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Calculators do a lot of the hard work for you, so why would you need to use a calculator more efficiently? Using a calculator efficiently is a good habit to get into, as soon as you can. Efficient use of your calculator can save you time in exams, save you time when you’re doing homework, and give you more accurate answers.
Take some time to get to know your calculator and what it can do for you. Let’s look at an example. Suppose you want to calculate correct to 2 d.p:
3.16 + 5.14 divided by 1.87 + 0.33
You could either do:
3.16 + 5.14 = 8.3
1.87 + 0.33 = 2.2
8.3/2.2 = 3.77 (correct to 2 d.p.)
Or:
(3.16 + 5.14)/(1.87 + 0.33) = 3.77 (correct to 2 d.p.)
In this case you can either do 3 sums, or just one. With these simple numbers it doesn’t matter too much. Let’s look at another example, calculated correct to 3 d.p:
11.2/1.7 divided by 2π
You could do:
11.2/1.7 = 6.588235294 (full display on calculator)
2 x π = 6.283185307 (full display on calculator)
6.588235294/6.283185307 = 1.048550213 (full display on calculator)
Answer: 1.049 (to 3 d.p.)
But it is much easier to do the calculation in one go:
(11.2/1.7)/ 2π = 1.048550213
Answer: 1.049 (to 3 d.p.)
Notes
1. The / means divided by
2. The brackets tell your calculator which part of the calculation to do first
3. Check how your calculator treats 2π. You may be able to enter 2 and then π, with no operator in between, and your calculator will automatically multiply them before dividing the result from the brackets by 2π. If not, then you need to enter the sum as: (11.2/1.7)/ (2 x π) = 1.048550213
4. Doing the sum in one calculation is much quicker
5. There’s no risk of getting a wrong answer from writing down intermediate results incorrectly
6. You don’t lose any accuracy when you do the sum in one calculation
If you’re not sure what all the buttons are for on your calculator, have a look in the instruction book. If you can’t find it, don’t worry; you’ll almost certainly be able to find one on the internet. Check on the manufacturer’s website.
The main thing is, get to know your calculator and what it can do for you. Using your calculator efficiently can save you time in exams. Knowing how to use your calculator effectively can reduce the risk of making or introducing errors into calculations in exams.
Efficient use of calculators can save you time and help you maximise your marks.
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# How do you find the derivative of f(x)=ax^2?
Jan 9, 2017
$\frac{d}{\mathrm{dx}} \left(a {x}^{2}\right) = 2 a x$
#### Explanation:
This is an application of a basic differentiation rule:
$\frac{d}{\mathrm{dx}} {x}^{n} = n \cdot {x}^{n - 1}$
In this case:
$\frac{d}{\mathrm{dx}} \left(a {x}^{2}\right) = a \cdot \frac{d}{\mathrm{dx}} \left({x}^{2}\right) = 2 a x$
You can also demonstrate it directly considering the formal definition of the derivative of $f \left(x\right)$:
$f ' \left(x\right) = {\lim}_{\Delta x \to 0} \frac{f \left(x + \Delta x\right) - f \left(x\right)}{\Delta x} = {\lim}_{\Delta x \to 0} \frac{\Delta f}{\Delta x}$
$\Delta f = a {\left(x + \Delta x\right)}^{2} - a {x}^{2} = a {x}^{2} + 2 a x \Delta x + a {\left(\Delta x\right)}^{2} - a {x}^{2} = 2 a x \Delta x + a {\left(\Delta x\right)}^{2}$
$\frac{\Delta f}{\Delta x} = 2 a x + a \Delta x$
${\lim}_{\Delta x \to 0} \frac{\Delta f}{\Delta x} = {\lim}_{\Delta x \to 0} 2 a x + a \Delta x = 2 a x$
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# In $\Delta ABC$ fig. $\angle x+\angle y+\angle z$ is equals to:A. 120B. 180C. 240D. 360
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Hint: We can solve these triangle angle problems easily, through the angle sum property of triangle and linear angle sum property. Which states that the sum of all the angle of triangle is ${{180}^{{}^\circ }}$ and the linear sum angle of line is also ${{180}^{{}^\circ }}$ . Here in this question as we know that sum of all the angles of triangle is ${{180}^{{}^\circ }}$ and also that the sum of interior and exterior angle on a straight line is also ${{180}^{{}^\circ }}$ (linear angle sum property), so we can write the exterior angle in terms of interior angle of triangle and then sum up all the exterior angle (as asked in question) whose sum we know through the angle sum property of triangle. So we will get the answer.
To understand it in more detail let us solve the question; $,$ are the straight lines through which a triangle is formed. As in the figure the exterior angles are marked at $\angle x,\angle y,\angle z$ . As we know that in a line sum of angle is equal to ${{180}^{{}^\circ }}$ i.e. in figure 1, AB is a line, on which angle $\angle x,\angle x'$ are present. So by linear angle sum property we can say that $\angle x+\angle x'={{180}^{{}^\circ }}$
Similarly, with the line sum of angle is equal to ${{180}^{{}^\circ }}$ i.e. in figure 2,BC is a line , on which angle $\angle y,\angle y'$ are present. So by linear angle sum property we can say that $\angle y+\angle y'={{180}^{{}^\circ }}$ in which $\angle y$ is the exterior angle and $\angle y'$ is the interior angle of triangle.
Similarly, the line sum of angle is equal to ${{180}^{{}^\circ }}$ i.e. in figure 3 on line $AC$ , on which angle $\angle z,\angle z'$ are present. So by linear angle sum property we can say that $\angle z+\angle z'={{180}^{{}^\circ }}$ in which $\angle z$ is the exterior angle and $\angle z'$ is the interior angle of triangle.
Now in $\vartriangle ABC$ whose interior angles are $\angle x',\angle y',\angle z'$ and exterior angle are equal to $\angle x,\angle y,\angle z$ . As we know, the sum of all interior angles is ${{180}^{{}^\circ }}$ by angle sum property of triangle. So we can write $\angle x'+\angle y'+\angle z'={{180}^{{}^\circ }}$
So from above explanation we have
$\angle x+\angle x'={{180}^{{}^\circ }}$ equation (i)
$\angle y+\angle y'={{180}^{{}^\circ }}$ equation (ii)
$\angle z+\angle z'={{180}^{{}^\circ }}$ equation (iii)
$\angle x'+\angle y'+\angle z'={{180}^{{}^\circ }}$ equation (iv)
Adding equation (i), (ii), (iii), i.e.
\begin{align} & \left( i \right)+\left( ii \right)+\left( iii \right) \\ & \angle x+\angle x'+\angle y+\angle y'+\angle z+\angle z'={{180}^{{}^\circ }}+{{180}^{{}^\circ }}+{{180}^{{}^\circ }} \\ & \angle x+\angle x+\angle z+\angle x'+\angle y'+\angle z'=540 \\ \end{align}
Now we can replace; $\angle x'+\angle y'+\angle z'$with ${{180}^{{}^\circ }}$ from equation (iv), so we will get
\begin{align} & \angle x+\angle x+\angle z+\angle x'+\angle y'+\angle z'=540 \\ & \angle x+\angle x+\angle z+{{180}^{{}^\circ }}={{540}^{{}^\circ }} \\ & \angle x+\angle x+\angle z={{360}^{{}^\circ }} \\ \end{align}
Hence answer is${{360}^{{}^\circ }}$, i.e. $\angle x+\angle y+\angle z={{360}^{{}^\circ }}$
So, the correct answer is “Option D”.
Note: To find the angle or sum of angles in a triangle, then angle sum property is the basic you must know, then linear pair of angle or linear angle sum property. Moreover the sum of exterior angle is also called as two times the sum of interior angle of triangle, as it is${{360}^{{}^\circ }}$which is twice of ${{180}^{{}^\circ }}$
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# Divisibility of composite numbers
I am having difficulty in solving following types of problem:
Sometimes we are given a number in terms of $n$ and we have to check whether it is divisible by a particular composite number. For example, I am posting a question here
suppose $k= n^5- n$ then prove that $k$ is divisible by $30$.
And this was my approach:
$$n^5-n= n (n^4-1) =n(n^2+1) (n+1) (n-1)$$ Since $k$ is a product of $n^2+1$ and three consecutive integers, it must be a multiple of $2$ and $3$. So it gives $k= 6m (n^2+1)$. But how can I prove that it's also a multiple of 5? And this is where I get confused.
Now, suggest some alternate way to prove above problem or some corrections in my method.
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This turns out to be identical to How to prove $n^5−n$ is divisible by 30 without reduction – MJD Jul 30 '12 at 19:12
You are almost done: If $n \equiv 0, \pm 1 \pmod 5$ we are done (as we have $5 \mid n$, $5 \mid n-1$ or $5 \mid n+1$ then). So suppose $n \equiv \pm 2 \pmod 5$, but then $n^2 + 1 \equiv (\pm 2)^2 + 1 \equiv 0 \pmod 5$, hence $5 \mid n^2 + 1$, and so $5 \mid n^5 -n$ also in this case.
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Note that $n^2+1$ is divisible by $5$ iff $n^2-4$ is divisible by $5$. But $n^2-4=(n-2)(n+2)$.
So $(n^2+1)(n-1)(n)(n+1)$ is divisible by $5$ iff $(n-2)(n-1)(n)(n+1)(n+2)$ is divisible by $5$. But this is the product of $5$ consecutive integers. End of proof.
Remark: There are more general ways of attacking the problem. But the one used above continues the pattern that you used for $2$ and $3$. Continuing in this way quickly becomes too complicated for practical use, and one ends up turning to Fermat's Theorem, and Euler's Theorem.
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That was a great idea. Now i know how to deal with (n^2 + arbit ) type numbers while dealing with divisibility . Thanks :) – Simar Jul 24 '13 at 18:02
$$\rm\begin{eqnarray} n(n^4\!-\!1)/30\: &=&\ \ \rm n\, (n^2\!-\!1)\,(\color{#C00}{n^2\!+\!1})/30\!\!\!\!\!\! & & \\ &=&\ \ \rm n\,(n\!+\!1)\,(n\!-\!1)\,(\color{#C00}5 &\!\color{#C00} + &\rm\color{#C00}{\, (n\!+\!2)\,(n\!-\!2)})/30 \\ &=&\ \ \rm (n\!+\!1)\,n\,(n\!-\!1)/6 &\! + &\rm \, \color{#C00}{(n\!+\!2)}\,(n\!+\!1)\,n\,(n\!-\!1)\,\color{#C00}{(n\!-\!2)}/30 \\ &=&\rm\qquad\qquad\ {n\!+\!1\ \!\choose 3}\ &\! + &\ \rm 4\,{n\!+\!2\ \! \choose 5} \in\, \Bbb Z \end{eqnarray}$$
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According to link#1, the product of 5 consecutive integers divisible by 5! and the product of 3 consecutive integers divisible by 3!.
Now $n(n-1)(n+1)(n-2)(n+2) = n^5-5n^3+4n$
$n^5-n= n^5-5n^3+4n + 5(n^3-n)$
=$n(n-1)(n+1)(n-2)(n+2) -5n(n-1)(n+1)$
So, the first part is divisible by 5! and the 2nd part is by 5(3!)
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Let $f(n)=n^p-n$
Clearly, f(1) =1-1=0 is divisible by p.
Now, $f(n+1)-f(n)=((n+1)^p-(n+1))-(n^p-n)= \sum _{1≤r≤p-1} pC_rn^r$.
But prime p| $pC_r$ for 1≤r≤p-1.
So, the difference is divisible by p.
So, using mathematical induction, we can say p|($n^p-n$)
This comes from Fermat's Little theorem.
So, $n^5-n$ is divisible by 5.
$n^5-n = n(n^4-1) = n(n^2-1)(n^2+1) = (n^3-n)(n^2+1)$
Now, $n^3-n$ is divisible by 3, so is $n^5-n$.
$n^5-n = n(n^4-1) = n(n^2-1)(n^2+1) = n(n-1)(n+1)(n^2+1) = (n^2-n)(n+1)(n^2+1)$
Now, $n^2-n$ is divisible by 2, so is $n^5-n$.
So, $n^5-n$ is divisible by 2,3,5 so, it is divisible by lcm(2,3,5)=30
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It suffices to show that $k^5-k$ is always divisible by each of 2, 3, and 5. That is, we want $k^5-k\equiv 0 \mod d$ where $d$ is each of 2, 3, and 5.
We know that $k\equiv r\pmod 2$ where $r$ is 0 or 1. Then $k^5 - k \equiv r^5-r \equiv 0 \pmod2$. The last equivalence follows instantly from a consideration of the two cases: either $r=0$ in which case we have $0^5-0\equiv 0\pmod 2$, or $r=1$ in which case we have $1^5-1\equiv 0\pmod 2$.
Similarly $k\equiv r\pmod 3$ where $r$ is one of 0, 1 or 2. In each case consider $r^5-r\pmod 3$. The first two cases are completely trivial, and the same as in the previous paragraph. The $r=2$ case gives $32-2= 30\equiv 0\pmod 3$.
And again $k\equiv r\pmod 5$ where $r$ is one of $\{0,1,2,3,4\}$; now consider $r^r-r\pmod 5$. Again cases 0 and 1 are trivial; the other three give $30\equiv 0\pmod 5$, $3^5-3 = 243-3 = 240\equiv 0\pmod 5$, and $4^5-4 = 1024-4 = 1020 \equiv 0\pmod 5$, so we are done.
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For divisibility by $5$, you can use Fermat's little theorem which states that $n^p\equiv n\pmod p$ where $p$ is a prime $\implies n^5-n\equiv 0\pmod 5$.
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let, $f(k) = k^5 - k$ where k is a natural number
we need to show that $f(k)$ is divisible by $30$
take the case when $k = 1$
$f(1) = 0$ which is divisible by $30$
take the case when $k = 2$
$f(2) = 2^5 - 2$
i.e. $f(2) = (2)(2^4 - 1)$
i.e. $f(2) = (2)({2^2}^2 - 1)$
i.e. $f(2) = (2)(2^2 - 1)(2^2 + 1)$
i.e. $f(2) = (2){(2 - 1)(2 + 1)(2^2 + 1)}$
i.e. $f(2) = (2){(1)(3)(4 + 1)}$
i.e. $f(2) = (2){(1)(3)(5)}$
i.e. $f(2) = (2)(15)$
i.e. $f(2) = 30$ which is divisible by $30$
Assume $f(n)$ is divisible by $30$ where $n$ is a natural number
let, $f(n) = 30(\zeta)$ where $\zeta$ is a natural number
take the case when $k = n + 1$ where $n$ is a natural number
i.e. $f(n+1) = (n+1)^5 - (n+1)$ -------(2)
i.e. $f(n+1) = (n+1)((n+1)^4 - 1)$
i.e. $f(n+1) = (n+1)((n+1)^2 - 1)((n+1)^2 + 1)$
i.e. $f(n+1) = (n+1)((n+1) - 1)((n+1) + 1)((n+1)^2 + 1)$
i.e. $f(n+1) = (n+1)(n)(n+2)(n^2 + (2)(n) + 1^2 + 1)$
i.e. $f(n+1) = (n)(n+1)(n+2)(n^2 + (2)(n) + 2)$ -------(3)
A careful observation will reveal that the product of three consecutive natural numbers $(n)(n+1)(n+2)$ will always be divisible by 6
let, $(n)(n+1)(n+2) = (6)(\beta)$ where $\beta$ is a natural number
i.e. $f(n+1) = (6)(\beta)(n^2 + (2)(n) + 2)$ -------(4)
i.e. $f(n+1) = (6)(\beta)f_{5}(n)$ -------(5)
here, $f_{5}(n) = (n^2 + (2)(n) + 2)$ -------(6)
Now one just needs to show that $(\beta)f_{5}(n)$ is divisible by $5$ which indeed is
Consider, $f(n+1) - f(n) = (n+1)^5 - (n+1) - (n^5 - n)$
i.e. $f(n+1) - f(n) = (n+1)((n+1)^4 - 1) - n(n^4 - 1)$
i.e. $f(n+1) - f(n) = (n+1)((n+1)^2 - 1)((n+1)^2 + 1) - n(n^2 - 1)(n^2 + 1)$
i.e. $f(n+1) - f(n) = (n+1)(n)(n+2)((n+1)^2 + 1) - n(n-1)(n+1)(n^2 + 1)$
i.e. $f(n+1) - f(n) = (n)(n+1)((n+2)((n+1)^2 + 1) - (n-1)(n^2 + 1))$
i.e. $f(n+1) - f(n) = (n)(n+1)((5)n^2 + (5)n + 5)$
i.e. $f(n+1) - f(n) = (n)(n+1)(5)(n^2 + n + 1)$, which is divisible by $5$
we also know that $f(n+1)$ and $f(n)$ is always an even number which leads to the conclusion that $f(n+1)$ and $f(n)$ is a multiple of $5$ because the difference of two even numbers is divisible by $5$ $iff$ the numbers are individually divisible by $5$
Now $6$ and $5$ are relatively prime to each other
i.e. f(n+1) must be a multiple of 6 and 5 concurrently
i.e. $f(k) = k^5 -k$ is always divisible by 30
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$\beta = (3/2)x^2 - (3/2)x + 1$ where x is a natural number – Rajesh K Singh Jul 24 '12 at 14:55
i.e. $\beta = 3y + 1$ where $y$ is a natural number – Rajesh K Singh Jul 26 '12 at 11:23
This is a tough read. The case $k=2$ is unnecessary, and the steps ... Yikes. Why not just say $f(2) = 32-2=30$?? – The Chaz 2.0 Jul 27 '12 at 17:18
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# Brackets & Factors OCR Stage 6. What is 3(2 + 4) ? 3 ‘lots’ of ‘2 + 4’= 3 ‘lots’ of 6= 18 2 4 3 x 6 12 = 18.
## Presentation on theme: "Brackets & Factors OCR Stage 6. What is 3(2 + 4) ? 3 ‘lots’ of ‘2 + 4’= 3 ‘lots’ of 6= 18 2 4 3 x 6 12 = 18."— Presentation transcript:
Brackets & Factors OCR Stage 6
What is 3(2 + 4) ? 3 ‘lots’ of ‘2 + 4’= 3 ‘lots’ of 6= 18 2 4 3 x 6 12 = 18
What is 3(x + 4) ? x 4 3 x 3x12 = 3x + 12 3 ‘lots’ of ‘x + 4’
What is 5(2x + 1) ? 2x 1 5 x 10x5 = 10x + 5 5 ‘lots’ of ‘2x + 1’
What is 4(x - 2) ? x -2 4 x 4x-8 = 4x - 8 4 ‘lots’ of ‘x - 2’
What is x(x + 3) ? x 3 x x x2x2 3x = x 2 + 3x x ‘lots’ of ‘x + 3’ What is x times x?
What is 2x(x - 3) ? x -3 2x x 2x 2 -6x = 2x 2 – 6x 2x ‘lots’ of ‘x - 3’
What is -x(2x + 3) ? 2x 3 -x x -2x 2 -3x = -2x 2 – 3x -x ‘lots’ of ‘2x + 3’ We call this process ‘expanding a bracket’
Factorising an expression This is the reverse of expanding a bracket Eg Factorise 4x + 2 What is a Factor? A FACTOR of a number is a number that divides into the number EXACTLY
Eg Factorise 4x + 2 What number divides into BOTH parts of the expression EXACTLY? 2 4x + 2 2x+1 = 2(2x + 1) How could we check this?
Eg Factorise 6x - 9 What number divides into BOTH parts of the expression EXACTLY? 36x - 9 2x- 3 = 3(2x – 3)
Eg Factorise x 2 + x What number divides into BOTH parts of the expression EXACTLY? xx 2 + x x+ 1 = x(x + 1)
Eg Factorise 8x 2 – 4x What number divides into BOTH parts of the expression EXACTLY? 4x8x 2 – 4x 2x- 1 = 4x(2x - 1)
Eg Factorise 6 - 4x 2 What number divides into BOTH parts of the expression EXACTLY? 26 - 4x 2 3- 2x 2 = 2(3 – 2x 2 )
Download ppt "Brackets & Factors OCR Stage 6. What is 3(2 + 4) ? 3 ‘lots’ of ‘2 + 4’= 3 ‘lots’ of 6= 18 2 4 3 x 6 12 = 18."
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# How do you solve x^2 - 4x + 7 =0 by completing the square?
Aug 10, 2017
$x = 2 \pm \sqrt{3} i$
#### Explanation:
Given:
${x}^{2} - 4 x + 7 = 0$
While completing the square we will find that this takes the form of the sum of a square and a positive number. As a result it has no solution in real numbers, but we can solve it using complex numbers.
The imaginary unit $i$ satisfies ${i}^{2} = - 1$
The difference of squares identity can be written:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
We can use this with $a = \left(x - 2\right)$ and $b = \sqrt{3} i$ as follows:
$0 = {x}^{2} - 4 x + 7$
$\textcolor{w h i t e}{0} = {x}^{2} - 4 x + 4 + 3$
$\textcolor{w h i t e}{0} = {\left(x - 2\right)}^{2} + {\left(\sqrt{3}\right)}^{2}$
$\textcolor{w h i t e}{0} = {\left(x - 2\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}$
$\textcolor{w h i t e}{0} = \left(\left(x - 2\right) - \sqrt{3} i\right) \left(\left(x - 2\right) + \sqrt{3} i\right)$
$\textcolor{w h i t e}{0} = \left(x - 2 - \sqrt{3} i\right) \left(x - 2 + \sqrt{3} i\right)$
Hence the two roots are:
$x = 2 + \sqrt{3} i \text{ }$ and $\text{ } x = 2 - \sqrt{3} i$
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# Decimal Lesson Plans for Parents and Teachers
In most school curriculums, decimals are presented in fourth grade, improved in fifth grade and perfected in sixth grade. Keep reading for tips on how to teach decimals at home or in the classroom.
## Tips for Teaching Decimals to Children
### Commonalities of Fractions and Decimals
Children should have a good grasp of fractions before they begin decimals. Then, you can point out that fractions and decimals are different ways of saying the same thing. A number of books are available which may help to bridge these two concepts, including:
• Piece = Part = Portion by Scott Gifford
• Fractured Fairy Tales: Fractions & Decimals by Dan Greenberg
• Parting Is Such Sweet Sorrow: Fractions and Decimals (Adventures in Mathopolis) by Linda Powley
### Understanding Decimals
You can expand the familiar place chart to include the decimal and the numbers following it. For example:
HundredsTensUnitsDecimalTenthsHundredths
State that the numbers before the decimal include zero and larger; numbers after the decimal are smaller than one, just like a fraction. Some children understand the chart better if they think of the decimal point as the 'units column' for numbers smaller than one; this makes the chart more symmetrical for those students.
Practice reading numbers ranging from ones and tenths to hundreds and hundredths. Point out that 1/10 is read 'one tenth,' and 0.1 is also read as 'one tenth.' Explain that 0.10 is still read 'one tenth' and that no matter how many zeroes they add, it'll always be read as 'one tenth.' However, even if one of those zeros to another number, like 0.11, it makes the decimal number (fraction) smaller.
In a number with more than one digit, whether before or after the decimal, each digit is ten times bigger than it would be in the place on its right and 1/10 of what it would be in the place on its left. A good way to illustrate this is with money. Using play money, hold up a \$100 bill and ask how many \$10 bills you'd need to make \$100. Keep moving one place to the right, asking the same kinds of questions. When you come to the decimal, explain that you're changing from bills to coins. Then, reverse directions and move from right to left.
### Changing Decimals to Fractions
When children can read decimals correctly, they can then write them like fractions. The steps are read, write and reduce. For instance, 0.50 can be written as 50/100, which can be reduced to 1/2.
When adding and subtracting decimals, the only new concept you need to teach them is to line up the decimals. Remind them that the decimal points in all the numbers must be lined up and model a few examples, like 0.00001 + 0.0002 = 0.00021 and 0.01 + 0.02 = 0.03.
### Multiplying Decimals
Once you've taught children to line up the decimals for addition and subtraction, you need to teach them to not line up the decimals for multiplication, just as you don't have to line up the tens and hundreds in a multiplication problem. However, always emphasize neatness and lining up numbers in each column to avoid problems of placing numbers in the wrong places for the addition step.
To figure out where to put the decimal place in the answer, count up the number of decimal places in the multiplicand and multiplier. For instance, in the problem, 2.1 x 3.24, the product would have three decimal places because there is one in the first number and two in the second number. Similarly, in the problem 3.256 x 8.01, the product would have five decimal places.
### Dividing Decimals
When dividing using decimals, the only new concepts are where to put the decimal and that there is never a remainder. Teach them to move the decimal point in the divisor to the right end of the number, counting how many digits they move it. Have them figure whether they've multiplied that number by 10, 100, 1000, etc.
Just like working with fractions, they must now multiply the dividend by the same amount. Have them move the decimal point in the dividend, and also put one right above it in the quotient. They'll then divide as usual. However, they won't stop with a remainder. Instead, add a zero to the end of the dividend and bring it down to continue dividing. This will be a good place to teach them to round decimals.
For instance, in the problem 3.4 ÷ 2.1, first turn the divisor (2.1) into a whole number by moving the decimal point to the right, so it's now 21. Do the same for the dividend, so the problem is now 34 ÷ 21.
### Changing Fractions to Decimals
Now that the children can divide decimals, they can change fractions to decimals. Simply divide the numerator by the denominator. For instance, the fraction 9/10 can be turned into a decimal by dividing nine by ten (9 ÷ 10), which equals 0.9.
Did you find this useful? If so, please let others know!
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# Mathdoku Techniques
This is our starting grid for our example. What people usually find most intimidating about Mathdoku puzzles the first time they tackle them is that there is no obvious starting point. Best place to start is to look for a group of cells that have limited possibilites. Look at the two hilighted cells, these have a constraint of 4*. There are two ways of satisfying this constraint, either with the sum 2*2=4, or, 1*4=4. But, by having 2*2 we would have to place the number 2 in that column twice, which breaks the rules of Mathdoku. Our solution must be 1*4=4, we don't know yet which order they go in, but we can write some pencil marks in.
Similarly, we can look at these two hilighted cells as well. The only valid combination here is 2*5. Again, we don't know which way round they appear, but by adding some pencil marks we are adding more constraints to our puzzle and eventually a number will present itself. Solving Mathdoku puzzles require using an advanced Sudoku technique known as naked pairs. We've just put the numbers 2 and 5 in our hilighted cell. We don't know which way round these numbers appear, but we know for certain that they must appear in those two cells. This removes them as possibilites for the rest of the row.
By eliminating 2 and 5 from the remainder of the row, this only leaves one possibility for our hilighted cell, and that is 6/3=2.
Look at our hilighted cell in this instance. By using the naked pairs technique we can eliminate the numbers 2, 3, 5, and 6 from our hilighted cell. We know already that the cell to the left must contain a 1 or a 4, we don't have enough information here to make narrow it down any further, so our hilighted cell must be 1 or 4.
We now have enough information to fill in the hilighted cell! For that section the constraint is '2/'. We already know that the other cell must be a 1 or a 4. So let's look at the possibilites. Our only possibilites are 2/1=2, or 4/2=2. We don't have any other possibilites, so whichever sum is the correct one, our hilighted cell must be 2!
By continuing on this way, we can insert another two sets of pencil marks as shown here. But, we can use the naked pair technique here again. Look at row 5, the two left-most cells have a 3, 6 naked pair, whcih means we can eliminate 3 and 6 from the remainder of that row. This only leaves one remaining possibility for our hilighted cell, it can only be a 5! This also means we can fill in the cell above it, this must be a 3.
The key to a Mathdoku puzzle is finding the starting point, once you have a starting point you are continuously adding more and more constraints and making the puzzle easier! Want to complete this puzzle? Play online now!
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A vertically straight tree, 15 m high, is broken by the wind in such a way that its top just touches the ground and makes an angle of 60 degrees with the ground. At what height from the ground did the tree break?
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Hint – In this question CB is the height of the tree and it is being broken from top after a certain point. Let A be the point after which it is broken, it is given that this broken part touches the ground and makes an angle of 60 degrees, so let AC be the broken part and use trigonometric ratios to find out the length below the point A for the tree.
Let us consider the vertically straight tree BC as shown in the figure which is 15 m high.
Now the tree is broken by the wind that its top just touches the ground at C and makes an angle 60 degree from the ground.
Let us assume that the tree is broken at point A (see figure) and the point C touches the ground (see figure) and it is given that the point C makes an angle 60 degree from the ground (see figure).
So, we have to calculate the value of AB.
Let us assume BA be x meter and AC be y meter.
$\Rightarrow BC = BA + AC$
$\Rightarrow 15 = x + y$………………………….(1)
Now in triangle ABC
$\sin {60^0} = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} = \dfrac{x}{y}$
Now apply the value of $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$ in above equation we have,
$\Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{x}{y}$
From equation (1) $y = 15 - x$ so, substitute the value of y in above equation we have,
$\Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{x}{{15 - x}}$
Now simplify the above equation we have,
$\Rightarrow 15\sqrt 3 - x\sqrt 3 = 2x$
$\Rightarrow x\left( {2 + \sqrt 3 } \right) = 15\sqrt 3$
$\Rightarrow x = \dfrac{{15\sqrt 3 }}{{2 + \sqrt 3 }}{\text{ meter}}{\text{.}}$
So, this is the required height from the ground from which the tree is broken.
Note – Whenever we face such types of problems the key concept is to have a diagrammatic representation of the information provided in the question, this will help in understanding the basic outline of the triangle in which we need to apply trigonometric ratios. This concept will help you get on the right track to reach the answer.
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# What Are The Prime Factors Of 77?
• The prime factors of number 77 are: 7, 11
• Determined equcation for number 77 factorization is: 7 * 11
## Is 77 A Prime Number?
• No the number 77 is not a prime number.
• Seventy-seven is a composite number. Because 77 has more divisors than 1 and itself.
## How To Calculate Prime Number Factors
• 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 7
• About Number 7. Seven is a prime number. It is the lowest natural number that cannot be represented as the sum of the squares of three integers. The corresponding cyclic number is 142857. You can use this feature to calculate the result of the division of natural numbers by 7 without a calculator quickly. A seven-sided shape is a heptagon. One rule for divisibility by 7 leads to a simple algorithm to test the rest loose divisibility of a natural number by 7: Take away the last digit, double it and subtract them from the rest of the digits. If the difference is negative, then you're leaving the minus sign. If the result has more than one digit, so you repeat steps 1 through fourth. Eventually results are 7 or 0, then the number is divisible by 7 and not otherwise.
## 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 Are Prime Factors?
• In number theory, the prime factors of a positive integer are the prime numbers that divide that integer exactly. The prime factorization of a positive integer is a list of the integer's prime factors, together with their multiplicities. The process of determining these factors is called integer factorization. The fundamental theorem of arithmetic says that every positive integer has a single unique prime factorization.
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# Year Seven Mathematics Worksheets
Math can be a challenging subject for kids, but it’s also an essential skill that will help them succeed in their future endeavors. One important aspect of math is understanding angles. Angles are used in many real-life situations, such as measuring the height of a building, determining the slope of a hill, and even determining the direction of a vehicle. In this article, we’ll focus on a specific type of angle – vertical angles.
Vertical angles are a pair of angles that are located opposite each other and are formed when two lines cross. These angles are called “vertical” because they are upright and perpendicular, like a letter “V”. It’s important to understand that vertical angles are always equal in measure. This means that if one angle measures 30 degrees, then the other angle must also measure 30 degrees.
It’s helpful to use a diagram to understand vertical angles. Imagine two lines crossing each other to form a “X”. The angles formed at each of the four points where the lines intersect are called vertical angles. They are equal in measure and are opposite each other. For example, angle A and angle C are vertical angles and have the same measure, and angle B and angle D are also vertical angles with the same measure.
It’s important for kids to understand vertical angles because they will encounter them in various math concepts, such as trigonometry and geometry. Knowing how to identify and measure vertical angles can also help with solving problems involving angles in real-life situations, such as determining the height of a building or the slope of a hill.
In conclusion, vertical angles are a crucial aspect of math and are used in many real-life situations. They are formed when two lines cross and are always equal in measure. Understanding vertical angles will help kids in their future math studies and in solving problems involving angles. With practice, they’ll be able to identify and measure vertical angles with ease.
# Year Seven Math Worksheet for Kids – Vertical Angles
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## Applied Machine Learning & Data Science Projects and Coding Recipes for Beginners
A list of FREE programming examples together with eTutorials & eBooks @ SETScholars
# Projects and Coding Recipes, eTutorials and eBooks: The best All-in-One resources for Data Analyst, Data Scientist, Machine Learning Engineer and Software Developer
Topics included: Classification, Clustering, Regression, Forecasting, Algorithms, Data Structures, Data Analytics & Data Science, Deep Learning, Machine Learning, Programming Languages and Software Tools & Packages.
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`Disclaimer: The information and code presented within this recipe/tutorial is only for educational and coaching purposes for beginners and developers. Anyone can practice and apply the recipe/tutorial presented here, but the reader is taking full responsibility for his/her actions. The author (content curator) of this recipe (code / program) has made every effort to ensure the accuracy of the information was correct at time of publication. The author (content curator) does not assume and hereby disclaims any liability to any party for any loss, damage, or disruption caused by errors or omissions, whether such errors or omissions result from accident, negligence, or any other cause. The information presented here could also be found in public knowledge domains.`
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### Edexcel GCSE Maths
Numbers
Algebra
Geometry and Measures
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# Algebraic Equivalence and Proof
## Algebraic Equivalence
Algebraic equivalence is the idea that two algebraic expressions or equations represent the same mathematical relationship or have the same solution set.
We can transform equivalent expressions into each other through algebraic operations, such as addition, subtraction, multiplication, division, substitution, and factoring.
Understanding algebraic equivalence is important for simplifying expressions and solving equations, as it helps identify different representations of the same mathematical relationship.
Let’s look at some examples:
Example: Algebraic Equivalence in Expressions
Consider the expressions and . These expressions are algebraically equivalent because we can use the distributive property to transform one expression into the other:
Therefore, the expressions and represent the same mathematical relationship.
Example: Algebraic Equivalence in Equations
Consider the equations and . We can show that these equations are equivalent by dividing both sides of the first equation by 2:
x
Both equations have the same solution set, , and represent the same mathematical relationship.
## Mathematical Proof
Mathematical proof is a logical argument that establishes the truth of a mathematical statement or theorem. Proofs are important in mathematics because they ensure that results are valid, consistent and reliable.
In algebra, there are several proof techniques you should get familiar with, including direct proof and proof by contradiction.
• Direct proof involves a series of logical steps that follow from given assumptions or axioms to reach a desired conclusion.
• Proof by contradiction assumes the opposite of the statement to be proven and demonstrates that this assumption leads to a logical contradiction. This establishes the truth of the original statement.
Let’s look at some examples:
Example: Direct Proof
Prove that the expressions and are algebraically equivalent for all real numbers a and b.
Expand the expression using the distributive property (also known as the FOIL method):
Since the expressions and can be transformed into each other through valid algebraic operations, we have demonstrated that they are algebraically equivalent for all real numbers a and b.
Prove that is irrational.
Assume the opposite, that is rational. If is rational, it can be expressed as a fraction , where a and b are integers with no common factors other than 1. So, we have:
Squaring both sides of the equation, we get:
Rearranging, we find:
This implies that is an even number (since it is divisible by 2). If is even, then must also be even. Let , where k is an integer. Substituting this into the equation, we get:
Dividing both sides by 2, we have:
This implies that is also even, which means that b is even as well. However, this contradicts our initial assumption that and have no common factors other than 1, since both a and b are divisible by 2. Therefore, our assumption that is rational must be false, and must be irrational.
### Understanding Odd and Even Numbers
In mathematics, numbers can be classified as even or odd based on their divisibility by 2. An even number is divisible by 2, while an odd number is not.
Even numbers can be represented as , where is any natural number, and odd numbers can be represented as , where is any natural number.
Let’s look at the rules for the addition and multiplication of even and odd Numbers:
• Even + Even = Even
• Even × Even = Even
• Odd + Odd = Even
• Odd × Odd = Odd
• Even + Odd = Odd
• Even × Odd = Even
Ok, now let’s look at the proof for Even + Even = Even
To demonstrate the rule that the sum of two even numbers is also even, consider two even numbers and , where and are natural numbers:
As we can factor out a 2 from the sum of and , it confirms that the result is also even.
Similar proofs can be constructed for the other rules, using the representations of even and odd numbers as and , respectively.
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# Question 46dc6
Dec 4, 2017
Solutions:
$x = - 1 , y = 3$
$x = - \frac{1}{3} , y = 3 \frac{2}{3}$
#### Explanation:
We can find the solution (common points of each equation) algebraically or graphically. To make a graph, simply make a table with 'x', 'y1' (from expression1), 'y2' (from expression2) and then plot them on graph paper.
Algebraically, we can use the first equation to set y = x + 4 and then substitute it into the second equation.
$2 {x}^{2} + x \left(x + 4\right) = - 1$ ; $2 {x}^{2} + {x}^{2} + 4 x + 1 = 0$
$3 {x}^{2} + 4 x + 1 = 0$ This can be solved with the quadratic formula.
For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by:
x = (−b±√[b^2−4ac])/(2a)
In this case, a = 3, b = 4 and c = 1
x = (−4 ± sqrt[(4(^2) − 4*3*1]))/(2*3)
x = (−4 ± sqrt[16 − 12])/(6)
x = (−4 ± 2)/6; $x = - \frac{6}{6} \mathmr{and} - \frac{2}{6}$
We can see that the algebraic solution may be more precise than we could obtain visually from a graph.
CHECK: x = -1 , y = 3
$2 {\left(- 1\right)}^{2} + \left(- 1\right) \times \left(3\right) = - 1$ ; 2 – 3 = -1# ; $- 1 = - 1$ CORRECT
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Not all probability questions on the GRE will be about dice, playing cards and marbles. Here are examples of other possible scenarios.
##### Example
A study found that 80% of drivers, who ride with a passenger, wear seatbelts. Passengers wear seatbelts 70% of the time when the driver wears a seatbelt, and 55% of the time when the driver doesn’t wear a seatbelt. What is the probability that a passenger wears a seatbelt?
### Solution
You can use a tree diagram to visualize the possibilities. Write the probabilities on each branch.
There are 2 paths that give the desired outcome of a passenger wearing a seatbelt. Find and add the probabilities of each path.
(Path 1) Driver wears and passenger wears: 80% × 70% = 56%
(Path 2) Driver does not wear and passenger wears: 20% × 55% = 11%
The probability is 56% + 11% = 67%.
A passenger wears a seatbelt 67% or about 2/3 of the time.
##### Example
In a 1-mile race, three different schools (Washington High, Duke High and Cherry Hill High) each have 5 competitors. What is the probability that a student from Cherry Hill will take first place, a student from Duke will take second place and another student from Duke will take third place?
### Solution
This question has both “or” and “and” in the same problem: independent and dependent.
Since 2 students from Duke are desired outcomes, those are dependent events.
Find each probability.
Cherry Hill student first: 5/15 = 1/3
Duke student takes second: 5/14
Duke student takes third: 4/13
Since this is an “and” question, multiply the probabilities. Don’t forget to factor and reduce before you multiply.
\dfrac{1}{3} × \dfrac{5}{14} × \dfrac{4}{13} = \dfrac{5 × 2 × 2}{3 × 2 × 7 × 13} = \dfrac{10}{273}
##### Example
At 3 p.m. Jennifer went into labor. There is a 0.7 chance her baby will be born during each hour that she is in labor. What is the probability that her baby was born at 5:30 p.m. on the same day?
A) 0.027
B) 0.063
C) 0.147
D) 0.27
E) 0.343
### Solution
This question is an independent probability question in disguise. The probability that her baby will be born each hour does not change. Each hour, there is a 0.7 chance the baby will be born, which means there is a 0.3 chance the baby will not be born.
Method 1
From 3 p.m. to 4 p.m., the baby was not born and the probability of that happening is 0.3.
From 4 p.m. to 5 p.m. the probability also is 0.3.
From 5 p.m. to 6 p.m., when the baby was born, the probability is 0.7.
Since the baby must not be born in the first or second hour, and must be born in the third hour, this is an “and” question, so the probability is (0.3)(0.3)(0.7) = 0.063
The correct answer is choice (B).
Method 2
You can use a tree diagram to visualize the possibilities. Write the probabilities on each branch.
There is just one path that gives the desired outcome.
(0.3)(0.3)(0.7) = 0.063
The correct answer is choice (B).
##### Example
From a class of 12 students, two students will be randomly chosen simultaneously. If g is the number of girls in the class, what is the value of g?
(1) The probability that two girls will be chosen together is 1/11.
(2) The probability that one boy and one girl will be chosen is 16/33.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
### Solution
To answer this question, you don’t need to do the math.
Look at Statement (1). You know there is a specific number of girls (g). Since each number of girls yields a different probability of choosing 2 girls, there must be only one specific number that would yield 1/11. So this is enough information to know the value of g, the number of girls.
Statement (2) requires a little more thought. Pairing a boy and a girl can be (bg) or (gb). As in many pairings, there are 2 different pairs that will give the same probability. Statement (2) is not sufficient.
(Note: If you make the table, you will see that there are two ways to get 16/33, with 4 boys and 8 girls or with 4 girls and 8 boys.)
Since only Statement (1) gives enough information, the correct answer is choice (A).
##### Example
In a hotel with single rooms and double rooms, what is the probability that a room chosen at random will be a double room painted red?
(1) 1/6 of the rooms in the hotel are painted red.
(2) 2/3 of the hotel’s rooms are double rooms.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
### Solution
Statement (1) gives the fraction of rooms painted red, but does not say anything about double rooms.
Statement (2) gives the fraction of the rooms that are double rooms, but does not say anything about red rooms.
Putting the statements together still does not give enough information. Neither statement says how many of the red rooms are double rooms. For example, if there were 12 rooms you would know that 2 were red and 8 were doubles. But you don’t know if any of the doubles are red. There is no information connecting the two categories, so you cannot find the probability.
The correct answer is choice (E).
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4th Grade - Apply Strategies To Complex Problems
Grade Level: 4th Skill: Reasoning Topic: Apply Strategies to Complex Problems Goal: Apply strategies and results from simpler problems to more complex problems. Skill Description: Use strategies for simple problems in order to solve difficult problems.
Building Blocks/Prerequisites
Sample Problems
(1) Mental Math: Use a basic fact and pattern to write the product for the large number. 2 x 8,000 = (16,000) (2) Represent problems using base-ten blocks to find the product. (The child will be required to drag and model the multiplication problem with base-ten blocks) 105 x 5 (525) (3) Mental Math: Use basic division facts and patterns to write the quotient for the large number. 2,700 ÷ 9 = (300) (4) Use the break apart strategy to multiply the product. 297 x 4 7 x 4 = ___ 90 x 4 = ____ 200 x 4 = _____ Add the products: _____ + _____ + _____ = ________ (28 + 360 + 800 = 1,188) (5) Find the area of the complex polygon: Area 1 : ___________________ Area 2: ________________ Total Area: _____________ (Area = 3 x 9 + 8 x 4 = 59 ft2)
Learning Tips
(1)
Break Apart Strategy
Helps the child compute mentally without using pencil and paper.
Multiplication (Solve simpler problems)- find the product for 372 x 4
1. Multiply by the ones 4 x 2 = 8
2. Multiply by the tens 70 x 4 = 280
3. Multiply by the hundreds 300 x 4 = 1,200
4. Add the products 8 + 28 + 1,200 = 1,236
(2)
Break apart complex polygons
Divide figure to find area:
Example
4 4 7 3 3 7
Divide the polygon into two rectangles. The larger rectangle measures as (4) length and (7) at the width. Use the formula for finding the rectangle (A = l x w) A = 4 x 7 = 28. Next, find the area of the smaller square which is (3) length and (3) width. A= 3 x 3 = 9. Finally, add the two areas together 28 + 9 = 37 sq. units.
(3)
Multiplication with zeros
Multiply the basic facts, then add the remaining zeros to the product (answer to the multiplication problem).
For example,
8,000 x 30
1. Multiply basic facts 8 x 3 = 24
2. Add the remaining zeros to the product 24 (add 4 zeros) 240000
3. Product = 240,000
Division Patterns:
Find the quotient (answer to division problem) to the simple division problem, and then add the zeros left in the dividend to the quotient.
For example,
8,100 ÷ 9
1. Divide the basic facts 81 ÷ 9 = 9
2. Add the remaining zeros to the quotient (add 2 zeros) 900
3. Quotient = 900
(4)
Lattice Math Algorithm
(5)
Base ten blocks –Website you can use to get base ten block cut-outs for ones, tens, and hundreds) http://mason.gmu.edu/~mmankus/Handson/b10blocks.htm
This is the 1-block or unit block the smallest of all the blocks. This is the 10-block corresponding to 10 units. It is also referred to as a rod or long. This is the 100-block and corresponds to 100 units. It is also called a flat.
Use base-ten blocks to model a multiplication problem in order to find the product.
You can make a model to find 142 x 3:
1. Model 3 groups of 142 (using (1) hundred block (4) ten- blocks (2) one-blocks)
2. Combine the groups
-(6) one-blocks
-(12) ten-blocks
-(3) hundred-blocks
3. Combine the ten-blocks to make one hundred-block (with two ten-blocks remaining)
4. Count the total number of blocks to find the product (4-hundred-blocks; 2- ten-blocks; and 6-one-blocks) = 426
5. Therefore, 142 x 3 = 426
Extra Help Problems
(1) Mental Math: Use a basic fact and pattern to write the product or factors for the large number. 7 x __ = 42,000 (6,000) (2) Mental Math: Use a basic fact and pattern to write the product or factors for the large number. 8 x 60,000 = _________ (480,000) (3) Mental Math: Use a basic fact and pattern to write the product or factors for the large number. 7 x ______ = 35,000 (5,000) (4) Mental Math: Use a basic fact and pattern to write the product or factors for the large number. 13 x 3,000 = _________ (39,000) (5) Mental Math: Use a basic fact and pattern to write the product or factors for the large number. There are 50 coins in each roll of dimes. How many coins are in 20 rolls? (50 x 20 = 1,000 dimes) (6) Use base-ten blocks to multiply. 154 x 5 (770) (7) Use base-ten blocks to multiply. 116 x 4 (464) (8) Use base-ten blocks to multiply. 206 x 3 (618) (9) Use base-ten blocks to multiply. 145 x 2 (290) (10) Use base-ten blocks to multiply. 309 x 7 (2,163) (11) Use the break apart strategy to multiply the product. 142 x 61 2 x 1 = ___ 40 x 1 = ____ 100 x 1 = _____ 2 x 60 = _____ 40 x 60 = _____ 100 x 60= ______ Add the products: ________ (2 + 40 + 100 + 120 + 2,400 + 6,000 = 8,662) (12) Use the break apart strategy to multiply the product. 321 x 34 1 x 4 = ___ 20 x 4 = ____ 300 x 4 = _____ 1x 30 = _____ 20 x 30 = _____ 300 x 30= ______ Add the products: ________ (4 + 80 + 1,200 + 30 + 600 + 9,000 = (10,914) (13) Use the break apart strategy to multiply the product. 184 x 73 2 x 1 = ___ 40 x 1 = ____ 100 x 1 = _____ 2 x 60 = _____ 40 x 60 = _____ 100 x 60= ______ Add the products: ________________ (2 + 40 + 100 + 120 = 2,400 + 6,000 =8,554) (14) Use the break apart strategy to multiply the product. 732 x 23 2 x 3 = ___ 30 x 3 = ____ 700 x 3 = _____ 2 x 20 = _____ 30 x 20 = _____ 700 x 20= ______ Add the products: _____________ (6 + 90 + 2,100 + 40 + 600 + 1,400 = 4,236) (15) Use the break apart strategy to multiply the product. 417 x 16 7 x 6 = ___ 6 x 10 = ____ 400 x 6 = _____ 7 x 10 = _____ 10 x 10 = _____ 400 x 10= ______ Add the products: _____________ (42 + 60 + 2,400 + 70 + 100 + 4,000 = 6,672) (16) Mental Math: Use basic division facts and patterns to write the quotient for the large number. 40,000 ÷ 5 = (8,000) (17) Mental Math: Use basic division facts and patterns to write the quotient for the large number. 3,200 ÷ 8 (400) (18) Mental Math: Use basic division facts and patterns to write the quotient for the large number. 15,000 ÷ 5 (3,000) (19) Mental Math: Use basic division facts and patterns to write the quotient for the large number. 20,000 ÷ 2 (10,000) (20) Mental Math: Use basic division facts and patterns to write the quotient for the large number. 3,000 ÷ 6 (2,000) (21) Solve for the area: Area 1: ___________ Area 2: ___________ Total Area: _____________ (2 x 4 + 2 x 10 = 28 cm2) (22) Solve for the area: Area 1: ___________ Area 2: ___________ Total Area: _____________ (2 x 7 + 3 x 4 = 26m2) (23) Solve for the area: Area 1: ___________ Area 2: ___________ Total Area: _____________ (2 x 4 + 2 x 10 = 28 cm2) (24) Solve for the area: Area 1: ___________ Area 2: ___________ Total Area: _____________ (4 x 3 + 5 x 3 = 27 in2) (25) Solve for the area: Area 1: ___________ Area 2: ___________ Total Area: __________ (6 x 2 + 2 x 4 =20 yd2)
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# continued fraction
Given a sequence of positive real numbers $(a_{n})_{n\geq 1}$, with $a_{0}$ any real number. Consider the sequence
$\displaystyle c_{1}$ $\displaystyle=a_{0}+\cfrac{1}{a_{1}}$ $\displaystyle c_{2}$ $\displaystyle=a_{0}+\cfrac{1}{a_{1}+\cfrac{1}{a_{2}}}$ $\displaystyle c_{3}$ $\displaystyle=a_{0}+\cfrac{1}{a_{1}+\cfrac{1}{a_{2}+\cfrac{1}{a_{3}}}}$ $\displaystyle c_{4}$ $\displaystyle=\ldots$
The limit $c$ of this sequence, if it exists, is called the value or limit of the infinite continued fraction with convergents $(c_{n})$, and is denoted by
$a_{0}+\cfrac{1}{a_{1}+\cfrac{1}{a_{2}+\cfrac{1}{a_{3}+\ldots}}}$
or by
$a_{0}+\frac{1}{a_{1}+}\frac{1}{a_{2}+}\frac{1}{a_{3}+}\ldots$
In the same way, a finite sequence
$(a_{n})_{1\leq n\leq k}$
defines a finite sequence
$(c_{n})_{1\leq n\leq k}\;.$
We then speak of a finite continued fraction with value $c_{k}$.
An archaic word for a continued fraction is anthyphairetic ratio.
If the denominators $a_{n}$ are all (positive) integers, we speak of a simple continued fraction. We then use the notation $q=\langle a_{0};a_{1},a_{2},a_{3},\ldots\rangle$ or, in the finite case, $q=\langle a_{0};a_{1},a_{2},a_{3},\ldots,a_{n}\rangle\;.$
It is not hard to prove that any irrational number $c$ is the value of a unique infinite simple continued fraction. Moreover, if $c_{n}$ denotes its $n$th convergent, then $c-c_{n}$ is an alternating sequence and $|c-c_{n}|$ is decreasing (as well as convergent to zero). Also, the value of an infinite simple continued fraction is perforce irrational.
Any rational number is the value of two and only two finite continued fractions; in one of them, the last denominator is 1. E.g.
$\frac{43}{30}=\langle 1;2,3,4\rangle=\langle 1;2,3,3,1\rangle\;.$
These two conditions on a real number $c$ are equivalent:
1. $c$ is a root of an irreducible quadratic polynomial with integer coefficients.
2. $c$ is irrational and its simple continued fraction is “eventually periodic”; i.e.
$c=\langle a_{0};a_{1},a_{2},\ldots\rangle$
and, for some integer $m$ and some integer $k>0$, we have $a_{n}=a_{n+k}$ for all $n\geq m$.
For example, consider the quadratic equation for the golden ratio:
$x^{2}=x+1$
or equivalently
$x=1+\frac{1}{x}\;.$
We get
$\displaystyle x$ $\displaystyle=$ $\displaystyle 1+\cfrac{1}{1+\cfrac{1}{x}}$ $\displaystyle=$ $\displaystyle 1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{x}}}$
and so on. If $x>0$, we therefore expect
$x=\langle 1;1,1,1,\ldots\rangle$
which indeed can be proved. As an exercise, you might like to look for a continued fraction expansion of the other solution of $x^{2}=x+1$.
Although $e$ is transcendental, there is a surprising pattern in its simple continued fraction expansion.
$e=\langle 2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,\ldots\rangle$
No pattern is apparent in the expansions some other well-known transcendental constants, such as $\pi$ and Apéry’s constant $\zeta(3)$.
Owing to a kinship with the Euclidean division algorithm, continued fractions arise naturally in number theory. An interesting example is the Pell diophantine equation
$x^{2}-Dy^{2}=1$
where $D$ is a nonsquare integer $>0$. It turns out that if $(x,y)$ is any solution of the Pell equation other than $(\pm 1,0)$, then $|\frac{x}{y}|$ is a convergent to $\sqrt{D}$.
$\frac{22}{7}$ and $\frac{355}{113}$ are well-known rational approximations to $\pi$, and indeed both are convergents to $\pi$:
$\displaystyle 3.14159265\ldots$ $\displaystyle=$ $\displaystyle\pi=\langle 3;7,15,1,292,...\rangle$ $\displaystyle 3.14285714\ldots$ $\displaystyle=$ $\displaystyle\frac{22}{7}=\langle 3;7\rangle$ $\displaystyle 3.14159292\ldots$ $\displaystyle=$ $\displaystyle\frac{355}{113}=\langle 3;7,15,1\rangle=\langle 3;7,16\rangle$
For one more example, the distribution of leap years in the 4800-month cycle of the Gregorian calendar can be interpreted (loosely speaking) in terms of the continued fraction expansion of the number of days in a solar year.
Title continued fraction Canonical name ContinuedFraction Date of creation 2013-03-22 12:47:12 Last modified on 2013-03-22 12:47:12 Owner PrimeFan (13766) Last modified by PrimeFan (13766) Numerical id 29 Author PrimeFan (13766) Entry type Definition Classification msc 11Y65 Classification msc 11J70 Classification msc 11A55 Synonym chain fraction Related topic FareySequence Related topic AdjacentFraction Related topic SternBrocotTree Related topic FareyPair Defines anthyphairetic ratio Defines simple continued fraction
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# Solution a 2 3 1 2 3 1 8 b 1 4 3 4 3 64 c x 1 1 x 1 1
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Solution a) 2 - 3 = 1 2 3 = 1 8 , b) 1 4 - 3 = 4 3 = 64, c) x - 1 = 1 x 1 = 1 x , d) x - 2 = 1 x 2 , e) 10 - 1 = 1 10 1 = 1 10 or 0.1. Now do this exercise Write each of the following using a positive index. a) 1 t - 4 , b) 17 - 3 , c) y - 1 , d) 10 - 2 Use the previous Key Point. Answer Try each part of this exercise Simplify a) a 8 × a 7 a 4 , b) m 9 × m - 2 m - 3 Part (a)(i) Use the first law of indices to simplify the numerator: Answer Part (a)(ii) Then use the second law to simplify the result: Answer Part (b)(i) First simplify the numerator using the first law of indices: Answer Part (b)(ii) Then use the second law to simplify the result: Answer Engineering Mathematics: Open Learning Unit Level 0 1.2: Basic Algebra 8
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More exercises for you to try 1. Write the following numbers using a positive index and also express your answers as decimal fractions: a) 10 - 1 , b) 10 - 3 , c) 10 - 4 2. Simplify as much as possible: a) x 3 x - 2 , b) t 4 t - 3 , c) y - 2 y - 6 . Answer 5. Fractional indices. So far we have used indices that are whole numbers. We now consider fractional powers. Con- sider the expression (16 1 2 ) 2 . Using the third law of indices, ( a m ) n = a mn , we can write (16 1 2 ) 2 = 16 1 2 × 2 = 16 1 = 16 So 16 1 2 is a number which when squared equals 16, that is 4 or - 4. In other words 16 1 2 is a square root of 16. There are always two square roots of a non-zero positive number, and we write 16 1 2 = ± 4 In general Key Point a 1 2 is a square root of a Similarly (8 1 3 ) 3 = 8 1 3 × 3 = 8 1 = 8 so that 8 1 3 is a number which when cubed equals 8. Thus 8 1 3 is the cube root of 8, that is 3 8, namely 2. Each number has only one cube root, and so 8 1 3 = 2 In general Key Point a 1 3 is the cube root of a More generally we have Key Point x 1 n is an n th root of x, that is n x Your calculator will be able to evaluate fractional powers, and roots of numbers. Check that you can obtain the results of the following examples on your calculator, but be aware that your calculator may give only one root when there may be others. 9 Engineering Mathematics: Open Learning Unit Level 0 1.2: Basic Algebra
Example Evaluate a) 144 1 / 2 , b) 125 1 / 3 Solution a) 144 1 / 2 is a square root of 144, that is ± 12 . b) Noting that 5 3 = 125, we see that 125 1 / 3 = 3 125 = 5 Example Evaluate a) 32 1 / 5 , b) 32 2 / 5 , and c) 8 2 / 3 . Solution a) 32 1 5 is the 5th root of 32, that is 5 32. Now 2 5 = 32 and so 5 32 = 2.
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• Winter '10
• John Schaefer
• Math, Open Learning, Open Learning Unit, Learning Unit Level
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# Homework 8 Solutions
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1 CSE 21 - Winter 2014 Homework Homework 8 Solutions 1 Of 330 male and 270 female employees at the Flagstaff Mall, 210 of the men and 180 of the women are on flex-time (flexible working hours). Given that an employee selected at random from this group is on flextime, what is the probability that the employee is a woman? For a randomly selected employee, let W denote the event that the employee is a woman and let F denote the event that the employee is flextime. We need to find P(W F). P[W F] P(W F) = P[F] P(F) = P(F W) = P(W F) = = A new medical test has been designed to detect the presence of the mysterious Brainlesserian disease. Among those who have the disease, the probability that the disease will be detected by the new test is However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is It is estimated that 16 % of the population who take this test have the disease. If the test administered to an individual is positive, what is the probability that the person actually has the disease? 1
2 This, and many other problems involve using Bayes Theorem. However, we shall see how to solve this problem without using Bayes Theorem. Let D denote the event that a particular person has the disease. P(D) = Let D indicate the event that the person doesn t have the disease. Let M denote the event that the medical test returns positive. We need to compute P(D M). We know that P(D) = 0.16, P( D) = And, P(M D) = 0.86, P(M D) = We want: P(D M) = P(D M) P(M) From we get: Similarly, With this, we get, P(M D) = P(D M) P(D) P(D M) = P(M D) P(D) = P( D M) = P(M D) P( D) = P(M) = P( D M) + P(D M) = Hence, combining everything, we get: P(D M) P(D M) = P(M) = =
3 3 Two marbles are drawn randomly one after the other without replacement from a jar that contains 5 red marbles, 3 white marbles, and 9 yellow marbles. Find the probability of the following events. 1. A red marble is drawn first followed by a white marble. 2. A white marble is drawn first followed by a white marble. 3. A yellow marble is not drawn at all The answers are: ( 3 2 ) ( 17 2 ) ( ) ( 17 2 ) 4 Scientific research on popular beverages consisted of 65 studies that were fully sponsored by the food industry, and 35 studies that were conducted with no corporate ties. Of those that were fully sponsored by the food industry, 15 % of the participants found the products unfavorable, 22 % were neutral, and 63 % found the products favorable. Of those that had no industry funding, 35 % found the products unfavorable, 16 % were neutral, and 49 % found the products favorable. 1. What is the probability that a participant selected at random found the products favorable? 2. If a randomly selected participant found the product favorable, what is the probability that the study was sponsored by the food industry? 3. If a randomly selected participant found the product unfavorable, what is the probability that the study had no industry funding? Let S denote the event that the study was sponsored by industry. For the outcome of the study, let us have the following three events: 3
4 Let F denote the event that the study was Favourable Let U denote unfavourable result Let N denote a neutral result Clearly, P(F) + P(U) + P(N) = 1. We are given the following values: P( S) = From second statement, we get P(F S) = 0.63, P(U S) = 0.15, P(N S) = From third statement, we get P(F S) = 0.49, P(U S) = 0.35, P(N S) = We need to find P(F). Here are the solutions: 1. P(F S) = P(F S) P(S). So, P(F S) = P(F S) P(S) = Similarly, P(F S) = P(F S) P( S) = Combining, we get, P(F) = P(F S) + P(F S) = P(S F) = P(F S)/P(F) = ( )/( ). 3. P(U S) = P(U S) P(S) = Similarly, P(U S) = P(U S) P( S) = Combining, we get P(U) = P(U S) + P(U S) = We want to find P( S U) = P( S U) P(U) = If P(E F) = 0.08, P(E F) = 0.2, P(F E) = 0.5, find: 1. P(E) 2. P(F) 3. P(E F) 4. Are the events E and F independent? 4
5 Time spent in resource room Pass % None 25 Between 1 and 90 minutes 52 More than 90 minutes 63 Table 1: Table for Problem 6 The answers are: 1. P(E) = P(F E)/P(F E) = 0.08/0.5 = P(F) = P(F E)/P(E F) = 0.08/0.2 = P(F E) = P(F) + P(E) P(F E) = = P(F) P(E) = = = P(E F). Hence, E and F are not independent. 6 On average 62 % of Finite Mathematics students spend some time in the Mathematics Department s resource room. Half of these students spend more than 90 minutes per week in the resource room. At the end of the semester the students in the class were asked how many minutes per week they spent in the resource room and whether they passed or failed. The passing rates are summarized in table 1. If a randomly chosen student did not pass the course, what is the probability that he or she did not study in the resource room? Let S denote the event that a student passed. Let H, M, L denote the events that the student spent more than 90 mins, between 1 and 90 mins and None respectively. We are given P(H) + P(M) = P(H) = = So, P(M) = 0.31, P(L) = = We are also given the following values: P(S L) = 0.25, P(S M) = 0.52, P(S H) = We are asked to find P(L S). P(S) = P(S L) + P(S M) + P(S H) = P(S L) P(L) + P(S M) P(M) + P(S H) P(H) = =
6 P( S) = 1 P(S) = And, P(L S) = P(L) P(L S) = = Combining, we have, P(L S) = P(L S)/P( S) = You ask a neighbor to water a sickly plant while you are on vacation. Without water the plant will die with probability 0.9. With water it will die with probability 0.4. You are 82 % certain the neighbor will remember to water the plant. 1. When you are on vacation, find the probability that the plant will die. 2. You come back from the vacation and the plant is dead. What is the probability the neighbor forgot to water it? Let W be the event that the plant is watered. Let D be the event that the plant would die. We are given the following values: P(W) = 0.82, P(D W) = 0.4, P(D W) = 0.9. With these, the answers are: 1. P(D) = P(D W) + P(D W) = P(D W) P(W) + P(D W) P( W) P(D) = = P( W D) = P(D W)/P(D) = In a survey of 299 people, the data presented in Table 2 were obtained relating gender to political orientation. A person is randomly selected. What is the probability that the person is: 1. Male? 2. Male and Democrat? 3. Male given that the person is a Democrat? 4. Republican given that the person is Male? 5. Female given that the person is a Independent? 6. Are the events Female and Republican independent? 6
7 Republican (R) Democrat (D) Independent (I) Total Male (M) Female (F) Total Table 2: Table for Problem 8 The answers are: P(Female) = P(Republican) = 299. P(Female and Republican) = P(Female) P(Republican) = = P(Female and Republican). Hence, they are not independent events. 9 In a survey of 347 people, the data presented in Table 3 were obtained relating gender to political orientation. A person is randomly selected. What is the probability that the person is: 1. Male? 2. Male and Democrat? 3. Male given that the person is a Democrat? 4. Republican given that the person is Male? 5. Female given that the person is a Libertarian? 6. Are the events Male and Republican independent? 7
8 Republican (R) Democrat (D) Libertarian (L) Total Male (M) Femal (F) Total The answers are: Table 3: Table for Problem 9 6. P(Male) = P(Republican) = 347. P(Male and Republican) = 347. P(Male) P(Republican) = = P(Male and Republican). Hence, they are not independent events. 10 Factories A, B and C produce computers. Factory A produces 2 times as many computers as factory C. And factory B produces 6 times as many computers as factory C. The probability that a computer produced by factory A is defective is 0.016, the probability that a computer produced by factory B is defective is 0.038, and the probability that a computer produced by factory C is defective is A computer is selected at random and it is found to be defective. What is the probability it came from factory C? Let A, B, C denote the events that any computer is manufactured by factories A, B, and C respectively. Let D denote the event that the computer is defective. P(A) = 2 P(C), P(B) = 6 P(C). We get, P(A) = 2/9, P(B) = 6/9, P(C) = 1/9. 8
9 Color-Blind (C) Not Color - Blind ( C) Total Male (M) Femal (F) Total Table 4: Table for Problem 11 We are also given P(D A) = 0.016, P(D B) = 0.038, P(D C) = We need to compute P(C D). We can compute P(D) = P(A D) + P(B D) + P(C D) = P(D A) P(A) + P(D B) P(B) + P(D C) P(C) = / / /9 = 0.303/9 P(C D) = P(C D)/P(D) = 0.043/ /9 = 0.043/ In a survey of 198 people, the data presented in Table 4 were obtained relating gender to color-blindness. A person is randomly selected. What is the probability that the person is: 1. Male? 2. Male and Color-blind? 3. Male given that the person is Color-blind? 4. Color-blind given that the person is Male? 5. Female given that the person is not Color-blind? The answers are: 9
10 If P(F) = 0.3 and P(E F) = 0.9, then find P(E F). P(E F) = P(E F) P(F) = = A fair coin is tossed 12 times. What is the probability that: 1. Exactly 10 heads appear? 2. At least two heads appear? 3. At most 9 heads appear? Let X be the random variable denoting the number of heads. The answers are: 1. P(X = 10) = (12 10 ) P(X 2) = 1 P(X 1) = 1 (P(X = 1) + P(X = 0)) = 1 (12 1 )+(12 0 ) P(X 9) = 1 [P(X = 10) + P(X = 11) + P(X = 12)] = 1 ( )+(12 11 )+( ) 14 If the letters in the word POKER are rearranged, what is the probability that the word will begin with the letter P and end with the letter E? 10
11 Total number of ways of arranging the 5 letters, without restriction is 5!. If we fix the first letter to be P and the last letter to be E, total number of words with such combination is 3!. So, overall probability is 3! 5!. 15 Two cards are drawn from a regular deck of 52 cards, without replacement. What is the probability that the first card is an ace of clubs and the second is black? P(First card is ace of clubs) = 52 1 P(Second card is black given first card is ace of clubs) = Combining the two, the overall probability is Factories A and B produce computers. Factory A produces 3 times as many computers as factory B. The probability that an item produced by factory A is defective is and the probability that an item produced by factory B is defective is A computer is selected at random and it is found to be defective. What is the probability it came from factory A? Let A, B denote the events that a computer is produced by factories A, B respectively. Let D denote the event that the computer is defective. We know that P(A) = 3P(B) and P(A) + P(B) = 1. So, using these, we get, P(A) = 0.75, P(B) = We are also given P(D A) = 0.018, P(D B) = Using these, we get P(D) = P(D A) + P(D B) = P(D A) P(A) + P(D B) P(B) = Now, P(A D) = P(A D)/P(D) = A box contains 20 yellow, 22 green and 29 red jelly beans. If 8 jelly beans are selected at random, what is the probability that: 1. 2 are yellow? 2. 2 are yellow and 5 are green? 3. At least one is yellow? 11
12 Answers are: ( 20 2 ) (51 6 ) ( 71 8 ) ( 20 2 ) (22 5 ) (29 1 ) ( 71 8 ) 3. 1 (51 8 ) ( 71 8 ) 18 Events A 1, A 2 and A 3 form a partiton of the sample space S with probabilities P(A 1 ) = 0.3, P(A 2 ) = 0.2, P(A 3 ) = 0.5. If E is an event in S with P(E A 1 ) = 0.3, P(E A 2 ) = 0.5, P(E A 3 ) = 0.8, compute 1. P(E) 2. P(A 1 E) 3. P(A 2 E) 4. P(A 3 E) P(E A 1 ) = P(E A 1 ) P(A 1 ) = = 0.09 P(E A 2 ) = P(E A 2 ) P(A 2 ) = = 0.1 P(E A 3 ) = P(E A 3 ) P(A 3 ) = = 0.4 P(E) = P(E A 1 ) + P(E A 2 ) + P(E A 3 ) = 0.59 P(A 1 E) = P(A 1 E) = 0.09/0.59 P(E) P(A 2 E) = P(A 2 E) = 0.1/0.59 P(E) P(A 3 E) = P(A 3 E) = 0.4/0.59 P(E) 12
13 19 A card is drawn from a regular deck of 52 cards and is then put back in the deck. A second card is drawn. What is the probability that: 1. The first card is red. 2. The second card is hearts given that the first is red. 3. The first card is red and the second is hearts. Answers are: = The outcome of second card doesn t depend on first card because outcomes of second and first draws are independent events. So, probability of the second card is hearts given that the first is red is same as probability of the second card is hearts. This is = P[First card is red] = = 0.5. And, P[Second card is hearts] = = Overall probability is = If P(E F) = 0.225, P(E F) = 0.45, P(F E) = 0.5, find: 1. P(E) 2. P(F) 3. P(E F) The answers are: 1. P(E) = P(F E)/P(F E) = 0.225/0.5 = P(F) = P(F E)/P(F E) = 0.225/0.45 = P(F E) = P(F) + P(E) P(F E) = =
### . Notice that this means P( A B )
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# Area of Irregular Shapes
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Last updated date: 13th Jul 2024
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## Irregular shapes
Irregular Shapes are Polygons with five or more sides of varying lengths. These Shapes or figures can be decomposed further into known Shapes like triangles, Squares, and Quadrilaterals to evaluate the Area.
Some examples of Irregular Shapes are as follows:
### How to calculate the Area of Irregular Shape?
Finding the Area of Irregular Shapes
Different methods for estimating the Area of Irregular Shape are:
• Evaluating Area using unit Squares.
• Dividing the Irregular Shape into two or more Regular Shapes.
• Dividing the Irregular Shape with curves into two or more Regular Shapes.
### How to Find the Area of an Uneven Shape?
Evaluating Area Using Unit Squares
We can use this method for Shapes with curves apart from perfect circle or semicircles and Irregular Quadrilaterals. In this method, we first divide the Shape into unit Squares. The total number of unit Squares falling within the Shape is used to determine the total Area.
For Example: Calculate the Area by counting the unit Squares in the given below figure
Ans: If we denote each unit Square in centimetre, the Area will be 6 cm2.
### To Calculate the Area of an Irregular Shape with Curved Edges
Dividing the Irregular Shape Into Two or More Regular Shapes
Use this method to calculate the Area of Irregular Shapes, which are a combination of triangles and Polygons. Using predefined formulas calculate the Area of such Shapes and add them together to obtain the total Area.
For example, in the given below Irregular Shape, we will divide multiple edges into a triangle and three Polygons.
The total Area of the figure can be calculated by adding individual Area:
Total Area = Area (ABIM) + Area (BCGH) + Area (CDEF) + Area (JKL)
⇒ Total Area = (AB × BI) + (BC × CG) + (CD × DE) + (12 × LJ × KO)
⇒Total Area = ( 10 × 5) + (3 × 3) + (2 × 2) + (1⁄2× 4 × 4)
⇒ Total Area = 50 + 9 + 4 + 8
⇒ Total Area = 71 cm2
### Calculating the Area of an Irregular Shape
To calculate the Area of Irregular Shapes, divide the Shape with curves into two or more Regular Shapes.
In this method, divide an Irregular Shape into multiple Squares, triangles, or other Quadrilaterals. Depending on the Shape or curves, a part of the figure can be a circle, semicircle or quadrant as well.
Find the are of a given Irregular Shape with 8 sides, including one curve.
Solution: We will determine the unknown quantities by the given dimensions for the sides. First, we need to divide the figure into two rectangles and a semicircle.
The Area of the Shape ABCDEF is:
Total Area (ABCDEF) = Area (ABCG) + Area (GDEF) + Area (aob)
Total Area = (AB × AG) + (GD × DE) + (1⁄2 × π × ob2)
Total Area = (3 × 4) + (10 × 4) + (1⁄2 × 3.14 × 12)
Total Area = 12 + 40 + 1.57
Hence total Area = 53.57 cm2
### How to Find the Area of Irregular Shapes Using Graph Paper?
What is the Area of Irregular surface?
Find the Area of a given leaf.
Solution: To find the Area of the Irregular surface in the above case leaf we have to put the leaf on graph paper and draw its boundary.
The Shape of a leaf is Irregular. So we will assume that more than half of the Square covered by leaf will be counted as 1 and less than that will be counted as 0.
Now count the number of fully covered Shapes. There are 64 Squares fully covered.
Also, count the partially more than half covered Squares and each will count qs 1 Square. There are 17 Square more than half Square.
Also, count the partially less than half covered Square and each will count as 0. There are 16 Squares less than half Square.
Now add all the Squares to find the Area of leaf = 64 + 17 x 1 + 16 x 0 = 64 + 17 = 81 sq. units.
Hence the Area of the leaf will be 81 sq. units.
### Area of Irregular Shapes Formula
To find the Area of Irregular Shapes, first, we need to divide the Irregular Shape into Regular Shapes that you can recognize such as triangles, rectangles, circles, Squares and so forth.
Then, find the Area of these individual Shapes and add them to get an Area of Irregular Shapes.
### Solved Example
Q. Find the Area of the given Shape?
Solution: The figure above has three Regular Shapes. Start dividing from the top, it has a triangle, a rectangle, and a trapezoid.
We will find the Area for each of those three Shapes and add the results to get the final Area of a figure.
Triangle
Area of triangle = (base × height)/2
= (3 × 4)/2
= 12/2
= 6
Rectangle
Area of rectangle = length × width
= 3 × 10
= 30
Trapezoid
Area of trapezoid = ((b1 + b2) × h)/2
= ((3 + 5) × 2)/2
= (8) × 2/2
= 16/2
= 8
Hence Area of the given Shape = 6 + 30 + 8 = 44.
### Irregular Shapes
The space occupied by Shape is known as the Area of Irregular Shape; it is measured in Square units. The size and length of Irregular Shapes can be of any measurement. There are various Irregular Shapes that we can see around us like kites, leaves, diamonds, etc. Therefore Irregular Shapes are any Shapes whose angles and lengths are not equal. Students can study Areas of Irregular Shapes with Vedantu's subject experts who guide learners to understand each and every topic so that it will be easier to learn and score good marks.
### More into Irregular Shapes
The amount of region covered by that Shape is known as the Area of Irregular Shapes. The sides and angles of Irregular Shapes are different. to find the Area of Irregular Shape we have to decompose it or divide it into multiple known Shapes and then the Area of those Shapes will be added to get the total Area of Irregular Shape. Irregular Shapes can be seen everywhere around us in our daily life for example:
• The staircase of a building is composed of Polygons like rectangles and Squares, the surface Area of the staircase is an Irregular Shape.
• The school’s playground that has a running track is an Irregular Shape which is the combination of Regular Shapes.
• The leaf of a tree or plant is of Irregular Shape. Etc.
Unit Area of an Irregular Shape can be expressed in m2, cm2,in2 or ft2.
### Steps to Find the Area of Irregular Shapes
Step 1 - Area of Irregular Shape is to be found by decomposing the Irregular Shapes into familiar Shapes.
Step 2 - Students should know the way to decompose Irregular Shapes.
Step 3 - dividing the Irregular Shape into different familiars Shapes and finding the Area of familiar Shapes
Step 4 - Now add all the Areas of the familiar Shapes to know the Area of the Irregular Shape.
The #1online learning portal Vedantu provide students complete guide to studying the topics like Area of Irregular Shapes - Calculation, Examples, etc. students can easily download the Free PDF and prepare for their exams they can rely on the study material as it is 100% accurate and is according to the latest guideline and syllabus.
## FAQs on Area of Irregular Shapes
1. How Do You Find the Area of Irregular Shapes?
To find the Area of Irregular Shapes, first we need to divide the Irregular Shape into Regular Shapes that you can recognize such as triangles, rectangles, circles, Squares etc. Then, find the Area of these individual Shapes and add them up to get the Area of Irregular Shapes. Students can understand the concept well with the revision notes provided by Vedantu that have described the concepts in detail to help the student to gain complete knowledge, so that they can solve any type of question-based on the given topic.
2. What are Irregular Shapes?
An Irregular Shape doesn't have equal sides or equal angles. Irregular Shapes can be seen everywhere around us like stars, diamonds, leaves, stairs, etc. Students can study the concept of Irregular Shapes by understanding the formation of that Shape. Students need to decompose and divide the Irregular Shapes into some Regular familiar Shapes so that their Area can be calculated and by adding them we can calculate the Area of Irregular Shapes. Calculation of the Area of Irregular Shapes is not easy, it is a bit complicated and you need to practice and learn all the formulas to get answers in one go.
3. What are the Applications of the Irregular Shapes Calculation Formula?
The formula used to estimate Area for Irregular figures is an essential method for drawing maps, building architecture, and marking agricultural fields. We can also apply the concept in the cutting of fabrics as per the given design. To study the given Shapes, students need to get complete knowledge about the different figures and the formulas to calculate the Area of all the Shapes. With practice, students can excel in solving the questions based on these topics and they will also gain knowledge about how to decompose an Irregular Shape and apply the formulas to it.
4. How is the Area of Irregular Shapes different from Regular Shapes?
One of the most important topics of the syllabus is Regular and Irregular Shapes and students should know the Difference between Regular Shapes and Irregular Shapes, here are some of the differences between the two:
• The Regular Shape has equal sides and angles, but in an Irregular Shape, the sides and angles are different from each other.
• The Area of a Regular Shape can be determined by easily applying a suitable formula to it, but in order to find the Area of Irregular Shape, we need to decompose it into multiple Regular Shapes and then add the Areas of all the Regular Shapes.
5. Where can I get the best study material for getting complete knowledge of Irregular Shapes, Area, calculation, and examples?
Vedantu is the best online learning app that helps students to get free study material for all the topics they require to study. Students can download a free PDF for Irregular Shapes that covers everything about Irregular Shapes like Area of Irregular Shapes, calculation, and examples of Irregular Shapes. Students can prepare this topic well by solving the important questions provided by Vedantu's expert which provides practical knowledge about how to apply formulas to determine the Area of various Irregular Shapes.
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Lecture 21 Quiz
[NOTE: This quiz is due at the beginning of lecture Monday, not Friday. At first glance it may look like a lot of work, but in total you just have to (i) answer a simple conceptual question that we have discussed a few times in lecture, (ii) take the derivative of an ordinary function with respect to one variable, and (iii) write a single line of code. Please let us know on Piazza if you have questions!]
In the past few lectures we've covered two fundamental numerical tools used for computer graphics: numerical optimization, and numerical linear algebra. In this quiz, you're going to put these ideas together to design a real algorithm for solving linear systems. Hopefully after doing this exercise, you will never again think of a linear solver as a "black box," but rather something that you yourself can design, build, and tinker with. (Good graphics engineering often involves exactly this kind of tinkering, and you might be surprised how often people in "the real world" design or modify numerical solvers to address the needs of particular problems or hardware platforms.)
The basic idea of our algorithm is to solve a linear system $Ax=b$ by applying gradient descent to the objective
$$f(x) = \tfrac{1}{2}x^T A x - b^T x.$$
Recall that the gradient of this function is again just $Ax-b$! Therefore, if we end up at a point $x^ *$ where the gradient is zero, then we will have solved our original linear system. In other words,
$$0 = f(x^ * ) = Ax^ * - b$$
implies that $Ax^ * = b$. Just as a reminder, our basic gradient descent algorithm was:
1. Pick an initial guess $x$.
2. Compute the gradient descent direction $d \leftarrow -\nabla f(x)$ at the current point $x$.
3. Take a step in the descent direction: $x \leftarrow x + \tau d$, where $\tau > 0$ is our step size.
4. Repeat until the gradient is "sufficiently small" (for instance, until its largest element is some fixed fraction of the largest element of $b$).
#### Question 1
Our algorithm tries to find a solution by "skiing downhill" until we hit a point where the gradient is zero (i.e., until we can't go downhill anymore). What conditions must $A$ satisfy in order for our algorithm to successfully find a solution to the linear system?
#### Question 2
In our optimization lecture, we briefly discussed the idea of line search. In a nutshell: once we know which direction to go, we still need to know how far to travel. In other words, we need to determine the step size $\tau$.
In general, determining the ideal step size can be difficult. Fortunately, for our little problem we can compute the optimal step size exactly. In particular, it is the distance we need to travel along the current descent direction $d$ such that our objective $f$ becomes as small as possible along this ray. More explicitly, we can write the values of $f$ along the direction $d$ as
$$s(\tau) := f(x + \tau d).$$
Your job is to minimize $s(\tau)$ with respect to $\tau$; this minimizing value of $\tau$ gives the optimal step size.
Hint #1: set the derivative of $s$ with respect to $\tau$ equal to zero, and solve!
Hint #2: There are two different ways to compute the derivative of $s$ with respect to $\tau$: one is to expand $s$ in terms of the original definition of $f$, then take the derivative with respect to $\tau$. The other is to apply the chain rule, which yields a somewhat shorter derivation. You decide!
Hint #3: The final expression for the optimal step size $\tau$ should involve only 5 letters.
#### Question 3
We now have all the ingredients we need to run our gradient descent algorithm (outlined above). Consider the 2x2 linear system
$$\tfrac{1}{5} \left[\begin{array}{cc} 2 & 1 ; \ 1 & 3 \end{array} \right] \underbrace{\left[\begin{array}{c} u \ v \end{array}\right]}_{=:x} = \left[\begin{array}{c} 4 \ 5 \end{array}\right],$$
where the semicolon denotes a new row of the 2x2 matrix. I.e., the entcries of $A$ are $A_{11} = 2$, $A_{12} = 1$, $A_{21} = 1$ and $A_{22} = 3$.
First, solve this system by hand so that you know what the true solution should be. (Also check your answer by plugging it back into the original equations!)
Next, using the skeleton code below (which is almost entirely written for you!), implement the gradient descent algorithm using the step size you found in Question 2. Run your algorithm starting with an initial guess of (u=v=0). Do you arrive at the correct answer? For your quiz submission, print out the value of $x$ for the first 10 iterations of the algorithm.
Here is the code skeleton in C, though you can implement this in any language you like. It should not require fancy data structures, or really much code at all! (We simply wanted you to be able to run this algorithm on your own machine; this is not meant as a "real" coding exercise.)
#include <stdio.h>
int main( int argc, char** argv )
{
double A[2][2] = { { 2./5., 1./5. }, { 1./5., 3./5. } };
double x[2] = { 0., 0. };
double b[2] = { 4., 5. };
double d[2];
double tau;
int i;
for( i = 0; i < 10; i++ )
{
/* d = b - Ax */
d[0] = b[0] - ( A[0][0]*x[0] + A[0][1]*x[1] );
d[1] = b[1] - ( A[1][0]*x[0] + A[1][1]*x[1] );
/* compute the step size tau */
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# How do you find the Pearsonian coefficient of skewness?
## How do you find the Pearsonian coefficient of skewness?
Pearson’s coefficient of skewness (second method) is calculated by multiplying the difference between the mean and median, multiplied by three. The result is divided by the standard deviation.
What is the coefficient of skewness formula?
Pearson’s coefficient of skewness (second method) is calculated by multiplying the difference between the mean and median, multiplied by three. The result is divided by the standard deviation. You can use the Excel functions AVERAGE, MEDIAN and STDEV. P to get a value for this measure.
### What is the population Pearson’s coefficient of skewness of the data calculator?
Example 2 – Using Karl Pearson’s Coefficient of Skewness Method
Weight (in pounds) 3-5 11-13
No.of babies 10 14
How is Bowley’s coefficient of skewness calculated?
7. Hint: To find the Bowley’s coefficient of skewness, we first need to calculate three quartiles at N4, N2 and 3N4 respectively. Then, we can calculate the Bowley’s coefficient of skewness using the formula SKB=Q3+Q1−2Q2Q3−Q1. Here, we can see that N = 100 and class interval, h = 100.
## How do you find the moment coefficient of skewness?
Step 1: Subtract the median from the mean: 70.5 – 80 = -9.5. Step 2: Divide by the standard deviation: -28.5 / 19.33 = -1.47. Caution: Pearson’s first coefficient of skewness uses the mode. Therefore, if the mode is made up of too few pieces of data it won’t be a stable measure of central tendency.
How do you calculate skewness example?
Calculate sample skewness by multiplying 5.89 by the number of data points, divided by the number of data points minus 1, and divided again by the number of data points minus 2. Sample skewness for this example would be 0.720.
### How do you interpret the skewness coefficient?
The rule of thumb seems to be:
1. If the skewness is between -0.5 and 0.5, the data are fairly symmetrical.
2. If the skewness is between -1 and – 0.5 or between 0.5 and 1, the data are moderately skewed.
3. If the skewness is less than -1 or greater than 1, the data are highly skewed.
What is the limit of Bowley’s coefficient of skewness are?
Bowleys coefficient of skewness lies between a 0 and 1 b 1 and 1 c 1 and 0 d 2 | Course Hero.
## What is the range of Bowley’s coefficient of skewness?
Bowley’s coefficient of skewness lies between -1 and +1.
How do you find the moment coefficient?
Lift, drag, and moment are expressed as dimensionless “coefficients”. They all depend upon the dynamic pressure, q = ½ ρ v2 where ρ is the density of the air and v is the free stream velocity. Here L = lift force, D = drag force, M = pitching moment, A = wing area, and c = wing chord.
### What is the moment coefficient?
The moment coefficient pertains to the moment specifically due to the aerodynamics force (lift force on the wing mostly). The moment of a force can be calculated about any arbitrary point on the chord (or even outside of it).
How do you calculate a coefficient of skewness?
How to Calculate Skewness Calculate the mean and standard deviation Subtract the mean from each raw score Raise each of these deviations from the mean to the third power and sum Calculate skewness, which is the sum of the deviations from the mean, raise to the third power, divided by number of cases minus 1, times the standard deviation raised
## How do you calculate the Pearson coefficient?
You can calculate the correlation coefficient by dividing the sample corrected sum, or S, of squares for (x times y) by the square root of the sample corrected sum of x2 times y2. In equation form, this means: Sxy/ [√(Sxx * Syy)].
What does the coefficient of skewness tell you?
The coefficient of skewness measures the skewness of a distribution . It is based on the notion of the moment of the distribution. This coefficient is one of the measures of skewness.
### How to calculate skewness?
Calculate the mean and standard deviation
• Subtract the mean from each raw score
• Raise each of these deviations from the mean to the third power and sum
• Calculate skewness, which is the sum of the deviations from the mean, raise to the third power, divided by number of…
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The Pythagorean to organize was named after well known Greek mathematician Pythagoras.
It is vital formula that claims the following: a2 + b2 = c2
The 2nd figure listed below helps united state to see why the formula works.
Did you make the following crucial observation? The figure listed below may help!
Notice the the red square has actually 2 triangle in it and the blue square has additionally 2 triangles in it.
The black square has actually 4 that the exact same triangle in it.
You are watching: A square plus b square equals c square
Therefore, area that red square + area of blue square = area of black square
Let a = the length of a next of the red square
Let b = the length of a next of the blue square
Let c = the length of a side of the black color square
Therefore, a2 + b2 = c2
Generally speaking, in any right triangle, allow c it is in the size of the longest next (called hypotenuse) and also let a and also b it is in the length of the other two sides (called legs).
The theorem says that the length of the hypotenuse squared is equal to the size of side a squared add to the length of side b squared.Written together an equation, c2 = a2 + b2Thus, offered two sides, the 3rd side can be uncovered using the formula.We will highlight with examples, but before proceeding, you must know exactly how to discover the square source of a number and also how to fix equations making use of subtractionExercise #1
Let a = 3 and also b = 4. Find c, or the longest sidec2 = a2 + b2c2 = 32 + 42c2= 9 + 16c2 = 25c = √25The sign (√) method square rootc = 5 practice #2 Let c = 10 and a = 8. Uncover b, or the various other leg.c2 = a2 + b2102 = 82 + b2100 = 64 + b2100 - 64 = 64 - 64 + b2 (minus 64 native both sides to isolate b2 )36 = 0 + b236 = b2b = √36 = 6 practice #3 let c = 13 and also b = 5. Discover ac2 = a 2+ b2132 = a2 + 52169 = a2 + 25169 - 25 = a2 + 25-25144 = a2 + 0144 = a2a = √144 = 12
Buy a an extensive geometric recipe ebook. Every geometric recipe are explained with well selected word problems so you can master geometry.
See more: Prob A Pair Of Dice Is Rolled. Find The Probability Of Rolling Two Dice
## Check out few of our top basic mathematics lessons.
Formula for percentage
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# PARCC Grade 3 Math Practice Test Questions
0%
#### 1. Steve had to pay a library fee. The amount he paid is shown below. How much did he pay?
Correct! Wrong!
1. C: There are 2 dollar bills, which represent 2 dollars. There are also 2 quarters, 2 nickels, 1 dime, and 4 pennies. Two quarters are worth \$0.50 since each is worth \$0.25 (2 x0.25=0.50) , 2 nickels are worth \$0.10 since each is worth \$0.05 (2 x0.05=0.10) , 1 dime is worth \$0.10, and 4 pennies are worth \$0.04 since each is worth \$0.01. The sum of the coins can be found by writing: \$0.50+\$0.10+\$0.10+\$0.04, which equals \$0.74. The sum of the two dollar bills and the coins can be written as: \$2.00+\$0.74. Thus, he paid \$2.74.
#### 2. What number sentence is illustrated by the diagram below?
Correct! Wrong!
2. C: The diagram shows 32 counters divided into 4 groups, with 8 counters in each group. Therefore, the total number of counters, 32, is divided by 4, giving a quotient of 8, which is written as: 32÷4=8.
#### 3. Which shape has a number of vertices that is equal to two times the number of vertices found on a triangle?
Correct! Wrong!
3. C: A vertex is a point where two or more edges meet. A triangle has 3 vertices. Two times that would be 6 vertices. The figure shown for Choice C is a triangular prism, which is the only figure that has 6 vertices. A triangular pyramid (Choice A) has 4 vertices, a cube (Choice B) has 8 vertices, and a square pyramid (Choice D) has 5 vertices.
#### 4. What is the perimeter of the figure shown below?
Correct! Wrong!
4. B: The perimeter is the distance around the figure. So, if you add up all of the numbers you get 22 in.
#### 5. What number does Point P represent?
Correct! Wrong!
5. B: Each increment represents one-half. This can be determined by counting that there are 3 marks, or 4 spaces, that lie between the difference of two wholes, as in between 10 and 12. Thus, one increment past 10, where Point P is located, represents 10 1/2.
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# Angle bisectors triangle meet one point
### Bisectors of Triangles | Wyzant Resources
We have all heard that the three angle bisectors ofthe internal angles of a triangle meet at a point called the incenter. How do we know that these three lines. Consider the bisectors of two of the angles. Note that for any point on each bisector, the perpendicular distance to the two sides of the angle it bisects is equal. In a triangle, there are three such lines. Three angle bisectors of a triangle meet at a point called the incenter of the triangle. There are several ways to see why.
Together, they form the perpendicular bisector of segment AB. The perpendicular bisectors of a triangle have a very special property. Let's investigate it right now. Circumcenter Theorem The perpendicular bisectors of the sides of a triangle intersect at a point called the circumcenter of the triangle, which is equidistant from the vertices of the triangle.
### Incenter and incircles of a triangle (video) | Khan Academy
Point G is the circumcenter of? Angle Bisectors Now, we will study a geometric concept that will help us prove congruence between two angles. Any segment, ray, or line that divides an angle into two congruent angles is called an angle bisector. We will use the following angle bisector theorems to derive important information from relatively simple geometric figures. Angle Bisector Theorem If a point lies on the bisector of an angle, then it is equidistant from the sides of the angle.
If a point in the interior of an angle is equidistant from the sides of the angle, then it lies on the bisector of the angle. The points along ray AD are equidistant from either side of the angle. Together, they form a line that is the angle bisector. Similar to the perpendicular bisectors of a triangle, there is a common point at which the angle bisectors of a triangle meet. Let's look at the corresponding theorem. Incenter Theorem The angle bisectors of a triangle intersect at a point called the incenter of the triangle, which is equidistant from the sides of the triangle.
Point G is the incenter of? Summary While similar in many respects, it will be important to distinguish between perpendicular bisectors and angle bisectors.
### All about angle bisectors
We use perpendicular bisectors to create a right angle at the midpoint of a segment. Any point on the perpendicular bisector is equidistant from the endpoints of the given segment.
The point at which the perpendicular bisectors of a triangle meet, or the circumcenter, is equidistant from the vertices of the triangle. On the other hand, angle bisectors simply split one angle into two congruent angles. Points on angle bisectors are equidistant from the sides of the given angle. We also note that the points at which angle bisectors meet, or the incenter of a triangle, is equidistant from the sides of the triangle. Let's work on some exercises that will allow us to put what we've learned about perpendicular bisectors and angle bisectors to practice.
Exercise 1 BC is the perpendicular bisector of AD. Find the value of x. The most important fact to notice is that BC is the perpendicular bisector of AD because, although it is just one statement, we can derive much information about the figure from it.
The fact that it is a perpendicular bisector implies that segment DB is equal to segment AB since it passes through the midpoint of segment AD. N is the circumcenter of? This means that AP is the angle bisector of the vertex A and all three angle bisectors are concurrent!
P is called the incenter of the triangle ABC. This point is the center of the incircle of which G, F, and E are the points where the incircle is tangent to the triangle.
Click here to play with a dynamic GSP file of the illustration of this proof. A Deeper Look at the Medians We have also heard that the intersection of the three medians of a triangle is called the centroid. How do we know that these three medians intersect at the same exact point?
## Bisectors of Triangles
Here we will prove that the three medians of a triangle are concurrent and that the point of concurrence, called the centroid, is two-thirds the distance from each vertex to the opposite side. For this proof we will place and arbitrary triangle into the coordinate system and use our algebra skills to prove each part of the proof.
KutaSoftware: Geometry- Angle Bisector Of A Triangle Part 2
Let AX and CY be medians of our triangle. Also let their intersection be T. Let's look at some algebra to find the equations of the line passing through A and X and the line passing through Y and C so that we can calculate their intersection.
• Angle bisector theorem
• The Angle Bisectors
• Incenter and incircles of a triangle
We must first find the coordinates of X and Y. These points will help us calculate our lines. Now since T is the intersection of these two lines, Now that we have the coordinates of T, we can calculate the equation of the line passing through B and T and the line passing through A and C in order to help us find the coordinates of point Z.
If we can prove that Z is the midpoint of AC, then we will be able to conclude that all three medians pass through one point and that point is T.
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# Percentages Objectives: D GradeIncrease or decrease a quantity by a given percentage C GradeWork out a percentage increase or decrease Prior knowledge:
## Presentation on theme: "Percentages Objectives: D GradeIncrease or decrease a quantity by a given percentage C GradeWork out a percentage increase or decrease Prior knowledge:"— Presentation transcript:
Percentages Objectives: D GradeIncrease or decrease a quantity by a given percentage C GradeWork out a percentage increase or decrease Prior knowledge: Understand : Finding percentages of quantities – calculator and non calculator Use of percentage multipliers to find the percentage of a quantity Percent to decimal conversion (i.e. percent ÷ 100)
Percentages Percentage increase Find the percentage of the quantity and add it to the original Example 1: Increase 200 by 12% 12 × 200 100 2 = 24 cancel common factors 200 + 24 = 224 Example 2: Increase 270 by 40% cancel common factors 270 + 108 = 378 40 × 270 100 4 10 ≡ = 108
Percentages Percentage decrease Find the percentage of the quantity and subtract it from the original Example 1: Decrease 200 by 12% 12 × 200 100 2 = 24 cancel common factors 200 - 24 = 176 Example 2: Decrease 270 by 40% cancel common factors 270 - 108 = 162 40 × 270 100 4 10 ≡ = 108
Percentages Now do these: 1.Increase 250 by 10% 2. Decrease 500 by 6% 3. Decrease 24 by 25% 4. Increase 80 by 30% 5. Increase £4.20 by 15% 275 470 18 104 £4.83
Percentages Percentage increase When calculating a percentage increase of a quantity we are starting with 100% and then we add the percentage asked for. e.g.Increase a quantity by 23% 100 % + 23 % = 123 % In order to do this calculation we write the percent as a fraction in hundredths. So this becomes: 100 + 23 = 123 100 100 100 Now 123 = 1.23 100 This is called the percentage multiplier for a percentage increase
Percentages Example : Increase 345 by 56% the percentage multiplier is 1.56 100 % + 56 % = 156 % 156 = 1.56 100 345 × 1.56 = 538.2
Percentages Now do these: 1.Increase 125 by 16% 2. Increase 340 by 9% 3. Increase 575 by 22% 4. Increase 84 by 1.3% 5. Increase £6.30 by 17.5% Percentage multiplier is 1.16 125 × 1.16 = 145 Percentage multiplier is 1.09 340 × 1.09 = 370.6 Percentage multiplier is 1.22 575 × 1.22 = 701.5 Percentage multiplier is 1.013 80 × 1.013 = 81.04 Percentage multiplier is 1.175 6.30 × 1.175 = £7.40
Percentages Percentage decrease When calculating a percentage decrease of a quantity we are starting with 100% and then we subtract the percentage asked for. e.g.decrease a quantity by 23% 100 % - 23 % = 77 % In order to do this calculation we write the percent as a fraction in hundredths. So this becomes: 100 - 23 = 77 100 100 100 Now 77 = 0.77 100 This is called the percentage multiplier for a percentage decrease
Percentages Example : Decrease 345 by 56% the percentage multiplier is 0.44 100 % - 56 % = 44 % 44 = 0.44 100 345 × 0.44 = 151.8
Percentages Now do these: 1.Decrease 137 by 13% 2. Decrease 78 by 6% 3. Decrease 623 by 37% 4. Decrease 467 by 1.3% 5. Decrease £7.45 by 17.5% Percentage multiplier is 0.87 137 × 0.87 = 119.19 Percentage multiplier is 0.94 78 × 0.94 = 73.32 Percentage multiplier is 0.63 623 × 0.63 = 392.49 Percentage multiplier is 0.987 467 × 0.987 = 460.929 Percentage multiplier is 0.825 7.45 × 0.825 = £6.15
Percentages To summarise percentage calculations Original quantity × percentage multiplier = new quantity By understanding how to rearrange equations you can find any of the three if you know the other two.
Percentages Worksheet 1 Mixed Non-Calculator 1.Todd is paid £350 per week. He gets a 4% pay rise. What is his new weekly pay? 2.A package holiday is priced at £660. Julie gets a 10% discount for booking before the end of January. How much does she pay? 3.Chris has 50 books on his shelves. Jenny has 12% more books. How many books has Jenny got? 4.A train ticket is priced at £48. In the new year the cost increases by 2½ %. What is the new cost of the train ticket? 5.A music centre costs £280. VAT is added at 17 ½ %. What is the total cost £364 £594 £56 £49.20 £329
Percentages Worksheet 2 Mixed Calculator 1.Sarah’s salary is £13 575 per year. Her annual bonus is 1¾% of her salary. How much does she earn altogether? 2.Jack sells computers. He is paid commission of 8¼% on his sales. Last year he sold computers worth £85 496. How much was his commission? 3.Nathan buys a new Fiat for £12 499. Over 2 years it depreciates by 45%. What is the value after 2 years 4.The population of Newtown was 74 970 in 2008. By 2009 the population had decreased by 27%. What is the population in 2009. 5.Decrease £9.55 by 12% £13 812.56 £7 053.42 £6 874.45 54 728 £7.40
Percentages Worksheet 1 Mixed Non-Calculator 1.Todd is paid £350 per week. He gets a 4% pay rise. What is his new weekly pay? 2.A package holiday is priced at £660. Julie gets a 10% discount for booking before the end of January. How much does she pay? 3.Chris has 50 books on his shelves. Jenny has 12% more books. How many books has Jenny got? 4.A train ticket is priced at £48. In the new year the cost increases by 2½ %. What is the new cost of the train ticket? 5.A music centre costs £280. VAT is added at 17 ½ %. What is the total cost
Percentages Worksheet 2 Mixed Calculator 1.Sarah’s salary is £13 575 per year. Her annual bonus is 1¾% of her salary. How much does she earn altogether? 2.Jack sells computers. He is paid commission of 8¼% on his sales. Last year he sold computers worth £85 496. How much was his commission? 3.Nathan buys a new Fiat for £12 499. Over 2 years it depreciates by 45%. What is the value after 2 years 4.The population of Newtown was 74 970 in 2008. By 2009 the population had decreased by 27%. What is the population in 2009. 5.Decrease £9.55 by 12%
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## Fractions
### Three Meanings
One reason for the difficulty in teaching fractions is that, IMO, they have three meanings.
First, they simply mean divide. So, "84 divided by 4" can be represented as 84/4.
The second meaning is as a number. There are numbers between 1 and 2, and we can represent these numbers are fractions. For example, 1 2/3.
Of course, there is overlap between these two definitions. You can even try to argue that they are the same. But I think you will get in trouble if you do, because they are different.
In the first, we are moreso talking about a process than a result. For example, we could say that (x*x + 2*x + 1)/(x+1) = x + 1. We are not talking about numbers here. In the second, the concern is with the result of the process, not the process itself. Here, 1 1/2 is the same as 1.5, and both are the same as 3/2, 6/4, and the number midway between 1 and 2.
In the first, we could have numerators bigger than demoninators, or as noted, variable expressions. In the second we are primarily going to be using mixed fractions, where the fraction is proper (numerator smaller than the denominator). We would rarely use 3/2 as indicating a length in inches.
### The Third Meaning
There is a third meaning to fractions, as proportions. Suppose we find out that 2/3 of the children in a class are boys. This tells us something. But it does not tell us how many boys are in the class.
In this third sense, fractions are just like percentages or probabilities. The lowest value is zero and means none; the highest value is one and that means all.
Of course, this is the form we need for making decisions. If you want to decide whether to get a flu shot, you need to the probability of getting the flu without the shot, which is (roughly) the fraction/proportion of people getting the flu out of all the people not getting a flu shot. The actual number of people who get the flu without getting a flu shot is irrelevant.
I ask my class to differentiate these two sentences. "I ate half of a banana" and "I ate half of the bananas". In the first, half is being used as a number; in the second, half is being used as a proportion. "I ate one and a half bananas" makes sense; "I ate one and a half of the bananas" doesn't make sense.
I also happened to ask if you could add fractions, and the answer from some students was no. From the third definition, this is actually somewhat reasonable. Suppose you knew that 1/2 the people in a class were male, and 1/3 had their birthdays in the summer. There is no question for which these two numbers can be meaningfully added. Of course, as numbers or lengths, it makes sense to add fractions.
### Are the Fractions Numbers?
I ask classes to list numbers. They are prone to just mention the counting numbers. On a bad day, they won't remember zero; on a good day, you might even get the negative numbers. You are unlikly to get fractions.
Then I ask if there are numbers between 1 and 2. Then I am likely to get fractions. Then I take a vote -- are the fractions numbers?
Usually the fractions lose the vote. One time they won and then the teacher later told the class that they weren't numbers. Actually, with no definition of "number", there is no right or wrong answer. And I am sympathetic to the claim that proportions are not numbers, though I think the fractions do point to real numbers and hence "are" true numbers.
If the class votes that fractions are not numbers, then you are back to the question of whether they are numbers between 1 and 2. Or to ask the question in a more concrete way, are there lengths between 1 inch and 2 inches?
One third-grade class voted overwhelmingly that fractions were numbers. I asked what a fraction was, which is a good exercise by itself, and they said, I think correctly, something to the effect of one number in the numerator and one number in the demoninator. (This left out the concept of division, which is central to the first definition of fraction but not essentially to the other definitions; as will be discussed below, students are exposed to fractions as part of a circle.)
So I asked if 1/2 / 6 was a fraction. They really didn't like calling that a fraction. But the standard way to get out of it was to say that 1/2 wasn't a number, and they had voted that fractions were numbers. (I gave them the word "integer", but they didn't use it in their definition construction.)
### The Ubiquitous Circle
Fractions apparently are first defined as parts of a circle. For example, the fraction 1/3 means to divide the circle into 3 parts, then color in one of the parts. As noted above, while the circle is divided into three parts to represent the fraction 2/3, two is never divided by three. So this representation does not fit the first meaning of fraction.
If the circle is taken as representing the number 1, then the circle fits the second definition. But I sincerely doubt that it how the circle is meant. The circle perfectly fits the third definition. If there are six bananas and I eat six of them, then we divide the circle into six parts, color in three, and see that I have eating 3/6 of the bananas. In this representation, the circle in a way is six bananas, and it has no connection to the number 1.
The circle concept can be bent to fill the other definitions. One third-grader could see that 1/2 / 3 was a number, using the circle concept -- she divided her circle into three parts and then imagined coloring in 1/2 of a part. Similarly, when faced with 84/4, she asked how that was possible -- you can't divide a circle up into 4 parts and then color in 84 of them. I said you would need 21 circles, and she seemed to understand that.
### A Beginning Exercise
Have some unit of measure. The "paper", equal to the width of a piece of standard paper, is very convenient. (You can make a lot of them very quickly.)
Now, there is a problem whenever you measure something, in that it won't be exactly some whole number of papers. You can have them measure.
Then once they see this problem, try to communicate a length to them. Tell them to draw a line that is this long: Divide the paper into 3 equal parts, then draw the line equal to two of these parts. You can use this to draw lines of any length.
I sometimes say that in the imaginary country of Riki-Tiki, they would say that something was 14 + 5 parts out of 7 papers long. This was then shorted to 14 5/7 and everyone knew what that meant.
### Other Meanings
The beginning exercise communicates the second meaning of fractions, as representing numbers. Ways of trying to communicate the third meaning are discussed elsewhere.
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# Difference Quotient: Definition & Example
## What is a Difference Quotient?
The difference quotient is one way to find a derivative or slope of a function. The formula is:
It might look intimidating, but it’s easier to solve than it looks, because many of the terms cancel out.
It’s called a “difference quotient” because the formula has two parts: a difference (subtraction) in the numerator and a quotient (division) between the two parts:
## Example
Find the difference quotient for the following function: f(x) = 3x + 2
Step 1: Insert your function into the first part of the formula. In this step, I’m replacing the “f(x+h)” in the left hand part of the numerator with the actual given function, 3x + 2:
Essentially, I just changed the “x” part of 3x + 2 with “x + h”.
Step 2: Replace the “f(x)” in the formula’s numerator. This part is easy, because you’re just copying the formula exactly as written in the question and pasting it in:
Step 3: Use algebra to simplify. In most cases, things will start to cancel out, making the formula a little less ugly.
For this particular formula, the first step of simplifying is to remove the parentheses, to get:
From here, you should be able to see that the following terms cancel out:
• 3x – 3x = 0
• 2 – 2 = 0
Leaving 3h / h = 3.
The difference quotient (aka the derivative or slope) is 3.
## Origins of the Difference Quotient
The formula is derived from the “rise/run” slope formula you’re probably familiar with from algebra:
It’s the same formula, with a few substitutions. For example, instead of “y2” in the slope formula, you have “f(x + h)” in the difference quotient formula.
The full list of substitutions:
• f(x + h) → y2
• f(x) → y1
• x → x1
• x + h → x2
• h → change in x
• f(x + h) – f(x) → (y2 – y1)
Substituting those in to the slope formula, you get:
## References
Difference Quotient. Retrieved October 13, 2019 from: https://www.csusm.edu/mathlab/documents/differencequotient-r6.pdf
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# Brian Austin Green – Future Predictions (11/10/2019)
How will Brian Austin Green fare on 11/10/2019 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is not at all guaranteed – don’t get too worked up about the result. I will first find the destiny number for Brian Austin Green, and then something similar to the life path number, which we will calculate for today (11/10/2019). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology people.
PATH NUMBER FOR 11/10/2019: We will consider the month (11), the day (10) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 10 we do 1 + 0 = 1. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 1 + 12 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 11/10/2019.
DESTINY NUMBER FOR Brian Austin Green: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Brian Austin Green we have the letters B (2), r (9), i (9), a (1), n (5), A (1), u (3), s (1), t (2), i (9), n (5), G (7), r (9), e (5), e (5) and n (5). Adding all of that up (yes, this can get tedious) gives 78. This still isn’t a single-digit number, so we will add its digits together again: 7 + 8 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the destiny number for Brian Austin Green.
CONCLUSION: The difference between the path number for today (6) and destiny number for Brian Austin Green (6) is 0. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is of questionable accuracy. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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# How many times does a simple cell divide mitotically to form 32 cells?
Contents
Five mitotic divisions are required to form 32 cells from a single cell.
## How many generations are required to produce 32 cells?
5 successive generations of mitosis must occur to produce 32 cells.
## How many mitotic divisions and number of generations respectively are required to form 32 cells from a cell?
For a single cell to produce 32 cells, 31 mitotic divisions are required. From the first cell, 2 cells are produced (1 mitotic division). Then these 2 cells are divided into 4 cells (2 mitotic division). Then from these 4 cells, by mitosis, 8 cells are produced (4 mitotic division).
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## How many times a cell will have to divide Mitotically?
Complete answer: The cell will have to divide 7 times mitotically, to form 128 cells. The number of times a cell will have to divide mitotically is calculated by the formula 2n, where n is the number of times the cell is dividing.
## How many number of generations are required to produce 32 cells from the two cells by mitotic division?
NEET Question
1-2-4-8-16-32-64-128 so total 7 generation req. Since Mitosis is an equational division process and it produces two daughter cells from the single parent cells. Clearly, the formula (2)^n would answer the number of cells produced, where n would be the number of divisions.
## How many generations are required to produce 256 cells?
Every mitotic division will lead to the formation of twice the number of mother cells, hence to form 256 cells from a root tip cell, 8 generations of mitotic divisions must occur (28=256 cells).
## How many generations of mitosis are required to produce 64 cells from one cell?
How many mitotic division and generation respectively are required to form 64 cells from a cell? 1 6,6.
## How many mitotic divisions are needed for 100 cells?
Number of generations (n) of mitosis for producing ‘x’ cells is 2^n. First formula is used when the question is saying “Find the number of mitotic divisions required to produce say 100 cells” so for example, to produce 100 cells, 100-1 = 99 mitosis are required.
## How many mitotic divisions are needed for a single cell?
Mitosis is an equational division where after division each cell produces two daughter cells, therefore after 7 divisions one cell will give 128 cells in case of mitosis.
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## How many times mitotic divisions are needed for a single cell to make 128 cells?
Hence 7 mitotic divisions cell needed for a single cell to make 128 cells.
## How many cells will be produced if a cell divides mitotically 6 times?
12 cells can be formed if cell divide by 6 times through mitosis process.
## How many times a cell will have to divide mitotically to form 256 cells?
Mitotic cells division is equational division where a mother cell divides to form 2 daughter cells. So to produce 256 cells from a single cell, 8 mitotic divisions will occur which can be represented as 1→2→4→8→16→64→128→256.
## How many times a cell can divide?
The Hayflick Limit is a concept that helps to explain the mechanisms behind cellular aging. The concept states that a normal human cell can only replicate and divide forty to sixty times before it cannot divide anymore, and will break down by programmed cell death or apoptosis.
## How many meiotic divisions are necessary for formation of 80 sperms?
Answer: The number of ‘meiotic divisions’ that are necessary for the production of 80 sperms is 20. Explanation: During the process of spermatogenesis, the haploid sperms are produced by the meiotic division that occurs.
## How many meiotic divisions are required to produce?
The meiotic division is involved in the formation of the two types of gametes in an organism, one male gamete, and the other female gamete. The formation of the male gamete requires only one meiotic division which results in four male gametes while the formation of four female gametes requires four meiotic divisions.
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## How many Meiosiss are needed for 50 seeds?
Thus, in order to produce 50 mature ovules, 50 megaspore mother cells will undergo 50 meiotic divisions, one meiosis per megaspore mother cell. This takes the requirement of 50 (in megaspore mother cells) + 13 (in microspore mother cells), i.e., 63 meiotic divisions.
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# Cross Multiplication Calculator
Created by Maciej Kowalski, PhD candidate
Reviewed by Steven Wooding
Last updated: Jan 18, 2024
Table of contents:
Welcome to Omni's cross multiplication calculator, where we'll be solving for $x$ with fractions. The expressions we tackle here are often called proportions, and there's an easy algorithmic way to deal with them: cross multiply the fractions. In fact, whichever of the four values is unknown, once we cross multiply and divide, we're sure to get our result, no strings attached.
But before we get ahead of ourselves, let's slow down a bit and learn how to do cross multiplication nice and easy.
## Solving for x with fractions
Typically, we use cross multiplication when we have one-variable equations including fractions. For instance, it involves expressions of the form:
$\begin{split} \frac{2}{x}& = \frac{5}{7}\\ \\[1.5em] -\frac{1}{3.4} &= \frac{9x}{10}\\ \\[1.5em] \frac{2}{3}x &= \frac{21}{8} \end{split}$
Note how we can have negative numbers or decimals in the numerators or denominators. Also, in the last example, $x$ is outside of the fraction, but we can easily get it inside by following the basic rules that tell us how to calculate the fraction multiplication:
$\frac{2}{3}x=\frac{2}{3}\times\frac{x}{1}=\frac{2\times x}{3\times1}=\frac{2x}{3}$
Such equations, even if they look fancy, must follow the same rules as any other. In particular, we can add or subtract any number, and we can multiply) or divide by any non-zero value as long as we do it on both sides of the equality sign. For our purposes, the latter pair of arithmetic operations prove crucial: we'll cross multiply and divide to find the value of $x$.
## How to cross multiply fractions
The clue is in the name "cross multiplication". We'll calculate the product of the values in a cross pattern:
$\scriptsize \text{num}_\text{left}\times\text{den}_{\text{right}}=\text{num}_\text{right}\times\text{den}_{\text{left}}$
Where $\text{num}$ and $\text{den}$ are, respectively, numerator and denominator. And if we were to use the symbols appearing in Omni's cross multiply calculator, i.e., take the equation:
$\frac{A}{B}=\frac{C}{D}$
we'd get:
$A\times D = B\times C$
In fact, that's all there is to solving for $x$ with fractions. After all, once we cross multiply, we don't have fractions anymore, so we can turn to other well-known methods of dealing with equations. For instance, if we wanted to find $A$ from the formula above, it'd be enough to divide both sides by $D$:
$\begin{split} \frac{A\times D}{D}&=\frac{B\times C}{D}\\ \\[1.5 em] A&=\frac{B\times C}{D} \end{split}$
Note how whichever letter we need (i.e., whichever numerator or denominator), the procedure would still be the same: cross multiply and divide. The difference is only in what we divide by in the second step. To be precise, we divide by:
• $D$ when seeking $A$;
• $C$ when seeking $B$;
• $B$ when seeking $C$; and
• $A$ when seeking $D$.
Furthermore, recall that we can always exchange the sides of an equation. Therefore, we can always change the formula:
$\scriptsize \text{num}_\text{left}\times\text{den}_{\text{right}}=\text{num}_\text{right}\times\text{den}_{\text{left}}$
into:
$\scriptsize \text{num}_\text{right}\times\text{den}_{\text{left}}=\text{num}_\text{left}\times\text{den}_{\text{right}}$
However, observe that the pairs stay the same: we need to preserve the cross multiplication pattern.
Alright, the instructions on how to do cross multiplication seem easy enough, don't you think? It's high time we move from symbols and theory to numbers and practice. And, to kill two birds with one stone, we'll take the opportunity to let our cross multiply calculator shine.
## Example: using the cross multiplication calculator
Suppose that you're constructing an aircraft model. After a few hours of meticulous gluing, the plane is ready: it will look awesome displayed on the shelf for everyone to admire. But now that the DIY part is over, why don't we learn something about it? Let's calculate how large the real-life equivalent is.
The box says that the model is done on a $1:100$ scale. You grab a ruler and check that your creation is $3.5$ inches long. Believe it or not, that's all we need to find the answer. The trick is in a good understanding of how scaling works: the unit rate on the box is proportional to the ratio of the model and real-life lengths. In our case, this means that the $3.5$ inches is to the actual aircraft's length what $1$ is to $100$:
$\frac{3.5}{x}=\frac{1}{100}$
where $x$ denotes the value we seek. However, before we rush to solve it ourselves, let's see how easy the task is with Omni's cross multiplication calculator at hand.
At the top of our tool, we see the formula:
$\frac{A}{B}=\frac{C}{D}$
and four variable fields corresponding to the letters. As the cross multiplication calculator states, it's enough to input three of the values, so we look back at our problem and write:
• $A = 3.5$, $C = 1$, and $D = 100$.
The moment we input the third number, the tool will spit out the answer. Note how it also provides a step-by-step explanation underneath: the same that we'll give right now.
We begin by doing what the above section taught us: we cross multiply the fractions:
$\frac{3.5}{x}=\frac{1}{100}$
Hence:
$x\times 1 = 3.5\times 100$
Which gives:
$x = 350$
Normally, we'd still need to divide the result by the number standing in front of $x$. However, in our case, that number turned out to be $1$, so there's no need.
We got our answer! The real-life aircraft is $350$ inches long. Well, we should probably move on from the cross multiplication calculator to a length converter to get a more reasonable answer, don't you think?
Models are often built in different scales: $1:72$, $1:48$, and so on. Learn how to quickly pass from one to the other with our scale calculator.
## FAQ
### How do I solve for x with fractions?
To solve for x with fractions, you need to:
1. Transform both sides into quotients.
2. Cross multiply the fractions.
3. Simplify the two expressions.
4. Divide by what's in front of x.
5. Enjoy having solved for x with fractions.
### How do I cross multiply fractions?
To cross multiply fractions, you need to:
1. Make sure you have only a fraction on each side.
2. Multiply the numerator of the first by the denominator of the second.
3. Multiply the numerator of the second by the denominator of the first.
4. Combine steps 2-3 into an equation.
5. If needed, solve the resulting equation with basic methods.
6. Enjoy having cross multiplied the fractions.
### Why does cross multiplication work?
Cross multiplication is, in fact, simple multiplication done twice.
Firstly, we multiply both sides by the left side's denominator, which leaves only the numerator on the left (according to fraction simplification rules), and multiplies the right numerator (according to fraction multiplication rules). Next, we multiply both sides by the right side's denominator, which gives a product on the left and kills the denominator on the right.
All in all, since we can always multiply both sides of an equation by a non-zero number, cross multiplication indeed works.
### How does cross multiplication work?
To use cross multiplication, you need to:
1. Make sure you have only a fraction on each side.
2. Multiply the numerator of the first by the denominator of the second.
3. Multiply the numerator of the second by the denominator of the first.
4. Combine steps 2-3 into an equation.
5. If needed, solve the resulting equation with basic methods.
6. Enjoy having used cross multiplication.
### How do I compare fractions using cross multiplication?
To compare fractions using cross multiplication, you need to:
1. Make sure you have only a fraction on each side.
2. Multiply the numerator of the first by the denominator of the second.
3. Multiply the numerator of the second by the denominator of the first.
4. Compare values from steps 2 and 3.
5. If the one in step 2 was:
• Smaller, then the first fraction is smaller; or
• Larger, then the first fraction is larger.
6. If one of the multipliers was negative, change the relation to its opposite.
7. Enjoy having compared fractions using cross multiplication.
### How do I solve proportions using cross multiplication?
To solve proportions using cross multiplication, you need to:
1. Make sure you only have a fraction on each side.
2. Multiply the numerator of the first by the denominator of the second.
3. Multiply the numerator of the second by the denominator of the first.
4. Combine steps 2-3 into an equation.
5. Solve the resulting equation with basic methods.
6. Enjoy having solved a proportion using cross multiplication.
Maciej Kowalski, PhD candidate
Enter any three numbers:
A
B
C
D
Would you like to learn more about proportionality? If so, check out the ratio calculator next!
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To use it replace square root sign with letter r. Rationalizing the denominator is the process of moving any root or irrational number cube roots or square roots out of the bottom of the fraction denominator and to top of the fraction numerator. It can rationalize denominators with one or two radicals. Estimating an n th Root. Square root calculator and perfect square calculator. Tap for more steps... Raise to the power of . Thus, = . This calculator eliminates radicals from a denominator. To rationalize a denominator, multiply the fraction by a "clever" form of 1--that is, by a fraction whose numerator and denominator are both equal to the square root in the denominator. For example, to rationalize the denominator of , multiply the fraction by : × = = = . Calculating n th roots can be done using a similar method, with modifications to deal with n.While computing square roots entirely by hand is tedious. Expand the denominator using the FOIL method. For example, with a square root, you just need to get rid of the square root. Combine fractions. From rationalize the denominator calculator with steps to power, we have every aspect discussed. In this tutorial, we learn how to rationalize square roots. 1 How to Add and Subtract Square Root. Also tells you if the entered number is a perfect square. To use it, replace square root sign ( â ) with letter r. Example: to rationalize $\frac{\sqrt{2}-\sqrt{3}}{1-\sqrt{2/3}}$ type r2-r3 for numerator and 1-r(2/3) for denominator. From here, this will make the square root go away, so your equation will be normal numbers. Table of Contents. 1.1 Addition and subtraction Can Be Done When the Numbers in a Radical Symbol Are Same; 1.2 Match the Numbers in the Root Symbol by Prime Factorization; 1.3 After Rationalizing the Denominator, Making the Common Denominator and Calculate It; 2 Square Root Multiplication and Division Using the Distributive Property You cannot have square roots in the denominator of an equation. Calculate the positive principal root and negative root of positive real numbers. For example, However, you canât fall for the trap of rationalizing a fraction by squaring the numerator and the denominator. This calculator eliminates radicals from a denominator. Come to Algebra-equation.com and understand linear systems, adding and subtracting rational and lots of additional algebra subject areas You can do this by multiplying the top and bottom of the equation by the bottom denominator. Find the square root, or the two roots, including the principal root, of positive and negative real numbers. Estimating higher n th roots, even if using a calculator for intermediary steps, is significantly more tedious. You need to multiply so the square root goes away. Rationalizing the denominator is accomplished by multiplying top and bottom by the square root found in the bottom. Normally, the best way to do that in an equation is to square both sides. Square roots and exponents, middle school pre algebra simplifying radicals, calculator to square a binomial, rationalizing denominator worksheet. We rationalize! The process is super easy to follow, and we can use the process of rationalizing ⦠In essence, we are merely multiplying by a form of 1. Simplify the numerator. Dividing negative fractions, clock problems in algebra, pre algebrathe area of a polygon formula. Simplify. 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Section10.6Applications Involving Densities
Subsection10.6.1Overview
Having developed the theory of definite integrals and functions defined as the accumulation on increments, we turn our attention to applications of these ideas. One of the most common mathematical applications is the calculation of area of regions bounded by curves. Physically, this is closely related to the calculation of total mass and center of mass. The same calculations are used in statistics to calculate probabilities and averages.
The general setup for many applications involving definite integrals is to think of the total quantity as a sum of the parts. If a region is cut into separate pieces, for example, then the area of the total region should be the sum of the areas of each region measured separately. Quantities that have this property are called extensive. A definite integral can be used to compute extensive quantities if we can consider the total as a sum of small increments over a partition of an independent variable.
In this context, the rate of accumulation function $f$ used to compute the quantity $Q$ as an integral is often called the density of $Q$ with respect to $x\text{.}$
This section will explore the application of definite integrals to compute various extensive quantities. This will require imagining a partition, identifying the independent variable and its corresponding interval for integration, determining the appropriate function used as the density, and setting up the definite integral. Because our emphasis will be on identifying the appropriate definite integral, we will use technology to compute the the resulting value.
Subsection10.6.2Area of Regions in the Plane
When we developed the definite integral of a function $f$ on an interval $[a,b]\text{,}$ we noted that the integral represented the total signed area over the interval. It was signed because the accumulation of negative values when the function was below the axis was subtracting area. Because area is an extensive quantity, it is an ideal example of a quantity that can be computed using integration.
Example10.6.2
Find the area of the region bounded between $y=x^2-2x$ and $y=x\text{.}$
Solution
Start by identifying a convenient variable to partition. The region is determined by where the graphs $y=x^2-2x$ and $y=x$ intersect. Using substitution and solving, the intersection occurs at the solution to $x^2-2x=x\text{.}$
\begin{gather*} x^2-2x=x\\ x^2-3x=0\\ x(x-3)=0 \end{gather*}
The region is completely contained between $x=0$ and $x=3\text{.}$ We choose our independent variable to be $x$ over the interval $[0,3]\text{.}$ Now, imagine a partition of the interval and consider the increments of area over each subinterval.
As the increments $\Delta x$ are smaller and smaller, the increment of area $\Delta A$ will be closely approximated by the width $\Delta x$ times the vertical distance between $y=x$ and $y=x^2-2x\text{,}$
\begin{equation*} \Delta A \approx \big(x - (x^2-2x)\big) \Delta x\text{.} \end{equation*}
The total area is the sum of the increments, so we can use a definite integral. The height $h(x) = x-(x^2-2x) = 3x-x^2$ between the curves acts as the density of area,
\begin{equation*} A = \int_0^3 \big(x - (x^2-2x)\big) \, dx = \int_0^3 3x-x^2 \, dx. \end{equation*}
The density is a simple polynomial so that we can compute the value using the elementary accumulation formulas.
\begin{align*} \int_0^3 3x-x^2 \, dx &= 3 \int_0^3 x \, dx - \int_0^3 x^2 \, dx \\ &=3\big(\frac{1}{2}(3)^2\big) - \big(\frac{1}{3}(3)^3\big)\\ &=\frac{27}{2} - 9\\ &=\frac{9}{2} \end{align*}
Except for such simple problems, we can use technology to compute or approximate the value of the integrals. The SageMath engine attempts to compute integrals exactly using the integrate command, which uses the following syntax.
When SageMath is unable to do the exact calculation, we can still do a numerical approximation using the numerical_integral command.
The result gives an approximate answer along with an estimated error bound. In this case, we find $A = 4.5 \pm 4.996 \times 10^{-14}\text{.}$
Example10.6.3
Express the area of a circle with center $(0,0)$ and radius $r=2$ as a definite integral.
Solution
The circle is clearly between the lines $x=-2$ and $x=2$ so that we can imagine a partition with variable $x$ on the interval $[-2,2]\text{.}$ The increments of area $\Delta A$ correspond to thin vertical strips with width $\Delta x$ (from the partition) and a height computed as the distance from the top of the circle to the bottom of the circle.
The equation of the circle is
\begin{equation*} x^2+y^2=4. \end{equation*}
To find the height of the increments, we need to know the two $y$-values for each $x$-value,
\begin{equation*} y = \pm \sqrt{4-x^2}\text{.} \end{equation*}
The height is the difference between the values,
\begin{equation*} h(x) = (\sqrt{4-x^2}) - (-\sqrt{4-x^2}) = 2\sqrt{4-x^2}\text{.} \end{equation*}
Consequently, the area of the circle is defined by the integral
\begin{equation*} A = \int_{-2}^{2} h(x) \, dx = \int_{-2}^2 2 \sqrt{4-x^2}. \end{equation*}
Computational tools can compute this value, which is consistent with the known formula
\begin{equation*} A=\pi r^2 = \pi(2)^2 = 4\pi\text{.} \end{equation*}
When curves cross multiple times, we may need to compute the area of individual regions.
Example10.6.4
Find the area bounded by the graphs $y=x$ and $y=x^3-4x\text{.}$
Solution
We start by identifying the points of intersection of the curves by solving the equation $x^3-4x = x\text{.}$
\begin{gather*} x^3-4x = x\\ x^3-5x = 0\\ x(x^2-5) = 0 \end{gather*}
One solution is $x=0\text{,}$ corresponding to the intersection at $(x,y)=(0,0)\text{.}$ Two other solutions come from $x^2=5$ at $x = \pm \sqrt{5}\text{.}$ The total area consists of two regions, the first with $x \in [-\sqrt{5},0]$ and the second with $x \in [0,\sqrt{5}]\text{.}$
On the first interval $[-\sqrt{5},0]\text{,}$ the height of increments is given by
\begin{equation*} h(x) = (x^3-4x) - x = x^3-5x \end{equation*}
because the cubic polynomial is the top curve. So the area over the interval $[-\sqrt{5},0]$ is computed by
\begin{equation*} A_1 = \int_{-\sqrt{5}}^{0} x^3-5x \, dx. \end{equation*}
In order to use the elementary accumulation formulas, the integral needs to start at $x=0\text{,}$ so we reverse the order of integration and change the sign.
\begin{align*} A_1 &= \int_{-\sqrt{5}}^{0} x^3-5x \, dx\\ &= -\int_{0}^{-\sqrt{5}} x^3-5x \, dx\\ &= -\int_{0}^{-\sqrt{5}} x^3 \, dx + 5 \int_{0}^{-\sqrt{5}} x \, dx\\ &= -\left( \frac{1}{4}(-\sqrt{5})^4\right) + 5 \left(\frac{1}{2}(-\sqrt{5})^2\right)\\ &= -\frac{25}{4} + \frac{25}{2}\\ &= \frac{25}{4} \end{align*}
On the second interval $[0,\sqrt{5}]\text{,}$ the height of increments is given by
\begin{equation*} h(x) = x - (x^3-4x) = 5x-x^3 \end{equation*}
because now the cubic polynomial is the bottom curve. The corresponding area is
\begin{equation*} A_2 = \int_{0}^{\sqrt{5}} 5x- x^3 \, dx \end{equation*}
which has a value
\begin{align*} A_2 &= \int_{0}^{\sqrt{5}} 5x- x^3 \, dx\\ &= 5\int_{0}^{\sqrt{5}} x \, dx - \int_{0}^{\sqrt{5}} x^3 \, dx\\ &= 5\left( \frac{1}{2}(\sqrt{5})^2\right) - \left(\frac{1}{4}(\sqrt{5})^4\right)\\ &= \frac{25}{2} -\frac{25}{4}\\ &= \frac{25}{4} \end{align*}
As we should expect from symmetry, the two areas are equal $A_1=A_2\text{.}$
The total area of the region is
\begin{equation*} A = A_1 + A_2 = \frac{25}{4} + \frac{25}{4} = \frac{25}{2}\text{.} \end{equation*}
If we were to think of the distance between the two curves in terms of the absolute value, the integral could be computed over a single interval,
\begin{equation*} A = \int_{-\sqrt{5}}^{\sqrt{5}} \left| x - (x^3-4x)\right| \, dx = \int_{-\sqrt{5}}^{\sqrt{5}} \left| 5x - x^3\right| \, dx\text{.} \end{equation*}
Using SageMath, the absolute value prevents an exact integral.
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# 5.07 Interpreting proportional relationships
Lesson
We've already learned about proportional relationships, where two quantities vary in such a way that one is a constant multiple of the other. In other words, they always vary by the same constant.
Remember!
Proportional relationships are a special kind of linear relationship that can be written generally in the form $y=kx$y=kx and always pass through the origin $\left(0,0\right)$(0,0).
To interpret information from a graph, we need to look at pairs of coordinates. Coordinates tell us how one variable relates to the other. Each pair has an $x$x value and a $y$y value in the form $\left(x,y\right)$(x,y).
• The $x$x-value tells us the value of the variable on the horizontal axis.
• The $y$y-value tells us the value of the variable on the vertical axis.
It doesn't matter what labels we give our axes, this order is always the same.
Let's look at some examples and see this in action.
#### Worked example
##### Question 1
The number of eggs farmer Joe's chickens produce each day are shown in the graph.
What does the point $\left(6,3\right)$(6,3) represent on the graph?
Think: The first coordinate corresponds to the values on the $x$x-axis (which in this problem would represent the number of days) and the second coordinate corresponds to values on the $y$y-axis ( which in this problem would represent the number of eggs).
Do:
Using the given information in context, we can interpret this point to mean that in $6$6 days the chickens will produce $3$3 eggs.
Reflect: How many days does it take for the chickens to lay $1$1 egg?
#### Practice questions
##### question 2
The number of liters of gas used by a fighter jet over a certain number of seconds is shown in the graph. What does the point on the graph represent?
1. $6$6 liters of gas are used by the fighter jet every $12$12 seconds.
A
$12$12 liters of gas are used by the fighter jet every $6$6 seconds.
B
$6$6 liters of gas are used by the fighter jet every $12$12 seconds.
A
$12$12 liters of gas are used by the fighter jet every $6$6 seconds.
B
##### question 3
The number of cupcakes eaten at a party is shown on the graph.
1. What does the point on the graph represent?
$2$2 cupcakes are eaten by $4$4 guests.
A
$4$4 cupcakes are eaten by $2$2 guests.
B
$2$2 cupcakes are eaten by $4$4 guests.
A
$4$4 cupcakes are eaten by $2$2 guests.
B
2. Danielle is having a party, and expects to have $10$10 guests. According to the rate shown on the graph, how many cupcakes should she buy?
### Outcomes
#### 7.RP.A.2
Recognize and represent proportional relationships between quantities.
#### 7.RP.A.2d
Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate
|
# A contour map is shown for a function f on the square r 0 6 0 6
## Introduction
A contour map is a topographical map that shows the relief of an area, and is usually drawn to show a particular landform or feature. In this case, the contour map shows the function f on the square r 0 6 0 6 . The contour lines represent points of equal elevation, and the space between them indicates the steepness of the terrain.
## Theorem
A contour map is a two-dimensional graph of a function. In this case, the function is graphed on the square r 0 6 0 6 . The theorem states that if f is a continuous function on r 0 6 0 6 , then f has a contour map.
## Proof
To prove that this contour map is accurate, we will use the definition of a contour map. A contour map is a two-dimensional cross section of a three-dimensional surface. In other words, it is a way of representting a three-dimensional surface on a two-dimensional plane.
To prove that this contour map is accurate, we will need to find the points on the square where the function f is equal to 0, 1, 2, and 3. We can do this by plugging in these values for x and y into the equation for f.
For x = 0 and y = 0, we get f(0,0) = 0.
For x = 0 and y = 1, we get f(0,1) = 1.
For x = 1 and y = 0, we get f(1,0) = 2.
For x = 1 and y = 1, we get f(1,1) = 3.
These points are plotted on the contour map, and they all fall on the correct contours. This means that the contour map is accurate.
## Examples
-A contour map is shown for a function f on the square r 0 6 0 6 – (a contour map is shown for a function f on the square r 0 6 0 6)
-The following example shows a contour map for the function f(x,y)=x^2+y^2 on the square [-1,1] x [-1,1].
## Conclusion
From the contour map, we can see that the function f has a minimum value at (0,6) and a maximum value at (6,0).
|
# Fraction calculator
This calculator divides fractions. The first step makes the reciprocal value of the second fraction - exchange numerator and denominator of 2nd fraction. Then multiply both numerators and place the result over the product of both denominators. Then simplify the result to the lowest terms or a mixed number.
## The result:
### 3/4 ÷ 1/3 = 9/4 = 2 1/4 = 2.25
The spelled result in words is nine quarters (or two and one quarter).
### How do we solve fractions step by step?
1. Divide: 3/4 : 1/3 = 3/4 · 3/1 = 3 · 3/4 · 1 = 9/4
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 1/3 is 3/1) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - three quarters divided by one third is nine quarters.
### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right.
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# chapter 9 name: monday tuesday wednesday thursday ... ... monday tuesday wednesday thursday friday 4
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• 1
Unit 10B – Quadratics - Chapter 9 Name:
The calendar and all assignments are subject to change. Students will be notified of any changes
during class, so it is their responsibility to pay attention and make any necessary changes.
All assignments are due the following class period unless indicated otherwise.
Monday Tuesday Wednesday Thursday Friday
4
MapTesting
5
Section 9.1
6
Section 9. 1b
Properties of
7
Section 9.3
8
Quick Quiz
Section 9.4 Completing
the Square
11
Section 9.4b
12
Section9.4c/ 9.5a
13
Section 9.5 b
14
9.5c
15
18
19
Test on Solving
Section Page Assignment
9.1 p. 485
9.1b p. 486
9.3 p. 501
9.4 p. 511
9.4b
9.5a p. 521
9.5b
Chapter 9 Review
• 2
Lesson 9.1 – Properties of Radicals Algebra 1
Essential Question How can you multiply and divide square roots?
For each operation with square roots, compare the results obtained using the two indicated orders of
operations. What can you conclude?
a. Square Roots and Multiplication
Is 4 9 equal to 4 9?
In general, is a b equal to ?a b Explain your reasoning.
b. Square Roots and Division
Is 100
4 equal to
100
4
In general, is a
b equal to ?
a
Core Concepts
Product Property of Square Roots
Words The square root of a product equals the product of the square roots of the
factors.
Numbers 9 • 5 9 • 5 3 5
Algebra • , where , 0ab a b a b
1. Simplifying Radicals with the Product Property
Simplify each radical into its simplest form.
(a) (b) (c) (d) (e)
(f) (g)
50 125 48 300 72
128 288
1 EXPLORATION: Operations with Square Roots
• 3
Quotient Property of Square Roots
Words The square root of a quotient equals the quotient of the square roots of the
numerator and denominator.
Numbers 3 3 3
4 24 Algebra , where 0 and 0
a a a b
b b
2. Simplifying Radicals with the Quotient Property
Simplify each radical into its simplest form.
(a) (b) (c) (d) (e)
3. Simplify each radical into its simplest form.
(a)
6162g (b) 7512h (c) 4
196
r (d)
3
2
49
64
x
y
Lesson 9.1 – Properties of Radicals – Day 2 Algebra 1
A radical is simplest form when 3 conditions are met:
No radicals appear in the…
9
25
25
18
6
72
90
40
50
32
• 4
(a) (b) (c)
(d) (e) ba
b
(f) 22
42
2. 7
4 3.
5
5 4.
2
10 5.
48
3
6. 75
9
7.
210
6
8.
37
12
2
5
14
7
35
6
23
4
• 5
Lesson 9.1 Worksheet Name _____________________________
I. Simplify each expression completely.
1. √ 2. √ 3. √
4. √ 5. √ √ 6.
7. √
8. √
9. 7√
10. √
11. √
12.
13. √
√ 14. √
15.
• 6
Lesson 9.3 – Solving Quadratic Equations Using Square Roots Algebra 1
Essential Question How can you determine the number of solutions of a quadratic equation of the form ax2 + c = 0?
1. Finding Square Roots Evaluate.
(a) (b) (c) (d)
2. What is the square root of 36?
3. Solving Quadratics Equations in the Form:
(a) (b) (c)
4. Solving Quadratics Equations in the Form:
(a) (b) (c)
(d) ( e) 2
2 196x (f) 2
2 7 49x
81 100 64 9
cx 2
812 x 172 x 52 x
02 cax
01222 x 012 x 0273 2 x
06012 2 x
• 7
5. Application Exercise – Falling Object Model
You have entered Mr. Esbrook’s first annual “egg dropping
contest”. The goal is to create a container for an egg so it can be
dropped from a height of 32 feet without breaking the egg. In you
quest to become egg-dropping champion, you have asked your
Algebra class to determine the time it will take for the egg
container to hit the ground. About how long will it take for the
egg’s container to hit the ground? Assume there is no air
Algebra: Section 9.1/ 9.3 Worksheet
Name______________________________________________Date____________________Hour______
Solve each equation using square roots. Simplify square roots if necessary. There should be no decimal answers.
1. x2 = 100 2. 2y2 = 32 3. ¼ a2 = -6
4. 10 – 2x2= 4 5. 7y2 + 14 = 0 6. 3b2 – 6 = 9
7. ½ x2 – 7 = 1 8. 2x2 + 5 = 9 9. -4x2 + 6 = -394
10. 6 – 3x2 = 27 11. 2
81 3 1 49x 12. 2
16 3 25x
• 8
simplify.
√ 13. √ 14. √
15. √
16. √
17. -3√
18. -5 √
√ 19. Find the area of a square with side length √
20. Find the area of a triangle with height √ and base √ .
21. Using the Falling Object Model found in section 9.1. Determine approximately how long it will take a I-
phone to hit the ground when dropped from a window 25 feet above the ground.
Lesson 9.4(a) – Completing the Square (w/ Leading Coefficient 1) Algebra 1
1. Completing the Square Complete the square so that each expression is a perfect square trinomial. Then factor the trinomial.
(a) (b) (c)
xx 122 xx 82 xx 302
Completing the Square To complete the square for any quadratic
expression in the form
add ____________ of the second coefficient (b)
_____________ to the end.
• 9
2. Solving Quadratic Equations by Completing the Square Solve each quadratic equation by completing the square.
(a) (b)
(c) (d)
Lesson 12.4(b) – Completing the Square (w/ Leading Coefficient 1) Algebra 1
Warm-Up Exercise Solve by Completing the Square.
(a) (b) Solving Quadratic Equations by Completing the Square Solve each quadratic equation by completing the square.
1. 2.
24102 xx 07262 xx
0422 xx 03 2
12 xx
01422 xx 7142 xx
41082 2 xx 0848 2 xx
• 10
3. 4. 210 13 9 0x x
5. 23 6 1 0x x 6.
212 8 2 0x x
In Exercises 7-12, determine whether the quadratic function has a maximum or minimum value.
Then find the value.
7. 2 4 3y x x 8. 2 6 10y x x 9. 2 8 2y x x
10. 2 10 8y x x 11. 23 3 1y x x 12. 24 8 12y x x
0263 2 xx
• 11
13. Choosing a Method to Solve a Quadratic Determine the easiest method to solve the quadratics below. Tell which method you used, and then solve the equation.
(a) (b)
(c)
Lesson 9.5(a) –The Quadratic Formula Algebra 1
Essential Question How can you derive a formula that can be used to write the solutions of any quadratic equation in standard form?
The following steps show a method of solving 2 0.ax bx c
|
# What is 201/338 as a decimal?
## Solution and how to convert 201 / 338 into a decimal
201 / 338 = 0.595
Fraction conversions explained:
• 201 divided by 338
• Numerator: 201
• Denominator: 338
• Decimal: 0.595
• Percentage: 0.595%
The basis of converting 201/338 to a decimal begins understanding why the fraction should be handled as a decimal. Both represent numbers between integers, in some cases defining portions of whole numbers But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. Now, let's solve for how we convert 201/338 into a decimal.
201 / 338 as a percentage 201 / 338 as a fraction 201 / 338 as a decimal
0.595% - Convert percentages 201 / 338 201 / 338 = 0.595
## 201/338 is 201 divided by 338
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 201 is being divided into 338. Think of this as our directions and now we just need to be able to assemble the project! The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We use this as our equation: numerator(201) / denominator (338) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. This is how we look at our fraction as an equation:
### Numerator: 201
• Numerators represent the number of parts being taken from a denominator. 201 is one of the largest two-digit numbers you'll have to convert. 201 is an odd number so it might be harder to convert without a calculator. Large numerators make converting fractions more complex. Time to evaluate 338 at the bottom of our fraction.
### Denominator: 338
• Denominators represent the total number of parts, located below the vinculum or fraction bar. 338 is one of the largest two-digit numbers to deal with. And it is nice having an even denominator like 338. It simplifies some equations for us. Have no fear, large two-digit denominators are all bark no bite. Now it's time to learn how to convert 201/338 to a decimal.
## How to convert 201/338 to 0.595
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 338 \enclose{longdiv}{ 201 }$$
We will be using the left-to-right method of calculation. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 338 \enclose{longdiv}{ 201.0 }$$
Because 338 into 201 will equal less than one, we can’t divide less than a whole number. So that means we must add a decimal point and extend our equation with a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 338 into 201 + 0 or 2010.
### Step 3: Solve for how many whole groups you can divide 338 into 2010
$$\require{enclose} 00.5 \\ 338 \enclose{longdiv}{ 201.0 }$$
How many whole groups of 338 can you pull from 2010? 1690 Multiply by the left of our equation (338) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.5 \\ 338 \enclose{longdiv}{ 201.0 } \\ \underline{ 1690 \phantom{00} } \\ 320 \phantom{0}$$
If there is no remainder, you’re done! If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 201/338 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. Same goes for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 201/338 and 0.595 bring clarity and value to numbers in every day life. Here are just a few ways we use 201/338, 0.595 or 59% in our daily world:
### When you should convert 201/338 into a decimal
Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.59 per hour and not$20 and 201/338.
### When to convert 0.595 to 201/338 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 201/338 = 0.595 what would it be as a percentage?
• What is 1 + 201/338 in decimal form?
• What is 1 - 201/338 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.595 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 338 Denominator What is 201/328 as a decimal? What is 191/338 as a decimal? What is 201/329 as a decimal? What is 192/338 as a decimal? What is 201/330 as a decimal? What is 193/338 as a decimal? What is 201/331 as a decimal? What is 194/338 as a decimal? What is 201/332 as a decimal? What is 195/338 as a decimal? What is 201/333 as a decimal? What is 196/338 as a decimal? What is 201/334 as a decimal? What is 197/338 as a decimal? What is 201/335 as a decimal? What is 198/338 as a decimal? What is 201/336 as a decimal? What is 199/338 as a decimal? What is 201/337 as a decimal? What is 200/338 as a decimal? What is 201/338 as a decimal? What is 201/338 as a decimal? What is 201/339 as a decimal? What is 202/338 as a decimal? What is 201/340 as a decimal? What is 203/338 as a decimal? What is 201/341 as a decimal? What is 204/338 as a decimal? What is 201/342 as a decimal? What is 205/338 as a decimal? What is 201/343 as a decimal? What is 206/338 as a decimal? What is 201/344 as a decimal? What is 207/338 as a decimal? What is 201/345 as a decimal? What is 208/338 as a decimal? What is 201/346 as a decimal? What is 209/338 as a decimal? What is 201/347 as a decimal? What is 210/338 as a decimal? What is 201/348 as a decimal? What is 211/338 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 338 Denominator 202/338 as a percentage 201/339 as a percentage 203/338 as a percentage 201/340 as a percentage 204/338 as a percentage 201/341 as a percentage 205/338 as a percentage 201/342 as a percentage 206/338 as a percentage 201/343 as a percentage 207/338 as a percentage 201/344 as a percentage 208/338 as a percentage 201/345 as a percentage 209/338 as a percentage 201/346 as a percentage 210/338 as a percentage 201/347 as a percentage 211/338 as a percentage 201/348 as a percentage
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# Problem
Given an integer N, find its prime factorization.
For example, $12 = 2 \times 2 \times 3 = 2^2 \times 3$.
It is somewhat difficult to factorize an integer. If $N$ is small enough, then it is possible to factorize it using prime generate by Sieve of Eratosthenes.
# Trial Division Method
Suppose we are trying to factorize $980$. How would we attempt to factorize it with pen and paper? We will try to extract the smallest possible prime factors from it and make it smaller. That is we will try to divide the number with prime numbers and keep track which is capable of dividing it.
The first prime we have is $2$. Can we divide $980$ with $2$? Yes. $980=2 \times 490$. Now we need to factorize $490$ which is smaller and thus easier.
Again, let’s try to factorize $490$ with $2$. It is possible, so $980=2 \times 2 \times 245= 2^2 \times 245$. It is not possible to extract another $2$ from $245$ so we move on to next prime.
$3$ fails to divide $245$, so we move on to $5$. By repeating this process, we find out that $980=2^2 \times 5 \times 7^2$.
This process is kind of laborious, but we have computers to do our bidding. So for this trial division to work, all we need is a list of primes so that we can try dividing $N$ one by one from that list. How will we get that list? Using Sieve of Eratosthenes.
But before we start generating prime, we need to answer another question. Sieve of Eratosthenes can generate prime up to a certain number. So up to which value should, we generate prime numbers to factorize N?
# Limit for Sieve of Eratosthenes
Well, if we are trying to factorize $N$, there is no point in generating primes that are larger than $N$. So we can generate up to $N$ and factorize using the list with primes less than or equal to $N$. But we can do even better by observing a special property of prime factors of $N$.
$N$ can have many prime factors. Some of the prime numbers will be $< \sqrt{N}$ and some $\ge \sqrt{N}$. Exactly how many primes factors can $N$ have that are greater than $\sqrt{N}$?
Notice that if $N$ can never have two prime factors $>\sqrt{N}$ since, $if : A > \sqrt{N}, then : A \times A > N$.
So if we generate primes up to $\sqrt{N}$ and use that list to factorize $N$ then at the end, we will either have 1 or a number greater than $\sqrt{N}$ which will be a prime for sure.
For example, $N=42$ and $\sqrt{N}=6.4 \approx 6$. So let’s try to factorize using primes less than or equal to $6$ only, that is using only $[2,3,5]$. What do we get? $N = 42 = 2 \times 3 \times 7$. Since $7$ can no longer be divided with any prime from our list $[2,3,5]$ we stop. $7$ is our last missing prime factor.
So, in order to factorize $N$, we need to generate primes up to $\sqrt{N}$ only.
# Code for Sieve of Eratosthenes
vector <int> factors;
void factorize( int n ) {
int sqrtn = sqrt ( n );
for ( int i = 0; i < prime.size() && prime[i] <= sqrtn; i++ ) {
if ( n % prime[i] == 0 ) {
while ( n % prime[i] == 0 ) {
n /= prime[i];
factors.push_back(prime[i]);
}
sqrtn = sqrt ( n );
}
}
if ( n != 1 ) {
factors.push_back(n);
}
}
Let me explain a few things about the code. In line 4, inside the for loop I wrote the following condition, i < prime.size() && prime[i] <= sqrtn. The first condition ensures that we don't go array over-bound. In order to understand it, imagine what will happen if we try to factorize a prime number $>\sqrt{N}$. Without that condition, you might get RTE.
The second condition ensures we are always trying to factorize using prime less than $\sqrt{N}$.
In line 5, we check if it is possible to divide the current $N$ with prime[i]. If so we start a loop in line 6 and keep on extracting until we cannot anymore.
In line 10, we are optimizing our code. Since, after extraction of prime factors, $N$ has become smaller, we now need to only factorize with a smaller amount of prime numbers. Due to this single line, the code converges really quickly.
In line 13, we are checking if the residual number after factorization is 1 or not. If not, it is the only prime factor which is greater than $\sqrt{N}$. So we add it to our prime factor list.
## Time Complexity
There are two loops in the code for factorize(). The first one is at line $4$. This loop runs once for every prime $\leq \sqrt{N}$. From "Prime Number Theorem", $\pi (N) \approx \frac{N}{ln(N)}$. So there are approximately $\frac{ \sqrt{N} }{ ln ( \sqrt{N} ) }$ primes $\leq \sqrt{N}$.
The second loop at line $6$ runs once for every prime factor $N$ has. So it cannot run more than $log_2(N)$ \times.
So the complexity is $O(\frac{ \sqrt{N} }{ ln ( \sqrt{N} ) } + log_2(N) : )$. In practice, if $N$ is not a prime then due to line $10$, $N$ becomes small quickly. So it is much faster than what we calculated.
## Optional Optimization
There is an optional optimization we can perform only when it is possible to generate sieve up to N. Just insert a single statement in line 5.
for ( int i = 0; i < prime.size() && prime[i] <= sqrtn; i++ ) {
if ( sieve[n] == 0 ) break; /*Checks if n is prime or not*/
if ( n % prime[i] == 0 ) {
The new line added simply checks if the number we are trying to factorize is a prime or not. If it is prime, then obviously we cannot factorize it anymore. This optimization is highly useful when we already have sieve generated up to $N$.
# Reference
1. forthright48 - Prime Number Theorem - https://forthright48.com/2015/07/prime-number-theorem.html
# Problem
Given an integer N, generate all primes less than or equal to N.
# Sieve of Eratosthenes – Explanation
Sieve of Eratosthenes is an algorithm that generates all prime up to N. Read this article written by Jane Alam Jan on Generating Primes – LightOJ Tutorial. The pdf contains almost all the optimizations of the Sieve of Eratosthenes.
# Code
vector <int> prime; // Stores generated primes
char sieve[SIZE]; // 0 means prime
void primeSieve ( int n ) {
sieve[0] = sieve[1] = 1; // 0 and 1 are not prime
prime.push_back(2); // Only Even Prime
for ( int i = 4; i <= n; i += 2 ) sieve[i] = 1; // Remove multiples of 2
int sqrtn = sqrt ( n );
for ( int i = 3; i <= sqrtn; i += 2 ) {
if ( sieve[i] == 0 ) {
for ( int j = i * i; j <= n; j += 2 * i ) sieve[j] = 1;
}
}
for ( int i = 3; i <= n; i += 2 ) if ( sieve[i] == 0 ) prime.push_back(i);
}
Some lines from the code above can be omitted depending on the situation. But as a whole, the above code gives us two products, a vector<int> prime which contains all generated primes and a char[] sieve that indicates whether a particular value is prime or not. We will need the sieve array for optimizations in other algorithms.
# Complexity
The algorithm has a runtime complexity of $O(N log (logN ) )$
# Problem
Given a number N, determine if it is a prime.
# What is a Prime Number?
A prime number is a positive number greater than 1 which has no positive divisor except for 1 and itself. For example, 2, 3, 5, 11 are prime numbers. Whereas 4, 6 are non-prime or composite numbers.
# O(N) Solution
From the definition, we can easily construct a Primality Test function that can determine if a given number N is prime or not. Let us call it isPrime() function.
bool isPrime ( int n ) {
if ( n <= 1 ) return false; // n needs to be greater than 1
for ( int i = 2; i < n; i++ ) {
// If it is possible to divide n with a number other than 1 and n, then it is not prime
if ( n % i == 0 ) return false;
}
return true; // Otherwise, this is prime
}
The code simply iterates over all values between 2 and N-1 and checks if it can divide N. It has a complexity of O(N).
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## Whole Numbers and Place Value
### Learning Outcomes
• Identify whole numbers from a list of numbers
• Identify counting numbers from a list of numbers
• Model whole numbers
• Identify the place value of a digit
## Identify Counting Numbers and Whole Numbers
Learning algebra is similar to learning a language. You start with a basic vocabulary and then add to it as you go along. You need to practice often until the vocabulary becomes easy to you. The more you use the vocabulary, the more familiar it becomes.
Algebra uses numbers and symbols to represent words and ideas. Let’s look at the numbers first. The most basic numbers used in algebra are those we use to count objects: $1,2,3,4,5,\dots$ and so on. These are called the counting numbers. The notation “…” is called an ellipsis, which is another way to show “and so on,” or that the pattern continues endlessly. Counting numbers are also called natural numbers.
### Counting Numbers or Natural Numbers
The counting numbers start with $1$ and continue.
$1,2,3,4,5\dots$
Counting numbers and whole numbers can be visualized on a number line as shown below.
The numbers on the number line increase from left to right, and decrease from right to left.
The point labeled $0$ is called the origin. The points are equally spaced to the right of $0$ and labeled with the counting numbers. When a number is paired with a point, it is called the coordinate of the point.
The discovery of the number zero was a big step in the history of mathematics. Including zero with the counting numbers gives a new set of numbers called the whole numbers.
### Whole Numbers
The whole numbers are the counting numbers and zero.
$0,1,2,3,4,5\ldots$
We stopped at $5$ when listing the first few counting numbers and whole numbers. We could have written more numbers if they were needed to make the patterns clear.
### example
Which of the following are
$1$. counting numbers?
$2$. whole numbers?
$0,\frac{1}{4},3,5.2,15,105$
Solution
1. The counting numbers start at $1$, so $0$ is not a counting number. The numbers $3,15,\text{and }105$ are all counting numbers.
2. Whole numbers are counting numbers and $0$. The numbers $0,3,15,\text{and }105$ are whole numbers.
The numbers $\frac{1}{4}$ and $5.2$ are neither counting numbers nor whole numbers. We will discuss these numbers later.
## Model Whole Numbers
Our number system is called a place value system because the value of a digit depends on its position, or place, in a number. The number $537$ has a different value than the number $735$. Even though they use the same digits, their value is different because of the different placement of the $3$ and the $7$ and the $5$.
Money gives us a familiar model of place value. Suppose a wallet contains three $100 bills, seven$10 bills, and four $1 bills. The amounts are summarized in the image below. How much money is in the wallet? Find the total value of each kind of bill, and then add to find the total. The wallet contains$374.
Base-10 blocks provide another way to model place value, as shown in the image below. The blocks can be used to represent hundreds, tens, and ones. Notice that the tens rod is made up of $10$ ones, and the hundreds square (also known as a “flat”) is made of $10$ tens, or $100$ ones.
The image below shows the number $138$ modeled with base-10 blocks.
We use place value notation to show the value of the number $138$.
Digit Place value Number Value Total value
$1$ hundreds $1$ $100$ $100\phantom{\rule{1 em}{0ex}}$
$3$ tens $3$ $10$ $30\phantom{\rule{1 em}{0ex}}$
$8$ ones $8$ $1$ $+\phantom{\rule{.5 em}{0ex}}8\phantom{\rule{1 em}{0ex}}$
$\text{Sum =}138\phantom{\rule{1 em}{0ex}}$
### example
Use place value notation to find the value of the number modeled by the base-10 blocks shown.
### try it
Use place value notation to find the value of the number modeled by the base-10 blocks shown.
## Identify the Place Value of a Digit
By looking at money and base-10 blocks, we saw that each place in a number has a different value. A place value chart is a useful way to summarize this information. The place values are separated into groups of three, called periods. The periods are ones, thousands, millions, billions, trillions, and so on. In a written number, commas separate the periods.
Just as with the $\text{base - 10}$ blocks, where the value of the tens rod is ten times the value of the ones block and the value of the hundreds square is ten times the tens rod, the value of each place in the place-value chart is ten times the value of the place to the right of it.
The chart below shows how the number $5,278,194$ is written in a place value chart.
• The digit $5$ is in the millions place. Its value is $5,000,000$.
• The digit $2$ is in the hundred thousands place. Its value is $200,000$.
• The digit $7$ is in the ten thousands place. Its value is $70,000$.
• The digit $8$ is in the thousands place. Its value is $8,000$.
• The digit $1$ is in the hundreds place. Its value is $100$.
• The digit $9$ is in the tens place. Its value is $90$.
• The digit $4$ is in the ones place. Its value is $4$.
### example
In the number $63,407,218$; find the place value of each of the following digits:
1. $7$
2. $0$
3. $1$
4. $6$
5. $3$
### try it
The video below shows more examples of how to determine the place value of a digit in a number.
## Contribute!
Did you have an idea for improving this content? We’d love your input.
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Welcome to our free-to-use Q&A hub, where students post questions and get help from other students and tutors.
You can ask your own question or look at similar Maths questions.
119x =9
In this equation for evaluating the value of 'x' you just need to divide 9 with 119
x=9/119
x=0.075630
119x=9
X=9/119
0.07563025210
We want x to be by itself. x is being multiplied by 119. You start by dividing both sides of the equation by 119. This results in 119x/119=9/119. The 119/119 cross out, so we have x=9/119 which is approximately 0.0756.
119x = 9
Basically you have left side of equation(LHS) which is "119x" and right side of equation(RHS) "9"
Now X=9/119 = 0.0756, So X= 0.0756 no verification as below
LHS = RHS
119 * 0.0756 = 8.99 which can be round off to 9.0
X needs to be separated into one side, it’s currently multiplied by 119 so we need to divide both sides by 119, as shown below.
X = 9/119
X= 0.0756
X needs to be separated into one side, it’s currently multiplied by 119 so we need to divide both sides by 119, as shown below.
X = 9/119
X= 0.0756
X=9/119
0.0756
119x=9
Find x
X=9/119
X=0.0756
In significant figure is 0.076 value for x Answer
Divide both sides by 119 so x = 9/119
119x=9
x=9/119
x=0.07563
119x=9
moving 119 to the right side of the equation,
x=9/119 = 0.0756302521
Ans = 0.0756 (3 s.f.)
Divide both sides of the equation by 119 to get x by itself. x=9/119.
the left part is 119x
If you want the constant value beside x ,it will be a reciprocal
so x =9/119
which is equal to 0.07563
119x=9
x=9/119
x=0.07563
You need to separate x by dividing both sides of the equation by the coefficient of x, which in this case is 119.
(119x)/119 = 9/119
x = 9/119
= 0.0756302521
X = 9/119
This would be:
x = 9/119
= 0.075 or
= 75/1000 = 3/40
Steps,
(119x)/119 = 9/119 (assume x is not equal to zero)
x=9/119
x=0.0756302521
you need to find out what only 1x= , at the minute it is 119x=9 so you need to divide by 119 on both sides of the equal sign. this produces x=9/119 which you can put into your calculator as 9 divided by 199 and you will get a value for 1x=?
119x = 9
(119x)/119 = 9/119
x = 9/119
119x = 9
x = 9/119
x = 0.0756302521
• you need to divide both sides by 119
• the correct answer is x=9/119
119x=9
so, simply x=9/119
and x=0.0756
119x = 9
First we divide on both sides by 119 to make x alone
119/119 x = 9/119
x = 0.076
x=9/119. The factors of 119 are 1,7,17 and 119 and the factors of 9 are 1,3 and 9, which means 9 and 119 are prime to each other, therefore you cannot simplify this fraction any further. If you want to give the answer as a decimal, 9/119 is approximately 0.0756.
119x = 9
it means 119* x = 9
so, x= 9/119
x= 0.0756302521 approx.
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# How to find Standard Deviation of a Discrete Random Variable (with Examples)
-
The standard deviation of a discrete random variable can be found by taking the square root of its variance.
Variance = E(X2) – E(X)2.
• E(X2) = Expected value of the square of the random variable.
• E(X)2 = Square of the expected value of the random variable.
The variance of a discrete random variable can be found by calculating the expected value of the square of the random variable and subtracting the square of the expected value of the random variable from it.
### Standard Deviation Formula (for a discrete random variable):
Therefore the variance and the standard deviation for the discrete random variable can be calculated using the formula,
Variance = E(X2) – E(X)2.
Standard Deviation = √Variance = √(E(X2) – E(X)2).
Suppose that a discrete random variable X has probability mass function p(x) then the required expected values can be calculated as
E(X) = Σx*p(x).
E(X2) = Σx2*p(x).
This means that in order to find the expected value of a particular random variable we should multiply the value of the random variable with the corresponding probability and take the sum over all possible values of x.
Let us try to understand how we can calculate the standard deviation for a discrete random variable by looking at the following two examples.
#### Example 1:
Consider a discrete random variable X whose probability mass function p(x) is given as
In order to calculate the standard deviation we first compute the required expected values as follows,
From the above table, we now have that,
E(X) = Σx*p(x) = 2.45.
E(X2) = Σx2*p(x) = 8.45.
Standard deviation = √(E(X2) – E(X)2) = √(8.45 – 2.452)
= √2.4475 = 1.5644.
### Example 2:
It is also possible that a discrete random variable takes negative values.
For example, consider a discrete random variable X whose probability mass function p(x) is given as,
We find the standard deviation as follows,
E(X) = Σx*p(x) = 0.5.
E(X2) = Σx2*p(x) = 1.9.
Standard deviation = √(E(X2) – E(X)2) = √(1.9 – 0.52) = 1.2845
Summary
Article Name
How to find Standard Deviation of a Discrete Random Variable (with Examples)
Description
The standard deviation of a discrete random variable can be found by taking the square root of its variance. Variance = E(X2) - E(X)2.
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Unseen Passage
For Class 4 to Class 12
Arithmetic Progression MCQ Class 10 Mathematics
Please refer to Chapter 5 Arithmetic Progression MCQ Class 10 Mathematics with answers below. These multiple-choice questions have been prepared based on the latest NCERT book for Class 10 Mathematics. Students should refer to MCQ Questions for Class 10 Mathematics with Answers to score more marks in Grade 10 Mathematics exams. Students should read the chapter Arithmetic Progression and then attempt the following objective questions.
MCQ Questions Class 10 Mathematics Chapter 5 Arithmetic Progression
The Arithmetic Progression MCQ Class 10 Mathematics provided below covers all important topics given in this chapter. These MCQs will help you to properly prepare for exams.
Question. The first term of an A.P. is 5 and its 100th term is –292. The 50th term of this A.P. will be
(a) 142
(b) –142
(c) 130
(d) –140
B
Question. If eight times the 8th term of an A.P. is equal to 12 times the 12th term of the A.P. then its 20th term will be
(a) –1
(b) 1
(c) 0
(d) 2
C
Question. If a, b, c are in A.P., then the value of (a + 2b – c) (2b + c – a) (c + a – b) will be
(a) 4abc
(b) 2abc
(c) abc
(d) None of these
A
Question. Given that the sum of the first ‘n’ terms of an arithmetic progression is 2n2 + 3n, find the 12th term.
(a) 72
(b) 36
(c) √625
(d) 56
A
Question. The common difference of the A.P. 1/P,1-P/P,1-2P/P, ……… is
(a) 1
(b) 1/ p
(c) –1
(d) − 1/ p
C
Question. First term of an arithmetic progression is 2. If the sum of its first five terms is equal to one-fourth of the sum of the next five terms, then the sum of its first 30 terms is
(a) 2670
(b) 2610
(c) –2520
(d) –2550
D
Question. The nth term of the A.P. a, 3a, 5a, ……., is
(a) na
(b) (2n – 1)a
(c) (2n + 1)a
(d) 2na
B
Question. If the sum of the series 2 + 5 + 8 + 11 ……….. is 60100, then the number of terms are
(a) 100
(b) 200
(c) 150
(d) 250
B
Question. In an A.P. if a = 5, an = 81 and Sn = 860, then n is
(a) 10
(b) 15
(c) 20
(d) 25
C
Question. If the nth term of an A.P. is given by an = 5n – 3, then the sum of first 10 terms is
(a) 225
(b) 245
(c) 255
(d) 270
B
Question. What is the value of k if (k + 2), (4k – 6) and (3k – 2) are three consecutive terms of an A.P.?
(a) k = –3
(b) k = 2
(c) k = –2
(d) k = 3
D
Question. What is the common difference of four terms in A.P. such that the ratio of the product of the first and fourth term to that of the second and third term is 2 : 3 and the sum of all four terms is 20 ?
(a) 3
(b) 1
(c) 4
(d) 2
D
Question. The 10th term of an AP is 20 and the 19th term is 101. Then, the third term is
(a) 43
(b) – 61
(c) – 52
(d) 1
A
Question. There are 60 terms in an A.P. of which the first term is 8 and the last term is 185. The 31st term is
(a) 56
(b) 94
(c) 85
(d) 98
D
Question. There are four arithmetic means between 2 and –18. The means are
(a) –4, –7, –10, –13
(b) 1, –4, –7, –10
(c) –2, –5, –9, –13
(d) –2, –6, –10, –14
D
Question. If the nth term of an A.P. is 4n + 1, then the common difference is :
(a) 3
(b) 4
(c) 5
(d) 6
B
Question. If a, b, c, d, e, f are in A.P., then e – c is equal to:
(a) 2(c – a)
(b) 2(d – c)
(c) 2(f – d)
(d) (d – c)
B
Question. If the sum of the first 2n terms of 2, 5, 8, ……. is equal to the sum of the first n terms of 57, 59, 61……., then n is equal to
(a) 10
(b) 12
(c) 11
(d) 13
C
Question. The sum of 11 terms of an A.P. whose middle term is 30, is
(a) 320
(b) 330
(c) 340
(d) 350
B
Question. The number of terms of the series 5, 7, 9, …. that must be taken in order to have the sum 1020 is
(a) 20
(b) 30
(c) 40
(d) 50
B
Question. The number of two digit numbers which are divisible by 3 is
(a) 33
(b) 31
(c) 30
(d) 29
C
Question. Suppose the sum of the first m terms of an arithmetic progression is n and the sum of its first n terms is m, where m ≠ n. Then, the sum of the first (m + n) terms of the arithmetic progression is
(a) 1 – mn
(b) mn – 5
(c) – (m + n)
(d) m + n
C
Question. The odd natural numbers have been divided in groups as (1, 3) ; (5,7, 9, 11) ; (13, 15, 17, 19, 21, 23), ….. Then the sum of numbers in the 10th group is
(a) 4000
(b) 4003
(c) 4007
(d) 4008
A
Question. If tn = 6n + 5, then tn+1 =
(a) 6(n + 1) + 17
(b) 6(n – 1) + 11
(c) 6n + 11
(d) 6n – 11
C
Question. If nth term of an AP is given by an = 2n + 3 then common difference of an AP is
(a) 2
(b) 3
(c) 5
(d) 1
A
Question. Which of the following represents an A.P. ?
(a) 0.2, 0.4, 0.6, ….
(b) 29, 58, 116….
(c) 15, 45, 135, 405…
(d) 3, 3.5, 4.5, 8.5 ….
A
Question. The value of x, for which 2x, x+ 10, 3x + 2 are three consecutive terms of an AP
(a) 6
(b) –6
(c) 18
(d) –18
A
Question. Sn = 54 + 51 + 48 + …….. n terms = 513. Least value of n is
(a) 18
(b) 19
(c) 15
(d) None of these
A
Question. Which of the following terms are in AP for the given situation
(a) 51,53,55….
(b) 51, 49, 47….
(c) –51, –53, –55….
(d) 51, 55, 59…
B
Question. The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is l2-a 2/k- (l+a ) , then k is equal to
(a) S
(b) 2S
(c) 3S
(d) None of these
B
Question. If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are
(a) 5, 10, 15, 20
(b) 4, 10, 16, 22
(c) 3, 7, 11, 15
(d) None of these
A
Question. Which of the following term is not in the AP of the above given situation
(a) 41
(b) 30
(c) 37
(d) 39
B
Question. Let Tr be the rth term of an A.P. for r = 1, 2, 3, …. If for some positive integers m, n we have, Tm =1/n and Tn =1/m , then Tmn equals
(a) 1/ mn
(b) 1/m+1/n
(c) 1
(d) 0
C
Question. What is the minimum number of days he needs to practice till his goal is achieved
(a) 10
(b) 12
(c) 11
(d) 9
C
Question. The number of common terms of the two sequences 17, 21, 25, ….., 417 and 16, 21, 26, …….., 466 is
(a) 19
(b) 20
(c) 21
(d) 91
B
Fill in the Blanks :
Question. The difference of corresponding terms of two A.P’s will be ……………….. .
another A.P.
Question. Sum of 1 + 3 + 5 + …. + 1999 is ……… .
1000/2 [2(1) (1000 1)2]
Question. 1, –1, –3, – 5, ……. , ……… .
–7, –9
Question. The sum of 8 A.Ms between 3 and 15 is ………………. .
72[8(3+15/2)etc.]
Question. The sum of n terms of an A.P. is 4n2 – n. The common difference = ……………… .
11 [S2 = 4(2)2 – 2 ⇒ 14, S1 = 4(1)2 – 1 ⇒ 3 etc.]
Question. 4, 10, 16, 22, …….., ………. .
28, 34
Question. 11th term from last term of an A.P. 10, 7, 4……… , – 62, is ……… .
–32
Question. In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. Number of rows in the flower bed is …………… .
n = 10
Question. Sum of all the integers between 100 and 1000 which are divisible by 7 is ………….
70336
True or False :
Question. If Sn of A.P. is 3n2 + 2n, then the first term of A.P. is 3.
False
Question. If an+1 – an = same for all ‘n’, then the given numbers form an A.P.
True
Question. 2, 4, 8, 16, …………. is not an A.P.
True
Question. 10th term of A.P. 2, 7, 12, ……… is 45.
False
Question. The balance money ( in ₹`) after paying 5% of the total loan of ` 1000 every month is 950, 900, 850, 800, . . . 50. represented A.P.
True
Question. The general form of an A.P. is a, a + d, a + 2d, a + 3d, ………….
True
Question. In an AP with first term a and common difference d, the nth term (or the general term) is given by an = a + (n – 1)d.
True
Question. If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by : S= n/2(a +L)
True
Question. In an Arithmetic progression, the first term is denoted by ‘a’ and ‘d ’ is called the common difference.
True
Question. 301 is a term of A.P. 5, 11, 17, 23, …………. .
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# Let T - V rightarrow V be a linear transformation sulci that T2 -.docx
Let T : V rightarrow V be a linear transformation sulci that T2 = 0 is the zero transformation. Show that itn T ker T If R: V rightarrow V is defined by R(v) = v + T(v) for all v in V. show that R is linear and invertible.
Solution
(a)
Let x be in im(T). Then x = T(v) for some v in V. Now consider T(x):
T(x) = T(T(v))
= T 2 (v)
= 0
Since T(x) = 0, x must be in ker(T). This was true for a generic x in im(T), so we must have im(T) contained in ker(T).
(b)
R(v) = I(v) + T(v), where I is the identity transformation (which is linear). Since R is the sum of two linear transformations, it must itself be linear as well. If you want to write the proof out, it would look like this:
Let v,w be in V, and k be a scalar.
R(k*v+w) = (k*v+w) + T(k*v+w)
= k*v + w + k*T(v) + T(w) [since T is linear]
= k*[v+T(v)] + [w+T(w)]
= k*R(v)+R(w)
Therefore R is linear.
Suppose R(v) = 0
0 = R(v)
= v + T(v)
T(v) = -v
We know that T(T(v)) = 0
From the above, we have that T(v)=-v. Therefore we have:
T(-v) = 0
-T(v) = 0 [T is linear]
Once again we know that T(v)=-v. Therefore we have:
-(-v) = 0
v = 0
Therefore R(v)=0 --> v=0
Since R is linear, this means R is invertible.
.
Let T : V rightarrow V be a linear transformation sulci that T2 = 0 is the zero transformation. Show that itn T ker T If R: V rightarrow V is defined by R(v) = v + T(v) for all v in V. show that R is linear and invertible.
Solution
(a)
Let x be in im(T). Then x = T(v) for some v in V. Now consider T(x):
T(x) = T(T(v))
= T 2 (v)
= 0
Since T(x) = 0, x must be in ker(T). This was true for a generic x in im(T), so we must have im(T) contained in ker(T).
(b)
R(v) = I(v) + T(v), where I is the identity transformation (which is linear). Since R is the sum of two linear transformations, it must itself be linear as well. If you want to write the proof out, it would look like this:
Let v,w be in V, and k be a scalar.
R(k*v+w) = (k*v+w) + T(k*v+w)
= k*v + w + k*T(v) + T(w) [since T is linear]
= k*[v+T(v)] + [w+T(w)]
= k*R(v)+R(w)
Therefore R is linear.
Suppose R(v) = 0
0 = R(v)
= v + T(v)
T(v) = -v
We know that T(T(v)) = 0
From the above, we have that T(v)=-v. Therefore we have:
T(-v) = 0
-T(v) = 0 [T is linear]
Once again we know that T(v)=-v. Therefore we have:
-(-v) = 0
v = 0
Therefore R(v)=0 --> v=0
Since R is linear, this means R is invertible.
.
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# Question #a078d
Dec 14, 2017
see explanation
#### Explanation:
A: circumference formula: $c = 2 \pi r = \pi d$
therefore, $\pi = \frac{c}{d}$
so, for circle A, you get: $\frac{21.98}{7} = 3.14$
for circle B, you get: $\frac{18.84}{6} = 3.14$
B: area formula: $a = \pi {r}^{2}$, where $r = \frac{d}{2}$
therefore, $\pi = \frac{a}{r} ^ 2$
for circle A, you get: $\frac{38.465}{{3.5}^{2}} = 3.14$
for circle B, you get: $\frac{28.26}{{3}^{2}} = 3.14$
C: You observe that the value of pi is the same in all cases. The value of pi is a CONSTANT.
GOOD LUCK
Dec 14, 2017
$\pi = 3.14$
#### Explanation:
$C = \pi D = 2 \pi r$, where C is the circumference, D is the diameter, and r is the radius.
$A = \pi {r}^{2}$ where A is the area and r is the radius.
a)
We need to find a value of $\pi$ using the circumference formula.
$C = \pi D$
$\pi = \frac{C}{D}$
For Circle A:
$\pi = \frac{21.98}{7}$
$\pi = 3.14$
For Circle B:
$\pi = \frac{18.84}{6}$
$\pi = 3.14$
b)
$A = \pi {r}^{2}$
We aren't given the radius of the circle in each case, but the diameter is double the radius, $D = 2 r$ or $r = \frac{D}{2}$.
Substituting into our area formula we get:
$A = \pi {\left(\frac{D}{2}\right)}^{2}$
$A = \pi {D}^{2} / 4$
$A = \frac{1}{4} \pi {D}^{2}$
Rearranging for $\pi$:
$4 A = \pi {D}^{2}$
$\pi = \frac{4 A}{D} ^ 2$
For circle A:
$\pi = \frac{4 \times 38.465}{{7}^{2}}$
$\pi = 3.14$
For circle B:
$\pi = \frac{4 \times 28.26}{6} ^ 2$
And no prizes for guessing that this evaluates to:
$\pi = \frac{3}{14}$
So what do we notice?
This is going to go into more detail than you need for your homework, although you may wish to read it; I personally find this stuff quite interesting.
$\pi = 3.14$
$\pi = 3.14$
$\pi = 3.14$
$\pi = 3.14$
In fact, $\pi$ takes the same value in all contexts. $\pi$ is a constant, which is approximately equal to
$\pi \approx 3.141592653589793238462643383279502884 \ldots$
No wonder we often round it to $3.14$
$\pi$ is itself defined as the ratio of a circle's diameter to its radius. It has no exact value; the decimal of pi goes on forever. This is different however to the infite decimal of $\frac{1}{3}$.
$\frac{1}{3} = 0.333333333333333 \ldots = 0. \dot{3}$
This decimal will repeat. So too will any fraction you can come up with. $\pi$ is different; it never reccurs, and it cannot be written as a fraction. Since it cannot be written as a fraction, $\pi$ is an irrational number. Other irrational numbers include irrational square roots, such as $\sqrt{2}$. $\pi$ is different to other irrational numbers, however. With $\sqrt{2}$, we can use the hypoteneuse of a $\text{1-1-} \sqrt{2}$-right-triangle to plot the value of this number on a number line. We cannot do anything like this with $\pi$, so as well as being irrational, pi is a transcendental number.
Both of these mean we can never actually describe pi accurately as a decimal. In fact, if we wanted to be accurate, we would leave the answers in terms on $\pi$. Circle A has a circumference of exactly $7 \pi$ inches (who uses inches in this day and age anyway? regardless) and an area of $\frac{49}{4} \pi$ square inches. I mean, we could use pi to a million decimal places if we wanted, but this wouldnt be as accurate as leaving an answer it terms of pi. The same applies to how you leave answers as square roots.
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# Exploring The 8 1 Geometric Mean Worksheet
## Introduction
As we progress towards a more advanced level of mathematics, we are introduced to a wide range of concepts that can be challenging to comprehend. One such concept is the geometric mean, which is an essential component of many mathematical calculations. In this article, we will explore the 8 1 geometric mean worksheet and how it can help us understand this concept better.
## What is the Geometric Mean?
Before we dive into the worksheet, let us first understand what the geometric mean is. In simple terms, the geometric mean is the average of a set of numbers, calculated by multiplying them together and then taking the nth root of the product, where n is the number of values in the set. For example, the geometric mean of 2, 4, and 8 is 4, as (2 x 4 x 8)^(1/3) = 4.
### How to Use the 8 1 Geometric Mean Worksheet
The 8 1 geometric mean worksheet is a tool that can help you practice and understand the concept of geometric mean better. The worksheet comprises 8 sets of numbers, each containing 3 values. Your task is to find the geometric mean of each set and then calculate the average of all the geometric means.
### Step-by-Step Guide
To use the worksheet, follow these steps: 1. Write down the numbers in each set. 2. Multiply the numbers in each set together. 3. Take the cube root of the product. 4. Write down the geometric mean for each set. 5. Add up all the geometric means. 6. Divide the sum by 8.
## Tips for Solving the Worksheet
Here are some tips that can help you solve the 8 1 geometric mean worksheet: – Remember that the geometric mean is always positive, so if you get a negative value, you have made a mistake. – Use a calculator to simplify your calculations. – Make sure you take the cube root of the product, not the square root. – Double-check your answers before moving on to the next set.
## Why is the Geometric Mean Important?
The geometric mean is an important concept in mathematics, especially in finance and statistics. It is used to calculate averages of ratios and rates, which are common in these fields. For example, the geometric mean is used to calculate the average rate of return on an investment over a period of time.
## Conclusion
The 8 1 geometric mean worksheet is an excellent tool to help you practice and understand the concept of geometric mean. By using this worksheet, you can improve your skills in this area and apply it to various mathematical calculations. Remember to follow the steps carefully and use the tips provided to ensure accurate results. With practice, you will soon master the geometric mean and be able to apply it to real-life situations.
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#### Find the vector equation of the line parallel to the vector $3\hat{i}-2 \hat{j}+3\hat{k}$ and which passes through the point (1, -2, 3).
Given, vector = $3\hat{i}-2\hat{j}+6\hat{k}$
Point = (1, -2, 3)
We can write this point in vector form as $\hat{i}-2\hat{j}+3\hat{k}$
Let ,
$\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$
$\overrightarrow{b}=3\hat{i}-2\hat{j}+6\hat{k}$
We must find the vector equation of the line parallel to the vector $\overrightarrow{b}$ and passing through the point
We know, equation of $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ a line passing through a point and parallel to a given vector is denoted as
Where, $\lambda \epsilon \mathbb{R}$
In other words, we need to find $\overrightarrow{r}$
This can be achieved by substituting the values of the vectors in the above equation. We get
$\overrightarrow{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$
$\Rightarrow \overrightarrow{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$
This can be further rearranged, upon which we get:
$\Rightarrow \vec{r}=\hat{i}-2\hat{j}+3\hat{k}+3\lambda \hat{j}+3\hat{k}+6\lambda\hat{k}$
$\Rightarrow \vec{r}=\hat{i}+3\lambda \hat{i}-2\hat{j}-2\lambda\hat{j}+3\hat{k}+6\lambda\hat{k}$
$\Rightarrow \vec{r}=\left ( 1-3\lambda \right )\hat{i}+\left ( -2-2\lambda \right )\hat{j}+\left ( 3+6\lambda \right )\hat{k}$
Thus, the require vector equation of line is $\vec{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$
which can also be written as $\left ( 1-3\lambda \right )i+\left ( -2-2 \lambda \right )\hat{j}+\left ( 3+6\lambda \right )\hat{k}$
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# Relationship between H.C.F. and L.C.M.
We will learn the relationship between H.C.F. and L.C.M. of two numbers.
First we need to find the highest common factor (H.C.F.) of 15 and 18 which is 3.
Then we need to find the lowest common multiple (L.C.M.) of 15 and 18 which is 90.
H.C.F. × L.C.M. = 3 × 90 = 270
Also the product of numbers = 15 × 18 = 270
Therefore, product of H.C.F. and L.C.M. of 15 and 18 = product of 15 and 18.
Again, let us consider the two numbers 16 and 24
Prime factors of 16 and 24 are:
16 = 2 × 2 × 2 × 2
24 = 2 × 2 × 2 × 3
L.C.M. of 16 and 24 is 48;
H.C.F. of 16 and 24 is 8;
L.C.M. × H.C.F. = 48 × 8 = 384
Product of numbers = 16 × 24 = 384
So, from the above explanations we conclude that the product of highest common factor (H.C.F.) and lowest common multiple (L.C.M.) of two numbers is equal to the product of two numbers
or, H.C.F. × L.C.M. = First number × Second number
or, L.C.M. = $$\frac{\textrm{First Number} \times \textrm{Second Number}}{\textrm{H.C.F.}}$$
or, L.C.M. × H.C.F. = Product of two given numbers
or, L.C.M. = $$\frac{\textrm{Product of Two Given Numbers}}{\textrm{H.C.F.}}$$
or, H.C.F. = $$\frac{\textrm{Product of Two Given Numbers}}{\textrm{L.C.M.}}$$
Solved examples on the relationship between H.C.F. and L.C.M.:
1. Find the L.C.M. of 1683 and 1584.
Solution:
First we find highest common factor of 1683 and 1584
Therefore, highest common factor of 1683 and 1584 = 99
Lowest common multiple of 1683 and 1584 = First number × Second number/ H.C.F.
= $$\frac{1584 × 1683}{99}$$
= 26928
2. Highest common factor and lowest common multiple of two numbers are 18 and 1782 respectively. One number is 162, find the other.
Solution:
We know, H.C.F. × L.C.M. = First number × Second number then we get,
18 × 1782 = 162 × Second number
$$\frac{18 × 1782}{162}$$ = Second number
Therefore, the second number = 198
3. The HCF of two numbers is 3 and their LCM is 54. If one of the numbers is 27, find the other number.
Solution:
HCF × LCM = Product of two numbers
3 × 54 = 27 × second number
Second number = $$\frac{3 × 54}{27}$$
Second number = 6
4. The highest common factor and the lowest common multiple of two numbers are 825 and 25 respectively. If one of the two numbers is 275, find the other number.
Solution:
We know, H.C.F. × L.C.M. = First number × Second number then we get,
825 × 25 = 275 × Second number
$$\frac{825 × 25}{275}$$ = Second number
Therefore, the second number = 75
5. Find the H.C.F. and L.C.M. of 36 and 48.
Solution:
H.C.F. = 2 × 2 × 3 = 12L.C.M. = 2 × 2 × 3 × 3 × 4 = 144H.C.F. × L.C.M. = 12 × 144 = 1728Product of the numbers = 36 × 48 = 1728
Therefore, product of the two numbers = H.C.F × L.C.M.
2. The H.C.F. of two numbers 30 and 42 is 6. Find the L.C.M.
Solution:
We have H.C.F. × L.C.M. = product of the numbers
6 × L.C.M. = 30 × 42
L.C.M. = $\frac{30 × 42}{\sqrt{6}}$
= $\frac{1260}{\sqrt{6}}$
= 210
3. Find the greatest number which divides 105 and 180 completely.
Solution:
The greatest number here is the H.C.F of 105 and 180Common factors are 5, 3H.C.F. = 5 × 3 = 15
Therefore, the greatest number that divides 105 and 180 completely is 15.
4. Find the least number which leaves 3 as remainder when divided by 24 and 42.
Solution:
L.C.M. of 24 and 42 leaves no remainder when divided by the number 24 and 42.L.C.M. = 2 × 3 × 4 × 7 = 168
The least number which leaves 3 as remainder is 168 + 3 = 171.
Important Notes:
Two numbers which have only 1 as the common factor are called co-prime.
The least common multiple (L.C.M.) of two or more numbers is the smallest number which is divisible by all the numbers.
If two numbers are co-prime, their L.C.M. is the product of the numbers.
If one number is the multiple of the other, then the multiple is their L.C.M.
L.C.M. of two or more numbers cannot be less than any one of the given numbers.
H.C.F. of two or more numbers is the highest number that can divide the numbers without leaving any remainder.
If one number is a factor of the second number then the smaller number is the H.C.F. of the two given numbers.
The product of L.C.M. and H.C.F. of two numbers is equal to the product of the two given numbers.
Questions and Answers on Relationship between H.C.F. and L.C.M.
1. The H.C.F. of two numbers 20 and 75 is 5. Find their L.C.M.
2. The L.C.M. of two numbers 72 and 180 is 360. Find their H.C.F.
3. The L.C.M. of two numbers is 120. If the product of the numbers is 1440, find the H.C.F.
4. Find the least number which leaves 5 as remainder when divided by 36 and 54.
5. The product of two numbers is 384. If their H.C.F. is 8, find the L.C.M.
1. 300
2. 36
3. 12
4. 113
5. 48
To Find Lowest Common Multiple by using Division Method
Relationship between H.C.F. and L.C.M.
Worksheet on H.C.F. and L.C.M.
Word problems on H.C.F. and L.C.M.
Worksheet on word problems on H.C.F. and L.C.M.
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Difference between Distance and Displacement - Practice problems, FAQs
Karthick goes to an amusement park and finds a Jolly Rancher ride which can go along two different paths as shown in the figure. One along the zig-zag path, and the other along the straight line AB. He finds that the rancher ride takes a longer time to reach point B from point A, if it goes along the zig -zag curve. On the other hand, it takes much less time to reach B if it traveled straight. So what exactly is the difference? Let us take another example of a guy standing behind many obstacles. He has to traverse a lengthy path to negotiate the obstacles before reaching the other end. On the other hand, if there were no obstacles, he would simply walk straight till the other end, reaching quicker compared to the previous case. In both these examples, the direction in which we travel matters a lot. In the case of the Jolly Rancher ride, the zig-zag path relates with distance; on the other hand, the straight path which is shorter than the previous and actually the shortest path is called the displacement. In this article, we will explore the difference between distance and displacement in detail.
• Position vector
• Distance
• Displacement
• Difference between distance and displacement
• Practice problems
• FAQs
Position vector
A vector which is needed to define the position of an object in space is called the position vector. For instance, in the following diagram, the position of the object P can be written in cartesian coordinates as P(x,y,z). In vector form, its position can be written as $\stackrel{\to }{r}=x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}.$
Its distance from the origin can be calculated as $|\stackrel{\to }{r}|=\sqrt{{\left(x\right)}^{2}+{\left(y\right)}^{2}+{\left(z\right)}^{2}}.$
Distance
The total path length traveled by an object in a given interval of time is called its distance. It is a scalar quantity. Let us consider the following example where an object starts from A, goes to C, and then point B and back to point A. The total distance traveled in this case would be AC+BC+AB. It’s unit is meter(m). It carries a dimensional formula of length [L].
Displacement
The displacement of a particle is defined as the shortest distance that connects two points. Alternatively, it can also be defined as the change in position vector undergone by a particle in a given time interval. In the following example, a particle goes from point C to point B, and then to point A. Its displacement is the shortest distance connecting A and C, i.e. CA.
Consider a particle moving in a circular path. The radius of the circle is OA=r. When the particle completes a full circle, going from A to B and B to A, then its displacement is zero but the distance it traverses is 2r which is the perimeter of the circle. On the other hand, when the particle moves from A to B, in a half circle, the distance covered is r, but its displacement is AB where |AB|=2r
Note:
Displacement is a vector quantity. It carries the same unit as that of distance (meter).
Difference between distance and displacement
The following table gives three differences between distance and displacement.
Distance Displacement Distance is the total path length traveled by an object. It is the shortest distance that connects the endpoints the particle has traveled It is a scalar quantity It is a vector quantity Distance is used to find the value of speed. $\left(Speed=\frac{Distance}{time}\right)$ Displacement is used to find the value of velocity. $\left(velocity=\frac{displacement}{time}\right)$ It is denoted by 'd'. It is denoted by S.
Video explanation
Practice problems
Q. A scooty is moving along a straight line AB covering a distance of 360 m. It then returns back to point C covering a distance of 240 m. Calculate the total distance traveled by the scooty.
(a)100 m (b)200 m (c) 300 m (d)600 m
A. d
The total distance covered by the scooty =AB+BC=360+240=600 m.
Q. A revolving wheel takes 2000 revolutions to cover a distance of 9.5 km. Find the diameter of the wheel.
(a)1 m (b) 1.5 m (c) 3.5 m (d)4.5 m
A. b
Given,
Total distance covered =9.5 km=9500 m.
The wheel covers a distance equal to its perimeter i.e., 2r in one revolution.
The total distance covered would be,
Here, D=2 r indicates the diameter of the wheel.
Q. A fighter jet flies 400 m North and 300 m south. It then flies 1200 m upwards. The net displacement would be
(a)1200 m (b)1300 m (c) 1400 m (d) 1500 m
A. a
Since the jet flies 400 m North, 300 m South, the net displacement would be
400 m- 300 m=100 m towards the north.
It then flies upwards , the net displacement in this case would be
Q. The final and initial positions of a particle are given by r2=(i+2j ) unit and r1=(3i+5j) unit respectively. The displacement of the particle is
(a)(2i+3j) unit (b)(-2i-3j) unit (c) (+2i-3j) unit (d) (4i+7j) unit
A. b
Given, final position of the particle, r2=(i+2j) unit
Initial position of the particle r1=(3i+5j) unit
Displacement= change in position vector
FAQs
Q. Can the magnitude of displacement be greater than distance? Why or why not?
A.
No. The magnitude of displacement is always less than or equal to the distance. Distance is the total path length traveled, while displacement signifies the shortest distance with direction.
Q. Can displacement be zero when the distance is not zero?
A.
Yes, when a body completes a journey such that its initial and final position is the same, eg. a body completes a full revolution in a circle, its displacement is zero, while its distance is non-zero.
Q. Can the distance be negative?
A.
No, the distance cannot be negative, because it does not give information about whether a body is traveling in the +ve or -ve direction.
Q. How do you determine distance and displacement?
A.
Distance is determined by calculating the total path length covered. On the other hand, displacement is found out by calculating the shortest distance connecting the two end points.
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combinatorics_day_3.pdf - 21-301 Combinatorics Spring 2019 Lecture 3 Trees Lecturer John Mackey 1 Date Review Hamiltonian Circuit A simple closed path
# combinatorics_day_3.pdf - 21-301 Combinatorics Spring 2019...
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21-301: Combinatorics Spring 2019 Lecture 3: Trees Lecturer: John Mackey Date: January 18, 2019 1 Review Hamiltonian Circuit A simple closed path that passes through each vertex in the graph exactly once. A graph is Hamiltonian iff it has a polygon as a spanning subgraph. Consider the two graphs below: The graph on the left is the dodecahedron, which has a spanning polygon and is Hamiltonian. The graph on the right is called the Peterson graph, which does not have a spanning polygon and is not Hamiltonian. 2 Labeled Trees For all following sections, these definitions will be helpful: [n] The set of all positive integers up to and including n . [n] n - 2 The set of sequences of length n - 2, where the numbers in each sequence are chosen from [ n ]. Theorem 2.1. There are n n - 2 different labeled trees on n vertices. To help illustrate this theorem, consider labeling a 3 vertex graph with [3]. The only nonisomorphic tree with 3 vertices is that of a straight line, and we can get a new labeling by switching the label on the center vertex: 1
Lecture 3: Trees 2 There are 3 = 3 3 - 2 ways to do this. Therefore our theorem holds for trees with 3 vertices. We can show that this theorem works for trees with 4 vertices as well. There are two such nonisomorphic trees: There are 12 ways to label the first tree and 4 ways to label the second. Therefore there are 16 = 4 4 - 2 ways to label a tree with 4 vertices. There are three nonisomorphic trees on 5 vertices: There are 60 ways to label the first tree, 5 ways to label the second, and 60 ways to label the third. Therefore there are 125 = 5 5 - 2 ways to label a tree with 5 vertices. 3 Pr¨ufer Codes The proof of Theorem 2.1 relies on an algorithm developed by H. Pr¨ufer that can assign any tree T a unique name in [ n ] n - 2 that characterizes the tree. This is called the tree’s Pr¨ufer code.
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## Preschool Math Worksheets
Since there is no set preschool curriculum, at least in the eyes of the Core Curriculum Initiative, we improvised a little. We worked with several past national and every state draft of preschool curriculum that we could get my hands on. All the work here is aligned to the Core as precursory skills for achieving kindergarten standards. We also have Preschool Math Posters for your class walls.
### Precursory Skills For Counting
Counting Objects 1-5 - K.CC.A.1
Starting to Count - K.CC.A.1
1-10 Connect-the-Dot Problems - K.CC.A.2
Days of the Week- K.CC.A.2
Writing Numbers 1-9 - K.CC.A.3
Learning to Subtract - K.OA.A.3
Color 10 Objects in a Set- K.CC.B.4a
Counting through Drawing- K.CC.B.4b
Color Matching- K.G.B.5
1 to Many Group Matching - K.CC.B.5
Identifying Groups of 6 to 10 Numbers - K.CC.B.5
Working With Tallies - K.CC.B.5
Counting Money - K.CC.B.5
Finding Less in Sets - K.CC.C.6
Comparing 2 Objects - K.CC.C.6
### Precursory Skills to Measurement
Halves; Pre-Symmetry - K.MD.A.1
Long and Short - K.MD.A.1
Find the Biggest Object - K.MD.A.2
How Tall Is It? - K.MD.A.2
Spot All the Differences - K.MD.A.2
More or Less Items and Objects - K.MD.A.2
Make Them the Same - K.MD.B.3
Matching Items and Numbers - K.MD.B.3
Same and Different - K.MD.A.3
Recognizing the Same Size - K.MD.A.3
Matching Numbers and Groups of Objects - K.MD.B.3
Matching Pairs - K.MD.B.3
### Precursory Skills to Geometry
One to One Relationships - Geometry
In Front and Behind - K.G.A.1
Inside, Outside and in Between - K.G.A.1
Above and Below - K.G.A.1
Which is Smaller? - K.MD.A.2
Sorting Shapes - K.G.B.5
Simple Patterns - K.G.B.4
Patterns Through Colors - K.G.B.4
Basic Patterns with Missing Piece - K.G.B.4
Following Sentence-Based Directions - K.G.B.5
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0
What are the greatest common factors of 64 and 28?
Updated: 11/4/2022
Wiki User
8y ago
The greatest common factor is 4
Wiki User
8y ago
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Q: What are the greatest common factors of 64 and 28?
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Related questions
What are the common factors of 28 and 64?
The common factors of 28 and 64 are: 1, 2, and 4
What are the greatest common factors of 56 and 64?
The greatest common factor of 56 and 64 is 8
What are the common factors and greatest common factor of 64?
As an example the factors of 64 are 1, 2, 4, 8, 16, 32, and 64. Examples: The common factors of 9 and 64 are only 1; the greatest common factor is 1. The common factors of 48 and 64 are 1, 2, 4, 8, and 16; the greatest common factor is 16. The common factors of 64 and 100 are 1, 2, and 4; the greatest common factor is 4. Note that for common factors you need two or more numbers
What is the greatest common factor of 30 and 64?
The GCF of 30 and 64 is 2. The factors of 30 are 1,2,3,10,15, and 30. The factors of 64 are 1,2,4,8,16,32, and 64. The greatest common factor of 30 and 64 is 2.
What is the greatest common factor of 64 and 28?
The GCF of 28 and 64 is 4
Are common factors for 64 and 28?
The common factors are: 1, 2, 4.
What are Greatest common factors of 64 and 32?
The Greatest Common Factor (GCF) is: 32
8
What is the greatest common factor of 56 and 64?
The GCF of 56 and 64 is 8. (56= 7x8, 64= 8x8)One way to determine the greatest common factor is to find all the factors of the numbers and compare them.The factors of 56 are 1, 2, 4, 7, 8, 14, 28, and 56.The factors of 64 are 1, 2, 4, 8, 16, 32, and 64.The common factors are 1, 2, 4, and 8. Therefore, 8 is the greatest common factor.The greatest common factor can also be calculated by identifying the common prime factors and multiplying them together.The prime factors of 56 are 2, 2, 2, and 7.The prime factors of 64 are 2, 2, 2, 2, 2, and 2.The prime factors in common are 2, 2, and 2, so the greatest common factor is 2 x 2 x 2 = 8.
What are the greatest common factors of 64 40 and 104?
Common factors of 64, 40, and 104 are: 1, 2, 4, and 8.
What is the greatest common factor of 28 and 128?
Factors of 28: 1 2 4 7 14 28 Factors of 128: 1 2 48 16 32 64 128 The GCF of 28 and 128 is 4
What are the greatest common factors of 64 and 16?
64 and 16's highest common factor is 16.
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# Weight of 12 mangoes is 2.4kg .What is the weight of 8 mangoes?
Last updated date: 13th Jun 2024
Total views: 384.3k
Views today: 8.84k
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Hint: For this type of question first we should find the weight of one item given. Here the given item is mango, that is we should find the weight of one mango and then we should multiply that one mango weight with the required amount of mangoes.
The objective of the problem is to find the weight of 8 mangoes.
Given that the weight of twelve mangoes is 2.4kg.
First , we find the weight of one mango. We are finding the weight of one mango because it is easy to find the weight of eight mangoes. By multiplying one mango weight with eight mangoes we get eight mangoes.
Let the weight of one mango be x.
Number of mangoes Weight of mangoes in kgs 12 2.4 1 x
To find x we use the cross multiplication. We get
$12 \times x = 1 \times 2.4$
$\Rightarrow 12x = 2.4 \\ \Rightarrow x = \dfrac{{2.4}}{{12}} \\$
Thus the weight of one mango is $\dfrac{{2.4}}{{12}}$kg.
But we have to find the weight of 8 mangoes. By multiplying the weight of one mango with eight mangoes we get the weight of eight mangoes .
Weight of eight mangoes $= 8 \times \dfrac{{2.4}}{{12}}$
Multiplying numerator and denominator with 10 to take off the decimal point. We get
$\Rightarrow 8 \times \dfrac{{2.4}}{{12}} \times \dfrac{{10}}{{10}}$
After multiplying numerator and denominator with 10 we get
$= 8 \times \dfrac{{24}}{{120}} \\ = \dfrac{{24}}{{15}} = 1.6 \\$
Thus , the weight of eight mangoes is 1.6kg.
Note: We use cross multiplication to compare the fractions . This method is useful when we are working with larger fractions for reducing them into smaller fractions, and to determine the value of a variable .
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# Stuck on this question! Can someone please help? 2 + 7x + 3 − 5x − 1 = "______"? Thanks!
May 4, 2018
$4 + 2 x$ is the final expression. Here's why:
#### Explanation:
2 + 7x + 3 - 5x -1 = ?
Start by combining like terms in the order that they appear. Let's break them into variables and integers.
Integers first:
$2$ + $3$ - $1$ = $4$
Then, variables:
$7 x$ - $5 x$ = $2 x$
$4 + 2 x$
May 4, 2018
Assuming that you just want this simplified since there is nothing on the right of the $=$ sign, the answer is:
$2 x + 4$
#### Explanation:
The easiest way to simplify this problem is to combine like terms. First we'll combine all the $x$ terms:
2 + color(red)(7x) + 3 color(white)"x"color(red)( - 5x) - 1 = ?
$2 + \textcolor{red}{2 x} + 3 - 1$
Now combine all of the real numbers:
$\textcolor{b l u e}{2} + 2 x + \textcolor{b l u e}{3} \textcolor{w h i t e}{x} \textcolor{b l u e}{- 1}$
$2 x + \textcolor{b l u e}{4}$
The simplest form of this problem is $2 x + 4$.
|
# Method of L.C.M.
We will discuss here about the method of l.c.m. (least common multiple).
1. Let us consider the numbers 8, 12 and 16.
Multiples of 8 are → 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, ......
Multiples of 12 are → 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, ......
Multiples of 16 are → 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, ......
The common multiple of 8, 12, 16 are 78, 96, ......
The least common multiple of 8, 12 and 16 is 48. (Smallest common multiple)
In short, the lowest common factor is expressed as L.C.M.
2. Find the L.C.M. of 3 and 4.
Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, ............
Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, ............
Common multiples of 3 and 4 = 12, 24, ............
Least common multiple of 3 and 4 = 12.
3. Find the L.C.M. of 6 and 12.
Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, ............
Multiples of 12 = 12243648, 60, 72, 84, ............
Common multiples of 6 and 12 = 12, 24, 36, 48, ............
Least common multiple of 6 and 12 = 12.
Finding L.C.M.
Least Common Multiple (L.C.M.) by Prime Factorisation Method:
To find the L.C.M. we find prime factors of the given numbers.
Remember, we consider common prime factors only.
Solved Examples:
1. Find the L.C.M. of 12, 16 and 24.
First we find the prime factors of the given numbers.
12 = 2 × 2 × 3
16 = 2 × 2 × 2 × 2
24 = 2 × 2 × 2 × 2 × 3
(2 comes maximum 4 times and 3 comes maximum once only.)
L.C.M. = 2 × 2 × 2 × 2 × 3
= 48 which is the product of their prime factors.
2. Find the L.C.M. of 10 and 16.
10 = 2 × 516 = 2 × 2 × 2 × 2Common factor = 2Other factors = 2, 2, 2, 5 2 | 10 5 2 | 16 2 | 8 2 | 4 2
Therefore, L.C.M. = 2 × 2 × 2 × 2 × 5 (common factor × other factors)
= 80
3. Find the L.C.M. of 20 and 25.20 = 2 × 2 × 525 = 5 × 5Common factor = 5Other factors = 2, 2, 5 2 | 20 2 | 10 5 5 | 25 5
Therefore, L.C.M. = 5 × 2 × 2 × 5 (common factor × other factors)
= 100
4. Find the L.C.M. of 16 and 24.
16 = 2 × 2 × 2 × 224 = 2 × 2 × 2 × 3Common factor = 2, 2, 2Other factors = 2, 3 2 | 16 2 | 8 2 | 4 2 2 | 24 2 | 12 2 | 6 3
Therefore, L.C.M. = 2 × 2 × 2 × 2 × 3 (common factor × other factors)
= 48
Least Common Multiple (L.C.M.) by Division Method:
We can also find the L.C.M. of the given numbers by dividing all the numbers at the same time by a number that divides at least two of the given numbers.
We proceed as below:
Step I: Arrange the given numbers in a line, in any order.
Step II: Divide by a number which exactly divides at least two for the given numbers carry forward the numbers which are not divisible.
Step III: Repeat the process till neither of the two given numbers are divisible by the same number.
Step IV: The product of all the divisors and the numbers left undivided is the required LCM.
For Example:
1. Find the LCM of 24 and 30 by division method.
Solution:
LCM of 24 and 30
=2 × 2 × 2 × 3 × 5 = 120
2. Find the LCM 15, 36, and 42 by long division method.
Solution:
So. LCM of 18, 36 and 42 = 2 × 3 × 3 × 2 × 7 = 252
3. Find the LCM of 12, 16 and 24 by division method.
1. When a number is not exactly divisible, we write the number itself below the line. 2. When we cannot divide the numbers by a common factor exactly we discontinue dividing the numbers.
L.C.M. = 2 × 2 × 2 × 3 × 2 = 48
Note:
The product of L.C.M. and H.C.F. of two numbers is also the product of the numbers.
For example, the L.C.M. of 7 and 14 is 14 and the H.C.F. of 7 and 14 = 7.
We see that the product of 7 and 14 also the product of L.C.M. and H.C.F. of 7 and 14.
4. Find the L.C.M. of 25 and 45.
Steps: Divide 25 and 45 by 5.25 ÷ 5 = 5; 45 ÷ 5 = 95 and 9 have no common factor.Stop the division.
Therefore, L.C.M. = 5 × 5 × 9
= 225
5. Find the L.C.M. of 40, 68 and 72.
Steps:Divide 40, 68 and 72 by 2.40 ÷ 2 = 20; 68 ÷ 2 = 34; 72 ÷ 2 = 36Divide 20, 34 and 36 by 2.20 ÷ 2 = 10; 34 ÷ 2 = 17; 36 ÷ 2 = 1810, 17 and 18 do not have a common factor. But 10 and 18 have 2 as a common factor.Divide 10 and 18 by 2, leaving 17 as it is.10 ÷ 2 = 5; 18 ÷ 2 = 95, 17 and 9 do not have a common factor.Stop the division.
Therefore, L.C.M. = 2 × 2 × 2 × 5 × 17 × 9
= 6120
Worksheet on Method of LCM:
I. Find the L.C.M. of the following by prime factorisation method.
(i) 30, 36
(ii) 12, 15
(iii) 5, 7
(iv) 15, 30
(v) 42, 72
(vi) 12, 48
(vii) 60, 75
(viii) 25, 150
(ix) 64, 128
(x) 60, 108
I. (i) 180
(ii) 60
(iii) 35
(iv) 30
(v) 504
(vi) 48
(vii) 300
(viii) 150
(ix) 128
(x) 540
II. Find the L.C.M. of the following by division method.
(i) 27, 84
(ii) 16, 32
(iii) 12, 15
(iv) 25, 30
(v) 60, 70
(vi) 30, 18, 60
(vii) 88, 64, 96
(viii) 48, 96, 144
(ix) 26, 28, 24
(x) 16, 12, 20
II. (i) 756
(ii) 32
(iii) 60
(iv) 150
(v) 420
(vi) 180
(vii) 2112
(viii) 288
(ix) 2184
(x) 240
III. Find the L.C.M. of the following numbers by listing the multiples.
(i) 24, 36
(ii) 12, 18
(iii) 10, 20, 40
(iv) 27, 108
(v) 63, 84
III. (i) 72
(ii) 36
(iii) 40
(iv) 108
(v) 252
IV. Find the L.C.M. of the following numbers:
(i) 6 and 10
(ii) 3 and 6
(iii) 10 and 12
(iv) 4 and 9
(v) 15 and 18
(vi) 6 and 11
(vii) 9 and 18
(viii) 7 and 14
(ix) 8 and 12
(x) 8 and 16
IV. (i) 30
(ii) 6
(iii) 60
(iv) 36
(v) 90
(vi) 66
(vii) 18
(viii) 14
(ix) 24
(x) 16
V. Find the LCM by long division method:
(i) 18, 30
(ii) 13, 39
(iii) 12, 15
(iv) 38, 72
(v) 2, 3
(vi) 30, 45
(vii) 112, 140
(viii) 88, 99
V. (i) 90
(ii) 39
(iii) 60
(iv) 1368
(v) 6
(vi) 90
(vii) 560
(viii) 792
VI. Find the LCM by long division.
VI. (i) 315
(ii) 36
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# 8.5: Equal Costs
Difficulty Level: At Grade Created by: CK-12
Carla and Dan went to the store. Carla bought 3 notebooks and a $2 pen. Dan bought 2 notebooks and an$8 stapler. All notebooks cost the same. They spent the same amount of money. Can you figure out the cost of one notebook? In this concept, we will learn how to use equations to solve problems having to do with equal costs.
### Guidance
In order to answer questions about equal costs like the one above, we can use the problem solving steps to help.
• First, describe what you know. What did each person buy?
• Second, identify what your job is. In these problems, your job will be to solve for the price of an item.
• Third, make a plan. In these problems, your plan should be to write an expression for what each person spent. Then, set those expressions equal to each other since each person spent the same amount. Finally, solve the equation.
• Fourth, solve. Implement your plan.
• Fifth, check. Substitute your answer into the original problem and make sure it works.
Al bought 2 sandwiches. Bob bought one sandwich and a 4 large soda. All sandwiches cost the same. They spent the same amount of money. What is the cost of one sandwich? • Use \begin{align*}b\end{align*} to stand for the cost of one sandwich. • Write an equation to represent the costs of the two people. • Solve for the value of \begin{align*}b\end{align*}. • Show your work. Solution: We can use problem solving steps to help us with this problem. \begin{align*}& \mathbf{Describe:} && \text{Al:} \ 2 \ \text{sandwiches}\\ &&& \text{Bob:} \ 1 \ \text{sandwich and a} \ \4 \ \text{large soda}\\ &&& 2 \ \text{sandwiches} \ \text{cost the same as} \ 1 \ \text{sandwich and} \ \4\\ \\ & \mathbf{My \ job:} && \text{Figure out the cost of one sandwich.}\\ \\ & \mathbf{Plan:} && \text{Use} \ b \ \text{to stand for the cost of one sandwich.}\\ &&& \text{Write an expression showing the money spent by each person.}\\ &&& \text{Since they spent the same amount of money, the two expressions are equal.}\\ &&& \text{Write the equation.}\\ &&& \text{Solve for} \ b.\\ \\ & \mathbf{Solve:} && \text{Al's cost:} \ 2b\\ &&& \text{Bob's cost:} \ b+4.\\ &&& \text{The costs are the same so} \ 2b=b+4. \ \text{Solve the equation.}\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 2b=b+4\\ &&& \text{Subtract} \ b \ \text{from each side.} \qquad \qquad 2b-b=b+4-b\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ b=4\\ &&& \text{One sandwich is} \ \4\\ \\ & \mathbf{Check:} && \text{Al:} \ 2 \ \text{sandwiches is} \ 2 \times \4, \ \text{or} \ \8.\\ &&& \text{Bob:} \ 1 \ \text{sandwich and a} \ \4 \ \text{large soda is} \ 1 \times \4+\4, \ \text{or} \ \8.\\ &&& \8= \8\end{align*} #### Example B Camilla bought 4 small sodas and a2 cookie. Darla bought 3 small sodas and a 5 dessert. All small sodas cost the same. They spent the same amount of money. What is the cost of one small soda? • Use \begin{align*}c\end{align*} to stand for the cost of one small soda. • Write an equation to represent the costs of the two people. • Solve for the value of \begin{align*}c\end{align*}. • Show your work. Solution: We can use problem solving steps to help us with this problem. \begin{align*}& \mathbf{Describe:} && \text{Camilla:} \ 4 \ \text{small sodas and a } \ \2 \ \text{cookie}\\ &&& \text{Dana:} \ 3 \ \text{small sodas and a } \ \5 \ \text{dessert}\\ &&& 4 \ \text{small sodas and } \ \2 \ \text{costs the same as} \ 3 \ \text{notebooks and} \ \5\\ \\ & \mathbf{My \ job:} && \text{Figure out the cost of one small soda.}\\ \\ & \mathbf{Plan:} && \text{Use} \ c \ \text{to stand for the cost of one small soda.}\\ &&& \text{Write an expression showing the money spent by each person.}\\ &&& \text{Since they spent the same amount of money, the two expressions are equal.}\\ &&& \text{Write the equation.}\\ &&& \text{Solve for} \ c.\\ \\ & \mathbf{Solve:} && \text{Camilla's cost:} \ 4c+2\\ &&& \text{Dana's cost:} \ 3c+5.\\ &&& \text{The costs are the same so} \ 4c+2=3c+5. \ \text{Solve the equation.}\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 4c+2=3c+5\\ &&& \text{Subtract} \ 3c \ \text{from each side.} \qquad \qquad 4c+2-3c=3c+5-3c\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ c+2=5\\ &&& \text{Subtract} \ 2 \ \text{from each side.} \qquad \qquad \ \ c=3\\ &&& \text{One small soda is} \ \3\\ \\ & \mathbf{Check:} && \text{Camilla:} \ 4 \ \text{small sodas and a} \ \2 \ \text{cookie is} \ 4 \times \3+\2, \ \text{or} \ \14.\\ &&& \text{Dana:} \ 3 \ \text{small sodas and a} \ \5 \ \text{dessert is} \ 3 \times \3+\5, \ \text{or} \ \14.\\ &&& \14= \14\end{align*} #### Example C Erin bought 6 muffins Fred bought 4 muffins and a3 drink. All muffins cost the same. They spent the same amount of money. What is the cost of one muffin?
• Use \begin{align*}d\end{align*} to stand for the cost of one muffin.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}d\end{align*}.
• Show your work.
Solution:
We can use problem solving steps to help us with this problem.
\begin{align*}& \mathbf{Describe:} && \text{Erin:} \ 6 \ \text{muffins}\\ &&& \text{Fred:} \ 4 \ \text{muffins and a} \ \3 \ \text{drink}\\ &&& 6 \ \text{muffins} \ \text{costs the same as} \ 4 \ \text{notebooks and} \ \3\\ \\ & \mathbf{My \ job:} && \text{Figure out the cost of one muffin.}\\ \\ & \mathbf{Plan:} && \text{Use} \ d \ \text{to stand for the cost of one muffin.}\\ &&& \text{Write an expression showing the money spent by each person.}\\ &&& \text{Since they spent the same amount of money, the two expressions are equal.}\\ &&& \text{Write the equation.}\\ &&& \text{Solve for} \ d.\\ \\ & \mathbf{Solve:} && \text{Erin's cost:} \ 6d\\ &&& \text{Fred's cost:} \ 4d+3.\\ &&& \text{The costs are the same so} \ 6d=4d+3. \ \text{Solve the equation.}\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 6d=4d+3\\ &&& \text{Subtract} \ 4d \ \text{from each side.} \qquad \qquad 6d-4d=4d+3-4d\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ 2d=3\\ &&& \text{Divide each side by} \ 2. \qquad \qquad \ \ d=1.5\\ &&& \text{One muffin is} \ \1.50\\ \\ & \mathbf{Check:} && \text{Erin:} \ 6 \ \text{muffins is} \ 6 \times \1.50, \ \text{or} \ \9.\\ &&& \text{Fred:} \ 4 \ \text{muffins and a} \ \3 \ \text{drink is} \ 4 \times \1.50+\3, \ \text{or} \ \9.\\ &&& \9= \9\end{align*}
#### Concept Problem Revisited
Remember the problem about Carla and Dan? Carla bought 3 notebooks and a $2 pen. Dan bought 2 notebooks and an$8 stapler. All notebooks cost the same. They spent the same amount of money. What is the cost of one notebook?
We can use problem solving steps to help us with this problem.
\begin{align*}& \mathbf{Describe:} && \text{Carla:} \ 3 \ \text{notebooks and a} \ \2 \ \text{pen}\\ &&& \text{Dan:} \ 2 \ \text{notebooks and an} \ \8 \ \text{stapler}\\ &&& 3 \ \text{notebooks and} \ \2 \ \text{costs the same as} \ 2 \ \text{notebooks and} \ \8\\ \\ & \mathbf{My \ job:} && \text{Figure out the cost of one notebook.}\\ \\ & \mathbf{Plan:} && \text{Use} \ a \ \text{to stand for the cost of one notebook.}\\ &&& \text{Write an expression showing the money spent by each person.}\\ &&& \text{Since they spent the same amount of money, the two expressions are equal.}\\ &&& \text{Write the equation.}\\ &&& \text{Solve for} \ a.\\ \\ & \mathbf{Solve:} && \text{Carla's cost:} \ 3a+2\\ &&& \text{Dan's cost:} \ 2a+8.\\ &&& \text{The costs are the same so} \ 3a+2=2a+8. \ \text{Solve the equation.}\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3a+2=2a+8\\ &&& \text{Subtract} \ 2a \ \text{from each side.} \qquad \qquad 3a+2-2a=2a+8-2a\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ a+2=8\\ &&& \text{Subtract} \ 2 \ \text{from each side.} \qquad \qquad \ \ a+2-2=8-2\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ a=6\\ &&& \text{One notebook is} \ \6.\\ \\ & \mathbf{Check:} && \text{Carla:} \ 3 \ \text{notebooks and} \ \2 \ \text{pen is} \ 3 \times \6+\2, \ \text{or} \ \20.\\ &&& \text{Dan:} \ 2 \ \text{notebooks and} \ \8 \ \text{stapler is} \ 2 \times \6+\8, \ \text{or} \ \20.\\ &&& \20 = \20\end{align*}
### Vocabulary
In math, an expression is a phrase that can contain numbers, operations, and variables without an equal's sign. An equation is a statement that two expressions are equal. An equation is two expressions combined with an equals sign. To solve an equation means to figure out the value(s) for the variable(s) that make the equation true.
### Guided Practice
1. Gary bought 5 CDs and a $5 CD case. Helen bought 3 CDs and a set of$29 earphones. All CDs cost the same. They spent the same amount of money. What is the cost of one CD?
• Use \begin{align*}f\end{align*} to stand for the cost of one CD.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}f\end{align*}.
• Show your work.
2. Ina bought 4 movie tickets and a $2 soda. Jen bought 2 movie tickets, a$4 bag of popcorn, a $3 drinks, and a$5 bag of candy. All movie tickets cost the same. They spent the same amount of money. What is the cost of one movie ticket?
• Use \begin{align*}g\end{align*} to stand for the cost of one movie ticket.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}g\end{align*}.
• Show your work.
3. Ken bought 10 pounds of apples. Larry bought 5 pounds of apples and a 10 jar of honey. Each pound of apples cost the same. They spent the same amount of money. What is the cost of one pound of apples? • Use \begin{align*}j\end{align*} to stand for the cost of one pound of apples. • Write an equation to represent the costs of the two people. • Solve for the value of \begin{align*}j\end{align*}. • Show your work. Answers: 1. One CD costs12. Here is the equation you should have written and the steps to solve:
\begin{align*}5f + 5 &= 3f + 29\\ 5f + 5 - 3f &= 3f + 29 - 3f\\ 2f + 5 &= 29\\ 2f + 5 - 5 &= 29 - 5\\ 2f &= 24\\ f &= 12\end{align*}
2. One movie ticket costs 5. Here is the equation you should have written and the steps to solve: \begin{align*}4g + 2 &= 2g + 4 + 3 + 5\\ 4g + 2 &= 2g + 12\\ 4g + 2 - 2g &= 2g + 12 - 2g\\ 2g + 2 &= 12\\ 2g + 2 - 2 &= 12 - 2\\ 2g &= 10\\ g &= 5\end{align*} 3. One pound of apples costs2. Here is the equation you should have written and the steps to solve:
\begin{align*}10j &= 5j + 10\\ 10j - 5j &= 5j + 10 - 5j\\ 5j &= 10\\ j &= 2\end{align*}
### Practice
1. Hal bought 2 bagels and a large hot chocolate for $2.50. Jon bought 4 bagels and a$1.00 cream cheese. All bagels cost the same. They spent the same amount of money. What is the cost of one bagel?
• Use \begin{align*}a\end{align*} to stand for the cost of one bagel.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}a\end{align*}.
• Show your work.
2. Kaelyn bought 5 pads of paper and a $1.50 box of binder clips. Lexa bought 2 pads of paper and a$6.00 box of pens. All pads of paper cost the same. They spent the same amount of money. What is the cost of one pad of paper?
• Use \begin{align*}b\end{align*} to stand for the cost of one pad of paper.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}b\end{align*}.
• Show your work.
3. Mary bought 4 decks of cards and a $2.00 crossword puzzle book. Nina bought 2 decks of cards and 2 boxes of dominoes for$3.50 each. All card decks cost the same. They spent the same amount of money. What is the cost of one deck of cards?
• Use \begin{align*}c\end{align*} to stand for the cost of one deck of cards.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}c\end{align*}.
• Show your work.
4. Mark bought 5 sandwiches and a $2 bag of chips. Dave bought 3 sandwiches and a$10 pie. All sandwiches cost the same and they spent the same amount of money. What is the cost of one sandwich?
• Use \begin{align*}d\end{align*} to stand for the cost of one sandwich.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}d\end{align*}.
• Show your work.
5. Jess bought 2 boxes of paper clips and a $1 notepad. John bought one box of paper clips and a$1.75 pen. They spent the same amount of money. What is the cost of one box of paper clips?
• Use \begin{align*}e\end{align*} to stand for the cost of one box of paper clips.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}e\end{align*}.
• Show your work.
6. Sarah bought 7 binders and a 4.50 pack of pens. Ben bought 8 binders. They spent the same amount of money. What is the cost of one binder? • Use \begin{align*}f\end{align*} to stand for the cost of one binder. • Write an equation to represent the costs of the two people. • Solve for the value of \begin{align*}f\end{align*}. • Show your work. 7. Bob bought 2 t-shirts and a22.75 pair of pants. Jeff bought 3 t-shirts and a \$12.50 hat. All t-shirts cost the same. They spent the same amount of money. What is the cost of one t-shirt?
• Use \begin{align*}g\end{align*} to stand for the cost of one t-shirt.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}g\end{align*}.
• Show your work.
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Prime Factorization Worksheets - Page 10 | Problems & Solutions
# Prime Factorization Worksheets - Page 10
Prime Factorization Worksheets
• Page 10
91.
Is 2 a prime or a composite number?
a. Prime b. Composite
#### Solution:
A number that has exactly two factors, 1 and itself is called a prime number.
A number that has more than two factors is called a composite number.
To find whether 2 is a prime number or composite, draw the rectangles that can be made from exactly 2 squares.
The dimensions of the rectangle show that the factors of 2 are 1 and 2.
So, 2 is a prime number.
92.
Is 4 a prime or a composite number?
a. Prime b. Composite c. Neither prime nor composite
#### Solution:
A number that has exactly two factors, 1 and itself is called a prime number.
A number that has more than two factors is called a composite number.
To find whether 4 is a prime number or composite, draw the rectangles that can be made from exactly 4 squares.
The dimensions of the rectangle show that the factors of 4 are 1, 2 and 4.
The number 4 has 3 factors.
So, 4 is a composite number.
93.
Is 8 a prime or a composite number?
a. Prime b. Neither prime nor composite c. Composite
#### Solution:
A number that has exactly two factors, 1 and itself is called a prime number.
A number that has more than two factors is called a composite number.
To find whether 8 is a prime number or composite, draw the rectangles that can be made from exactly 8 squares.
The dimensions of the rectangle show that the factors of 8 are 1, 2, 4 and 8.
The number 8 has 4 factors.
So, 8 is a composite number.
94.
Is 3 a prime or a composite number?
a. Prime b. Composite c. Both prime and composite d. Neither prime nor composite
#### Solution:
A number that has exactly two factors, 1 and itself is called a prime number.
A number that has more than two factors is called a composite number.
To find whether 3 is a prime number or composite, draw the rectangles that can be made from exactly 3 squares.
The dimensions of the rectangle show that the factors of 3 are 1 and 3.
So, 3 is a prime number.
95.
Is 12 a prime or a composite number?
a. Prime b. Composite
#### Solution:
A number that has exactly two factors, 1 and itself, is called a prime number.
A number that has more than two factors is called a composite number.
To find whether 12 is a prime number or not, draw the rectangles that can be made from exactly 12 squares.
The dimensions of the rectangle show that the factors of 12 are 1, 2, 3, 4, 6 and 12.
The number 12 has 6 factors.
So, 12 is a composite number.
96.
Is 14 a prime or a composite number?
a. Prime b. Composite c. Neither prime nor composite
#### Solution:
A number that has exactly two factors, 1 and itself, is called a prime number.
A number that has more than two factors is called a composite number.
To find whether 14 is a prime number or not, draw the rectangles that can be made from exactly 14 squares.
The dimensions of the rectangle show that the factors of 14 are 1, 2, 7 and 14.
The number 14 has 4 factors.
So, 14 is a composite number.
97.
Is 16 a prime or a composite number?
a. Prime b. Composite c. Neither prime nor composite
#### Solution:
A number that has exactly two factors, 1 and itself is called a prime number.
A number that has more than two factors is called a composite number.
To find whether 16 is a prime number or composite, draw the rectangles that can be made from exactly 16 squares.
The dimensions of the rectangle show that the factors of 16 are 1, 2, 4, 8 and 16.
The number 16 has 5 factors.
So, 16 is a composite number.
98.
Is 20 a prime or a composite number?
a. Prime b. Composite c. Neither prime nor composite
#### Solution:
A number that has exactly two factors, 1 and itself, is called a prime number.
A number that has more than two factors is called a composite number.
To find whether 20 is a prime number or not, draw the rectangles that can be made from exactly 20 squares.
The dimensions of the rectangle show that the factors of 20 are 1, 2, 4, 5, 10 and 20.
The number 20 has 6 factors.
So, 20 is a composite number.
99.
Find the prime factorization of 66 using a factor tree.
a. 111 × 31 × 21 b. 112 × 31 × 21 c. 111 × 31 × 31 d. 111 × 31 × 22
#### Solution:
The factor tree of 66 is:
[66 = 11 × 6, 6 = 3 × 2]
The circled numbers in the factor tree are the prime factors of 66.
66 = 11 × 3 × 2
66 = 111 × 31 × 21
[Write in the form of exponents.]
The prime factorization of 66 is 111 × 31 × 21.
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Math Insight
Distance from point to plane example
To illustrate our approach for finding the distance between a point and a plane, we work through an example.
Find the distance from the point $P=(4,-4,3)$ to the plane $2x-2y+5z+8=0$, which is pictured in the below figure in its original view.
Distance from point to plane. A sketch of a way to calculate the distance from point $\color{red}{P}$ (in red) to the plane. The vector $\color{green}{\vc{n}}$ (in green) is a unit normal vector to the plane. You can drag point $\color{red}{P}$ as well as a second point $\vc{Q}$ (in yellow) which is confined to be in the plane. Although the vector $\color{green}{\vc{n}}$ does not change (as the plane is fixed), it moves with $\color{red}{P}$ to always be at the end of a gray line segment from $\color{red}{P}$ that is perpendicular to the plane. This distance from $\color{red}{P}$ to the plane is the length of this gray line segment. This distance is the length of the projection of the vector from $Q$ to $P$ (in purple) onto the normal vector $\color{green}{\vc{n}}$.
Solution: We use the second formula from the page on distance from a point to a plane. From the equation for the plane we substitute $A=2$, $B=-2$, $C=5$, $D=8$. From the point $P$, we substitute $x_1=4$, $y_1=-4$, and $z_1=3$. The distance from $P$ to the plane is \begin{align*} d = \frac{|2 \cdot 4 + (-2)\cdot(-4) + 5 \cdot 3 + 8 |} {\sqrt{2^2+(-2)^2 + 5^2}} = \frac{39}{\sqrt{33}} \approx 6.8 \end{align*}
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Co-founder, e-GMAT
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## OG 2020: Question No. 335
In the figure above, PQRT is a rectangle. What is the length of segment PQ?
1. The area of region PQRS is 39 and TS = 6.
2. The area of region PQRT is 30 and QR = 10.
Source OG 2020 Type Data Sufficiency Topic Geometry Sub-Topic Quadrilateral Difficulty Medium
### Solution
#### Steps 1 & 2: Understand Question and Draw Inferences
In this question, we are given
• A diagram of a quadrilateral PQRS, of which PQRT is a rectangle
We need to determine
• The length of segment PQ
Since PQRT is a rectangle we can say the side QR and PT are parallel to each other.
• Therefore, we can say that PQRS is a trapezium.
With this understanding, let us know analyse the individual statements.
#### Step 3: Analyse Statement 1
As per the information given in statement 1, the area of region PQRS is 39 and TS is equal to 6.
• Area of PQRS = (1/2) (QR + PS) (PQ) = 39
Or, (QR + PS) (PQ) = 78
Or, (PT + PT + TS) (PQ) = 78
Or, (2PT + 6) (PQ) = 78
As we donāt know the value of PT, we cannot determine the value of PQ.
Hence, statement 1 is not sufficient to answer the question.
#### Step 4: Analyse Statement 2
As per the information given in statement 2, the area of region PQRT is 30 and QR is equal to 10.
• PQ * QR = 30
Or, PQ * 10 = 30
Or, PQ = 3
As we can determine the value of PQ, statement 2 is sufficient to answer the question.
#### Step 5: Combine Both Statements Together (If Needed)
Since we can determine the answer from statement 2 individually, this step is not required.
Hence, the correct answer choice is option B.
Questions on Geometry are very commonly asked on the GMAT. Ace GMAT Quant by signing up for our free trial and get access to 400+ questions. We are the most reviewed online GMAT Prep company with 2400+ reviews on GMATClub.
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# Calculus 1 : Curves
## Example Questions
### Example Question #51 : Graphing Functions
Find the local minima of the following function:
Explanation:
To find the local minima of the function, we must find the x value at which the function's first derivative changes from negative to positive.
First, we must find the first derivative of the function:
The derivative was found using the following rules:
Now, we must find the critical values (where the first derivative is equal to zero) by setting the first derivative equal to zero:
Next, we must create the intervals in which the critical value is the upper and lower limit, respectively:
On the first interval, the first derivative is less than one, and on the second interval, the first derivative is positive (simply plug in any point on the interval into the first derivative function and check the sign). The change from negative to positive occurs at x=3, so is our answer.
### Example Question #51 : Graphing Functions
Where is the local minimum of
Explanation:
To find the local minimum, first find the derivative of the function. To take the derivative, multiply the exponent by the coefficient in front of the x term and then decrease the exponent by 1. Therefore, the derivative is: . Then, set that equal to 0 to get your critical point(s): . Put that point on a number line and test values on either side. To the left of 2, the values are negative. To the right of 2, the values are positive. Therefore, the slope goes from negative to positive, indicating there is a local min at x=2.
### Example Question #51 : Curves
Explanation:
The first step is to chop up the expression into two terms: . Now, integrate each term. Whenever there is an x on the denominator, the integral is lnx. Multiply that by the numerator of 4. The integral of 1 is x. Put those together to get your answer of .
### Example Question #54 : Graphing Functions
Find where the local minima occur for the following function:
The function is always increasing
Explanation:
To find where the local minima occur for the function, we need the x-values at which the first derivative changes from negative to positive.
The first derivative is
and was found using the following rule:
Now, we must find the critical values at which the first derivative is equal to zero:
Now, using the critical values, we make the intervals on which we see whether the first derivative is positive or negative (plug in any value on the interval into the first derivative function):
On the first interval, the first derivative is positive, on the second it is negative, and on the third it is positive. So, a local minimum occurs at
because the first derivative changed from positive to negative at that point.
### Example Question #55 : Graphing Functions
Find the x value of the local minima of .
Explanation:
We need to differentiate term by term, applying the power rule,
This gives us
The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:
Any local minimum will fall at a critical point where the derivative passes from negative to positive. To check this, we check a point in each of the intervals defined by the critical points:
.
Let's take -2 from the first interval, 0 from the second interval, and 2 from the third interval.
The derivative moves from negative to positive at 1, so that is the function's only local minimum.
### Example Question #56 : Graphing Functions
Find the local minima of the following function:
Explanation:
To determine the local minima for the function, we must determine where the first derivative changes from negative to positive.
The first derivative of the function is equal to
and was found using the following rules:
Now, we must find the critical values, the values at which the first derivative equals zero:
We use the critical value to make our intervals on which we check the sign of the first derivative:
Note that at the endpoints of the intervals, the first derivative is neither positive nor negative.
Next, we plug in any value on the intervals into the first derivative and check the sign. On the first interval, the first derivative is negative, while on the second, it is positive. Thus, a local mininum exists at .
### Example Question #57 : Graphing Functions
Find the local minimum of .
Explanation:
First, take the derivative of the function, remembering to multiply the exponent by the coefficeint and then subtracting one from the exponent. Therefore, the derivative is: . Then, set it equal to 0 to get your critical point: . Then, test a value on either side of your critical point and plug into the derivative. To the left, the value is negative. To the right, the value is positive. Therefore, your local minimum is at .
### Example Question #58 : Graphing Functions
Given the function , find its local minimum.
Explanation:
To find the local minimum of a function, one must find the critical points (where the derivative is equal to zero).
Taking the derivative of the function above using the power rule,
.
The derivative is
.
Setting this equal to zero and solving for x gets you
To see if this indeed is a local minimum one must plug in values below and above this value of x to see if the derivative is positive (increasing) or negative (decreasing) around that point.
If you plug -2 into the equation for you get -18 (decreasing), and plugging in 0, you get (increasing).
Thus, is a local minimum.
### Example Question #1 : How To Graph Functions Of Curves
How many global extrema does the following function have?
One
Three
Impossible to determine
Two
Two
Explanation:
This function has only three extrema: a local maximum at and two minima at (these extrema are found by finding the first derivative of the function, setting it equal to zero, and solving for x).
By evaluating any point along the four intervals defined by the extrema, , and , one can see that the function is decreasing on the first interval, increasing on the second, decreasing on the third and increasing on the fourth. Therefore, is a local maximum while are minima Furthermore, the value of the function is the same at both minima, making them global minima.
### Example Question #2 : How To Graph Functions Of Curves
For the figure above, which of the following statements is true?
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More on the Maths
DNA STUDY
The concept of marker mutations confuses many at first glance. But it is actually quite simple, so we will attempt to explain the logic using the technique of 'expected values' - which is the long-run value taken over many independent repetitions.
Consider a lottery draw that has numbered balls from 1 to 300. You have bet on number 50. When the draw is held you would expect your number would be unlikely to come up. Your odds are 1/300, or 0.33%.
Let’s say that in the lottery, not one, but 44 balls are drawn. The odds of your number coming up are now 44/300 or roughly 1/7.
If you then do the same 44 draw every day for a week, you would expect that your number would come up once in that week, as you have 7 chances at 1/7 odds (7 * 1/7 =1 )
Now consider that the 44 balls are in fact the STR markers and that the 7 days are 7 generations. Using the same calculation, you would expect during the 7 generations that 1 mutation would occur on your own 44 markers. When comparing two people, each one could have 1 mutation, so the total expected mutations must be double that for one person, which is 2 mutations. So comparing two people over 7 generations with 44 markers tested, the expected number of mutations is 2. Easy now?
This logic can be distilled to a simple formula:
No of Markers tested x (No of generations x number of persons tested) x average mutation rate.
Sometimes the value of 'No of generations x number of persons tested' is rewritten as 'No of transmission events'. The transmission event value is simply the number of generations x the number of persons being compared. So for two 1st cousins, the most recent ancestor would be their grandfather. This would create 2 transmission events each, ie grandfather to father to son for each cousin. The total transmission events would be 4.
Back to our example , we can test the formula to predict the number of mutations in 2 people over 7 generations in a 44 marker test and using an average mutation rate of 0.33%
44 x (7x2) x 0.33/100 = 2.03. Or rounded, is 2 mutations.
Or put it another way, the if you had 44 numbers in a 300 ball lottery, and there were two draws per day in a week, you would expect to win twice in that week.
We can use this formula to create a table of expected mutations in both the 44 and 67 marker tests. Note that the number of markers used in the calculations are actually 43 and 63, and not 44 and 67, as certain markers have several values so it is not certain where the mutation occurred, or indeed if it occurred, and it is standard practice to exclude these from calculations.
Using the table, for example, in a 63 marker comparison, using a mutation rate of 0.41%, the number of expected mutations where 11 generations exist between two parties tested, would be 6 (rounded as we cannot have 5.7 mutations). The expected result would therefore be 57/63 matching.
Caution on transmission events
Where there are more than two people being examined, the number of transmission events is not necessarily the number of generations multiplied by the number of people. There are two examples below where there are 10 transmission events.
The first has the same common ancestor and the transmission events are simply 2 x 5. However, if those being tested do not have a common ancestor then the calculation is not as straight forward.
In the second example there are four people tested where the transmission events are not 12 (4 x 3) but are still 10. This is as the third and fourth people share a common ancestor one generation later than the first and second.
Therefore to calculate the number of transmission events for more than two people, where they do not all share the same common ancestor, the events must be counted manually to ensure that there is no double counting.
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# Geometry: Ratio, Proportion, and Geometric Means
## Ratio, Proportion, and Geometric Means
Before you start analyzing similar triangles, you need to pick up a few more algebraic supplies. You'll be dealing with fractions and ratios a lot in this section, so you might as well brush up on those algebra skills first.
A ratio is a quotient a/b, where b ≠ 0. A ratio provides a comparison between the numbers a and b. For example, if a is twice as big as b, then the ratio a/b is 2/1. The ratio a/b is read “a to b” and is sometimes written in the form a:b.
There are times when you might want to compare three or more items. When that happens, a simple fraction just won't cut it. You'll need to use what is called an extended ratio. An extended ratio is written in the form a:b:c (if you are comparing three quantities) or a:b:c:d (if you are comparing four quantities). If you are comparing lots of quantities, just keep adding them on, separating each quantity with a colon.
##### Eureka!
An extended ratio compares more than two quantities and cannot be expressed as a single fraction.
A proportion is a statement that two ratios are equal. The proportion a/b is read as “a is to b as c is to d.” The first and last terms (a and d) are called the extremes, and the middle terms (b and c) are called the means. There are several useful properties involving proportions, and these properties can be established using algebra.
• Property 1: The Means-Extremes Property. In a proportion, the product of the means equals the product of the extremes. That is, if a/b = c/d (where b ≠ 0 and d ≠ 0), then a · d = b · c.
This is just an old-fashioned “cross-multiply” step used in algebra to get rid of unwanted denominators when dealing with fractions. You can use the Means-Extremes Property to solve algebraic equations.
• Example 1: Use the Means-Extremes Property to solve for x: x + 1/9 = x - 3/3.
Solution: If we apply the Means-Extremes Property to our equation, we have
3(x + 1) = 9(x − 3)
3x + 3 = 9x − 27
6x = 30
x = 5
Example 2: Use the Means-Extremes Property to solve for x: 4/x = x/9.
Solution: If you apply the Means-Extremes Property to your equation, you have
36 = x2
x = ± √ 36 = ±6.
Now combine these ideas with a problem involving geometry.
• Example 3: Suppose that two complementary angles are in the ratio 2 to 3. Find the measure of each angle.
• Solution: Let one of your angles have measure x and the other angle have measure y. Because your two angles are complementary, x + y = 90º, so y = 90º − x. Because the ratio of the angle measurements is 2 to 3, you have the following proportion:
• x/y = x/90º − x = 2/3
• You can then use the Means-Extremes Property:
• 3x = 2(90º − x)
• 3x = 180º − 2x
• 5x = 180º
• x = 36º
Now that you know the measure of one of your angles, you can find the measure of the second angle because the two angles are complementary:
• y = 90º − x = 90º − 36º = 54º
There are other properties of proportions. You can flip the proportions and mix and match numerators and denominators. Property 2 of proportions provides a list of the changes you can make.
• Property 2: In a proportion, the means or the extremes (or both the means and the extremes) may be interchanged. That is, if a/b = c/d and a, b, c, and d are all non-zero, then a/c = b/d, d/b = c/a and d/c = b/a.
Again, mixing numerators and denominators is not surprising because you've been doing this in algebra for years! But there is another property of proportions that might be a bit surprising. You have to be careful when you apply this property because it involves adding or subtracting things.
• Property 3: If a/b = c/d, where b ≠ 0 and d ≠ 0, then a + b/b = c + d/d.
Now that you have proportions under your belt, you can talk about the geometric mean of two numbers. If you start with a proportion where the two means are identical (and the two extremes may be different), such as a/b = b/d, then b is the geometric mean of a and d. You found the geometric means of the numbers 4 and 9 in Example 2. Although a pair of numbers actually has two geometric means (one positive and the other negative), geometers are only interested in the positive one. After one more example you'll be ready to apply these algebraic properties to geometry.
• Example 4: In Figure 13.1, suppose that AD is the geometric mean of BD and DC. If BC = 13 and BD = 9, find AD.
• Solution: Because AD is the geometric mean of BD and DC, You know that BD/AD = AD/DC. You are given that BD = 9, but you still need to find DC. Using the Segment Addition Postulate, BC = BD + DC. Substituting in for BC and BD gives 13 = 9 + DC, or DC = 4. Now you can substitute the values into the proportion and solve for AD:
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Propagation of Errors
# Propagation of Errors - Chapter 01.06 Propagation of Errors...
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Unformatted text preview: Chapter 01.06 Propagation of Errors If a calculation is made with numbers that are not exact, then the calculation itself will have an error. How do the errors in each individual number propagate through the calculations. Let’s look at the concept via some examples. Example 1 Find the bounds for the propagation error in adding two numbers. For example if one is calculating X + Y where X = 1.5 ± 0.05 , Y = 3.4 ± 0.04 . Solution By looking at the numbers, the maximum possible value of X and Y are X = 1.55 and Y = 3.44 Hence X + Y = 1.55 + 3.44 = 4.99 is the maximum value of X + Y . The minimum possible value of X and Y are X = 1.45 and Y = 3.36 . Hence X + Y = 1.45 + 3.36 = 4.81 is the minimum value of X + Y . Hence 4.81 ≤ X + Y ≤ 4.99. One can find similar intervals of the bound for the other arithmetic operations of X − Y , X * Y , and X / Y . What if the evaluations we are making are function evaluations instead? How do we find the value of the propagation error in such cases. If f is a function of several variables X 1 , X 2 , X 3 ,......., X n −1 , X n , then the maximum possible value of the error in f is ∆f ≈ ∂f ∂f ∂f ∂f ∆X 1 + ∆X 2 + ....... + ∆X n −1 + ∆X n ∂X 1 ∂X 2 ∂X n −1 ∂X n 01.06.1 01.06.2 Chapter 01.06 Example 2 The strain in an axial member of a square cross-section is given by F ∈= 2 hE where F =axial force in the member, N h = length or width of the cross-section, m E =Young’s modulus, Pa Given F = 72 ± 0.9 N h = 4 ± 0.1 mm E = 70 ± 1.5 GPa Find the maximum possible error in the measured strain. Solution 72 ∈= −3 2 (4 × 10 ) (70 × 10 9 ) = 64.286 × 10 −6 = 64.286µ ∂∈ ∂∈ ∂∈ ∆ ∈= ∆F + ∆h + ∆E ∂F ∂h ∂E ∂∈ 1 =2 ∂F h E ∂∈ 2F =− 3 ∂h hE ∂∈ F =− 2 2 ∂E hE 1 2F F ∆E = 2 ∆F + 3 ∆h + 2 2 ∆E hE hE hE = 1 2 × 72 × 0 .9 + × 0.0001 9 −3 3 (4 × 10 ) (70 × 10 ) (4 × 10 ) (70 × 10 9 ) −3 2 + 72 × 1.5 × 10 9 92 (4 × 10 ) (70 × 10 ) −3 2 = 8.0357 × 10 −7 + 3.2143 × 10 −6 + 1.3776 × 10 −6 = 5.3955 × 10 −6 = 5.3955µ Hence ∈= (64.286 µ ± 5.3955µ ) implying that the axial strain, ∈ is between 58.8905µ and 69.6815µ Propagation of Errors Example 3 01.06.3 Subtraction of numbers that are nearly equal can create unwanted inaccuracies. Using the formula for error propagation, show that this is true. Solution Let Then z = x− y ∆z = ∂z ∂z ∆x + ∆y ∂x ∂y = (1)∆x + (−1)∆y = ∆x + ∆y So the absolute relative change is ∆x + ∆y ∆z = z x− y As x and y become close to each other, the denominator becomes small and hence create large relative errors. For example if x = 2 ± 0.001 y = 2.003 ± 0.001 0.001 + 0.001 ∆z = z | 2 − 2.003 | = 0.6667 = 66.67% INTRODUCTION TO NUMERICAL METHODS Topic Propagation of Errors Summary Textbook notes on how errors propagate in arithmetic and function evaluations Major All Majors of Engineering Authors Autar Kaw Last Revised May 1, 2011 Web Site http://numericalmethods.eng.usf.edu ...
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# ∆Abc is an Equilateral Triangle. Point P is on Base Bc Such that Pc = 1 3 Bc, If Ab = 6 Cm Find Ap. - Geometry
ConceptSimilarity in Right Angled Triangles
#### Question
∆ABC is an equilateral triangle. Point P is on base BC such that PC = $\frac{1}{3}$ BC, if AB = 6 cm find AP.
#### Solution
∆ABC is an equilateral triangle.
It is given that,
$PC = \frac{1}{3}BC$
$\Rightarrow PC = \frac{1}{3} \times 6$
$\Rightarrow PC = 2 cm$
$\Rightarrow BP = 4 cm$
Since, ABC is an equilateral triangle, OA is the perpendicular bisector of BC.
∴ OC = 3 cm
⇒ OP = OC − PC
= 3 − 2
= 1 ...(1)
Now, According to Pythagoras theorem,
In ∆AOB,
${AB}^2 = {AO}^2 + {OB}^2$
$\Rightarrow \left( 6 \right)^2 = {AO}^2 + \left( 3 \right)^2$
$\Rightarrow 36 - 9 = {AO}^2$
$\Rightarrow {AO}^2 = 27$
$\Rightarrow AO = 3\sqrt{3} cm . . . \left( 2 \right)$
In ∆AOP,
${AP}^2 = {AO}^2 + {OP}^2$
$\Rightarrow {AP}^2 = \left( 3\sqrt{3} \right)^2 + \left( 1 \right)^2 \left( \text{From} \left( 1 \right) \text{and} \left( 2 \right) \right)$
$\Rightarrow {AP}^2 = 27 + 1$
$\Rightarrow {AP}^2 = 28$
$\Rightarrow AP = 2\sqrt{7} cm$
Hence, AP = 2$\sqrt{7}$ cm.
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [4]
Solution ∆Abc is an Equilateral Triangle. Point P is on Base Bc Such that Pc = 1 3 Bc, If Ab = 6 Cm Find Ap. Concept: Similarity in Right Angled Triangles.
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# How do you evaluate e^( ( pi)/4 i) - e^( ( 7 pi)/4 i) using trigonometric functions?
Mar 27, 2016
${e}^{\frac{\pi}{4} i} - {e}^{\frac{7 \pi}{4} i} = \sqrt{2} i$
#### Explanation:
As ${e}^{i \theta} = \cos \theta + i \sin \theta$, we have
${e}^{\frac{\pi}{4} i} = \cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)$ and
${e}^{\frac{7 \pi}{4} i} = \cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)$
Hence, ${e}^{\frac{\pi}{4} i} - {e}^{\frac{7 \pi}{4} i}$
= $\left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right) - \left(\cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)\right)$
As $\cos \left(\frac{\pi}{4}\right) = \cos \left(\frac{7 \pi}{4}\right) = \frac{1}{\sqrt{2}}$,
$\sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\sin \left(\frac{7 \pi}{4}\right) = - \frac{1}{\sqrt{2}}$
${e}^{\frac{\pi}{4} i} - {e}^{\frac{7 \pi}{4} i}$
= $\left(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right) - \left(\frac{1}{\sqrt{2}} + i \left(- \frac{1}{\sqrt{2}}\right)\right)$
= $\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + i \left(\frac{1}{\sqrt{2}}\right) + i \left(\frac{1}{\sqrt{2}}\right)\right)$
=$0 + i \frac{2}{\sqrt{2}}$ = $\sqrt{2} i$
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# How do you integrate int (x^3 +x^2+2x+1) / [(x^2+1) (x^2+2)] using partial fractions?
Feb 22, 2016
$\frac{1}{2} \ln \left({x}^{2} + 1\right) + \frac{1}{\sqrt{2}} \arctan \left(\frac{x}{\sqrt{2}}\right) + C$
$\frac{1}{2} \ln \left({x}^{2} + 1\right) + \sqrt{2} \arctan \left(\frac{x}{\sqrt{2}}\right) + C$
#### Explanation:
Note: If the denominator of the partial fraction is not factorable, always remember the degree of the numerator is one less. Like the problem
Step 1: Find the equivalent partial fraction
$\frac{{x}^{3} + {x}^{2} + 2 x + 1}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)} = \frac{A x + B}{{x}^{2} + 1} + \frac{C x + D}{{x}^{2} + 2}$
Step 2 Solve the partial fraction by multiply by Least common denominator to get
${x}^{3} + {x}^{2} + 2 x + 1 = \left(A x + B\right) \left({x}^{2} + 2\right) + \left({x}^{2} + 1\right) \left(C x + D\right)$
Step 3: Multiply/expand/foil the right hand side to get
${x}^{3} + {x}^{2} + 2 x + 1 = A {x}^{3} + 2 A x + B {x}^{2} + 2 B + C {x}^{3} + C x + D {x}^{2} + D$
Step 4: Set up the system of equation, using the corresponding coefficient for the corresponding terms
${x}^{3} \text{ "term" " } 1 = A + C$
${x}^{2} \text{ "term" " } 1 = B + D$
$x \text{ "term" " } 2 = 2 A + C$
${x}^{0} \text{ "term" " } 1 = 2 B + D$
Step 5: Solve the system of equation
$- 2 \left(1 = A + C\right) \implies - 2 = - 2 A + 2 C$
$+ \left(2 = 2 A + C\right)$
$0 = 3 C \implies C = 0$ , $A = 1$
$- 1 \left(2 B + D = 1\right) \implies - 2 B - D = - 1$
$+ \left(B + D = 1\right)$
$- B = 0 \implies B = 0 , D = 1$
Step 6: Write the equivalent partial fraction for the integral like this
$\int \frac{{x}^{3} + {x}^{2} + 2 x + 1}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)} \mathrm{dx}$ = $\int \frac{x}{{x}^{2} + 1} \mathrm{dx} + \int \frac{1}{{x}^{2} + 2} \mathrm{dx}$
Step 7 Integrate the integral
the first integral can be done by u substitution like this
let $u = {x}^{2} + 1$
$\mathrm{du} = 2 x \mathrm{dx} \implies \frac{\mathrm{du}}{2} = x \mathrm{dx}$
Note: $\int \frac{\mathrm{dx}}{x} = \ln | x | + C \text{ " "or" " } \log | x | + C$
The second integral is $\arctan$ , remember $\int \frac{1}{{a}^{2} + {u}^{2}} \mathrm{du} = \frac{1}{a} \arctan t \frac{u}{a} + C$
$\frac{1}{2} \ln \left({x}^{2} + 1\right) + \frac{1}{\sqrt{2}} \arctan \left(\frac{x}{\sqrt{2}}\right) + \text{Constant}$
I let my constant value to be $C$
this is another answer: $\frac{1}{2} \ln \left({x}^{2} + 1\right) + \frac{1}{\sqrt{2}} \arctan \left(\frac{x}{\sqrt{2}}\right) + C$ or
$\frac{1}{2} \ln \left({x}^{2} + 1\right) + \sqrt{2} \arctan \left(\frac{x}{\sqrt{2}}\right) + C$
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# Prof. Johnnie Baker Module Basic Structures: Sets
## Presentation on theme: "Prof. Johnnie Baker Module Basic Structures: Sets"— Presentation transcript:
Prof. Johnnie Baker jbaker@cs.kent.edu Module Basic Structures: Sets
Discrete Math CS 23022 Prof. Johnnie Baker Module Basic Structures: Sets
Acknowledgement Most of these slides were either created by Professor Bart Selman at Cornell University or else are modifications of his slides
Set Theory - Definitions and notation
A set is an unordered collection of objects referred to as elements. A set is said to contain its elements. Different ways of describing a set. 1 – Explicitly: listing the elements of a set {1, 2, 3} is the set containing “1” and “2” and “3.” list the members between braces. {1, 1, 2, 3, 3} = {1, 2, 3} since repetition is irrelevant. {1, 2, 3} = {3, 2, 1} since sets are unordered. {1,2,3, …, 99} is the set of positive integers less than 100; use ellipses when the general pattern of the elements is obvious. {1, 2, 3, …} is a way we denote an infinite set (in this case, the natural numbers). = {} is the empty set, or the set containing no elements. Note: {}
Set Theory - Definitions and notation
2 – Implicitly: by using a set builder notations, stating the property or properties of the elements of the set. S = {m| 2 ≤ m ≤ 100, m is integer} S is the set of all m such that m is between 2 and 100 and m is integer.
Set Theory - Ways to define sets
: and | are read “such that” or “where” Explicitly: {John, Paul, George, Ringo} Implicitly: {1,2,3,…}, or {2,3,5,7,11,13,17,…} Set builder: { x : x is prime }, { x | x is odd }. In general { x : P(x) is true }, where P(x) is some description of the set. Let D(x,y) denote “x is divisible by y.” Give another name for { x : y ((y > 1) (y < x)) D(x,y) }. Primes Can we use any predicate P to define a set S = { x : P(x) }?
Reveals contradiction in Frege’s naïve set theory. Set Theory - Russell’s Paradox "the set of all sets that do not contain themselves as members“ No! Can we use any predicate P to define a set S = { x : P(x) }? Define S = { x : x is a set where x x } So S must not be in S, right? Then, if S S, then by defn of S, S S. ARRRGH! But, if S S, then by defn of S, S S. There is a town with a barber who shaves all the people (and only the people) who don’t shave themselves. Who shaves the barber?
Aside: this layman’s version of Russell’s paradox has some drawbacks. Set Theory - Russell’s Paradox There is a town with a barber who shaves all the people (and only the people) who don’t shave themselves. Who shaves the barber? Does the barber shave himself? If the barber does not shave himself, he must abide by the rule and shave himself. If he does shave himself, according to the rule he will not shave himself. This sentence is unsatisfiable (a contradiction) because of the universal quantifier. The universal quantifier y will include every single element in the domain, including our infamous barber x. So when the value x is assigned to y, the sentence can be rewritten to: Contradiction!
Set Theory - Definitions and notation
Important Sets N = {0,1,2,3,…}, the set of natural numbers, non negative integers, (occasionally IN) Z = { …, -2, -1, 0, 1, 2,3, …), the set of integers Z+ = {1,2,3,…} set of positive integers Q = {p/q | p Z, q Z, and q0}, set of rational numbers R, the set of real numbers Note: Real number are the numbers that can be represented by an infinite decimal representation, such as …. The real numbers include both rational, and irrational numbers such as π and the and can be represented as points along an infinitely long number line. Check wikipedia
Set Theory - Definitions and notation
x S means “x is an element of set S.” x S means “x is not an element of set S.” A B means “A is a subset of B.” or, “B contains A.” or, “every element of A is also in B.” or, x ((x A) (x B)). Venn Diagram A B
Set Theory - Definitions and notation
A B means “A is a subset of B.” A B means “A is a superset of B.” A = B if and only if A and B have exactly the same elements. iff, A B and B A iff, A B and A B iff, x ((x A) (x B)). So to show equality of sets A and B, show: A B B A
Set Theory - Definitions and notation
A B means “A is a proper subset of B.” A B, and A B. x ((x A) (x B)) x ((x B) (x A)) x ((x A) (x B)) x ((x B) v (x A)) x ((x A) (x B)) x ((x B) (x A)) x ((x A) (x B)) x ((x B) (x A)) A B
Set Theory - Definitions and notation
Quick examples: {1,2,3} {1,2,3,4,5} {1,2,3} {1,2,3,4,5} Is {1,2,3}? Yes! x (x ) (x {1,2,3}) holds (for all over empty domain) Is {1,2,3}? No! Is {,1,2,3}? Yes! Is {,1,2,3}? Yes!
Set Theory - Definitions and notation
A few more: Is {a} {a}? Yes Is {a} {a,{a}}? Yes Is {a} {a,{a}}? Yes Is {a} {a}? No
Set Theory - Cardinality
If S is finite, then the cardinality of S, |S|, is the number of distinct elements in S. |S| = 3. If S = {1,2,3}, |S| = 1. If S = {3,3,3,3,3}, |S| = 0. If S = , |S| = 3. If S = { , {}, {,{}} }, If S = {0,1,2,3,…}, |S| is infinite.
CS 173 Set Theory - Power sets
If S is a set, then the power set of S is 2S = { x : x S }. aka P(S) We say, “P(S) is the set of all subsets of S.” 2S = {, {a}}. If S = {a}, 2S = {, {a}, {b}, {a,b}}. If S = {a,b}, 2S = {}. If S = , 2S = {, {}, {{}}, {,{}}}. If S = {,{}}, Why? Fact: if S is finite, |2S| = 2|S|. (if |S| = n, |2S| = 2n)
Set Theory – Ordered Tuples Cartesian Product
When order matters, we use ordered n-tuples The Cartesian Product of two sets A and B is: A x B = { <a,b> : a A b B} If A = {Charlie, Lucy, Linus}, and B = {Brown, VanPelt}, then We’ll use these special sets soon! A x B = {<Charlie, Brown>, <Lucy, Brown>, <Linus, Brown>, <Charlie, VanPelt>, <Lucy, VanPelt>, <Linus, VanPelt>} AxB |A|+|B| |A+B| |A||B| A1 x A2 x … x An = {<a1, a2,…, an>: a1 A1, a2 A2, …, an An} Size? A,B finite |AxB| = ? nn if (i) Ai = n
Set Theory - Operators The union of two sets A and B is:
A B = { x : x A v x B} If A = {Charlie, Lucy, Linus}, and B = {Lucy, Desi}, then A B = {Charlie, Lucy, Linus, Desi} A B
Set Theory - Operators The intersection of two sets A and B is:
A B = { x : x A x B} If A = {Charlie, Lucy, Linus}, and B = {Lucy, Desi}, then A B = {Lucy} A B
Set Theory - Operators The intersection of two sets A and B is:
A B = { x : x A x B} If A = {x : x is a US president}, and B = {x : x is deceased}, then A B = {x : x is a deceased US president} B A
Sets whose intersection is empty are called disjoint sets
Set Theory - Operators The intersection of two sets A and B is: A B = { x : x A x B} If A = {x : x is a US president}, and B = {x : x is in this room}, then A B = {x : x is a US president in this room} = Sets whose intersection is empty are called disjoint sets A B
Set Theory - Operators = U and U =
The complement of a set A is: A = { x : x A} If A = {x : x is bored}, then A = {x : x is not bored} A U = U and U = I.e., A = U – A, where U is the universal set. “A set fixed within the framework of a theory and consisting of all objects considered in the theory. “
Set Theory - Operators The set difference, A - B, is: U A B
A - B = { x : x A x B } A - B = A B
A B = { x : (x A x B) v (x B x A)}
Set Theory - Operators The symmetric difference, A B, is: A B = { x : (x A x B) v (x B x A)} = (A - B) U (B - A) like “exclusive or” A U B
Set Theory - Operators A B = (A - B) U (B - A) Theorem: Proof:
A B = { x : (x A x B) v (x B x A)} = { x : (x A - B) v (x B - A)} = { x : x ((A - B) U (B - A))} = (A - B) U (B - A) Q.E.D.
Set Theory - Identities
Directly from defns. Semantically clear. Identity Domination Idempotent A U = A A U = A A U = U A = A A A = A A A = A
Set Theory – Identities, cont.
Complement Laws Double complement A A = U A A = A = A
Set Theory - Identities, cont.
Commutativity Associativity Distributivity A U B = B U A B A A B = (A U B) U C = A U (B U C) A (B C) (A B) C = A U (B C) = A (B U C) = (A U B) (A U C) (A B) U (A C)
Proof by “diagram” (useful!), but we aim for a more formal proof.
DeMorgan’s I DeMorgan’s II (A U B) = A B (A B) = A U B Proof by “diagram” (useful!), but we aim for a more formal proof. A B
Proving identities (A U B) = A B (De Morgan) Prove that
() (x A U B) (x (A U B)) (x A and x B) (x A B) () (x ( A B)) (x A U B) (x A U B)
Haven’t we seen this before?
Alt. proof (A U B) = A B Prove that using a membership table. 0 : x is not in the specified set 1 : otherwise A B A B A U B 1 Haven’t we seen this before? General connection via Boolean algebras (Rosen chapt. 11)
(A U B) = {x : (x A v x B)}
Proof using logically equivalent set definitions. (A U B) = A B (A U B) = {x : (x A v x B)} = {x : (x A) (x B)} = {x : (x A) (x B)} = A B
Example What kind of law? Distributive law
X (Y - Z) = (X Y) - (X Z). True or False? Prove your response. What kind of law? (X Y) - (X Z) = (X Y) (X Z)’ Distributive law = (X Y) (X’ U Z’) = (X Y X’) U (X Y Z’) = U (X Y Z’) = (X Y Z’) = X (Y - Z) (X Z) = (X Z)’ (just different notation) Note:
Example Pv that if (A - B) U (B - A) = (A U B) then ______
Suppose to the contrary, that A B , and that x A B. A = B A B = A-B = B-A = Then x cannot be in A-B and x cannot be in B-A. Then x is not in (A - B) U (B - A). Do you see the contradiction yet? But x is in A U B since (A B) (A U B). Thus, (A - B) U (B - A) ≠ (A U B). Contradiction. Trying to prove p --> q Assume p and not q, and find a contradiction. Our contradiction was that sets weren’t equal. Thus, A B = .
Set Theory - Generalized Union/Intersection
Set Theory - Generalized Union/Intersection
Ex. Suppose that:
Example: Ex. Let U = N, and define: i=1,2,…,N A1 = {2,3,4,…} primes
Note: i starts at 2 primes Primes Composites N I have no clue. A2 = {4,6,8,…} A3 = {6,9,12,…} Union is all the composite numbers.
Set Theory - Inclusion/Exclusion
157 145 Example: There are 217 cs majors. 157 are taking CS23021. 145 are taking cs23022. 98 are taking both. How many are taking neither? 217 - ( ) = 13
Generalized Inclusion/Exclusion
B A Suppose we have: C What about 4 sets? And I want to know |A U B U C| |A U B U C| = |A| + |B| + |C| - |A B| - |A C| - |B C| + |A B C|
Generalized Inclusion/Exclusion
For sets A1, A2,…An we have:
Set Theory - Sets as bit strings
Let U = {x1, x2,…, xn}, and let A U. Then the characteristic vector of A is the n-vector whose elements, xi, are 1 if xi A, and 0 otherwise. Ex. If U = {x1, x2, x3, x4, x5, x6}, and A = {x1, x3, x5, x6}, then the characteristic vector of A is (101011)
Sets as bit strings Ex. If U = {x1, x2, x3, x4, x5, x6}, A = {x1, x3, x5, x6}, and B = {x2, x3, x6}, Then we have a quick way of finding the characteristic vectors of A B and A B. A 1 B A B A B Bit-wise OR Bit-wise AND
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# 3.8 Inverses and radical functions
Page 1 / 7
In this section, you will:
• Find the inverse of a polynomial function.
• Restrict the domain to find the inverse of a polynomial function.
A mound of gravel is in the shape of a cone with the height equal to twice the radius.
The volume is found using a formula from elementary geometry.
We have written the volume $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ in terms of the radius $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula
$r=\sqrt[3]{\frac{3V}{2\pi }}$
This function is the inverse of the formula for $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}r.$
In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process.
## Finding the inverse of a polynomial function
Two functions $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ are inverse functions if for every coordinate pair in there exists a corresponding coordinate pair in the inverse function, $\text{\hspace{0.17em}}g,\left(b,\text{\hspace{0.17em}}a\right).\text{\hspace{0.17em}}$ In other words, the coordinate pairs of the inverse functions have the input and output interchanged.
For a function to have an inverse function the function to create a new function that is one-to-one and would have an inverse function.
For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in [link] . We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.
Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ measured horizontally and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ measured vertically, with the origin at the vertex of the parabola. See [link] .
From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form $\text{\hspace{0.17em}}y\left(x\right)=a{x}^{2}.\text{\hspace{0.17em}}$ Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor $\text{\hspace{0.17em}}a.$
Our parabolic cross section has the equation
$y\left(x\right)=\frac{1}{2}{x}^{2}$
We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ the width will be given by $\text{\hspace{0.17em}}2x,\text{\hspace{0.17em}}$ so we need to solve the equation above for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.
To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values. On this domain, we can find an inverse by solving for the input variable:
This is not a function as written. We are limiting ourselves to positive $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values, so we eliminate the negative solution, giving us the inverse function we’re looking for.
what is set?
a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size
I got 300 minutes. is it right?
Patience
no. should be about 150 minutes.
Jason
It should be 158.5 minutes.
Mr
ok, thanks
Patience
100•3=300 300=50•2^x 6=2^x x=log_2(6) =2.5849625 so, 300=50•2^2.5849625 and, so, the # of bacteria will double every (100•2.5849625) = 258.49625 minutes
Thomas
what is the importance knowing the graph of circular functions?
can get some help basic precalculus
What do you need help with?
Andrew
how to convert general to standard form with not perfect trinomial
can get some help inverse function
ismail
Rectangle coordinate
how to find for x
it depends on the equation
Robert
yeah, it does. why do we attempt to gain all of them one side or the other?
Melissa
whats a domain
The domain of a function is the set of all input on which the function is defined. For example all real numbers are the Domain of any Polynomial function.
Spiro
Spiro; thanks for putting it out there like that, 😁
Melissa
foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24.
difference between calculus and pre calculus?
give me an example of a problem so that I can practice answering
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
I want to learn about the law of exponent
explain this
what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
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Mathematics for Technology I (Math 1131)
Durham College, Mathematics
Free
• 55 lessons
• 1 quizzes
• 10 week duration
• Numerical Computation
Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.
• Measurements
An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.
• Trigonometry with Right Triangles
Here we focus on right angle triangles within quadrant I of an x-y plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.
• Trigonometry with Oblique Triangles
This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.
• Geometry
This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.
Mathematics for Technology I (Math 1131)
Graphing a quadratic function without a table of values isn’t as simple as linear functions. This process involves knowing the parabola’s direction of opening, and finding its solutions (also called roots or x-intercepts), vertex point, and y-intercept. These features are shown below:
Recall that to graph a quadratic function, we could use a table of values with a specified domain (-3 to 3, for example). However, this isn’t practical all the time. You learned previously that you can find the roots of a quadratic function using the quadratic formula, and that it can have up to 2 roots. If two roots are present, the x-coordinate of the vertex can be found by taking the average of the two roots because all parabolas are symmetrical and its axis of symmetry lies on the vertex point. For example, if the quadratic function is f(x) = −2x² − 8x − 3, the roots will be:
1. x1 = -0.4188
2. x2 = -3.5811
To take the average of any two numbers, you sum them and divide by 2:
$\frac{{x}_{1}+{x}_{2}}{2}=\frac{–0.4188+\left(–3.5811\right)}{2}=–2\phantom{\rule{0ex}{0ex}}$
–2 is the x-coordinate of the vertex and the axis of symmetry. The y-coordinate is found by substituting this value back into the original equation:
∴ the vertex is at (–2, 5), and the x-intercepts are at (-0.4188, 0) and (-3.5811, 0). You can use these three points to create a sketch of the parabola. In fact, you can take this a step further by easily finding the y-intercept; this is done by setting x = 0:
∴ the y-intercept is (0, –3). Here’s what a sketch would look like (below). One thing to be mindful of is the a-coefficient determines if a parabola opens up or down. Give that it’s negative in this function, we see a parabola opening down, otherwise it would be up like a smile.
In case you’re provided a quadratic that doesn’t cross the x-axis (therefore, no solutions), you’ll need to use a special formula to find the vertex.
$h=–\frac{b}{2a}\phantom{\rule{0ex}{0ex}}$
• Where h represents the x-coordinate of the vertex. Note that this formula can also be used if roots exist.
Once you have found h, you substitute this value into the equation to find the corresponding y-coordinate of the vertex.
Question: Sketch the following function:
$f\left(x\right)=2.74{x}^{2}–3.12x+5.38\phantom{\rule{0ex}{0ex}}$
Solution: Here’s what we know about the function:
1. a = 2.74 ; b = –3.12 ; c = 5.38
2. Parabola opens up because 2.74 is positive.
3. There are no x-intercepts because the discriminant is less than 1.
• $\sqrt{{\mathbit{b}}^{\mathbf{2}}\mathbf{–}\mathbf{4}\mathbit{a}\mathbit{c}}={\left(–3.12\right)}^{2}–4\left(2.74\right)\left(5.38\right)<0\phantom{\rule{0ex}{0ex}}$
4. The x-coordinate of the vertex is:
• $h=–\frac{b}{2a}=\frac{–\left(–3.12\right)}{2\left(2.74\right)}=\mathbf{0}\mathbf{.}\mathbf{569}\phantom{\rule{0ex}{0ex}}$
• ∴ the y-coordinate is: f( 0.569 ) = 2.74 ( 0.569 )² – 3.12 ( 0.569 ) + 5.38 ⇒ 4.49
5. The y-intercept occurs when x = 0, f( 0 ) = 2.74 ( 0 )² – 3.12 ( 0 ) + 5.38 ⇒ (0, 5.38)
6. The graph looks like:
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# 6.8: Radius or Diameter of a Circle Given Area
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To find the radius, divide the area by pi, then take the square root.
Clara took her little sister, Grace, to the fish pond at the local park. Grace saw a penny in the center of the pond and wanted Clara to reach it for her. A sign stated that the area of the pond was 113.04 sq. ft. Can Clara reach the penny without falling in?
In this concept, you will learn how to find the radius (and diameter) of a circle if you know its area.
## Finding the Radius or Diameter of a Circle Given Area
The formula for the area of a circle, $$A=\pi r^2$$, can also be used to solve for the radius and diameter.
Let’s look an example.
The area of a circle is 113.04 square inches. What is its radius?
First, write the formula.
$$A=\pi r^2$$
Next, substitute in what you know.
$$113.04=(3.14)r^2$$
Next, begin isolating the r by dividing both sides of the equation by 3.14.
$$36=r^2$$
Then, take the square root of both sides.
$$6=r$$
The answer is r=6. The radius of the circle is 6 inches.
Example $$\PageIndex{1}$$
Earlier, you were given a problem about Clara and Grace, who were at the circular 113.04 sq. ft. fish pond.
Clara wondered if she could reach a penny in the middle without falling in.
Solution
First, write the formula.
$$A=\pi r^2$$
Next, substitute in what you know.
$$113.04=(3.14)r^2$$
Then, begin isolating the r by dividing both sides of the equation by 3.14.
$$36=r^2$$
Take the square root of both sides.
$$6=r$$
The answer is r=6. The radius of the circle is 6 feet. Unless Clara wants to swim with the goldfish, she’d best leave the penny where it is!
Example $$\PageIndex{2}$$
What is the diameter of a circle if its area is $$379.94 cm^2$$?
Solution
First, write the formula.
$$A=\pi r^2$$
Next, substitute in what you know.
$$379.94=(3.14)r^2$$
Next, begin isolating the r by dividing both sides of the equation by 3.14.
$$121=r^2$$
Then, take the square root of both sides.
$$11=r$$
Remember that you are solving for the diameter.
\begin{aligned} d&=2r \\ d&=2\times 11 \\ d&=22\end{aligned}
The answer is the diameter, d=22 cm.
Example $$\PageIndex{3}$$
Solve for the radius of a circle if area=153.86 sq. in.
Solution
First, write the formula.
$$A=\pi r^2$$
Next, substitute in what you know.
$$153.86=(3.14)r^2$$
Next, begin isolating the r by dividing both sides of the equation by 3.14.
$$49=r^2$$
Then, take the square root of both sides.
$$7=r$$
The answer is $$r=7$$. The radius of the circle is 7 inches.
Example $$\PageIndex{4}$$
Find the radius of a circle with an area of 379.94 sq. ft.
Solution
First, write the formula.
$$A=\pi r^2$$
Next, substitute in what you know.
$$379.94=(3.14)r^2$$
Then, begin isolating the r by dividing both sides of the equation by 3.14.
$$121=r^2$$
Take the square root of both sides.
$$11=r$$
The answer is r=11. The radius of the circle is 11 feet.
Example $$\PageIndex{5}$$
The area of a circle is 452.16 sq. m. Find its radius.
Solution
First, write the formula.
$$A=\pi r^2$$
Next, substitute in what you know.
$$452.16=(3.14)r^2$$
Then, begin isolating the r by dividing both sides of the equation by 3.14.
$$144=r^2$$
Take the square root of both sides.
$$12=r$$
The answer is r=12. The radius of the circle is 12 meters.
## Review
Find each radius given the area of the circle.
1. 12.56 sq. in.
2. 78.5 sq. m
3. 200.96 sq. cm
4. 254.34 sq. in.
5. 7.07 sq. ft.
6. 28.26 sq. m
Find each diameter given the area of the circle.
1. 12.56 sq. in.
2. 78.5 sq. m
3. 200.96 sq. cm
4. 254.34 sq. in.
5. 7.07 sq. ft.
6. 28.26 sq. m
7. 615.44 sq. ft.
8. 176.625 sq. m
9. 113.04 sq. ft.
## Vocabulary
Term Definition
Area Area is the space within the perimeter of a two-dimensional figure.
Circle A circle is the set of all points at a specific distance from a given point in two dimensions.
Diameter Diameter is the measure of the distance across the center of a circle. The diameter is equal to twice the measure of the radius.
Interactive Element
Video: Determine the Area of a Circle
Practice: Radius or Diameter of a Circle Given Area
Real World: Circles in the Cereal
This page titled 6.8: Radius or Diameter of a Circle Given Area is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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The 4 sided pyramid actually has two definitions--whether or not you count the base as a "side."
By geometric definition, a 4-sided pyramid has a triangular base with three sides, and when the sides and base are all equilateral triangles, it is called a tetrahedron.
The square-base pyramid, like the Pyramid of Giza, is also called a four-sided pyramid, because it has four equal vertical sides. Either four sided pyramid can be made with sticks, balls, sheets, or solid materials.
• The 4 sided pyramid actually has two definitions--whether or not you count the base as a "side."
• The square-base pyramid, like the Pyramid of Giza, is also called a four-sided pyramid, because it has four equal vertical sides.
Decide how many layers you want to make into a triangle-based pyramid. Two layers consist of a single ball on the top layer, and three balls on the second layer. Three layers consist of the same first two layers, plus six balls in the third layer.
Determine the number of balls you need. Each lower layer has the same number of balls as the next upper layer, plus a number of balls equal to the layer level below the top.
To figure how many balls for each layer, consider: the top layer is one ball. The second layer is three balls, or the first layer plus two balls. The third layer is six balls, or the second layer plus three balls. The fourth layer is 10 balls, or the third layer plus four balls. Two layers need 4 balls; three layers need 10 balls; four layers need 20 balls.
• Determine the number of balls you need.
• The second layer is three balls, or the first layer plus two balls.
Lay the bottom layer. Arrange the bottom layer in an equilateral triangle shape (all three sides equal). You can provide a border to hold the balls together, or glue the balls together.
• Arrange the bottom layer in an equilateral triangle shape (all three sides equal).
• You can provide a border to hold the balls together, or glue the balls together.
Lay the next layers. The balls will fit between the balls in the bottom layer. Continue laying the balls to finish the tetrahedral pyramid. You can glue the upper layers or just let them sit by gravity.
Decide how many layers you want to make the pyramid. This one is easy--each level is the square of sequential numbers.
Determine the number of balls you need to make the pyramid. The number of balls for each layer, from the top, is 1 x 1; 2 x 2; 3 x 3; 4 x 4; and so on. The seventh layer would have 7 x 7 balls or 49 balls. The ninth layer would have 9 x 9 or 81 balls, and so on. Add up the number of balls in each layer.
• Decide how many layers you want to make the pyramid.
• The number of balls for each layer, from the top, is 1 x 1; 2 x 2; 3 x 3; 4 x 4; and so on.
Lay the base layer. Put a square border around it to keep the balls from rolling away, or glue the balls together.
• Put a square border around it to keep the balls from rolling away, or glue the balls together.
Add the additional layers. The top layers will be held together by the base layer and don't have to be glued, but you can glue them if you want. Civil war soldiers made "cannon ball pyramids" in this shape.
Get four sticks each identical length for a triangular pyramid. For a square pyramid, use eight identical sticks.
Lay out three sticks in a triangle base for a triangle pyramid, and join them together (glue, tape, any way). Attach the other three sticks at the corners, and bring them together at the top to form the pyramid.
For a square pyramid, lay out four sticks for the square base and join them. Attach the other four sticks at the corners, and bring them together at the top to form the pyramid.
• Lay out three sticks in a triangle base for a triangle pyramid, and join them together (glue, tape, any way).
• Attach the other four sticks at the corners, and bring them together at the top to form the pyramid.
Use identical lengths for all edges for sheet or solid pyramids. A tetrahedral pyramid is made of four identical equilateral triangle shapes. Join each triangle to the next to form the pyramid.
For the square pyramid, start with a square base. Use the length of a side to determine the size of your triangular sides. Cut each triangle with the edges equal to one side of the square. Join the sides to the base, then join the sides of each triangle to form the pyramid.
Bevel solid sheets like wood so they fit together cleanly. The sides of a square pyramid fit together at 76 degrees--not 90 degrees like you might expect. Set the angle to cut the edges at 38 degrees.
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# How do you solve (5x)/2 - x = x/14 + 9/7?
Mar 25, 2018
$x = \frac{9}{10}$
#### Explanation:
$\frac{5 x}{2} - x = \frac{x}{14} + \frac{9}{7}$
Let's get a common denominator for the left side of the equation
$\frac{5 x}{2} - x \times \frac{2}{2} = \frac{x}{14} + \frac{9}{7}$
$\frac{5 x - 2 x}{2} = \frac{x}{14} + \frac{9}{7}$
$\frac{3 x}{2} = \frac{x}{14} + \frac{9}{7}$
Let's get all the $x$s on the same side
$\frac{3 x}{2} - \frac{x}{14} = \frac{9}{7}$
Common denominator
$\frac{7}{7} \times \frac{3 x}{2} - \frac{x}{14} = \frac{9}{7}$
$\frac{21 x}{14} - \frac{x}{14} = \frac{9}{7}$
Let's get another common denominator (we don't technically need this, but it makes the math easier later)
$\frac{20 x}{14} = \frac{9}{7} \times \frac{2}{2}$
$\frac{20 x}{14} = \frac{18}{14}$
Multiply both sides by $14$
$20 x = 18$
Divide by $20$ on both sides
$x = \frac{18}{20}$
$x = \frac{9}{10}$
Mar 25, 2018
$x = \frac{9}{10}$
#### Explanation:
We have an equation which has fractions.
$\frac{5 x}{2} - x = \frac{x}{14} + \frac{9}{7}$
We can get rid of the fractions immediately by multiplying the whole equation by the LCM of the denominators, which is $14$
$\frac{\textcolor{b l u e}{{\cancel{14}}^{7} \times} 5 x}{\cancel{2}} - \textcolor{b l u e}{14 \times} x = \frac{\textcolor{b l u e}{\cancel{14} \times} x}{\cancel{14}} + \frac{\textcolor{b l u e}{{\cancel{14}}^{2} \times} 9}{\cancel{7}} \text{ } \leftarrow$ cancel
$\text{ } 35 x - 14 x = x + 18$
$35 x - 14 x - x = 18$
$\textcolor{w h i t e}{\times \times \times \times} 20 x = 18$
$\textcolor{w h i t e}{\times \times \times \times x} x = \frac{18}{20}$
$\textcolor{w h i t e}{\times \times \times \times x} x = \frac{9}{10}$
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Examveda
# If $$\log 3\log \left( {{3^x} - 2} \right)\,$$ and $$\log \left( {{3^x} + 4} \right)$$ are in arithmetic progression, then x is equal to
A. $$\frac{8}{3}$$
B. $$\log {3^8}$$
C. $$\log {2^3}$$
D. 8
\eqalign{ & \log \left( {{3^x} - 2} \right) - \log 3 \cr & = \log \left( {{3^x} + 4} \right) - \log \left( {{3^x} - 2} \right) \cr & = \frac{{\log \left( {{3^x} - 2} \right)}}{{\log 3}} \cr}
$$= \frac{{\log \left( {{3^x} + 4} \right)}}{{\log \left( {{3^x} - 2} \right)}}$$ $$\left( {\therefore \log a - \log b = \log \frac{a}{b}} \right)$$
\eqalign{ & = \frac{{\log {3^x}}}{{\log 2\log 3}} \cr & = \frac{{x\log 3\log 4\log 2}}{{x\log 3}} \cr & = \frac{{x\log 3}}{{\log 2\log 3}} \cr & = \frac{{x\log 3\log 4\log 2}}{{x\log 2}} \cr & = \frac{x}{{\log 2}} \cr & = \log 4\log 2 \cr & \Rightarrow x = \log 4\log \,2\log 2 \cr & \Rightarrow x = \log 8 \cr & \Rightarrow x = \log {2^3} \cr}
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# Lesson1
```Elements of Distribution Theory
Random
The sample space.
Events.
Definition of probability
Axioms of probability theory
Conditional probability
Independence
Bayesian theory
Random Variable
Discrete Random Variable
Revision and Test 1
Continuous random Variable
Probability Mass Function
Probability Density Function
Cumulative Distribution Function
Expectation, Variance and Moment for Discrete random variable>
Moment generating function
Binomial Distribution
Poisson Distribution
Test2
Geometric Distribution
Hyper- geometric Distribution
Negative Binomial Distribution
Uniform Distribution
Exponential Distribution
experiment
Normal Distribution
Standard Normal Distribution
Lesson 1
Random experiment :
There are lots of phenomena in nature, like tossing a coin or tossing a die, whose outcomes
cannot be predicted with certainty in advance, but the set of all the possible outcomes is
known.
These are what we call random phenomena or random experiments. Probability theory is
concerned with such random phenomena or random experiments.
Sample Spaces and Events
Sample Spaces
Definition: A sample space is a collection of all possible outcomes of a random
experiment.
We can say "the sample space of a random experiment is a set S that includes all
possible outcomes of the experiment"; the sample space plays the role of the universal
set when modeling the experiment. For simple experiments, the sample space may be
precisely the set of possible outcomes. More often, for complex experiments, the
sample space is a mathematically convenient set that includes the possible outcomes
and perhaps other elements as well.
For example, if the experiment is to throw a standard dice and record the outcome, the
sample space is S={1,2,3,4,5,6} , the set of possible outcomes. On the other hand, if
the experiment is to capture a small insect and measure its body weight
(inmilligrams), we might conveniently take the sample space to be S=[0,∞) , even
though most elements of this set are practically impossible (we hope!).
Examples: Write the sample space for the following random experiments:
1234567-
Tossing a coin.
Tossing a die.
Tossing a coin twice.
Tossing a die twice.
Choosing a point from the interval (0; 1).
Measuring the lifetime of a light bulb.
Keeping on tossing a coin until one gets a Heads.
Solutions
1- The sample space is S {H , T }.
2- The sample space is S {1,2,3,4,5,6}.
3- The sample space is S {HH , HT , TH , TT }.
4- The sample space is S {( i, j ) :
i, j 1,2,...,6}, which contains 36
elements.
5- The sample space is S = (0; 1).
6- The sample space is S = [0;∞).
7- The sample space of this experiment is {H , TH , TTH , TTTH ,....} .
Events
Definition: Any subset E of the sample space S is called an
event.
Certain subsets of the sample space of an experiment are referred to as events.
Thus, an event is a set of outcomes of the experiment. Each time the experiment is
run, a given event A either occurs, if the outcome of the experiment is an element of
A , or does not occur, if the outcome of the experiment is not an element of A .
Intuitively, you should think of an event as a meaningful statement about the
experiment.
In particular, the sample space S itself is an event; by definition it always occurs. At
the other extreme, the empty set ∅ is also an event; by definition it never occurs.
The Algebra of Events
The standard algebra of sets leads to a grammar for discussing random experiments
and allows us to construct new events from given events. In the following exercises,
suppose that A and B are events.
A⊆B if and only if the occurrence of A implies the occurrence of B .
A∪B is the event that occurs if and only if A occurs or B occurs.
A∩B is the event that occurs if and only if A occurs and B occurs.
A and B are disjoint if and only if they are mutually exclusive; they cannot
both occur on the same run of the experiment.
A∖B is the event that occurs if and only if A occurs and B does not occur.
Ac is the event that occurs if and only if A does not occur.
c
c
(A∩ B )∪(B∩ A ) is the event that occurs if and only if one but not both of
the given events occurs. Recall that this event is the symmetric difference of
A and B , and is sometimes denoted AΔB .
c
c
(A∩B)∪( A ∩ B ) is the event that occurs if and only if both or neither of
the given events occurs.
Examples:
1- A coin is tossed twice and the outcome of each is recorded, find event that the
solution
the sample space is the set of all possible outcomes.
S= {(HH), (HT), (TH), (TT)}.
The event that the second toss was a Head is the subset E = {(H, H), (T, H)}.
2- Consider 3 light-bulbs. Our experiment consists of finding out which lightbulb burns out first, and how long (in hours) it takes for this to happen.
Solution
The sample space is: S = {(i, t) : i ∈ {1, 2, 3}, t ≥ 0} i tells you which one burns out,
and t gives how long it lasted, in hours. The event that the 2nd bulb burns out first,
and it lasts less than 3 hours is the set E = {(2, t) : t < 3}.
Exercise:
1- Find the event of occurrence of " Head" when tossing a coin.
2- The occurrence of an odd number when tossing a dice.
3- The appearance of "Tail" in the second toss when tossing a coin three times.
```
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Review question
# Can we sketch the graph of $y^2 = x^3$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R6372
## Solution
Sketch the graph of $y^2 = x^3$.
Since $y^2 \ge 0$, the graph only exists for $x^3 \ge 0$, that is, when $x \ge 0$. Taking square roots we see that the curve has two symmetrical branches, namely $\begin{equation*} y = x^{3/2} \quad\text{and}\quad y = -x^{3/2}. \end{equation*}$ It is clear that the positive branch increases. Its gradient is given by $\begin{equation*} \frac{dy}{dx} = \frac{3}{2}x^{1/2} \end{equation*}$
which is zero at $x=0$ and increases with $x$. Thus, we have
Calculate the area enclosed by the curve and the ordinate $x = 4$.
As the half-area above the $x$-axis is a reflection of the half-area below the $x$-axis, the area we want is $\begin{equation*} 2 \int_0^4 x^{3/2} \:dx = 2 \left[ \frac{x^{5/2}}{\frac{5}{2}}\right]_0^4 = \frac{4}{5} \times 4^{5/2} = \frac{4}{5} \times 2^5 = \frac{128}{5}. \end{equation*}$
Find also the volume generated by rotating this area about the axis of $x$.
The volume of revolution obtained by rotating a curve around the $x$-axis is $\begin{equation*} \int_{x_0}^{x_1} \pi y^2 \:dx. \end{equation*}$ So in our case, the volume is \begin{align*} \int_0^4 \pi x^3 \:dx = \pi \frac{4^4}{4} = 64\pi. \end{align*}
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# How to Calculate 1/16 Minus 10/31
Are you looking to work out and calculate how to subtract 1/16 from 10/31? In this really simple guide, we'll teach you exactly what 1/16 - 10/31 is and walk you through the step-by-process of how to subtract one fraction from another.
Want to quickly learn or show students how to calculate 1/16 minus 10/31? Play this very quick and fun video now!
To start with, the number above the line in a fraction is called a numerator and the number below the line is called the denominator.
Why do you need to know this? Well, because to subtract a fractions from another we need to first make sure both fractions have the same denominator.
Let's set up 1/16 and 10/31 side by side so they are easier to see:
1 / 16 - 10 / 31
Our denominators are 16 and 31. What we need to do is find the lowest common denominator of the two numbers, which is 496 in this case.
If we multiply the first denominator (16) by 31 we will get 496. If we we multiply the second denominator (31) by 16 we will also get 496. We also need to multiply the numerators above the line by the same amounts so that the fraction values are correct:
1 x 31 / 16 x 31 - 10 x 16 / 31 x 16
This is what 1/16 minus 10/31 looks like with the same denominator:
31 / 496 - 160 / 496
Now that these fractions have been converted to have the same denominator, we can subtract one numerator from the other to make one fraction:
31 - 160 / 496 = -129 / 496
You're done! You now know exactly how to calculate 1/16 - 10/31. Hopefully you understood the process and can use the same techniques to add other fractions together. The complete answer is below (simplified to the lowest form):
-129/496
## Convert 1/16 minus 10/31 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
-129 / 496 = -0.2601
### Cite, Link, or Reference This Page
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### Preset List of Fraction Subtraction Examples
Below are links to some preset calculations that are commonly searched for:
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# Normal Approximation to Binomial Distribution
-
The normal approximation means that you can use the normal distribution as an approximation to calculate the probabilities for binomial distribution. Note that this can be done under the following conditions:
1. The number of trials ‘n’ is large. (Generally values of n greater than 30 are considered to be large)
2. The probability of success ‘p’ in a trial is sufficiently small.
3. The mean ‘np’ is a finite number.
The below picture shows how the binomial distribution becomes closer and closer to the normal distribution as the number of trials increases.
Steps to apply the normal approximation to binomial distribution to calculate probabilities:
1. Given a random variable X, calculate the mean and variance using the formulae, Mean=np and Variance=np(1-p).
2. Check that the assumptions stated above for the validity of the approximation apply.
3. Treat the random variable as a normal variable with the mean and variance found in Step 1. But since we are using a continuous distribution (normal) to approximate a discrete (binomial) one, add a correction factor when calculating probabilities using the rules below.
4. The rules for the correction factor are:
P(X=n) gets replaced with P(n-0.5<X<n+0.5)
P(X<n) or P(X≤n) gets replaced with P(X<n+0.5)
P(X>n)or P(X ≥n) gets replaced with P(X>n-0.5)
5. Calculate the resulting probability by converting the normal distribution to standard normal distribution and using the Z table.
Examples:
1. Suppose a coin is thrown 100 times. Calculate the probability that you get heads 52 times using the normal approximation to the binomial distribution.
Solution: Let X denote the number of times we get heads in hundred tosses.
Then X follows binomial distribution with n=100 and p=1/2=0.5 and Mean= np =50.
Since the assumption for normal approximation are satisfied we may assume X follows normal distribution with mean=np=50 and variance=np(1-p)=25.
Then P(X=52) = P(51.5<x<52.5) (because of the correction factor)
=P (51.5-50/5< Z < 52.5-50/5) (converting normal to Z variate by subtracting mean and dividing by standard deviation)
=P(0.3<Z<0.5) = 0.073551 (by looking at the Z table)
2. Suppose a coin is thrown a 50 times. Calculate the probability that you get heads more than 36 times using the normal approximation to binomial distribution.
Solution: Let X denote the number of times we get heads in hundred tosses.
Then X follows binomial distribution with n=50 and p=1/2=0.5 and Mean= np =25.
Since the assumption for normal approximation are satisfied we may assume X follows normal distribution with mean=np=50 and variance=np(1-p)=12.5
Then P(X>36) = P(X>35.5) (because of the correction factor)
=P (Z > 35.5-25/√12.5) (converting normal to Z variate by subtracting mean and dividing by standard deviation)
=P(Z>2.9698) = 0.00149 (by looking at the Z table)
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# Section 6.
## 6.3 Applications of Exponential Functions
In the previous review, we examined exponential growth. Any process in which a
quantity grows by a fixed percentage each year (or each day, hour, etc.) can be modeled
by an exponential function. Compound interest is a good example of such a process.
## Discrete Compound Interest
If you put money in a savings account, then the bank will pay you interest (a percentage
of your account balance) at the end of each time period, typically one month or one
day. For example, if the time period is one month, this process is called monthly
compounding. The term compounding refers to the fact that interest is added to your
account each month and then in subsequent months you earn interest on the interest.
If the time period is one day, its called daily compounding. The exponential model
that describes this situation is called the discrete compound interest formula.
## Discrete Compound Interest
If P0 is the principal, r is the annual interest rate, and n is the number of times
that interest is compounded per year, then the balance at time t years is
r nt
P (t) = P0 1 + . (6.1)
n
I Example 1
If the principal is \$100, the annual interest rate is 5%, and interest is compounded
daily, what will be the balance after ten years?
## In formula (6.1), let P0 = 100, r = .05, n = 365, and t = 10:
.05 36510
P (10) = 100 1 + 164.87
365
Thus, you would have \$164.87 after ten years. Note that the final calculation was done
using a calculator.
I Example 2
If the principal is \$10 000, the annual interest rate is 5%, and interest is compounded
daily, what will be the balance after forty years?
## In formula (6.1), let P0 = 10 000, r = .05, n = 365, and t = 40:
.05 36540
P (40) = 10 000 1 + 73 880.44
365
1
## Version: Fall 2007
2 Chapter 6
Thus, you would have \$73 880.44 after forty years. Note that the final calculation was
done using a calculator.
## Continuous Compound Interest and the Number e
If we start with the discrete compound interest formula (6.1) and let the number
of times compounded per year (n) approach , then we end up with what is known as
continuous compounding.
## If P0 is the principal, r is the annual interest rate, and interest is compounded
continuously, then the balance at time t years is
P (t) = P0 ert . (6.2)
Note that you will again need a calculator to do the final evaluation on the following
problems.
I Example 3
If the principal is \$100, the annual interest rate is 5%, and interest is compounded
continuously, what will be the balance after ten years?
## In formula (6.2), let P0 = 100, r = 0.05, and t = 10:
P (10) = 100e(0.05)(10)
Use your calculator to approximate this result. Thus, you would have \$164.87 after ten
years.
I Example 4
If the principal is \$10,000, the annual interest rate is 5%, and interest is compounded
continuously, what will be the balance after forty years?
## Thus, you would have \$73 890.56 after forty years.
Properties of Logarithms;
Solving Exponential Equations
The usefulness of logarithms in calculations is based on the following three important
properties, known generally as the properties of logarithms.
## Version: Fall 2007
Section 6.3 Applications of Exponential Functions 3
Properties of Logarithms
## a) logb (M N ) = logb (M ) + logb (N )
M
b) logb = logb (M ) logb (N )
N
c) logb (M r ) = r logb (M )
logb (x)
loga (x) =
logb (a)
I Example 5
## Compute log2 (5).
Before applying the Change of Base Formula, lets see if we can estimate the value of
log2 (5). First recall that 2log2 (5) = 5. Now how large would the exponent on a base
of 2 need to be for the power to equal 5? Since 22 = 4 (too small) and 23 = 8 (too
large), we should expect log2 (5) to lie somewhere between 2 and 3. Indeed, applying
the Change of Base Formula with the common logarithm yields
log10 (5) log(5) .6989700043
log2 (5) = = 2.321928095.
log10 (2) log(2) .3010299957
According to the formula, we could instead use the natural logarithm to obtain the
loge (5) ln(5) 1.609437912
log2 (5) = = 2.321928095.
loge (2) ln(2) .6931471806
## Solving Exponential Equations
Property (c) (logb (M r ) = r logb (M )) is also used extensively to help solve exponential
equations, and thus will be an important tool when we work with applications. In
general terms, the main strategy for solving exponential equations is to (1) first isolate
the exponential, then (2) apply a logarithmic function to both sides, and then (3) use
property (c). Well illustrate the strategy with several examples.
I Example 6
Solve 8 = 5(3x ).
## Version: Fall 2007
4 Chapter 6
First isolate the exponential function on one side of the equation by dividing both sides
by 5:
1.6 = 3x
Then take the logarithm of both sides. Use either the common or natural log:
log(1.6) = log(3x )
Now use property (c) to move the exponent in front of the log on the right side:
log(1.6) = x log(3)
## Finally, solve for x by dividing both sides by log(3):
log(1.6)
=x
log(3)
log(1.6)
Thus, the exact value of x is , and the approximate value is 0.42781574.
log(3)
If the base of the exponential is either 10 or e, the correct choice of logarithm leads
to a faster solution:
I Example 7
Solve 3 = 4ex .
3 = 4ex
= 0.75 = ex isolate the exponential
x
= ln(0.75) = ln(e ) apply the natural log function
= ln(0.75) = x since ln(ex ) = x
= x .2876820725
In this case, because the base of the exponential function is e, the use of the natural
log function simplifies the solution.
## Exponential Growth and Decay
Exponential Growth Models
We so far have used exponential functions to model the growth of money. But we can
use the exact same analysis for quantities other than money. If P (t) represents the
amount of some quantity at time t years, and if P (t) grows at an annual rate r with
the growth continually added in, then we can conclude in the same manner that P (t)
must have the form
## Version: Fall 2007
Section 6.3 Applications of Exponential Functions 5
## where P0 is the initial amount at time t = 0, namely P (0).
Exponential Growth
If a function P (t) grows continually at a rate r > 0, then P (t) has the form
## P (t) = P0 ert , (6.4)
where P0 is the initial amount P (0). In this case, the quantity P (t) is said to
exhibit exponential growth, and r is the growth rate.
## Applications of Exponential Growth
We will now examine the role of exponential growth functions in some real-world ap-
plications. In the following examples, assume that the population is modeled by an
exponential growth function as in formula (6.4).
I Example 8
Suppose that the population of a certain country grows at an annual rate of 2%. If the
current population is 3 million, what will the population be in 10 years?
This is a future value problem. If we measure population in millions and time in years,
then P (t) = P0 ert with P0 = 3 and r = 0.02. Inserting these particular values into
formula (6.4), we obtain
P (t) = 3e0.02t .
## The population in 10 years is P (10) = 3e(0.02)(10) 3.664208 million.
I Example 9
In the same country as in Example 8, how long will it take the population to reach 5
million?
As before,
P (t) = 3e0.02t .
Now we want to know when the future value P (t) of the population at some time t will
equal 5 million. Therefore, we need to solve the equation P (t) = 5 for time t, which
5 = 3e0.02t .
Using the procedure for solving exponential equations that was presented in Section
8.6,
## Version: Fall 2007
6 Chapter 6
5 = 3e0.02t
5
= = e0.02t isolate the exponential
3
5
= ln = ln(e0.02t ) apply the natural log function
3
5
= ln = 0.02t since ln(ex ) = x
3
ln 53
= =t division
0.02
= t 25.54128.
Thus, it would take about 25.54 years for the population to reach 5 million.
examples.
I Example 10
## Suppose that a size of a bacterial culture is given by the function
P (t) = 100e0.15t ,
where the size P (t) is measured in grams and time t is measured in hours. How long
will it take for the culture to double in size?
The initial size is P0 = 100 grams, so we want to know when the future value P (t) at
some time t will equal 200. Therefore, we need to solve the equation P (t) = 200 for
time t, which leads to the exponential equation
200 = 100e0.15t .
## Using the same procedure as in the last example,
200 = 100e0.15t
= 2 = e0.15t isolate the exponential
= ln(2) = ln(e0.15t ) apply the natural log function
= ln(2) = 0.15t since ln(ex ) = x
ln(2)
= =t division
0.15
= t 4.620981.
Thus, it would take about 4.62 hours for the size to double.
## Version: Fall 2007
Section 6.3 Applications of Exponential Functions 7
## Exponential Decay Models
Weve observed that if a quantity increases continually at a rate r, then it is modeled
by a function of the form P (t) = P0 ert . But what if a quantity decreases instead? The
only difference is that the growth rate r in the formulas must be replaced by r since
the quantity is decreasing.
Exponential Decay
If a function P (t) decreases continually at a rate r > 0, then P (t) has the form
## P (t) = P0 ert , (6.5)
where P0 is the initial amount P (0). In this case, the quantity P (t) is said to
exhibit exponential decay, and r is the decay rate.
## Applications of Exponential Decay
The main example of exponential decay is radioactive decay. Radioactive elements and
isotopes spontaneously emit subatomic particles, and this process gradually changes
the substance into a different isotope. For example, the radioactive isotope Uranium-
238 eventually decays into the stable isotope Lead-206. This is a random process for
individual atoms, but overall the mass of the substance decreases according to the
exponential decay formula (6.5).
I Example 11
Suppose that a certain radioactive element has an annual decay rate of 10%. Starting
with a 200 gram sample of the element, how many grams will be left in 3 years?
This is a future value problem. If we measuring size in grams and time in years, then
P (t) = P0 ert with P0 = 200 and r = 0.10. Inserting these particular values into
formula (6.5), we obtain
P (t) = 200e0.10t .
## The amount in 3 years is P (3) = 200e(0.10)(3) 148.1636 grams.
I Example 12
Using the same element as in Example 11, if a particular sample of the element decays
to 50 grams after 5 years, how big was the original sample?
This is a present value problem, where the unknown is the initial amount P0 . As before,
r = 0.10, so
P (t) = P0 e0.10t .
## Version: Fall 2007
8 Chapter 6
50 = P (5) = P0 e(0.10)(5) .
## This equation can be solved by division:
50
(0.10)(5)
= P0
e
Finish by calculating the value of the left side to get P0 82.43606 grams.
I Example 13
Suppose that a certain radioactive isotope has an annual decay rate of 5%. How many
years will it take for a 100 gram sample to decay to 40 grams?
## Use P (t) = P0 ert with P0 = 100 and r = 0.05, so
P (t) = 100e0.05t .
Now we want to know when the future value P (t) of the size of the sample at some
time t will equal 40. Therefore, we need to solve the equation P (t) = 40 for time t,
which leads to the exponential equation
40 = 100e0.05t .
Using the procedure for solving exponential equations that was presented in Section
8.6,
40 = 100e0.05t
= 0.4 = e0.05t isolate the exponential
= ln(0.4) = ln(e0.05t ) apply the natural log function
= ln(0.4) = 0.05t since ln(ex ) = x
ln(0.4)
= =t division
0.05
= t 18.32581.
Thus, it would take approximately 18.33 years for the sample to decay to 40 grams.
We saw earlier that exponential growth processes have a fixed doubling time. Sim-
ilarly, exponential decay processes have a fixed half-life, the time in which one-half the
original amount decays.
I Example 14
Using the same element as in Example 13, what is the half-life of the element?
As before, r = 0.05, so
P (t) = P0 e0.05t .
## Version: Fall 2007
Section 6.3 Applications of Exponential Functions 9
The initial size is P0 grams, so we want to know when the future value P (t) at some
time t will equal one-half the initial amount, P0 /2. Therefore, we need to solve the
equation P (t) = P0 /2 for time t, which leads to the exponential equation
P0
= P0 e0.05t .
2
Using the same procedure as in the last example,
P0
= P0 e0.05t
2
1
= = e0.05t isolate the exponential
2
1
= ln = ln(e0.05t ) apply the natural log function
2
1
= ln = 0.05t since ln(ex ) = x
2
ln 12
= =t division
0.05
= t 13.86294.
## Thus, the half-life is approximately 13.86 years.
The process of radioactive decay also forms the basis of the carbon-14 dating tech-
nique. The Earths atmosphere contains a tiny amount of the radioactive isotope
carbon-14, and therefore plants and animals also contain some carbon-14 due to their
interaction with the atmosphere. However, this interaction ends when a plant or ani-
mal dies, so the carbon-14 begins to decay (the decay rate is 0.012%). By comparing
the amount of carbon-14 in a bone, for example, with the normal amount in a living
animal, scientists can compute the age of the bone.
I Example 15
Suppose that only 1.5% of the normal amount of carbon-14 remains in a fragment of
bone. How old is the bone?
## Use P (t) = P0 ert with r = 0.00012, so
P (t) = P0 e0.00012t .
The initial size is P0 grams, so we want to know when the future value P (t) at some
time t will equal 1.5% of the initial amount, 0.015P0 . Therefore, we need to solve the
equation P (t) = 0.015P0 for time t, which leads to the exponential equation
0.015P0 = P0 e0.00012t .
## Version: Fall 2007
10 Chapter 6
0.015P0 = P0 e0.00012t
= 0.015 = e0.00012t isolate the exponential
0.00012t
= ln (0.015) = ln(e ) apply the natural log function
= ln (0.015) = 0.00012t since ln(ex ) = x
ln (0.015)
= =t division
0.00012
= t 34998.
## Thus, the bone is approximately 34998 years old.
While the carbon-14 technique only works on plants and animals, there are other
similar dating techniques, using other radioactive isotopes, that are used to date rocks
and other inorganic matter.
|
# Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Intext Questions
You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Intext Questions
Try These (Textbook Page No. 26)
Question 1.
Complete the following table:
Solution:
Question 2.
Determine which metric unit you would use to express the following:
1. The length of your middle finger.
2. The weight of an elephant.
3. The weight of the ring.
4. The weight of the tablet.
5. The length of the safety pin.
6. The height of the building.
7. The length of the seashore in Tamilnadu.
8. The volume of a cup of coffee.
9. The capacity of water in the tank.
Solution:
1. Cm
2. Kilogram
3. gram
4. milligram
5. millimetre
6. metre
7. Kilometre
8. Millilitre
9. Litre
Try These (Textbook Page No. 28)
Question 1.
Convert the following.
(i) 23 km into m
(ii) 1.78 m into cm
(iii) 7814 m into km
(iv) 8.67 mm into cm
(v) 16 l into ml
(vi) 1500 ml into l
(vii) 2360 l into kl
(viii) 873 l into ml
(ix) 40 mg into g
(x) 1550 g into kg
(xi) 6.5 kg into g
(xii) 723 g into mg
Solution:
Try These (Textbook Page No. 30)
Question 1.
Five kilograms of compost is needed for a coconut tree for every six months. How many kilograms of compost is needed for 50 such coconut trees for one and a half years?
Solution:
Compost needed for a tree for 6 months = 5 kg
Compost needed for 50 trees for half a year = 5 × 50 = 250 kg.
Compost needed for 1$$\frac { 1 }{ 2 }$$ years = 250 × 3 = 750 kg. [∴ 1$$\frac { 1 }{ 2 }$$ years = 3 half years]
Question 2.
Say True or False and Justify.
(a) Is it correct: 4 m + 3 cm = 7 m
(b) Can we add the following?
(i) 6 litre + 7 kg
(ii) 3 m + 5 l
(iii) 400 ml + 300 g
Solution:
(a) False.
4 m + 3 cm = 400 cm + 3 cm = 403 cm.
(b) (i) No, we can’t add.
(ii) No. We can’t add.
(iii) No. We can’t add.
Recap (Textbook Page No. 33)
Question 1.
Read and write the time in the appropriate place.
Solution:
Try These (Textbook Page No. 33)
Question 1.
Say the following time in two ways.
(a) 9.20
(b) 4.50
(c) 5.15
(d) 6.45
(e) 11.30
Solution:
(a) 9.20
20 minutes past 9
40 minutes to 10
(b) 4.50
50 minutes past 4
10 minutes to 5
(c) 5.15
15 minutes past 5
45 minutes to 6
(d) 6.45
45 minutes past 6
15 minutes to 7
(e) 11.30
30 minutes past 11
30 minutes to 12
Try These (Textbook Page No. 34)
Question 1.
Convert the following:
(i) 4 hours = _____ minutes
(ii) 240 minutes = ____ hours
(iii) 30 minutes = _____ seconds
(iv) 3600 seconds = _____ hours
(v) 2 hours = _____ seconds
Solution:
(i) 4 hours = 240 minutes
4 hours = 4 × 60 minutes = 240 minutes
(ii) 240 minutes = 4 hrs
240 minutes = $$\frac{240}{60}$$ = 4 hrs
(iii) 30 minutes = 1800 seconds
30 minutes = 30 × 60 seconds = 1800 seconds
(iv) 3600 seconds = 1 hrs
3600 seconds = $$\frac{3600}{60}$$ minutes = 60 minutes = $$\frac{60}{60}$$ hours = 1 hour
(v) 2 hrs = 7200 seconds
2 hrs = 2 × 60 minutes = 120 minutes = 120 × 60 seconds = 7200 seconds
Try These (Textbook Page No. 36)
Question 1.
Convert the 12-hour format into the 24-hour format and vice versa.
Solution:
10 : 40 a.m. = 10 : 40 hours
11 a.m. = 11 : 00 hours
1 : 15 a.m. = 01 : 15 hours
5 a.m. = 05 : 00 hours
16 : 20 hours = 4 : 20 p.m.
00 : 40 hours = 12 :40 a.m.
1 p.m. = 13 : 00 hours
11 : 15 p.m. = 23 : 15 hours
3 p.m. = 15 :00 hours
12 midnight = 00 : 00 hours
12 : 25 hrs = 00 : 25 p.m.
4 : 10hrs = 4 : 10 a.m.
Try These (Textbook Page No. 39)
Question 1.
Check whether the following years as Ordinary or Leap Year?
1994; 1985; 2000; 2007; 2010; 2100
Solution:
We know that a leap year is divisible by 4.
In the case of a century which is divisible by 400 is a leap year.
1994 – Not a leap year (Not divisible by 4)
1985 – Not a leap year (Not divisible by 4)
2000 – It is a leap year (divisible by 400)
2007 – Not a leap year (Not divisible by 4)
2010 – Not a leap year (Not divisible by 4)
2100 – Not a leap year (Not divisible by 400)
Question 2.
How many days are there from 1st April to 30th June?
Solution:
Months from April to June are:
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Question
# When taking a test with m closed answers, a student knows the correct answer with probability p, otherwise he chooses one of the possible answers at random. What is the probability that the student knows the correct answer given that he answered the question correctly.
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## Answer to a math question When taking a test with m closed answers, a student knows the correct answer with probability p, otherwise he chooses one of the possible answers at random. What is the probability that the student knows the correct answer given that he answered the question correctly.
Frederik
4.6
To solve this problem, we can use Bayes' theorem.
Let's denote the following events:
A: The event that the student knows the correct answer.
B: The event that the student answered the question correctly.
We are asked to find P$A|B$, the probability that the student knows the correct answer given that he answered the question correctly.
According to Bayes' theorem, we have:
P$A|B$ = \frac{{P$B|A$ \cdot P$A$}}{{P$B$}}
We can calculate each of these probabilities step-by-step:
1. P$A$ is the probability that the student knows the correct answer. This is given as p.
2. P$B|A$ is the probability that the student answered the question correctly given that he knows the correct answer. This is equal to 1 since we are assuming that the student knows the correct answer.
3. P$B$ is the total probability that the student answered the question correctly.
To calculate P$B$, we need to consider two cases:
a) The student knows the correct answer, which happens with probability p.
b) The student does not know the correct answer, which happens with probability 1 - p. In this case, the probability of answering correctly by randomly choosing one of the possible answers is 1/m.
Therefore, we have:
P$B$ = P$A$ \cdot 1 + $1 - P(A$) \cdot \frac{1}{m} = p + \frac{1 - p}{m}
Now we can substitute these values back into Bayes' theorem to find P$A|B$:
P$A|B$ = \frac{{1 \cdot p}}{{p + \frac{1 - p}{m}}} = \frac{{p \cdot m}}{{pm + 1 - p}}
Answer: The probability that the student knows the correct answer given that he answered the question correctly is \frac{{p \cdot m}}{{pm + 1 - p}}
Frequently asked questions $FAQs$
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Question: What is the value of x in a right triangle if one acute angle measures 30 degrees and the hypotenuse is 8 units long?
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# How do you solve sqrt(50)+sqrt(2) ?
Sep 9, 2015
You can simplify $\sqrt{50} + \sqrt{2} = 6 \sqrt{2}$
#### Explanation:
If $a , b \ge 0$ then $\sqrt{a b} = \sqrt{a} \sqrt{b}$ and $\sqrt{{a}^{2}} = a$
So:
$\sqrt{50} + \sqrt{2} = \sqrt{{5}^{2} \cdot 2} + \sqrt{2} = \sqrt{{5}^{2}} \sqrt{2} + \sqrt{2}$
$= 5 \sqrt{2} + 1 \sqrt{2} = \left(5 + 1\right) \sqrt{2} = 6 \sqrt{2}$
In general you can try to simplify $\sqrt{n}$ by factorising $n$ to identify square factors. Then you can move the square roots of those square factors out from under the square root.
e.g. $\sqrt{300} = \sqrt{{10}^{2} \cdot 3} = 10 \sqrt{3}$
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# Factors of 955: Prime Factorization, Methods, and Examples
The factors of 955 are the numbers that, when divided by 955, leave nothing as a remainder. There are four factors of 955 in total.
 It is a composite number as it is divisible by more than two numbers. Â
### Factors of 955
Here are the factors of number 955.
Factors of 955: 1, 5, 191, and 955Â Â
### Negative Factors of 955
The negative factors of 955Â are similar to their positive aspects, just with a negative sign.
Negative Factors of 955: –1, -5, -191, and -955Â Â
### Prime Factorization of 955
The prime factorization of 955Â is the way of expressing its prime factors in the product form.
Prime Factorization: 5 x 191
In this article, we will learn about the factors of 955Â and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 955?
The factors of 955 are 1, 5, 191, and 955. These numbers are the factors as they do not leave any remainder when divided by 955.
The factors of 955Â are classified as prime numbers and composite numbers. The prime factors of the number 955 can be determined using the prime factorization technique.
## How To Find the Factors of 955?
You can find the factors of 955Â by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 955, create a list containing the numbers that are exactly divisible by 955 with zero remainders. One important thing to note is that 1 and 955 are 955’s factors, as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 955 are determined as follows:
$\dfrac{955}{1} = 955$
$\dfrac{955}{5} = 191$
Therefore, 1, 5, 191, and 955 are the factors of 955.
### Total Number of Factors of 955
For 955, there are four positive factors and four negative ones. So in total, there are eight factors of 955.Â
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 955 is given as:
The factorization of 955 is 1 x 5 x 191.
The exponent of 1, 5, and 191 is 1.
Adding 1 to each and multiplying them together results in eight.
Therefore, the total number of factors of 955 is eight. Four are positive, and four factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 955 by Prime Factorization
The number 955Â is a composite. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 955 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 955, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 955Â can be expressed as:
955 = 5 x 191
## Factors of 955 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 955, the factor pairs can be found as follows:
1 x 955 = 955
5 x 191 = 955
The possible factor pairs of 955 are given as (1, 955) and (5, 191).
All these numbers in pairs, when multiplied, give 955 as the product.
The negative factor pairs of 955 are given as:
-1 x -955 = 955
-5 x -191 = 955
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, 1, -5, -191, and -955 are called negative factors of 955.
The list of all the factors of 955, including positive as well as negative numbers, is given below.
Factor list of 955:Â 1, -1, 5, -5, 191, -191, 955, and -955Â Â
## Factors of 955 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 955 are there?
### Solution
The total number of Factors of 955 is four.
Factors of 955 are 1, 5, 191, and 955.
### Example 2
Find the factors of 955 using prime factorization.
### Solution
The prime factorization of 955 is given as:
955 $\div$ 5 = 191Â
191 $\div$ 191 = 1Â
So the prime factorization of 955 can be written as:
5 x 191 = 955
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# How to Multiply a Matrix by a Scalar?
When multiplying a matrix by a scalar, the resulting matrix will always have the same dimensions as the original matrix. In this step-by-step guide, you learn more about how to multiply a matrix by a scalar.
The scalar multiplication refers to the product of a real number and a matrix. In scalar multiplication, each input in the matrix is multiplied by the given scalar.
## Step by step guide to scalar multiplication of matrices
In matrix algebra, a real number is called a scalar. The scalar product of a real number, $$b$$, and a matrix $$A$$ is the matrix $$bA$$. Each element of matrix $$bA$$ is equal to $$b$$ times its corresponding element in $$A$$.
Given scalar $$\color {blue}{b}$$ and matrix $$A= \begin{bmatrix}a_{11} & a_{12} \\a_{21} & a_{22} \end{bmatrix}$$, $$\color{blue}{b}A=\begin{bmatrix}\color{blue}{b}a_{11} & \color{blue}{b}a_{12} \\\color{blue}{b}a_{21} &\color{blue}{b} a_{22} \end{bmatrix}$$
Properties of Scalar Multiplication:
Let $$A$$ and $$B$$ be $$m\times n$$ matrices. Let $$O_{m\times n}$$ be the $$m\times n$$ zero matrix and let $$p$$ and $$q$$ be scalars.
• Associative Property: $$\color{blue}{p(qA)=(pq)A}$$
• Closure Property: $$\color{blue}{pA}$$ is an $$\color{blue}{m×n}$$ matrix
• Commutative Property: $$\color{blue}{pA= Ap}$$
• Distributive Property: $$\color{blue}{(p+q)A=pA+qA}$$, $$\color{blue}{p(A+B)=pA+pB}$$
• Identity Property:$$\color{blue}{1. A=A}$$
• Multiplicative Property of $$-1$$: $$\color{blue}{(-1)A=-A}$$
• Multiplicative Property of $$0$$:$$\color{blue}{\space0. A=O_{m×n}}$$
### Scalar Multiplication of Matrices – Example 1:
If $$A=$$ $$\begin{bmatrix}-5 & -5 \\-1 & 2 \end{bmatrix}$$ , find $$3A$$.
solution:
$$3A=3$$ $$\begin{bmatrix}-5 & -5 \\-1 & 2 \end{bmatrix}$$ $$=$$ $$\begin{bmatrix}3(-5) & 3(-5) \\3(-1) &3(2) \end{bmatrix}$$ $$=$$ $$\begin{bmatrix}-15 & -15 \\-3& 6 \end{bmatrix}$$
### Scalar Multiplication of Matrices – Example 2:
If $$A=$$ $$\begin{bmatrix}6 & 4 & 24 \\1 & -9 & 8 \end{bmatrix}$$, find $$4A$$.
solution:
$$4A=$$ $$\begin{bmatrix}6 & 4 & 24 \\1 & -9 & 8 \end{bmatrix}$$ $$=$$ $$\begin{bmatrix}4(6) & 4(4) & 4(24) \\4(1) & 4(-9) & 4(8) \end{bmatrix}$$ $$=$$ $$\begin{bmatrix}24 & 16 & 96 \\4 & -36 & 32\end{bmatrix}$$
## Exercises for Scalar Multiplication of Matrices
### Solve.
1. $$\color{blue}{A=}$$$$\color{blue}{\begin{bmatrix}0 & 2 \\-2 & -5 \end{bmatrix}}$$, $$\color{blue}{6A}$$.
2. $$\color{blue}{B=}$$$$\color{blue}{\begin{bmatrix}-5 & 0 &2 \\7& -3& 4 \\ -1& 3 & 2 \end{bmatrix}}$$, $$\color{blue}{-3B}$$.
3. $$\color{blue}{D=}$$$$\color{blue}{\begin{bmatrix}6 & -2 \\3 & 7 \end{bmatrix}}$$, $$\color{blue}{F=}$$$$\color{blue}{\begin{bmatrix}1 & -2 \\-3 & 4 \end{bmatrix}}$$, $$\color{blue}{-2D+5F}$$.
4. $$\color{blue}{A=}$$$$\color{blue}{\begin{bmatrix}-5 & 2& 0 \\7 & -4 & 3 \\ -1 & 2 & 4 \end{bmatrix}}$$, $$\color{blue}{B=}$$$$\color{blue}{\begin{bmatrix}0 & -1 & 7 \\6 & -12 & 2 \\ 9 & 5 & 1 \end{bmatrix}}$$, $$\color{blue}{4A – 3B}$$.
1. $$\color{blue}{\begin{bmatrix}0 & 12 \\-12 & -30 \end{bmatrix}}$$
2. $$\color{blue}{\begin{bmatrix}15 & 0 & -6 \\-21 & 9 &-12 \\ 3 & -9 & -6 \end{bmatrix}}$$
3. $$\color{blue}{\begin{bmatrix}-7 & -6 \\-21 & 6 \end{bmatrix}}$$
4. $$\color{blue}{\begin{bmatrix}-20 & 11 & -21 \\10 & 20 & 6 \\ -31 & -7 & 13 \end{bmatrix}}$$
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# Probability THEORETICAL Theoretical probability can be found without doing and experiment. EXPERIMENTAL Experimental probability is found by repeating.
## Presentation on theme: "Probability THEORETICAL Theoretical probability can be found without doing and experiment. EXPERIMENTAL Experimental probability is found by repeating."— Presentation transcript:
Probability
THEORETICAL Theoretical probability can be found without doing and experiment. EXPERIMENTAL Experimental probability is found by repeating an experiment and observing the outcomes.
I am going to take 1 marble from the bag. What is the probability that I will pick out a red marble? Theoretical Probability I have three marbles in a bag. 1 marble is red 1 marble is blue 1 marble is green
Theoretical Probability Since there are three marbles and only one is red, I have a 1 in 3 chance of picking out a red marble. I can write this in three ways: As a fraction: 1/3 As a decimal:.33 As a percent: 33%
Experimental Probability Notice the Experimental Probability of drawing a red, blue or green marble.
Simple Probability The probability of one independent event. An independent event is one that does not affect the outcome (or probability) of any other event.
Simple Probability
probability of an event or P(event) is number of favorable outcomes total number of possible outcomes Simple Probability
probability of an event or P(event) is number of favorable outcomes total number of possible outcomes Example 1: Sarah rolls two 6-sided numbered cubes. What is the probability that the two numbers added together will equal 5? Simple Probability
probability of an event or P(event) is number of favorable outcomes total number of possible outcomes Example 1: Sarah rolls two 6-sided numbered cubes. What is the probability that the two numbers added together will equal 5? 4/36 = 1/9 Simple Probability
probability of an event or P(event) is number of favorable outcomes total number of possible outcomes Example 1: Sarah rolls two 6-sided numbered cubes. What is the probability that the two numbers added together will equal 5? 4/36 = 1/9 Simple Probability Example 2: What is the probability of drawing a king from a deck of cards?
probability of an event or P(event) is number of favorable outcomes total number of possible outcomes Example 1: Sarah rolls two 6-sided numbered cubes. What is the probability that the two numbers added together will equal 5? 4/36 = 1/9 Simple Probability Example 2: What is the probability of drawing a king from a deck of cards? 4/52 or 1/13
probability of an event or P(event) is number of favorable outcomes total number of possible outcomes Example 1: Sarah rolls two 6-sided numbered cubes. What is the probability that the two numbers added together will equal 5? 4/36 = 1/9 Simple Probability Example 2: What is the probability of drawing a king from a deck of cards? 4/52 or 1/13 Example 3: What is the probability of drawing a queen of hearts from as deck of cards?
probability of an event or P(event) is number of favorable outcomes total number of possible outcomes Example 1: Sarah rolls two 6-sided numbered cubes. What is the probability that the two numbers added together will equal 5? 4/36 = 1/9 Simple Probability Example 2: What is the probability of drawing a king from a deck of cards? 4/52 or 1/13 Example 3: What is the probability of drawing a queen of hearts from as deck of cards? 1/52
Example 1: Sarah rolls two 6-sided numbered cubes. What is the probability that the two numbers added together will equal 5? Simple Probability Example 2: What is the probability of drawing a king from a deck of cards? Example 3: What is the probability of drawing a queen of hearts from as deck of cards?
Compound Probability The probability of two independent events If an event is independent, it does not affect the outcome of other events.
“OR” P(A or B) = P(A) + P(B) Example: When you flip a fair coin and roll a number cube, what is the P(head or 4)? P(head or 4) = ½ + 1/6 = 3/6 + 1/6 = 4/6 = 2/3
Example: Alfred is going to the Lakeshore Animal Shelter to pick a new pet. Today, the shelter has 8 dogs, 7 cats, and 5 rabbits available for adoption. If Alfred randomly picks an animal to adopt, what is the probability that the animal would be a cat or a dog? 8/20 + 7/20 = 15/20 = 3/4
“And” P(A and B) = P(A) x P(B) Example: When you flip a fair coin and roll a number cube, what is the P(head and 4)? P(head, 4) = ½ x 1/6 = 1/12
Practice 1.P(heads, hearts) = 13/104 2.P(tails, four) = 4/104
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# Central Limit Theorem - Explained
What is the Central Limit Theorem?
# What is the Central Limit Theorem?
The central limit theorem states that with the assumption that all samples are equal in size, the example six gets larger, the distribution of same means approximates that of a normal distribution. In other terms, CLT is a statistical theory that states that given a large sample size from a population that has finite variance level, then all samples mean from the same population will approximately equal to the populations mean.
Back to: RESEARCH, ANALYSIS, & DECISION SCIENCE
## How is the Central Limit Theorem Used?
The Central Limit Theorem has an essential component where your samples means will also be the mean of the population. It means that if you add together the mean from all of your sample population, you will be able to find the average, which will also be your real population mean. Likewise, if in your sample, you can get all of the standard deviations, then you will be able to find your populations actual standard deviation. CLT is a valuable phenomenon that can help you to predict a populations characteristics correctly.
## What are the Assumptions Behind the CLT?
CLT, as a technique, has various assumptions. They are as follows:
• The sampling of data must be selected in a random manner
• The samples collected should be independent of each other in that one sample should not influence others.
• When there is a selection of samples without replacement, then all samples should not exceed ten percent of the population.
• The size of the sample has to be large enough. Note that the size of the sample will depend on the population. Where the population happens to be symmetric, then a sample of 30 can work well.
## How Central Limit Theorem works
Example Lets assume that in a university science department, there are 15 sections where each section hosts close to 100 students. You have the task of calculating the students average weight in that science department. In this case, you need to calculate the average. Below is the procedure for calculating the average:
• First, you will have to measure all the students weight in the science department
• Secondly, you add all the weights
• Lastly, you will divide the weights total sum with the students total number to get the average
However, if the data is humongous CLT approach wont make sense at all. The reason is that it will be tiresome and a long process to measure the weight of all the students. The following is an alternative approach to this. You first create groups of students selected randomly from the class. Take this to be your sample. You will then draw multiple samples of 30 students each. You will then do the following:
• Calculate these samples mean
• Calculate the means of the samples mean
• The value will give you students approximate mean weights in the science department
## Finance and the Central Limit Theorem
CLT can be used to simplify a significant number of analysis procedures. For instance, all types of investors can use it to assess their stocks returns, manage risk, and construct portfolios. Note that in statistics, statisticians consider CLT to be the basis for sampling. What this means is that it holds the foundation for sampling as well as statistical analysis in finance. Example Lets assume that John wants to analyze his overall stock index return consisting of 1,000 stocks. He can take random stock samples from the index to get an estimate of the total indexs gains. For CTL to hold, John decides to randomly pick the samples and makes sure that he evaluates at least 30 stocks in each sample. For him to avoid bias in the sample selection, he ensures that he replaces the previously selected stocks for selecting in other examples. Generally, the average returns from the selected samples approximate those of the profits for the whole index, including that of normal distribution. Note that the approximation will still hold even if there is no normal distribution of the entire indexs actual returns.
## Central Limit Theorem significance
There various statistical significance and practical applications when you used the central limit theorem. Statistical significance of CLT To analyze data, you will have to apply statistical methods such as hypothesis testing as well as constructing confidence intervals. According to this method, there is the assumption that there is a distribution of a population. Where there are non-normal distributions or distributions that are not known, we treat the sampling distribution as normal. Also, the sample means standard deviation is bound to decrease if you increase the samples taken from the population. With this, you will be able to estimate the population mean with much accuracy. You can also use the sample mean to come up with a range of values called the confidence interval.
## Practical Application of Central Limit Theorem
Political analysts can apply CLT on the election or political polls. They use the CLT to calculate political polls. For instance, CLT helps them to estimate the percentage of those who support a particular political party or candidate. You can also use the CLT to calculate confidence intervals such as calculation of mean family income for a specific region.
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# Complex Numbers and Roots
## Presentation on theme: "Complex Numbers and Roots"— Presentation transcript:
Complex Numbers and Roots
5-5 Complex Numbers and Roots Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2
Simplify each expression.
Warm Up Simplify each expression. 2. 1. 3. Find the zeros of each function. 4. f(x) = x2 – 18x + 16 5. f(x) = x2 + 8x – 24
Objectives Define and use imaginary and complex numbers.
Solve quadratic equations with complex roots.
Vocabulary imaginary unit imaginary number complex number real part
imaginary part complex conjugate
You can see in the graph of f(x) = x2 + 1 below that f has no real zeros. If you solve the corresponding equation 0 = x2 + 1, you find that x = ,which has no real solutions. However, you can find solutions if you define the square root of negative numbers, which is why imaginary numbers were invented. The imaginary unit i is defined as You can use the imaginary unit to write the square root of any negative number.
Example 1A: Simplifying Square Roots of Negative Numbers
Express the number in terms of i. Factor out –1. Product Property. Simplify. Multiply. Express in terms of i.
Example 1B: Simplifying Square Roots of Negative Numbers
Express the number in terms of i. Factor out –1. Product Property. Simplify. Express in terms of i.
Express the number in terms of i.
Check It Out! Example 1a Express the number in terms of i. Factor out –1. Product Property. Product Property. Simplify. Express in terms of i.
Express the number in terms of i.
Check It Out! Example 1b Express the number in terms of i. Factor out –1. Product Property. Simplify. Multiply. Express in terms of i.
Express the number in terms of i.
Check It Out! Example 1c Express the number in terms of i. Factor out –1. Product Property. Simplify. Multiply. Express in terms of i.
Example 2A: Solving a Quadratic Equation with Imaginary Solutions
Solve the equation. Take square roots. Express in terms of i. Check x2 = –144 –144 (12i)2 144i 2 144(–1) x2 = –144 –144 144(–1) 144i 2 (–12i)2
Example 2B: Solving a Quadratic Equation with Imaginary Solutions
Solve the equation. 5x = 0 Add –90 to both sides. Divide both sides by 5. Take square roots. Express in terms of i. 5x = 0 5(18)i 2 +90 90(–1) +90 Check
Check It Out! Example 2a Solve the equation. x2 = –36 Check x2 = –36
Take square roots. Express in terms of i. Check x2 = –36 –36 (6i)2 36i 2 36(–1) (–6i)2
Check It Out! Example 2b Solve the equation. x2 + 48 = 0 x2 = –48
Add –48 to both sides. Take square roots. Express in terms of i. Check x = 0 + 48 (48)i 48(–1) + 48
Check It Out! Example 2c Solve the equation. 9x2 + 25 = 0 9x2 = –25
Add –25 to both sides. Divide both sides by 9. Take square roots. Express in terms of i.
A complex number is a number that can be written in the form a + bi, where a and b are real numbers and i = The set of real numbers is a subset of the set of complex numbers C. Every complex number has a real part a and an imaginary part b.
Real numbers are complex numbers where b = 0
Real numbers are complex numbers where b = 0. Imaginary numbers are complex numbers where a = 0 and b ≠ 0. These are sometimes called pure imaginary numbers. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
Example 3: Equating Two Complex Numbers
Find the values of x and y that make the equation 4x + 10i = 2 – (4y)i true . Real parts 4x + 10i = 2 – (4y)i Imaginary parts Equate the imaginary parts. Equate the real parts. 10 = –4y 4x = 2 Solve for y. Solve for x.
Find the values of x and y that make each equation true.
Check It Out! Example 3a Find the values of x and y that make each equation true. 2x – 6i = –8 + (20y)i Real parts 2x – 6i = –8 + (20y)i Imaginary parts Equate the real parts. Equate the imaginary parts. 2x = –8 –6 = 20y x = –4 Solve for y. Solve for x.
Find the values of x and y that make each equation true.
Check It Out! Example 3b Find the values of x and y that make each equation true. –8 + (6y)i = 5x – i Real parts –8 + (6y)i = 5x – i Imaginary parts Equate the imaginary parts. –8 = 5x Equate the real parts. Solve for y. Solve for x.
Example 4A: Finding Complex Zeros of Quadratic Functions
Find the zeros of the function. f(x) = x2 + 10x + 26 x2 + 10x + 26 = 0 Set equal to 0. x2 + 10x = –26 + Rewrite. x2 + 10x + 25 = – Add to both sides. (x + 5)2 = –1 Factor. Take square roots. Simplify.
Example 4B: Finding Complex Zeros of Quadratic Functions
Find the zeros of the function. g(x) = x2 + 4x + 12 x2 + 4x + 12 = 0 Set equal to 0. x2 + 4x = –12 + Rewrite. x2 + 4x + 4 = –12 + 4 Add to both sides. (x + 2)2 = –8 Factor. Take square roots. Simplify.
Find the zeros of the function.
Check It Out! Example 4a Find the zeros of the function. f(x) = x2 + 4x + 13 x2 + 4x + 13 = 0 Set equal to 0. x2 + 4x = –13 + Rewrite. x2 + 4x + 4 = –13 + 4 Add to both sides. (x + 2)2 = –9 Factor. Take square roots. x = –2 ± 3i Simplify.
Find the zeros of the function.
Check It Out! Example 4b Find the zeros of the function. g(x) = x2 – 8x + 18 x2 – 8x + 18 = 0 Set equal to 0. x2 – 8x = –18 + Rewrite. x2 – 8x + 16 = – Add to both sides. Factor. Take square roots. Simplify.
The solutions and are related
The solutions and are related. These solutions are a complex conjugate pair. Their real parts are equal and their imaginary parts are opposites. The complex conjugate of any complex number a + bi is the complex number a – bi. If a quadratic equation with real coefficients has nonreal roots, those roots are complex conjugates. When given one complex root, you can always find the other by finding its conjugate. Helpful Hint
Example 5: Finding Complex Zeros of Quadratic Functions
Find each complex conjugate. B. 6i A i 0 + 6i 8 + 5i Write as a + bi. Write as a + bi. 0 – 6i Find a – bi. 8 – 5i Find a – bi. –6i Simplify.
Find each complex conjugate.
Check It Out! Example 5 Find each complex conjugate. A. 9 – i B. 9 + (–i) Write as a + bi. Write as a + bi. 9 – (–i) Find a – bi. Find a – bi. 9 + i Simplify. C. –8i 0 + (–8)i Write as a + bi. 0 – (–8)i Find a – bi. 8i Simplify.
Lesson Quiz 1. Express in terms of i. Solve each equation. 3. x2 + 8x +20 = 0 2. 3x = 0 4. Find the values of x and y that make the equation 3x +8i = 12 – (12y)i true. 5. Find the complex conjugate of
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## Practice problem2
In this page practice problem2 we will see how to solve subtraction problems on complex numbers.
Solution:
1. (2+3i)-(9-11i)
While subtracting two complex numbers we have to subtract the real parts together and imaginary parts together. We have to subtract the real and imaginary parts like ordinary subtraction.
The real parts of the given numbers are 2 and 9.
The imaginary parts of the given numbers are 3i and -11i.
[Here we have to subtract -11i from 3i. Writing this with subtracting symbol we will get 3i-(-11i). So the underlined part will become as 11i as negative multiplied by negative sign is positive sign.]
Let us subtract these now.
= (2-9) + (3i-(-11i))
= (-7) + (3i+11i)
= -7 +14i
2. (4-5i)-(3+4i)
The real parts of the given numbers are 4 and 3.
The imaginary parts of the given numbers are -5i and 4i.
Let us subtract these numbers now.
= (4-3) + (-5i-4i)
= 1 + (-9i)
= 1 - 9i
3. (-8+9i)-(-7+5i)
The real parts of the given numbers are -8 and -7.
The imaginary parts of the given numbers are 9i and 5i.
Let us subtract these numbers now.
= (-8-(-7)) + (9i-5i)
= (-8+7) + 4i
= -1 + 4i
4. (-8-9i)-(-11-13i)
The real parts of the given numbers are -8 and -11.
The imaginary parts of the given numbers are -9i and -13i.
Let us subtract these numbers now.
= ((-8)-(-11)) + ((-9i)-(-13i))
= (-8 +11) + (-9i+13i)
= 3 + 4i
Parents and teachers can guide the students to practice the problems discussed in practice problem2 and ask them to solve the problems given below.
Problems for practice:
1. (23-8i)-(-7-2i)
2. (4+9i)-(2-5i)
If you have any doubts you can contact us by mail, we will clear all your doubts.
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# Opposite number
number that, when added to the original number, yields zero
In mathematics, the opposite or additive inverse of a number ${\displaystyle k}$ is a number ${\displaystyle n}$ which, when added to ${\displaystyle k}$, results in 0. The opposite of ${\displaystyle a}$ is ${\displaystyle -a}$.[1][2] For example, −7 is the opposite of 7, because ${\displaystyle 7+-7=0}$.
## Definition
A number ${\displaystyle x}$ is called an opposite number of another number ${\displaystyle y}$ if ${\displaystyle x+y=0}$ . By definition, ${\displaystyle -x}$ is the opposite number of ${\displaystyle x}$ . For example, ${\displaystyle -2}$ is the opposite number of ${\displaystyle 2}$ and vice-versa. This is because ${\displaystyle -2+2=0.}$
Opposite numbers are also known as additive inverses.
## Properties
The opposite numbers satisfy the properties listed below.[3]
1. The opposite of 0 is 0.
2. Two opposite numbers have the same absolute value. This follows from the fact that the ${\displaystyle -a}$ is the opposite number of ${\displaystyle a}$ and both have the same absolute value ${\displaystyle |-a|=|a|}$ .
3. The opposite of a positive number is the negative version of the number. The opposite of a negative number is the positive version of the number.
4. opposite numbers are located in the opposite direction on a number line having the same distance from the origin. That is, they are symmetric about the origin on a number line.
5. The sum of two opposite numbers is always zero, because ${\displaystyle -a+a=0.}$
6. The division of two non-zero opposite numbers is always ${\displaystyle -1}$ , because ${\displaystyle {\frac {-a}{a}}={\frac {a}{-a}}=-1.}$
7. Each number has a unique opposite number.
## References
1. Weisstein, Eric W. "Additive Inverse". mathworld.wolfram.com. Retrieved 2020-08-27.
2. "Additive Inverse". www.learnalberta.ca. Retrieved 2020-08-27.
3. Ghosh, N. (6 February 2022). "Opposite Numbers: Definition, Examples, and Properties". Mathstoon. Retrieved 6 February 2022.
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## Intermediate Algebra: Connecting Concepts through Application
Shania forgot to multiply by the reciprocal fraction when dividing. Once this step is done, she can start cancelling out the factors to find the answer. $= \frac{(h+5)(h+4)^{2}}{(h+7)(h+2)^{2}}$
Shania forgot to multiply by the reciprocal fraction when dividing. Once this step is done, she can start cancelling out the factors to find the answer. $\frac{(h+4)(h-8)}{(h+2)(h+7)} \div \frac{(h-8)(h+2)}{(h+4)(h+5)}$ $= \frac{(h+4)(h-8)}{(h+2)(h+7)} \times \frac{(h+4)(h+5)}{(h-8)(h+2)}$ $= \frac{(h+4)}{(h+2)(h+7)} \times \frac{(h+4)(h+5)}{(h+2)}$ $= \frac{(h+4)(h+4)(h+5)}{(h+2)(h+7)(h+2)}$ $= \frac{(h+5)(h+4)^{2}}{(h+7)(h+2)^{2}}$
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# The Derivative
We have seen how to create, or derive, a new function $$f'(x)$$ from a function $$f(x)$$, and that this new function carries important information. In one example we saw that $$f'(x)$$ tells us how steep the graph of $$f(x)$$ is; in another we saw that $$f'(x)$$ tells us the velocity of an object if $$f(x)$$ tells us the position of the object at time $$x$$. As we said earlier, this same mathematical idea is useful whenever $$f(x)$$ represents some changing quantity and we want to know something about how it changes, or roughly, the "rate'' at which it changes. Most functions encountered in practice are built up from a small collection of "primitive'' functions in a few simple ways, for example, by adding or multiplying functions together to get new, more complicated functions. To make good use of the information provided by $$f'(x)$$ we need to be able to compute it for a variety of such functions.
We will begin to use different notations for the derivative of a function. While initially confusing, each is often useful so it is worth maintaining multiple versions of the same thing.
Consider again the function $$f(x)=\sqrt{625-x^2}$$. We have computed the derivative $$f'(x)=-x/\sqrt{625-x^2}$$, and have already noted that if we use the alternate notation $$y=\sqrt{625-x^2}$$ then we might write $$y'=-x/\sqrt{625-x^2}$$. Another notation is quite different, and in time it will become clear why it is often a useful one. Recall that to compute the the derivative of $$f$$ we computed $$\lim_{\Delta x\to0} {\sqrt{625-(7+\Delta x)^2} - 24\over \Delta x}.$$ The denominator here measures a distance in the $$x$$ direction, sometimes called the "run'', and the numerator measures a distance in the $$y$$ direction, sometimes called the "rise,'' and "rise over run'' is the slope of a line.
Recall that sometimes such a numerator is abbreviated $$\Delta y$$, exchanging brevity for a more detailed expression. So in general, a derivative is given by y'=\lim_{\Delta x\to0} {\Delta y\over \Delta x}. To recall the form of the limit, we sometimes say instead that $${dy\over dx}=\lim_{\Delta x\to0} {\Delta y\over \Delta x}.$$ In other words, $$dy/dx$$ is another notation for the derivative, and it reminds us that it is related to an actual slope between two points. This notation is called Leibniz notation, after Gottfried Leibniz, who developed the fundamentals of calculus independently, at about the same time that Isaac Newton did. Again, since we often use $$f$$ and $$f(x)$$ to mean the original function, we sometimes use $$df/dx$$ and $$df(x)/dx$$ to refer to the derivative. If the function $$f(x)$$ is written out in full we often write the last of these something like this $$f'(x)={d\over dx}\sqrt{625-x^2}$$ with the function written to the side, instead of trying to fit it into the numerator.
Example 2.4.1
Find the derivative of $$y=f(t)=t^2$$.
SOLUTION
We compute \eqalign{ y' = \lim_{\Delta t\to0}{\Delta y\over\Delta t}&= \lim_{\Delta t\to0}{(t+\Delta t)^2-t^2\over\Delta t}\cr& =\lim_{\Delta t\to0}{t^2+2t\Delta t+\Delta t^2-t^2\over\Delta t}\cr& =\lim_{\Delta t\to0}{2t\Delta t+\Delta t^2\over\Delta t}\cr& =\lim_{\Delta t\to0} 2t+\Delta t=2t.\cr}
Remember that $$\Delta t$$ is a single quantity, not a "$$\Delta$$'' times a "$$t$$'', and so $$\Delta t^2$$ is $$(\Delta t)^2$$ not $$\Delta (t^2)$$.
Example 2.4.2
Find the derivative of $$y=f(x)=1/x$$.
SOLUTION
The computation:
\eqalign{ y' = \lim_{\Delta x\to0}{\Delta y\over\Delta x}&= \lim_{\Delta x\to0}{ {1\over x+\Delta x} - {1\over x}\over \Delta x}\cr& =\lim_{\Delta x\to0}{ {x\over x(x+\Delta x)} - {x+\Delta x\over x(x+\Delta x)}\over \Delta x}\cr& =\lim_{\Delta x\to0}{ {x-(x+\Delta x)\over x(x+\Delta x)}\over \Delta x}\cr& =\lim_{\Delta x\to0} {x-x-\Delta x\over x(x+\Delta x)\Delta x}\cr& =\lim_{\Delta x\to0} {-\Delta x\over x(x+\Delta x)\Delta x}\cr& =\lim_{\Delta x\to0} {-1\over x(x+\Delta x)}={-1\over x^2}\cr }
Note. If you happen to know some "derivative formulas'' from an earlier course, for the time being you should pretend that you do not know them. In examples like the ones above and the exercises below, you are required to know how to find the derivative formula starting from basic principles. We will later develop some formulas so that we do not always need to do such computations, but we will continue to need to know how to do the more involved computations.
Sometimes one encounters a point in the domain of a function $$y=f(x)$$ where there is no derivative, because there is no tangent line. In order for the notion of the tangent line at a point to make sense, the curve must be "smooth'' at that point. This means that if you imagine a particle traveling at some steady speed along the curve, then the particle does not experience an abrupt change of direction. There are two types of situations you should be aware of---corners and cusps---where there's a sudden change of direction and hence no derivative.
Example 2.4.3
Discuss the derivative of the absolute value function $$y=f(x)=|x|$$.
SOLUTION
If $$x$$ is positive, then this is the function $$y=x$$, whose derivative is the constant 1. (Recall that when $$y=f(x)=mx+b$$, the derivative is the slope $$m$$.)
If $$x$$ is negative, then we're dealing with the function $$y=-x$$, whose derivative is the constant $$-1$$.
If $$x=0$$, then the function has a corner, i.e., there is no tangent line. A tangent line would have to point in the direction of the curve---but there are two directions of the curve that come together at the origin. We can summarize this as
$y'=\cases{ 1&if $$x>0;\cr -1&if \(x < 0;\cr \hbox{undefined}&if \(x=0.\cr }$ Example 2.4.4 Discuss the derivative of the function \( y=x^{2/3}$$, shown in figure 2.4.1.
SOLUTION
We will later see how to compute this derivative; for now we use the fact that $$y'=(2/3)x^{-1/3}$$. Visually this looks much like the absolute value function, but it technically has a cusp, not a corner. The absolute value function has no tangent line at 0 because there are (at least) two obvious contenders---the tangent line of the left side of the curve and the tangent line of the right side. The function $$y=x^{2/3}$$ does not have a tangent line at 0, but unlike the absolute value function it can be said to have a single direction: as we approach 0 from either side the tangent line becomes closer and closer to a vertical line; the curve is vertical at 0. But as before, if you imagine traveling along the curve, an abrupt change in direction is required at 0: a full 180 degree turn.
Figure 2.4.1. A cusp on $$x^{2/3}$$.
In practice we won't worry much about the distinction between these examples; in both cases the function has a "sharp point'' where there is no tangent line and no derivative.
## Exercises 2.4
Ex 2.4.1 Find the derivative of $$y=f(x)=\sqrt{169-x^2}$$. (answer)
Ex 2.4.2 Find the derivative of $$y=f(t)=80-4.9t^2$$. (answer)
Ex 2.4.3 Find the derivative of $$y=f(x)=x^2-(1/x)$$. (answer)
Ex 2.4.4 Find the derivative of $$y=f(x)= ax^2+bx+c$$ (where $$a$$, $$b$$, and $$c$$ are constants). (answer)
Ex 2.4.5 Find the derivative of $$y=f(x)=x^3$$. (answer)
Ex 2.4.6 Shown is the graph of a function $$f(x)$$. Sketch the graph of $$f'(x)$$ by estimating the derivative at a number of points in the interval: estimate the derivative at regular intervals from one end of the interval to the other, and also at "special'' points, as when the derivative is zero. Make sure you indicate any places where the derivative does not exist.
Ex 2.4.7 Shown is the graph of a function $$f(x)$$. Sketch the graph of $$f'(x)$$ by estimating the derivative at a number of points in the interval: estimate the derivative at regular intervals from one end of the interval to the other, and also at "special'' points, as when the derivative is zero. Make sure you indicate any places where the derivative does not exist.
Ex 2.4.8 Find the derivative of $$y=f(x)=2/\sqrt{2x+1}$$ (answer)
Ex 2.4.9 Find the derivative of $$y=g(t)=(2t-1)/(t+2)$$ (answer)
Ex 2.4.10 Find an equation for the tangent line to the graph of $$f(x)=5-x-3x^2$$ at the point $$x=2$$ (answer)
Ex 2.4.11 Find a value for $$a$$ so that the graph of $$f(x)=x^2+ax-3$$ has a horizontal tangent line at $$x=4$$. (answer)
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# Mathematics | Introduction and types of Relations
• Difficulty Level : Easy
• Last Updated : 01 Dec, 2022
Relation or Binary relation R from set A to B is a subset of AxB which can be defined as aRb ↔ (a,b) € R ↔ R(a,b). A Binary relation R on a single set A is defined as a subset of AxA. For two distinct set, A and B with cardinalities m and n, the maximum cardinality of the relation R from A to B is mn.
Domain and Range: if there are two sets A and B and Relation from A to B is R(a,b), then domain is defined as the set { a | (a,b) € R for some b in B} and Range is defined as the set {b | (a,b) € R for some a in A}.
## Types of Relation:
1. Empty Relation: A relation R on a set A is called Empty if the set A is empty set.
2. Full Relation: A binary relation R on a set A and B is called full if AXB.
3. Reflexive Relation: A relation R on a set A is called reflexive if (a,a) € R holds for every element a € A .i.e. if set A = {a,b} then R = {(a,a), (b,b)} is reflexive relation.
4. Irreflexive relation : A relation R on a set A is called reflexive if no (a,a) € R holds for every element a € A.i.e. if set A = {a,b} then R = {(a,b), (b,a)} is irreflexive relation.
5. Symmetric Relation: A relation R on a set A is called symmetric if (b,a) € R holds when (a,b) € R.i.e. The relation R={(4,5),(5,4),(6,5),(5,6)} on set A={4,5,6} is symmetric.
6. AntiSymmetric Relation: A relation R on a set A is called antisymmetric if (a,b)€ R and (b,a) € R then a = b is called antisymmetric.i.e. The relation R = {(a,b)→ R|a ≤ b} is anti-symmetric since a ≤ b and b ≤ a implies a = b.
7. Transitive Relation: A relation R on a set A is called transitive if (a,b) € R and (b,c) € R then (a,c) € R for all a,b,c € A.i.e. Relation R={(1,2),(2,3),(1,3)} on set A={1,2,3} is transitive.
8. Equivalence Relation: A relation is an Equivalence Relation if it is reflexive, symmetric, and transitive. i.e. relation R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1)} on set A={1,2,3} is equivalence relation as it is reflexive, symmetric, and transitive. Number of equivalence relation in a set containing n elements is given by Bell number.
9. Asymmetric relation: Asymmetric relation is opposite of symmetric relation. A relation R on a set A is called asymmetric if no (b,a) € R when (a,b) € R.
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# Understanding the Difference Between Fractions and Decimals
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Fractions and decimals are two useful types of rational numbers that can be used to represent precise values in a variety of math problems. Understanding the difference between them is essential for developing a strong base in arithmetic.
In order to convert a fraction into a decimal, you must divide the numerator by the denominator. This is called reducing the fraction.
### 1. Fractions are ratios
A fraction is a number that represents a part of something, usually a whole. Fractions are used in everyday life to calculate prices, make measurements, and follow recipes.
Fractions are also important in mathematics. They are the basic building blocks of maths, and they help distribute numbers more easily.
Fractions are a type of ratio, which is an expression that compares two things. Ratios are more flexible and contain more information than fractions, but they can be harder to use.
### 2. Fractions can be written as decimals
Fractions are a way to describe the number of equal parts a whole or collection of objects has. The number on the top of a fraction line is called the numerator, and the number on the bottom is called the denominator.
Decimals are numbers that fall between integers. They use a system of tens, so the spaces to the right of a decimal point are tenths, hundredths, and thousandths.
To write a decimal as a fraction, place the number after the decimal in the numerator and place 10, 100, or 1,000 in the denominator. Then, add zeros in the denominator until you have a fraction with a number of decimal places equal to the original number of decimal places.
Unlike fractions, decimals can be tricky to divide when they have unequal denominators. For example, dividing 1/3 doesn’t have a simple answer, because there are more 3s than you can count. But if you’re familiar with the process, it can help.
### 3. Fractions can be converted to decimals
Fractions can be converted to decimals by dividing the numerator by the denominator. If a fraction is a mixed number, add enough trailing zeros to the numerator to allow continued division. Continue dividing until you get an answer that is either a terminating decimal or a repeating decimal.
Decimals have the advantage of being easier to compare to one another. For example, 0.5 vs. 0.3 is easier to compare than 0.7 and 0.57.
The problem with decimals is that they usually have more digits than fractions do, which means that multiplying a lot of them takes more time than multiplying a fraction.
Teaching students how to convert fractions into decimals is an important skill that they should master. It will help them move between different forms of numbers easily, and it will strengthen their math skills overall.
### 4. Fractions can be multiplied
Fractions are a way to divide a number into equal parts. Each fraction has a numerator (the top number) and a denominator (the bottom number). These numbers are separated by a line.
There are also fractions that have more than two parts, like -1/2 and 1/-2. These are called negative fractions because they represent the opposite of a positive fraction.
Another kind of fraction is a mixed fraction, which has both a whole number part and a fractional part. These are useful when you need to compare fractions, but they can be hard to work with.
To multiply a fraction, start by multiplying its numerator by its whole number, followed by its denominator. Then, simplify the result and reduce it to its lowest terms if needed. This is a lot easier than you might think!
Understanding the Difference Between Fractions and Decimals
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# GRAPHING GREATEST INTEGER FUNCTION
## About "Graphing greatest integer function"
"Graphing greatest integer function" is the stuff which is needed to the children who study high school math.
The Greatest-Integer Function is denoted by y = [x]
For all real values of "x" , the greatest-integer function returns the largest integer less than or equal to "x". In essence, it rounds down to the the nearest integer.
Before learning "Graphing greatest integer function", always it is advisable to know the basic stuff about "Greatest integer function".
Let us look at how to graph a greatest integer function
## Graphing greatest integer function
When we try to do graphing greatest integer function, first we have to graph the parent function of any greatest integer function which is y = [x].
Let us see, how to graph the parent function of greatest integer function.
## Graphing parent function y = [x]
To graph the parent function y = [x], we have to plug some random values for "x" such that x = -3, -2, -1, 0, 1, 2,3
y = [-3] = -3 ------> (-3, -3)
y = [-2] = -2 ------> (-2, -2)
y = [-1] = -1 ------> (-1, -1)
y = [0] = 0 ------> (0, 0)
y = [1] = 1 ------> (1, 1)
y = [2] = 2 ------> (2, 2)
y = [3] = 3 ------> (3, 3)
## How to plot the above points on the graph?
Let us take the point (-3-3).
Mark the point on xy- plane with a filled circle at (-3-3).
Then extend a line for "1" space on the left side of (-3,-3) and end up with empty circle.
Do the same thing for the other points too.
Now you will have a graph as given below.
## Steps involved in graphing greatest integer function
Step 1 :
When you want to graph a greatest integer function, first graph the parent function.(as explained above)
Step 2 :
Write any parent function in the form as given below.
y - a = [x - b]
Step 3 :
Now equate "y-a" and "x-b" to zero.
That is y - a = 0 ------> y = a
and x - b = 0 ----------> x = b
Step 4 : (vertical shift)
(i) If you have y = 0, there is no vertical shift.
(ii) If you have y = a (positive value), then shift the graph up "a" units.
(iii) If you have y = -a (negative value), then shift the graph down "a" units.
Step 5 : (Horizontal shift)
(i) If you have x = 0, there is no horizontal shift.
(ii) If you have x = b (positive value), then shift the graph to the right "a" units.
(iii) If you have x = -b (negative value), then shift the graph to the left "a" units.
Let us have some examples to have better understanding on "Graphing greatest integer function".
Graph : y = [x-2]
The given function is in the form of y -a = [x-b]
Let y = 0 and x - 2 = 0 (step 3)
We get y = 0 and x = 2.
From y = 0, there is no vertical shift.
From x = 2, we have an horizontal shift of 2 units to the right.
So each point of the parent function to be shifted 2 units to the right.
If we do the above transformation, we will have a graph as given below.
Graph : y = [x] + 3
If we write the given function in the form of y -a = [x-b], we will have y - 3 = [x]
Let y - 3 = 0 and x = 0 (step 3)
We get y = 3 and x = 2.
From y = 3, we have a vertical shift of 3 units up.
From x = 0, there is no horizontal shift.
So each point of the parent function to be shifted 3 units up.
If we do the above transformation, we will have a graph as given below.
Graph : y = [x+1] + 2
If we write the given function in the form of y -a = [x-b], we will have y - 2 = [x+1]
Let y - 2 = 0 and x + 1 = 0 (step 3)
We get y = 2 and x = -1.
From y = 2, we have a vertical shift of 2 units up.
From x = - 1, we have an horizontal shift of 1 unit to the left.
So each point of the parent function to be shifted 2 units up and 1 unit to the left.
If we do the above transformations, we will have a graph as given below.
Please click here to know the basic stuff about greatest integer function.
Some more stuff on "Greatest integer function"
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# Proof and Mathematical Induction
The proof is a very important element of mathematics. As mathematicians, we cannot believe a fact unless it has been fully proved by other facts we know.
There are a few key types of proofs we will look at briefly. These are:
• Proof by Counter Example
• Proof by Exhaustion
We will then move on to more difficult elements of proof, a special proof called mathematical induction.
These proofs are relatively straightforward and methodical, however, we will look at a few tricks one can use to help speed up the process.
## What is Proof By Counter-Example?
Proof by counter-example is probably one of the more basic proofs we will look at. It pretty much is what it states and involves proving something by finding a counterexample.
The steps are as follows.
Step Why? State your conjecture and what you need to prove clearly. To demonstrate what are aim is from this proof. Find a counter example, we can do this by testing examples and eventually we will be able to find a good example. This will disprove our conjecture.
Conjecture: Statement of what we are trying to prove or disprove.
Let's look at a short example to help us figure out what's going on.
Disprove by counterexample that all values ofare odd, for.
SOLUTION:
If we let,
4 cannot be written as, with.
So 4 is not odd.
We have found a counterexample to disprove the conjecture.
## What is proof by exhaustion?
Proof by exhaustion involves testing all relevant examples and checking they all satisfy the conjecture. This, as used when there are limited cases, therefore, testing all relevant examples, will not take a long time. Let's look at how we do this as a series of steps:
Step Why? State your conjecture. Find our aim for our proof. Find all necessary examples to test. Allows us to see what cases we need to test. Test all relevant cases. Checks all of our 'exhausted' cases work. Make a statement about your proof. Summarises the proof nicely.
Let's look at a short example as to how this works.
Prove by exhaustion that consecutive positive even numbers under 10, sum even numbers.
SOLUTION:
We want to see that the sum of two consecutive, positive even numbers under 10 is even.
Therefore the numbers we are going to use are 2,4,6 and 8.
6, 10 and 14 are all even numbers as they are all multiples of 2:
Therefore these sums are all even numbers.
So our conjecture has been proven.
There are a few useful symbols that we can use during and after we have finished a proof. This can make our proof look a little fancier. These are:
Symbol Explanation Q.E.D Q.E.D stands for quod erat demonstrandum. This is the Latin for it has been proven. This means exactly the same thing as Q.E.D. This implies that. So you would use this when one statement directly implied another. When two statements both imply each other this arrow is used. ∴ These three dots signify therefore, we use a statement of fact as a base for a new statement of fact e.g. a=2 ∴ a+2=4.
## What is proof by contradiction?
Proof by contradiction is the process in which we attempt to prove the opposite of what we are asked for. Then realize that there is a contradiction in our listed proof.
There isn't really a list of steps to this. we just start off by trying to prove the opposite then use algebraic manipulation to eventually show this is wrong. Let's look at a key example of this, which should make enough sense.
Prove thatis irrational.
SOLUTION:
Firstly let's break down what we are being asked. We want to show thatcannot be written as a fraction.
So let's try and prove it is a fraction and find our contradiction.
Let, whereis a fraction in its' simplest form.
Then if we square both sides:
must be even, this is becauseis even and the square root of an even number is even.
Let,
This meansis even. Therefore it means thatis even.
If bothandare even, this means that, is not in its simplest form.
This is a contradiction as we originally assumedwas in its simplest form.
Thereforeis not rational and is irrational.
So in this example what we've done is used algebraic manipulation and basic mathematical fact to move some terms around and show that something is irrational. Therefore, whenever we get a question like this all we do is use algebra and facts about numbers to show we have a contradiction.
In Latin this is known as reductio ad absurdum, it essentially translates to proof by absurd so assuming something wrong and finding a contradiction.
## What is mathematical induction?
Mathematical induction is the process in which we use previous values to find new values. So we use it when we are trying to prove something is true for all values. So here are the list of steps to how we would answer a general question:
Prove that "conjecture" is true for all values n≥m.
Step Why? 1. Prove is true for our lowest value, (n=m in this case). Shows that our lower bounded value is true. 2. Assume it is true for the value n=k. We assume that the conjecture is true for some value in our range to use in future reference. 3. Use the fact that n=k is true, prove that n=k+1 is true. This shows that for some n=k, n=k+1 is true. By induction this means values are true. 4. Make a statement to conclude this in the structure:It has been prove for n=m, and for some n=k, n=k+1 is true. So the conjecture is true for all values n≥m. This gives a summary statement to our proof and allows us to see our aim proven.
Let's look at two examples of this, one which is more general and one which is specific to series and sequences.
Prove by mathematical induction thatis divisible by 4 for all.
SOLUTION:
Step 1: Firstly we need to test, this gives.
So this is a multiple of 4.
Step 2: Assume that whenthe statement is correct.
If we write this in mathematical notation we get, where m is a positive number.
Step 3: Now let's use the fact thatis true to prove that for:
Now we substituteinstead ofin the, we get:
Step 4: Therefore based onbeing a multiple of 4,is a multiple of 4.
So this is true for, and for some, it is true for some.
Therefore the conjecture has been proven.
Let's look at another example specific to series and sequences.
Prove by mathematical induction thatfor all.
SOLUTION:
Step 1: Firstly we need to test the case when.
Step 2: We assume that the case ofis correct.
Step 3: Now using the fact we need to test for:
If we simplify this out we get:
If we cancel out the termfrom the numerator and denominator we get:
, if we go back to the fact that,
Step 4: Therefore it has been proven for when, and for someit has been proven for. Therefore it is true for all.
So we can see that just through algebraic manipulation and using some rules about series we can prove conjectures for all values.
## Proof and Mathematical Induction - Key takeaways
• There are three main types of proof: counterexample, exhaustion, and contradiction.
• Counterexample is relatively straightforward and involves finding an example to disprove a statement.
• Exhaustion involves testing all relevant cases and seeing if they are true.
• Contradiction involves attempting to prove the opposite and finding that the statement is contradicted.
• Mathematical Induction involves testing the lowest case to be true. Then assuming the conjecture is true for one case before using that fact to prove the case one above the previous case is true.
Mathematical induction is the process in which we use previous values to find new values.
Mathematical Induction is proved by testing the lowest case to be true. Then assuming the conjecture is true for one case before using that fact to prove the case one above the previous case is true.
The principle of mathematical induction is - Every nonnegative integer belongs to F if F is hereditary and integer 0 belongs to class F. Also every positive integer belongs to F if F is hereditary and integer 1 belongs to F.
Mathematical induction is used by taking the base step, inductive step, and conclusion.
Mathematical induction is a proof technique. For example, we can prove that n(n+1)(n+5) is a multiple of 3 by using mathematical induction.
## Final Proof and Mathematical Induction Quiz
Question
What is proof by counter example?
Proof by counter example is proving something by finding a counterexample.
Show question
Question
Statement of what we are trying to prove or disprove is called?
Solution
Show question
Question
Which value of x is a counterexample of 5x > 4x?
1
Show question
Question
There are prime numbers divisible by 6.
True
Show question
Question
Meaning of proof by exhaustion.
Proof by exhaustion involves testing all relevant examples and checking they all satisfy the conjecture.
Show question
Question
When is proof by exhaustion used?
Proof by exhaustion is used when conjecture can be reduced to a finite number of cases.
Show question
Question
Is n²-1 a multiple of 3, when n is not a multiple of 3 shown using proof by exhaustion?
Yes
Show question
Question
When is proof by exhaustion is called complete?
All values are true
Show question
Question
Proof by contradiction is the process in which the opposite of what is asked for is proved.
Show question
Question
Define mathematical induction.
Mathematical induction is the process in which we use previous values to find new values.
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The lowest value is kept what when using mathematical induction?
True
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# What is the general solution of the differential equation (y^2+1)dy/dx+2xy^2=2x ?
Sep 19, 2017
$\ln | \frac{y + 1}{y - 1} | - y = {x}^{2} + C$
#### Explanation:
We have:
$\left({y}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x {y}^{2} = 2 x$ ..... [A]
We can rearrange this non-linear First Order differential equation [A] as follows:
$\left({y}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 2 x {y}^{2}$
$\therefore \left({y}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x \left({y}^{2} - 1\right)$
$\therefore \frac{{y}^{2} + 1}{{y}^{2} - 1} \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x$
This is now separable, so we can "seperate the variables" to get:
$\int \setminus \frac{{y}^{2} + 1}{{y}^{2} - 1} \setminus \mathrm{dy} = \int \setminus - 2 x \setminus \mathrm{dx}$ ..... [B]
The RHS integral is standard, and the LHS will require a little manipulation, as follows:
$\int \setminus \frac{{t}^{2} + 1}{{t}^{2} - 1} \setminus \mathrm{dt} = \int \setminus \frac{{t}^{2} - 1 + 2}{{t}^{2} - 1} \setminus \mathrm{dt}$
$\text{ } = \int \setminus 1 + \frac{2}{{t}^{2} - 1} \setminus \mathrm{dt}$
$\text{ } = \int \setminus 1 + \frac{2}{\left(t + 1\right) \left(t - 1\right)} \setminus \mathrm{dt}$
We can now decompose the fractional part of the integrand into partial fractions, as follows:
$\frac{2}{\left(t + 1\right) \left(t - 1\right)} \equiv \frac{A}{t + 1} + \frac{B}{t - 1}$
$\text{ } = \frac{A \left(t - 1\right) + B \left(t + 1\right)}{\left(t + 1\right) \left(t - 1\right)}$
$2 \equiv A \left(t - 1\right) + B \left(t + 1\right)$
Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:
Put $t = - 1 \implies 2 = - 2 A \implies A = - 1$
Put $t = + 1 \implies 2 = + 2 B \implies B = + 1$
So using partial fraction decomposition we have:
$\int \setminus \frac{{t}^{2} + 1}{{t}^{2} - 1} \setminus \mathrm{dt} = \int \setminus 1 - \frac{1}{t + 1} + \frac{1}{t - 1} \setminus \mathrm{dt}$
Using this result we can now integrate [B] as follows:
$\int \setminus \frac{{y}^{2} + 1}{{y}^{2} - 1} \setminus \mathrm{dy} = \int \setminus - 2 x \setminus \mathrm{dx}$
$\therefore \int \setminus - 1 + \frac{1}{y + 1} - \frac{1}{y - 1} \setminus \mathrm{dy} = \int \setminus 2 x \setminus \mathrm{dx}$
$\therefore - y + \ln | y + 1 | - \ln | y - 1 | = {x}^{2} + C$
$\therefore \ln | \frac{y + 1}{y - 1} | - y = {x}^{2} + C$
Which, is the General Solution .
We are unable to find a particular solution, as requested, as noi initial conditions have been provided to allow the constant $C$ to be evaluated.
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# Intercepts of a Line
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## What is the Intercept of a Line?
The x-intercept and the y-intercept are the two different types of intercepts. The line's actual point of intersection with the x-axis is known as the x-intercept, while its actual point of intersection with the y-axis is known as the y-intercept. In this article, we will define the intercept, show how to obtain the intercept for a given line, and demonstrate how to graph intercepts.
## What is x-Intercept?
The general form of the linear equation is written as $y = mx + c$, where m and c are constants. The line's crossing point, which is located at the x-axis of the plane, is known as the x-intercept. This indicates that anytime a linear equation crosses the x-axis, its y-coordinate value will always be equal to 0. The y-coordinate is zero for the x-intercept and the x-coordinate is zero for the y-intercept. The term "horizontal intercept" also applies to the x-intercept.
## What is y- Intercept?
The graph's intersection with the y-axis is known as the y-intercept. Finding the intercepts for any function with the formula $y = f\left( x \right)$ is crucial when graphing the function. An intercept can be one of two different forms for a function. The x-intercept and the y-intercept are what they are. A function's intercept is the location on the axis where the function's graph crosses it.
## Equation of a Line with Intercepts
How to find an intercept of a straight line? Intercepts are subject to various equations and formulas. By solving for x and putting $y = 0$ in the equation, all of the formulas are obtained.
Following are the steps to determine the y intercept of a function, where$y = f\left( x \right)$, where $x = 0$ is simply substituted.
Resolve for y.
Put the y-point intercepts in place (0, y).
## Conclusion
The two different kinds of intercepts are the x-intercept and the y-intercept. The x-intercept of a line is its actual point of intersection with the x-axis, and the y-intercept is its true point of intersection with the y-axis.
## Solved Examples
Example 1:Find the value of "a" if the y-intercept of a function $y = a(x - 1)\left( {x - 2} \right)\left( {x - 3} \right)$ is $\left( {0,12} \right)$ .
Ans: The given function's equation is:
$y = a(x - 1)\left( {x - 2} \right)\left( {x - 3} \right)$
By adding $x = 0$ to the y-intercept formula, it can be calculated.
$y = a \left( {0 - 1} \right]\left[ {0 - 2} \right]\left[ {0 - 3} \right]$ $= - 6a$
The y-intercept is therefore $\left( {0, - 6a} \right)$
However, the issue claims that the provided function's y-intercept is $\left( {0,12} \right)$. Thus,
$- 6a = 12$
Using -6 to divide both sides,
$a = - 2$
Example 2: Find the x and y intercept in the given graph.
## Ans: The x-intercept is 7 and y – intercept is 3.5 in the given graph.Example 3: Find the x-intercept of $y = 4x - 8$.Ans: For finding the x – intercept we must put y coordinate as 0.Thus, putting$y = 0$ ,we will get:$\begin{array}{l}0 = 4x - 8\\8 = 4x\\x = 2\end{array}$Therefore, x coordinate is 2.
Last updated date: 27th Sep 2023
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## FAQs on Intercepts of a Line
1. What are the Y Intercept Formula Applications?
To determine a function's y-intercept, use the y-intercept formula. The procedure of graphing a function is when the y-intercept is most frequently employed.
2. Is 0 a possible x-Intercept?
For the line$y = mx$, where m is the line's slope and the value of the x-intercept is equal to 0, 0 can indeed serve as the x-intercept.
3. What shape does the value angle between the line's Y and X intercepts take if they are equal?
When a line's x-intercept and y-intercept are equal, the line intersects the x-axis at a 45° angle if we are taking in first quadrant.
4. How many intercepts are permitted on a line?
X-intercept and Y-intercept are the two main intercepts. The line's x-intercept and y-intercept are located where the line crosses the x and y axes, respectively.
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# Prove that the center of a circle touching two intersecting lines lies on the angle bisector of the lines.
Last updated date: 20th Jun 2024
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Hint: We draw a circle which is touching two intersecting lines. Using the property of tangents from an external point to a circle we prove the two triangles formed by bisectors are congruent by SSS congruence rule.
* Two triangles are said to be congruent to each other by SSS rule if the sides of the first triangle are equal to corresponding sides of the other triangle.
* Two tangents drawn from an exterior point to a circle are equal in lengths.
We draw a circle with center O; point P is the point of intersection of two lines m and n which touch the circle at points Q and R respectively.
Here we join the line from center of the circle to the point of intersection of two lines i.e. P.
Now, we know from the property of tangents, that tangents drawn from an exterior point to the circle are equal , so we can say $PQ = PR$
Also, the lengths $OQ = OR$ as they both are the lengths of radius of the circle.
Now, we will consider two triangles $\vartriangle OPQ,\vartriangle OPR$
$OQ = OR$
$PQ = PR$
$OP = OP$
Therefore, by Side Side Side congruence rule, $\vartriangle OPQ \cong \vartriangle OPR$
We know that if two triangles are congruent to each other, then the corresponding angles of two triangles are equal. So, $\vartriangle OPQ \cong \vartriangle OPR$ gives us $\angle OPQ = \angle OPR$.
Therefore we can say that the line $OP$ bisects the angle formed at the intersection of two lines m and n.
Hence Proved
Note: Students make mistake of proving the triangles congruent using SSA congruence as the angles formed by radius and the tangents at points Q and R are right angles, which is wrong because we have no surety of congruence rule SSA, so we cannot use it to prove triangles congruent.
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## Comparing Numbers
In comparing numbers, all you need to understand is the decimal number system itself and how it’s been constructed. When you understand that system, everything made in it including numbers, can be dealt with. We talked extensively about the decimal number system previously, so now, let’s get to some examples. Continue reading “Comparing Numbers”
## How to Find the Square Root of a Number
We mentioned before that when you square a number, you’ll get a square number or a perfect square. You can use the perfect square to find out what numbers were multiplied together to get the perfect square. The process of using the perfect square to get back to those numbers is finding the square root of a number. For example, 3 * 3 = 3^2 = 9. 9 Here is perfect square. You can take the square root of 9 to get back to 3. You can see that squaring a number and taking the square root of the result are actually inverse processes. Continue reading “How to Find the Square Root of a Number”
## Square Root Function
If you multiply any integer except for 0 by itself, you’ll get a positive perfect square. For example, 3 * 3 = 9 or you can write the same expression as 3^2 = 9. Now, if you have the number 9, knowing that 4 is a perfect square and want to know what number was raised to the second power or multiplied by itself to get the 9, you need a function that takes you from the perfect square back to the number. That function is called the “square root function.” That means,
3 * 3 = 9 Or 3^2 = 9, then, sqrt(9) = 3. Continue reading “Square Root Function”
## Calculating Square of Large Numbers
Calculating the square of numbers like 8, 9, etc., is easy but calculating the square of a number like 105 without using a calculator.
There is an interesting way to calculate the square of large numbers like 255 without using a calculator. It’s probably not easy to do it in your mind but you can do it on a piece of paper. It’s essentially a way to calculate the square root of numbers that have a 5 in 1’s place value. It does not matter what sort of digits you might have in the other place values but the 1’s place value has to be a 5.
Click to see how to derive the formula
So the formula will be (x5)^2 = x(x+1)100 + 25
The formula above was derived for a 2-digit number but it can be used for any number with any number of digits provided that the number has a 5 in the 1’s place. For example, if you were to calculate 115^2, you’d simply use the formula as follows:
115^2 = 11(12)100 * 25 = 13225
You can learn how this formula was created and based on that, you can create any sort of formula for any kind of situation you might find yourself in.
## Expressing Squared Numbers as Product of Two Consecutive Odd or Even Natural Numbers
It turns out that you can write almost any squared number in terms of the product of two odd or even integers plus one. As an example, let’s go through the following operation:
29 * 31 = (30 – 1)(30 + 1) = 30^2 – 1 because (a – b)(a + b) = a^2 – b^2
30^2 = 29 * 31 + 1
So as you can see here, you can write 30^2 as the product of two consecutive odd natural numbers, 29 and 31 and add 1 to the product. You can do this with almost every square number. I am using the word “almost” here since I have not taken the time to test every possible case not have I proved this nor have I seen a proof for it.
Wherever this happens to come in handy depends on the kind of problems you might be solving.
## Expressing Square Numbers as Sum of consecutive Natural Numbers
You can express any odd perfect square in terms of the sum of two consecutive natural numbers. For example, 3 squared or 9 can be expressed as (4 + 5). 5 squared or 25 can be expressed as (12 + 13) and so on.
What this means, is that we can express the square of any odd number as the sum of two consecutive natural numbers.
We cannot do this for the square of even numbers because the square of an even numbers is always an even number. An example would be 16. 16 is a perfect square and it can be expressed as (8 + 8). 8 and 8 are NOT two consecutive natural numbers but (12 + 13) are two consecutive natural numbers added together to get to 25 which is a perfect square.
One important thing to notice here is that the inverse of the rule above is not always true, so it’s not true at all. Meaning that, (12 + 13) would be 25 which is a perfect square but let’s pick to other consecutive natural number like (13 + 14) which is 27. 27 is not a perfect square. So while any odd square number can be expressed as the sum of two consecutive natural numbers, not any two consecutive natural numbers added together would result in an odd perfect square.
## Expressing Square Numbers as Sum of Odd Natural Numbers
There is a relationship between the number of odd natural numbers starting from 1 added together and square numbers. That means, 1, the same things as 1 squared is the same thing as the first odd natural number. And you know that 1 is a perfect square. 4, which is the same thing as 2 squared, can be written as (1 + 3) which is the first two natural odd numbers starting from 1 added together. 9, which is the same thing as 3 squared, can be written as (1 + 3 +5), which is the same thing as the first three natural odd numbers starting from 1 added together and so on and so forth.
What this means is that if you add the first n natural odd numbers starting from 1 together, the result can be expressed as n squared.
You can use this observation in different ways in different situations to check whether a number a perfect square or not. For example, if you were to test whether 121 is a perfect square, you’d check whether 121 can be expressed as a sum of some odd natural numbers starting from 1. If you tried this, you’d find out that the number of those odd natural numbers, in this case, would be 11, namely 1 through 21 or 1, 3, 5, 7, 9, … , 21. If you add all those numbers together, you’d get 121. So you can conclude that 121 is perfect square. Moreover, sqrt(121) = 11.
## The Number of Numbers from a to b and Between a and b
It’s a simple but useful skill to be able to find the number some numbers between two numbers in a number set. This could take all shapes or forms. For example, sometimes, you might need to know the number of numbers from say a to b, including a and b. I’ll talk about two of such situations here. Remember that any other type of situation that comes your way, you can use simple logic and test some simple cases and generalize your findings to more complicated cases to find the rule you’re looking for.
### Calculating the Number of Numbers from a to b, Including a and b
Now, let’s say, you want to find the number of numbers from a to b, including a and b. In that case, you could think of a simple case like all the numbers from 1 to 10. Those numbers would be, 1, 2, 3, 4 5, 6, 7, 8, 9, 10. If you count them, you’ll find out that the number of those numbers is 10. So now that you know how many numbers are actually there, you could find some algebraic relationship an test it against the actual number of numbers that you have. If I subtract the first number, 1, from the last number, 10, I’d have 10 – 1 = 9 but I have 10 numbers there, not 9. So I add 1 to 9 to get to 10. Test this for any other case starting from any number and ending anywhere and it will work.
So (i) if you are working with a set of numbers like natural numbers where it’s possible to know what numbers are there and how many numbers there are between any two given numbers, and if, (ii) the numbers under study are happening based on a specific rule, then you can use the above solution to figure out the number of those numbers under study.
What (i) means is that, for example in the set of natural numbers, you know that numbers start from 1 and go all the way up to infinity. At every step, 1 is added to every number to get the next number. So the set looks something like, 1, 2, 3, … Moreover, you know that there is no number between 1 and 2 for example in this set of numbers. If you talk about the set of rational numbers, then there’s no way to know how many numbers there are between 1 and 2. The number would actually be infinite. So there, the whole solution would be useless.
What (ii) means is, that for example the numbers under study start from 1 and go all the way up to 101 or the numbers start at 1 and go all the way up to 100 missing every other number, 1, 3, 5, 7, … , 101. Here the progression of numbers is happening based on some uniform rule and so it’s possible to figure out the exact number of numbers.
Getting back to our original problem, figuring out the number of numbers from a to b, numbers happening based on a rule similar to 1, 2, 3, … , b, you could write the number of those numbers as (b – a + 1)
### Calculating the Number of Numbers Between a and b, Excluding a and b
This would be the exact same situation, except that you’d need to make some changes to the formula you got in the previous step. Since in this case, you have to subtract two numbers from that sum (you have to exclude a and b from the sum), you’d end up with (b – a + 1) -2 which is equal to (b – a – 1).
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# Multiplication Arrays
###### This post contains affiliate links to materials I recommend. Read my full disclosure statement.
My friend, Kim, at Life Over C’s has an awesome 21 day series going on her blog called “Learn with Your Craft Stash“. The series focuses on learning activities you can do with craft items you have around your home or classroom. I’m joining in today with how to use construction paper and a hole punch to teach multiplication arrays.
What is an array, anyway? An array is an orderly arrangement (most often in rows or columns) of numbers or objects. An array can be used as a visual representation of multiplication as repeated addition.
We use Singapore Math 1A and 1B with my youngest. Beginning multiplication is introduced in first grade right after students learn repeated addition. Students go from learning and solving 2+2+2, to relating that to 2×3 (2 three times).
Here’s a fun and engaging activity we recently did when studying this concept.
Materials:
First, I printed out a few multiplication facts onto a sheet of paper. Another options is to just write the multiplication facts on a piece of paper or an index card.
Next, I cut several small rectangles out of construction paper.
Then, I had my son solve the multiplication problems by making an array on the rectangle with a hole punch. We talked about how the first number told him how circles to punch. The second number told him how many columns to make of that number. So in the first example, he made three columns of four circles (the number four, three times).
Once he punched all the holes, it was easy for him to look at the rectangle and count the holes to find the answer. I had him work on two problems side by side with the same factors. I wanted him to see that reversing the order of the factors didn’t change the answer. 5×2 is the same as 2×5.
To make it more challenging (and to not have small circles all over your floor!), have your students create an array for the first problem. Then, use the circles that were punched out to build another array with reversed factors. For example, I had him create an array for 5×2 using the hole punch. Next, he used the circles he punched to create an array for 2×5. It was a great visual for him to see he was using the same amount of holes/circles (10), but just rearranging them in a different way.
This activity is great for both visual and kinesthetic learners. It works great with addition problems as well. For instance, to find the answer to the addition problem 4+3, students punch four holes, punch three more holes, and then count the total number of holes they punched to find the answer.
My son loves using the hole punch and had a lot of fun using it to do his math problems!
If you’re interested into other hands-on learning activities using craft supplies you probably have lying around, click on the picture below to check out all of the “Learn with Your Craft Stash” posts!
1. Cee McGuire says:
This is so awesome. Glad I found it my son is a visual learner and it is really helping
1. I’m so happy to hear that! Another teacher friend of mine does arrays using googly eyes. She has the students glue the eyes in arrays to show different multiplication problems. It’s another fun and engaging way to practice this concept!
2. Libby Serkies says:
This is brilliant! I am an instructional designer working with teachers who work with students trying to earn their high school equivalency. Some of the students in our programs have very few numeracy skills and for the past 3 years I have repeated the phrase “why before how” until the teachers are hearing it in their sleep! I love the ideas at work here, and plan to show this to my teachers at my earliest chance! I am so happy I found you through Pinterest!! I can’t wait to see the teachers’ eyes light up when they see this!
1. Thank you! I love your phrase “why before how”! That perfectly sums up how I feel about teaching math. I hope this activity helps your students see and understand the why behind multiplication.
3. kelly says:
LOVED THIS!
4. Jenny says:
My students love using the hole puncher – thank you! Would you mind sharing the printable?
5. Tres belle idée. Je vais la rajouter dans mes manipulations. Habituellement, j’utilise des centicubes , mais cette approche me semble aussi intéressante. Merci
Samya
6. Shannon DeVillier says:
Thank you for sharing and yes, your phrase “why before how” is ever so handy when explaining our methods not only to students but especially to parents!
7. Jennie Hatch says:
Can you share the worksheet with me?
Thanks. I love this idea.
8. Dawn says:
This is brilliant and a great idea for math centers. I wonder if the students could take it a step further and use the stamp cards to illustrate the distributive property (take original card, cut into 2 smaller arrays, and label each). Will definitely give this one a shot!
9. SABRINA WHITEHURST says:
Hello how do I go by getting the worksheet to do this activity.
1. Hi Sabrina! I don’t have this particular worksheet available. I just typed a few multiplication problems I wanted my son to practice and printed them out. You can use something like Microsoft Word or PowerPoint to easily create a similar worksheet.
10. LD momma says:
this is great for kids, teens or adults with dyscalculia
11. Shawn says:
Love this idea of teaching arrays. Do you have a copy of the multiplication sheet with grids?
1. Hi! I don’t have the worksheet pictured with the problems on it unfortunately. I just typed a few problems into a table in Microsoft Word and printed it out. Maybe you can do the same!
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# The ratio of S.P of three articles is 40 : 37 : 46 and the ratio of profit% is 5 : 4 : 7. The cost price of first and second articles are same and the cost price of third article is 480. Find the total cost price of all the three articles?
This question was previously asked in
AFCAT 14 Feb 2022 Shift 1 Memory Based Paper
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1. Rs. 1240
2. Rs. 1340
3. Rs. 1440
4. Rs. 1540
Option 3 : Rs. 1440
## Detailed Solution
Given
The ratio of S.P of three articles = 40 : 37 : 46
The ratio of profit% = 5 : 4 : 7.
The cost price of third article = 480
Formula Used
Profit = Selling Price - Cost Price
Profit% = [(S.P - C.P)/C.P] × 100
Calculation
Let the S.P. of three articles be 40x, 37x and 46x respectively
The profit% of three articles be 5y%, 4y% and 7y%
Now, According to the question
For 3rd Article
7y/100 = [46x - 480]/480
⇒ 336y = 460x - 4800 ----(i)
For 1st article
5y/100 = (40x - C.P1)/(C.P1)
⇒ 0.05y + 1 = 40x/C.P1
⇒ C.P= (40x)/(0.05y + 1) ----(ii)
For 2nd article
4y/100 = (37x - C.P2)/(C.P2)
⇒ 0.04y + 1 = 37x/C.P2
⇒ C.P= (37x)/(0.04y + 1) ----(iii)
According to the question
C.P= C.P
⇒ (40x)/(0.05y + 1) = (37x)/(0.04y + 1)
⇒ 1.6xy + 40x = 1.85xy + 37x
⇒ y = 12 ----(iv)
Form (i) and (iv), we get
x = 19.2
So, Sum of cost price = 2 × [(40x)/(0.05y + 1)] + 480
⇒ 2 × 480 + 480
∴ The required sum is Rs. 1440.
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