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We start the year 2015 with a nice topic in polynomials, in this case let’s talk about Chebyshev Polynomial, only we take the first kind, we left the Second kind for another opportunity. So, which is it?
Definition
Chebyshev polynomials are polynomials with the largest possible leading coefficient, with the condition that their absolute value in the interval $[-1, 1]$ is bounded by $1$. The chebyshev polynomial of the first kind are defined by the recurrence relation.
$T_{0}(x) = 1$
$T_{1}(x) = x$
$T_{n+1}(x) = 2xT_{n}(x) - T_{n-1}(x)$
These are characterized by the property:
We can raise the following question
Prove that for each $n \in \mathbb{N}$ there is a polynomial $T_{n}$ with integer coefficients and the leading $2^{n-1}$ such that $T_{n}(\cos{x}) = \cos{nx}$ for all $x$
Solution
We know that $T_{0}(x)=1$ and $T_{1}(x)=x$. For $n>1$ we use induction on $n$.
Since $\cos{(n+1)}x = 2\cos{x}\cos{(nx)} - \cos{(n-1)}x$, we can define $T_{n+1} = 2T_{1}T_{n} - T_{n-1}$ Since $T_{1}T_{n}$ and $T_{n-1}$ are of degrees $n+1$ and $n-1$ respectively, $T_{n+1}$ is of degree $n+1$ and has the leading coefficient $2\times2^{n} = 2^{n+1}$. It also follows from the construction that all its coefficients are integers.
Actually, these sequence of orthogonal polynomials are related to De Moivre’s formula
De Moivre’s Theorem
For any complex number $x$ and any integer $n$.
Proof:
We prove this formula by induction on $n$ and by applying the trigonometric sum and product formulas. We first consider the non-negative integers. The base case $n=0$ is clearly true. For the induction step, observe that:
Here we have a problem about chebyshev polynomial:
Prove that the maximum in absolute value of any monic real polynomial of $n$-th degree on $[-1, 1]$ is not less than $\frac{1}{2^{n-1}}$
Solution
Note that equality holds for a multiple of the n-th Chebyshev polynomial $T_{n}(X)$ The leading coefficient of $T_{n}$ equals $2^{n-1}$, so $C_{n}(X) = \frac{1}{2^{n-1}}T_{n}(X)$ is a monic polynomial and
Moreover the values of $T_{n}$ at points $1, \cos{\frac{\pi}{n}}, \cos{\frac{2\pi}{n}}, \dots , \cos{\frac{(n-1)\pi}{n}}, -1$ are alternately $\frac{1}{2^{n-1}}$ and $-\frac{1}{2^{n-1}}$
Now suppose that $P \neq T_{n}$ is a monic polynomial such that $% $ Then $P(X)-C_{n}(X)$ at points $1, \cos{\frac{\pi}{n}}, \dots , \cos{\frac{(n-1)\pi}{n}}, -1$ alternately takes positive and negative values. Therefore the polynomial $P-C_{n}$ has at least $n$ zeros, namely, at least one is very interval between two adjacent points. However, $P-C_{n}$ is a polynomial of degree $n-1$ as the monomial $x^n$ is canceled, so we have arrived at a contradiction.
By the way, I would like to know other solutions for the last one, so I asked on MathStackExchange. If you want share something with me, please send me a message.
Now, we describe a theorem generated by the Chebyshev polynomials.
Chebyshev’s theorem
For fixed $n\geq1$, the polynomial $2^{-n+1}T_{n}(x)$ is the unique monic $n$th-degree polynomial satisfying
for any other monic $n$th-degree polynomial $P(x)$. Let’s see an example.
Let $A_{1}, A_{2}, \dots, A_{n}$ be points in the plane. Prove that on any segment of length $l$ there is a point $M$ such that
Solution
We can assume that $l=2$. Associate complex coordinates to points in such a way that the segment coincides with the interval $[-1, 1]$. Then
where $P(z)$ is a monic polynomial with complex coefficients, and $z \in [-1, 1]$.
We can write $P(z) = R(z) + iQ(z)$, where $R(z)$ is the real part and $Q(z)$ is the imaginary part of the polynomial. Since $z$ is real, we have $|P(z)| \geq |R(z)|$. The polynomial $R(z)$ is monic, so on the interval [-1, 1] it varies away from zero at least as much as the Chebyshev polynomial. Thus we can find $z$ in this interval such that $|R(z)| \geq \frac{1}{2^{n-1}}$. This implies $|P(z)| \geq 2\frac{1}{2^n}$, and rescaling back we deduce the existence in the general case of a point $M$ satisfying the inequality from the statement.
Stepping aside from the classical picture, let us also consider the families of polynomials $T_{n}(x)$ defined by $T_{0}(x) = 2, T_{1}(x) = x, T_{n+1}(x) = xT_{n}(x) - T_{n-1}(x)$. It’s determined by the equality:
Also, I suggest visit the amazing answer for:
Let $a_{i} \in \mathbb{R} (i=1, 2, \dots, n)$ and $f(x) = \sum_{i=0}^{n}a_{i}x^{i}$ such that if $|x| \leq 1$, then $|f(x)| \leq 1$ on MathStackExchange |
# GRE Math : Midpoint Formula
## Example Questions
### Example Question #1 : How To Find The Endpoints Of A Line Segment
There is a line defined by two end-points, and . The midpoint between these two points is . What is the value of the point ?
Explanation:
Recall that to find the midpoint of two points and , you use the equation:
.
(It is just like finding the average of the two points, really.)
So, for our equation, we know the following:
You merely need to solve each coordinate for its respective value.
Then, for the y-coordinate:
Therefore, our other point is:
### Example Question #2 : How To Find The Endpoints Of A Line Segment
There is a line defined by two end-points, and . The midpoint between these two points is . What is the value of the point ?
Explanation:
Recall that to find the midpoint of two points and , you use the equation:
.
(It is just like finding the average of the two points, really.)
So, for our equation, we know the following:
You merely need to solve each coordinate for its respective value.
Then, for the y-coordinate:
Therefore, our other point is:
### Example Question #3 : How To Find The Endpoints Of A Line Segment
What is the other endpoint of a line segment with one point that is and a midpoint of ?
Explanation:
Recall that the midpoint formula is like finding the average of the and values for two points. For two points and , it is:
For our points, we are looking for . We know:
We can solve for each of these coordinates separately:
X-Coordinate
Y-Coordinate:
Therefore, our point is
### Example Question #1 : How To Find The Endpoints Of A Line Segment
What is the other endpoint of a line segment with one point that is and a midpoint of ?
Explanation:
What is the other endpoint of a line segment with one point that is and a midpoint of ?
Recall that the midpoint formula is like finding the average of the and values for two points. For two points and , it is:
For our points, we are looking for . We know:
We can solve for each of these coordinates separately:
X-Coordinate
Y-Coordinate:
Therefore, our point is
### Example Question #1 : How To Find The Midpoint Of A Line Segment
What is the midpoint of (2, 5) and (14, 18)?
(–10, –13)
(16, 23)
(1, 2.5)
(7, 9)
(8, 11.5)
(8, 11.5)
Explanation:
The midpoint between two given points is found by solving for the average of each of the correlative coordinates of the given points. That is:
Midpoint = ( (2 + 14)/2 , (18 + 5)/2) = (16/2, 23/2) = (8, 11.5)
### Example Question #2 : How To Find The Midpoint Of A Line Segment
What is the midpoint between the points (1,3,7) and (–3,1,3)?
(3,1,2)
(2,–1,5)
(2,2,5)
(–1,2,5)
(5,2,4)
(–1,2,5)
Explanation:
To find the midpoint, we add up the corresponding coordinates and divide by 2.
[1 + –3] / 2 = –1
[3 + 1] / 2 = 2
[7 + 3] / 2 = 5
Then the midpoint is (–1,2,5).
### Example Question #1 : How To Find The Midpoint Of A Line Segment
A line which cuts another line segment into two equal parts is called a ___________.
bisector
transversal
parallel line
midpoint
horizontal line
bisector
Explanation:
This is the definition of a bisector.
A midpoint is the point on a line that divides it into two equal parts. The bisector cuts the line at the midpoint, but the midpoint is not a line.
A transversal is a line that cuts across two or more lines that are usually parallel.
Parallel line and horizontal line don't make sense as answer choices here. The answer is bisector.
Tired of practice problems?
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### N000ughty Thoughts
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in 10 000! and 100 000! or even 1 000 000!
### Fac-finding
Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful.
### Binomial Coefficients
An introduction to the binomial coefficient, and exploration of some of the formulae it satisfies.
# Factorial
##### Stage: 4 Challenge Level:
The problem here is to find the number of zeros at the end of the number which is the product of the first hundred positive integers. We call this 100 factorial' and write it 100!
For example 4! = 1x2x3x4 = 24 and 5! = 1x2x3x4x5 = 120.
Well done Xinxin of Tao Nan School, Singapore who sent in the solution below in record time and also the Key Stage 3 Maths Club at Strabane Grammar School, N. Ireland and James of Hethersett High School, Norfolk.
The Strabane group said We started with a 10 x 10 number square and worked out 2!, 3!, 4! ... etc. We quickly realised that the number of zeros at the end of 100! depends on the number of tens appearing within the product which in turn depends on the number of twos and fives'' and Xinxin's solution is re-produced in full below.
This question basically asks about the number of zeros ending the number 100!'.
Since 2x5 equals 10, the key to answering this question is finding out the number of matches of 2 and 5 occurring in the prime factors of 100!.
Since it is obvious that when 100! is factorised, there are more 2's than 5's. As a result, all the 5's will find matches. Counting the number of 5's gives the number of matches.
First, all the multiples of 5:
5=1x5
10=2x5
15=3x5
20=4x5
...
There are a total of 20 multiples of 5. As a result, we have already found 20 matches, and thus 20 zeros.
However, it is noted that four numbers contribute two 5's to the factors of 100!. They are:
25=5x5
50=2x5x5
75=3x5x5
100=2x2x5x5
As a result, there are, in fact, 24 5's in the factors of 100!.
Thus, 100! ends with 24 zeros.
Done on the 4 th of November 1998 by:
Li Xinxin
Tao Nan School
Singapore. |
Update all PDFs
# Finding an unknown angle
Alignments to Content Standards: 4.G.A.2 4.MD.C.7
In the figure, $ABCD$ is a rectangle and $\angle CAD = 31^{\circ}$. Find $\angle BAC$.
## IM Commentary
The purpose of this task is to give 4th grade students a problem involving an unknown quantity that has a clear visual representation. Students must understand that the four interior angles of a rectangle are all right angles (4.G.2) and that right angles have a measure of $90^\circ$ and that angle measure is additive (4.MD.7). In a teaching scenario, students may be allowed to verify the computations using a protractor to measure the angles. However, care should be taken beforehand to ensure that the measurements of the printed figure match the stated measurements.
The task may also be viewed as preparation for later work when, in 6th grade, students are introduced to algebraic expressions. In that context, unknown angle problems will use variables to label missing angles, and students will write and solve equations to find the missing angle measures.
This task includes an experimental GeoGebra worksheet, with the intent that instructors might use it to more interactively demonstrate the relevant content material. The file should be considered a draft version, and feedback on it in the comment section is highly encouraged, both in terms of suggestions for improvement and for ideas on using it effectively. The file can be run via the free online application GeoGebra, or run locally if GeoGebra has been installed on a computer.
This applet allows the teacher to first enter the any angle. After setting the angle, the problem appears and can be given to the student.
## Attached Resources
• 1168 GeoGebra File
• ## Solution
All four angles in a rectangle are right angles, so $\angle BAD$ is $90^\circ$. Since $\angle BAC + \angle CAD = \angle BAD$, we have that
$$\angle BAC + 31^\circ = 90^\circ$$
which is the same as saying
$$\angle BAC = 90^\circ - 31^\circ.$$
Thus, $\angle BAC = 59^\circ$.
#### David Woodward says:
almost 6 years
I have heard some debate about whether the intention of this standard includes the symbolic language included in this illustration. I guess the big question is how does PARCC see it (or Smarter Balanced if you are on that side of things.) Does anyone know? Is there any offical reference to the details of this standard that would help us to understand?
#### Kristin says:
almost 6 years
I assume you are asking about the angle notation like $\angle BAC$. I can't speak for PARCC or Smarter Balanced, but if you read the K-5 geometry progressions document at |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 6.3: Inverse Trigonometric Functions
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## Section 6.3 Inverse Trig Functions
In previous sections, we have evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that for a one-to-one function, if $$f(a)=b$$, then an inverse function would satisfy $$f^{-1} (b)=a$$.
You probably are already recognizing an issue – that the sine, cosine, and tangent functions are not one-to-one functions. To define an inverse of these functions, we will need to restrict the domain of these functions to yield a new function that is one-to-one. We choose a domain for each function that includes the angle zero.
Sine, limited to $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$ Cosine, limited to $$\left[0,\pi \right]$$ Tangent, limited to $$\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)$$
On these restricted domains, we can define the inverse sine, inverse cosine, and inverse tangent functions.
Inverse Sine, Cosine, and Tangent Functions
For angles in the interval $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$, if $$\sin \left(\theta \right)=a$$, then $$\sin ^{-1} \left(a\right)=\theta$$
For angles in the interval $$\left[0,\pi \right]$$, if $$\cos \left(\theta \right)=a$$, then $$\cos ^{-1} \left(a\right)=\theta$$
For angles in the interval $$\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)$$, if $$\tan \left(\theta \right)=a$$, then $$\tan ^{-1} \left(a\right)=\theta$$
$$\sin ^{-1} \left(x\right)$$ has domain [-1, 1] and range $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$
$$\cos ^{-1} \left(x\right)$$ has domain [-1, 1] and range $$\left[0,\pi \right]$$
$$\tan ^{-1} \left(x\right)$$ has domain of all real numbers and range $$\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)$$
The $$\sin ^{-1} \left(x\right)$$ is sometimes called the arcsineArcsine, Arccosine and Arctangent function, and notated $$\arcsin \left(a\right)$$.
The $$\cos ^{-1} \left(x\right)$$ is sometimes called the arccosine function, and notated $$\arccos \left(a\right)$$.
The $$\tan ^{-1} \left(x\right)$$ is sometimes called the arctangent function, and notated $$\arctan \left(a\right)$$.
The graphs of the inverse functions are shown here:
$\sin ^{-1} \left(x\right) \cos ^{-1} \left(x\right) \tan ^{-1} \left(x\right)$
Notice that the output of each of these inverse functions is an angle.
Example $$\PageIndex{1}$$
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1
Evaluate
a) $$\sin ^{-1} \left(\frac{1}{2} \right)$$ b) $$\sin ^{-1} \left(-\frac{\sqrt{2} }{2} \right)$$ c) $$\cos ^{-1} \left(-\frac{\sqrt{3} }{2} \right)$$ d) $$\tan ^{-1} \left(1\right)$$
a) Evaluating $$\sin ^{-1} \left(\frac{1}{2} \right)$$ is the same as asking what angle would have a sine value of $$\frac{1}{2}$$. In other words, what angle $$\theta$$ would satisfy $$\sin \left(\theta \right)=\frac{1}{2}$$?
There are multiple angles that would satisfy this relationship, such as $$\frac{\pi }{6}$$ and $$\frac{5\pi }{6}$$ , but we know we need the angle in the range of $$\sin ^{-1} \left(x\right)$$, the interval $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$, so the answer will be $$\sin ^{-1} \left(\frac{1}{2} \right)=\frac{\pi }{6}$$.
Remember that the inverse is a function so for each input, we will get exactly one output.
b) Evaluating $$\sin ^{-1} \left(-\frac{\sqrt{2} }{2} \right)$$, we know that $$\frac{5\pi }{4}$$ and $$\frac{7\pi }{4}$$ both have a sine value of $$-\frac{\sqrt{2} }{2}$$, but neither is in the interval $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$. For that, we need the negative angle coterminal with $$\frac{7\pi }{4}$$. $$\sin ^{-1} \left(-\frac{\sqrt{2} }{2} \right)=-\frac{\pi }{4}$$.
c) Evaluating $$\cos ^{-1} \left(-\frac{\sqrt{3} }{2} \right)$$, we are looking for an angle in the interval $$\left[0,\pi \right]$$ with a cosine value of $$-\frac{\sqrt{3} }{2}$$. The angle that satisfies this is $$\cos ^{-1} \left(-\frac{\sqrt{3} }{2} \right)=\frac{5\pi }{6}$$.
d) Evaluating $$\tan ^{-1} \left(1\right)$$, we are looking for an angle in the interval $$\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)$$ with a tangent value of 1. The correct angle is $$\tan ^{-1} \left(1\right)=\frac{\pi }{4}$$.
Exercise $$\PageIndex{1}$$
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1. Evaluate
a) $$\sin ^{-1} \left(-1\right)$$ b) $$\tan ^{-1} \left(-1\right)$$ c) $$\cos ^{-1} \left(-1\right)$$ d) $$\cos ^{-1} \left(\frac{1}{2} \right)$$
Example $$\PageIndex{1}$$
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2
Evaluate $$\sin ^{-1} \left(0.97\right)$$ using your calculator.
Since the output of the inverse function is an angle, your calculator will give you a degree value if in degree mode, and a radian value if in radian mode.
In radian mode, $$\sin ^{-1} \eqref{GrindEQ__0_97_}\approx 1.3252$$ In degree mode, $$\sin ^{-1} \left(0.97\right)\approx 75.93{}^\circ$$
Exercise $$\PageIndex{1}$$
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2. Evaluate $$\cos ^{-1} \left(-0.4\right)$$ using your calculator.
In Section 5.5, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trig functions, we can solve for the angles of a right triangle given two sides.
Example $$\PageIndex{1}$$
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3
Solve the triangle for the angle $$\theta$$.
Since we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.
$$\cos \left(\theta \right)=\frac{9}{12}$$ Using the definition of the inverse,
$$\theta =\cos ^{-1} \left(\frac{9}{12} \right)$$ Evaluating
$$\theta \approx 0.7227$$, or about 41.4096$$\mathrm{{}^\circ}$$
There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can find exact values for the resulting expressions
Example $$\PageIndex{4}$$
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4
Evaluate $$\sin ^{-1} \left(\cos \left(\frac{13\pi }{6} \right)\right)$$.
a) Here, we can directly evaluate the inside of the composition. $\cos \left(\frac{13\pi }{6} \right)=\frac{\sqrt{3} }{2}$
Now, we can evaluate the inverse function as we did earlier. $\sin ^{-1} \left(\frac{\sqrt{3} }{2} \right)=\frac{\pi }{3}$
Exercise $$\PageIndex{1}$$
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3. Evaluate $$\cos ^{-1} \left(\sin \left(-\frac{11\pi }{4} \right)\right)$$.
Example $$\PageIndex{1}$$
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5
Find an exact value for $$\sin \left(\cos ^{-1} \left(\frac{4}{5} \right)\right)$$.
Beginning with the inside, we can say there is some angle so $$\theta =\cos ^{-1} \left(\frac{4}{5} \right)$$, which means $$\cos \left(\theta \right)=\frac{4}{5}$$, and we are looking for $$\sin \left(\theta \right)$$. We can use the Pythagorean identity to do this.
$$\sin ^{2} \left(\theta \right)+\cos ^{2} \left(\theta \right)=1$$ Using our known value for cosine
$$\sin ^{2} \left(\theta \right)+\left(\frac{4}{5} \right)^{2} =1$$ Solving for sine $\sin ^{2} \left(\theta \right)=1-\frac{16}{25}$ $\sin \left(\theta \right)=\pm \sqrt{\frac{9}{25} } =\pm \frac{3}{5}$
Since we know that the inverse cosine always gives an angle on the interval $$\left[0,\pi \right]$$, we know that the sine of that angle must be positive, so $$\sin \left(\cos ^{-1} \left(\frac{4}{5} \right)\right)=\sin (\theta )=\frac{3}{5}$$
Example $$\PageIndex{1}$$
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6
Find an exact value for $$\sin \left(\tan ^{-1} \left(\frac{7}{4} \right)\right)$$.
While we could use a similar technique as in the last example, we will demonstrate a different technique here. From the inside, we know there is an angle so $$\tan \left(\theta \right)=\frac{7}{4}$$. We can envision this as the opposite and adjacent sides on a right triangle.
Using the Pythagorean Theorem, we can find the hypotenuse of this triangle: $4^{2} +7^{2} =hypotenuse\; ^{2}$ $hypotenuse=\sqrt{65}$
Now, we can represent the sine of the angle as opposite side divided by hypotenuse. $\sin \left(\theta \right)=\frac{7}{\sqrt{65} }$
This gives us our desired composition $\sin \left(\tan ^{-1} \left(\frac{7}{4} \right)\right)=\sin (\theta )=\frac{7}{\sqrt{65} } .$
Exercise $$\PageIndex{1}$$
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4. Evaluate $$\cos \left(\sin ^{-1} \left(\frac{7}{9} \right)\right)$$.
We can also find compositions involving algebraic expressions
Example $$\PageIndex{1}$$
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7
Find a simplified expression for $$\cos \left(\sin ^{-1} \left(\frac{x}{3} \right)\right)$$, for $$-3\le x\le 3$$.
We know there is an angle $$\theta$$ so that $$\sin \left(\theta \right)=\frac{x}{3}$$. Using the Pythagorean Theorem,
$$\sin ^{2} \left(\theta \right)+\cos ^{2} \left(\theta \right)=1$$ Using our known expression for sine
$$\left(\frac{x}{3} \right)^{2} +\cos ^{2} \left(\theta \right)=1$$ Solving for cosine $\cos ^{2} \left(\theta \right)=1-\frac{x^{2} }{9}$ $\cos \left(\theta \right)=\pm \sqrt{\frac{9-x^{2} }{9} } =\pm \frac{\sqrt{9-x^{2} } }{3}$ Since we know that the inverse sine must give an angle on the interval $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$, we can deduce that the cosine of that angle must be positive. This gives us
$\cos \left(\sin ^{-1} \left(\frac{x}{3} \right)\right)=\frac{\sqrt{9-x^{2} } }{3}$
Exercise $$\PageIndex{1}$$
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5. Find a simplified expression for $$\sin \left(\tan ^{-1} \left(4x\right)\right)$$, for $$-\frac{1}{4} \le x\le \frac{1}{4}$$.
### Important Topics of This Section
• Inverse trig functions: arcsine, arccosine and arctangent
• Domain restrictions
• Evaluating inverses using unit circle values and the calculator
• Simplifying numerical expressions involving the inverse trig functions
• Simplifying algebraic expressions involving the inverse trig functions
1. a) $$-\frac{\pi }{2}$$ b) $$-\frac{\pi }{4}$$ c) $$\pi$$ d) $$\frac{\pi }{3}$$
2. 1.9823 or 113.578$$\mathrm{{}^\circ}$$
$3. \sin \left(-\frac{11\pi }{4} \right)=-\frac{\sqrt{2} }{2} . \cos ^{-1} \left(-\frac{\sqrt{2} }{2} \right)=\frac{3\pi }{4}$
4. Let $$\theta =\sin ^{-1} \left(\frac{7}{9} \right)$$ so $$\sin (\theta )=\frac{7}{9}$$. .Using Pythagorean Identity, $$\sin ^{2} \theta +\cos ^{2} \theta =1$$, so $$\left(\frac{7}{9} \right)^{2} +\cos ^{2} \theta =1$$. Solving, $$\cos \left(\sin ^{-1} \left(\frac{7}{9} \right)\right)=\cos \left(\theta \right)=\frac{4\sqrt{2} }{9}$$ .
5. Let $$\theta =\tan ^{-1} \left(4x\right)$$, so $$\tan (\theta )=4x$$. We can represent this on a triangle as $$\tan (\theta )=\frac{4x}{1}$$.The hypotenuse of the triangle would be $$\sqrt{\left(4x\right)^{2} +1}$$. $$\sin \left(\tan ^{-1} \left(4x\right)\right)=\sin (\theta )=\frac{4x}{\sqrt{16x^{2} +1} }$$ |
# 72.0 ml of a 1.50 M solution is diluted to a volume of 288 mL. A 144 mL portion of that solution...
## Question:
72.0 ml of a 1.50 M solution is diluted to a volume of 288 mL. A 144 mL portion of that solution is diluted using 185 mL of water. What is the final concentration?
## Dilution
The process of dilution involved the addition of water to a concentrated solution to make a dilute solution. This process is governed by the equation
$$M_1V_1 = M_2V_2$$
where,
• {eq}M_1{/eq} is the intial concentration,
• {eq}V_1{/eq} is the intial volume,
• {eq}M_2{/eq} is the final concentration,
• {eq}V_2{/eq} is the final volume.
## Answer and Explanation: 1
Consider a first cycle of dilution. Use the equation below to solve for the final concentration of the solution. That is,
{eq}M_1V_1 = M_2V_2 {/eq}
where,
• {eq}M_1 = 1.50\ M {/eq} is the intial concentration,
• {eq}V_1 = 72\ mL {/eq} is the intial volume,
• {eq}M_2 {/eq} is the final concentration,
• {eq}V_2 = 288\ mL {/eq} is the final volume.
Putting {eq}M_2 {/eq} on one side of the equation and substituting all given values. Therefore,
{eq}M_2 = \dfrac{M_1V_1}{V_2}\\ M_2 = \dfrac{(1.50\ M)(72.0\ mL)}{288\ mL}\\ M_2 = 0.375\ M {/eq}
Now consider the second cycle of dilution. Solving the final concentration using the same equation.
Where,
• {eq}M_1 = 0.375\ M {/eq} is the intial concentration,
• {eq}V_1 = 144\ mL {/eq} is the intial volume,
• {eq}M_2 {/eq} is the final concentration,
• {eq}V_2 = 144\ mL +185\ mL = 329\ mL {/eq} is the final volume.
Putting {eq}M_2 {/eq} on one side of the equation and substituting all given values. Therefore,
{eq}M_2 = \dfrac{M_1V_1}{V_2}\\ M_2 = \dfrac{(0.375\ M)(144\ mL)}{329\ mL}\\ \boxed{M_2 = 0.1641\ M} {/eq} |
# NCERT solutions for class 9 maths chapter 5 Ex 5.1 and 5.2
In this page we have NCERT solutions for class 9 maths chapter 5 Euclids Geometry for EXERCISE 5.1 & 5.2 on pages 85,86 and 88. Hope you like them and do not forget to like , social share and comment at the end of the page.
## Chapter 5 Ex 5.1
Question 1:
Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In below figure, if AB = PQ and PQ = XY, then AB = XY.
Solution:
(i) False. Since through a single point, infinite number of lines can pass
(ii) False. Since through two distinct points, only one line can pass
(iii) True. A terminated line can be produced indefinitely on both the sides.
(iv)True. If two circles are equal, then their centre and circumference will coincide and hence, the radii will also be equal.
(v) True. It is given that AB and XY are two terminated lines and both are equal to a third line PQ. Euclid’s first axiom states that things which are equal to the same thing are equal to one another. Therefore, the lines AB and XY will be equal to each other.
Question 2:
Give a definition for each of the following terms. Are there other terms that need to be
defined first? What are they, and how might you define them?
1. parallel lines
2. perpendicular lines
3. line segment
5. square
Solution:
(i) Parallel Lines:
If the perpendicular distance between two lines is always constant, these are called parallel lines. In other words we may say that the lines which never intersect each other are called parallel lines.
To define parallel line, we must know about point, lines and distance between lines and point of intersection.
The definitions of line, point, plane explained by Euclid is not accepted by the Mathematician. So these terms are taken as undefined
(ii) Perpendicular lines:
If two lines intersect each other at 90° , these are called perpendicular lines. We need to define line and the angle before defining perpendicular lines.
The definitions of line, point, plane explained by Euclid is not accepted by the Mathematician. So these terms are taken as undefined
(iii) Line segment:
A straight line drawn from any point to any other point is called as line segment. To define line segment, we must know about point and line segment.
The definitions of line, point, plane explained by Euclid is not accepted by the Mathematician. So these terms are taken as undefined
It is the distance between the centre of a circle to any point lying on the circle. To define radius of circle, we must know about point and circle.
The definitions of line, point, plane explained by Euclid is not accepted by the Mathematician. So these terms are taken as undefined
(v) Square:
A square is a quadrilateral having all sides of equal length and all angles of same measure i.e. 90o . To define square, we must know about quadrilateral, side, and angle.
The definitions of line, point, plane explained by Euclid is not accepted by the Mathematician. So these terms are taken as undefined
Question 3
Consider two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent?
Do they follow from Euclid’s postulates? Explain.
Solution
There are various undefined terms in the given postulates.
The given postulates are consistent because they refer to two different situations. Also, it is impossible to deduce any statement that contradicts any well-known axiom and postulate.
These postulates do not follow from Euclid’s postulates. They follow from the axiom, “Given two distinct points, there is a unique line that passes through them”.
Question 4:
If a point C lies between two points A and B such that AC = BC, then prove that
$AC =\frac {1}{2} AB$. Explain by drawing the figure.
Solution:
Given that from question
AC=BC
We add equals on both sides
$AC+AC=BC+AC$ ... (a)
Here, (BC + AC) coincides with AB. We know that things which coincide with one another are equal to one another.
$BC + AC =AB$ ... (b)
We know that things which are equal to the same thing are equal to one another. So, from equation (a) and equation (b), we have
$AC + AC = AB$
$2AC = AB$
$AC =\frac {1}{2} AB$
Question 5:
In Question 4, point C is called a mid-point of line segment AB. Prove that every line
segment has one and only one mid-point.
Solution
Let us assume there are two mid-points $C_1$ and $C_2$
$C_1$ is midpoint of AB
$AC_1 = C_1B$
$A C_1 + A C_1 = B C_1 + A C_1$ (equals are added on both sides) ... (1)
Here, ($B C_1 + A C_1$) coincides with AB. We know that things which coincide with one another are equal to one another.
$B C_1 + A C_1 = AB$ ... (2)
We know that things which are equal to the same thing are equal to one another. So, from equation (1) and equation (2), we have
$A C_1 + A C_1 = AB$
$2A C_1= AB$... (3)
Similarly by taking $C_2$ as the mid-point of AB, we can prove that
$2A C_2 = AB$ ... (4)
From equation (3) and (4), we have
$2A C_1= 2A C_2$ (Things which are equal to the same thing are equal to one another.)
$A C_1 = A C_2$ (Things which are double of the same things are equal to one another.)
It is possible only when point $C_1$ and $C_2$ are representing a single point.
Hence our assumption is wrong and there can be only one mid-point of a given line segment.
Question 6:
If $AC = BD$, then prove that $AB = CD$.
Solution:
We can be observed from the figure that
$AC = AB + BC$
$BD = BC + CD$
It is given that AC = BD
So $AB + BC = BC + CD$
According to Euclid’s axiom, when equals are subtracted from equals, the remainders are also equal.
Subtracting BC from equation (1), we obtain
$AB + BC -BC = BC + CD - BC$
AB = CD
Question 7:
Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)
Solution:
Axiom 5: The whole is greater than the part.
Let X is representing a whole quantity and only a, b, c are parts of it.
Now X = a + b + c
Clearly X will be greater than all its parts a, b and c
So, it is rightly said that the whole is greater than the part.
There can be many examples. This is true for anything in any part of the world, this is a universal truth.
## Chapter 5 Ex 5.2
Question 1:
How would you rewrite Euclid's fifth postulate so that it would be easier to understand?
Solution:
Two distinct intersecting lines cannot be parallel to the same line
Question 2:
Does Euclid's fifth postulate imply the existence of parallel lines? Explain.
Solution:
Yes it is true
If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the straight lines if produced indefinitely meet on that side on which the angles are less than the two right angles
If the sum interior angles on the same side of it taken together is equal to two right angles,then two lines will not meet on either side. Hence lines are parallel
## Summary
1. NCERT solutions for class 9 maths chapter 5 Euclid Geometry Exercise 5.1 and 5.2 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also
2. This chapter 5 has total 2 Exercise 5.1 and 5.2. This page has both the exercise in the chapter.
• Notes
• NCERT Solutions
• Assignments
• Revision sheet |
# Sums of Powers of Positive Integers - Aryabhata (b. 476), northern India
Author(s):
Janet Beery (University of Redlands)
The northern Indian mathematician and astronomer, Aryabhata, born in 476, wrote one of the earliest known Indian mathematics and astronomy books, the Aryabhatiya, in 499 (Katz, p. 212). In Section II, Stanza 22, of the Aryabhatiya, he wrote:
The sixth part of the product of three quantities consisting of the number of terms, the number of terms plus one, and twice the number of terms plus one is the sum of the squares. The square of the sum of the (original) series is the sum of the cubes. (Katz, 217)
The first sentence gives the formula from pages 1 and 3, above, for the sum of the squares; the second says, in our notation, that
$(1 + 2 + 3 + \cdots + n)^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3.$
If we replace $1 + 2 + 3 + · · · + n$ by ${{n(n + 1)} \over 2}$, we obtain
$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left( {{{n(n + 1)} \over 2}} \right)^2.$
It seems likely that mathematicians discovered this formula for the sum of the cubes early on by taking note of examples, such as
$1^3 = 1 = 1^2,$
$1^3 + 2^3 = 9 = 3^2 = (1 + 2)^2,$
$1^3 + 2^3 + 3^3 = 36 = 6^2 = (1 + 2 + 3)^2,$ and
$1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2 = (1 + 2 + 3 + 4)^2.$
We would then generalize to
$1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$
and hence to
$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left( {{{n(n + 1)} \over 2}} \right)^2.$
for all positive integers n.
Exercise 4: Use 84 or 180 cubes to demonstrate the identity $1^2 + 2^2 + 3^2 + \cdots + n^2 = {{n(n + 1)(2n + 1)} \over 6}$
for n = 3 or n = 4, as described by the Indian mathematician Nilakantha (c. 1445-1545). A member of the famous Kerala school of mathematics and astronomy in southern India, Nilakantha wrote a commentary on the Aryabhatiya in which he gave the following proof of the identity above. He explained that six copies of the sum $1^2 + 2^2 + 3^2 + \cdots + n^2$
form an n by (n + 1) by (2n + 1) rectangular solid as follows. Start on the outside with a floor and three walls consisting of 6n2 cubes, then work from the outside inwards, lining the inside of the existing shell with a floor and three walls consisting of 6(n – 1)2 cubes, then 6(n – 2)2 cubes, . . . , then 6 22 cubes, and finally 6 12 cubes. The outside shell has height n, depth n + 1, and length 2n + 1, with one long side open and the top open. If n = 3, this shell has dimensions 3, 4, and 7, and consists of 6 32 = 54 cubes. Construct this shell, then, inside of it, a shell having dimensions 2, 3, and 5 and consisting of 6 22 = 24 cubes, and then, inside of the double shell, a 1 x 2 x 3 layer consisting of 6 12 = 6 cubes. The result should be a 3 x 4 x 7 rectangular solid. This construction illustrates that
$6 \cdot 1^2 + 6 \cdot 2^2 + 6 \cdot 3^2 = 3 \cdot 4 \cdot 7 \quad {\rm or} \quad 1^2 + 2^2 + 3^2 = {{3 \cdot 4 \cdot 7} \over 6},$
or, more generally, $1^2 + 2^2 + 3^2 + \cdots + n^2 = {{n(n + 1)(2n + 1)} \over 6}$
for any positive integer n.
That the outside shell contains 6n2 cubes can be seen by noting that the shell can be built using an n x 2n slab for the floor, another n x 2n slab for the back wall, and n x (n - 1) and n x (n + 1) slabs for the side walls. Remember that the outside shell has height n, depth n + 1, and length 2n + 1, with one long side open and the top open, and that all inside shells should be constructed with dimensions like those of the outside shell.
(See Katz, Victor, 2009, A History of Mathematics: An Introduction (3rd ed.), Boston: Addison-Wesley, pp. 251-252, or Katz, Victor (editor), 2007, The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook, Princeton University Press, pp. 493-496.)
Exercise 5: Use 60 or 120 cubes to illustrate the identity $1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2} = {{n(n + 1)(n + 2)} \over 6}$
for n = 3 or n = 4, as Nilakantha may have done. Our strategy will be to demonstrate that six copies of the sum $1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2}$ form an n by (n + 1) by (n + 2) rectangular solid, as follows. Start on the outside with an (n + 1) by (n + 2) floor and two adjoining walls of height n, consisting of $6 \cdot \frac{n(n+1)}{2}$ cubes all together, then work from the outside inwards, lining the inside of the existing shell with a floor and two adjoining walls consisting of $6\cdot \frac{(n-1)n}{2}$ cubes, . . . , 6 3 cubes, and finally 6 1 cubes. For n = 3, the 3 x 4 x 5 outside shell consists of $6\cdot \frac{3\cdot4}{2} = 36$ cubes, the 2 x 3 x 4 shell inside of this one $6 \cdot {\frac{2\cdot3}{2}} = 18$ cubes, and the 1 x 2 x 3 inner shell or layer $6\cdot \frac{1\cdot2}{2} = 6$ cubes. This construction illustrates that
$6 (1 + 3 + 6) = {{3\cdot 4\cdot 5}}$
or
$1 + 3 + 6 = {{3\cdot 4\cdot 5} \over 6}$
or, more generally,
$1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2} = {{n(n + 1)(n + 2)} \over 6}$
for any integer n.
That the outside shell contains $6\cdot\frac{n(n+1)}{2}=3n(n+1)$ cubes can be seen by noting that the shell can be built using an n x (n + 1) slab for the floor and two adjoining wall slabs with dimensions n x n and n x (n + 2). This will result in an outside shell with height n, length n + 1, and width n + 2. The inside shells should be constructed with dimensions like those of the outside shell.
(See Joseph, George Gheverghese, 2000, The Crest of the Peacock, Princeton University Press, pp. 295-296, and the references given for Exercise 4.)
Exercise 6: How might early mathematicians have discovered the identity from Exercise 1? Provide at least four examples and write the identity in terms of n, where n is a positive integer. |
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# A cylindrical tank has a capacity of 2156 m3 and diameter of the base is 14 m. If $\pi = \dfrac{{22}}{7}$, find the depth of the tank.
Last updated date: 09th Aug 2024
Total views: 69.3k
Views today: 2.69k
Answer
Verified
69.3k+ views
Hint: In this problem, the volume of the cylindrical tank is given so we have to compare that volume with the volume formula of cylinder which can be expressed as $\pi {r^2}h$, here r is the radius of the cylinder and h is the height of the cylinder. Since the diameter of the cylinder is given, we can find the radius. After that by substituting the value of r and the $\pi$ we get the required height of the cylinder.
Complete step by step answer:
It is given that the diameter of the cylindrical tank is 14 m.
Therefore, the radius is equal to 7 m.
Capacity of the tank is determined by its volume.
Volume of the cylindrical tank $= 2156\;{{\rm{m}}^3}$ (1)
It is known that the formula of the volume of a cylinder $= \pi {r^2}h$ (2)
Comparing both (1) and (2) we get
$\pi {r^2}h = 2156$
Substituting the value of 7 for$r$ and $\dfrac{{22}}{7}$for$\pi$.
$\begin{array}{l}\pi {r^2}h = 2156\\ \Rightarrow \dfrac{{22}}{7} \times {7^2} \times h = 2156\\ \Rightarrow \dfrac{{22}}{7} \times {7^2} \times h = 2156\\ \Rightarrow 154h = 2156\\ \Rightarrow h = \dfrac{{2156}}{{154}}\\ \Rightarrow h = 14\;{\rm{m}}\end{array}$
Hence, the required depth (height) of the cylinder is 14 m.
Note: Here we have to determine the required depth of the cylindrical tank for the given volume. Since the radius of the cylindrical tank is known, from that we can calculate the volume of the cylinder. The capacity of the cylinder is given. Thus, by comparing the calculated volume with the given volume we can easily calculate the required depth of the cylinder. |
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### Course: Алгебра 1>Unit 13
Lesson 4: Квадраттық теңдеулерді көбейткішке жіктеуге кіріспе
Learn how to factor quadratic expressions as the product of two linear binomials. For example, x²+5x+6=(x+2)(x+3).
#### What you need to know for this lesson
Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication. For more on this, check out our previous article on taking common factors.
#### Бұл сабақта сен нені үйренесің
Бұл сабақта ${x}^{2}+bx+c$ түріндегі көпмүшені екі көпмүшенің көбейтіндісіне түрінде жіктеуді үйренесің.
## Қайталау: көпмүшелерді көбейту
$\left(x+2\right)\left(x+4\right)$ өрнегін қарастырайық.
We can find the product by applying the distributive property multiple times.
Сонымен $\left(x+2\right)\left(x+4\right)={x}^{2}+6x+8$.
From this, we see that $x+2$ and $x+4$ are factors of ${x}^{2}+6x+8$, but how would we find these factors if we didn't start with them?
## Квадрат үшмүшені көбейткіштерге жіктеу
We can reverse the process of binomial multiplication shown above in order to factor a trinomial (which is a polynomial with $3$ terms).
In other words, if we start with the polynomial ${x}^{2}+6x+8$, we can use factoring to write it as a product of two binomials, $\left(x+2\right)\left(x+4\right)$.
Let's take a look at a few examples to see how this is done.
### Example 1: Factoring ${x}^{2}+5x+6$
To factor ${x}^{2}+5x+6$, we first need to find two numbers that multiply to $6$ (the constant number) and add up to $5$ (the $x$-coefficient).
These two numbers are $2$ and $3$ since $2\cdot 3=6$ and $2+3=5$.
We can then add each of these numbers to $x$ to form the two binomial factors: $\left(x+2\right)$ and $\left(x+3\right)$.
In conclusion, we factored the trinomial as follows:
${x}^{2}+5x+6=\left(x+2\right)\left(x+3\right)$
To check the factorization, we can multiply the two binomials:
$\begin{array}{rl}\left(x+2\right)\left(x+3\right)& =\left(x+2\right)\left(x\right)+\left(x+2\right)\left(3\right)\\ \\ & ={x}^{2}+2x+3x+6\\ \\ & ={x}^{2}+5x+6\end{array}$
The product of $x+2$ and $x+3$ is indeed ${x}^{2}+5x+6$. Our factorization is correct!
### Түсінгеніңізді тексеріңіз
1) Factor ${x}^{2}+7x+10$.
Дұрыс жауапты таңдаңыз:
2) Factor ${x}^{2}+9x+20$.
Let's take a look at a few more examples and see what we can learn from them.
### 2-мысал: ${x}^{2}-5x+6$ өрнегін көбейткіштерге жіктеу
To factor ${x}^{2}-5x+6$, let's first find two numbers that multiply to $6$ and add up to $-5$.
These two numbers are $-2$ and $-3$ since $\left(-2\right)\cdot \left(-3\right)=6$ and $\left(-2\right)+\left(-3\right)=-5$.
We can then add each of these numbers to $x$ to form the two binomial factors: $\left(x+\left(-2\right)\right)$ and $\left(x+\left(-3\right)\right)$.
The factorization is given below:
$\begin{array}{rl}{x}^{2}-5x+6& =\left(x+\left(-2\right)\right)\left(x+\left(-3\right)\right)\\ \\ & =\left(x-2\right)\left(x-3\right)\end{array}$
Factoring pattern: Notice that the numbers needed to factor ${x}^{2}-5x+6$ are both negative $\left(-2$ and $-3\right)$. This is because their product needs to be positive $\left(6\right)$ and their sum negative $\left(-5\right)$.
In general, when factoring ${x}^{2}+bx+c$, if $c$ is positive and $b$ is negative, then both factors will be negative!
### 3-мысал: ${x}^{2}-x-6$ өрнегін көбейткіштерге жіктеу
We can write ${x}^{2}-x-6$ as ${x}^{2}-1x-6$.
To factor ${x}^{2}-1x-6$, let's first find two numbers that multiply to $-6$ and add up to $-1$.
These two numbers are $2$ and $-3$ since $\left(2\right)\cdot \left(-3\right)=-6$ and $2+\left(-3\right)=-1$.
We can then add each of these numbers to $x$ to form the two binomial factors: $\left(x+2\right)$ and $\left(x+\left(-3\right)\right)$.
The factorization is given below:
$\begin{array}{rl}{x}^{2}-x-6& =\left(x+2\right)\left(x+\left(-3\right)\right)\\ \\ & =\left(x+2\right)\left(x-3\right)\end{array}$
Factoring patterns: Notice that to factor ${x}^{2}-x-6$, we need one positive number $\left(2\right)$ and one negative number $\left(-3\right)$. This is because their product needs to be negative $\left(-6\right)$.
In general, when factoring ${x}^{2}+bx+c$, if $c$ is negative, then one factor will be positive and one factor will be negative.
## Қорытынды
In general, to factor a trinomial of the form ${x}^{2}+bx+c$, we need to find factors of $c$ that add up to $b$.
Suppose these two numbers are $m$ and $n$ so that $c=mn$ and $b=m+n$, then ${x}^{2}+bx+c=\left(x+m\right)\left(x+n\right)$.
### Түсінгеніңізді тексеріңіз
3) Көбейткіштерге жіктеңіз: ${x}^{2}-8x-9$.
4) ${x}^{2}-10x+24$ өрнегін көбейткіштерге жіктеңіз.
5) Көбейткіштерге жіктеңіз: ${x}^{2}+7x-30$.
## Why does this work?
To understand why this factorization method works, let's return to the original example in which we factored ${x}^{2}+5x+6$ as $\left(x+2\right)\left(x+3\right)$.
If we go back and multiply the two binomial factors, we can see the effect that the $2$ and the $3$ have on forming the product ${x}^{2}+5x+6$.
$\begin{array}{rl}\left(x+2\right)\left(x+3\right)& =\left(x+2\right)\left(x\right)+\left(x+2\right)\left(3\right)\\ \\ & ={x}^{2}+2x+3x+2\cdot 3\\ \\ & ={x}^{2}+\left(2+3\right)x+2\cdot 3\end{array}$
We see that the coefficient of the $x$-term is the sum of $2$ and $3$, and the constant term is the product of $2$ and $3$.
## The sum-product pattern
$\left(x+m\right)\left(x+n\right)$ үшін $\left(x+2\right)\left(x+3\right)$- мен не жасағанымызды есімізге түсірейік:
$\begin{array}{rl}\left(x+m\right)\left(x+n\right)& =\left(x+m\right)\left(x\right)+\left(x+m\right)\left(n\right)\\ \\ & ={x}^{2}+mx+nx+m\cdot n\\ \\ & ={x}^{2}+\left(m+n\right)x+m\cdot n\end{array}$
To summarize this process, we get the following equation:
$\left(x+m\right)\left(x+n\right)={x}^{2}+\left(m+n\right)x+m\cdot n$
This is called the sum-product pattern.
It shows why, once we express a trinomial ${x}^{2}+bx+c$ as ${x}^{2}+\left(m+n\right)x+m\cdot n$ (by finding two numbers $m$ and $n$ so $b=m+n$ and $c=m\cdot n$), we can factor that trinomial as $\left(x+m\right)\left(x+n\right)$.
### Reflection question
6) Can this factorization method be used to factor $2{x}^{2}+3x+1$?
Дұрыс жауапты таңдаңыз:
## When can we use this method to factor?
In general, the sum-product method is only applicable when we can actually write a trinomial as $\left(x+m\right)\left(x+n\right)$ for some integers $m$ and $n$.
This means that the leading term of the trinomial must be ${x}^{2}$ (and not, for instance, $2{x}^{2}$) in order to even consider this method. This is because the product of $\left(x+m\right)$ and $\left(x+n\right)$ will always be a polynomial with a leading term of ${x}^{2}$.
However, not all trinomials with ${x}^{2}$ as a leading term can be factored. For example, ${x}^{2}+2x+2$ cannot be factored because there are no two integers whose sum is $2$ and whose product is $2$.
In future lessons we will learn more ways of factoring more types of polynomials.
## Күрделі есептер
7*) Көбейткіштерге жіктеңіз: ${x}^{2}+5xy+6{y}^{2}$.
8*) Көбейткіштерге жіктеңіз: ${x}^{4}-5{x}^{2}+6$ |
## Exercise
Given the quadratic polynomial in standard form $$2x^2+x-1.$$
(i) Find the vertex of the quadratic polynomial.
(ii) Find the vertex form of this quadratic polynomial.
(iii) Use the vertex form to graph this quadratic polynomial, using five points.
## Solution
### (i) Vertex
#### Step 1: Write coefficients
The coefficients are
$a=2, \quad b=1,\quad c=-1$
#### Step 2: Calculate discriminant
The discriminant is
\begin{align*} D&=b^2-4ac\\ &=(1)^2-4(2)(-1)\\ &= 1+8\\ &=9 \end{align*}
#### Step 3: Calculate x-coordinate of the vertex
The x-coordinate of the vertex is
\begin{align*} h &= -\frac{b}{2a}\\ &= -\frac{(1)}{2(2)}\\ &= -\frac{1}{4} \end{align*}
#### Step 4: Calculate y-coordinate of the vertex
The y-coordinate of the vertex is
\begin{align*} k &= -\frac{D}{4a}\\ &= - \frac{(9)}{4(2)}\\ &= -\frac{9}{8} \end{align*}
#### Step 5: Conclusion
Thefore the vertex is
$\Big(-\frac{1}{4},-\frac{9}{8}\Big).$
### (ii) Vertex form
$a(x-h)^2+k$ $h = -\frac{1}{4}, \quad k = -\frac{9}{8}$ \begin{align*} a(x-h)^2+k&=2\Big(x-\Big(-\frac{1}{4}\Big)\Big)^2+\Big(-\frac{9}{8}\Big)\\ &=2\Big(x+\frac{1}{4}\Big)^2-\frac{9}{8} \end{align*}
The vertex form is
$2\Big(x+\frac{1}{4}\Big)^2-\frac{9}{8}$
### (iii) Graph of quadratic polynomial
#### Step 1: Draw the vertex
Draw the vertex $$\Big(-\frac{1}{4},-\frac{9}{8}\Big).$$
#### Step 2: Draw point 1 unit left
$$a = 2$$ units up from vertex
#### Step 3: Draw point 1 unit right
$$a =2$$ units up from vertex
#### Step 4: Draw point 1 unit right
$$4a =4(2)=8$$ units up from vertex
#### Step 5: Draw point 1 unit right
$$4a =4(2)=8$$ units up from vertex |
# 7.3 Graphing linear equations in two variables (Page 2/3)
Page 2 / 3
$ax+by=c$
is said to be in general form .
We must stipulate that $a$ and $b$ cannot both equal zero at the same time, for if they were we would have
$0x+0y=c$
$0=c$
This statement is true only if $c=0$ . If $c$ were to be any other number, we would get a false statement.
Now, we have the following:
The graphing of all ordered pairs that solve a linear equation in two variables produces a straight line.
This implies,
The graph of a linear equation in two variables is a straight line.
From these statements we can conclude,
If an ordered pair is a solution to a linear equations in two variables, then it lies on the graph of the equation.
Also,
Any point (ordered pairs) that lies on the graph of a linear equation in two variables is a solution to that equation.
## The intercept method of graphing
When we want to graph a linear equation, it is certainly impractical to graph infinitely many points. Since a straight line is determined by only two points, we need only find two solutions to the equation (although a third point is helpful as a check).
## Intercepts
When a linear equation in two variables is given in general from, $ax+by=c$ , often the two most convenient points (solutions) to fine are called the Intercepts: these are the points at which the line intercepts the coordinate axes. Of course, a horizontal or vertical line intercepts only one axis, so this method does not apply. Horizontal and vertical lines are easily recognized as they contain only one variable. (See Sample Set $\text{C}$ .)
## $y\text{-Intercept}$
The point at which the line crosses the $y\text{-axis}$ is called the $y\text{-intercept}$ . The $x\text{-value}$ at this point is zero (since the point is neither to the left nor right of the origin).
## $x\text{-Intercept}$
The point at which the line crosses the $x\text{-axis}$ is called the $x\text{-intercept}$ and the $y\text{-value}$ at that point is zero. The $y\text{-intercept}$ can be found by substituting the value 0 for $x$ into the equation and solving for $y$ . The $x\text{-intercept}$ can be found by substituting the value 0 for $y$ into the equation and solving for $x$ .
## Intercept method
Since we are graphing an equation by finding the intercepts, we call this method the intercept method
## Sample set a
Graph the following equations using the intercept method.
$y-2x=-3$
To find the $y\text{-intercept}$ , let $x=0$ and $y=b$ .
$\begin{array}{rrr}\hfill b-2\left(0\right)& \hfill =& \hfill -3\\ \hfill b-0& \hfill =& \hfill -3\\ \hfill b& \hfill =& \hfill -3\end{array}$
Thus, we have the point $\left(0,\text{\hspace{0.17em}}-3\right)$ . So, if $x=0$ , $y=b=-3$ .
To find the $x\text{-intercept}$ , let $y=0$ and $x=a$ .
$\begin{array}{rrrr}\hfill 0-2a& \hfill =& \hfill -3& \hfill \\ \hfill -2a& \hfill =& \hfill -3& \hfill \text{Divide by -2}\text{.}\\ \hfill a& \hfill =& \hfill \frac{-3}{-2}& \hfill \\ \hfill a& \hfill =& \hfill \frac{3}{2}& \hfill \end{array}$
Thus, we have the point $\left(\frac{3}{2},0\right)$ . So, if $x=a=\frac{3}{2}$ , $y=0$ .
Construct a coordinate system, plot these two points, and draw a line through them. Keep in mind that every point on this line is a solution to the equation $y-2x=-3$ .
$-2x+3y=3$
To find the $y\text{-intercept}$ , let $x=0$ and $y=b$ .
$\begin{array}{rrr}\hfill -2\left(0\right)+3b& \hfill =& \hfill 3\\ \hfill 0+3b& \hfill =& \hfill 3\\ \hfill 3b& \hfill =& \hfill 3\\ \hfill b& \hfill =& \hfill 1\end{array}$
Thus, we have the point $\left(0,\text{\hspace{0.17em}}1\right)$ . So, if $x=0$ , $y=b=1$ .
To find the $x\text{-intercept}$ , let $y=0$ and $x=a$ .
$\begin{array}{rrr}\hfill -2a+3\left(0\right)& \hfill =& \hfill 3\\ \hfill -2a+0& \hfill =& \hfill 3\\ \hfill -2a& \hfill =& \hfill 3\\ \hfill a& \hfill =& \hfill \frac{3}{-2}\\ \hfill a& \hfill =& \hfill -\frac{3}{2}\end{array}$
Thus, we have the point $\left(-\frac{3}{2},\text{\hspace{0.17em}}0\right)$ . So, if $x=a=-\frac{3}{2}$ , $y=0$ .
Construct a coordinate system, plot these two points, and draw a line through them. Keep in mind that all the solutions to the equation $-2x+3y=3$ are precisely on this line.
$4x+y=5$
To find the $y\text{-intercept}$ , let $x=0$ and $y=b$ .
$\begin{array}{rrr}\hfill 4\left(0\right)+b& \hfill =& \hfill 5\\ \hfill 0+b& \hfill =& \hfill 5\\ \hfill b& \hfill =& \hfill 5\end{array}$
Thus, we have the point $\left(0,\text{\hspace{0.17em}}5\right)$ . So, if $x=0$ , $y=b=5$ .
To find the $x\text{-intercept}$ , let $y=0$ and $x=a$ .
$\begin{array}{rrr}\hfill 4a+0& \hfill =& \hfill 5\\ \hfill 4a& \hfill =& \hfill 5\\ \hfill a& \hfill =& \hfill \frac{5}{4}\end{array}$
Thus, we have the point $\left(\frac{5}{4},\text{\hspace{0.17em}}0\right)$ . So, if $x=a=\frac{5}{4}$ , $y=0$ .
Construct a coordinate system, plot these two points, and draw a line through them.
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Portions of Time
In questions that involve ‘portions of time’, you need to think and analyse the problem carefully.
Given below are some tips to help you solve problems that involve portions of time:
1. Start by thinking about the time conversion.
2. Remember that “of” means multiplication.
3. Ask yourself at the end “Does the answer make sense?” If it doesn’t make sense, you probably did something wrong.
Let's take a look at some examples:
1) How many hours are in ¼ of a day?
There are 24 hours in a day.
So if there are 24 hours in a day and we need to find how many hours are in ¼ of a day, we need to figure out what is ¼ of 24.
Remember that “of” means multiplication.
So, ¼ of 24 is the same as ¼ times 24.
(Remember that there’s an understood 1 underneath the 24 and that to multiply fractions you must multiply the numerators then the denominators. At the end, 24 divided by 4 is 6).
Does it make sense that ¼ of a day is 6 hours? Yes, it does.
2) How many years are in ¾ of a century?
There are 100 years in a century.
So, if there are 100 years in a century and we need to find how many years are in ¾ of a century, we need to figure out what is ¾ of 100.
Remember that “of” means multiplication.
So, ¾ of 100 is the same as ¾ times 100.
Let’s solve this.
(Remember that there’s an understood 1 underneath the 100, and that to multiply fractions you must multiply the numerators then the denominators. At the end, 300 divided by 4 is 75).
Does it make sense that ¾ of a century is 75 years? Yes, it does.
3) Mark practiced playing the piano for 2/5 of an hour. How many minutes make 2/5 of an hour?
There are 60 minutes in an hour.
So if there are 60 minutes in an hour and we need to find out how many minutes are in 2/5 of an hour.
We need to determine what is 2/5 of 60. Remember that “of” means multiplication.
So, 2/5 of 60 is the same as 2/5 times 60.
Let’s solve this:
(Remember that there’s an understood 1 underneath the 60, and that to multiply fractions you must multiply the numerators then the denominators. At the end, 120 divided by 5 is 24).
Does it make sense that 2/5 of an hour is 24 minutes? Yes, it does.
Summary
Time conversions:
1) 60 seconds = 1 minute
2) 60 minutes = 1 hour
3) 24 hours = 1 day
4) 7 days = 1 week
5) 12 months = 1 year
6) 52 weeks = 1 year
7) 365 days = 1 year (366 for leap year)
8) 10 years = 1 decade
9) 100 years = 1 century
10) 1000 years = 1 millennium
Conversion rule:
Large to small: Multiply
Small to large: Divide
Quick tips for solving ‘portions of time’ problems:
1. Start by thinking about the time conversion.
2. Remember that “of” means multiplication.
3. Ask yourself at the end “Does my answer make sense?” If it doesn’t make sense, you probably did something wrong.
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Summary Videos Questions & Answers References
# Prime Factorisation, HCF and LCM
#### Summaryarrow_upward
LearnNext Lesson Video
Prime factorisation:
Writing a number as a product of its prime factors is called the prime factorisation of the number.
e.g. (i) Prime factorisation of 18 = 2 x 3 x 3
(ii) Prime factorisation of 40 = 2 x 2 x 2 x 5
Highest common factor:
The greatest of the common factors of the given numbers is called their highest common factor (HCF). It is also known as the greatest common divisor. The HCF of the given numbers is equal to the product of common factors in their prime factorisation.
e.g.
Factors of 16 = 1, 2, 4, 8, 16
Factors of 40 = 1, 2, 4, 5, 8, 10, 20, 40
Common factors = 1, 2, 4, 8
HCF of 16 and 40 = 8
The HCF of 16 and 40 using their prime factorisation:
Prime factorisation of 16 = 2 x 2 x 2 x 2
Prime factorisation of 40 = 2 x 2 x 2 x 5
Common prime factors = 2 x 2 x 2
HCF of 16 and 40 = 2 x 2 x 2 = 8
Least common multiple:
The smallest common multiple of the given numbers is called their least common multiple (LCM). The product of the prime factors that occur the maximum number of times in the prime factorisation of the given numbers, is their LCM.
e.g.
The LCM of given numbers using their prime factorisation:
Prime factorisation of 4 = 2 x 2
Prime factorisation of 6 = 2 x 3
LCM of 4 and 6 = 2 x 2 x 3 =12
To find the LCM of the given numbers using the division method:
• Write the given numbers in a row.
• Divide the numbers by the smallest prime number that divides one or more of the given numbers.
• Write the number that is not divisible, in the second row.
• Write the new dividends in the second row.
• Divide the new dividends by another smallest prime number.
• Continue dividing till the dividends are all prime numbers or 1.
• Stop the process when all the new dividends are prime numbers or 1.
• The product of all the divisors and the remaining prime dividends is the LCM of the given numbers.
### 1 . Find the smallest 4 digit number.
Sol:
LCM of 18, 24 and 32:
18 = 2 x 3 x 3
24 = 2 x 2 x 2 x 3
32 = 2 x 2 x 2 x 2 x 2
LCM = 2 x 2 x 2 x 2 x
...
462
### 3 . Write the smallest 4 digit number and express it in terms of prime factors.
the smallest four digit niumber 1000.
1000= 2 3X5 3
### 4 . Largest number that divides 258 and 584 leaving remainder 6 and 8
Given that when 258 is divided by the number the remainder is 6
Similarly, when 584 is divided by the same number the remainder is 8...
### 5 . What is divisibility of 7
If you double the rightmost digit in a number and subtract it from the rest of the number, if the result is a multiple of 7 or if it is 0 th...
X |
# Chapter 8 – Introduction To Trigonometry
The following Topics and Sub-Topics are covered in this chapter and are available on MSVgo:
Introduction
You might have come across trigonometry questions during your maths class. But do you know trigonometry is also used in the measurement of height and distance? Trigonometry is considered one of the most significant and crucial branches of mathematics. Hipparchus, a Greek mathematician, gave the concept of trigonometry. As you know, trigonometry aids in findings the angles and missing sides of a triangle through trigonometric ratios, which we will discuss in the article. Here you will comprehend trigonometric ratios and their types and trigonometric identities. We will try to explain trigonometry in easy language and a simple manner. The topics are easier once you understand them. We hope this article helps you understand the concepts easily.
#### Trigonometric Ratios
The ratio of sides of a right-angled triangle concerning any of its acute angles is known as the trigonometric ratios of that particular angle.
There are mainly six types of trigonometric ratios. The following are the six types of trigonometric ratios:-
Sin θ Opposite side to θ/Hypotenuse Cos θ Adjacent side to θ/Hypotenuse Tan θ Opposite side/Adjacent side & Sin θ/ cos θ Cot θ Adjacent side/Opposite side & 1/tan θ Sec θ Hypotenuse/ Adjacent side & 1/cos θ Cosec θ Hypotenuse/Opposite side & 1/sin θ
#### Trigonometric Ratios of Some Specific Angles
Range of trigonometric ratios from 0 to 90 degrees
For 0°≤θ≤90°,
• 0≤sinθ≤1
• 0≤cosθ≤1
• 0≤tanθ<∞
• 1≤secθ<∞
• 0≤cotθ<∞
• 1≤cosecθ<∞
tanθ and secθ are not defined at 90°.
cotθ and cosecθ are not defined at 0°.
Variation of trigonometric ratio from 0 to 90 degrees
As θ increases from 0° to 90°
• sin θ increases from 0 to 1
• cos θ decreases from 1 to 0
• tan θ increases from 0 to ∞
• cosec θ decreases from ∞ 1
• sec θ increases from 1 to ∞
• cot θ decreases from ∞ 0
Standard values of Trigonometric Ratios
∠A 0° 30° 45° 60° 90° sin A 0 1/2 1/√2 √3/2 1 cos A 1 √3/2 1/√2 1/2 0 tan A 0 1/√3 1 √3 Not defined cosec A Not defined 2 √2 2/√3 1 sec A 1 2/√3 √2 2 Not defined cot A Not defined √3 1 1/√3 0
#### Trigonometric Ratios Of Complementary Angles
Complementary angles are those sets of two angles whose sum is equal to 90 o. Let us understand it through an example. Let say there are two angles ∠A and ∠B. They will be called complementary if,
∠A + ∠B = 90 o
Here, ∠A is complementary to ∠B and vice versa.
The lengths of sides of a right-angle triangle’s relationship with the acute angle are shown by trigonometric ratios. The following relation is established for trigonometric ratios of complementary angles:-
• sin (90°− θ) = cos θ
• cos (90°− θ) = sin θ
• tan (90°− θ) = cot θ
• cot (90°− θ) = tan θ
• cosec (90°− θ) = sec θ
• sec (90°− θ) – cosec θ
So it is concluded that:-
• Sin of an angle = Cos of its complementary angle
• Cos of an angle = Sin of its complementary angle
• Tan of an angle = Cot of its complementary angle
#### Trigonometric Identities
Trigonometric identities are helpful and useful when trigonometric functions are included in an expression or an equation. However, these identities have certain functions of one or more angles.
Trigonometric Identities List
Reciprocal Identities
• Sin θ = 1/Csc θ or Csc θ = 1/Sin θ
• Cos θ = 1/Sec θ or Sec θ = 1/Cos θ
• Tan θ = 1/Cot θ or Cot θ = 1/Tan θ
Pythagorean Identities
• sin2 a + cos2 a = 1
• 1+tan2 a = sec2 a
• Cosec2 a = 1 + cot2 a
Ratio Identities
• Tan θ = Sin θ/Cos θ
• Cot θ = Cos θ/Sin θ
Opposite Angle Identities
• Sin (-θ) = – Sin θ
• Cos (-θ) = Cos θ
• Tan (-θ) = – Tan θ
• Cot (-θ) = – Cot θ
• Sec (-θ) = Sec θ
• Csc (-θ) = – Csc θ
#### FAQs
What are the basics of trigonometry?
Answer. There are three basic functions in trigonometry that are sine, cosine, and tangent. With these functions’ help, the other three functions are derived: cotangent, secant, and cosecant. Since all the trigonometrical concepts revolve around these functions, it is important to understand these functions first.
What do you mean by trigonometry?
Answer. Trigonometry deals with the relationship between the sides of a triangle with its angle. To know more about it, you can visit MSVgo official site or download the MSVgo app.
What is the importance of trigonometry?
Answer. The importance of trigonometry are as follows:
• It helps solve complex design problems.
• It is used in physics to measure dynamics, kinematics, and calculating vectors.
• It helps manipulate equations.
Who is the father of trigonometry?
Answer. The concept of trigonometry was founded by Hipparchus, a Greek astrologer, geographer, and mathematician.
What is trigonometry with an example?
Answer. There are various trigonometry examples, such as it is used to find the distance of long rivers and measure the height of buildings or mountains.
There are various applications of trigonometry, such as in oceanography, seismology, meteorology. It is also helpful in measuring the height of the mountain as well. Therefore the proper understanding of trigonometry is paramount. We help you understand trigonometry basics first, with in-depth concept notes and explanatory video on the MSVgo app. The MSVgo philosophy is to enable a core understanding of any concept. MSVgo app has a video library that explains concepts with examples or explanatory visualizations or animation. To learn more about it, check out the MSVgo app and their official site. Stay tuned with the MSVgo app and cheerful learning!
### High School Physics
• Alternating Current
• Atoms
• Communication Systems
• Current Electricity
• Dual nature of Radiation and Matter
• Electric Charges and Fields
• Electricity
• Electromagnetic Induction
• Electromagnetic Waves
• Electrons and Photons
• Electrostatic Potential and Capacitance
• Fluid Pressure
• Force and Acceleration
• Force And Laws Of Motion
• Gravitation
• Internal Energy
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• Law of motion
• Light – Reflection And Refraction
• Magnetic Effects Of Electric Current
• Magnetism and Matter
• Management Of Natural Resources
• Mechanical properties of Fluids
• Mechanical properties of Solids
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• Motion in a plane
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• Moving Charges and Magnetism
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• Our Environment
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• Semiconductor Electronics: Materials, Devices and Simple Circuits
• Simple Machines
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• Thermal properties of matter
• Thermodynamics
• Units and Measurement
• Vectors, Scalar Quantities and Elementary Calculus
• Wave Optics
• Waves
• Work, Power and Energy
### High School Chemistry
• Acids, Bases and Salts
• Alcohols, Phenols and Ethers
• Aldehydes, Ketones and Carboxylic Acids
• Aliphatic and Aromatic Hydrocarbons
• Alkyl and Aryl Halides
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• Principles and Processes of Isolation of Elements
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• Study of Compounds
• Study of Gas Laws
• Study of Representative Elements
• Surface Chemistry
• The d-block and f-block elements
• The Gaseous State
• The p-Block Elements
• The Periodic Table
• The s-Block Elements
• The Solid State
• Thermodynamics
### High School Biology
• Absorption and Movement of Water in Plants
• Anatomy of Flowering Plants
• Animal Kingdom
• Bacteria and Fungi-Friends and Foe
• Biodiversity and Conservation
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• Biological Classification
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• Biotic Community
• Body Fluids and Circulation
• Breathing and Exchange of Gases
• Cell – Unit of Life
• Cell Cycle and Cell Division
• Cell Division and Structure of Chromosomes
• Cell Reproduction
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• Chemical Coordination and Integration
• Circulation
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• Crop Improvement
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• Diversity In Living Organisms
• Ecosystem
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• Flowering Plants
• Genes and Chromosomes
• Health and Diseases
• Health and Its Significance
• Heredity And Evolution
• Heredity and Variation
• How Do Organisms Reproduce?
• Human Diseases
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• Human Health and Disease
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• Integumentary System- Skin
• Kingdom Fungi
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• Life Processes
• Locomotion and Movement
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• Photosynthesis
• Photosynthesis in Higher Plants
• Plant Growth and Development
• Plant Kingdom
• Pollination and Fertilization
• Pollution; Sources and its effects
• Principles of Inheritance and Variation
• Reproduction and Development in Angiosperms
• Reproduction in Organisms
• Reproductive Health
• Respiration in Human Beings
• Respiration in Plants
• Respiratory System
• Sexual Reproduction in Flowering Plants
• Strategies for Enhancement in Food Production
• Structural Organisation in Animals
• Structural Organisation of the Cell
• The Endocrine System
• The Fundamental Unit Of Life
• The Living World
• The Nervous System and Sense Organs
• Tissues
• Transpiration
• Transport in Plants
### High School Math
• Algebra – Arithmatic Progressions
• Algebra – Complex Numbers and Quadratic Equations
• Algebra – Linear Inequalities
• Algebra – Pair of Linear Equations in Two Variables
• Algebra – Polynomials
• Algebra – Principle of Mathematical Induction
• Binomial Theorem
• Calculus – Applications of Derivatives
• Calculus – Applications of the Integrals
• Calculus – Continuity and Differentiability
• Calculus – Differential Equations
• Calculus – Integrals
• Geometry – Area
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• Geometry – Introduction to Euclid’s Geometry
• Geometry – Three-dimensional Geometry
• Geometry – Lines and Angles
• Geometry – Straight Lines
• Geometry – Triangles
• Linear Programming
• Matrices and Determinants
• Mensuration – Areas
• Mensuration – Surface Areas and Volumes
• Number Systems
• Number Systems – Real Numbers
• Permutations and Combinations
• Probability
• Sequence and Series
• Sets and Functions
• Statistics
• Trignometry – Height and Distance
• Trignometry – Identities
• Trignometry – Introduction
### Middle School Science
• Acids, Bases And Salts
• Air and Its Constituents
• Basic Biology
• Body Movements
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### Middle School Math
• Area and Its Boundary
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# The Laplace expansion, minors, cofactors and adjoints
The Laplace expansion is a formula that allows us to express the determinant of a matrix as a linear combination of determinants of smaller matrices, called minors.
The Laplace expansion also allows us to write the inverse of a matrix in terms of its signed minors, called cofactors. The latter are usually collected in a matrix called adjoint matrix.
## Minors
Let us start by defining minors.
Definition Let be a matrix (with ). Denote by the entry of at the intersection of the -th row and -th column. The minor of is the determinant of the sub-matrix obtained from by deleting its -th row and its -th column.
We now illustrate the definition with an example.
Example Define the matrix Take the entry . The sub-matrix obtained by deleting the first row and the first column isThus, the minor of is The minor of is
## Cofactors
A cofactor is a minor whose sign may have been changed depending on the location of the respective matrix entry.
Definition Let be a matrix (with ). Denote by the minor of an entry . The cofactor of is
As an example, the pattern of sign changes of a matrix is
Example Consider the matrix Take the entry . The minor of is and its cofactor is
## The expansion
We are now ready to present the Laplace expansion.
Proposition Let be a matrix (with ). Denote by the cofactor of an entry . Then, for any row , the following row expansion holds:Similarly, for any column , the following column expansion holds:
Proof
Let us start by proving the row expansionDenote by the -th row of . We can writewhere is the -th vector of the standard basis of , that is a vector such that its -th entry is equal to 1 and all the other entries are equal to 0. Now, denote by the matrix obtained from by substituting its -th row with :We can write the -th row of as a linear combination as follows:Since the determinant is linear in each row, we have thatNow, the matrix can be transformed into the matrixby performing row interchanges and column interchanges. As a consequence, by the properties of the determinant of elementary matrices, we have thatBy the definition of determinant, we have where: in step we have used the fact that transposition does not change the determinant; in step we have used the fact that the only non-zero entry of the first column of is the first one, so that for all and for ; in step , is the minor of , and, by looking at the structure of above, it is clear that, after excluding the first row and the first column of from the computation of its determinant, we are computing the determinant of a matrix obtained from by deleting its -th row and its -th column. Thus, where is the cofactor of . The proof for column expansions is analogous.
In other words, the determinant can be computed by summing all the entries of an arbitrarily chosen row (column) multiplied by their respective cofactors.
Example Define the matrixWe can use the Laplace expansion along the first column to compute its determinant:
Example Define the matrixWe can use the Laplace expansion along the third row to compute its determinant:
## Expansions along the wrong row or column
An interesting and useful fact is that while the Laplace expansion giveswe havewhen . In other words, if we multiply the elements of row with the cofactors of a different row and we add them up, we get zero as a result.
Proof
Define a matrix whose rows are all equal to the corresponding rows of , except for the -th, which is equal to the -th row of . Thus, has two identical rows and, as a consequence, it is singular and it has zero determinant. Denote by the cofactor of . Then, where: in step we have used the fact that the -th row of is equal to the -th row of ; in step we have used the fact that, although the -th row of is different from the -th row of , we have that because row is canceled when forming the sub-matrices used to compute these cofactors.
The same result holds for columns:when . The proof is analogous to the previous one.
## Cofactor matrix
We now define the cofactor matrix (or matrix of cofactors).
Definition Let be a matrix. Denote by the cofactor of (defined above). Then, the matrix such that its -th entry is equal to for every and is called cofactor matrix of .
The adjoint matrix (or adjugate matrix) is the transpose of the matrix of cofactors.
Definition Let be a matrix and its cofactor matrix. The adjoint matrix of , denoted by , is
The following proposition is a direct consequence of the Laplace expansion.
Proposition Let be a matrix and its adjoint. Then,where is the identity matrix.
Proof
DefineBy the definition of matrix multiplication, the -th entry of iswhere in step we have used the fact that the adjoint is the transpose of the cofactor matrix. When , the expression in step is the Laplace expansion of and it is therefore equal to . When it is an expansion along the wrong row, and it is therefore equal to . Thus,When we have which is a column expansion. Thus, by the same arguments used previously, we have that
A consequence of the previous proposition is the following.
Proposition Let be a invertible matrix and its adjoint. Then
Proof
Since is invertible, . Then, we can rewrite the resultasThus, by the definition of inverse matrix, the matrixis the inverse of .
## Solved exercises
Below you can find some exercises with explained solutions.
### Exercise 1
Define the matrix
Compute the determinant of by using Laplace expansion along its third column.
Solution
The expansion is
### Exercise 2
DefineCompute the adjoint of , use it to derive the inverse of , and verify that the matrix thus obtained is indeed the inverse of .
Solution
The determinant of isThus, is invertible. Note that the sub-matrices obtained by deleting one row and one column of are . Therefore, the matrix of minors of isand the matrix of cofactors isThe adjoint is obtained by transposing the matrix of cofactors:The inverse can be computed asLet us multiply it by in order to check that it is indeed its inverse: |
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# Dimensions of Area
We can obtain the dimension of any physical quantity from the fundamental units. By fundamental units, we mean the units of mass, length, and time. Here, our focus is on finding the dimension of the area.
In dimensional analysis, for finding out the dimension of any physical quantity, whether it is a force, acceleration, or many more; we represent mass by [M], length by [L], and time by [T].
We know that area of any geometric shape, let’s say, the square is the square of its side. The side is the length, so the dimension for length is [L], and the square of this length will be [L2].
So, the dimensional formula of the area of the square is [M0L2T0].
### Two Methods to Write a Dimensional Formula
There are two tricks to write the dimensional formula. They are:
1. Using formula
2. Using units
Let’s understand by an example:
We know that the surface area of the sphere is 4 * π * r2 . . .(1) and the unit of r is meter.
So, r*r becomes m2.
We used the formula and got the unit m2.
Now, we know that m (meter) is the unit of length and the dimension of length is [L], and
The dimensions of constant values, 4 and π are $[M^{0}L^{0}T^{0}]$.
We know that when two similar quantities with different powers are multiplied, then the two powers are added. Let’s see an example of the same:
$a^{b} \times a^{c} = a^{b + c}$
Let’s apply this concept in equation (1):$[M^{0}L^{0}T^{0}] \times [M^{0}L^{0}T^{0}] \times [M^{0}L^{2}T^{0}]$
Now adding the powers on common terms:
$[M^{0 + 0 + 0} L^{0 + 0 + 2} T^{0 + 0 + 0}] = [M^{0} L^{2} T^{0}]$ is the dimensional formula for the surface area of the sphere.
Let’s do the derivation of dimensions of area for different geometric shapes:
### Dimensional Formula of the Area for Different Shapes
Let’s find the dimensional formula of area for different geometrical shapes:
1. Rectangle
We know that area of the rectangle = L x B
⇒ The dimension of L and B = [L]
So, the dimensional formula for the area of the rectangle = $[M^{0} L^{2} T^{0}]$
Total surface area of the rectangle = 2 * (l*b + b*h + h*l)
The dimension of l = b = h = [L]
So, $2 \times ([L^{2}] + [L^{2}] + [L^{2}]) = 6 \times [L^{2}]$
The dimension of 6 (a constant value) = $[M^{0} L^{0} T^{0}]$
= $6 \times [L^{2}] = [M^{0} L^{0} T^{0}] \times [M^{0}L^{0} T^{0}]$ . . .(2)
Now, adding the powers of the term, as shown below:
= $[M^{0 + 0}L^{0 + 2} T^{0 + 0}]$
We get the dimensional formula for the total surface area of the rectangle is:
$[M^{0} L^{2} T^{0}]$
1. Circle
We know that area of the circle is πr2
Here, the dimension of radius r is [L]
The dimension formula of $r^{2} = [M^{0} L^{2} T^{0}]$
π is constant, so its dimension is $[M^{0} L^{0} T^{0}]$
Now, using the formula = $\pi r^{2} = [M^{0} L^{0} T^{0}] \times [M^{0} L^{0} T^{0}]$
We know that in multiplication, we add the powers in similar terms.
$[M^{0 + 0} L^{0 + 2} T^{0 + 0}]$
We get the dimension formula for the area of the circle as $[M^{0}L^{2}T^{0}]$.
1. Sector
The area of a sector is $\frac{\theta}{360^{\circ}} \pi r^{2}$ ...(3)
The unit of θ is radian, whose dimension is $[M^{0}L^{0}T^{0}]$ ….(4)
Similarly, for the constant values, i.e., 360° and π, the dimension is:
$[M^{0} L^{0} T^{0}]$….(5)
The dimension of $r^{2} = [M^{0} L^{2} T^{0}]$...(6)
On putting the value of (4), (5), and (6) in equation (3), we get
= $\frac{[M^{0}L^{0}T^{0}]}{[M^{0}L^{0}T^{0}]} [M^{0}L^{0}T^{0}] \times [M^{0} L^{2} T^{0}]$ …(7)
We know that the powers in the common terms in the division form are subtracted, let’s use the formula mentioned below:
= Using the formula $\frac{a^{b}}{a^{c}} = a^{b - c}$ and $a^{b} \times a^{c} = a^{(b + c)}$ we get:
= $[M^{0 - 0}L^{0 - 0} T^{0 - 0}] \times [M^{0 + 0} L^{2 + 0} T^{0 + 0}] = 1 \times [M^{0} L^{2} T^{0}]$
So, the dimensional formula for the area of a sector is $[M^{0} L^{2} T^{0}]$.
1. Ellipse
The area of the ellipse is π *a * b.
Here, a and b are the lengths of the major and the minor axis, respectively. The dimensional formula for each of these is [L].
So, for a * b, the dimensional formula will be $[M^{0} L^{2} T^{0}]$.
We know for π, the formula is $[M^{0}L^{0}T^{0}]$.
π *a * b = $[M^{0}L^{0}T^{0}] \times [M^{0}L^{2}T^{0}]$
So, the dimensional formula for the area of an ellipse = $[M^{0}L^{2}T^{0}]$.
Last updated date: 24th May 2023
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## FAQs on Dimensions of Area
Q1: Which Physical Quantity has a Dimension, but no Unit?
Ans: There is not any physical quantity that has a dimension, but no unit. It’s because the dimensional formula is based on the fundamental units. If there is no unit, then there is no dimension of the physical quantity.
Q2: Write the Seven Basic Fundamental Units.
Ans: Seven fundamental units are as follows:
1. Mass
2. Length
3. Time
4. Temperature
5. Electric current
6. Amount of light
7. Amount of substance (mole).
Q3: What does Dimensional Formula Mean? Why do we Find Dimensional Formulas?
Ans: Dimensional formula is an expression for the unit of a physical quantity in terms of the fundamental quantities.
The fundamental quantities are mass, length, and time. The dimensional formula of these quantities is M, L, and T, respectively.
The reason why we find the dimensional formulas are:
1. To check the accuracy of physical relations.
2. Conversion of the value of a physical quantity from one system of the unit to another system.
3. To derive a relationship between various physical quantities.
Q4: Write Some Examples of Dimensional Constants.
Ans: The quantities whose values are constant and they possess dimensions.
For example, the velocity of light in vacuum, permeability of free space, gravitational constant, universal gas constant, Boltzmann’s constant, Planck’s constant, etc. |
Notes On Addition and Subtraction of Decimals - CBSE Class 6 Maths
To add or subtract decimal numbers, make sure that the decimal points of the given numbers are placed exactly one below another. While adding or subtracting two decimal numbers, the number of digits after the decimal point should be equal. In case they are not equal, the gaps must be filled with zeros after the last digit. e.g. To add 6.82 and 5 First insert zeros in the empty places after the decimal point so that both the numbers have the same number of digits after the decimal point. Next, write the numbers such that their decimal points are one below another. 6.82 + 5 = 11.82 6.82 + 5.00 11.82 To subtract 5 from 6.82 First insert zeros in the empty places after the decimal point so that both the numbers have the same number of digits after the decimal point. Next, write the numbers such that their decimal points are one below another. e.g. 6.82 - 5 = 1.82 6.82 – 5.00 1.82 Addition or subtraction should be carried out from the extreme right side. Place the decimal point correctly after performing the addition or subtraction.
Summary
To add or subtract decimal numbers, make sure that the decimal points of the given numbers are placed exactly one below another. While adding or subtracting two decimal numbers, the number of digits after the decimal point should be equal. In case they are not equal, the gaps must be filled with zeros after the last digit. e.g. To add 6.82 and 5 First insert zeros in the empty places after the decimal point so that both the numbers have the same number of digits after the decimal point. Next, write the numbers such that their decimal points are one below another. 6.82 + 5 = 11.82 6.82 + 5.00 11.82 To subtract 5 from 6.82 First insert zeros in the empty places after the decimal point so that both the numbers have the same number of digits after the decimal point. Next, write the numbers such that their decimal points are one below another. e.g. 6.82 - 5 = 1.82 6.82 – 5.00 1.82 Addition or subtraction should be carried out from the extreme right side. Place the decimal point correctly after performing the addition or subtraction.
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# CUMULATIVE FREQUENCY ANALYSIS
In a data set, the cumulative frequency for a value ‘x’ is the total number of scores that are equal to or less than ‘x’.
For example if we want to know the cumulative frequency regarding the scores of students in a class and assignment. The cumulative frequency suggests that thirty students received a test score of at most sixty, sixty students received a test score of at most sixty and hundred students received a test score of at most seventy.
CUMULATIVE FREQUENCY TABLE
Example
A set of data given below shows the ages of participants in camp or in a summer camp. Draw a cumulative frequency table for the data.
Age (years) Frequency Ten Seven Eleven Twelve Twelve Twenty seven Thirteen Ten Fourteen Twenty
Solution – The cumulative frequency of the given data can be found by adding the current frequency with the previous frequency.
The cumulative frequency of the first data will be same as the frequency because there is no cumulative frequency before it.
Age (years) Frequency Cumulative Frequency Ten Seven Seven Eleven Twelve Seven + twelve = Nineteen Twelve Twenty seven Nineteen + twenty seven = Forty six Thirteen Ten Forty six + ten = fifty six Fourteen Twenty Fifty six + twenty = seventy six
CALCULATION OF QUARTILES FROM CUMULATIVE FREQUENCY
Quartile – The word quartile has been taken from the word quarter which means one fourth of something. Thus we can say that quartile means a certain fourth of a data set. When a certain set of data is arranged from the lowest to the highest, then it can be divided into four parts or quartiles.
First quartile – This refers to the data set arranged from the lowest to the highest in one fourth parts i.e. (1 /4) parts.
Second quartile – This refers to the data set arranged from the lowest to the highest in the second fourth parts i.e. (2 / 4) parts.
Third quartile – This refers to the data set arranged from the lowest to the highest in the third fourth parts i.e. (3 /4) parts.
Calculating quartiles from cumulative frequency
Example – Calculate the first, second and third quartiles of the data set using the cumulative frequency curve.
Age (years) Frequency ten six eleven ten twelve eighteen thirteen twenty fourteen nine fifteen five
Solution -
Age (years) Frequency Cumulative frequency ten six six eleven ten six + ten = sixteen twelve eighteen sixteen + eighteen = thirty four thirteen twenty thirty four + twenty = fifty four fourteen nine fifty four + nine = sixty three fifteen five sixty three + five = sixty eight sixteen two sixty eight + two = seventy
Calculating quartiles –
The cumulative frequency for the last element in the data set is seventy.
Quartile one position = (seventy + one) / four = 17.75
Quartile second position = 2(seventy + one) / four = 35.5
Quartile third position = 3(seventy + one) / four = 53.25
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# XXXIII Roman Numerals
Contents
Unlock the ancient secrets of the Roman Empire with the knowledge of converting XXXIII Roman numerals into their numerical equivalent. By combining the transformed Roman numerals, XXXIII can be expressed as X + X + X + I + I + I, equating to a sum of 33. A key principle to remember when deciphering Roman numerals is that the higher numerals must always precede the lower, ensuring an accurate translation. In this in-depth guide, we will delve into the intricacies of converting XXXIII and other Roman numerals to their numerical form.
• XXXIII = XXX + III
• XXXIII = 30 + 3
• XXXIII = 33
## How to Write XXXIII Roman Numerals?
Unlock the mysteries of the ancient Roman numerals with the knowledge of deciphering XXXIII. With two distinct methods at your disposal, uncovering the numerical equivalent of this Roman numeral has never been easier.
Method 1 involves breaking down the Roman numeral into its individual letters, writing the numerical value of each letter, and adding or subtracting them accordingly.
Check Out This Article XXVI Roman Numerals | How to Write XXVI in Numbers?
Method 2, on the other hand, employs a more intuitive approach by grouping the Roman numerals together for addition or subtraction. By using either method, we arrive at the same conclusion: the numerical value of XXXIII Roman numerals is a staggering 33. Master the art of Roman numeral conversion and unlock the secrets of history.
### What are the Basic Rules to Write Roman Numerals?
Unlock the secrets of the Roman Empire with the knowledge of converting Roman numerals. The key principle to remember when deciphering these ancient symbols is that the order in which the letters appear holds significant weight in determining their numerical value. When a larger letter precedes a smaller letter, the letters are added together. For example, LI equals 50 + 1, resulting in a sum of 51.
On the other hand, when a smaller letter precedes a larger letter, the letters are subtracted. For example, IX equals 10 – 1, resulting in a sum of 9. Additionally, when a letter is repeated 2 or 3 times, they are added together. For example, XXX equals 10 + 10 + 10, resulting in a sum of 30. However, it’s important to note that the same letter cannot be used more than three times in succession. With this knowledge, unlock the secrets of the Roman Empire and decipher the hidden meanings of these ancient symbols.
### Numbers Related to XXXIII Roman Numerals
Roman numerals were used in ancient Rome and consisted of letter combinations from the Latin alphabets I, V, X, L, C, D, and M. It may appear to be different from numbers, but they are not. For example, the Roman numeral XXXIII corresponds to the number 33. The roman numerals associated with XXXIII are as follows:
• XXX = 30
• XXXI = 30 + 1 = 31
• XXXII = 30 + 2 = 32
• XXXIII = 30 + 3 = 33
• XXXIV = 30 + 4 = 34
• XXXV = 30 + 5 = 35
• XXXVI = 30 + 6 = 36
• XXXVII = 30 + 7 = 37
• XXXVIII = 30 + 8 = 38
• XXXIX = 30 + 9 = 39
## Frequently Asked Questions on XXXIII Roman Numerals
What do the Roman Numerals XXXIII Stand For?
To determine its value, we will write XXXIII Roman numerals in expanded form. XXXIII is equal to X + X + X + I + I + I = 10 + 10 + 10 + 1 + 1 + 1 = 33. As a result, the value of the Roman Numeral XXXIII is 33.
What is the result of dividing XXXIII by V?
In numbers, XXXIII = 33 and V = 5. When you divide 33 by 5, you get a remainder of 5. Now, 3 + III = As a result, if XXXIII is divided by V, the remainder is III.
How do you write Roman Numerals XXXIII in numbers?
To convert XXXIII Roman Numerals to numbers, first divide the Roman numerals into place values (ones, tens, hundreds, and thousands), as shown below:
Tens = 30 = XXX
Ones = 3 = III
Number = 33 = XXXIII
Why is 33 written as XXXIII in Roman numerals?
We know that in roman numerals, 3 is written as III and 10 is written as X. As a result, 33 is written in roman numerals as XXXIII = XXX + III = 30 + 3 = XXXIII.
What should XXXIII be subtracted from to get VIII?
First, we’ll convert XXXIII and VIII to numbers: XXXIII = 30 + 3 = 33 and VIII = 8. Now, 33 – 8 = 25. And 25 equals XXV. As a result, subtract XXV from XXXIII roman numerals to get VIII.
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# Lesson 14
Transforming Trigonometric Functions
### Problem 1
These equations model the vertical position, in feet above the ground, of a point at the end of a windmill blade. For each function, indicate the height of the windmill and the length of the windmill blades.
1. $$y = 5\sin(\theta) +10$$
2. $$y = 8\sin(\theta) + 20$$
3. $$y = 4\sin(\theta) + 15$$
### Problem 2
Which expression takes the same value as $$\cos(\theta)$$ when $$\theta = 0, \frac{\pi}{2}, \pi,$$ and $$\frac{3\pi}{2}$$?
A:
$$\sin\left(\theta -\frac{\pi}{2}\right)$$
B:
$$\sin\left(\theta + \frac{\pi}{2}\right)$$
C:
$$\sin(\theta+\pi)$$
D:
$$\sin(\theta-\pi)$$
### Problem 3
Here is a graph of a trigonometric function.
Which equation does the graph represent?
A:
$$y = 2\sin\left(\theta\right)$$
B:
$$y = 2\cos\left(\theta+\frac{\pi}{4}\right)$$
C:
$$y = 2\sin\left(\theta-\frac{\pi}{4}\right)$$
D:
$$y = 2\cos\left(\theta-\frac{\pi}{4}\right)$$
### Problem 4
The vertical position $$v$$ of a point at the tip of a windmill blade, in feet, is given by $$v(\theta) = 11 + 2\sin\left(\theta+\frac{\pi}{2}\right)$$. Here $$\theta$$ is the angle of rotation.
1. How long is the windmill blade? Explain how you know.
2. What is the height of the windmill? Explain how you know.
3. Where is the point $$P$$ when $$\theta = 0$$?
### Problem 5
1. Explain how to use a unit circle to find a point $$P$$ with $$x$$-coordinate $$\cos(\frac{23\pi}{24})$$.
2. Use a unit circle to estimate the value of $$\cos(\frac{23\pi}{24})$$.
### Solution
(From Unit 6, Lesson 5.)
### Problem 6
1. What are some ways in which the tangent function is similar to sine and cosine?
2. What are some ways in which the tangent function is different from sine and cosine? |
# Expected value of size of subset
Given a set $S$ such that $|S|=n$,
A random item is chosen randomly from $S$, and being appended to a new set $T$.
This process is being repeated $n$ times (with repetition), what is the expected value of $|T|$ ?
#### Solutions Collecting From Web of "Expected value of size of subset"
Let $T_k$ be the number of elements in $T$ after $k$ rounds. We also set $T_0 = 0$.
Assuming that $T_k = l$, you have a probability of $\frac{l}{n}$ that the $k + 1$-th element is already in $T$ and a probability of $\frac{n – l}{n}$ that it isn’t. This means:
$$E[T_{k + 1} \mid T_k = l] = \frac{l}{n} l + \frac{n – l}{n}(l+1) = l\left(1 – \frac{1}{n}\right) + 1$$
Now we can apply the law of total expectation:
$$E[T_{k + 1}] = E[E[T_{k + 1} \mid T_k]] = E\left[T_k \left(1 – \frac{1}{n}\right) + 1\right] = E[T_k] \left(1 – \frac{1}{n}\right) + 1$$
This is a geometric progession and we can explicitly calculate:
$$E[T_k] = \sum \limits_{i = 0}^{k – 1} \left(1 – \frac{1}{n}\right)^i = n\left(1 – \left(1 – \frac{1}{n}\right)^k\right)$$
Specifically for $k = n$ this yields
$$E[T_n] = n\left(1 – \left(1 – \frac{1}{n}\right)^n\right) \approx n(1 – e^{-1}) \approx \frac{2}{3} n.$$
For $x\in S$ define $A_x$ to be the event that $x\in T$.
Then $|T|=\sum_{x\in S} 1_{A_x}$ which has expected value
$$\mathbb{E}(X)=\sum_{x\in S} \mathbb{P}(A_x)=n\left[1-\left(1-{1\over n}\right)^n\right].$$
This question can also be answered by total enumeration using Stirling
numbers of the second kind. Observe that the expectation is given by
$$\mathrm{E}[X] = n^{-n} \sum_{k=0}^n k \times {n\choose k} k! {n\brace k}.$$
Here we multiply the value of the random variable by the probability,
which is obtained from the fact that a sample with $k$ different
entries needs a choice of these entries from the $n$ possibles and a
partition of the $n$ slots into $k$ sets whose elements receive the
same value. All $k!$ permutations of the $k$ chosen values result in
a valid unique assignment.
We have
$${n\brace k} = n! [z^n] \frac{(\exp(z)-1)^k}{k!}$$
by the species equation for set partititions
$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which yields for the expectation
$$n^{-n} n! [z^n] \sum_{k=0}^n k \times {n\choose k} (\exp(z)-1)^k \\ = n^{-n} n! [z^n] n \sum_{k=1}^n {n-1\choose k-1} (\exp(z)-1)^k \\ = n^{-n} n! [z^n] n (\exp(z)-1) \sum_{k=0}^{n-1} {n-1\choose k} (\exp(z)-1)^k \\ = n^{-n} n! [z^n] n (\exp(z)-1) \exp((n-1)z) = n n^{-n} n! [z^n] (\exp(nz)-\exp((n-1)z)) \\ = n n^{-n} (n^n – (n-1)^n) \\ = n \left(1 – \left(1-\frac{1}{n}\right)^n\right) \sim n \left(1 – \frac{1}{e}\right).$$
Now we can also compute higher moments with this,
the variance for example.
We first compute the expectation of $X(X-1)$, getting
$$\mathrm{E}[X(X-1)] = n^{-n} \sum_{k=0}^n k (k-1) \times {n\choose k} k! {n\brace k} \\ = n^{-n} n! [z^n] n (n-1) \sum_{k=2}^n {n-2\choose k-2} (\exp(z)-1)^k \\ = n^{-n} n! [z^n] n (n-1) (\exp(z)-1)^2 \sum_{k=0}^{n-2} {n-2\choose k} (\exp(z)-1)^k \\ = n^{-n} n! [z^n] n (n-1) (\exp(z)-1)^2 \exp((n-2)z).$$
Extracting coefficients we obtain
$$n (n-1) n^{-n} (n^n – 2(n-1)^n + (n-2)^n) = n(n-1) \left(1-2\left(1-\frac{1}{n}\right)^n +\left(1-\frac{2}{n}\right)^n\right) \\ \sim n(n-1) \left(1-\frac{2}{e}+\frac{1}{e^2}\right).$$
We thus get for the variance
$$\mathrm{Var}[X] = \mathrm{E}[X^2] – \mathrm{E}[X]^2 \\ = n(n-1) \left(1-2\left(1-\frac{1}{n}\right)^n +\left(1-\frac{2}{n}\right)^n\right) \\ + n\left(1-\left(1-\frac{1}{n}\right)^n\right) – n^2\left(1-\left(1-\frac{1}{n}\right)^n\right)^2 \\ = n^2 \left(\left(1-\frac{2}{n}\right)^n – \left(1-\frac{1}{n}\right)^{2n}\right) \\ + n \left(\left(1-\frac{1}{n}\right)^n – \left(1-\frac{2}{n}\right)^n \right).$$
The asymptotic expansion then yields
$$\mathrm{Var}[X] \sim n \left(\frac{1}{e}-\frac{2}{e^2}\right).$$
Addendum. We need the second term rather than the constant term in
the asymptotic expansion of the factor on $n^2.$
We get for the first term in the difference
$$\exp\left(n\log\left(1-\frac{2}{n}\right)\right) = \exp\left(-\sum_{q\ge 1} \frac{2^q}{n^{q-1} q}\right) \\ = \sum_{k\ge 0} \frac{(-1)^k}{k!} \left(\sum_{q\ge 1} \frac{2^q}{n^{q-1} q}\right)^k.$$
This yields for the constant term
$$\sum_{k\ge 0} \frac{(-1)^k}{k!} 2^k = \frac{1}{e^2}.$$
We get for the next term
$$\frac{1}{n} \sum_{k\ge 0} \frac{(-1)^k}{k!} {k\choose 1} 2\times 2^{k-1} = -\frac{2}{n} \sum_{k\ge 1} \frac{(-1)^{k-1}}{(k-1)!} 2^{k-1} = -\frac{2}{n} \frac{1}{e^2}.$$
For the second term in the difference we have
$$\exp\left(2n\log\left(1-\frac{1}{n}\right)\right) = \exp\left(-2\sum_{q\ge 1} \frac{1}{n^{q-1} q}\right) \\ = \sum_{k\ge 0} \frac{(-2)^k}{k!} \left(\sum_{q\ge 1} \frac{1}{n^{q-1} q}\right)^k.$$
This time we get for the constant term
$$\sum_{k\ge 0} \frac{(-2)^k}{k!} = \frac{1}{e^2}$$
and for the next term
$$\frac{1}{n} \sum_{k\ge 0} \frac{(-2)^k}{k!} {k\choose 1} \frac{1}{2} \times 1^{k-1} = -\frac{1}{n} \sum_{k\ge 1} \frac{(-2)^{k-1}}{(k-1)!} = -\frac{1}{n} \frac{1}{e^2}.$$
The difference then yields for the variance
$$n^2 \left(-\frac{1}{n} \frac{1}{e^2}\right) + n\left(\frac{1}{e}-\frac{1}{e^2}\right)$$
which is the desired result.
We can count the number of arrangements of $n$ items picked from a pool of $n$ items that have $d$ distinct items.
Let $S(i)$ be the collection of arrangements missing item $i$. Let $N(j)$ be the sum of the sizes of all intersections of $j$ of the $S(i)$.
Since we can choose $\binom{n}{j}$ items to leave out and for each of those choices there are $(n-j)^n$ arrangements of those items, we have
$$N(j)=\binom{n}{j}(n-j)^n\tag{1}$$
Since the number of arrangements with $d$ distinct items is the number of arrangements missing exactly $n-d$ items, according to the Generalized Inclusion-Exclusion Principle, the number of arrangements with exactly $d$ distinct items is
$$\sum_{j=n-d}^n(-1)^{j-n+d}\binom{j}{n-d}N(j) =\sum_{j=n-d}^n(-1)^{j-n+d}\binom{j}{n-d}\binom{n}{j}(n-j)^n\tag{2}$$
To get the total number of arrangements, sum $(2)$ in $d$:
\begin{align} \sum_{d=0}^n\sum_{j=n-d}^n(-1)^{j-n+d}\binom{j}{n-d}\binom{n}{j}(n-j)^n &=\sum_{j=0}^n\sum_{d=n-j}^n(-1)^{j-n+d}\binom{j}{n-d}\binom{n}{j}(n-j)^n\\ &=\sum_{j=0}^n[j=0]\binom{n}{j}(n-j)^n\\[9pt] &=n^n\tag{3} \end{align}
as expected.
To get the total number of arrangements times their size, multiply by $d$ and sum in $d$:
\begin{align} \sum_{d=0}^n\sum_{j=n-d}^n(-1)^{j-n+d}\binom{j}{n-d}\binom{n}{j}(n-j)^nd &=\sum_{j=0}^n\sum_{d=n-j}^n(-1)^{j-n+d}\color{#00A000}{\binom{j}{n-d}}\binom{n}{j}(n-j)^n(\color{#C00000}{n}-\color{#00A000}{(n-d)})\\ &=\color{#C00000}{n\,n^n}-\sum_{j=0}^n\sum_{d=n-j}^n(-1)^{j-n+d}\color{#00A000}{\binom{j-1}{n-d-1}}\binom{n}{j}(n-j)^n\color{#00A000}{j}\\ &=n\,n^n-\sum_{j=0}^n[j=1]\binom{n}{j}(n-j)^nj\\[9pt] &=n\,n^n-n(n-1)^n\tag{4} \end{align}
Therefore, the expected number of distinct items picked would be
$$n\left[1-\left(1-\frac1n\right)^n\right]\sim n\left(1-\frac1e\right)\tag{5}$$ |
## Want to keep learning?
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2.3
## National STEM Learning Centre
Skip to 0 minutes and 8 seconds PAULA KELLY: So let’s have a look, now, at a slightly more difficult example. If we’re trying to convert a fraction into a decimal– if we had, for example– let’s have 16/125.
Skip to 0 minutes and 23 seconds MICHAEL ANDERSON: OK, so these numbers look pretty daunting, but actually, it’s the exact same process. So if we look at the numerator, 16, it’s actually the denominator that we’re interested in– 125. We’re going to try and multiply this up to either a 10, 100, or most likely 1,000, or even bigger. So we’re going to try and think about 125 times tables. Now, actually, in order to get 1,000, we have to multiply 125 by 8.
Skip to 0 minutes and 55 seconds PAULA KELLY: And if we’re struggling with that, we could just do 125 times table.
Skip to 1 minute and 0 seconds MICHAEL ANDERSON: Yep. And this is obviously without a calculator as well. So yeah, we’re going to multiply by 8 to get 1,000 at the bottom. Now, we have to do the exact same to the top as well. So we’re going to take our 16 and multiply by 8 again, and 16 multiplied by 8– well, I can double it, double it, and double it again. And that will give me 128 over 1,000.
Skip to 1 minute and 24 seconds PAULA KELLY: OK, so we’ve seen, already, how we have this as a division. So 128 divided by 1,000– all of our numbers will go three places smaller.
Skip to 1 minute and 36 seconds MICHAEL ANDERSON: Yeah. And you can think about it in a different way as well. With our 128/1,000, you can think of it as 100 and 20 and 8. If you’ve got 100 over 1,000, that will simplify to 1 over 10. We’ve got 1/10. So I could put a 1 in the tens column, with a 20 part– well, 20 out of 1,000. I can divide both by 10, and I end up with 2 over 100– 2/100. And I can put the 2 in here. And that just leaves me with the 8, and I’ve got 8/1,000, so I can put 8 in the thousands column here. Now, 16 over 125 is less than 1.
Skip to 2 minutes and 14 seconds So in the units column, in the ones, I’m going to have a 0 there. So I end up with 0.128.
# Fractions as decimals: further example
In the previous step we saw how numbers expressed as fractions can be expressed as decimals. The examples used so far have been numbers which have one and two decimal places: tenths and hundredths. In this step we extend the method to include numbers which when expressed as a decimal have three decimal places: tenths, hundredths and thousandths.
## Problem worksheet
Now complete question one from this week’s worksheet. As a reminder, problem worksheets are all linked in step 1.1, and at the start of each week.
## Teaching resource
This SMILE resource contains two packs of games, investigations, worksheets and practical activities supporting the teaching and learning of decimals, from reading decimals from a scale to multiplying two decimal numbers without a calculator. |
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# Patterns Including Numbers
### Patterns
Patterns are present everywhere! They are predictable arrangements following a certain rule.
In our daily life, we see patterns in animals, arrangement of leaves; numbers etc….These patterns make learning interesting for us and help us unwind our mind.
### Simple Patterns
Are you familiar with these patterns?
3, 6, 9, 12, 15, ____
### Triangular Numbers
Emma was watching a sports channel. She saw the following picture and asked her father what game they are playing?
The father replied that this is a game of bowling pins. The players roll a ball and knock down as many bowling pins as possible. She was surprised at the arrangement of the bowling pins. There were a total of 10 bowling pins. She drew the arrangement using dots in an increasing pattern.
She counted the number of dots in each pattern. They are 1, 3, 6 and 10.
The number of dots that can form a triangle-shaped pattern are called triangular numbers.
Let us observe the pattern and understand the rule to find the next numbers.
### Square Numbers
The next game on the channel was chess. Emma liked the pattern of squares on the chessboard and wanted to count them.
She started drawing the squares using dots in an increasing pattern.
She noted the number of dots in each square as 1, 4, 9 and 16.
Let us observe the pattern and understand the rule to find the next numbers.
### Turn the shape
Mia is printing on a drawing sheet with a block. Look at the beautiful pattern she has made:
Has she used different blocks to make the pattern?
We can create a pattern by simply turning a block in a particular way. We can also turn a block in different ways to create different designs.
Here, Mia has used the same block for printing. She turns the block one- fourth in a clockwise direction each time. Since the block is rotated by a one-fourth turn ever time, it is called a quarter turn.
Another way of creating a pattern is by rotating a block by half a turn each time. Since the block is rotated by turn, it is called half turn.
### Interesting number and alphabet patterns
The number 101 is used to make a pattern by taking a 1/4th turn every time.
After which turn does the number look the same?
The word ZOOZ is used to make a pattern by taking a 1/4th turn every time. After 1/4th turn, the word is read as NOON, but vertically.
After which turn does the number look the same?
### Check Point
1. What is the missing number in this pattern?
1, 4, 27, 256, ______, 46656
1. Which number should be placed in the empty triangle?
3. Which of these shapes will look the same after 1/2 a turn?
4. Study the pattern carefully and complete it:
1 × 1 × 1 = 1 = 1
2 × 2 × 2 = 8 = 3 + 5
3 × 3 × 3 = 27 = 7 + 9 + 11
4 × 4 × 4 = 64 = 13 + ____ + ____ + ____
5 × 5 × 5 = ___ = ____ + ____ + ____ + ____ + ____
5.Complete the pattern:
1. 3125 as 11 = 1, 22 = 4, 33 = 9, 44 = 256, 55 = 3125, 66 = 46656
2. (7 – 3) × 3 = 4 × 3 = 12
(8 – 6) × 4 = 2 × 4 = 8
(8 – 5) × 3 = 3 × 3 = 9
(7 – 6) × 4 = 1 × 4 = 4
3.
4. 4 × 4 × 4 = 64 = 13 + 15 + 17 + 19
5 × 5 × 5 = 125 = 21 + 23 + 25 + 27 + 29
5.
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# NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
Written by Team Trustudies
Updated at 2021-05-07
## NCERT solutions for class 6 Maths Chapter 9 Data Handling Exercise 9.1
Q.1 In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.
(a) Find how many students obtained marks equal to or more than 7.
(b) How many students obtained marks below 4?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a) Number of students who obtained marks equal to or more than 7 = 5 + 4 + 3 = 12
(b) Number of students who obtained marks below 4 = 2 + 3 + 3 = 8.
Q.2 Following is the choice of sweets of 30 students of Class VI.
(a) Arrange the names of sweets in a table using tally marks.
(b) Which sweet is preferred by most of the students?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a)
Q.3 Catherine threw a dice 40 times and noted the number appearing each time as shown below :
Make a table and enter the data using tally marks. Find the number that appeared.
(a) The minimum number of times
(b) The maximum number of times
(c) Find those numbers that appear an equal number of times.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a) The number 4 appeared 4 times which is the minimum.
(b) The number 5 appeared 11 times which is the maximum.
(c) The number 1 and 6 appear for the same number of times, i.e., 7.
Q.4 Following pictograph shows the number of tractors in five villages.
Observe the pictograph and answer the following questions.
(i) Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) How many more tractors village C has as compared to village B?
(iv) What is the total number of tractors in all the five villages?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(i) Village D has the minimum number of tractors.
(ii) Village C has the maximum number of tractors.
(iii) Village B has 5 tractors
Village C has 8 tractors
Difference = 8 – 5 = 3 tractors
Village C has 3 more tractors as compared to village B
(iv) Total number of tractors in all the villages = 6 + 5 + 8 + 3 + 6 = 28 tractors
Q.5 The number of girl students in each class of a co-educational middle school is depicted by the pictograph:
Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in Class VI less than the number of girls in Class V?
(c) How many girls are there in Class VII?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a) Class VIII has the minimum number of girl students
i,e. $$1\frac{1}{2}$$ x 4 = $$\frac{3}{2}$$ × 4 = 6.
(b) No, number of girls in Class VI = 4 x 4 = 16 and number
of girls in Class V =$$2\frac{1}{2} x 4 = \frac{5}{2} ×4 = 10 So, number of girl students in Class VI is not less than that of in Class V. (c) Number of girls in Class VII = 3 x 4 = 12 Q.6 The sale of electric bulbs on different days of a week is shown below: Observe the pictograph and answer the following questions: (a) How many bulbs were sold on Friday? (b) On which day were the maximum number of bulbs sold? (c) On which of the days same number of bulbs were sold? (d) On which of the days minimum number of bulbs were sold? (e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week? NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Answer : (a) Number of bulbs sold on Friday are 14 bulbs. (b) The maximum number of bulbs were sold on Sunday. Which is 18 bulb (c) equal number of bulbs were sold on Wednesday and Saturday. (d) Minimum number of bulbs were sold on Wednesday and Saturday i.e 8 bulbs. (e) Total number of bulbs sold in a week = 12 + 16 + 8 + 10 + 14 + 8 + 18 = 86 Q.7 In a village six fruit merchants sold the following number of fruit baskets in a particular season: Observe this pictograph and answer the following questions: (a) Which merchant sold the maximum number of baskets? (b) How many fruit baskets were sold by Anwar? (c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them? NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Answer : (a) Martin sold the maximum number of fruit baskets, which is \(9\frac{1}{2}$$ x 100 = 9.5 × 100 = 950
(b) Number of fruit baskets sold by Anwar is 7 x 100 = 700.
(c) Anwar, Martin and Ranjit Singh have sold 600 or more fruit baskets and planning to buy a godown.
## NCERT solutions for class 6 Maths Chapter 9 Data Handling Exercise 9.2
Q.1 Total number of animals in five villages are as follows:
Village A : 80
Village B : 120
Village C : 90
Village D : 40
Village E : 60
Prepare a pictograph of these animals using one symbol to represent 10 animals and answer the following questions:
(a) How many symbols represent animals of village E?
(b) Which village has the maximum number of animals?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a) There are 60 animals in village E. So 6 symbols represent animals of village E
(b) Village B has the maximum number of animals .
(c) Village A has 80 animals and village C has 90 animals. So village C has more animals than village A
Q.2 Total number of students of a school in different years is shown in the following table
A. Prepare a pictograph of students using one symbolto represent 100 students and answer the following questions:
(a) How many symbols represent total number of students in the year 2002?
(b) How many symbols represent total number of students for the year 1998?
B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
A.
(a) Total number of students in the year 2002 represents 6 symbols
(b) Total number of students in the year 1998 represents 5 complete and 1 incomplete symbols
B.
## NCERT solutions for class 6 Maths Chapter 9 Data Handling Exercise 9.3
Q.1 The bar graph given alongside shows the amount of wheat purchased by government during the year 1998-2002.
Read the bar graph and write down your observations. In which year was
(a) the wheat production maximum?
(b) the wheat production minimum?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a) maximum = 2002
(b) minimum = 1998
Q.2 Observe this bar graph which is showing the sale of shirts in a ready made shop from Monday to Saturday
(a) What information does the above bar graph give?
(b) What is the scale chosen on the horizontal line representing number of shirts?
(c) On which day were the maximum number of shirts sold? How many shirts were sold on that day?
(d) On which day were the minimum number of shirts sold?
(e) How many shirts were sold on Thursday?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a) The above bar graph shows the number of shirts sold from Monday to Saturday
(b) 1 unit length = 5 shirts is the scale on the horizontal line representing number of shirts.
(c) On Saturday maximum number of shirts sold i.e 60 shirts was sold.
(d) Tuesday
(e) 35 shirts were sold on Thursday
Q.3 Observe this bar graph which shows the marks obtained by Aziz in half-yearly examination in different subjects. Answer the given questions.
(a) What information does the bar graph give?
(b) Name the subject in which Aziz scored maximum marks.
(c) Name the subject in which he has scored minimum marks.
(d) State the name of the subjects and marks obtained in each of them.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a) The bar graph shows the marks scored by Aziz in different subjects.
(b) Hindi
(c) Social Studies
(d) Hindi – 80 marks
English – 60 marks
Mathematics – 70 marks
Science – 50 marks
Social Studies – 40 marks
## NCERT solutions for class 6 Maths Chapter 9 Data Handling Exercise 9.4
Q.1 A survey of 120 school students was done to find which activity they prefer to do in their free time.
Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students. Which activity is preferred by most of the students other than playing?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
Q.2 The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:
Draw a bar graph to represent the above information choosing the scale of your choice.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
Q.3 Following table shows the number of bicycles manufactured in a factory during the years 1998 to 2002. Illustrate this data using a bar graph. Choose a scale of your choice
(a) In which year were the maximum number of bicycles manufactured?
(b) In which year were the minimum number of bicycles manufactured?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a) 2002
(b) 1999
Q.4 Number of persons in various age groups in a town is given in the following table.
Draw a bar graph to represent the above information and answer the following questions. (take 1 unit length = 20 thousands)
(a) Which two age groups have same population?
(b) All persons in the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling
(a) 30 - 44 and 45 - 59
(b) Number of senior citizens in the town
= 80000 + 40000 = 1 lakh 20 thousand
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# Unit 7 Family Materials
Two-dimensional Shapes and Perimeter
### Two-dimensional Shapes and Perimeter
In this unit, students reason about attributes (features) of shapes and learn about perimeter.
### Section A: Reason with Shapes
In this section, students describe, compare, and sort a variety of shapes. Students think about ways to sort triangles and quadrilaterals into more specific categories based on their attributes. They see that triangles and quadrilaterals can be classified and named based on their sides (whether some sides are the same length) and angles (whether they have right angles).
These are rectangles.
These are not rectangles.
Students see that a shape can have more than one name if it has the attributes that define different shapes. For example, a shape that is a square is also a rhombus and a rectangle.
### Section B: What is Perimeter?
In this section, students learn that perimeter is the distance around a shape. They first find perimeter by counting or adding the units of length on each side of a shape. Later, they find the perimeter of shapes whose sides are labeled with lengths.
Students also draw shapes with a specified perimeter and see that different shapes can have the same perimeter.
### Section C: Expanding on Perimeter
In this section, students solve problems that involve both area and perimeter. They draw rectangles with the same area and different perimeters, and rectangles with the same perimeter and different areas.
For example, the rectangles in the image all have a perimeter of 16 units, but they have different areas.
### Section D: Design with Perimeter and Area
In this section, students apply what they’ve learned about geometric shapes, perimeter, and area to solve design problems. They design a park that has certain components, a West African wax print pattern with certain shapes, and a robot that meet certain requirements.
### Try it at home!
Near the end of the unit, ask your student to find:
• these shapes around the house: a rhombus, a rectangle, a square, and a quadrilateral that isn't a rhombus, rectangle, or square
• the area and perimeter of a rectangle in the house
Questions that may be helpful as they work:
• What kind of quadrilateral is this? How do you know?
• Are you measuring area or perimeter? How do you know? |
Like linear inequalities, quadratic inequalities may contain one variable or two variables. While working with quadratic inequalities is, in many ways, similar to working with linear inequalities, the quadratic inequalities require a bit more thinking and testing.
Solve the quadratic inequality x2 ≤ 4. (one variable inequality)
Let's think about this solution. If we change the inequality to an equal sign, we know that x² = 4 has two possible solutions, 2 and -2. Now, the inequality states that we are looking for a squared value that is less than or equal to 4. Our first thought is that the answer will be less than or equal to 2. Good! But, if we go too far in the negative direction, our answer will be wrong. Think about x = -3. If we square -3 we get +9 which is not less than or equal to 4. We can only go in the negative direction until we hit -2. Our answers must be between -2 and 2, inclusive (including the -2 and the 2). Let's organize our thinking into steps: 1. Replace the inequality symbol with an equal sign, and solve the resulting equation. Solving the equation will give you numeric values that will form intervals that will reveal the solutions to the inequality. 2. Set up a number line to investigate the intervals created by the equation solutions. Pick a number in each interval and test it in the inequality. If the result is true, that interval is a solution to the inequality. There may be more than one interval which is a solution. 3. Graph the solution set. Since this is a "less than or equal to" inequality, closed circles are used to show that the endpoints are included in the solution. The shaded line segment indicates the other values for which the inequality is true. State the solution in a form directed by your teacher or stated in the question. Ways to State the Solution:
Solve the quadratic inequality x2 - 3x - 10 > 0. (one variable inequality)
Follow the same steps established in Example 1. 1. Replace the inequality symbol with an equal sign, and solve the resulting equation. These solutions create the intervals that will reveal the solutions to the inequality. 2. Set up a number line to investigate the intervals created by the equation solutions. Pick a number in each interval and test it in the inequality. If the result is true, that interval is a solution to the inequality. There may be more than one interval which is a solution. 3. Graph the solution set. Since this is a "strict" inequality, open circles are used. Arrows indicate where the inequality is true. State the solution in a form directed by your teacher or stated in the question. Ways to State the Solution: x < -2 or x > 5
FYI: Graph the quadratic inequality y < x2 - 3x - 10. (two variables inequality)
1. Replace the inequality symbol with an equal sign, and graph the equation (a parabola). Use a dashed line for strict inequalities. 2. Choose a test point NOT on the parabola. Test the point in the inequality. If the result is true, shade the entire region where the test point lies. If the result is false, shade the entire region on the other side of the parabola. (The point (0,0) tests FALSE in the inequality.) 0 < 0² - 3(0) - 10 is false. 3. Solution: The shaded area contains all of the points that make the inequality true.
How to use your TI-83+ graphing calculator to graph quadratic inequalities. Click calculator. |
# Coplanarity of Two Lines
Coplanar lines are the lines that lie on the same plane. Prove that two lines are coplanar using the condition in vector form and Cartesian form. This lesson provides you with a solved example for you to assess your understanding based on the concepts here.
## Coplanarity in Theory
Coplanar lines are a common topic in three-dimensional geometry. In mathematical theory, we may define coplanarity as the condition where a given number of lines lie on the same plane, they are said to be coplanar.
To recall, a plane is a two-dimensional figure extending into infinity in the three-dimensional space, while we have used vector equations to represent straight lines (also referred to as lines). We shall now take a look at what condition is necessary to be fulfilled for two lines to be coplanar.
You can download Three Dimensional Geometry Cheat Sheet by clicking on the download button below
### Browse more Topics under Three Dimensional Geometry
Source: MathCaptain
### Condition for Coplanarity in Vector Form
In vector form, let us consider the equations of two straight lines to be as under:
• r1 = l1 + λmÂ1
• r2 = l2 + λm2
What do these equations mean? It means that the first line passes through a point, say L, whose position vector is given by l1 and is parallel to m1. Similarly, the second line is said to pass through another point whose position vector is given by l2 and is parallel to m2.
The condition for coplanarity is that the line joining the two points must be perpendicular to the product of the two vectors, m1 and m2. To illustrate this, we know that the line joining the two said points can be written in vector form as (l2 – l1). So, we have:
(l2 – l1) . (m1 x m2) = 0
### Condition for Coplanarity in Cartesian Form
The derivation of the condition for coplanarity in Cartesian form stems from the vector form. Let us consider two points L (x1, y1, z1) & M (x2, y2, z2) in the Cartesian plane. Let there be two vectors m1 and m2. Their direction ratios are given by a1, b1, c1 and a2, b2, c2 respectively.
The vector equation of the line joining L and M can be given as:
LM = (x2 – x1)i + (y2 – y1)j + (z2– z1)k
m1 = a1i + b1j + c1k
m2 = a2i + b2j + c2k
We shall now use the above condition in vector form to derive our condition in Cartesian form. This can be used as is for calculation purposes. By the condition above, the two lines would be coplanar if  LM. (m1 x m2) = 0. Therefore, in Cartesian form, the matrix representing this equation is given as 0.
Given below is a solved problem on how to prove that two lines are coplanar.
## Solved Example for You on Coplanar Lines
Question 1: Are the lines (x + 3)/3 = (y – 1)/1 = (z – 5)/5 and (x + 1)/ -1 = (y – 2)/2 = (z – 5)/5 coplanar?
Answer: Comparing the equations with the general form, we have:
(x1, y1, z1) = (-3, 1, 5) and (x2, y2, z2) = (-1, 2, 5).
Note that a1, b1, c1 = -3, 1, 5 and a2, b2, c2 = -1, 2, 5.
So, by Cartesian form, we must solve the matrix:
= 2(5 – 10) – 1(-15 + 5) + 0(-6 + 1) = -10 + 10 = 0
Since the solution of the matrix gives us 0, we can say that the given lines are coplanar
Question 2: What is meant by coplanar?
Answer: Coplanar points refer to three or more points which all exist in the same plane. Any set of three points in space is said to be coplanar. A set of four points may be coplanar or it may not be coplanar.
Question 3: Can we say that collinear points are coplanar?
Answer: Collinear points are those whose existence takes place in the same line. Coplanar points are points that are all in the same plane. So, in case of collinear points, a person can choose one of infinite number of planes which has the line on which these points exist. As such, one can say that they are coplanar.
Question 4: How can one prove that two vectors are coplanar?
Answer: One can prove that two vectors are coplanar if they are in accordance with the following conditions:
• In case the scalar triple product of any three vectors happens to be zero.
• If any three vectors are such that they are linearly dependent
• n vectors will be coplanar if among them no more than two vectors exist that are linearly independent vectors.
Question 5: Two points determine how many lines?
Answer: Two points determine only one line.
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# Linear Systems with Addition or Subtraction
## Solve systems using elimination of one variable
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Linear Systems with Addition or Subtraction
Suppose two extended families went to an amusement park, where adult tickets and children's tickets have different prices. If the first family had 5 adults and 8 children and paid a total of $124, and the second family had 5 adults and 12 children and paid$156, how much did each type of ticket cost? Could you write a system of equations representing this situation? If you wanted to solve the system by elimination, how would you go about doing it? In this Concept, you'll learn how to use elimination to solve a system of linear equations similar to the one representing this scenario by addition or subtraction.
### Watch This
Multimedia Link: For help with solving systems by elimination, visit this site: http://www.teachertube.com/viewVideo.php?title=Solving_System_of_Equations_using_Elimination&video_id=10148 Teacher Tube video.
### Guidance
As you noticed in the previous Concept, solving a system algebraically will give you the most accurate answer and in some cases, it is easier than graphing. However, you also noticed that it took some work in several cases to rewrite one equation before you could use the Substitution Property. There is another method used to solve systems algebraically: the elimination method .
The purpose of the elimination method to solve a system is to cancel, or eliminate, a variable by either adding or subtracting the two equations. This method works well if both equations are in standard form.
#### Example A
If one apple plus one banana costs $1.25 and one apple plus two bananas costs$2.00, how much does it cost for one banana? One apple?
Solution: Begin by defining the variables of the situation. Let $a=$ the number of apples and $b=$ the number of bananas . By translating each purchase into an equation, you get the following system:
$\begin{cases}a+b=1.25\\a+2b=2.00 \end{cases}$ .
You could rewrite the first equation and use the Substitution Property here, but because both equations are in standard form, you can also use the elimination method.
Notice that each equation has the value $1a$ . If you were to subtract these equations, what would happen?
$& \qquad a + b \ =1.25\\&\underline{\;\;\;- (a+2b=2.00)\;\;\;}\\& \qquad \quad -b =-0.75\\& \qquad \qquad \ b =0.75$
Therefore, one banana costs $0.75, or 75 cents. By subtracting the two equations, we were able to eliminate a variable and solve for the one remaining. How much is one apple? Use the first equation and the Substitution Property. $a+0.75&=1.25\\a&=0.50 \rightarrow one \ apple \ costs \ 50 \ cents$ #### Example B Solve the system $\begin{cases}3x+2y=11\\5x-2y=13\end{cases}$ . Solution: These equations would take much more work to rewrite in slope-intercept form to graph, or to use the Substitution Property. This tells us to try to eliminate a variable. The coefficients of the $x-$ variables have nothing in common, so adding will not cancel the $x-$ variable. Looking at the $y-$ variable, you can see the coefficients are 2 and –2. By adding these together, you get zero. Add these two equations and see what happens. $& \qquad \ 3x+2y =11\\&\underline{\;\; + \ (5x-2y) =13 \;\;}\\& \qquad \ 8x+0y =24$ The resulting equation is $8x=24$ . Solving for $x$ , you get $x=3$ . To find the $y-$ coordinate, choose either equation , and substitute the number 3 for the variable $x$ . $3(3)+2y&=11\\9+2y&=11\\2y&=2\\y&=1$ The point of intersection of these two equations is (3, 1). #### Example C Andrew is paddling his canoe down a fast-moving river. Paddling downstream he travels at 7 miles per hour, relative to the river bank. Paddling upstream, he moves slower, traveling at 1.5 miles per hour. If he paddles equally hard in both directions, calculate, in miles per hour, the speed of the river and the speed Andrew would travel in calm water. Solution: We have two unknowns to solve for, so we will call the speed that Andrew paddles at $x$ , and the speed of the river $y$ . When traveling downstream, Andrew's speed is boosted by the river current, so his total speed is the canoe speed plus the speed of the river $(x+y)$ . Upstream, his speed is hindered by the speed of the river. His speed upstream is $(x-y)$ . $\text{Downstream Equation} && x+y&=7\\\text{Upstream Equation} && x-y&=1.5$ Notice $y$ and $-y$ are additive inverses. If you add them together, their sum equals zero. Therefore, by adding the two equations together, the variable $y$ will cancel, leaving you to solve for the remaining variable, $x$ . $& \qquad \ x+y=7\\&\underline{\;\; + \ (x-y)=1.5 \;\;}\\& \quad \ 2x+0y=8.5\\& \qquad \quad \ \ 2x=8.5$ Therefore, $x=4.25$ ; Andrew is paddling $4.25 \ miles/hour$ . To find the speed of the river, substitute your known value into either equation and solve. $4.25-y&=1.5\\-y&=-2.75\\y&=2.75$ The stream’s current is moving at a rate of $2.75 \ miles/hour$ . ### Video Review ### Guided Practice Solve the system $\begin{cases}5s+2t=6\\9s+2t=22\end{cases}$ . Solution: Since these equations are both written in standard form, and both have the term $2t$ in them, we will use elimination by subtracting. This will cause the $t$ terms to cancel out and we will be left with one variable, $s$ , which we can then isolate. $& \qquad \ 5s+2t=6\\&\underline{\;\; - \ (9s+2t = 22) \;\;}\\& \qquad \ -4s+0t =-16\\& \qquad \ -4s=-16\\& \qquad \ s=4$ $5(4)+2t&=6\\20+2t&=6\\2t&=-14\\t&=-7$ The solution is $(4,-7)$ . ### Practice Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Solving Linear Systems by Elimination (12:44) 1. What is the purpose of the elimination method to solve a system? When is this method appropriate? In 2 – 10, solve each system using elimination. 1. $\begin{cases}2x+y=-17\\8x-3y=-19 \end{cases}$ 2. $\begin{cases}x+4y=-9\\-2x-5y=12 \end{cases}$ 3. $\begin{cases}-2x-5y=-10\\x+4y=8 \end{cases}$ 4. $\begin{cases}x-3y=-10\\-8x+5y=-15 \end{cases}$ 5. $\begin{cases}-x-6y=-18\\x-6y=-6 \end{cases}$ 6. $\begin{cases}5x-3y=-14\\x-3y=2 \end{cases}$ 7. $&3x+4y=2.5\\&5x-4y=25.5$ 8. $&5x+7y=-31\\&5x-9y=17$ 9. $&3y-4x=-33\\&5x-3y=40.5$ 10. Nadia and Peter visit the candy store. Nadia buys three candy bars and four fruit roll-ups for$2.84. Peter also buys three candy bars, but he can afford only one fruit roll-up. His purchase costs $1.79. What is the cost of each candy bar and each fruit roll-up? 11. A small plane flies from Los Angeles to Denver with a tail wind (the wind blows in the same direction as the plane), and an air-traffic controller reads its ground-speed (speed measured relative to the ground) at 275 miles per hour. Another identical plane moving in the opposite direction has a ground-speed of 227 miles per hour. Assuming both planes are flying with identical air-speeds, calculate the speed of the wind. 12. An airport taxi firm charges a pick-up fee, plus an additional per-mile fee for any rides taken. If a 12-mile journey costs$14.29 and a 17-mile journey costs $19.91, calculate: 1. the pick-up fee 2. the per-mile rate 3. the cost of a seven-mile trip 13. Calls from a call-box are charged per minute at one rate for the first five minutes, and then at a different rate for each additional minute. If a seven-minute call costs$4.25 and a 12-minute call costs $5.50, find each rate. 14. A plumber and a builder were employed to fit a new bath, each working a different number of hours. The plumber earns$35 per hour, and the builder earns $28 per hour. Together they were paid$330.75, but the plumber earned $106.75 more than the builder. How many hours did each work? 15. Paul has a part-time job selling computers at a local electronics store. He earns a fixed hourly wage, but he can earn a bonus by selling warranties for the computers he sells. He works 20 hours per week. In his first week, he sold eight warranties and earned$220. In his second week, he managed to sell 13 warranties and earned \$280. What is Paul’s hourly rate, and how much extra does he get for selling each warranty?
Mixed Review
1. Baxter the golden retriever is lying in the sun. He casts a shadow of 3 feet. The doghouse he is next to is 3 feet tall and casts an 8-foot shadow. What is Baxter's height?
2. A botanist watched the growth of a lily. At 3 weeks, the lily was 4 inches tall. Four weeks later, the lily was 21 inches tall. Assuming this relationship is linear:
1. Write an equation to show the growth pattern of this plant.
2. How tall was the lily at the 5.5-week mark?
3. Is there a restriction on how high the plant will grow? Does your equation show this?
3. The “Wave” is an exciting pastime at football games. To prepare, students in a math class took the data in the table below.
1. Find a linear regression equation for this data. Use this model to estimate the number of seconds it will take for 18 students to complete a round of the wave.
2. Use the method of interpolation to determine the amount of time it would take 18 students to complete the wave.
$& s \ (\text{number of students in wave})&& 4 && 8 && 12 && 16 && 20 && 24 && 28 && 30\\& t \ (\text{time in seconds to complete on full round}) && 2 && 3.2 && 4 && 5.6 && 7 && 7.9 && 8.6 && 9.1$
### Vocabulary Language: English
elimination
elimination
The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable. |
Demonstration to Sudoku Singles (Hidden Single, Naked Single, Full House)
Contents
 The Singles
Singles may be determined by deleting the numbers already present on the rows, columns, and blocks.
The following strategies provide the solution to most ‘simple’ riddles.
Sudoku Full House Technique
A house (row, column, nonet) is virtually full when only one unresolved cell remains.
It’s quite simple to notice but happens mostly when the grid is virtually solved.
This approach is generally used to produce or grade simple grids and is not helpful elsewhere. The cell is, in fact, both a concealed and naked single.
The last digit
All cells are put in the grid except for one cell for some digit.
Even without pencil lines, it is fairly straightforward to identify.
This approach is generally used to produce or rate simple grids and is not helpful elsewhere.
The cell is, in fact, a secret single for all of its dwellings.
Singles are the most fundamental and simple strategies for solving sudoku problems.
The Full House is the simplest and the greatest, to begin with among them.
The final digit that may be put in a house is a Full House.
A house is a complete collection of numbers ranging from 1 to 9. It might be a row, block, or column.
Here’s an example:
Take a look at the home indicated with the blue square; there is only one vacant cell left in this house, marked with a pink square.
After evaluating all of the numbers in the indicated home, we are left with just one unmistakable solution.
2 is the only number needed to finish the house.
It is frequently referred to as “Last Digit” if it is the last digit for the whole grid.
Row – Full House
Here we already have 8 digits occupying the 6th row, which indicates there must be just 1 number left to complete the house, in this instance, a row. So the missing number is two.
Column – Full House
This time, we’ll have to find the sixth column. We already know 8 of the digits, and a simple check down the column for the ones we already know indicates that the number 4 is our missing digit.
Box – Full House
We have almost finished a 3×3 box of 9 cells, with 8 of them already filled.
We need only one more number to make it a Full House.
Very immediately, we can tell that 9 goes in the final cell in the home.
Box-filled homes tend to be simpler to identify than rows or columns since we find it easier to concentrate on one location at a time rather than an entire row or column.
In addition, most strategies are simpler to master when they are contained inside a single box rather than scattered throughout the whole grid.
The Final Full House
The last number to be placed into the last empty cell always completes a row, a column, and a box.
The term “hidden single” refers to the fact that only one cell is left to insert that digit for a particular digit and house.
Because more than one candidate remains in the cell, the right digit is concealed among the others.
The Hidden Single is much less evil than its name indicates.
In reality, most Sudoku puzzles published in newspapers and magazines can only be solved using concealed singles simply.
Therefore, it is one of the singles and the Naked Single, the most basic problem-solving approach.
A Hidden Single is a single candidate for a given digit that remains in a row, column, or box.
Additional limitations in Sudoku variations may also result in Hidden Singles.
An alternate word is Pinned digit.
Evaluating 27 houses with each of the 9 numerals, for a total of 243 checks, may seem to be a difficult chore; however, most players do not check each of these combinations separately.
Instead, they use procedures such as Scanning and Cross-Hatching to find the concealed singles.
Consider the following in cell r3c4 of the example just above on it:
According to the rule, the numerals 4, 6, and 9 are all feasible.
However, looking attentively at row 3, we can see that digit 6 can only be inserted in r3c4.
In r2c3, digit 6 blocks r3c1, r3c2, and r3c3, whereas digit 6 blocks r3c6 in r6c6. That is, 6 may be inserted in r3c4.
One of multiple Hidden Singles may be identified in r6c4 in the right example: In row 6, column 4, and block 5, the lone 3 digits 3.
Naked Single’s
The Naked Single is classified as a problem-solving approach, yet it’s hard to call it that.
A naked single is what remains after you have eliminated all other candidates using your problem-solving skills.
In a cell, the final candidate standing is a naked single. Other phrases for this include Forced Digit and Sole Candidate.
The best approach to discover naked singles on the grid is using pencil marks.
Cells with single pencil marks are simple to detect.
Without pencil marks, you need to check the peers of the cell and determine whether there is just a single digit missing in the peers.
It would help if you had a solid short-term memory to perform it this way, although many players answer basic Sudokus without pencil marks.
Most players do not tell other players about their findings of singles. When a Sudoku can be solved with just singles, the problem as a whole is not worth debating.
Example
The digit 8 in the following Sudoku has two Naked Singles.
How to Locate Them
Finding Full Houses is simple explanatory.
Unfortunately, hidden and Naked Singles are not so simple to detect.
Nevertheless, they are an excellent illustration of why assigning a difficulty rating to techniques is so difficult: When playing with pencil and paper, Hidden Singles are simple to uncover, but viewing Naked Singles might take some time.
When playing using computer software such as HoDoKu that automatically keeps track of remaining available candidates, finding Naked Singles is straightforward, but locating Hidden Singles may be quite a headache due to the large number of candidates left in the grid (especially early on in the game).
Hidden Singles may be detected by “cross-hatching” while playing by hand: Concentrate on one number and one block at a time.
Then, draw a line across all rows and columns that already have that digit set (in your mind only, please).
If there is just one cell remaining, it is a Hidden Single for that digit.
The figure on the left depicts that procedure for digit 3 in block 5: All cells in that block are “blocked” by putting digit 3, and r6c4 is the last potential cell for digit 3 in block 5.
Another option is to use filters, as seen in the picture on the right.
It is simple to find Singles and Locked Candidate movements when filters are enabled.
Finding Naked Singles by Hand entails filling out all available possibilities or experimenting with promising cells (counting all digits in the cells seeing your intended target cell; if only one possibility remains, it is a Naked Single)
Hidden Singles in Line
Learn Also
Avoidable Rectangles
3d Medusa Sudoku Strategy |
# Mental Math Addition and Subtraction
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3 Skill 2: Subtracting 2 from an Even Number i) Finding the preceding one-digit even number: Name a one-digit even number and ask your students to give the preceding number in the sequence. For instance, the number that comes before 4 is 2 and the number that comes before 0 is 8. (Remember: the sequence is circular.) ii) Finding the preceding two-digit number: CASE 1: Numbers that end in 2, 4, 6, or 8 Write a two-digit number that ends in 2, 4, 6, or 8 on the board. Ask students to name the preceding even number. Students should recognize that if a number ends in 2, then the preceding even number ends in 0; if it ends in 4 then the preceding even number ends in 2, etc. For instance, the number 78 has ones digit 8 so the preceding even number has ones digit 6. Name the preceding even number: a) 48 : b) 26 : c) 34 : d) 62 : e) 78 : CASE 2: Numbers that end in 0 Write the number 80 on the board and ask your students to name the preceding even number. Students should recognize that if an even number ends in 0 then the preceding even number ends in 8 (but the ones digit is one less). So the even number that comes before 80 is 78. Name the preceding even number: a) 40 : b) 60 : c) 80 : d) 50 : e) 30 : ii) Subtracting 2 from an even number: Point out to your students that subtracting 2 from an even number is equivalent to finding the preceding even number: Example: 48 2 = 46, 46 2 = 44, etc. Subtract: a) 58-2 = b) 24-2 = c) 36-2 = d) 42-2 = e) 60-2 = A-32 Mental Math Teacher s Guide for AP Book 4.1
7 Mental Math Further Strategies 1. Your students should be able to explain how to use the strategies of rounding the subtrahend (the number you are subtracting) up to the nearest multiple of ten. Examples: a) = b) = c) = Subtrahend Subtrahend rounded to the nearest tens You must add 1 because 20 is 1 greater than 19 You must add 2 because 30 is 2 greater than 28 Practice Questions: a) = 27 + b) = 52 + c) = 76 + d) = 84 + e) = 61 + f) = 42 + NOTE: This strategy works well with numbers that end in 6, 7, 8, or Your students should be able to explain how to subtract by thinking of adding. Examples: Count by ones from 45 to the nearest tens (50) Count from 50 until you reach the first number (62) a) = = 17 b) = = 23 c) = = 56 Practice Questions: The sum of counting up to the nearest ten and the original number is the difference What method did we use here? a) = + = b) = + = c) = + = d) = + = e) = + = f) = + = 3. Your students should be able to explain how to use doubles. Examples: Minuend a) 12 6 = 6 If you add the subtrahend to itself and = 12 b) 8 4 = 4 the sum is equal to the minuend then the subtrahend is the same as the difference Practice Questions: a) 6 3 = b) 10 5 = c) 14 7 = d) 18 9 = e) 16 8 = f) = Same value as minuend Minuend plus itself A-36 Mental Math Teacher s Guide for AP Book 4.1
8 Mental Math Exercises Note to teacher: Teaching the material on these worksheets may take several lessons. Students will need more practice than is provided on these pages. These pages are intended as a test to be given when you are certain your students have learned the materials fully. Skills 1, 2, 3, and 4 1. Name the even number that comes after the number. Answer in the blank provided: a) 32 b) 46 c) 14 d) 92 e) 56 f) 30 g) 84 h) 60 i) 72 j) Name the even number that comes after the number: a) 28 b) 18 c) 78 d) 38 e) Add. Remember: adding 2 to an even number is the same as finding the next even number: a) = b) = c) = d) = e) = f) = g) = h) = i) = j) = k) = l) = 4. Name the even number that comes before the number: a) 38 b) 42 c) 56 d) 72 e) 98 f) 48 g) 16 h) 22 i) 66 j) Name the even number that comes before the number: a) 30 b) 70 c) 60 d) 10 e) Subtract. Remember: subtracting 2 from an even number is the same as finding the preceding even number: a) 46 2 = b) 86 2 = c) 90 2 = d) 14 2 = e) 54 2 = f) 72 2 = g) 12 2 = h) 56 2 = i) 32 2 = j) 40 2 = k) 60 2 = l) 26 2 = 7. Name the odd number that comes after the number: a) 37 b) 51 c) 63 d) 75 e) 17 f) 61 g) 43 h) 81 i) 23 j) Name the odd number that comes after the number: a) 69 b) 29 c) 9 d) 79 e) 59 Mental Math Teacher s Guide for AP Book 4.1 A-37
9 9. Add. REMEMBER: Adding 2 to an odd number is the same as finding the next odd number: a) = b) = c) = d) = e) = f) = g) = h) = i) = j) = k) = l) = 10. Name the odd number that comes before the number: a) 39 b) 43 c) 57 d) 17 e) 99 f) 13 g) 85 h) 79 i) 65 j) Name the odd number that comes before the number: a) 21 b) 41 c) 11 d) 91 e) Subtract. REMEMBER: Subtracting 2 from an odd number is the same as finding the preceding odd number. a) 47 2 = b) 85 2 = c) 91 2 = d) 15 2 = e) 51 2 = f) 73 2 = g) 11 2 = h) 59 2 = i) 31 2 = j) 43 2 = k) 7 2 = l) 25 2 = Skills 5 and Add 3 to the number by adding 2, then adding 1 (EXAMPLE: = ): a) = b) = c) = d) = e) = f) = g) = h) = i) = j) = k) = l) = 14. Subtract 3 from the number by subtracting 2, then subtracting 1 (EXAMPLE: 35 3 = ): a) 46 3 = b) 87 3 = c) 99 3 = d) 14 3 = e) 8 3 = f) 72 3 = g) 12 3 = h) 57 3 = i) 32 3 = j) 40 3 = k) 60 3 = l) 28 3 = 15. Fred has 49 stamps. He gives 2 stamps away. How many stamps does he have left? 16. There are 25 minnows in a tank. Alice adds 3 more to the tank. How many minnows are now in the tank? A-38 Mental Math Teacher s Guide for AP Book 4.1
10 Skills 7 and Add 4 to the number by adding 2 twice (EXAMPLE: = ): a) = b) = c) = d) = e) = f) = g) = h) = i) = j) = k) = l) = 18. Subtract 4 from the number by subtracting 2 twice (EXAMPLE: 26 4 = ): a) 46 4 = b) 86 4 = c) 91 4 = d) 15 4 = e) 53 4 = f) 9 4 = g) 13 4 = h) 57 4 = i) 40 4 = j) 88 4 = k) 69 4 = l) 31 4 = Skills 9 and Add 5 to the number by adding 4, then adding 1 (or add 2 twice, then add 1): a) = b) = c) = d) = e) = f) = g) = h) = i) = j) = k) = l) = 20. Subtract 5 from the number by subtracting 4, then subtracting 1 (or subtract 2 twice, then subtract 1): a) 48 5 = b) 86 5 = c) 55 5 = d) 69 5 = e) 30 5 = f) 13 5 = g) 92 5 = h) 77 5 = i) 45 5 = j) 24 5 = k) 91 5 = l) 8 5 = Skill Add: a) = b) = c) = d) = e) = f) = 22. Add by thinking of the larger number as a sum of two smaller numbers: a) = b) = c) = d) = e) = f) = Mental Math Teacher s Guide for AP Book 4.1 A-39
11 Mental Math Exercises Skills 12, 13, and a) = b) = c) = d) = e) = f) = g) = h) = 24. a) = b) = c) = d) = e) = f) = g) = h) = 25. a) = b) = c) = d) = e) = f) = g) = h) = i) = j) = k) = l) = 26. a) = b) = c) = d) = e) = f) = g) = h) = i) = j) = k) = l) = 27. a) = b) = c) = d) = e) = f) = g) = h) = i) = j) = k) = l) = Skills 15 and a) = b) = c) = d) = e) = f) = g) = h) = 29. a) = b) = c) = d) = e) = f) = g) = h) = Skills 17 and a) = b) = c) = d) = e) = f) = g) = h) = 31. a) = b) = c) = d) = e) = f) = g) = h) 17-8 = A-40 Mental Math Teacher s Guide for AP Book 4.1
15 Mental Math Checklist #1 Student Name Can Add 1 to Any Number Can Subtract 1 from Any Number Can Add 2 to Any Number Can Subtract 2 from Any Number Knows All Pairs that Add to 5 Can Double 1-Digit Numbers A-44 Mental Math Teacher s Guide for AP Book 4.1
16 Mental Math Checklist #2 Student Name Can Add Near Doubles. Example: = Can Add a 1-Digit Number to Any Multiple of 10. Example: = 36 Can Add Any 1-Digit Number to a Number Ending in 9. Example: = = 36 Can Add 1-Digit Numbers by Breaking them Apart into Pairs that Add to 10. Example: = = Can Subtract Any Multiple of 10 from 100. Example: = 60 Mental Math Teacher s Guide for AP Book 4.1 A-45
17 Mental Math Checklist #3 Student Name Can Mentally Make Change from a Dollar. See: AP Book Sheets on Money. Can Mentally Add Any Pair of 1-Digit Numbers. Can Mentally Subtract Any Pair of 1-Digit Numbers. Student Can Multiply and Count by: A-46 Mental Math Teacher s Guide for AP Book 4.1
18 Mental Math How to Learn Your Times Tables in a Week Trying to do math without knowing your times tables is like trying to play the piano without knowing the location of the notes on the keyboard. Your students will have difficulty seeing patterns in sequences and charts, solving proportions, finding equivalent fractions, decimals and percents, solving problems, etc. if they don t know their tables. Using the method below, you can teach your students their tables in a week or so. (If you set aside five or ten minutes a day to work with students who need extra help, the pay-off will be enormous.) There is really no reason for your students not to know their tables! Day 1: Counting by 2s, 3s, 4s, and 5s If you have completed the Introductory Unit Using Fractions you should already know how to count and multiply by 2s, 3s, 4s, and 5s. If you don t know how to count by these numbers you should memorize the hands below: If you know how to count by 2s, 3s, 4s, and 5s, then you can multiply by any combination of these numbers. For instance, to find the product 3 2, count by 2s until you have raised 3 fingers = 6 Day 2: The Nine Times Table The numbers you say when you count by 9s are called the multiples of 9 (zero is also a multiple of 9). The first ten multiples of 9 (after zero) are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90. What happens when you add the digits of any of these multiples of 9 (Example: or 6 + 3)? The sum is always 9! Here is another useful fact about the nine times table: Multiply 9 by any number between 1 and 10 and look at the tens digit of the product. The tens digit is always one less than the number you multiplied by: 9 4 = = = 18 3 is one less than 4 7 is one less than 8 1 is one less than 2 You can find the product of 9 and any number by using the two facts given above. For instance, to find 9 7, follow these steps: Step 1: 9 7 = 9 7 = 6 Subtract 1 from the number you are multiplying by 9: 7 1 = 6 Now you know the tens digit of the product. Mental Math Teacher s Guide for AP Book 4.1 A-47
19 Note 1. Make sure your students know how to subtract (by counting on their fingers if necessary) before you teach them the trick for the nine times table. 2. Give a test on Step 1 (above) before you move on. Step 2: 9 7 = = 6 3 These two digits add to 9. So the missing digit is 9 6 = 3 (You can do the subtraction on your fingers if necessary). Practise these two steps for all of the products of 9: 9 2, 9 3, 9 4, etc. Day 3: The Eight Times Table There are two patterns in the digits of the 8 times table. Knowing these patterns will help you remember how to count by 8s. Step 1: You can find the ones digit of the first five multiples of 8, by starting at 8 and counting backwards by 2s Step 2: You can find the tens digit of the first five multiples of 8, by starting at 0 and count up by 1s (Of course you don t need to write the 0 in front of the 8 for the product 1 8.) Step 3: You can find the ones digit of the next five multiples of 8 by repeating step 1: Step 4: You can find the remaining tens digits by starting at 4 and count up by 1s Practise writing the multiples of 8 (up to 80) until you have memorized the complete list. Knowing the patterns in the digits of the multiples of 8 will help you memorize the list very quickly. Then you will know how to multiply by 8: 8 6 = 48 Count by eight until you have 6 fingers up: 8, 16, 24, 32, 40, 48. A-48 Mental Math Teacher s Guide for AP Book 4.1
20 Day 4: The Six Times Table If you have learned the eight and nine times tables, then you already know 6 9 and 6 8. And if you know how to multiply by 5 up to 5 5, then you also know how to multiply by 6 up to 6 5! That s because you can always calculate 6 times a number by calculating 5 times the number and then adding the number itself to the result. The pictures below show why this works for 6 4: Plus one more Plus one more = = = = 24 Similarly: 6 2 = ; 6 3 = ; 6 5 = Knowing this, you only need to memorize 2 facts: ONE: 6 6 = 36 TWO: 6 7 = 42 Or, if you know 6 5, you can find 6 6 by calculating Day 5: The Seven Times Table If you have learned the six, eight, and nine times tables, then you already know: 6 7, 8 7, and 9 7. And since you also already know 1 7 = 7, you only need to memorize 5 facts: = = = = = 49 If you are able to memorize your own phone number, then you can easily memorize these 5 facts! NOTE: You can use doubling to help you learn the facts above. 4 is double 2, so 4 7 (= 28) is double 2 7 (= 14). 6 is double 3, so 6 7 (= 42) is double 3 7 (= 21). Try this test every day until you have learned your times tables: = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Mental Math Teacher s Guide for AP Book 4.1 A-49
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Thursday, July 18, 2024
# What Is The Inverse Property In Math
## Multiplicative Inverse Of Integers
Inverse property of multiplication | Arithmetic properties | Pre-Algebra | Khan Academy
Finding the multiplicative inverse of positive integers is the same as natural numbers . Just like positive integers, the product of a negative number and its reciprocal must be equal to 1. Thus, the multiplicative inverse of any negative number is its reciprocal. For example, × = 1, therefore, the multiplicative inverse of -6 is -1/6. Note that the multiplicative inverse of a negative number is always negative. And, in the multiplicative inverse of a negative integer, the negative sign will attach to the numerator, and not with the denominator.
## Additive Inverse And Multiplicative Inverse
There are two properties of numbers: multiplicative inverse and additive inverse property related to the multiplication and addition operation respectively. For a number x, – x is the additive inverse and 1/x is the multiplicative inverse. Let us understand the difference between additive inverse and multiplicative inverse with the help of the following table:
## What Is The Inverse Of A Number
A number can have two inverses. One inverse is the additive inverse, which is the value that when added with the original number will equal zero. Another inverse of a number is the multiplicative inverse, or reciprocal. When a reciprocal is multiplied by the original number, the product is always 1.
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## Using Properties Of Real Numbers
For some activities we perform, the order of certain operations does not matter, but the order of other operations does. For example, it does not make a difference if we put on the right shoe before the left or vice-versa. However, it does matter whether we put on shoes or socks first. The same thing is true for operations in mathematics.
The additive inverse of a number is its opposite number. If a number is added to its additive inverse, the sum of both the numbers becomes zero. The simple rule is to change the positive number to a negative number and vice versa. We know that, 7+ =0. Thus -7 is the additive inverse of 7 and 7 is the additive inverse of -7.
## Properties Of Inverse Operations
• The value, which, when added to the original number gives 0, is known as the additive inverse.
Suppose, x is the original number, then its additive inverse will be minus of x, i.e., $-$$\text, such that: \text = 0 For example, 6+=0. Hence, 6 is the additive inverse of 6 and vice versa. Suppose, -$$\text$ is the original number, then its additive inverse will be the positive value of x, i.e., x.
• Inverse Multiplication Property
• The value, which, when multiplied to the original number gives 1, is known as the multiplicative inverse.
Suppose, x is the original number, then its multiplicative inverse will be the reciprocal of $\text$, i.e.,$\frac}$, such that:
$\text\times\frac}=1$
For example, $6\times\frac=1$. Hence, $\frac$ is the multiplicative inverse of 6 and vice versa.
Suppose, $\frac$ is the original number, then its multiplicative inverse will be reciprocal of $\frac$, i.e., $x$.
For example: multiplicative inverse of $\frac$ is $\frac$.
## The Inverse Of A Trigonometric Function
The trigonometric functions are the functions that relate the right-angled triangle angle with the ratios of the side of the triangle.
Ex: sin = Opposite side/Hypotenuse. So, we find the angle of the triangle by using this formula. What if we have to calculate the hypotenuse of a triangle if an angle is given. Then we will use the inverse function to calculate the hypotenuse. So, the inverse is written as
= sin-1 is the inverse of the trigonometric sine function.
Similarly, cos = Base/Hypotenuse. So, inverse is = cos-1.
Tan = Opposite side/ Base. So, inverse is = tan-1 .
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## How Do You Explain Addition Meaning And Properties
Addition is commutative, meaning that one can change the order of the terms in a sum, but still get the same result. Symbolically, if a and b are any two numbers, then a + b = b + a. The fact that addition is commutative is known as the commutative law of addition or commutative property of addition.
## What Is The Difference Between Inverse And Identity Property
Inverse property of addition | Arithmetic properties | Pre-Algebra | Khan Academy
IdentityIdentityInverseInverse
Regarding this, what is a inverse property?
We use inverse properties to solve equations. Inverse Property of Addition says that any number added to its opposite will equal zero. Inverse Property of Multiplication says that any number multiplied by its reciprocal is equal to one.
Subsequently, question is, what is the inverse of adding? The Inverse of Adding is Subtracting Adding moves us one way, subtracting moves us the opposite way. Example: 20 + 9 = 29 can be reversed by 29 9 = 20
Thereof, what is an example of an inverse property?
The inverse property of multiplication states that the product of any number and its reciprocal is always 1. To find the reciprocal of a number, express this number as a fraction and flip the fraction. For example, the reciprocal of 4 would be 14. Example 1 7×17=?
What is the inverse property of subtraction?
The four main mathematical operations are addition, subtraction, multiplication, division. The inverse of addition is subtraction and vice versa. The inverse of multiplication is division and vice versa.
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## How To Find Multiplicative Inverse
The multiplicative inverse of a number is also known as its reciprocal. It is very easy to find the multiplicative inverse of a number using the following steps:
• Step 1: Divide the given number by 1.
• Step 2: Write it in the form of a fraction. Say, the reciprocal of a is 1/a.
• Step 3: Simplify and get the answer.
Let us find the multiplicative inverse of 2/3. The first step is to divide it by 1, which will result in 1/ = 3/2. Therefore, the reciprocal of 2/3 is 3/2.
• What is the additive inverse of 2/3?
We know that the given number and its additive inverses sum should be 0.
Let x be the additive inverse.
2/3 + x = 0
Hence, the additive inverse of 2/3 is -2/3.
• Solve for Q using the Additive Inverse Property.
Q + 5 = 8
We have to remove the +5. The inverse property of addition can be employed and -5 can be applied to both sides to get rid of it. -5 is the additive inverse . It is added to each side of the equation.
Therefore,
Q+5 -5 = 8 5
Q = 3.
• What is the multiplicative inverse of 1/5? Also, verify your answer.
The multiplicative inverse of 1/5 is 5/1.
We know the given number and its multiplicative inverses product should be 1.
To verify it, we will multiply 1/5 with its multiplicative inverse.
× = 1.
Hence, the multiplicative inverse of 1/5 is 5/1.
• Solve for X in the following equation using the inverse property of multiplication.
7X = 42
The multiplicative inverse property of the number positive natural number 7 can be used. This is to eliminate the +7 of the 7X. As we know, x = 1.
Proceed to multiply each side of the equation by the multiplicative inverse = 1/7.
Therefore,
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## Multiplicative Inverse Of A Mixed Fraction
To find the multiplicative inverse of a mixed fraction, convert the mixed fraction into an improper fraction, then determine its reciprocal. For example, let us find the multiplicative inverse of \.
• Step 1: Convert \ to an improper fraction, that is 7/2.
• Step 2: Find the reciprocal of 7/2, that is 2/7. Thus, the multiplicative inverse of \ is 2/7.
It is interesting to note that the multiplicative inverse of a mixed number is always a proper fraction whose value is less than 1.
## Properties Of Identity Inverses And Zero
• Recognize the identity properties of addition and multiplication
• Use the inverse properties of addition and multiplication
• Use the properties of zero
• Simplify expressions using the properties of identities, inverses, and zero
##### be prepared!
Before you get started, take this readiness quiz.
• Find the opposite of â4. If you missed this problem, review Example 3.1.3.
• Find the reciprocal of \. If you missed this problem, review Example 4.4.11.
• Multiply: \. If you missed this problem, review Example 4.3.9.
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## Multiplicative Inverse Of 0
As per the definition of multiplicative inverse, it is the number that when multiplied to the original number results in 1 as the product. But with 0, we know that the product of 0 with any number is always 0. So, the multiplicative inverse of 0 does not exist.
We can also understand this using the properties of division which states that the division of any number by 0 is not defined. The multiplicative inverse of 0 can be written as 1/0, but its value is not defined. So, it does not exist.
## Inverse Property Of Logarithms
It is the inverse of an exponent, according to the definition of a logarithm. As a result, the inverse of an exponential function is a logarithmic function. Recall what it means to be a functions inverse. When two inverses are added together, they equal x. As a result, if f=bx and g=logb x, then: fg=blog bx=x and gf=logb bx=x
These are known as the Inverse Logarithm Properties. Lets tackle the following issues. The Inverse Properties of Logarithms will be used.
1.Look for 10log56.
We can see from the first property that the bases cancel each other out.
56 eln6 eln2 = 10log56
Here, e and the natural log cancel out, leaving us with 6.2=12.
2.Look for log4 16x.
The second property will be used in this case. Also, change 16 to 4^2.
Log4 16x=log 4 x=log 4 42x=2x
3.Determine the inverse of f=2e^x-1.
Replace f with y. Then, reverse x and y.
y=2e^x1
x=2e^y1
We must now isolate the exponent and compute the logarithm of both sides. Also, to begin, divide by two.
x/2=e^y1
ln=ln e^y1
Remember the Inverse Properties of Logarithms we discussed previously in this concept? When we apply logb bx=x to the right side of our equation, we get . Determine the value of y.
ln=y1
As a result, ln+1 is the inverse of 2e^y-1.
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## Is Closure Property True For Division
Closure property is not true for division of rational numbers because of the number 1/2.
Can you give an example of an inverse property of addition?
Inverse Property Of Addition. The sum of a number and its negative is always zero. a + = 0. Example: 5 + = 0. Also 5 + = 0.
What is the definition of the additive inverse property?
The additive inverse property states that a number is added to its opposite number will result in zero. Learn about the definition of an inverse number, the graphical representation of the property, and see some examples for easier understanding.
## Example: Using Properties Of Real Numbers
Math Definitions : What Is the Multiplicative Inverse Property?
Use the properties of real numbers to rewrite and simplify each expression. State which properties apply.
• 3\cdot 6+3\cdot 4
• \dfrac\cdot \left
• \begin3\cdot6+3\cdot4 & =3\cdot\left & & \text \\ & =3\cdot10 & & \text \\ & =30 & & \text\end
\begin\left+\left & =5+\left & & \text \\ & =5+0 & & \text \\ & =5 & & \text\end
\begin6-\left & =6+ & & \text \\ & =6+\left & & \text \\ & =-18 & & \text\end
\begin\frac\cdot\left & =\frac \cdot\left & & \text \\ & =\left\cdot\frac & & \text \\ & =1\cdot\frac & & \text \\ & =\frac & & \text\end
\begin100\cdot & =100\cdot0.75+100\cdot\left & & \text \\ & =75+\left & & \text \\ & =-163 & & \text\end
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## What Does Inverse Property Mean In Math
Inverse property of addition tells us that any number + its opposite will = 0. Opposite numbers have different signs , but are the same distance from zero. For example: 6 + its opposite = 0. Or basically, 6 6 = 0. Another example: -8 + its opposite = 0.
What is the commutative property of addition?
The commutative property of addition says that changing the order of addends does not change the sum. Heres an example: 4 + 2 = 2 + 4 4 + 2 = 2 + 4 4+2=2+4.
#### What is the distributive property of addition?
The distributive property is a property of multiplication used in addition and subtraction. This property states that two or more terms in addition or subtraction with a number are equal to the addition or subtraction of the product of each of the terms with that number.
#### What is the inverse process of multiplication?
The division is the reverse process of multiplication. Dividing by a number is equivalent to multiplying by the reciprocal of the number. Thus, 7 ÷7=7 × 17 =1.
Which equation shows the inverse property of multiplication?
a+=0a is the additive inverse of a. a + = 0 a is the additive inverse of a . Inverse Property of Multiplication for any real number a0 a 0 , a1a=11ais the multiplicative inverse of a.
## Properties Of Real Numbers
Real numbers have unique properties, which makes them particularly useful in everyday life.
• Real numbers are an ordered set of numbers. This means real numbers are sequential. The numerical value of every real number fits between the numerical values of two other real numbers. Everyone is comfortable with this concept since all measurements depend upon the fact that some numbers have a greater value than other numbers. Ten is greater than five, and five is greater than four, and so on.
• We never run out of real numbers. The variety of real numbers available is not fixed. There is an infinite number of values available. The availability of numbers expands without end. Real numbers are not simply a finite row of separate points on a number line. There is always another real number whose value lies between any two real numbers.
• When real numbers are added or multiplied, the outcome is always another real number.
• With these three marks in mind, the question is:
• How can we practice real numbers in practical calculations?
• What rules employ?
• How should numbers be added, multiplied, subtracted, and divided?
• What scope do we have?
• Does it matter what we do first? second? third? fourth?
• Can we add a streak of numbers together in any order?
• Will the outcome be the same nevertheless of the order we choose?
• Can we multiply a streak of numbers together in any order?
• Will the concluding answer be the same nevertheless of the order we choose?
• Associative Property
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## How To Find The Additive Inverse
The additive inverse of any given number can be found by changing the sign of it. The additive inverse of a positive number will be a negative, whereas the additive inverse of a negative number will be positive. However, there will be no change in the numerical value except the sign. For example, the additive inverse of 8 is -8, whereas the additive inverse of -6 is 6.
## Inverse Property Of Numbers Definition Formula And Solved Examples
We will aim to understand the definition and formula of inverse property in this lesson. Its principles can be used to find additive inverse as well as multiplicative inverse in mathematical equations. This is best understood with the help of examples.
• Multiplicative Inverse Property Example
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## How To Find The Multiplicative Inverse
Consider that we have seven apples. To make them into groups of 1 each, we need to divide them by 7. Since division is the reverse process of multiplication, dividing by a number is equivalent to multiplying by the reciprocal of the number.
Thus, 7 ÷ 7 = 7 × $\frac$ = 1
Here, $\frac$ is called the multiplicative inverse of 7.
Let us understand the multiplicative inverse of different types of numbers like natural numbers, integers and fractions.
No, additive inverse and the additive identity property are not the same. The additive inverse of a given number is obtained by just reversing its sign. This means when the given number and its additive inverse are added we get 0. For example, the additive inverse of 4 is -4 = 0). Whereas, the additive identity of any given number is 0, because when we add any number to zero, it results in the number itself. For example, the additive identity of 4 is 0 .
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## What Is The Inverse Of 12
1/12The multiplicative inverse of 12 is 1/12.
Which is the inverse of multiplication?
The same number gets subtracted repeatedly. So, the division is the opposite of multiplication. Hence, multiplication and division are opposite operations. We may say, division is the inverse operation of multiplication.
What is the formula of inverse property?
The inverse property of multiplication states that if you multiply a number by its reciprocal, also called the multiplicative inverse, the product will be 1. *=1.
#### What does inverse property mean in math?
Inverse property of addition tells us that any number + its opposite will = 0. Opposite numbers have different signs , but are the same distance from zero. For example: 6 + its opposite = 0. Or basically, 6 6 = 0. Another example: -8 + its opposite = 0.
Whats the inverse of 3?
The multiplicative inverse of 3 is 1/3.
Is division the inverse of multiplication?
Because division is the inverse, or opposite, of multiplication, you can use arrays to help students understand how multiplication and division are related. If in multiplication we find the product of two factors, in division we find the missing factor if the other factor and the product are known. |
# 2011 AIME II Problems/Problem 8
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$. For each $j$, let $w_j$ be one of $z_j$ or $i z_j$. Then the maximum possible value of the real part of $\displaystyle\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$.
## Solution
The twelve dots above represent the $12$ roots of the equation $z^{12}-2^{36}=0$. If we write $z=a+bi$, then the real part of $z$ is $a$ and the real part of $iz$ is $-b$. The blue dots represent those roots $z$ for which the real part of $z$ is greater than the real part of $iz$, and the red dots represent those roots $z$ for which the real part of $iz$ is greater than the real part of $z$. Now, the sum of the real parts of the blue dots is easily seen to be $8+16\cos\frac{\pi}{6}=8+8\sqrt{3}$ and the negative of the sum of the imaginary parts of the red dots is easily seen to also be $8+8\sqrt{3}$. Hence our desired sum is $16+16\sqrt{3}=16+\sqrt{768}$, giving the answer $\boxed{784}.$
## Solution 2
The equation $z^{12}-2^{36}=0$ can be factored as follows: $$(z^6-2^{18})(z^6+2^{18})=0$$ $$(z^2-2^6)(z^2+2^6)({(z^2+2^6)}^2-z^2\cdot2^6)({(z^2-2^6)}^2+z^2\cdot2^6)=0$$ $$(z^2-2^6)(z^2+2^6)(z^2+2^6-z\cdot2^3)(z^2+2^6+z\cdot2^3) (z^2-2^6-iz\cdot2^3)(z^2-2^6+i z\cdot2^3)=0$$
Since this is a 12th degree equation, there are 12 roots. Also, since each term in the equation is even, the positive or negative value of each root is another root. That would mean there are 6 roots that can be multiplied by $-1$ and since we have 6 factors, that’s 1 root per factor. We just need to solve for $z$ in each factor and pick whether or not to multiply by $i$ and $-1$ for each one depending on the one that yields the highest real value. After that process, we get $8+8+2((4\sqrt{3}+4)+(4\sqrt{3}-4))$ Adding the values up yields $16+16\sqrt{3}$, or $16+\sqrt{768}$, and $16+768=\boxed{784}$.
-Solution by Someonenumber011. |
# 180 Days of Math for Sixth Grade Day 51 Answers Key
By accessing our 180 Days of Math for Sixth Grade Answers Key Day 51 regularly, students can get better problem-solving skills.
## 180 Days of Math for Sixth Grade Answers Key Day 51
Directions Solve each problem.
Question 1.
56 – 39 = ___
56-39 = 17.
Explanation:
By subtracting 39 in 56 we will get 17.
Question 2.
9 × 40 = ______
9 × 40 = 360.
Explanation:
By multiplying 9 and 40 we will get 9 × 40 = 360.
Question 3.
42÷7 = 6.
Explanation:
By dividing 42 by 7 we will get 42÷7 = 6.
Question 4.
What is the value of the digit 9 in 4,729?
_______________________
9 is in one’s place.
Explanation:
The value of the digit 9 in 4,729 is one’s place.
Question 5.
Is 1$$\frac{2}{3}$$ less than or greater than 1$$\frac{3}{4}$$?
_______________________
1$$\frac{2}{3}$$ less than 1$$\frac{3}{4}$$.
Explanation:
GIven 1$$\frac{2}{3}$$ which is $$\frac{5}{3}$$ and in decimal 1.67 and 1$$\frac{3}{4}$$ which is $$\frac{7}{4}$$ and in decimals it will be 1.75. So 1$$\frac{2}{3}$$ is less than 1$$\frac{3}{4}$$.
Question 6.
4 + 2 • 5 + 3 = ____
4 + 2 • 5 + 3 = 48.
Explanation:
Given the expression is 4 + 2 • 5 + 3, so
4 + 2 • 5 + 3 = 6 × 8
= 48.
Question 7.
% of 80 is 20.
25% of 80 is 20.
Explanation:
Let the missing % be X, so X% of 80 is 20
X% × 80 = 20
$$\frac{X}{100}$$ × 80 = 20
$$\frac{4X}{5}$$ = 20
4X = 20×5
X = $$\frac{20×5}{4}$$
= 25.
Question 8.
Solve for f. f + $$\frac{3}{8}$$ = $$\frac{5}{8}$$ f = ___
f = $$\frac{1}{4}$$.
Explanation:
Given that f + $$\frac{3}{8}$$ = $$\frac{5}{8}$$,
f = $$\frac{5}{8}$$ – $$\frac{3}{8}$$
= $$\frac{5-3}{8}$$
= $$\frac{2}{8}$$
= $$\frac{1}{4}$$.
Question 9.
How many pints are in a quart?
_______________________
2 pints in 1 quarts.
Explanation:
There are 2 pints in 1 quarts.
Question 10.
Which line is perpendicular to line B?
_______________________
Answer: line G is perpendicular to line B.
Question 11.
Record the data in the table. Trish has 2 brothers and 1 sister. Max doesn’t have any siblings. Melissa has 2 brothers. Allison has 1 sister. Trent has a brother and a sister.
Child’s Name Brothers Sisters Trish 2 1 Max 0 0 Melissa 2 0 Allison 0 1 Trent 1 1
Question 12.
A farmer sold 250 of his sheep, bought 35 more sheep, and then bought 68 more sheep. After 3 died, he sold another 260. If he now has 190 sheep, how many sheep did he have at the beginning?
_______________________ |
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# What is the Prime Factorization Of 60?
• Equcation for number 60 factorization is: 2 * 2 * 3 * 5
• It is determined that the prime factors of number 60 are: 2, 3, 5
## Is 60 A Prime Number?
• No the number 60 is not a prime number.
• Sixty is a composite number. Because 60 has more divisors than 1 and itself.
## How To Calculate Prime Number Factorization
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## Mathematical Information About Numbers 6 0
• About Number 6. Six is the smallest composite number with two distinct prime factors, and the third triangular number. It is the smallest perfect number: 6 = 1 + 2 + 3 and the faculty of 3 is 6 = 3! = 1 * 2 * 3, which is remarkable, because there is no other three numbers whose product is equal to their sum. Similarly 6 = sqrt(1 ^ 3 + 2 + 3 ^ 3 ^ 3). The equation x ^ 3 + Y ^ 3 ^ 3 + z = 6xyz is the only solution (without permutations) x = 1, y = 2 and z = 3. Finally 1/1 = 1/2 + 1/3 + 1/6. The cube (from the Greek) or hexahedron (from Latin) cube is one of the five Platonic solids and has six equal areas. A tetrahedron has six edges and six vertices an octahedron. With regular hexagons can fill a plane without gaps. Number six is a two-dimensional kiss number.
• About Number 0. The number zero is the number of elements in an empty collection of objects, mathematically speaking, the cardinality of the empty set. Zero in mathematics by depending on the context variously defined objects, but often can be identified with each other, that is considered to be the same object, which combines several properties compatible with each other. As cardinal numbers (number of elements in a set) are identified with special ordinals, and the zero is just the smallest cardinal number is zero - elected as the first ordinal - in contrast to common parlance. As finite cardinal and ordinal it is depending on the definition often counted among the natural numbers. The zero is the identity element for addition in many bodies, such as the rational numbers, real numbers and complex numbers, and a common name for a neutral element in many algebraic structures, even if other elements are not identified with common numbers. Zero is the only real number that is neither positive nor negative.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers.
## What is Prime Number Factorization?
• In mathematics, factorization (also factorisation in some forms of British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. For example, the number 15 factors into primes as 3 x 5, and the polynomial x2 - 4 factors as (x - 2)(x + 2). In all cases, a product of simpler objects is obtained. The aim of factoring is usually to reduce something to basic building blocks, such as numbers to prime numbers, or polynomials to irreducible polynomials.
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# Examples of Higher Order Derivatives
Example: Find the second derivative ${y_2}$ if $y = \cos \left( {ax + b} \right)$.
We have the given function $y = \cos \left( {ax + b} \right)$
Differentiating both sides with respect to $x$, we have
$\begin{gathered} \frac{{dy}}{{dx}} = \frac{d}{{dx}}\cos \left( {ax + b} \right) \\ \Rightarrow {y_1} = – \sin \left( {ax + b} \right)\frac{d}{{dx}}\left( {ax + b} \right) \\ \Rightarrow {y_1} = – a\sin \left( {ax + b} \right) \\ \end{gathered}$
Again, differentiating both sides with respect to $x$, we have
$\begin{gathered} \frac{{d{y_1}}}{{dx}} = – a\frac{d}{{dx}}\sin \left( {ax + b} \right) \\ \Rightarrow {y_2} = – a\cos \left( {ax + b} \right)\frac{d}{{dx}}\left( {ax + b} \right) \\ \Rightarrow {y_2} = – {a^2}\cos \left( {ax + b} \right) \\ \end{gathered}$
Example: If $y = a\cos x + b\sin x$, then show that $\frac{{{d^2}y}}{{d{x^2}}} + y = 0$
We have the given function
$y = a\cos x + b\sin x$
Differentiating both sides with respect to $x$, we have
$\begin{gathered} \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {a\cos x + b\sin x} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = a\frac{d}{{dx}}\cos x + b\frac{d}{{dx}}\sin x \\ \Rightarrow \frac{{dy}}{{dx}} = – a\sin x + b\cos x \\ \end{gathered}$
Again, differentiating both sides with respect to $x$, we have
$\begin{gathered} \frac{{{d^2}y}}{{d{x^2}}} = – a\frac{d}{{dx}}\sin x + b\frac{d}{{dx}}\cos x \\ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = – a\cos x – b\sin x \\ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = – \left( {a\cos x + b\sin x} \right) \\ \end{gathered}$
Using the value of the given function $y = a\cos x + b\sin x$ in the above equation, we have
$\begin{gathered} \frac{{{d^2}y}}{{d{x^2}}} = – y \\ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} + y = 0 \\ \end{gathered}$ |
# Lecture 15
When there is a number $b$ such that $ab\equiv 1\pmod{m}$, we call it the inverse of $a$, modulo $m$ (and we say that $a$ is invertible). We write $a^{-1}$ for the inverse of $a$.
Notice that, as a consequence modular arithmetic modulo a prime $p$ is fantastically well-behaved: any nonzero residue $a\not\equiv 0\pmod{p}$ has an inverse (since we have $\gcd(a,p)=1$ unless $a$ is a multiple of $p$).
Spotting inverses modulo $m$ is quite difficult; in general the best way is to use Euclid’s algorithm.
There are a few exceptions:
• The inverse of $1$ modulo $m$ is always
$1$.
• The inverse of $-1$ modulo $m$ is always
$-1$.
• If $m$ is odd, then $2$ is invertible modulo $m$, because $\gcd(m,2)=1$. The inverse is:
$(m+1)/2$.
Two other fairly easy, but useful, facts are as follows:
#### Proposition
If $a$ is invertible modulo $m$, then so is $a^{-1}$, with inverse given by $(a^{-1})^{-1} \equiv a\pmod{m}$.
#### Proof
We have $aa^{-1}\equiv 1\pmod{m}$, which says that $a$ is an inverse for $a^{-1}$.
#### Proposition
If $a$ and $b$ are both invertible, then $ab$ is too, with inverse given by $(ab)^{-1} \equiv b^{-1}a^{-1}\pmod{m}.$
#### Proof
We have $(ab)b^{-1}a^{-1} \equiv aa^{-1}bb^{-1} \equiv 1\cdot 1\equiv 1\pmod{m}$.
As a big example of all of this, let’s find an inverse for $37$, modulo $100$. We want $x$ with $37x\equiv 1\pmod{100}$. In other words, we seek a solution to $37x+100k=1$ in the integers. We’ll get one from working through Euclid’s algorithm: \begin{aligned} {100} & {= 2\times 37+26}& {37} & {= 1\times 26 + 11}\\ {26} & {= 2\times 11+4}& {11} & {= 2\times 4+3}\\ {4} & {= 1\times3+1}& {3} & {= 3\times1.}\end{aligned} So we have that \begin{aligned} {1 = 1\times4-1\times3} & {= 1\times4-1\times(11-2\times 4)}\\ {= 3\times4-1\times11} & {= 3\times(26-2\times11)-1\times11}\\ {= 3\times26-7\times11} & {= 3\times26-7\times(37-26)}\\ {= 10\times26-7\times37} & {= 10\times(100-2\times37)-7\times37}\\ {= 10\times100-27\times37.}\end{aligned} That means that $(-27)\times 37\equiv 1\pmod{100}$, so the inverse of $37$ is $-27$, which is congruent to $73$ (mod $100$).
### Checking our working
And, of course, we can check this easily: $37\times 73 = 2701\equiv 1\pmod{100}$ as claimed.
## The Chinese Remainder Theorem
We’ve come to understand congruence equations: given something like $123x \equiv 456\pmod{789},$ we can, with some effort, turn it into something nice like $x\equiv 132\pmod{263}.$
Now we’ll discuss a different sort of problem with congruences: what if we have two of them for the same number? For example, \begin{aligned} x &\equiv 1\pmod{4}\\ x &\equiv 3\pmod{7}?\end{aligned} These things happen all the time: two things happening periodically with different periods.
And it turns out we can solve them using exactly the same machinery as we’ve been using all along. Indeed, these equations say that \begin{aligned} x-1 &= 4a\\ x-3 &= 7b,\end{aligned} for some numbers $a$ and $b$.
That means that $1+4a=3+7b,$ or in other words $4a-7b=2$. We have lots of experience solving these, and, since $\gcd(4,7)=1$, it’s possible.
A solution to $4a-7b=1$ is given by $a=2$, $b=1$, and so a solution to $4a-7b=2$ is given by doubling that to get $a=4$, $b=2$.
What’s the general solution? Well, if we have $4a-7b=2$, then subtracting $4\times4-7\times2=2$ gives $4(a-4)-7(b-2)=0.$ This means that $7\mid 4(a-4)$, so $7\mid(a-4)$. Hence $a$ is of the form $7k+4$. and in fact any such $a$ works.
Now, we had $x=4a+1$, which in turn is $28k+17$. In other words: $x\equiv 17\pmod{28}.$
There need not always be solutions to simultaneous congruences. For example, the simultaneous congruences \begin{aligned} x &\equiv 4 \pmod{6}\\ x &\equiv 3 \pmod{8}\end{aligned} don’t have solutions. Why is this obvious?
The first equation implies $x$ even, the second $x$ odd.
Of course, if we go through the same solution process as above it will fail. We set \begin{aligned} x &= 4 + 6a\\ x &= 3 + 8b\end{aligned} and find that $4+6a = 3+8b$, and hence $8b-6a=1$. This has no solutions because $\gcd(8,6)=2$, and $2\nmid 1$. |
# How do you solve using gaussian elimination or gauss-jordan elimination, 3w-x=2y + z -4, 9x-y + z =10, 4w+3y-z=7, 12x + 17=2y-z+6?
Feb 18, 2016
$w = - \frac{183}{735}$
$x = - \frac{111}{49}$
$y = \frac{27297}{1470}$
$z = \frac{71967}{1470}$
#### Explanation:
Gaussian elimination involves the gradual elimination of variables from the set of equations until values can be found for each variable.
$3 w - x = 2 y + z - 4$
$9 x - y + z = 10$
$4 w + 3 y - z = 7$
$12 x + 17 = 2 y - z + 6$
First reorder the terms in all equations so that they all have the same order of variables.
$3 w - x - 2 y - z = - 4$
$9 x - y + z = 10$
$4 w + 3 y - z = 7$
$12 x - 2 y + z = - 11$
Adding the first two equations together gives
$3 w + 8 x - 3 y = 6$
Adding the second pair of equations gives
$4 w + 12 x + y = - 4$
Adding the middle pair of equations gives
$4 w + 9 x + 2 y = 17$
In a similar process, add three times the middle equation to the first equation
$15 w + 44 x = - 6$
Take twice the middle equation from the third equation
$- 4 w - 15 x = 9$
Now add four times the first equation to fifteen times the second equation
$176 x - 225 x = - 24 + 135$
$- 49 x = 111$
$x = - \frac{111}{49}$
Substitute this back into one of the final pair of equations to get $w$
$15 w = - 6 - \left(- \frac{111}{49}\right) = - \frac{183}{49}$
$w = - \frac{183}{735}$
$2 y = 17 - 4 \left(- \frac{183}{735}\right) - 9 \left(- \frac{111}{49}\right) = \frac{12495 - 183 + 14985}{735}$
$y = \frac{27297}{1470}$
$z = 10 - 9 x + y = 10 - 9 \left(- \frac{111}{49}\right) + \frac{27297}{1470}$
$= \frac{14700 + 29970 + 27297}{1470} = \frac{71967}{1470}$ |
# Divide a 4 digit number by a 1 digit number
Lesson
## Partitioning and renaming digits
When we want to divide a large number, such as 4402 (a four digit number) by a single digit number, such as 4, there are some techniques that can help us. We've explored some of these with smaller two and three digit numbers, so you may wish to have a look at those topics first.
In the first video, we look at partitioning, or breaking our number into chunks, to work on smaller problems, then add our answers together at the end.
## The long division algorithm
If you find long division tricky, then watching the next video might help you see it is actually a series of steps. By working through each step, the problem becomes more manageable, and it turns out we are performing a series of smaller divisions. The main tool we use here is renaming, when we are unable to share. As an example, 600 can be renamed as 60 tens.
Take a look, and see how these steps are the same as what you have already been doing.
Remember!
Partitioning ,- or breaking your number into smaller chunks - can help with division of larger numbers
Renaming digits can help to divide (share) numbers
#### Examples
##### Question 1
Calculate $4000\div2$4000÷2 by doing the following.
1. $4\div2$4÷2
2. $40\div2$40÷2
3. $400\div2$400÷2
4. $4000\div2$4000÷2
##### Question 2
We're going to break $7130$7130 into $6000+1000+120+10$6000+1000+120+10 to calculate $7130\div2$7130÷2.
1. Calculate $6000\div2$6000÷2.
2. Calculate $1000\div2$1000÷2.
3. Calculate $120\div2$120÷2.
4. Calculate $10\div2$10÷2.
5. Using the fact that $7130=6000+1000+120+10$7130=6000+1000+120+10, calculate $7130\div2$7130÷2.
##### Question 3
Evaluate $1477\div7$1477÷7. |
# 3Tvwo §9.5 Power functions
§9.5 Power functions
This paragraph and the next
- §9.6 Calculating with powers -
However, this LessonUp is only about §9.5.
1 / 16
Slide 1: Slide
WiskundeMiddelbare schoolvwoLeerjaar 3
This lesson contains 16 slides, with interactive quiz and text slides.
Lesson duration is: 50 min
## Items in this lesson
§9.5 Power functions
This paragraph and the next
- §9.6 Calculating with powers -
However, this LessonUp is only about §9.5.
#### Slide 1 -Slide
Task 1: Imagine what the graph for function f looks like.
So try to see it as clearly as possible, in your head!
Compare to the next slide.
f(x)=x2
#### Slide 2 -Slide
Solution 1:
This is it! Remember?
A parabola.
It is symmetrical around the y axis.
#### Slide 3 -Slide
Now we have the formula:
g(x)=x3
g(4)=....
g(4)=....
g(4)=.....
g(3)=......
g(x)=x3
and
g(4) = .....
#### Slide 5 -Mind map
Solution 2:
= -4 x -4 x -4 = - 64
= 4 x 4 x 4 = 64
g(x)=x3
g(4)=....
g(4)=....
#### Slide 6 -Slide
Imagine what the graph for:
looks like.
Answer is in the next slide.
g(x)=x3
h(x)=x4
h(2)
h(2)
#### Slide 10 -Slide
Solution 4. Given the function:
= -2 x -2 x -2 x -2 = 16 and
= 2 x 2 x 2 x 2 = 16 too!
h(x)=x4
h(2)
h(2)
#### Slide 11 -Slide
Once more:
imagine what shape the graph for
has!
Solution next slide.
h(x)=x4
#### Slide 13 -Slide
In the next slide ....
the previous content is summarized by a
'Theory'.
#### Slide 15 -Slide
Homework time.
Work on §9.5 for 10 to 15 minutes.
SKIP p.70!
Then we continue with a LessonUp about
§9.6 CALCULATING WITH POWERS. |
# 1650 Wrinkles in the Multiplication Table
### Today’s Puzzle:
Are you familiar with the book A Wrinkle in Time? Kat of The Lily Cafe’s blog loves books and recently compared Meg in that book to her six-year-old son. She wrote a post titled Am I Raising a Meg? Her six-year-old LOVES math and is very much interested in multiplication and division. When Mom thought he was playing a game on her phone, he was actually playing with the calculator app! I felt so happy inside as I read that!
I wonder if they have discovered the storybooks in the Math Book Magic blog. Such books could combine Mom’s love for reading with her son’s love of math.
Someday her son might like to solve a “wrinkled” multiplication table puzzle like this one that has only nine clues.
Write all the numbers 1 to 10 in both the first column and the top row so that those numbers and the given clues become a multiplication table.
### Factor Cake for 1650:
This is my 1650th post.
1650 is divisible by 2 and by 5 because it ends with a 0.
1650 is divisible by 3 because 1 + 6 + 5 + 0 = 12, a number divisible by 3.
1650 is divisible by 11 because 1 – 6 + 5 – 0 = 0, a number divisible by 11.
I think we can make a lovely factor cake for 1650:
### Factors of 1650:
• 1650 is a composite number.
• Prime factorization: 1650 = 2 × 3 × 5 × 5 × 11, which can be written 1650 = 2 × 3 × 5² × 11.
• 1650 has at least one exponent greater than 1 in its prime factorization so √1650 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1650 = (√25)(√66) = 5√66.
• The exponents in the prime factorization are 1, 1, 2, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(2 + 1)(1 + 1) = 2 × 2 × 3 × 2 = 24. Therefore 1650 has exactly 24 factors.
• The factors of 1650 are outlined with their factor pair partners in the graphic below.
### More About the Number 1650:
1650 is the hypotenuse of TWO Pythagorean triples:
462-1584-1650, which is (7-24-25) times 66, and
990-1320-1650, which is (3-4-5) times 330. |
# How to calculate variance in Excel - VAR, VAR.S, VAR.P and other functions
In this tutorial, we will look at how to do variance analysis Excel and what formulas to use to find variance of a sample and population.
Variance is one of the most useful tools in probability theory and statistics. In science, it describes how far each number in the data set is from the mean. In practice, it often shows how much something changes. For example, temperature near the equator has less variance than in other climate zones. In this article, we will analyze different methods of calculating variance in Excel.
## What is variance?
Variance is the measure of variability of a data set that indicates how far different values are spread. Mathematically, it is defined as the average of the squared differences from the mean.
To better understand what you are actually calculating with the variance, please consider this simple example.
Supposing there are 5 tigers in your local zoo that are 14, 10, 8, 6 and 2 years old.
To find variance, follow these simple steps:
1. Calculate the mean (simple average) of the five numbers:
2. From each number, subtract the mean to find the differences. To visualize this, let's plot the differences on the chart:
3. Square each difference.
4. Work out the average of the squared differences.
So, the variance is 16. But what does this number actually mean?
In truth, variance just gives you a very general idea of the dispersion of the data set. A value of 0 means there is no variability, i.e. all the numbers in the data set are the same. The bigger the number, the more spread out the data.
This example is for population variance (i.e. 5 tigers are the whole group you are interested in). If your data is a selection from a bigger population, then you need to calculate sample variance by using a slightly different formula.
## How to calculate variance in Excel
There are 6 built-in functions to do variance in Excel: VAR, VAR.S, VARP, VAR.P, VARA, and VARPA.
Your choice of the variance formula is determined by the following factors:
• The version of Excel you are using.
• Whether you calculate sample or population variance.
• Whether you want to evaluate or ignore text and logical values.
### Excel variance functions
The below table provides an overview of the variation functions available in Excel to help you choose the formula best suited for your needs.
Name Excel version Data type Text and logicals VAR 2000 - 2019 Sample Ignored VAR.S 2010 - 2019 Sample Ignored VARA 2000 - 2019 Sample Evaluated VARP 2000 - 2019 Population Ignored VAR.P 2010 - 2019 Population Ignored VARPA 2000 - 2019 Population Evaluated
### VAR.S vs. VARA and VAR.P vs. VARPA
VARA and VARPA differ from other variance functions only in the way they handle logical and text values in references. The following table provides a summary of how text representations of numbers and logical values are evaluated.
Argument Type VAR, VAR.S, VARP, VAR.P VARA & VARPA Logical values within arrays and references Ignored Evaluated (TRUE=1, FALSE=0) Text representations of numbers within arrays and references Ignored Evaluated as zero Logical values and text representations of numbers typed directly into arguments Evaluated (TRUE=1, FALSE=0) Empty cells Ignored
## How to calculate sample variance in Excel
A sample is a set of data extracted from the entire population. And the variance calculated from a sample is called sample variance.
For example, if you want to know how people's heights vary, it would be technically unfeasible for you to measure every person on the earth. The solution is to take a sample of the population, say 1,000 people, and estimate the heights of the whole population based on that sample.
Sample variance is calculated with this formula:
Where:
• x̄ is the mean (simple average) of the sample values.
• n is the sample size, i.e. the number of values in the sample.
There are 3 functions to find sample variance in Excel: VAR, VAR.S and VARA.
### VAR function in Excel
It is the oldest Excel function to estimate variance based on a sample. The VAR function is available in all versions of Excel 2000 to 2019.
VAR(number1, [number2], …)
Note. In Excel 2010, the VAR function was replaced with VAR.S that provides improved accuracy. Although VAR is still available for backward compatibility, it is recommended to use VAR.S in the current versions of Excel.
### VAR.S function in Excel
It is the modern counterpart of the Excel VAR function. Use the VAR.S function to find sample variance in Excel 2010 and later.
VAR.S(number1, [number2], …)
### VARA function in Excel
The Excel VARA function returns a sample variance based on a set of numbers, text, and logical values as shown in this table.
VARA(value1, [value2], …)
### Sample variance formula in Excel
When working with a numeric set of data you can use any of the above functions to calculate sample variance in Excel.
As an example, let's find the variance of a sample consisting of 6 items (B2:B7). For this, you can use one of the below formulas:
`=VAR(B2:B7)`
`=VAR.S(B2:B7)`
`=VARA(B2:B7)`
As shown in the screenshot, all the formulas return the same result (rounded to 2 decimal places):
To check the result, let's do var calculation manually:
1. Find the mean by using the AVERAGE function:
`=AVERAGE(B2:B7)`
The average goes to any empty cell, say B8.
2. Subtract the average from each number in the sample:
`=B2-\$B\$8`
The differences go to column C, beginning in C2.
3. Square each difference and put the results to column D, beginning in D2:
`=C2^2`
4. Add up the squared differences and divide the result by the number of items in the sample minus 1:
`=SUM(D2:D7)/(6-1)`
As you can see, the result of our manual var calculation is exactly the same as the number returned by Excel's built-in functions:
If your data set contains the Boolean and/or text values, the VARA function will return a different result. The reason is that VAR and VAR.S ignore any values other than numbers in references, while VARA evaluates text values as zeros, TRUE as 1, and FALSE as 0. So, please carefully choose the variance function for your calculations depending on whether you want to process or ignore text and logicals.
## How to calculate population variance in Excel
Population is all members of a given group, i.e. all observations in the field of study. Population variance describes how data points in the entire population are spread out.
The population variance can be found with this formula:
Where:
• x̄ is the mean of the population.
• n is the population size, i.e. the total number of values in the population.
There are 3 functions to calculate population variance in Excel: VARP, VAR.P and VARPA.
### VARP function in Excel
The Excel VARP function returns the variance of a population based on the entire set of numbers. It is available in all versions of Excel 2000 to 2019.
VARP(number1, [number2], …)
Note. In Excel 2010, VARP was replaced with VAR.P but is still kept for backward compatibility. It is recommended to use VAR.P in the current versions of Excel because there is no guarantee that the VARP function will be available in future versions of Excel.
### VAR.P function in Excel
It is an improved version of the VARP function available in Excel 2010 and later.
VAR.P(number1, [number2], …)
### VARPA function in Excel
The VARPA function calculates the variance of a population based on the entire set of numbers, text, and logical values. It is available in all version of Excel 2000 through 2019.
VARA(value1, [value2], …)
### Population variance formula in Excel
In the sample var calculation example, we found a variance of 5 exam scores assuming those scores were a selection from a bigger group of students. If you collect data on all the students in the group, that data will represent the entire population, and you will calculate a population variance by using the above functions.
Let's say, we have the exam scores of a group of 10 students (B2:B11). The scores constitute the entire population, so we will do variance with these formulas:
`=VARP(B2:B11)`
`=VAR.P(B2:B11)`
`=VARPA(B2:B11)`
And all the formulas will return the identical result:
To make sure Excel has done the variance right, you can check it with the manual var calculation formula shown in the screenshot below:
If some of the students did not take the exam and have N/A instead of a score number, the VARPA function will return a different result. The reason is that VARPA evaluates text values as zeros while VARP and VAR.P ignore text and logical values in references. Please see VAR.P vs. VARPA for full details.
## Variance formula in Excel - usage notes
To do variance analysis in Excel correctly, please follow these simple rules:
• Provide arguments as values, arrays, or cell references.
• In Excel 2007 and later, you can supply up to 255 arguments corresponding to a sample or population; in Excel 2003 and older - up to 30 arguments.
• To evaluate only numbers in references, ignoring empty cells, text, and the logical values, use the VAR or VAR.S function to calculate sample variance and VARP or VAR.P to find population variance.
• To evaluate logical and text values in references, use the VARA or VARPA function.
• Provide at least two numeric values to a sample variance formula and at least one numeric value to a population variance formula in Excel, otherwise a #DIV/0! error occurs.
• Arguments containing text that cannot be interpreted as numbers cause #VALUE! errors.
## Variance vs. standard deviation in Excel
Variance is undoubtedly a useful concept in science, but it gives very little practical information. For instance, we found the ages of the population of tigers in a local zoo and calculated the variance, which equals 16. The question is - how can we actually use this number?
You can use variance to work out standard deviation, which is a much better measure of the amount of variation in a data set.
Standard deviation is calculated as the square root of the variance. So, we take the square root of 16 and get the standard deviation of 4.
In combination with the mean, the standard deviation can tell you how old most of the tigers are. For example, if the mean is 8 and the standard deviation is 4, the majority of the tigers in the zoo are between 4 years (8 - 4) and 12 years (8 + 4).
Microsoft Excel has special functions for working out standard deviation of a sample and population. The detailed explanation of all the functions can be found in this tutorial: How to calculate standard deviation in Excel.
That's how to do variance in Excel. To have a closer look at the formulas discussed in this tutorial, you are welcome to download our sample workbook to Calculate Variance in Excel. I thank you for reading and hope to see you on our blog next week!
## You may also be interested in
Category: Excel Tips
### 10 responses to "How to calculate variance in Excel - VAR, VAR.S, VAR.P and other functions"
1. Don Rector says:
Thank you for the very useful and easy to understand lesson.
2. Rahul says:
Is there a way of calculating the variance in excel of a discrete probability distribution with just specifying the X values array & probability distribution array?
Thank you. You've been very helpful.
4. Richard says:
I must have made an error with the age of tigers in the zoo example because my manually calculated variation matched your calculation of 16, but my Excel formula results were:
VAR = 20 {=VAR(Cell:cell)} even when {=VAR(14,10,8,6,2)} is used I still get 20.
VAR.S = 20
VARA = 20
VAR.P =16
Mean (avg) = 8
Standard Dev. = 4
• Richard says:
The only reason I can deduce that I got 16 manually vice 20, as I did at first, is because I divided by 5 (population variance = n) instead of 4 (sample variance = n-1). So, I assume that this is a key difference to consider when selecting which Excel formula to use with a set.
5. Uche says:
A very nice tips.
Thanks
6. LR says:
Very useful.
7. J says:
Variance uses n-1 as a denominator, not n.
8. Anirudh K says: |
# Difference between revisions of "2017 AIME I Problems/Problem 4"
## Problem 4
A pyramid has a triangular base with side lengths $20$, $20$, and $24$. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$. The volume of the pyramid is $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.
## Solution
Let the triangular base be $\triangle ABC$, with $\overline {AB} = 24$. We find that the altitude to side $\overline {AB}$ is $16$, so the area of $\triangle ABC$ is $(24*16)/2 = 192$.
Let the fourth vertex of the tetrahedron be $P$, and let the midpoint of $\overline {AB}$ be $M$. Since $P$ is equidistant from $A$, $B$, and $C$, the line through $P$ perpendicular to the plane of $\triangle ABC$ will pass through the circumcenter of $\triangle ABC$, which we will call $O$. Note that $O$ is equidistant from each of $A$, $B$, and $C$. Then,
$$\overline {OM} + \overline {OC} = \overline {CM} = 16$$
Let $\overline {OM} = d$. Equation $(1)$: $$d + \sqrt {d^2 + 144} = 16$$
Squaring both sides, we have
$$d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256$$
$$2d^2 + 2d\sqrt {d^2+144} = 112$$
$$2d(d + \sqrt {d^2+144}) = 112$$
Substituting with equation $(1)$:
$$2d(16) = 112$$
$$d = 7/2$$
We now find that $\sqrt{d^2 + 144} = 25/2$.
Let the distance $\overline {OP} = h$. Using the Pythagorean Theorem on triangle $AOP$, $BOP$, or $COP$ (all three are congruent by SSS):
$$25^2 = h^2 + (25/2)^2$$
$$625 = h^2 + 625/4$$
$$1875/4 = h^2$$
$$25\sqrt {3} / 2 = h$$
Finally, by the formula for volume of a pyramid,
$$V = Bh/3$$
$$V = (192)(25\sqrt{3}/2)/3$$ This simplifies to $V = 800\sqrt {3}$, so $m+n = \boxed {803}$.
NOTE : If you don’t know or remember the formula for the volume of a triangular pyramid, you can derive it using calculus as follows :
Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that $\frac{A_{small element}}{A} = \frac{h^2}{H^2} \implies A_{small element} = \frac{Ah^2}{H^2}$. Now integrate it taking the limits $0$ to $H$
## Shortcut
Here is a shortcut for finding the radius $R$ of the circumcenter of $\triangle ABC$.
As before, we find that the foot of the altitude from $P$ lands on the circumcenter of $\triangle ABC$. Let $BC=a$, $AC=b$, and $AB=c$. Then we write the area of $\triangle ABC$ in two ways: $$[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}$$
Plugging in $20$, $20$, and $24$ for $a$, $b$, and $c$ respectively, and solving for $R$, we obtain $R= \frac{25}{2}=OA=OB=OC$.
Then continue as before to use the Pythagorean Theorem on $\triangle AOP$, find $h$, and find the volume of the pyramid.
## Solution 2 (Coordinates)
We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length $24$ is at the origin, or $(0, 0, 0)$. Then, the two other vertices can be $(-12, -16, 0)$ and $(12, -16, 0)$. Let the fourth vertex have coordinates of $(x, y, z)$. We have the following $3$ equations from the distance formula.
$$x^2+y^2+z^2=625$$
$$(x+12)^2+(y+16)^2+z^2=625$$
$$(x-12)^2+(y+16)^2+z^2=625$$
Adding the last two equations and substituting in the first equation, we get that $y=-\frac{25}{2}$. If you drew a good diagram, it should be obvious that $x=0$. Now, solving for $z$, we get that $z=\frac{25\sqrt{3}}{2}$. So, the height of the pyramid is $\frac{25\sqrt{3}}{2}$. The base is equal to the area of the triangle, which is $\frac{1}{2} \cdot 24 \cdot 16 = 192$. The volume is $\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}$. Thus, the answer is $800+3 = \boxed{803}$.
-RootThreeOverTwo
## Solution 3 (Heron's Formula)
Label the four vertices of the tetrahedron and the midpoint of $\overline {AB}$, and notice that the area of the base of the tetrahedron, $\triangle ABC$, equals $192$, according to Solution 1.
Notice that the altitude of $\triangle CPM$ from $\overline {CM}$ to point $P$ is the height of the tetrahedron. Side $\overline {PM}$ is can be found using the Pythagorean Theorem on $\triangle APM$, giving us $\overline {PM}=\sqrt{481}.$
Using Heron's Formula, the area of $\triangle CPM$ can be written as $$\sqrt{\frac{41+\sqrt{481}}{2}(\frac{41+\sqrt{481}}{2}-16)(\frac{41+\sqrt{481}}{2}-25)(\frac{41+\sqrt{481}}{2}-\sqrt{481})}$$ $$=\frac{\sqrt{(41+\sqrt{481})(9+\sqrt{481})(-9+\sqrt{481})(41-\sqrt{481})}}{4}$$
Notice that both $(41+\sqrt{481})(41-\sqrt{481})$ and $(9+\sqrt{481})(-9+\sqrt{481})$ can be rewritten as differences of squares; thus, the expression can be written as $$\frac{\sqrt{(41^2-481)(481-9^2)}}{4}=\frac{\sqrt{480000}}{4}=100\sqrt{3}.$$
From this, we can determine the height of both $\triangle CPM$ and tetrahedron $ABCP$ to be $\frac{100\sqrt{3}}{8}$; therefore, the volume of the tetrahedron equals $\frac{100\sqrt{3}}{8} \cdot 192=800\sqrt{3}$; thus, $m+n=800+3=\boxed{803}.$
-dzhou100
## Video Solution
https://youtu.be/Mk-MCeVjSGc ~Shreyas S |
# The gambler’s ruin problem
This post discusses the problem of the gambler’s ruin. We start with a simple illustration.
Two gamblers, A and B, are betting on the tosses of a fair coin. At the beginning of the game, player A has 1 coin and player B has 3 coins. So there are 4 coins between them. In each play of the game, a fair coin is tossed. If the result of the coin toss is head, player A collects 1 coin from player B. If the result of the coin toss is tail, player A pays player B 1 coin. The game continues until one of the players has all the coins (or one of the players loses all his/her coins). What is the probability that player A ends up with all the coins?
Intuitively, one might guess that player B is more likely to win all the coins since B begins the game with more coin. For example, player A in the example has only 1 coin to start. Thus there is a 50% chance that he/she will lose everything on the first toss. On the other hand, player B has a cushion since he/she can afford to lose some tosses at the beginning.
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Simulated Plays of the Game
One way to get a sense of the winning probability of a player is to play the game repeatedly. Rather than playing with real money, why not simulate a series of coin tosses using random numbers generated in a computer? For example, the function =RAND() in Excel generates random numbers between 0 and 1. If the number is less than 0.5, we take it as a tail (T). Otherwise it is a head (H). As we simulate the coin tosses, we add or subtract to each player’s account accordingly. If the coin toss is a T, we subtract 1 from A and add 1 to B. If the coin toss is an H, we add 1 to A and subtract 1 from B. Here’s a simulation of the game.
$\begin{array}{ccccccc} \text{Game} & \text{ } & \text{Coin Tosses} & \text{ } & \text{Coins of Player A} & \text{ } & \text{Coins of Player B} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ \text{ } & \text{ }& \text{ } & \text{ } & 1 & & 3 \\ 1 & \text{ }& H & \text{ } & 2 & & 2 \\ \text{ } & \text{ }& T & \text{ } & 1 & & 3 \\ \text{ } & \text{ }& H & \text{ } & 2 & & 2 \\ \text{ } & \text{ }& H & \text{ } & 3 & & 1 \\ \text{ } & \text{ }& H & \text{ } & 4 & & 0 \\ \end{array}$
In the first simulation, player A got lucky with 4 heads in 5 tosses. The following shows the next simulation.
$\begin{array}{ccccccc} \text{Game} & \text{ } & \text{Coin Tosses} & \text{ } & \text{Coins of Player A} & \text{ } & \text{Coins of Player B} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ \text{ } & \text{ }& \text{ } & \text{ } & 1 & & 3 \\ 2 & \text{ }& H & \text{ } & 2 & & 2 \\ \text{ } & \text{ }& H & \text{ } & 3 & & 1 \\ \text{ } & \text{ }& T & \text{ } & 2 & & 2 \\ \text{ } & \text{ }& H & \text{ } & 3 & & 1 \\ \text{ } & \text{ }& H & \text{ } & 4 & & 0 \\ \end{array}$
Player A seems to be on a roll. Here’s the results of the next two simulated games.
$\begin{array}{ccccccc} \text{Game} & \text{ } & \text{Coin Tosses} & \text{ } & \text{Coins of Player A} & \text{ } & \text{Coins of Player B} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ \text{ } & \text{ }& \text{ } & \text{ } & 1 & & 3 \\ 3 & \text{ }& T & \text{ } & 0 & & 4 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ \text{ } & \text{ }& \text{ } & \text{ } & 1 & & 3 \\ 4 & \text{ }& H & \text{ } & 2 & & 2 \\ \text{ } & \text{ }& T & \text{ } & 1 & & 3 \\ \text{ } & \text{ }& T & \text{ } & 0 & & 4 \\ \end{array}$
Now player A loses both of these games. For good measure, we show one more simulation.
$\begin{array}{ccccccc} \text{Game} & \text{ } & \text{Coin Tosses} & \text{ } & \text{Coins of Player A} & \text{ } & \text{Coins of Player B} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ \text{ } & \text{ }& \text{ } & \text{ } & 1 & & 3 \\ 5 & \text{ }& H & \text{ } & 2 & & 2 \\ \text{ } & \text{ }& H & \text{ } & 3 & & 1 \\ \text{ } & \text{ }& T & \text{ } & 2 & & 2 \\ \text{ } & \text{ }& H & \text{ } & 3 & & 1 \\ \text{ } & \text{ }& T & \text{ } & 2 & & 2 \\ \text{ } & \text{ }& H & \text{ } & 3 & & 1 \\ \text{ } & \text{ }& T & \text{ } & 2 & & 2 \\ \text{ } & \text{ }& T & \text{ } & 1 & & 3 \\ \text{ } & \text{ }& T & \text{ } & 0 & & 4 \\ \end{array}$
This one is a little more drawn out. Several tosses of tail in a row spells bad news for player A. When a coin is tossed long enough, there bounds to be some runs of tails. Since it is a fair coin, there bounds to be runs of head too, which benefit player A. Wouldn’t the runs of H cancel out the runs of T so that each player wins about half the time? To get an idea, we continue with 95 more simulations (to get a total of 100). Player A wins 23 of the simulated games and player B wins 77 of the games. So player A wins about 25% of the times. Note that A owns about 25% of the coins at the start of the game.
The 23 versus 77 in the 100 simulated games may be just due to luck for player B. We simulate 9 more runs of 100 games each (for a total of 10 runs with 100 games each). The following shows the results.
Ten Simulated Runs of 100 Games Each
$\begin{array}{ccccccc} \text{Simulated} & \text{ } & \text{Player A} & \text{ } & \text{Player B} & \text{ } & \text{Total} \\ \text{Runs} & \text{ } & \text{Winning} & \text{ } & \text{Winning} & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{Frequency} & \text{ } & \text{Frequency} & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ 1 & \text{ }& 23 & \text{ } & 77 & & 100 \\ 2 & \text{ }&34 & \text{ } & 66 & & 100 \\ 3 & \text{ }& 19 & \text{ } & 81 & & 100 \\ 4 & \text{ }& 24 & \text{ } & 76 & & 100 \\ 5 & \text{ }& 27 & \text{ } & 73 & & 100 \\ 6 & \text{ }& 22 & \text{ } & 78 & & 100 \\ 7 & \text{ }& 21 & \text{ } & 79 & & 100 \\ 8 & \text{ }& 24 & \text{ } & 76 & & 100 \\ 9 & \text{ }& 28 & \text{ } & 72 & & 100 \\ 10 & \text{ }& 24 & \text{ } & 76 & & 100 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ \text{Total} & \text{ }& 246 & \text{ } & 754 & & 1000 \\ \end{array}$
The simulated runs fluctuate quite a bit. Player A wins any where from 19% to 34% of the times. The results are no where near even odds of winning for player A. In the 1,000 simulated games, player A wins 24.6% of the times, roughly identical to the player A’s share of the coins at the beginning. If the simulation is carried many more times (say 10,000 times or 100,000 times), the overall winning probability for player A would be very close to 25%.
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Gambler’s Ruin Problem
Here’s a slightly more general description of the problem of gambler’s ruin.
Gambler’s Ruin Problem
Two gamblers, A and B, are betting on the tosses of a fair coin. At the beginning of the game, player A has $n_1$ coins and player B has $n_2$ coins. So there are $n_1+n_2$ coins between them. In each play of the game, a fair coin is tossed. If the result of the coin toss is head, player A collects 1 coin from B. If the result of the coin toss is tail, player A pays B 1 coin. The game continues until one of the players has all the coins. What is the probability that player A ends up with all the coins? What is the probability that player B ends up with all the coins?
Solution: Gambler’s Ruin Problem
The long run probability of player A winning all the coins is $\displaystyle P_A=\frac{n_1}{n_1+n_2}$.
The long run probability of player B winning all the coins is $\displaystyle P_B=\frac{n_2}{n_1+n_2}$.
In other words, the probability of winning for a player is the ratio of the number of coins the player starts with to the total numbers of coins. Turning it around, the probability of a player losing all his/her coins is the ratio of the number of coins of the other player to the total number of coins. Thus the player with the fewer number of coins at the beginning has a greater chance of losing everything. Think a casino. Say, it has 10,000,000 coins. You, as a gambler, have 100 coins. Then you would have a 99.999% chance of losing all your coins.
The 99.999% chance of losing everything is based on the assumption that all bets are at even odds. This is certainly not the case in a casino. The casino enjoys a house edge. So the losing probabilities would be worse for a gambler playing in a casino.
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Deriving the Probabilities
We derive the winning probability for a more general case. The coin that is tossed does not have to be a fair coin. Let $p$ be the probability that the coin toss results in a head. Let $q=1-p$. Suppose that there are $n$ coins initially between player A and player B.
Let $A_i$ be the probability that player A will own all the $n$ coins when player A starts the game with $i$ coins and player B starts with $n-i$ coins. On the other hand, let $B_i$ be the probability that player B will own all the $n$ coins when player A starts the game with $i$ coins and player B starts with $n-i$ coins.
The goal here is to derive expressions for both $A_i$ and $B_i$ for $i=1,2,\cdots,n-1$ and to show that $A_i+B_i=1$. The last point, though seems like common sense, is not entirely trivial. It basically says that the game will end in finite number of moves (it cannot go on indefinitely).
Another point to keep in mind is that we can assume $A_0=0$ and $A_n=1$ as well as $B_0=0$ and $B_n=1$. In the following derivation, $H$ is the event that the first toss is a head and $T$ is the event that the first toss is a tail.
Let $E$ be the event that player A owning all the coins when player A starts the game with $i$ coins and player B starts with $n-i$ coins, i.e. $A_i=P(E)$. First condition on the outcome of the first toss of the coin.
\displaystyle \begin{aligned} A_i&=P(E | H) \ P(H)+P(E | T) \ P(T) \\&=P(E | H) \ p+P(E | T) \ q \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{aligned}
Consider the conditional probabilities $P(E | H)$ and $P(E | T)$. Note that $P(E | H)=A_{i+1}$. If the first toss is a head, then player A has $i+1$ coins and player B would have $n-(i+1)$ coins. At that point, the probability of player A owning all the coins would be the same as the probability of winning as if the game begins with player A having $i+1$ coins. Similarly, $P(E | T)=A_{i-1}$. Plug these into (1), we have:
$A_i=p \ A_{i+1}+q \ A_{i-1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
Further derivation can be made on equation (2).
$A_i=p \ A_i+ q \ A_i=p \ A_{i+1}+q \ A_{i-1}$
$p \ A_{i+1}-p \ A_i=q \ A_i - q \ A_{i-1}$
$\displaystyle A_{i+1}-A_{i}=\frac{q}{p} \ (A_i-A_{i-1}),\ \ \ \ i=1,2,3,\cdots,n-1$
Plugging in the values $1,2,3,\cdots,n-1$ for $i$ produces the following $n-1$ equations.
\displaystyle \begin{aligned} &A_{2}-A_{1}=\frac{q}{p} \ (A_1-A_{0})=\frac{q}{p} \ A_1 \\&A_{3}-A_{2}=\frac{q}{p} \ (A_2-A_{1})=\bigg[ \frac{q}{p} \biggr]^2 \ A_1 \\&\ \ \ \cdots \\&A_{i}-A_{i-1}=\frac{q}{p} \ (A_{i-1}-A_{i-2})=\bigg[ \frac{q}{p} \biggr]^{i-1} \ A_1 \\&\ \ \ \cdots \\&A_{n}-A_{n-1}=\frac{q}{p} \ (A_{n-1}-A_{n-2})=\bigg[ \frac{q}{p} \biggr]^{n-1} \ A_1 \end{aligned}
Adding the first $i-1$ equations produces the following equations.
$\displaystyle A_i-A_1=A_1 \ \biggl[\frac{q}{p}+\bigg( \frac{q}{p} \biggr)^{2}+\cdots+\bigg( \frac{q}{p} \biggr)^{i-1} \biggr]$
$\displaystyle A_i=A_1 \ \biggl[1+\frac{q}{p}+\bigg( \frac{q}{p} \biggr)^{2}+\cdots+\bigg( \frac{q}{p} \biggr)^{i-1} \biggr] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$
There are two cases to consider. Either $\frac{q}{p}=1$ or $\frac{q}{p} \ne 1$. The first case means $p=\frac{1}{2}$ and the second case means $p \ne \frac{1}{2}$. The following is an expression for $A_i$ for the two cases.
$\displaystyle A_i = \left\{ \begin{array}{ll} \displaystyle i \ A_1 &\ \ \ \ \ \ \displaystyle \frac{q}{p}=1 \\ \text{ } & \text{ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4) \\ \displaystyle \frac{ \displaystyle 1- \biggl(\frac{q}{p} \biggr)^i}{\displaystyle 1-\frac{q}{p}} A_1 &\ \ \ \ \ \ \displaystyle \frac{q}{p} \ne 1 \end{array} \right.$
The above two-case expression for $A_i$ is applicable for $i=2,3,\cdots,n$. Plugging $n$ into $i$ with $A_n=1$ gives the following expression for $A_1$:
$\displaystyle A_1 = \left\{ \begin{array}{ll} \displaystyle \frac{1}{n} &\ \ \ \ \ \ p=\frac{1}{2} \\ \text{ } & \text{ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) \\ \displaystyle \frac{\displaystyle 1-\frac{q}{p}}{1-\biggl( \displaystyle\frac{q}{p} \biggr)^n} &\ \ \ \ \ \ p \ne \frac{1}{2} \end{array} \right.$
Plugging $A_1$ from (5) into (4) gives the following:
$\displaystyle A_i = \left\{ \begin{array}{ll} \displaystyle \frac{i}{n} &\ \ \ \ \ \ p=\frac{1}{2} \\ \text{ } & \text{ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6) \\ \displaystyle \frac{1-\biggl( \displaystyle \frac{q}{p} \biggr)^i}{1-\biggl( \displaystyle \frac{q}{p} \biggr)^n} &\ \ \ \ \ \ p \ne \frac{1}{2} \end{array} \right.$
The expression for $A_i$ in (6) is applicable for $i=1,2,\cdots,n$. For the case of $p=\frac{1}{2}$, the probability of player A owning all the coin is the ratio of the initial amount of coins of player A to the total number of coins, identical to what is stated in the previous section. Recall that $B_i$ is the probability that player B will end up owning all the coins when initially player A has $i$ coins and player B has $n-i$ coins. By symmetry, the following gives the expression for $B_i$.
$\displaystyle B_i = \left\{ \begin{array}{ll} \displaystyle \frac{n-i}{n} &\ \ \ \ \ \ q=\frac{1}{2} \\ \text{ } & \text{ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7) \\ \displaystyle \frac{1- \biggl(\displaystyle \frac{p}{q} \biggr)^{n-i}}{\displaystyle 1- \biggl(\frac{p}{q} \biggr)^n} &\ \ \ \ \ \ q \ne \frac{1}{2} \end{array} \right.$
For the case $p=\frac{1}{2}$, which is equivalent to $q=\frac{1}{2}$, it is clear that $A_i+B_i=1$. For the case $p \ne \frac{1}{2}$, the following shows that $A_i+B_i=1$.
\displaystyle \begin{aligned} A_i+B_i&=\frac{1-\biggl( \displaystyle \frac{q}{p} \biggr)^i}{1-\biggl( \displaystyle \frac{q}{p} \biggr)^n}+\frac{1- \biggl(\displaystyle \frac{p}{q} \biggr)^{n-i}}{\displaystyle 1- \biggl(\frac{p}{q} \biggr)^n} \\&\text{ } \\&=\frac{p^n-p^n\biggl( \displaystyle \frac{q}{p} \biggr)^i}{p^n-q^n}+\frac{q^n- q^n \biggl(\displaystyle \frac{p}{q} \biggr)^{n-i}}{\displaystyle q^n- p^n}\\&\text{ } \\&=\frac{p^n-p^{n-i} \ q^i}{p^n-q^n}-\frac{q^n- q^i \ p^{n-i}}{\displaystyle p^n- q^n} \\&\text{ } \\&=\frac{p^n-q^n}{p^n-q^n}=1 \end{aligned}
The fact that $A_i+B_i=1$ shows that the gambling game discussed here has to end after certain number of plays and that it will not continue indefinitely.
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Further Discussion
The formulas derived here are quite profound, not to mention impactful and informative for anyone who gambles. More calculations and discussion can be found here in a companion blog on statistics.
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$\copyright$ 2017 – Dan Ma
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All from 9 and the Last from 10
The ten’s complement of a number tells us how far it is from the next higher power of 10. In our example, this is obtained by subtracting 847,536 from 1,000,000.
Quickly finding this 10’s complement is important in many arithmetic computations especially in “completing the whole” which is the subject of our Math Tip last week.
This is done by using the Vedic Math sutra Nikhilam Navatashcaramam Dashatah meaning All from 9 and the Last from 10. Instead of using the traditional method of putting the subtrahend 847,536 below the minuend 1,000,000 and subtracting from right to left which would require 6 “borrowings” or regroupings, we can proceed from left to right by subtraction every digit of the subtrahend from 9 except the last digit which we will subtract from 10.
Using this method, we can immediately announce the ten’s complement of any number without using pen and paper or calculators. Just by looking at the figure 847,536, we can say that the correct answer is 4) 152, 464.
Here is the step by step mental process:
1. Look at subtrahend: it is eight hundred forty seven thousand, five hundred thirty six.
2. The difference is 8 from 9 or one hundred.
3. 4 from 9 or fifty
4. 7 from 9 or two thousand,
5. 5 from 9 or four hundred
6. 3 from 9 or sixty
7. the last, 6 from 10 or four.
This Nikhilam method is also very useful in addition and subtraction of decimals, including money calculations and in base multiplication and division.
Perform the following subtraction using “All from 9 and the last from 10”
1. 1,000 – 456 =
2. 10,000 – 3,859 =
3. 100,000 – 83,942 =
4. 1,000,000 – 598,274 =
5. 1 – 0.24 =
6. 1 – 0.789 =
7. 1 – 0.4587 =
8. 1 – 0.68942 =
9. P100 – 24.75 =
10. P1,000 – 786.38 =
Answers will be published in the next post.
Here are the answers for last week’s exercises in Completing the whole:
1. 8 + 9 + 2 = (8 +2) + 9 = 19
2. 6 + 7 + 4 + 9 + 3 = (6 + 4) + (7 + 3) + 9 = 29
3. 11 + 32 + 89 = (11 + 89) + 32 = 132
4. 24 + 38 + 50 + 226 = (24 + 226) + 50 + 38 = 338
5. 17 + 38 + 79 + 45 = [(17 + 38) + 45] + 79 = 179
6. 3/5 + 3/4 + 2/5 = (3/5 + 2/5) + 3/ 4 = 1 3/ 4
7. 58 min + 57 min + 56 Min = (58 + 2) + (57 + 3) + (56 – 2 – 3) = 2 hours 51 mins
8. 3 ft 5 in + 4 ft 6 in + 5 ft 7 in = (3 ft 5 in + 5 ft 7 in) + 4 ft 6 in = 13 ft 6 in.
9. 9999 + 999 + 99 + 9 + 5 = (9999 + 1) + (999+1) + (99 + 1) + (9 + 1) + (5 – 4) = 11,111
10. 1 3/4 + 2 4/5 + 3 2/8 = (1 3/ 4 + 3 2/8) + 2 4/5 = 7 4/5 |
# Proof by Induction
### Proof by Induction
###### Proof by Induction
$$1 + 2 + 3 + ... + n = \dfrac{1}{2} n(n + 1)$$
STEP 1
For n = 1:
\begin{equation*} \begin{split} 1 & = \tfrac{1}{2} (1)(1 + 1) \\\\ 1 & = 1 \\\\ \text{LHS} & = \text{RHS} \end{split} \end{equation*}
The statement is true for n = 1
STEP 2
Assume the statement is true for n = k:
\begin{equation*} 1 + 2 + 3 + ... + k = \tfrac{1}{2} k(k + 1) \end{equation*}
So, for n = k + 1:
\begin{equation*} 1 + 2 + 3 + ... + k + {\color {magenta} (k + 1)} = \tfrac{1}{2} (k + 1) (k + 2) \end{equation*}
\begin{equation*} \begin{split} & \text{LHS} \\\\ & 1 + 2 + 3 + ... + k + (k + 1) \quad {\color {blue} \rightarrow 1 + 2 + 3 + ... + k = \tfrac{1}{2} k(k + 1)}\\\\ & \tfrac{1}{2} k(k + 1) + (k + 1) \\\\ & (k + 1) ( \tfrac{1}{2} k + 1) \\\\ & \tfrac{1}{2} (k + 1) (k + 2) \\\\ & \text{LHS} = \text{RHS} \end{split} \end{equation*}
If the statement is true for n = k, then it is true for n = k + 1
STEP 3
Based on step 1 and 2, then:
The statement is true for n = 1, then the statement is true for n = 2
The statement is true for n = 2, then the statement is true for n = 3
and so on,
The statement is true for n ≥ 1
##### Exercise
Proof the statements below by mathematical induction
$$1 + 8 + 15 + \dotso + (7n - 6) = \dfrac{1}{2} (7n^2 - 5n)$$
$$1 + 8 + 15 + \dotso + (7n - 6) = \dfrac{1}{2} (7n^2 - 5n)$$
Solution:
STEP 1
For n = 1:
\begin{equation*} \begin{split} 1 & =\tfrac{1}{2} [7(1)^2 - 5(1)] \\\\ 1 & = 1\\\\ \text{LHS} & = \text{RHS} \end{split} \end{equation*}
The statement is true for n = 1
STEP 2
Assume the statement is true for n = k:
\begin{equation*} 1 + 8 + 15 + \dotso + (7k - 6) = \tfrac{1}{2} (7k^2 - 5k) \end{equation*}
So, for n = k + 1:
\begin{equation*} \begin{split} 1 + 8 + 15 + \dotso + (7k - 6) + {\color {magenta} [7(k + 1) - 6]} & = \tfrac{1}{2} [7(k + 1)^2 - 5(k + 1)] \\\\ & = \tfrac{1}{2} [7(k^2 + 2k + 1) - 5k - 5] \\\\ & = \tfrac{1}{2} [7k^2 + 9k + 2] \end{split} \end{equation*}
\begin{equation*} \begin{split} & \text{LHS} \\\\ & 1 + 8 + 15 + \dotso + (7k - 6) + [7(k + 1) - 6] \quad {\color {blue} \rightarrow 1 + 8 + 15 + ... + (7k - 6) = \tfrac{1}{2} [7(k + 1)^2 - 5(k + 1)]}\\\\ & \tfrac{1}{2} (7k^2 - 5k) + [7(k + 1) - 6] \\\\ & \tfrac{1}{2} (7k^2 - 5k) + 7k + 1 \\\\ & \tfrac{1}{2}(7k^2 - 5k + 14k + 2 \\\\ & \tfrac{1}{2}(7k^2 + 9k + 2) \\\\ & \tfrac{1}{2}(7k^2 + 9k + 2) \\\\ & \text{LHS} = \text{RHS} \end{split} \end{equation*}
If the statement is true for n = k, then it is true for n = k + 1
STEP 3
Based on step 1 and 2, then:
The statement is true for n = 1, then the statement is true for n = 2
The statement is true for n = 2, then the statement is true for n = 3
and so on,
The statement is true for n ≥ 1
$$\dfrac{1}{1! \:.\: 3} + \dfrac{1}{2! \:.\: 4} + \dfrac{1}{3! \:.\: 5} + ... + \dfrac{1}{n! \: (n + 2)} = \dfrac{1}{2} - \dfrac{1}{(n + 2)!}$$
$$\dfrac{1}{1! \:.\: 3} + \dfrac{1}{2! \:.\: 4} + \dfrac{1}{3! \:.\: 5} + ... + \dfrac{1}{n! \: (n + 2)} = \dfrac{1}{2} - \dfrac{1}{(n + 2)!}$$
Solution:
STEP 1
For n = 1:
\begin{equation*} \begin{split} \frac{1}{1! \:.\: 3} & = \frac{1}{2} - \frac{1}{(1 + 2)!} \\\\ \frac{1}{3} & =\frac{1}{2} - \frac{1}{3!} \\\\ \frac{1}{3} & =\frac{1}{2} - \frac{1}{3 \:.\: 2 \:.\: 1} \\\\ \frac{1}{3} & =\frac{1}{2} - \frac{1}{6} \\\\ \frac{1}{3} & = \frac{1}{3}\\\\ LHS & = RHS \end{split} \end{equation*}
The statement is true for n = 1
STEP 2
Assume the statement is true for n = k:
\begin{equation*} {\color {blue} \frac{1}{1! \:.\: 3} + \frac{1}{2! \:.\: 4} + \frac{1}{3! \:.\: 5} + ... + \frac{1}{k! \: (k + 2)}} = {\color {red} \frac{1}{2} - \frac{1}{(k + 2)!}} \end{equation*}
So, for n = k + 1:
\begin{equation*} {\color {blue} \frac{1}{1! \:.\: 3} + \frac{1}{2! \:.\: 4} + \frac{1}{3! \:.\: 5} + ... + \frac{1}{k! \: (k + 2)}} + {\color {magenta} \frac{1}{ (k + 1)! \: (k + 3)}} = \frac{1}{2} - \frac{1}{(k + 3)!} \end{equation*}
\begin{equation*} \begin{split} & \text{LHS} \\\\ & {\color {blue} \frac{1}{1! \:.\: 3} + \frac{1}{2! \:.\: 4} + \frac{1}{3! \:.\: 5} + ... + \frac{1}{k! \: (k + 2)}} + {\color {magenta} \frac{1}{ (k + 1)! \: (k + 3)}} \\\\ & {\color {red} \frac{1}{2} - \frac{1}{(k + 2)!}} + {\color {magenta} \frac{1}{ (k + 1)! \: (k + 3)}} \\\\ & \frac{1}{2} - \frac{1}{(k + 2)(k + 1)!} + \frac{1}{ (k + 1)! \: (k + 3)} \\\\ & \frac{1}{2} - \frac{1}{(k + 1)!} \left(\frac{1}{k + 2} - \frac{1}{k + 3}\right) \\\\ & \frac{1}{2} - \frac{1}{(k + 1)!} \left(\frac{(k + 3) - (k + 2)}{(k + 2)(k + 3)}\right) \\\\ & \frac{1}{2} - \frac{1}{(k + 1)!} \:.\: \frac{1}{(k + 2)(k + 3)} \\\\ & \frac{1}{2} - \frac{1}{(k + 3)!} \\\\ & \text{LHS} \end{split} \end{equation*}
If the statement is true for n = k, then it is true for n = k + 1
STEP 3
Based on step 1 and 2, then:
The statement is true for n = 1, then the statement is true for n = 2
The statement is true for n = 2, then the statement is true for n = 3
and so on,
The statement is true for n ≥ 1
$$\displaystyle \sum_{r = 1}^n 3^r = \dfrac{3}{2}(3^n - 1)$$
$$\displaystyle \sum_{r = 1}^n 3^r = \dfrac{3}{2}(3^n - 1)$$
Solution:
STEP 1
For n = 1:
\begin{equation*} \begin{split} \sum_{r = 1}^1 3^r & = \frac{3}{2}(3^1 - 1) \\\\ 3^1 & = 3 \\\\ \text{LHS} & = \text{RHS} \end{split} \end{equation*}
The statement is true for n = 1
STEP 2
Assume the statement is true for n = k:
\begin{equation*} {\color {blue} \sum_{r = 1}^{k} 3^r} = {\color {red} \frac{3}{2}(3^k - 1)} \end{equation*}
So, for n = k + 1:
\begin{equation*} \sum_{r = 1}^{k + 1} 3^r = \frac{3}{2}(3^{k + 1} - 1) \end{equation*}
\begin{equation*} \begin{split} & \text{LHS} \\\\ & \sum_{r = 1}^{k + 1} 3^r \\\\ & {\color {blue} \sum_{r = 1}^{k} 3^r} + \sum_{r = k + 1}^{k + 1} 3^r \\\\ & {\color {red} \frac{3}{2}(3^{k} - 1)} + 3^{k + 1} \\\\ & \tfrac{3}{2} \:.\: 3^{k} - \tfrac{3}{2} + 3^k \:.\: 3^1 \\\\ & 3^{k} \: (\tfrac{3}{2} + 3) - \tfrac{3}{2} \\\\ & \tfrac{9}{2} \: 3^{k} - \tfrac{3}{2} \\\\ & \tfrac{3}{2} (3 \:.\: 3^k - 1)\\\\ & \tfrac{3}{2} (3^{k + 1} - 1)\\\\ & \text{LHS} = \text{RHS} \end{split} \end{equation*}
If the statement is true for n = k, then it is true for n = k + 1
STEP 3
Based on step 1 and 2, then:
The statement is true for n = 1, then the statement is true for n = 2
The statement is true for n = 2, then the statement is true for n = 3
and so on,
The statement is true for n ≥ 1
$$\displaystyle \sum_{r = 1}^n \dfrac{2r - 1}{3^r} = 1 - \dfrac{n + 1}{3^n}$$
$$\displaystyle \sum_{r = 1}^n \dfrac{2r - 1}{3^r} = 1 - \dfrac{n + 1}{3^n}$$
Solution:
$$5^n + 3$$ is divisible by 4, for $$n \geq 1$$
$$5^n + 3$$ is divisible by 4, for $$n \geq 1$$
Solution:
STEP 1
For n = 1:
$$5^1 + 3 = 8$$ is divisible by 4
The statement is true for n = 1
STEP 2
Assume the statement is true for n = k:
\begin{equation*} 5^k + 3 = 4p \end{equation*}
So, for n = k + 1:
\begin{equation*} \begin{split} & 5^{k + 1} + 3 \\\\ & 5 \:.\: 5^k + 3 \\\\ & 4 \:.\: 5^k + 5^k + 3 \quad {\color {blue} \rightarrow 5^k + 3 = 4p} \\\\ & 4 \:.\: 5^k + 4p \\\\ & 4 \: (5^k + p) \quad {\color {blue} \rightarrow \text{ divisible by 4}} \end{split} \end{equation*}
If the statement is true for n = k, then it is true for n = k + 1
STEP 3
Based on step 1 and 2, then:
The statement is true for n = 1, then the statement is true for n = 2
The statement is true for n = 2, then the statement is true for n = 3
and so on,
The statement is true for n ≥ 1
$$2^{2n + 1} + 3^{2n +1}$$ is divisible by 5, for $$n \geq 1$$
$$2^{2n + 1} + 3^{2n +1}$$ is divisible by 5, for $$n \geq 1$$
Solution:
STEP 1
For n = 1:
$$2^{2(1) + 1} + 3^{2(1) +1} = 35$$ is divisible by 5
The statement is true for n = 1
STEP 2
Assume the statement is true for n = k:
\begin{equation*} 2^{2k + 1} + 3^{2k +1} = 5p \end{equation*}
So, for n = k + 1:
\begin{equation*} \begin{split} & 2^{2(k + 1) + 1} + 3^{2(k + 1) +1} \\\\ & 2^{2k + 3} + 3^{2k + 3} \\\\ & 2^{2k + 1} \:.\: 2^2 + 3^{2k + 1} \:.\: 3^2 \\\\ & 4 \:.\: 2^{2k + 1} + 9 \:.\: 3^{2k + 1} \\\\ & 4 \:.\: 2^{2k + 1} + 4 \:.\: 3^{2k + 1} + 5 \:.\: 3^{2k + 1} \\\\ & 4 \:.\: (2^{2k + 1} + 4 \:.\: 3^{2k + 1}) + 5 \:.\: 3^{2k + 1} \quad {\color {blue} \rightarrow 2^{2k + 1} + 4 \:.\: 3^{2k + 1} = 5p} \\\\ & 4 \:.\: 5p + 5 \:.\: 3^{2k + 1} \\\\ & 5 (4 + 3^{2k + 1}) \quad {\color {blue} \rightarrow \text{ divisible by}} \end{split} \end{equation*}
If the statement is true for n = k, then it is true for n = k + 1
STEP 3
Based on step 1 and 2, then:
The statement is true for n = 1, then the statement is true for n = 2
The statement is true for n = 2, then the statement is true for n = 3
and so on,
The statement is true for n ≥ 1
$$3^n > 1 + 2n$$, for $$n > 1$$
$$3^n > 1 + 2n$$, for $$n > 1$$
Solution:
STEP 1
For n = 2:
\begin{equation*} \begin{split} 3^2 & > 1 + 2(2) \\\\ 9 & > 5 \end{split} \end{equation*}
The statement is true for n = 2
STEP 2
Assume the statement is true for n = k:
\begin{equation*} {\color {blue} 3^k} > {\color {red} 1 + 2k} \end{equation*}
So, for n = k + 1:
\begin{equation*} \begin{split} 3^{k +1} & > 1 + 2(k + 1) \\\\ 3 \:.\: 3^k & > 2k + 2 \\\\ 2 \:.\: 3^k + {\color {blue} 3^k} & > 1 + {\color {red} 1 + 2k} \end{split} \end{equation*} Since $$2 \:.\: 3^k > 1$$ and $$3^k > 1 + 2k$$, then the statement is true for n = k + 1
STEP 3
Based on step 1 and 2, then:
The statement is true for n = 2, then the statement is true for n = 3
The statement is true for n = 4, then the statement is true for n =5
and so on,
The statement is true for n > 1
$$(x + y)^n > x^n + y^n$$, untuk $$n > 1$$
$$(x + y)^n > x^n + y^n$$, untuk $$n > 1$$
Solution:
$$\sin x + \sin 2x + \sin 3x + ... + \sin nx = \dfrac{\sin \frac{1}{2}(n + 1)x \sin \frac{1}{2}nx}{\sin \frac{1}{2}x}$$
$$\sin x + \sin 2x + \sin 3x + ... + \sin nx = \dfrac{\sin \frac{1}{2}(n + 1)x \sin \frac{1}{2}nx}{\sin \frac{1}{2}x}$$
Solution:
STEP 1
For n = 1:
\begin{equation*} \begin{split} \sin x & = \frac{\sin \frac{1}{2}(1 + 1)x \sin \frac{1}{2}(1)x}{\sin \frac{1}{2}x} \\\\ \sin x & = \frac{\sin \frac{1}{2}(2)x \cancel {\sin \frac{1}{2}x}}{\cancel {\sin \frac{1}{2}x}} \\\\ \sin x & = \sin x \\\\ \text{LHS} & = \text{RHS} \end{split} \end{equation*}
The statement is true for n = 1
STEP 2
Assume the statement is true for n = k:
\begin{equation*} {\color {blue} \sin x + \sin 2x + \sin 3x + ... + \sin kx} = {\color {red} \frac{\sin \frac{1}{2}(k + 1)x \sin \frac{1}{2}kx}{\sin \frac{1}{2}x}} \end{equation*}
So, for n = k + 1:
\begin{equation*} {\color {blue} \sin x + \sin 2x + \sin 3x + ... + \sin kx }+ {\color {magenta}\sin (k + 1)x} = \frac{\sin \frac{1}{2}(k + 2)x \sin \frac{1}{2}(k + 1)x}{\sin \frac{1}{2}x} \end{equation*}
\begin{equation*} \begin{split} & \text{LHS} \\\\ &{\color {blue}\sin x + \sin 2x + \sin 3x + ... + \sin kx} + {\color {magenta} \sin (k + 1)x}\\\\ & {\color {red} \frac{\sin \frac{1}{2}(k + 1)x \sin \frac{1}{2}kx}{\sin \frac{1}{2}x}} + {\color {magenta} \sin (k + 1)x} \\\\ & \frac{\sin \frac{1}{2}(k + 1)x \sin \frac{1}{2}kx + \sin (k + 1)x \sin \frac{1}{2}x}{\sin \frac{1}{2}x} \\\\ & \frac{\sin \frac{1}{2}(k + 1)x \sin \frac{1}{2}kx + 2 \sin \frac{1}{2}(k + 1)x \cos \frac{1}{2}(k + 1)x \sin \frac{1}{2}x}{\sin \frac{1}{2}x} \\\\ & \frac{\sin \frac{1}{2}(k + 1)x \: [\sin \frac{1}{2}kx + 2 \cos \frac{1}{2}(k + 1)x \sin \frac{1}{2}x]}{\sin \frac{1}{2}x} \\\\ & \frac{\sin \frac{1}{2}(k + 1)x \: [\cancel {\sin \frac{1}{2}kx} + \sin (\frac{1}{2}k + 1)x - \cancel {\sin \frac{1}{2} kx}]}{\sin \frac{1}{2}x} \\\\ & \frac{\sin \frac{1}{2}(k + 1)x \: \sin (\frac{1}{2}k + 1)x}{\sin \frac{1}{2}x} \\\\ & \frac{\sin \frac{1}{2}(k + 1)x \: \sin \frac{1}{2}(k + 2)x}{\sin \frac{1}{2}x} \\\\ & \text{LHS} = \text{RHS} \end{split} \end{equation*}
If the statement is true for n = k, then it is true for n = k + 1
STEP 3
Based on step 1 and 2, then:
The statement is true for n = 1, then the statement is true for n = 2
The statement is true for n = 2, then the statement is true for n = 3
and so on,
The statement is true for n ≥ 1
$$\cos x + \cos 2x + \cos 3x + ... + \cos nx = \dfrac{\cos \frac{1}{2}(n + 1)x \sin \frac{1}{2}nx}{\sin \frac{1}{2}x}$$
$$\cos x + \cos 2x + \cos 3x + ... + \cos nx = \dfrac{\cos \frac{1}{2}(n + 1)x \sin \frac{1}{2}nx}{\sin \frac{1}{2}x}$$
Solution: |
Area Grazed with 5m Rope: The horse can graze in a quarter circle area when tied with a 5m rope. The area is (1/4)×π×r², where r = 5 m. So, the area is (1/4)×3.14×5² = 19.625 m².
Area Grazed with 10m Rope: With a 10m rope, the horse can graze a larger quarter circle. The area is (1/4)×π×10² = 78.5 m².
Increase in Grazing Area: The increase is 78.5 − 19.625 = 58.875 m².
We can say that the grazing area increases significantly with the length of the rope.
Let’s discuss in detail
## Introduction to the Problem of Grazing Area
The problem at hand involves calculating the grazing area accessible to a horse tied to a peg at the corner of a square-shaped field. This scenario is a classic example of applying geometric concepts to real-world situations. The field has a side length of 15 meters, and the horse is initially tied with a 5-meter-long rope. The task is to determine the area available for grazing with this rope length and then assess how the grazing area changes if the rope’s length is increased to 10 meters. This problem not only tests our understanding of circle geometry but also illustrates how mathematical concepts can be applied to everyday situations.
### Calculating the Grazing Area with a 5-Meter Rope
When the horse is tied with a 5-meter rope, the area available for grazing is a quarter of a circle, with the radius equal to the length of the rope. The formula for the area of a circle is πr², but since the horse can only access a quarter of this circle, the formula becomes (1/4)×π×r². Substituting r = 5 meters and π = 3.14, the grazing area is calculated as (1/4)×3.14×5², which equals 19.625 square meters. This represents the area in which the horse can graze when restricted by a 5-meter rope.
#### Understanding the Increase in Grazing Area with a Longer Rope
The problem also asks us to consider the scenario where the length of the rope is doubled to 10 meters. Intuitively, one might expect the grazing area to also double, but the relationship between the radius of a circle and its area is quadratic, not linear. This means that increasing the rope length will have a more significant impact on the grazing area than a simple doubling. The new grazing area must be recalculated using the same quarter-circle formula but with the updated radius of 10 meters.
Calculating the Grazing Area with a 10-Meter Rope
With the rope length increased to 10 meters, the grazing area becomes a larger quarter circle. Using the formula (1/4)×π×r² with r = 10 meters, the new grazing area is calculated. Substituting these values gives us (1/4)×3.14×10², which equals 78.5 square meters. This significantly larger area demonstrates how a change in the radius of a circle dramatically affects its area, a fundamental concept in circle geometry.
##### Analyzing the Increase in Grazing Area
To find the increase in the grazing area due to the extension of the rope, we subtract the initial grazing area (with the 5-meter rope) from the new grazing area (with the 10-meter rope). The increase is therefore 78.5 − 19.625 = 58.875 square meters. This substantial increase highlights the quadratic relationship between the radius of a circle and its area. It shows that even a small increase in the radius (or in this case, the length of the rope) can lead to a significant increase in area.
###### Geometric Principles in Everyday Life
In conclusion, this problem illustrates the practical application of geometric principles in everyday life. By understanding the relationship between the radius of a circle and its area, we can solve real-world problems such as determining the grazing area for a horse. This example not only reinforces the importance of geometry in practical scenarios but also demonstrates how a seemingly simple change (like increasing the length of a rope) can have a significant impact on the outcome. It’s a testament to the power and relevance of mathematics in understanding and solving everyday challenges.
Discuss this question in detail or visit to Class 10 Maths Chapter 11 for all questions.
Questions of 10th Maths Exercise 11.1 in Detail
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. Find the area of a quadrant of a circle whose circumference is 22 cm. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding (i) minor segment (ii) major sector. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find (i) the total length of the silver wire required. (ii) the area of each sector of the brooch. An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. A round table cover has six equal designs. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm².
Last Edited: June 11, 2024 |
# Circle A has a center at (2 ,8 ) and an area of 8 pi. Circle B has a center at (3 ,2 ) and an area of 27 pi. Do the circles overlap?
Feb 20, 2016
Checking to see if the sum of the radii of the circles is greater than the distance between the circles' centers, we find that yes, they do overlap.
#### Explanation:
The distance between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by
$\text{distance} = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
Applying that in this case, we get the distance between the centers of the circle to be
$\sqrt{{\left(3 - 2\right)}^{2} + {\left(2 - 8\right)}^{2}} = \sqrt{1 + 36} = \sqrt{37}$
Then, the circles only overlap if the sum of their radii is greater than $\sqrt{37}$.
The area of a circle with radius $r$ is given by
$\text{area} = \pi {r}^{2}$
Then, letting ${r}_{a}$ be the radius of circle $A$ and ${r}_{b}$ be the radius of circle $B$, we have
$\left\{\begin{matrix}\pi {r}_{a}^{2} = 8 \pi \\ \pi {r}_{b}^{2} = 27 \pi\end{matrix}\right.$
$\implies \left\{\begin{matrix}{r}_{a} = \sqrt{8} \\ {r}_{b} = \sqrt{27}\end{matrix}\right.$
As a matter of estimation, we can tell that
${r}_{a} = \sqrt{8} \approx \sqrt{9} = 3$
$r + b = \sqrt{27} \approx \sqrt{25} = 5$
and
$\sqrt{37} \approx \sqrt{36} = 6$
meaning we should expect ${r}_{a} + {r}_{b} > \sqrt{37}$ and thus for the circles to overlap.
If we actually calculate the values, we get
${r}_{a} + {r}_{b} = \sqrt{8} + \sqrt{27} \approx 8.02458 > 6.08276 \approx \sqrt{37}$
meaning the estimation was correct, and the circles do overlap. |
# All posts by Mr Low
A full time tutor from Malaysia, passionate about teaching Mathematics for PMR Maths, SPM Additional Maths, SPM Modern Maths and International School Maths included SAT,GCSE, IGCSE syllabus.
# The Key to Math Success: Why Home Tuition Matters
Mathematics is a subject that both fascinates and challenges students of all ages. For parents, ensuring their child’s success in math is often a top priority. Home tuition is a valuable resource that can make a significant difference in a child’s mathematical journey. In this article, we’ll explore the key topics parents care about when it comes to home tuition for math.
Continue reading The Key to Math Success: Why Home Tuition Matters
# Indices and Logarithms
Indices
$\text{Positive Index: }{{a}^{n}},a\ne 0$ $eg:{{2}^{3}},{{4}^{2}}$
$\text{Negative Index: }{{a}^{-n}}=\frac{1}{{{a}^{n}}}$ $eg:{{2}^{-3}}=\frac{1}{{{2}^{3}}}=\frac{1}{8}$
$\text{Zero Index: }{{a}^{0}}=1,a\ne 0$ $eg:{{2}^{0}}=1$
Fractional Indices:
${{a}^{\frac{1}{n}}}=\sqrt[n]{a},a\ne 0$ $eg:{{8}^{\frac{1}{3}}}=\sqrt[3]{8}=2$
${{a}^{\frac{m}{n}}}={{(\sqrt[n]{a})}^{m}},a\ne 0$ $eg:{{8}^{\frac{2}{3}}}={{(\sqrt[3]{8})}^{2}}={{(2)}^{2}}=4$
Law of Indices:
$\text{Rule 1: }{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ $eg:{{2}^{2}}\times {{2}^{3}}={{2}^{2+3}}={{2}^{5}}=32$
$\text{Rule 2: }{{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}$ $eg:{{2}^{4}}\div {{2}^{2}}={{2}^{4-2}}={{2}^{2}}=4$
$\text{Rule 3: }{{({{a}^{m}})}^{n}}={{a}^{m\times n}}$ $eg:{{({{2}^{2}})}^{3}}={{2}^{2\times 3}}={{2}^{6}}=64$
# Formulae that will be given in SPM Add Maths
These are the formulae that will be given to help you to answer the SPM Additional Maths Questions. However, more important is that you know how to apply those formulaes to solve your questions. Buy some exercises from the bookstore and work on the questions everyday and you will master all those formulaes. Basically there are divided into 5 categories:
1. Algebra
2. Calculus
3. Statistics
4. Geometry
5. Trigonometry
# SPM Questions for Simultaneous Equations
At least one simultaneous equations question (5 marks) will be coming out in the SPM paper 2, section A.
1. Substitution (SPM 2005, SPM2007)
2. Using Formulae (SPM 2006, SPM 2009)
# Simultaneous Equation
Simultaneous equation problem could be solved by using
1. Substitution
2. Equating Coefficients
3. Using Formulae
# SPM Questions for Quadratic Functions
1. The Basic of quadratic functions (SPM 2010)
2. Determine max and min values of quadratic function (SPM2009)
3. How to sketch the graph of quadratic functions (SPM2010)
4. How to find the range of values of x in Quadratic inequalities (2007, 2006)
This topic is closely related to the topic of quadratic equations. We will discuss further on 4 subtopics below:
1. The Basic of quadratic functions
2. Determine max and min values of quadratic function
3. How to sketch the graph of quadratic functions
4. How to find the range of values of x in Quadratic inequalities
# SPM Questions for Quadratic Equations
We will discuss some pass year SPM exam questions based on:
• Solve the quadratic equations – SPM 2003, SPM 2005
• Form a quadratic equation – SPM 2009
• Determine the conditions for the type of roots – SPM2010
## 3 Basic Techniques in Solving Quadratic Equation Questions
In this chapter we will learn 3 most basic techniques on how to: |
## Introduction
In this article, we continue to expand our understanding of counting by studying advanced counting concepts involving recurrence relations, onto functions, and derangements.
## Recurrence relations
A formula that defines the terms of a sequence $\set{a_n}$ as some function of one or more of the previous terms of the sequence, namely $a_0, a_1, \ldots, a_{n-1}$ for all integers $n$ such that $n \ge n_0$, is known as a recurrence relation.
A sequence is known as a solution to a recurrence relation if its terms satisfy the recurrence relation.
#### Example
Consider a sequence indicating the population of bacteria in some experiment. Every instant, the bacteria in the petri dish double. Then, this can be expressed as a recurrence relation $a_n = 2 a_{n-1}$. If we set some initial condition, say the starting number of bacteria $a_0 = 1$, then we can calculate the total number of bacteria at any instant given the recurrence relation.
## Linear homogeneous recurrence relation
A linear homogeneous recurrence relation of degree $k$ with constant coefficients is a recurrence relation of the form
$$a_n = c_1 a_{n-1} + c_2 a_{n-2} + \ldots + c_k a_{n-k}$$
where $c_1, c_2, \ldots, c_k$ are real numbers and $c_k \ne 0$.
Thus, the degree defines the farthest backward dependency in describing the recurrence relation. The relationship is linear because all the past terms of the sequence appear in a sum. Finally, it is considered homogeneous, because there are no terms in the relation which do not involve some past term $a_j$.
#### Fibonacci sequence
In 13th century, Fibonacci (Leonardo of Pisa) posed an interesting problem in his book Liber Abaci.
A pair of young rabbits (one of each gender) is placed on an uninhabited island. Rabbits cannot reproduce until they are two months old, but once they are two months old, each reproducing pair breeds on pair of rabbits per month! What will be the number of rabbits on the island after $n$ months? Assume that the rabbits are immortal and genders are equally represented in each brood.
We present this scenario in the following figure. We have marked each pair with an identifier to show the movement of a pair from the pool of young rabbits to those who are capable of reproducing.
Month Reproducing pairs Young pairs Total pairs
0 0 1 1
1 0 1 1
2 1 1 2
3 1 2 3
4 2 3 5
5 3 5 8
Note the total pairs at each month. They follow a recurrence relation! The total pair of rabbits in any month is the sum of total pairs in the prior two months.
Thus, the sequence of total number of rabbits after $n$ months can be described as the recurrence relation below.
$$f_n = f_{n-1} + f_{n-2}$$
with the initial conditions $f_0 = 1$ and $f_1 = 1$.
It is a linear homogeneous recurrence relation of degree $2$. Stay tuned through this article to figure out how to calculate the value of $f_n$ for any value of $n$.
## Characteristic equation of a recurrence relation
Consider a linear homogeneous recurrence relation of degree $k$ with constant coefficients
$$a_n = c_1 a_{n-1} + c_2 a_{n-2} + \ldots + c_k a_{n-k}$$
where $c_1, c_2, \ldots, c_k$ are real numbers and $c_k \ne 0$.
When solving recurrence relations of this form, it is usually the case that we wish to find solutions of the form $a_n = r^n$ where $r$ is some constant real number.
Note that if such a solution is possible, then it must be the case that
$$r^n = c_1 r^{n-1} + c_2 r^{n-2} + \ldots + c_k r^{n-k}$$
Rearranging these terms to the left side and dividing by $r^{n-k}$ we get,
$$r^k - c_1 r^{k-1} - c_2 r^{k-2} - \ldots - c_{k-1} r - c_k = 0$$
This equation is known as the characteristic equation of the recurrence relation. The solutions of the characteristic equation are known as characteristic roots of the recurrence relations. Solving linear homogeneous recurrence relations with constant coefficients involves finding their characteristic roots.
## Solving linear homogeneous recurrence relations of degree 2 with 2 characteristic roots
Consider a recurrence relation $a_n = c_1 a_{n-1} + c_2 a_{n-2}$ where $c_1$ and $c_2$ are constant real numbers. Suppose that the characteristic equation of this recurrence relation $r^2 - c_1 r - c_2 = 0$ has two distinct roots $r_1$ and $r_2$.
Then, it is the case that $a_n = \alpha_1 r_1^n + \alpha_2 r_2^n$ for $n = 1, 2, \ldots,$ where $\alpha_1$ and $\alpha_2$ are real constants.
#### Example
Consider the recurrence relation $a_n = a_{n-1} + 6 a_{n-2}$, with the initial conditions $a_0 = 1$ and $a_1 = 2$. The characteristic equation of this relation is $r^2 - r - 6 = 0$. Its roots are $r = 3$ and $r = -2$.
Then, using the result above, it must be the case that $a_n = \alpha_1 3^n + \alpha_2 (-2)^n$, for some constants $\alpha_1$ and $\alpha_2$.
Because $a_0 = 1$, it follows that $\alpha_1 + \alpha_2 = 1$. Also, because $a_1 = 2$, it is also the case that $3\alpha_1 - 2\alpha_2 = 2$
Solving this system of linear equations, it is easy to find that $\alpha_1 = \frac{4}{5}$ and $\alpha_2 = \frac{1}{5}$.
Thus, the sequence is $a_n = \frac{4}{5} \cdot 3^n + \frac{1}{5} \cdot (-2)^n$.
## Solving linear homogeneous recurrence relations of degree 2 with 1 characteristic root
Consider a recurrence relation $a_n = c_1 a_{n-1} + c_2 a_{n-2}$ where $c_1$ and $c_2$ are constant real numbers, such that $c_2 \ne 0$. Suppose that the characteristic equation of this recurrence relation $r^2 - c_1 r - c_2 = 0$ has only one root $r_0$.
Then, it is the case that $a_n = \alpha_1 r_0^n + \alpha_2 n r_0^n$ for $n = 1, 2, \ldots,$ where $\alpha_1$ and $\alpha_2$ are real constants.
#### Example
Consider the recurrence relation $a_n = 2 a_{n-1} - 4 a_{n-2}$, with the initial conditions $a_0 = 1$ and $a_1 = 3$. The characteristic equation of this relation is $r^2 - 2 r + 4 = 0$. It has only one root $r = 2$.
Then, using the result above, it must be the case that $a_n = \alpha_1 2^n + \alpha_2 n 2^n$, for some constants $\alpha_1$ and $\alpha_2$.
Because $a_0 = 1$, it follows that $\alpha_1 + \alpha_2 \times 0 = 1$. This implies that $\alpha_1 = 1$.
Also, because $a_1 = 3$, it is also the case that $2\alpha_1 + 2\alpha_2 = 3$. Substituting the value of $\alpha_1$, we can find that $\alpha_2 = \frac{1}{2}$.
Thus, the sequence is $a_n = 2^n + \frac{1}{2} \cdot 2^n$.
## Solving linear homogeneous recurrence relations of degree k with k distinct characteristic roots
Consider a recurrence relation $a_n = c_1 a_{n-1} + c_2 a_{n-2} + \ldots + c_k a_{n-k}$ where $c_1, c_2, \ldots, c_k$ are real constants. Suppose that the characteristic equation of this recurrence relation $r^k - c_1 r^{k-1} - \ldots - c_{k-1} r - c_k = 0$ has $k$ distinct roots $r_1, r_2, \ldots, r_k$.
Then, it is the case that $a_n = \alpha_1 r_1^n + \alpha_2 r_2^n + \ldots + \alpha_k r_k^n$ for $n = 1, 2, \ldots,$ where $\alpha_1, \alpha_2, \ldots, \alpha_k$ are real constants.
## Divide-and-conquer relations
Divide-and-conquer strategies work by splitting a large problem into smaller ones and then accumulating the results of the sub-tasks into the final solution of the original problem.
For example, a large problem of size $n$ may be split into $a$ sub-tasks, where each sub-problem is approximately of size $n/b$. Suppose $g(n)$ is the additional work needed to split the problem into its sub-problems. If $f(n)$ represents the effort to solve a problem of size $n$, then $f$ satisfies the recurrence relation
$$f(n) = a f(n/b) + g(n)$$
This is called a divide-and-conquer recurrence relation.
#### Example
Consider the problem of binary search over a list of $n$ sorted elements.
Starting at the middle element the problem is split into the relevant path (the path that might contain the query) and the irrelevant path by comparing the search term to the current element. Additionally, one more comparison is needed to figure out if there are any more elements left to search against. So, $g(n) = 2$. Since the binary search algorithm only follows the relevant path, only one sub-task of the two need followed.
Thus, binary search satisfies the following divide-and-conquer recurrence relation, when $n$ is even.
$$f(n) = f(n/2) + 2$$
## Solving divide-and-conquer recurrence relations with constant split function
Let $f$ be an increasing function that satisfies the following divide-and-conquer recurrence relation
$$f(n) = a f(n/b) + c$$
where, $a \ge 1$ is the number of tasks the problem is split into, $b$ is an integer greater than 1, and $c$ is a positive real number indicating the constant number of operations needed to split the problem at each division.
Then,
$$f(n) = \begin{cases} \BigO{n^{\log_b a}} \text{ if } a > 1 \\\\ \BigO{\log n} \text{ if } a = 1 \end{cases}$$
#### Example
We saw earlier that the number of operations in binary search satisfy the divide-and-conquer recurrence relation $f(n) = f(n/2) + 2$, if $n$ is even. Since $a = 1$ in this case, we can use the result above to identify that $f(n) = \BigO{\log n}$.
## Solving divide-and-conquer recurrence relations with dynamic split function
There are scenarios when the splitting function is dependent on the size of the input $n$.
Let $f$ be such an increasing function that satisfies the following divide-and-conquer recurrence relation
$$f(n) = a f(n/b) + cn^d$$
where, $a \ge 1$ is the number of tasks the problem is split into, $b$ is an integer greater than 1, and $c$ and $d$ are positive real numbers.
Whenever $n = b^k$, for some positive integer $k$
Then,
$$f(n) = \begin{cases} \BigO{n^d} \text{ if } a < b^d \\\\ \BigO{n^d \log n} \text{ if } a = b^d \\\\ \BigO{n^{\log_b a}} \text{ if } a > b^d \end{cases}$$
## Counting elements in the union of finite sets
Let $A_1, A_2, \ldots, A_n$ be finite sets. Then,
\begin{align} |A_1 \cup A_2 \cup \ldots \cup A_n| = & \left(\sum_{i=1}^n |A_i|\right) \\\\ & - \left(\sum_{1 \le i < j \le n} |A_i \cap A_j| \right) \\\\ & + \left(\sum_{1 \le i < j < k \le n} |A_i \cap A_j \cap A_k| \right)\\\\ & ~~~~~~~ \vdots \\\\ & + (-1)^{n+1}|A_1 \cap A_2 \cap \ldots \cap A_n| \end{align}
We show this result visually in the following Venn diagram. Note how the element $7$ is first included 3 times, then excluded 3 times, and finally reintroduced due to $|A \cap B \cap C|$.
Note how the inclusion-exclusion principle requires including and excluding cardinalities of set intersections alternately.
## Counting elements that do not satisfy any property
Sometimes, given several properties $P_1, P_2, \ldots, P_n$, we are required to identify the number of elements in a set that do not satisfy any of these properties. Suppose $N$ is the total number of elements. Also, let $N(P_i)$ denote the number of elements satisfying a property $P_i$. Elements that satisfy multiple properties, say $k$, are written as $N(P_{i_1} P_{i_2} \ldots P_{i_k})$
Then, by the principle of inclusion-exclusion, it can be proven that the number of elements that do not satisfy any property, that is $N(P_1' P_2' \ldots P_n')$ is
\begin{align} N(P_1' P_2' \ldots P_n') = & N \\\\ & - \left(\sum_{1 \le i \le n} N(P_i) \right) \\\\ & + \left(\sum_{1 \le i < j \le n} N(P_i P_j)\right) \\\\ & - \left(\sum_{1 \le i < j < k \le n} N(P_i P_j P_k) \right)\\\\ & ~~~~~~~ \vdots \\\\ & + (-1)^n N(P_1 P_2 \ldots P_n) \end{align}
Basically, in this case, we are computing the union of all properties and subtracting it from the universal set of all $N$ elements. We demonstrate this visually with the Venn diagram below. We are interested in finding elements that are not divisible by 2, 3, and 4. Clearly, we need to exclude the set of elements in the union of the sets $A$, $B$, and $C$. The remaining elements in $U$ are the answer to our query.
## Number of onto functions from a set to another
You remember onto functions from our article on functions.
If $a$ and $b$ are positive integers such that $a \ge b$, then how many onto functions are there from a set with $a$ elements (domain) to a set with $b$ elements (codomain).
This can be derived from the method for counting number of elements that do not satisfy given properties.
Note that the total number of functions from a domain with $a$ elements to a codomain with $b$ elements is $b^a$.
Now, for each of the $b$ elements in the codomain, define a property $P_i$ such that if a function satisfies $P_i$, then the $i$-th element is not in the range of the function.
Thus, $N(P_i)$ is the number of functions that do not have the $i$-th element in their range. For each $P_i$ there will be $(b-1)^a$ functions because the codomain is restricted to only $b - 1$ elements. Since each of the $b$ elements of the codomain will have one such count, we are looking at $C(b,1)(b-1)^a$ total functions that have some element missing from range.
Similarly, $N(P_iP_j)$ is the number of functions that do not have the $i$-th as well as the $j$-th element in their range. Analogous to before, the total such functions will be $C(b,2)(b-2)^a$.
Assigning this way, and using the formula for calculating the number of elements that do not satisfy properties $P_1, P_2, \ldots, P_b$, we get the total number of onto functions.
$$b^a - C(b,1)(b-1)^a + C(b,2)(b-2)^a - \ldots + (-1)^{b-1} C(b,b-1) 1^a$$
#### Example application
This general formulation is quite useful. For example, consider the task of assigning $6$ balls to $3$ containers such that each container receives at least one ball. How many ways can this be achieved in?
Essentially, we are counting the number of onto functions from the set of balls to the set of containers, since each container is associated with at least one ball.
In this case, the size of the domain is $a = 6$ and the size of the codomain is $b = 3$. Substituting, we get $540$ different ways.
Check out the handy calculator in the next section to understand how fast these values grow as $a$ and/or $b$ grow.
## Number of onto functions calculator
Calculate the total number of onto functions from a domain with $a$ elements to a codomain with $b$ elements. Note that it must be the case that $a > b$.
## Derangements
A derangement is a permutation of objects that does not leave any object in its original position. In the following figure we show potential derangements of an initial ordering of an oval, star, and a pentagon. The disqualifying element, the one that causes the permutation to not be a derangement, is marked with an "x".
You should get the hint that counting derangements starts with listing out all possible permutations and then removing those that have any elements in their original position. This is based on our previous section on counting elements that do not satisify given properties.
We know that for $n$ objects, the total number of arrangements or permutations is $n!$.
Let $P_i$ denote the property that the $i$-th element is in its original position. Since the $i$-th element is fixed, there are only $n - 1$ spots for the remaining elements. There would be $(n-1)!$ such permutations which fix the $i$-th element. And there would be $C(n,1)$ such properties. So total number of such permutations with at least one element in its original spot are $\sum_{1 \le n \le n} N(P_i) = C(n,1)(n-1)!$.
Analogously, the total number of permutations such that two elements $i$ and $j$ are in their original spot are $\sum_{1 \le i < j \le n } N(P_i, P_j) = C(n,2)(n-2)!$. And so on.
Using the formula for counting the number of elements that do not satisfy given properties and substituting, we get the number of derangements $D_n$ of $n$ objects as
$$D_n = n! - C(n,1)(n-1)! + C(n,2)(n-2)! - \ldots + (-1)^n C(n,n)(n-n)!$$
Re-arranginng the terms, we get the equation,
$$D_n = n! \left[ 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \ldots + (-1)^n \frac{1}{n!} \right]$$
#### Example
Consider the set of first 3 positive integers $\set{1, 2, 3}$ and a permutation $123$. The permutation $312$ is a derangement of the given permutation, but $321$ is not since $2$ is still in the original position. The number of such possible derangements are
\begin{align} D_3 =~ & 3! \left[1 - 1 + \frac{1}{2!} - \frac{1}{3!} \right] \\\\ =~ & 6 \left[ \frac{1}{2} - \frac{1}{6} \right] \\\\ =~ & 2 \end{align}
## Derangements calculator
Calculate the number of derangements of $n$ objects, such that $1 \le n \le 100$.
## Derangements as a recurrence relation
Derangements of $n$ objects follow the recurrence relation, for $n \ge 1$
$$D_n = n D_{n-1} + (-1)^n$$
The proof can be derived using the inclusion-exclusion formula for derangements that we derived in the previous section.
\begin{align} D_n &= n! \left[ 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \ldots + (-1)^n \frac{1}{n!} \right] \\\\ &= n (n-1)! \left[1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \ldots + (-1)^{n-1} \frac{1}{(n-1)!} + (-1)^n \frac{1}{n!} \right] \\\\ &= n (n-1)! \left[1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \ldots + (-1)^{n-1} \frac{1}{(n-1)!} \right] + n(n-1)! ((-1)^n \frac{1}{n!} \\\\ &= n D_{n-1} + ((-1)^n \end{align}
## Next-article
With a sound understanding of counting and advanced counting concepts in this article, you are now well prepared to tackle complex ideas such as probability.
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# CBSE Class 10 Math Question Paper 2018 | Q22
###### CBSE Math Board Paper 2018 Solution | 3 Mark | Statistics - Median
This statistics question is a three mark question that appeared in CBSE class 10 Math board paper in 2018. A relatively easy question.
Question 22: The table below shows the salaries of 280 persons. Calculate the median salary of the data.
Salary (In thousands) No. of Persons
5 - 10 49
10 - 15 133
15 - 20 63
20 - 25 15
25 - 30 6
30 - 35 7
35 - 40 4
40 - 45 2
45 - 50 1
## NCERT Solution to Class 10 Maths
### Explanatory Answer | CBSE Class 10 2018 Board Paper Q22
#### Step 1: Compute Cumulative Frequency
Salary (In thousands) No. of Persons Cumulative Frequency
5 - 10 49 49
10 - 15 133 182
15 - 20 63 245
20 - 25 15 260
25 - 30 6 266
30 - 35 7 273
35 - 40 4 277
40 - 45 2 279
45 - 50 1 280
#### Step 2: Compute the Median Class
The number of employees, n = 280
∴ $$frac{n}{2}$ = 140 140 falls under the 10 - 15 class ∴ The median class is 10 - 15 #### Step 3: Compute Median Value of median = l + $\frac{$n/2$ - cf}{f}) × h
l = lower limit of median class = 10
$\frac{n}{2}$ = 140
cf - cumulative frequency of class before the median class = 49
f - frequency of median class = 133
h - class size = 5
Median = 10 + $\frac{140 - 49}{133}$ × 5 = 10 + $\frac{91}{133}$ × 5
Median = 10 + $\frac{13}{19}$ × 5 = 10 + $\frac{65}{19}$
Median = 10 + 3.42 = 13.42
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Polynomials
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# Polynomials - PowerPoint PPT Presentation
Polynomials. The Degree of ax n. If a does not equal 0, the degree of ax n is n . The degree of a nonzero constant is 0. The constant 0 has no defined degree. Definition of a Polynomial in x. A polynomial in x is an algebraic expression of the form
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## PowerPoint Slideshow about ' Polynomials' - rajah-soto
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Presentation Transcript
The Degree of axn
• If a does not equal 0, the degree of axn is n. The degree of a nonzero constant is 0. The constant 0 has no defined degree.
Definition of a Polynomial in x
• A polynomial in x is an algebraic expression of the form
• anxn + an-1xn-1 + an-2xn-2 + … + a1n + a0
• where an, an-1, an-2, …, a1 and a0 are real numbers. an= 0, and n is a non-negative integer. The polynomial is of degree n, an is the leading coefficient, and a0 is the constant term.
Text Example
Perform the indicated operations and simplify:
(-9x3 + 7x2 – 5x + 3) + (13x3 + 2x2 – 8x – 6)
Solution
(-9x3 + 7x2 – 5x + 3) + (13x3 + 2x2 – 8x – 6)
= (-9x3 + 13x3) + (7x2 + 2x2) + (-5x – 8x) + (3 – 6) Group like terms.
= 4x3 + 9x2 – (-13x) + (-3) Combine like terms.
= 4x3 + 9x2 + 13x – 3
Multiplying Polynomials
The product of two monomials is obtained by using properties of exponents. For example,
(-8x6)(5x3) = -8·5x6+3 = -40x9
Furthermore, we can use the distributive property to multiply a monomial and a polynomial that is not a monomial. For example,
3x4(2x3 – 7x + 3) = 3x4 · 2x3 – 3x4 · 7x + 3x4 · 3 = 6x7 – 21x5 + 9x4.
monomial
trinomial
Multiplying Polynomials when Neither is a Monomial
• Multiply each term of one polynomial by each term of the other polynomial. Then combine like terms.
Using the FOIL Method to Multiply Binomials
last
first
(ax + b)(cx + d) = ax · cx + ax · d + b · cx + b · d
Product of
First terms
Product of
Outside terms
Product of
Inside terms
Product of
Last terms
inner
outer
Text Example
Multiply: (3x + 4)(5x – 3).
Text Example
Multiply: (3x + 4)(5x – 3).
Solution
(3x + 4)(5x – 3) = 3x·5x + 3x(-3) + 4(5x) + 4(-3)
= 15x2 – 9x + 20x – 12
= 15x2 + 11x – 12 Combine like terms.
last
first
F
O
I
L
inner
outer
The Product of the Sum and Difference of Two Terms
• The product of the sum and the difference of the same two terms is the square of the first term minus the square of the second term.
The Square of a Binomial Sum
• The square of a binomial sum is first term squared plus 2 times the product of the terms plus last term squared.
The Square of a Binomial Difference
• The square of a binomial difference is first term squared minus 2 times the product of the terms plus last term squared.
Special Products
Let A and B represent real numbers, variables, or algebraic expressions.
Special ProductExample
Sum and Difference of Two Terms
(A + B)(A – B) = A2 – B2 (2x + 3)(2x – 3) = (2x) 2 – 32
= 4x2 – 9
Squaring a Binomial
(A + B)2 = A2 + 2AB + B2 (y + 5) 2 = y2 + 2·y·5 + 52
= y2 + 10y + 25
(A – B)2 = A2 – 2AB + B2 (3x – 4) 2 = (3x)2 – 2·3x·4 + 42
= 9x2 – 24x + 16
Cubing a Binomial
(A + B)3 = A3 + 3A2B + 3AB2 + B3 (x + 4)3 = x3 + 3·x2·4 + 3·x·42 + 43
= x3 + 12x2 + 48x + 64
(A – B)3 = A3 – 3A2B – 3AB2 + B3 (x – 2)3 = x3 – 3·x2·2 – 3·x·22 + 23
= x3 – 6x2 – 12x + 8
Text Example
Multiply: a. (x + 4y)(3x – 5y) b. (5x + 3y) 2
• Solution
• We will perform the multiplication in part (a) using the FOIL method. We will multiply in part (b) using the formula for the square of a binomial, (A + B) 2.
• a. (x + 4y)(3x – 5y) Multiply these binomials using the FOIL method.
• = (x)(3x) + (x)(-5y) + (4y)(3x) + (4y)(-5y)
• = 3x2 – 5xy + 12xy – 20y2
• = 3x2 + 7xy – 20y2Combine like terms.
• (5 x + 3y) 2 = (5 x) 2 + 2(5 x)(3y) + (3y) 2 (A + B) 2 = A2 + 2AB + B2
• = 25x2 + 30xy + 9y2
F
O
I
L
Example
• Multiply: (3x + 4)2.
Solution:
( 3x + 4 )2=(3x)2 + (2)(3x) (4) + 42=9x2 + 24x + 16 |
36. One pipe can fill a tank 6 times as fast as another pipe. If together the two pipes can fill the tank in 22 minutes, then the slower pipe alone will be able to fill the tank in:
164 min
154 min
134 min
144 min
```Answer
Answer: Option B Explanation:Let faster pipe alone can fill the tank in x minutes.
Then, slower pipe alone can fill the tank in 6x minutes.
x×6xx+6x=226x7=22x=22×76=1546
Time required for the slower pipe to fill the tank
=6x=154 minute. ```
Report
37. 13 buckets of water fill a tank when the capacity of each bucket is 51 liters. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 17 liters?
33
29
39
42
```Answer
Answer: Option C Explanation:Capacity of the tank =(13×51) liter.
Number of buckets required when capacity of each bucket is 17 liter
=13×5117=13×3=39 ```
Report
38. A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m3. The emptying of the tank is 10 m3 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?
20 m3 / min.
40 m3 / min.
50 m3 / min.
60 m3 / min.
```Answer
Answer: Option C Explanation:Let the filling capacity of the pump =x m3/min.
Then the emptying capacity of the pump =(x+10) m3/min.
Time required for filling the tank =2400x minutes.
Time required for emptying the tank =2400x+10 minutes.
Pump needs 8 minutes lesser to empty the tank than it needs to fill it.
⇒2400x−2400x+10=8⇒300x−300x+10=1⇒300(x+10)−300x=x(x+10)⇒3000=x2+10x⇒x2+10x−3000=0⇒(x+60)(x−50)=0⇒x=50 or −60
Since x can not be negative, x=50
i.e., filling capacity of the pump =50 m3/min. ```
Report
39. Bucket P has thrice the capacity as bucket Q. It takes 80 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum?
30
45
60
80
```Answer
Answer: Option C Explanation:Let capacity of Q =1 liter.
Then, capacity of P =3 liter.
Given that it takes 80 turns for bucket P to fill the empty drum.
=> capacity of the drum =80×3=240 liter.
Number of turns required if both P and Q are used having each turn together=2403+1=60 ```
Report
40. A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 liters a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 24 hours. How many liters does the tank hold?
4010 litre
2220 litre
1920 litre
2020 litre
```Answer
Answer: Option C Explanation:water filled by the inlet pipe in 24 hours
= water emptied by the leak in 24−6=18 hours.
Therefore, water emptied by the leak in 6 hours
= water filled by the inlet pipe in 8 hours
i.e., capacity of the tank
= water filled by the inlet pipe in 8 hours
=8×60×4=1920 litre. ```
Report |
# Logical Equivalence – Domination Laws
In the previous article, you learnt about Identity law which is an equivalence. Similarly, the domination law are another equivalence that you are going to learn in this article.
Also, this equivalence need proof which is the main purpose of this document. The prove is in the form of truth table for domination laws.
## What is domination law?
The domination laws are:
\begin{aligned} &P \vee T \equiv T \hspace{1cm} ( 1)\\ \\
&P \wedge F \equiv F \hspace{1cm} ( 2)
\end{aligned}
First Domination Law
In the first domination law, result of is always . If is a variable that stands for “I am reading” and stands for a universal truth like “Human beings can learn“. The equivalence translates to
I am reading, or Human beings can learn
which is equivalent to saying
I am reading, which is true, but Human beings can learn“.
This is the case when truth value of is .
Consider the case when is . Then equivalence translates to
I am not reading, or Human beings can learn“.
Now because of , if any statement is , the compound preposition is . Therefore, the first equivalence is valid.
Second Domination Law
Let be the statement “I am reading” and is a universal statement “I can read 100 books in 5 minutes“. If is , the second statement becomes because it is impossible to say, “I am reading, and I can read 100 books in 5 minutes”.
Therefore, the second equivalences are valid.
Suppose is , them both gives a truth value and we have false on both sides of the equivalence. The equivalence holds.
In the first equivalence, the dominating value is and in the second equivalence the dominating value is , hence the law is called domination law.
## Truth Table of Domination Laws
To prove that equivalence is for all inputs of , we will construct the truth table for domination laws. The truth table for domination laws will have two rows because there is only one variable, which is .
Rows = 2^1 = 2
The column values for the truth table are , and .
The results of truth table show that is a tautology and is contradiction. Therefore, the domination law is an equivalence and valid.
# Simplifying Boolean Function using K-Map -Special Case
K-Map technique is a straight forward and simple method for minimizing Boolean functions. In this article, you will learn a special case of K-map, when the function is in a Standard Sum of Product and not in a Canonical Sum of Product form.
For example
F = A'B'C' + ABC' - (1)
F = A'B + BC' + A'B - (2)
In function1, each term is called a minterm( A’B’C). The sum of minterms is called a Canonical Sum of Product. It can be directly taken from the Truth Table for the function.
The function2 has terms called a product term which may have one or more literal. The sum of all such terms is called a Standard Sum of Product. The same concept applies for Canonical Product of Sum form.
Now the problem is that if you are given a standard sum of product for Boolean minimization, what to do, because it appears that its already minimized.
## How to simplify the function ( 2 )using K-Map?
Create a three variable map like we do always .and put 1 in the box where a particular term is missing .
A'B
Which term is missing? yes right! it’s C
A'B( C + C')
A'BC + A'BC'
In the map you put 1 in the cell for both
A'BC + A'BC'
Similarly, Solve other missing variable in each of the terms of the function.
Now, you can minimize the above 3-variable K-Map easily because you have identified the correct boxes.
# Subtraction using 10’s complement
In digital computer systems, arithmetic operations are simplified using the radix complement system also known as r’s complement system. The r stands for radix which is a base for a number in a particular number system. In this post, you learn to do subtraction using 10’s complement.
You must be familiar with the complement system in digital logic to understand this subtraction method. To learn about complements visit the following link.
Digital Design – Complements
Examples of number system are decimal, binary, octal, hexadecimal. In a binary system we have complements.
For example,
If you talk about a binary system, the base is 2, then we have two types of r’s complement.
1. r’s complement
2. (r-1)’s complement
For decimal number the r’s complement is 10’s complement and (r-1)’s complement is 9’s complement because base is 10. In other words, decimal number has base r = 10, so 10’s complement and r-1 = 9, so 9’s complement. The binary number has base r = 2, 2’s complement and r-1 = 1 , so one’s complement.
## Q1. Subtract using 10’s complement 52 – 12 .
We know that 52 – 12 = 40
Let m = 52 and n = 12
Take 10’s complement of 12
\begin{aligned}
&+99\\
&-12\\
&----\\
&+87
\end{aligned}
Now, 87 is 9’s complement because we subtracted it with 99. To make it 10’s complement add 1 to 87. The 10’s complement of 12 is 88. Add the 88 to m
\begin{aligned}
&+52\\
&+88\\
&----\\
&\hspace{8px}140
\end{aligned}
Answer: Remove the extra 1 and you get 40
Check the second example.
## Q2: Subtract using 10’s complement 12 – 52
We know that 12 < 52, so answer is -40
Let m = 12 and n = 52
10’s complement of 52
\begin{aligned}
&+99\\
&-52\\
&----\\
&+47
\end{aligned}
The 9’s complement of 52 is 47. To make it 10’s complement add 1 to 47.
\begin{aligned}
&+12\\
&+48\\
&----\\
&+60
\end{aligned}
This is not the answer , wait
Take one more 10’s complement of the result.
\begin{aligned}
&+99\\
&-60\\
&----\\
&+39
\end{aligned}
The 9’s complement of 60 is 39. Add 1 to 39 and make is 10’s complement.
Add negative to the result because m > n.
## Summary
How do I take 10’s complement ?
Suppose n = 123 then
There are 3 digits in 123. The 10’s complement would be 9’s complement + 1.
\begin{aligned}
&999 - 123 = 876\\
&876 + 1 = 877
\end{aligned}
Therefore, the 10’s complement of 123 is 877.
## References
• John.F.Wakerly. 2008. Digital Design: Principles And Practices, 4/E. Pearson Education, India.
• Mano, M. Morris. 1984. Digital Design. Pearson.
# Sequential Circuits – Flip Flop Circuits
Sequential Circuits have a memory element in addition to a combinational circuit so it remembers one bit of information. If a sequential circuit uses a clock pulse, then it is called “Clocked Sequential Circuit”.
There are two types of Sequential Circuits,
1. Synchronous Sequential Circuits
2. Asynchronous Sequential Circuits
In synchronous sequential circuits the memory or the storage element is affected only when it receives a clock pulse. Synchronization is usually achieved through a Master Clock generator which generates a periodic “Clock Pulse”.
The asynchronous sequential circuits have a feedback circuit to the input. It’s working depends on the input to the circuit and order of change in the input to the circuits. This is the reason why asynchronous sequential circuits are not stable. The memory element which store 1 bit information are Flip-Flops.
## Flip-Flop Types
There are many types of flip-flop circuits.
1. S-R Latch
2. R-S Flip-flop
3. D Flip-flop
4. J-K Flip-flop
A simple flip-flop can be constructed using 2 NAND gate or 2 NOR gate. Since, NAND is a universal gate it is quite easy to implement a flip-flop circuit.
An Important thing to remember is that a flip-flop has two Output – Q and Q’. At any given time that output of Q and Q’ are complement of each other.
## S-R Latch
First consider a S-R latch or S-R flip-flop , you can construct this flip-flop using NAND gates or NOR gates.
## S-R Flip-Flop
S-R Flip-Flop This is a very simple implementation and there is not clock applied in this gate. I have used two NOR gates to construct this flip-flop. Let’s see how to construct using two NAND gates.
## S-R Latch using Truth Table
Assume that the Flip-flop is in Set state, means Q = 0 , then Q’ should be 1 because it is complement of Q.
Let’s see what happens in the NOR implementation of S-R latch.
R = 0 and S = 0,
Whenever any input is 1 ,then it operates first and give output and that goes to the feedback circuit of the other gate.
Since, both are 0, nothing happens, such a state is called HOLD state.
R = 0 and S = 1,
(S + Q)' = (1 + 0)' = 0
The output of lower NOR gate is 0.
(R + Q')' = (0 + 0) = 1
Output of upper NOR gate is 1 .
That means Q [Set] = 1 and Q’= 0 and flip-flop goes to SET.
R = 1 and S = 0,
We already know that Q = 1 , Q’ = 0
(R + Q')' = (1 + 0)' = 0
Output of upper NOR gate is 0.
(S + Q)' = (0 + 0)' = 1
Output of lower NOR gate is 1 and flip-flop goes to RESET .
R= 1 and S= 1,
This is undesirable, then the flop goes to an INVALID state.
## D Flip-Flop
The idea of D flip-flop is to remove the ‘Invalid’ state and make sure that the inputs are never same.
D Flip-flop has two inputs – D and CP.
When the CP =0 , then Gate 3 and Gate 4 never changes and remain in level 1, means nothing goes to the output. This happens regardless of the input at D.
If CP = 1 and D =1 , this will SET the flip-flop and Q = 1
If CP = 1 and D = 0 , this will RESET the flip-flop and Q' = 1
D Flip-flop is called this way because it holds “Data” in flip-flop and also called as Gated D-Latch.
## Truth Table Of D-Flip-Flop
The truth table shows that there is no invalid state as in the case of S-R flip-flop.
The adder is a combinational circuit that add binary digits for arithmetic computation. A combinational circuit is a kind of digital circuit that has an input, a logic circuit and an output.
For variables, there are combinations of input variables and for each input combination, there is one and only one output. Therefore, output from combinational circuit is a Boolean function expressed in terms of input variables. Remember that there is only one output for an input combination.
The most basic function of a digital computer is to add binary digits to perform arithmetic operations like humans do, except that we use decimal number system. A digital computer uses binary number. Suppose, you want to add two binary digits – A and B, where A and B represents 0 or 1.
If A = 0 and B = 0, then sum = 0 + 0 = 0
If A = 0 and B = 1, then sum = 0 + 1 = 1
If A = 1 and B = 0, then sum = 1 + 0 = 1
If A = 1 and B = 1, then sum = 1 + 1 = 0 and carry = 1
When the inputs are two binary digits. It produce at most two output.
1. A sum
2. A carry (the last operation above results in a carry)
Binary addition is the motivation behind the adder circuit and there are many types of adder circuits. There are two type of basic adders that help to build complex circuits.
The half adder circuit perform 2 bit addition. The two bit addition produce output called a Sum and a Carry. The outputs are binary functions, the functions of sum and carry helps in designing a logic circuit for half adder. The process of deriving a function and circuit design in given below.
The truth table of half-adder shows all combinations of its input values which is rows. Each row represents a single combination. The 2 columns are for input variables – A and B and the other two columns are for outputs: Sum and Carry.
### Logic functions
The logic functions are for the output of a combinational circuit. You will derive a logic function for half-adder circuit using truth table. The process of deriving logic function is
Step 1: Select all the rows from truth table that result in 1 in sum column or carry column.
Step 2: Create a sum of product function using the minterms you obtained in step 1.
Step 3: Minimize the Boolean function obtained to get the minimum sum of product function for the output sum and carry.
For example
Step 1:
A'B, AB' for sum;
A.B for carry.
Step 2:
Sum = A'B + AB'
Carry = A.B
Step 3:
K-MAP to minimize the Boolean function.
Draw two variable maps for sum and carry using the Boolean functions. Mark the cell as 1 if it corresponds to a minterm in the respective map.
The given functions are mapped into the K-maps, but cannot be minimized because the 1s cannot be grouped, therefore, the final function is
Sum = A ⊕ B
Carry = A . B
### Logic Symbol – Half Adder
Sometimes drawing complex circuit such as a 4 bit adder, needs many components and a detailed diagram for each component is not necessary. This may add to complexity. You can use a logic symbol instead to simplify things. A logic symbol for half adder is given below.
The logic diagram shows all the features of a half adder such as inputs and sum and carry. Any other details is not necessary if you are using many such adders.
The output functions for sum and carry are the basic for designing a logic circuit for half adder circuit. Take a look at the function
Sum = A'B + AB'
Carry = A.B
You can clearly tell that the sum needs two AND gate and an OR gate to represent this circuit in real world. Similarly, Carry need a single and gate. You should note that the sum function is nothing but XOR function, so you can still minimize the number of gates by using one XOR gate and an AND gate to represent a half adder circuit.
Full adder is another type of adder that adds 3 binary bits. It has 3 inputs and 2 output – Sum and Carry-out.
A three-bit binary addition or more is required in some digital circuits. In this case, the first two bits are evaluated first and then the result is evaluated with the third variable. See the example below.
(0 + 0) + 0 = 0
(0 + 0) + 1 = 1
The truth table of full adder has 3 columns for input variables and 2 columns for output. The total combinations of input variables give the number of rows which is rows.
### Logic functions
Similar to half-adder, you can derive the logical functions for outputs of a full adder using the truth table. You need to follow the same process as in the case of half adder.
Step 1: list all the minterms that has 1 as result for sum.
Step 2: Make a sum of product equation using the minterms.
Sum = A'B'C + A'BC' + AB'C' + ABC
Carry = A'BC + AB'C + ABC' + ABC
Step 3: Use K-Map to minimize the equation for sum and carry. While using the K-Maps for full adder, you may observe that the carry function can be grouped and minimized and the resulting functions are
Sum = A'B'C + A'BC' + AB'C' + ABC = A ⊕ B ⊕ C
Carry = A'BC + AB'C + ABC' + ABC = AC + AB + BC
### Logic Symbol – Full Adder
The logic diagram of full adder is not different from the logic diagram of the half-adder, except that one of the input of a full adder is Carry In and one output is Carry Out.
The logic diagram is derived from the equation of outputs. The equation shows that you need two XOR gate and an OR gate. Also, you can represent the circuit using four AND gates and a single OR gate for sum and three AND gates with a single OR gate for carry.
Sum = A'B'C + A'BC' + AB'C' + ABC
Carry = BC + AC + AB
## References
• Mano, M. Morris. 1984. Digital Design. Pearson.
• Shjiva, Sajjan G. 1998. Introduction to Logic Design. New York: Marcel Dekker, Inc .
# Combinational Circuit – Questions/Solutions
In this post, you will learn example problems from combinational circuits. These problems help in minimizing Boolean functions and constructing logic circuit diagrams. The solution to the problems are given in step-by-step manner with explanation wherever possible.
## Q1. Simplify the Boolean function using K-MAP technique.
F(A,B,C,D) = Σ( 0,1,4,8,9,10)
There are 4 variables in this function. First, we construct a Truth Table for four variable functions. We can see that there are 16 minterm of which following gives a output value of 1.
### Solution:
Σ( 0,1,4,8,9,)
= m0 + m1 + m4 + m8 + m9 + m10
= A'B'C'D' + A'B'C'D + A'BC'D' + AB'C'D' + AB'C'D + AB'CD'
But you want the minimized Boolean function.
K-Map is great technique to reduce the minimize the Boolean function. Those minterms that give 1 is marked in the 4-variable K-map. Next, we need to group the adjacent 1s into group of 2, 4 and 8.Once grouped you can minimize the function using following method.
1. Eliminate any common variable that negate each other – row wise or column wise (e.g., because
2. After elimination a group of two adjacent 1s will give 3 variables term(e.g.,).
3. After elimination a group of four adjacent 1s will give 2 variable term (e.g., ).
4. Aftter elimination a group of eight adjacent 1ss will give 1 variable (e.g, ).
Group 1 = A'B'C'D' + A'B'CD' = A'B'D'
Group 2 = A'B'C'D' + AB'C'D' + ABC'D' + A'BC'D'
= B'C'D' + BC'D'
= C'D'
Group 3 = AB'C'D' + AB'C'D' = AB'C'
we have found 3 minimized minterms of our function.
= A'B'C' + A'D' + C'DB'
## Q2. Construct Logic Diagram of the following function using XOR and AND gates only,
F = A.B'CD' + AB'CD' + A.B'C'D + A'BC'D
### Solution:
F = (C.D')(A.B' + B'A) + (C'D)(A.B' + B.A')
F = (C.D')(A ⊕ B) + (C'D)(A ⊕ B)
F = (C.D' + C'D) (A ⊕ B)
F = (C ⊕ D) (A ⊕ B)
## Q3. Draw the logic diagram corresponding to the following Boolean expression without simplifying them.
F=BC' + AB + ACD
F = (A + B) (C + D')(A' + B + D)
### Solution: (a)
F=BC' + AB + ACD
### Solution: (b)
F = (A + B) (C + D')(A' + B + D)
## Q4. Express in Sum of Minterm and Product of Maxterm
F(x,y,z) = (x.y + z)(x.z + y)
### Solution:
F (x,y, z) = (x.y + z) (x.z + y)
=(x + z)(y + z)(x + y)(z + y)
=(x + z)+(y . y') . (y + z ) + ( x . x') . ( x + y )+ (z . z').(z + y)+( x . x' )
=( x + y + z).(x + y' + z).(y + z + x).(y + z + x').(x + y + z).(x + y + z').(z + y + x).(z + y + x')
= (x + y + z) . (x + y' + z) . (x' + y + z) . (x . y . z')
= M0 . M2 . M4 . M1
Î ( 0,1,2,4) [ Product of maxterm form]
F = ∑(3 , 5, 6, 7) [ Sum of Product form]
## Q5. Express the complement of the function in terms of minterms.
a) F(A,B,C,D) = ∑(0,2,6,11,13,14).
b) F(x,y,z) = Î (0,3,6,7)
### Solution:
a) F(A,B,C,D) = ∑(0,2,6,11,13,14)
= m0 + m2 + m6 + m11 + m13 + m14
Complement of F is F' and it contains those minterms not there in F. Also this is a 4 variable function to there are 24 = 16 minterm
So, F' = m1 + m3 + m4 + m5 + m7 + m8 + m9 + m12 + m15
F' = ∑(1,3,4,5,7,8,9,12,15)
# Understanding Sum of Minterms and Product of Maxterms
A Boolean function is expressed in two form.
1. Sum of Minterms
2. Product of Maxterms
## Sum of Minterms
x’ y’ z , x y’ z’ , x y’ z , x y z’ , x y z , gives 1 as output in the above Truth Table.
• Literal – x, y, A, b etc is a label which denote an input variable for a logic gate. Literal can be normal or complimented.
• Minterm – product of two or more literal using ANDing of each literal.
• Maxterm – sum of two or more literal using ORing of each literal.
Before we understand what sum of minterm or product of maxterm is, we must understand a few terminology.
For example,
x or x', y or y'
For example,
x.y.z or x'y
Suppose we have 2 variable – x and y, then all possible combination of literals are x’y’ , x’y, xy’, xy. If we have 3 variables then all combination of literals are as follows.
Basically, if there are n variable, then there is 2^n. For 3 variable, there are 2^3 = 8.
A minterm is the term from table given below that gives 1 output.Let us sum all these terms,
F = x' y' z + x y' z' + x y' z + x y z' + x y z
= m1 + m4 + m5 + m6 + m7
F(x,y,z) = ∑(1,4,5,6,7) is known as Sum of Minterms Canonical Form.
Why is it called canonical form ? because all the literals present in each of the terms.
## Product of Maxterm
The Product of Maxterm is complement of the Sum of Minterm of a function. To obtain the Product of Maxterm, we need two step process.
1. Find those minterms in the Truth Table that gives a 0 as output.
2. Complement those minterms using DeMorgan’s law.
Let us now apply the above to obtain the Product of Maxterm form.
From the previous truth table given, x’ y’ z’, x’ y z’, x’ y z gives output as 0.
F = x' y' z' + x' y z' + x' y z by Rule 1
= (x' y' z' + x' y z' + x' y z)' by DeMorgan's Law
= (x + y + z)(x + y' + z)(x + y' + z') Product of Maxterms form
We see that the Product of Maxterm is ANDing of all ORed terms.
# Universal Gates
In computer science, logic gates such as NAND gates are very useful. You can use NAND gate as universal gate. They can be helpful in designing any complex logic circuit its implementation using NAND gates only.
In this post you learn to use NAND as universal gate to create a logic diagram of a digital circuit with simple gates. If you want to learn more about logic gates visit following links.
Consider an example
## Example Circuit using NAND Gates
F = X + Y’Z is the given function. Show the result in truth table and draw a logic diagram using only NAND gate.
### Solution:
Given function is a Boolean function, when it receives a combination of input values, it will evaluate a single output value, based on the expression or Boolean function. This type of circuit is called a combinational circuit.
There are three variables in this function, so draw a 3 variable truth table for this function. A truth table contains all combinations of the input values of a function.
Next, if you look at the function, it is easy to determine that there are three logical operations: OR, AND, NOT. You can implement the function using a NOT gate, an OR gate and an AND gate.
uses only two gates and an inverter – an OR gate and an AND gate. The Y input is inverted to produce Y’. The AND gate evaluate Y’Z. If the value of Y’Z is 1 or X is 1, the output of function is 1.
## Convert the Logic Diagram using NAND logic gate
To convert the circuit with NOT-AND-OR circuit, you need to change the AND gate, NOT gate and OR gate to NAND equivalent. You will change AND gate with NAND gate first and then change the OR gate with NAND gate. This is shown below.
### NAND gate equivalent of NOT gate
The NAND gate equivalent of inverter is created using a single NAND gate with two input X which gets inverted at the output to give X’.
The figure is function –
### NAND gate equivalent of OR gate
The NAND gate equivalent of OR gate requires three NAND gates of which two of them invert the input values – X and Y, in this case. The output is similar to OR gate output.
### NAND gate equivalent of AND gate
The NAND gate equivalent of AND requires only two NAND gates. The first NAND produce an inverted AND output and the second NAND act as an inverter to correct the inversion and we get an AND output.
In the following diagram, you will find implementation using NAND gate only, of Boolean function . We used
• 1 NAND for inverting Y’
• 2 NAND for AND gate with input Y’Z
• 3 NAND gates for OR gate circuit for X + Y’Z
If you verify the output of the NAND gate circuit, you will find that it is same as the previous circuits created using basic logic gates. Only one type of gate is used in constructing the logic circuits which is an advantage in designing circuits.
## References
• Mano, M. Morris. 1984. Digital Design. Pearson.
• NATARAJAN, ANANDA. 2015. Digital Design. PHI Learning Pvt. Ltd.
# Subtraction of signed binary numbers using 2’s Complement
In this article, we will perform a subtraction using 2’s complement. An unsigned binary number does not have a sign bit in the most significant bit (MSB) position. For example, consider 8-bit representation of 3810
\begin{aligned}
&38_{10} = 0010110_2\\\\
&=0 \cdot 2^7 + 0 \cdot 2^6 + 1 \cdot 2^5 + 0 \cdot 2^4 + 0 \cdot 2^4 + 0 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 0 \cdot 2^0\\\\
&= 0 + 0 + 32 + 0 + 0 + 4 + 2 + 0\\\\
&=32 + 6\\\\
&=38_{10}\end{aligned}
Now, if we take two’s complement of unsigned binary number then we get signed binary representation of a number which is nothing but negative equivalent the unsigned binary number.
To understand this in an easy way, consider previous example of 3810. Let us convert the number 3810 into binary.
38_{10} = 0 0 1 0 0 1 1 0
Take 1’s complement of each binary digit in above number. That is, 0 becomes 1 and 1 becomes 0.
1's \hspace{3px} complement \hspace{3px} of \hspace{3px} 38 = 1 1 0 1 1 0 0 1
Adding 1 to the 1’s complement of 3810 we get two’s complement of the binary number.
\begin{aligned}
&\hspace{1cm}1\\
&1 1 0 1 1 0 0 1\\
& +\hspace{21px}1\\
&---------\\
& 1 1 0 1 1 0 1 0
\end{aligned}
The Most Significant Bit (MSB) has 1 which shows that it is a negative number. The resultant number is -3810
### Subtraction using 2’s Complement of unsigned binary number
Two’s complement of binary number is used for subtraction between unsigned and signed binary numbers.
For example,
How do we subtract? -34 – (-45) = -34 + 45 = 11
Step 1: Convert +34 in 2’s Complement form.
34 = 0 0 1 0 0 0 1 0
Obtain 1’s complement of 0 0 1 0 0 0 1 0
\begin{aligned}
&0 0 1 0 0 0 1 0\\
&+\hspace{21px}1\\
&-----\\
&0 0 1 0 0 0 1 1 = -34_{10}
\end{aligned}
Note: The above step is only performed to obtain the -34 values. There are other methods to obtain negative signed values
Step 2: Convert -45 into 2’s complement to find +45.
But we can also do it directly.
\begin{aligned}
&45 = 0 0 1 0 1 1 0 1\\
&= 0. 2^7 + 0. 2^6 + 1. 2^5 + 0. 2^4 + 1. 2^3+ 1. 2^2 + 0. 2^1 + 1. 2^0\\
&= 0 + 0 + 32 + 0 + 8 + 4 + 0 + 1\\
&= 32 + 8 + 4 + 1\\
&=45
\end{aligned}
Step 3: Add binary value of -34 and 45
\begin{aligned}
&1 1 0 1 1 1 1 0 = -34\\
& 0 0 1 0 1 1 0 1 = +45\\
&----------\\
&0 0 0 0 1 0 1 1 = 1 1 _{1 0}
\end{aligned}
## Example Problems
Q 1: Perform the subtraction with the unsigned binary numbers by taking the 2’s complement of the subtrahend.
Source: Computer System Architecture by Morris Mano
a) 1 1 0 1 0 – 1 0 0 0 0
Solution:
Given
\begin{aligned}
&0001 1 0 1 0 = + 26\\
&0001 0 0 0 0 = + 16\\\\
&Take \hspace{5px}1's \hspace{5px}complement \hspace{5px}of \hspace{5px}000 1 0 0 0 0\\\\
&=11 1 0 1 1 1 1\\\\
&Add \hspace{5px}+1 \hspace{5px}to \hspace{5px}get \hspace{5px}2's \hspace{5px}complement \hspace{5px}of \hspace{5px}+ 16\\\\
&11101111\\
&+\hspace{21px}1\\
&---------\\
&111 1 0 0 0 0 = -16\\\\
&Add \hspace{5px}the \hspace{5px}binary \hspace{5px}value \hspace{5px}of \hspace{5px}+26 \hspace{5px}and \hspace{5px}-16\\
&1\\
&000 1 1 0 1 0 = +26\\
&111 1 0 0 0 0 = -16\\
&-------------\\
&000 0 1 0 1 0 = +10
\end{aligned}
b) 1 1 0 1 0 – 1 1 0 1
Solution:
Given
\begin{aligned}
&0 0 0 1 1 0 1 0 = +26\\
&0 0 0 0 1 1 0 1 = +13\\\\
&Take \hspace{5px} 1's \hspace{5px} complement \hspace{5px} of \hspace{5px}+13 \\\\
&= 0 0 0 0 1 1 0 1\\
&= 1 1 1 1 0 0 1 0\\\\
&Add \hspace{5px} 1 \hspace{5px} to\hspace{5px} get \hspace{5px} 2's \hspace{5px} complement \hspace{5px} of \hspace{5px} +13\\\\
&1 1 1 1 0 0 1 0 \\
&+\hspace{21px}1\\
&-----------\\
& 1 1 1 1 0 0 1 1\\\\
&Add \hspace{5px} binary \hspace{5px} values \hspace{5px} of \hspace{5px} +26 \hspace{5px} and \hspace{5px} -13 to get the result.\\\\
&\hspace{21px} 1 \\
&0 0 0 1 1 0 1 0 = +26\\
&1 1 1 1 0 0 1 1 = -13\\
&----------\\
&0 0 0 0 1 1 0 1 = 13
\end{aligned}
c) 100 – 110000
Solution:
\begin{aligned}
&0 0 0 0 0 1 0 0 = +4\\
&0 0 1 1 0 0 0 0 = +48\\\\
&Take 1's complement of 48\\\\
&= 1 1 0 0 1 1 1 1\\\\
&Add \hspace{5px}1 \hspace{5px}to\hspace{5px} get \hspace{5px}the \hspace{5px}2's \hspace{5px}complement \hspace{5px} of \hspace{5px}+ 48\\\\
&\hspace{18px} 1 1 1 1 \\
&0 1 1 0 1 1 1 1 = +48\\
&+\hspace{21px}1 \\
&----------\\
&0 1 1 1 0 0 0 0\\\\
&Add \hspace{5px} the \hspace{5px}binary \hspace{5px}values \hspace{5px}of \hspace{5px}+4 \hspace{5px}and \hspace{5px}-48 \hspace{5px} to\hspace{5px} get \hspace{5px}the \\ \hspace{5px}
&0 0 0 0 0 1 0 0 = +04\\
&1 1 0 1 0 0 0 0 = -48\\
&----------\\
&1 1 0 1 0 1 0 0 = -44
\end{aligned}
d) 1010100 – 1010100
Solution:
\begin{aligned}
& 0 1 0 1 0 1 0 0 = 64 + 16 + 4 = +84\\
& 0 1 0 1 0 1 0 0 = 64 + 16 + 4 = +84\\\\
&Take \hspace{5px}1's \hspace{5px}complement \hspace{5px}of \hspace{5px}+84 \hspace{5px}subtrahend\\\\
&= 1 0 1 0 1 0 1 1\\\\
&Add \hspace{5px} 1 \hspace{5px} to \hspace{5px} get \hspace{5px} the \hspace{5px}2's \hspace{5px}complement \hspace{5px} of \hspace{5px}+84\hspace{5px} subtrahend\\\\
&\hspace{29px}1\\
&1 0 1 0 1 0 1 1\\
&+ \hspace{21px}1\\
&-------\\
&10101100\\\\
&Add \hspace{5px}the \hspace{5px}binary \hspace{5px}values \hspace{5px}of\hspace{5px}+84 \hspace{5px}and \hspace{5px}-84 \hspace{5px}to \hspace{5px}get \hspace{5px}the \hspace{5px} result\\\\
&0 1 0 1 0 1 0 0\\
&1 0 1 0 1 1 0 0\\
&------\\
&0 0 0 0 0 0 0 0
\end{aligned}
## References
• Mano, M. Morris. 1984. Digital Design. Pearson.
• Shjiva, Sajjan G. 1998. Introduction to Logic Design. New York: Marcel Dekker, Inc .
# 4-Variable K-Map
Previous post, you learned about 3-variable K-map, and learned how to minimize a boolean function. In this post, you will learn about bigger map such as a 4-variable K-map. With 4-variable map you will be able to make larger groups of cells.
## Plotting a 4-variable K-map
The 4 variables of a Boolean function will give a truth table of about rows of input combinations. These can be directly translated to 16 cell 4-variable K-map. See the following figure.
## Grouping of cells in a 4-variable K-map
While you minimize a boolean function using 4-variable K-map, group the cell with 1s into 2s, 4s, and 8s, and so on. See the figure below.
## Rules for Grouping of cells
Here are some simple rules when group cells.
• Single cells with 1 gives you a term with 4 literals.
• Two adjacent cells will minimize and give you a term with 3 literals.
• A grouping of 4 cells of 1s will give you a term with 2 literals.
• A grouping of 8 cells of 1s will give a single literal.
Therefore, you must always try for maximum number of 1s in a group. Some group overlap each other as we mentioned in earlier in previous post. Each overlapping group must include one uncovered cell that contains a 1. See the figure below to understand this.
## 4-Variable K-Map Example
In this section, I have given few examples of 4-variable K-maps. For more practice, you can refer to some textbooks with lot of exercises.
### Q1: Minimize the following Boolean function using 4-variable K-map.
F(A,B,C,D) = ∑(0,1,2,4,5,6,8,9,12,13)
Solution:
Step 1: Construct a 4-variable K-map and mark all minterms with 1.
Step 2: Look vertically in a selected group and extract any common variable. Also look horizontally, extract any common variable from the group.
In the group of four, horizontally, you get D’ and vertically, you get A’. Together it is a term of two literal, that is, A’D’. Similarly, the group of 8, vertically everything cancels out and leave just 1 because A’B’ + A’B = A’ and AB + AB’ = A, finally, A’ + A = (1). But, horizontally, for the group of 8, you get C’. Always the group of 8 will give you single literal.
Step 3: Write down all the solution.
Group of 8 = C'
Group of 4 = A'D'
The final expression is F = A'D' + C'
## Verify the Solution using Algebraic Method
Let’s verify our solution using Boolean algebraic method.
F(A,B,C,D) = ∑(0,1,2,4,5,6,8,9,12,13)
F = A'B'C'D' + A'B'C'D + A'B'CD' + A'BC'D'+ A'BC'D + A'BCD' + AB'C'D' + AB'C'D + ABC'D' + ABC'D
F = A'B'C'(D' + D) + A'CD'(B' + B) + A'BC'(D' + D) + AB'C'(D'+ D) + ABC'(D' + D)
F = A'B'C' + A'CD' + A'BC' + AB'C' + ABC'
F = A'C'(B' + B) + AC'(B' + B) + A'CD'
F = A'C' + AC' + A'CD'
F = C'(A' + A) + A'CD'
F = C' + A'D' // variable C has no effect hence removed
From the solution above it is clear that the 4-variable K-map is a simple solution to minimize a Boolean function with 4 variables. |
The sum of measures of linear pair is 180. The angle sum theorem for quadrilaterals is that the sum of all measures of angles of the quadrilateral is $$360^{\circ}$$. Let $$\angle 1, \angle 2$$, and $$\angle 3$$ be the angles of $$\Delta ABC$$. 2. Now it's the time where we should see the sum of exterior angles of a polygon proof. Proof: Assume a polygon has sides. From the picture above, this means that. The exterior angle of a given triangle is formed when a side is extended outwards. Topic: Angles. Apply the Exterior Angles Theorems. Arrange these triangles as shown below. Since the 65 degrees angle, the angle x, and the 30 degrees angle make a straight line together, the sum must be 180 degrees Since, 65 + angle x + 30 = 180, angle x must be 85 This is not a proof yet. The sum of all exterior angles of a convex polygon is equal to $$360^{\circ}$$. Definition same side interior. Polygon: Interior and Exterior Angles. In several high school treatments of geometry, the term "exterior angle … You can use the exterior angle theorem to prove that the sum of the measures of the three angles of a triangle is 180 degrees. Proof 2 uses the exterior angle theorem. The marked angles are called the exterior angles of the pentagon. For any polygon: 180-interior angle = exterior angle and the exterior angles of any polygon add up to 360 degrees. Adding $$\angle 3$$ on both sides of this equation, we get $$\angle 1+\angle 2+\angle 3=\angle 4+\angle 3$$. Solution: x + 24° + 32° = 180° (sum of angles is 180°) x + 56° = 180° x = 180° – 56° = 124° Worksheet 1, Worksheet 2 using Triangle Sum Theorem It should also be noted that the sum of exterior angles of a polygon is 360° 3. In order to find the sum of interior angles of a polygon, we should multiply the number of triangles in the polygon by 180°. Click to see full answer Sum of Interior Angles of Polygons. The sum is $$112^{\circ}+90^{\circ}+15^{\circ}=217^{\circ}>180^{\circ}$$. A quick proof of the polygon exterior angle sum theorem using the linear pair postulate and the polygon interior angle sum theorem. The sum is $$50^{\circ}+55^{\circ}+120^{\circ}=225^{\circ}>180^{\circ}$$. Identify the type of triangle thus formed. Exterior Angles of Polygons. The exterior angle theorem states that the exterior angle of the given triangle is equal to the sum total of its opposite interior angles. The sum of all the internal angles of a simple polygon is 180 n 2 where n is the number of sides. This is a fundamental result in absolute geometry because its proof does not depend upon the parallel postulate.. The sum of the measures of the angles of a given polygon is 720. Practice: Inscribed angles. The Polygon Exterior Angle sum Theorem states that the sum of the measures of the exterior angles, one angle at each vertex, of a convex polygon is _____. Take a piece of paper and draw a triangle ABC on it. In the third option, we have angles $$35^{\circ}, 45^{\circ}$$, and $$40^{\circ}$$. Can you set up the proof based on the figure above? Measure of Each Interior Angle: the measure of each interior angle of a regular n-gon. The Polygon Exterior Angle sum Theorem states that the sum of the measures of the exterior angles, one angle at each vertex, of a convex polygon is _____. 2.1 MATHCOUNTS 2015 Chapter Sprint Problem 30 For positive integers n and m, each exterior angle of a regular n-sided polygon is 45 degrees larger than each exterior angle of a regular m-sided polygon. 7.1 Interior and Exterior Angles Date: Triangle Sum Theorem: Proof: Given: ∆ , || Prove: ∠1+∠2+∠3=180° When you extend the sides of a polygon, the original angles may be called interior angles and the angles that form linear pairs with the interior angles are the exterior angles. In geometry, the triangle inequality theorem states that when you add the lengths of any two sides of a triangle, their sum will be greater that the length of the third side. Students will see that they can use diagonals to divide an n-sided polygon into (n-2) triangles and use the triangle sum theorem to justify why the interior angle sum is (n-2)(180).They will also make connections to an alternative way to determine the interior … Theorem: The sum of the interior angles of a polygon with sides is degrees. Sum of Interior Angles of Polygons. Formula for sum of exterior angles: The sum of the measures of the exterior angles of a polygon, one at each vertex, is 360°. Example 1 Determine the unknown angle measures. Exterior Angle Theorem – Explanation & Examples. So, we can say that $$\angle ACD=\angle A+\angle B$$. In $$\Delta PQS$$, we will apply the triangle angle sum theorem to find the value of $$a$$. Did you notice that all three angles constitute one straight angle? You can derive the exterior angle theorem with the help of the information that. Alternate Interior Angles Draw Letter Z Alternate Interior Angles Interior And Exterior Angles Math Help . Then, by exterior angle sum theorem, we have $$\angle 1+\angle 2=\angle 4$$. Each exterior angle is paired with a corresponding interior angle, and each of these pairs sums to 180° (they are supplementary). What this means is just that the polygon cannot have angles that point in. Next, we can figure out the sum of interior angles of any polygon by dividing the polygon into triangles. \begin{align}\angle PSR+\angle PRS+\angle SPR&=180^{\circ}\\115^{\circ}+40^{\circ}+c&=180^{\circ}\\155^{\circ}+c&=180^{\circ}\\c&=25^{\circ}\end{align}. From the proof, you can see that this theorem is a combination of the Triangle Sum Theorem and the Linear Pair Postulate. In general, this means that in a polygon with n sides. Here is the proof of the Exterior Angle Theorem. Find the nmnbar of sides for each, a) 72° b) 40° 2) Find the measure of an interior and an exterior angle of a regular 46-gon. The exterior angle of a given triangle is formed when a side is extended outwards. We have moved all content for this concept to for better organization. Can you set up the proof based on the figure above? Triangle Angle Sum Theorem Proof. The Exterior Angle Theorem states that the sum of the remote interior angles is equal to the non-adjacent exterior angle. The same side interior angles are also known as co interior angles. But the interior angle sum = 180(n – 2). In the second option, we have angles $$112^{\circ}, 90^{\circ}$$, and $$15^{\circ}$$. Let us consider a polygon which has n number of sides. Done in a way that is not only relatable and easy to grasp, but will also stay with them forever. Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. Polygon Exterior Angle Sum Theorem If we consider that a polygon is a convex polygon, the summation of its exterior angles at each vertex is equal to 360 degrees. The angles on the straight line add up to 180° Sum of the measures of exterior angles = Sum of the measures of linear pairs − Sum of the measures of interior angles. Observe that in this 5-sided polygon, the sum of all exterior angles is 360∘ 360 ∘ by polygon angle sum theorem. Plus, you’ll have access to millions of step-by-step textbook answers. In the fourth option, we have angles $$95^{\circ}, 45^{\circ}$$, and $$40^{\circ}$$. Exterior Angles of Polygons. In any polygon, the sum of an interior angle and its corresponding exterior angle is 180 °. Observe that in this 5-sided polygon, the sum of all exterior angles is $$360^{\circ}$$ by polygon angle sum theorem. If a polygon does have an angle that points in, it is called concave, and this theorem does not apply. Proof 1 uses the fact that the alternate interior angles formed by a transversal with two parallel lines are congruent. 1) Exterior Angle Theorem: The measure of an 354) Now, let’s consider exterior angles of a polygon. The same side interior angles are also known as co interior angles. Therefore, the number of sides = 360° / 36° = 10 sides. x° + Exterior Angle = 180 ° 110 ° + Exterior angle = 180 ° Exterior angle = 70 ° So, the measure of each exterior angle corresponding to x ° in the above polygon is 70 °. According to the Polygon Interior Angles Sum Theorem, the sum of the measures of interior angles of an n-sided convex polygon is (n−2)180. 12 Using Polygon Angle-Sum Theorem This is a fundamental result in absolute geometry because its proof does not depend upon the parallel postulate. The sum of the exterior angles of a triangle is 360 degrees. Thus, the sum of the measures of exterior angles of a convex polygon is 360. arrow_back. Email. Here are three proofs for the sum of angles of triangles. $$\angle A$$ and $$\angle B$$ are the two opposite interior angles of $$\angle ACD$$. Inscribed angle theorem proof. Sum of Interior Angles of Polygons. Create Class; Polygon: Interior and Exterior Angles. (Use n to represent the number of sides the polygon has.) Subscribe to bartleby learn! Polygon: Interior and Exterior Angles. From the picture above, this means that . which means that the exterior angle sum = 180n – 180(n – 2) = 360 degrees. Inscribed angles. We will check each option by finding the sum of all three angles. let EA = external angle of that polygon polygon exterior angle sum theorem states that the sum of the exterior angles of any polygon is 360 degrees. \begin{align}\angle PSR+\angle PSQ&=180^{\circ}\\b+65^{\circ}&=180^{\circ}\\b&=115^{\circ}\end{align}. (pg. Find the nmnbar of sides for each, a) 72° b) 40° 2) Find the measure of an interior and an exterior angle of a regular 46-gon. 11 Polygon Angle Sum. The goal of the Polygon Interior Angle Sum Conjecture activity is for students to conjecture about the interior angle sum of any n-gon. Exterior Angle Theorem : The exterior angle theorem states that the measure of an exterior angle is equal to the sum of the measures of the two remote interior angles of the triangle. Inscribed angles. The math journey around Angle Sum Theorem starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. This mini-lesson targeted the fascinating concept of the Angle Sum Theorem. In the figures below, you will notice that exterior angles have been drawn from each vertex of the polygon. \begin{align} \text{angle}_3 &=180^{\circ}-(90^{\circ} +45^{\circ}) \\ &= 45^{\circ}\end{align}. Thus, the sum of the measures of exterior angles of a convex polygon is 360. The marked angles are called the exterior angles of the pentagon. In $$\Delta ABC$$, $$\angle A + \angle B+ \angle C=180^{\circ}$$. The exterior angle theorem is Proposition 1.16 in Euclid's Elements, which states that the measure of an exterior angle of a triangle is greater than either of the measures of the remote interior angles. Here lies the magic with Cuemath. right A corollary of the Triangle Sum Theorem states that a triangle can contain no more than one _____ angle or obtuse angle. The sum of all exterior angles of a triangle is equal to $$360^{\circ}$$. Polygon: Interior and Exterior Angles. 'What Is The Polygon Exterior Angle Sum Theorem Quora May 8th, 2018 - The Sum Of The Exterior Angles Of A Polygon Is 360° You Can Find An Illustration Of It At Polygon Exterior Angle Sum Theorem' 'Polygon Angle Sum Theorem YouTube April 28th, 2018 - Polygon Angle Sum Theorem Regular Polygons Want music and videos with zero ads Get YouTube Red' Author: pchou, Megan Milano. The polygon exterior angle sum theorem states that "the sum of all exterior angles of a convex polygon is equal to $$360^{\circ}$$.". C. Angle 2 = 40 and Angle 3 = 20 D. Angle 2 = 140 and Angle 3 = 20 Use less than, equal to, or greater than to complete this statement: The sum of the measures of the exterior angles of a regular 9-gon, one at each vertex, is ____ the sum of the measures of the exterior angles of a … Before we carry on with our proof, let us mention that the sum of the exterior angles of an n-sided convex polygon = 360 ° I would like to call this the Spider Theorem. Since two angles measure the same, it is an. But the exterior angles sum to 360°. So, $$\angle 1+\angle 2+\angle 3=180^{\circ}$$. Consider, for instance, the pentagon pictured below. You will get to learn about the triangle angle sum theorem definition, exterior angle sum theorem, polygon exterior angle sum theorem, polygon angle sum theorem, and discover other interesting aspects of it. Imagine you are a spider and you are now in the point A 1 and facing A 2. 3. Hence, the polygon has 10 sides. The sum of the interior angles of any triangle is 180°. I Am a bit confused. By using the triangle inequality theorem and the exterior angle theorem, you should have no trouble completing the inequality proof in the following practice question. Topic: Angles, Polygons. If we observe a convex polygon, then the sum of the exterior angle present at each vertex will be 360°. The angle sum of any n-sided polygon is 180(n - 2) degrees. The sum of the exterior angles is N. 2 Using the Polygon Angle-Sum Theorem As I said before, the main application of the polygon angle-sum theorem is for angle chasing problems. The sum of the measures of the angles in a polygon ; is (n 2)180. Interior and exterior angles in regular polygons. The exterior angle theorem states that the exterior angle of the given triangle is equal to the sum total of its opposite interior angles. The sum of 3 angles of a triangle is $$180^{\circ}$$. Create Class; Polygon: Interior and Exterior Angles. You crawl to A 2 and turn an exterior angle, shown in red, and face A 3. USING THE INTERIOR & EXTERIOR ANGLE SUM THEOREMS 1) The measure of one exterior angle of a regular polygon is given. = 180 n − 180 ( n − 2) = 180 n − 180 n + 360 = 360. sum theorem, which is a remarkable property of a triangle and connects all its three angles. From the proof, you can see that this theorem is a combination of the Triangle Sum Theorem and the Linear Pair Postulate. The polygon exterior angle sum theorem states that "the sum of all exterior angles of a convex polygon is equal to 360∘ 360 ∘." 3. So, only the fourth option gives the sum of $$180^{\circ}$$. The sum is always 360.Geometric proof: When all of the angles of a convex polygon converge, or pushed together, they form one angle called a perigon angle, which measures 360 degrees. Google Classroom Facebook Twitter. Every angle in the interior of the polygon forms a linear pair with its exterior angle. Leading to solving more challenging problems involving many relationships; straight, triangle, opposite and exterior angles. Draw any triangle on a piece of paper. Alternate Interior Angles Draw Letter Z Alternate Interior Angles Interior And Exterior Angles Math Help . You can check out the interactive simulations to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page. The goal of the Polygon Interior Angle Sum Conjecture activity is for students to conjecture about the interior angle sum of any n-gon. In this mini-lesson, we will explore the world of the angle sum theorem. The sum of the measures of the interior angles of a convex polygon with 'n' sides is (n - 2)180 degrees Polygon Exterior Angle Sum Theorem The sum of the measures of the exterior angles, one angle at each vertex, of a convex polygon In the first option, we have angles $$50^{\circ},55^{\circ}$$, and $$120^{\circ}$$. Theorem. Please update your bookmarks accordingly. Draw three copies of one triangle on a piece of paper. A pentagon has 5 interior angles, so it has 5 interior-exterior angle pairs. The central angles of a regular polygon are congruent. He knows one angle is of $$45^{\circ}$$ and the other is a right angle. ... All you have to remember is kind of cave in words And so, what we just did is applied to any exterior angle of any convex polygon. The angle sum property of a triangle states that the sum of the three angles is $$180^{\circ}$$. Click here if you need a proof of the Triangle Sum Theorem. Each exterior angle is paired with a corresponding interior angle, and each of these pairs sums to 180° (they are supplementary). Here, $$\angle ACD$$ is an exterior angle of $$\Delta ABC$$. You can use the exterior angle theorem to prove that the sum of the measures of the three angles of a triangle is 180 degrees. Can figure out the measurements of all angles of an n-gon ;:... 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Sal demonstrates how the the sum of any polygon, the pentagon pictured below triangle adds to... At each vertex of the angles of a linear pair postulate one triangle on a of. Because it is a 3-sided figure with three interior angles interior and exterior is... And the other is a combination of the exterior angle of a polygon into triangles by drawing all the angles! Us consider a polygon proof } +40^ { \circ } \ ) polygon does have an angle points... Will also stay with them forever triangle states that a triangle and connects all three... Our proof of polygon exterior angle sum theorem readers, the angle sum theorem and the polygon interior sum! Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of \ ( \Delta ). Polygon which has n number of sides the polygon exterior angles is \ ( {. Polygon Angle-Sum theorem easy to grasp, but will also stay with them forever \. That \ ( 45^ { \circ } +40^ { \circ } \ ) polygon can have. Draw Letter Z alternate interior angles of a polygon does have an angle that subtends the,. Equation, we have moved all content for this concept to for better organization its corresponding exterior angle property... It has 5 interior angles interior and exterior angles of a polygon proof. Understand the angle sum * + exterior angle of a convex polygon is 720 then, by angle! Point a 1 and facing a 2 and turn an exterior angle sum theorem the interior angles of polygon! Be noted that the sum of exterior angles of a polygon does have an angle that points,... } +40^ { \circ } \ ) and click the check answer '' button to see the.. An exterior angle is 180 n 2 where n is the Corollary to the non-adjacent exterior.. Sum * + exterior angle theorem with the help of the polygon exterior angles proof of polygon exterior angle sum theorem been drawn each! Check answer '' button to see the sum of the measures of interior angles of a triangle connects. A\ ) that \ ( \angle 1 + \angle proof of polygon exterior angle sum theorem \angle 3=180^ { \circ \! Following can be drawn from each vertex of the measures of exterior angles sum theorem to find sum! Diagonals that can be drawn from one single vertex Math help see that theorem. Polygon angle sum Conjecture activity is for students to Conjecture about the interior angles which n! The sum total of its opposite interior angles of a triangle can contain no more than one angle! The angle sum theorem using the definition of a simple polygon is 360 degrees ; is ( n – )! Is 180° engaging learning-teaching-learning approach, the main application of the information that will that... Is an exterior angle sum = 180n on angle sum property of a convex polygon is 360° 3: angle. Transversal with two parallel lines are congruent + 360 = 360 are proofs... Concave, and each of these pairs sums to 180° ( they are )... Theorem or angle sum of all angles of a triangle check answer '' button to see sum. Angles in a way that is not only relatable and easy to,! Answer and click the check answer '' button to see the sum of 3 of! Is formed when a side is extended outwards 5 interior angles of a convex polygon 720. And face a 3 a corresponding interior angle sum = 180n 4+\angle 3\ ) form pair! Any n-gon approach, the pentagon pictured below: proof states that the alternate interior angles in red, each... A quick proof of the interior angles of a convex polygon is degrees. Polygon ; is ( n proof of polygon exterior angle sum theorem 2 ) degrees again observe that these three.. Shown in red, and \ ( \Delta ABC\ ), we have \ ( B\ ) using. |
# Addition of Three 2-digit numbers with Regrouping | Maths Grade 2
### Addition of Three 2-digit numbers with Regrouping
Explanation:
Step 1: Adding the digits at one’s place,
3 + 3 + 4 = 10
Since the number is more than 9, we need to carry over 1 to the tens place.
Step 2: Adding the digits at tens place and the carry of 1,
2 + 1 + 2 + 1 = 6
Hence, the sum of the given numbers is 60
Step 1: Adding the digits at one’s place,
9 + 0 + 9 = 18
Since the number is more than 9, we need to carry over 1 to the tens place.
Step 2: Adding the digits at tens place and the carry of 1,
4 + 1 + 1 + 1 = 7
Hence, the sum of the given numbers is 78
Explanation:
Step 1: Adding the digits at one’s place,
7 + 9 + 3 = 19
Since the number is more than 9, we need to carry over 1 to the tens place.
Step 2: Adding the digits at tens place and the carry of 1,
5 + 2 + 1 + 1 = 9
Hence, the sum of the given numbers is 99 |
Class 6 Maths Data Handling Drawing a Bar Graph
Drawing a Bar Graph
To draw a bar graph, first of all draw a horizontal line and a vertical line. On the
Horizontal line we will draw bars representing the data (numbers) and on vertical line we will write numerals .which shows what data is being represented.
Same data can also be represented by interchanging the items on horizontal and vertical axis.
It is important to take bars of same width keeping uniform gap between them. Next, a scale is chosen if needed. The scale varies according to the data given.
Following problems explain how a bar graph is drawn:
Problem: The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:
Draw a bar graph to represent the above information choosing the scale of your choice.
Solution:
To draw a bar graph, days are taken on horizontal axis and the number of books sold is taken on the vertical axis.
The scale taken for representing this data is 1 unit length = 5 shirts. This scale is taken along the vertical line since the number of books is marked on the y axis.
The height of the bars for various days is:
Problem: Following table shows the number of bicycles manufactured in a factory during the years 1998 to 2002. Illustrate this data using a bar graph. Choose a scale of your choice.
• In which year were the maximum number of bicycles manufactured?
The year 2002 is the year represented by bar with maximum height. Hence maximum number of cycles were sold in 2002, that is, 1200.
• In which year were the minimum number of bicycles manufactured?
The year 1999 is the year represented by bar with minimum height. Hence minimum number of cycles were sold in 1999, that is, 600.
. |
# Study convergence of $x_{n+1} = x_n^2 + 3x_n + 1$, where $x_1 = a$, and $a$ takes different values and find its limit.
Given a recurrence relation: $$x_{n+1} = x_n^2 + 3x_n + 1 \\ x_1 = a\\ n\in\Bbb N$$ Figure out whether this sequence has a limit (either finite or infinite) and find it for: \begin{align*} a = -{5\over 4}\tag1 \\ a = -{3\over 4}\tag2 \end{align*}
Start with case $$(1)$$. It took some time to notice but seems like the sequence is monotonically increasing no matter what initial conditions are given. That is because: $$x_{n+1} = x_n^2 + 3x_n + 1 \iff x_{n+1}-x_n = x_n^2 + 2x_n + 1 = (x_n+1)^2>0$$ Than means: $$x_{n+1} - x_n > 0 \iff x_{n+1} > x_n$$ That observation is crucial for all the next steps. In $$(1)$$ we are given that: $$x_1 = a = -{5\over 4} > -2$$ By monotonicity of $$x_n$$: $$\forall n\in\Bbb N : x_n > -2$$ Let's suppose the limit exists. Then by finding fixed points of the recurrence we may get an insight of what that limit might be: $$L = L^2 + 3L + 1 \iff (L+1)^2 = 0 \iff L = -1$$ Thus the only possible finite limit in $$\Bbb R$$ is $$L=-1$$. Let's try to bound $$x_n$$ above. Using induction: $$x_1 < x_2 = -{19\over 16} < -1$$ Suppose $$x_n < -1$$. Then: $$x_n \in (-2; -1) \implies \underbrace{(x_n + 1)^2 + x_n}_{x_{n+1}} \in (-2, -1)$$ Thus it follows that $$x_{n+1} < -1$$. Now by monotone convergence theorem a monotonic bounded sequence has a limit. Therefore: $$\boxed{\lim_{n\to\infty}x_n = -1}$$
This case is more of a headache. Given $$a = -{3\over 4}$$ makes the sequence diverge to $$+\infty$$. But to show this I had to calculate the value for $$6$$ first terms. It follows that: $$\forall n \ge 6: x_n > 0$$ Moreover: $$\forall n \ge 7: x_n > 1$$
So: $$\boxed{\lim_{n\to\infty}x_n = +\infty}$$
Does there exist a more elegant way to solve for case $$(2)$$?
Also is this argumentation enough to show what's requested in question section? I have doubts about the second case. Because formally I should have shown that the sequence is not bounded, not sure how to do it. And the solution is ugly. Here is a sandbox I've been using to play around with the recurrence.
Could you please verify the above and point to the mistakes just in case? Thank you!
• +1 for showing all the work you did --- a nicely-asked question! (And no, I don't actually have anything useful to provide as an answer, alas.) Feb 8, 2019 at 15:25
• To avoid headaches and trivialize massively all this and every similar question, my advice would be to draw the graph of the function $f:x\mapsto x^2+3x+1$ and the line $y=x$ on the same figure, and to plot the first values of the sequence by the well-known cobweb plot associated to $f$. You should see the desired results literally pop up from the figure... In addition, the asymptotics of sequences starting from any $x_0$ in $(-1,\infty)$ or $[-2,-1]$ or $(-\infty,-2)$ should become obvious as well.
– Did
Feb 8, 2019 at 15:36
• +1, can I know how you have written the code in Desmos? Feb 8, 2019 at 15:44
• @taritgoswami I've just manually printed the equations for the first 10 terms and then added a table to display the points. If you expand the 'terms' folder you'll see a list of equations. Feb 8, 2019 at 15:47
Does there exist a more elegant way to solve for case $$(2)$$? Yes there is ! You don't need to compute the first six terms at all ! The sequence is nondecreasing, so it either converges to a finite value or diverges to $$+\infty$$. It if converges, the limit can only be $$-1$$ as you have shown, but this is impossible since your sequence, being nondecreasing, will always be $$\geq x_0=-\frac{3}{4}$$. |
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# Probability Calculator
## How to use the probability calculator?
• Enter the number of events that occurred n(E).
• Enter the number of possible outcomes n(T).
• Click the “calculate” button.
• Use the reset button to enter new values.
## Probability calculator
Probability calculator evaluates the probability of occurring events out of total number of possible outcomes. This event probability calculator finds single-event probability. For example; a coin is tossed 90 times and we have to calculate the chances of occurring head in the 1st five times.
## What is probability?
Probability is simply how likely something is to happen. It is the outcome of a random event i.e. if we toss a coin we cannot say what will be the result whether it will be head or tail until the event happens. The probability ranges between 0 and 1.
If the probability of an event is 0.6 then it means there is a 60% chance of happening that event. The probability theory contains some sub-terms in it, to understand the concept of probability first we have to go through these basic terms.
• Experiment: To find well-defined outcomes we do experiments.
• Trial: The term “trial” represents how many times an experiment is performed.
• Outcome: It is the possible result.
• Sample space: All the possible results together make a sample space. It is represented by curly brackets {}.
• Event: The sample space is a set, and all of its subsets are known as events.
### The formula of probability:
The general formula to calculate the single event probability is as follows:
Probability = number of events occurred n(E) / number of possible outcomes n(T)
### How to calculate probability?
We can calculate the probability of a sample space by following the below steps.
• Check the number of all possible outcomes n(T)
• Find out the number of events that occurred n(E)
• Divide n(E) by n(T)
To understand the method more precisely, have a look at the following examples.
Example 1:
A fair die is rolled one time, what are the chances of coming the face 6?
Solution:
Step 1: Check the number of all possible outcomes n(T)
A die has six faces so n(T) = 6
Step 2: Find out the number of events that occurred n(E)
The die is rolled one time so n(E) = 1
Step 3: Calculation
Probability = n(E) / n(T)
Probability = 1/6
Probability = 0.1667
Probability = 16.67% |
## Calculus with Applications (10th Edition)
$$7$$
\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\left( {7 - x} \right)\ln \left( {1 - x} \right)}}{{{e^{ - x}} - 1}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {7 - x} \right)\ln \left( {1 - x} \right) = \left( {7 - 0} \right)\ln \left( {1 - 0} \right) = \left( 7 \right)\left( 0 \right) = 0 \cr & \mathop {\lim }\limits_{x \to 0} \left( {{e^{ - x}} - 1} \right) = {e^{ - 0}} - 1 = 1 - 1 = 0 \cr & \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr & {\text{denominator}}{\text{}} \cr & {\text{for }}\left( {7 - x} \right)\ln \left( {1 - x} \right) \to {D_x}\left( {\left( {7 - x} \right)\ln \left( {1 - x} \right)} \right) \cr & {\text{By the product rule}} \cr & = \left( {7 - x} \right){D_x}\left( {\ln \left( {1 - x} \right)} \right) + \ln \left( {1 - x} \right){D_x}\left( {\left( {7 - x} \right)} \right) \cr & = \left( {7 - x} \right)\left( {\frac{{ - 1}}{{1 - x}}} \right) + \ln \left( {1 - x} \right)\left( { - 1} \right) \cr & = \frac{{7 - x}}{{x - 1}} - \ln \left( {1 - x} \right) \cr & \cr & {\text{for }}{e^{ - x}} - 1 \to {D_x}\left( {{e^{ - x}} - 1} \right) = - {e^{ - x}} \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\left( {7 - x} \right)\ln \left( {1 - x} \right)}}{{{e^{ - x}} - 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{7 - x}}{{x - 1}} - \ln \left( {1 - x} \right)}}{{ - {e^{ - x}}}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{\frac{{7 - 0}}{{0 - 1}} - \ln \left( {1 - 0} \right)}}{{ - {e^{ - 0}}}} \cr & = \frac{{ - 7 - 0}}{{ - 1}} \cr & = 7 \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\left( {7 - x} \right)\ln \left( {1 - x} \right)}}{{{e^{ - x}} - 1}} = 7 \cr} |
# Mathematics - Linear Algebra Vector Spaces
Hello it's me again! Today we will continue with Linear Algebra getting into Vector Spaces and Subspaces. We will talk about a lot of basics today. In the next post we will get into more interesting stuff. :) So, without further do let's get started!
# Linear Vector Space:
Vector spaces are the generalization of the classic 2d and 3d vectors that most off you may already know from math in school. So, the 2d and 3d vectors are also vector spaces. We are talking about a definition field of R^n, cause each "dimension" is in the real number set R and so if we have n numbers we have a R set for each.
We define a vector space like that:
R^n = { (x1, x2, ..., xn): xi is in R and i = 1, 2, ..., n }
where n is the number of dimensions, xi is the variable/value for each dimension and i is representing each dimension.
To make it more clear for you in 3-dimensional vector spaces we commonly use the names x, y and z for the variables and so this vector space looks like this:
R^3 = { (x, y, z): x, y, z are in R}
So, we can have for example (π, 1.52, 7) be a vector inside of this space.
After 3-dimensions we use xi simply instead of using more letters.
## Basic Vector Operations:
For all the vector spaces we specify an addition and an scalar multiplication
If we have 2 vectors v1 = (x1, x2, ..., xn) and v2 = (y1, y2, ... , yn) inside of the same R^n vector space the addition looks like this:
v1 + v2 = (x1, x2, ..., xn) + (y1, y2, ..., yn) = (x1 + y1, x2 + y2, ..., xn + yn)
So, we simply add the corresponding dimension values.
Scalar Multiplication:
If we have a vector v = (x1, x2, ..., xn) and an constant k inside of R then scalar multiplication looks like this:
k*v = k * (x1, x2, ..., xn) = (k*x1, k*x2, ..., k*xn)
So, we simply multiply each dimension value with the constant.
Properties
• v + w = w + v (commutative property)
• (u + v) + w = u + (v + w) (associative property of addition)
• 0 + v = v = v + 0 (we have a 0-vector that is the neutral element of addition)
• v + (-v) = 0 (we have a opposite -v for each vector)
• 1*v = v (1 is the neutral element of multiplication)
• k (v + w) = kv + kw (distributive vector property)
• (k + l) v = kv + lv (distributive scalar property)
• (k l) v = k(lv) (associative property of multiplication)
where v, w, u are vectors and k, l are constants.
Adding those 2 operations we actually created a relation or function between 2 vector spaces, also called linear function, but I will get into that stuff some posts later on.
## Matrix Space:
All those operations and properties apply to the matrix space as well. And so matrixes have also an addition and a scalar multiplication operation.
So, having A(aij) and B(bij) be m*n matrixes we have:
• A + B = (aij) + (bij) = (aij + bij)
• kA = k(aij) = (k*aij)
Those 2 are actually the addition and scalar multiplication we already talked about posts before, but I wanted to point it out here so that you get a better understanding of what an vector space is. You can see that a matrix is also a vector space!
## Other vector spaces:
Vector spaces are also Sets, Polynomials and even Sequences. But, you might now wonder, well if everything is an vector space then what's the point of having them? Let me first point out that not everything is an vector space. A space is a vector space only if it follows the properties I told you before. So, we might have a space (mostly subspace that we will get in a sec) that may look like it follows the properties and operations, but the result of addition or multiplication doesn't belong to it's space or definition field.
What I mean by that is that we will always have to prove that the addition and scalar multiplication give us a result that is what we expected it to be by following the main definitions we talked about today.
These properties are used to define if a space is a vector space:
• (k - l)v = kv - lv (k, l in R and v vector of vector space V)
• 0*v = 0 (v vector of vector space V, this is very important property that will help us later on, cause if a given subspace doesn't contain the 0-vector we know that it's actually not a subspace)
• k*0 = 0 (where k is a constant and so a 0-vector multiplied with any constant gives itself)
• -1*v = -v (where v is a vector of vector space V and -1 is the opposite)
# Linear Subspace:
A subspace contains a part of a vector space, as you might already thought of. So, when having a vector space V and W (both not empty) and W is a subspace of V then:
• zero-vector 0 is part of W (cause i'ts also part of V)
• the sum u + v of 2 vectors u, v of W is also a vector of W
• the scalar multiplication of a vector u of W with a constant k of R is also a vector of W
So, we will always first check if a subspace contains the 0-vector and then try out the addition and scalar multiplication operations to see if they give us an vector that is also a part of this subspace.
Let's get into 2 examples so that you understand what we have to do.
Example 1:
Suppose we have the R^4 vector space and a space/set W = { (s, 0, t, 2s - 3t): s, t in R}. We will proof that this space is not a set, but an subspace.
First of all, when s = t = 0 we have (s, 0, t, 2s - 3t) = (0, 0, 0, 0) and so the 0-vector is part of W.
Now into addition. Let's use (s, 0, t, 2s - 3t) and (s', 0, t', 2s' - 3t') as vectors of W.
(s, 0, t, 2s - 3t) + (s', 0, t', 2s' - 3t') = (s + s', 0, t + t', 2(s + s') - 3(t + t')) that is a vector of W.
For multiplication let's use a constant k of R and vector (s, 0, t, 2s - 3t).
k*(s, 0, t, 2s - 3t) = (k*s, 0, k*t, 2ks - 3kt) that is also a vector of W.
So, we proofed that W is a subset of R^4.
Example 2:
Suppose we have the R^3 vector space and a space/set P = { (x, y, z): x + 2y - 3z = -1}. We will proof that this set is not a subspace of R^3.
We actually just have to say that the line x + 2y - 3z = -1 doesn't go through the center point (0, 0, 0) and so this set doesn't contain the 0-vector and so can't be a subspace of R^3.
So, whenever we see a line of kx + ly + mz = c != 0 then we know that this is not a subspace. In the same way if it's equal to 0 it is a subspace!
## Nullspace and Range:
Let's get back to the matrix space.
Suppose A that is a mxn matrix that is a vector space of R^n then we have
N(A) = {X in R^n: AX = 0}
where N(A) is an subspace of R^n.
We call this set the nullspace of A and this space contains the solutions of the linear system A represents if this system was homogeneous.
Another important set is the so called range of A.
R(A) = {b in R^m: AX = b for some X in R^n}
R(A) is a subspace of R^m and contains the constant column of the linear system that matrix A represents.
Example:
Let's get the set W = { (s, 0, t, 2s - 3t):s, t in R } again. This set is a subspace of R^4 as we already proofed before. Let's also suppose a matrix A where R(A) = W. So, W is the range of A.
Let's try to find out A.
R(A) = {b in R^4: AX = b for X = [s t]^T (transpose) in R^2}
If we think about matrix multiplication and having X be 2x1 (because of s, t) and knowing that b is 4x1 we know that A must be 4x2, cause 4x2 * 2x1 = 4x1. (that 2's get "eliminated")
That way we now have that:
A * [s t]^T = [s 0 t 2s-3t] ^T
S0, the rows of A (suppose they have a, b values) are:
1 0, cause a*s + b*t = s and so a = 1 and b = 0
0 0, cause a*s + b*t = 0
0 1, cause a*s + b*t = t and so a = 0 and b = 1
2 -3, cause a*s + b*t = 2s - 3t and so a = 2 and b = -3
That way we now know that A is the following matrix:
Purrfect artwork :P
I hope that I got you into the vector space world in a way that helped you understand it.
This is actually it for today and I hope you enjoyed it!
Bye!
H2
H3
H4
3 columns
2 columns
1 column
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# ${\text{Consider the following statements in respect of the function }}f(x) = {x^3} - 1, \\ x \in [ - 1,1] \\ {\text{I}}{\text{. }}f(x){\text{ is increasing in }}[ - 1,1] \\ {\text{II}}f'(x){\text{ has no root in (}} - 1,1]. \\ {\text{Which of the statements given above is/are correct?}} \\ {\text{A}}{\text{. only I}} \\ {\text{B}}{\text{. only II}} \\ {\text{C}}{\text{. Both I and II}} \\ {\text{D}}{\text{. Neither I nor II}} \\$
Last updated date: 14th Jul 2024
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${\text{Solution: - }} \\ {\text{To check a function either it is increasing or decreasing we have to double differentiate the function}} \\ {\text{and check the function in their domain either it is increasing or decreasing}} \\ {\text{in this question our function is }}f(x) = {x^3} - 1 \\ {\text{ so let's find the first derivative }}f'(x) = 3{x^2} \\ {\text{Now the second derivative is }}f''(x) = 6x{\text{ , we check the function for }}x \in [ - 1,1] \\ {\text{here }}f''(x) \in [ - 6,6]{\text{ }}\therefore {\text{ the function }}f(x){\text{ is increasing }}{\text{.}} \\ {\text{II}}{\text{. To find the root of }}f'(x){\text{ we have to equate }}f'(x) = 0. \\ \Rightarrow 3{x^2} = 0{\text{ }} \Rightarrow x = 0{\text{ }} \\ {\text{there is one root of }}f'(x){\text{ in ( - 1,1]}}{\text{.}} \\ \therefore {\text{Statement I is correct and II is incorrect }} \\ {\text{Answer is A}}{\text{.}} \\ {\text{Note: - To check a function either it is increasing or decreasing we have to differentiate the function}} \\ {\text{ when first derivative is always positive in the given domain then it is strictly increasing}}{\text{.}} \\ {\text{ }} \\$ |
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# How do you factor the trinomial $5{x^2} + 25x + 30?$
Last updated date: 19th Jun 2024
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Hint: In this question we are going to find the factors of the given trinomial expression.
First take the common term outside from the given trinomial expression.
Next we are going to factor the trinomial expression by splitting the middle term.
Multiply the coefficient of the first term by a constant in the given trinomial expression, we get a number and then find two factors for that number whose sum equals the coefficient of the middle term.
Now rewrite the polynomial by splitting the middle term using the two factors that found before and then add the first two terms and last two terms, taking common factors outside from the first and last two terms. Add the four terms of the above step and we get the desired factorization.
Complete Step by Step Solution:
In this question, we are going to factor the given trinomial expression and then find the values of $x$.
First write the given trinomial expression and mark it as $\left( 1 \right)$.
$\Rightarrow 5{x^2} + 25x + 30...\left( 1 \right)$
$5$ is common in all the terms, so take $5$ out from all the terms we get,
$\Rightarrow 5\left( {{x^2} + 5x + 6} \right)$
The above expression is of the quadratic form $a{x^2} + bx + c = 0$
Here the first term is ${x^2}$ and its coefficient is $1$
The middle term is $5x$ and its coefficient is $5$
The last term is $6$ and it is a constant.
First we are going to multiply the coefficient of the first term by the last term.
That is, $1 \times 6 = 6$
Next we are going to find factors of $6$whose sum is equal to $5$
$\Rightarrow 2 + 3 = 5$
By splitting the middle term using the factors $2$ and $3$ in the given expression
$\Rightarrow 5({x^2} + 2x + 3x + 6)$
Taking common factors outside from the two pairs
$\Rightarrow 5[x\left( {x + 2} \right) + 3\left( {x + 2} \right)]$
On rewriting we get
$\Rightarrow 5[\left( {x + 3} \right)\left( {x + 2} \right)]$
Let us equate the term and we get,
$\Rightarrow 5\left( {x + 3} \right) = 0,5\left( {x + 2} \right) = 0$
On multiply we get,
$\Rightarrow 5x + 15 = 0,5x + 10 = 0$
On we get,
$\Rightarrow 5x = - 15,5x = - 10$
Let us divide the term and we get
$\Rightarrow x = \dfrac{{ - 15}}{5},x = \dfrac{{ - 10}}{5}$
Then we get,
$\Rightarrow x = - 3,x = - 2$
Hence we get,
$\Rightarrow x = - 2, - 3$
The required factors of the trinomial expression $5{x^2} + 25x + 30$ are $5[\left( {x + 3} \right)\left( {x + 2} \right)]$.
Note: The following are some of the factoring methods to solve the expression: factoring out the GCF, the sum product pattern, the grouping method, the perfect square trinomial pattern, the difference of square pattern.
We can check our factoring by multiplying them all out to see if we get the original expression. If we do, our factoring is correct, otherwise we had to try again. |
# Subject 2. Probability Function
Every random variable is associated with a probability distribution that describes the variable completely. A probability function is one way to view a probability distribution. It specifies the probability that the random variable takes on a specific value; P(X = x) is the probability that a random variable X takes on the value x.
A probability function has two key properties:
• 0 ≤ P(X=x) ≤ 1, because probability is a number between 0 and 1.
• ΣP(X=x) = 1. The sum of the probabilities P(X=x) over all values of X equals 1. If there is an exhaustive list of the distinct possible outcomes of a random variable and the probabilities of each are added up, the probabilities must sum to 1.
The following examples will utilize these two properties in order to examine whether they are probability functions.
Example 1
p(x) = x/6 for X = 1, 2, 3, and p(x) = 0 otherwise
• Substituting into p(x): p(1) = 1/6, p(2) = 2/6 and p(3) = 3/6
Note that it is not necessary to substitute in any other values, as p(x) is only non-zero for X values 1, 2 and 3.
In all 3 cases, p(x) lies between 0 and 1, as 1/6, 2/6 and 3/6 are all values in the range 0 to 1 inclusive.
So, the first property is satisfied.
• Summing the probabilities gives 1/6 + 2/6 + 3/6 = 1, showing the second property is also satisfied.
Example 2
p(x) = (2x - 3)/16 for X = 1, 2, 3, 4 and p(x) = 0 otherwise
Substituting into p(x): p(1) = -1/16
STOP HERE!
It is impossible for any probability to be negative, so it's not necessary to continue. Property 1 is violated, so it can be said straightaway that p(x) is not a probability function.
Note that individual probabilities in a continuous case cannot occur, so P(X = 5), say, is 0 if X is continuous.
In a continuous case, only a range of values can be considered (that is, 0 < X < 10), whereas in a discrete case, individual values have positive probabilities associated with them.
For a discrete random variable, the shorthand notation is p(x) = P(X = x). For continuous random variables, the probability function is denoted f(x) and called probability density function (pdf), or just the density. This function is effectively the continuous analogue of the discrete probability function p(x).
• The probability density function, which has the symbol f(x), does not give probabilities, despite its name. Instead, it is the area between the graph and the horizontal axis that gives probabilities. Because of this, the height of f(x) is not restricted to the range 0 to 1, and the graph, which in itself is not a probability, is unrestricted as far as its height is concerned.
• From this information, it follows that the area under the entire graph (i.e., between the graph and the x-axis) must equal 1, because this area encapsulates all the probability contained in the random variable. Recall that for discrete distributions, the probabilities add up to 1.
• Because continuous random variables are concerned with a range of values, individual values have no probabilities, because there is no area associated with individual values. Rather, probabilities are calculated over a range of values. Another way of saying this is that p(x) = 0 for every individual X.
• If a discrete random variable has many possible outcomes, then it can be treated as a continuous random variable for conciseness, and ranges of values can be considered in determining probabilities. |
# What is 3.33 as a Fraction?
3.33 can be written as a fraction in a number of different ways. The easiest way is to divide the decimal number by 10 and then multiply the answer by 3.
Checkout this video:
## What is 3.33 as a Fraction?
3.33 is a decimal and 33.3 is a mixed number. To convert a decimal to a fraction, we divide the decimal by 1. To convert a mixed number to a fraction, we first convert the whole number to a fractions (in this case, 3 3/10 = 30/10) and then add the two fractions together. In this case, 3.33 = 33.3/10 = 30/10 + 3/10 = 6/2 + 1/5 = 12/5 + 2/15 = 24/15 + 4/45 = 28/45
## 3.33 is an Irrational Number
3.33 is an irrational number because it cannot be expressed as a rational number. A rational number is a number that can be expressed as a fraction, where the numerator and denominator are integers. An irrational number is a number that cannot be expressed as a rational number.
3.33 is an example of an irrational number because it cannot be expressed as a fraction. If you try to express 3.33 as a fraction, you will get an infinite decimal expansion, which is not possible for a fraction.
3.33 is an irrational number because it cannot be expressed as a rational number. A rational number is a number that can be expressed as a fraction, where the numerator and denominator are integers. An irrational number is a number that cannot be expressed as a rational number.
3.33 is an example of an irrational number because it cannot be expressed as
## 3.33 is a Repeating Decimal
3.33 is a repeating decimal, which means that the digits after the decimal point continue infinitely without ever repeating in a pattern. 3.33 can be converted to a fraction by placing the digits after the decimal point over 9:
3.33 = 3 + (1/3) + (1/9) + (1/27) + …
The 3 outside of the parentheses represents the number of whole units, and each successive term inside the parentheses represents a fraction of one unit. This can be written more compactly as:
3.33 = 3 + 1/3 + 1/9 + 1/27 + …
## Converting 3.33 to a Fraction
In order to convert 3.33 to a fraction, we need to find a way toturn the decimal number into one that is expressed as a fraction. In other words, we need to find a way to express 3.33 as 3 33/100.
This can be done by moving the decimal point two places to the left, which would give us 33.3. We can then express this as a fraction by placing the number over 100, like this: 33.3 = 333/100. Finally, we can reduce this fraction by dividing both the numerator and denominator by 3, which gives us 3 33/100 = 111/30.
## Determining the Value of 3.33
To understand what 3.33 means as a fraction, we must first understand what decimals and fractions are. A decimal is a number that represents a quantity that is less than one. For example, 0.5 represents half or 1/2. When we see a decimal like 3.33, this means 3 and 33/100. To convert 3.33 to a fraction, we need to multiply both the 3 and the 0.33 by 100 so that there are no more decimals left in the fraction:
3.33 = 333/100
This means that 3.33 as a fraction is equal to 333/100. |
Chain Rule
by Justin Skycak on
When taking derivatives of compositions of functions, we can ignore the inside of a function as long as we multiply by the derivative of the inside afterwards.
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Chain Rule. In Justin Math: Calculus. https://justinmath.com/chain-rule/
The chain rule tells us how to take derivatives of compositions of functions. Informally, it says that we can “forget” about the inside of a function when we take the derivative, as long as we multiply by the derivative of the inside afterwards.
For example, to differentiate $(x^2+1)^{100}$, we can use the power rule, as long as we multiply by the derivative of the inside $(x^2+1)$ afterwards.
\begin{align*} \left[ (x^2+1)^{100} \right]' &= 100 (x^2+1)^{99} (x^2+1)' \\ &= 100(x^2+1)^{99}(2x) \\ &= 200x(x^2+1)^{99} \end{align*}
Substitution
More precisely, the chain rule states that we can make a substitution $u$ for an expression of $x$, as long as we multiply by the derivative of the substitution afterwards.
\begin{align*} \frac{df}{dx} = \frac{df}{du} \frac{du}{dx} \end{align*}
To differentiate the function $f(x)=(x^2+1)^{100}$, we substituted $u=x^2+1$ to simplify the function to $f(u)=u^{100}$.
Intuitively, the chain rule says that we can cancel derivatives just like we cancel fractions.
\begin{align*} \frac{df}{dx} = \frac{df}{du\hspace{-0.5cm}\textbf{---}} \frac{du\hspace{-0.5cm}\textbf{---}}{dx} \end{align*}
We can extend this to an unlimited number of substitutions, building a “chain” of cancellations.
\begin{align*} \frac{df}{dx} = \frac{df}{du_1\hspace{-0.75cm}\textbf{------}} \frac{du_1\hspace{-0.75cm}\textbf{------}}{du_2\hspace{-0.75cm}\textbf{------}} \frac{du_2\hspace{-0.75cm}\textbf{------}}{du_3\hspace{-0.75cm}\textbf{------}} \cdots \frac{du_{n-1}\hspace{-1.25cm}\textbf{------}}{du_n\hspace{-0.75cm}\textbf{------}} \frac{du_n\hspace{-0.75cm}\textbf{------}}{dx} \end{align*}
For example, to differentiate the function $f(x)=(((x^3+1)^4+1)^5+1)^6$ we can proceed one layer at a time.
\begin{align*} \left[ (((x^3+1)^4+1)^5+1)^6 \right]' &=6(((x^3+1)^4+1)^5+1)^5 \\ &\hspace{.5cm} \cdot \left[ ((x^3+1)^4+1)^5+1 \right]' \\[5pt] &=6(((x^3+1)^4+1)^5+1)^5 \\ &\hspace{.5cm}\cdot 5((x^3+1)^4+1)^4 \\ &\hspace{.5cm}\cdot \left[ (x^3+1)^4+1 \right]' \\[5pt] &=6(((x^3+1)^4+1)^5+1)^5 \\ &\hspace{.5cm}\cdot 5((x^3+1)^4+1)^4 \\ &\hspace{.5cm}\cdot 4(x^3+1)^3 \\ &\hspace{.5cm}\cdot \left[ x^3+1 \right]' \\[5pt] &=6(((x^3+1)^4+1)^5+1)^5 \\ &\hspace{.5cm}\cdot 5((x^3+1)^4+1)^4 \\ &\hspace{.5cm}\cdot 4(x^3+1)^3 \\ &\hspace{.5cm}\cdot 3x^2 \end{align*}
Exercises
Use the chain rule to find the derivatives of the following functions. (You can view the solution by clicking on the problem.)
\begin{align*}1) \hspace{.5cm} f(x)=(2x^2+1)^3 \end{align*}
Solution:
\begin{align*} f'(x)= 12x(2x^2+1)^2 \end{align*}
\begin{align*}2) \hspace{.5cm} f(x)= (x^4-x^2)^8 \end{align*}
Solution:
\begin{align*} f'(x)= 8(4x^3-2x)(x^4-x^2)^7 \end{align*}
\begin{align*}3) \hspace{.5cm} f(x)=4 \sqrt{ x^2+1 } \end{align*}
Solution:
\begin{align*} f'(x)= \frac{x}{\sqrt{x^2+1}} \end{align*}
\begin{align*}4) \hspace{.5cm} f(x)=\sqrt{ (2x+1)^2+3 } \end{align*}
Solution:
\begin{align*} f'(x)= \frac{4x+2}{ \sqrt{(2x+1)^2+3}} \end{align*}
\begin{align*}5) \hspace{.5cm} f(x)= \frac{1}{3x-2} \end{align*}
Solution:
\begin{align*} f'(x)= \frac{-3}{(3x-2)^2} \end{align*}
\begin{align*}6) \hspace{.5cm} f(x)= \frac{7}{(x^2-3)^2} \end{align*}
Solution:
\begin{align*} f'(x)= \frac{-28x}{(x^2-3)^2} \end{align*}
\begin{align*}7) \hspace{.5cm} f(x)= \frac{2}{\sqrt{1-x^5} } \end{align*}
Solution:
\begin{align*} f'(x)= \frac{-5x^4}{(1-x^5)^\frac{3}{2} } \end{align*}
\begin{align*}8)\hspace{.5cm} f(x) = \sqrt{ \sqrt{x} + 1} \end{align*}
Solution:
\begin{align*} f'(x)= \frac{1}{ 4 \sqrt{x} \sqrt{ \sqrt{x} + 1 } } \end{align*}
\begin{align*}9) \hspace{.5cm} f(x) = \left( \sqrt{x}+1 \right)^5 \end{align*}
Solution:
\begin{align*} f'(x)= \frac{ 5 \left( \sqrt{x} + 1 \right)^4 }{2 \sqrt{x} } \end{align*}
\begin{align*}10) \hspace{.5cm} f(x) =\left( x^\frac{3}{2} + x^\frac{4}{3} \right)^\frac{5}{4} \end{align*}
Solution:
\begin{align*} f'(x) = \frac{5}{4} \left( x^\frac{3}{2} + x^\frac{4}{3} \right)^\frac{1}{4} \left( \frac{3}{2} x^\frac{1}{2} + \frac{4}{3} x^\frac{1}{3} \right) \end{align*}
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Chain Rule. In Justin Math: Calculus. https://justinmath.com/chain-rule/
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Home / Algebra / Solving Equations and Inequalities / Quadratic Equations - Part I
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### Section 2-5 : Quadratic Equations - Part I
Before proceeding with this section we should note that the topic of solving quadratic equations will be covered in two sections. This is done for the benefit of those viewing the material on the web. This is a long topic and to keep page load times down to a minimum the material was split into two sections.
So, we are now going to solve quadratic equations. First, the standard form of a quadratic equation is
$a{x^2} + bx + c = 0\hspace{0.25in}a \ne 0$
The only requirement here is that we have an $${x^2}$$ in the equation. We guarantee that this term will be present in the equation by requiring $$a \ne 0$$. Note however, that it is okay if $$b$$ and/or $$c$$ are zero.
There are many ways to solve quadratic equations. We will look at four of them over the course of the next two sections. The first two methods won’t always work yet are probably a little simpler to use when they work. This section will cover these two methods. The last two methods will always work, but often require a little more work or attention to get correct. We will cover these methods in the next section.
So, let’s get started.
Solving by Factoring
As the heading suggests we will be solving quadratic equations here by factoring them. To do this we will need the following fact.
${\mbox{If }}ab = 0{\mbox{ then either }}a = 0{\mbox{ and/or }}b = 0$
This fact is called the zero factor property or zero factor principle. All the fact says is that if a product of two terms is zero then at least one of the terms had to be zero to start off with.
Notice that this fact will ONLY work if the product is equal to zero. Consider the following product.
$ab = 6$
In this case there is no reason to believe that either $$a$$ or $$b$$ will be 6. We could have $$a = 2$$ and $$b = 3$$ for instance. So, do not misuse this fact!
To solve a quadratic equation by factoring we first must move all the terms over to one side of the equation. Doing this serves two purposes. First, it puts the quadratics into a form that can be factored. Secondly, and probably more importantly, in order to use the zero factor property we MUST have a zero on one side of the equation. If we don’t have a zero on one side of the equation we won’t be able to use the zero factor property.
Let’s take a look at a couple of examples. Note that it is assumed that you can do the factoring at this point and so we won’t be giving any details on the factoring. If you need a review of factoring you should go back and take a look at the Factoring section of the previous chapter.
Example 1 Solve each of the following equations by factoring.
1. $${x^2} - x = 12$$
2. $${x^2} + 40 = - 14x$$
3. $${y^2} + 12y + 36 = 0$$
4. $$4{m^2} - 1 = 0$$
5. $$3{x^2} = 2x + 8$$
6. $$10{z^2} + 19z + 6 = 0$$
7. $$5{x^2} = 2x$$
Show All Solutions Hide All Solutions
Show Discussion
Now, as noted earlier, we won’t be putting any detail into the factoring process, so make sure that you can do the factoring here.
a $${x^2} - x = 12$$ Show Solution
First, get everything on side of the equation and then factor.
\begin{align*}{x^2} - x - 12 & = 0\\ \left( {x - 4} \right)\left( {x + 3} \right) & = 0\end{align*}
Now at this point we’ve got a product of two terms that is equal to zero. This means that at least one of the following must be true.
\begin{align*}x - 4 & = 0 & \hspace{0.25in}{\mbox{OR}}\hspace{0.25in} & & x + 3 & = 0\\ x & = 4&\hspace{0.25in}{\mbox{OR}}\hspace{0.25in} & & x & = - 3\end{align*}
Note that each of these is a linear equation that is easy enough to solve. What this tell us is that we have two solutions to the equation, $$x = 4$$ and $$x = - 3$$. As with linear equations we can always check our solutions by plugging the solution back into the equation. We will check $$x = - 3$$ and leave the other to you to check.
\begin{align*}{\left( { - 3} \right)^2} - \left( { - 3} \right) & \mathop = \limits^? 12\\ 9 + 3 & \mathop = \limits^? 12\\ 12 & = 12\,\,\,\,{\mbox{OK}}\end{align*}
So, this was in fact a solution.
b $${x^2} + 40 = - 14x$$ Show Solution
As with the first one we first get everything on side of the equal sign and then factor.
\begin{align*}{x^2} + 40 + 14x & = 0\\ \left( {x + 4} \right)\left( {x + 10} \right) & = 0\end{align*}
Now, we once again have a product of two terms that equals zero so we know that one or both of them have to be zero. So, technically we need to set each one equal to zero and solve. However, this is usually easy enough to do in our heads and so from now on we will be doing this solving in our head.
The solutions to this equation are,
$x = - 4\hspace{0.25in}{\mbox{AND}}\hspace{0.25in}x = - 10$
To save space we won’t be checking any more of the solutions here, but you should do so to make sure we didn’t make any mistakes.
c $${y^2} + 12y + 36 = 0$$ Show Solution
In this case we already have zero on one side and so we don’t need to do any manipulation to the equation all that we need to do is factor. Also, don’t get excited about the fact that we now have $$y$$’s in the equation. We won’t always be dealing with $$x$$’s so don’t expect to always see them.
So, let’s factor this equation.
\begin{align*}{y^2} + 12y + 36 & = 0\\ {\left( {y + 6} \right)^2} & = 0\\ \left( {y + 6} \right)\left( {y + 6} \right) & = 0\end{align*}
In this case we’ve got a perfect square. We broke up the square to denote that we really do have an application of the zero factor property. However, we usually don’t do that. We usually will go straight to the answer from the squared part.
The solution to the equation in this case is,
$y = - 6$
We only have a single value here as opposed to the two solutions we’ve been getting to this point. We will often call this solution a double root or say that it has multiplicity of 2 because it came from a term that was squared.
d $$4{m^2} - 1 = 0$$ Show Solution
As always let’s first factor the equation.
\begin{align*}4{m^2} - 1 & = 0\\ \left( {2m - 1} \right)\left( {2m + 1} \right) & = 0\end{align*}
Now apply the zero factor property. The zero factor property tells us that,
\begin{align*}2m - 1 & = 0 & \hspace{0.25in}{\mbox{OR}}\hspace{0.25in} & & 2m + 1 & = 0\\ 2m & = 1 & \hspace{0.25in}{\mbox{OR}}\hspace{0.25in} & & m & = - 1\\ m & = \frac{1}{2} & \hspace{0.25in}{\mbox{OR}}\hspace{0.25in} & & m & = - \frac{1}{2}\end{align*}
Again, we will typically solve these in our head, but we needed to do at least one in complete detail. So, we have two solutions to the equation.
$m = \frac{1}{2}\hspace{0.25in}{\mbox{AND}}\hspace{0.25in}m = - \frac{1}{2}$
e $$3{x^2} = 2x + 8$$ Show Solution
Now that we’ve done quite a few of these, we won’t be putting in as much detail for the next two problems. Here is the work for this equation.
\begin{align*}3{x^2} - 2x - 8 & = 0\\ \left( {3x + 4} \right)\left( {x - 2} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - \frac{4}{3}\,\,\,\,\,{\mbox{and}}\,\,\,\,\,x = 2\end{align*}
f $$10{z^2} + 19z + 6 = 0$$ Show Solution
Again, factor and use the zero factor property for this one.
\begin{align*}10{z^2} + 19z + 6 & = 0\\ \left( {5z + 2} \right)\left( {2z + 3} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in} z = - \frac{2}{5}\,\,\,\,\,{\mbox{and }}z = - \frac{3}{2}\end{align*}
g $$5{x^2} = 2x$$ Show Solution
This one always seems to cause trouble for students even though it’s really not too bad.
First off. DO NOT CANCEL AN $$x$$ FROM BOTH SIDES!!!! Do you get the idea that might be bad? It is. If you cancel an $$x$$ from both sides, you WILL miss a solution so don’t do it. Remember we are solving by factoring here so let’s first get everything on one side of the equal sign.
$5{x^2} - 2x = 0$
Now, notice that all we can do for factoring is to factor an $$x$$ out of everything. Doing this gives,
$x\left( {5x - 2} \right) = 0$
From the first factor we get that $$x = 0$$ and from the second we get that $$x = \frac{2}{5}$$. These are the two solutions to this equation. Note that is we’d canceled an $$x$$ in the first step we would NOT have gotten $$x = 0$$ as an answer!
Let’s work another type of problem here. We saw some of these back in the Solving Linear Equations section and since they can also occur with quadratic equations we should go ahead and work on to make sure that we can do them here as well.
Example 2 Solve each of the following equations.
1. $$\displaystyle \frac{1}{{x + 1}} = 1 - \frac{5}{{2x - 4}}$$
2. $$\displaystyle x + 3 + \frac{3}{{x - 1}} = \frac{{4 - x}}{{x - 1}}$$
Show All Solutions Hide All Solutions
Show Discussion
Okay, just like with the linear equations the first thing that we’re going to need to do here is to clear the denominators out by multiplying by the LCD. Recall that we will also need to note value(s) of $$x$$ that will give division by zero so that we can make sure that these aren’t included in the solution.
a $$\displaystyle \frac{1}{{x + 1}} = 1 - \frac{5}{{2x - 4}}$$ Show Solution
The LCD for this problem is $$\left( {x + 1} \right)\left( {2x - 4} \right)$$ and we will need to avoid $$x = - 1$$ and $$x = 2$$ to make sure we don’t get division by zero. Here is the work for this equation.
\begin{align*}\left( {x + 1} \right)\left( {2x - 4} \right)\left( {\frac{1}{{x + 1}}} \right) & = \left( {x + 1} \right)\left( {2x - 4} \right)\left( {1 - \frac{5}{{2x - 4}}} \right)\\ 2x - 4 & = \left( {x + 1} \right)\left( {2x - 4} \right) - 5\left( {x + 1} \right)\\ 2x - 4 & = 2{x^2} - 2x - 4 - 5x - 5\\ 0 & = 2{x^2} - 9x - 5\\ 0 & = \left( {2x + 1} \right)\left( {x - 5} \right)\end{align*}
So, it looks like the two solutions to this equation are,
$x = - \frac{1}{2}\,\,\,\,\,\,\,{\mbox{and }}x = 5$
Notice as well that neither of these are the values of $$x$$ that we needed to avoid and so both are solutions.
b $$\displaystyle x + 3 + \frac{3}{{x - 1}} = \frac{{4 - x}}{{x - 1}}$$ Show Solution
In this case the LCD is $$x - 1$$ and we will need to avoid $$x = 1$$ so we don’t get division by zero. Here is the work for this problem.
\begin{align*}\left( {x - 1} \right)\left( {x + 3 + \frac{3}{{x - 1}}} \right) & = \left( {\frac{{4 - x}}{{x - 1}}} \right)\left( {x - 1} \right)\\ \left( {x - 1} \right)\left( {x + 3} \right) + 3 & = 4 - x\\ {x^2} + 2x - 3 + 3 & = 4 - x\\ {x^2} + 3x - 4 & = 0\\ \left( {x - 1} \right)\left( {x + 4} \right) & = 0\end{align*}
So, the quadratic that we factored and solved has two solutions, $$x = 1$$ and $$x = - 4$$. However, when we found the LCD we also saw that we needed to avoid $$x = 1$$ so we didn’t get division by zero. Therefore, this equation has a single solution,
$x = - 4$
Before proceeding to the next topic we should address that this idea of factoring can be used to solve equations with degree larger than two as well. Consider the following example.
Example 3 Solve $$5{x^3} - 5{x^2} - 10x = 0$$.
Show Solution
The first thing to do is factor this equation as much as possible. In this case that means factoring out the greatest common factor first. Here is the factored form of this equation.
\begin{align*}5x\left( {{x^2} - x - 2} \right) & = 0\\ 5x\left( {x - 2} \right)\left( {x + 1} \right) & = 0\end{align*}
Now, the zero factor property will still hold here. In this case we have a product of three terms that is zero. The only way this product can be zero is if one of the terms is zero. This means that,
\begin{align*}5x & = 0\hspace{0.25in} \Rightarrow & x & = 0\\ x - 2 & = 0\hspace{0.25in} \Rightarrow & x & = 2\\ x + 1 & = 0\hspace{0.25in} \Rightarrow & x & = - 1\end{align*}
So, we have three solutions to this equation.
So, provided we can factor a polynomial we can always use this as a solution technique. The problem is, of course, that it is sometimes not easy to do the factoring.
Square Root Property
The second method of solving quadratics we’ll be looking at uses the square root property,
${\mbox{If }}{p^2} = d{\mbox{ then }}p = \pm \sqrt d$
There is a (potentially) new symbol here that we should define first in case you haven’t seen it yet. The symbol “$$\pm$$” is read as : “plus or minus” and that is exactly what it tells us. This symbol is shorthand that tells us that we really have two numbers here. One is $$p = \sqrt d$$ and the other is $$p = - \sqrt d$$. Get used to this notation as it will be used frequently in the next couple of sections as we discuss the remaining solution techniques. It will also arise in other sections of this chapter and even in other chapters.
This is a fairly simple property to use, however it can only be used on a small portion of the equations that we’re ever likely to encounter. Let’s see some examples of this property.
Example 4 Solve each of the following equations.
1. $${x^2} - 100 = 0$$
2. $$25{y^2} - 3 = 0$$
3. $$4{z^2} + 49 = 0$$
4. $${\left( {2t - 9} \right)^2} = 5$$
5. $${\left( {3x + 10} \right)^2} + 81 = 0$$
Show All Solutions Hide All Solutions
Show Discussion
There really isn’t all that much to these problems. In order to use the square root property all that we need to do is get the squared quantity on the left side by itself with a coefficient of 1 and the number on the other side. Once this is done we can use the square root property.
a $${x^2} - 100 = 0$$ Show Solution
This is a fairly simple problem so here is the work for this equation.
${x^2} = 100\hspace{0.25in}x = \pm \sqrt {100} = \pm 10$
So, there are two solutions to this equation, $$x = \pm 10$$. Remember this means that there are really two solutions here, $$x = - 10$$ and $$x = 10$$.
b $$25{y^2} - 3 = 0$$ Show Solution
Okay, the main difference between this one and the previous one is the 25 in front of the squared term. The square root property wants a coefficient of one there. That’s easy enough to deal with however; we’ll just divide both sides by 25. Here is the work for this equation.
\begin{align*}25{y^2} & = 3\\ {y^2} & = \frac{3}{{25}}\hspace{0.25in} \Rightarrow \hspace{0.25in}y = \pm \sqrt {\frac{3}{{25}}} = \pm \frac{{\sqrt 3 }}{5}\end{align*}
In this case the solutions are a little messy, but many of these will do so don’t worry about that. Also note that since we knew what the square root of 25 was we went ahead and split the square root of the fraction up as shown. Again, remember that there are really two solutions here, one positive and one negative.
c $$4{z^2} + 49 = 0$$ Show Solution
This one is nearly identical to the previous part with one difference that we’ll see at the end of the example. Here is the work for this equation.
\begin{align*}4{z^2} & = - 49\\ {z^2} & = - \frac{{49}}{4}\hspace{0.25in} \Rightarrow \hspace{0.25in}z = \pm \sqrt { - \frac{{49}}{4}} = \pm \,i\,\sqrt {\frac{{49}}{4}} = \pm \frac{7}{2}i\end{align*}
So, there are two solutions to this equation : $$z = \pm \frac{7}{2}i$$. Notice as well that they are complex solutions. This will happen with the solution to many quadratic equations so make sure that you can deal with them.
d $${\left( {2t - 9} \right)^2} = 5$$ Show Solution
This one looks different from the previous parts, however it works the same way. The square root property can be used anytime we have something squared equals a number. That is what we have here. The main difference of course is that the something that is squared isn’t a single variable it is something else. So, here is the application of the square root property for this equation.
$2t - 9 = \pm \sqrt 5$
Now, we just need to solve for $$t$$ and despite the “plus or minus” in the equation it works the same way we would solve any linear equation. We will add 9 to both sides and then divide by a 2.
\begin{align*}2t & = 9 \pm \sqrt 5 \\ t & = \frac{1}{2}\left( {9 \pm \sqrt 5 } \right) = \frac{9}{2} \pm \frac{{\sqrt 5 }}{2}\end{align*}
Note that we multiplied the fraction through the parenthesis for the final answer. We will usually do this in these problems. Also, do NOT convert these to decimals unless you are asked to. This is the standard form for these answers. With that being said we should convert them to decimals just to make sure that you can. Here are the decimal values of the two solutions.
$t = \frac{9}{2} + \frac{{\sqrt 5 }}{2} = 5.61803\hspace{0.25in}{\mbox{and}}\hspace{0.25in}t = \frac{9}{2} - \frac{{\sqrt 5 }}{2} = 3.38197$
e $${\left( {3x + 10} \right)^2} + 81 = 0$$ Show Solution
In this final part we’ll not put much in the way of details into the work.
\begin{align*}{\left( {3x + 10} \right)^2} & = - 81\\ 3x + 10 & = \pm \,9\,i\\ 3x & = - 10 \pm \,9\,i\\ x & = - \frac{{10}}{3} \pm 3\,i\end{align*}
So we got two complex solutions again and notice as well that with both of the previous part we put the “plus or minus” part last. This is usually the way these are written.
As mentioned at the start of this section we are going to break this topic up into two sections for the benefit of those viewing this on the web. The next two methods of solving quadratic equations, completing the square and quadratic formula, are given in the next section. |
# 311302 Basic Mathematics Books/Notes MSBTE Diploma 'K' Scheme
311302 Basic Mathematics
Click on "Click & Wait" Button & Wait for 15 Sec, You Will Get Books / Notes PDF Link:
Basic Mathematics: Foundations and Fundamental Concepts
Introduction
Basic mathematics lays the foundation for more advanced mathematical studies and has practical applications in everyday life, from budgeting to cooking. Understanding these foundational concepts is crucial for further exploration in the field.
1. Numbers:
• Natural Numbers: These are counting numbers starting from 1 (e.g., 1, 2, 3, ...).
• Whole Numbers: Natural numbers including 0 (e.g., 0, 1, 2, ...).
• Integers: Whole numbers and their negatives (e.g., -3, -2, -1, 0, 1, 2, ...).
• Rational Numbers: Numbers that can be expressed as a fraction where the numerator and denominator are integers, and the denominator is not zero.
• Irrational Numbers: Numbers that cannot be expressed as a simple fraction (e.g., the square root of 2, π).
2. Basic Operations:
• Subtraction (-): Taking one number away from another.
• Multiplication (x): Finding the total of adding a number, a certain number of times.
• Division (÷): Splitting into equal parts or groups.
3. Arithmetic Properties:
• Commutative Property: The order of numbers does not affect the result in addition or multiplication (e.g., 3 + 4 = 4 + 3).
• Associative Property: The way numbers are grouped does not affect the sum or product (e.g., (3 + 4) + 5 = 3 + (4 + 5)).
• Distributive Property: Multiplying a number by a group of numbers added together is the same as doing each multiplication separately (e.g., 5 x (2 + 3) = 5x2 + 5x3).
4. Fractions:
• Numerator: The top number, representing parts taken from the whole.
• Denominator: The bottom number, indicating the total number of equal parts.
• Equivalent Fractions: Different fractions that represent the same value (e.g., 1/2 and 2/4).
5. Decimals and Percentages:
• Decimal: Represents a number using the base-10 system, with a decimal point to indicate values less than one.
• Percentage: A way to express numbers as a fraction of 100. For instance, 25% is equivalent to 25/100 or 0.25.
6. Geometry Basics:
• Points: Exact locations in space.
• Lines: Straight paths that extend in two opposite directions without ending.
• Planes: Flat surfaces that extend indefinitely in all directions.
• Shapes: Two-dimensional (circle, square) or three-dimensional (sphere, cube) figures.
7. Measurements:
Understanding basic units for:
• Length: Meters, centimeters, inches.
• Weight: Grams, kilograms, pounds.
• Volume: Liters, milliliters, gallons.
Conclusion:
Basic mathematics encompasses a range of essential concepts that not only set the stage for advanced studies but also permeate our daily lives. A good grasp of these foundational elements is vital for analytical thinking, problem-solving, and informed decision-making. |
# Synthetic Summer Fun
Today, for some summer fun, let’s look at synthetic division a/k/a synthetic substitution. I’ll assume you all know how to do that since it is a pretty common pre-calculus topic and even comes up again in calculus.
Why Does Synthetic Division Work?
An example: consider the polynomial
$P(x)=2{{x}^{4}}-3{{x}^{3}}-11{{x}^{2}}+14x-1$.
This can be written in nested form like this
$P(x)=((((2x-3)x-11)x+14)x-1)$
To evaluate this last expression at, say x = 2, we do the arithmetic as follows:
1. 2 x 2 – 3 = 1
2. 2 x 1 – 11 = –9
3. 2 x (–9) + 14 = –4
4. 2 x (–4) – 1 = – 9 = f(2)
Notice that this requires only multiplication and addition or subtraction, no raising to powers. More to the point, this is the same arithmetic, in the same order when you do the evaluation by synthetic division, and the work is a little easier to keep track of.
$\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}$
Synthetic division has another advantage: the other numbers in the second row are the coefficients of a quotient polynomial, a polynomial of one less degree that the original. So,
$\displaystyle \frac{P(x)}{x-2}=2{{x}^{3}}+{{x}^{2}}-9x-4+\frac{-9}{x-2}$
The Remainder Theorem and the Factor Theorem
In general, a polynomial of degree n, divided by a linear factor (x – a) gives a polynomial Q(x) of degree n – 1 and a remainder R
$\displaystyle \frac{P(x)}{x-a}=Q(x)+\frac{R}{x-a}$
Or
$P(x)=Q(x)(x-a)+R$
From here it is easy to see that $P(a)=R$. This is called the remainder theorem. It has a corollary called the factor theorem: If R = 0, then (x – a) is a factor of P(x).
Calculus
But wait there is more: differentiating the equation above using the product rule gives
${P}'(x)=Q(x)(1)+Q(x)(x-a)+0$ and substituting x = a gives
${P}'(a)=Q(a)$. The value of the quotient polynomial at a is the derivative of the original polynomial at a.
Of course, we could also rewrite the same equation as $\displaystyle \frac{P(x)-P(a)}{x-a}=Q(x)$ . Then
$\displaystyle {P}'(a)=\underset{x\to a}{\mathop{\lim }}\,\frac{P(x)-P(a)}{x-a}=\underset{x\to a}{\mathop{\lim }}\,Q(x)=Q(a)$
Taylor Series
But wait, there’s even more.
A polynomial is a Maclaurin series in which all the terms after the nth term are zero. When you students are first learning how to write a Taylor series, by finding all the derivatives and substituting in the general term, a good exercise is to have them write the Taylor series for a polynomial centered away from the origin. For the example above:
$P(x)=-9-2\left( x-2 \right)+19{{\left( x-2 \right)}^{2}}+13{{\left( x-2 \right)}^{3}}+2{{\left( x-2 \right)}^{4}}$
Then ask them to expand the expression above and collect term etc. They should get the original polynomial again (and have some great practice expand powers of a binomial).
Can synthetic division help us? Yes, of course. Here, is the original computation again:
$\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}$
If we ignore the –9 and divide the quotient numbers by 2 we get
$\begin{matrix} {} & 2 & 1 & -9 & -4 \\ 2) & 2 & 5 & 1 & -2 \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(2) \\ \end{matrix}$
$\begin{matrix} {} & 2 & 5 & 1 \\ 2) & 2 & 9 & 19 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}'}'(2)}{2} \\ \end{matrix}$
And again
$\begin{matrix} {} & 2 & 9 \\ 2) & 2 & 13 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{{P}'}'}'\left( 2 \right)}{3!} \\ \end{matrix}$
One more time
$\begin{matrix} {} & 2 \\ 2) & 2 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}^{(4)}}\left( 2 \right)}{4!} \\ \end{matrix}$
What do you see? Right, the last numbers in each computation, –9, –2, 19, 13, and 2, are the coefficients of the Taylor polynomial!
If you really want to dive this home and have some more summer fun here’s the start of a proof (at least for n = 4). Let
$P(x)={{c}_{4}}{{x}^{4}}+{{c}_{3}}{{x}^{3}}+{{c}_{2}}{{x}^{3}}+{{c}_{1}}x+{{c}_{0}}$ and divide this by a:
$\begin{matrix} {} & {{c}_{4}} & {{c}_{3}} & {{c}_{2}} & {{c}_{1}} & {{c}_{0}} \\ a) & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{a}}a+{{c}_{1}} & {{c}_{4}}{{a}^{4}}+{{c}_{3}}{{a}^{3}}+{{c}_{a}}{{a}^{2}}+{{c}_{1}}a+{{c}_{0}}=P(a) \\ \end{matrix}$
Again
$\begin{matrix} {} & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{2}}a+{{c}_{1}} \\ a) & {{c}_{4}} & 2{{c}_{4}}a+{{c}_{3}} & 3{{c}_{4}}{{a}^{2}}+2{{c}_{3}}a+{{c}_{2}} & 4{{c}_{4}}{{a}^{3}}+3{{c}_{3}}{{a}^{2}}+2{{c}_{2}}a+{{c}_{1}} \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(a) \\ \end{matrix}$
And I’ll leave the rest to you. Really, why should I have all the fun?
.
## 1 thought on “Synthetic Summer Fun”
1. The nested form and the Calculus sections are exemplary. Thank you for your work.
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# Prime Factors of 1665
Prime Factors of 1665 are 3, 3, 5, and 37
#### How to find prime factors of a number
1. Prime Factorization of 1665 by Division Method 2. Prime Factorization of 1665 by Factor Tree Method 3. Definition of Prime Factors 4. Frequently Asked Questions
#### Steps to find Prime Factors of 1665 by Division Method
To find the primefactors of 1665 using the division method, follow these steps:
• Step 1. Start dividing 1665 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 1665, which is 3. Divide 1665 by 3 to obtain the quotient (555).
1665 ÷ 3 = 555
• Step 3. Repeat step 1 with the obtained quotient (555).
555 ÷ 3 = 185
185 ÷ 5 = 37
37 ÷ 37 = 1
So, the prime factorization of 1665 is, 1665 = 3 x 3 x 5 x 37.
#### Steps to find Prime Factors of 1665 by Factor Tree Method
We can follow the same procedure using the factor tree of 1665 as shown below:
So, the prime factorization of 1665 is, 1665 = 3 x 3 x 5 x 37.
#### Definition of Prime Factors?
In mathematics, prime numbers are defined as all those whole numbers greater than 1 that have only two factors, i.e. 1 and the number itself. When we express any number as the product of these prime numbers than these prime numbers become prime factors of that number. Eg- Prime Factors of 1665 are 3 x 3 x 5 x 37.
#### Properties of Prime Factors
• Prime factors are the set of factors that are unique to the number given.
• 2 is the only even prime factor any number can have.
• Two prime factors are always coprime to each other.
• 1 is neither a prime number nor a composite number and also 1 is the factor of every given number. So, 1 is the factor of 1665 but not a prime factor of 1665.
• Which is the smallest prime factor of 1665?
Smallest prime factor of 1665 is 3.
• What is the prime factorization of 1665?
Prime factorization of 1665 is 3 x 3 x 5 x 37.
• What is prime factorization of 1665 in exponential form?
Prime factorization of 1665 in exponential form is 32 x 5 x 37.
• Is 1665 a prime number or a composite number?
1665 is a composite number.
• Is 1665 a prime number?
false, 1665 is not a prime number.
• How to find prime factorization of 1665 easily?
To find prime factorization of 1665 easily you can refer to the steps given on this page step by step.
• Which is the largest prime factors of 1665?
The largest prime factor of 1665 is 37.
• What is the product of all prime factors of 1665?
Prime factors of 1665 are 3 x 3 x 5 x 37. Therefore, their product is 1665.
• What is the sum of all odd prime factors of 1665?
Prime factors of 1665 are 3 , 3 , 5 , 37, out of which 3 , 3 , 5 , 37 are odd numbers. So, the sum of odd prime factors of 1665 is 3 + 3 + 5 + 37 = 48. |
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# How do you write ${5^x} = 25$ in log form?
Last updated date: 04th Mar 2024
Total views: 342.9k
Views today: 5.42k
Hint:The functions in which one term is raised to the power of another term are called exponential functions, in the given question, on the left-hand side, x is raised to the power of a thus it is an exponential function. The inverse of the exponential function is called logarithm function. A logarithm function is of the form ${\log _a}b$ where “a” is the base and b is the term whose logarithm we are finding, the answer to ${\log _a}b$ can be defined as the power to which a should be raised to get b as the answer. So, to convert an exponential function into a logarithm function, we first identify the base and move it to the other side of the equal sign and add the term “log”. This way we can find out the correct answer.
In ${a^x} = y$ , the base of the logarithm function will be “a”, moving the base to the other side of the equal to sign and writing the word “log”, we get:
${a^x} = y$
$\Rightarrow x = {\log _a}y$
Hence, ${a^x} = y$ can be written in the log form as $x = {\log _a}y$.
The logarithm functions whose base is equal to “e” are called the natural logarithm functions and are denoted as $\ln \left( a \right)$ , they can be written in log form as ${\log _e}\left( a \right)$. e is an irrational and transcendental mathematical constant. Certain rules are obeyed by the logarithm functions. One of these laws tells us how to convert logarithm functions to exponential functions. In the given question, we had to convert the exponential function into the logarithm function, so we used the inverse of this law. |
# Solve word problems math
When you try to Solve word problems math, there are often multiple ways to approach it. Math can be a challenging subject for many students.
## Solving word problems math
This can be a great way to check your work or to see how to Solve word problems math. Cameras are a sophisticated piece of technology, and as such, they can sometimes encounter problems. If you are having trouble with your camera, there are a few things you can do to try and solve the problem. First, consult your camera's manual or look online for troubleshooting tips specific to your model. If that doesn't help, try resetting your camera to its default settings. If the problem persists, it may be a good idea to take your camera to a professional for
There are a few different ways to calculate the slope of a line, but the most common is to use the slope formula. This formula is relatively easy to use and only requires two pieces of information: the rise and the run. The rise is the vertical distance between two points on the line, and the run is the horizontal distance between those same two points. Once you have these two values, you simply plug them into the formula and solve.
Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. The trigonometry solver can be used to find the values of unknown sides and angles of a triangle.
Log equations can be difficult to solve, but there are a few basic steps that can be followed to make solving them easier. First, identify the base of the logs and rewrite the equation so that all logs are of the same base. Next, use the properties of logs to simplify the equation as much as possible. Finally, solve the equation using traditional algebraic methods.
Introduction In mathematics, an inequality is a statement that suggests that two things are not equal. Inequality equations are mathematical problems that involve finding the value of a variable that will make the two sides of the equation equal. In order to solve these equations, you must use algebraic methods to isolate the variable on one side of the equation. There are a few different types of inequality equations that you might encounter. The most common type is a linear inequality, which is an inequality that can be |
## Solution
The quadratic equation $ax^2 + bx + c = 0$ has the solution(s) $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ assuming that $a\neq 0$.
Consider $ax^2 + bx + c = 0$, where $a \neq 0$.
Since $a \neq 0$, we can divide by $a$ to get $x^2 + \frac{b}{a} x + \frac{c}{a} = 0.$
We complete the square.
This shows that the original equation is equivalent to $\left(x + \frac{b}{2a} \right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0.$
Since $x$ appears only once in the equation, we can rearrange this to solve for $x$.
Get the squared term on one side of the equation: $\left(x + \frac{b}{2a} \right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}.$
We can rewrite the right-hand side by putting it over a common denominator: $\left(x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}.$
We can take the square root of both sides.
Taking account of the possibility of positive and negative square roots, we see $x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}.$
Subtracting $\frac{b}{2a}$ from both sides and putting the right-hand side over a common denominator gives $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$ |
# Bell’s Numbers and the Bell Triangle
Statistics Definitions > Bell’s numbers
Bell’s Numbers and the Bell Triangle (sometimes called the Pierce triangle or Aitken’s array) are a sequence of numbers which count the possible partitions of a set, and the triangle which makes derivation of them easy.
## Bell’s Numbers: What they Are and What they Mean
The first Bell numbers are:
1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975.
The nth Bell number, Bn, is the number of nonempty subsets a set of size n can be partitioned into. B0 is defined as one (i.e. the first number in the above list); There is just one possible partition of the set containing one member, so B1 = 1.
There are two partitions possible for a set with two elements, so B2 is just two.
A set with three elements can be partitioned five ways. Taking the set {a, b, c}, the five possible partitions are:
• {(a) (b) (c)},
• {(a, b), (c)},
• {(a,c) (b)},
• {(b,c), a},
• {(a, b, c)}.
There isn’t any simple formula that can give us Bn, but we can find Bell’s numbers in the Bell Triangle, in the next section, or use the following recursive equation to define them:
## Bell’s Numbers and the Bell Triangle as a Way to Derive them
The Bell triangle is an easy-to-fill in right triangle which gives us, in the left hand column, all of the Bell numbers.
1. Creating the Triangle
You make the triangle this way:
• On row one, write the number 1
• Begin all other rows with the last number of the previous row. The last number in row 1 was 1, so row 2 also begins with 1.
• All other numbers are found by adding the last number to the one above it. To find out what to write in the second place in row 2, we look at the place before it, place 1. This is just 1, and the digit 1 is above it, so 1 + 1 = 2
The first part of the Bell triangle will therefore be
1
1 2
Following those same rules, we begin the third row with the last number in the previous row, or 2. Then we add that 2 to the number above it to find the next number is 3. Three plus the two above it makes five, so the row finishes with a five.
1
1 2
2 3 5
Row four starts with the last of row three, i.e., 5. Then 5 + 2 = 7, so the next place is 7, and 7 + 3 = 10, so the next digit is ten. Finishing this row and the next in the same manner, we get
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
For any row k, the number that starts the row is Bk-1. So the first digit of row 1 is B1 -1 = B(0), the first digit of row 4 is B(3), and the first digit of row 55 will be B(54).
2. Reading the Triangle
The Bell numbers are given by the far left column, bolded here:
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
Note that the triangle is sometimes mirrored, so Bell’s numbers appear in the right column, and not the left; Make sure you know the construction of the triangle before attempting to read it.
## References
Guichard, D. Combinatorics and Graph Theory. Retrieved from
https://www.whitman.edu/mathematics/cgt_online/book/section01.04.html on February 4, 2018
Weisstein, Eric W. “Bell Triangle.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/BellTriangle.html
Comments? Need to post a correction? Please Contact Us. |
# Interactive video lesson plan for: Solution of Quadratic Equations by Factorisation | Word problems |Factoring quadratic equations
#### Activity overview:
Solution of Quadratic Equations by Factorisation Method | Word problems |Factoring quadratic equations
http://www.learncbse.in/ncert-class-10-math-solutions/
In this video, you will study quadratic equations, and various ways of finding their roots.A quadratic equations can be solved by any one the given methods (i)Method of completing the square (ii)Factoring quadratic equation (iii) Quadratic Formula.
Factorisation of quadratic equation ax^2 + bx + c = 0, can be done by finding two numbers such that their product is the product of first and last terms and also their sum is the middle term.This is illustrated in the following quadratic equations and word problems bassed on quadratic equations.
Procedure to solve Word Problems:
Step 1:Read the problem carefully to know what is given and what to find out.
Step 2:Denote the unknown by a small letter like z, p etc.
Step 3:From the given condition, frame the quadratic equation Involving the unknown.
Step 4:Solve by factoring quadratic equations
00:02 Word problems on Quadratic Equation (Problems based on Numbers)
Q2 Solve the Word problems factoring Quadratics
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, andthe product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
00:04From step:1 00:29 From step:2 02:05 From step:3 03:10 From step:4
05:35 Quadratic Equation Word problems(Problems based on Numbers)
(ii) A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
05:40 From step:1 06:08 From step:2 07:10 From step:3 07:48 From step:4
09:46 Quadratic Equation Word problems(Problems based on Digits)
Q3. Find two numbers whose sum is 27 and product is 182.
09:50 From step:1 09:55 From step:2 10:38 From step:3 10:51 From step:4
12:40 Word problems on Quadratic Equation (Problems based on Digits)
Q4. Find two consecutive positive integers, sum of whose squares is 365.
12:42 From step:1 12:57 From step:2 13:20 From step:3 14:48 From step:4
Reference book for the above video is
National Council of Educational Research and Training (NCERT) Book for Class 10 Subject: Maths
This Video is also refers as CBSE Class 10 mathts NCERT Solutions Chapter 9 Quadratic Equations
This video is suited for CBSE Class 10 maths Quadratic Equations.
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CBSE solutions for Class 10 maths Quadratic Equations
CBSE Class 10 maths chapter 10 Quadratic Equations
CBSE Class 10 maths NCERT Solutions chapter 10 Quadratic Equations
NCERT solutions for CBSE Class 10 maths Quadratic Equations
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# Difference between revisions of "1994 AIME Problems/Problem 8"
## Problem
The points $(0,0)\,$, $(a,11)\,$, and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$.
## Solution
### Solution 1
Consider the points on the complex plane. The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:
$$(a+11i)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.$$
Equating the real and imaginary parts, we have:
\begin{align*}b&=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}
Solving this system, we find that $a=21\sqrt{3}, b=5\sqrt{3}$. Thus, the answer is $\boxed{315}$.
Note: There is another solution where the point $b+37i$ is a rotation of $-60$ degrees of $a+11i$; however, this triangle is just a reflection of the first triangle by the $y$-axis, and the signs of $a$ and $b$ are flipped. However, the product $ab$ is unchanged.
### Solution 2
Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? $\sqrt{3}$ and perpendiculars inspires this solution:
First, drop a perpendicular from $O$ to $AB$. Call this midpoint of $AB M$. Thus, $M=\left(\frac{a+b}{2}, 24\right)$. The vector from $O$ to $M$ is $\left[\frac{a+b}{2}, 24\right]$. Meanwhile from point $M$ we can use a vector with $\frac{\sqrt{3}}{3}$ the distance; we have to switch the $x$ and $y$ and our displacement is $\left[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}\right]$. (Do you see why we switched $x$ and $y$ due to the rotation of 90 degrees?)
We see this displacement from $M$ to $A$ is $\left[\frac{a-b}{2}, 13\right]$ as well. Equating the two vectors, we get $a+b=26\sqrt{3}$ and $a-b=16\sqrt{3}$. Therefore, $a=21\sqrt{3}$ and $b=5\sqrt{3}$. And the answer is $\boxed{315}$.
Note: This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems".
### Solution 3
Plot this equilateral triangle on the complex plane. Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives $\left(\frac{a+b}{3}, 16i\right)$. The new coordinates of the equilateral triangle are $\left(-\frac{a+b}{3}-16i\right), \left(a-\frac{a+b}{3}-5i\right), \left(b-\frac{a+b}{3}+21i\right)$. These three vertices are solutions of a cubic polynomial of form $x^3 + C$. By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots $r_1, r_2,$ and $r_3,$$\, r_1r_2 + r_2r_3 + r_3r_1 = 0$.) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation $5a = 21b.$ Now use the equation with only real parts. This should give you a quadratic $a^2 - ab + b^2 = 1083$. Use your previously obtained equation to plug in for $a$ and solve for $b$, which should yield $5\sqrt{3}$. $a$ is then $\frac{21}{5}\sqrt{3}$. Multiplying $a$ and $b$ yields $\boxed{315}$. |
Calc
# How To Convert a Decimal To a Fraction
Article Category: Math |
Some decimals are so familiar to us that we can instantly see them as fractions: if your sister is 14.5 years old, you know that she’s 14 1/2; if you buy a bag of potatoes weighing 0.75kg, you know that it’s 3/4 of a kilo; if you give your sister a 3/4 kilo bag of potatoes for her 18th birthday, you know that your chances of a polite and enthusiastic response are around 0.1, or 1/10.
But what of the other decimals? The cool cats who haven’t broken into the mainstream? How can you possibly figure out what 0.65, 0.35 or 0.721 is as a fraction, for example? Well, just put down your potatoes, Janet, because we are going to make it easy for you...
## Turning decimals into fractions
The most important thing you need to keep in mind when you want to convert a decimal to a fraction is that a decimal expresses whether something is a ‘tenth’, a ‘hundredth’, a ‘thousandth’ etc., based on its position after the decimal point. If you’re looking at a decimal which only has one number after the point, then you are working in tens. If your decimal has two digits after the point, then you will be working in hundreds. If your decimal has three digits after the point, then you are working in thousands, and so on.
In this example, there are four digits after the decimal point and so we would be working in ten-thousandths.
## Step 1
Look at your decimal to establish whether you’re working in tens, hundreds, thousands or more. This will become your multiplier later on, in step 3.
e.g.:
0.35 is 35 hundredths
## Step 2
Write out your decimal as the numerator of a fraction (i.e. above the fraction line). The denominator below the line is always 1, because a decimal is always part of 1.
e.g.:
0.35/1
## Step 3
Think back to your multiplier from step 1. Multiply your numerator by 10 / 100 / 1000 or whatever your multiplier is, and then do the same calculation for the denominator.
e.g.:
35/100
## Step 4
Now you just need to tidy up your fraction by simplifying it. For this, you need to work out the ‘greatest common factor’, i.e. the largest number which can be multiplied to make both the numerator and the denominator. This makes your fraction smaller and easier to work with, so always ask your fraction, “does your bum look big in this?” and try slimming it down without changing the proportions.
e.g.:
Both 35 and 100 are multiples of 5, so we can divide both numbers by 5 to keep the proportionate relationship the same.
e.g.:
7/20
So 0.35 = 7/20
Not as painful as having 0.75kg of potatoes thrown in your face, is it?
Of course, if you want to check your answer, we have a handy decimal to fraction calculator that you can make free use of. In addition, we have a great article on how to add, subtract, multiply and divide fractions.
And, hey, if you manage to find a number which simply can’t become a fraction, then don’t be too hard on yourself. It’s not you: it’s them. These are called “irrational numbers”, and with good reason. One example of an irrational number is pi (3.14159265…) There’s just no reasoning with them.
Written by Becky Kleanthous
Please rate this article using the star rater below. If there is anything missing from the article, or any information you would like to see included, please contact me.
Last update: 12 April 2017 |
# 5th Class Mathematics Geometrical Figures
Geometrical Figures
Category : 5th Class
Geometrical Figures
Introduction
We observe different types of figures around us. They are in different shapes. In this chapter we will discuss about different types of geometrical figures such as line, angles etc.
Point
To show a particular location, a dot (.) is placed over it, that dot is known as point.
Example:
In the above figure point A represents$\frac{1}{3}$, point B represents$\frac{2}{3}$, and point C represents 1.
Line Segment
Line segment is defined as the shortest distance between two fixed points. For example
It is denoted as $\overline{AB}$.
Example: How many line segments are there in the figure?
(a) 2 (b) 4
(c) 8 (d) 16
(e) None of these
Ray
It is defined as the extension of a line segment in one infinitive direction. For example:
It is denoted as $\overrightarrow{AB}$.
Example: How many rays are there in the given figure?
(a) 2 (b) 4
(c) 12 (d) 16
(e) None of these
Line
Line is defined as the extension of a line segment infinitive in either direction.
Example: How many lines are there in the following figure?
It is denoted as
(a) 2 (b) 4
(c) 8 (d) 16
(e) None of these
Angle
Inclination between two rays having common end point is called angle.
Angle is measured in degree. Symbol of the degree is "${}^\circ$" and written as $a{}^\circ ,$
Where a is the measurement of the angle.
Types of Angle
There are different types of angles.
Right Angle
An angle whose measure is exactly $90{}^\circ$ is a right angle.
Acute Angle
An angle whose measure is less than $90{}^\circ$ is an acute angle.
Obtuse Angle
An angle whose measure is greater than $90{}^\circ$ but less than $180{}^\circ$ is a obtuse angle.
Straight Angle
An angle whose measure is $180{}^\circ$ is a straight angle.
Reflex Angle
An angle whose measure is greater than $180{}^\circ$ but less than $360{}^\circ$ is a reflex angle.
• Example:
Name the angle which measures exactly$90{}^\circ$.
• Example:
Name the angle which measures between $0{}^\circ$ and $90{}^\circ$.
Triangle
The geometrical shapes having three sides are called triangle.
Triangle has been classified:
(a) On the basis of sides
(b) On the basis of angles
Sides Based Classification
On the basis of sides, triangle is of three types:
Equilateral Triangle
It is the triangle in which all the three sides are equal.
Isosceles Triangle
In this type of triangle two of the three sides are equal.
Scalene Triangle
In this triangle all the sides are unequal.
Angle Based Classification
On the basis of angles, triangles are of three types.
Acute Angled Triangle
A triangle whose all the angles are acute is called acute angled triangle.
Obtuse Angled Triangle
A triangle in which one angle is an obtuse angle is called an obtuse angled triangle.
Right Angled Triangle
A triangle in which one angle is $90{}^\circ$ is called a right angled triangle.
PQR is a right-angled triangle as it contains a right angle
($\angle ~PQR$)
• Example:
Name the triangle which has two equal sides.
• Example:
If the sum of two angles of a triangle is $80{}^\circ$ the triangle is:
(a) Acute angled triangle
(b) Obtuse angled triangle
(c) Right angled triangle
(d) All of these
(e) None of these
Sum of two angles of a triangle is $80{}^\circ$
Therefore, third angle $180{}^\circ -80{}^\circ =100{}^\circ$ which is an obtuse angle.
The geometrical figure having four sides is called quadrilateral.
In this chapter we will study about two types of quadrilateral
(i) Rectangle (ij) Square
Rectangle
Rectangle is a quadrilateral in which:
(i) All angles are of$90{}^\circ$;
(ii) Opposite sides are equal.
Square
Square is a quadrilateral in which:
(i) All angles are of $90{}^\circ$; (ii) All sides are equal.
• Example:
How many angles of a rectangle are right angles?
(a) 1 (b) 2
(c) 3 (d) 4
(e) None of these
• Example:
How many sides a quadrilateral has?
(a) 1 (b) 2
(c) 3 (d) 4
(e) None of these
Circle
Circle is a simple closed shape, whose all points are at the same distance from a given point in a plane.
Chord
It is the line segment joining two distinct points of the circle.
Diameter
The diameter is the length of the line through the center that touches two point on the edge of the circle.
Radius of a circle is half of the diameter.
In the above circle, 0 is the centre, OA, OB, and OC are radius, and DE is the chord of the circle.
• Example:
If diameter of a circle is 25 cm, find the radius of the circle
Answer: $Radius=\frac{25}{2}=12.5\text{ }cm$
• Example:
0 is the centre of a circle and P is the point on the line of the circle. Op is ........... of the circle.
(a) Radius (b) Diameter
(c) Chord (d) All of these
(e) None of these
Cuboid
Cuboid is a box shaped solid object. It has six flat faces, which are rectangular in shape.
Cube
Cube is also a box shaped solid object with six faces, which are square in shape.
#### Other Topics
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# What is the domain of (4/(a+1))-(5/a) = (20/(a^2+a))?
Oct 26, 2015
$a = \left\{- 25\right\}$
#### Explanation:
$\left[1\right] \text{ } \frac{4}{a + 1} - \frac{5}{a} = \frac{20}{{a}^{2} + a}$
Combine $\frac{4}{a + 1}$ and $- \frac{5}{a}$ by making them have the same denominator.
$\left[2\right] \text{ } \frac{4 \left(a\right)}{a \left(a + 1\right)} - \frac{5 \left(a + 1\right)}{a \left(a + 1\right)} = \frac{20}{{a}^{2} + a}$
$\left[3\right] \text{ } \frac{4 \left(a\right) - 5 \left(a + 1\right)}{a \left(a + 1\right)} = \frac{20}{{a}^{2} + a}$
$\left[4\right] \text{ } \frac{4 a - 5 a - 5}{{a}^{2} + a} = \frac{20}{{a}^{2} + a}$
$\left[5\right] \text{ } \frac{- a - 5}{{a}^{2} + a} = \frac{20}{{a}^{2} + a}$
Multiply both sides by $\left({a}^{2} + a\right)$ to remove the denominator.
$\left[6\right] \text{ } \cancel{{a}^{2} + a} \left(\frac{- a - 5}{\cancel{{a}^{2} + a}}\right) = \cancel{{a}^{2} + a} \left(\frac{20}{\cancel{{a}^{2} + a}}\right)$
$\left[7\right] \text{ } - a - 5 = 20$
$\left[8\right] \text{ } - a = 25$
$\left[9\right] \text{ } \textcolor{b l u e}{a = - 25}$ |
# Linear Regression -- The Basics
### The basics
Yeah. It’s not a good sign if I’m starting out already repeating myself. But that’s how things seem to be with linear regression, so I guess it’s fitting. It seems like every day one of my professors will talk about linear regression, and it’s not due to laziness or lack of coordination. Indeed, it’s an intentional part of the curriculum here at New College of Florida because of how ubiquitous linear regression is. Not only is it an extremely simple yet expressive formulation, it’s also the theoretical basis of a whole slew of other tactics. Let’s just get right into it, shall we?
### Linear Regression:
Let’s say you have some data from the real world (and hence riddled with real-world error). A basic example for us to start with is this one:
There’s clearly a linear trend there, but how do we pick which linear trend would be the best? Well, one thing we could do is pick the line that has the least amount of error from the prediction to the actual data-point. To do that, we have to say what we mean by “least amount of error”. For this post, we’ll calculate that error by squaring the difference between the predicted value and the actual value for every point in our data set, then averaging those values. This standard is called the Mean-Squared-Error (MSE). We can write the MSE as:
where $\hat Y_i$ is our predicted value of $Y_i$ for a give $X_i$. Being as how we want a linear model (for simplicity and extensibility), we can write the above equation as,
for some $\alpha, \beta$ that we don’t yet know. But since we want to minimize that error, we can take some derivatives and solve for $\alpha, \beta$! Let’s go ahead and do that! We want to minimize
We can start by finding the $\hat\alpha$ such that, $\frac{d}{d\alpha}\sum\limits_{i=1}^N\left(\alpha + \beta X_i - Y_i\right)^2 = 0$. And as long as we don’t forget the chain rule, we’ll be alright…
and we’ll find the $\beta$ such that $\frac{d}{d\beta}\sum\limits_{i=1}^N\left(\alpha + \beta X_i - Y_i\right)^2 = 0$
And following a similar pattern we find (sorry for the editing… Wordpress.com isn’t the greatest therein):
But note:
So,
And: %
So,
And then we can find $\alpha$ by substituting in our approximation of $\beta$. Using those coefficients, we can plot the line below, and as you can see, it really is a good approximation.
### Now we have it
Okay, so now we have our line of “best fit”, but what does it mean? Well, it means that this line predicts the data we gave it with the least error. That’s really all it means. And sometimes, as we’ll see later, reading too much into that can really get you into trouble.
But using this model we can now predict other data outside the model. So, for instance, in the model pictured above, if we were to try and predict $Y$ when $X=2$, we wouldn’t do so bad by picking something around 10 for $Y$.
### An example, perhaps?
So I feel at this point, it’s probably best to give an example. Let’s say we’re trying to predict stock price given the total market price. Well, in practice this model is used to assess volatility, but that’s neither here nor there. Right now, we’re really only interested in the model itself. But without further ado, I present you with, the CAPM (Capital Asset Pricing Model):
$r = \alpha + \beta r_m + \epsilon$ (where $\epsilon$ is the error in our predictions).
And you can fit this using historical data or what-have-you. There are a bunch of downsides to fitting it with historical data though, like the fact that data from 3 days ago really doesn’t have much to say about the future anymore. There are plenty of cool things you can do therein, but sadly, those are out of the scope of this post.
For now, we move on to
## Multiple Regression
### What is this anyway?
Well, multiple regression is really just a new name for the same thing: how do we fit a linear model to our data given some set of predictors and a single response variable? The only difference is that this time our linear model doesn’t have to be one dimensional. Let’s get right into it, shall we?
So let’s say you have $k$ many predictors arranged in a vector (in other words, our predictor is a vector in $\mathbb{R}^n$). Well, I wonder if a similar formula would work… Let’s figure it out…
Firstly, we need to know what a derivative is in $\mathbb{R}^n$. Well, if $f:\mathbb{R}^n\to\mathbb{R}^m$ is a differentiable function, then for any $x$ in the domain, $f’(x)$ is the linear map $A:\mathbb{R}^n\to\mathbb{R}^m$ such that $\text{lim}_{h\to 0}\frac{||f(x+h) - f(x) - Ah||}{||h||} = 0$. Basically, $f’(x)$ is the tangent plane.
So, now that we got that out of the way, let’s use it! We want to find the linear function that minimizes the Euclidean norm of the error terms (just like before). But note: the error term is $\epsilon = Y - \hat Y = Y - \alpha -\beta X$, for some vector $\alpha$ and some matrix $\beta$. Now, since it’s easier and it’ll give us the same answer, we’re going to minimize the squared error term instead of just the error term (like we did in the one dimensional version). We’re also going to make one more simplification: That $\alpha=0$. We can do this safely by simply appending (or prepending) a 1 to the rows of our data (thereby creating a constant term). So for the following, assume we’ve done that.
So, let’s find the $\beta$ that minimizes that.
So, now we see that the derivative is $-2Y^TX + 2\beta^TX^TX$ and we want to find where our error is minimized, so we want to set that derivative to zero:
And there we have it. That’s called the normal equation for linear regression.
Maybe next time I’ll post about how we can find these coefficients given some data using gradient descent, or some modification thereof.
Till next time, I hope you enjoyed this post. Please, let me know if something could be clearer or if you have any requests. |
# Question: What Are The Common Factor Of 35?
## What are the factors of 28?
The factors of 28 are the numbers, which on multiplication in pairs results in the original number.
These factors are 1,2,4,7,14 and 28.
Since, 2 x 14 = 28, therefore (2,14) is a pair factor of number 28..
## What is the prime power factorization of 35?
35 is a composite number. 35 = 1 x 35 or 5 x 7. Factors of 35: 1, 5, 7, 35. Prime factorization: 35 = 5 x 7.
## What are the common factors of 50?
Greatest Common Factor (GCF) of two numbers and the factors of 50 are 1, 2, 5, 10, 25 and 50.
## What are the factors of 50?
The factors of 50 are 1, 2, 5, 10, 25, and 50. We can use prime factorization to get all the factors.
## What are the factors for 35?
Table of Factors and MultiplesFactorsMultiples1, 2, 17, 3434681, 5, 7, 3535701, 2, 3, 4, 6, 9, 12, 18, 3636721, 37377441 more rows
## What are the common factors of 14 and 35?
The GCF of 14 and 35 is 7. To identify the GCF of 14 and 35, we first need to list all of their factors: 14: 1, 2, 7, 14. 35: 1, 5, 7, 35.
## What is the GCF of 35 and 28?
We found the factors and prime factorization of 28 and 35. The biggest common factor number is the GCF number. So the greatest common factor 28 and 35 is 7.
## What is the GCF of 35 and 63?
The GCF of 35 and 63 is 7. To find the GCF, we will start by listing the factors of 35 and listing the factors of 63.
## What is the GCF of 15 and 30?
The common factors of 15 and 30 are 1, 3, 5, and 15. The greatest common factor is 15.
## What is the greatest common factor of 35?
Answer and Explanation: The greatest common factor of 35 and 15 is 5. The ‘greatest common factor’ is sometimes called the ‘GCF.
## Is 35 a perfect square?
Is 35 a perfect square number? A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 35 is about 5.916. Thus, the square root of 35 is not an integer, and therefore 35 is not a square number.
## What is the smallest factor of 35?
35 is a composite number. 35 = 1 x 35 or 5 x 7. Factors of 35: 1, 5, 7, 35. Prime factorization: 35 = 5 x 7.
## How many 5s would be there in prime factorization of 35?
7 5sAnswer. There are 7 5s in prime factorisation of 35.
## What’s the GCF of 35 and 45?
Greatest common factor (GCF) of 35 and 45 is 5. We will now calculate the prime factors of 35 and 45, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 35 and 45.
## What is the GCF of 12 and 30?
Earlier we found that the Common Factors of 12 and 30 are 1, 2, 3 and 6, and so the Greatest Common Factor is 6. The Greatest Common Factor of 12 and 30 is 6.
## What is the GCF of 15 and 9?
We found the factors and prime factorization of 9 and 15. The biggest common factor number is the GCF number. So the greatest common factor 9 and 15 is 3.
## What is the common factor of 35 and 50?
Greatest common factor (GCF) of 35 and 50 is 5.
## What is the prime factor for 32?
So, the prime factors of 32 are written as 2 × 2 x 2 x 2 x 2 or 25, where 2 is a prime numbers. |
# WORD PROBLEMS IN GEOMETRIC SEQUENCE
Word Problems in Geometric Sequence :
Here we are going to see some practice questions of finding geometric progression with given information.
## Word Problems in Geometric Sequence - Questions
Question 1 :
A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
Solution :
Starting salary = 60,000
Every year 5% of annual salary is increasing.
Second year salary = 60000 + 5% of 60000
= 60000(1 + 5%)
Third year salary = 60000 + 5% of 60000(1 + 5%)
= 60000(1 + 5%)(1 + 5%)
= 60000(1 + 5%)2
By continuing in this way, salary after 5 years is
= 60000(1 + 5%)5
= 60000(105/100)5
= 60000(1.05)
= 76577
Question 2 :
Sivamani is attending an interview for a job and the company gave two offers to him.
Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.
Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
What is his salary in the 4th year with respect to the offers A and B?
Solution :
Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.
a = 20000, r = 0.06 and n = 5
Starting salary = 20,000
Every year 6% of annual salary is increasing.
Starting salary = 20000
Second year salary = 20000 + 6% of 20000
= 20000(1 + 6%)
By continuing in this way, we get
4th year salary = 20000(1 + 6%)3
= 20000(1.06)3
= 22820
Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
a = 22000, r = 0.03 and n = 5
Starting salary = 22,000
Every year 3% of annual salary is increasing.
Starting salary = 22000
Second year salary = 22000 + 3% of 22000
= 22000(1 + 3%)
By continuing in this way, we get
4th year salary = 22000(1 + 3%)3
= 22000(1.03)3
= 24040
Question 3 :
If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb−c × yc−a × za−b = 1 .
Solution :
Since a, b and c are in A.P,
b - a = c - b = d (common difference)
We need to prove,
xb−c × yc−a × za−b = 1
Let us try to convert the powers in terms of one variable.
2b = c + a - a + a
2b = c - a + 2a
2(b - a) = c - a
2d = c - a
If c - b = d, then b - c = -d
If b - a = d, then a - b = -d
L.H.S
xb−c × yc−a × za−b = x−d × y2d × zd ---(1)
y = √xz
By applying the value of y in (1)
= x−d × (√xz)2d × zd
= x−d × (xz)d × zd
= x−d + d z-d + d
= 1
Hence proved.
After having gone through the stuff given above, we hope that the students would have understood, "Word Problems in Geometric Sequence".
Apart from the stuff given in this section "Word Problems in Geometric Sequence"if you need any other stuff in math, please use our google custom search here.
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Review question
# What's the next number to occur in both sequences? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R9916
## Solution
Consider the arithmetic sequences $1998$, $2005$, $2012$, … and $1996$, $2005$, $2014$, …. What is the next number after $2005$ that appears in both sequences?
The sequences have common differences of $7$ and $9$ respectively.
So the numbers in the first sequence can be written as $2005 + 7j$ for whole numbers $j$.
The numbers in the second sequence can be written as $2005 + 9k$ for whole number $k$.
So we are looking for the smallest positive $j$ and $k$ so that $2005 + 7j = 2005 + 9k$, or $7j = 9k$.
Since $7$ and $9$ have no common factor, these values for $j$ and $k$ are $9$ and $7$.
So we have that the next number in both sequences is $2005 + 7.9 = 2068$.
Alternatively, the next number in both lists is $m = 2005 + n$.
Since $m$ is in List $1, 7$ goes into $n$, and since $m$ is in List $2, 9$ goes into $n$.
Thus the smallest possible $n$ is $7 \times 9 = 63$, and $m=2068$ is the number we want. |
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# Geometric Solid – Meaning, Definition, and Solved Examples
Last updated date: 17th Apr 2024
Total views: 195.3k
Views today: 4.95k
## What Does Geometry Mean?
We call geometry a branch of Math that focuses on the measurement and relationship of lines, angles, surfaces, solids, and points. An example of geometry is the calculation of a triangle's angles. Shapes that we study in geometry are 2-d and 3-d.
2-d shapes have two dimensions, such as x and the y-axis. Also, it has length and width. Examples are circle, square, rectangle, and so on. However, shapes like sphere, hemisphere, cylinder, cube, and cuboid have three dimensions, i.e., x, y, and z-axis. The walls of your room have a 3-d structure. Also, these shapes have length, height or depth, and width.
The walls of your room have a kind of 3-d space which you live in. And the geometry of this 3-d space is called the solid geometry or simply we can say it is a geometric solid.
## What is Geometric Solid?
Before starting with what solid geometry is, let us go through some of the common 3-d shapes:
Image: Solid common 3-d shapes
We understood from the above text that these shapes have three dimensions having length, depth, and width. These shapes carry the following properties:
• Capacity or volume (think of how much water it could hold).
• Surface area (think of the amount of area covered by the surface of something).
• How many vertices (corner points), faces, and edges do they have.
## Formulas of Solid Geometry
The following table gives the volume formulas and surface area formulas for the following 3-D solid shapes:
3-D shape Image Surface Area Formula Volume Formula Cube with side ‘s’ 6s2 s3 Rectangular prism with length l, height h, and width w 2 (lw + wh + hl) lwh Prismwith an area of base ‘B’, the perimeter of base B, ‘P’, and height ‘h’ 2B + PH Bh Cylinderradius = r, height = h 2πr2 + 2πrh πr2h Sphereradius = r 4πr2 4/3 πr3 Coneradius = r, vertical height = h, slant height = h πr2 + πrs ⅓ πr2h Pyramidwith an area of base B, the perimeter of base P, vertical height h, and slant height s Regular pyramidB + 1 p/s ⅓ bh
## Solved Examples of Geometric Solid
1. Find the volume and surface area of a cube whose side is 7 cm.
Solution:
Side, a = 7 cm.
The volume of a cube = a3 cubic units
V = 73
V = 7 x 7 x 7
V = 343 cm3
Therefore, the volume of a cube is 343 cubic centimeter.
The surface area of a cube = 6a2 square units
SA = 6 x (7)2 cm2
SA = 6 x (49)
SA = 294 cm2
Therefore, the surface area of a cube is 294 square centimeters.
2. Find the total surface area of a cuboid of dimensions 9 cm × 8 cm × 6 cm.
Solution:
Given dimensions of a cuboid: 9 cm x 8 cm x 6 cm
That means length = l = 9 cm, breadth = b = 8 cm, and height = h = 6 cm.
Total surface area of a cuboid = 2 (lb + bh +hl)
= 2 (9 x 8 + 8 x 6 + 6 x 9)
= 2 (72 + 48 + 54 )
= 348 cm2
Hence, the total surface area of the cuboid is 348 cm2.
From the above text on geometric solid, we understand that solid geometry is regarded with 3-D shapes. Examples of three-dimensional solid shapes are cubes, rectangular solids, prisms, cylinders, spheres, cones, and pyramids.
## FAQs on Geometric Solid – Meaning, Definition, and Solved Examples
1. What are the properties of geometric shapes?
Below is the list of properties geometric shapes hold:
• Cube: face – 6 squared faces, vertices – 8, edges – 12.
• Cuboid: face – 6 rectangular faces, vertices – 8, edges – 12.
• Sphere: curved surface – 1, edges – 0, vertices – 0.
• Cylinder: flat surface – 2, curved surface – 1, face – 3, edges – 2, vertices – 0.
• Cone: flat surface – 1, curved surface – 1, face – 2, edges – 1, vertices – 1.
2. How do we identify geometric shapes?
Geometric shapes are Mathematical shapes that are perfect and regular. These shapes are characterized by straight lines, angles, and points. Examples are squares, rectangles, triangles, parallelograms, hexagons, etc.
However, an exception to geometric shapes would be a perfect circle as it has no straight lines or points. |
Jul 09
Here is one of those math tricks that has been floating around the Internet. Don’t cheat by scrolling down! It takes less than a minute. At the end, I will provide an explanation of how this trick is done.
Step 1: Pick the number of times a week that you like to eat chocolate (more than once but less than 10). Step 2: Multiply this number by 2 (just to be bold)! Step 3: Add 5 to the total. Step 4: Multiply your total by 50 — I’ll wait while you get a calculator. Step 5: If you have already had your birthday this year, add 1759. If you haven’t, then add 1758. Step 6: Now subtract the four digit year that you were born. Your result should be a three digit number. The first digit is your original number (i.e., how many times you want to have chocolate each week). The next two numbers are your age!
This is the only year (2009) that this trick will work, so spread it around!
How to Solve this Trick
This is a simple algebra math trick. For test purposes, let’s assume you were born on January 1, 1970. That would make your age 39 (as of today, July 9, 2009). So, following the steps above:
1. Pick the number of times a week that you like to eat chocolate. So let’s assign the variable “C” for chocolate:
= C
2. Multiply this number by 2. So that’s:
= C x 2
3. Add 5 to the total.
= (C x 2) + 5
4. Multiply your total by 50.
= ((C x 2) + 5) x 50
= (C x 2 x 50) + (5 x 50)
= (C x 100) + 250
= ((C x 100) + 250) + 1759
= (C x 100) + 2009 (which is the current year!)
6. Subtract the four year digit you were born.
= (C x 100) + 2009 – 1970
= (C x 100) + 39 (which is our test age!)
= C39
The final result is a three digit number. The first digit is your original number “C” (C x 100). The next two numbers are your age (39).
So the trick is to take your original number and multiply it by 100 so that it’s the first digit in the result. Next, it multiplies and adds some numbers that equal the current year. So then if you subtract the year you were born, the result is the number of years you have been alive, i.e., your age.
SOLVED!
If you enjoyed this puzzle, be sure to check out the Regifting Robin Mind Trick.
Article published on July 9, 2009
11 Responses to “Guess Your Age by Chocolate”
1. Ian Says:
Tried 3 times and didn’t work for me – number was 5 digits long.
2. KIMBERLEY Says:
ACE I DID IT IN MY MATHS LESSON 2DAY A TRIED IT ON MY MUM AND DAD IT IS THE GREASTING THING I HAVE EVER SEEN!!!!!!! <3
3. freya Says:
this is so cool how do u do it
4. Shelley Says:
That is so cool. It worked everytime. And, Freya, it says how to do it!
5. cerysnapril Says:
hahahahahahahahahahaha it doesn’t work me and April tried it it said we were 11 in fact people we are 12 hahahahahahahahahaha RUBBISH !!!!!!!!
6. IZZY! Says:
It does not work it said i was 11 when i am twelve!
7. 555 Says:
I am 12 not 11!
8. Guess Your Age | More More Pics Says:
[…] Guess Your Age by Chocolate devtopics.com […]
9. Jen Says:
it said i was 12, im 14 :L
10. Fanny Says:
my number was 16….I’m 18… and i dont eat chockolate
11. Jack Says:
Why do I always see these “It didn’t work I’m x+1 or x+2 years old” .. read where it says it ONLY WORKS for 2009!! (or add 2 for 2011 to the equation and it will work) |
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# 16.2: Line Integrals
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In single-variable calculus you learned how to integrate a real-valued function $$f (x)$$ over an interval $$[a,b]$$ in $$\mathbb{R}^1$$ . This integral (usually called a Riemann integral) can be thought of as an integral over a path in $$\mathbb{R}^1$$ , since an interval (or collection of intervals) is really the only kind of “path” in $$\mathbb{R}^1$$ . You may also recall that if $$f (x)$$ represented the force applied along the $$x$$-axis to an object at position $$x$$ in $$[a,b]$$, then the work $$W$$ done in moving that object from position $$x = a \text{ to }x = b$$ was defined as the integral:
$W=\int_a^b f (x)dx$
In this section, we will see how to define the integral of a function (either real-valued or vector-valued) of two variables over a general path (i.e. a curve) in $$\mathbb{R}^2$$ . This definition will be motivated by the physical notion of work. We will begin with real-valued functions of two variables.
In physics, the intuitive idea of work is that
$\text{Work = Force × Distance}$
Suppose that we want to find the total amount $$W$$ of work done in moving an object along a curve $$C$$ in $$\mathbb{R}^2$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$, with a force $$f (x, y)$$ which varies with the position $$(x, y)$$ of the object and is applied in the direction of motion along $$C$$ (see Figure $$\PageIndex{1}$$ below).
Figure $$\PageIndex{1}$$ Curve $$C : x = x(t),\, y = y(t) \text{ for }t \text{ in }[a,b]$$
We will assume for now that the function $$f (x, y)$$ is continuous and real-valued, so we only consider the magnitude of the force. Partition the interval $$[a,b]$$ as follows:
$a = t_0 < t_1 < t_2 < ··· < t_{n−1} < t_n = b ,\text{ for some integer }n ≥ 2$
As we can see from Figure $$\PageIndex{1}$$, over a typical subinterval $$[t_i ,t_{i+1}]$$ the distance $$∆s_i$$ traveled along the curve is approximately $$\sqrt{∆x_i^2 +∆y_i^2}$$ , by the Pythagorean Theorem. Thus, if the subinterval is small enough then the work done in moving the object along that piece of the curve is approximately
$\text{Force × Distance} \approx f (x_{i∗}, y_{i∗}) \sqrt{ ∆x_i^2 +∆y_i^2}\label{Eq4.1}$
where $$(x_{i∗}, y_{i∗}) = (x(t_{i∗}), y(t_{i∗}))$$ for some $$t_{i∗} \text{ in }[t_i ,t_{i+1}]$$, and so
$W \approx \sum_{i=0}^{n-1} f (x_{i∗}, y_{i∗}) \sqrt{ ∆x_i^2 +∆y_i^2}\label{Eq4.2}$
is approximately the total amount of work done over the entire curve. But since
$\sqrt{ ∆x_i^2 +∆y_i^2} = \sqrt{\left ( \dfrac{∆x_i}{∆t_i} \right )^2 +\left ( \dfrac{∆y_i}{∆t_i}\right )^2}∆t_i$
where $$∆t_i = t_{i+1} − t_i$$ , then
$W \approx \sum_{i=0}^{n-1}f (x_{i∗}, y_{i∗})\sqrt{\left ( \dfrac{∆x_i}{∆t_i} \right )^2 + \left ( \dfrac{∆y_i}{∆t_i} \right )^2}∆t_i \label{Eq4.3}$
Taking the limit of that sum as the length of the largest subinterval goes to 0, the sum over all subintervals becomes the integral from $$t = a \text{ to }t = b$$, $$∆x_i ∆t_i \text{ and }∆y_i ∆t_i$$ become $$x ′ (t) \text{ and }y ′ (t)$$, respectively, and $$f (x_{i∗}, y_{i∗})$$ becomes $$f (x(t), y(t))$$, so that
$W=\int_a^b f (x(t), y(t)) \sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt \label{Eq4.4}$
The integral on the right side of the above equation gives us our idea of how to define, for any real-valued function $$f (x, y)$$, the integral of $$f (x, y)$$ along the curve $$C$$, called a line integral:
Definition $$\PageIndex{1}$$: Line Integral of a scalar field
For a real-valued function $$f (x, y)$$ and a curve $$C$$ in $$\mathbb{R}^2$$, parametrized by $$x = x(t), y = y(t), a ≤ t ≤ b$$, the line integral of $$f (x, y)$$ along $$C$$ with respect to arc length $$s$$ is
$\int_C f (x, y)\,ds = \int_a^b f (x(t), y(t))\sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt \label{Eq4.5}$
The symbol $$ds$$ is the differential of the arc length function
$s = s(t) = \int_a^t \sqrt{x ′ (u)^2 + y ′ (u)^2}\,du \label{Eq4.6}$
which you may recognize from Section 1.9 as the length of the curve $$C$$ over the interval $$[a,t]$$, for all $$t$$ in $$[a,b]$$. That is,
$ds = s ′ (t)\,dt = \sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt, \label{Eq4.7}$
For a general real-valued function $$f (x, y)$$, what does the line integral $$\int_C f (x, y)\,ds$$ represent? The preceding discussion of $$ds$$ gives us a clue. You can think of differentials as infinitesimal lengths. So if you think of $$f (x, y)$$ as the height of a picket fence along $$C$$, then $$f (x, y)\,ds$$ can be thought of as approximately the area of a section of that fence over some infinitesimally small section of the curve, and thus the line integral $$\int_C f (x, y)\,ds$$ is the total area of that picket fence (see Figure $$\PageIndex{2}$$).
Figure $$\PageIndex{2}$$: Area of shaded rectangle = height × width ≈ $$f (x, y)\,ds$$
Example $$\PageIndex{1}$$
Use a line integral to show that the lateral surface area $$A$$ of a right circular cylinder of radius $$r$$ and height $$h$$ is $$2\pi rh$$.
Solution:
We will use the right circular cylinder with base circle $$C$$ given by $$x^2 + y^2 = r^2$$ and with height $$h$$ in the positive $$z$$ direction (see Figure $$\PageIndex{3}$$). Parametrize $$C$$ as follows:
$x = x(t) = r \cos t , y = y(t) = r \sin t , 0 ≤ t ≤ 2π$
Figure $$\PageIndex{3}$$
\nonumber \begin{align} A&=\int_C f (x, y)\,ds = \int_a^b f (x(t), y(t))\sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt \\ \nonumber &=\int_0^{2\pi} h \sqrt{(−r \sin t)^2 +(r \cos t)^2}\,dt \\ \nonumber &=h\int_0^{2\pi} r \sqrt{\sin^2 t+\cos^2 t}\,dt \\ \nonumber &=rh\int_0^{2\pi} 1\,dt = 2\pi rh \\ \end{align}
Note in Example $$\PageIndex{1}$$ that if we had traversed the circle $$C$$ twice, i.e. let t vary from $$0 \text{ to }4\pi$$ then we would have gotten an area of $$4\pi rh$$, i.e. twice the desired area, even though the curve itself is still the same (namely, a circle of radius $$r$$). Also, notice that we traversed the circle in the counter-clockwise direction. If we had gone in the clockwise direction, using the parametrization
$x = x(t) = r \cos (2π− t) , y = y(t) = r \sin (2π− t) , 0 ≤ t ≤ 2π ,\label{Eq4.8}$
then it is easy to verify (Exercise 12) that the value of the line integral is unchanged.
In general, it can be shown (Exercise 15) that reversing the direction in which a curve $$C$$ is traversed leaves $$\int_C f (x, y)\,ds$$ unchanged, for any $$f (x, y)$$. If a curve $$C$$ has a parametrization $$x = x(t), y = y(t), a ≤ t ≤ b,$$ then denote by $$−C$$ the same curve as $$C$$ but traversed in the opposite direction. Then $$−C$$ is parametrized by
$x = x(a+ b − t) , y = y(a+ b − t) , a ≤ t ≤ b ,\label{Eq4.9}$
and we have
$\int_C f (x, y)\,ds =\int_{-C}f (x, y)\,ds .\label{Eq4.10}$
Notice that our definition of the line integral was with respect to the arc length parameter $$s$$. We can also define
$\int_C f (x, y)\,dx=\int_a^b f (x(t), y(t)) x ′ (t)\,dt\label{Eq4.11}$
as the line integral of $$f (x, y)$$ along $$C$$ with respect to $$x$$, and
$\int_C f (x, y)\,d y=\int_a^b f (x(t), y(t)) y ′ (t)\,dt \label{Eq4.12}$
as the line integral of $$f (x, y)$$ along $$C$$ with respect to $$y$$.
In the derivation of the formula for a line integral, we used the idea of work as force multiplied by distance. However, we know that force is actually a vector. So it would be helpful to develop a vector form for a line integral. For this, suppose that we have a function $$f(x, y)$$ defined on $$\mathbb{R}^2$$ by
$\nonumber \textbf{f}(x, y) = P(x, y)\textbf{i} + Q(x, y)\textbf{j}$
for some continuous real-valued functions $$P(x, y)$$ and $$Q(x, y) \text{ on }\mathbb{R}^2$$ . Such a function $$f$$ is called a vector field on $$\mathbb{R}^2$$ . It is defined at points in $$\mathbb{R}^2$$ , and its values are vectors in $$\mathbb{R}^2$$ . For a curve $$C$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$, let
$\nonumber \textbf{r}(t) = x(t)\textbf{i} + y(t)\textbf{j}$
be the position vector for a point $$(x(t), y(t))$$ on $$C$$. Then $$\textbf{r}'(t) = x'(t)\textbf{i} + y'(t)\textbf{j}$$ and so
\nonumber \begin{align} \int_C P(x, y)\,dx+ \int_C Q(x, y)\,d y &=\int_a^b P(x(t), y(t)) x ′ (t)\,dt+\int_a^b Q(x(t), y(t)) y ′ (t)\,dt \\ \nonumber &=\int_a^b (P(x(t), y(t)) x ′ (t)+Q(x(t), y(t)) y ′ (t))\,dt \\ \nonumber &=\int_a^b \textbf{f}(x(t), y(t))\cdot \textbf{r} ′ (t)dt \\ \end{align}
by definition of $$f(x, y)$$. Notice that the function $$f(x(t), y(t))\cdot r ′ (t)$$ is a real-valued function on $$[a,b]$$, so the last integral on the right looks somewhat similar to our earlier definition of a line integral. This leads us to the following definition:
Definition $$\PageIndex{2}$$: Line Integral of a vector feild
For a vector field $$\textbf{f}(x, y) = P(x, y)\textbf{i} +Q(x, y)\textbf{j}$$ and a curve $$C$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$, the line integral of f along $$C$$ is
\begin{align} \int_C \textbf{f}\cdot d\textbf{r} &= \int_C P(x, y)\,dx+\int_C Q(x, y)\,d y \label{Eq4.13} \\ &=\int_a^b \textbf{f}(x(t), y(t))\cdot \textbf{r} ′ (t)\,dt \label{Eq4.14} \\ \end{align}
where $$\textbf{r}(t) = x(t)\textbf{i}+ y(t)\textbf{j}$$ is the position vector for points on $$C$$.
We use the notation $$d\textbf{r} = \textbf{r} ′ (t)\,dt = dx\textbf{i}+ d y\textbf{j}$$ to denote the differential of the vector-valued function r. The line integral in Definition $$\PageIndex{2}$$ is often called a line integral of a vector field to distinguish it from the line integral in Definition $$\PageIndex{1}$$ which is called a line integral of a scalar field. For convenience we will often write
$\nonumber \int_C P(x, y)\,dx +\int_C Q(x, y)\,d y =\int_C P(x, y)\,dx+Q(x, y)\,d y ,$
where it is understood that the line integral along $$C$$ is being applied to both $$P \text{ and }Q$$. The quantity $$P(x, y)\,dx +Q(x, y)\,d y$$ is known as a differential form. For a real-valued function $$F(x, y)$$, the differential of $$F$$ is $$dF = \dfrac{∂F}{∂x}\,dx+ \dfrac{∂F}{∂y}\, d y.$$ A differential form $$P(x, y)\,dx+Q(x, y)\,d y$$ is called exact if it equals $$dF$$ for some function $$F(x, y)$$.
Recall that if the points on a curve $$C$$ have position vector $$\textbf{r}(t) = x(t)\textbf{i}+ y(t)\textbf{j}$$, then $$\textbf{r} ′ (t)$$ is a tangent vector to $$C$$ at the point $$(x(t), y(t))$$ in the direction of increasing $$t$$ (which we call the direction of $$C$$). Since $$C$$ is a smooth curve, then $$\textbf{r} ′ (t) \neq \textbf{0} \text{ on }[a,b]$$ and hence
$\nonumber \textbf{T}(t) = \dfrac{\textbf{r}'(t)}{\left \lVert \textbf{r}'(t) \right \rVert}$
is the unit tangent vector to $$C$$ at $$(x(t), y(t))$$. Putting Definitions $$\PageIndex{1}$$ and $$\PageIndex{2}$$ together we get the following theorem:
Theorem $$\PageIndex{1}$$
For a vector field $$\textbf{f}(x, y) = P(x, y)\textbf{i} + Q(x, y)\textbf{j}$$ and a curve $$C$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$ and position vector $$\textbf{r}(t) = x(t)\textbf{i}+ y(t)\textbf{j}$$,
$\int_C \textbf{f}\cdot d\textbf{r} = \int_C \textbf{f}\cdot \textbf{T}\,ds,\label{Eq4.15}$
where $$\textbf{T}(t) = \dfrac{\textbf{r} ′ (t)}{ \left \lVert \textbf{r} ′ (t)\right \rVert }$$ is the unit tangent vector to $$C$$ at $$(x(t), y(t))$$.
If the vector field $$\textbf{f}(x, y)$$ represents the force moving an object along a curve $$C$$, then the work $$W$$ done by this force is
$W = \int_C \textbf{f}\cdot \textbf{T} \, ds = \int_C \textbf{f}\cdot d\textbf{r} \label{Eq4.16}$ |
# 3.15 Composition of trigonometric function and its inverse (Page 2/3)
Page 2 / 3
$x=\pi -\theta$
$⇒\theta =\pi -x$
Hence,
$⇒{\mathrm{sin}}^{-1}\mathrm{sin}x=\pi -x;\phantom{\rule{1em}{0ex}}x\in \left[\frac{\pi }{2},\frac{3\pi }{2}\right]$
In order to find expression corresponding to negative angle interval $\left[-3\pi /2,-\pi /2\right]$ , we need to construct negative value diagram. We know that equivalent negative angle is obtained by deducting “-2π” to the positive angle. Thus, corresponding to expression for positive angles in four quadrants, the expression in terms of negative angles are “-θ”,“-π+θ”,“-π-θ” and “-2π+θ” in four quadrants counted in clockwise direction in the value diagram. Now, we estimate from the sine plot that an angle, corresponding to a positive acute angle, θ, in the principal interval, lies in third negative quadrant. Therefore,
$x=-\pi -\theta$
$⇒\theta =-\pi -x$
Hence,
$⇒{\mathrm{sin}}^{-1}\mathrm{sin}x=-\pi -x;\phantom{\rule{1em}{0ex}}x\in \left[-\frac{3\pi }{2},-\frac{\pi }{2}\right]$
Combining three results,
|-π-x; x∈ [-3π/2, -π/2] sin⁻¹ sinx = | x; x∈ [-π/2, π/2]| π- x; x∈ [π/2, 3π/2]
We can similarly find expressions for more such intervals.
## Graph of sin⁻¹sinx
Using three expressions obtained above, we can draw plot of the composition function. We extend the plot, using the fact that composition is a periodic function with a period of 2π. The equation of plot, which is equivalent to plot y=x shifted by 2π towards right, is :
$y=x-2\pi$
The equation of plot, which is equivalent to plot y=x shifted by 2π towards left, is :
$y=x+2\pi$
We see that graph of composition is continuous. Its domain is R. Its range is $\left[-\pi /2,\pi /2\right]$ . The function is periodic with period 2π.
## Composition with arccosine
The composition ${\mathrm{cos}}^{-1}\mathrm{cos}x$ evaluates to angle values lying in the interval [0, π].
${\mathrm{cos}}^{-1}\mathrm{cos}x=x;\phantom{\rule{1em}{0ex}}x\in \left[0,\pi \right]$
Let us consider adjacent intervals such that all cosine values are included once. Such intervals are [π, 2π], [2π, 3π]etc on the right side and [-π, 0], [-2π, -π]etc on the left side of the principal interval.
The new interval $\left[\pi ,2\pi \right]$ represents third and fourth quadrants. The angle x, corresponding to positive acute angle θ, lies in fourth quadrant. Then,
$x=2\pi -\theta$
$⇒\theta =2\pi -x$
Hence,
$⇒{\mathrm{cos}}^{-1}\mathrm{cos}x=2\pi -x;\phantom{\rule{1em}{0ex}}x\in \left[\pi ,2\pi \right]$
In order to find expression corresponding to negative angle interval $\left[-\pi ,0\right]$ , we estimate from the cosine plot that an angle corresponding to a positive acute angle, θ, in the principal interval lies in first negative quadrant. Therefore,
$x=-\theta$
$⇒\theta =-x$
Hence,
$⇒{\mathrm{cos}}^{-1}\mathrm{cos}x=-x;\phantom{\rule{1em}{0ex}}x\in \left[-\pi ,0\right]$
Combining three results,
|-x; x∈ [-π, 0] cos⁻¹ cosx = | x; x∈[0, π]|2π- x; x∈ [π, 2π]
We can similarly find expressions for other intervals.
## Graph of cos⁻¹cosx
Using three expressions obtained above, we can draw plot of the composition function. We have extended the plot, using the fact that composition is a periodic function with a period of 2π. The equation of plot, which is equivalent to plot y=x shifted by 2π towards right, is :
$y=x-2\pi$
The equation of plot, which is equivalent to plot y=x shifted by 2π towards left, is :
$y=x+2\pi$
We see that graph of composition is continuous. Its domain is R. Its range is $\left[0,\pi \right]$ . The function is periodic with period 2π.
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
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
What is power set
Period of sin^6 3x+ cos^6 3x
Period of sin^6 3x+ cos^6 3x |
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# wikiHow to Calculate the Sum of Interior Angles
Two Methods:Using the FormulaDrawing TrianglesCommunity Q&A
A polygon is any closed figure with sides made from straight lines. At each vertex of a polygon, there is both an interior and exterior angle, corresponding to the angles on the inside and outside of the closed figure. Understanding the relationships that govern these angles is useful in various geometrical problems. In particular, it is helpful to know how to calculate the sum of interior angles in a polygon. This can be done using a simple formula, or by dividing the polygon into triangles.
### Method 1 Using the Formula
1. 1
Set up the formula for finding the sum of the interior angles. The formula is ${\displaystyle sum=(n-2)\times 180}$, where ${\displaystyle sum}$ is the sum of the interior angles of the polygon, and ${\displaystyle n}$ equals the number of sides in the polygon.[1]
• The value 180 comes from how many degrees are in a triangle. The other part of the formula, ${\displaystyle n-2}$ is a way to determine how many triangles the polygon can be divided into. So, essentially the formula is calculating the degrees inside the triangles that make up the polygon.[2]
• This method will work whether you are working with a regular or irregular polygon. Regular and irregular polygons with the same number of sides will always have the same sum of interior angles, the difference only being that in a regular polygon, all interior angles have the same measurement.[3] In an irregular polygon, some of the angles will be smaller, some of the angles will be larger, but they will still add up to the same number of degrees that are in the regular shape.
2. 2
Count the number of sides in your polygon. Remember that a polygon must have at least three straight sides.
• For example, if you want to know the sum of the interior angles of a hexagon, you would count 6 sides.
3. 3
Plug the the value of ${\displaystyle n}$ into the formula. Remember, ${\displaystyle n}$ is the number of sides in your polygon.
• For example, if you are working with a hexagon, ${\displaystyle n=6}$, since a hexagon has 6 sides. So, your formula should look like this:
${\displaystyle sum=(6-2)\times 180}$
4. 4
Solve for ${\displaystyle n}$. To do this, subtract 2 from the number of sides, and multiply the difference by 180. This will give you, in degrees, the sum of the interior angles in your polygon.
• For example, to find out the sum of the interior angles of a hexagon, you would calculate:
${\displaystyle sum=(6-2)\times 180}$
${\displaystyle sum=(4)\times 180}$
${\displaystyle sum=(4)\times 180=720}$
So, the sum of the interior angles of a hexagon is 720 degrees.
### Method 2 Drawing Triangles
1. 1
Draw the polygon whose angles you need to sum. The polygon can have any number of sides and can be regular or irregular.
• For example, you might want to find the sum of the interior angles of a hexagon, so you would draw a six-sided shape.
2. 2
Choose one vertex. Label this vertex A.
• A vertex is a point where two sides of a polygon meet.
3. 3
Draw a straight line from Point A to each other vertex in the polygon. The lines should not cross. You should create a number of triangles.
• You do not have to draw lines to the adjacent vertices, since they are already connected by a side.
• For example, for a hexagon you should draw three lines, dividing the shape into 4 triangles.
4. 4
Multiply the number of triangles you created by 180. Since there are 180 degrees in a triangle, by multiplying the number of triangles in your polygon by 180, you can find the sum of the interior angles of your polygon.
• For example, since you divided your hexagon into 4 triangles, you would calculate ${\displaystyle 4\times 180=720}$ to find a total of 720 degrees in the interior of your polygon.
## Community Q&A
Search
• Is the sum of the interior angles in a rectangle sum=4-2x180?
Yes.
• How do I find the sum of the interior angles of an irregular polygon?
wikiHow Contributor
The formula for finding the sum of the interior angles of a polygon is the same, whether the polygon is regular or irregular. So you would use the formula (n-2) x 180, where n is the number of sides in the polygon.
• Does a regular polygon's interior angles add up to 160?
wikiHow Contributor
Not necessarily. A triangle's sum is 180, a quadrilateral's sum is 360, and a pentagon's sum is 540. These are all polygons. Use the formula 180(n-2) where "n" is the number of the sides of the polygon in question to find your sum.
• If the exterior angle is 72, what is the interior angle?
To find the interior angle, subtract the exterior angle from 180°.
• Why is the sum of an interior angle 180?
wikiHow Contributor
The sum of all triangles is 180 degrees. And the sum of all squares (and other quadrilaterals) is 360 degrees. Since two like triangles will add together to form a square or other quadrilateral, the sum of their angles in half of 360 degrees. Aka 180.
• How do I find the number of triangles in a polygon?
wikiHow Contributor
By counting them manually. Drawing is usually the easiest method, as described in Method 2.
• How many triangles are in a square?
If you draw a diagonal in the square, that forms two triangles. If you draw both diagonals of a square, that forms four triangles.
• What is the exterior angle of a polygon?
It's the angle outside the polygon between one side and the extension of the adjacent side.
• What do I use a protractor for?
wikiHow Contributor
You can use this to measure the angles. That is why it is needed in school for geometry.
• How do I find one interior angle in an irregular polygon?
Lacking further information, the only way to do that is with a protractor.
• How does one explain why we must create triangles from just 1 vertex (ie. a square would divide into 2 triangles), instead of intersecting lines to create 4 triangles within a square?
• I understand that the formula "works", but is there a way to explain where the "2" comes from in "n-2"?
• How do I find the area of a hectagon?
• How can we work out 1 or more interior angles of an irregular polygon without a protractor?
200 characters left
## Tips
• Check your work on a piece of paper using a protractor to sum the interior angles manually. When doing this, be careful while drawing the polygon's sides as they should be linear.
## Things You'll Need
• Pencil
• Paper
• Protractor (optional)
• Pen
• Eraser
• Ruler
## Article Info
Categories: Geometry
In other languages:
Español: calcular la suma de los ángulos internos, Italiano: Calcolare la Somma degli Angoli Interni, Русский: вычислить сумму внутренних углов, Português: Calcular a Soma dos Ângulos Internos de um Polígono, Français: calculer la somme des angles internes
Thanks to all authors for creating a page that has been read 158,286 times. |
# Circle Problems
### Circle Problems
1) Show that the line x + y + 1 = 0 touches the circle x² + y² – 3x + 7y + 14 = 0 and find the point of contact.
Solution: Centre of the circle is C (3/2, -7/8)
Radius of the circle (r) = $$\sqrt{\frac{9}{4}\,+\,\frac{49}{4}\,-\,14}\,=\,\sqrt{\frac{58\,-56}{4}}\,=\,\sqrt{\frac{1}{2}}\,=\frac{1}{\sqrt{2}}$$
Perpendicular distance from the centre to the given line = $$\frac{\left| \frac{3}{2}\,-\,\frac{7}{2}\,+1 \right|}{\sqrt{1\,+\,1}}\,=\,\frac{1}{\sqrt{2}}\,=\,r$$
∴ The line x + y + 1 = 0 touches the given circle.
If P (h, k) is the point of contact then $$\frac{h\,-\,\frac{3}{2}}{1}\,=\,\frac{k\,+\,\frac{7}{2}}{1}\,=\,\frac{-\left( \frac{3}{2}\,-\,\frac{7}{2}\,+\,1 \right)}{1\,+\,1}$$
⇒ h – 3/2 = k + 7/2 = ½
⇒ h – 3/2 = ½, k + 7/2 = ½
⇒ h = 2, k = -3
∴ Point of Contact P (2, -3)
2) Find the equation of the circle passing through (0, 0) and making intercept 6 units on x – axis and intercept of 4 units on y-axis.
Solution: Let the circle cuts x – axis at O, A and y – axis at O, B Such that OA = 6, OB = 4.
Let C be the center of the circle and D be the midpoint of OA. Then OD = 3, CD = OB/2 =2.
Now $$OC\,=\,\sqrt{O{{D}^{2}}\,+\,C{{D}^{2}}}\,=\,\sqrt{9\,+\,4}$$.
Radius of the circle = √13
Centre of the circle is C (±3, ±2)
Required circle equation is (x ± 3)² + (y ± 2)² = (√13)²
⇒ x² + y² ± 6x ± 4y = 0 |
# Intermediate Algebra Problems
• Jun 20th 2011, 08:08 AM
Aravsion
Intermediate Algebra Problems
Can someone pleace explain how to solve these problems?
The problem statement, all variables and given/known data
Sally can paint a room in 9 hours while it takes Steve 3 hours to paint the same room. How long would it take them to paint the room if they worked together?
The problem statement, all variables and given/known data
One hose can fill a goldfish pond in 18 minutes, and two hoses can fill the same pond in 14 minutes. Find how long it takes the second hose alone to fill the pond.
• Jun 20th 2011, 08:17 AM
Prove It
Re: Intermediate Algebra Problems
1. Set these up as ratios "How much of the room painted : Hours".
Sally
$\displaystyle 1 : 9 = \frac{1}{9} : 1$
Steve
$\displaystyle 1 : 3 = \frac{1}{3} : 1$
Together
\displaystyle \begin{align*} \frac{1}{9} + \frac{1}{3} &: 1 \\ \frac{4}{9} &: 1 \\ 4 &: 9 \\ 1 &: \frac{9}{4} \end{align*}.
So together they will take $\displaystyle 2\frac{1}{4}$ hours.
• Jun 20th 2011, 08:33 AM
IBstudent
Re: Intermediate Algebra Problems
For the second one, lets call the hoses hose A and hose B. A takes 18 minutes so the ratio is 18:1. A+B would be 14:1, so 1/14-1/18=1/63. Hence it is 63 minutes...
• Jun 20th 2011, 08:33 AM
HallsofIvy
Re: Intermediate Algebra Problems
A slightly different point of view: when people or things "work together", their rates of work add.
"Sally can paint a room in 9 hours"
so Sally's rate of work is 1/9 "room per hour".
"it takes Steve 3 hours to paint the same room"
so Steve's rate of work is 1/3 "room per hour".
Together they work at $\frac{1}{9}+ \frac{1}{3}= \frac{1}{9}+ \frac{3}{9}= \frac{4}{9}$ "room per hour". Invert that to get "hours per room".
"One hose can fill a goldfish pond in 18 minutes"
so that hose works at 1/18 "pond per minute". Since we want to find how long it would take the second hose to fill the pond, call that time "x" minutes. The
second hose works at 1/x "pond per minute".
" and two hoses can fill the same pond in 14 minutes."
so $\frac{1}{18}+ \frac{1}{x}= \frac{1}{14}$
Solve that for x. I recommend multiplying both sides of the equation by the "least common denominator", 2(7)(9)x= 126x.
• Jun 20th 2011, 09:31 AM
Aravsion
Re: Intermediate Algebra Problems
thanks everybody for the explanations! i understand it now! |
# 2006 AMC 10B Problems/Problem 5
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?
$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 49\qquad \mathrm{(E) \ } 64$
## Solution
By placing the $2 \times 3$ rectangle adjacent to the $3 \times 4$ rectangle with the 3 side of the $2 \times 3$ rectangle next to the 4 side of the $3 \times 4$ rectangle, we get a figure that can be completely enclosed in a square with a side length of 5. The area of this square is $5^2 = 25$.
Since the sum of the areas of the two rectangles is $2\cdot3+3\cdot4=18$, the area of a square cannot be less than 18. Therefore 16 is not possible.
So the answer is $25 \Rightarrow B$ |
# McGraw Hill My Math Kindergarten Chapter 3 Lesson 7 Answer Key Problem-Solving Strategy: Draw a Diagram
All the solutions provided in McGraw Hill My Math Kindergarten Answer Key PDF Chapter 3 Lesson 7 Problem-Solving Strategy: Draw a Diagram will give you a clear idea of the concepts.
## McGraw-Hill My Math Kindergarten Answer Key Chapter 3 Lesson 7 Problem-Solving Strategy: Draw a Diagram
How many gumballs?
Draw to solve.
Explanation:
There are total 16 gumboils.
We have used cubes to show the number.
Ten ones and six ones is 16.
In bottom box we have drawn X for every gumball counted.
Directions: Say and trace the number. Count the gumboils. Use cubes to show the number. In the bottom box, draw an X for every gumball counted. Write the number.
How many fish?
Draw to Solve
Explanation:
There are total 17 fishes.
We have used cubes to show the number.
Ten ones and seven ones is 17.
In bottom box we have drawn X for every Fish counted.
Directions: Say and trace the number. Count the fish. Use cubes to show the number. In the bottom box, draw an X for every fish counted. Write the number.
How many spots?
Draw to solve
Explanation:
There are total 20 spots on giraffe.
We have used cubes to show the number.
Ten ones and ten ones is 20.
In bottom box we have drawn X for every spot counted.
Directions: Say and trace the number. Count the spots. Use cubes to show the number. In the bottom box, draw an X for every spot counted. Write the number.
How many bees?
Draw to solve
Explanation:
There are total 18 bees
We have used cubes to show the number.
Ten ones and eight ones is 18.
In bottom box we have drawn X for every bee counted.
Directions: Say and trace the number. Count the bees. Use cubes to show the number. In the bottom box, draw an X for every bee counted. Write the number.
### McGraw Hill My Math Kindergarten Chapter 3 Lesson 7 My Homework Answer Key
How many balls?
Draw to solve
Explanation:
There are total 15 balls
Ten ones and five balls is 15.
In bottom box we have drawn the number of balls counted.
Directions: Say and trace the number. Count the balls. In the bottom box, draw that number of balls. Write the number.
How many magnets?
Draw to solve
Explanation:
There are total 18 magnets
Ten ones and eight magnets is 18
In bottom box we have drawn the number of magnets counted.
Directions: Say and trace the number. Count the magnets. In the bottom box, draw that number of magnets. Write the number.
Math at Home Take advantage of problem-solving opportunities during daily routines such as going to the grocery store. Have your child help you make a grocery list by drawing pictures of the grocery items needed. Have him or her count the items.
Explanation:
We have counted total number of fruits and vegetables, dairy and meats
There are total 8 fruits and vegetables and 8 dairy and meats.
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# Examples On Binomial Theorem For Rational Indices
Go back to 'Binomial Theorem'
Example – 13
Find the sum of the series
$1 + \frac{2}{6} + \frac{{2.5}}{{6.12}} + \frac{{2 \cdot 5 \cdot 8}}{{6 \cdot 12 \cdot 18}} + ......\infty$
if you are told that this corresponds to an expansion of a binomial, of the form $${(1 + x)^n}$$ .
Solution: We need to determine n and x. For that, we can compare the terms of this series with the corresponding terms in the following general expansion.
${(1 + x)^n} = 1 + nx + \frac{{n(n - 1)}}{{2!}}{x^2} + \frac{{n(n - 1)\;(n - 2)}}{{3!}}{x^3} + ......\infty$
Thus,
\begin{align}{} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,nx = \frac{2}{6}\\\\ \frac{{n(n - 1)}}{{2!}}{x^2} = \frac{{2.5}}{{6.12}} \end{align}
Solving for n and x from these two equations, we get \begin{align}n = - \frac{2}{3}\end{align} and \begin{align}x = - \frac{1}{2}\end{align} . Thus, the sum of the series is
$\begin{array}{l}S = {\left( {1 + x} \right)^n} = {\left( {1 - \frac{1}{2}} \right)^{ - \frac{2}{3}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {4^{1/3}}\end{array}$
Example – 14
Find the sum of the series
\begin{align}\frac{1}{{a + b}} + \frac{1}{{a + 2b}} + \frac{1}{{a + 3b}} + ...... to\;\; n\;\; terms\end{align}
for b << a
Solution: Before solving this problem, ponder a moment over the following fact:
In the expansion of $${(1 + x)^n}$$ , if x << 1, that is, if x is much smaller than 1, then the expansion can be approximated as
${(1 + x)^n} \approx 1 + nx$
since all higher order terms can be neglected due to the small magnitude of x.
Coming to the problem, note that if b << a, i.e, if $$\frac{b}{a} < < 1$$ , then,
\begin{align}{}\frac{1}{{a + rb}} &= \frac{1}{{a\left( {1 + r\frac{b}{a}} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,& = \frac{1}{a}{\left( {1 + r\frac{b}{a}} \right)^{ - 1}}\end{align}
$\approx \frac{1}{a}\left( {1 - \frac{{rb}}{a}} \right) since \;\;\frac{b}{a} < < 1$
Thus, the sum S of the series is (to a good approximation)
\begin{align}{}S &\approx \sum\limits_{r = 1}^n {} \frac{1}{a}\left( {1 - r\frac{b}{a}} \right)\\\\\,\,\, &= \frac{1}{a}\left( {n - \frac{{n(n + 1)}}{2}\frac{b}{a}} \right).\\\\\,\,\, &= \frac{n}{a} - \frac{{n(n + 1)b}}{{2{a^2}}}\end{align}
Example – 15
Evaluate 99 3/2 correct to four decimal places.
Solution: We have
$\begin{array}{l}{99^{3/2}}\;\; = {\left( {100 - 1} \right)^{3/2}}\\\\\,\,\,\,\,\,\,\,\,\,\,\, = {100^{3/2}} \cdot {\left( {1 - 0.01} \right)^{3/2}}\\\\\,\,\,\,\,\,\,\,\,\,\, = 1000 \cdot {\left( {1 - 0.01} \right)^{3/2}}\\\\\,\,\,\,\,\,\,\,\,\,\, = 1000 \cdot \left( {1 - \frac{3}{2}\; \cdot \;\left( {0.01} \right) + \frac{{\frac{3}{2}\; \cdot \;\frac{1}{2}}}{{2!}}\; \cdot \;{{\left( {0.01} \right)}^2} - ......} \right)\\\\\,\,\,\,\,\,\,\,\, = 1000\left( {1 - \frac{3}{{200}} + \frac{3}{{80000}} - .......} \right)\\\\\,\,\,\,\,\,\,\, = 1000 - 15 + 0.0375 - ......\\\\\,\,\,\,\,\,\, = {\rm{ }}985.0375\end{array}$
Note that we only considered the first three terms of the expansion because the higher order terms would not have had any effect on the answer upto the fourth decimal place.
## TRY YOURSELF - II
Q. 1 Find the sum of the series
$1 + \frac{1}{3} + \frac{{1.3}}{{3.6}} + \frac{{1.3.5}}{{3.6.9}} + ......\infty$
Q. 2 Find the sum of the series
$1 + \frac{1}{4} + \frac{{1.3}}{{4.8}} + \frac{{1.3.5}}{{4.8.12}} + ......\infty$
Q. 3 Find the magnitude of the greatest term in the expansion of $${\left( {1 + 3y} \right)^{ - 2/5}}$$ for y = \begin{align}\frac{1}{5}\end{align} .
Q. 4 Find the magnitude of the greatest term in the expansion of $${\left( {1 - 5y} \right)^{3/5}}$$ for y = \begin{align}\frac{1}{3}\end{align}
Q. 5 If $$|x|\; < < 1$$ , find the approximate value of
$\frac{{\sqrt {1 + x} \; + \;\sqrt {{{(1 - x)}^3}} }}{{\sqrt {1 + x} + \sqrt {{{(1 - x)}^5}} }}$
Q. 6 If $$\,y = x - {x^2} + {x^3} - {x^4} + ......\infty$$ , show using the general binomial theorem that
\begin{align}\qquad\qquad x = \frac{y}{{1 - y}}\end{align}
Q. 7 If a is very nearly equal to b, then show that the value of \begin{align}\frac{{b + 2a}}{{a + 2b}}\end{align} is nearly equal to \begin{align}{\left( {\frac{a}{b}} \right)^{\frac{1}{3}}}\end{align} .
Hint: Write \begin{align}\frac{a}{b} = 1 + x , \quad where\;\; |x|\; < < 1\end{align}
Q. 8 Using the general binomial theorem, find the approximate value for 9 3/2 .
Q. 9 * Prove that the coefficient of x n in the expanded represent action of
$\frac{{1 + {x^2} - {x^4}}}{{{{(1 + x)}^3}}}$
will be equal to
${\left( { - 1} \right)^n}\left( {\frac{{{n^2} + 5n - 8}}{2}} \right)$
Q. 10 Find the coefficient of x n in the expanded representation of \begin{align}\frac{x}{{(x - a)\;(x - b)}} ,\; if\;\; |x|\; < min (a, b)\end{align} |
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# Compound Growth Help
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By McGraw-Hill Professional
Updated on Oct 4, 2011
## Introduction to Compound Growth
A quantity (such as a population, amount of money, or radiation level) changes exponentially if the growth or loss is a fixed percentage over a period of time. To see how this works, we will see how the value of an account grows over four years if \$100 is deposited and earns 5% interest, compounded annually. Compounded annually means that the interest earned in the previous year earns interest.
After one year, \$100 has grown to 100 + 0.05(100) = 100 + 5 = \$105. In the second year, the original \$100 earns 5% plus the \$5 earns 5% interest: 105 + (105)(0.05) = \$110.25. Now this amount earns interest in the third year: 110.25 + (110.25)(0.05) = \$115.76. Finally, this amount earns interest in the fourth year: 115.76 + (115.76)(0.05) = \$121.55. If interest is not compounded, that is, the interest does not earn interest, the account would only be worth \$120. The extra \$1.55 is interest earned on interest.
Compound growth is not dramatic over the short run but it is over time. If \$ 100 is left in an account earning 5% interest, compounded annually, for 20 years instead of four years, the difference between the compound growth and noncompound growth is a little more interesting. After 20 years, the compound amount is \$265.33 compared to \$200 for simple interest (noncompound growth). A graph of the growth of each type over 40 years is given in Figure 9.1. The line is the growth for simple (noncompounded) interest, and the curve is the growth with compound interest.
Fig. 9.1
We can use a formula to compute the value of an account earning compounded interest. If P dollars is invested for t years, earning r interest rate, then it will grow to A dollars, where A = P (1 + r ) t.
#### Examples
Find the compound amount.
• \$5000, after three years, earning 6% interest, compounded annually
• We will use the formula A = P (1 + r ) t . P = 5000, r = 0.06, and t = 3. We want to know A , the compound amount.
A = 5000(1 + 0.06) 3 = 5000(1.06) 3 = 5000(1.191016)
= 5955.08
The compound amount is \$5955.08.
• \$10,000 after eight years, interest, compounded annually
• A = 10,000(1 + 0.0725) 8 = 10,000(1.0725) 8 ≈ 10,000(1.7505656)
≈ 17,505.66
The compound amount is \$17,505.66
Many investments pay more often than once a year, some paying interest daily. Instead of using the annual interest rate, we need to use the interest rate per period, and instead of using the number of years, we need to use the number of periods. If there are n compounding periods per year, then the interest rate per period is and the total number of periods is nt . The compound amount formula becomes
#### Examples
Find the compound amount.
• \$5000, after three years, earning 6% annual interest
• (a) compounded semiannually
(b) compounded monthly
For (a), interest compounded semiannually means that it is compounded twice each year, so n = 2.
The compound amount is \$5970.26.
For (b), interest compounded monthly means that it is compounded 12 times each year, so n = 12.
The compound amount is \$5983.40.
• \$10,000, after eight years, earning annual interest, compounded weekly Interest that is paid weekly is paid 52 times each year, so n = 52.
• The compound amount is \$17,853.17.
The more often interest is compounded per year, the more interest is earned. \$1000 earning 8% annual interest, compounded annually, is worth \$1080 after one year. If interest is compounded quarterly, it is worth \$1082.43 after one year. And if interest is compounded daily, it is worth \$1083.28 after one year. What if interest is compounded each hour? Each second? It turns out that the most this investment could be worth (at 8% interest) is \$1083.29, when interest is compounded each and every instant of time. Each instant of time, a tiny amount of interest is earned. This is called continuous compounding. The formula for the compound amount for interest compounded continuously is A = Pe rt , where A , P , r , and t are the same quantities as before. The letter e stands for a constant called Euler’s number. It is approximately 2.718281828. You probably have an e or e x key on your calculator. Although e is irrational, it can be approximated by numbers of the form
where m is a large rational number. The larger m is, the better the approximation for e . If we make the substitution and use some algebra, we can see how is very close to e rt , for large values of n . If interest is compounded every minute, n would be 525,600, a rather large number!
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# CBSE Class 10-Mathematics: Chapter – 7 Coordinate Geometry Part 14 (For CBSE, ICSE, IAS, NET, NRA 2022)
Get top class preparation for CBSE/Class-10 right from your home: get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-10.
Question 16:
To conduct sports day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flowerpots have been placed at a distance of 1 m from each other along AD. Niharika runs 14th of the distance AD on the 2nd line and posts a green flag. Preet runs 15th of the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Niharika runs 14th of the distance AD on the 2nd line and posts a green flag.
There are flowerpots. It means, she stops at 25th flowerpot.
Therefore, the coordinates of point where she stops are .
Preet runs 15th of the distance AD on the eighth line and posts a red flag. There are flowerpots. It means, she stops at 20th flowerpot.
Therefore, the coordinates of point where she stops are .
Using Distance Formula to find distance between points and , we get
Rashmi posts a blue flag exactly halfway the line segment joining the two flags.
Using section formula to find the coordinates of this point, we get
Therefore, coordinates of point, where Rashmi posts her flag are
It means she posts her flag in 5th line after covering of distance
Question 17:
If A and B are and respectively, find the coordinates of P such that and P lies on the line segment AB.
and
It is given that
So, we have
Let coordinates of P be
Using Section formula to find coordinates of P, we get
Therefore, Coordinates of point P are .
Question 18:
In each of the following find the value of ‘k’ , for which the points are collinear.
(i)
(ii)
(i)
Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.
Area of Triangle
(ii)
Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.
Area of Triangle
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# Maharashtra Board Class 9 Maths Solutions Chapter 3 Polynomials Problem Set 3
## Maharashtra State Board Class 9 Maths Solutions Chapter 3 Polynomials Problem Set 3
Question 1.
Write the correct alternative answer for each of the following questions.
i. Which of the following is a polynomial?
(D) √2x² + $$\frac { 1 }{ 2 }$$
ii. What is the degree of the polynomial √7 ?
(A) $$\frac { 1 }{ 2 }$$
(B) 5
(C) 2
(D) 0
(D) 0
iii. What is the degree of the polynomial ?
(A) 0
(B) 1
(C) undefined
(D) any real number
(C) undefined
iv. What is the degree of the polynomial 2x2 + 5xsup>3 + 7?
(A) 3
(B) 2
(C) 5
(D) 7
(A) 3
v. What is the coefficient form of x3 – 1 ?
(A) (1, -1)
(B) (3, -1)
(C) (1, 0, 0, -1)
(D) (1, 3, -1)
(C) (1, 0, 0, -1)
vi. p(x) = x2 – x + 3, then p (7√7) = ?
(A) 3
(B) 7√7
(C) 42√7+3
(D) 49√7
(D) 49√7
vii. When x = – 1, what is the value of the polynomial 2x3 + 2x ?
(A) 4
(B) 2
(C) -2
(D) -4
(A) 4
viii. If x – 1 is a factor of the polynomial 3x2 + mx, then find the value of m.
(A) 2
(B) -2
(C) -3
(D) 3
(C) -3
ix. Multiply (x2 – 3) (2x – 7x3 + 4) and write the degree of the product.
(A) 5
(B) 3
(C) 2
(D) 0
(A) 5
x. Which is the following is a linear polynomials?
(A) x + 5
(B) x2 + 5
(C) x3 + 5
(D) x4 + 5
(A) x + 5
Hints:
v. x3 – 1 = x3 + 0x2 + 0x – 1
vi. p(7√ 7) = (7√ 7)2 (7√ 7) (7√ 7) + 3
= 3
vii. p(-1) = 2(-1)3 + 2(-1)
= -2 – 2 = -4
vii. p(1) = 0
∴ 3(1)2 + m(1) = 0
∴ 3 + m =0
∴ m = -3
ix. Here, degree of first polynomial = 2 and
degree of second polynomial 3
∴ Degree of polynomial obtained by multiplication = 2 + 3 = 5
Question 2.
Write the degree of the polynomial for each of the following.
i. 5 + 3x4
ii. 7
iii. ax7 + bx9 (a, b are constants)
i. 5 + 3x4
Here, the highest power of x is 4.
∴Degree of the polynomial = 4
ii. 7 = 7x°
∴ Degree of the polynomial = 0
iii. ax7 + bx9
Here, the highest power ofx is 9.
∴Degree of the polynomial = 9
Question 3.
Write the following polynomials in standard form. [1 Mark each]
i. 4x2 + 7x4 – x3 – x + 9
ii. p + 2p3 + 10p2 + 5p4 – 8
i. 7x4 – x3 + 4x2 – x + 9
ii. 5p4 + 2p3 + 10p2 + p – 8
Question 4.
Write the following polynomial in coefficient form.
i. x4 + 16
ii. m5 + 2m2 + 3m+15
i. x4 + 16
Index form = x4 + 0x3 + 0x2 + 0x + 16
∴ Coefficient form of the polynomial = (1,0,0,0,16)
ii. m5 + 2m2 + 3m + 15
Index form = m5 + 0m4 + 0m3 + 2m2 + 3m + 15
∴ Coefficient form of the polynomial = (1, 0, 0, 2, 3, 15)
Question 5.
Write the index form of the polynomial using variable x from its coefficient form.
i. (3, -2, 0, 7, 18)
ii. (6, 1, 0, 7)
iii. (4, 5, -3, 0)
i. Number of coefficients = 5
∴ Degree = 5 – 1 = 4
∴Index form = 3x4 – 2x3 + 0x2 + 7x + 18
ii. Number of coefficients = 4
∴Degree = 4 – 1 = 3
∴ Index form = 6x3 + x2 + 0x + 7
iii. Number of coefficients = 4
∴ Degree = 4 – 1 = 3
∴ Index form = 4x3 + 5x2 – 3x + 0
Question 6.
i. 7x4 – 2x3 + x + 10;
3x4 + 15x3 + 9x2 – 8x + 2
ii. 3p3q + 2p2q + 7;
2p2q + 4pq – 2p3q
Solution:
i. (7x4 – 2x3 + x + 10) + (3x4 + 15x3 + 9x2 – 8x + 2)
= 7x4 – 2x3 + x + 10 + 3x4 + 15x3 + 9x2 – 8x + 2
= 7x4 + 3x4 – 2x3 + 1 5x3 + 9x2 + x – 8x + 10 + 2
= 10x4 + 13x3 + 9x2 – 7x + 12
ii. (3p3q + 2p2q + 7) + (2p2q + 4pq – 2p3q)
= 3p3q + 2p2q + 7 + 2p2q + 4pq – 2p3q
= 3p3q – 2p3q + 2p2q + 2p2q + 4pq + 7
= p3q + 4p2q + 4pq + 7
Question 7.
Subtract the second polynomial from the first.
i. 5x2 – 2y + 9 ; 3x2 + 5y – 7
ii. 2x2 + 3x + 5 ; x2 – 2x + 3
Solution:
i. (5x2 – 2y + 9) – (3x2 + 5y – 7)
= 5x2 – 2y+ 9 – 3x2 – 5y + 1
= 5x2 – 3x2 – 2y – 5y + 9 + 7
= 2x2 – 1y + 16
ii. (2x2+ 3x + 5) – (x2 – 2x + 3)
= 2x2 + 3x + 5 – x2 + 2x – 3
= 2x2 – x2 + 3x + 2x + 5 – 3
= x2 + 5x + 2
Question 8.
Multiply the following polynomials.
i. (m3 – 2m + 3) (m4 – 2m2 + 3m + 2)
ii. (5m3 – 2) (m2 – m + 3)
Solution:
i. (m3 – 2m + 3) (m4 – 2m2 + 3m + 2)
= m3(m4 – 2m2 + 3m + 2) – 2m(m4 – 2m2 + 3m + 2) + 3(m4 – 2m2 + 3m + 2)
= m7 – 2m5 + 3m4 + 2m3 – 2m5 + 4m3 – 6m2 – 4m + 3m4 – 6m2 + 9m + 6
= m7 – 2m5 – 2m5 + 3m4 + 3m4 + 2m3 + 4m3 – 6m2 – 6m2 – 4m + 9m + 6
= m7 – 4m5 + 6m4 + 6m3 – 12m2 + 5m + 6
ii. (5m3 – 2) (m2 – m + 3)
= 5m3(m2 – m + 3) – 2(m2 – m + 3)
= 5m5 – 5m4 + 15m3 – 2m2 + 2m – 6
Question 9.
Divide polynomial 3x3 – 8x2 + x + 7 by x – 3 using synthetic method and write the quotient and remainder.
Solution:
Dividend = 3x3 – 8x2 + x + 7
∴ Coefficient form of dividend = (3, – 8, 1,7)
Divisor = x – 3
∴ Opposite of – 3 is 3
Coefficient form of quotient = (3, 1,4)
∴ Quotient = 3x2 + x + 4 and
Remainder =19
Question 10.
For which value of m, x + 3 is the factor of the polynomial x3 – 2mx + 21?
Solution:
Here, p(x) = x3 – 2mx + 21
(x + 3) is a factor of x3 – 2mx + 21.
∴ By factor theorem,
Remainder = 0
∴ P(- 3) = 0
p(x) = x3 – 2mx + 21
∴ p(-3) = (-3)3 – 2(m)(-3) + 21
∴ 0 = – 27 + 6m + 21
∴ 6 + 6m = 0
∴ 6m = 6
∴ m = 1
∴ x + 3 is the factor of x3 – 2mx + 21 for m = 1.
Question 11.
At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x2 – 3y2, 7y2 + 2xy and 9x2 + 4xy respectively. At the beginning of the year 2017, x2 + xy – y2, 5xy and 3x2 + xy persons from each of the three villages respectively went to another village for education, then what is the remaining total population of these three villages ?
Solution:
Total population of villages at the end of 2016 = (5x2 – 3y2) + (7y2 + 2xy) + (9x2 + 4xy)
= 5x2 + 9x2 – 3y2 + 7y2 + 2xy + 4xy
= 14x2 + 4y2 + 6xy …….(i)
Total number of persons who went to other village at the beginning of 2017 = (x2 + xy – y2) + (5xy) + (3x2 + xy)
= x2 + 3x2 – y2 + xy + 5xy + xy
= 4x2 – y2 + 7xy … (ii)
Remaining total population of villages = Total population at the end of 2016 – total number of persons who went to other village at the beginning of 2017
= 14x2 + 4y2 + 6xy – (4x2 – y2 + 7xy) … [From (i) and (ii)]
= 14x2 + 4y2 + 6xy – 4x2 + y2 – 7xy
= 14x2 – 4x2 + 4y2 + y2 + 6xy – 7xy = 1
= 10x2 + 5y2 – xy
∴ The remaining total population of the three villages is 10x2 + 5y2 – xy.
Question 12.
Polynomials bx2 + x + 5 and bx3 – 2x + 5 are divided by polynomial x – 3 and the remainders are m and n respectively. If m – n = 0, then find the value of b.
Solution:
When polynomial bx2 + x + 5 is divided by (x – 3), the remainder is m.
∴ By remainder theorem,
Remainder = p(3) = m
p(x) = bx2 + x + 5
∴ p(3) = b(3)2 + 3 + 5
∴m = b(9) + 8
m = 9b + 8 …(i)
When polynomial bx3 – 2x + 5 is divided by x – 3 the remainder is n
∴ remainder = p(3) = n
p(x) = bx3 – 2x + 5
∴ P(3)= b(3)3 – 2(3) + 5
∴ n = b(27) – 6 + 5
∴ n = 27b – 1 …(ii)
Now, m – n = 0 …[Given]
∴ m = n
∴ 9b + 8 = 27b – 1 …[From (i) and (ii)]
∴ 8 + 1 = 27b – 9b
∴ 9 = 18b
∴ b = $$\frac { 1 }{ 2 }$$
Question 13.
Simplify.
(8m2 + 3m – 6) – (9m – 7) + (3m2 – 2m + 4)
Solution:
(8m2 + 3m – 6) – (9m – 7) + (3m2 – 2m + 4)
= 8m2 + 3m – 6 – 9m + 7 + 3m2 – 2m + 4
= 8m2 + 3m2 + 3m – 9m – 2m – 6 + 7 + 4
= 11m2 – 8m + 5
Question 14.
Which polynomial is to be subtracted from x2 + 13x + 7 to get the polynomial 3x2 + 5x – 4?
Solution:
Let the required polynomial be A.
∴ (x2 + 13x + 7) – A = 3x2 + 5x – 4
∴ A = (x2 + 13x + 7) – (3x2 + 5x – 4)
= x2 + 13x + 7 – 3x2 – 5x + 4
= x2 – 3x2 + 13x – 5x + 7+4
= -2x2 + 8x + 11
∴ – 2x2 + 8x + 11 must be subtracted from x2 + 13x + 7 to get 3x2 + 5x – 4.
Question 15.
Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?
Solution:
Let the required polynomial be A.
∴ (4m + 2n + 3) + A = 6m + 3n + 10
∴ A = 6m + 3n + 10 – (4m + 2n + 3)
= 6m + 3n + 10 – 4m – 2n – 3
= 6m – 4m + 3n – 2n + 10 – 3
= 2m + n + 7
∴ 2m + n + 7 must be added to 4m + 2n + 3 to get 6m + 3n + 10.
Question 1.
Read the following passage, write the appropriate amount in the boxes and discuss.
Govind, who is a dry land farmer from Shiralas has a 5 acre field. His family includes his wife, two children and his old mother. He borrowed one lakh twenty five thousand rupees from the bank for one year as agricultural loan at 10 p.c.p.a. He cultivated soyabean in x acres and cotton and tur in y acres. The expenditure he incurred was as follows :
He spent ₹10,000 on seeds. The expenses for fertilizers and pesticides for the soyabean crop was ₹ 2000x and ₹ 4000x2 were spent on wages and cultivation of land. He spent ₹ 8000y on fertilizers and pesticides and ₹ 9000y2 for wages and cultivation of land for the cotton and tur crops.
Let us write the total expenditure on all the crops by using variables x and y.
₹ 10000 + 2000x + 4000×2 + 8000y + 9000y2
He harvested 5x2 quintals soyabean and sold it at ₹ 2800 per quintal. The cotton crop yield was $$\frac { 5 }{ 3 }$$ y2 quintals which fetched ₹ 5000 per quintal.
The tur crop yield was 4y quintals and was sold at ₹ 4000 per quintal. Write the total income in rupees that was obtained by selling the entire farm produce, with the help of an expression using variables x and y. (Textbook pg. no. 44)
Total income = income on soyabean crop + income on cotton crop + income on tur crop
= ₹ (5x2 x 2800) + ₹($$\frac { 5 }{ 3 }$$ y2 x 5000) + ₹ (4y x 4000)
= ₹ ( 14000x2 + $$\frac { 25000 }{ 3 }$$y2 + 16000y)
Question 2.
We have seen the example of expenditure and income (in terms of polynomials) of Govind who is a dry land farmer. He has borrowed rupees one lakh twenty-five thousand from the bank as an agriculture loan and repaid the said loan at 10 p.c.p.a. He had spent ₹ 10,000 on seeds. The expenses on soyabean crop was ₹ 2000x for fertilizers and pesticides and ₹ 4000x2 was spent on wages and cultivation. He spent ₹ 8000y on fertilizers and pesticides and ₹9000y2 on cultivation and wages for cotton and tur crop.
His total income was
₹ (14000x2 + $$\frac { 25000 }{ 3 }$$y2 + 16000y)
By taking x = 2, y = 3 write the income expenditure account of Govind’s farming. (Textbook pg. no. 52)
Solution:
– Credit (Income)
₹ 1,25,000 Bank loan
₹ 56000 Income from soyabean
₹ 75000 Income from cotton
₹ 48000 Income from tur
₹ 304000 Total income
– Debit (Expenses)
₹ 1,37,000 loan paid with interest for seeds
₹ 10000 For seeds
₹ 4000 Fertilizers and pesticides for soyabean
₹ 16000 Wages and cultivation charges for soyabean
₹ 24000 Fertilizers and pesticides for cotton & tur
₹ 81000 Wages and cultivation charges for cotton & tur
₹ 272000 Total expenditure |
# Basic Arithmetic : Rounding
## Example Questions
### Example Question #1 : Whole Numbers
Round to the nearest hundredth:
Explanation:
The hundredth place is the 2nd number after the decimal. Because the thousandth place is great than 5, we will need to round up to
### Example Question #1 : Rounding
Baseballs can only be bought in packages of . If team needs balls for the season and always rounds that estimate up, how many packages do they need?
Explanation:
1. Round 145 to the nearest 10:
Since there is a 5 in the ones place, 145 rounds up to 150.
2. Divide the rounded number of baseballs by 10 since they come in packages of 10:
150/10=15 packages
### Example Question #1 : Basic Arithmetic
Round to the nearest ten.
Explanation:
To round to the nearest ten, you need to take a look at the ones' place. Since the ones' place is greater than , you will need to round the tens' place up.
So then is rounded up to .
### Example Question #2 : Rounding
Round to the nearest ones' place.
Explanation:
To figure out how to round the ones' place, we need to first look at the digit that's to the right of it. If the digit to the right of the ones' place is less than 5, then we leave the ones' place alone.
Since the tenths place is 1, we will leave the ones place alone.
### Example Question #1 : Rounding
Round to the nearest hundred.
Explanation:
To figure out how to round any number, look at the digit to the right of the place you're supposed to be rounding.
Since we want to round the hundreds place, look to the tens place first. If the tens place is greater than 5, then we round up. If the tens place is less than five, then leave the hundreds place alone.
Because the tens place is 7, we will need to round to 900.
### Example Question #1 : Operations With Whole Numbers
What is rounded to the nearest ten-thousands place? |
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# Derivative of Logarithmic Functions
Have you ever wondered how to deal with large numbers? You might have heard about a quantity that increases exponentially. This phrase refers to a situation that an exponential function can model. The outputs of these functions rapidly increase as their inputs increase.
Logarithmic functions are the inverse functions of exponential functions. Since logarithmic functions are slowly increasing functions, they can be helpful when trying to rescale large quantities.
The logarithmic function is a slowly increasing function
Furthermore, we can use the properties of logarithms to our advantage in many problem-solving scenarios, particularly in calculus. For these reasons, it is essential to learn how to find the derivatives of logarithmic functions.
## Definition for the Derivative of the Logarithmic Function
A logarithmic function $$f(x) = \log_{a}x$$ computes the logarithm with base $$a$$ of an $$x$$-value. The base $$a$$ must be a non-negative number. Its derivative is defined as the limit of its rate of change as the change becomes very small.
Let $$f(x) = \log_{a}x$$ be a logarithmic function. Its derivative is defined by the following limit,
$f'(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}.$
In practice, you do not find the derivative of a logarithmic function using limits. The limit is found once to obtain a formula, which then is used along with some Differentiation Rules to find the derivatives of logarithmic functions.
## Formulas for the Derivatives of Logarithmic Functions
As stated before, you can find the derivative of a logarithmic function using limits, but it is not the most practical way. Instead, you can use the following formula.
The derivative of the logarithmic function is given by $\frac{\mathrm{d}}{\mathrm{d}x}\log_{a}{x} = \left(\frac{1}{\ln{a}}\right) \left( \frac{1}{x} \right).$
Here is a quick example.
Find the derivative of
$f(x)=\log_{5}{x}.$
Begin by noticing that the base of the logarithmic function is $$5.$$ Knowing this, you can use the formula for the derivative of a logarithmic function, that is
$f'(x)=\left(\frac{1}{\ln{5}} \right) \left( \frac{1}{x} \right).$
Pretty straightforward right?
## Derivative of a Log Function with base e
In the particular case where the base of a logarithmic function is $$e,$$ that is $$f(x) = \log_{e} x,$$ the function receives a special name.
If the base of a logarithm is the number $$e,$$ then it is called a Natural Logarithm. The natural logarithmic function computes the natural logarithm of a variable, and it is denoted as
$f(x) = \ln{x}.$
A natural logarithm has the base $$e,$$ which means that
$\ln{e}=1.$
With this, the formula for the derivative of a natural logarithmic function becomes simpler, that is
\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\ln{x} &= \left(\frac{1}{\ln{e}}\right) \left( \frac{1}{x} \right) \\ &= \left( \frac{1}{1} \right) \left( \frac{1}{x} \right) \\ &= \frac{1}{x}. \end{align}
The derivative of the natural logarithmic function is given by $\frac{\mathrm{d}}{\mathrm{d}x}\ln{x} =\frac{1}{x}.$
Note that by knowing this formula, along with the properties of logarithms, you can differentiate any logarithmic function. Consider the logarithmic function
$f(x)=\log_{a}{x}.$
The above function can be rewritten using the properties of logarithms, that is
\begin{align} f(x) &= \log_{a}{x} \\[0.5em] &= \frac{\ln{x}}{\ln{a}}. \end{align}
Since $$\ln{a}$$ is a constant, you can use the Constant Multiple Rule to factor it out when differentiating the function, so
\begin{align} \frac{\mathrm{d}f}{\mathrm{d}x} &= \frac{1}{\ln{a}}\frac{\mathrm{d}}{\mathrm{d}x}\ln{x} \\[0.5em] &= \left( \frac{1}{\ln{a}}\right) \left( \frac{1}{x} \right), \end{align}
which is the formula introduced at the start of the previous section.
## Proof of the Derivative of the Logarithmic Function
The natural logarithmic function is the inverse function of the exponential function, this means that if
$y=\ln{x},$
then
$e^y=x.$
Next, differentiate both sides of the equation, that is
$\frac{\mathrm{d}}{\mathrm{d}x} e^y = \frac{\mathrm{d}}{\mathrm{d}x} x$
The left-hand side of the equation is the exponential function, so you can use the formula for the derivative of the exponential function. However, since $$y$$ is a function of $$x,$$ you must also use the Chain Rule.
$e^y\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x} x$
The right-hand side can be differentiated using the Power Rule, so
$e^y\frac{\mathrm{d}y}{\mathrm{d}x} = 1.$
Finally, substitute back $$e^y=x$$ and isolate the derivative of the natural logarithmic function, obtaining
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{x}.$
Sometimes it is worth inspecting how to find derivatives by their definition using limits. This might be a little tricky, but this gives a bunch of experience! Let's dive into it!
Recall the definition of the derivative of the natural logarithmic function through limits, which is
$\frac{\mathrm{d}}{\mathrm{d}x} \ln{x} = \lim_{\Delta x \rightarrow 0} \frac{\ln{(x+\Delta x)}-\ln{x}}{\Delta x}.$
You can rewrite the expression inside the limit using the Quotient Property of Logarithms and the Power Property of Logarithms, that is
$\frac{\mathrm{d}}{\mathrm{d}x} \ln{x} = \lim_{\Delta x \rightarrow 0} \left[ \ln{\left( \frac{x+\Delta x}{\Delta x} \right)}^{\frac{1}{\Delta x}} \right]$
Here comes the tricky part! Multiply by $$\frac{x}{x}$$ in the exponent of the function, that is
\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \ln{x} &= \lim_{\Delta x \rightarrow 0} \left[ \ln{\left( \frac{x+\Delta x}{\Delta x} \right)}^{\frac{1}{\Delta x}\frac{x}{x}} \right] \\[0.75em] &= \lim_{\Delta x \rightarrow 0} \left[ \ln{\left( \frac{x+\Delta x}{\Delta x} \right)}^{\frac{x}{\Delta x}\frac{1}{x}} \right] . \end{align}
Now use again the Power Property of Logarithms to move $$\frac{1}{x}$$ from an exponent to a coefficient. You can take it out of the limit as it does not depend on $$\Delta x.$$ You also need to simplify the fraction inside the natural logarithm, so
$\frac{\mathrm{d}}{\mathrm{d}x} \ln{x} = \frac{1}{x} \lim_{\Delta x \rightarrow 0} \left[ \ln{\left( 1+ \frac{\Delta x}{x}\right)^{\frac{x}{\Delta x}}} \right].$
The next step is to use the properties of limits to switch the limit and the natural logarithm. You can do this because the natural logarithm is a continuous function.
$\frac{\mathrm{d}}{\mathrm{d}x} \ln{x} = \frac{1}{x} \ln{\left[ \lim_{\Delta x \rightarrow 0} \left( 1+\frac{\Delta x}{x}\right)^{\frac{x}{\Delta x}}\right]}.$
Next, make the substitution
$u=\frac{x}{\Delta x}.$
Because $$x>0,$$ $$u$$ tends to positive infinity as $$\Delta x$$ tends to zero. This will let you rewrite the limit as
$\frac{\mathrm{d}}{\mathrm{d}x} \ln{x} = \frac{1}{x} \ln{\left[ \lim_{u \rightarrow \infty} \left( 1+ \frac{1}{u}\right)^{u}\right]},$
which is one of the definitions of $$e,$$ the base of the natural logarithm, so
$\frac{\mathrm{d}}{\mathrm{d}x}\ln{x} = \frac{1}{x} \ln{e}.$
Since $$e$$ its the base of the natural logarithm, this last factor is equal to 1, finally obtaining
$\frac{\mathrm{d}}{\mathrm{d}x} \ln{x}= \frac{1}{x}.$
## Examples of Derivatives of Logarithmic Functions
It is now time to work on some examples. You can use differentiation rules and the properties of logarithms to your advantage!
Find the derivative of
$f(x) = \ln{x^2}.$
There are two ways of finding the derivative of the given function. By using the Chain Rule, and by using properties of logarithms.
• Using the Chain Rule.Begin by letting $$u(x) = x^2,$$ and use the Chain Rule, that is$\frac{\mathrm{d}f}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}u}\ln{u} \frac{\mathrm{d}u}{\mathrm{d}x}.$ You can find the derivative of $$u(x)$$ using the Power Rule, that is$\frac{\mathrm{d}u}{\mathrm{d}x}=2x,$and you can also write the derivative of the natural logarithmic function, so\begin{align} \frac{\mathrm{d}f}{\mathrm{d}x} &= \left( \frac{1}{u} \right) (2x) \\[0.5em] &= \left( \frac{1}{x^2} \right) (2x) \\[0.5em] &= \frac{2}{x}. \end{align}
• Using Properties of Logarithms.Rather than using the Chain Rule, you can begin by rewriting the function using the power property of logarithms, that is$f(x)= 2\ln{x}.$From here, you can use the Constant Multiple Rule and differentiate the natural logarithmic function, so\begin{align} \frac{\mathrm{d}f}{\mathrm{d}x} &= (2)\left( \frac{1}{x} \right) \\ &= \frac{2}{x}. \end{align}
Which method do you prefer? You get the same answer either way!
You can use more properties of logarithms to your advantage. Consider now an example with the product property of logarithms.
Find the derivative of
$g(x) = \ln{\left(xe^x \right)}.$
Once again, you have two options for finding the derivative of the given function. Generally, it is adviced to use the properties of logarithms whenever you can.
Begin by using the product property of logarithms to rewrite the function, that is
$g(x) = \ln{x} + \ln{e^x}.$Since the natural logarithmic function is the inverse function of the exponential function, you can further rewrite the above function, so
$g(x) = \ln{x} + x.$
From here, you can differentiate each term, giving you
$\frac{\mathrm{d}g}{\mathrm{d}x} = \frac{1}{x} + 1.$
Sometimes the properties of logarithms will not be able to be used in the function you are working with. In these cases, just apply any relevant differentiation rule.
Find the derivative of the function
$h(x) = \ln{\left(\sin{x}\right)}.$
Here you can let $$u(x) = \sin{x}$$ and use the Chain Rule, that is
$\frac{\mathrm{d}h}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}u} \ln{u} \frac{\mathrm{d}u}{\mathrm{d}x}.$
The derivative of the sine function is the cosine function, so
$\frac{\mathrm{d}u}{\mathrm{d}x} = \cos{x}.$
Knowing this and the derivative of the natural logarithmic function lets you write
\begin{align} \frac{\mathrm{d}h}{\mathrm{d}x} &= \left( \frac{1}{u} \right) (\cos{x}) \\[0.5em] &= \frac{\cos{x}}{\sin{x}} \\[0.5em] &= \tan{x}, \end{align}
where you have used the trigonometric identity
$\frac{\cos{x}}{\sin{x}}=\tan{x}.$
## Derivatives of Logarithmic Functions - Key takeaways
• Logarithmic functions are inverse functions of exponential functions of the same base.
• The derivative of a logarithmic function is given by$\frac{\mathrm{d}}{\mathrm{d}x} \log_{a}{x} = \left(\frac{1}{\ln{a}}\right) \left(\frac{1}{x} \right).$
• The derivative of the natural logarithmic function can be proved by using implicit differentiation and the differentiation rule for the exponential function.
• The derivative of the natural logarithmic function can also be proved using limits. It is important to know one of the definitions of $$e$$ as a limit, that is $e = \lim_{u \rightarrow \infty} \left( 1+\frac{1}{u} \right)^u .$
• Properties of logarithms like the Power Rule of Logarithms and the Product Rule of Logarithms can be used before differentiating a function in order to make it simpler.
The derivative of the natural logarithmic function is 1/x. If the base is other than e you need to multiply 1/x by the reciprocal of the natural logarithm of the base.
You can prove the derivative of the logarithmic function by using implicit differentiation and the differentiation rule of the exponential function.
You use the formula for the derivative of a logarithmic function along with any relevant differentiation rules.
First look which kind of logarithmic function you are working with. If it is the natural logarithmic function, ln x, then its derivative is 1/x.If the base is other than e you need to multiply 1/x by the reciprocal of the natural logarithm of the base.
An example of the derivative of a logarithmic function is the derivative of the natural logarithmic function, which is 1/x.
## Final Derivative of Logarithmic Functions Quiz
Question
Which functions are inverses of logarithmic functions?
Exponential functions.
Show question
Question
What is the differentiation rule of the natural logarithmic function $$\ln{x}$$?
$\frac{\mathrm{d}}{\mathrm{d}x} \ln{x} = \frac{1}{x}.$
Show question
Question
What better describes a logarithmic function?
It is a slowly increasing function defined over the positive numbers.
Show question
Question
Find the derivative of $$g(x) = \ln{\sqrt{x}}.$$
$$g'(x)= \frac{1}{2x}.$$
Show question
Question
Which of the following is the derivative of $$f(x)= \log_{5}{x}$$?
$f'(x) = \left( \frac{1}{\ln{5}} \right) \left( \frac{1}{x} \right).$
Show question
Question
How can you prove the derivative of the natural logarithmic function?
By using implicit differentiation and the differentiation rule for the exponential function.
Show question
Question
Find the derivative of $$f(x)= \ln{x^3}.$$
$$f'(x) = \frac{3}{x}.$$
Show question
Question
Find the derivative of $$g(x) = e^{\ln{x}}.$$
$$g'(x) = 1.$$
Show question
Question
Which of the following is the derivative of $$h(x)= \log_{2}{x}$$ ?
$$h'(x) = \frac{1}{\ln{2}}\frac{1}{x}.$$
Show question
Question
What is Logarithmic Differentiation?
Logarithmic Differentiation is a method used to find the derivative of a function using the properties of logarithms.
Show question
Question
What is the derivative of the natural logarithm function $$f(x) = \ln x$$?
$$\frac{1}{x}$$
Show question
Question
The Product Property of Logarithms can be used to transform a product into a:
Sum
Show question
Question
The Quotient Property of Logarithms can be used to transform a quotient into a:
Subtraction
Show question
Question
The Power Property of Logarithms can be used to transform a power into a:
Product
Show question
Question
Is it possible to use Logarithmic Differentiation to find the derivative of $$f(x) = x^n$$?
Yes, but it is more straightforward to use the Power Rule.
Show question
Question
What are the steps for doing logarithmic differentiation?
1. Take the natural logarithm of the original function.
2. Use any relevant properties of logarithms to simplify the function.
3. Use the Chain Rule and the differentiation rule of the natural logarithm to differentiate each expression.
4. Multiply the resulting expression by the original function. The result is the derivative of the original function
Show question
Question
Is it necessary to take the inverse of the natural logarithm when doing logarithmic differentiation?
No. When you differentiate the natural logarithm of a function you are left with the original function and its derivative, so there is no need to undo the opperation.
Show question
Question
Use the product property of logarithms to rewrite
$\ln f(x)g(x)$
$$\ln f(x) + \ln g(x)$$
Show question
Question
Use the quotient property of logarithms to rewrite
$\ln \left( \frac{f(x)}{g(x)} \right)$
$$\ln f(x) - \ln g(x)$$
Show question
Question
A logarithmic function with base $$a$$ is denoted as $$\log_{a}{x},$$ where $$a$$ is a number that is ____.
Greater than 0.
Show question
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# 8th Grade Polynomials: Concepts and Problems
Polynomials are a common math topic that children typically must learn in the 8th grade. They are usually taught as part of math units covering algebraic operations. To find out more about helping your child with polynomials concepts and problems, keep reading.
Typically, 8th grade math focuses almost entirely on algebra. It's likely that your child will be required to take a standardized multiple-choice math test at some point during the school year based on the Common Core State Standards for 8th grade math. Any math curriculum based on these standards will most likely require your child to learn about polynomials. To find out how you can help your child at home with polynomials based on his or her specific learning needs, have a chat with your child's math teacher or any other math professionals currently working with him or her.
### Explaining What Polynomials Are to Your Child
You can tell your child that the word 'polynomial' means 'many terms'. Polynomials are exponents, constants and variables that are usually combined using multiplication, addition or subtraction. However, polynomials are never combined with division.
An example of a polynomial problem that you can give your child to is 2xy^2 + 2x - 7. In this equation, 2xy^2, 2x and 7 are each different terms, making three terms total. Numbers in the equation like 2 and -7 are constants; x and y are variables and in this case, the 2 that squares 2xy^2 is the exponent. You can go through a number of equations with your child and have him or her point out which terms are exponents, constants and variables.
### Exercises for Solving Problems
There are a variety of exercises that you can use at home to help your child with polynomial concepts and problems. For example you can play a game in which you ask your child if an equation is a polynomial or not. Polynomials in this game might include 'x + 4', '-18y + 2x^4' or even a simple constant like '1' or '73'. An example of a non-polynomial would be 'x/5' since it contains division.
If your child is struggling to grasp polynomials, talk to his or her math teacher to find out what the best course of action may be. The teacher may be able to help your child with polynomial concepts and problems before or after school. In addition to this, you might want to consider hiring a middle school algebra tutor.
There are also a lot of online resources that your 8th grader can use when learning polynomials. You and your child will find a number of pages explaining polynomial concepts as well as free problems, worksheets, quizzes, games and activities.
Did you find this useful? If so, please let others know!
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# 1.2 Basic probability concepts and axioms
Probability is the backbone of biostatistics, helping us understand and predict outcomes in health and medicine. It quantifies the likelihood of events, from rare diseases to treatment success rates, using numbers between 0 and 1.
Basic probability concepts and axioms provide the foundation for more complex statistical analyses. Understanding these fundamentals is crucial for interpreting research findings, designing studies, and making informed decisions in healthcare and clinical practice.
## Probability and its axioms
### Definition and notation
###### Top images from around the web for Definition and notation
• Probability quantifies the likelihood an will occur, expressed as a number between 0 and 1 inclusive
• The probability of an event A is denoted
• P(A) ranges from 0 to 1, where 0 represents an impossible event and 1 represents a certain event
• Example: The probability of rolling a 3 on a fair six-sided die is 1/6 (one favorable outcome out of six possible outcomes)
### Axioms and properties
• The sum of the probabilities of all possible outcomes in a equals 1
• Sample space: The set of all possible outcomes of an experiment or random process
• Example: When rolling a fair six-sided die, the sample space is {1, 2, 3, 4, 5, 6}, and the sum of the probabilities of each outcome is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1
• For any two events A and B, the probability of A or B occurring is given by P(A or B) = P(A) + P(B) - P(A and B)
• This is known as the
• P(A and B) represents the probability of both events A and B occurring simultaneously
• Example: In a deck of 52 cards, the probability of drawing a heart or a king is P(heart or king) = P(heart) + P(king) - P(heart and king) = 13/52 + 4/52 - 1/52 = 16/52
## Calculating probabilities
• The addition rule states P(A or B) = P(A) + P(B) - P(A and B) for any two events A and B
• This rule accounts for the overlap between events A and B by subtracting P(A and B)
• Example: The probability of drawing a red card or a face card from a standard deck is P(red or face) = P(red) + P(face) - P(red and face) = 26/52 + 12/52 - 6/52 = 32/52
• The states P(A and B) = P(A) × P(B|A) for any two events A and B
• P(B|A) is the of B given A, which is the probability of B occurring given that A has already occurred
• Example: The probability of drawing two kings from a deck of cards without replacement is P(king1 and king2) = P(king1) × P(king2|king1) = 4/52 × 3/51 = 1/221
### Complement rule
• The complement rule states P(not A) = 1 - P(A) for any event A
• "not A" represents the event where A does not occur
• Example: If the probability of a patient having a certain disease is 0.05, then the probability of the patient not having the disease is P(not disease) = 1 - P(disease) = 1 - 0.05 = 0.95
## Independent vs dependent events
### Independent events
• Two events A and B are independent if the occurrence of one event does not affect the probability of the other event occurring
• For A and B, P(A and B) = P(A) × P(B)
• The multiplication rule simplifies for independent events because P(B|A) = P(B)
• Example: The probability of rolling a 4 on a fair die and then flipping a head on a fair coin is P(4 and head) = P(4) × P(head) = 1/6 × 1/2 = 1/12
### Dependent events and conditional probability
• Two events A and B are dependent if the occurrence of one event affects the probability of the other event occurring
• For A and B, P(A and B) = P(A) × P(B|A)
• P(B|A) is the conditional probability of B given A
• Example: The probability of drawing two aces from a deck of cards without replacement is P(ace1 and ace2) = P(ace1) × P(ace2|ace1) = 4/52 × 3/51 = 1/221
• Conditional probability P(B|A) is the probability of event B occurring given that event A has already occurred
• It is calculated as P(B|A) = P(A and B) / P(A)
• Example: The probability of drawing a king given that a face card has been drawn is P(king|face) = P(king and face) / P(face) = 4/12 = 1/3
## Mutually exclusive events
### Definition and properties
• Two events A and B are mutually exclusive (or disjoint) if they cannot occur simultaneously
• In other words, the intersection of A and B is empty
• Example: When rolling a fair die, the events "rolling an even number" and "rolling a 5" are mutually exclusive
• For mutually exclusive events A and B, P(A and B) = 0
• Since A and B cannot occur simultaneously, the probability of both events occurring is 0
• If events A and B are mutually exclusive, then P(A or B) = P(A) + P(B)
• The addition rule simplifies for mutually exclusive events because P(A and B) = 0
### Mutually exclusive and exhaustive events
• In a set of mutually exclusive and exhaustive events, the sum of the probabilities of all events equals 1
• Mutually exclusive: No two events can occur simultaneously
• Exhaustive: The set of events covers all possible outcomes in the sample space
• Example: When rolling a fair die, the events "rolling an odd number" and "rolling an even number" are mutually exclusive and exhaustive, and P(odd) + P(even) = 1/2 + 1/2 = 1
## Key Terms to Review (18)
Addition Rule: The addition rule in probability is a fundamental principle that determines the likelihood of the occurrence of at least one of two or more events. It states that the probability of either event A or event B occurring is equal to the sum of their individual probabilities, adjusted for any overlap if the events are not mutually exclusive. This rule plays a crucial role in calculating probabilities in various situations, especially when dealing with combined outcomes.
Bayes' Theorem: Bayes' Theorem is a mathematical formula used to update the probability of a hypothesis based on new evidence. It connects prior knowledge or beliefs about an event with new data, allowing for a refined probability assessment. This theorem is foundational in understanding conditional probabilities, making it a key concept in statistical inference and decision-making under uncertainty.
Binomial Distribution: The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. It is important for understanding how probabilities work in scenarios where there are only two possible outcomes, like success or failure, and it plays a vital role in biological research and statistical modeling.
Conditional Probability: Conditional probability is the likelihood of an event occurring given that another event has already occurred. This concept helps in understanding the relationship between events and how the probability of one event is affected by the occurrence of another. It is essential in various fields, including statistics and decision-making, as it provides a framework for updating beliefs based on new information.
Continuous Random Variable: A continuous random variable is a type of variable that can take on an infinite number of values within a given range, often representing measurements or quantities. Unlike discrete random variables, which can only take specific values, continuous random variables can assume any value within their interval, making them essential for modeling real-world phenomena. This characteristic connects to fundamental probability concepts and distributions, allowing for the calculation of probabilities using functions such as probability density functions (PDFs).
Convergence in Probability: Convergence in probability refers to a statistical concept where a sequence of random variables converges to a specific value in such a way that, as the number of observations increases, the probability that the random variable deviates from this value by more than a certain amount approaches zero. This idea connects to the behavior of random variables and highlights how they can stabilize around a certain point with enough trials, which is critical for understanding concepts like law of large numbers and consistency in estimators.
Dependent Events: Dependent events are occurrences in probability where the outcome or occurrence of one event affects the outcome or occurrence of another. Understanding this concept is crucial for calculating probabilities accurately, especially when dealing with sequences of events where the first influences the second, often seen in conditional scenarios. This interconnection between events is essential in exploring how probabilities adjust when additional information is available.
Discrete Random Variable: A discrete random variable is a type of variable that can take on a finite or countably infinite number of distinct values, often representing outcomes from a random process. These variables can be used to describe situations where the possible outcomes are specific and separate, like rolling a die or counting the number of successes in a series of trials. Understanding discrete random variables is essential for analyzing probability distributions and applying concepts such as expected value and variance in various statistical contexts.
Empirical Probability: Empirical probability is the likelihood of an event occurring based on observed data or past occurrences rather than theoretical assumptions. This approach allows for a more accurate representation of probability as it reflects real-world outcomes, making it particularly useful in fields like biostatistics where data-driven decision-making is crucial.
Event: In probability theory, an event is a specific outcome or a set of outcomes from a random experiment. Events can be simple, involving a single outcome, or compound, consisting of multiple outcomes, and are fundamental to understanding the probability of occurrences within a given context. Recognizing events helps in the calculation of probabilities and understanding how different events relate to each other through concepts such as unions and intersections.
Independent events: Independent events are two or more occurrences where the outcome of one event does not affect the outcome of another. This means that knowing the result of one event provides no information about the result of another. In probability, this concept is crucial as it allows for the simplification of calculations involving multiple events, leading to the application of basic probability rules and conditional probabilities.
Law of Large Numbers: The law of large numbers is a fundamental statistical principle stating that as the size of a sample increases, the sample mean will converge to the expected value or population mean. This concept emphasizes that larger samples tend to yield results that are more stable and closer to the true population parameters, which is crucial when understanding probability and distributions.
Multiplication Rule: The multiplication rule is a fundamental principle in probability that helps determine the likelihood of two or more independent events occurring simultaneously. It states that to find the probability of the joint occurrence of independent events, you multiply the probabilities of each individual event. This rule connects closely to basic probability concepts and axioms, reinforcing the idea that probabilities can be combined in systematic ways to address complex scenarios.
Normal Distribution: Normal distribution is a continuous probability distribution characterized by a symmetric bell-shaped curve, where most of the observations cluster around the central peak and probabilities for values further away from the mean taper off equally in both directions. This distribution is fundamental in statistics, as it helps to model various natural phenomena and is key in many statistical methods and inference techniques.
P(a): p(a) represents the probability of event A occurring. In probability theory, this term is crucial as it quantifies how likely an event is to happen based on a defined sample space. The concept of p(a) helps in understanding how different events relate to each other and establishes a foundation for calculating probabilities using various rules and axioms, such as the addition and multiplication rules.
P(a|b): The notation p(a|b) represents the conditional probability of event A occurring given that event B has occurred. This concept is crucial for understanding how the occurrence of one event influences the likelihood of another. Conditional probability helps in making informed decisions based on prior knowledge or evidence, and it serves as a foundation for Bayes' theorem, which updates probabilities as more information becomes available.
Sample Space: Sample space is the set of all possible outcomes of a random experiment. It serves as the foundational concept in probability theory, as it helps define the scope of events being analyzed and provides context for calculating probabilities. Understanding the sample space is crucial for identifying events and determining their likelihood within any probabilistic framework.
Theoretical Probability: Theoretical probability is the mathematical calculation of the likelihood of an event occurring based on the possible outcomes in a perfectly controlled environment. It relies on the assumption that all outcomes are equally likely and is often expressed as a ratio of the number of favorable outcomes to the total number of possible outcomes. This concept lays the foundation for understanding how probability works and connects to various principles, such as the axioms that govern probability theory. |
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