what kind of information is in that column.
pet tail length (inches)
dog 22
cat 12
mouse 2 |
# Calculate the derivative of the function.f(x)=(x-3)/(x^2+2)
sciencesolve | Certified Educator
You need to use quotient rule to evaluate the derivative of the given function, such that:
`f'(x) = ((x - 3)'(x^2 + 2) - (x - 3)(x^2 + 2)')/((x^2 + 2)^2)`
`f'(x) = (1*(x^2 + 2) - (x - 3)(2x))/((x^2 + 2)^2)`
`f'(x) = (x^2 + 2 - 2x^2 + 6x)/((x^2 + 2)^2)`
`f'(x) = (-x^2 + 6x + 2)/((x^2 + 2)^2)`
Hence, evaluating the derivative of teh function, using the quotient rule, yields `f'(x) = (-x^2 + 6x + 2)/((x^2 + 2)^2).`
giorgiana1976 | Student
We'll use the limit method to calculate the value of derivative of a function in a given point.
lim [f(x) - f(1)]/(x-1) = lim [(x-3)/(x^2+2) + 2/3]/(x-1)
lim [(x-3)/(x^2+2) + 2/3]/(x-1) = lim (3x-9+2x^2+4)/(x-1)
lim (3x-9+2x^2+4)/(x-1) = lim (3x-5+2x^2)/(x-1)
We'll substitute x by 1:
lim (3x-5+2x^2)/(x-1) = (3-5+2)/(1-1) = 0/0
Since we've obtained an indeterminacy, we'll apply L'Hospital rule:
lim (3x-5+2x^2)/(x-1) = lim (3x-5+2x^2)'/(x-1)'
lim (3x-5+2x^2)'/(x-1)' = lim (3+4x)/1
We'll substitute x by 1:
lim (3+4x)/1 = (3+4)/1
f'(1) = lim (3x-5+2x^2)/(x-1) = 7
f'(1) = 7 |
# What is 51/74 as a decimal?
## Solution and how to convert 51 / 74 into a decimal
51 / 74 = 0.689
Fraction conversions explained:
• 51 divided by 74
• Numerator: 51
• Denominator: 74
• Decimal: 0.689
• Percentage: 0.689%
To convert 51/74 into 0.689, a student must understand why and how. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). Now, let's solve for how we convert 51/74 into a decimal.
51 / 74 as a percentage 51 / 74 as a fraction 51 / 74 as a decimal
0.689% - Convert percentages 51 / 74 51 / 74 = 0.689
## 51/74 is 51 divided by 74
Converting fractions to decimals is as simple as long division. 51 is being divided by 74. For some, this could be mental math. For others, we should set the equation. Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 51 divided by 74. We must divide 51 into 74 to find out how many whole parts it will have plus representing the remainder in decimal form. This is how we look at our fraction as an equation:
### Numerator: 51
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. Any value greater than fifty will be more difficult to covert to a decimal. The bad news is that it's an odd number which makes it harder to covert in your head. Values closer to one-hundred make converting to fractions more complex. So how does our denominator stack up?
### Denominator: 74
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. Larger values over fifty like 74 makes conversion to decimals tougher. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Overall, two-digit denominators are no problem with long division. Now it's time to learn how to convert 51/74 to a decimal.
## How to convert 51/74 to 0.689
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 74 \enclose{longdiv}{ 51 }$$
Use long division to solve step one. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 74 \enclose{longdiv}{ 51.0 }$$
Uh oh. 74 cannot be divided into 51. Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 74 into 510.
### Step 3: Solve for how many whole groups you can divide 74 into 510
$$\require{enclose} 00.6 \\ 74 \enclose{longdiv}{ 51.0 }$$
Since we've extended our equation we can now divide our numbers, 74 into 510 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiple this number by our furthest left number, 74, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.6 \\ 74 \enclose{longdiv}{ 51.0 } \\ \underline{ 444 \phantom{00} } \\ 66 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 51/74 into 0.689 If you have a remainder over 74, go back. Your solution will need a bit of adjustment. If you have a number less than 74, continue!
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. They each bring clarity to numbers and values of every day life. And the same is true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But 51/74 and 0.689 bring clarity and value to numbers in every day life. Here are just a few ways we use 51/74, 0.689 or 68% in our daily world:
### When you should convert 51/74 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.689 to 51/74 as a fraction
Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same.
### Practice Decimal Conversion with your Classroom
• If 51/74 = 0.689 what would it be as a percentage?
• What is 1 + 51/74 in decimal form?
• What is 1 - 51/74 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.689 + 1/2?
### Convert more fractions to decimals
From 51 Numerator From 74 Denominator What is 51/64 as a decimal? What is 41/74 as a decimal? What is 51/65 as a decimal? What is 42/74 as a decimal? What is 51/66 as a decimal? What is 43/74 as a decimal? What is 51/67 as a decimal? What is 44/74 as a decimal? What is 51/68 as a decimal? What is 45/74 as a decimal? What is 51/69 as a decimal? What is 46/74 as a decimal? What is 51/70 as a decimal? What is 47/74 as a decimal? What is 51/71 as a decimal? What is 48/74 as a decimal? What is 51/72 as a decimal? What is 49/74 as a decimal? What is 51/73 as a decimal? What is 50/74 as a decimal? What is 51/74 as a decimal? What is 51/74 as a decimal? What is 51/75 as a decimal? What is 52/74 as a decimal? What is 51/76 as a decimal? What is 53/74 as a decimal? What is 51/77 as a decimal? What is 54/74 as a decimal? What is 51/78 as a decimal? What is 55/74 as a decimal? What is 51/79 as a decimal? What is 56/74 as a decimal? What is 51/80 as a decimal? What is 57/74 as a decimal? What is 51/81 as a decimal? What is 58/74 as a decimal? What is 51/82 as a decimal? What is 59/74 as a decimal? What is 51/83 as a decimal? What is 60/74 as a decimal? What is 51/84 as a decimal? What is 61/74 as a decimal?
### Convert similar fractions to percentages
From 51 Numerator From 74 Denominator 52/74 as a percentage 51/75 as a percentage 53/74 as a percentage 51/76 as a percentage 54/74 as a percentage 51/77 as a percentage 55/74 as a percentage 51/78 as a percentage 56/74 as a percentage 51/79 as a percentage 57/74 as a percentage 51/80 as a percentage 58/74 as a percentage 51/81 as a percentage 59/74 as a percentage 51/82 as a percentage 60/74 as a percentage 51/83 as a percentage 61/74 as a percentage 51/84 as a percentage |
# The two shorter sides of a right triangle have the same length. The area of the right triangle is 7.07 square, what is the length of the triangle?
##### 1 Answer
Oct 2, 2016
Each of the equal sides has a length of 3.760 and the hypotenuse has a length of 5.318
#### Explanation:
Uh, I don't get what exact measurement of the triangle you want so I'll give you the length of all 3 sides.
In your case you've described an isosceles right triangle. Meaning the angle between the two equal sides is 90 degrees. Since we know that the formula for the area of a triangle is $\frac{b a s e \cdot h e i g h t}{2}$ and we can use each of the equal sides for base and height, we get:
$\frac{s i \mathrm{de} \cdot s i \mathrm{de}}{2} = 7.07$ so:
$s i {\mathrm{de}}^{2} = 14.14$
$s i \mathrm{de} = \sqrt{14.14}$
$s i \mathrm{de} = 3.76$
Now to find the hypotenuse, we can use Pythagoras' Theorem that states that $s i \mathrm{de} {1}^{2} + s i \mathrm{de} {2}^{2} = h y p o t e n$$u s {e}^{2}$ or as more commonly seen ${a}^{2} + {b}^{2} = {c}^{2}$
In our case a and b are equal and we know that ${a}^{2} = 14.14$ so:
$14.14 + 14.14 = {c}^{2}$
$c = \sqrt{28.28}$
$c = 5.318$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Geometry Concepts Go to the latest version.
# Chapter 7: Similarity
Difficulty Level: At Grade Created by: CK-12
## Introduction
In this chapter, we will start with a review of ratios and proportions. Second, we will introduce the concept of similarity. Two figures are similar if they have the same shape, but not the same size. We will apply similarity to polygons, quadrilaterals and triangles. Then, we will extend this concept to proportionality with parallel lines and dilations. Finally, there is an extension about self-similarity, or fractals, at the end of the chapter.
## Summary
This chapter is all about proportional relationships. It begins by introducing the concept of ratio and proportion and detailing properties of proportions. It then focuses on the geometric relationships of similar polygons. Applications of similar polygons and scale factors are covered. The AA, SSS, and SAS methods of determining similar triangles are presented and the Triangle Proportionality Theorem is explored. The chapter wraps up with the proportional relationships formed when parallel lines are cut by a transversal, similarity and dilated figures, and self-similarity.
### Chapter Keywords
• Ratio
• Proportion
• Means
• Extremes
• Cross-Multiplication Theorem
• Similar Polygons
• Scale Factor
• AA Similarity Postulate
• Indirect Measurement
• SSS Similarity Theorem
• SAS Similarity Theorem
• Triangle Proportionality Theorem
• Triangle Proportionality Theorem Converse
• Transformation
• Rigid Transformation
• Non-rigid Transformation
• Dilation
• Self-Similar
• Fractal
### Chapter Review
1. Solve the following proportions.
1. x+33=102\begin{align*}\frac{x+3}{3}=\frac{10}{2}\end{align*}
2. 85=2x1x+3\begin{align*}\frac{8}{5}=\frac{2x-1}{x+3}\end{align*}
2. The extended ratio of the angle in a triangle are 5:6:7. What is the measure of each angle?
3. Rewrite 15 quarts in terms of gallons.
Determine if the following pairs of polygons are similar. If it is two triangles, write why they are similar.
1. Draw a dilation of A(7,2),B(4,9),\begin{align*}A(7, 2), B(4, 9),\end{align*} and C(1,4)\begin{align*}C(-1, 4)\end{align*} with k=32\begin{align*}k=\frac{3}{2}\end{align*}.
Algebra Connection Find the value of the missing variable(s).
### Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9692.
Aug 09, 2013 |
# MAT 101: Symmetry and Transformations II
In this notes, we study the remaining transformations of a function, namely horizontal shifting, vertical shifting, and dilation (stretching and shrinking). We also discuss how those transformations can be used to sketch the graph of a complicated function from its most basic form.
Horizontal Shifting (translation to the right or left):
Example. Let us plot $y=x^2$, $y=(x-3)^2$, and $y=(x+3)^2$ all together. As you can see clearly, the graph of $y=(x-3)^2$ is 3 units translation to the right of the graph of $y=x^2$, while the graph of $y=(x+3)^2$ is 3 units translations to the left of the graph of $y=x^2$.
Let $a>0$. Then:
The graph of $y=f(x-a)$ is $a$ units translation to the right of the graph of $y=f(x)$;
The graph of $y=f(x+a)$ is $a$ units translation to the left of the graph of $y=f(x)$.
Vertical Shifting (translation upward or downward):
Example. Let us plot $y=x^2$, $y=x^2+3$, and $y=x^2-3$ all together. As you can see clearly, the graph of $y=x^2+3$ is 3 units translation upward of the graph of $y=x^2$, while the graph of $y=x^2-3$ is 3 units translations downward of the graph of $y=x^2$.
Let $a>0$. Then:
The graph of $y=f(x)+a$ is $a$ units translation to upward of the graph of $y=f(x)$;
The graph of $y=f(x)-a$ is $a$ units translation to downward of the graph of $y=f(x)$.
Stretching and Shrinking:
Example. Let us compare the graphs of $y=x^2$, $y=2x^2$ and $y=\frac{1}{2}x^2$. The graph of $y=2x^2$ looks thinner than the graph of $y=x^2$, while the graph of $y=\frac{1}{2}x^2$ looks wider than the graph of $y=x^2$. The graph of $y=2x^2$ can be considered as a (vetical) stretching of the graph of $y=x^2$. Similarly, the graph of $=\frac{1}{2}x^2$ can be considered as a (vetical) shrinking of the graph of $y=x^2$.
If $a>0$, the graph of $y=af(x)$ is a (vertical) stretching of the graph of $y=f(x)$;
If $0<a<1$, the graph of $y=af(x)$ is a (vertical) shrinking of the graph of $y=f(x)$.
Graphing functions using transformations: Sometimes we can sketch graph of a function by analyzing transformations involved such as reflection, stretching/shrinking, horizontal/vertical shifting.
Example. Sketch the graph of $f(x)=\frac{1}{3}x^3+2$ using transformations.
Solution. The most basic form of the function is $y=x^3$. So we obtain the graph of $f(x)$ from $y=x^3$
1. first by shrinking $y=\frac{1}{3}x^3$
2. and then translate the resultign function 2 units upward $y=\frac{1}{3}x^3+2$.
Example. Sketch the graph of $g(x)=-(x-3)^2+5$ using transformations.
Solution. The most basic form of the function is $y=x^2$. We obtain the graph of $g(x)$ from $y=x^2$
1. first by reflection $y=-x^2$
2. next translate the resulting function 3 units to the right
3. and finally translate the resulting function 5 units upward. |
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# Laws of indices
Algebra uses symbols or letters to represent quantities; for example I = PRT
I is used to stand for interest, P for principle, R for rate, and T for time.
A quantity made up of symbols together with operations ($+ - \times \div$) is called an algebraic expression. We use the laws of indices to simplify expressions involving indices.
Expand the following boxes for the laws of indices. The videos show why the laws are true.
### The first law: multiplication
If the two terms have the same base (in this case $x$) and are to be multiplied together their indices are added.
In general: $x^m \times x^n = x^{m+n}$
Here is an explanation of why this is true:
### The second law: division
If the two terms have the same base (in this case $x$) and are to be divided their indices are subtracted.
In general: $\dfrac{x^m}{x^n}=x^{m-n}$
Here is an explanation of why this is true:
### The third law: brackets
If a term with a power is itself raised to a power then the powers are multiplied together.
In general: $(x^m)^n = x^{m \times n}$
Here is an explanation of why this is true:
### Negative powers
Consider this example: $\dfrac{a^2}{a^6} = a^{2-6} = a^{-4}$
Also we can show that: $\dfrac{a^2}{a^6} = \dfrac{1}{a^4}$
So a negative power can be written as a fraction.
In general: $x^{-m} = \dfrac{1}{x^m}$
Here is an explanation of why this is true:
### Power of zero
The second law of indices helps to explain why anything to the power of zero is equal to one.
We know that anything divided by itself is equal to one. So $\dfrac {x^3}{x^3} = 1$
Also we know that $\dfrac {x^3}{x^3} = x^{3-3} = x^0 = 1$
Therefore, we have shown that $\dfrac {x^3}{x^3} = x^0 = 1$
Here is an explanation of why this is true:
### Fractional powers
Both the numerator and denominator of a fractional power have meaning.
The bottom of the fraction stands for the type of root; for example, $x^{\frac{1}{3}}$ denotes a cube root $\sqrt[3]{x}$
The top line of the fractional power gives the usual power of the whole term.
For example: $x^{\frac{2}{3}} = (\sqrt[3]{x})^2$
In general: $x^{\frac{m}{n}} = (\sqrt[n]{x})^m$
This explanation show why a root is shown as a fractional power:
## Further information
• Press the Printer Friendly button at the bottom left-hand corner to download a printable handout
• RMIT University has online videos to review the use of indices including fractional indices, brackets and logarithms. |
# How do you find the base if 11 1/2% of the base is 46?
Jun 6, 2017
The base is $400$
#### Explanation:
Calculations with percent can always be done by direct proportion.
Write equivalent fractions comparing the values with the matching percent. (The full amount is always 100%
In this question:
11.5% represents 46% and we want to know the value for 100%
11.5/46 = 100/x" "(larr"percent")/(larr"value")
$x = \frac{46 \times 100}{11.5} \text{ } \leftarrow$ cross-multiply to get $x$
$x = 400$
Check: 11 1/2% xx 400 = 46 |
The trigonometric functions are defined in two related approaches:
(i) Right Triangle Trigonometry.
(ii) Unit Circle Trigonometry.
(i) In right triangle trigonometry, six functions are defined with acute angles, right triangles, and the theory of similar of triangles: sine, cosine, tangent, cotangent, secant and cosecant.
In most courses on triangle trigonometry, the main attention is paid to the sine, cosine, and tangent functions. These three functions are the most frequently used for the study of triangles- especially with problems of solving triangles using the laws of sines and cosines and the use of tangents to measure inclination of an angle.
In the study of trigonometry applied to general triangles, treating obtuse angles becomes relevant. The definitions of the trigonometric functions are extended to angles of degree measures between 0 and 180. Making sense of the laws of sines and cosines and solving triangles are central applications for this extension.
For our treatment of triangle trigonometry we consider all six functions as "core".
Note: The theory of similar right triangle shows that the value of any one of these functions will determine the angle and thus the other five functions, we could consider the sine function as the core of all trigonometric functions for right triangles.
Details: Right Triangle Trigonometric Functions
(ii) The unit circle approach to the trigonometric functions comes from considering central angles in the geometry of circles- in particular looking at a unit circle centered at the point \$O=(0,0)\$ in cartesian coordinate plane.
Central angles with the positive horizontal axis as the initial ray are determined by an arc length, \$t\$- the radian measure of the angle. This arc also determines a right triangle by the center \$O\$, the endpoint of the arc, \$P(t)=(x(t), y(t)\$, and the point on the horizontal axis, \$C(t)=(x(t),0)\$. The right angle of the triangle is at the point \$C(t)\$.
Since the triangle \$OC(t)P(t)\$ has a unit length for the measure of its hypotenuse, the right triangle trigonometric functions of cosine and sine correspond to the coordinates of \$P(t)\$. This correspondence is used to extend the definition of these trigonometric functions to all real numbers. The use of the coordinates related to the unit circle also gives a sensible way to visualize the values of all the trigonometric functions as lengths of line segments, with the easiest line segment visualizations being connected to the sine, cosine, and tangent functions.
Details: Unit Circle Trigonometric Functions |
secord.doc - Second-order Systems Definition the transfer function contains terms in s2(and often but not necessarily s as well but none in s3 or above
# secord.doc - Second-order Systems Definition the transfer...
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Second-order Systems Definition -- the transfer function contains terms in s 2 (and often, but not necessarily, s as well!) but none in s 3 or above. What would it be in terms of the differential equation? To see what the problems might be, let us determine how systems having the following transfer functions will respond to a unit step input. 1. 10/(s 2 + 5s + 4) 2. 10/(s 2 + 2s + 4) 3. 10/(s 2 + 4s + 4) System 1 The transform of the unit step is 1/s, so the transform of the output will be: 10 ---------------- s(s 2 + 5s + 4) We have, unfortunately, no direct conversion for this one in the tables. We will need to use Partial Fractions. Fortunately, we note that (s 2 + 5s + 4) will factorise as (s + 1)(s + 4). So we have: A B C ----- + ------ + -- s + 1 s + 4 s We put it all over its common denominator as usual: As(s + 4) + Bs(s + 1) + C(s 2 + 5s + 4) ----------------------------------------------------------------------- s(s + 1)(s + 4) The numerator of the fraction turns out to be: s 2 (A + B + C) + s(4A + B + 5C) + (4C) and it all has to equal the 10 (+ 0s + 0s 2 ) we started out with. So 4C = 10, and C = 2.5. The s2 terms then give us A + B + 2.5 = 0, whilst the s terms give us 4A + B + 12.5 = 0. If we subtract the first equation from the second, we obtain: 3A + 10 = 0, so A = -10/3 = -3.333. Putting this back into A + B + 2.5 = 0, we obtain: -3.333 + B + 2.5 = 0, giving B = 0.833. The Laplace Transform of the output is therefore: 2.5/s - 3.333/(s + 1) + 0.833/(s + 4) and the time response becomes: 2.5 - 3.333e -t + 0.833e -4t .
We will sketch the graph in the lecture but it is particularly noteworthy that the response does not overshoot the steady-state value. System 2 On the face of it, this one looks like a repeat performance - but there are actually important differences! The output transform is 10/[s(s 2 + 2s + 4)]. Unfortunately, s 2 + 2s + 4 does not factorise ... we will have to adopt another approach. The tables do give us the following guidance: w/[(s + a) 2 + w 2 ] = e -at sin(wt) (s + a)/[(s + a) 2 + w 2 ] = e -at cos(wt) so we will have to resort to partial fractions again.
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# 009 chapter ii
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### 009 chapter ii
1. 1. Chapter II<br />This chapter centers on the complex numbers which is a number comprising a real number and an imaginary number. Under this, we have the number i, the complex plane where the points are plotted and the 4 arithmetic operations such as addition and subtraction, multiplication and division of complex numbers. To round up the chapter, simple equation involving complex numbers will be studied and solved.<br /> TARGET SKILLS: <br />At the end of this chapter, students are expected to:<br />• identify complex numbers;<br />• differentiate the real pat and imaginary part of complex numbers; and<br />• explore solving of the 4 arithmetic operations on the complex numbers.<br />Lesson 2<br />Defining Complex Numbers<br />OBJECTIVES:<br />At the end of this lesson, students are expected to:<br /><ul><li>identify complex numbers;
2. 2. differentiate the real number and standard imaginary unit; and
3. 3. extend the ordinary real number.</li></ul>A complex number, in mathematics, is a number comprising a real number and an imaginary number; it can be written in the form a + bi, where a and b are real numbers, and i is the standard imaginary unit, having the property that i2 = −1.[1] The complex numbers contain the ordinary real numbers, but extend them by adding in extra numbers and correspondingly expanding the understanding of addition and multiplication.<br />Equation 1: x2 - 1 = 0.<br />Equation 1 has two solutions, x = -1 and x = 1. We know that solving an equation in x is equivalent to finding the x-intercepts of a graph; and, the graph of y = x2 - 1 crosses the x-axis at (-1,0) and (1,0).<br />Equation 2: x2 + 1 = 0<br />Equation 2 has no solutions, and we can see this by looking at the graph of y = x2 + 1.<br />Since the graph has no x-intercepts, the equation has no solutions. When we define complex numbers, equation 2 will have two solutions.<br />-438785-684530Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br />Solve each equation and graph.<br />x² + 4 = 0<br /> _____________________________________________<br />2x² + 18 = 0<br /> _____________________________________________<br />2x² + 14 = 0<br /> _____________________________________________<br />3x² + 27 = 0<br /> _____________________________________________<br />x² - 3 = 0 <br /> _____________________________________________<br />x² + 21 = 0<br />-466725-741103 _____________________________________________<br />3x² - 5 = 0<br /> _____________________________________________<br />5x² + 30 = 0<br /> _____________________________________________<br />2x² + 3 = 0<br /> _____________________________________________<br /> x² + 50 = 0<br /> _____________________________________________<br /> x² - 2 = 0<br /> _____________________________________________<br /> 3x² - 50 = 0<br /> _____________________________________________<br /> x² - 3 = 0<br /> _____________________________________________<br />-514350-683895 x² + 4 = 0<br /> _____________________________________________<br /> 2x² + 14 = 0<br /> _____________________________________________<br />Lesson 3<br />The Number i<br />OBJECTIVES:<br />At the end of this lesson, students are expected to:<br /><ul><li>recognize the property of the number i;
4. 4. discuss the powers of i; and
5. 5. solve the high powers of imaginary unit.</li></ul>Consider Equations 1 and 2 again.<br />Equation 1Equation 2 x2 - 1 = 0.x2 + 1 = 0. x2 = 1.x2 = -1. <br />Equation 1 has solutions because the number 1 has two square roots, 1 and -1. Equation 2 has no solutions because -1 does not have a square root. In other words, there is no number such that if we multiply it by itself we get -1. If Equation 2 is to be given solutions, then we must create a square root of -1.<br />The imaginary unit i is defined by<br />The definition of i tells us that i2 = -1. We can use this fact to find other powers of i.<br />Example<br />i3 = i2 * i = -1*i = -i.<br />i4 = i2 * i2 = (-1) * (-1) = 1.<br />Exercise:<br />Simplify i8 and i11. <br />We treat i like other numbers in that we can multiply it by numbers, we can add it to other numbers, etc. The difference is that many of these quantities cannot be simplified to a pure real number.<br />For example, 3i just means 3 times i, but we cannot rewrite this product in a simpler form, because it is not a real number. The quantity 5 + 3i also cannot be simplified to a real number.<br />However, (-i)2 can be simplified. (-i)2 = (-1*i)2 = (-1)2 * i2 = 1 * (-1) = -1.<br />Because i2 and (-i)2 are both equal to -1, they are both solutions for Equation 2 above.<br />-485775-683953Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br />Instruction: Express each number in terms of i and simplify.<br /><ul><li>-64</li></ul> ______________________________________________________<br /><ul><li>-9 4</li></ul> _____________________________________________________<br /><ul><li>-50</li></ul> ______________________________________________________<br /><ul><li>2-18</li></ul> ______________________________________________________<br /><ul><li>4-45</li></ul> ______________________________________________________<br /><ul><li>-514350-693420-16 49</li></ul> ______________________________________________________<br /><ul><li>3-25 16</li></ul> ______________________________________________________<br /><ul><li>-7-4</li></ul> ______________________________________________________<br /><ul><li>40-425</li></ul> ______________________________________________________<br /><ul><li> -100</li></ul> ______________________________________________________<br /><ul><li> -48</li></ul> ______________________________________________________<br /><ul><li> 5-18100</li></ul> ______________________________________________________<br /><ul><li>-534789-636213 4-2527</li></ul> ______________________________________________________<br /><ul><li> -649</li></ul> ______________________________________________________<br /><ul><li> --75</li></ul> ______________________________________________________<br />Lesson 4<br />The Complex Plane<br />OBJECTIVES:<br />At the end of this lesson, students are expected to:<br /><ul><li>distinguish the points on the plane;
6. 6. differentiate the real and imaginary part; and
7. 7. draw from memory the figure form by the plot points on the complex plane.</li></ul>A complex number is one of the form a + bi, where a and b are real numbers. a is called the real part of the complex number, and b is called the imaginary part.<br />Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. I.e., a+bi = c+di if and only if a = c, and b = d.<br />Example.<br />2 - 5i.<br />6 + 4i.<br />0 + 2i = 2i.<br />4 + 0i = 4.<br />The last example above illustrates the fact that every real number is a complex number (with imaginary part 0). Another example: the real number -3.87 is equal to the complex number -3.87 + 0i.<br />It is often useful to think of real numbers as points on a number line. For example, you can define the order relation c < d, where c and d are real numbers, by saying that it means c is to the left of d on the number line.<br />We can visualize complex numbers by associating them with points in the plane. We do this by letting the number a + bi correspond to the point (a,b), we use x for a and y for b.<br />Exercises: Represent each of the following complex number by a point in the plane. <br /><ul><li>3 + 2i
8. 8. 1 – 4i
9. 9. 4 + 3i
10. 10. 2 – 5i
11. 11. 4 – 3i</li></ul>-485775-683953Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br />Instruction: Represent each of the following Complex Numbers by a point in the plane.<br /><ul><li>4 + 3i</li></ul> ______________________________________________________<br /><ul><li>5 – 7i</li></ul> ______________________________________________________<br /><ul><li>5i</li></ul> ______________________________________________________<br /><ul><li>0</li></ul> ______________________________________________________<br /><ul><li>3</li></ul> ______________________________________________________<br /><ul><li>32-2i
12. 12. _____________________________________________________
13. 13. 12</li></ul> ______________________________________________________<br /><ul><li>3 + 2i</li></ul> ______________________________________________________<br /><ul><li>-504826-6838955 – 3i2</li></ul> _____________________________________________________<br /><ul><li>1 – 4i</li></ul> ______________________________________________________<br /><ul><li>32+ 4i</li></ul> _____________________________________________________<br /><ul><li>4+3i</li></ul> ______________________________________________________ <br /><ul><li>4-3i</li></ul> ______________________________________________________<br /><ul><li>2-5i</li></ul> ______________________________________________________<br /><ul><li>53+2i</li></ul> ______________________________________________________<br />Lesson 5<br />Complex Arithmetic<br />OBJECTIVES:<br />At the end of this lesson, students are expected to:<br /><ul><li>define the four arithmetic operations on complex numbers;
14. 14. comply with the steps in solving the different operations; and
15. 15. solve the four arithmetic operations.</li></ul>When a number system is extended the arithmetic operations must be defined for the new numbers, and the important properties of the operations should still hold. For example, addition of whole numbers is commutative. This means that we can change the order in which two whole numbers are added and the sum is the same: 3 + 5 = 8 and 5 + 3 = 8.<br />We need to define the four arithmetic operations on complex numbers.<br />Addition and Subtraction<br />To add or subtract two complex numbers, you add or subtract the real parts and the imaginary parts.<br />(a + bi) + (c + di) = (a + c) + (b + d)i.(a + bi) - (c + di) = (a - c) + (b - d)i.<br />Example <br />(3 - 5i) + (6 + 7i) = (3 + 6) + (-5 + 7)i = 9 + 2i.<br />(3 - 5i) - (6 + 7i) = (3 - 6) + (-5 - 7)i = -3 - 12i.<br />Note<br />These operations are the same as combining similar terms in expressions that have a variable. For example, if we were to simplify the expression (3 - 5x) + (6 + 7x) by combining similar terms, then the constants 3 and 6 would be combined, and the terms -5x and 7x would be combined to yield 9 + 2x.<br />The Complex Arithmetic applet below demonstrates complex addition in the plane. You can also select the other arithmetic operations from the pull down list. The applet displays two complex numbers U and V, and shows their sum. You can drag either U or V to see the result of adding other complex numbers. As with other graphs in these pages, dragging a point other than U or V changes the viewing rectangle.<br />Multiplication<br />The formula for multiplying two complex numbers is<br />(a + bi) * (c + di) = (ac - bd) + (ad + bc)i.<br />You do not have to memorize this formula, because you can arrive at the same result by treating the complex numbers like expressions with a variable, multiply them as usual, then simplify. The only difference is that powers of i do simplify, while powers of x do not.<br />Example <br />(2 + 3i)(4 + 7i)= 2*4 + 2*7i + 4*3i + 3*7*i2= 8 + 14i + 12i + 21*(-1)= (8 - 21) + (14 + 12)i= -13 + 26i.<br />Notice that in the second line of the example, the i2 has been replaced by -1.<br />Using the formula for multiplication, we would have gone directly to the third line.<br />Exercise <br />Perform the following operations.<br />(a) (-3 + 4i) + (2 - 5i)<br />(b) 3i - (2 - 4i)<br />(c) (2 - 7i)(3 + 4i)<br />(d) (1 + i)(2 - 3i)<br />Division<br />The conjugate (or complex conjugate) of the complex number a + bi is a - bi.<br />Conjugates are important because of the fact that a complex number times its conjugate is real; i.e., its imaginary part is zero.<br />(a + bi)(a - bi) = (a2 + b2) + 0i = a2 + b2.<br />Example <br />NumberConjugateProduct2 + 3i2 - 3i4 + 9 = 133 - 5i3 + 5i9 + 25 = 344i-4i16<br />Suppose we want to do the division problem (3 + 2i) ÷ (2 + 5i). First, we want to rewrite this as a fractional expression .<br />Even though we have not defined division, it must satisfy the properties of ordinary division. So, a number divided by itself will be 1, where 1 is the multiplicative identity; i.e., 1 times any number is that number.<br />So, when we multiply by , we are multiplying by 1 and the number is not changed.<br />Notice that the quotient on the right consists of the conjugate of the denominator over itself. This choice was made so that when we multiply the two denominators, the result is a real number. Here is the complete division problem, with the result written in standard form.<br />Exercise:<br />Write (2 - i) ÷ (3 + 2i) in standard form. <br />We began this section by claiming that we were defining complex numbers so that some equations would have solutions. So far we have shown only one equation that has no real solutions but two complex solutions. In the next section we will see that complex numbers provide solutions for many equations. In fact, all polynomial equations have solutions in the set of complex numbers. This is an important fact that is used in many mathematical applications. Unfortunately, most of these applications are beyond the scope of this course. See your text (p. 195) for a discussion of the use of complex numbers in fractal geometry.<br />-485775-683953Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br />Instruction: Perform the indicated operations and express the result in the form a +bi.<br /><ul><li>(8+2i)+(4+7i)</li></ul>_____________________________________________________<br /><ul><li>3-i)+(-4+2i)</li></ul>_____________________________________________________<br /><ul><li>(-7+4i)+(5-6i)</li></ul>_____________________________________________________<br /><ul><li>(-8+6i)-(-3+2i)</li></ul>_____________________________________________________<br /><ul><li>(1.2+3i)+(3.88+1.6i)
16. 16. _____________________________________________________
17. 17. (6-i)+(2-5i)</li></ul>_____________________________________________________<br /><ul><li>(5+3i)-(3-2i)</li></ul>_____________________________________________________<br /><ul><li>-533399-6266888-15i)+(-2+10i)</li></ul>_____________________________________________________<br /><ul><li>(8-2i)-(10-5i)</li></ul>_____________________________________________________<br /><ul><li> (6-5i)-(6-4i)</li></ul>_____________________________________________________<br /><ul><li> (2+i)(3+i)</li></ul>_____________________________________________________<br /><ul><li> (4-3i)(2+3i)</li></ul>_____________________________________________________<br /><ul><li> (1-2i)(5-2i)</li></ul>_____________________________________________________<br /><ul><li> (3+2i)(-7+5i)</li></ul>_____________________________________________________<br /><ul><li> (8-7i)(2i+3i)</li></ul>_____________________________________________________<br />Define and/or describe each of the following terms.<br />Imaginary part<br />Real number<br />Complex number<br />Complex plane<br />Imaginary unit<br />Commutative property<br />Complex conjugate<br />1. Simplify: <br />i15 <br />i25<br />i106<br />i207<br />i21<br />2. Perform the indicated operation and express each answer.<br /><ul><li>a. ¯9 + ¯25
18. 18. b. ¯16 + ¯49
19. 19. c. ¯100 + ¯81
20. 20. d. ¯169 + ¯225
21. 21. e. ¯450 + ¯162</li></ul>f. ¯147 + ¯48<br />3. Represent each complex numbers by a point in the plane.<br />a. 3 – i<br />b. -2 + 4i<br />c. -3 + 3i<br />d. 4 + 5i<br />e. -3 + 5i<br />4. Give the real part and the imaginary part of each complex numbers in #3.<br />5. Perform the indicated operations.<br />a. (3 – 2i) + (-7 + 3i)<br />b. (-4 + 7i) + (9 – 2i)<br />c. (14 – 9i) + (7 – 6i)<br />d. (5 + i) – (3 + 2i)<br />e. (7 – 2i) – (4 – 6i)<br />f. (8 + 3i) – (-4 – 2i)<br />g. (3 – 2i) (3 +2i)<br />h. (5 + 3i) (4 – i)<br />i. (11 + 2i)2 (5 – 2i)<br />j. (5 + 4i) / (3 – 2i)<br />k. (4 + i) (3 – 5i) / (2 – 3i)<br />l. (7 + 3i) / (3 – 3i / 4)<br /> |
# How do you find the quotient of (m^4-2m^3+m^2+12m-6)/(m-2) using synthetic division?
Jan 31, 2018
Quotient $\textcolor{g r e e n}{Q = {m}^{2} + m + 14}$, Remainder $\textcolor{g r e e n}{R = \frac{22}{m - 2}}$
#### Explanation:
$\textcolor{w h i t e}{a a} 2 \textcolor{w h i t e}{a a} | \textcolor{w h i t e}{a a} 1 \textcolor{w h i t e}{a a} - 2 \textcolor{w h i t e}{a a} 1 \textcolor{w h i t e}{a a} 12 \textcolor{w h i t e}{a a} - 6$
$\textcolor{w h i t e}{a a a a a} | \textcolor{w h i t e}{a} \downarrow \textcolor{w h i t e}{a a a a} 2 \textcolor{w h i t e}{a a} 0 \textcolor{w h i t e}{a a a} 2 \textcolor{w h i t e}{a a a} 28$
$\textcolor{w h i t e}{a a a a a} - - - - - - - - - -$
$\textcolor{w h i t e}{a a a a a a a a a} 1 \textcolor{w h i t e}{a a a a} 0 \textcolor{w h i t e}{a a} 1 \textcolor{w h i t e}{a a a} 14 \textcolor{w h i t e}{a a a} 22$
Quotient $\textcolor{g r e e n}{Q = {m}^{2} + m + 14}$, Remainder $\textcolor{g r e e n}{R = \frac{22}{m - 2}}$ |
How about putting all the knowledge you’ve gained about the ASVAB Mathematics Knowledge subtest to the test? Here are ten questions that are very similar to those you’re likely to see when you take the actual test.
## Sample questions
1. Which of the following fractions is the smallest?
• (A)3/4
• (B)14/17
• (C)4/7
• (D)5/8
2. Solve: 2x – 3 = x + 7.
• (A)10
• (B)6
• (C)21
• (D)–10
3. A circle has a radius of 15 feet. What is most nearly its circumference?
• (A)30 feet
• (B)25 feet
• (C)94 feet
• (D)150 feet
4. At 3 p.m., the angle between the hands of the clock is
• (A)90 degrees
• (B)180 degrees
• (C)120 degrees
• (D)360 degrees
5. If 3 + y ≥ 13, what is the value of y?
• (A)Greater than or equal to 10
• (B)Less than or equal to 10
• (C)10
• (D)6
6. y3 × y2 × y–3 =
• (A)y2
• (B)y–18
• (C)y8
• (D)x23
7. 14 yards + 14 feet =
• (A)16 yards
• (B)15 yards
• (C)28 feet
• (D)56 feet
8. What is 35 percent of 85?
• (A)33.2
• (B)65.32
• (C)21.3
• (D)29.75
9. What is most nearly the average of 37, 22, 72, and 44?
• (A)43.8
• (B)5.2
• (C)175
• (D)77.1
Use this answer key to score the practice Mathematics Knowledge questions.
1. C. One method of comparing fractions is called the cross-product method.
The cross-products of the first fraction and the second fraction are 3 × 17 = 51 and 14 × 4 = 56. The first fraction is smaller.
The cross-products of the first fraction and the third fraction are 3 × 7 = 21 and 4 × 4 = 16. The third fraction is smaller.
The cross-products of the third fraction and the fourth fraction are 4 × 8 = 32 and 5 × 7 = 35. The third fraction, Choice (C), is still smaller, so it’s the smallest of all the fractions.
2. A. Rearrange the equation and solve as follows:
3. C. The circumference of a circle is ð × diameter; the diameter equals two times the radius; and ð is approximately 3.14. Therefore, 30 × 3.14 ≈ 94.
The ≈ sign means approximately equals. It’s used here because the answer, 94, is a rounded number.
4. A. At 3 p.m., one hand is on the 12, and the other is on the 3. This setup creates a right angle — a 90-degree angle.
5. A. Solve the inequality the same way you’d solve an algebraic equation:
6. A. When you multiply powers with the same base, add the exponents: y3 × y2 × y–3 = y3 + 2 + (–3) = y2.
7. D. Convert the yards to feet by multiplying by 3: 14 × 3 = 42 feet. Add this to 14 feet:42 + 14 = 56 feet.
8. D. Multiply 85 by the decimal equivalent of 35 percent, or 0.35: 0.35 × 85 = 29.75.
9. A. Add the numbers and then divide by the number of terms: 37 + 22 + 72 + 44 = 175, and 175 ÷ 4 = 43.75. Round this number up to 43.8. |
## Remainder of One
In this lesson, children will arrange them 25 cubes into groups of two, three, four and five. The children will examine the different groups and learn about the concept of a remainder.
### Lesson for:
Toddlers/Preschoolers
(See Step 5: Adapt lesson for toddlers or preschoolers.)
Algebra
### Learning Goals:
This lesson will help toddlers and preschoolers meet the following educational standards:
• Understand numbers, ways of representing numbers, relationships among numbers and number systems
• Understand patterns, relations and functions
• Understand meanings of operations and how they relate to one another
### Learning Targets:
After this lesson, toddlers and preschoolers should be more proficient at:
• Counting with understanding and recognizing “how many” in sets of objects
• Developing a sense of whole numbers and representing and using them in flexible ways, including relating, composing and decomposing numbers
• Understanding situations that entail multiplication and division, such as equal groupings of objects and sharing equally
• Sorting, classifying and ordering objects by size, number and other properties
• Recognizing, describing and extending patterns such as sequences of sounds and shapes or simple numeric patterns and translating from one representation to another
## Remainder of One
### Lesson plan for toddlers/preschoolers
#### Step 1: Gather materials.
• The book, A Remainder of One by Elinor J. Pinczes
• Paper lunch bags with 25 counting blocks in each bag (Each child should have one bag with 25 blocks.)
• Chart paper and markers
Note: Small parts pose a choking hazard and are not appropriate for children age five or under. Be sure to choose lesson materials that meet safety requirements.
#### Step 2: Introduce activity.
1. Explain that the queen in the book, A Remainder of One, is having a problem and she needs the children’s help. The queen needs her squadron, her army, to line up into even and equal lines. She has 25 members in her bug army.
2. Explain that the children are going to receive bags that have 25 counting blocks in them to help them solve the problem. The children are to use the blocks as we read the book.
#### Step 3: Engage children in lesson activities.
1. Begin reading the book. Stop after reading: “The troop had divided by two for the show.” Have the children take out their blocks and make two equal lines/groups. The children will be able to make two groups of 12 but will have one left over. Have the children share their findings. Record their findings on the chart paper. Read the rest of the page: “Each bug had a partner, except soldier Joe.” Ask the children: “There are 25 bug soldiers. Can they make two even and equal lines? No.” Introduce the concept of division. “So, we can say: 25 bugs divided into two groups has 12 bugs in each line with one bug left over. Is this right?” Write the division equation on the chart paper.
2. Continue reading the book. Stop after reading: “The troop had divided by three for the show.” Have the children take out their blocks and make three equal lines/groups. The children will be able to make three groups of eight, but will have one left over. Have the children share their findings. Record their findings on the chart paper. Read the rest of the page: “Each line seemed perfect. Then someone spied Joe.” Ask the children: “There are 25 bug soldiers. Can they make three even and equal lines? No. So, we can say: 25 bugs divided into three groups has eight bugs in each line with one bug left over. Is this right?” Write the division equation on the chart paper.
3. Continue reading the book. Stop after reading: “The troop had divided by four for the show.” Have the children take out their blocks and make four equal lines/groups. The children will be able to make four groups of six, but will have one left over. Have the children share their findings. Record their findings on the chart paper. Read the rest of the page: “The lines all looked even, till they spotted Joe.” Ask the children: “There are 25 bug soldiers. Can they make four even and equal lines? No. So, we can say: 25 bugs divided into four groups has six bugs in each line with one bug left over. Is this right?” Write the division equation on the chart paper.
4. Continue reading the book. Stop after reading: “Five lines of soldiers….” Again, have the children take out their blocks and make five equal lines/groups. The children will be able to make five groups of five with no remainders. Have the children share their findings. Record their findings on the chart paper. Read the rest of the page: “…with 5 in each row… perfect at last—and that’s counting Joe.” Ask the children: “There are 25 bug soldiers. Can they make five even and equal lines? Yes. So, we can say: 25 bugs divided into five groups has five bugs in each line, with no bugs left over. Is this right?” Write the division equation on the chart paper. “Did we do our job and help the queen?”
• Present different scenarios and numbers for the children to group and divide. Use larger or smaller numbers, depending on the children’s abilities: 18, 24, 30.
#### Step 4: Vocabulary.
• Remainder: Amount left over after dividing a number (e.g.,”The children will be able to make five groups of five with no remainders.”)
• Divide: Sharing or grouping a number into equal parts (e.g.,”So, we can say: 25 bugs divided into four groups has six bugs in each line with one bug left over.”)
#### Step 5: Adapt lesson for toddlers or preschoolers.
###### Toddlers may:
• Still be working with one-to-one correspondence, counting and grouping
###### Child care providers may:
• Have the children work with smaller numbers (nine is a good number to work with because there are several ways in which the children can group the number)
###### Preschoolers may:
• Have a command of bigger numbers and can easily work with grouping, sorting and identifying the relationships and patterns among numbers
• Understand the concept of division—grouping numbers into equal parts
###### Child care providers may:
• Increase the number of blocks that the children divide (61 is a good number)
• Make the connection between multiplication and division. (If division is grouping numbers into equal parts, multiplication is groups of a number. Make these connections without necessarily using the terms “division” or “multiplication.” Use “groups of” to describe the inverse relation.
### Suggested Books
• A Remainder of One by Elinor J. Pinczes (Houghton Mifflin Harcourt, 2002)
### Outdoor Connections
Go outside to act out the book with the children. Work with 25 children or a number with several factors and add one more child. Have the children line up into even rows and see if there are any children who cannot be incorporated into one of the lines. Do this until all of the children are evenly grouped into an equal number of lines.
### Web Resources
• A game about fair shares
• The Jelly Bean Game (practice number sense concepts such as counting, more and less, estimation, algebraic thinking and missing addend equations, as well as addition and subtraction math facts) |
COMPOUNDED CONTINUOUSLY FORMULA
Compounded continuously formula :
In compound interest, we would have heard about the terms like "compounded annually" or "compounded semi annually" or "compounded quarterly" or compounded monthly".
But, always we have question about compounded continuously.
To understand "compounded continuously", let us consider the example given below.
When we invest some money in a bank, it will grow continuously. That is, at any instant the balance is changing at a rate that equals "r" (rate of interest per year) times the current balance.
Compounded continuously formula
Many real world phenomena are being modeled by functions which describe how things grow continuously at any instance.
Formula :
The formula given below is related to compound interest formula and represents the case where interest is being compounded continuously.
That is, at any instant the balance is changing at a rate that equals "r" times the current balance.
We use this formula, when it is given "compounded continuously"
A ---> Ending amount
---> Beginning amount
---> Growth rate
---> Time
Compounded continuously formula- Examples
Example 1 :
You invest \$2500 in bank which pays 10% interest per year compounded continuously. What will be the value of the investment after 10 years ?
Solution :
We have to use the formula given below to know the value of the investment after 3 years.
Here,
A = Final value of the deposit
P = 2500, r = 10% or 0.1, t = 10, e = 2.71828 and also
rt = 0.1x10 = 1
A = 2500(2.71828)¹
A = 6795.70
Hence, the value of the investment after 10 years is \$6795.70
Example 2 :
If David invests \$500 at annual rate of 20% compounded continuously, calculate the final amount that David will have after 5 years.
Solution :
We have to use the formula given below to know the final amount that David will have after 5 years .
Here,
A = Final value of the deposit
P = 500, r = 20% or 0.2, t = 5, e = 2.71828 and also
rt = 0.2x5 = 1
A = 500(2.71828)¹
A = 1359.14
Hence, the final amount that David will have after 5 years is \$1359.14
After having gone through the stuff given above, we hope that the students would have understood "continuous compounding formula". |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
## Categories of quadrilaterals based on sides and angles.
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What if you were given the coordinates of four points that form a quadrilateral? How could you determine if that quadrilateral qualifies as one of the special four-sided figures you learned about in the previous Concepts? After completing this Concept, you'll be able to make such a determination.
### Watch This
CK-12 Classifying Quadrilaterals in the Coordinate Plane
### Guidance
In order to be successful in this concept you need to already be familiar with the definitions and properties of the following quadrilaterals: parallelograms, rhombuses, rectangles, squares, kites and trapezoids. The definitions for each are provided in the vocabulary section as a resource, and further information can be found by searching on those topic words.
When working in the coordinate plane, you will sometimes want to know what type of shape a given shape is. You should easily be able to tell that it is a quadrilateral if it has four sides. But how can you classify it beyond that?
First you should graph the shape if it has not already been graphed. Look at it and see if it looks like any special quadrilateral. Do the sides appear to be congruent? Do they meet at right angles? This will give you a place to start.
Once you have a guess for what type of quadrilateral it is, your job is to prove your guess. To prove that a quadrilateral is a parallelogram, rectangle, rhombus, square, kite or trapezoid, you must show that it meets the definition of that shape OR that it has properties that only that shape has.
If it turns out that your guess was wrong because the shape does not fulfill the necessary properties, you can guess again. If it appears to be no type of special quadrilateral then it is simply a quadrilateral.
The examples below will help you to see what this process might look like.
#### Example A
Determine what type of parallelogram TUNE\begin{align*}TUNE\end{align*} is: T(0,10),U(4,2),N(2,1)\begin{align*}T(0, 10), U(4, 2), N(-2, -1)\end{align*}, and E(6,7)\begin{align*}E(-6, 7)\end{align*}.
This looks like a rectangle. Let’s see if the diagonals are equal. If they are, then TUNE\begin{align*}TUNE\end{align*} is a rectangle.
EU=(64)2+(72)2=(10)2+52=100+25=125TN=(0+2)2+(10+1)2 =22+112 =4+121 =125
If the diagonals are also perpendicular, then TUNE\begin{align*}TUNE\end{align*} is a square.
Slope of EU=7264=510=12Slope of TN=10(1)0(2)=112\begin{align*}\text{Slope of}\ EU = \frac{7 - 2}{-6 - 4} = -\frac{5}{10} = -\frac{1}{2} \quad \text{Slope of}\ TN = \frac{10 - (-1)}{0-(-2)} = \frac{11}{2}\end{align*}
The slope of EU\begin{align*}EU \neq\end{align*} slope of TN\begin{align*}TN\end{align*}, so TUNE\begin{align*}TUNE\end{align*} is a rectangle.
#### Example B
A quadrilateral is defined by the four lines y=2x+1\begin{align*}y=2x+1\end{align*}, y=x+5\begin{align*}y=-x+5\end{align*}, y=2x4\begin{align*}y=2x-4\end{align*}, and y=x5\begin{align*}y=-x-5\end{align*}. Is this quadrilateral a parallelogram?
To check if its a parallelogram we have to check that it has two pairs of parallel sides. From the equations we can see that the slopes of the lines are 2\begin{align*}2\end{align*}, 1\begin{align*}-1\end{align*}, 2\begin{align*}2\end{align*} and 1\begin{align*}-1\end{align*}. Because two pairs of slopes match, this shape has two pairs of parallel sides and is a parallelogram.
#### Example C
Determine what type of quadrilateral RSTV\begin{align*}RSTV\end{align*} is.
This looks like a kite. Find the lengths of all the sides to check if the adjacent sides are congruent.
RS=(52)2+(76)2=(7)2+12=50ST=(25)2+(6(3))2 =(3)2+92 =90
RV=(5(4))2+(70)2=(1)2+72=50VT=(45)2+(0(3))2 =(9)2+32 =90
From this we see that the adjacent sides are congruent. Therefore, RSTV\begin{align*}RSTV\end{align*} is a kite.
CK-12 Classifying Quadrilaterals in the Coordinate Plane
-->
### Guided Practice
1. A quadrilateral is defined by the four lines y=2x+1\begin{align*}y=2x+1\end{align*}, y=2x+5\begin{align*}y=-2x+5\end{align*}, y=2x4\begin{align*}y=2x-4\end{align*}, and y=2x5\begin{align*}y=-2x-5\end{align*}. Is this quadrilateral a rectangle?
2. Determine what type of quadrilateral ABCD\begin{align*}ABCD\end{align*} is. A(3,3),B(1,5),C(4,1),D(1,5)\begin{align*}A(-3, 3), B(1, 5), C(4, -1), D(1, -5)\end{align*}.
3. Determine what type of quadrilateral EFGH\begin{align*}EFGH\end{align*} is. E(5,1),F(11,3),G(5,5),H(1,3)\begin{align*}E(5, -1), F(11, -3), G(5, -5), H(-1, -3)\end{align*}
1. To be a rectangle a shape must have four right angles. This means that the sides must be perpendicular to each other. From the given equations we see that the slopes are 2\begin{align*}2\end{align*}, 2\begin{align*}-2\end{align*}, 2\begin{align*}2\end{align*} and 2\begin{align*}-2\end{align*}. Because the slopes are not opposite reciprocals of each other, the sides are not perpendicular, and the shape is not a rectangle.
2. First, graph ABCD\begin{align*}ABCD\end{align*}. This will make it easier to figure out what type of quadrilateral it is. From the graph, we can tell this is not a parallelogram. Find the slopes of BC¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} and AD¯¯¯¯¯¯\begin{align*}\overline{AD}\end{align*} to see if they are parallel.
Slope of BC¯¯¯¯¯=5(1)14=63=2\begin{align*}\overline{BC} = \frac{5-(-1)}{1-4} = \frac{6}{-3} = -2\end{align*}
Slope of AD¯¯¯¯¯¯=3(5)31=84=2\begin{align*}\overline{AD} = \frac{3-(-5)}{-3-1} = \frac{8}{-4} = -2\end{align*}
BC¯¯¯¯¯AD¯¯¯¯¯¯\begin{align*}\overline{BC} \| \overline{AD}\end{align*}, so ABCD\begin{align*}ABCD\end{align*} is a trapezoid. To determine if it is an isosceles trapezoid, find AB\begin{align*}AB\end{align*} and CD\begin{align*}CD\end{align*}.
AB=(31)2+(35)2=(4)2+(2)2=20=25ST=(41)2+(1(5))2 =32+42 =25=5
ABCD\begin{align*}AB \neq CD\end{align*}, therefore this is not an isosceles trapezoid.
3. We will not graph this example. Let’s find the length of all four sides.
EFGH=(511)2+(1(3))2=(6)2+22=40=(5(1))2+(5(3))2=62+(2)2=40 FG=(115)2+(3(5))2 =62+22=40HE=(15)2+(3(1))2 =(6)2+(2)2=40
All four sides are equal. This quadrilateral is either a rhombus or a square. Let’s find the length of the diagonals.
EG=(55)2+(1(5))2=02+42=16=4 FH=(11(1))2+(3(3))2=122+02=144=12
The diagonals are not congruent, so EFGH\begin{align*}EFGH\end{align*} is a rhombus and not a square.
### Explore More
Determine what type of quadrilateral ABCD\begin{align*}ABCD\end{align*} is.
1. A(2,4),B(1,2),C(3,1),D(4,3)\begin{align*}A(-2, 4), B(-1, 2), C(-3, 1), D(-4, 3)\end{align*}
2. \begin{align*}A(-2, 3), B(3, 4), C(2, -1), D(-3, -2)\end{align*}
3. \begin{align*}A(1, -1), B(7, 1), C(8, -2), D(2, -4)\end{align*}
4. \begin{align*}A(10, 4), B(8, -2), C(2, 2), D(4, 8)\end{align*}
5. \begin{align*}A(0, 0), B(5, 0), C(0, 4), D(5, 4)\end{align*}
6. \begin{align*}A(-1, 0), B(0, 1), C(1, 0), D(0, -1)\end{align*}
7. \begin{align*}A(2, 0), B(3, 5), C(5, 0), D(6, 5)\end{align*}
\begin{align*}SRUE\end{align*} is a rectangle and \begin{align*}PRUC\end{align*} is a square.
1. What type of quadrilateral is \begin{align*}SPCE\end{align*}?
2. If \begin{align*}SR = 20\end{align*} and \begin{align*}RU = 12\end{align*}, find \begin{align*}CE\end{align*}.
3. Find \begin{align*}SC\end{align*} and \begin{align*}RC\end{align*} based on the information from part b. Round your answers to the nearest hundredth.
For questions 11-14, determine what type of quadrilateral \begin{align*}ABCD\end{align*} is. If it is no type of special quadrilateral, just write quadrilateral.
1. \begin{align*}A(1, -2), B(7, -5), C(4, -8), D(-2, -5)\end{align*}
2. \begin{align*}A(6, 6), B(10, 8), C(12, 4), D(8, 2)\end{align*}
3. \begin{align*}A(-1, 8), B(1, 4), C(-5, -4), D(-5, 6)\end{align*}
4. \begin{align*}A(5, -1), B(9, -4), C(6, -10), D(3, -5)\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 6.8.
### Vocabulary Language: English Spanish
isosceles trapezoid
isosceles trapezoid
An isosceles trapezoid is a trapezoid where the non-parallel sides are congruent.
midsegment (of a trapezoid)
midsegment (of a trapezoid)
A line segment that connects the midpoints of the non-parallel sides.
Rectangle
Rectangle
A rectangle is a quadrilateral with four right angles.
Rhombus
Rhombus
A rhombus is a quadrilateral with four congruent sides.
square
square
A parallelogram is a square if and only if it has four right angles and four congruent sides. {{Inline image |source=Image:geo-0603-04b.png|size=100px}}
Trapezoid
Trapezoid
A trapezoid is a quadrilateral with exactly one pair of parallel opposite sides.
Kite
Kite
Parallelogram
Parallelogram
A parallelogram is a quadrilateral with two pairs of parallel sides. |
# How do you write an equation for a circle with center (2,4) and passes through point (5, 9)?
Apr 16, 2016
${\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} = 34$
#### Explanation:
The radius can be found using Pythagoras Theorem
$r = \sqrt{{\left(5 - 2\right)}^{2} + {\left(9 - 4\right)}^{2}} = \sqrt{34}$
For a circle with radius $r$ and centered at $\left(a , b\right)$, its cartesian equation is
${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
Therefore. for a circle with radius $\sqrt{34}$ and centered at $\left(2 , 4\right)$. its cartesian equation is
${\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} = 34$ |
## Engage NY Eureka Math 5th Grade Module 4 Lesson 4 Answer Key
### Eureka Math Grade 5 Module 4 Lesson 4 Problem Set Answer Key
Question 1.
Draw a tape diagram to solve. Express your answer as a fraction. Show the multiplication sentence to check your answer. The first one is done for you.
a. 1 ÷ 3 = $$\frac{1}{3}$$
b. 2 ÷ 3 =
2/3
Explanation:
c. 7 ÷ 5 =
1 2/5
Explanation:
d. 14 ÷ 5 =
2 4/5
Explanation:
Question 2.
Fill in the chart. The first one is done for you.
Question 3.
Greg spent $4 on 5 packs of sport cards. a. How much did Greg spend on each pack? Answer: Greg spent$80 on each pack.
Explanation:
5 units = $4 1 unit =$4 ÷ 5 = 4/5
4/5 of $1=$80
b. If Greg spent half as much money and bought twice as many packs of cards, how much did he spend on each pack? Explain your thinking.
He spent $20 dollars on each pack. Explanation: 10 units =$2
1 unit = 2 ÷ 10 = 2/10 = 1/5
1/5 of $1 =$20. Hence he spent \$20 on each pack.
Question 4.
Five pounds of birdseed is used to fill 4 identical bird feeders.
a. What fraction of the birdseed will be needed to fill each feeder?
1/4
Explanation:
There are 4 identical bird feeders, so 1/4 of the birdseed would be needed to fill each feeder.
b. How many pounds of birdseed are used to fill each feeder? Draw a tape diagram to show your thinking.
1 1/4
4 units = 5lb
1 unit = 5lb ÷ 4
= 5/4
= 1 1/4
Explanation:
1 1/4 lb of birdseed is used to fill each feeder.
c. How many ounces of birdseed are used to fill three bird feeders?
1 lb = 16 oz
1 1/4 = x
= 1 1/4 * 16 oz
= 16 oz + 4 oz
= 20 oz
1 unit = 20ounces
3 units = 3*20 = 60 ounces
Explanation:
60 ounces of birdseed are used to fill three birdfeeders.
### Eureka Math Grade 5 Module 4 Lesson 4 Exit Ticket Answer Key
Matthew and his 3 siblings are weeding a flower bed with an area of 9 square yards. If they share the job equally, how many square yards of the flower bed will each child need to weed? Use a tape diagram to show your thinking.
Each child will need to weed =2 1/4 square yards
Explanation:
Matthew and his 3 siblings are weeding a flower bed with an area of 9 square yards.
Matthew and his 3 siblings = 1 + 3 = 4
The Total area given is = 9 square yards.
Hence, Each child will need to weed = 9/4 square yards
= 2 1/4 square yards
### Eureka Math Grade 5 Module 4 Lesson 4 Homework Answer Key
Question 1.
a. 1 ÷ 4 = $$\frac{1}{4}$$
b. 4 ÷ 5 =
c. 8 ÷ 5 =
8/5 = 1 3/8
Explanation:
d. 14 ÷ 3 =
2 4/5
Explanation:
Question 2.
Fill in the chart. The first one is done for you.
Question 3.
Jackie cut a 2-yard spool into 5 equal lengths of ribbon.
a. What is the length of each ribbon in yards? Draw a tape diagram to show your thinking.
2/5 yard
Explanation:
2 ÷ 5 = 2/5 yards
b. What is the length of each ribbon in feet? Draw a tape diagram to show your thinking.
1 1/5 ft
Explanation:
1 yd * 0.4 / 3ft * 0.4 = 2/5 yard / ? feet
3 * 2/5 = 6/5 = 1 1/5
Question 4.
Baa Baa, the black sheep, had 7 pounds of wool. If he separated the wool equally into 3 bags, how much wool would be in 2 bags?
4 2/3 pounds
Explanation:
2 bags = 2 1/3 + 2 1/3
= 4 2/3 pounds
Question 5.
An adult sweater is made from 2 pounds of wool. This is 3 times as much wool as it takes to make a baby sweater. How much wool does it take to make a baby sweater? Use a tape diagram to solve. |
## What is Bezier interpolation?
In general, our Bezier curve interpolation is the fixed interpolation which means that the shape of the interpolating curve is fixed for the given interpolating data and control polygon, since the interpolating function is unique for the given control points.
## What is Bezier geometry?
A Bézier curve (/ˈbɛz. i. eɪ/ BEH-zee-ay) is a parametric curve used in computer graphics and related fields. The curves, which are related to Bernstein polynomials, are named after French engineer Pierre Bézier, who used it in the 1960s for designing curves for the bodywork of Renault cars.
## What is a cubic Bezier?
The cubic-bezier() functional notation defines a cubic Bézier curve. As these curves are continuous, they are often used to smooth down the start and end of the interpolation and are therefore sometimes called easing functions. A cubic Bézier curve is defined by four points P0, P1, P2, and P3.
## How does a Bezier curve work?
By combining multiple curves together you can create any shape you want. The Bézier curve is the fundamental primitive of curved shapes. Being defined as a polynomial means we can do interesting things with it, like evaluate the equation at any point we want, then make objects move along the point.
## How is Bezier curve calculated?
Maths
1. The formula for a 2-points curve: P = (1-t)P1 + tP2
2. For 3 control points: P = (1−t)2P1 + 2(1−t)tP2 + t2P3
3. For 4 control points: P = (1−t)3P1 + 3(1−t)2tP2 +3(1−t)t2P3 + t3P4
## When you change the shape of a Bezier What does it do?
Bezier handles are two-directional controls that change the curve of the line segment between the handle and the next point on either side. The farther you pull a handle from its keyframe (center point), the more the line bends or curves.
## What is a Nurbs curve?
NURBS, Non-Uniform Rational B-Splines, are mathematical representations of 3D geometry that can accurately describe any shape from a simple 2D line, circle, arc, or curve to the most complex 3D organic free-form surface or solid.
## What is quadratic Bezier curve?
Quadratic Bezier curve is a point-to-point linear interpolation of two Linear Bezier Curves. For given three points P0, P1 and P2, a quadratic bezier curve is a linear interpolation of two points, got from Linear Bezier curve of P0 and P1 and Linear Bezier Curve of P1 and P2.
## What is a cubic Bézier transition?
A Cubic Bezier curve is defined by four points P0, P1, P2, and P3. P0 and P3 are the start and the end of the curve and, in CSS these points are fixed as the coordinates are ratios. The cubic-bezier() function can be used with the animation-timing-function property and the transition-timing-function property.
## How do you plot a Bezier curve?
To draw a line using this equation, one can divide the curve into smaller segments, calculate the end points of each segment using the Bezier cubic equation and draw the line for the segment. For instance, one can draw a line between the points defined by t = 0 and t = 0.01, then t = 0.01 and t = 0.02, and so on.
## Is it possible to reduce the degree of Bezier curve?
In contrast to other methods, ours minimizes the L_2-error for the whole composite curve instead of minimizing the L_2-errors for each segment separately. As a result, an additional optimization is possible.
## Which curves allows local control of curve?
B-spline allows the local control over the curve surface because each vertex affects the shape of a curve only over a range of parameter values where its associated basis function is nonzero. The curve exhibits the variation diminishing property. The curve generally follows the shape of defining polygon. |
A Realistic Numbers Worksheet might help your child be a little more knowledgeable about the concepts behind this percentage of integers. With this worksheet, students will be able to resolve 12 various issues related to reasonable expressions. They are going to learn how to increase two or more phone numbers, group of people them in couples, and figure out their items. They will likely also process simplifying logical expression. As soon as they have mastered these concepts, this worksheet will certainly be a valuable device for continuing their studies. Multiplying Rational Numbers Worksheet 7th Grade Answers.
## Logical Numbers certainly are a ratio of integers
The two main varieties of phone numbers: irrational and rational. Rational figures are considered entire figures, whereas irrational figures will not perform repeatedly, and get an limitless number of digits. Irrational figures are no-zero, non-terminating decimals, and square beginnings that are not best squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To outline a realistic number, you need to realize just what a rational number is. An integer is a complete quantity, plus a realistic amount is a rate of two integers. The rate of two integers is the amount at the top separated by the amount at the base. For example, if two integers are two and five, this would be an integer. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They can be produced in a fraction
A reasonable quantity includes a denominator and numerator that are not no. This means that they are often indicated like a small percentage. Together with their integer numerators and denominators, logical phone numbers can in addition have a bad benefit. The negative benefit should be located left of along with its definite benefit is its length from no. To easily simplify this instance, we will claim that .0333333 is a fraction that may be published as being a 1/3.
Together with bad integers, a reasonable variety can even be manufactured into a small percentage. For instance, /18,572 is actually a realistic quantity, when -1/ is just not. Any small fraction consisting of integers is rational, given that the denominator is not going to contain a and might be published for an integer. Similarly, a decimal that leads to a stage is another rational variety.
## They can make perception
Even with their name, realistic phone numbers don’t make very much sense. In mathematics, they are solitary organizations with a distinctive span about the quantity range. Which means that once we count up anything, we are able to buy the size and style by its ratio to its authentic quantity. This holds correct even if you can find limitless logical numbers in between two particular figures. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To get the period of a pearl, for instance, we could count up its size. One particular pearl weighs in at ten kilos, which is a rational number. In addition, a pound’s body weight means twenty kgs. Thus, we should certainly split a lb by twenty, without having be concerned about the length of an individual pearl.
## They can be expressed as a decimal
If you’ve ever tried to convert a number to its decimal form, you’ve most likely seen a problem that involves a repeated fraction. A decimal variety can be composed as being a several of two integers, so 4x 5 various is the same as seven. The same dilemma necessitates the repeated small percentage 2/1, and both sides must be divided by 99 to get the right response. But how can you have the conversion? Here are a few good examples.
A rational amount can also be designed in great shape, which includes fractions plus a decimal. A great way to stand for a reasonable quantity in the decimal is to separate it into its fractional counterpart. There are actually 3 ways to split a logical number, and all these techniques yields its decimal counterpart. One of these approaches is to break down it into its fractional comparable, and that’s what’s called a terminating decimal. |
# How do you evaluate 2^4-6-:2?
Feb 23, 2017
You will use PEMDAS, the order of operations.
#### Explanation:
PEMDAS stands for:
• Parentheses
• Exponents
• Multiplication
• Division
• Subtraction
PEMDAS tells us the order in which we should solve equations. We can apply it to the current problem. First, we need to solve everything in parentheses. However, there are no parentheses in the problem. So we move on to exponents. There is an exponent in the expression, so we solve it like this:
${2}^{4} = 2 \cdot 2 \cdot 2 \cdot 2 = 16$
Then, we continue through PEMDAS. Multiplication is next, but there is none of that in the problem. We move on to the division, which is solved like this:
$\frac{6}{2} = 3$
So, we now have:
$16 - 3$
Addition comes next in PEMDAS, but we have none of that, so we do the final PEMDAS step, the subtraction.
$16 - 3 = 13$ |
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# Simplify the given Expression : $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$.
Last updated date: 16th Jul 2024
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Hint:The given problem is related to simplification by factorization. Express each term as a product of its factors and then simplify the expression.
We will proceed with the solution by taking each term, factoring it, and then substituting the factored form in the given expression. Then we will simplify the given expression.
The given expression is $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$. The first term of the expression is ${{a}^{2}}+10a+21$. It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $10a$ into two terms, such as their sum is equal to $10a$ and their product is equal to $21{{a}^{2}}$ .
We can write $10a$ as $7a+3a$ . Here, the sum of $7a$ and $3a$ is $10a$ and the product of $7a$ and $3a$ is $21{{a}^{2}}$ . So, ${{a}^{2}}+10a+21$can be written as ${{a}^{2}}+7a+3a+21$ .
$\Rightarrow {{a}^{2}}+10a+21=a\left( a+7 \right)+3\left( a+7 \right)$
$\Rightarrow {{a}^{2}}+10a+21=\left( a+3 \right)\left( a+7 \right)$
So, we can write ${{a}^{2}}+10a+21$ as $\left( a+3 \right)\left( a+7 \right)$ .
Now, the second term is ${{a}^{2}}+6a-7$ . It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $6a$ into two terms, such as their sum is equal to $6a$ and their product is equal to $-7{{a}^{2}}$ .
We can write $6a$ as $7a-a$ . Here, the sum of $7a$ and $-a$ is $6a$ and the product of $7a$ and $-a$ is $-7{{a}^{2}}$ . So, ${{a}^{2}}+6a-7$can be written as ${{a}^{2}}+7a-a-7$ .
$\Rightarrow {{a}^{2}}+6a-7=a\left( a+7 \right)-1\left( a+7 \right)$
$\Rightarrow {{a}^{2}}+6a-7=\left( a-1 \right)\left( a+7 \right)$
So, we can write ${{a}^{2}}+6a-7$ as $\left( a-1 \right)\left( a+7 \right)$ .
Now, the third term is ${{a}^{2}}-1$ . We know, we can write ${{a}^{2}}-1$ as ${{a}^{2}}-{{1}^{2}}$. So, ${{a}^{2}}-1={{a}^{2}}-{{1}^{2}}=\left( a+1 \right)\left( a-1 \right)$ . So, we can write ${{a}^{2}}-1$ as $\left( a+1 \right)\left( a-1 \right)$
Now, the fourth term is $\left( a+3 \right)$ . It is already in its simplest form. So, we don’t need to factorize it.
Now, the expression $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ can be written as $\dfrac{\left( a+7 \right)\left( a+3 \right)}{\left( a+7 \right)\left( a-1 \right)}\times \dfrac{\left( a-1 \right)\left( a+1 \right)}{\left( a+3 \right)}$ .
$\Rightarrow \dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}=a+1$
Hence, the simplified value of $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ is $a+1$ .
Note: While making substitutions, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong. |
Jessica Wesaquate and Andrea Rogers
Strand: Number Grade Level: Three Students will be able to view how a tipi raising is performed. Students will be able to create estimations based on their observations of the tipi raising video clips. Materials: tipi-raising videos, graph paper, pencils, math logs/duo tangs/journals Video Clips: As the teacher you may choose to use the video clips that demonstrate the tipi-raising done with Elder Glen Anaquod using a Saulteaux perspective. Or you may choose the video clips that demonstrate the tipi raising done with Tim Haywahe using a Nakota perspective. Depending on your area, it may be appropriate to choose one over the other. If time permits, showing them both tipi raisings is a good opportunity to compare and contrast different traditions and teachings. Introduction: As a class, you are going to show the students the tipi raising videos. Start with showing them the clip that demonstrates them measuring the first poles on the canvas. As they are viewing the videos, have students make estimations to how many people they think could fit comfortably in this tipi? They should record this in their math logs/duo tangs/journals. Step Two: Play the next clips. Pause the clip when the canvas is 1/2 to ¾ around the poles. Allow students to re-think their estimation. Do they still want to go with their original estimation or make a new estimation? Have them record their estimation and state whether it is new or if it remained the same. Step Three: Watch the remainder of the videos and pause the last slide where the students can see the complete tipi. This is their last opportunity to either keep their estimation, or create a new one, and record. Introduce or review what mean, median and mode are with your students. Choose five students’ numbers to work with and record these on the board (students should have recorded three numbers each). Have students work individually to determine what the mean, median and mode are of those numbers. Review as a class. Mean: Average Median: arrange the numbers from lowest to highest and the median is the middle number Activity: On construction paper, have students trace their hand (fingers together and thumb close in) and then cut it out. This will be used as a non-standard measuring tool for items around the classroom. If you have cultural items available to measure, great, but if not this can be used for basically anything. For example, if you have a rain stick in your classroom have students measure how many hand prints long it is. Remember to share background on the meaning of the rain stick so they understand its significance. Other items you can measure are things like their desks, tables, drawers, windows, and etcetera. You can set up stations for this activity so that students are not trying to measure the same thing all at the same time, also so that you can take anecdotal records on the way they measure items*, behaviors, other. *Do they measure items with the hand vertically or horizontally, do they use the width of their hand cut-out or the length of it? Mode: the number that occurs most often Optional Activities: Have the students do some graphing with this information. Students can create a line or bar graph. Have them note where they see the most common guesses. Have the class make a consensus on how many people can fit into this tipi. Find a place outdoors around the school where you can measure a diameter of anywhere for 10 to 25 feet using broken branches, or other objects Mother Nature provides us with (to represent the tipi) to explore their estimations. You can also use masking tape if needed. Have students sit inside the circle, how many fit comfortably? |
Chapter 2: Linear Equations
# 2.2 Solving Linear Equations
When working with questions that require two or more steps to solve, do the reverse of the order of operations to solve for the variable.
Example 2.2.1
Solve $4x+16=-4$ for $x.$
$\begin{array}{rrrrrl} 4x& +& 16 &=&-4& \\ &&-16&& -16&\text{Subtract 16 from each side} \\ \hline &&\dfrac{4x}{4}& =& \dfrac{-20}{4}&\text{Divide each side by 4}\\ \\ &&x& =& -5 & \text{Solution} \end{array}$
In solving the above equation, notice that the general pattern followed was to do the opposite of the equation. $4x$ was added to 16, so 16 was then subtracted from both sides. The variable $x$ was multiplied by 4, so both sides were divided by 4.
Example 2.2.2
1. $\begin{array}[t]{rrrrr} 5x &+& 7 &=& 7 \\ &-&7&&-7 \\ \hline &&\dfrac{5x}{5} &=& \dfrac{0}{5}\\ \\ &&x& =& 0 \end{array}$
2. $\begin{array}[t]{rrrrr} 4 &- &2x& =& 10 \\ -4&&&&-4\\ \hline &&\dfrac{-2x}{-2}& = &\dfrac{6}{-2}\\ \\ &&x& =& -3 \end{array}$
3. $\begin{array}[t]{rrrrr} -3x& -& 7& =& 8 \\ &+&7& = & + 7 \\ \hline &&\dfrac{-3x}{-3}& =& \dfrac{15}{-3} \\ \\ &&x &= &-5 \end{array}$
# Questions
For questions 1 to 20, solve each linear equation.
1. $5 + \dfrac{n}{4} = 4$
2. $-2 = -2m + 12$
3. $102 = -7r + 4$
4. $27 = 21 - 3x$
5. $-8n + 3 = -77$
6. $-4 - b = 8$
7. $0 = -6v$
8. $-2 + \dfrac{x}{2} = 4$
9. $-8 = \dfrac{x}{5} - 6$
10. $-5 = \dfrac{a}{4} - 1$
11. $0 = -7 + \dfrac{k}{2}$
12. $-6 = 15 + 3p$
13. $-12 + 3x = 0$
14. $-5m + 2 = 27$
15. $\dfrac{b}{3} + 7 = 10$
16. $\dfrac{x}{1} - 8 = -8$
17. $152 = 8n + 64$
18. $-11 = -8 + \dfrac{v}{2}$
19. $-16 = 8a + 64$
20. $-2x - 3 = -29$ |
# Quadratic Equations Class 10 Notes Maths Chapter 4
##### CHAPTER AT A GLANCE
Standard from of the quadratic equation in the variable x is an equation of the form ax2+bx+c= 0, where a, b, c are real numbers, a not= 0.
Any equation of the form P(x)= 0, where P(x) is a polynomial of degree 2, is a quadratic equation.
A real number a is said to be a root of the quadratic equation ax2 +bx+ c = 0, a not= 0 ifa a2 + ba + c = 0.
We can say that x = a, is a solution of the quadratic equation or that a satisfies the quadratic equation.
The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the equation ax2 + bx + c = 0 are same.
A quadratic equation has at most two roots/zeroes.
3. Relation between Zeroes and Co-efficient of a Quadratic Equation
4. Methods of Solving Quadratic Equation
Following are the methods which are used to solve quadratic equations:
(i) Factorisation.
(ii) Completing the square.
5. Methods of Factorisation
In this method we find the roots of a quadratic equation (ax2 +bx+ c = 0) by factorising LHS it into two linear factors and equating each factor to zero:
6. Method of Completing the Square
This is the method of converting L.H.S. of a quadratic equation which is not a perfect square into the sum or difference of a perfect square and a constant by adding and subtracting the suitable constant terms.
Consider a quadratic equation: ax2+bx+c = 0.
If b2-4ac => 0,then the roots of the above equation are given by:
8. Nature of Roots
(a not=0),value of (b2-4ac) is called discriminant of the equation and denoted as D.
:. D=b2-4ac
Discriminant is very important in finding nature of the roots.
(i) If D = 0, then roots are real and equal.
(ii) If D > 0, then roots are real and unequal
(iii) If D < 0, then roots are not real.
Related Articles: |
finding out Objectives Graphing position as a duty of Time exercise Problems inspect Your knowledge
### Learning Objectives
By the end of this section, girlfriend will have the ability to do the following:
Explain the meaning of steep in place vs. Time graphsSolve troubles using position vs. Time graphsSection vital Terms
dependent variable independent variable tangent
### Graphing place as a role of Time
A graph, like a picture, is worth a thousands words. Graphs not only contain number information, they also reveal relationships between physical quantities. In this section, we will investigate kinematics by analyzing graphs of place over time.
You are watching: The slope at a point on a position versus time graph of an object is
Graphs in this text have perpendicular axes, one horizontal and also the other vertical. Once two physical amounts are plotted versus each other, the horizontal axis is usually considered the live independence variable, and the upright axis is the dependent variable. In algebra, friend would have referred to the horizontal axis together the x-axis and the upright axis as the y-axis. Together in figure 2.11, a straight-line graph has actually the general type y=mx+by=mx+b.
Here m is the slope, identified as the rise split by the operation (as seen in the figure) that the directly line. The letter b is the y-intercept which is the allude at i beg your pardon the line the cross the vertical, y-axis. In regards to a physical instance in the real world, these quantities will take on a certain significance, together we will see below. (Figure 2.11.)
Figure 2.11 The diagram mirrors a straight-line graph. The equation for the right line is y amounts to mx + b.
In physics, time is typically the live independence variable. Other quantities, such as displacement, are claimed to rely upon it. A graph of place versus time, therefore, would have position top top the vertical axis (dependent variable) and also time on the horizontal axis (independent variable). In this case, come what would the slope and also y-intercept refer? stop look ago at our original instance when examining distance and also displacement.
The journey to institution was 5 kilometres from home. Let’s assume it took 10 minute to do the drive and also that your parental was driving in ~ a constant velocity the whole time. The position versus time graph because that this ar of the trip would look choose that displayed in number 2.12.
Figure 2.12 A graph of place versus time because that the drive to college is shown. What would certainly the graph look choose if we added the return trip?
As we said before, d0 = 0 due to the fact that we contact home our O and start calculating indigenous there. In number 2.12, the line starts in ~ d = 0, together well. This is the b in ours equation for a directly line. Our initial position in a position versus time graph is constantly the location where the graph crosses the x-axis in ~ t = 0. What is the slope? The rise is the adjust in position, (i.e., displacement) and also the run is the adjust in time. This partnership can likewise be composed
This relationship was how we identified average velocity. Therefore, the steep in a d matches t graph, is the mean velocity.
Sometimes, as is the case where we graph both the trip to school and the return trip, the actions of the graph looks various during various time intervals. If the graph looks choose a collection of straight lines, then you have the right to calculate the median velocity because that each time interval by looking in ~ the slope. If you then desire to calculation the median velocity for the whole trip, you deserve to do a weight average.
Let’s look at at another example. Number 2.13 mirrors a graph of position versus time because that a jet-powered auto on a very flat dry lake bed in Nevada.
Figure 2.13 The diagram shows a graph of position versus time for a jet-powered vehicle on the Bonneville Salt Flats.
Using the relationship between dependent and independent variables, we check out that the slope in the graph in number 2.13 is average velocity, vavg and the intercept is displacement in ~ time zero—that is, d0. Substituting this symbols right into y = mx + b offers
2.5d=vt+d0d=vt+d0
or
2.6d=d0+vt.d=d0+vt.
Thus a graph of place versus time gives a basic relationship among displacement, velocity, and also time, and also giving detailed numerical information around a particular situation. From the number we deserve to see that the car has a position of 400 m at t = 0 s, 650 m at t = 1.0 s, and also so on. And we have the right to learn around the object’s velocity, as well.
Graphing Motion
In this activity, you will certainly release a round down a ramp and also graph the ball’s displacement vs. Time.
Choose an open ar with too many of space to spread out so there is much less chance for tripping or falling because of rolling balls.
1 ball1 board2 or 3 books1 stopwatch1 tape measure6 pieces of masking tape1 piece of graph paper1 pencil
Build a ramp by put one end of the plank on top of the ridge of books. Change location, as necessary, till there is no obstacle follow me the right line course from the bottom that the ramp till at the very least the following 3 m. Mark distances of 0.5 m, 1.0 m, 1.5 m, 2.0 m, 2.5 m, and 3.0 m indigenous the bottom that the ramp. Write the ranges on the tape.Have one human take the duty of the experimenter. This person will relax the sphere from the optimal of the ramp. If the round does no reach the 3.0 m mark, then increase the incline that the ramp by adding another book. Repeat this action as necessary.Have the experimenter release the ball. Have a 2nd person, the timer, begin timing the trial once the ball reaches the bottom of the ramp and also stop the timing once the sphere reaches 0.5 m. Have a 3rd person, the recorder, record the time in a data table.Repeat action 4, protecting against the time at the distances of 1.0 m, 1.5 m, 2.0 m, 2.5 m, and 3.0 m indigenous the bottom of the ramp.Use your measurements of time and the displacement to make a position vs. Time graph that the ball’s motion.Repeat actions 4 through 6, through different people taking ~ above the roles of experimenter, timer, and recorder. Perform you get the same measurement values regardless of that releases the ball, steps the time, or records the result? Discuss feasible causes the discrepancies, if any.
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Grasp Check
True or False: The average speed that the ball will be much less than the median velocity that the ball.
True False
Solving troubles Using position vs. Time Graphs
So how do we usage graphs to resolve for points we desire to recognize like velocity? |
# LCM of 10 and 12
Welcome to our blog on the least common multiple (LCM) of 10 and 12 In this blog, we will be diving into the different methods for finding the LCM, including the prime factorization method, the listing multiples method, and the division method. Learn how to find the LCM and make your life a little bit easier in scheduling or managing time.
## LCM of 10 and 12
The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both numbers. To find the LCM of 10 and 12, you can list out the multiples of each number and look for the smallest one that they have in common.
A more efficient method is to use the formula for the LCM, which is:
LCM(a,b) = (a x b) / GCD(a,b)
Where GCD is the greatest common divisor of a and b.
In this case, GCD of 10 and 12 is 2
so, LCM(10,12) = (10 x 12) / 2 = 60
Hence, LCM of 10 and 12 is 60
## LCM of 10 and 12 though Listing Multiples
The method of listing multiples to find the LCM of two numbers involves listing out the multiples of each number and looking for the smallest one that they have in common.
Multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, …
Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180, …
As we can see the smallest common multiple of 10 and 12 is 60, which is the LCM of 10 and 12.
Hence, LCM of 10 and 12 by Listing Multiples method is 60.
## LCM of 10 and 12 by Prime Factorization
The prime factorization method for finding the LCM of two numbers involves breaking down each number into its prime factors, then multiplying the highest powers of each prime factor together.
To find the prime factorization of 10, we can start by dividing it by the smallest prime number, 2.
10 / 2 = 5
5 is a prime number, so the prime factorization of 10 is 2 * 5
To find the prime factorization of 12, we can start by dividing it by the smallest prime number, 2.
12 / 2 = 6
6 is not a prime number, it can be further divided by 2 to get 3.
so the prime factorization of 12 is 2^2 * 3
To find the LCM, we take the highest powers of each prime factor and multiply them together.
LCM(10, 12) = (2^2) * (3^1) * (5^1)
= 4 * 3 * 5
= 60
Hence, LCM of 10 and 12 by Prime Factorization method is 60.
## LCM of 10 and 12 by Division Method
The division method for finding the LCM of two numbers involves dividing the product of the two numbers by their greatest common divisor (GCD).
The steps to find the LCM of two numbers using the division method are:
Multiply the two numbers together, in this case 10 x 12 = 120
Find the GCD of the two numbers, in this case GCD(10,12) = 2
Divide the product of the two numbers by the GCD.
LCM = (10 x 12) / GCD(10,12) = 120/2 = 60
Hence, LCM of 10 and 12 by Division Method is 60
Categories LCM |
## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.6
Exercise 1.6 Class 9 Maths Samacheer Question 1.
(i) If n(A) = 25, n(B) = 40, n(A ∪ B) = 50 and n(B’) = 25 , find n(A ∩ B) and ii(U).
(ii) If n(A) = 300, n(A ∪ B) = 500, n(A ∩ B) = 50 and n(B’) = 350, find n(B) and n(U).
Solution:
(i) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
n(A ∩ B) = 25 + 40 – 50 = 65 – 50 = 15
n(U) = n(B) + n(B’) = 40 + 25 = 65
(ii) n(U) = n(B) + n(B’)
n(A ∩ B) = n(A) + n(B) – n(A B)
n(B) = n(A ∪ B) + n(A ∩ B) – n(A) = 500 + 50 – 300 = 250
n(U) = 250 + 350 = 600
9th Maths Exercise 1.6 Question 2.
If U = {x : x ∈ N, x ≤ 10}, A = {2, 3, 4, 8, 10) and b = {1, 2, 5, 8, 10}, then verify that n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Solution:
n(A) = 5, n(B) = 5
A ∪ B = {1, 2, 3, 4, 5, 8, 10}, A ∩ B= {2, 8, 10}
n(A ∪ B) = 7, n(A ∩ B) = 3
L.H.S n(A ∪ B) = 7
R.H.S = n(A) + n(B) – n(A ∩ B) = 5 + 5 – 3 = 7
∴ L.H.S = R.H.S proved.
9th Standard Maths Exercise 1.6 Question 3.
Verify n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) for the following sets.
(i) A = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f}
(ii) A = {1, 3, 5} B = {2, 3, 5, 6} and C = {1, 5, 6, 7}.
Solution:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
(i) A = {a, c, e, f, h}, B = {c, d, e, f}, C = {a, b, c, f}
n (A) = 5, n (B) = 4, n (C) = 4
n( A ∩ B) =3
n(B ∩ C) = 2
n( A ∩ C) =3
n( A ∩ B ∩ C) = 2
A ∩ B = {c, e, f}
B ∩ C = {c, f}
A ∩ C = {a, c, f}
A ∩ B ∩ C = {c, f}
A ∪ B ∪ C = {a, c, d, e, f, b, h}
∴ n(A ∪ B ∪ C) = 7 ……………. (1)
n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 5 + 4 + 4 – 3 – 2 – 3 + 2 = 15 – 8 = 7 ……………… (2)
∴ (1) =(2)
⇒ n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
Hence it is verified.
(ii) A = {1, 3, 5}, B = {2, 3, 5, 6 }, C = {1, 5, 6,7} = 3, n (B) = 4, n (C) = 4
n(A ∩ B) = 2
n(B ∩ C) = 2
n(C ∩ A) = 2
n(A ∩ B ∩ C) = 1
n(A ∪ B ∪ C) = 6
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
6 = 3 + 4 + 4 – 2 – 2 – 2 + 1 = 12 – 6 = 6
Hence it is verified.
9th Maths 1.6 Question 4.
In a class, all students take part in either music or drama or both. 25 students take part in music, 30 students take part in drama and 8 students take part in both music and drama. Find
(i) The number of students who take part in only music.
(ii) The number of students who take part in only drama.
(iii) The total number of students in the class.
Solution:
Let the number of students take part in music is M.
Let the number of students take part in drama is D.
By using venn diagram
(i) The number of students take part in only music is 17.
(ii) The number of students take part in only drama is 22.
(iii) The total number of students in the class is 17 + 8 + 22 = 47.
Samacheer Kalvi Guru 9th Maths Question 5.
In a party of 45 people, each one likes tea or coffee or both. 35 people like tea and 20 people like coffee. Find the number of people who
(i) like both tea and coffee.
(ii) do not like tea.
(iii) do not like coffee.
Solution:
Let the people who like tea be T.
Let the people who like coffee be C
By using formula
n( A ∪ B) = n(A) + n(B) – n(A ∩ B)
(i) n(T ∩ C) = n(T) + n(C) – n(T ∪ C) = 35 + 20 – 45 = 55 – 45 = 10
The number of people who like both coffee and tea = 10.
(ii) The number of people who do not like Tea
n(T) = n(U) – n(T) = 45 – 35 = 10
(iii) The number of people who do not like coffee
n(C’) = n(U) – n(C) = 45 – 20 = 25.
Samacheer Kalvi 9th Standard Maths Question 6.
In an examination 50% of the students passed in Mathematics and 70% of students passed in Science while 10% students failed in both subjects. 300 students passed in at least one subjects. Find the total number of students who appeared in the examination, if they took examination in only two subjects.
Solution:
Let the students who appeared in the examination be 100%.
Let the percentage of students who failed in mathematics be M.
Let the percentage of students who failed in science be S.
Failed in Maths = 100 % – Pass% = 100% – 50% = 50%
Failed in Science% 100% – 70% = 30%
Failed in both% = 10%
n(M ∪ S) = n(M) + n(S) – n(M ∩ S)
= 50% + 30% – 10% = 70%
% of students failed in atleast one subject = 70%
∴ The % of students who have passed in atleast one subject = 100% – 70% = 30%
30% = 300
∴ $$100 \%=\frac{100 \times 300}{30}=1000$$
∴ The total number of students who appeared in the examination = 1000 students.
Samacheer Kalvi 9th Maths Solutions Pdf Question 7.
A and B are two sets such that n(A – B) = 32 + x, n(B – A) = 5x and n(A ∩ B) = x. Illustrate the information by means of a venn diagram. Given that n(A) = n(B), calculate the value of x.
Solution:
n (A – B) = 32 +x
n(B – A) = 5x
n(A ∩ B) = x
n( A) = n(B)
32 + x + x = 5x + x
32 + 2x = 6x
4x = 32
x = 8
Samacheer Kalvi 9th Maths Question 8.
Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this data correct?
Solution:
n( A ∪ B) = n(A) + n(B) – n( A ∩ B)
n(A ∪ B) = 500 (given) …………. (1).
n( A) = 400
n(B) = 200
n( A ∩ B) =50
∴ n(A ∩ B) = 400 + 200 – 50 = 550 …………… (2)
1 ≠ 2
∴ This data is incorrect.
9th Maths Samacheer Kalvi Question 9.
In a colony, 275 families buy Tamil newspaper, 150 families buy English newspaper, 45 families buy Hindi newspaper, 125 families buy Tamil and English newspapers, 17 families buy English and Hindi newspapers, 5 families buy Tamil and Hindi newspapers and 3 families buy all the three newspapers. If each family buy atleast one of these newspapers then find
(i) Number of families buy only one newspaper
(ii) Number of families buy atleast two newspapers
(iii) Total number of families in the colony.
Solution:
(i) Tamil Newspaper buyers n(A) = 275
English Newspaper buyers n(B) = 150
Hindi Newspaper buyers n(C) = 45
Tamil and English Newspaper buyers n(A ∩ B) = 125
English and Hindi Newspaper buyers n(B ∩ C) = 17
Hindi and Tamil Newspaper buyers n(C ∩ A) = 5
All the three Newspaper buyers n(A ∩ B ∩ C) = 3
(i) Number of families buy only one newspaper = 148 + 11 + 26 = 185
(ii) Number of families buy atleast two news papers = 122 + 14 + 2 + 3 = 141
(iii) Total number of families in the colony = 148 + 11 + 26 + 122 + 14 + 2 + 3 = 326
Question 10.
A survey of 1000 farmers found that 600 grew paddy, 350 grew ragi, 280 grew corn, 120 grew paddy and ragi, 100 grew ragi and corn, 80 grew paddy and corn. If each farmer grew atleast any one of the above three, then find the number of farmers who grew all the three.
Solution:
a = 600 – (120 – x + x + 80 – x)
= 600 – (200 – x)
= 600 – 200 + x
= 400 + x
b = 350 – (120 – x + x + 100 – x)
= 350 – (220 – x)
= 350 – 230 + x
= 130 + x
c = 280 – (80 – x + x + 100 – x)
= 2800 – (180 – x) = 280 – 180 + x = 100 + x
Each farmer grew atleast one of the above three, the number of farmers who grew all the three is x.
= a + b + c + 120 – x + 100 – x + 80 – x + x = 1000
400 + x + 130 + x + 100 + x + 120 – x + 100 – x + 80 – x + x= 1000
∴ 930 + x = 1000
x = 1000 – 930 = 70
∴ 70 farmers grew all the three crops
9th Standard Maths Samacheer Kalvi Question 11.
In the adjacent diagram, if n(U) = 125,y is two times of x and z is 10 more than x, then find the value of x, y and z.
Solution:
n(U) = 125
y = 2x
z = x + 10
∴ x + y + z + 4 + 17 + 6 + 3 + 5 = 125
x + 2x + x + 10 + 35 = 125
4x + 45 = 125
4x = 125 – 45
4x = 80
x = 20
∴ y = 2x = 2 × 20 = 40
z = x + 10 = 20 + 10 = 30
Hence x = 20 ; y = 40; z = 30
Samacheer Kalvi Guru Maths 9th Question 12.
Each student in a class of 35 plays atleast one game among chess, carrom and table tennis. 22 play chess, 21 play carrom, 15 play table tennis, 10 play chess and table tennis, 8 play carrom and table tennis and 6 play all the three games. Find the number of students who play (i) chess and carrom but not table tennis (ii) only chess (iii) only carrom (Hint: Use Venn diagram)
Solution:
A – Chess
B – Carrom
C – Table Tennis
n(A) = 22
n(B) = 21
n(C) = 15
n(A ∩ C) = 10
n(B ∩ C) = 8
n(A ∩ B ∩ C) = 6
(i) y = 22 – (x + 6 + 4) = 22 – (x + 10)
= 22 – x – 10
= 12 – x
z = 21 – (x + 6 + 2) = 21 – (8 + x)
21 – 8 – x = 13 – x
y + z + 3 + x + 2 + 4 + 6 = 35
12 – x + 13 – x + 15 + x = 35
40 – x = 35
x = 40 – 35 = 5
(i) Number of students who pay only chess and Carrom but not table tennis = 5
(ii) Number of students who play only chess = 12 – x = 12 – 5 = 7
(iii) Number of students who play only carrom = 13 – x = 13 – 5 = 8
9th Class Maths Exercise 1.6 Solution Question 13.
In a class of 50 students, each one come to school by bus or by bicycle or on foot. 25 by bus, 20 by bicycle, 30 on foot and 10 students by all the three. Now how many students come to school exactly by two modes of transport?
Solution:
A – by bus
B – by bicycle
C – on foot
n(A) = 25
n(B) = 20
n(C) = 30
n(A ∩ B ∩ C ) = 10
n(A ∪ B ∪ C ) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
50 = 25 + 20 + 30 – (10 + x) – (10 + y) – (10 + z) + 10
50 = 75 – 10 – x – 10 – y – 10 – z + 10
= 75 – 20 – (x + y + z)
= 55 – (x + y + z)
x + y + z = 55 – 50 = 5
∴ The number of students who come to school exactly by two modes of transport = 5 |
# How to Find the Area of a Rectangle
Baca Cepat
## Unlocking the Secrets to Calculating Rectangle Areas
Welcome, dear reader! Whether you are a student, a professional, or just an enthusiast, youβve come to the right place. In this article, we will explore the ins and outs of finding the area of a rectangle. With our easy-to-follow guide, you will no longer be puzzled by this essential math concept.
## π What is a Rectangle?
A rectangle is a type of quadrilateral with four sides and four right angles. In simpler terms, it is a shape with a length and a width. The opposite sides of a rectangle are equal, and the adjacent sides are perpendicular to each other.
### π Properties of a Rectangle:
Property Description
Length The longer side of a rectangle
Width The shorter side of a rectangle
Perimeter The sum of all sides of a rectangle
Area The amount of space inside a rectangle
## π How to Find the Area of a Rectangle?
The formula for finding the area of a rectangle is:
Area = Length x Width
### π Steps to Calculate the Area of a Rectangle:
Step 1: Measure the length of the rectangle
Step 2: Measure the width of the rectangle
Step 3: Multiply the length by the width
Step 4: The result is the area of the rectangle
### π Example:
Letβs say we have a rectangle with a length of 6 inches and a width of 4 inches:
Area = 6 inches x 4 inches = 24 square inches
## π Tips and Tricks for Finding the Area of a Rectangle
### π Tip #1: Use Units of Measurement
Always include the units of measurement in your answer. For example, if the length and width are measured in inches, the area should be expressed in square inches.
### π Tip #2: Check Your Units of Measurement
Make sure the units of measurement for the length and width are the same. For example, if the length is measured in feet, the width should also be measured in feet.
### π Tip #3: Use Estimation
Estimation is a useful technique when you need a quick answer. Round off the length and width to the nearest whole number and multiply them.
### π Tip #4: Break Up Complicated Shapes
If a shape is too complicated, divide it into smaller rectangles, find the area of each rectangle, and add them together.
### π Tip #5: Practice
The more you practice, the easier it becomes. Try solving different problems and exercises to get a better understanding of finding the area of a rectangle.
## π FAQ
### π Q1. What is the difference between perimeter and area?
The perimeter is the distance around the outside of a shape, and the area is the space inside the shape.
### π Q2. Can you find the area of a rectangle with only one measurement?
No, you need both the length and width to find the area of a rectangle.
### π Q3. Can you find the area of a rectangle with decimal measurements?
Yes, you can. Just multiply the decimal length by the decimal width.
### π Q4. What if the rectangle is a square?
If the length and width are the same, itβs a square. The formula for finding the area of a square is:
Area = Length x Length = LengthΒ²
### π Q5. Can you find the area of a rectangle if the length and width are not equal?
Yes, you can. Just multiply the length and width together.
### π Q6. Is it possible for a rectangle to have a negative area?
No, itβs not possible. The area of a shape cannot be negative.
### π Q7. What if the sides of a rectangle are measured in different units?
Convert one of the measurements to the other unit before multiplying them.
## π Conclusion
In conclusion, finding the area of a rectangle is a simple process if you follow the steps and understand the formula. Remember to include the units of measurement and practice solving different problems. We hope this article has been helpful to you, and youβve learned something new today. Now, go out there and conquer rectangles like a pro! |
# 8.2.2 Sequences & Linear Functions
8
Subject:
Math
Strand:
Algebra
Standard 8.2.2
Recognize linear functions in real-world and mathematical situations; represent linear functions and other functions with tables, verbal descriptions, symbols and graphs; solve problems involving these functions and explain results in the original context.
Benchmark: 8.2.2.1 Represent Linear Functions
Represent linear functions with tables, verbal descriptions, symbols, equations and graphs; translate from one representation to another.
Benchmark: 8.2.2.2 Graphs of Lines
Identify graphical properties of linear functions including slopes and intercepts. Know that the slope equals the rate of change, and that the y-intercept is zero when the function represents a proportional relationship.
Benchmark: 8.2.2.3 Coefficients & Lines
Identify how coefficient changes in the equation f (x) = mx + b affect the graphs of linear functions. Know how to use graphing technology to examine these effects.
Benchmark: 8.2.2.4 Arithmetic Sequences
Represent arithmetic sequences using equations, tables, graphs and verbal descriptions, and use them to solve problems.
For example: If a girl starts with $100 in savings and adds$10 at the end of each month, she will have 100 + 10x dollars after x months.
Benchmark: 8.2.2.5 Geometric Sequences
Represent geometric sequences using equations, tables, graphs and verbal descriptions, and use them to solve problems.
For example: If a girl invests $100 at 10% annual interest, she will have 100(1.1x) dollars after x years. ## Overview Big Ideas and Essential Understandings Standard 8.2.2 Essential Understandings In this standard, it is essential for students to be able to move fluidly between different representations of linear functions. Students move from their prior learning of linear relationships that are proportional to all linear functions. Tables, graphs and equations are used to find and interpret solutions to real-world linear situations. When students identify a situation as linear, it is essential that they are able to identify and make meaning of the slope and y-intercept. Students can understand more about linear when they compare linear function to other functions. As students explore other functions such as inverse and exponential functions, it is important that the tables, graphs, equations and situations are continuously compared to what they know about linear functions. All Standard Benchmarks 8.2.2.1 Represent linear functions with tables, verbal descriptions, symbols, equations and graphs; translate from one representation to another. 8.2.2.2 Identify graphical properties of linear functions including slopes and intercepts. Know that the slope equals the rate of change, and that the y-intercept is zero when the function represents a proportional relationship. Example: Coordinates used for determining slope must contain integer values. 8.2.2.3 Identify how coefficient changes in the equation f (x) = mx + b affect the graphs of linear functions. Know how to use graphing technology to examine these effects. 8.2.2.4 Represent arithmetic sequences using equations, tables, graphs and verbal descriptions, and use them to solve problems. Example: If a girl starts with$100 in savings and adds $10 at the end of each month, she will have 100 + 10x dollars after x months. 8.2.2.5 Represent geometric sequences using equations, tables, graphs and verbal descriptions, and use them to solve problems. Example: If a girl invests$100 at 10% annual interest, she will have 100(1.1x) dollars after x years.
Benchmark Cluster
8.2.2 Sequences and Functions
8.2.2.1 Represent linear functions with tables, verbal descriptions, symbols, equations and graphs; translate from one representation to another.
8.2.2.2 Identify graphical properties of linear functions including slopes and intercepts. Know that the slope equals the rate of change, and that the y-intercept is zero when the function represents a proportional relationship.
Example: Coordinates used for determining slope must contain integer values.
8.2.2.3 Identify how coefficient changes in the equation f (x) = mx + b affect the graphs of linear functions. Know how to use graphing technology to examine these effects.
8.2.2.4 Represent arithmetic sequences using equations, tables, graphs and verbal descriptions, and use them to solve problems.
Example: If a girl starts with $100 in savings and adds$10 at the end of each month, she will have 100 + 10x dollars after x months.
8.2.2.5 Represent geometric sequences using equations, tables, graphs and verbal descriptions, and use them to solve problems.
Example: If a girl invests $100 at 10% annual interest, she will have 100(1.1x) dollars after x years. What students should know and be able to do [at a mastery level] related to these benchmarks: • Recognize linear relationships as they are expressed in a variety of formats; • Given one form of a linear function, such as a table, words, equation, or graph, be able to transfer to any other form; • Identify the slope and y -intercept of a linear function; • Interpret the slope and y -intercept in the context of the given situation; • Know how changing the coefficient changes the line on the graph; • Know how to solve a proportional situation differently than a linear function that is not proportional; • Students should be able to describe the pattern of a sequence by stating what is repeatedly being added or repeatedly being multiplied in the sequence; • Students should be able to connect the sequence to the function that represents the sequence; • Students will be able to translate between the equation, table, graph and verbal descriptions of the sequences; • Students should make the connection between arithmetic sequences and linear functions as well as geometric sequences and exponential functions; • Students will be able to use the different representations of arithmetic/geometric sequences to solve problems. Work from previous grades that supports this new learning includes: • Know how to represent proportional relationships with tables, verbal descriptions, symbols, equations and graphs; • Know how to calculate the constant of proportionality (unit rate or slope) given a proportional relationship; • Use of exponents and exponential form; • Understand linear functions (equations, tables, graphs). Correlations NCTM Standards • Represent, analyze, and generalize a variety of patterns with tables, graphs, words, and, when possible, symbolic rules; • Relate and compare different forms of representation for a relationship; • Identify functions as linear or nonlinear and contrast their properties from tables, graphs, or equations. • Represent and analyzemathematical situations and structures using algebraic symbols: • Explore relationships between symbolic expressions and graphs of lines, paying particular attention to the meaning of intercept and slope; • Use symbolic algebra to represent situations and to solve problems, especially those that involve linear relationships. • Use mathematical models to represent and understand quantitative relationships: • Model and solve contextualized problems using various representations, such as graphs, tables, and equations. • Analyze change in various contexts: • Use graphs to analyze the nature of changes in quantities in linear relationships. Common Core State Standards (CCSS) • 8.F (Functions) Define, evaluate, and compare functions. • 8.F.2. Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a linear function represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change. • 8.F (Functions)Use functions to model relationships between quantities. • 8.F.4. Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values. • 8.EE (Expressions and Equations)Understand the connections between proportional relationships, lines, and linear equations. • 8.EE.5. Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed. • F-LE (Linear, Quadratic, & Exponential Models)Construct and compare linear, quadratic, and exponential models and solve problems. • F-LE.2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table). ## Misconceptions Student Misconceptions • When given a table representation of a linear function with the first entry pair of the table not being "when x = 0," students sometimes give the first y value given as the y-intercept and not the y-value associated to x = 0. • When calculating slope from a graph, students sometimes will not pay attention to the "direction" of the rise or the run which will sometimes cause the sign of the slope to be wrong. • When given a table that doesn't have consecutive x-values, students sometimes will calculate the slope wrong. They will only pay attention to the change in y. • Students get confused about which value is the y-intercept in the sequence. • Students sometimes jump to conclusions and try to determine the pattern of the sequence by just looking at the first two terms. This leads students to sometimes mistake an arithmetic sequence for a geometric sequence (or vice versa) and then predict incorrect terms. For example, 1, 4 could be an arithmetic sequence (1, 4, 7, 10...) or a geometric sequence (1, 4, 16, 64...). • In geometric sequences such as 16, 8, 4, 2..., students will want to say the pattern is divide by 2 instead of multiplying by 0.5. ## Vignette In the Classroom In this vignette, students use a sequence of figures of toothpick triangles and a table of numbers as two ways to determine an algebraic expression. Teacher: Today we are going to work with patterns and see what comes next. Look at this pattern of figures. Who came come to the board and draw what they think the fourth figure would look like and explain the process. Student: I can, it's easy. (Student draws the fourth figure). All we need to do is add one more triangle so that is three toothpicks, but one of the sides will overlap so I will subtract one. Student: I thought of it a different way. I thought that each figure adds a triangle but because one line is already there I only need two to make the triangle. Teacher: So who is right? Student: I think they are both right since basically they are both adding two toothpicks each time to get the next figure. Teacher: So let's say that bn denotes the nth box number. For example, b3 = 7 since figure three has 7 toothpicks. Let's see if we can come up with some sort of equation that will represent this pattern. Let students work on their own for a bit. Student: I think it's y = x + 2. Teacher: I know you are probably used to using y = but here we are going to use different notation and use bn and n as our variables. And let's give your equation a try: bn = n + 2. So in Figure 3, n = 3. Substitute that into the equation and bn = 5, is that what we were supposed to get? Student: No, we should have gotten 7. I see why he is using 2 in his equation because the pattern is adding two but I think that n should be multiplied by 2 instead of added to 2. Student: So you think the equation is bn = 2n? But if I plug in 3 like the last time, I get 6 and we should have gotten 7. What if we say bn = 2n + 1? Would that work for all of them? Teacher: Well, let's try one. What about Figure 5? How many toothpicks would be there? Student: I just made it, and Figure 5 had 11 toothpicks. And it worked in the equation bn = 2n + 1. Student: The equation I came up with works, but it's not the same as yours. I said that bn = 2(n - 1) + 3. I did this because I had three toothpicks in the first figure and then I kept adding on two for every figure after that. And it works. I plugged in 5 for n and bn = 11. Teacher: Could it be possible that both work? The teacher takes this time to revisit the distributive property if the students didn't catch on right away. Student: I think they both work because if you simplify the second equation, you end up with exactly the same as the first. Teacher: OK, now let's look at a different problem without the drawing. Here, the information is given in a table. n 1 2 3 4 5 an 11 23 Student: So with this one, Figure 2 has 11 toothpicks and figure 5 has 23? Teacher: Yes, that is what this table is saying. See if you can figure out the missing numbers in the table. The teacher gives students time to work. Student: Well, I noticed that the change was 4. Teacher: How did you know that? Student: Isn't it obvious? Teacher: Well, not for some. Can you be a little more specific? Student: Well, the bottom number went up 12 when the top number changed 3, so 12 / 3 equals +4 per change. Teacher: This change is a sequence of numbers that has a constant rate of change is called the common difference. So I hear you say that the $\text{common difference}=\frac{\text{increase in sequence values}}{\text{the number of steps it takes}}=\frac{23-11}{5-2}=4$. Student: So last time with the picture drawing of the toothpicks, the equation was bn = 2n + 1, and the common difference was 2, here it's 4? Teacher: Yes, so where did the 1 come from in the last equation and what would need to be in its place for this table? Student: I noticed before that if we would have gone backwards and drawn a figure 0 there would have been 1 toothpick. Could that be what happens? Teacher: Well, let's try it. If we went to where n = 0 in the table what value would be there? n 0 1 2 3 4 5 an 7 11 15 19 23 Student: I think the number in the 0 spot should be 3. So then the equation would be an = 3 + 4n? Student: I think she is right. It works for n = 5. 3 + 4*5 does equal 23. Essentially I think you start with 3 and take n steps forward at 4 toothpicks per step. This gives us an = 3 + 4n. Note: Students have essentially just derived the slope formula and found the equation of a line given two points. All that is left to be done is transform the knowledge about sequences to correct language of linear equations. ## Resources Instructional Notes Teacher Notes • Students may need support in further development of previously studied concepts and skills. • It is important for students to be able to translate between all representations of a function. Whenever students are working with a situation, they need to make connections between the context language, math language, table, graph and equation. When given one representation, students should be constantly creating the other representations of that same function. • Most curricula will not address arithmetic and geometric sequences. They will address linear and exponential functions. The standard implies knowledge of exponential function, but never explicitly states "exponential." It is essential for students to learn how to write equations, make tables and graphs for exponential functions, and make the connection to geometric sequences. • Students need to understand that rate of change/slope is what the dependent variable (output) changes when the independent variable (input) changes by one. Remind students to not only look at what the y-value changes by, but also what the x-value changes by in a table. Point out that sometimes to truly assess whether or not students understand the concept of linear functions, the question writers will make a test question where there is a linear relationship, but the x-value isn't changing by a constant amount so it's hard to distinguish linear vs. non-linear just by examining the y values. So it is important to expose students to tables that are not always in order and/or increase by a constant value. This makes students actually think about slope rather than just assume slope by looking at the patterns. For example: x f(x) -2 3 2 5 4 6 10 9 • When converting a sequence to a function, it is common that the independent variable is the position of the number in the sequence. Therefore, the first value would be considered the first term. In order to find the y-intercept, students need to work backwards to find the previous term. For example, if the sequence is 3, 5, 7, 9..., the y-intercept would be 1. • To convert from a sequence list to a table, have the students label the first column "term" and the second column "value." • Some math books/resources show how to find the nth term in an arithmetic sequence by using the formula an = a0 + (n - 1)d. The term an is the term value we are searching for, a0 is the first term, n is the term being calculated and d is the common difference. Other math books will direct students to find the value of the zero term (y-intercept) and use y = mx + b. In this equation, y is the term value we are searching for, m is the common difference (slope), x is the term number being calculated and b is value of the "zero" term. • The study of sequences lays the foundation for linear equations. Finding the patterns in arithmetic sequences and using them to find the next terms in the sequence prepares students for finding slope of a line. • When students are finding the slope of a line, encourage them to go back to the graph and look to see if the line is increasing or decreasing. This will help them avoid making a sign error with the slope. • This lesson compares linear patterns to exponential patterns: Watch It Grow! Suppose you are offered these two methods to be paid for 30 days of work. • Plan 1-You receive$1000 the first day, and for each day following, you get $100 more than the day before. This means on day 2 you get$1100, on day 3 you get $1200 and so on. This continues for 30 days. • Plan 2-You receive$1 the first day, and for each day following, your wage doubles. This means on day 2, you get $2, on day 3 you get$4 and so on. This continues for 30 days.
• Without doing any calculating, which plan sounds better? Do you think it will be a lot better or a little better?
• After you have made your preliminary choice, work with your group to calculate and compare the earnings under both plans. Decide how to display the information and report conclusions.
• You are then offered two additional plans.
• Plan 3-You receive $1000 the first day, and for each day following, you get$1000 more. How does this plan compare with Plans 1 & 2?
• Plan 4-You receive $1000 the first day, and an additional$10,000 for each day following. How does this plan compare with the other plans?
• How would you rank the four plans from best to worst? What variables might influence the order of your list?
• How does your ranking change if the number of days is changed? (Examine both fewer days and more days.)
• This activity will help students learn to appreciate the concept of geometric growth and how it differs from linear growth.
• The preliminary guesses can be kept private or discussed. Let students choose whether to share their guesses.
• Groups may want to use calculators or graphing calculators or the computer to help organize and display their comparisons of the two plans.
• Encourage groups to write equations to represent what they see happening.
• Which plan is better if the job suddenly ends after 5 days? 10 days? 15 days? 20 days?
• At what point does the "best plan" change from one plan to another? What happens to the graphs at that point?
• What factors - other than choosing the pay plan that yields the most money - affect your choice of plan?
Adapted from the 97 Frameworks Document.
Instructional Resources
Effects of m and b on a linear graph
Movement with functions
A common problem when students learn about the slope-intercept equation y = mx + b is that they mechanically substitute for m and b without understanding their meaning. This lesson is intended to provide students with a method for understanding that m is a rate of change and b is the value when x = 0. This kinesthetic activity allows students to form a physical interpretation of slope and y-intercept by running across a football field. Students will be able to verbalize the meaning of the equation to reinforce understanding and discover that slope (or rate of movement) is the same for all sets of points given a set of data with a linear relationship.
Linear Relationships: Tables, Equations, and Graphs
This website offers good ideas that model real world linear data using tables, graphs, rules and expressions.
Slope slider
This activity allows students or teachers to move a slider to examine what happens to the graph of a linear function if the slope is changed or if the y-intercept is changed.
Linear function resources
This website provides a list of resources to use focused on representing linear functions. It includes the connection to NCTM Focal Points in Grade 8.
Effects of changing slope or y-intercept
This TI-83/84 calculator and navigator activity explores the effects on the graph of changing the coefficient or constant in a linear equation.
Math in the real world for real!
This lesson engages students in a relevant discussion and exploration of linear functions based on the cost of the iPad. Teacher notes and student pages are included.
New Vocabulary
linear function: a function whose rule may be written in the form f(x) = mx + b where m and b are real numbers. The term m represents the slope and b represents the y-intercept of the function. A linear function has a constant rate of change that results in a straight line graph.
arithmetic sequence: a sequence of numbers of the form: a, a + b, a + 2b, a + 3b, ... , a + (n - 1)b. There is a number that is constantly being added to each term to get the next term.
Example: 4, 7, 10, 13, 16, 19, 22, 25, ...
geometric sequence: a sequence of numbers of the form: a, ar, ar2, ar3,....., ar(n-1). Each term is being multiplied by a constant number to get the next term.
Example: 2, 4, 8, 16, 32, 64, ....
non-linear function: any function that does not follow a linear pattern of a constant rate of change, a straight line graph and an equation of the form f(x)=mx + b.
slope: the ratio of the vertical change to the horizontal change of a line on a graph. Slope represents the constant rate of change of a linear function. Given two points on a line slope is the ratio of the change in y to the change in x.
$\text{Slope}=\frac{\text{vertical change}}{\text{horizontal change}}=\frac{\Delta y}{\Delta x}=\frac{\text{rise}}{\text{run}}=\frac{y2-y1}{x2-x1}$
y-intercept: the value on the y-axis where a graph crosses the y-axis.
domain: the set of x-coordinates of the set of points on a graph; the set of x-coordinates of a given set of ordered pairs; the value that is the input in a function or relation.
range: the y-coordinates of the set of points on a graph; also, the y-coordinates of a given set of ordered pairs. The range is the output in a function or a relation.
independent variable: a variable whose value determines the value of other variables. Example: In the formula for the area of a circle, A = πr2, r is the independent variable, as its value determines the value of the area (A).
dependent variable: a variable whose value is determined by the value of an independent variable.
Example: In the formula for the area of a circle, A =πr2, A is the dependent variable, as its value depends on the value of the radius (r).
coefficient: the number multiplies times a product of variables or powers of variables in a term.
Example: 123 is the coefficient in the term 123x3y.
constant: a term or expression with no variables.
arithmetic sequence: a sequence of numbers in which the difference between any two consecutive terms is the same. In other words, an arithmetic sequence occurs when you add the same number each time as you move from one term to the next term in the sequence. This fixed number is called the common difference for the sequence.
geometric sequence: a sequence of numbers in which the ratio between any two consecutive terms is the same. In other words, you multiply by the same number each time to get the next term in the sequence. This fixed number is called the common ratio for the sequence.
exponential function: a function of the form y = a*bx where a represents the y-intercept and b represents the growth factor (the number multiplied by).
Professional Learning Communities
Reflection - Critical Questions regarding the teaching and learning of these benchmarks
• Why is it important for students to understand the different possible representations of a linear relationship: a verbal description, a numerical description (table or a set of ordered pairs), a geometrical description (graph) and an algebraic description (equation)? What are the strengths of each representation?
Taken from Focal Points: Focus In Grade 8, p.16.
• What is the ultimate goal of having students do activities in which they explore equations, slopes and y-intercepts? What prior background do students bring to this discussion that can be built on?
Taken from Focal Points: Focus in Grade 8, p.32
• How did students demonstrate understanding of the materials presented?
• Did students make the connection between slope and rate of change?
• How did students communicate that they understand the meaning of the slope-intercept equation?
• What were some of the ways in which students illustrated that they were actively engaged in the learning process?
References
• Interactivate: Slope Slider. (n.d.). Shodor: A National Resource for Computational Science Education. Retrieved June 20, 2011, from this source.
• Algebra 1 Graphing Linear Equations. (n.d.). Glencoe Mathematics Online Study Tools. Retrieved June 20, 2011, from this source.
• Illuminations: Movement with Functions. (n.d.). Illuminations. Retrieved June 20, 2011, from this source.
• Integrated Algebra Practice A.A.34#1. (n.d.). Jefferson Math Project. Retrieved June 20, 2011, from this source.
• Interpreting the Effects of Changing Slope and Y-intercept. (n.d.). Welcome to Mr. Livingston's Algebra Assignments Page. Retrieved June 20, 2011, from this source.
• Linear Relationships: Tables, Equations, and Graphs. (n.d.). Utah Education Network. Retrieved June 20, 2011, from this source.
• Principles and standards for school mathematics. (2000). Reston, VA: National Council of Teachers of Mathematics.
• Schielack, J. F. (2010). Focus in grade 8: Teaching with curriculum focal points. Reston, VA: National Council of Teachers of Mathematics.
• Classroom Activities: TAKS: Effects of changing slope or y-intercept - Texas Instruments - US and Canada. (n.d.). Calculators and Education Technology by Texas Instruments - US and Canada. Retrieved June 20, 2011, from this source.
• iCost: Mathalicious. (n.d.). Mathalicious. Retrieved June 20, 2011, from this source.
## Assessment
1.
DOK Level: 2
Taken from MCA III item sampler
2.
DOK Level: 2
Taken from MCA III item sampler
3.
DOK Level: 2
Taken from FCAT Math released test booklet
4.
DOK Level: 2
Taken from FCAT Math released test booklet
5.
DOK Level: 2
Taken from Massachusetts Comprehensive Assessment Released Exam
6.
DOK Level: 2
Taken from 2009 Texas TAKS test
7.
DOK Level: 2
Taken from 2009 Texas TAKS test
8.
DOK Level: 2
Taken from Minnesota MCA III item sampler
9.
DOK Level: 2
Taken from Minnesota MCA III item sampler
10.
DOK Level: 2
Taken from Massachusetts Comprehensive Assessment Released Exam
11.
DOK Level: 2
Taken from Massachusetts Comprehensive assessment
The following documents include assessment questions regarding these benchmarks.
Some questions regarding Interpreting the Effects of Changing Slope and y-intercept
Document 2
Document 3
Document 4
## Differentiation
Struggling Learners
• Give students multiple opportunities to experience the translation between graph, table, equation and situation. For example:
•
• Do a matching activity to put together all the representations of the same function. Use a table to organize. Then compare the different forms.
• Use a video to remind students of steps to translate from one representation to another.
• Use partners to describe their process.
• Look through magazines or catalogs to find real world examples of slope and linear functions. Compare the examples.
• Look at families of functions. For example, give students a group of three different functions that have something in common. Ask students to discuss how the functions compare to each other. Give students a chance to compare families of each representation.
• Ask students to define the word pattern. Where do they see patterns in their world? Continually go back to their understanding of pattern to help them represent linear patterns.
• Representing Patterns in Multiple Ways
This link provides some guided lesson plans to guide students through the process of understanding how to represent linear (arithmetic) patterns. Many organization maps are used including the Frayer model.
English Language Learners
• Give students multiple opportunities to experience the translation between graph, table, equation and situation. For example:
• Use a table to show multiple linear functions in all different forms.
• Use a video to remind students of steps to translate from one representation to another.
• Use partners to describe their process.
• Look through magazines or catalogs to find real world examples of slope and linear functions. Compare the examples.
• Look at families of functions. For example, give students a group of three different functions that have something in common. Ask students to discuss how the functions compare to each other. Give students a chance to compare families of each representation.
• Ask students to define the word pattern. Where do they see patterns in their world? Continually go back to their understanding of pattern to help them represent linear patterns.
Representing Patterns in Multiple Ways
This link provides some guided lesson plans to guide students through the process of understanding how to represent linear (arithmetic) patterns. Many organization maps are used including the Frayer model.
Extending the Learning
• Introduce representations of non-linear models, and ask students to compare them to linear models.
• Give students more open-ended problems that involve linear representations. |
# What’s just a scalar in math?
Nicely, it’s really a concept that permits the calculations expressed along with to be symbolized using factors which were defined as singulars or like a number of factors. For example, if your square is loosely defined and also this is a example of one variable, then a scalar could be utilised to be a symbol of different aspects of the sq feet.
Scalar refers paper writers for this only factor. Perhaps one of the uses of the scalar is to earn comparisons. By way of example, if you are comparing apples to apples and are attempting to find out that ones will be bigger, you have to make use of a scalar. Although this can be challenging for some, the simple fact we employ a scalar to evaluate things means that they are sometimes generalized to any or all kinds of comparison.
When you find a scalar in math, you will find a way to tell the two variables. Two numbers that are not the same may be represented by a different scalar. 1 case of a scalar in mathematics would be a integer. https://web.iastate.edu/students The scalar could be known to be itself some thing that is multiplied on it’s own over again before one is reached by it.
You may multiply two integers with each other in order to find the sum of your own worth employing the scalar, or in different words to come across the integer that reflects it. In case you aren’t sure the best way to do this, then strive to understand how to put in them and you will see you are helped by the scalar .
If you are trying to find how many variables for a given number can represent, you can use the scalar in math to find the total number of variables in the number. A good example of a scalar in math would be the square root of two.
So what is a scalar in math? It is just a type of number, similar to an integer. It is basically a variable, or a number https://expert-writers.net/thesis-statement-help that represents a variable.
Try to imagine a calculator where you could use the different values for each of the different variables. In addition to this, you would be able to enter the number of variables for that particular calculation and the calculator would automatically display all the results and let you know if the input is correct or not. If you were able to compute the square root of two using just a scalar, then you would not need to go through the process of multiplying the input numbers. This would save you time and would help you focus on other more important calculations. |
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The x-coordinate of the parabola is given by the equation: -b/2a In this case b = 6 and a = 1 So x = -6/2 x = -3 Now to find the y-coordinate that corresponds to that x -value, we will plug it into the quadratic function: y= -3^2 + 6(-3) -2 y = 9 -18 -2 y = -11 Therefore the vertex is at (-3,-11) .
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First distribute the 2 to the terms in parenthesis: 10x + 2x - 16 = 0 Combine Like Terms: 12x - 16 = 0 Move the 16 to the other side: 12x = 16 Divide by 12 to isolate the x: x = 16/12 simplify by dividing the numerator and denominator by 4: x = 4/3
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Solve for x in the first equation by first adding the -7y over to the other side: x = 7y - 11 Substitute the equation above into x in the second equation: 5( 7y - 11 ) + 2y = -18 Distribute the 5: 35y - 55 + 2y = -18 Combine like terms: 37y = 37 Divide by 37 to solve for y: y = 1 Plug y = 1 into the first equation in the system: x - 7 = -11 Add the 7 over: x = -4 Therefore, the coordinate pair that satisfies this system is (-4 , 1).
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Analytical reasoning tests a child’s ability to analyse data and use logic to find patterns or draw conclusions.
In the real world, we use analysis to scrutinise sequences, patterns, diagrams, charts, and graphs in order to gather the most pertinent data.
Kids with strong analytical skills will think about how key elements in that information interact with one another, and they will be more likely to notice important patterns and details.
The topics for analytical and logical reasoning are listed below, along with questions and answers. For a better understanding of the questions, explanations have been added. To improve your child’s critical thinking and problem-solving abilities, help them practise these questions.
## Logical Reasoning and Analytical Ability for Kids
As parents, we want to see our children’s cognitive abilities develop quickly during their early years. We must engage in logical and analytical exercises that contribute to their cognitive development because children develop their skills by observing and interacting.
The goal of the logical and analytical exercise for kids is to help your child remember what he or she has learned. Physical and mental exercises are both possible. These activities help students improve their academic performance while also providing an outlet for their energy.
Simple logical and analytical exercises can help your child stay sharp, improve their intelligence, and perform better at school or at home.
### Logical Ability Questions and Answers for Kids
#Due to old age issues, a man dies on his 24th birthday. How is this logical?
Solution: He was born on the 29th of February.
Explanation: February 29th occurs every four years; he would age each year but only celebrate birthdays every four years.
#On a table, there are three matches. Make a four with them without adding any more match sticks. None of the matches should be broken.
Solution: IV.
Explanation: Make the shape of a roman number four IV with three matchsticks.
#In town, there is a clothing store. The suit will set you back Rs 24, the socks will set you back Rs 30, the cap will set you back Rs 18, and a pair of trousers will set you back Rs 48. Guess the price of a hoodie using the above method.
Solution: 36 rupees.
Explanation: Pay attention to the prices and the number of letters in each word. For each letter in the word, the owner charges 6 rs. Because the word ‘Suit’ has four letters, 6 x 4 equals 24. Similarly, because the word ‘Hoodie’ has six letters, 6 x 6 = 36.
#Which month has a total of 28 days?
Solution: Every month.
Explanation: The first thing that comes to mind when the question is posed is February, but we forget that every month has 28 days.
#Consider a three-digit number in which the second digit is four times the first and the last digit is three times a second. Try to figure out the three-digit number.
Solution: 141
Explanation: The second digit in the number 141 is four times the first, 4 x 1, and the last digit is three less than the second, 4 – 3 = 1.
### Analytical Ability Questions for Kids
What does a snowman eat first thing in the morning?
When it comes to turkeys, which side has the most feathers?
What animal always wears its shoes to bed?
When you add two letters to a word, it becomes shorter.
What can you keep after you’ve given it away?
In the middle of April and March, what can be seen?
What, no matter how hard it rains, never gets any wetter?
Last year, how many men were born?
Answer: No one, only babies were born
How long does it take you to answer this question?
What can you keep but not share because once you share it, you can no longer keep it?
What always brings everything to a close?
What are two things you won’t be able to eat for dinner?
### What Are Logical Ability and Reasoning Questions?
#### Concentration Exercises
Concentration games assist children in focusing their attention in order to efficiently memorise the names of objects. Put toys or other playthings in front of the child and give him 30 seconds to remember them all in this game. It will be difficult to recall the names of toys or objects in a short period of time.
#### The Child Becomes Sharper and More Intelligent
This logical and reasoning activity encourages your child to think creatively and analytically, making him or her sharper and smarter. The memory of the child improves as a result of this exercise, and they learn faster than other children.
#### Increases Confidence and Self-Esteem in Children
Logical and analytical exercises improve a child’s self-esteem as well as their level of confidence. Creativity, organisational skills, analytical skills, coordination, and concentration are all enhanced by puzzle games.
#### Creativity is Boosted
These questions aid in the development of creativity in children. In addition to creativity, children develop coordination and self-reliance at a young age.
#### Enhances Communication Abilities
This logical activity and reasoning improve communication skills as well as story-telling techniques for effectively presenting ideas. Researchers have discovered that children with better communication skills have an easier time succeeding and developing leadership qualities.
When compared to reading a picture book or listening to a story, telling a story improves brain development.
##### Conclusion
So, to sum up, any exercise and questions that increase brain power is necessary for making the child a better person rather than turning him into a machine as is the case with traditional methods.
The questions and benefits listed above demonstrate how important logical and analytical questions are for your child’s brain development.
As a result, the Real School Of Montessori was founded with the goal of improving children’s fundamentals and providing them with the best education possible for a brighter future. Rather than teaching children using traditional teaching methods, The Real School Of Montessori focuses on laying a strong foundation of knowledge for them. |
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You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 7 Go to the latest version.
# 7.5: Simplify Products or Quotients of Single Variable Expressions
Difficulty Level: At Grade Created by: CK-12
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Practice Simplify Products or Quotients of Single Variable Expressions
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Have you ever had a stamp collection?
Marc has twice as many stamps in his collection as his Grandfather has in his. Write an expression to represent $m$ , the number of stamps in his Grandfather's collection.
To solve this problem, you will need to know how to write a single variable expression. Pay attention to this Concept, and you will know how to do this by the end of the Concept.
### Guidance
Recall that when you add and subtract terms in an expression, you can only combine like terms.
However, you can multiply or divide terms whether they are like terms or not.
For example, $6a$ and $3a$ are like terms because both terms include the variable $a$ . We can multiply them to simplify an expression like this.
$6a \times 3a= 18 \times a \times a=18a^2$ .
However, even though $6a$ and 3 are not like terms, we can still multiply them, like this.
$6a \times 3=18a$ .
The Commutative and Associative Properties of Multiplication may help you understand how to multiply expressions with variables. Remember, the Commutative property states that factors can be multiplied in any order. The Associative property states that the grouping of factors does not matter.
Let's apply this information.
$6a(3a)$
We can take these two terms and multiply them together.
First, we multiply the number parts.
$6 \times 3 = 18$
Next, we multiply the variables.
$a \cdot a= a^2$
Our answer is $18a^2$ .
Here is another one.
$5x(8y)$
Even though these two terms are different, we can still multiply them together.
First, we multiply the number parts.
$5 \times 8 = 40$
Next, we multiply the variables.
$x \cdot y=xy$
Our answer is $40xy$ .
Find the product $4z \times \frac{1}{2}$ .
$4z$ and $\frac{1}{2}$ are not like terms, however, you can multiply terms even if they are not like terms.
Use the commutative and associative properties to rearrange the factors to make it easier to see how they can be multiplied.
According to the commutative property, the order of the factors does not matter.
So, $4z \times \frac{1}{2}=\frac{1}{2}\times 4z$ .
According to the associative property, the grouping of the factors does not matter. Group the factors so that the numbers are multiplied first.
So, $\frac{1}{2} \times 4z=\frac{1}{2} \times 4 \times z=\left(\frac{1}{2} \times 4\right) \times z$ .
Now, multiply.
$\left(\frac{1}{2} \times 4\right) \times z=\left(\frac{1}{2} \times \frac{4}{1}\right) \times z=\frac{4}{2}\times z=2 \times z=2z.$
The product is $2z$ .
Remember that the word PRODUCT means multiplication and the word QUOTIENT means division.
Here is one that uses division.
Find the quotient $42c \div 7$ .
It may help you to rewrite the problem like this $\frac{42c}{7}$ . Then separate out the numbers and variables like this.
$\frac{42c}{7}=\frac{42 \cdot c}{7}=\frac{42}{7} \cdot c$
Now, divide 42 by 7 to find the quotient.
$\frac{42}{7} \cdot c=6 \cdot c=6c$
The quotient is $6c$ .
Now it's your turn. Find each product or quotient.
#### Example A
$6a(9a)$
Solution: $54a^2$
#### Example B
$\frac{15b}{5b}$
Solution: $3$
#### Example C
$\frac{20c}{4}$
Solution: $5c$
Here is the original problem once again.
Marc has twice as many stamps in his collection as his Grandfather has in his. Write an expression to represent $m$ , the number of stamps in his Grandfather's collection.
To write this, we simply use the variable and the fact that Marc has twice as many stamps.
$2m$
This term represents Marc's stamps.
### Vocabulary
Here are the vocabulary words in this Concept.
Expression
a number sentence without an equal sign that combines numbers, variables and operations.
Simplify
to make smaller by combining like terms
Product
the answer in a multiplication problem.
Quotient
the answer in a division problem.
Commutative Property of Multiplication
states that the product is not affected by the order in which you multiply factors.
Associative Property of Multiplication
states that the product is not affected by the groupings of the numbers when multiplying.
### Guided Practice
Here is one for you to try on your own.
Find the quotient $50 g \div 10 g$ .
It may help you to rewrite the problem like this $\frac{50g}{10g}$ . Then separate out the numbers and variables like this.
$\frac{50 g}{10 g}=\frac{50 \cdot g}{10 \cdot g}=\frac{50}{10} \cdot \frac{g}{g}$
Now, divide 50 by 10 and divide $g$ by $g$ to find the quotient. Since any number over itself is equal to 1, you know that $\frac{g}{g}=1$ .
$\frac{50}{10} \cdot \frac{g}{g}=5.1=5$
The quotient is 5.
### Video Review
Here is a video for review.
### Practice
Directions: Simplify each product or quotient.
1. $6a(4a)$
2. $9x(2)$
3. $14y(2y)$
4. $16a(a)$
5. $22x(2x)$
6. $18b(2)$
7. $\frac{21a}{7}$
8. $\frac{22b}{2b}$
9. $\frac{25x}{x}$
10. $\frac{45a}{5a}$
11. $\frac{15x}{3x}$
12. $\frac{18y}{9}$
13. $\frac{22y}{11y}$
14. $\frac{15x}{3y}$
15. $\frac{82x}{2x}$
### Vocabulary Language: English
Associative property
Associative property
The associative property states that the order in which three or more values are grouped for multiplication or addition will not affect the product or sum. For example: $(a+b) + c = a + (b+c) \text{ and\,} (ab)c = a(bc)$.
Commutative Property
Commutative Property
The commutative property states that the order in which two numbers are added or multiplied does not affect the sum or product. For example $a+b=b+a \text{ and\,} (a)(b)=(b)(a)$.
Expression
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Product
Product
The product is the result after two amounts have been multiplied.
Quotient
Quotient
The quotient is the result after two amounts have been divided.
Simplify
Simplify
To simplify means to rewrite an expression to make it as "simple" as possible. You can simplify by removing parentheses, combining like terms, or reducing fractions.
Nov 30, 2012
Mar 22, 2015 |
# Logarithms Quiz Set 020
### Question 1
What is x if logx$(3/5)$ = $1/2$?
A
${9/25}$.
B
${3/5}$.
C
$√{3/5}$.
D
$√{5/3}$.
Soln.
Ans: a
logx$(3/5)$ = $1/2$, by definition, gives $x^{1/2}$ = $3/5$. So x = $(3/5)^2$ = ${9/25}$.
### Question 2
What is the value of log2(128)?
A
7.
B
$√7$.
C
0.
D
$√128$.
Soln.
Ans: a
Let log2(128) = P. By definition, we have 2P = 128 = 27, which gives 7 as the answer.
### Question 3
Which of these is correct?
A
$\text"log"(1 + 2 + 3)$ = $\text"log"(1 × 2 × 3)$.
B
$\text"log"_4(4)$ = 4.
C
$\text"log"_6(6)$ = 36.
D
$\text"log"_7(1)$ = 7.
Soln.
Ans: a
Speaking factually, log(m + n + p) = log(m × n × p) is possible only if m × n × p = m + n + p, hence the answer, because 1 × 2 × 3 = 1 + 2 + 3. Expressions of the form $\text"log"_m(n) = p$ are same as mp = n. We can see that none of the other options makes it correct.
### Question 4
Which of these is correct?
A
$\text"log"_6(8)$ = $1/{\text"log"_8(6)}$.
B
$\text"log"_8(8)$ = 8.
C
$\text"log"_4(4)$ = 16.
D
$\text"log"(6 + 8 + 4)$ = $\text"log"(192)$.
Soln.
Ans: a
Speaking factually, $\text"log"_m(n)$ = $1/{\text"log"_n(m)}$, hence the answer. Expressions of the form $\text"log"_m(n) = p$ are same as mp = n. We can see that none of the options makes it correct. Also, log(m + n + p) = log(m × n × p) is possible only if m × n × p = m + n + p.
### Question 5
What is the value of $\text"log"_3(√3)$?
A
$1/2$.
B
$2$.
C
$\text"log"_√3(3)$.
D
0.
Soln.
Ans: a
From the theory of logarithms, we know that $\text"log"_m(√m)$ = $1/2$ × $\text"log"_m(m)$ = $1/2$ × 1 = 1/2. |
# What is a Multiple?
Any concept of math is easier to understand when it’s done with the help of practical examples. After all, I have yet to see a student who could learn by blindly following theories! So let’s get into what is a multiple with this practical example.
## Explanation
Take a single-digit non-zero number for example the number 2 and add it any number of times you want to. Let’s go ahead and add the number 2, five times. We can write this as 2 + 2 + 2 + 2 + 2 and the answer to that problem would be 10. We will call this new number the product of 2 times 5.
Similarly, if we took the number 5, and added it two times, we would get 10. 10 is the product of 5 times 2.
The number 10 is a multiple of 2 (since 2 x 5 = 10) and is also a multiple of 5 (since 5 x 2 = 10).
If you are with us so far, then it’s going to be very easy for you to understand multiples and all its concepts which are included in this article. As mentioned before, there will be no shortage of examples and questions for you to practice along. The answer to each practice question will be given at the end of the article.
## What is a Multiple?
To define it simply, a multiple of a number is the product of that number and another non-zero whole number.
For example, 12 is a product of 2 and 6. Both 2 and 6 are non-zero whole numbers.
When you keep finding the product of a number by multiplying it with a series of whole numbers in any numerical order, you will get a multiplication table of that number.
For example:
For this, we are going to continue with the number 7 and make up a multiplication table for it.
Here, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 are all multiples of 7. We stopped the table at 10, but in reality, it can go up to infinity. You did a good job of understanding the basic concept of multiples. Let’s add a few rules or properties to make it even easier.
## Properties of Multiple
• Any number that is a multiple of 2 is an even number for e.g., 2, 4, 6, 8…. etc
• All the non 2 multiple numbers are called odd numbers for e.g., 1, 3, 7, 9….. etc
• There can be infinite, i.e., no limit, number of multiples.
• Every integer is a multiple of itself.
• Smallest multiple is the number itself; in the above example, the smallest multiple was 7.
• A product is a multiple of all its factors.
• 0 is a multiple of anything.
• If a and b are multiples of x, then a + b and a – b are also multiples of x.
## Factors Are Not Multiples
Often easily confused, factors are not like multiples at all. Essentially, factors are what we multiply to get a number, and multiple is the result of that multiplication when calculated with an integer.
While there are infinite numbers of multiples, factors are finite. They are also less than or equal to the integer or number, whereas a multiple of a number is always equal or greater than its numbers.
For Example 2 x 2 = 4, Here 2 is the factor of 4, and 4 is the multiple of 2.
## Methods of Finding out Multiples
### Least Common Multiple (LCM)
Finding out the least common multiple of any two or more numbers is a fun and sometimes necessary way of solving any problem. There are different ways of finding out the least common multiplies of two or more numbers. Here are the four most common ways:
• The listing method
• The tree factor method
• LCM of mixed fraction through the ladder method
#### 1. Listing Method
In the listing method, we find the least common multiple that exists in the listing of all the numbers available.
For example:
If we want to find out the LCM of 4 and 5 using the listing method, we have to write the multiple listing of both the numbers. This would go something like this:
The multiple listing of 4 is: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44…
The multiple listing of 5 is: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55…
Notice how both 20 and 40 are repeated in both the listings? These are the common multiples of 4 and 5, but we are looking for the least common multiple and that is 20.
As soon as you spot a common multiple in both the lists, you can stop the listing. It’s a quick way to spot the least common multiple. However, to do this, you would need to memorize all the multiples of a single number in a numeric, as shown above.
#### 2. Tree Factor
The tree factor, the factor tree, or prime factorization is a useful way of finding a factor of any integer until we have reached the prime number and cannot factor anymore.
Finding the least common multiple through the tree factor is a fun way of breaking down or decomposing a large number and building it back up. Through the tree factor method, you can conduct a prime factorization of any two or more numbers. It is a useful tool for breaking down large and difficult problems.
For example:
We are going to find the least common multiple of 18 and 12 using the tree factor method.
So we write down “18” and “12” separately and start writing down the factors of it beneath the numbers.
Here, for 18, the lowest factors are 2, 3, and 3. For the number 12, we have 2, 2, and 3. So we can write it like this:
18 = 2 x 3 x 3 12 = 2 x 2 x 3
Now here is the super important part so pay close attention. We need to list the prime numbers as many times as they occur most often for each of the numbers. For the number 18, we only have one “2” and two “3”. For the number 12, we have two “2” and only one “3”. So we will take the number 3 (two times) since that occurs the most often in the number 18, and we will take number 2 (two times) since that occurs the most often in the number 12.
Then multiply. So 3 x 3 x 2 x 2 = 36. 36 is the least common multiple of both 12 and 18.
Fun fact: knowing how to find the prime factors of any problem is especially useful if you want a career in encryption or cryptology.
The ladder or grid is a method of finding out the least common multiple of any two or more numbers. You can find the LCM of larger numbers in a much faster way. The ladder or grid looks like a tic-tac-toe grid. It is a set of two horizontal parallel lines that intersect each other at a vertical angle.
You start writing the numbers on the top row and move down as you factor down each number until you reach a prime number.
For example:
For the ladder method, we are going to find the LCM of 18 and 12 again.
We have the prime numbers 2 and 3 on the vertical line and 3 and 2 on the horizontal line. These numbers are multiplied with each other to get the common multiple of both 18 and 12 that is 36.
2 x 3 x 3 x 2 = 36
As mentioned above, 36 is the multiple of 18 and 12.
#### 4. LCM Of Mixed Fractions
The best way to find a common denominator of a proper or improper fraction is through the LCM method for mixed fractions. In this method, you use the ladder LCM method for both the denominator number of the fraction to come up with a common number.
For example:
If you want to add two fractions that have a different denominator, you will have to find a common multiple or a common denominator first. Assume that the denominators of two fractions are 8 and 12. You need to find out the lease common multiple between the numbers 8 and 12. Here the common multiple is 24. 8 times 3 is 24, and 12 times 2 is 24. However, 24 isn’t the only common multiple with both 8 and 12. We also have 72, i.e., 8 times 9 is 72, and 12 times 6 is 72.
So, what makes 24 special? It is the least common multiple of both denominators.
### Grid Method Multiplication
The grid method, also known as the grammar school or box method, is a good place to start learning multiples and the calculations with larger numbers that are more than 10. This method breaks the process of manual multiplication into three steps.
For example:
Here we will find out the multiples of 15 with 37.
The first step is to make up a grid or table with the first column and first row being your numbers to be multiplied. The first box will remain empty. Like this:
The second step is to multiply all the numbers in the rows and the columns with each other. Like this:
The third step is a revelation of addition. When you add up 300 + 150 + 70 + 35, you will get a sum of 555. When you calculate 15 x 37, you will get exactly 555.
This method makes it easy to break down and simplify the factors of that number.
## Bottom Line:
This was a fun and informative part of math. We got to know what a multiple is, how we can multiply factors and the different methods of finding out the least common multiple. Multiples are incredibly fun and useful. |
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Scientific Notations of Real Numbers and Logarithms
Express Large Numbers in Standard Form and Vice-versa
Let us suppose we are given 3 numbers: 2, 3 and 9.
Now, we know that 32 = 9
Also, $\sqrt{9}=3$
The above two expressions are formed by combining 2 and 3, and 2 and 9 respectively to get the third number.
Is there an expression wherein we can combine 3 and 9 to get 2?
3 and 9 can be combined to get 2 as:
Here, ‘log’ is the abbreviated form of a concept called ‘Logarithms’.
The expression can be read as ‘logarithm of 9 to the base 3 is equal to 2’.
In general, if a is any positive real number (except 1), n is any rational number such that , then n is called the logarithm of b to the base a, and is written as.
Thus, if and only if .
is called the exponential form and is called the logarithmic form.
The following are the properties of logarithms.
1. Since a is any positive real number (except 1), an is always a positive real number for every rational number n, i.e., b is always a positive real number.
Thus, logarithms are only defined for positive real numbers.
2. Since
Thus, and
where, a is any positive real number except 1
3. If
Then, and
x = y
Thus,
x = y
4. Logarithms to the base 10 are called common logarithms.
5. If no base is given, the base is always taken as 10.
For example, log 5 = log10 5
Let us consider the following example.
Convert the following into logarithmic form.
(i) 53 = 125
(ii)
There are three standard laws of logarithms.
(i) Product Law
In general,
(ii) Quotient Law
(iii) Power Law
On the basis of the above laws, we have
For a and b two positive numbers, ${\mathrm{log}}_{b}a=\frac{1}{{\mathrm{log}}_{a}b}$.
Also, we know that, log of a number at the same base is 1 i.e ${\mathrm{log}}_{a}a=1$.
$⇒x{\mathrm{log}}_{a}a=x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{log}}_{a}{a}^{x}=x$
Example 1:
Solve for x.
(i) log7 343 = 5x − 4
(ii) logx 216 = 3
Solution:
(i)
(ii)
Example 2:
If what is x?
Solution:
Now,
Example: 3
Solve for x.
(i)
(ii)
${\mathrm{log}}_{5}{5}^{9}=x$
Solution:
(i)
We know that,
For a and b two positive numbers, ${\mathrm{log}}_{b}a=\frac{1}{{\mathrm{log}}_{a}b}$.
Therefore,
(ii)
We know that, log of a number at the same base is 1 i.e ${\mathrm{log}}_{a}a=1$.
$⇒x{\mathrm{log}}_{a}a=x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{log}}_{a}{a}^{x}=x$
Therefore,
${\mathrm{log}}_{5}{5}^{9}=x=9\phantom{\rule{0ex}{0ex}}⇒x=9$
Common Logarithms
We know:
101 = 10 and 102 = 100
∴ log10 10 = 1 and log10 100 = 2
For 10 < n < 100, 1 < log10 n < 2.
Example: log10 20 = 1.3010 and log10 25 = 1.3979
Here, the integral part 1 is called the characteristic of the logarithm and the fractional part is called the mantissa of the logarithm.
For any positive number N,
log10 N = Characteristic + Mantiss…
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# What is wrong with my solution to the system of congruences?
In the following system of congruences, $n_1$ and $n_2$ are relatively prime.
\begin{align} \newcommand{\mod}{\text{mod }} a &\equiv a_1 \; (\mod n_1) \\ a &\equiv a_2 \; (\mod n_2) \\ \end{align}
Using the Chinese Remainder System described in [CLRS: Introduction to Algorithms (Theorem 31.27, Page 951, the 3rd edition)], we have
$$m_1 = n_2, m_2 = n_1$$ \begin{align} c_1 = m_1(m_1^{-1} \;\mod n_1) = n_2 (n_2^{-1} \;\mod n_1) \\ c_2 = m_2(m_2^{-1} \;\mod n_2) = n_1 (n_1^{-1} \;\mod n_2) \end{align}
Then, we have ($n = n_1 n_2$ in the following)
\begin{align} a &= (a_1 c_1 + a_2 c_2) \; (\mod n) \\ &= \left(a_1 n_2 (n_2^{-1} \;\mod n_1) + a_2 n_1 (n_1^{-1} \;\mod n_2)\right) \; (\mod n) \end{align}
Another method is to solve the congruences directly as follows.
$$a \equiv a_1 \; (\mod n_1) \Rightarrow \exists x \in \mathbb{Z}, a = n_1 x + a_1$$
Substitute $a$ into the second congruence, we get $$n_1 x \equiv a_2 - a_1 \; (\mod n_2)$$
Solving the congruence above, we get $$x = (a_2 - a_1) (n_1^{-1} \; \mod n_2)$$
Therefore, \begin{align} a &= n_1 x + a_1 \\ &= n_1 \left( (a_2 - a_1) (n_1^{-1} \; \mod n_2) \right) + a_1 \\ &= a_2 n_1 (n_1^{-1} \;\mod n_2) - a_1 n_1 (n_1^{-1} \;\mod n_2) + a_1 \end{align}
However, the result is not the same as the one obtained by applying the Chinese Remainder Theorem. What is wrong with it?
• Maybe the expressions are both correct (although they appear quite different)! If I did not make an error in the calculation, both solutions seem to satisfy the required congruences. Note that the expressions only have to be congruent modulo $n$ Mar 30, 2017 at 16:02
• Really, how do you know they are not the same? $2+2$ and $12\over3$ look different as well, but there is a catch. Mar 30, 2017 at 20:30
Hint $\ {\rm mod}\ (n_1,n_2)\!:\!\!\! \underbrace{(\color{#0a0}{a_1,0})\,+\,(0,a_2)} _{\Large \begin{array}{r} \color{#0a0}{n_2(n_2^{-1}a_1\bmod n_1})\\ +\ n_1(n_1^{-1}a_2\bmod n_2) \end{array} }\, \!\!\!\!\!\!\!\!\equiv\, (a_1,a_2)\, \equiv\,\underbrace{(\color{#c00}{a_1,a_1})\, +\, (0,\,a_2-a_1)}_{\Large \color{#c00}{a_1} +\: n_1(n_1^{-1}(a_2-a_1)\bmod n_2)\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! }$
i.e. $\ \color{#0a0}{n_2(n_2^{-1}a_1\bmod n_1) \equiv a_1}\!\pmod {\!n_1}\,$ and $\equiv \color{#0a0}0\pmod {\!n_2},\,$ i.e. it's $\equiv\color{#0a0}{(a_1,0)}\pmod{\!n_1,n_2}$
Remark $\$ These "vector" operations will be arithmetically clarified when one learns about the ring-theoretic view of CRT = Chinese Remainder Theorem. |
CBSE Class 8 Maths Worksheets for Rational Numbers
# CBSE Class 8 Maths Worksheets for Rational Numbers
Rational numbers are a fundamental concept in mathematics, and understanding them is crucial for success in higher-level math classes. In this worksheet, we will explore the basics of rational numbers, including their definition, properties, and operations.
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Infinity Learn offers a treasure trove of free Class 8 Maths worksheets Rational Numbers, meticulously crafted by experienced teachers to support students in mastering mathematical concepts effectively. These worksheets are designed to align seamlessly with the CBSE Syllabus curriculum, ensuring relevance and accuracy in content.
## What are Rational Numbers?
A rational number is a number that can be expressed as the ratio of two integers, i.e., p/q, where p and q are integers and q is non-zero. For example, 3/4, 22/7, and -1/2 are all rational numbers.
Properties of Rational Numbers:
1. Closure Property: The sum, difference, product, and quotient of two rational numbers is always a rational number.
2. Commutative Property: The order of the numbers does not change the result when adding or multiplying rational numbers.
3. Associative Property: The order in which we add or multiply rational numbers does not change the result.
4. Distributive Property: Rational numbers follow the distributive property of multiplication over addition.
Operations on Rational Numbers:
### Addition and Subtraction of Rational Numbers
To add or subtract rational numbers, we need to follow these steps:
1. Find the least common multiple (LCM) of the denominators.
2. Convert both numbers to have the LCM as the denominator.
3. Add or subtract the numerators.
4. Simplify the result, if possible.
Example 1: Add 2/3 and 3/4.
Solution: LCM of 3 and 4 is 12. Convert both numbers to have 12 as the denominator: 2/3 = 8/12 and 3/4 = 9/12. Add the numerators: 8 + 9 = 17. So, 2/3 + 3/4 = 17/12.
### Multiplication of Rational Numbers
To multiply rational numbers, we multiply the numerators and denominators separately.
Example 2: Multiply 2/3 and 3/4.
Solution: Multiply the numerators: 2 × 3 = 6. Multiply the denominators: 3 × 4 = 12. So, 2/3 × 3/4 = 6/12 = 1/2.
### Division of Rational Numbers
To divide rational numbers, we invert the second number (i.e., flip the numerator and denominator) and then multiply.
Example 3: Divide 2/3 by 3/4.
Solution: Invert the second number: 3/4 = 4/3. Multiply: 2/3 × 4/3 = 8/9. So, 2/3 ÷ 3/4 = 8/9.
Exercises:
2. Subtract 3/4 from 5/6.
3. Multiply 2/5 by 3/7.
4. Divide 3/8 by 2/5.
5. Simplify the rational number 12/16.
1. 1/2 + 2/3 = 7/6
2. 5/6 – 3/4 = 1/12
3. 2/5 × 3/7 = 6/35
4. 3/8 ÷ 2/5 = 15/16
5. 12/16 = 3/4
Conclusion
Rational numbers are an essential part of mathematics, and understanding their properties and operations is crucial for success in higher-level math classes. Practice the exercises provided to reinforce your understanding of rational numbers.
## Class 8 Maths Rational Numbers FAQs
### What are rational numbers, and why are they important in mathematics?
Rational numbers are numbers that can be expressed as a fraction of two integers. They are crucial in mathematics as they help us represent quantities that are not whole numbers, such as fractions and decimals. Understanding rational numbers is essential for various mathematical operations and real-world applications.
### How can I add and subtract rational numbers effectively?
To add or subtract rational numbers, ensure the denominators are the same by finding a common denominator. Add or subtract the numerators while keeping the denominator constant. Simplify the result if possible. Practice with different examples to improve your skills in adding and subtracting rational numbers.
### What are the properties of rational numbers?
Rational numbers exhibit properties such as closure under addition, subtraction, multiplication, and division. They follow the commutative, associative, and distributive properties. Rational numbers also have an additive identity (0) and a multiplicative identity (1). Understanding these properties helps in manipulating rational numbers effectively.
### How can I multiply and divide rational numbers accurately?
To multiply rational numbers, multiply the numerators and denominators separately. For division, invert the second number (reciprocal) and then multiply. Simplify the result if needed. Pay attention to signs and ensure proper handling of negative numbers. Practice various multiplication and division exercises to enhance your skills.
### How can I practice and improve my understanding of rational numbers through worksheets?
To enhance your understanding of rational numbers, regularly practice worksheets that include a variety of problems on addition, subtraction, multiplication, and division of rational numbers. Challenge yourself with word problems and real-life scenarios involving rational numbers. Seek feedback on your solutions and review any mistakes to learn from them.
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# Discrete Mathematics: An Open Introduction, 3rd edition
## Section2.3Polynomial Fitting
###### Investigate!
A standard $$8 \times 8$$ chessboard contains 64 squares. Actually, this is just the number of unit squares. How many squares of all sizes are there on a chessboard? Start with smaller boards: $$1\times 1\text{,}$$ $$2 \times 2\text{,}$$ $$3\times 3\text{,}$$ etc. Find a formula for the total number of squares in an $$n\times n$$ board.
So far we have seen methods for finding the closed formulas for arithmetic and geometric sequences. Since we know how to compute the sum of the first $$n$$ terms of arithmetic and geometric sequences, we can compute the closed formulas for sequences which have an arithmetic (or geometric) sequence of differences between terms. But what if we consider a sequence which is the sum of the first $$n$$ terms of a sequence which is itself the sum of an arithmetic sequence?
Before we get too carried away, let’s consider an example: How many squares (of all sizes) are there on a chessboard? A chessboard consists of $$64$$ squares, but we also want to consider squares of longer side length. Even though we are only considering an $$8 \times 8$$ board, there is already a lot to count. So instead, let us build a sequence: the first term will be the number of squares on a $$1 \times 1$$ board, the second term will be the number of squares on a $$2 \times 2$$ board, and so on. After a little thought, we arrive at the sequence
\begin{equation*} 1,5,14,30, 55,\ldots\text{.} \end{equation*}
This sequence is not arithmetic (or geometric for that matter), but perhaps it’s sequence of differences is. For differences we get
\begin{equation*} 4, 9, 16, 25, \ldots\text{.} \end{equation*}
Not a huge surprise: one way to count the number of squares in a $$4 \times 4$$ chessboard is to notice that there are $$16$$ squares with side length 1, 9 with side length 2, 4 with side length 3 and 1 with side length 4. So the original sequence is just the sum of squares. Now this sequence of differences is not arithmetic since it’s sequence of differences (the differences of the differences of the original sequence) is not constant. In fact, this sequence of second differences is
\begin{equation*} 5, 7, 9, \ldots\text{,} \end{equation*}
which is an arithmetic sequence (with constant difference 2). Notice that our original sequence had third differences (that is, differences of differences of differences of the original) constant. We will call such a sequence $$\Delta^3$$-constant. The sequence $$1, 4, 9, 16, \ldots$$ has second differences constant, so it will be a $$\Delta^2$$-constant sequence. In general, we will say a sequence is a $$\Delta^k$$-constant sequence if the $$k$$th differences are constant.
### Example2.3.1.
Which of the following sequences are $$\Delta^k$$-constant for some value of $$k\text{?}$$
1. $$2, 3, 7, 14, 24, 37,\ldots\text{.}$$
2. $$1, 8, 27, 64, 125, 216, \ldots\text{.}$$
3. $$1,2,4,8,16,32,64,\ldots\text{.}$$
Solution.
1. This is the sequence from Example 2.2.6, in which we found a closed formula by recognizing the sequence as the sequence of partial sums of an arithmetic sequence. Indeed, the sequence of first differences is $$1,4,7, 10, 13,\ldots\text{,}$$ which itself has differences $$3,3,3,3,\ldots\text{.}$$ Thus $$2, 3, 7, 14, 24, 37,\ldots$$ is a $$\Delta^2$$-constant sequence.
2. These are the perfect cubes. The sequence of first differences is $$7, 19, 37, 61, 91, \ldots\text{;}$$ the sequence of second differences is $$12, 18, 24, 30,\ldots\text{;}$$ the sequence of third differences is constant: $$6,6,6,\ldots\text{.}$$ Thus the perfect cubes are a $$\Delta^3$$-constant sequence.
3. If we take first differences we get $$1,2,4,8,16,\ldots\text{.}$$ Wait, what? That’s the sequence we started with. So taking second differences will give us the same sequence again. No matter how many times we repeat this we will always have the same sequence, which in particular means no finite number of differences will be constant. Thus this sequence is not $$\Delta^k$$-constant for any $$k\text{.}$$
The $$\Delta^0$$-constant sequences are themselves constant, so a closed formula for them is easy to compute (it’s just the constant). The $$\Delta^1$$-constant sequences are arithmetic and we have a method for finding closed formulas for them as well. Every $$\Delta^2$$-constant sequence is the sum of an arithmetic sequence so we can find formulas for these as well. But notice that the format of the closed formula for a $$\Delta^2$$-constant sequence is always quadratic. For example, the square numbers are $$\Delta^2$$-constant with closed formula $$a_n= n^2\text{.}$$ The triangular numbers (also $$\Delta^2$$-constant) have closed formula $$a_n = \frac{n(n+1)}{2}\text{,}$$ which when multiplied out gives you an $$n^2$$ term as well. It appears that every time we increase the complexity of the sequence, that is, increase the number of differences before we get constants, we also increase the degree of the polynomial used for the closed formula. We go from constant to linear to quadratic. The sequence of differences between terms tells us something about the rate of growth of the sequence. If a sequence is growing at a constant rate, then the formula for the sequence will be linear. If the sequence is growing at a rate which itself is growing at a constant rate, then the formula is quadratic. You have seen this elsewhere: if a function has a constant second derivative (rate of change) then the function must be quadratic.
This works in general:
### Finite Differences.
The closed formula for a sequence will be a degree $$k$$ polynomial if and only if the sequence is $$\Delta^k$$-constant (i.e., the $$k$$th sequence of differences is constant).
This tells us that the sequence of numbers of squares on a chessboard, $$1, 5, 14, 30, 55, \ldots\text{,}$$ which we saw to be $$\Delta^3$$-constant, will have a cubic (degree 3 polynomial) for its closed formula.
Now once we know what format the closed formula for a sequence will take, it is much easier to actually find the closed formula. In the case that the closed formula is a degree $$k$$ polynomial, we just need $$k+1$$ data points to “fit” the polynomial to the data.
### Example2.3.2.
Find a formula for the sequence $$3, 7, 14, 24,\ldots\text{.}$$ Assume $$a_1 = 3\text{.}$$
Solution.
First, check to see if the formula has constant differences at some level. The sequence of first differences is $$4, 7, 10, \ldots$$ which is arithmetic, so the sequence of second differences is constant. The sequence is $$\Delta^2$$-constant, so the formula for $$a_n$$ will be a degree 2 polynomial. That is, we know that for some constants $$a\text{,}$$ $$b\text{,}$$ and $$c\text{,}$$
\begin{equation*} a_n = an^2 + bn + c\text{.} \end{equation*}
Now to find $$a\text{,}$$ $$b\text{,}$$ and $$c\text{.}$$ First, it would be nice to know what $$a_0$$ is, since plugging in $$n = 0$$ simplifies the above formula greatly. In this case, $$a_0 = 2$$ (work backwards from the sequence of constant differences). Thus
\begin{equation*} a_0 = 2 = a\cdot 0^2 + b \cdot 0 + c\text{,} \end{equation*}
so $$c = 2\text{.}$$ Now plug in $$n =1$$ and $$n = 2\text{.}$$ We get
\begin{equation*} a_1 = 3 = a + b + 2 \end{equation*}
\begin{equation*} a_2 = 7 = a4 + b 2 + 2\text{.} \end{equation*}
At this point we have two (linear) equations and two unknowns, so we can solve the system for $$a$$ and $$b$$ (using substitution or elimination or even matrices). We find $$a = \frac{3}{2}$$ and $$b = \frac{-1}{2}\text{,}$$ so $$a_n = \frac{3}{2} n^2 - \frac{1}{2}n + 2\text{.}$$
### Example2.3.3.
Find a closed formula for the number of squares on an $$n \times n$$ chessboard.
Solution.
We have seen that the sequence $$1, 5, 14, 30, 55, \ldots$$ is $$\Delta^3$$-constant, so we are looking for a degree 3 polynomial. That is,
\begin{equation*} a_n = an^3 + bn^2 + cn + d\text{.} \end{equation*}
We can find $$d$$ if we know what $$a_0$$ is. Working backwards from the third differences, we find $$a_0 = 0$$ (unsurprisingly, since there are no squares on a $$0\times 0$$ chessboard). Thus $$d = 0\text{.}$$ Now plug in $$n = 1\text{,}$$ $$n =2\text{,}$$ and $$n =3\text{:}$$
\begin{align*} 1 = \amp a + b + c\\ 5 = \amp 8a + 4b + 2c\\ 14 = \amp 27a + 9b + 3c\text{.} \end{align*}
If we solve this system of equations we get $$a = \frac{1}{3}\text{,}$$ $$b = \frac{1}{2}$$ and $$c = \frac{1}{6}\text{.}$$ Therefore the number of squares on an $$n \times n$$ chessboard is $$a_n = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n = \frac{1}{6}n(n+1)(2n+1)\text{.}$$
Note: Since the squares-on-a-chessboard problem is really asking for the sum of squares, we now have a nice formula for $$\d\sum_{k=1}^n k^2\text{.}$$
Not all sequences will have polynomials as their closed formula. We can use the theory of finite differences to identify these.
### Example2.3.4.
Determine whether the following sequences can be described by a polynomial, and if so, of what degree.
1. $$\displaystyle 1, 2, 4, 8, 16, \ldots$$
2. $$\displaystyle 0, 7, 50, 183, 484, 1055, \ldots$$
3. $$\displaystyle 1,1,2,3,5,8,13,\ldots$$
Solution.
1. As we saw in Example 2.3.1, this sequence is not $$\Delta^k$$-constant for any $$k\text{.}$$ Therefore the closed formula for the sequence is not a polynomial. In fact, we know the closed formula is $$a_n = 2^n\text{,}$$ which grows faster than any polynomial (so is not a polynomial).
2. The sequence of first differences is $$7, 43, 133, 301, 571,\ldots\text{.}$$ The second differences are: $$36, 90, 168, 270,\ldots\text{.}$$ Third difference: $$54, 78, 102,\ldots\text{.}$$ Fourth differences: $$24, 24, \ldots\text{.}$$ As far as we can tell, this sequence of differences is constant so the sequence is $$\Delta^4$$-constant and as such the closed formula is a degree 4 polynomial.
3. This is the Fibonacci sequence. The sequence of first differences is $$0, 1, 1, 2, 3, 5, 8, \ldots\text{,}$$ the second differences are $$1, 0, 1, 1, 2, 3, 5\ldots\text{.}$$ We notice that after the first few terms, we get the original sequence back. So there will never be constant differences, so the closed formula for the Fibonacci sequence is not a polynomial.
### ExercisesExercises
#### 1.
Use polynomial fitting to find the formula for the $$n$$th term of the sequence $$(a_n)_{n \ge 0}$$ which starts,
$${0, -1, 4, 15, 32, 55}, \ldots$$
$$a_n =$$
#### 2.
Use polynomial fitting to find the formula for the $$n$$th term of the sequence $$(a_n)_{n \ge 0}$$ which starts,
$${-4, 5, 20, 41, 68, 101}, \ldots$$
$$a_n =$$
#### 3.
Use polynomial fitting to find the formula for the $$n$$th term of the sequence $$(a_n)_{n \ge 0}$$ which starts,
\begin{equation*} {2, 7, 30, 89, 202, 387}, \ldots \end{equation*}
$$a_n =$$
#### 4.
Use polynomial fitting to find the formula for the $$n$$th term of the sequence $$(a_n)_{n \ge 1}$$ which starts,
$${6, 9, 16, 33, 66}, \ldots$$ Note the first term above is $$a_1\text{,}$$ not $$a_0\text{.}$$
$$a_n =$$
#### 5.
Make up sequences that have
1. 3, 3, 3, 3, … as its second differences.
2. 1, 2, 3, 4, 5, … as its third differences.
3. 1, 2, 4, 8, 16, … as its 100th differences.
#### 6.
Consider the sequence $$1, 3, 7, 13, 21, \ldots\text{.}$$ Explain how you know the closed formula for the sequence will be quadratic. Then “guess” the correct formula by comparing this sequence to the squares $$1, 4, 9, 16, \ldots$$ (do not use polynomial fitting).
#### 7.
Use a similar technique as in the previous exercise to find a closed formula for the sequence $$2, 11, 34, 77, 146, 247,\ldots\text{.}$$
#### 8.
Suppose Suppose $$a_n = {n^{2}+5n-3}\text{.}$$ Find a closed formula for the sequence of differences by computing $$a_n - a_{n-1}\text{.}$$ Simplify your answer as much as posible.
$$a_n - a_{n-1} =$$
#### 9.
Generalize Exercise 2.3.8: Find a closed formula for the sequence of differences of $$a_n = an^2 + bn + c\text{.}$$ That is, prove that every quadratic sequence has arithmetic differences.
#### 10.
Can you use polynomial fitting to find the formula for the $$n$$th term of the sequence 4, 7, 11, 18, 29, 47, …? Explain why or why not.
#### 11.
Will the $$n$$th sequence of differences of $$2, 6, 18, 54, 162, \ldots$$ ever be constant? Explain.
#### 12.
In their down time, ghost pirates enjoy stacking cannonballs in triangular based pyramids (aka, tetrahedrons), like those pictured here:
Note, these are solid tetrahedrons, so there will be some cannonballs obscured from view (the picture on the right has one cannonball in the back not shown in the picture, for example)
The pirates wonder how many cannonballs would be required to build a pyramid 15 layers high (thus breaking the world cannonball stacking record). Can you help?
1. Let $$P(n)$$ denote the number of cannonballs needed to create a pyramid $$n$$ layers high. So $$P(1) = 1\text{,}$$ $$P(2) = 4\text{,}$$ and so on. Calculate $$P(3)\text{,}$$ $$P(4)$$ and $$P(5)\text{.}$$
2. Use polynomial fitting to find a closed formula for $$P(n)\text{.}$$ Show your work.
3. Answer the pirate’s question: how many cannonballs do they need to make a pyramid 15 layers high?
4. Bonus: Locate this sequence in Pascal’s triangle. Why does that make sense?
#### 13.
Use polynomial fitting to find the formula for the $$n$$th term of the sequence $$(a_n)_{n \ge 1}$$ which starts,
$${3, 8, 18, 36, 65}, \ldots$$ Note the first term above is $$a_1\text{,}$$ not $$a_0\text{.}$$
$$a_n =$$ |
usagirl007A
2021-09-18
At one point in a pipeline the water’s speed is 3.00 m/s and the gauge pressure is $5.00×{10}^{4}Pa$. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.
Tuthornt
Step 1
The diameters of the two pipe cross-sections are ${d}_{1}$ and ${d}_{2}=2{d}_{1}$. The area of the first is ${A}_{1}=\pi {d}_{1}^{2}$ and of the the second ${A}_{2}=\pi {d}_{2}^{2}=\pi 4{d}_{1}^{2}=4A1$
Step 2
Let's first determine the relation between the velocities based on the continuity equation ${v}_{1}{A}_{1}={v}_{2}{A}_{2}$ and the fact that ${A}_{2}=4{A}_{1}$ from which we can write
${v}_{2}=\frac{{v}_{1}{A}_{1}}{{A}_{2}}$
${v}_{2}=\frac{{v}_{1}}{4}$
Step 3
Since the atmospheric pressures are the same we can write Bernoulli's equation as
${p}_{1}+\rho g{y}_{1}+\frac{1}{2}\rho {v}_{1}^{2}={p}_{2}+\rho g{y}_{2}+\frac{1}{2}\rho {v}_{2}^{2}$ where ${p}_{1}$ and ${p}_{2}$ are gauge pressures. We can now insert ${v}_{2}=\frac{{v}_{1}}{4}$ to get
${p}_{2}={p}_{1}+\rho g\left({y}_{1}-{y}_{2}\right)+\frac{1}{2}\rho \left({v}_{1}^{2}-\frac{{v}_{1}^{2}}{16}\right)$
${p}_{2}={p}_{1}+\rho g\left({y}_{1}-{y}_{2}\right)+\frac{15}{32}\rho {v}_{1}^{2}$
Step 4
At this point we can insert all the given values to get the following
${p}_{2}=5×{10}^{4}+{10}^{3}×9.81×11+{10}^{3}×\frac{15}{32}×{3}^{2}$
${p}_{2}=162kPa$
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# Proportionality (mathematics)
(Redirected from )
For other uses, see Proportionality.
Variable y is directly proportional to the variable x.
In mathematics, two variables are proportional if a change in one is always accompanied by a change in the other, and if the changes are always related by use of a constant multiplier. The constant is called the coefficient of proportionality or proportionality constant.
• If one variable is always the product of the other and a constant, the two are said to be directly proportional. x and y are directly proportional if the ratio y/x is constant.
• If the product of the two variables is always a constant, the two are said to be inversely proportional. x and y are inversely proportional if the product xy is constant.
To express the statement "y is (directly) proportional to x" mathematically, we write an equation y = cx, for some real constant, c. Symbolically, this is written yx.
To express the statement "y is inversely proportional to x" mathematically, we write an equation y = c/x. We can equivalently write "y is proportional to 1/x", which y = c/x would represent.
If a linear function transforms 0, a and b into 0, c and d, and if the product a b c d is not zero, we say a and b are proportional to c and d. An equality of two ratios such as a/c = b/d, where no term is zero, is called a proportion.
## Direct proportionality
Given two variables x and y, y is directly proportional to x (x and y vary directly, or x and y are in direct variation)[1] if there is a non-zero constant k such that
${\displaystyle y=kx.\,}$
U+221D ∝ PROPORTIONAL TO (HTML ∝ · ∝) U+007E ~ TILDE (HTML ~) U+223C ∼ TILDE OPERATOR (HTML ∼ · ∼) U+223A ∺ GEOMETRIC PROPORTION (HTML ∺) See also: Equals sign
The relation is often denoted, using the ∝ or ~ symbol, as
${\displaystyle y\propto x}$
and the constant ratio
${\displaystyle k={\frac {y}{x}}\,}$
is called the proportionality constant, constant of variation or constant of proportionality.
### Examples
• If an object travels at a constant speed, then the distance traveled is directly proportional to the time spent traveling, with the speed being the constant of proportionality.
• The circumference of a circle is directly proportional to its diameter, with the constant of proportionality equal to π.
• On a map drawn to scale, the distance between any two points on the map is directly proportional to the distance between the two locations that the points represent, with the constant of proportionality being the scale of the map.
• The force acting on a certain object due to gravity is directly proportional to the object's mass; the constant of proportionality between the mass and the force is known as gravitational acceleration.
### Properties
Since
${\displaystyle y=kx}$
is equivalent to
${\displaystyle x=\left({\frac {1}{k}}\right)y,}$
it follows that if y is directly proportional to x, with (nonzero) proportionality constant k, then x is also directly proportional to y with proportionality constant 1/k.
If y is directly proportional to x, then the graph of y as a function of x is a straight line passing through the origin with the slope of the line equal to the constant of proportionality: it corresponds to linear growth.
## Inverse proportionality
Inverse proportionality with a function of y = 1/x.
The concept of inverse proportionality can be contrasted against direct proportionality. Consider two variables said to be "inversely proportional" to each other. If all other variables are held constant, the magnitude or absolute value of one inversely proportional variable decreases if the other variable increases, while their product (the constant of proportionality k) is always the same.
Formally, two variables are inversely proportional (also called varying inversely, in inverse variation, in inverse proportion, in reciprocal proportion) if each of the variables is directly proportional to the multiplicative inverse (reciprocal) of the other, or equivalently if their product is a constant.[2] It follows that the variable y is inversely proportional to the variable x if there exists a non-zero constant k such that
${\displaystyle y=\left({\frac {k}{x}}\right)}$
(Also sometimes written as: ${\displaystyle xy=k}$)
The constant can be found by multiplying the original x variable and the original y variable.
As an example, the time taken for a journey is inversely proportional to the speed of travel; the time needed to dig a hole is (approximately) inversely proportional to the number of people digging.
The graph of two variables varying inversely on the Cartesian coordinate plane is a rectangular hyperbola. The product of the x and y values of each point on the curve equals the constant of proportionality (k). Since neither x nor y can equal zero (if k is non-zero), the graph never crosses either axis.
## Hyperbolic coordinates
The concepts of direct and inverse proportion lead to the location of points in the Cartesian plane by hyperbolic coordinates; the two coordinates correspond to the constant of direct proportionality that locates a point on a ray and the constant of inverse proportionality that locates a point on a hyperbola.
## Exponential and logarithmic proportionality
A variable y is exponentially proportional to a variable x, if y is directly proportional to the exponential function of x, that is if there exist non-zero constants k and a such that
${\displaystyle y=ka^{x}.\,}$
Likewise, a variable y is logarithmically proportional to a variable x, if y is directly proportional to the logarithm of x, that is if there exist non-zero constants k and a such that
${\displaystyle y=k\log _{a}(x).\,}$[dubious ]
## Notes
1. ^ Weisstein, Eric W. "Directly Proportional." MathWorld -- A Wolfram Web Resource
2. ^ Weisstein, Eric W. "Inversely Proportional." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverselyProportional.html
## References
• Ya.B. Zeldovich, I.M. Yaglom: Higher math for Beginners. pp. 34-35
• Brian Burell: Merriam-Webster's Guide to Everyday Math: A Home and Business Reference. Merriam-Webster, 1998, ISBN 9780877796213, pp. 85-101
• Lanius, Cynthia S.; Williams Susan E.: PROPORTIONALITY: A Unifying Theme for the Middle Grades. Mathematics Teaching in the Middle School 8.8 (2003), pp. 392-96 (JSTOR)
• Seeley, Cathy; Schielack Jane F.: A Look at the Development of Ratios, Rates, and Proportionality. Mathematics Teaching in the Middle School, 13.3, 2007, pp. 140-42 (JSTOR)
• Van Dooren, Wim; De Bock Dirk; Evers Marleen; Verschaffel Lieven : Students' Overuse of Proportionality on Missing-Value Problems: How Numbers May Change Solutions. Journal for Research in Mathematics Education, 40.2, 2009, pp. 187-211 (JSTOR) |
Finger Math Multiplication
I was surprised when I came across with an article on Polish hand magic, a strategy for multiplication using fingers. I remember us doing it when we were in the first grade, but in a slightly different way. Here in our country, we call it Finger Math or Finger Multiplication.
Finger multiplication is a strategy for multiplying numbers from 6 to 10. It is used by pupils in the early grades who have not memorized the multiplication table yet. The idea is to assign the numbers from 6 to 10 to each finger on both hands (see Figure 1): 6 to the pinkie, 7 to the ring finger, 8 to the middle finger, 9 to the index finger, and 10 to the thumb.
Figure 1
To multiply, do the following (see Figure 2):
(1) Connect the fingers assigned to the numbers that you want to multiply. For example, in multiplying 8 and 7, connect the middle finger and the ring finger (see Figure 2).
(2) Next, we count by 10s the connected finger and all the fingers below them, then find their sum. In the figure, we have 50.
(3) Next, count the fingers above the connected fingers and multiply the number of fingers on the left hand to the number of fingers on the right hand. In the second figure, we have 2 x 3 = 6.
(4) Add the sum obtained in (2) and the product obtained in (3): 50 + 6 = 56.
Figure 2
In the second example (see Figure 3), we multiply 7 and 9. We do this by connecting the index finger and the ring finger. Again, we count by 10s the connected fingers below and all the fingers below them. This gives us 60. Next, we count by 1 the fingers above the connected fingers and multiply the number of fingers the left by the number of fingers at the right which 3 x 1 = 3. We add up: 60 + 3 = 63.
Figure 3
Now, try multiplying the following if it works: (1) 8 and 6, (2) 9 and 6, and (3) 10 and 0.
Amazed?
In the next post, you will be more amazed when we discuss why the strategy above works.
***
Photo Credit: Mi Mano vy Vimayr
7 thoughts on “Finger Math Multiplication”
• Thanks Roman.
1. Thank you for sharing this strategy with us. Many young children learning to multiply struggle to learn the not so friendly factors of 6, 7,8 and 9, and this finger multiplication strategy is a great tool for young children to have. Some adults like me might also benefit from knowing this strategy, it is so fast and easy that I truly wonder why my teachers did not share this tool with me when I was memorizing my multiplication tables. I will be saving this strategy in my math toolbox for future use, and it will soon be added to many more toolboxes. |
# How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of f(x)=x^4+2x^3-12x^2-40x-32?
Sep 23, 2017
Possible rational zeros:
$\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16 , \pm 32$
Descartes gives us that $f \left(x\right)$ has one positive and $3$ or $1$ negative zeros.
Actual zeros: $- 2 , - 2 , - 2 , 4$
#### Explanation:
Given:
$f \left(x\right) = {x}^{4} + 2 {x}^{3} - 12 {x}^{2} - 40 x - 32$
Rational roots theorem
By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 32$ and $q$ a divisor of the coefficient $1$ of the leading term.
That means the the only possible rational zeros are:
$\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16 , \pm 32$
Descartes' Rule of Signs
The pattern of signs of the coefficients of $f \left(x\right)$ is $+ + - - -$. With one change of signs, Descartes' Rule of Signs tells us that $f \left(x\right)$ has exactly one positive real zero.
The pattern of signs of coefficients of $f \left(- x\right)$ is $+ - - + -$. With $3$ changes of sign, Descartes' Rule of Signs allows us to deduce that $f \left(x\right)$ has $3$ or $1$ negative real zero.
Bonus - Find the actual zeros
Note that the coefficient of ${x}^{4}$ is odd, but all the other coefficients are even. Therefore any integer zero of $f \left(x\right)$ must be even.
We find:
$f \left(- 2\right) = {\left(- 2\right)}^{4} + 2 {\left(- 2\right)}^{3} - 12 {\left(- 2\right)}^{2} - 40 \left(- 2\right) - 32 = 16 - 16 - 48 + 80 - 32 = 0$
So $- 2$ is a zero and $\left(x + 2\right)$ a factor:
${x}^{4} + 2 {x}^{3} - 12 {x}^{2} - 40 x - 32 = \left(x + 2\right) \left({x}^{3} - 12 x - 16\right)$
We find that $- 2$ is also a zero of the remaining cubic expression:
${\left(- 2\right)}^{3} - 12 \left(- 2\right) - 16 = - 8 + 24 - 16 = 0$
So $\left(x + 2\right)$ is a factor again:
${x}^{3} - 12 x - 16 = \left(x + 2\right) \left({x}^{2} - 2 x - 8\right)$
Finally, to factor the remaining quadratic, note that $4 \cdot 2 = 8$ and $4 - 2 = 2$, so we find:
${x}^{2} - 2 x - 8 = \left(x - 4\right) \left(x + 2\right)$
So:
$f \left(x\right) = {\left(x + 2\right)}^{3} \left(x - 4\right)$
has zeros $- 2$ with multiplicity $3$ and $4$ with multiplicity $1$. |
# Lesson 8
Document Sample
``` Circular Motion
I. Circular Motion and Polar Coordinates
A. Consider the motion of ball on a circle from point A to point B as shown below. We could
describe the path of the ball in Cartesian coordinates or by polar coordinates. In Cartesian
coordinate system, we see that both coordinates change!! This makes the problem 2-
dimensional.
(XB,YB)
r (XA,YA)
B
If we use polar coordinates, the radius is constant and only the angle theta changes. This
simplifies the system to a 1-dimensional problem and makes the math simpler.
We will deal with this in more detail in Chapter 9 when we study rotation!
B. Tangential Velocity
By its definition in terms of the derivative of the position vector, the velocity vector is
Tangent to the Curve every point on the ball's path. Thus, we call it the
Tangential velocity.
This is the same velocity that we dealt with in 1-dimensional and projectile motion problems.
For a rigid body composed of many particles traveling in circles of different radii, it is convenient
to also define another type of velocity (angular velocity) based upon polar coordinates.
C. Acceleration
1. Acceleration is defined as the Time Rate of Change of the Velocity
Vector.
A Velocity Vector has two parts: Magnitude (Speed) and Direction.
If either part changes then the object is undergoing Acceleration.
Thus, it is often convenient to break the acceleration into components based upon the change
in speed (magnitude) or direction of the velocity vector instead of x and y directions. This is
again an example of using polar coordinates to represent the motion.
2. Tangential Acceleration
This is the acceleration an object feels due to a change in the object’s Speed.
The magnitude of the tangential acceleration is usually either specified in the problem statement
or found using trigonometry.
The tangential acceleration is the only acceleration possible for straight line motion. We
can use this to help us find the direction of the acceleration vector.
Direction
Same as Velocity if Speeding Up
Opposite of Velocity if Slowing Down
Speeding Up Slowing Down
Centripetal means Center Seeking
This tells you that the centripetal acceleration always points to the Center of
the Circle. It is therefore Perpendicular to the Tangential
Acceleration.
Centripetal acceleration is due to the change in the Direction of the
Velocity Vector.
Any object traveling in a Curved Path MUST HAVE Centripetal
Acceleration.
Furthermore, notice that the Centripetal Acceleration is always
Perpendicular to the Tangential Acceleration.
This is why the moon can be accelerating toward the Earth while not moving toward the Earth!!
The magnitude of the centripetal acceleration vector can be found by the formula:
v2
a 2 r
r
This is a very useful formula for solving problems and can be derived directly from the
definition of acceleration using Calculus. This derivation is usually reserved for students in either
Engineering Principles I (Dynamics) or the Junior Level Mechanics class for Physics and
Engineering Physics Majors. You will probably prefer just to memorize the equation.
4. Total Acceleration
The total acceleration of an object traveling in a circle is the vector sum of the
tangential acceleration and the centripetal acceleration.
Example: A car is slowing down at a rate of 6.00 m/s2 while traveling counter clockwise on a circular
track of radius 100.0 m. What is the total acceleration on the car when it has slowed to 20.0 m/s as shown
below:
II. Uniform Circular Motion
An object that is traveling in a circle at constant speed is said to be traveling in
uniform circular motion.
This is just a special case of circular motion where the object has no tangential
acceleration. It does have centripetal acceleration.
III. General Curve-linear Motion In A Plane
As our concept question shows, any curve-linear motion can be seen at every instant as circular
motion a circle whose radius is the radius of curvature of the trajectory at that particular point. In
the case of straight line motion, the radius of curvature is infinity so their is no centripetal
acceleration. In the case of circular motion, the radius of curvature is constant!
So why didn't we treat projectile motion using the concepts of centripetal and tangential
acceleration?
Because it makes the math harder to perform!! In Cartesian coordinates, the
acceleration has only one component (vertical) and it is constant in magnitude. Thus, we can use
the kinematic equations.
In polar form, both the tangential and centripetal acceleration components vary in direction and
magnitude. Thus, we couldn't use the kinematic equations with either component.
We Use Different Coordinate Systems and Define New Quantities To
Make The Math Simpler For Solving Problems.
This Doesn't Mean That There Is New Physics!!
Concept Question
Consider the case of projectile motion from the last lesson: A cannon ball is
fired out of a cannon and follows a parabolic path before hitting the ground.
What type(s) of acceleration does the cannon ball have during its flight at
point X?
X
A. Tangential Acceleration
B. Centripetal Acceleration
C. Both Tangential and Centripetal Acceleration
D. Neither Centripetal or Tangential Acceleration
```
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