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# The diameter of a circle is 10 cm , then find the length of the arc when the corresponding central angle is ${180^ \circ }(\pi = 3.14)$ ?A. 15.7B. 16C. 3.14D. 18
Last updated date: 14th Sep 2024
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Hint: Using the given diameter we can get the radius of the circle as the radius is half of the diameter and when the central angle is given the length of the arc is given by the formula $Length = \dfrac{\theta }{{360}}\times 2\pi r$ and using the given values we get the required length.
Complete step by step solution:
We are given a circle and it is given that the diameter of the circle is 10 cm
We know that the radius of the circle is given by half of its diameter.
Hence the radius of the given circle is given by
$\Rightarrow r = \dfrac{d}{2} = \dfrac{{10}}{2} = 5cm$
Now we have the radius of the circle
Now we are asked to find the length of the arc corresponding to the central angle
Whenever the central angle is given the length of the arc is given by the formula
$\Rightarrow Length = \dfrac{\theta }{{360}}\times 2\pi r$
Where $\theta$ is the central angle and r is its radius
Here the central angle is ${180^ \circ }$ and radius is 5 cm
$\Rightarrow Length = \dfrac{{180}}{{360}}\times 2\times 3.14\times 5 \\ \Rightarrow Length = \dfrac{1}{2}\times 2\times 3.14\times 5 \\ \Rightarrow Length = 3.14\times 5 = 15.7 \\$
Hence we get the length of the arc to be 15.7 units
Therefore the correct answer is option A.
Note :
1) A part of a curve or a part of a circumference of a circle is called Arc.
2) In general, the length of a curve is called the arc length.
3) An arc length is measured by taking a part of the whole circle. |
# How do you find the product of (2t^2+t+3)(4t^2+2t-2)?
Apr 26, 2017
#### Explanation:
The method for performing the multiplication will become obvious when we write it in a form where each term of the first factor multiplies the second factor:
$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$
$2 {t}^{2} \left(4 {t}^{2} + 2 t - 2\right) +$
$t \left(4 {t}^{2} + 2 t - 2\right) +$
$3 \left(4 {t}^{2} + 2 t - 2\right)$
Distribute the first term of the first factor into the second factor:
$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$
$8 {t}^{4} + 4 {t}^{3} - 4 {t}^{2} +$
$t \left(4 {t}^{2} + 2 t - 2\right) +$
$3 \left(4 {t}^{2} + 2 t - 2\right)$
Distribute the second term of the first factor into the second factor:
$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$
$8 {t}^{4} + 4 {t}^{3} - 4 {t}^{2} +$
$\textcolor{w h i t e}{\text{.........}} 4 {t}^{3} + 2 {t}^{2} - 2 t +$
$3 \left(4 {t}^{2} + 2 t - 2\right)$
Distribute the third term of the first factor into the second factor:
$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$
$8 {t}^{4} + 4 {t}^{3} - 4 {t}^{2} +$
$\textcolor{w h i t e}{\text{.........}} 4 {t}^{3} + 2 {t}^{2} - 2 t +$
$\textcolor{w h i t e}{\text{.................}} 12 {t}^{2} + 6 t - 6$
I have aligned the columns so that they can be added:
$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$
$8 {t}^{4} + 4 {t}^{3} - 4 {t}^{2} +$
$\textcolor{w h i t e}{\text{.........}} 4 {t}^{3} + 2 {t}^{2} - 2 t +$
$\underline{\textcolor{w h i t e}{\text{.................}} 12 {t}^{2} + 6 t - 6}$
$8 {t}^{4} + 8 {t}^{3} + 10 {t}^{2} + 4 t - 6$
Because multiplication is commutative, we could have multiplied the first factor by each term of the second factor and obtained the same result. |
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# 2.3: The General Planar Truss
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
Let us now consider something that resembles the mechanical prospection problem introduced in the introduction to matrix methods to matrix methods for mechanical systems. In the figure below we offer a crude mechanical model of a planar tissue, say, e.g., an excised sample of the wall of a vein.
Elastic fibers, numbered 1 through 20, meet at nodes, numbered 1 through 9. We limit our observation to the motion of the nodes by denoting the horizontal and vertical displacements of node j by $$x_{2j-1}$$ (horizontal) and $$x_{2j}$$ (vertical), respectively. Retaining the convention that down and right are positive we note that the elongation of fiber 1 is
$e_{1} = x_{2}-x_{8} \nonumber$
while that of fiber 3 is
$e_{3} = x_{3}-x_{1} \nonumber$.
As fibers 2 and 4 are neither vertical nor horizontal their elongations, in terms of nodal displacements, are not so easy to read off. This is more a nuisance than an obstacle however, for noting our discussion of elongation in the small planar truss module, the elongation is approximately just the stretch along its undeformed axis. With respect to fiber 2, as it makes the angle $$-\frac{\pi}{4}$$ with respect to the positive horizontal axis, we find
$e_{2} = x_{9} \cos(-\frac{\pi}{4})-x_{10} \sin(-\frac{\pi}{4}) = \frac{x_{9}-x_{1}+x_{2}-x_{10}}{\sqrt{22}} \nonumber$
Similarly, as fiber 4 makes the angle $$-\frac{3 \pi}{4}$$ with respect to the positive horizontal axis, its elongation is
$e_{4} = x_{7} \cos(-\frac{3\pi}{4})-x_{8} \sin(-\frac{3\pi}{4}) = \frac{x_{3}-x_{7}+x_{4}-x_{8}}{\sqrt{22}} \nonumber$
These are both direct applications of the general formula
$e_{j} = x_{2n-1} \cos(\theta_{j})-x_{2n} \sin(\theta_{j}) \nonumber$
for fiber j Figure below, connecting node mm to node nn and making the angle $$\theta_{j}$$ with the positive horizontal axis when node $$m$$ is assumed to lie at the point $$(0, 0)$$. The reader should check that our expressions for $$e_{1}$$ and $$e_{3}$$ indeed conform to this general formula and that $$e_{2}$$ and $$e_{4}$$ agree with ones intuition. For example, visual inspection of the specimen suggests that fiber 2 can not be supposed to stretch (i.e., have positive $$e_{2}$$) unless $$x_{9} > x_{1}$$ and/or $$x_{2} > x_{10}$$. Does this jive with Equation?
Applying Equation to each of the remaining fibers we arrive at $$\textbf{e} = A\textbf{x}$$ where $$A$$ is 20-by-18, one row for each fiber, and one column for each degree of freedom. For systems of such size with such a well defined structure one naturally hopes to automate the construction. We have done just that in the accompanying M-file and diary. The M-file begins with a matrix of raw data that anyone with a protractor could have keyed in directly from Figure 1.:
This data is precisely what Euqation requires in order to know which columns of $$A$$ receive the proper $$\cos$$ or $$\sin$$. The final $$A$$ matrix is displayed in the diary.
The next two steps are now familiar. If $$K$$ denotes the diagonal matrix of fiber stiffnesses and $$\textbf{f}$$ denotes the vector of nodal forces then
$\begin{array} {ccc} {\textbf{y} = K \textbf{e}}&{and}&{A^{T} \textbf{y} = \textbf{f}} \nonumber \end{array}$
and so one must solve $$S \textbf{x} = \textbf{f}$$ where $$S = A^{T}KA$$. In this case there is an entire three--dimensional class of $$\textbf{z}$$ for which $$A \textbf{z} = \textbf{0}$$ and therefore $$S \textbf{z} = \textbf{0}$$ e.g., two translations and a rotation. As a result $$S$$ is singular and x = S\f in MATLAB will get us nowhere. The way out is to recognize that $$S$$ has $$18-3 = 15$$ stable modes and that if we restrict $$S$$ to 'act' only in these directions then it 'should' be invertible. We will begin to make these notions precise in discussions on the Fundamental Theorem of Linear Algebra. For now let us note that every matrix possesses such a pseudo-inverse and that it may be computed in MATLAB via the pinv command. Supposing the fiber stiffnesses to each be one and the edge traction to be of the form
$\textbf{f} =\begin{pmatrix} {-1}&{1}&{0}&{1}&{1}&{1}&{-1}&{0}&{0}&{0}&{1}&{0}&{-1}&{-1}&{0}&{-1}&{1}&{-1} \end{pmatrix}^T \nonumber$,
we arrive at $$\textbf{x}$$ via x=pinv(S)*f and offer below its graphical representation. |
# Question b9e5a
Feb 22, 2017
The Father is now $40$ and Mary is $10$.
#### Explanation:
Let us say the current Mary is $M$ and her current father is $F$.
1) In the present:
$F = 4 M$
Mary’s father is four times as old as Mary.
2) Using the terms above, the equation of their age $5$ years ago is this:
$F - 5 = 7 \left(M - 5\right)$
The father is $5$ years younger and is $7$ times older than Mary was $5$ years ago.
3) Remove the brackets from the previous equation, simplify it and bring it into the terms of $F$.
i) " "F - 5 = 7M - 35
ii) " "F = 7M - 30#
4) Piece it all together.
$7 M - 30 = 4 M$
$3 M - 30 = 0$
$3 M = 30$
$M = 10$
We already know that $F = 4 M$, and we just worked out that F also $= 7 M - 30$. This means $4 M$ and $7 M - 30$ is the same. After working it all out, we see $M = 10$.
5) Double check by substituting.
$M = 10$
$F = 4 M = 40$
$F - 5 = 35$
$M - 5 = 5$
$5 \times 7 = 35$
It all works out...
6) All Done! I hope I helped.
Jul 29, 2017
She is $10$ now and he is $40$
#### Explanation:
There are two people: Mary and her Father
and two periods of time: present and $5$ years ago.
Write an expression for each person for each period of time:
Mary is younger, so let her present age be $x$
Her father is $4$ times older, so his present age is $4 x$
Five years ago, they were both younger by $5$ years
Mary's age was $x - 5$
Her father was $4 x - 5$
However, $5$ years ago, he was $7$ times as old as she was.
Write this as an equation.
$7 \left(x - 5\right) = 4 x - 5$
$7 x - 35 = 4 x - 5$
$7 x - 4 x = 35 - 5$
$3 x = 10$
$x = 10$
She is $10$ now and he is $40$
$5$ years ago .... She was $5$ and he was $35$
$7 \times 5 = 35$ |
## Algebra 1: Common Core (15th Edition)
$(1+5)^{2}-(18\div3)$ -To solve this correctly you must follow PEMDAS (Parenthesis, Exponents, Multiplication, Division, Subtraction). 1. The first step is to solve what is inside the parenthesis. $(1+5)=6$ and $(18\div3)=6$. This simplifies the equation to $6^{2}-(6)$. 2. Next, we must solve the exponents. $6^{2}=36$. This further simplifies the equation to $36-6$. 3. The final step is to subtract $36-6$, leaving us with $30$. The correct answer to this problem is $30$. If you do not follow the order of operations, you will get a different answer. For example, if you were to distribute the exponent before solving the parenthesis, you would get $(1+25)-(18\div3)$. This would significantly change your final answer as you would get $(1+25)-6$, leaving you with an incorrect answer of $20$. This proves that solving a problem while applying PEMDAS in a different order will leave you with an incorrect answer. |
# Elimination Method
Follow the steps to solve the system of linear equations by using the elimination method:
(i) Multiply the given equation by suitable constant so as to make the coefficients of the variable to be eliminated equal.
(ii) Add the new equations obtained if the terms having the same coefficient are opposite signs and subtract if they are of the same sign.
(iii) Solve the equation thus obtained.
(iv) Substitute the value found in any one the given equations.
(v) Solve it to get the value of the other variable.
Worked-out examples on elimination method:
1. Solve the system of equation 2x + y = -4 and 5x – 3y = 1 by the method of elimination.
Solution:
The given equations are:
2x + y = -4 …………… (i)
5x – 3y = 1 …………… (ii)
Multiply equation (i) by 3, we get;
{2x + y = -4} …………… {× 3}
6x + 3y = -12 …………… (iii)
Adding (ii) and (iii), we get;
or, x = -11/11
or, x = -1
Substituting the value of x = -1 in equation (i), we get;
2 × (-1) + y = -4
-2 + y = -4
-2 + 2 + y = -4 + 2
y = -4 + 2
y = -2
Therefore, x = -1 and y = -2 is the solution of the system of equations 2x + y = -4 and 5x – 3y = 1
2. Solve the system of equation 2x + 3y = 11, x + 2y = 7 by the method of elimination.
Solution:
The given equations are:
2x + 3y = 11 …………… (i)
x + 2y = 7 …………… (ii)
Multiply the equation (ii) by 2, we get
{x + 2y = 7} …………… (× 2)
2x + 4y = 14 …………… (iii)
Subtract equation (i) and (ii), we get
Substituting the value of y = 3 in equation (i), we get
2x + 3y = 11
or, 2x + 3 × 3 = 11
or, 2x + 9 = 11
or, 2x + 9 – 9 = 11 – 9
or, 2x = 11 – 9
or, 2x = 2
or, x = 2/2
or, x = 1
Therefore, x = 1 and y = 3 is the solution of the system of the given equations.
3. Solve 2a – 3/b = 12 and 5a – 7/b = 1
Solution:
The given equations are:
2a – 3/b = 12 …………… (i)
5a – 7/b = 1 …………… (ii)
Put 1/b = c, we have
2a – 3c = 12 …………… (iii)
5a – 7c = 1 …………… (iv)
Multiply equation (iii) by 5 and (iv) by 2, we get
10a – 15c = 60 …………… (v)
10a + 14c = 2 …………… (vi)
Subtracting (v) and (vi), we get
or, c = 58 /-29
or, c = -2
But 1/b = c
Therefore, 1/b = -2 or b = -1/2
Subtracting the value of c in equation (v), we get
10a – 15 × (-2) = 60
or, 10a + 30 = 60
or, 10a + 30 - 30= 60 - 30
or, 10a = 60 – 30
or, a = 30/10
or, a = 3
Therefore, a = 3 and b = 1/2 is the solution of the given system of equations.
4. x/2 + 2/3 y = -1 and x – 1/3 y = 3
Solution:
The given equations are:
x/2 + 2/3 y = -1 …………… (i)
x – 1/3 y = 3 …………… (ii)
Multiply equation (i) by 6 and (ii) by 3, we get;
3x + 4y = -6 …………… (iii)
3x – y = 9 …………… (iv)
Solving (iii) and (iv), we get;
or, y = -15/5
or, y = -3
Subtracting the value of y in (ii), we get;
x - 1/3̶ × -3̶ = 3
or, x + 1 = 3
or, x = 3 – 1
or, x = 2
Therefore, x = 2 and y = -3 is the solution of the equation.
x/2 + 2/3 y = -1 and x - y/3 = 3
Simultaneous Linear Equations
Simultaneous Linear Equations
Comparison Method
Elimination Method
Substitution Method
Cross-Multiplication Method
Solvability of Linear Simultaneous Equations
Pairs of Equations
Word Problems on Simultaneous Linear Equations
Word Problems on Simultaneous Linear Equations
Practice Test on Word Problems Involving Simultaneous Linear Equations
Simultaneous Linear Equations - Worksheets
Worksheet on Simultaneous Linear Equations
Worksheet on Problems on Simultaneous Linear Equations |
## Intermediate Algebra (12th Edition)
$\left( 3,\infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $-(4+r)+2-3r\lt-14 ,$ use the properties of inequality. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Using the Distributive Property and the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -4-r+2-3r\lt-14 \\\\ -r-3r\lt-14+4-2 \\\\ -4r\lt-12 .\end{array} Dividing both sides by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} r\gt\dfrac{-12}{-4} \\\\ r\gt3 .\end{array} The red graph is the graph of the solution set. In interval notation, the solution set is $\left( 3,\infty \right) .$ |
# Math website that shows steps
In this blog post, we discuss how Math website that shows steps can help students learn Algebra. Our website can solving math problem.
## The Best Math website that shows steps
Best of all, Math website that shows steps is free to use, so there's no reason not to give it a try! There are a number of ways to solve equations involving synthetic division, but one of the most popular is to use a synthetic division solver. This tool can be found online or in many math textbooks, and it can be a great help in solving complex equations. Synthetic division solvers work by breaking down an equation into smaller pieces, which makes it easier to solve. In addition, they often include step-by-step instructions that can make the process of solving an equation much simpler. If you're struggling with an equation that involves synthetic division, a synthetic division solver can be a valuable resource.
How to solve radicals can be a tricky process, but there are a few steps that can help. First, rationalize the denominator by multiplying by an accessory root. This will eliminate any fractions in the denominator. Next, extract any perfect square roots from the radical. For example, if the radical is 4√5, you would take out the 2√5. Finally, simplify the radical by using absolute value signs and grouping like terms. How to solve radicals may seem complicated at first, but with some practice it can become second nature.
There are a variety of methods that can be used to solve mathematical equations. One of the most common is known as elimination. This method involves adding or subtracting terms from both sides of the equation in order to cancel out one or more variables. For example, consider the equation 2x + 3y = 10. To solve for x, we can add 3y to both sides of the equation, which cancels out y and leaves us with 2x = 10. We can then divide both sides by 2 in order to solve for x, giving us a final answer of x = 5. While elimination may not always be the easiest method, it can be very effective when used correctly.
To solve for the square root of a number, we can use a few different methods. One method is to use factor trees. Another method is to use the long division method. Lastly, we can use estimation methods to approximate the answer. No matter which method we use, being able to solve by square roots is a valuable skill to have!
## We cover all types of math issues
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# How to Calculate 1/5 Times 3/5
Are you looking to work out and calculate how to multiply 1/5 by 3/5? In this really simple guide, we'll teach you exactly what 1/5 times 3/5 is and walk you through the step-by-process of how to multiply two fractions together.
Just a quick reminder here that the number above the fraction line is called the numerator and the number below the fraction line is called the denominator.
Want to quickly learn or refresh memory on factor multiplication, play this quick and informative video developed by visualfractions.com!For more educational and explainer videos on math and numbers including fractions, percentage calculation, conversions and much more visit visualfractions.com Youtube channel.
To multiply two fractions, all we need to do is multiply the numerators and the denominators together and then simplify the fraction if we can.
Let's set up 1/5 and 3/5 side by side so they are easier to see:
1 / 5 x 3 / 5
The next step is to multiply the numerators on the top line and the denominators on the bottom line:
1 x 3 / 5 x 5
From there, we can perform the multiplication to get the resulting fraction:
1 x 3 / 5 x 5 = 3 / 25
You're done! You now know exactly how to calculate 1/5 x 3/5. 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):
3/25
## Convert 1/5 times 3/5 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:
3 / 25 = 0.12
### Cite, Link, or Reference This Page
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "How to Calculate 1/5 times 3/5". VisualFractions.com. Accessed on November 30, 2023. http://visualfractions.com/calculator/multiply-fractions/what-is-1-5-times-3-5/.
• "How to Calculate 1/5 times 3/5". VisualFractions.com, http://visualfractions.com/calculator/multiply-fractions/what-is-1-5-times-3-5/. Accessed 30 November, 2023.
• How to Calculate 1/5 times 3/5. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/multiply-fractions/what-is-1-5-times-3-5/.
### Preset List of Fraction Multiplication Examples
Below are links to some preset calculations that are commonly searched for: |
800score.comhttp://www.800score.com/forum/ GMAT Coordinate Geometryhttp://www.800score.com/forum/viewtopic.php?f=3&t=9 Page 1 of 1
Author: questioner [ Tue Jun 08, 2010 3:57 pm ] Post subject: GMAT Coordinate Geometry In a rectangular coordinate system, what is the area of a quadrilateral whose vertices have the coordinates (2,-2), (2, 6), (15, 2), (15,-4)?A. 91B. 95C. 104D. 117E. 182(A) First, we should make a rough sketch of the figure to determine its general shape. Its left side and right side are parallel, with the left side having a length of 8 and the right side having a length of 6. The distance between these two sides is 13.This figure is a trapezoid. A trapezoid is any quadrilateral that has one set of parallel sides, and the formula for the area of a trapezoid is:Area = (1/2) × (Base 1 + Base 2) × (Height), where the bases are the parallel sides.We can now determine the area of the quadrilateral:Area = 1/2 × (8 + 6) × 13 = 1/2 × 14 × 13 = 7 × 13 = 91.The correct answer is choice (A).Alternate Method (Breaking the figure apart):Without the formula for the area of a trapezoid, we can still solve the problem. We can draw two horizontal lines through the figure, one at y = 2 and one at y = -2 to divide the trapezoidinto an upper triangle, a rectangle, and a lower triangle.The upper triangle has an area of (1/2) × 4 × 13 = 26.The rectangle has an area of 4 × 13 = 52.The lower triangle has an area of 1/2 × 2 × 13 = 13.Adding these areas, we get the area for the quadrilateral:52 + 26 + 13 = 91.Again, we see that the correct answer is choice (A).-------------If the y coordinates are -2 and 6 then the length of the left side is 9 and the right side is 7 (you need to count the 0)! Area = 1/2 × (9+7) × 13 = 104(and not 91)
Author: Gennadiy [ Wed Jun 09, 2010 4:55 pm ]
Post subject: Re: GMAT Coordinate Geometry
It may be easier if you break each segment in two.
Segment that connects points (2, -2) and (2, 6) we can break into
(2, 0) - (2, 6) segment, length 6
&
(2, -2) - (2, 0) segment, length 2.
The total length is 8.
The same works for the right side.
Attachments: xy_plane_2.gif [6.51 KiB] Downloaded 8 times
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# What is 5/3 divided by 5
So you want to divide your fraction 5/3 by your whole number 5, right? You're in the right place. In this simple walkthrough guide, we'll show you exactly what you need to do to divide any fraction by a whole number (it's super simple). Keep reading to find out!
If you've ready any of our fraction walkthroughs before, you'll know we always kick the show off with a quick recap for the kids. The number above the dividing line is the numerator, and the number below the line is the denominator. Simple stuff but sometimes we can all get a little forgetful!
To visualize the question we are trying to solve, let's put 5/3 and 5 side-by-side so it's easier to see:
5 / 3 ÷ 5
So here is the incredibly easy way to figure out what 5/3 divided by 5 is. All we need to do here is keep the numerator exactly the same (5) and multiple the denominator by the whole number:
5 / 3 x 5 = 5 / 15
Can it possibly be that simply to divide a fraction by a whole number? Yup. I hate to disappoint you but this might be the easiest problem you've had to solve all day long!
In this particular case there is one slight adjustment we can make. The new fraction we have (5/15) can actually be simplified down to a smaller fraction:
1 / 3
You're done! You now know exactly how to calculate 5/3 divided by 5. Hopefully you understood the process and can use the same techniques to divide other fractions by whole numbers.
Want to quickly learn or refresh memory on how to divide fractions by whole numbers play this quick and informative video now!
## Convert 5/3 divided by 5 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. Once you have your final fraction, just divide the numerator by the denominator to get your answer in decimal form:
1 / 3 = 0.3333
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 5/3 divided by 5". VisualFractions.com. Accessed on April 18, 2024. http://visualfractions.com/calculator/fraction-divided-by-whole/what-is-5-3-divided-by-5/.
• "What is 5/3 divided by 5". VisualFractions.com, http://visualfractions.com/calculator/fraction-divided-by-whole/what-is-5-3-divided-by-5/. Accessed 18 April, 2024. |
# What is the vertex form of y=−x^2 − 7x + 1 ?
The Vertex Form ${\left(x - - \frac{7}{2}\right)}^{2} = - \left(y - \frac{53}{4}\right)$ with vertex at $\left(- \frac{7}{2} , \frac{53}{4}\right)$
#### Explanation:
We start from the given and do the "Completing the Square Method"
$y = - {x}^{2} - 7 x + 1$
factor out the $- 1$ first
$y = - 1 \cdot \left({x}^{2} + 7 x\right) + 1$
Compute the number to be added and subtracted using the numerical coefficient of x which is the 7. Divide the 7 by 2 and square the result,...that is ${\left(\frac{7}{2}\right)}^{2} = \frac{49}{4}$
$y = - 1 \cdot \left({x}^{2} + 7 x\right) + 1$
$y = - 1 \cdot \left({x}^{2} + 7 x + \frac{49}{4} - \frac{49}{4}\right) + 1$
the first three terms inside the parenthesis forms a PST-perfect square trinomial.
$y = - 1 \cdot \left({x}^{2} + 7 x + \frac{49}{4} - \frac{49}{4}\right) + 1$
$y = - 1 \cdot \left(\left({x}^{2} + 7 x + \frac{49}{4}\right) - \frac{49}{4}\right) + 1$
$y = - 1 \cdot \left({\left(x + \frac{7}{2}\right)}^{2} - \frac{49}{4}\right) + 1$
simplify by multiplying the -1 back and removing the grouping symbol
$y = - 1 {\left(x + \frac{7}{2}\right)}^{2} + \frac{49}{4} + 1$
$y = - 1 {\left(x + \frac{7}{2}\right)}^{2} + \frac{53}{4}$
$y - \frac{53}{4} = - 1 {\left(x + \frac{7}{2}\right)}^{2}$
Let us form the Vertex Form
${\left(x - h\right)}^{2} = \pm 4 p \left(y - k\right)$
${\left(x - - \frac{7}{2}\right)}^{2} = - \left(y - \frac{53}{4}\right)$
Kindly see the graph
graph{(x- -7/2)^2=-(y-53/4)[-30,30,-15,15]}
God bless ....I hope the explanation is useful. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Linear Programming
## Maximize and minimize quantities using linear inequality systems
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Practice Linear Programming
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Applications of Systems of Inequalities
The following diagram shows a feasible region solution for a system of linear inequalities.
If \begin{align*}z=2x+3y\end{align*}, at what point on the graph is the value of \begin{align*}z\end{align*} the largest?
### Guidance
A system of linear inequalities is often used to determine the maximum or minimum values of a situation with multiple constraints. For example, you might be determining how many of a product should be produced to maximize a profit.
In order to solve this type of problem using linear inequalities, follow these steps:
• Step 1: Make a table to organize the given information.
• Step 2: List the constraints of the situation. Write an inequality for each constraint.
• Step 3: Write an equation for the quantity you are trying to maximize (like profit) or minimize (like cost).
• Step 4: Graph the constraints as a system of inequalities.
• Step 5: Find the exact coordinates for each vertex from the graph or algebraically.
• Step 6: Use the Vertex Theorem. Test all vertices of the feasible region in the equation and see which point is the maximum or minimum.
Look at the examples below to see how this process works. Example A allows you to see Steps 5 and 6. Examples B and C allow you to see all the steps in the process.
#### Example A
Evaluate the expression \begin{align*}z=3x+4y\end{align*} for the given feasible region to determine the point at which \begin{align*}z\end{align*} has a maximum value and the point at which \begin{align*}z\end{align*} has a minimum value.
Solution:
The maximum value of \begin{align*}z\end{align*} occurred at the vertex (6, 3). The minimum value of \begin{align*}z\end{align*} occurred at the vertex (–4, 0).
Note: Using the vertices of the feasible region to determine the maximum or the minimum value is a branch of mathematics known as linear programming. Linear programming is a technique used by businesses to solve problems. The types of problems that usually employ linear programming are those where the profit is to be maximized and those where the expenses are to be minimized. However, linear programming can also be used to solve other types of problems. The solution provides the business with a program to follow to obtain the best results for the company.
#### Example B
A company that produces flags makes two flags for Nova Scotia-the traditional blue flag and the green flag for Cape Breton. To produce each flag, two types of material, nylon and cotton, are used. The company has 450 units of nylon in stock and 300 units of cotton. The traditional blue flag requires 6 units of nylon and 3 units of cotton. The Cape Breton flag requires 5 units of nylon and 5 units of cotton. Each blue flag that is made realizes a profit of $12 for the company, whereas each Cape Breton flag realizes a profit of$15. For the nylon and cotton that the company currently has in stock, how many of each flag should the company make to maximize their profit?
Solution: Let ‘\begin{align*}x\end{align*}’ represent the number of blue flags. Let ‘\begin{align*}y\end{align*}’ represent the number of green flags.
Step 1: Transfer the information presented in the problem to a table.
Units Required per Blue Flag Units Required Per Green Flag Units Available
Nylon 6 5 450
Cotton 3 5 300
Profit(per flag) $12$15
The information presented in the problem identifies the restrictions or conditions on the production of the flags. These restrictions are known as constraints and are written as inequalities to represent the information presented in the problem.
Step 2: From the information (now in the table), list the constraints.
• The number of blue flags that are produced must be either zero or greater than zero. Therefore, the constraint is .
• The number of green flags that are produced must be either zero or greater than zero. Therefore, the constraint is .
• The total number of units of nylon required to make both types of flags cannot exceed 450. Therefore, the constraint is .
• The total number of units of cotton required to make both types of flags cannot exceed 300. Therefore, the constraint is .
Step 3: Write an equation to identify the profit.
Step 4: Graph the listed constraints to identify the feasible region.
The feasible region is the area shaded in teal blue.
Step 5: Algebraically, determine the exact point of intersection between the constraints. Also, the \begin{align*}x\end{align*}-intercept of the feasible region must be calculated. Write the constraints as linear equations and solve the system by elimination.
The \begin{align*}x\end{align*}-intercept for the inequality \begin{align*}6x+5y \le 450\end{align*} must be calculated. Write the inequality as a linear equation. Set ‘\begin{align*}y\end{align*}’ equal to zero and solve the equation for ‘\begin{align*}x\end{align*}’.
The \begin{align*}x\end{align*}-intercept of the feasible region is (75, 0).
The \begin{align*}y\end{align*}-intercept is (0, 60). This point was plotted when the inequalities were put into slope-intercept form for graphing.
The following graph shows the vertices of the polygon that encloses the feasible region.
Step 6: Calculate the profit, using the profit equation, for each vertex of the feasible region:
The maximum profit occurred at the vertex (50, 30). This means, with the supplies in stock, the company should make 50 blue flags and 30 green flags to maximize their profit.
#### Example C
A local smelting company is able to provide its customers with iron, lead and copper by melting down either of two ores, A or B. The ores arrive at the company in railroad cars. Each railroad car of ore A contains 3 tons of iron, 3 tons of lead and I ton of copper. Each railroad car of ore B contains 1 ton of iron, 4 tons of lead and 3 tons of copper. The smelting receives an order for 7 tons of iron, 19 tons of lead and 8 tons of copper. The cost to purchase and process a carload of ore A is $7000 while the cost for ore B is$6000. If the company wants to fill the order at a minimum cost, how many carloads of each ore must be bought?
Solution: Let ‘\begin{align*}x\end{align*}’ represent the number of carloads of ore A to purchase. Let ‘\begin{align*}y\end{align*}’ represent the number of carloads of ore B to purchase.
Step 1: Transfer the information presented in the problem to a table.
One Carload of ore A One Carload of ore B Number of tons to fill the order
Tons of Iron 3 1 7
Tons of Lead 3 4 19
Tons of Copper 1 3 8
Step 2: From the information, list the constraints.
• The number of carloads of ore A that must be bought is either zero or greater than zero. Therefore, the constraint is .
• The number of carloads of ore B that must be bought is either zero or greater than zero. Therefore, the constraint is .
• The total number of tons of iron from ore A and ore B must be greater than or equal to the 7 tons needed to fill the order. Therefore, the constraint is .
• The total number of tons of lead from ore A and ore B must be greater than or equal to the 20 tons needed to fill the order. Therefore, the constraint is .
• The total number of tons of copper from ore A and ore B must be greater than or equal to the 8 tons needed to fill the order. Therefore, the constraint is .
Step 3: Write an equation to represent the cost in dollars of \begin{align*}x\end{align*} carloads of ore A and y carloads of ore B.
Step 4: Graph the listed constraints to identify the feasible region.
The feasible region shows that there are an infinite number of ways to fill the order. The feasible region is the large shaded area that is sitting above the graphed lines.
Step 5: Algebraically, determine the exact point of intersection between the constraints. Also, the \begin{align*}x\end{align*}-intercept of the feasible region must be calculated. Write the constraints as linear equations and solve the system by elimination.
The \begin{align*}x\end{align*}-intercept for the inequality \begin{align*}x+3y \ge 8\end{align*} must be calculated. Write the inequality as a linear equation. Set ‘\begin{align*}y\end{align*}’ equal to zero and solve the equation for ‘\begin{align*}x\end{align*}’.
The \begin{align*}x\end{align*}-intercept of the feasible region is (8, 0).
The \begin{align*}y\end{align*}-intercept is (0, 7). This point was plotted when the inequalities were put into slope-intercept form for graphing.
The following graph shows the vertices of the region borders the feasible region.
Step 6: Calculate the cost, using the cost equation, for each vertex of the feasible region:
The minimum cost is located at the vertex (1, 4). Therefore the company should buy one carload of ore A and four carloads of ore B.
#### Concept Problem Revisited
The following diagram shows a feasible region that is within a polygonal region.
The linear function \begin{align*}z=2x+3y\end{align*} will now be evaluated for each of the vertices of the polygon.
To evaluate the value of ‘\begin{align*}z\end{align*}’ substitute the coordinates of the point into the expression for ‘\begin{align*}x\end{align*}’ and ‘\begin{align*}y\end{align*}’.
The value of \begin{align*}z=2x+3y\end{align*}, for each of the vertices, remains constant along any line with a slope of \begin{align*}-\frac{2}{3}\end{align*}. This is obvious on the following graph.
As the line moved away from the origin, the value of \begin{align*}z=2x+3y\end{align*} increased. The maximum value for the shaded region occurred at the vertex (9, 4) while the minimum value occurred at the vertex (0, 0). These statements confirm the vertex theorem for a feasible region:
If a linear expression is to be evaluated for all points of a convex, polygonal region, then the maximum value of \begin{align*}z\end{align*}, if one exists, will occur at one of the vertices of the feasible region. Also, the minimum value of \begin{align*}z\end{align*}, if one exists, will occur at one of the vertices of the feasible region.
### Vocabulary
Constraint
A constraint is a restriction or condition presented in a real-world problem. The constraints are written as inequalities and are used to solve the problem.
Linear Programming
Linear programming is a branch of mathematics that uses systems of linear inequalities to solve real-world problems. The vertex theorem of regions is applied to the vertices to determine the best solution to the problem.
Vertex Theorem for Regions
The vertex theorem for regions states that if a linear expression \begin{align*}z=ax+by+c\end{align*} is to be evaluated for all points of a convex, polygonal region, then the maximum value of \begin{align*}z\end{align*}, if one exists, will occur at one of the vertices of the feasible region. Also, the minimum value of \begin{align*}z\end{align*}, if one exists, will occur at one of the vertices of the feasible region.
### Guided Practice
1. For the following graphed region and the expression \begin{align*}z=5x+7y-1\end{align*}, find a point where ‘\begin{align*}z\end{align*}’ has a maximum value and a point where ‘\begin{align*}z\end{align*}’ has a minimum value.
2. The following table shows the time required on three machines for a company to produce Super 1 and Super 2 coffee percolators. The table also shows the amount of time that each machine is available during a one hour period. The company is trying to determine how many of each must be made to maximize a profit if they make $30 on each Super 1 model and$35 on each Super 2 model. List the constraints and write a profit statement to represent the information.
Super 1 Super 2 Time Available
Machine A 1 minute 3minutes 24 minutes
Machine B 3 minutes 2minutes 36 minutes
Machine C 3 minutes 4 minutes 44 minutes
3. A local paint company has created two new paint colors. The company has 28 units of yellow tint and 22 units of red tint and intends to mix as many quarts as possible of color X and color Y. Each quart of color X requires 4 units of yellow tint and 1 unit of red tint. Each quart of color Y requires 1 unit of yellow tint and 4 units of red tint. How many quarts of each color can be mixed with the units of tint that the company has available? List the constraints, complete the graph and determine the solution using linear programming.
1. The vertices of the polygonal region are (–7, –1); (2, 5); (6, 1); and (0, –4).
The maximum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (2, 5). The minimum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (–7, –1).
2. Let ‘\begin{align*}x\end{align*}’ represent the number of Super 1 coffee percolators. Let ‘\begin{align*}y\end{align*}’ represent the number of Super 2 coffee percolators.
• The number of Super 1 coffee percolators that are made must be either zero or greater than zero. Therefore, the constraint is .
• The number of Super 2 coffee percolators that are made must be either zero or greater than zero. Therefore, the constraint is .
• The total amount of time that both a Super 1 and a Super 2 model can be processed on Machine A is less than or equal to 24 minutes. Therefore, the constraint is .
• The total amount of time that both a Super 1 and a Super 2 model can be processed on Machine B is less than or equal to 36 minutes. Therefore, the constraint is .
• The total amount of time that both a Super1 and a Super 2 model can be processed on Machine C is less than or equal to 44 minutes. Therefore, the constraint is .
• The profit equation is
3. Table:
Color X Color Y Units Available
Yellow Tint 4 units 1 unit 28
Red Tint 1 unit 4 units 22
Constraints: Let ‘\begin{align*}x\end{align*}’ represent the number of quarts of Color X paint to be made. Let ‘\begin{align*}y\end{align*}’ represent the number of quarts of Color Y paint to be made.
• The number of quarts of Color X paint that are mixed must be either zero or greater than zero. Therefore, the constraint is .
• The number of quarts of Color Y paint that are mixed must be either zero or greater than zero. Therefore, the constraint is .
• The total amount of yellow tint that is used to mix Color X and Color Y must be less than or equal to 28. Therefore, the constraint is .
• The total amount of red tint that is used to mix Color X and Color Y must be less than or equal to 22. Therefore, the constraint is .
Equation: The company wants to mix as many quarts as possible. They want to maximize the value of \begin{align*}Q\end{align*} given by \begin{align*}Q=x+y\end{align*}.
Graph:
Vertices:
The three points in the feasible region are \begin{align*}(6, 4), (7,0), (0, 5.5)\end{align*}. The company wants to maximize \begin{align*}Q=x+y\end{align*}. The point that produces the maximum value is \begin{align*}(6, 4)\end{align*}.
The company should mix 6 quarts of Color X paint and 4 quarts of Color Y paint.
### Practice
For each graphed region and corresponding equation, find a point at which ‘\begin{align*}z\end{align*}’ has a maximum value and a point at which ‘\begin{align*}z\end{align*}’ has a minimum value.
A small manufacturing company makes $125 on each DVD player it produces and$100 profit on each color TV set it makes. Each DVD player and each TV must be processed by a cutting machine (A), a fitting machine (B) and a polishing machine (C). Each DVD player must be processed on Machine A for one hour, on Machine B for one hour and on Machine C for four hours. Each TV set must be processed on Machine A for two hours, on Machine B for one hour and on Machine C for one hour. Machines A, B, and C are available for 16, 9, and 24 hours per day respectively. How many DVD players and TV sets must be made each day to maximize the profit?
1. List the constraints and state the profit equation.
2. Create a graph and identify the feasible region.
3. Determine what the company must do to maximize their profit.
April has a small business during the winter months making hats and scarves. A hat requires 2 hours on Machine A, 4 hours on Machine B and 2 hours on Machine C. A scarf requires 3 hours on Machine A, 3 hours on Machine B and 1 hour on Machine C. Machine A is available 36 hours each week, Machine B is available 42 hours each week and Machine C is available 20 hours each week. The profit on a hat is $7.00 and the profit on a scarf is$4.00. How many of each should be made each week to maximize the profit?
1. List the constraints and state the profit equation.
2. Create a graph and identify the feasible region.
3. Determine what the April must do to maximize her profit.
Beth is knitting mittens and gloves. Each pair must be processed on three machines. Each pair of mittens requires 2 hours on Machine A, 2 hours on Machine B and 4 hours on Machine C. Each pair of gloves requires 4 hours on Machine A, 2 hours on Machine B and 1 hour on Machine C. Machine A, B, and C are available 32, 18 and 24 minutes each day respectively. The profit on a pair of mittens is $8.00 and on a pair of gloves is$10.00. How many pairs of each should be made each day to maximize the profit?
1. List the constraints and state the profit equation.
2. Create a graph and identify the feasible region.
3. Determine what the Beth must do to maximize her profit.
A patient is prescribed a pill that contains vitamins A, B and C. These vitamins are available in two different brands of pills. The first type is called Brand X and the second type is called Brand Y. The following table shows the amount of each vitamin that a Brand X and a Brand Y pill contain. The table also shows the minimum daily requirement needed by the patient. Each Brand X pill costs \begin{align*}0.32\end{align*} and each Brand Y pill costs \begin{align*} 0.29\end{align*}. How many pills of each brand should the patient take each day to minimize the cost?
Brand X Brand Y Minimum Daily Requirement
Vitamin A 2mg 1mg 5mg
Vitamin B 3mg 3mg 12mg
Vitamin C 25mg 50mg 125mg
1. List the constraints and state the cost equation.
2. Create a graph and identify the feasible region.
3. Determine what the patient must do to minimize his/her cost.
A local smelting company is able to provide its customers with lead, copper and iron by melting down either of two ores, X or Y. The ores arrive at the company in railroad cars. Each railroad car of ore X contains 5 tons of lead, 1 ton of copper and 1 ton of iron. Each railroad car of ore Y contains 1 ton of lead, 1 ton of copper and 2 tons of iron. The smelting company receives an order and must make at least 20 tons of lead, 12 tons of copper and 20 tons of iron. The cost to purchase and process a carload of ore X is $6000 while the cost for ore Y is$5000. If the company wants to fill the order at a minimum cost, how many carloads of each ore must be bought?
1. List the constraints and state the cost equation.
2. Create a graph and identify the feasible region.
3. Determine what the company must do to minimize their cost.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 5.7.
### Vocabulary Language: English
Constraint
Constraint
A constraint is a limitation.
Feasible region
Feasible region
The feasible region is the inner region within a set of four inequalities on a graph.
Linear Programming
Linear Programming
Linear programming uses the vertices of the feasible region (the area of overlap between multiple inequalities) to determine a maximum or minimum value.
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### Explore More
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# How to do 3 Digit Multiplication?
We can solve the problem of multiplication by simply repeated addition also, but at times when we have larger digits, this method of repeated addition does not work. Here we will discuss how to do 3 digit multiplication.
For multiplying a three digit number with another number of one digit, we will first place the digits of the multiplicand under their respective place values and in the next row we write the digit of the multiplier. Thus every digit of the multiplicand is multiplied by the multiplier and in case the number is greater than 9, i.e. in two digits, then the digit at the tens place is transferred to the next place value and added their with the product. Let us look at the following example: 125 * 5,
H t o
1 2 5
X 5
5 2 5
Now we will learn how a three digit number is multiplied by a two digit number. In this case we will first write the number in two rows at their proper place so that the product of 204 * 12 looks as follows:
H T O
2 0 4
X 1 2
---------------
4 0 8 (Here we will multiply the multiplicand with the ones place of the multiplier i.e. 2. Now we will write 0 at the ones place in the next row).
2 0 4 x (and then multiply 204 by 1, which is at the tens place of the number 12)
------------------
2 4 4 8 ( this is the product of 204 and 12, which we will get by adding the digits in the two rows. )
In this way we get the product of the three digit number with a two digit number, and here we observe 204 X 12 = 2448. |
## A ratio compares values .
A ratio says how much of one thing there is compared to another thing.
Ratios can be shown in different ways:
A ratio can be scaled up:
## Try it Yourself
Using ratios.
The trick with ratios is to always multiply or divide the numbers by the same value .
## Example: A Recipe for pancakes uses 3 cups of flour and 2 cups of milk.
So the ratio of flour to milk is 3 : 2
To make pancakes for a LOT of people we might need 4 times the quantity, so we multiply the numbers by 4:
3 ×4 : 2 ×4 = 12 : 8
In other words, 12 cups of flour and 8 cups of milk .
The ratio is still the same, so the pancakes should be just as yummy.
## "Part-to-Part" and "Part-to-Whole" Ratios
The examples so far have been "part-to-part" (comparing one part to another part).
But a ratio can also show a part compared to the whole lot .
## Example: There are 5 pups, 2 are boys, and 3 are girls
Part-to-Part:
The ratio of boys to girls is 2:3 or 2 / 3
The ratio of girls to boys is 3:2 or 3 / 2
Part-to-Whole:
The ratio of boys to all pups is 2:5 or 2 / 5
The ratio of girls to all pups is 3:5 or 3 / 5
## Try It Yourself
We can use ratios to scale drawings up or down (by multiplying or dividing).
## Example: To draw a horse at 1/10th normal size, multiply all sizes by 1/10th
This horse in real life is 1500 mm high and 2000 mm long, so the ratio of its height to length is
1500 : 2000
What is that ratio when we draw it at 1/10th normal size?
We can make any reduction/enlargement we want that way.
"I must have big feet, my foot is nearly as long as my Mom's!"
But then she thought to measure heights, and found she is 133cm tall, and her Mom is 152cm tall.
In a table this is:
The "foot-to-height" ratio in fraction style is:
We can simplify those fractions like this:
And we get this (please check that the calcs are correct):
"Oh!" she said, "the Ratios are the same".
"So my foot is only as big as it should be for my height, and is not really too big."
You can practice your ratio skills by Making Some Chocolate Crispies
## 5.4 Ratios and Proportions
Learning objectives.
After completing this section, you should be able to:
• Construct ratios to express comparison of two quantities.
• Use and apply proportional relationships to solve problems.
• Determine and apply a constant of proportionality.
• Use proportions to solve scaling problems.
Ratios and proportions are used in a wide variety of situations to make comparisons. For example, using the information from Figure 5.15 , we can see that the number of Facebook users compared to the number of Twitter users is 2,006 M to 328 M. Note that the "M" stands for million, so 2,006 million is actually 2,006,000,000 and 328 million is 328,000,000. Similarly, the number of Qzone users compared to the number of Pinterest users is in a ratio of 632 million to 175 million. These types of comparisons are ratios.
## Constructing Ratios to Express Comparison of Two Quantities
Note there are three different ways to write a ratio , which is a comparison of two numbers that can be written as: a a to b b OR a : b a : b OR the fraction a / b a / b . Which method you use often depends upon the situation. For the most part, we will want to write our ratios using the fraction notation. Note that, while all ratios are fractions, not all fractions are ratios. Ratios make part to part, part to whole, and whole to part comparisons. Fractions make part to whole comparisons only.
## Example 5.28
Expressing the relationship between two currencies as a ratio.
The Euro (€) is the most common currency used in Europe. Twenty-two nations, including Italy, France, Germany, Spain, Portugal, and the Netherlands use it. On June 9, 2021, 1 U.S. dollar was worth 0.82 Euros. Write this comparison as a ratio.
Using the definition of ratio, let a = 1 a = 1 U.S. dollar and let b = 0.82 b = 0.82 Euros. Then the ratio can be written as either 1 to 0.82; or 1:0.82; or 1 0.82 . 1 0.82 .
Example 5.29, expressing the relationship between two weights as a ratio.
The gravitational pull on various planetary bodies in our solar system varies. Because weight is the force of gravity acting upon a mass, the weights of objects is different on various planetary bodies than they are on Earth. For example, a person who weighs 200 pounds on Earth would weigh only 33 pounds on the moon! Write this comparison as a ratio.
Using the definition of ratio, let a = 200 a = 200 pounds on Earth and let b = 33 b = 33 pounds on the moon. Then the ratio can be written as either 200 to 33; or 200:33; or 200 33 . 200 33 .
Using and applying proportional relationships to solve problems.
Using proportions to solve problems is a very useful method. It is usually used when you know three parts of the proportion, and one part is unknown. Proportions are often solved by setting up like ratios. If a b a b and c d c d are two ratios such that a b = c d , a b = c d , then the fractions are said to be proportional . Also, two fractions a b a b and c d c d are proportional ( a b = c d ) ( a b = c d ) if and only if a × d = b × c a × d = b × c .
## Example 5.30
Solving a proportion involving two currencies.
You are going to take a trip to France. You have $520 U.S. dollars that you wish to convert to Euros. You know that 1 U.S. dollar is worth 0.82 Euros. How much money in Euros can you get in exchange for$520?
Step 1: Set up the two ratios into a proportion; let x x be the variable that represents the unknown. Notice that U.S. dollar amounts are in both numerators and Euro amounts are in both denominators.
Step 2: Cross multiply, since the ratios a b a b and c d c d are proportional, then a × d = b × c a × d = b × c .
You should receive 426.4 426.4 Euros ( 426.4 € ) ( 426.4 € ) .
Example 5.31, solving a proportion involving weights on different planets.
A person who weighs 170 pounds on Earth would weigh 64 pounds on Mars. How much would a typical racehorse (1,000 pounds) weigh on Mars? Round your answer to the nearest tenth.
Step 1: Set up the two ratios into a proportion. Notice the Earth weights are both in the numerator and the Mars weights are both in the denominator.
Step 2: Cross multiply, and then divide to solve.
So the 1,000-pound horse would weigh about 376.5 pounds on Mars.
Example 5.32, solving a proportion involving baking.
A cookie recipe needs 2 1 4 2 1 4 cups of flour to make 60 cookies. Jackie is baking cookies for a large fundraiser; she is told she needs to bake 1,020 cookies! How many cups of flour will she need?
Step 1: Set up the two ratios into a proportion. Notice that the cups of flour are both in the numerator and the amounts of cookies are both in the denominator. To make the calculations more efficient, the cups of flour ( 2 1 4 ) ( 2 1 4 ) is converted to a decimal number (2.25).
Step 2: Cross multiply, and then simplify to solve.
Jackie will need 38.25, or 38 1 4 38 1 4 , cups of flour to bake 1,020 cookies.
Part of the definition of proportion states that two fractions a b a b and c d c d are proportional if a × d = b × c a × d = b × c . This is the "cross multiplication" rule that students often use (and unfortunately, often use incorrectly). The only time cross multiplication can be used is if you have two ratios (and only two ratios) set up in a proportion. For example, you cannot use cross multiplication to solve for x x in an equation such as 2 5 = x 8 + 3 x 2 5 = x 8 + 3 x because you do not have just the two ratios. Of course, you could use the rules of algebra to change it to be just two ratios and then you could use cross multiplication, but in its present form, cross multiplication cannot be used.
## People in Mathematics
Eudoxus was born around 408 BCE in Cnidus (now known as Knidos) in modern-day Turkey. As a young man, he traveled to Italy to study under Archytas, one of the followers of Pythagoras. He also traveled to Athens to hear lectures by Plato and to Egypt to study astronomy. He eventually founded a school and had many students.
Eudoxus made many contributions to the field of mathematics. In mathematics, he is probably best known for his work with the idea of proportions. He created a definition of proportions that allowed for the comparison of any numbers, even irrational ones. His definition concerning the equality of ratios was similar to the idea of cross multiplying that is used today. From his work on proportions, he devised what could be described as a method of integration, roughly 2000 years before calculus (which includes integration) would be fully developed by Isaac Newton and Gottfried Leibniz. Through this technique, Eudoxus became the first person to rigorously prove various theorems involving the volumes of certain objects. He also developed a planetary theory, made a sundial still usable today, and wrote a seven volume book on geography called Tour of the Earth , in which he wrote about all the civilizations on the Earth, and their political systems, that were known at the time. While this book has been lost to history, over 100 references to it by different ancient writers attest to its usefulness and popularity.
## Determining and Applying a Constant of Proportionality
In the last example, we were given that 2 1 4 2 1 4 cups of flour could make 60 cookies; we then calculated that 38 1 4 38 1 4 cups of flour would make 1,020 cookies, and 720 cookies could be made from 27 cups of flour. Each of those three ratios is written as a fraction below (with the fractions converted to decimals). What happens if you divide the numerator by the denominator in each?
The quotients in each are exactly the same! This number, determined from the ratio of cups of flour to cookies, is called the constant of proportionality . If the values a a and b b are related by the equality a b = k , a b = k , then k k is the constant of proportionality between a a and b b . Note since a b = k , a b = k , then b = a k . b = a k . and b = a k . b = a k .
One piece of information that we can derive from the constant of proportionality is a unit rate. In our example (cups of flour divided by cookies), the constant of proportionality is telling us that it takes 0.0375 cups of flour to make one cookie. What if we had performed the calculation the other way (cookies divided by cups of flour)?
In this case, the constant of proportionality ( 26.66666 … = 26 2 3 ) ( 26.66666 … = 26 2 3 ) is telling us that 26 2 3 26 2 3 cookies can be made with one cup of flour. Notice in both cases, the "one" unit is associated with the denominator. The constant of proportionality is also useful in calculations if you only know one part of the ratio and wish to find the other.
## Example 5.33
Finding a constant of proportionality.
Isabelle has a part-time job. She kept track of her pay and the number of hours she worked on four different days, and recorded it in the table below. What is the constant of proportionality, or pay divided by hours? What does the constant of proportionality tell you in this situation?
To find the constant of proportionality, divide the pay by hours using the information from any of the four columns. For example, 87.5 7 = 12.5 87.5 7 = 12.5 . The constant of proportionality is 12.5, or $12.50. This tells you Isabelle's hourly pay: For every hour she works, she gets paid$12.50.
Example 5.34, applying a constant of proportionality: running mph.
Zac runs at a constant speed: 4 miles per hour (mph). One day, Zac left his house at exactly noon (12:00 PM) to begin running; when he returned, his clock said 4:30 PM. How many miles did he run?
The constant of proportionality in this problem is 4 miles per hour (or 4 miles in 1 hour). Since a b = k , a b = k , where k k is the constant of proportionality, we have
a miles b hours = k a miles b hours = k
a 4 .5 = 4 a 4 .5 = 4 (30 minutes is ½ ½ , or 0.5 0.5 , hours)
a = 4 ( 4.5 ) a = 4 ( 4.5 ) , since from the definition we know a = k b a = k b
a = 18 a = 18
Zac ran 18 miles.
Example 5.35, applying a constant of proportionality: filling buckets.
Joe had a job where every time he filled a bucket with dirt, he was paid $2.50. One day Joe was paid$337.50. How many buckets did he fill that day?
The constant of proportionality in this situation is $2.50 per bucket (or$2.50 for 1 bucket). Since a b = k , a b = k , where k k is the constant of proportionality, we have
a dollars b buckets = k 337.50 b = 2.50 a dollars b buckets = k 337.50 b = 2.50
Since we are solving for b b , and we know from the definition that b = a k : b = a k :
b = 337.50 2.50 b = 135 b = 337.50 2.50 b = 135
Joe filled 135 buckets.
Example 5.36, applying a constant of proportionality: miles vs. kilometers.
While driving in Canada, Mabel quickly noticed the distances on the road signs were in kilometers, not miles. She knew the constant of proportionality for converting kilometers to miles was about 0.62—that is, there are about 0.62 miles in 1 kilometer. If the last road sign she saw stated that Montreal is 104 kilometers away, about how many more miles does Mabel have to drive? Round your answer to the nearest tenth.
The constant of proportionality in this situation is 0.62 miles per 1 kilometer. Since a b = k , a b = k , where k k is the constant of proportionality, we have
a miles b kilometers = k a 104 = 0.62 a = 0.62 ( 104 ) a = 64.48 a miles b kilometers = k a 104 = 0.62 a = 0.62 ( 104 ) a = 64.48
Rounding the answer to the nearest tenth, Mabel has to drive 64.5 miles.
Using proportions to solve scaling problems.
Ratio and proportions are used to solve problems involving scale . One common place you see a scale is on a map (as represented in Figure 5.16 ). In this image, 1 inch is equal to 200 miles. This is the scale. This means that 1 inch on the map corresponds to 200 miles on the surface of Earth. Another place where scales are used is with models: model cars, trucks, airplanes, trains, and so on. A common ratio given for model cars is 1:24—that means that 1 inch in length on the model car is equal to 24 inches (2 feet) on an actual automobile. Although these are two common places that scale is used, it is used in a variety of other ways as well.
## Example 5.37
Solving a scaling problem involving maps.
Figure 5.17 is an outline map of the state of Colorado and its counties. If the distance of the southern border is 380 miles, determine the scale (i.e., 1 inch = how many miles). Then use that scale to determine the approximate lengths of the other borders of the state of Colorado.
When the southern border is measured with a ruler, the length is 4 inches. Since the length of the border in real life is 380 miles, our scale is 1 inch = 95 = 95 miles.
The eastern and western borders both measure 3 inches, so their lengths are about 285 miles. The northern border measures the same as the southern border, so it has a length of 380 miles.
Example 5.38, solving a scaling problem involving model cars.
Die-cast NASCAR model cars are said to be built on a scale of 1:24 when compared to the actual car. If a model car is 9 inches long, how long is a real NASCAR automobile? Write your answer in feet.
The scale tells us that 1 inch of the model car is equal to 24 inches (2 feet) on the real automobile. So set up the two ratios into a proportion. Notice that the model lengths are both in the numerator and the NASCAR automobile lengths are both in the denominator.
This amount (216) is in inches. To convert to feet, divide by 12, because there are 12 inches in a foot (this conversion from inches to feet is really another proportion!). The final answer is:
The NASCAR automobile is 18 feet long.
Check your understanding, section 5.4 exercises.
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Home / United States / Math Classes / 7th Grade Math / Writing and Interpreting Ratios
## Writing and Interpreting Ratios
We use ratios to compare two quantities having the same unit. Learn how to write ratios, interpret comparisons from real -life problems, the concept of equivalent ratios, and how ratios can be expressed using a ratio table. ...Read More Read Less
• What are Ratios?
• How do we express Ratios?
## Application of Ratios
Equivalent ratios and ratio table, solved examples.
## What are Ratios ?
A ratio is a comparison of quantities having the same unit. A ratio helps us indicate how big or how small a quantity is when compared to another quantity. Two known quantities can be compared by dividing them. We can use division to compare the quantities $$a$$ and $$b$$: $$\frac{a}{b}$$ . Here, $$a$$ is the dividend and $$b$$ is the divisor. Ratio is an efficient method of representing this division operation in a different manner.
## How do we express Ratios ?
A ratio can be expressed in three ways:
• A ratio is denoted using the ‘:’ symbol. A ratio is expressed by writing the ‘:’ symbol in the middle of the two quantities that are being compared. In the ratio $$a : b$$ , $$a$$ and $$b$$ are together known as the terms of the ratio. Also, $$a$$ is known as the antecedent or the first term, and $$b$$ is known as the consequent or the second term.
• But ratios needn’t always be represented using the ‘:’ symbol. As ratios are basically fractions, a fraction like $$\frac{a}{b}$$ can also be interpreted as a ratio.
• A ratio can also be expressed in words as $$a$$ to $$b$$ .
∴ $$a : b$$ = $$\frac{a}{b}$$ = $$a$$ to $$b$$
As discussed earlier, ratios can be used to compare quantities in our day-to-day lives. You can easily compare the number of boys and the number of girls in your classroom by taking the ratio.
Ratios can help make decision-making easier while performing activities like grocery shopping and cooking easier.
Suppose you need 3.5 ounces of ice cream and 10 ounces of milk while making a milkshake for one person. Ratios make it easy to scale up this recipe if we need to make this milkshake for multiple people. Let’s see how ratios can be scaled up by performing operations on them.
Two ratios that describe the same relationship are known as equivalent ratios. Equivalent ratios can be represented using a ratio table by adding or subtracting quantities in equivalent ratios, or by multiplying and dividing with the same values.
In the previous example of the recipe of milkshakes, a ratio table can be used to scale up the number of servings.
Ingredients for 1 milkshake: 3.5 ounces of ice cream for 10 ounces of milk.
With this information in hand, it is possible to scale up the recipe to the required amount.
Missing values in a ratio table can be calculated by performing addition, subtraction, multiplication, or division.
Example 1: If there are 25 students in a class and 15 of them are girls, find the simplest ratio of the number of boys to the number of girls.
Total number of students = 25
Number of girls = 15
Number of boys = 25 – 15 = 10
Ratio of number of boys to the number of girls: 15 : 10
Divide both sides by 5.
$$\frac{15 \div 5}{10 \div 5} = \frac{3}{2}$$
The simplest form of the ratio 15 : 10 is 3 : 2.
Example 2: You have to mix $$\frac{1}{3}$$ cup of yellow paint for every $$\frac{1}{2}$$ cup of red paint to make 10 cups of orange paint. Find the required number of cups of yellow paint.
We can use a ratio table to find the number of cups of yellow paint required to get 10 cups of orange paint,
By looking at the ratio table, we can conclude that we need 4 cups of yellow paint to get a cup of orange paint.
Example 3: Find the missing values from the ratio table.
Solution: In this case, we can perform math operations to find the missing values of the ratio table.
Note that during multiplication or division, the same operation was done on both sides. But during addition or subtraction, the values used for the operation are different on both sides: they are equivalent ratios.
## Should the quantities in a ratio have the same unit?
The quantities or terms in a ratio should be of the same unit. If we want to compare two quantities of different units, we use rate instead of ratios.
## Are ratios and fractions the same?
Ratios and fractions are mathematically the same: 1 : 2 is the same as $$\frac{1}{2}$$. But in most cases, they carry a different meaning. For example, $$\frac{1}{2}$$ of a sandwich means half of a sandwich. But a ratio is a comparison of two different quantities, i.e. 1 : 2 represents the ratio of non-vegetarian sandwiches to vegetarian sandwiches, which means there are two vegetarian sandwiches for every non-vegetarian sandwich.
## Can we multiply or divide any values in the ratio table?
The values in the ratio table can be multiplied or divided by any values. We just need to make sure that we use the same value while multiplying or dividing both sides.
## Can we add or subtract any values in the ratio table?
While multiplication and division operations can be done with any values, this is not possible with addition and subtraction. In the case of addition and subtraction, the operation is done with equivalent ratios
Level 1 - 10
## Writing and Using Ratios
A ratio is a comparison of quantities. For example, for most mammals, the ratio of legs to noses is $$4:1,$$ but for humans, the ratio of legs to noses is $$2:1.$$
## Writing Ratios
Using ratios to problem solve.
Ratios can be written using the word "to," a colon, or a fraction.
For example, in a group of 3 girls and 5 boys, the ratio of girls to boys can be written as $$3 \text{ to } 5,$$ $$3:5,$$ or $$\dfrac{3}{5}.$$
A bowl contains 3 apples, 5 oranges, and 9 kiwi. What is the ratio of kiwi to apples? ANSWER There are 9 kiwi and 3 apples, so the ratio of kiwi to apples is $$9:3,$$ or in simplified form, $$3:1.$$
A bowl contains 3 apples, 5 oranges, and 9 kiwi. What is the ratio of oranges to total pieces of fruit? ANSWER There are 5 oranges and $$3+5+9=17$$ total pieces of fruit, so the ratio of oranges to total pieces of fruit is $$5:17.$$
Marshall has 18 bananas. Joey comes up to Marshall and says that he has 30 apples. What is the ratio of apples to bananas?
In a parking lot, the ratio of white cars to blue cars is 2:3, and the ratio of blue cars to silver cars is 2:5. What is the least number of cars in the parking lot?
Assume that there is at least 1 car in the parking lot.
Ratios are often easiest to use if written in fraction form. To find an unknown value in a ratio, we can use two equivalent ratios.
For example, if the ratio $$3:4$$ is equivalent to the ratio $$18:x,$$ then $$\frac{3}{4}=\frac{18}{x}$$ and $$x=24$$ because the values in the fraction on the right are six times greater than the corresponding values in the fraction on the left.
If $$6: 15 = 10 : x,$$ what is $$x\,?$$ ANSWER Expressing the ratios as fractions, we get $$\frac{6}{15} = \frac{10}{x} .$$ The fraction $$\frac{6}{15}$$ simplifies to $$\frac{2}{5},$$ so now we have $$\frac{2}{5} = \frac{10}{x} .$$ The values in the ratio on the right are five times greater than the values in the ratio on the left, so $$x=25.$$ We can also rewrite the ratios as fractions and cross multiply to solve. \begin{align} \frac{6}{15} &= \frac{10}{ x} \\ 6x &= (10)(15) \\ 6x &= 150 \\ x &= 25.\end{align}
If Calvin paid 5 for 7 pencils, how much would he pay for 56 pencils? ANSWER Let $$x$$ be the price of $$56$$ pencils. Since the price of a single pencil does not change, we have \begin{align} \frac{\5}{7} &= \frac{x}{56} \\ 7x &= \5 \times 56 \\ x &= \40. \end{align} Hence, Calvin would pay $$\40$$ for $$56$$ pencils. The ratio of Alice's pay to Bob's pay is $$\frac{5}{4}$$. The ratio of Bob's pay to Charlie's pay is $$10:9$$. If Alice is paid75, how much is Charlie paid? ANSWER Since the ratio of Alice's pay to Bob's pay is $$5:4$$, Bob's pay must be $$b$$, where $$\frac{5}{4}=\frac{75}{b}$$. Cross-multiplying by the denominators, we get $$5b = 4(75)$$, so $$b = 60$$. Continuing in the same way, we compare Bob to Charlie: $\frac{10}{9}=\frac{60}{c} \implies 10c = 9(60) \implies c = 54.$ Thus, Charlie is paid $54. $$_\square$$ The ratio of boys to girls in Mr. John's math class is 2 : 3 . If there are 4030 students in the class, how many more girls than boys are in the class? Ten years ago, the ratio of the ages of two brothers was 1:4. Fourteen years from now, the ratio will be 4:7. What is the ratio of their present ages? Problem Loading... Note Loading... Set Loading... • How to Read and Write Ratios Are you ready to embark on a mathematical journey? If so, let's explore together the world of ratios. This concept, while seemingly simple, forms the basis of many mathematical calculations and real-world applications. • What is a Ratio? Let's start at the very beginning. In the simplest terms, a ratio is a way of comparing quantities. It tells us how much of one thing there is compared to another thing. Ratios can be written in different ways, but they all serve the same purpose of comparison. ## How to Write Ratios? There are three main ways to write ratios. You can write them using the word "to", using a colon (:), or as a fraction. For example, if there are four apples and three oranges, we can express the ratio of apples to oranges as "4 to 3", "4:3", or "4/3". All three methods represent the same ratio and can be used interchangeably. ## How to Read Ratios? When reading ratios, the order is very important. The number before the "to" or the colon is always compared to the number after. So, in the ratio 4:3, we say "four is to three". This means there are four parts of the first quantity for every three parts of the second quantity. ## The Concept of Equivalent Ratios Equivalent ratios are ratios that express the same relationship between numbers. They are essentially the same fraction but multiplied or divided by the same number. For example, the ratios 2:1, 4:2, 6:3 are all equivalent because they all represent the same relationship "two is to one". ## Famous Mathematicians and Ratios Many famous mathematicians have made significant contributions to the understanding and development of ratios. For instance, Pythagoras, a Greek mathematician, discovered a profound relationship between ratios and musical harmony. His findings laid the groundwork for our understanding of music theory today. Similarly, Euclid, another ancient Greek mathematician, wrote extensively about ratios in his work "Elements", one of the most influential works in the history of mathematics. He used ratios to develop the concept of proportion, which has wide-ranging applications in fields such as geometry, physics, and engineering. ## Important Considerations When Working with Ratios When working with ratios, it's important to remember that they are a form of comparison, not an absolute measure. A ratio tells us about the relationship between quantities, not their absolute values. For example, a ratio of 2:1 tells us that there are twice as many of one quantity as there are of another, but it doesn't tell us the exact amounts. Another important point is that ratios must always be simplified. Similar to fractions, ratios should be expressed in their simplest form for clarity and ease of understanding. For example, the ratio 8:4 should be simplified to 2:1. ## Practical Applications of Ratios Understanding ratios can be incredibly useful in a variety of practical situations. From cooking recipes to financial analysis, from art and design to health and fitness, ratios help us quantify relationships and make informed decisions. Hopefully, this tutorial has shed some light on how to read and write ratios. This fundamental mathematical concept, introduced by pioneering mathematicians like Pythagoras and Euclid, plays a crucial role in many areas of study and everyday life. Reading and writing ratios correctly is a skill that can be easily mastered with practice. Remember, ratios are simply a way of comparing quantities and should be expressed in their simplest form. As we move forward in our journey of understanding ratios, always remember their origin and importance. The concept of ratios has been at the heart of many great mathematical discoveries and continues to be a powerful tool for understanding the world around us. ## Introduction to Ratios Tutorials If you found this ratio information useful then you will likely enjoy the other ratio lessons and tutorials in this section: • Examples of Ratios in Everyday Life • Identify the Ratios: Exercise Let's use this illustration of shapes to learn more about ratios. How can we write the ratio of squares to circles, or 3 to 6? The most common way to write a ratio is as a fraction, 3/6. We could also write it using the word "to," as "3 to 6." Finally, we could write this ratio using a colon between the two numbers, 3:6. Be sure you understand that these are all ways to write the same number. Which way you choose will depend on the problem or the situation. • ratio of squares to circles is 3/6 • ratio of squares to circles is 3 to 6 • ratio of squares to circles is 3:6 There are still other ways to make the same comparison, by using equal ratios. To find an equal ratio, you can either multiply or divide each term in the ratio by the same number (but not zero). For example, if we divide both terms in the ratio 3:6 by the number three, then we get the equal ratio, 1:2. Do you see that these ratios both represent the same comparison? Some other equal ratios are listed below. To find out if two ratios are equal, you can divide the first number by the second for each ratio. If the quotients are equal, then the ratios are equal. Is the ratio 3:12 equal to the ratio 36:72? Divide both, and you discover that the quotients are not equal. Therefore, these two ratios are not equal. Some other equal ratios: 3:6 = 12:24 = 6:12 = 15:30 Are 3:12 and 36:72 equal ratios? Find 3÷12 = 0.25 and 36÷72 = 0.5 The quotients are not equal —> the ratios are not equal. You can also use decimals and percents to compare two quantities. In our example of squares to circles, we could say that the number of squares is "five-tenths" of the number of circles, or 50%. Here is a chart showing the number of goals made by five basketball players from the free-throw line, out of 100 shots taken. Each comparison of goals made to shots taken is expressed as a ratio, a decimal, and a percent. They are all equivalent, which means they are all different ways of saying the same thing. Which do you prefer to use? ## What Is a Ratio? Definition and Examples How to use ratios in math Larry Washburn / Getty Images • Math Tutorials • Pre Algebra & Algebra • Exponential Decay • Worksheets By Grade • B.B.A., Finance and Economics, University of Oklahoma Ratios are a helpful tool for comparing things to each other in mathematics and real life, so it is important to know what they mean and how to use them. These descriptions and examples will not only help you to understand ratios and how they function but will also make calculating them manageable no matter what the application. ## What Is a Ratio? In mathematics, a ratio is a comparison of two or more numbers that indicates their sizes in relation to each other. A ratio compares two quantities by division, with the dividend or number being divided termed the antecedent and the divisor or number that is dividing termed the consequent . Example: you have polled a group of 20 people and found that 13 of them prefer cake to ice cream and 7 of them prefer ice cream to cake. The ratio to represent this data set would be 13:7, with 13 being the antecedent and 7 the consequent. A ratio might be formatted as a Part to Part or Part to Whole comparison. A Part to Part comparison looks at two individual quantities within a ratio of greater than two numbers, such as the number of dogs to the number of cats in a poll of pet type in an animal clinic. A Part to Whole comparison measures the number of one quantity against the total, such as the number of dogs to the total number of pets in the clinic. Ratios like these are much more common than you might think. ## Ratios in Daily Life Ratios occur frequently in daily life and help to simplify many of our interactions by putting numbers into perspective. Ratios allow us to measure and express quantities by making them easier to understand. Examples of ratios in life: • The car was traveling 60 miles per hour, or 60 miles in 1 hour. • You have a 1 in 28,000,000 chance of winning the lottery. Out of every possible scenario, only 1 out of 28,000,000 of them has you winning the lottery. • There were enough cookies for every student to have two, or 2 cookies per 78 students. • The children outnumbered the adults 3:1, or there were three times as many children as there were adults. ## How to Write a Ratio There are several different ways to express a ratio. One of the most common is to write a ratio using a colon as a this-to-that comparison such as the children-to-adults example above. Because ratios are simple division problems, they can also be written as a fraction . Some people prefer to express ratios using only words, as in the cookies example. In the context of mathematics, the colon and fraction format are preferred. When comparing more than two quantities, opt for the colon format. For example, if you are preparing a mixture that calls for 1 part oil, 1 part vinegar, and 10 parts water, you could express the ratio of oil to vinegar to water as 1:1:10. Consider the context of the comparison when deciding how best to write your ratio. ## Simplifying Ratios No matter how a ratio is written, it is important that it be simplified down to the smallest whole numbers possible, just as with any fraction. This can be done by finding the greatest common factor between the numbers and dividing them accordingly. With a ratio comparing 12 to 16, for example, you see that both 12 and 16 can be divided by 4. This simplifies your ratio into 3 to 4, or the quotients you get when you divide 12 and 16 by 4. Your ratio can now be written as: • 0.75 (a decimal is sometimes permissible, though less commonly used) ## Practice Calculating Ratios With Two Quantities Practice identifying real-life opportunities for expressing ratios by finding quantities you want to compare. You can then try calculating these ratios and simplifying them into their smallest whole numbers. Below are a few examples of authentic ratios to practice calculating. • What is the ratio of apples to the total amount of fruit? (answer: 6:8, simplified to 3:4) • If the two pieces of fruit that are not apples are oranges, what is the ratio of apples to oranges? (answer: 6:2, simplified to 3:1) • What is the ratio of cows to horses that she treated? (answer: 12:16, simplified to 3:4. For every 3 cows treated, 4 horses were treated) • What is the ratio of cows to the total number of animals that she treated? (answer: 12 + 16 = 28, the total number of animals treated. The ratio for cows to total is 12:28, simplified to 3:7. For every 7 animals treated, 3 of them were cows) ## Practice Calculating Ratios With Greater Than Two Quantities Use the following demographic information about a marching band to complete the following exercises using ratios comparing two or more quantities. Instrument type • 160 woodwinds • 84 percussion • 127 freshmen • 63 sophomores 1. What is the ratio of boys to girls? (answer: 2:3) 2. What is the ratio of freshmen to the total number of band members? (answer: 127:300) 3. What is the ratio of percussion to woodwinds to brass? (answer: 84:160:56, simplified to 21:40:14) 4. What is the ratio of freshmen to seniors to sophomores? (answer: 127:55:63. Note: 127 is a prime number and cannot be reduced in this ratio) 5. If 25 students left the woodwind section to join the percussion section, what would be the ratio for the number of woodwind players to percussion? (answer: 160 woodwinds – 25 woodwinds = 135 woodwinds; 84 percussionists + 25 percussionists = 109 percussionists. The ratio of the number of players in woodwinds to percussion is 109:135) • Sixth-Grade Lesson Plan: Ratios • How to Calculate a Male to Female Ratio and Other Quantities • Empirical Formula: Definition and Examples • Mole Ratio: Definition and Examples • Math Glossary: Mathematics Terms and Definitions • Learn About Molecular and Empirical Formulas • Definition and Examples of Colons • Learn to Calculate Percent Change • How Are Odds Related to Probability? • Probabilities for Rolling Three Dice • Example of an ANOVA Calculation • Exponents and Bases • Computations With Fractions • What Is the Median? • Histogram Classes • Calculate Simplest Formula From Percent Composition Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Unit 1: Ratio, Rate, & Proportion ## Topic A: Writing Ratios Introduction to ratios. Ratio is a comparison of one number or quantity with another number or quantity. Ratio shows the relationship between the quantities. Ratio is pronounced “rā’shō” or it can be pronounced “rā’shēō.” Check out this YouTube video to listen to someone pronouce the word: How to Pronounce Ratio . You often use ratios, look at these examples: • Scores in games are ratios. For example, the Penguins won 4 to 3 or the Canucks lost 1 to 5. • Directions for mixing can be ratios. For example, use 1 egg to each cup of milk or mix 25 parts gas to 1 part oil for the motorcycle. • Betting odds are given as ratios. For example, Black Jade is a 3 to 1 favourite or the heavyweight contender is only given a 2 to 5 chance to win. • The scale at the bottom of maps is a ratio. For example, 1 centimetre represents 10 kilometres. • Prices are often given as ratios. For example, 100 grams for$0.79 or 2 cans for $1.85. For ratios to have meaning you must know what is being compared and the units that are being used. Read these examples of ratios and the units that are used. A general ratio may say “parts” for the units. • It rained four days and was sunny for three days last week. The ratio of rainy days to sunny days was $4:3$. ($4:3$ is properly read “4 is compared to 3” but is often read “4 to 3”). • The class has 12 men and 15 women registered. The ratio of men to women in the class is $12:15$. • At the barbeque, 36 hot dogs and 18 hamburgers were eaten. The ratio of hot dogs eaten to hamburgers eaten is $36:18$. • The class spends 3 hours on English and 2 hours on math each day. The ratio of time spent on English compared to math is $3:2$. Write the ratios asked for in these questions using the $:$ symbol (for example, $4:1$). Write the units and what is being compared beside the ratio. • Powdered milk is mixed using 1 part of milk powder to 3 parts of water. Write a ratio to compare the milk powder to the water. Answer: $1:3$ — 1 part of milk powder to 3 parts of water • One kilogram of ground beef will make enough hamburger for 5 people. Write a ratio to express the amount of ground beef for hamburgers to the number of people. • Seventy-five vehicles were checked by the police. Fifteen vehicles did not meet the safety standards, but 60 of them did. Write a ratio comparing the unsafe vehicles to the safe vehicles. • The recipe says to roast a turkey according to its weight. For every kilogram, allow 40 minutes of cooking. Write a ratio comparing time to weight. • The 4-litre pail of semi-transparent oil stain should cover 24 square metres of the house siding if the wood is smooth. Write the ratio comparing quantity of stain to the smooth wood surface area. • The same 4 L of stain will only cover 16 square metres of the house siding if the wood is rough. Write that ratio. Answers to Exercise 1 • $1:5$ 1 kg of beef to 5 people • $15:60$ 15 unsafe vehicles to 60 safe vehicles • $40:1$ 40 minutes to 1 kg of turkey • $4:24$ 4 L of stain to 24 m 2 of smooth wood • $4:16$ 4 L of stain to 16 m 2 of rough wood The numbers that you have been using to write the ratios are called the terms of the ratio. The order that you use to write the terms is very important. Read a ratio from left to right and the order must match what the numbers mean. For example, 3 scoops of coffee to 12 cups of water must be written $3:12$ as a ratio because you are comparing the quantity of coffee to the amount of water. If you wish to talk about the amount of water compared to the coffee you have, you would say, “Use 12 cups of water for every 3 scoops of coffee” and the ratio would be written $12:3$. Ratios can be written 3 different ways: • Using the $:$ symbol — $2:5$ • The first number in the ratio is the numerator; the second number is the denominator. • Ratios written as a common fraction are read as a ratio, not as a fraction. Say “2 to 5,” not “two-fifths.” • Using the word “to” — 2 to 5 Use the ratios you wrote in Exercise 1 to complete the chart. Answers to Exercise 2 • $1:5$ or $\frac{1}{5}$ or 1 to 5 • $15:60$ or $\frac{15}{60}$ or 15 to 60 • $40:1$ or $\frac{40}{1}$ or 40 to 1 • $4:24$ or $\frac{4}{24}$ or 4 to 24 • $4:16$ or $\frac{4}{16}$ or 4 to 16 Answers to Exercise 3 • $9:6$ or $\frac{9}{6}$ or 9 to 6 • $7:9$ or $\frac{7}{9}$ or 7 to 9 • $6:9$ or $\frac{6}{9}$ or 6 to 9 ## Equivalent Ratios Like equivalent fractions, equivalent ratios are equal in value to each other. $10:100 = 1:10$ Ratios can be written as common fractions. It is convenient to work with ratios in the common fraction form. You can then easily: • Find equivalent ratios in higher terms • Find equivalent ratios in lower terms • Find a missing term Express $4:5$ in higher terms. $4:5=\dfrac{4}{5}\longrightarrow \dfrac{4}{5} \times \left(\dfrac{2}{2}\right) \longrightarrow\left(\dfrac{4\times 2}{5\times 2}\right)\longrightarrow\dfrac{8}{10}$ $4:5$ is equivalent to $8:10$ Express $3:6$ in lower terms. $3:6=\dfrac{3}{6}\longrightarrow \dfrac{3}{6} \div \left(\dfrac{3}{3}\right) \longrightarrow \left(\dfrac{3\div 3}{6\div 3}\right)\longrightarrow\dfrac{1}{2}$ $3:6$ is equivalent to $1:2$ To find equivalent ratios in higher terms, multiply each term of the ratio by the same number. To find equivalent ratios in lower terms, divide each term of the ratio by the same number. Write equivalent ratios in any higher term. You may want to write the ratio as a common fraction first. Ask your instructor to mark this exercise. • $5:6= \dfrac{5}{6} \times \left(\dfrac{3}{3}\right)=\left(\dfrac{5\times 3}{6\times 3}\right)=\dfrac{15}{18}=15:18$ • $4:3$ • $10:2$ • $50:1$ • $9:4$ • $3:5$ Answers to Exercise 4 See your instructor. Write these ratios in lowest terms—that is, simplify the ratios. • $4:12=\dfrac{4}{12}\div\left(\dfrac{4}{4}\right)=\left(\dfrac{4\div4}{12\div4}\right)=\dfrac{1}{3}=1:3$ • $10:5$ • $7:21$ • $20:5$ • $6:14$ • $2:4$ • $6:3$ • $16:8$ Answers to Exercise 5 Ratios written as a common fraction or using the word “to” will also be correct in this exercise. The terms must be the same. • $1:3$ • $2:1$ • $4:1$ • $3:7$ • $1:2$ Using a colon, write a ratio in lowest terms for the information given. • In the class of 25 students, only 5 are smokers. Write the ratio of smokers to non-smokers in the class. ( Note —you must first calculate the number of non-smokers.) • The police issued 12 roadside suspensions to drivers out of the 144 who were checked in the road block last Friday. Write the ratio of suspended drivers to the number checked. • Twenty-seven students registered for the course and 24 completed it. Write a ratio showing number of completions compared to number enrolled. • During an hour (60 minutes) of television viewing last night there were 14 minutes of commercials, so there were only 46 minutes of the actual program! Write the ratio of commercial time to program time. • A nickel to a dime $5:10=1:2$ • A nickel to a quarter • A nickel to a dollar • A dime to a nickel • A dime to a quarter • A dime to a dollar • A dollar to a dime Answers to Exercise 6 • $1:4$ • $1:12$ • $8:9$ • $1:5$ • $1:20$ • $2:5$ • $1:10$ • $10:1$ ## Topic A: Self-Test Mark /12 Aim 10/12 • Terms of the ratio • Equivalent ratios • The campground had three vacant campsites and 47 occupied sites. Write the ratio of occupied sites to vacant sites. Ratio: Read: • For every ten dogs in the city, only 2 have current dog licences. Write the ratio of licensed dogs to unlicensed dogs. (Find the number of unlicensed dogs first). Ratio: Read: • $9:12$ • $6:4$ • $500:1000$ • $2:9$ • $35:15$ ## Answers to Topic A Self-Test • A ratio is a comparison of one number or quantity with another number or quantity. Ratios show the relationship between the quantities or amounts. • Terms of a ratio are the numbers used in the ratio, the parts of the ratio. • Equivalent ratios are ratios of equal value to each other. • $47:3$ Read: ““47 occupied sites to 3 vacant sites.” • $1:4$ Read: “1 licensed dog to 4 unlicensed dogs.” • $3:4$ • $3:2$ • $7:3$ Adult Literacy Fundamental Mathematics: Book 6 - 2nd Edition Copyright © 2022 by Liz Girard and Wendy Tagami is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted. ## Share This Book Simplifying Ratios and Rates Learning Objective(s) · Write ratios and rates as fractions in simplest form. · Find unit rates. · Find unit prices. Introduction Ratios are used to compare amounts or quantities or describe a relationship between two amounts or quantities. For example, a ratio might be used to describe the cost of a month’s rent as compared to the income earned in one month. You may also use a ratio to compare the number of elephants to the total number of animals in a zoo, or the amount of calories per serving in two different brands of ice cream. Rates are a special type of ratio used to describe a relationship between different units of measure, such as speed, wages, or prices. A car can be described as traveling 60 miles per hour; a landscaper might earn$35 per lawn mowed; gas may be sold at $3 per gallon. Ratios compare quantities using division. This means that you can set up a ratio between two quantities as a division expression between those same two quantities. Here is an example. If you have a platter containing 10 sugar cookies and 20 chocolate chip cookies, you can compare the cookies using a ratio. The ratio of sugar cookies to chocolate chip cookies is: The ratio of chocolate chip cookies to sugar cookies is: You can write the ratio using words, a fraction, and also using a colon as shown below. Some people think about this ratio as: “For every 10 sugar cookies I have, I have 20 chocolate chip cookies.” You can also simplify the ratio just as you simplify a fraction. So we can also say that: Below are two more examples that illustrate how to compare quantities using a ratio, and how to express the ratio in simplified form. Often, one quantity in the ratio is greater than the second quantity. You do not have to write the ratio so that the lesser quantity comes first; the important thing is to keep the relationship consistent. Ratios can compare a part to a part or a part to a whole . Consider the example below that describes guests at a party. A rate is a ratio that compares two different quantities that have different units of measure. A rate is a comparison that provides information such as dollars per hour, feet per second, miles per hour, and dollars per quart, for example. The word “per” usually indicates you are dealing with a rate. Rates can be written using words, using a colon, or as a fraction. It is important that you know which quantities are being compared. For example, an employer wants to rent 6 buses to transport a group of 300 people on a company outing. The rate to describe the relationship can be written using words, using a colon, or as a fraction; and you must include the units. six buses per 300 people 6 buses : 300 people As with ratios, this rate can be expressed in simplest form by simplifying the fraction. This fraction means that the rate of buses to people is 6 to 300 or, simplified, 1 bus for every 50 people. Finding Unit Rates A unit rate compares a quantity to one unit of measure. You often see the speed at which an object is traveling in terms of its unit rate. For example, if you wanted to describe the speed of a boy riding his bike—and you had the measurement of the distance he traveled in miles in 2 hours—you would most likely express the speed by describing the distance traveled in one hour. This is a unit rate; it gives the distance traveled per one hour. The denominator of a unit rate will always be one. Consider the example of a car that travels 300 miles in 5 hours. To find the unit rate, you find the number of miles traveled in one hour. A common way to write this unit rate is 60 miles per hour. Finding Unit Prices A unit price is a unit rate that expresses the price of something. The unit price always describes the price of one unit, so that you can easily compare prices. You may have noticed that grocery shelves are marked with the unit price (as well as the total price) of each product. This unit price makes it easy for shoppers to compare the prices of competing brands and different package sizes. Consider the two containers of blueberries shown below. It might be difficult to decide which is the better buy just by looking at the prices; the container on the left is cheaper, but you also get fewer blueberries. A better indicator of value is the price per single ounce of blueberries for each container. Look at the unit prices—the container on the right is actually a better deal, since the price per ounce is lower than the unit price of the container on the left. You pay more money for the larger container of blueberries, but you also get more blueberries than you would with the smaller container. Put simply, the container on the right is a better value than the container on the left. So, how do you find the unit price? Imagine a shopper wanted to use unit prices to compare a 3-pack of tissue for$4.98 to a single box of tissue priced at $1.60. Which is the better deal? Since the price given is for 3 boxes, divide both the numerator and the denominator by 3 to get the price of 1 box, the unit price. The unit price is$1.66 per box.
The unit price of the 3-pack is $1.66 per box; compare this to the price of a single box at$1.60. Surprisingly, the 3-pack has a higher unit price! Purchasing the single box is the better value.
Like rates, unit prices are often described with the word “per.” Sometimes, a slanted line / is used to mean “per.” The price of the tissue might be written $1.60/box, which is read “$1.60 per box.”
The following example shows how to use unit price to compare two products and determine which has the lower price.
Ratios and rates are used to compare quantities and express relationships between quantities measured in the same units of measure and in different units of measure. They both can be written as a fraction, using a colon, or using the words “to” or “per”. Since rates compare two quantities measured in different units of measurement, such as dollars per hour or sick days per year, they must include their units. A unit rate or unit price is a rate that describes the rate or price for one unit of measure.
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## 4.1.1: Simplifying Ratios and Rates
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## Learning Objectives
• Write ratios and rates as fractions in simplest form.
• Find unit rates.
• Find unit prices.
## Introduction
Ratios are used to compare amounts or quantities or describe a relationship between two amounts or quantities. For example, a ratio might be used to describe the cost of a month’s rent as compared to the income earned in one month. You may also use a ratio to compare the number of elephants to the total number of animals in a zoo, or the amount of calories per serving in two different brands of ice cream.
Rates are a special type of ratio used to describe a relationship between different units of measure, such as speed, wages, or prices. A car can be described as traveling 60 miles per hour; a landscaper might earn $35 per lawn mowed; gas may be sold at$3 per gallon.
Ratios compare quantities using division. This means that you can set up a ratio between two quantities as a division expression between those same two quantities.
Here is an example. If you have a platter containing 10 sugar cookies and 20 chocolate chip cookies, you can compare the cookies using a ratio.
$$\ \frac{\text { sugar cookies }}{\text { chocolate chip cookies }}=\frac{10}{20}$$
$$\ \frac{\text { chocolate chip cookies }}{\text { sugar cookies }}=\frac{20}{10}$$
You can write the ratio using words, a fraction, and also using a colon as shown below.
$$\ \begin{array}{c} \text { ratio of } {\color{red}\text{sugar cookies to}}\\ \color{blue}\text{chocolate chip cookies}\\ {\color{red} 10} \text{ to } \color{blue}20\\ \frac{\color{red}10}{\color{blue}20}\\ {\color{red}10}:\color{blue}20 \end{array}$$
You can also simplify the ratio just as you simplify a fraction.
$$\ \frac{10}{20}=\frac{10 \div 10}{20 \div 10}=\frac{1}{2}$$
So we can also say that:
$$\ \begin{array}{c} \text { ratio of } {\color{red}\text{sugar cookies to}}\\ \color{blue}\text{chocolate chip cookies}\\ {\color{red} 1} \text{ to } \color{blue}2\\ \frac{\color{red}1}{\color{blue}2}\\ {\color{red}1}:\color{blue}2 \end{array}$$
## How to Write a Ratio
A ratio can be written in three different ways:
• with the word “to”: 3 to 4
• as a fraction: $$\ \frac{3}{4}$$
• with a colon: 3:4
A ratio is simplified if it is equivalent to a fraction that has been simplified.
Below are two more examples that illustrate how to compare quantities using a ratio, and how to express the ratio in simplified form.
A basketball player takes 50 jump shots during a practice. She makes 28 of them. What is the ratio of shots made to shots taken? Simplify the ratio.
The ratio of shots made to shots taken is $$\ \frac{14}{25}$$, 14:15, or 14 to 15.
Often, one quantity in the ratio is greater than the second quantity. You do not have to write the ratio so that the lesser quantity comes first; the important thing is to keep the relationship consistent.
Paul is comparing the amount of calories in a large order of French fries from his two favorite fast food restaurants. Fast Foodz advertises that an order of fries has 450 calories, and Beef Stop states that its fries have 300 calories. Write a ratio that represents the amount of calories in the Fast Foodz fries compared to the calories in Beef Stop fries.
The ratio of calories in Fast Foodz fries to Beef Stop fries is \frac{3}{2}, 3:2, or 3 to 2.
Ratios can compare a part to a part or a part to a whole . Consider the example below that describes guests at a party.
Luisa invites a group of friends to a party. Including Luisa, there are a total of 22 people, 10 of whom are women.
Which is greater: the ratio of women to men at the party, or the ratio of women to the total number of people present?
The ratio of women to men at the party, $$\ \frac{5}{6}$$, is greater than the ratio of women to the total number of people, $$\ \frac{5}{11}$$.
A poll at Forrester University found that 4,000 out of 6,000 students are unmarried. Find the ratio of unmarried to married students. Express as a simplified ratio.
• Incorrect. The ratio 3 to 2 compares the total number of students to the number of unmarried students. The correct answer is 2 to 1.
• Incorrect. The ratio 1 to 3 compares the number of married students to the total number of students. The correct answer is 2 to 1.
• Correct. If 4,000 students out of 6,000 are unmarried, then 2,000 must be married. The ratio of unmarried to married students can be represented as 4,000 to 2,000, or simply 2 to 1.
• Incorrect. The ratio 2 to 3 compares the number of unmarried students to the total number of students. The correct answer is 2 to 1.
A rate is a ratio that compares two different quantities that have different units of measure. A rate is a comparison that provides information such as dollars per hour, feet per second, miles per hour, and dollars per quart, for example. The word “per” usually indicates you are dealing with a rate. Rates can be written using words, using a colon, or as a fraction. It is important that you know which quantities are being compared.
For example, an employer wants to rent 6 buses to transport a group of 300 people on a company outing. The rate to describe the relationship can be written using words, using a colon, or as a fraction; and you must include the units.
$$\ \begin{array}{c} \text{six buses per }300 \text{ people}\\ 6 \text{ buses : }300 \text{ people}\\ \frac{6 \text{ buses}}{300 \text{ people}} \end{array}$$
As with ratios, this rate can be expressed in simplest form by simplifying the fraction.
$$\ \frac{6 \text { buses}{\div6 }}{300 \text { people}{\div6 }}=\frac{1 \text { bus }}{50 \text { people }}$$
This fraction means that the rate of buses to people is 6 to 300 or, simplified, 1 bus for every 50 people.
Write the rate as a simplified fraction: 8 phone lines for 36 employees.
The rate of phone lines for employees can be expressed as $$\ \frac{2 \text { phone lines }}{9 \text { employees }}$$.
Write the rate as a simplified fraction: 6 flight attendants for 200 passengers.
The rate of flight attendants to passengers is $$\ \frac{3 \text { flight attendants }}{100 \text { passengers }}$$.
Anyla rides her bike 18 blocks in 20 minutes. Express her rate as a simplified fraction.
• $$\ 18: 20$$
• $$\ \begin{array}{cc} 9 & \text { blocks } \\ \hline 10 & \text { minutes } \end{array}$$
• $$\ \frac{9 \text { minutes }}{10 \text { blocks }}$$
• $$\ \frac{18 \text { blocks }}{20 \text { minutes }}$$
Incorrect. Anyla’s trip compares quantities with different units, so it can be described as a rate. Since rates compare two quantities measured in different units of measurement, they must include their units. The correct answer is \ \begin{aligned} 9 & \text { blocks } \\ \hline 10 & \text { minutes } \end{aligned}.
Correct. Anyla’s trip compares quantities with different units (blocks and minutes), so it is a rate and can be written $$\ \begin{array}{cc} 18 & \text { blocks } \\ \hline 20 & \text { minutes } \end{array}$$. This fraction can be simplified by dividing both the numerator and the denominator by 2.
Incorrect. 18 blocks in 20 minutes is not equivalent to 10 blocks in 9 minutes. Check the units again in your answer. The correct answer is \ \begin{aligned} 9 & \text { blocks } \\ \hline 10 & \text { minutes } \end{aligned}.
Incorrect. Anyla’s trip compares quantities with different units, so it can be described as a rate. This is a correct representation and includes the units, but the fraction can be simplified. The correct answer is \ \begin{aligned} 9 & \text { blocks } \\ \hline 10 & \text { minutes } \end{aligned}.
## Finding Unit Rates
A unit rate compares a quantity to one unit of measure. You often see the speed at which an object is traveling in terms of its unit rate.
For example, if you wanted to describe the speed of a boy riding his bike—and you had the measurement of the distance he traveled in miles in 2 hours—you would most likely express the speed by describing the distance traveled in one hour. This is a unit rate; it gives the distance traveled per one hour. The denominator of a unit rate will always be one.
Consider the example of a car that travels 300 miles in 5 hours. To find the unit rate, you find the number of miles traveled in one hour.
$$\ \frac{300 \text { miles}{\div5 }}{5 \text { hours}{\div5 }}=\frac{60 \text { miles }}{1 \text { hour }}$$
A common way to write this unit rate is 60 miles per hour.
A crowded subway train has 375 passengers distributed evenly among 5 cars. What is the unit rate of passengers per subway car?
The unit rate of the subway car is 75 riders per subway car.
## Finding Unit Prices
A unit price is a unit rate that expresses the price of something. The unit price always describes the price of one unit, so that you can easily compare prices.
You may have noticed that grocery shelves are marked with the unit price (as well as the total price) of each product. This unit price makes it easy for shoppers to compare the prices of competing brands and different package sizes.
Consider the two containers of blueberries shown below. It might be difficult to decide which is the better buy just by looking at the prices; the container on the left is cheaper, but you also get fewer blueberries. A better indicator of value is the price per single ounce of blueberries for each container.
Look at the unit prices—the container on the right is actually a better deal, since the price per ounce is lower than the unit price of the container on the left. You pay more money for the larger container of blueberries, but you also get more blueberries than you would with the smaller container. Put simply, the container on the right is a better value than the container on the left.
So, how do you find the unit price?
Imagine a shopper wanted to use unit prices to compare a 3-pack of tissue for $4.98 to a single box of tissue priced at$1.60. Which is the better deal?
Find the unit price of the 3-pack: $$\ \frac{\ 4.98}{3 \text { boxes }}$$
Since the price given is for 3 boxes, divide both the numerator and the denominator by 3 to get the price of 1 box, the unit price. The unit price is $1.66 per box. The unit price of the 3-pack is$1.66 per box; compare this to the price of a single box at $1.60. Surprisingly, the 3-pack has a higher unit price! Purchasing the single box is the better value. Like rates, unit prices are often described with the word “per.” Sometimes, a slanted line / is used to mean “per.” The price of the tissue might be written$1.60/box, which is read "$1.60 per box." 3 pounds of sirloin tips cost$21. What is the unit price per pound?
The unit price of the sirloin tips is $7.00/pound. The following example shows how to use unit price to compare two products and determine which has the lower price. Sami is trying to decide between two brands of crackers. Which brand has the lower unit price? Brand A:$1.12 for 8 ounces
Brand B: $1.56 for 12 ounces The unit price of Brand A crackers is 14 cents/ounce and the unit price of Brand B is 13 cents/ounce. Brand B has a lower unit price and represents the better value. A shopper is comparing two packages of rice at the grocery store. A 10-pound package costs$9.89 and a 2-pound package costs $1.90. Which package has the lower unit price to the nearest cent? What is its unit price? • The 2-pound bag has a lower unit price of$.95/pound.
• The 10-pound bag has a lower unit price of $0.99/pound. • The 10-pound bag has a lower unit price of$.95/pound.
• The 2-pound bag has a lower price of $1.89/2pound. • Correct. The unit price per pound for the 2-pound bag is $$\ \ 1.90 \div 2=\ 0.95$$. The unit price per pound for the 10-pound bag is $$\ \ 9.89 \div 10=\ 0.989$$, which rounds to$0.99.
• Incorrect. $$\ \ 9.89 \div 10=\ 0.989$$, which rounds to $0.99. $$\ \ 1.90 \div 2=\ 0.95$$. The 2-pound bag has a lower unit price. The correct answer is A. • Incorrect. $$\ \ 9.89 \div 10=\ 0.989$$, which rounds to$0.99. The correct answer is A.
• Incorrect. A unit price is the price for one unit; in this case, you need to find the cost of one pound, not two pounds. The correct answer is A.
Ratios and rates are used to compare quantities and express relationships between quantities measured in the same units of measure and in different units of measure. They both can be written as a fraction, using a colon, or using the words “to” or “per”. Since rates compare two quantities measured in different units of measurement, such as dollars per hour or sick days per year, they must include their units. A unit rate or unit price is a rate that describes the rate or price for one unit of measure.
#### IMAGES
1. A ratio can be expressed 3 different ways:
2. How To Write A Ratio
3. How To Write A Ratio
4. How to write a ratio three different ways MGSE6.RP.1 Understand the concept of a ratio
5. Simplifying Ratios
6. What Are The Three Ways of Writing Ratios? Step-by-step explanation
#### VIDEO
1. Define Ratio #shorts
2. What are ratio and how to write them
3. 🔵 Ratio Pronunciation
4. Ratios
5. How to write ratio in a funny way #shorts #noor_artz
6. ##maths shortcut,Ratio##yt shorts##viral##trending##
1. How To Write A Ratio
Example 1: writing a ratio about an everyday life situation. Ms. Holly is looking after 7 children. She has 5 toys in her nursery. Write the ratio of children to toys. Identify the different quantities being compared and their order. There are 7 children and 5 toys. The order of the ratio is children to toys. 2 Write the ratio using a colon ...
2. Ratios
Ratios can be shown in different ways: Use the ":" to separate the values: 3 : 1 : Or we can use the word "to": 3 to 1 : Or write it like a fraction: 31: A ratio can be scaled up: Here the ratio is also 3 blue squares to 1 yellow square, even though there are more squares.
3. 5.4 Ratios and Proportions
Constructing Ratios to Express Comparison of Two Quantities. Note there are three different ways to write a ratio, which is a comparison of two numbers that can be written as: a a to b b OR a: b a: b OR the fraction a / b a / b. Which method you use often depends upon the situation. For the most part, we will want to write our ratios using the ...
4. Ratio review (article)
Ratio review. Google Classroom. Learn how to find the ratio between two things given a diagram. A ratio compares two different quantities. For example, those two quantities could be monkeys and bananas: Notice that there are 4 monkeys and 5 bananas. Here are a few different ways we can describe the ratio of monkeys to bananas:
5. 1.3: Ratios
Consider the ratio $$6 \colon 11$$. Write $$6 \colon 11$$ in two other ways. Write three equivalent ratios. Is the ratio $$6\colon 11$$ in lowest terms? Explain why or why not in a complete sentence. In a given pond, there are 22 rock fish for every 121 minnows. Express the ratio of minnows to rock fish in lowest terms.
6. 1.1: Topic A- Writing Ratios
Ratios can be written 3 different ways: Using the :: symbol — 2: 5 2: 5. As a common fraction — 25 2 5. The first number in the ratio is the numerator; the second number is the denominator. Ratios written as a common fraction are read as a ratio, not as a fraction. Say "2 to 5," not "two-fifths.".
7. Writing and Interpreting Ratios
A ratio can be expressed in three ways: A ratio is denoted using the ':' symbol. A ratio is expressed by writing the ':' symbol in the middle of the two quantities that are being compared. In the ratio \ (a : b\), \ (a\) and \ (b\) are together known as the terms of the ratio. Also, \ (a\) is known as the antecedent or the first term ...
8. 5.5: Ratios and Proportions
Constructing Ratios to Express Comparison of Two Quantities. Note there are three different ways to write a ratio, which is a comparison of two numbers that can be written as: /**/(\$1 =0.82\,{€})/**/, how many dollars should you receive? Round to the nearest cent if necessary. 9. Writing and Using Ratios Writing Ratios. Ratios can be written using the word "to," a colon, or a fraction. For example, in a group of 3 girls and 5 boys, the ratio of girls to boys can be written as 3 \text { to } 5, 3 to 5, 3:5, 3: 5, or \dfrac {3} {5}. 53. A bowl contains 3 apples, 5 oranges, and 9 kiwi. What is the ratio of kiwi to apples? 10. Writing a Ratio A ratio is a way to show a relationship two numbers.Ratios can be used to compare things of the same type. For example, we may use a ratio to compare the nu... 11. How to Read and Write Ratios Let's start at the very beginning. In the simplest terms, a ratio is a way of comparing quantities. It tells us how much of one thing there is compared to another thing. Ratios can be written in different ways, but they all serve the same purpose of comparison. How to Write Ratios? There are three main ways to write ratios. 12. Ratios and proportions It compares the amount of one ingredient to the sum of all ingredients. part: whole = part: sum of all parts. To write a ratio: Determine whether the ratio is part to part or part to whole. Calculate the parts and the whole if needed. Plug values into the ratio. Simplify the ratio if needed. 13. Ratios and Proportions Ratios can be written in several different ways: as a fraction, using the word "to", or with a colon. ... Finally, we could write this ratio using a colon between the two numbers, 3:6. Be sure you understand that these are all ways to write the same number. Which way you choose will depend on the problem or the situation. ratio of squares to ... 14. Ratios As you can see, the parts consist of the number of girls and boys which sum up to the whole or the total number of students. $3$ girls + $1$ boy = $4$ students. With this setup, it is now easy to come up with various kinds of ratios. Examples: Find the required ratios in three different formats. 15. Intro to ratios (video) Intro to ratios. Google Classroom. About. Transcript. The video explains ratios, which show the relationship between two quantities. Using apples and oranges as an example, it demonstrates how to calculate and reduce ratios (6:9 to 2:3) and how to reverse the ratio (9:6 to 3:2). Created by Sal Khan. Questions. 16. What Is a Ratio? Definition and Examples There are several different ways to express a ratio. One of the most common is to write a ratio using a colon as a this-to-that comparison such as the children-to-adults example above. Because ratios are simple division problems, they can also be written as a fraction. Some people prefer to express ratios using only words, as in the cookies ... 17. Topic A: Writing Ratios Ratios can be written 3 different ways: Using the :: symbol — 2: 5 2: 5. As a common fraction — 2 5 2 5. The first number in the ratio is the numerator; the second number is the denominator. Ratios written as a common fraction are read as a ratio, not as a fraction. Say "2 to 5," not "two-fifths.". 18. How to Write Ratios Using Different Notations Write a ratio comparing the number of circles to triangles using a symbol. Step 1: Determine the number of both items asked in the problem. The problem asks for circles and triangles. There are 2 ... 19. Simplifying Ratios and Rates Which is greater: the ratio of women to men at the party, or the ratio of women to the total number of people present? Identify the first relationship. Write a ratio comparing women to men. Since there are 22 people and 10 are women, 12 must be men. Simplify the ratio. 10 and 12 have a common factor of 2; the ratio of women to men at the party is . 20. 9.1: Introducing Ratios and Ratio Language A ratio is an association between two or more quantities. There are many ways to describe a situation in terms of ratios. For example, look at this collection: Figure 9.1.3 9.1. 3. Here are some correct ways to describe the collection: The ratio of squares to circles is 6: 3 6: 3. 21. Comparing Ratios: Definition, Methods, Examples, FAQs The ratio of cookies to cupcakes in the given image can be expressed in three different ways: 6 to 7; 6 : 7$\frac{6}{7}\$ Ratio is the quantitative relationship between two quantities or numbers. In the ratio a : b, the first quantity is called an antecedent and the second quantity is called consequent. ... Write the given ratios and their ...
22. 4.1.1: Simplifying Ratios and Rates
Consider the two other ways to write a ratio. You'll want to express your answer in a particular format if required. The ratio of shots made to shots taken is $$\ \frac{14}{25}$$, 14:15, or 14 to 15. ... A rate is a ratio that compares two different quantities that have different units of measure. A rate is a comparison that provides ... |
# Common Core: 6th Grade Math : Find Area of Polygons: CCSS.Math.Content.6.G.A.1
## Example Questions
← Previous 1 3 4 5 6
### Example Question #1 : Geometry
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
### Example Question #2 : Geometry
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
### Example Question #1 : Find Area Of Polygons: Ccss.Math.Content.6.G.A.1
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
### Example Question #4 : Geometry
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
### Example Question #5 : Geometry
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
### Example Question #6 : Geometry
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
### Example Question #2 : Find Area Of Polygons: Ccss.Math.Content.6.G.A.1
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
### Example Question #3 : Find Area Of Polygons: Ccss.Math.Content.6.G.A.1
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
### Example Question #9 : Geometry
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
### Example Question #10 : Geometry
What is the area of the right triangle in the following figure?
Explanation:
There are several different ways to solve for the area of a right triangle. In this lesson, we will transform the right triangle into a rectangle, use the the simpler formula for area of a rectangle to solve for the new figure's area, and divide this area in half in order to solve for the area of the original figure.
First, let's transform the triangle into a rectangle:
Second, let's remember that the formula for area of a rectangle is as follows:
Substitute in our side lengths.
Last, notice that our triangle is exactly half the size of the rectangle that we made. This means that in order to solve for the area of the triangle we will need to take half of the area of the rectangle, or divide it by .
Thus, the area formula for a right triangle is as follows:
or
← Previous 1 3 4 5 6 |
# Solving Reverse Fraction Problems
## Presentation on theme: "Solving Reverse Fraction Problems"— Presentation transcript:
Solving Reverse Fraction Problems
Mathematics Solving Reverse Fraction Problems
Lesson Objectives The aim of this powerpoint is to help you…
to find the original value when you have been given the increased or decreased value to revise dividing by fractions
Reversing Fractional Changes
You MUST use ‘method 2’ of fractional change Writing this method as a formula we have… New fraction × Original quantity = Changed Quantity To work backwards to find the original quantity we must do the opposite… Changed Quantity ÷ New fraction = Original quantity You will need to remember the method for dividing by fractions!
Example 1 In a sale, all the prices are marked down by 3 10 If the sale price of some jeans is £56, what was their original price? 1 – 3 10 £56 ÷ 7 10 × original = £56 56 1 × = 56×10 1×7 560 7 = 560 ÷ 7 = £80
Reversing Fractional Changes
Watch out for the increase or decrease being quoted rather than the new value. Writing this as a formula we have… Fraction × Original quantity = Increase or decrease To work backwards to find the original quantity we must do the opposite… Increase or Decrease ÷ Fraction = Original quantity You will still need to remember the method for dividing by fractions!
Example 2 Rosemary’s weekly wage increases by 2 5 This equates to an increase of £40. What did she originally earn each week? 2 5 × original = £40 £40 ÷ 2 5 40 1 × = 40×5 1×2 200 2 = 200 ÷ 2 = £100
Quick Practice A Increase 36 by 4 9 B Decrease 50 by 3 10
C A hat usually costs £36.99. In a sale a third is knocked off the price. What is the dale price of the hat? D Sarah earns £7.50 per hour. If this increases by 2 5 , what is her new hourly rate of pay? Work out your answers before clicking on to the next slide.
Answers A 4 9 of 36 = B 1 – 3 10 C 1 – 1 3 D 1 + 2 5
(36 ÷ 9) x 4 = 16 = 52 Using Method 1 Using Method 2… of 50 = (50 ÷ 10) x 7 = 35 2 3 of = (36.99 ÷ 3) x 2 = £24.66 7 5 of = (7.5 ÷ 5) x 7 = £10.50
What next? Print out the notes called Frac10. Read them and make sure you answer any questions If you need more practice, try the worksheet called RevFrac-S1.xlsx. You have now completed all the work in the Fractions Module, though you will come across FDP conversions in the Percentages module. Please attempt the Fractions Assessment. |
# Power rule in calculus
## Statement
The derivative of $x^r$ is $r * x^{r-1}$, where $r$ is a real number.
$$\tfrac{d}{dx}(x^r) = r * x^{r-1}, r \in \Reals$$
## Proof
Define function $f$.
$$f(x) = x^r$$
Rewrite the function using $e$ and $\ln$. Since they are inverse functions, it stays exactly the same.
$$f(x) = e^{\ln(x^r)}$$
Differentiate this function by the derivative of e^x, and mutliply by the chain rule.
$$f'(x) = e^{\ln(x^r)} * \tfrac{d}{dx}(\ln(x^r))$$
The first factor simplifies to $x^r$, since $e$ and $\ln$ are inverses.
$$f'(x) = x^r * \tfrac{d}{dx}(\ln(x^r))$$
For the second factor, bring down the exponent $r$ using the properties of logarithms. Then bring $r$ out of the $\frac{d}{dx}$ function.
$$f'(x) = x^r * r * \tfrac{d}{dx}(\ln(x))$$
The derivative of $\ln(x)$ is $\frac{1}{x}$.
$$f'(x) = x^r * r * \frac{1}{x}$$
Now simplify this function to the final result by using negative exponents and adding exponents.
$$f'(x) = r * x^r * x^{-1}$$
$$f'(x) = r * x^{r - 1}$$
## Proofs building upon this proof
### Exponent rule in calculus
This proofs show the derivative of a^x is a^x * ln(a).
### Quotient rule in calculus
This proof shows that the derivative for the quotient or fraction a/b is (a'b - ab') / b^2.
### The derivative of ln(x)
This proof shows that the derivative of ln(x) is 1/x. |
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# Online Algebra Tutoring: Simultaneous Equations 1
## Algebra Worksheets
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### Learn Simultaneous Equations 1 from Online Algebra Tutor
In this method, we eliminate one variable (say x) to find the value of the other variable (y) and then put back the value of y in the original equation to find the value of the x . This can be done by adding the equations if they have the opposites (like 2y and – 2y are opposites). Let us see how this method works with an example
### Example:
Solve system of linear equations by elimination method 2x + 3y = 15 equation 1 x – 3y = 3 equation 2 Step 1: These are two equations, we will first line up the variables (if they are not lined up) 2 x + 3y = 15 x – 3y = 3 Step 2: Here we have opposites 3y and – 3y , so we will simply add the equations to eliminate y variable 2 x + 3 y = 15 x – 3y = 3 3 x = 18 x = 6 Step 3: Putting this value back of x into any one original equation. We are putting in equation 2 x – 3y = 3 6 – 3y = 3 -3y = 3 – 6 -3y = -3 y = 1 Solution x = 6 , y = 1 or (6 , 1).
## Check Point
Solve these systems of Equations by Elimination method
1. 2x – y = 17, x – y = 10
2. –2x + y = 10, 4x – y = –14
3. –x – 3y = 18, 2x + 3y = –12
4. 2x + 7y = 21, 3x + 7y = 14
5. y = −4x – 18 , y = 4x + 22
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# 4.7: Newton's Method
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In many areas of pure and applied mathematics, we are interested in finding solutions to an equation of the form $$f(x)=0.$$ For most functions, however, it is difficult—if not impossible—to calculate their zeroes explicitly. In this section, we take a look at a technique that provides a very efficient way of approximating the zeroes of functions. This technique makes use of tangent line approximations and is behind the method used often by calculators and computers to find zeroes.
## Describing Newton’s Method
Consider the task of finding the solutions of $$f(x)=0.$$ If $$f$$ is the first-degree polynomial $$f(x)=ax+b$$, then the solution of $$f(x)=0$$ is given by the formula $$x=−\frac{b}{a}$$. If $$f$$ is the second-degree polynomial $$f(x)=ax^2+bx+c$$, the solutions of $$f(x)=0$$ can be found by using the quadratic formula. However, for polynomials of degree 3 or more, finding roots of $$f$$ becomes more complicated. Although formulas exist for third- and fourth-degree polynomials, they are quite complicated. Also, if f is a polynomial of degree 5 or greater, it is known that no such formulas exist. For example, consider the function
$f(x)=x^5+8x^4+4x^3−2x−7.$
No formula exists that allows us to find the solutions of $$f(x)=0.$$ Similar difficulties exist for nonpolynomial functions. For example, consider the task of finding solutions of $$tan(x)−x=0.$$No simple formula exists for the solutions of this equation. In cases such as these, we can use Newton’s method to approximate the roots.
Newton’s method makes use of the following idea to approximate the solutions of $$f(x)=0.$$ By sketching a graph of $$f$$, we can estimate a root of $$f(x)=0$$. Let’s call this estimate $$x_0$$. We then draw the tangent line to $$f$$ at $$x_0$$. If $$f′(x0)≠0$$, this tangent line intersects the $$x$$-axis at some point $$(x1,0)$$. Now let $$x_1$$ be the next approximation to the actual root. Typically, $$x_1$$ is closer than $$x_0$$ to an actual root. Next we draw the tangent line to $$f$$ at $$x_1$$. If $$f′(x1)≠0$$, this tangent line also intersects the $$x$$-axis, producing another approximation, $$x_2$$. We continue in this way, deriving a list of approximations: $$x_0,x_1,x_2,….$$ Typically, the numbers $$x_0,x_1,x_2,…$$ quickly approach an actual root $$x*$$, as shown in the following figure.
Figure $$\PageIndex{1}$$:The approximations $$x_0,x_1,x_2,…$$ approach the actual root $$x*$$. The approximations are derived by looking at tangent lines to the graph of $$f$$.
Now let’s look at how to calculate the approximations $$x_0,x_1,x_2,….$$ If $$x_0$$ is our first approximation, the approximation $$x_1$$ is defined by letting $$(x_1,0)$$ be the $$x$$-intercept of the tangent line to $$f$$ at $$x_0$$. The equation of this tangent line is given by
$y=f(x_0)+f′(x_0)(x−x_0). \nonumber$
Therefore, $$x_1$$ must satisfy
$f(x_0)+f′(x_0)(x_1−x_0)=0.\nonumber$
Solving this equation for $$x_1$$, we conclude that
$x_1=x_0−\frac{f(x_0)}{f'(x_0)}.\nonumber$
Similarly, the point $$(x_2,0)$$ is the $$x$$-intercept of the tangent line to $$f$$ at $$x_1$$. Therefore, $$x_2$$ satisfies the equation
$x_2=x_1−\frac{f(x_1)}{f'(x_1)}.\nonumber$
In general, for $$n>0,x_n$$ satisfies
$x_n=x_{n−1}−\frac{f(x_{n−1})}{f'(x_{n−1})}.\label{Newton}$
Next we see how to make use of this technique to approximate the root of the polynomial $$f(x)=x^3−3x+1.$$
Example $$\PageIndex{1}$$: Finding a Root of a Polynomial
Use Newton’s method to approximate a root of $$f(x)=x^3−3x+1$$ in the interval $$[1,2]$$. Let $$x_0=2$$ and find $$x_1,x_2,x_3,x_4$$, and $$x_5$$.
Solution
From Figure $$\PageIndex{2}$$, we see that $$f$$ has one root over the interval $$(1,2)$$. Therefore $$x_0=2$$ seems like a reasonable first approximation. To find the next approximation, we use Equation \ref{Newton}. Since $$f(x)=x^3−3x+1$$, the derivative is $$f′(x)=3x^2−3$$. Using Equation \ref{Newton} with $$n=1$$ (and a calculator that displays $$10$$ digits), we obtain
$x_1=x_0−\frac{f(x_0)}{f'(x_0)}=2−\frac{f(2)}{f'(2)}=2−\frac{3}{9}≈1.666666667.\nonumber$
To find the next approximation, $$x_2$$, we use Equation with $$n=2$$ and the value of $$x_1$$ stored on the calculator. We find that
$x_2=x_1=\frac{f(x1)}{f'(x1)}≈1.548611111.\nonumber$
Continuing in this way, we obtain the following results:
• $$x_1≈1.666666667$$
• $$x_2≈1.548611111$$
• $$x_3≈1.532390162$$
• $$x_4≈1.532088989$$
• $$x_5≈1.532088886$$
• $$x_6≈1.532088886.$$
We note that we obtained the same value for $$x_5$$ and $$x_6$$. Therefore, any subsequent application of Newton’s method will most likely give the same value for $$x_n$$.
Figure $$\PageIndex{2}$$: The function $$f(x)=x^3−3x+1$$ has one root over the interval $$[1,2].$$
Exercise $$\PageIndex{1}$$
Letting $$x_0=0$$, let’s use Newton’s method to approximate the root of $$f(x)=x^3−3x+1$$ over the interval $$[0,1]$$ by calculating $$x_1$$ and $$x_2$$.
Hint
Use Equation \ref{Newton}.
Answer
$x_1≈0.33333333,x_2≈0.347222222$
Newton’s method can also be used to approximate square roots. Here we show how to approximate $$\sqrt{2}$$. This method can be modified to approximate the square root of any positive number.
Example $$\PageIndex{2}$$: Finding a Square Root
Use Newton’s method to approximate $$\sqrt{2}$$ (Figure $$\PageIndex{3}$$). Let $$f(x)=x^2−2$$, let $$x_0=2$$, and calculate $$x_1,x_2,x_3,x_4,x_5$$. (We note that since $$f(x)=x^2−2$$ has a zero at $$\sqrt{2}$$, the initial value $$x_0=2$$ is a reasonable choice to approximate $$\sqrt{2}$$).
Figure $$\PageIndex{3}$$: We can use Newton’s method to find $$\sqrt{2}$$.
Solution
For $$f(x)=x^2−2,f′(x)=2x.$$ From \ref{Newton}, we know that
$$x_n=x_{n−1}−\frac{f(x_{n−1})}{f'(x_{n−1})}$$
$$=x_{n−1}−\frac{x^2_{n−1}−2}{2x_{n−1}}$$
$$=\frac{1}{2}x_{n−1}+\frac{1}{x_{n−1}}$$
$$=\frac{1}{2}(x_{n−1}+\frac{2}{x_{n−1}}).$$
Therefore,
$$x_1=\frac{1}{2}(x_0+\frac{2}{x_0})=\frac{1}{2}(2+\frac{2}{2})=1.5$$
$$x_2=\frac{1}{2}(x_1+\frac{2}{x_1})=\frac{1}{2}(1.5+\frac{2}{1.5})≈1.416666667.$$
Continuing in this way, we find that
$$x_1=1.5$$
$$x_2≈1.416666667$$
$$x_3≈1.414215686$$
$$x_4≈1.414213562$$
$$x_5≈1.414213562.$$
Since we obtained the same value for $$x_4$$ and $$x_5$$, it is unlikely that the value xn will change on any subsequent application of Newton’s method. We conclude that $$\sqrt{2}≈1.414213562.$$
Exercise $$\PageIndex{2}$$
Use Newton’s method to approximate $$\sqrt{3}$$ by letting $$f(x)=x^2−3$$ and $$x_0=3$$. Find $$x_1$$ and $$x_2$$.
Hint
For $$f(x)=x^2−3$$, Equation \ref{Newton} reduces to $$x_n=\frac{x_{n−1}}{2}+\frac{3}{2x_{n−1}}$$.
Answer
$$x_1=2,x_2=1.75$$
When using Newton’s method, each approximation after the initial guess is defined in terms of the previous approximation by using the same formula. In particular, by defining the function $$F(x)=x−[\frac{f(x)}{f′(x)}]$$, we can rewrite Equation as $$x_n=F(x_{n−1})$$. This type of process, where each $$x_n$$ is defined in terms of $$x_{n−1}$$ by repeating the same function, is an example of an iterative process. Shortly, we examine other iterative processes. First, let’s look at the reasons why Newton’s method could fail to find a root.
## Failures of Newton’s Method
Typically, Newton’s method is used to find roots fairly quickly. However, things can go wrong. Some reasons why Newton’s method might fail include the following:
1. At one of the approximations $$x_n$$, the derivative $$f′$$ is zero at $$x_n$$, but $$f(x_n)≠0$$. As a result, the tangent line of $$f$$ at $$x_n$$ does not intersect the $$x$$ -axis. Therefore, we cannot continue the iterative process.
2. The approximations $$x_0,x_1,x_2,…$$ may approach a different root. If the function $$f$$ has more than one root, it is possible that our approximations do not approach the one for which we are looking, but approach a different root (see Figure). This event most often occurs when we do not choose the approximation $$x_0$$ close enough to the desired root.
3. The approximations may fail to approach a root entirely. In Example, we provide an example of a function and an initial guess $$x_0$$ such that the successive approximations never approach a root because the successive approximations continue to alternate back and forth between two values.
Figure $$\PageIndex{4}$$: If the initial guess $$x_0$$ is too far from the root sought, it may lead to approximations that approach a different root.
Example $$\PageIndex{3}$$: When Newton’s Method Fails
Consider the function $$f(x)=x^3−2x+2$$. Let $$x_0=0$$. Show that the sequence $$x_1,x_2,…$$ fails to approach a root of $$f$$.
Solution
For $$f(x)=x^3−2x+2,$$ the derivative is $$f′(x)=3x^2−2$$.Therefore,
$x_1=x_0−\frac{f(x0)}{f′(x0)}=0−\frac{f(0)}{f′(0)}=−\frac{2}{−2}=1. \nonumber$
In the next step,
$x_2=x_1−\frac{f(x1)}{f'(x1)}=1−\frac{f(1)}{f′(1)}=1−\frac{1}{1}=0. \nonumber$
Consequently, the numbers $$x_0,x_1,x_2,…$$ continue to bounce back and forth between $$0$$ and $$1$$ and never get closer to the root of $$f$$ which is over the interval $$[−2,−1]$$ (Figure $$\PageIndex{5}$$). Fortunately, if we choose an initial approximation $$x_0$$ closer to the actual root, we can avoid this situation.
Figure $$\PageIndex{5}$$: The approximations continue to alternate between $$0$$ and $$1$$ and never approach the root of $$f$$.
Exercise $$\PageIndex{3}$$
For $$f(x)=x^3−2x+2,$$ let $$x_0=−1.5$$ and find $$x_1$$ and $$x_2$$.
Hint
Use Equation \ref{Newton}.
Answer
$$x_1≈−1.842105263,x_2≈−1.772826920$$
From Example, we see that Newton’s method does not always work. However, when it does work, the sequence of approximations approaches the root very quickly. Discussions of how quickly the sequence of approximations approach a root found using Newton’s method are included in texts on numerical analysis.
## Other Iterative Processes
As mentioned earlier, Newton’s method is a type of iterative process. We now look at an example of a different type of iterative process.
Consider a function $$F$$ and an initial number $$x_0$$. Define the subsequent numbers $$x_n$$ by the formula $$x_n=F(x_{n−1})$$. This process is an iterative process that creates a list of numbers $$x_0,x_1,x_2,…,x_n,….$$ This list of numbers may approach a finite number $$x*$$ as $$n$$ gets larger, or it may not. In Example, we see an example of a function $$F$$ and an initial guess $$x_0$$ such that the resulting list of numbers approaches a finite value.
Example $$\PageIndex{4}$$: Finding a Limit for an Iterative Process
Let $$F(x)=\frac{1}{2}x+4$$ and let $$x_0=0$$. For all $$n≥1$$, let $$x_n=F(x_{n−1})$$. Find the values $$x_1,x_2,x_3,x_4,x_5$$. Make a conjecture about what happens to this list of numbers $$x_1,x_2,x_3…,x_n,…$$ as $$n→∞$$. If the list of numbers $$x_1,x_2,x_3,…$$ approaches a finite number $$x*$$, then $$x*$$ satisfies $$x*=F(x*)$$, and (x*\) is called a fixed point of $$F$$.
Solution
If $$x_0=0$$, then
• $$x_1=\frac{1}{2}(0)+4=4$$
• $$x_2=\frac{1}{2}(4)+4=6$$
• $$x_3=\frac{1}{2}(6)+4=7$$
• $$x_4=\frac{1}{2}(7)+4=7.5$$
• $$x_5=\frac{1}{2}(7.5)+4=7.75$$
• $$x_6=\frac{1}{2}(7.75)+4=7.875$$
• $$x_7=\frac{1}{2}(7.875)+4=7.9375$$
• $$x_8=\frac{1}{2}(7.9375)+4=7.96875$$
• $$x _9=\frac{1}{2}(7.96875)+4=7.984375.$$
From this list, we conjecture that the values $$x_n$$ approach $$8$$.
Figure provides a graphical argument that the values approach $$8$$ as $$n→∞$$. Starting at the point $$(x_0,x_0)$$, we draw a vertical line to the point $$(x_0,F(x_0))$$. The next number in our list is $$x_1=F(x_0)$$. We use $$x_1$$ to calculate $$x_2$$. Therefore, we draw a horizontal line connecting $$(x_0,x_1)$$ to the point $$(x_1,x_1)$$ on the line $$y=x$$, and then draw a vertical line connecting $$(x_1,x_1)$$ to the point $$(x_1,F(x_1))$$. The output $$F(x_1)$$ becomes $$x_2$$. Continuing in this way, we could create an infinite number of line segments. These line segments are trapped between the lines $$F(x)=\frac{x}{2}+4$$ and $$y=x$$. The line segments get closer to the intersection point of these two lines, which occurs when $$x=F(x)$$. Solving the equation $$x=\frac{x}{2}+4,$$ we conclude they intersect at $$x=8$$. Therefore, our graphical evidence agrees with our numerical evidence that the list of numbers $$x_0,x_1,x_2,…$$ approaches $$x*=8$$ as $$n→∞$$.
Figure $$\PageIndex{6}$$: This iterative process approaches the value $$x*=8.$$
Exercise $$\PageIndex{4}$$
Consider the function $$F(x)=\frac{1}{3}x+6$$. Let $$x_0=0$$ and let $$x_n=F(x_{n−1})$$ for $$n≥2$$. Find $$x_1,x_2,x_3,x_4,x_5$$. Make a conjecture about what happens to the list of numbers $$x1_,x_2,x_3,…x_n,…$$ as $$n→∞.$$
Hint
Consider the point where the lines $$y=x$$ and $$y=F(x)$$ intersect.
Answer
$$x_1=6,x_2=8,x_3=\frac{26}{3},x_4=\frac{80}{9},x_5=\frac{242}{27};x*=9$$
Iterative Processes and Chaos
Iterative processes can yield some very interesting behavior. In this section, we have seen several examples of iterative processes that converge to a fixed point. We also saw in Example that the iterative process bounced back and forth between two values. We call this kind of behavior a 2-cycle. Iterative processes can converge to cycles with various periodicities, such as 2−cycles, 4−cycles (where the iterative process repeats a sequence of four values), 8-cycles, and so on.
Some iterative processes yield what mathematicians call chaos. In this case, the iterative process jumps from value to value in a seemingly random fashion and never converges or settles into a cycle. Although a complete exploration of chaos is beyond the scope of this text, in this project we look at one of the key properties of a chaotic iterative process: sensitive dependence on initial conditions. This property refers to the concept that small changes in initial conditions can generate drastically different behavior in the iterative process.
Probably the best-known example of chaos is the Mandelbrot set (see Figure), named after Benoit Mandelbrot (1924–2010), who investigated its properties and helped popularize the field of chaos theory. The Mandelbrot set is usually generated by computer and shows fascinating details on enlargement, including self-replication of the set. Several colorized versions of the set have been shown in museums and can be found online and in popular books on the subject.
Figure $$\PageIndex{7}$$: The Mandelbrot set is a well-known example of a set of points generated by the iterative chaotic behavior of a relatively simple function.
In this project we use the logistic map
$f(x)=rx(1−x)$
where $$x∈[0,1]$$ and $$r>0$$
as the function in our iterative process. The logistic map is a deceptively simple function; but, depending on the value of $$r$$, the resulting iterative process displays some very interesting behavior. It can lead to fixed points, cycles, and even chaos.
To visualize the long-term behavior of the iterative process associated with the logistic map, we will use a tool called a cobweb diagram. As we did with the iterative process we examined earlier in this section, we first draw a vertical line from the point $$(x_0,0)$$ to the point $$(x_0,f(x0))=(x_0,x_1)$$. We then draw a horizontal line from that point to the point $$(x_1,x_1),$$ then draw a vertical line to $$(x_1,f(x_1))=(x_1,x_2)$$, and continue the process until the long-term behavior of the system becomes apparent. Figure shows the long-term behavior of the logistic map when $$r=3.55$$ and $$x_0=0.2$$. (The first $$100$$ iterations are not plotted.) The long-term behavior of this iterative process is an $$8$$-cycle.
Figure $$\PageIndex{8}$$: A cobweb diagram for $$f(x)=3.55x(1−x)$$ is presented here. The sequence of values results in an 8-cycle.
1. Let $$r=0.5$$ and choose $$x_0=0.2$$. Either by hand or by using a computer, calculate the first $$10$$ values in the sequence. Does the sequence appear to converge? If so, to what value? Does it result in a cycle? If so, what kind of cycle (for example, $$2−cycle,4−cycle.$$)?
2. What happens when $$r=2$$?
3. For $$r=3.2$$ and $$r=3.5$$, calculate the first $$100$$ sequence values. Generate a cobweb diagram for each iterative process. (Several free applets are available online that generate cobweb diagrams for the logistic map.) What is the long-term behavior in each of these cases?
4. Now let $$r=4.$$ Calculate the first $$100$$ sequence values and generate a cobweb diagram. What is the long-term behavior in this case?
5. Repeat the process for $$r=4,$$ but let $$x_0=0.201.$$ How does this behavior compare with the behavior for $$x_0=0.2$$?
## Key Concepts
• Newton’s method approximates roots of $$f(x)=0$$ by starting with an initial approximation $$x_0$$, then uses tangent lines to the graph of $$f$$ to create a sequence of approximations $$x_1,x_2,x_3,….$$
• Typically, Newton’s method is an efficient method for finding a particular root. In certain cases, Newton’s method fails to work because the list of numbers $$x_0,x_1,x_2,…$$ does not approach a finite value or it approaches a value other than the root sought.
• Any process in which a list of numbers $$x_0,x_1,x_2,…$$ is generated by defining an initial number $$x_0$$ and defining the subsequent numbers by the equation $$x_n=F(x_{n−1})$$ for some function $$F$$ is an iterative process. Newton’s method is an example of an iterative process, where the function $$F(x)=x−[\frac{f(x)}{f′(x)}]$$ for a given function $$f$$.
## Glossary
iterative process
process in which a list of numbers $$x_0,x_1,x_2,x_3…$$ is generated by starting with a number $$x_0$$ and defining $$x_n=F(x_{n−1})$$ for $$n≥1$$
Newton’s method
method for approximating roots of $$f(x)=0;$$ using an initial guess x0; each subsequent approximation is defined by the equation $$x_n=x_{n−1}−\frac{f(x_{n−1})}{f'(x_{n−1})}$$
### Contributors
• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. |
# class 7 Maths Chapter-4 Linear equation
### Exercise-4.1
#### Question 1:
Complete the last column of the table.
(i) The equation we have is: x + 3 = 0
Note: To check whether x =3 satisfies the given equation, we put x = 3 in the equation.
L.H.S. = 3 + 3 = 6 ≠ RHS
Thus, the answer is No i.e. the equation is not satisfied.
(ii) The equation we have is: x + 3 = 0
Note: To check whether the x = 0 satisfies the given equation, we put x = 0 in the given equation.
L.H.S. = 0 + 3 = 3 ≠ RHS
Thus, the answer is No i.e. the equation is not satisfied.
(iii) The equation we have is: x + 3 = 0
Note: To check whether the x = -3 satisfies the given equation, we put x = -3 in the given equation.
L.H.S. = -3 + 3 = 0 = RHS
Thus, the answer is Yes i.e. the equation is satisfied.
(iv) The equation we have is: x – 7 = 1
Note: To check whether the x = 7 satisfies the given equation, we put x = 7 in the given equation.
L.H.S. = 7 – 7 = 0 ≠ RHS
Thus, the answer is No i.e. the equation is not satisfied.
(v) The equation we have is: x – 7 = 1
Note: To check whether the x = 8 satisfies the given equation, we put x = 8 in the given equation.
L.H.S. = 8 – 7 = 1 = RHS
Thus, the answer is Yes i.e. the equation is satisfied.
(vi) The equation we have is: 5x = 25
Note: To check whether the x = 0 satisfies the given equation, we put x = 0 in the given equation.
L.H.S. = 5(0) = 0 ≠ RHS
Thus, the answer is No i.e. the equation is not satisfied.
(vii) The equation we have is: 5x = 25
Note: To check whether the x = 5 satisfies the given equation, we put x = 5 in the given equation.
L.H.S. = 5(5) = 25 = RHS
Thus, the answer is Yes i.e. the equation is satisfied.
(viii) The equation we have is: 5x = 25
Note: To check whether the x = -5 satisfies the given equation, we put x = -25 in the given equation.
L.H.S. = 5(-5) = -25 = RHS
Thus, the answer is No i.e. the equation is not satisfied.
### Exercise-4.2
Question 1:
Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
Answer: (a) x – 1 = 0
Solution: We have to separate the variables
Thus, x – 1 = 0
Now,
Adding 1 on both side of the equation, We get,
x – 1 + 1 = 0 + 1
x = 1
(b) x + 1 = 0
Solution: Here, we have to separate the variables
We have,
x + 1 = 0
Now,
Subtracting 1 on both side of the equation,We get,
x – 1 + 1 = 0 – 1
x = -1
(c) x – 1 = 5
Solution: Here, we have to separate the variables
We have,
x – 1 = 5
Now,Adding 1 on both side of the equation,We get,
x – 1 + 1 = 5 + 1
x = 6
(d) x + 6 = 2
Solution: Here, we have to separate the variables
We have,
x + 6 = 2
Now, subtracting 6 on both side of the equation,
We get,
x + 6 – 6 = 2 – 6
x = -4
(e)y – 4 = – 7
Solution: Here, we have to separate the variables
We have,
y – 4 = -7
Now,
Adding 4 on both side of the equation,
We get,
x – 4 + 4 = -7 + 4
x = -3
(f) y – 4 = 4
Solution: Here, we have to separate the variables
We have,
y – 4 = 4
Now,Adding 4 on both side of the equation,We get,
x – 4 + 4 = 4 + 4
x = 8
(g) y + 4 = 4
Solution: Here, we have to separate the variables
Thus,
We have, y + 4 = 4
Now,
Subtracting 4 on both side of the equation, We get,
x + 4 – 4 = 4 – 4
x = 0
(h )y + 4 = – 4
Solution: Here,We have to separate the variables
Thus,We have,
y + 4 = -4
Now,
Subtracting 4 on both side of the equation,We get,
x + 4 – 4 = -4 – 4
x = -8
Question 2:
Check whether the value given in the brackets is a solution to the given equation or not:
(a)3l = 42
(b )b/2 = 6
(c)p/7 = 4
(f)
(h)20t = -10
The given parts of the question are solved below:
Here,
Thus,
Dividing both sides by 3, we get, l=423
We have to separate the variables
Multiplying both sides by 2, we get,
b = 12
(c) p/7 = 4
We have, p7=4
p/7×7 = 4 × 7
### Exercise-4.4
Question 1:
Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of
(f) Notebooks he has from 50, he finds the result to be 8.
(g) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(h) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
The parts of this given questions are solved below:
(a) Add 4 to eight times a number you get 60.We have to frame the equation on the basis of given statement
And then solve the so framed equation
Thus,
Let the number be a
We know that,
Eight times of a is 8a
Therefore,
As per the statement,
We can write the equation as,
8a + 4 = 60
8a = 60 – 48a = 56
Now,
Dividing 8 on both the sides, we get
8a/8=56/8
Therefore,
a = 7
(b) One-fifth of a number minus 4 gives 3.
We have to frame the equation on the basis of given statement
And then solve the so framed equation
Thus,
Let the number be a
We know that,
One fifth of a is x/5
Therefore,
As per the statement,
We can write the equation as,
x/5-4=3
x/5 = 3 + 4
x/5 = 7
Now,
Multiplying 5 to both the sides, we get
a/5×5=7×5
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# Monika answered $\dfrac{3}{4}$ of the questions on her quiz correctly. What percent of the questions did she answer correctly?$A)25\%$$B)34\%$$C)50\%$$D)75\%$
Last updated date: 24th Jul 2024
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Hint: To solve this question we need to know the concept of percentage. This question asks us to change the fraction into percentage, which means this fraction will be represented as the fraction of $100$.
Complete step by step solution:
The question asks us to find the percentage of correct answers, when the answers marked correctly is $\dfrac{3}{4}$ of the questions given. To solve, the fraction $\dfrac{3}{4}$ needs to be converted into the fraction of $100$, which means percentage form. So solving this, the first step will be ,
The fraction will be multiplied by $100$, to find the value in percentage
$\Rightarrow \dfrac{3}{4}\times 100$
Multiply the terms in the numerator and the terms in the denominator,
$\Rightarrow \dfrac{3\times 100}{4}$
On multiplying we get
$\Rightarrow \dfrac{300}{4}$
The fraction need to be reduced in its lowest term, thus on reducing we get
$\Rightarrow 75$
After the calculation of fraction in percentage we have got the value as$75$ but since the fraction which we have converted in percentage means the fraction of hundred therefore sign of percentage will be used ,$\%$ hence indicating the value to be in percentage.
$\therefore$ The percent of the question Monika answered correctly is Option $D)75\%$ when the answer marked correctly by her is $\dfrac{3}{4}$ of the question.
So, the correct answer is “Option D”.
Note: A number with a percentage sign means that the number is a fraction of 100. So here $75\%$ when written in fraction will be written as
$\Rightarrow \dfrac{75}{100}$
Now we will bring the fraction to it’s lowest form. We see that the common factor for both the numerator and the denominator is $25$. So on dividing the number by the common factor (same number) of both numerator and denominator which is $25$, we get:
$\Rightarrow \dfrac{3}{4}$
So, the fraction we get after converting the percentage to fraction is $\dfrac{3}{4}$. This mean the answer we obtained is correct. |
# Class 12 NCERT Solutions – Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 2
Last Updated : 04 Apr, 2024
### (x cos x)x + (x sin x)1/x
Solution:
Given: (x cos x)x + (x sin x)1/x
Let us considered y = u + v
Where, u = (x cos x)x and v = (x sin x)1/x
So, dy/dx = du/dx + dv/dx ………(1)
So first we take u = (x cos x)
On taking log on both sides, we get
log u = log(x cos x)
log u = xlog(x cos x)
Now, on differentiating w.r.t x, we get
[Tex]\frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}(\log x+\log(\cos x))+\log x+\log \cos x[/Tex]
[Tex]\frac{1}{u}\frac{du}{dx}=x(\frac{1}{x}+\frac{1}{\cos x}\frac{d}{dx}\cos x)+\log x+\log\cos x[/Tex]
[Tex]\frac{du}{dx}=u(x(\frac{1}{x}+\frac{-\sin x}{\cos x})+\log x+\log(\cos x))[/Tex]
[Tex]\frac{du}{dx}=(x\cos x)^x(1-x\tan x+\log x+\log(\cos x)) [/Tex] ………(2)
Now we take u =(x sin x)1/x
On taking log on both sides, we get
log v = log (x sin x)1/x
log v = 1/x log (x sin x)
log v = 1/x(log x + log sin x)
Now, on differentiating w.r.t x, we get
[Tex]\frac{1}{v}\frac{dv}{dx}=\frac{1}{x}\frac{d}{dx}(\log x+\log(\sin x)+\frac{d}{dx}(\frac{1}{x}).(\log x+\log(\sin x)))[/Tex]
[Tex]\frac{1}{v}.\frac{dv}{dx}=\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.\frac{d}{dx}\sin x)+(\frac{-1}{x^2})(\log x+\log(\sin x))[/Tex]
[Tex]\frac{dv}{dx}=v(\frac{1}{x}(\frac{1}{x}+\frac{\cos x}{\sin x})\frac{-1}{x^2}(\log x+\log(\sin x)))[/Tex]
[Tex]\frac{dv}{dx}=(x\sin x)^{1/2}.[(\frac{1}{x^2}+\frac{\cot x}{x})-\frac{\log x}{x^2}-\frac{\log(\sin x)}{x^2}] [/Tex] ………(3)
Now put all the values from eq(2) and (3) into eq(1)
[Tex]\frac{dy}{dx}=(x\cos)^x(1-x\tan x+\log x+\log(\cos x))+(x\sin x)^{\frac{1}{x}}.[\frac{xcotx+1-log(xsinx)}{x^2}][/Tex]
### Question 12. xy + yx = 1
Solution:
Given: xy + yx = 1
Let us considered
u = xy and v = yx
So,
[Tex]\frac{du}{dx}+\frac{dv}{dx}=0 [/Tex]………(1)
So first we take u = xy
On taking log on both sides, we get
log u = log(xy)
log u = y log x
Now, on differentiating w.r.t x, we get
[Tex]\frac{1}{u}.\frac{du}{dx}=y.\frac{d}{dx}\log x+\frac{dy}{dx}.\log x[/Tex]
[Tex]\frac{1}{u}\frac{du}{dx}=\frac{y}{x}+\frac{dy}{dx}\log x[/Tex]
[Tex]\frac{du}{dx}=x^4(\frac{y}{x}+\frac{dy}{dx}\log x) [/Tex] ………(2)
Now we take v = yx
On taking log on both sides, we get
log v = log(y)x
log v = x log y
Now, on differentiating w.r.t x, we get
[Tex]\frac{1}{v}.\frac{dv}{dx}=x\frac{d}{dx}(\log x)+\log y\frac{d}{dx}x[/Tex]
[Tex]\frac{dv}{dx}=v(x.\frac{1}{y}.\frac{dy}{dx}+\log y)[/Tex]
[Tex]\frac{dv}{dx}=y^x(\frac{x}{y}\frac{dy}{dx}+\log x) [/Tex] ………(3)
Now put all the values from eq(2) and (3) into eq(1)
[Tex]x^y(\frac{y}{x}+\frac{dy}{dx}\log x)+y^x(\frac{x}{y}\frac{dy}{dx}+\log y)=0[/Tex]
[Tex](x^y.\log x+xy^{x-1})\frac{dy}{dx}=-(yx^{y-1}+y^x\log y)[/Tex]
[Tex]\frac{dy}{dx}=\frac{-yx^{y-1}+y^x\log y}{x^y\log x+xy^{x-1}}[/Tex]
### Question 13. yx = xy
Solution:
Given: yx = xy
On taking log on both sides, we get
log(yx) = log(xy)
xlog y = y log x
Now, on differentiating w.r.t x, we get
[Tex]x\frac{dy}{dx}(\log y)+\log y(\frac{d}{dx}x)=y\frac{d}{dx}\log x+\log x\frac{d}{dx}y[/Tex]
[Tex]x.\frac{d}{dx}.y+\log y.1=y.\frac{1}{x}+\log x\frac{dy}{dx}[/Tex]
[Tex]\frac{x}{y}\frac{dy}{dx}+\log y=\frac{y}{x}+\log x\frac{dy}{dx}[/Tex]
[Tex](\frac{x}{y}-\log x)\frac{dy}{dx}=(\frac{y}{x}-\log y)[/Tex]
[Tex]\frac{dy}{dx}=\frac{\frac{y}{x}-\log y}{\frac{x}{y}-\log x}[/Tex]
[Tex]\frac{dy}{dx}=\frac{y}{x}(\frac{y-x\log y}{x-y\log x})[/Tex]
### Question 14. (cos x)y = (cos y)x
Solution:
Given: (cos x)y = (cos y)x
On taking log on both sides, we get
y log(cos x) = x log (cos y)
Now, on differentiating w.r.t x, we get
[Tex]y\frac{d}{dx}\log(\cos x)+\log(\cos x).\frac{dy}{dx}=x\frac{d}{dx}\log (\cos y)+\log(\cos y)\frac{dx}{dx}[/Tex]
[Tex]y\frac{1}{\cos x}\frac{d}{dx}\cos x+\log(\cos x)\frac{dy}{dx}=x\frac{1}{\cos y}\frac{d}{dx}\cos y+\log(\cos y).1[/Tex]
[Tex]\frac{y}{\cos x}.-\sin x+\log(\cos x).\frac{dy}{dx}=\frac{x}{\cos y}.(-\sin y).\frac{dy}{dx}+\log(\cos y)[/Tex]
[Tex](\log(\cos x)+x\tan y)\frac{dy}{dx}=\log(\cos y)+y \tan x[/Tex]
[Tex]\frac{dy}{dx}=\frac{\log(\cos y)+y\tan x}{\log(\cos x)+x\tan y}[/Tex]
### Question 15. xy = e(x – y)
Solution:
Given: xy = e(x – y)
On taking log on both sides, we get
log(xy) = log ex – y
log x + log y = x – y
Now, on differentiating w.r.t x, we get
[Tex]\frac{d}{dx}\log x+\frac{d}{dx}\log y=\frac{dx}{dx}-\frac{dy}{dx}[/Tex]
[Tex]\frac{1}{x}.+\frac{1}{y}.\frac{dy}{dx}=1-\frac{dy}{dx} [/Tex]
[Tex](\frac{1}{y}+1)\frac{dy}{dx}=(1-\frac{1}{x})[/Tex]
[Tex]\frac{dy}{dx}=\frac{(1-\frac{1}{x})}{(1+\frac{1}{y})} [/Tex]
[Tex]\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}[/Tex]
### Question 16. Find the derivative of the function given by f(x) = (x + 1)(x + x2)(1 + x4)(1 + x8) and hence find f'(1).
Solution:
Given: f(x) = (x + 1)(x + x2)(1 + x4)(1 + x8)
Find: f'(1)
On taking log on both sides, we get
log(f(x)) = log(1 + x) + log(1 + x2) + log(1 + x4) + log(1 + x8)
Now, on differentiating w.r.t x, we get
[Tex]\frac{1}{f(x)}.\frac{d}{dx}f{x}=\frac{1}{1+x}\frac{d}{dx}(1+x)+\frac{1}{1+x^2}(1+x)^2+\frac{1}{1+x^4}.\frac{d}{dx}(1+x^4)+\frac{1}{1+x^8}\frac{d}{dx}(1+x^8)\frac{f'(x)}{f(x)}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}[/Tex]
[Tex]f'(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)(\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4x^3}{1+x^2}+\frac{8x^7}{1+x^8})[/Tex]
∴ f'(1) = 2.2.2.2.[Tex](\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2})[/Tex]
[Tex]f'(1)=16.(\frac{15}{2})[/Tex]
f'(1) = 120
### Do they all give the same answer?
Solution:
(i) By using product rule
[Tex]\frac{d}{dx}(u.v)=v\frac{du}{dv}+u\frac{dv}{dx}[/Tex]
[Tex]\frac{dy}{dx}=(x^2-5x+8)\frac{d}{dx}(x^3+7x+9)+(x^3+7x+9).\frac{d}{dx}(x^2-5x+8)[/Tex]
dy/dx = (3x4 – 15x3 + 24x2 + 7x2 – 35x + 56) + (2x4 + 14x2 + 18x – 5x3 – 35x – 45)
dy/dx = 5x4 – 20x3 + 45x2 – 52x + 11
(ii) By expansion
y = (x2 – 5x + 8)(x3 + 7x + 9)
y = x5 + 7x3 + 9x2 – 5x4 – 35x2 – 45x + 8x3 + 56x + 72
y = x5 – 5x4 + 15x3 – 26x2 + 11x + 72
dy/dx = 5x4 – 20x3 + 45x2 – 52x + 11
(iii) By logarithmic expansion
Taking log on both sides
log y = log(x2 – 5x + 8) + log(x3 + 7x + 9)
Now on differentiating w.r.t. x, we get
[Tex]\frac{1}{y}.\frac{dy}{dx}=\frac{1}{x^2-5x+8}.\frac{d}{dx}(x^2-5x+8)+\frac{1}{x^3+7x+9}\frac{d}{dx}(x^3+7x+9)[/Tex]
[Tex]\frac{1}{y}.\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}[/Tex]
[Tex]\frac{1}{(x^2-5x+8)(x^3+7x+9)}\frac{dy}{dx}=\frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)}[/Tex]
dy/dx = 2x4 + 14x2 + 18x – 5x3 – 35x – 45 + 3x4 – 15x3 + 24x2 + 7x2 – 35x + 56
dy/dx = 5x4 – 20x3 + 45x2 – 52x + 11
Answer is always same what-so-ever method we use.
### [Tex]\frac{d}{dx}(u.v.w)=\frac{du}{dx}v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}[/Tex]
Solution:
Let y = u.v.w.
Method 1: Using product Rule
[Tex]\frac{dy}{dx}=u\frac{d}{dx}(v.w)+v.w\frac{d}{dx}u[/Tex]
[Tex]\frac{dy}{dx}=u.[v.\frac{dw}{dx}+w\frac{du}{dx}]+v.w.\frac{du}{dx}[/Tex]
[Tex]\frac{dy}{dx}=u.v.\frac{dw}{dx}+u.w.\frac{dv}{dx}+v.w\frac{du}{dx}[/Tex]
Method 2: Using logarithmic differentiation
Taking log on both sides
log y = log u + log v + log w
Now, Differentiating w.r.t. x
[Tex]\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx} [/Tex]
[Tex]\frac{dy}{dx}=(u.v.w)(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx})[/Tex]
[Tex]\frac{dy}{dx}=v.w\frac{du}{dx}+uw\frac{dv}{dx}+uv\frac{dw}{dx}[/Tex]
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# Taylor and maclaurin series - Sequence and Series
## Taylor Series and Maclaurin Series
In order to understand Taylor and Maclaurin Series, we need to first look at power series.
## What are Power Series?
A power series is basically a series with the variable x in it. Formally speaking, the power series formula is:
where $c_{n}$ are the coefficients of each term in the series and $a$ is a constant. Power series are important because we can use them to represent a function. For example, the power series representation of the function $f(x) = \frac{1}{(1-x)} (for |x|$ < $1)$ is:
where $a = 1$ and $c_{n} = 1$.However, what if I want to find a power series representation for the integral of $\frac{1}{(1-x)}$? All you have to do is integrate the power series.
## Find a Power Series Representation for the function
Question 1: Find a power series representation for the integral of the function
• Recall that earlier we said that:
• So if we integrate both sides, then we get:
• To find $c$, we set $x=0$. So we have:
• Notice that $\ln (1)$ is equal to 0. So we get:
• Hence we can conclude that:
• If you want, you can make the series start at $n=1$ instead, making the series become:
## Power Series to a Taylor Series
Now this is where Taylor and Maclaurin Series come in. Taylor Series and Maclaurin Series are very important when we want to express a function as a power series. For example, $e^{x}$ and $\cos x$ can be expressed as a power series! First, we will examine what Taylor Series are, and then use the Taylor Series Expansion to find the first few terms of the series. Then we will learn how to represent some function as a Taylor series, and even differentiate or integrate them. Lastly, we will look at how to derive Taylor Polynomials from Taylor Series, and then use them to approximate functions. Note that we will also look at Maclaurin Series.
## What is a Taylor Series
So what exactly are Taylor Series? If possible (not always), we can represent a function $f(x)$ about $x=a$ as a Power Series in the form:
where $f^{n} (a)$ is the $n^{th}$ derivative about $x = a$. This is the Taylor Series formula. If it is centred around $x = 0$, then we call it the Maclaurin Series. Maclaurin Series are in the form:
Here are some commonly used functions that can be represented as a Maclaurin Series:
We will learn how to use the Taylor Series formula later to get the common series, but first let’s talk about Taylor Series Expansion.
## Taylor Series Expansion
Of course if we expand the Taylor series out, we will get:
This is known as the Taylor Expansion Formula. We can use this to compute an infinite number of terms for the Taylor Series.
## Finding the First Few Terms
For example, let’s say I want to compute the first three terms of the Taylor Series $e^{x}$ about $x = 1$.
Question 2: Find the first three terms of the Taylor Series for $f(x) = e^{x}$.
• We will use the Taylor Series Expansion up to the third term. In other words, the first three terms are:
• Note that this is centred about $x = 1$, hence we know
• We also know that taking derivatives gives us:
• Hence plugging $a$ into each of these functions will give us:
• So we know that the first three terms are:
## Finding the Taylor Series
Instead of finding the first three terms of the Taylor series, what if I want to find all the terms? In other words, can I find the Taylor Series which can give me all the terms? This is possible; however it can be difficult because you need to notice the pattern. Let’s try it out!
Question 3: Find the Taylor Series of $f(x) = e^{x}$ at $x = 1$.
• Recall that the Taylor Expansion is:
• We know the first three terms, but we don’t know any terms after. In fact, there are an infinite amount of terms after the third term. So how is it possible to figure what the term is when $n$$\infty$? Well, we look for the pattern of the derivatives. If we are able to spot the patterns, then we will be able to figure out the $n^{th}$ derivative is. Let’s take a few derivatives first. Notice that:
• The more derivatives you take, the most you realize that you will just get $e^{x}$ back. Hence we can conclude that the $n^{th}$ derivative is:
• Furthermore we know at $a = 1$, hence
• Therefore plugging this in into the Taylor Series Formula gives:
• Notice that this Taylor Series for $e^{x}$ is different from the Maclaurin Series for $e^{x}$. This is because this one is centred at $x=1$, while the other is centred around $x=0$.You may have noticed that finding the $n^{th}$ derivative was really easy here. What if the $n^{th}$ derivative was not so easy to spot?
Question 4: Find the Taylor Series of $f(x) = \sin x$ centred around $a = 0$.
• Notice that if we take a few derivatives, we get:
• Now the $n^{th}$ derivative is not easy to spot here because the derivatives keep switching from cosine to sine. However, we do notice that the $4^{th}$ derivative goes back $\sin x$ again. This means that if we derive more after the $4^{th}$ derivative, then we are going to get the same things again. We may see the pattern, but it doesn’t tell us much about the $n^{th}$ derivative. Why don’t we plug $a = 0$ into the derivatives?
• Now we are getting something here. The values of the $n^{th}$ derivative are always going to be 0, -1, or 1. Let’s go ahead and find the first six terms of the Taylor Series using these derivative.
• If we are to add all the terms together (including term after the sixth term), we will get:
• This is the Taylor Expansion of $\sin x$. Notice that every odd term is 0. In addition, every second term has interchanging signs. So we are going to rewrite this equation to:
• Even though we have these three terms, we can pretty much see the patterns of where this series is going. The powers of $x$ are always going to be odd. So we can generalize the powers to be $2n+1$. The factorials are also always odd. So we can generalize the factorials to be $2n+1$. The powers of -1 always go up by 1, so we can generalize that to be $n$. Hence, we can write the Taylor Series $\sin x$ as
• which is a very common Taylor series. Note that you can use the same strategy when trying to find the Taylor Series for $y = \cos x$.
Question 5: Find the Taylor Series of f(x) = cosx centred around.
• Notice that if we take a few derivatives, we get:
• Again, the $n^{th}$ derivative is not easy to spot here because the derivatives keep switching from cosine to sine. However, we do notice that the $4^{th}$ derivative goes back $\cos x$ again. This means if we derive more after the $4^{th}$ derivative, then we are going to get a loop. Now plugging in $a=0$ we have
• Again, the values of the $n^{th}$ derivative are always going to be 0, -1, or 1. Let’s find the first six terms of the Taylor Series using the derivatives from above.
• If we are to add all the terms together (including term after the sixth term), we will get:
• Notice that this time all even terms are 0 and every odd term have interchanging signs. So we are going to rewrite this equation to:
• We pretty much know the pattern here. The powers of x are always even. So we can generalize the powers to be $2n$. The factorials are always even, so we can generalize them to be $2n$. Lastly, the powers of -1 goes up by 1. So we can generalize that to be n. Hence, we can write the Taylor Series $\cos x$ as:
## Taylor Expansion Relationship of cosx and sinx
• Notice that the Maclaurin Series of $\cos x$ and $\sin x$ are very similar. In fact, they only defer by the powers. If we were to expand the Taylor series of $\cos x$ and $\sin x$, we see that:
• We can actually find a relationship between these two Taylor expansions by integrating. Notice that we were to find the integral of $\cos x$, then
• See that if $x=0$, then
• So the integral of cosine is
• which is the Taylor Expansion of $\sin x$. Likewise, the integral of $\sin x$ gives:
• See if $x=0$, then,
• Then we are left with:
• Dividing both sides of the equation by -1 gives:
which is the Taylor Expansion of $\cos x$.
## Taylor Series of Harder Functions
Now that we know how to use the Taylor Series Formula, let’s learn how to manipulate the formula to find Taylor Series of harder functions.
Question 6:Find the Taylor Series of $f(x) = \frac{\sin x}{x}$.
• So we see that the function has $\sin x$ in it. We know that $\sin x$ has the common Taylor series:
• So if we were to divide both sides by $x$, then we will get:
• We can manipulate the right hand side so that:
and so we just found the Taylor series for $\frac{\sin x}{x}$. Let’s do a harder question.
Question 7: Find the Taylor Series of
• Notice that cosine is in the function. So we probably want to use the Taylor Series:
• See that inside the cosine is $3x^{4}$. So what were going to do is replace all the $x$’s, and make them into $3x^{4}$. In other words,
• Doing so gives us:
• Now we are going to multiply both sides of the equation by $2x^{3}$. This leads to:
• Now we are going to clean up the series a little bit so that everything is inside the general term.
Thus we are done and this is the Taylor Series of $2x^{3} \cos (3x^{4})$. If you want to do more practice problems, then I suggest you look at this link.
http://tutorial.math.lamar.edu/Problems/CalcII/TaylorSeries.aspx
Each question has a step-by-step solution, so you can check your work!
## Taylor Series Approximation
Note that the Taylor Series Expansion goes on as $n$→ \intfy, but in practicality we cannot go to infinity. As humans (or even computers) we cannot go on forever, so we have to stop somewhere. This means we need to alter the formula for us so that it is computable.
We alter the formula will be:
Notice that since we stopped looking for terms after n, we have to make it an approximation instead. This formula is known as the Taylor approximation. It is a well known formula that is used to approximate certain values.
Notice on the right hand side of the equation that it is a polynomial of degree n. We actually call this the Taylor polynomial $T_{n} (x)$. In other words, the Taylor polynomial formula is:
Let’s do an example of finding the Taylor polynomial, and approximating a value.
Question 8: Find the $3^{rd}$ degree Taylor Polynomial of $f(x) = \ln (x)$ centred at $a = 1$. Then approximate $\ln (2)$.
• If we are doing a Taylor Polynomial of degree 3 centred at $a = 1$, then use the formula up to the $4^{th}$ term:
• Notice that taking the derivatives gives us:
• We also know that $a = 1$, so:
• Just in case you forgot, $\ln 1$ gives us 0. That’s why $f(a) = 0$. Now plugging everything into the formula of the $3^{rd}$ degree Taylor polynomial gives:
• Now we have to approximate $\ln (2)$. In order to do this, we need to use the Taylor polynomial that we just found. Notice that according to the Taylor approximation:
• This means that:
• If we are to set $x = 2$, then we will see that:
• So $\ln (2)$ is approximately around $\frac{5}{6}$. See that $\frac{5}{6}$ in decimal form is 0.833333...
Now if you pull out your calculator, we are actually pretty close. The actual value of $\ln (2)$ is 0.69314718056....
## The Error Term
We know that Taylor Approximation is just an approximation. However, what if we want to know the difference between the actual value and the approximated value? We call the difference the error term, and it can be calculated using the following formula:
Keep in mind that the $z$ variable is a value that is between $a$ and $x$, which gives the largest possible error.
Let’s use the error term formula to find the error of our previous question.
Question 9: Find the error of $\ln (2)$.
• In order to find the error, we need to find
• Notice from our previous question that we found the Taylor polynomial of degree 3. So we set $n = 3$. This means we need to find:
• See that the fourth derivative of the function is:
• Now our function is in terms of $x$, but we need it in term of $z$. So we just set $z = x$. This means that:
• So plugging this into our error term formula gives us:
• Remember that we set $x=2$ and $a=1$, so we have:
• Since we are talking the error of our approximation, the negative sign doesn’t matter here. So realistically we are looking at:
• Now recall that $z$ is a number between $a$ and $x$ which makes the error term the largest value. In other words, $z$ must be:
because $a=1$, and $x=2$. Now what $z$ value must we pick so that our error term is the largest?
• Notice that the variable $z$ is in the denominator. So if we pick smaller values of $z$, then the error term will become bigger. Since the smallest value of $z$ we can pick is 1, then we set $z = 1$. Thus,
is our error.
## Taylor’s Theorem
Now think of it like this. If we were to add the error term and the approximated value together, wouldn’t I get the actual value? This is correct! In fact, we can say this formally. If the Taylor polynomial is the approximated function and $R_{n} (x)$ is the error term, then adding them gives the actual function. In other words,
This is known as Taylor Theorem.
### Taylor and maclaurin series
In this lesson, we will learn that most functions can be expressed as a Taylor Series. These are power series with a special form, and are centred at a point. If it is centred at 0, then it is called a Maclaurin Series. All of these series require the n'th derivative of the function at point a. We will first apply the Taylor Series formula to some functions. You may notice that trying to find a Taylor Series of a polynomial will just give us back the same polynomial, and not a power series. Then we will learn how to manipulate some of the formulas for some harder functions. Lastly, we will find the Taylor series for sine and cosine, which will requires us to recognize some patterns. |
# 16.2 Mathematics of waves
Page 1 / 11
By the end of this section, you will be able to:
• Model a wave, moving with a constant wave velocity, with a mathematical expression
• Calculate the velocity and acceleration of the medium
• Show how the velocity of the medium differs from the wave velocity (propagation velocity)
In the previous section, we described periodic waves by their characteristics of wavelength, period, amplitude, and wave speed of the wave. Waves can also be described by the motion of the particles of the medium through which the waves move. The position of particles of the medium can be mathematically modeled as wave function s , which can be used to find the position, velocity, and acceleration of the particles of the medium of the wave at any time.
## Pulses
A pulse can be described as wave consisting of a single disturbance that moves through the medium with a constant amplitude. The pulse moves as a pattern that maintains its shape as it propagates with a constant wave speed. Because the wave speed is constant, the distance the pulse moves in a time $\text{Δ}t$ is equal to $\text{Δ}x=v\text{Δ}t$ ( [link] ).
## Modeling a one-dimensional sinusoidal wave using a wave function
Consider a string kept at a constant tension ${F}_{T}$ where one end is fixed and the free end is oscillated between $y=\text{+}A$ and $y=\text{−}A$ by a mechanical device at a constant frequency. [link] shows snapshots of the wave at an interval of an eighth of a period, beginning after one period $\left(t=T\right).$
Notice that each select point on the string (marked by colored dots) oscillates up and down in simple harmonic motion, between $y=+A$ and $y=\text{−}A,$ with a period T . The wave on the string is sinusoidal and is translating in the positive x -direction as time progresses.
At this point, it is useful to recall from your study of algebra that if f ( x ) is some function, then $f\left(x-d\right)$ is the same function translated in the positive x -direction by a distance d . The function $f\left(x+d\right)$ is the same function translated in the negative x -direction by a distance d . We want to define a wave function that will give the y -position of each segment of the string for every position x along the string for every time t .
Looking at the first snapshot in [link] , the y -position of the string between $x=0$ and $x=\lambda$ can be modeled as a sine function. This wave propagates down the string one wavelength in one period, as seen in the last snapshot. The wave therefore moves with a constant wave speed of $v=\lambda \text{/}T.$
Recall that a sine function is a function of the angle $\theta$ , oscillating between $\text{+}1$ and $-1$ , and repeating every $2\pi$ radians ( [link] ). However, the y -position of the medium, or the wave function, oscillates between $\text{+}A$ and $\text{−}A$ , and repeats every wavelength $\lambda$ .
when I click on the links in the topics noting shows. what should I do.
can we regard torque as a force?
Torque is only referred a force to rotate objects.
SHREESH
thanks
Emmanuel
I need lessons on Simple harmonic motion
Emmanuel
what is the formulae for elastic modulus
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
At time to = 0 the current to the DC motor is reverse, resulting in angular displacement of the motor shafts given by angle = (198rad/s)t - (24rad/s^2)t^2 - (2rad/s^3)t^3 At what time is the angular velocity of the motor shaft zero
3s
Basit
what is angular velocity
In three experiments, three different horizontal forces are ap- plied to the same block lying on the same countertop. The force magnitudes are F1 " 12 N, F2 " 8 N, and F3 " 4 N. In each experi- ment, the block remains stationary in spite of the applied force. Rank the forces according to (a) the
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
AFLAX
ty
Sharath
CD=5.83 n direction is NE
Ark
state Hooke's law of elasticity
Hooke's law states that the extension produced is directly proportional to the applied force provided that the elastic limit is not exceeded. F=ke;
Shaibu
thanks
Aarti
You are welcome
Shaibu
thnx
Junaid
what is drag force
Junaid
A backward acting force that tends to resist thrust
Ian
solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
Ian
The weight inside the tank is lesser due to the buoyancy force by the water displaced. Weight of water displaced = His weight outside - his weight inside tank = 720 - 34.3 = 685.7N Now, the density of water = 997kg/m³ (this is a known value) Volume of water displaced = Mass/Density (next com)
Sharath
density or relative density
Shaibu
density
Ian
Upthrust =720-34.3=685.7N mass of water displayed = 685.7/g vol of water displayed = 685.7/g/997 hence, density of man = 720/g / (685.7/g/997) =1046.6 kg/m3
1046.8
R.d=weight in air/upthrust in water =720/34.3=20.99 R.d=density of substance/density of water 20.99=x/1 x=20.99g/cm^3
Shaibu
Kg /cubic meters
Shaibu
Upthrust = 720-34.3=685.7N vol of water = 685.7/g/density of water = 685.7/g/997 so density of man = 720/g /(685.7/g/997) =1046.8 kg/m3
is there anyway i can see your calculations
Ian
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
okay i got it thanks
Ian
u welcome
Shaibu
HELLO kindly assist me on this...(MATHS) show that the function f(x)=[0 for xor=0]is continuous from the right of x->0 but not from the left of x->0
I do not get the question can you make it clearer
Ark
Same here, the function looks very ambiguous. please restate the question properly.
Sharath
please help me solve this problem.a hiker begins a trip by first walking 25kmSE from her car.she stops and sets her tent for the night . on the second day, she walks 40km in a direction 60°NorthofEast,at which she discovers a forest ranger's tower.find components of hiker's displacement for each day
Take a paper. put a point (name is A), now draw a line in the South east direction from A. Assume the line is 25 km long. that is the first stop (name the second point B) From B, turn 60 degrees to the north of East and draw another line, name that C. that line is 40 km long. (contd.)
Sharath
Now, you know how to calculate displacements, I hope? the displacement between two points is the shortest distance between the two points. go ahead and do the calculations necessary. Good luck!
Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
Sharath
cheers!
Liteboho
cheers too
Duncan
hii
Lakshya
hii too
Siciid
haye
Siciid
yes
Siciid
yes
Lakshya
shggggg
Lakshya
you mean
Siciid
solution problems
Siciid
what is the definition of model
please is there any way that i can understand physics very well i know am not support to ask this kind of question....
matthew
yes
Duncan
prove using vector algebra that the diagonals of a rhombus perpendicular to each other.
A projectile is thrown with a speed of v at an angle of theta has a range of R on the surface of the earth. For same v and theta,it's range on the surface of moon will be
0
Keshav
what is soln..
Keshav
o
Duncan
Using some kinematics, time taken for the projectile to reach ground is (2*v*g*Sin (∆)) (here, g is gravity on Earth and ∆ is theta) therefore, on Earth, R = 2*v²*g*Sin(∆)*Cos(∆) on moon, the only difference is the gravity. Gravity on moon = 0.166*g substituting that value in R, we get the new R
Sharath
Some corrections to my old post. Time taken to reach ground = 2*v*Sin (∆)/g R = (2*v²*Sin(∆)*Cos(∆))/g I put the g in the numerator by mistake in my old post. apologies for that. R on moon = (R on Earth)/(0.166)
Sharath
state Newton's first law of motion
Every body will continue in it's state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external force.
Kumaga
if you want this to become intuitive to you then you should state it
Shii
changing the state of rest or uniform motion of a body
koffi
if a body is in rest or motion it is always rest or motion, upto external force appied on it. it explains inertia
Omsai
what is a vector
smith
a ship move due north at 100kmhr----1 on a River flowing be due east on at 25kmperhr. cal the magnitude of the resultant velocity of the ship.
The result is a simple vector addition. The angle between the vectors is 90 degrees, so we can use Pythagoras theorem to get the result. V magnitude = sqrt(100*100 + 25*25) = 103.077 km/hr. the direction of the resultant vector can be found using trigonometry. Tan (theta) = 25/100.
Kumar
103.077640640442km/h
Peter
state Newton's first law of motion
An object continues to be in its state of rest or motion unless compelled by some external force
Alem
First law (law of inertia)- If a body is at rest, it would remain at rest and if the body is in the motion, it would be moving with the same velocity until or unless no external force is applied on it. If force F^=0 acceleration a^=0 or v^=0 or constant.
Govindsingh |
# Applied Discrete Structures
## Section10.3Rooted Trees
In the next two sections, we will discuss rooted trees. Our primary foci will be on general rooted trees and on a special case, ordered binary trees.
### Subsection10.3.1Definition and Terminology
One can formally define the genealogical terms above. We define child here since it’s used in our formal definition of a rooted tree and leave the rest of the definitions as an exercise.
#### Definition10.3.4.Child of a Root.
Given a rooted tree with root $$v\text{,}$$ a child of $$v$$ is a vertex that is connected to $$v$$ by an edge of the tree. We refer to the root as the parent of each of its children.
#### Definition10.3.5.Rooted Tree.
1. A single vertex $$v$$ with no children is a rooted tree with root $$v\text{.}$$
2. Recursion: Let $$T_1, T_2,\ldots,T_r\text{,}$$ $$r\geq 1\text{,}$$ be disjoint rooted trees with roots $$v_1\text{,}$$ $$v_2, \ldots \text{,}$$ $$v_r\text{,}$$ respectively, and let $$v_0$$ be a vertex that does not belong to any of these trees. Then a rooted tree, rooted at $$v_0\text{,}$$ is obtained by making $$v_0$$ the parent of the vertices $$v_1\text{,}$$ $$v_2, \ldots\text{,}$$ and $$v_r\text{.}$$ We call $$T_1, T_2, \ldots, T_r$$ subtrees of the larger tree.
The level of a vertex of a rooted tree is the number of edges that separate the vertex from the root. The level of the root is zero. The depth of a tree is the maximum level of the vertices in the tree. The depth of a tree in Figure 10.3.3 is three, which is the level of the vertices $$L$$ and $$M\text{.}$$ The vertices $$E\text{,}$$ $$F\text{,}$$ $$G\text{,}$$ $$H\text{,}$$ $$I\text{,}$$ $$J\text{,}$$ and $$K$$ have level two. $$B\text{,}$$ $$C\text{,}$$ and $$D$$ are at level one and $$A$$ has level zero.
Figure 2.1.2 is a rooted tree with Start as the root. It is an example of what is called a decision tree.
One of the keys to working with large amounts of information is to organize it in a consistent, logical way. A data structure is a scheme for organizing data. A simple example of a data structure might be the information a college admissions department might keep on their applicants. Items might look something like this:
\begin{equation*} \begin{split} ApplicantItem & = (FirstName,MiddleInitial,LastName,StreetAddress,\\ & City,State,Zip,HomePhone,CellPhone,EmailAddress,\\ & HighSchool,Major,ApplicationPaid,MathSAT,VerbalSAT,\\ & Recommendation1,Recommendation2,Recommendation3) \end{split} \end{equation*}
This structure is called a “flat file”.
A spreadsheet can be used to arrange data in this way. Although a “flat file” structure is often adequate, there are advantages to clustering some the information. For example the applicant information might be broken into four parts: name, contact information, high school, and application data:
\begin{equation*} \begin{split} ApplicantItem & = ((FirstName,MiddleInitial, LastName),\\ &((StreetAddress,City,State,Zip),\\ & (HomePhone,CellPhone),EmailAddress),\\ &HighSchool,\\ &(Major,ApplicationPaid,(MathSAT,VerbalSAT),\\ &(Recommendation1,Recommendation2,Recommendation3)) \end{split} \end{equation*}
The first item in each ApplicantItem is a list $$(FirstName, MiddleInitial, LastName)\text{,}$$ with each item in that list being a single field of the original flat file. The third item is simply the single high school item from the flat file. The application data is a list and one of its items, is itself a list with the recommendation data for each recommendation the applicant has.
The organization of this data can be visualized with a rooted tree such as the one in Figure 10.3.8.
In general, you can represent a data item, $$T\text{,}$$ as a rooted tree with $$T$$ as the root and a subtree for each field. Those fields that are more than just one item are roots of further subtrees, while individual items have no further children in the tree.
### Subsection10.3.2Kruskal’s Algorithm
An alternate algorithm for constructing a minimal spanning tree uses a forest of rooted trees. First we will describe the algorithm in its simplest terms. Afterward, we will describe how rooted trees are used to implement the algorithm. Finally, we will demonstrate the SageMath implementation of the algorithm. In all versions of this algorithm, assume that $$G = (V, E, w)$$ is a weighted undirected graph with $$\lvert V\rvert = m$$ and $$\lvert E\rvert = n\text{.}$$
Step 1 can be accomplished using one of any number of standard sorting routines. Using the most efficient sorting routine, the time required to perform this step is proportional to $$n \log n\text{.}$$ The second step of the algorithm, also of $$n \log n$$ time complexity, is the one that uses a forest of rooted trees to test for whether an edge should be added to the spanning set.
#### Note10.3.11.
1. Since we start the Kruskal’s algorithm with $$m$$ trees and each addition of an edge decreases the number of trees by one, we end the algorithm with one rooted tree, provided a spanning tree exists.
2. The rooted tree that we develop in the algorithm is not the spanning tree itself.
### Subsection10.3.3SageMath Note - Implementation of Kruskal’s Algorithm
Kruskal’s algorithm has been implemented in Sage. We illustrate how the spanning tree for a weighted graph in can be generated. First, we create such a graph
We will create a graph using a list of triples of the form $$(\text{vertex},\text{vertex}, \text{label})\text{.}$$ The $$weighted$$ method tells Sage to consider the labels as weights.
edges=[(1, 2, 4), (2, 8, 4), (3, 8, 4), (4, 7, 5), (6, 8, 5), (1, 3, 6), (1, 7, 6), (4, 5, 6), (5, 10, 9), (2, 10, 7), (4, 6, 7), (2, 4, 8), (1,8, 9), (1, 9, 9), (5, 6, 9), (1, 10, 10), (2, 9, 10), (4, 9, 10), (5, 9, 10), (6, 9, 10)]
G=Graph(edges)
G.weighted(True)
G.graphplot(edge_labels=True,save_pos=True).show()
Next, we load the kruskal function and use it to generate the list of edges in a spanning tree of $$G\text{.}$$
from sage.graphs.spanning_tree import kruskal
E = kruskal(G, check=True);E
To see the resulting tree with the same embedding as $$G\text{,}$$ we generate a graph from the spanning tree edges. Next, we set the positions of the vertices to be the same as in the graph. Finally, we plot the tree.
T=Graph(E)
T.set_pos(G.get_pos())
T.graphplot(edge_labels=True).show()
### Exercises10.3.4Exercises
#### 1.
Suppose that an undirected tree has diameter $$d$$ and that you would like to select a vertex of the tree as a root so that the resulting rooted tree has the smallest depth possible. How would such a root be selected and what would be the depth of the tree (in terms of $$d$$)?
Locate any simple path of length $$d$$ and locate the vertex in position $$\lceil d/2\rceil$$ on the path. The tree rooted at that vertex will have a depth of $$\lceil d/2\rceil\text{,}$$ which is minimal.
#### 2.
Use Kruskal’s algorithm to find a minimal spanning tree for the following graphs. In addition to the spanning tree, find the final rooted tree in the algorithm. When you merge two trees in the algorithm, make the root with the lower number the root of the new tree.
#### 3.
Suppose that information on buildings is arranged in records with five fields: the name of the building, its location, its owner, its height, and its floor space. The location and owner fields are records that include all of the information that you would expect, such as street, city, and state, together with the owner’s name (first, middle, last) in the owner field. Draw a rooted tree to describe this type of record |
Question Video: Finding the Kinetic Energy of a Body Moving between Two Points given Their Coordinates and the Acting Force in Vector Form | Nagwa Question Video: Finding the Kinetic Energy of a Body Moving between Two Points given Their Coordinates and the Acting Force in Vector Form | Nagwa
# Question Video: Finding the Kinetic Energy of a Body Moving between Two Points given Their Coordinates and the Acting Force in Vector Form Mathematics
The coordinates of the points ๐ด and ๐ต are (โ8, โ8) and (9, โ3). A body of unit mass moved from ๐ด to ๐ต in the direction of ๐๐ under the action of the force ๐น, where ๐
= (6๐ข hat + 7๐ฃ hat) force units. Given that the body started moving from rest, use the workโenergy principle to find its kinetic energy at point ๐ต.
04:52
### Video Transcript
The coordinates of the points ๐ด and ๐ต are negative eight, negative eight and nine, negative three. A body of unit mass moved from ๐ด to ๐ต in the direction of ๐๐ under the action of the force ๐น, where ๐
is six ๐ข hat plus seven ๐ฃ hat force units. Given that the body started moving from rest, use the workโenergy principle to find its kinetic energy at point ๐ต.
The workโenergy principle equates the net work done on some object to the total change in kinetic energy of that object. The total change in kinetic energy of an object is the difference between its final kinetic energy and its initial kinetic energy. In this case though, our object starts from rest, so its initial kinetic energy is zero. But if the initial kinetic energy is zero, then the total change in kinetic energy is just the final kinetic energy, which is, in fact, the kinetic energy that weโre looking for. Okay, well, that means that to find the final kinetic energy that weโre looking for, all we need to do is evaluate the left-hand side of the workโenergy principle, that is, find the net work done to the object.
Work itself is the energy gained or lost by an object as it moves in the presence of some external force. The simplest expression for work is force times distance, which is valid as long as the force is parallel to the objectโs motion. In our situation though, weโre given a vector representing the force. And weโre also told the beginning and endpoints of the objectโs motion and also that the object moved in a straight line between these two points. So if we want to use force times distance, we need to make sure that the force vector and the direction of motion are indeed parallel. Here, we have a set of axes and weโve labeled the initial and final points ๐ด and ๐ต, respectively.
The vector ๐๐, with its tail at point ๐ด and head at point ๐ต, is the vector that the object follows as it moves. We call this the displacement vector. The displacement vector for any motion has its tail at the initial position and its head at the final position and represents the net motion of the object. Also, on this diagram is the force vector ๐
six ๐ข hat plus seven ๐ฃ hat. Just by looking at this diagram, which we have carefully drawn to scale, we can see that ๐
and ๐๐ are not parallel. This means that weโll need a different way to express the work. The formula weโll use involves the vector dot product of the force and the displacement, specifically work equals force dotted with displacement.
This formula is valid as long as the force is constant for the entire motion. And in fact, six ๐ข hat plus seven ๐ฃ hat is constant. Since force dot displacement gives us the work and work gives us the kinetic energy that weโre looking for, all we need to do is calculate the dot product of ๐
and ๐๐, the displacement of the motion. Of course, to carry out this dot product, we first need to find an expression for the vector ๐๐. To find the vertical and horizontal components of this vector, weโll need to consider the changes in the vertical and horizontal coordinates between the points ๐ด and ๐ต. The vertical change is a rise from negative eight to negative three. This is a total change of positive five, and it is in the ๐ฃ hat direction.
The total horizontal change is the change from negative eight to positive nine. This is a net change of 17, and it is in the ๐ข hat direction. So the vector ๐๐ itself expressed in the unit vectors ๐ข hat and ๐ฃ hat would be 17๐ข hat plus five ๐ฃ hat.
All right, now letโs calculate the dot product. To calculate the dot product of two vectors, we multiply the components with the same unit vector and then add those results together. So in this case, the dot product would be six times 17, the product of the ๐ข hat components, plus seven times five, the product of the ๐ฃ hat components. Six times 17 is 102, and seven times five is 35. 102 plus 35 is 137. And this is the numerical value of the total work done on the object. Now we need to give this answer units. The units that we have for force are the rather arbitrary force units, which just means whatever unit we use for force. Therefore, our answer will be 137 work units because itโs the calculation of work that we did from force measured in the arbitrary force units.
What this really means is just that the unit of force determines the unit of work. For example, if we decide that the force was measured in newtons, then the work unit would be joules. But if we used any other unit for force, we would still have the same numerical answer, just a different unit for the work. Now, remember that the work energy theorem tells us that the final kinetic energy that is what weโre looking for is exactly the net work done on the system. So the final kinetic energy is 137 work units. Itโs also worth mentioning that the work energy theorem guarantees that work units are exactly the same as energy units. |
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#### Question 1:
Define the terms:
(i) Data
(ii) Raw data
(iii) Array
(iv) Tabulation of data
(v) Observations
(vi) Frequency of an observation
(vii) Statistics
(i) Data: It refers to the information in the form of numerical figures.
The marks obtained by 5 students of a class in a unit test are 34, 45, 65, 67, 87.
We call it the data related to the marks obtained by 5 students of a class in a unit test.
(ii) Raw Data: Data obtained in the original form is called raw data.
(iii) Array: Arranging the numerical figures in an ascending or a descending order is called an array.
(iv) Tabulation of data: Arranging the data in a systematic form in the form of a table is called tabulation or presentation of the data.
(v) Observations: Each numerical figure in a data is called an observation.
(vi) Frequency of an observation: The number of times a particular observation occurs is called its frequency.
(viii) Statistics: It is the science that deals with the collection, presentation, analysis and interpretation of numerical data.
#### Question 2:
The number of children in 25 families of a colony are give below:
2, 0, 2, 4, 2, 1, 3, 3, 1, 0, 2, 3, 4, 3, 1, 1, 1, 2, 2, 3, 2, 4, 1, 2, 2.
Represent the above data in the form of a frequency distribution table.
#### Question 3:
The sale of shoes of various sizes at a shop on a particular day is given below:
6, 9, 8, 5, 5, 4, 9, 8, 5, 6, 9, 9, 7, 8, 9, 7, 6, 9, 8, 6, 7, 5, 8, 9, 4, 5, 8, 7.
Represent the above data in the form of a frequency distribution table.
#### Question 4:
Construct a frequency table for the following:
3, 2, 5, 4, 1, 3, 2, 2, 5, 3, 1, 2, 1, 1, 2, 2, 3, 4, 5, 3, 1, 2, 3.
#### Question 5:
Construct a frequency table for the following:
7, 8, 6, 5, 6, 7, 7, 9, 8, 10, 7, 6, 7, 8, 8, 9, 10, 5, 7, 8, 7, 6.
#### Question 6:
Fill in the blanks:
(i) Data means information in the form of ...... .
(ii) Data obtained in the ...... form is called raw data.
(iii) Arranging the numerical figures in an ascending or a descending order is called an ....... .
(iv) The number of times a particular observation occurs is called its ...... .
(v) Arranging the data in the form of a table is called ...... . |
# 2015 AIME I Problems/Problem 7
## Problem
7. In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$.
## Solution
We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\sqrt5$. Since angles $\angle DCE$ and $\angle FCJ$ are complimentary, we have that $\triangle CDE ~ \triangle JFC$ (and similarly the rest of the triangles are $1-2-\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$, giving us:
$JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}$ and $BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}$.
Since $BC = CJ + JC$,
$2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}$,
Solving for $b$ in terms of $a$ yields $b = \frac{4a\sqrt5}{7}$.
We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\sqrt{11}$. We also draw the perpendicular from E to ML and label the point of intersection P.
This gives that $AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}$ and $ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}$
Since $AE$ = $AM + ME$, we get
$2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a$
$\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a$
$\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a$
$\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}$
$\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a$
$\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}$
So our final answer is $(7\sqrt{11})^2 = \boxed{539}$ |
Multiplying Terms over Brackets
# Multiplying Terms over Brackets
GCSE(F), GCSE(H),
Multiplying out brackets is expanding an expression.
For example: expand 3a(4a - 3b + 6). Set out the terms within the brackets as a list, deal with them individually, then add them together:
4a x 3a = 12a2
-3b x 3a = -9ab
6 x 3a = 18a
Note that every term in the bracket is multiplied by the term outside the bracket, and that negative signs are taken into account. Put the answer together:
3a(4a - 3b + 6) = 12a2 - 9ab + 18a
## Examples
1. Expand and simplify 3(m + 2) + 4m.
Expand the bracket:
m x 3 = 3m
2 x 3 = 6
Put it all together. Include the 4m that was outside the bracket:
3m + 6 + 4m = 7m + 6
2. Expand and simplify: 6 - a(a - 2).
Answer: -a2 + 2a + 6
Expand the brackets, taking very careful note of the negative signs:
a x -a = -a2
-2 x -a = 2a
Putting it all together, with the leading 6:
-a2 + 2a + 6
It is normal to put a list of terms in descending order of their powers. |
Students / Subjects
# Calculating Percentage Change
What does the "percentage change" () element of our elasticity formula mean? We simply want to look at how much the quantity and price changes, and then express this as a percentage. It is important to note that there are two common ways that percentage change is calculated. The two different methods give slightly different answers, so it is important to know which method you are being asked to use. Both methods are presented here; the standard method and the mid-point method.
Standard Method
= New - Old Old 100
We can use this formula to calculate the percentage change between any two numbers or quantities. As applied to economics, we typically present percentage change as follows where...
P = New Price
P = Old Price
Q = New Quantity
Q = Old Quantity
Y = New Income
Y = Old Quantity
Percentage change in price:
price = (P - P)P 100
Percentage change in quantity:
quantity = (Q - Q)Q 100
Percentage change in income:
income = (Y - Y)Y 100
Example
Suppose the price of Atlanta Thrashers tickets falls from \$45 to \$40 and the quantity demanded for tickets increases from 18,000 tickets to 23,000 tickets. Find the price elasticity of demand.
price = (40 - 45)45 100
= 11%
quantity = (23,000 - 18,000)18,000 100
= 28%
Now that we have calculated our percentage change in price and quantity demanded, we can measure the price elasticity of demand.
E = change in quantity demanded
Change in Price
E = 11%28%
E = 39%
Midpoint Method
Calculating percentage change as done above is often sufficient. However, you may notice that if we calculate the percentage change in price as (45 - 40)/40 x 100 we find that the percentage change is (-12.5 percent). In other words, it makes a difference if we look at the change as a rise or a fall; this is "end-point problem". To get around this problem, economists use the average of the two values as shown in the formulas below.
price = (P - P) .5(P + P) 100
This formula is not as complicated as it may look; the only thing we have added is finding the average in the numerator to take care of the "end-point problem".
.5(P + P)
This gives us an average of the two values; we can use this for any two values we may need to average including price, quantity, and income.
Example
To demonstrate how to find percentage change using the mid-point method, consider the following situation. Bill's income increased from \$100,000 to \$108,000 and his consumption of cruises increased from one cruise to two. Find the Income Elasticity of Demand.
quantity = (2 - 1) .5(1 + 2) 100 = 67%
income = (\$108,000 - \$100,000) .5(\$100,000 + \$108,000) 100 = 8%
Income Elasticity = 67%8%
= 8.4 |
## Calculus (3rd Edition)
$\begin{array}{*{20}{c}} {Rectangular}&{Polar}\\ {\left( {x,y} \right)}&{\left( {r,\theta } \right)}\\ {\left( {1, - 3} \right)}&{\left( {\sqrt {10} ,5.034} \right)}\\ {\left( {3, - 1} \right)}&{\left( {\sqrt {10} ,5.961} \right)} \end{array}$
We have $\left( {x,y} \right) = \left( {1, - 3} \right)$. Using the conversion formula from rectangular coordinates to polar coordinates given by $r = \sqrt {{x^2} + {y^2}}$, ${\ \ \ }$ $\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$ we obtain the polar coordinates: $r = \sqrt {1 + {{\left( { - 3} \right)}^2}} = \sqrt {10}$, $\theta = {\tan ^{ - 1}}\left( {\frac{{ - 3}}{1}} \right) \simeq - 1.249$ ${\ \ }$ or ${\ \ }$ $\theta \simeq 2\pi - 1.249 \simeq 5.034$ $\left( {r,\theta } \right) = \left( {\sqrt {10} ,5.034} \right)$ Similarly for the point $\left( {x,y} \right) = \left( {3, - 1} \right)$. $r = \sqrt {{3^2} + {{\left( { - 1} \right)}^2}} = \sqrt {10}$, $\theta = {\tan ^{ - 1}}\left( {\frac{{ - 1}}{3}} \right) \simeq - 0.322$ ${\ \ }$ or ${\ \ }$ $\theta \simeq 2\pi - 0.322 \simeq 5.961$ $\left( {r,\theta } \right) = \left( {\sqrt {10} ,5.961} \right)$ In summary: $\begin{array}{*{20}{c}} {Rectangular}&{Polar}\\ {\left( {x,y} \right)}&{\left( {r,\theta } \right)}\\ {\left( {1, - 3} \right)}&{\left( {\sqrt {10} ,5.034} \right)}\\ {\left( {3, - 1} \right)}&{\left( {\sqrt {10} ,5.961} \right)} \end{array}$ |
Irrational numbers
An irrational number is a number that cannot be written in the form of a common fraction of two integers. It is part of the set of real numbers alongside rational numbers. It can also be defined as the set of real numbers that are not rational numbers.
When an irrational number is expanded in decimal form, it is a non-terminating decimal that does not repeat. Note that a non-terminating decimal that repeats is a rational number, not an irrational number.
Examples
The following are a few of the more commonly known irrational numbers:
π = 3.14159... e = 2.71828... = 1.41421...
No matter the number of decimal places we calculate these values to, there will always be another digit after it, hence the term non-terminating decimal.
Properties of irrational numbers
As a subset of real numbers, irrational numbers share the same properties as the real numbers. Below are some of the properties of irrational numbers as they relate to their rational counterpart.
• The sum of an irrational number and a rational number is irrational.
• The product of an irrational number and a rational number is irrational, as long as the rational number is not 0.
• Two irrational numbers may or may not have a least common multiple.
• Irrational numbers are not closed under addition, subtraction, multiplication, and division. This is in contrast to rational numbers which are closed under all these operations.
In regards to the last bullet point, the property of closure, this means that operations involving only the set of irrational numbers can result in numbers that are members of different sets, such as rational numbers:
Addition and subtraction
Addition and subtraction of irrational numbers can result in either an irrational number or a rational number. Whenever operations between two irrational numbers can result in a number that is not irrational, it is not closed under that operation.
Examples
Addition:
(rational)
Subtraction:
(rational)
Multiplication and division
Irrational numbers are also not closed under multiplication and division. In both cases, it is possible for irrational numbers undergoing these operations to result in a rational number.
Examples
Multiplication:
(rational)
Division:
(rational)
Did you know??
There are more irrational numbers than there are rational numbers. Even though there are an infinite number of both types of numbers, we still know that there are more irrational numbers than rational ones. One way to think about this is that within even the relatively small set of natural numbers, the square root of all natural numbers that are not perfect squares (1, 4, 9, 16, etc), are irrational numbers. Having only listed the first 4 perfect squares, we've already reached the natural number 16. Between 1 and 16, there are 12 natural numbers the square root of which are irrational numbers. Furthermore, irrational numbers are non-terminating and non-repeating, so imagine adding many decimal places to each natural number along with all the combinations of digits we can use for each of the decimal places, and you can start to imagine just how many more irrational numbers there are! |
# Mean Value Theorem Formula
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## Mean Value Theorem
Mean Value Theorem is an important theorem in calculus. The first form of the mean value theorem was proposed in the 14th century by Parmeshwara, a mathematician from Kerela, India. Further, a simpler version of this was proposed by Rolle in the 17th century: Rolle’s Theorem, which was proved only for polynomials and was not a part of the calculus. Finally, the present version of the Mean Value Theorem was proposed by Augustin Louis Cauchy in the year 1823.
The mean value theorem states that for a curve passing through two given points there is one point on the curve where the tangent is parallel to the secant passing through the two given points. Rolle’s theorem has been derived from this mean value theorem.
## What is Mean Value Theorem?
The mean value theorem states that for a curve f(x) passing through two given points (a, f(a)), (b, f(b)), there is at least one point (c, f(c)) on the curve where the tangent is parallel to the secant passing through the two given points. The mean value theorem is defined herein calculus for a function f(x): [a, b] → R, such that it is continuous and differentiable across an interval.
• The function f(x) is continuous across the interval [a, b].
• The function f(x) is differentiable across the interval (a, b). There exists a point c in (a, b).
Here we have proved that the tangent at c is parallel to the secant passing through the points (a, f(a)), (b, f(b)). This mean value theorem is used to prove a statement across a closed interval. Further, Rolle’s theorem is derived from this mean value theorem.
## Proof of Mean Value Theorem
Statement: The mean value theorem states that if a function f is continuous over the closed interval [a,b], and differentiable over the open interval (a,b), then there exists at least one point c in the interval (a,b) such that f(c) is the average rate of change of the function over [a,b] and it is parallel to the secant line over [a,b].
Proof: Let g(x) be the secant line to f(x) passing through (a, f(a)) and (b, f(b)). We know that the equation of the secant line is y-y1 = m (x- x1).
Thus the mean value theorem is proved.
Note: The result may not hold if the function is not differentiable, even at a single point.
### Graphical Representation of Mean Value Theorem
The graphical representation of the function f(x) helps in understanding the mean value theorem. Here we consider two distinct points (a, f(a)), (b, f(b)). The line connecting these points is the secant of the curve, which is parallel to the tangent cutting the curve at (c, f(c)). The slope of the secant of the curve joining these points is equal to the slope of the tangent at the point (c, f(c)). We know that the derivative of the tangent is the slope at that point.
Slope of the Tangent = Slope of the Secant
Here we observe that the point (c, f(c)), lies between the two points (a, f(a)), (b, f(b)).
### Difference Between Mean Value Theorem and Rolle’s Theorem
The difference between the mean value theorem and Rolle’s theorem helps in a better understanding of these theorems. Both the theorems define the function f(x) such that it is continuous across the interval [a, b], and it is differentiable across the interval (a, b). In the mean value theorem, the two referred points (a, f(a)), (b, f(b)) are distinct and f(a) ≠ f(b). In Rolle’s theorem, the points are defined such that f(a) = f(b).
The value of c in the mean value theorem is defined such that the slope of the tangent at the point (c, f(c)) is equal to the slope of the secant joining the two points. The value of c in Rolle’s theorem is defined such that the slope of the tangent at the point (c, f(c)) is equal to the slope of the
## Examples of Mean Value Theorem
f'(x) = 5
2x = 5
x = 2.5
Answer: Since 2.5 lies in the interval (1, 4), the function satisfies the mean value theorem.
Example 2: Find the value of c if the function f(x) = x2 – 4x + 3 satisfies mean value theorem in the interval (1, 4).
Solution:
The given function f(x) = x2 – 4x + 3 satisfies mean values theorem. Hence it is continuous in [1, 4] and is differentiable in (1, 4).
f'(x) = 2x – 4
f(1) = 1 – 4 + 3 = 0
f(4) = 42 – 4(4) + 3 = 16 – 16 + 3 = 3
f'(c) = 1
2c – 4 = 1
2c = 5
c = 5/2 = 2.5
c = 2.5 belongs to the interval (1, 4)
Example 3: For the function f(x) = x2 + 2x, find all the values of c that satisfy the mean value theorem, over the interval [-4,4].
Solution:
f(x) = x2 + 2x is a polynomial and hence it is continuous and differentiable over the given interval [4,-4]
f'(x) = 2x+ 2
f(4) =42 + 2(4) = 24
f(-4) = (-4)2 + 2(-4)= 8
Let us find C in (-4,4) such that f'(c) = 2
f'(x) = 2x+ 2
f'(2) = 2(2)+ 2 = 6
f'(x) = 2c+ 2 = 2
⇒ c = 0.
Answer: For the function f(x) = x2 + 2x, the value of C = 0 that satisfy the mean value theorem, over the interval [-4,4].
## FAQs on Mean Value Theorem
### What is the Mean Value Theorem?
The mean value theorem states that if a function f is continuous over the closed interval [a,b], and differentiable over the open interval (a,b), then there exists a point c in the interval (a,b) such that f(c) is the average rate of change of the function over [a,b] and it is parallel to the secant line over [a,b].
### What is the Mean Value Theorem Equation?
The mean value theorem is defined for a function f(x): [a, b]→ R, such that it is continuous in the interval [a, b], and differentiable in the interval (a,b). For a point c in (a, b), the equation for the mean value theorem is as follows.
### What Does Mean Value Theorem Mean?
The mean value theorem states that for a curve passing through two given points there is one point on the curve where the tangent is parallel to the secant passing through the two given points. Rolle’s theorem has been derived from this mean value theorem.
### What is the Hypothesis of Mean Value Theorem?
The hypothesis for the mean value theorem is that, for a continuous function f(x), it is continuous in the interval [a, b], and it is differentiable in the interval (a, b).
### How to Find the Values that Satisfy Mean Value Theorem?
The values satisfying the mean value theorem are calculated by finding the differential of the given function f(x). The given function is defined in the interval (a, b), and the value satisfying the mean value theorem is the point c, which belongs to the interval (a, b). And we can find its value from
### How to Find C for Mean Value Theorem in Integrals?
As per the mean value theorem for the function f(x) defined in the interval (a, b), the value of C belongs to (a, b), and is calculated using the slope of the secant connecting the points (a, f(a)), (b, f(b)). The value of c is calculated from the derivative formula of
### Mean Value Theorem Formula
The mean value theorem formula tells us about a point c that must exist in a function if it follows the following conditions:
Let f(x) be a function defined on [ab] such that
(i) It is continuous on [ab].
(ii) It is differentiable on (ab).
Then there exists a real number c∈(a,b), and tangent to the curve of the function at point c will be parallel to the secant line between the points (a, f(a)), and (b, f(b)).
Let us learn more about the mean value theorem formula using solved examples in the section given below.
### What is Mean Value Theorem Formula?
If a function follows all the above-mentioned conditions, then there will be a point c between (a, b) where the tangent to the curve of the function at point c will be parallel to the secant line between the points (a, f(a)), and (b, f(b)).
Mean value theorem formula is:
### Solved Examples Using Mean Value Theorem Formula
1. Example 1: Find the value of c, for which the following function 4x3−8x2+7x−2, satisfies the mean value theorem between the interval [2, 5]. Solution: To find: c
f(2) = 12
f(5) = 333
f′(x)=12×2−16x+7
Using the mean value theorem formula,
12c2−16c+7=107c=−2.29, and 3.629
Since only 3.629 falls under the given interval.
Answer: Hence, the value of c is 3.629.
Example 2:
Determine all the value of c, for the function f(x) = x2−5x+7
for the interval [-1, 3] which satisfies the mean value theorem.
Solution:
To find: c
f(-1) = 13
f(3) = 1
f′(x)=2x−5
Using the mean value theorem formula,
### Solved Example
Question: Evaluate f(x) = x+ 2 in the interval [1, 2] using mean value theorem.
Solution:
Given function is:
f(x) = x+ 2
Interval is [1, 2].
i.e. a = 1, b = 2
Mean value theorem is given by,
f(b) = f(2) = 22 + 2 = 6
f(a) = f(1) = 12 + 2 = 3
Arithmetic Mean Formula
Geometric Mean Formula ⭐️⭐️⭐️⭐️⭐
Sample Mean Formula ⭐️⭐️⭐️⭐️⭐️
Root Mean Square Formula ⭐️⭐️⭐️⭐️⭐️
Mean Deviation Formula ⭐️⭐️⭐️⭐️⭐️
Mean Value Theorem Formula ⭐️⭐️⭐️⭐️⭐️
HARMONIC MEAN FORMULA ⭐️⭐️⭐️⭐️⭐ |
MCAT Physical : Conservation of Energy
Example Questions
Example Question #1 : Conservation Of Energy
A ball with mass of 2kg is dropped from the top of a building this is 30m high. What is the approximate velocity of the ball when it is 10m above the ground?
14m/s
20m/s
10m/s
30m/s
25m/s
20m/s
Explanation:
Use conservation of energy. The gravitational potential energy lost as the ball drops from 30m to 10m equals the kinetic energy gained.
Change in gravitational potential energy can be found using the difference in mgh.
So 400 Joules are converted from gravitational potential to kinetic energy, allowing us to solve for the velocity, v.
Example Question #41 : Work, Energy, And Power
Consider a spring undergoing simple harmonic motion. When the spring is at its maximum velocity __________.
kinetic energy is at a minimum and potential energy is at a maximum
kinetic energy and potential energy are at a maximum
kinetic energy is at a maximum and potential energy is at a minimum
kinetic energy and potential energy are at a minimum
kinetic energy is at a maximum and potential energy is at a minimum
Explanation:
Kinetic energy is highest when the spring is moving the fastest. Conversely, potential energy is highest when the spring is most compressed, and momentarily stationary. When the force resulting from the compression causes the spring to extend, potential energy decreases as velocity increases.
Example Question #2 : Conservation Of Energy
A pendulum with a mass of 405kg reaches a maximum height of 2.4m. What is its velocity at the bottommost point in its path?
17m/s
48m/s
6.9m/s
4.5m/s
6.9m/s
Explanation:
First solve for the potential energy of the pendulum at the height of 2.4m.
PE = mgh
PE = (405kg)(10m/s2)(2.4m) = 9720J
This must be equal to the maximum kinetic energy of the object.
KE = ½mv2
9720J = ½mv2
Plug in the mass of the object (405 kg) and solve for v.
9720J = ½(405kg)v2
v = 6.9m/s
Example Question #239 : Mcat Physical Sciences
Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top.
They then decide to sled down the hill, but disagree about who will go first.
Scenario 1:
Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.
Scenario 2:
Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.
Scenario 3:
Unable to agree, Sam and Sally tether themselves with a rope and go down together.
At the bottom of a neighboring hill, a neighbor watches Sally and Sam come down the hill. Sally is traveling 15m/s and Sam is traveling 10m/s. From the moment the neighbor begins watching, to just after they both come to a stop, who has dissipated more heat in the form of friction? (Assume all friction is lost as heat).
Sally, because she has less momentum
They dissipate equal amounts because the coeffecients of friction are the same for both
Sam, because he has more momentum
Sally, because she has greater kinetic energy
Sam, because he has greater kinetic energy
Sally, because she has greater kinetic energy
Explanation:
Sally has greater kinetic energy in this example than does Sam. From the moment when the neighbor begins watching we can calculate the kinetic energy. Once stopped, all of the kinetic energy will have been dissipated.
Sally's KE = 1/2 (52kg) (15m/s)2 = 5850J
Sam's KE = 1/2 (60kg) (10m/s)2 = 3000J
All of this energy will be dissipated as friction before Sam and Sally come to a stop.
Example Question #1 : Conservation Of Energy
Ignoring air resistance, which of the following is true regarding the motion of a pendulum?
At the bottom of the oscillation potential energy is at a minimum and kinetic energy is at a minimum
At the bottom of the oscillation potential energy is at a maximum and kinetic energy is at a maximum
At the bottom of the oscillation potential energy is at a maximum and kinetic energy is at a minimum
At the bottom of the oscillation potential energy is at a minimum and kinetic energy is at a maximum
At the bottom of the oscillation potential energy is at a minimum and kinetic energy is at a maximum
Explanation:
Energy must be conserved through the motion of a pendulum. Let point 1 represent the bottom of the oscillation and point 2 represent the top. At point 1, there is no potential energy, using point 1 as our "ground/reference," thus all of the system energy is kinetic energy. At point 2, the velocity is zero; thus, the kinetic energy is zero and all of the system energy is potential energy. At the highest point in the swing, potential energy is at a maximum, and at the lowest point in the swing, kinetic energy is at a maximum.
Example Question #1 : Conservation Of Energy
A boulder drops from rest off of a cliff. Find its velocity at before impact.
Explanation:
Conservation of energy dictates that the initial energy and final energy will be equal.
In this case, the boulder starts with zero kinetic energy and ends with both kinetic and potential energy.
We can cancel the mass from each term and plug in the given values to solve for the velocity at a height of .
Example Question #1 : Conservation Of Energy
A block of wood is floating in space. A bullet is fired from a gun and hits the block, embedding itself in the wood and generating heat. Which of the following is conserved?
Mechanical energy
Temperature
Kinetic energy
Momentum
Momentum
Explanation:
Momentum is always conserved in a system, when not experiencing external forces.
An inelastic collision can be identified if the two objects stick together after the collision occurs, such as the bullet becoming embedded in the wood. In an inelastic collision, kinetic energy is not conserved. Since mechanical energy is the sum of kinetic and potential energy, mechanical energy is also not conserved. This lack of conservation is due to the conversion of some of the kinetic energy to heat and sound. The kinetic energy decreases as the heat energy increases, resulting in a non-constant temperature.
Example Question #1 : Conservation Of Energy
A rock is dropped from a given height and allowed to hit the ground. The velocity of the rock is measured upon impact with the ground. Assume there is no air resistance.
In order for the velocity of the rock to be doubled before impact, which of the following is necessary?
Halve the height from which the rock is dropped
Double the height from which the rock is dropped
Reduce by 25% the height from which the rock is dropped
Quadruple the height from which the rock is dropped
Reduce by 75% the height from which the rock is dropped
Quadruple the height from which the rock is dropped
Explanation:
We can compare height and velocity by comparing the equations for potential and kinetic energy. This is possible because the rock initially has no kinetic energy (velocity is zero) and has no potential energy upon impact (height is zero). Using conservation of energy will yield the comparison below.
Because velocity is squared in the equation for kinetic energy, it requires a quadrupling of height in order to double the velocity.
Example Question #1 : Conservation Of Energy
A stone of mass m sits atop a hill of height h. As it rolls downhill, which of the following is true?
1. Half way down the hill,
2. Half way down the hill,
3. Half way down the hill,
4. Half way down the hill, PE still equals mgh.
5. None of these is true.
3
4
2
5
1
1
Explanation:
4. Choice 1 is correct because initially, all of the mechanical energy in the stone was potential energy and none was kinetic energy: ME = KE + PE. PE is “stored” in the stone-hill system by rolling the stone up hill. It is obvious that it takes half as much energy to roll the stone half way up the hill, compared with rolling it to the top. At the bottom of the hill, all of the PE will have been converted into KE, given by the formula
Since the PE was ½ mgh when rolling the stone half way up hill, it is the same as it rolls down hill.
Example Question #1 : Conservation Of Energy
An empty mining cart has a mass of and is traveling down a track that has a slope of to the horizontal. The cart is traveling at a rate of when an operator notices a disturbance on the track ahead and locks the wheels of the cart. What is the speed of the cart after it has traveled with the wheels locked?
Explanation:
We need the equation for conservation of energy for this problem:
We can eliminate final potential energy if we set the final height to be zer. We are solving for final velocity, so let's rearrange for final kinetic energy.
Substituting our equations for each variable, we get:
Rearranging for final velocity we get:
If you derive this formula and are unsure of your work, simply check your units. Each term under the square root has units of , which will ultimately give us units of , which is what we want.
We have values for all variables except two: height and normal force.
Let's calculate the height. We know that the between the initial and final states, the cart has traveled 20 meters down a slope of 40 degrees. Therefore, we can calculate height with the formula:
Now we just need to find the normal force. The following diagram will help visualize this calculation.
If you are unsure whether to use sine or cosine, think about it practically. As the angle gets less and less, the normal force is going to get larger. This is characteristic of a cosine function.
Therefore, we can say that:
Now that we have all of our variables, it's time to plug and chug: |
# Chapter 2 Probability Review
We give a very brief review of some necessary probability concepts. As the treatment is less than complete, a list of references is given at the end of the chapter. For example, we ignore the usual recap of basic set theory and omit proofs and examples.
## 2.1 Probability Models
When discussing probability models, we speak of random experiments that produce one of a number of possible outcomes.
A probability model that describes the uncertainty of an experiment consists of two elements:
• The sample space, often denoted as $$\Omega$$, which is a set that contains all possible outcomes.
• A probability function that assigns to an event $$A$$ a nonnegative number, $$P[A]$$, that represents how likely it is that event $$A$$ occurs as a result of the experiment.
We call $$P[A]$$ the probability of event $$A$$. An event $$A$$ could be any subset of the sample space, not necessarily a single possible outcome. The probability law must follow a number of rules, which are the result of a set of axioms that we introduce now.
## 2.2 Probability Axioms
Given a sample space $$\Omega$$ for a particular experiment, the probability function associated with the experiment must satisfy the following axioms.
1. Nonnegativity: $$P[A] \geq 0$$ for any event $$A \subset \Omega$$.
2. Normalization: $$P[\Omega] = 1$$. That is, the probability of the entire space is 1.
3. Additivity: For mutually exclusive events $$E_1, E_2, \ldots$$ $P\left[\bigcup_{i = 1}^{\infty} E_i\right] = \sum_{i = 1}^{\infty} P[E_i]$
Using these axioms, many additional probability rules can easily be derived.
## 2.3 Probability Rules
Given an event $$A$$, and its complement, $$A^c$$, that is, the outcomes in $$\Omega$$ which are not in $$A$$, we have the complement rule:
$P[A^c] = 1 - P[A]$
In general, for two events $$A$$ and $$B$$, we have the addition rule:
$P[A \cup B] = P[A] + P[B] - P[A \cap B]$
If $$A$$ and $$B$$ are also disjoint, then we have:
$P[A \cup B] = P[A] + P[B]$
If we have $$n$$ mutually exclusive events, $$E_1, E_2, \ldots E_n$$, then we have:
$P\left[\textstyle\bigcup_{i = 1}^{n} E_i\right] = \sum_{i = 1}^{n} P[E_i]$
Often, we would like to understand the probability of an event $$A$$, given some information about the outcome of event $$B$$. In that case, we have the conditional probability rule provided $$P[B] > 0$$.
$P[A \mid B] = \frac{P[A \cap B]}{P[B]}$
Rearranging the conditional probability rule, we obtain the multiplication rule:
$P[A \cap B] = P[B] \cdot P[A \mid B] \cdot$
For a number of events $$E_1, E_2, \ldots E_n$$, the multiplication rule can be expanded into the chain rule:
$P\left[\textstyle\bigcap_{i = 1}^{n} E_i\right] = P[E_1] \cdot P[E_2 \mid E_1] \cdot P[E_3 \mid E_1 \cap E_2] \cdots P\left[E_n \mid \textstyle\bigcap_{i = 1}^{n - 1} E_i\right]$
Define a partition of a sample space $$\Omega$$ to be a set of disjoint events $$A_1, A_2, \ldots, A_n$$ whose union is the sample space $$\Omega$$. That is
$A_i \cap A_j = \emptyset$
for all $$i \neq j$$, and
$\bigcup_{i = 1}^{n} A_i = \Omega.$
Now, let $$A_1, A_2, \ldots, A_n$$ form a partition of the sample space where $$P[A_i] > 0$$ for all $$i$$. Then for any event $$B$$ with $$P[B] > 0$$ we have Bayes’ Rule:
$P[A_i | B] = \frac{P[A_i]P[B | A_i]}{P[B]} = \frac{P[A_i]P[B | A_i]}{\sum_{i = 1}^{n}P[A_i]P[B | A_i]}$
The denominator of the latter equality is often called the law of total probability:
$P[B] = \sum_{i = 1}^{n}P[A_i]P[B | A_i]$
Two events $$A$$ and $$B$$ are said to be independent if they satisfy
$P[A \cap B] = P[A] \cdot P[B]$
This becomes the new multiplication rule for independent events.
A collection of events $$E_1, E_2, \ldots E_n$$ is said to be independent if
$P\left[\bigcap_{i \in S} E_i \right] = \prod_{i \in S}P[E_i]$
for every subset $$S$$ of $$\{1, 2, \ldots n\}$$.
If this is the case, then the chain rule is greatly simplified to:
$P\left[\textstyle\bigcap_{i = 1}^{n} E_i\right] = \prod_{i=1}^{n}P[E_i]$
## 2.4 Random Variables
A random variable is simply a function which maps outcomes in the sample space to real numbers.
### 2.4.1 Distributions
We often talk about the distribution of a random variable, which can be thought of as:
$\text{distribution} = \text{list of possible} \textbf{ values} + \text{associated} \textbf{ probabilities}$
This is not a strict mathematical definition, but is useful for conveying the idea.
If the possible values of a random variables are discrete, it is called a discrete random variable. If the possible values of a random variables are continuous, it is called a continuous random variable.
### 2.4.2 Discrete Random Variables
The distribution of a discrete random variable $$X$$ is most often specified by a list of possible values and a probability mass function, $$p(x)$$. The mass function directly gives probabilities, that is,
$p(x) = p_X(x) = P[X = x].$
Note we almost always drop the subscript from the more correct $$p_X(x)$$ and simply refer to $$p(x)$$. The relevant random variable is discerned from context
The most common example of a discrete random variable is a binomial random variable. The mass function of a binomial random variable $$X$$, is given by
$p(x | n, p) = {n \choose x} p^x(1 - p)^{n - x}, \ \ \ x = 0, 1, \ldots, n, \ n \in \mathbb{N}, \ 0 < p < 1.$
This line conveys a large amount of information.
• The function $$p(x | n, p)$$ is the mass function. It is a function of $$x$$, the possible values of the random variable $$X$$. It is conditional on the parameters $$n$$ and $$p$$. Different values of these parameters specify different binomial distributions.
• $$x = 0, 1, \ldots, n$$ indicates the sample space, that is, the possible values of the random variable.
• $$n \in \mathbb{N}$$ and $$0 < p < 1$$ specify the parameter spaces. These are the possible values of the parameters that give a valid binomial distribution.
Often all of this information is simply encoded by writing
$X \sim \text{bin}(n, p).$
### 2.4.3 Continuous Random Variables
The distribution of a continuous random variable $$X$$ is most often specified by a set of possible values and a probability density function, $$f(x)$$. (A cumulative density or moment generating function would also suffice.)
The probability of the event $$a < X < b$$ is calculated as
$P[a < X < b] = \int_{a}^{b} f(x)dx.$
Note that densities are not probabilities.
The most common example of a continuous random variable is a normal random variable. The density of a normal random variable $$X$$, is given by
$f(x | \mu, \sigma^2) = \frac{1}{\sigma\sqrt{2\pi}} \cdot \exp\left[\frac{-1}{2} \left(\frac{x - \mu}{\sigma}\right)^2 \right], \ \ \ -\infty < x < \infty, \ -\infty < \mu < \infty, \ \sigma > 0.$
• The function $$f(x | \mu, \sigma^2)$$ is the density function. It is a function of $$x$$, the possible values of the random variable $$X$$. It is conditional on the paramters $$\mu$$ and $$\sigma^2$$. Different values of these parameters specify different normal distributions.
• $$-\infty < x < \infty$$ indicates the sample space. In this case, the random variable may take any value on the real line.
• $$-\infty < \mu < \infty$$ and $$\sigma > 0$$ specify the parameter space. These are the possible values of the parameters that give a valid normal distribution.
Often all of this information is simply encoded by writing
$X \sim N(\mu, \sigma^2)$
### 2.4.4 Several Random Variables
Consider two random variables $$X$$ and $$Y$$. We say they are independent if
$f(x, y) = f(x) \cdot f(y)$
for all $$x$$ and $$y$$. Here $$f(x, y)$$ is the joint density (mass) function of $$X$$ and $$Y$$. We call $$f(x)$$ the marginal density (mass) function of $$X$$. Then $$f(y)$$ the marginal density (mass) function of $$Y$$. The joint density (mass) function $$f(x, y)$$ together with the possible $$(x, y)$$ values specify the joint distribution of $$X$$ and $$Y$$.
Similar notions exist for more than two variables.
## 2.5 Expectations
For discrete random variables, we define the expectation of the function of a random variable $$X$$ as follows.
$\mathbb{E}[g(X)] \triangleq \sum_{x} g(x)p(x)$
For continuous random variables we have a similar definition.
$\mathbb{E}[g(X)] \triangleq \int g(x)f(x) dx$
For specific functions $$g$$, expectations are given names.
The mean of a random variable $$X$$ is given by
$\mu_{X} = \text{mean}[X] \triangleq \mathbb{E}[X].$
So for a discrete random variable, we would have
$\text{mean}[X] = \sum_{x} x \cdot p(x)$
For a continuous random variable we would simply replace the sum by an integral.
The variance of a random variable $$X$$ is given by
$\sigma^2_{X} = \text{var}[X] \triangleq \mathbb{E}[(X - \mathbb{E}[X])^2] = \mathbb{E}[X^2] - (\mathbb{E}[X])^2.$
The standard deviation of a random variable $$X$$ is given by
$\sigma_{X} = \text{sd}[X] \triangleq \sqrt{\sigma^2_{X}} = \sqrt{\text{var}[X]}.$
The covariance of random variables $$X$$ and $$Y$$ is given by
$\text{cov}[X, Y] \triangleq \mathbb{E}[(X - \mathbb{E}[X])(Y - \mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X] \cdot \mathbb{E}[Y].$
## 2.6 Likelihood
Consider $$n$$ iid random variables $$X_1, X_2, \ldots X_n$$. We can then write their likelihood as
$\mathcal{L}(\theta \mid x_1, x_2, \ldots x_n) = \prod_{i = i}^n f(x_i; \theta)$
where $$f(x_i; \theta)$$ is the density (or mass) function of random variable $$X_i$$ evaluated at $$x_i$$ with parameter $$\theta$$.
Whereas a probability is a function of a possible observed value given a particular parameter value, a likelihood is the opposite. It is a function of a possible parameter value given observed data.
Maximumizing likelihood is a common techinque for fitting a model to data.
## 2.7 Videos
The YouTube channel mathematicalmonk has a great Probability Primer playlist containing lectures on many fundamental probability concepts. Some of the more important concepts are covered in the following videos:
## 2.8 References
Any of the following are either dedicated to, or contain a good coverage of the details of the topics above. |
# Number Tricks. How to work out the product of two numbers that differ by 2 (mental math tricks).
Here I will show you a number trick that can be used to multiply two numbers that have a difference of 2. For example, 17 × 19, 22 ×20, 41 × 43... and so on. The only downside is that you need to know your square number well. Here are the first twenty square numbers to help you out:
1,4,9,16,25,36,49,64,81,100,121,144,169,196,225, 256, 289, 324, 361, 400...
All you need to do then is follow these two steps:
Step 1 Work out the number in the middle of the two numbers and square this number.
Let’s try these two steps out on some examples:
Example 1
Work out 11 × 13
Step 1 Work out the number in the middle of the two numbers and square this number.
The number in the middle of 11 and 13 is 12 and 12 squared is 144.
144 – 1 = 143
Example 2
Work out 16 × 14
Step 1 Work out the number in the middle of the two numbers and square this number.
The number in the middle of 14 and 16 is 15 and 15 squared is 225.
225 – 1 = 224
Example 3
Work out 17 × 19
Step 1 Work out the number in the middle of the two numbers and square this number.
The number in the middle of 17 and 19 is 18 and 18 squared is 324.
324 – 1 = 323
Example 4
Work out 79 × 81
Step 1 Work out the number in the middle of the two numbers and square this number.
The number in the middle of 79 and 81 is 80 and 80 squared is 6400.
6400 – 1 = 6399 |
# What Does Halving Mean in Math?
To divide by 2 means to give away half of what you have. Imagine that you have six pieces of candy and a friend comes over. To share the candy equally, you need to split the candy into two equally large piles and give one of the piles to your friend. When you have given one half to your friend, you will be left with half of what you had. You have divided your candy by 2.
Imagine that you have $6$ toy cars, but you have to share them with your sister. When you split these $6$ cars into two equal piles, there will be $3$ cars in each pile. You give one pile of cars to your sister. You went from having $6$ cars to having $3$ cars, which means you have divided the number of cars you have by 2.
Are you able to find out what the half of 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24 is?
• $2$ divided by 2 is $1$
• $4$ divided by 2 is $2$
• $6$ divided by 2 is $3$
• $8$ divided by 2 is $4$
• $10$ divided by 2 is $5$
• $12$ divided by 2 is $6$
• $14$ divided by 2 is $7$
• $16$ divided by 2 is $8$
• $18$ divided by 2 is $9$
• $20$ divided by 2 is $10$
• $22$ divided by 2 is $11$
• $24$ divided by 2 is $12$
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# Section 5.0A Factoring Part 1
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1 Section 5.0A Factoring Part 1 I. Work Together A. Multiply the following binomials into trinomials. (Write the final result in descending order, i.e., a + b + c ). ( 7)( + 5) ( + 7)( + ) ( + 7)( + 5) ( )( + 7) B. Where did the first term ( a ) of the trinomial come from in each of the problems above? (Use the words like product, sum, First terms, Outer terms, Inner terms, and Last terms) C. Where did the last term (c) of the trinomial come from in each of the problems above? (Use the words like product, sum, First terms, Outer terms, Inner terms, and Last terms) D. Where did the middle term (b) of the trinomial come from in each of the problems above? (Use the words like product, sum, First terms, Outer terms, Inner terms, and Last terms) E. Wrap Up Honors Algebra Page 1
2 II. Think and Discuss A. Factoring Reversing the FOILing process. Factor the following trinomials (i.e., break them apart to be two binomials again) = ( )( ) =. 10 ( )( ) + =. 1 ( )( ) Remember what you just discovered and what we discussed on the previous page (i.e., start with the first two questions and then verify with the last question). + =. ( )( ) = ( )( ) B. Factoring can also be the reverse of distributing (i.e., what is in common with all the terms?) Try the following 15 y 10 y = ( ) Remember what you do when you distribute ( y + ),. + y + = ( ) now do the opposite. C. What happens when both are together (reverse distribution and reverse FOILing)? Which do you do first? Always do reverse distribution first, then the reverse FOILing. Try the following r + r 5 r = ( )= ( )( ). r + r + 1 r = ( )= ( )( ). r 1 r ( )= ( )( = ) Homework: p. P 1 all Honors Algebra Page
3 Section 5.0B Factoring Part Objective: Factoring polynomials with or more terms. I. Work Together A. Multiply the following binomials. ( a)( y + b). ( + 7)( y + ). ( )( + ). ( + 7)( + 5) B. Looking at the binomials you just multiplied, can you figure out a way to factor the following? a + b + ay + by = ( )( ) y + + y = ( )( ). + 1 = ( )( ) = ( )( ) C. Wrap Up Honors Algebra Page
4 II. Think and Discuss A. Factor the following using the grouping method a + 5 a = ( )+( )= ( )+ ( )=( )( ). rs + st + r + t = ( )+( )= ( )+ ( )=( )( ). 5y 10 y + = ( )+( )= ( )+ ( )=( )( ) a + = ( )+( )= ( )+ ( )= ( )( ) = ( )+( )= ( )+ ( )=( )( ) B. What happens if none of the methods I use is able to factor the problem? Then the polynomial is prime. Try the following y Homework: p all and p all Honors Algebra Page
5 Section 5.1 Graphing Quadratics Objectives: Graph Quadratic Functions. Find the ais of symmetry and coordinates of the verte of a parabola.. Model data using a quadratic function. y = 5 I. Think and Discuss A. Quadratic Functions Form a) Quadratic term b) Linear Term c) Constant Term. Graph of a quadratic a) Symmetrical (1) Def () Ais of Symmetry (a) Def b) Verte (1) () () (b) Equation for a parabola: c) Direction of Opening (1) If a is positive, it opens () If a is negative, it opens B. Which points are on the graph of the function y = 5? (1, -7). (, 0). (0, -5). (-, 11) C. What do you think the greatest eponent is for a quadratic function? for a linear function? D. Try This Tell whether each function is linear or quadratic? a). y = ( )( ) b). y = ( + ) c). y = ( + 5 ) d). y = ( 5). Find the ais of symmetry, verte, and determine if the verte is a maimum or minimum point for the following: y = Use you calculator to verify your answers to problem. Homework: p all, 19, 7, - all, 1 Honors Algebra Page 5
6 Section 5.1B Modeling Real World Data Modeling Data A. The table shows the average temperature in Gatlinburg, TN, for each month. Plot the points on a graph. Would it be useful to represent this data with a linear model? Eplain. Month Temp Feb() 5 Apr() 7 Jun() Aug() Oct(10) 71 Nov(11) 5 B. Finding Equations to model Quadratic Functions Find an equation to model the data mentioned, using your graphing calculator. (Hint: It is done the same way you did the linear functions, ecept for one thing.). Use the equation you just found to predict the average temperature in September.. How close was it to the actual temp of 1? Homework: p all Honors Algebra Page
7 Section 5. Solving Quadratic Equations by Graphing Objectives: Solve quadratics by graphing I. Solving Quadratic Functions Graphically A. Solution (Root or Zero) Algebraic Definition. Graphical definition (What would happen if the graph didn't cross the -ais or just touched it?) B. Solve the given graph: y = 5. C. Solve the following using your graphing calculator. y = + y. = Homework: p. 1-1 all, 0- all, 9, 5 and p all Honors Algebra Page 7
8 Objective: Solve polynomials Section 5. Solving Quadratics by Factoring I. Work Together A. Solve for each variable in the following equations. = 0 y = 0 5y = 0 Why were these easy to solve? B. Solve for each variable in the following equations. ab = 1 5y = 75 Can I use the same method I did with the problems in section A above? Eplain C. Wrap Up II. Think and Discuss A. The Process... Honors Algebra Page
9 B. Try the following = 0 10 = 0 ( )( ) = 0 ( )( ) = 0 ( ) = 0 or ( ) = 0 ( ) = 0 or ( ) = 0 = or = = or = + = 0 1 = 0 ( )( ) = 0 ( )( ) = 0 ( ) = 0 or ( ) = 0 ( ) = 0 or ( ) = 0 = or = = or = + = 0 1 = 0 ( )( ) = 0 ( )( ) = 0 = 0 or ( ) = 0 or ( ) = 0 = 0 or ( ) = 0 or ( ) = 0 = or = or = = or = or = C. What is wrong with the following problems? Find the error; eplain the error in this person's thought process; and correct the problem. 5 = 15 + = 5 5 ( 9)( + 5) = 15 + = ( 9) = 15 or ( + 5) = 15 + = 5 = or = = 0 ( 9)( + ) = 0 ( 9) = 0 or ( + ) = 0 = 9 or = Homework: p all, - all, 7, 9-5 all, 59- all, 79 Honors Algebra Page 9
10 Section 5.A Square Roots I. General Information A. Definition of Square Roots B. Note: Square roots have more than one root. Eamples has two square roots.. The nonnegative root is called the principal root. a) is asking for the principle root. b) is asking for the opposite of the principle root. c) ± is asking for the both roots. C. Problems Find each root II. Radical Epressions A. Properties a b =. 11a b. a b = ± 19 B. Eamples on Simplifying a. ± y z y z. 0y y 0y 7. 1m n mn III. Rationalizing the Denominator A. Protocol B. Eamples Homework: p all Honors Algebra Page 10
11 Section 5.B Operations with Radical Epressions I. Adding and Subtracting Radicals A. Note: To add radicals they must be like radical epressions. Treat like they are. B. Eamples II. Multiplying Radicals Eamples ( + )( + ). ( + 5 )( + 5 ). ( 1 + )( 1 ). ( )( 5 7 ) III. Rationalizing the Denominator Eamples 5. 5 Homework: p all Honors Algebra Page 11
12 Section 5. Comple Numbers Goals: To simplify radicals containing negative radicands.. To multiply pure imaginary numbers.. To solve quadratics equations that has pure imaginary solutions.. To add, subtract, and multiply comple numbers. I. General Information A. Rene Descartes (00 yrs ago) came up with a way to solve Proposals: i = 1 where i is not a real number.. i = 1 B. Pure Imaginary Numbers: Eamples 1 = and 11 = =. Simplify: i i = and 5 0 =. Simplify: i 1 = and i 5 =. Solve: + 1 = 0 and a + = 7 0 II. Comple Numbers A. Comple Number Form Real part Imaginary part Eamples ( + 7 i) + ( i). (9 i) (1 i). ( + 5 i)( i). ( + i)(5 i) III. Comple Conjugates Eamples: + 7i 7i ( )( ). ( 9 7i)( 9 + 7i) IV. Comple Numbers in the Denominator Eamples Simplify i 5 i. + 5 i + 7i Homework: p all, 7,,, 50, 1, Honors Algebra Page 1
13 Section 5.5 Completing the Square Goals: To solve quadratic equations by completing the square. Work Together FOIL These ( ). ( + 5). ( 7) Making perfect squares + = ( ) = ( + ). 0 + = ( ) Solve the following problems without setting equations equal to zero. =. ( ) = 1. (1 + ) = = = Wrap Up Think and Discuss Completing the square is a method to solve quadratic equations when they do not factor. Method Get constant on one side of the equation.. Factor out the coefficient in front of the.. Make the variable side into a perfect square (reminder what ever you add to one side must be done to the other).. Square root both sides and simplify. Eamples = = = = 0 Homework: p. 5-1 all, 7, (0-)/, 5, 7-9, 75-7 Honors Algebra Page 1
14 Section 5. The Quadratic Formula and Discriminant Goals:. To solve quadratic equations using the quadratic formula.. To use the discriminant to determine the nature of the roots of the quadratic equation. Work Together Solve using the Complete the Square Method a + b + c = 0 Can you write a formula to solve for in all quadratics? Wrap Up Think and Discuss II. The Quadratic Formula is a method to solve quadratic equations when they do not factor. A. Formula: If a b c + + = 0, then ( ) ± ( ) ( )( ) ( ) ( ) = where 0 B. Eamples = = = = 0 III. The Discriminant (Determines Nature of the Roots) A. Formula: If a + b + c = 0, then. B. Translation If D > 0, then. If D = 0, then. If D < 0, then C. Eamples Determine the nature of the roots = = 5 Homework: p all, 1,,, 5-5 all, Honors Algebra Page 1
15 Section 5.B Sum and Product of Roots Goals:. To find the sum and product of the roots of a quadratic equation. 5. To find all possible integral roots of a quadratic equation.. To find a quadratic equation to fit a given condition. Work Together Solve the following quadratics 15 = = = 0 Find the sum and product of each problem s roots. Sum =. Sum =. Sum = Product = Product = Product = Can you make a conclusion? (Do you see a pattern with the original quadratic and the sum and products?) Wrap Up Think and Discuss Sum and Product Theorem: Formula: If the roots of a + b + c = 0 are r 1 and r, then Eamples Find the quadratic given its roots. Roots are and -. Roots are 5 and 1 *. One root is 5 + i *. One root is 1+ *Note: a + bi ( a + b c ) is a root iff a bi ( a b c ). Homework: p all Honors Algebra Page 15
16 Section 5.7 Transformations with Quadratic Functions Goals: To graph quadratic equations of the form y = a( h) + k and identify the verte, the ais of symmetry, and the direction of the opening.. To determine the equation of the parabola from given information about the graph. I. Terms A. Verte B. Ais of Symmetry C. Parent Graph: y = II. Dynamics of y = a( h) + k A. What does h do? Graph: y = ( ). Graph: 10 y = ( + ) B. What does k do? Graph: y = +. Graph: 10 y = C. What does a do? Graph: y 10 =. Graph: y = Re-graph these two equations but us a negative coefficient D. Wrap Up y = a( h) + k h. k Verte: Ais of Symmetry:. a Honors Algebra Page 1
17 E. Eamples: Name the verte, ais of symmetry and direction of opening. a) y = ( + 11) + b) y = ( ) +. Put the following quadratics into a) y = + + b) y = a( h) + k y = c) y = Graph the following: a) y = ( + ) 1 b) y = = ( 1) + c) y ( ) III. Find the equation of the parabola A. Eamples: (,1) (,-) (,-) - -. Parabola passes through the verte (5, ) and the point (, ). Homework: p all, (9-1)/, (-) /, 5-9 all, 1,, 59, 5, 9, 7 Honors Algebra Page 17
18 Section 5. Graphing and Solving Inequalities Goals: To graph quadratic inequalities.. To solve quadratic inequalities in one variable. I. Graphing Quadratic Inequalities A. Same linear functions. B. Eamples y < ( ) 1 y II. Solving quadratic inequalities A. Graphically: 0 > < B. Algebraically 0 > < 0 Homework: p all, 0, (-)/, 7, 5, 7, 71 Honors Algebra Page 1
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# How to Tell If a Number is Divisible by 2 or 3
Learn how to test if a number is divisible by 2 or 3, and find out why each of these divisibility tests actually work.
By
Jason Marshall, PhD,
March 25, 2011
Episode #056
There are many types of problems in life that require dividing one number by another. But sometimes, we don’t actually need to do the division—we just need to know if it’s even possible to do the division in the first place. In these situations, you can save yourself time and trouble by learning a few quick and dirty tips. So today we’re kicking things off by looking at how to quickly tell if a number is divisible by 2 or 3.
But first, we’re giving away \$100 this month! Visit http://stitcher.com/math to download the FREE Stitcher app to listen to podcasts on your iPhone, BlackBerry, Android, or Pre without downloading or syncing. You must use the code MATH to enter.
## What Does it Mean for a Number to be Divisible?
Before we get to testing for divisibility, let’s first make sure we understand what divisibility actually means. As you know, whenever you multiply two numbers together, you get another number. So, in a way, you can think of the process of multiplying numbers as a means of building new numbers. For example, in 3 x 7 = 21 the product of the numbers 3 and 7 “build” the number 21.
Whenever a number can be built from other numbers like this, it obviously must be true that we can also divide the number by one of those building numbers to get the other building number. In other words, in this case, we can divide 21 by 7 to get 3, or we can divide 21 by 3 to get 7. When a number can be divided evenly by another number like this, we say that the number is “divisible by” that number. So, in this case, 21 is evenly divisible by both 3 and 7 since 21/3 is exactly equal to the whole number 7 and 21/7 is exactly equal to the whole number 3.
## How to Tell if a Number is Divisible by 2
Now that we understand what we’re trying to test for, let’s turn our attention to actually doing the testing. Up first, let’s talk about how to quickly tell whether or not a number is divisible by 2. The quick and dirty tip is that all even numbers are divisible by 2. That means that any number whose final digit is a 0, 2, 4, 6, or 8 must be divisible by 2. So now right away—without doing any division—we can tell that the number 1,018 (which is even) must be divisible by 2, while the number 1,033 (which is odd) is not.
Why does this work? Well, in this case the answer is pretty intuitive; but let’s talk about it a little to help prepare us to understand the divisibility tests for bigger integers that we’ll soon see. The important thing to notice is that when we divide an even number by 2, the remainder is always 0; and when we divide an odd number by 2, the remainder is always 1. If you think about this for a minute, you’ll see that we can use that information to conclude that a number is only divisible by 2 if it is even—since it won’t have a remainder.
For the math whizzes out there, you’ll notice that we could just as well have summed this all up using the connection between modular arithmetic and remainders that we talked about in the past few weeks to say that a number is only divisible by 2 if that number modulo 2 is congruent to 0.
## How to Tell if a Number is Divisible by 3
Let’s now talk about how to test if a number is divisible by 3. The quick and dirty tip to check for divisibility by 3 is to see if the sum of all the digits in the number is divisible by 3. If so, the number itself must also be divisible by 3. For example, is 1,529 divisible by 3? Well, the sum of the digits of 1,529 is 1+5+2+9=17. Since 17 is not divisible by 3, we can conclude that 1,529 is also not divisible by 3. How about 1,530? Well, this time the sum is 1+5+3+0=9. Since 9 is divisible by 3, we know that 1,530 is divisible by 3 too.
## Why Does the Divisibility by 3 Test Work?
But why does this work? Well, to find out let’s take a look at the number 111. First, since 1+1+1=3 is divisible by 3, we can immediately tell that 111 must be divisible by 3 too. To help us see why this works, let’s write 111 in a funny way (you’ll see why in a minute):
111 = 100 + 10 + 1
Now, let’s do one more thing that’s even a little weirder. Let’s use the fact that 10 = 9+1 and that 100 = 99+1 to write this sum for 111 as
111 = 99+1 + 9+1 + 1
What does this do for us? Well, believe it or not, writing 111 like this is extremely useful since it allows us to look at the problem 111/3 in a very interesting way.
In particular, if you think about it for a minute, you’ll see that any number of 9s or any number of 99s can always be evenly divided by 3. So since we know that the 9s and 99s are divisible by 3, the only question is whether the sum of everything that’s not a 9 or a 99 is divisible by 3 too. If it is, then since all the parts are divisible by 3, the whole thing must be divisible by 3! And notice that the only things in the sum 111 = 99+1 + 9+1 + 1 that aren’t 9s or 99s are the three 1s—precisely the digits of the number 111.
Try breaking apart some other numbers like this for yourself and you’ll see that they can all be written as some number of 9s, plus some number of 99s, plus some number of 999s (for larger numbers), and so on, plus some other stuff that will always be the digits of the original number. That’s precisely where the quick and dirty tip comes from!
## How Modular Arithmetic Can Explain Divisibility by 3
[[AdMiddle]Once again, for the math superstars out there, notice that we could have written this whole explanation using the language of modular arithmetic that we talked about in previous articles. In particular, 111 is only divisible by 3 if 111 mod 3 is congruent to 0. And since going around a modulus 3 clock any number of 9s, 99s, 999s, and so on always leaves you exactly where you started, the only thing that matters is whether the sum of the digits in the original number also leave you where you started. If so, then the number is divisible by 3. Who knew modular arithmetic was so useful!
## Practice Problems
Okay, that’s all the time we have for today. But before we finish up, here are a few practice problems to help you test your divisibility testing skills:
1. Is 213 divisible by 2? ____ (Yes/No) By 3? ____ (Yes/No)
2. Is 2,023,182 divisible by 2? ____ (Yes/No) By 3? ____ (Yes/No)
3. Is 1,109 divisible by 2? ____ (Yes/No) By 3? ____ (Yes/No)
You can find the answers at the very end of this article. After checking them, feel free to leave a comment at the bottom of the page and let me know how you did.
## Wrap Up
If you have questions about how to solve these practice problems or any other math questions, please email them to me at mathdude@quickanddirtytips.com, send them via Twitter, or become a fan of the Math Dude on Facebook and get help from me and the other math fans there.
Until next time, this is Jason Marshall with The Math Dude’s Quick and Dirty Tips to Make Math Easier. Thanks for reading math fans!
## Practice Problem Solutions
1. 1. Is 213 divisible by 2? No. By 3? Yes.
2. 2. Is 2,023,182 divisible by 2? Yes. By 3? Yes.
3. 3. Is 1,109 divisible by 2? No. By 3? No. |
9
# THE ISOSCELES RIGHT TRIANGLE
AN ISOSCELES RIGHT TRIANGLE is one of two special triangles. (The other is the 30°-60°-90° triangle.) In each triangle the student should know the ratios of the sides.
(An isosceles triangle has two equal sides. See Definition 8 in Some Theorems of Plane Geometry. The theorems cited below will be found there.)
Theorem. In an isosceles right triangle the sides are in the ratio 1:1:.
Proof. In an isosceles right triangle, the equal sides make the right angle. They have the ratio of equality, 1 : 1.
To find the ratio number of the hypotenuse h, we have, according to the Pythagorean theorem,
h2 = 12 + 12 = 2.
Therefore,
h = .
(Lesson 26 of Algebra.) Therefore the three sides are in the ratio
1 : 1 : .
Since the triangle is isosceles, the angles at the base are equal. (Theorem 3.) Therefore each of those acute angles is 45°.
(For the definition of measuring angles by "degrees," see Topic 3.)
Example 1. Evaluate sin 45° and tan 45°.
Answer. For any problem involving 45°, the student should sketch the triangle and place the ratio numbers.
We see:
sin 45° = 1 = ½,
on rationalizing the denominator. (Lesson 26 of Algebra.)
tan 45° = 11 = 1.
Problem 1. Evaluate cos 45° and csc 45°.
cos 45° = 1 = ½.
Thus cos 45° is equal to sin 45°; they are complements.
csc 45° = 1 = .
Example 2. Solve the isosceles right triangle whose side is 6.5 cm.
Answer. To solve a triangle means to know all three sides and all three angles. Since this is an isosceles right triangle, the only problem is to find the hypotenuse.
In every isosceles right triangle, the sides are in the ratio 1 : 1 : , as shown on the right. Those two triangles are similar. The side 1 has been multiplied by 6.5. Therefore every side will be multiplied by 6.5. The hypotenuse will be 6.5. (The theorem of the same multiple.)
Whenever we know the ratio numbers, then to solve the triangle we use this method of similar figures; not the trigonometric functions.
(In Topic 6, we will solve right triangles the ratios of whose sides we do not know.)
Problem 2. In an isosceles right triangle, the hypotenuse is inches. How long are the sides?
The student should sketch the triangles and place the ratio numbers.
Again, those triangles are similar. The side has been multiplied by . Therefore, the remaining sides will be multiplied by . Each of the equal sides is 1 = .
Problem 3. Inspect the values of 30°, 60°, and 45° -- that is, look at the two triangles --
-- and in each equation, decide which of those three angles is the value of x.
a) sin x = cos x. x = 45°. b) tan x = 1. x = 45°. c) sin x = ½. x = 30° d) cos x = ½. x = 60°. e) sin x = ½. x = 60° f) cos x = ½. x = 45°.
x = 30°. h) csc x = 2. x = 30°.
45° = π4 . 60° = π3 . 30° = π6 .
Problem 5. Evaluate each of the following.
a) tan π4 = 1 b) cos π3 = ½ c) sin π6 = ½
d) sin π4 = ½ e) cos π4 = ½ f) csc π6 = 2
g) tan π3 = h) cot π6 = i) sec π4 =
Next Topic: Solving Right Triangles
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# How to Solve Division Problems
We hope that the ideas shared in this article will help you and your students to succeed in this potentially challenging topic. If you do find these tips helpful, check out what Happy Numbers offers its users by setting up your class and starting a free trial which is available only this week!
Learn the parts of a division problem and how to solve them in a few easy steps.
You have 20 cookies and 10 friends. How many cookies should you give each of your friends?
This is a basic division problem.
Division is one of the four basic operations: addition, subtraction, and multiplication are the other three.
Division is a simple operation in which a number is divided. It’s easiest to think of it as a number of objects being divided among a certain number of people, such as in the above example. Of course, you always want to give the same amount to each person to be fair! That’s basically how division works, you divide numbers into equal groups of numbers.
So, how can you solve a division problem? First, you have to know the parts of a division problem.
## Parts of a Division Problem
There are three main parts to a division problem: the dividend, the divisor, and the quotient.
The dividend is the number that will be divided. The divisor is the number of “people” that the number is being divided among. The quotient is the answer.
## How to Solve Division Problems
Solving simple division problems is closely linked to multiplication. In fact, to check your work, you’ll have to multiply the quotient by the divisor to see if it equals the dividend. If it doesn’t, you’ve solved incorrectly.
Let’s try solving one simple division problem. For example:
12 ÷ 2 =
In this problem, you can see how Happy Numbers helps children visualize the problem. There are 12 oranges. There are 2 placed in each box. How many boxes are there?
The answer is 6.
You can check the answer by multiplying the quotient, 6, by the divisor, 2, (6 x 2) which gives us 12. So, the answer is correct.
## What Is a Remainder in Math?
You may have heard of a remainder and wondered, what is a remainder in math?
A remainder in math is used when a division problem doesn’t come out evenly. For example:
11 ÷ 4 =
As you can see in the above example of tennis balls, first, the balls are divided into groups of 4. However, after making 2 groups of balls, 3 balls are left that can’t make a group of 4. These are the remainder. So, the quotient is 2 (2 groups of 4 can be made) and the remainder is 3.
To check the work, multiply the quotient, 2, times the divisor, 4. The answer is 8. Then, add the remainder of 3. The answer is 11, which was the original dividend, so the answer is correct.
Division can get more and more complicated as the numbers get bigger. Then, you must use strategies such as long division, estimation, and more to determine the answers. However, with these basic steps, you can solve just about any division problem.
These are just a few examples of how Happy Numbers incorporates the ideas of small- group instruction into its curriculum. To find out more and try them out, register as a teacher on the website to start a free trial period which is available only this week! |
# How To Count The Limits
## Video: How To Count The Limits
In textbooks on mathematical analysis, considerable attention is paid to techniques for calculating the limits of functions and sequences. There are ready-made rules and methods, using which, you can easily solve even relatively complex problems on the limits.
## Instructions
### Step 1
In mathematical analysis, there are the concepts of the limits of sequences and functions. When it is required to find the limit of a sequence, it is written as follows: lim xn = a. In such a sequence of the sequence, xn tends to a, and n tends to infinity. A sequence is usually represented as a series, for example:
x1, x2, x3…, xm,…, xn….
Sequences are subdivided into ascending and descending sequences. For example:
xn = n ^ 2 - increasing sequence
yn = 1 / n - decreasing sequence
So, for example, the limit of the sequence xn = 1 / n ^ 2 is:
lim 1 / n ^ 2 = 0
x → ∞
This limit is equal to zero, since n → ∞, and the sequence 1 / n ^ 2 tends to zero.
### Step 2
Usually, the variable x tends to a finite limit a, moreover, x is constantly approaching a, and the value of a is constant. This is written as follows: limx = a, while n can also tend to both zero and infinity. There are infinite functions, for which the limit tends to infinity. In other cases, when, for example, a function describes the deceleration of a train, we can talk about a limit tending to zero.
Limits have a number of properties. Typically, any function has only one limit. This is the main property of the limit. Their other properties are listed below:
* The sum limit is equal to the sum of the limits:
lim (x + y) = lim x + lim y
* The product limit is equal to the product of the limits:
lim (xy) = lim x * lim y
* The quotient limit is equal to the quotient of the limits:
lim (x / y) = lim x / lim y
* The constant multiplier is taken out of the limit sign:
lim (Cx) = C lim x
Given a function 1 / x with x → ∞, its limit is zero. If x → 0, the limit of such a function is ∞.
There are exceptions to these rules for trigonometric functions. Since the sin x function always tends to unity when it approaches zero, the identity holds for it:
lim sin x / x = 1
x → 0
### Step 3
In a number of problems, there are functions in the calculation of the limits of which an uncertainty arises - a situation in which the limit cannot be calculated. The only way out of this situation is to apply the L'Hôpital rule. There are two types of uncertainties:
* uncertainty of the form 0/0
* uncertainty of the form ∞ / ∞
For example, a limit of the following form is given: lim f (x) / l (x), moreover, f (x0) = l (x0) = 0. In this case, an uncertainty of the form 0/0 arises. To solve such a problem, both functions are subjected to differentiation, after which the limit of the result is found. For uncertainties of the form 0/0, the limit is:
lim f (x) / l (x) = lim f '(x) / l' (x) (as x → 0)
The same rule is valid for ∞ / ∞ uncertainties. But in this case the following equality is true: f (x) = l (x) = ∞
Using L'Hôpital's rule, you can find the values of any limits in which uncertainties appear. A prerequisite for
volume - no errors when finding derivatives. So, for example, the derivative of the function (x ^ 2) 'is 2x. From this we can conclude that:
f '(x) = nx ^ (n-1) |
UPSC > Set Theory - 2
# Set Theory - 2 Video Lecture - CSAT Preparation - UPSC
## CSAT Preparation
197 videos|151 docs|200 tests
## FAQs on Set Theory - 2 Video Lecture - CSAT Preparation - UPSC
1. What is set theory?
Ans. Set theory is a branch of mathematical logic that deals with the study of sets, which are collections of distinct objects. It provides a foundation for mathematical analysis and various other areas of mathematics.
2. What are the fundamental concepts of set theory?
Ans. The fundamental concepts of set theory include sets, elements, empty set, universal set, subset, power set, intersection, union, and complement. These concepts form the basis of all set operations and relationships.
3. How are sets represented in set theory?
Ans. In set theory, sets are typically represented using braces { } and commas to separate the elements. For example, a set containing the numbers 1, 2, and 3 can be represented as {1, 2, 3}. Sets can also be defined using set builder notation, which specifies the properties or conditions that define the elements of the set.
4. What is the difference between a subset and a proper subset?
Ans. A subset is a set that contains all the elements of another set, including the possibility of being equal to the other set. On the other hand, a proper subset is a subset that does not include all the elements of the other set. In other words, a proper subset is a subset that is not equal to the other set.
5. How are set operations performed in set theory?
Ans. Set operations in set theory include union, intersection, and complement. The union of two sets A and B is a set that contains all the elements that are in either A or B. The intersection of two sets A and B is a set that contains all the elements that are in both A and B. The complement of a set A with respect to a universal set U is a set that contains all the elements in U that are not in A. These operations can be performed using various mathematical symbols and notations.
## CSAT Preparation
197 videos|151 docs|200 tests
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### How do I solve Hannah’s sweet question?
This refers to a question in the Edexcel GCSE paper this year which took students to Twitter venting their frustration. The question is as follows:
‘There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow.
Hannah takes a random sweet from the bag. She eats the sweet.
Hannah then takes at random another sweet from the bag. She eats the sweet.
The probability that Hannah eats two orange sweets is 1/3.
Show that n² – n – 90 = 0’
Seemingly out of nowhere you’re asked to prove that a certain quadratic equation holds using the information provided. The first three lines set up the situation whilst the fourth line provides you with some extra information to use to obtain the answer. Intuition should tell you that you need to calculate the probability that Hannah eats two orange sweet using the first three lines and then apply what you’re given in the fourth line.
So let’s do that. What’s the probability that the first sweet she eats from the bag is orange? There are n sweets in the bag, 6 of which are orange. So the probability is 6/n.
What’s the probability that the second sweet she eats from the bag is orange? Now there are n-1 sweets in the bag, 5 of which are orange (since she has eaten an orange sweet!). So the probability is 5/(n-1).
These two events are separate from one another, so the probability that both happen (i.e. both the sweets are orange) are the two probabilities multiplied together: 6/n × 5/(n-1) = 30/n(n-1)
But you’re told that this probability is 1/3! So all you need to do is set the expression equal to 1/3, rearrange and (hopefully!) obtain the required quadratic equation.
30/n(n-1) = 1/3
⇒ 90/n(n-1) = 1 (multiplying both sides by 3)
⇒ 90 = n(n-1) (multiplying both sides by n(n-1))
⇒ n(n-1) – 90 = 0 (subtracting 90 from both sides)
⇒ n² – n – 90 = 0 (expanding the brackets)
Tah-dah. We’ve found the required equation and we’re done. This question was only worth three marks; a bit stingy in my opinion!
2 years ago
Answered by George, a GCSE Maths tutor with MyTutor
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#### Posts by George
Given that a and b are distinct positive numbers, find a polynomial P(x) such that the derivative of f(x) = P(x)e^(−x²) is zero for x = 0, x = ±a and x = ±b, but for no other values of x.
How do I solve Hannah’s sweet question?
Let P(z) = z⁴ + az³ + bz² + cz + d be a quartic polynomial with real coefficients. Let two of the roots of P(z) = 0 be 2 – i and -1 + 2i. Find a, b, c and d.
What values of θ between 0 and 2π satisfy the equation cosec(θ) + 5cot(θ) = 3sin(θ)?
#### Other GCSE Maths questions
What is completing the square and how do you do it?
A ladder 6·8m long is leaning against a wall. The foot of the ladder is 1·5m from the wall. Calculate the distance the ladder reaches up the wall. Give your answer to a sensible degree of accuracy.
Solve x^2 - 10x + 21 = 0 through the following methods: Factorisation, Completing the Square and using the Quadratic Formula
1/4 of a number is 20. What is 5 times the number?
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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 11 Maths Textbook Solution.
Answer: Function is continuous but not differentiable at x = c
Hint: To check whether the function f(x) is differentiable at point x=c, we need to check that $f'\left ( c^{-} \right )=f'\left ( c^{+} \right )$
Given: $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} (x-c) \cos \left(\frac{1}{x-c}\right), \mathrm{x} \neq \mathrm{c} \\ 0, x=c \end{array}\right.$
Solution:
It is given that, f(c) = 0.
Consider,
$\lim _{x \rightarrow c}(x-c) \cos \left(\frac{1}{x-c}\right)$
Substituting x – c = y, we get:
\begin{aligned} \lim _{y \rightarrow 0} y \cos \left(\frac{1}{y}\right) &=\lim _{y \rightarrow 0} y \cdot \lim _{y \rightarrow 0} \cos \left(\frac{1}{y}\right) \\ &=0 \cdot \lim _{y \rightarrow 0} \cos \left(\frac{1}{y}\right) \end{aligned}
$=0$
So,$\lim_{x\rightarrow c}f\left ( x \right )=f\left ( c \right )=0$
Thus, f(x) is continuous at x = c.
Now, we need to check differentiability at x=c.
Using the formula, we have:
$f'\left ( c \right )=\lim_{x\rightarrow c}\frac{f\left ( x \right )-f(c)}{x-c}$
So, LHD is given by,
$f'\left ( c \right )=\lim_{x\rightarrow c}\frac{f\left ( x \right )-f(c)}{x-c}$
$=\lim_{x\rightarrow c}\frac{\left ( x-c \right )\cos \left ( \frac{1}{x-c} \right )-0}{x-c}$
$=\lim_{x\rightarrow c}\cos \left ( \frac{1}{x-c} \right )$
Put x = c – h
$f'\left ( c \right )=\lim_{x\rightarrow c}\cos \left ( \frac{1}{c-h-c} \right )$
$=\lim_{x\rightarrow c}\cos \frac{1}{h}$ $\left [ \cos \left ( \Theta \right )=\cos \Theta \right ]$
Since the value of cos is going to infinity, its limit will oscillate between -1 and 1.
Now, R.H.D is given by,
\begin{aligned} \mathrm{f}^{\prime}\left(\mathrm{c}^{+}\right) &=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c} \\ &=\lim _{x \rightarrow c^{+}} \frac{(x-c) \cos \left(\frac{1}{x-c}\right)-0}{x-c} \\ &=\lim _{h \rightarrow 0} \cos \left(\frac{1}{c-h-c}\right) \\ &=\lim _{h \rightarrow 0} \cos \left(\frac{1}{h}\right) \end{aligned}
Since the value of cos is going to infinity, its limit will oscillate between -1 and 1. As the value of limit is not a finite value, the function is not differentiable. |
# MATHNASIUM #MATHTRICKS: PROBABILITY (PART 3)
Apr 25, 2023 | Woodmore
Welcome to Mathnasium’s Math Tricks series. Today we are calculating the conditional probability of one event occurring, given that another has already occurred.
Two events are considered dependent if the occurrence of one event affects the probability of the other event. For example, if we draw a red marble from a bag of multicolored marbles and do not replace it, the probabilities of drawing different colored marbles change; so, drawing marbles of particular colors from a bag without replacing them are dependent events.
The probability of a dependent event, B, occurring given that another event, A, has already occurred is called a conditional probability, and is denoted as: P(B | A). Conditional probabilities can change based on some event that has already occurred.
Follow the example below to find the conditional probability.
##### Example: A jar contains 8 quarters, 6 dimes, and 10 pennies. If a quarter is removed from the jar and not replaced, what is the probability of then randomly drawing another quarter?
Step 1: Identify the two dependent events.
The two dependent events are “drawing a quarter” and “drawing a quarter.” We want to find the probability of drawing a quarter given that we have already drawn one quarter from the jar:
P(quarter | quarter), or P(Q | Q).
Step 2: Find the probability of the first dependent event.
The probability of drawing a quarter is: P(Q) = 824 = 13.
Step 3: Find the conditional probability of the second event occurring, given that the first event occurred.
The probability of drawing a quarter, given that you already drew a quarter is: P(Q | Q) = 723. |
Error and Relative Error of Approximations
# Error and Relative Error of Approximations
In many applications of mathematics, we are often interested in knowing accuracy of an approximate value $x_A$ base on its error and relative error from the true value $x_T$. We will define these terms below.
Definition: The Error or Absolute Error between the true value $x_T$ and the approximate value $x_A$ is denoted $\mathrm{Error}(x_A) = x_T - x_A$. The Relative Error between the true value $x_T$ and the approximate value $x_A$ is denoted $\mathrm{Rel}(x_A) = \frac{x_T - x_A}{x_T}$ provided that $x_T \neq 0$.
Some people define the error $x_A$ and $x_T$ to be $\mathrm{Error}(x_A) = \mid x_T - x_A \mid$ and the relative error between $x_A$ and $x_T$ to be $\mathrm{Rel}(x_A) = \frac{\mid x_T - x_A \mid}{\mid x_T \mid}$.
It should be noted that the error of $x_A$ is going to be how far off $x_A$ is from $x_T$, meanwhile, the relative error of $x_A$ is going to be a percentage of how far off $x_A$ is relative to $x_T$. For example, suppose that $x_T = 4$ and $x_A = 3.5$. Then the error is $\mathrm{Error} \left ( 3.5 \right) = 0.5$, while the relative error is $\mathrm{Rel} \left ( 3.5 \right ) = \frac{0.5}{4} = 0.125$. Now consider the example $x_T = 100$ and $x_A = 99.5$. Then the error is $\mathrm{Error} \left ( 99.5 \right ) =0.5$, while the relative error is $\mathrm{Rel} \left ( 99.5 \right ) = \frac{0.5}{100} = 0.005$.
In the two examples above, notice that the errors between the true and actual values are equal, however, the relative errors are much different. Thus, the relative error gives us a more concise picture of how accurate $x_A$ is from $x_T$.
Let's look at some examples of finding the error and relative errors of approximate values $x_A$ from true values $x_T$.
## Example 1
Compute the error and relative error when $x_T = 0.3320$ and $x_A = 0.3321$.
Using the formula above, we have that the error of $x_A$ from $x_T$ is:
(1)
\begin{align} \quad \mathrm{Error} (x_A) = x_T - x_A = 0.3320 - 0.3321 = -0.0001 \end{align}
Now the relative error of $x_A$ from $x_T$ is:
(2)
\begin{align} \quad \mathrm{Rel} (x_A) = \frac{x_T - x_A}{x_T} = \frac{-0.0001}{0.3320} \approx -0.00030120... \end{align}
## Example 2
A polynomial $p(x)$ is used to approximate a function $f(x)$ that is difficult to compute at $x = 0$. We find that $p(0) = 9.1111334$ and the true value $f(0) = 9.1122114$. Find the error and relative error of $p(0)$ from $f(0)$.
Using the formula above, we have that the error of $p(0)$ (our approximate value) from $f(0)$ (the actual value) is:
(3)
\begin{align} \quad \mathrm{Error} (p(0)) = f(0) - p(0) = 9.1122114 - 9.1111334 = 0.001078 \end{align}
Now the relative error of $p(0)$ from $f(0)$ is:
(4)
\begin{align} \quad \mathrm{Rel} (p_0) = \frac{f(0) - p(0)}{f(0)} = \frac{0.001078}{9.1122114} \approx 0.0001183... \end{align}
Note that Example 2 above is not a realistic scenario. If we knew the value of $f$ at $x = 0$, then there would be no reason to approximate $f$ with a polynomial $p$. We will see later that sometimes we do not know $f$ directly, and instead, we may be able to determine the maximum error that an approximate can have from the true value. |
## Fractional Clothesline
• Lesson
6-8
1
In this lesson, a string will be stretched across the classroom and various points will be marked for 0, 1, 2, 3, and 4. This classroom number line will be used to show that all proper fractions are grouped between 0 and 1, and that improper fractions or mixed numbers are all grouped above 1. Students clip index cards with various proper fractions, improper fractions, and mixed numbers on the clothesline to visually see groupings. Students then play an estimation game with groups using the same principle. Encouraging students to look at fractions in various ways will help foster their conceptual fraction sense.
Preparing for the Lesson
Several practice fractions have been provided on the second page of the Fractional Clothesline activity sheet. Create a class set of the fractions by gluing each fraction to an index card. You may want to adapt and use different fractions that your class has been struggling with or that you wish to add to extend your number line.
Fractional Clothesline Activity Sheet
Additionally, create index cards for the number 0, 1, 2, 3, and 4, and stretch a string across the length of your classroom, attaching the ends to opposite walls with tape.
Student Number Line
Students should work in pairs or groups to allow for discussions. Pass out the 0, 1, 2, 3, and 4 index cards to any 5 students in your class. Have those students place their cards on the clothesline. Each of these whole numbers should be spaced an equal distance apart. As students place them, one or more of the cards may need to be adjusted to appear equally spaced. As a class, discuss whether the cards were spaced evenly and are agreed upon. The string and number line will cover the length of your classroom and should have several feet between the whole numbers. This will allow for other cards, representing fractions, to be placed on the number line.
Distribute the prepared fraction cards (one per student/pair/group). Once all the students have a card, tell them to arrange themselves in number line order, similar to the string. They will need to be in order from least to greatest, according to the value on their card. During this part, students will have to apply various strategies to compare and order themselves. You may remind students of strategies you’ve used to compare fractions. Several of the cards do not have common denominators and students will have to decide which is larger through common numerators, or based on decimal equivalents. Some students may organize themselves realizing that 3/4 is actually three 1/4's and figure out where to be. You may also have students use a calculator to verify their answers. Question students to help them explore their thinking. This portion of the lesson can be done quickly or, if you wish, extended to incorporate maximum comparison strategies, depending on students' experience with multiple fractions with different denominators.
In their groups, have students discuss briefly where the cards should be pinned on the clothesline. Once the groups have come to some conclusions, have the members of one group come up to the clothesline and pin up their index cards with paper clips or clothespins. For students who have equivalent numbers (e.g., an improper fraction that is equal to a mixed number), clip the cards together in a column so they are both in the same spot and still visible. Once all cards have been placed, have a whole-class discussion to decide as a group if any cards need to be moved, changed, or adjusted. Ask the class, “Do these cards look like they are in the proper spot? Are there any cards that need to be moved?”
As students are placing the cards, there are several ways to compare fractions. You can point out that fourths are dividing the distance from 0 to 1 into 4 equal lengths. This can also be done for thirds, fifths, etc. Then discuss how 2/3 compares to 3/4 or other fractions by using lengths of wholes instead of the usual circles and squares.
Ask students, "What strategies did you use to decide if a fractional card is closer to 0, 1/2, or a whole number?"
[Each multiple of the denominator is a whole number. If the numerator is close to the next multiple, it’s close to another whole value.]
Repeat this process for each group until all the index cards are placed. It is important to note that as more students begin to place cards, it will be come more obvious when previously placed cards are not in the correct spot. This is where discussion through the process can be invaluable. Lead your students in identifying whether some cards should be moved slightly so that the card currently being added will be placed accurately. Have students continue to adjust the cards on the clothesline until they are reasonable estimates.
Ask students, "What type of fractions are on the numberline that are greater than 1 whole?"
[Mixed numbers are greater than 1 whole and are a way of writing numbers so the whole and the part are seen separately.]
They may also notice that several mixed numbers and improper fractions overlap. It is important to have students use the words equivalent or equal when discussing this.
Clothesline Addition
The Clothesline Addition Game is a way for students to combine different amounts that are represented on the clothesline using estimation. To play the game, distribute the Fractional Clothesline Activity Sheet, and allow time for students to cut out the fraction cards on the page 2. Students should start the game by flipping over 2 cards from a face-down pile of fraction cards, and then estimating what they think the sum of the 2 cards will be. They should place a marker or penny on their sheet that shows where they think the sum is. For example, if they add 1/2 and 13/4, they will place their marker close to 21/4. Remind students to look at the lengths each fraction is represented by on the clothesline and to picture those lengths being placed end to end to create the sum. Once they agree on the answers, they repeat the process. If any students struggle, have them use calculators to double check their estimations. You may have students perform the operations according to a time limit or until the deck runs out of cards. As you move through the classroom, help students find exact answers for themselves if there are any disagreements.
Concluding the Lesson
After students have played several rounds of the Clothesline Addition Game, ask them to summarize the lesson and lead them in a discussion about their strategies for the game. Ask probing questions to find out how they thought of adding the fractions together.
Some key questions may include:
• How are you combining fractions with other fractions?
[Some students may see the fractions as lengths on the number line; some may see decimals; some may create equivalent fractions.]
• What strategy are you using when you have to add a mixed number with a proper fraction?
[Some may create improper fractions; some may estimate the fractions and whole numbers separately.]
• How can you tell if a proper fraction is closer to 1 whole or closer to 0?
[The closer the numerator value is to the denominator, the closer to 1 it becomes.]
• In an improper fraction, how can you tell which whole number it is closer to?
[Each multiple of the denominator is a whole number. If the numerator is closer to the next multiple, then it is closer to the next whole value. For example, in 5/4, the numerator is between 4 and 8, two multiples of the denominator. Since it is closer to 4 than to 8, it is closer to the whole number 1 than the whole number 2.]
• Are there any patterns you see in the placement of fractions on the number line?
[All proper fractions are greater than 0 and less than 1. Fractions beyond 1 involve a whole number and are actually mixed numbers or are improper fractions.]
Some students may think of using equivalent fractions while others may think of using lengths on the clothesline. Allow students to express their strategies and the patterns they noticed. It might also prove a useful tool to leave the clothesline up for a while following the lesson as a visible tool in the room.
• String long enough to stretch across classroom
• Index cards
• Paper clips or clothespins
• Calculator (optional)
• Scissors
• Pennies (or other markers)
• Envelopes to hold the cards
• Fractional Clothesline Activity Sheet
Assessments
1. As an entry task the next day, place the original 5 index cards on the clothesline. Give each student an index card and tell them to write a number on their card, that is not a whole number. Have students take turns placing another student's card on the line, discussing if it is in the correct spot.
2. In whole-class discussions, ask students to describe how they discovered their patterns. Encourage and validate a variety of appropriate responses.
3. Have students place the game cards in order from least to greatest on their desks.
4. Write on the board numbers that are not included in the fraction cards. Ask students to mark on their number lines where each number belongs.
5. Have students collaborate to write down situations in real life where they have used the numbers that are on the clothesline and relate how comparing or adding those numbers might apply to those situations.
Extensions
1. As students progress in their experiences, include decimals on the clothesline and in the game cards. As the year progresses, include percentages.
2. Create a clothesline for positive and negative integers. Have students play a game where they add positive and negative integers on a modified clothesline.
3. Place several fractions on the clothesline. Then, have students place the whole numbers.
4. Students can play the Fraction Game.
Fraction Game
This tool allows students to individually practice working with relationships among fractions and ways of combining fractions. The object of this game is to get all of the markers to the right side of the game board, using as few cards as possible. This activity is a good extension of today's lesson, as it relies on number lines divided into specific fractional values.
Questions for Students
1. What is a fraction?
[Fractions are parts of wholes, whether part of a circle, part of a square, or part of a straight line. There is an amount that represents a whole piece (the denominator) and a fraction is a part (the numerator) of that whole.]
2. What is an improper fraction?
[Improper fractions are fractions with a numerator greater than the denominator. They are more than 3. 1 whole and include fractions of another whole.]
3. What is a mixed number?
[Mixed numbers are fractions that have regrouped the fractions pieces into a whole number and a proper fraction, which is less than another whole piece.]
4. How can you combine 13/4 with 11/2?
[Some students may point out an algorithm and actually add. Others may use decimals or consider adding the lengths of the line segments on the clothesline.]
Teacher Reflection
• Was your lesson developmentally appropriate? If not, what was inappropriate? What would you do to change it to make it more/less challenging?
• How did your lesson address auditory, tactile, and visual learning styles?
• Did you find it necessary to make adjustments while teaching the lesson? If so, what adjustments did you make? Were they effective?
• What worked with classroom behavior management? What didn't work? How would you change what didn’t work?
• What are alternative ways you can have students look at fractions to help their overall conceptual understanding?
### Fraction Game
3-5, 6-8
This applet allows students to individually practice working with relationships among fractions and ways of combining fractions.
### Learning Objectives
Students will:
• Create a number line with fractions, improper fractions, mixed numbers, and integers.
• Use estimation to practice combining various numbers. |
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# Vectors Definition. Main information
## Vector definition
Definition. Vector is a directed line segment, ie the segment having a length and a definite direction. Graphically vector depicted as a directed line segments of a certain length. (Fig. 1)
Fig. 1
## Vector designation
The vector which has a beginning point A and end point B, denoted AB (Fig. 1). Also, the vector represent one small letter, for example a.a.
## Vector length
Definition. The length of the directed segment determines the numerical value of the vector and is called the length of the vector AB.
The length of the vector AB is denoted as: |AB|.
## Zero vector
Definition. Zero vector is a vector whose start and end points coincide.
The zero vector is usually is denoted as 0.
The length of the zero vector is zero.
## Collinear vectors
Definition. Vector parallel to one line or lying on one line are called collinear vectors (Fig. 2).
Fig. 2
## Codirected vectors
Definition. Two collinear vectors a and b are called codirected vectors if their directions are the same: a↑↑b (Fig. 3).
Fig. 3
## Oppositely directed vectors
Definition. Two collinear vectors a and b are called oppositely directed vectors if their directions are opposite: a↑↓b (Fig. 4).
Fig. 4
## Coplanar vectors
Definition. Vectors parallel to the same plane, or lie on the same plane are called coplanar vectors (Fig. 5).
Fig. 5
It is always possible to find a plane parallel to the two random vectors, in that any two vectors are always coplanar.
## Equal vectors
Definition. Vectors a and b is an equal vectors if they are in the same or parallel lines, their directions are the same and the lengths are equal (Fig. 6).
Fig. 6
Two vectors are equal if they are collinear, codirected and have the same length:
a = b, if a↑↑b and |a| = |b|.
## Unit vector
Definition. Unit vector or orth is a vector whose length is equal to one. |
# Mixed Word Problem5
In this page mixed word problem5 you can find the question as well as solution for this question with detailed explanation.Like this you can see every problems with detailed solution and explanation.
Question:
The total weight of 4 girls is 155 ⅕ . If three of them weighs 20 ⅕ kg,44 ½ kg, and 30 ⅓ kg.Find the weight o the remaining girl.
Basic idea:
In this problem we have total weight of 4 girls that is 155 ⅕ .If the weight of three girls is 20 ⅕ kg,44 ½ kg, and 30 ⅓ kg.Now we need to find the weight o the remaining girl.So let u consider the weight of the remaining girl is x.If we add the given three weights with x we will get 155 ⅕ .Then we can easily solve for x.
Total weight of 4 girls = 155 ¼
Sum of the weight of three girls = (20 ⅕ + 44 ½ + 30 ⅓ )
= (100+1)/5 + (88+1) /2 + (90+1)/3.
= 101/5 + 89/2 + 91/3
The denominators of the three fractions are not same.So we have to take L.C.M
Here L.C.M is 5 x 2 x 3 = 30
= (101/5) x (6/6) + (89/2) x (15/15) + (91/3) x(10/10)
= 606/30 + 1335/30 + 910/30
= (505 + 1335 + 910) / 30
= 2851 /30
Weight of remaining girl = 155 ⅕ - 2851 /30
= (620+1) /5 - 2851/30
= 621/5 - 2851/30
L.C.M = 30
= (606/5) x (6/6) -(2851/30)
= (3636 -2851)/30
= 785/30 |
Home » Math Theory » Graphs and Charts » Graphing Lines in Slope-Intercept Form
# Graphing Lines in Slope-Intercept Form
## Introduction
Graphing lines in slope-intercept form is a fundamental skill in mathematics, a way to visually represent linear equations and understand the relationship between two variables. We often encounter these lines in algebra and geometry, and knowing how to graph them is essential for success in these areas.
## Grade Appropriateness
Understanding and graphing lines in slope-intercept form is most appropriate for students in middle school, roughly between grades 6 and 8. However, these concepts can be introduced to more advanced students in grade 5 and revisited throughout high school math courses.
## Math Domain
The domain for this topic is primarily Algebra, but it also has applications in Geometry, Calculus, and even beyond in more advanced mathematical disciplines.
## Applicable Common Core Standards
Here are a few of the Common Core Standards this topic addresses:
CCSS.MATH.CONTENT.8.F.B.4: Construct a function to model a linear relationship between two quantities.
CCSS.MATH.CONTENT.8.SP.A.1: Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities.
## Definition of the Topic
The Slope Intercept Form of a line is a way to express a linear equation. Y = mx + b is the formula for the slope-intercept form, where m denotes the slope of the line and b denotes the y-intercept (the point at which the line crosses the y-axis).
## Key Concepts
Slope: The slope (m) measures how steep the line is. If the slope is positive, the line rises as it moves to the right. If the slope is negative, the line falls as it moves to the right.
Y-intercept: The point where the line crosses the y-axis is known as the y-intercept (b). It is y’s value when x is equal to 0.
## Discussion with Illustrative Examples
### Types of Slopes
The slope is rise over run written as a fraction. If slope (m) is a whole number, you can use one (1) as the denominator to form a fraction.
A positive slope means the line will go uphill from left to right. A negative slope means the line will go downhill from left to right. A zero slope means that the line is horizontal. An undefined slope means that the line is vertical.
### y-intercept
The y-intercept (b) is where the line crosses the y-axis. The point (0, b) are the coordinates for the y-intercept.
### Graphing Lines in Slope-Intercept Form
The graph of an equation in a slope-intercept form is a straight line. Lines can be horizontal, vertical, or diagonal.
The following are the steps to consider in graphing lines in slope-intercept form:
Identify the y-intercept (b) and locate the point on the y-axis.
Identify the slope (m), rise over run.
Count the rise over run from the point on the y-axis to identify another point.
Draw a line through the two points.
For example, we graph the line for the equation y =2x + 3.
The slope (m) is 2 or $\frac{2}{1}$ in fraction form, which means the line rises two units for every 1 unit it moves to the right. The y-intercept (b) is 3, so the line crosses the y-axis at the point (0,3).
We first plot the y-intercept at the point (0,3) to graph this line.
From there, we use the slope to determine where to plot the next point. We move two units up and 1 unit to the right and plot a point.
And then, we draw a line through the points.
## Examples with Solutions
Example 1
Graph the line for the equation y=$\frac{2}{3}$x-4.
Solution
The slope (m) is positive $\frac{2}{3}$, meaning that the line rises two units for every three units it moves to the right. The y-intercept is at the point (0,-4) or b=-4.
Example 2
Graph the line for the equation y = -x + 2.
Solution
The slope (m) is -1 or -$\frac{1}{1}$ in fraction form. Since the slope is negative, the line falls 1 unit for every 1 unit moving to the right. The y-intercept (b) is 2, so the line crosses the y-axis at the point (0,2).
First, plot the y-intercept at (0,2), then move down 1 unit and 1 unit to the right to plot the next point.
Example 3
Graph the line for the equation y = 3x+1.
Solution
The slope (m) is positive 3, so the line rises three units for every 1 unit it moves to the right.
The y-intercept (b) is 1, so the line crosses the y-axis at the point (0,1).
Plot the y-intercept at (0,1) first, then move three units up and one unit to the right to plot the following point.
## Real-life Application with Solution
Problem
John’s lemonade stand is earning money. He makes \$5 per glass sold. He already had \$10 to start. How much money does he have after selling x glasses of lemonade?
Solution
The problem can be modeled as a linear equation y = 5x + 10, where y is the total amount of money John has, and x is the number of glasses sold.
The slope (m) is 5, meaning he earns \$5 for each glass sold. The y-intercept (b) is 10, meaning he started with \$10.
If John sells four glasses of lemonade, he will have,
y = (5×4) + 10
y= 20+10
y= 30
## Frequently Asked Questions (FAQs)
### What if the y-intercept (b) is zero?
If the y-intercept is zero, the line crosses the y-axis at the origin (0,0).
### What does a slope of zero mean?
A slope of zero means the line is horizontal. No matter what value x is, y will always be the same.
### What if a linear equation does not have a y-term?
If the equation does not have a y-term, the line is vertical and has no defined slope.
### What happens when the slope is undefined?
When the slope is undefined, it usually means that we have a vertical line that does not intercept the y-axis at any point other than, possibly, y = 0.
### What is the relationship between the sign of the slope and the direction of the line?
If the slope is positive, the line goes up from left to right. If the slope is negative, the line goes down from left to right.
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
Solve for $y$ by dividing both sides of the equation by $3$: $\frac{3y}{3}=\frac{2}{3} \\y=\frac{2}{3}$ This means that equation $3y=2$ is equivalent to and has the same graph as $y=\frac{2}{3}$. RECALL: The graph of an equation of the form $y=k$ where $k$ is a real number is a horizontal line whose y-intercept is $(0, k)$. The line is parallel to the y-axis and each point on the line has a y-coordinate of $k$. Thus, the equation $y=\frac{2}{3}$ is a horizontal line whose y-intercept is $(0, \frac{2}{3})$. Every point on the line has a y-coordinate of $\frac{2}{3}$. This means the following points are on the line: $(0, \frac{2}{3})$, $(-2, \frac{2}{3})$, and $(2, \frac{2}{3})$ Plot the three points above then connect them using a line to complete the graph. (Refer to the graph in the answer part above.) |
# Kaplan GMAT Sample Problem: Advanced Rates
May 4, 2011 by
Try this advanced sample GMAT problem-solving question focusing on rates. Remember, as complicated as the problems get, the key is using the distance = rate x time formula for all of the specific distances travelled.
Problem:
Shannon and Maxine work in the same building and leave work at the same time. Shannon lives due north of work and Maxine lives due south. The distance between Maxine’s house and Shannon’s house is 60 miles. If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon. If Maxine rides her bike home at the rate of R miles per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine. How far does Maxine live from work?
A) 15
B) 20
C) 30
D) 40
E) 45
Solution:
To solve for the distance between Maxine’s house and the building in which she works, we must first translate our question into a series of equations. Remember that for each leg of a journey, we can use the distance = rate x time relationship. First, we know that the distance between Maxine and Shannon’s houses is 60 miles. If we call the distance between Maxine’s house & work DM, and the distance between Shannon’s house and work Ds, we can write the equation DM + DS = 60. We can also write equations for the three types of trips. For Shannon’s drive, we can say that DS = 2RT. For Maxine’s drive, we can say that DM = 2R(T + 2/3) – Shannon’s time, which is T, plus 2/3 of an hour. For Maxine’s bike ride, we can say DM = R(T + 2). From these equations we can solve for DM.
DM = R(T + 2) and DM = 2R(T + 2/3), so R(T + 2) = 2R(T + 2/3). This can be simplified to RT + 2R = 2RT + 4R/3, which in turn becomes 3RT + 6R = 6RT + 4R. We can then simplify even further to say 3RT = 2R, 3T = 2, T = 2/3. Therefore, we know that it takes Shannon 2/3 of an hour to drive home.
Next, we can use the fact that Ds = 60 – DM. This can be substituted into the equation Ds = 2RT, giving us 60 – Dm = 2RT. Because we know that T = 2/3, we know that 60 = DM + 2(2/3)R. We also know that DM = R(T + 2). Again keeping in mind that T = 2/3, we can substitute and say that 60 = R(2/3 + 2) + 2(2/3)R. We then solve for R:
60 = R(2/3 + 2) + 2(2/3)R
60 = 2R/3 + 2R + 4R/3
180 = 2R + 6R + 4R
180 = 12R
R = 15.
Once we know R is 15, we can plug back into the equation DM = R(T + 2), which becomes DM = 15(2/3 + 2) = 10 + 30 = 40. Therefore, choice (D) is the correct answer.
" " |
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## Transcription
1 cups cups cup Fractions are a form of division. When I ask what is / I am asking How big will each part be if I break into equal parts? The answer is. This a fraction. A fraction is part of a whole. The top part of the fraction is the numerator and the bottom part is the denominator. Since < this is a proper fraction because the numerator is less than the denominator. In an improper fraction, the numerator is greater than or equal to the denominator, such as /, or /. Proper fractions have values less than and improper fractions have values greater than or equal to. whole cup is divided into equal parts. cup is of those equal parts. cup cup cup cup
2 Writing a Fraction in Lowest Terms 8 Notice the same size box is split three different ways in equal parts: in parts, parts, or 8 parts. By looking at the pictures, each box has the same portion (or fraction) filled, so these portions are called equivalent fractions. Fundamental Property of Fractions Multiplying or dividing the numerator AND the denominator of a fraction by the same nonzero number (not 0) does not change the value of the fraction. The fractions are equivalent. a b a a x x a and b x x
3 A Fraction in Lowest Terms Means the numerator and denominator have no common factor other than. The denominator is the lowest it can be. /8 is not in lowest terms because and 8 have common factors: and. goes into both the numerator,, and the denominator, 8, and also goes into both and 8. / is not in lowest terms because and have a common factor:. goes into both the numerator,, and the denominator,. is in lowest terms because there is no number (factor) that goes into both the numerator,, and the denominator,, other than. Do self-check: Are the following fractions in lowest terms? a) / b) /8 c) 9/ d) /
4 Using Greatest Common Factor (GCF) to reduce a fraction to its lowest terms The Greatest Common Factor is the largest number that goes into two numbers. What is the GCF of 0 and? The largest number that goes into both 0 and (the largest divisor of both 0 and ) Both numbers are divisible by. Is the GCF? 0 = 0, and =. But 0 and have common factors, so is not the GCF of 0 and. A bigger number can go into both 0 and. is the GCF of 0 and. 0 =, and =. and have no common factors, so is the GCF. To reduce a fraction to its lowest terms, find the GCF of the numerator and the denominator, and divide both the numerator and denominator by the GCF. 0 0 so is 0 in its lowest terms.
5 Example If you write the prime factorization both the numerator and denominator, the fraction in lowest terms is easily found by cancelling out the common prime factors. Write 90 in lowest terms = = 90 You can only cancel out factors on the bottom (denominator) with corresponding factors on the top (numerator).
6 FRACTIONS A fraction is just a division problem. Numerator Denominator = Numerator Denominator A fraction is in LOWEST TERMS, also known as REDUCED or SIMPLEST FORM when the numerator and denominator have no common factors. 0 is not in lowest terms, is in lowest terms A fraction can be put in reduced form by dividing both the numerator and the denominator by the GCF. This can be done easily by putting both the numerator and denominator in prime-factored form and canceling the common factors between the numerator and denominator. 0 Mixed Number - A number, such as number and a fraction., consisting of an integer and a fraction. A mixed number is just the sum of a whole Improper Fraction - A fraction in which the numerator is larger than the denominator. Converting from mixed number to improper fraction: Mixed numbers must be converted to improper fractions or decimals before doing ANY MULTIPLICATION OR DIVISION OPERATIONS on them. denominato r numerator A fraction can be converted into a decimal by dividing: Numerator Denominator 0 When ADDING or SUBTRACTING FRACTIONS that must have the same denominator, then you just add the numerators and leave the denominator the same. However, consider 8
7 Expressing a fraction in higher terms: Later when we add fractions, we might have to express them in higher terms so that each fraction will have the same denominator. Example: Write / as an equivalent fraction with a denominator of 8.?? Example Write as an equivalent fraction with a denominator of. Is a fraction? We can make it one by writing it as.??
8 Multiplying fractions is easy: you multiply the top numbers and multiply the bottom numbers. For instance: When possible, you reduce. In this case, however, nothing reduces, because 8 and have no factors in common. If you're not sure, you can always do prime factorization: Nothing cancels. Copyright Often, though, something will cancel: Simplify Also, you can simplfiy by cancelling common factors between numerators and denominators. 9 9 Dividing fractions is just about as easy; there's just one extra step. When you divide by a fraction, the first thing you do is "flip-n-multiply". That is, you take the second fraction, flip it upside-down, and multiply it by the first fraction. (The upside-down fraction is called the reciprocal). For instance: Simplify 8
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Question Video: Finding the Unknown Elements of a Matrix given Its Rank | Nagwa Question Video: Finding the Unknown Elements of a Matrix given Its Rank | Nagwa
# Question Video: Finding the Unknown Elements of a Matrix given Its Rank Mathematics
If the matrix π΄ = (6, β9, 1 and π, 24, 4 and 15, 18, β11) has rank 2, what is the value of π?
04:09
### Video Transcript
If the matrix π΄ has rank two, what is the value of π?
In order to solve this question, we will use the fact that the question says that matrix π΄ has a rank of two. And since matrix π΄ is a three-by-three matrix, this tells us that the determinant of π΄ is equal to zero. And this is because if the determinant of π΄ is nonzero, then π΄ has a rank of three. And so we can use the fact that the determinant of π΄ is equal to zero to form an equation involving π, which can be solved to find the value of π.
Letβs now recall how to find the determinant of a three-by-three matrix. So given this matrix, we have that its determinant is equal to π lots of ππ minus πβ minus π lots of ππ minus ππ plus π lots of πβ minus ππ. Letβs quickly look at where this formula comes from.
When finding the determinant of a three-by-three matrix, we know that there will be three terms. And each of these terms will start with their respective entry in the first row of the matrix. And so in our case, thatβs π, π, and π. And we know that the signs of these terms will be positive and then negative and then positive. And so thatβs where we get the minus sign in front of the π from.
Now in order to work out what goes in the bracket of each term, we take the first part β so in the first term, thatβs π β and we cross out everything thatβs in the same row or column as π. And weβre left with a two-by-two matrix, which we then find the determinant of. And we multiply π by this determinant. So this gives us that ππ minus πβ in the brackets.
We then repeat this process for π. Crossing out everything in the same row and column as π leaves us with four entries, which form a two-by-two matrix. And we then find the determinant of this matrix and multiply it by π. And we then repeat this process for π. Crossing out the rows and columns, weβre left with this four-by-four matrix, which we find the determinant of, giving us then πβ minus ππ, which we multiply π by.
And so now weβre ready to use this formula to find the determinant of matrix π΄. And we get that the determinant of π΄ is equal to. We get six lots of 24 times minus 11 minus four times 18 minus negative nine lots of π timesed by negative 11 minus four times 15 plus one lot of π times 18 minus 24 times 15. And we can multiply inside the brackets to leave us with six lots of minus 264 minus 72 plus nine lots of minus 11 times π minus 60 plus 18π minus 360. And now we expand the brackets to leave us with. We get minus 1584 minus 432 minus 99π minus 540 plus 18π minus 360, which simplifies to minus 2916 minus 81π.
And now we have found an equation for the determinant of π΄. But we also have that the determinant of π΄ is equal to zero. So therefore, we can put our equation equal to zero. And then we can solve this for π. We obtain a solution of π is equal to negative 36.
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# When equations have infinite solutions?
An infinite solution has both sides equal. For example, 6x + 2y - 8 = 12x +4y - 16. If you simplify the equation using an infinite solutions formula or method, you'll get both sides equal, hence, it is an infinite solution.
## How do you know if an equation has infinite solutions?
We can identify which case it is by looking at our results. If we end up with the same term on both sides of the equal sign, such as 4 = 4 or 4x = 4x, then we have infinite solutions. If we end up with different numbers on either side of the equal sign, as in 4 = 5, then we have no solutions.
## What is an equation with infinite solutions?
No solution would mean that there is no answer to the equation. It is impossible for the equation to be true no matter what value we assign to the variable. Infinite solutions would mean that any value for the variable would make the equation true.
## What are infinitely many solutions?
Having infinitely many solutions means that you couldn't possibly list all the solutions for an equation, because there are infinite. Sometimes that means that every single number is a solution, and sometimes it just means all the numbers that fit a certain pattern.
## Are infinite solutions consistent?
A system of two linear equations can have one solution, an infinite number of solutions, or no solution. ... If a system has at least one solution, it is said to be consistent . If a consistent system has exactly one solution, it is independent . If a consistent system has an infinite number of solutions, it is dependent .
## One Solution, No Solution, or Infinitely Many Solutions - Consistent & Inconsistent Systems
26 related questions found
### Is it infinite solutions or all real numbers?
When you end up with a true statement like this, it means that the solution to the equation is “all real numbers”. Try substituting x = 0 into the original equation—you will get a true statement! Try , and it also will check! This equation happens to have an infinite number of solutions.
### What's the difference between all real numbers and infinite solutions?
There is no difference. The notation (−∞,∞) in calculus is used because it is convenient to write intervals like this in case not all real numbers are required, which is quite often the case.
### What equations have no solution?
The coefficients are the numbers alongside the variables. The constants are the numbers alone with no variables. If the coefficients are the same on both sides then the sides will not equal, therefore no solutions will occur.
### How do u know how many solutions an equation has?
If solving an equation yields a statement that is true for a single value for the variable, like x = 3, then the equation has one solution. If solving an equation yields a statement that is always true, like 3 = 3, then the equation has infinitely many solutions.
### Which of the following system of equations has infinitely many solutions?
A dependent system has infinitely many solutions. The lines are exactly the same, so every coordinate pair on the line is a solution to both equations.
### Is all real numbers infinity to infinity?
The misconception is in choosing to treat ∞ \infty ∞ as an integer or whole number or as one of the real numbers. ... Infinity is a "real" and useful concept. However, infinity is not a member of the mathematically defined set of "real numbers" and, therefore, it is not a number on the real number line.
### Is 0 0 infinite or no solution?
Since 0 = 0 for any value of x, the system of equations has infinite solutions.
### Can all real numbers be infinity?
Any non-zero real number is either negative or positive. ... The real numbers make up an infinite set of numbers that cannot be injectively mapped to the infinite set of natural numbers, i.e., there are uncountably infinitely many real numbers, whereas the natural numbers are called countably infinite.
Yes, it does.
### What graph has infinite solutions?
Recall that a linear equation graphs as a line, which indicates that all of the points on the line are solutions to that linear equation. There are an infinite number of solutions. If you have a system of linear equations, the solution for the system is the value that makes all of the equations true.
### How do you tell if a system of equations has no solution or infinitely many without graphing?
Two equations have parallel lines (no solution to the system) if the slopes are equal and and y-intercepts are not. Adding the equations gives an obviously false statement. This system of equations has no solution.
### How can a linear system have infinite solutions?
A system of linear equations has no solution when the graphs are parallel. Infinite solutions. A system of linear equations has infinite solutions when the graphs are the exact same line.
### Which equation has only one solution?
A quadratic equation only has one solution if the discriminant is zero.
### Is Pi an infinite?
We still call it Pi Day. ... No matter how big your circle, the ratio of circumference to diameter is the value of Pi. Pi is an irrational number---you can't write it down as a non-infinite decimal. This means you need an approximate value for Pi.
### Is Pi a number?
Regardless of the circle's size, this ratio will always equal pi. In decimal form, the value of pi is approximately 3.14. But pi is an irrational number, meaning that its decimal form neither ends (like 1/4 = 0.25) nor becomes repetitive (like 1/6 = 0.166666...). (To only 18 decimal places, pi is 3.141592653589793238.)
### Is infinity an element of R?
No. If you look up the definition of the real numbers, you will not find any of its elements called "infinity". However, the extended real numbers has two numbers called +∞ and −∞, which become the endpoints of the number line in the extended reals.
### When linear equations have infinitely many solutions lines are?
The system of an equation has infinitely many solutions when the lines are coincident, and they have the same y-intercept. If the two lines have the same y-intercept and the slope, they are actually in the same exact line.
### How many solutions might a system of inequalities have?
A system of linear inequalities can have none, one, or an infinite number of solutions; therefore, there are three. |
## Define and Simplify Rational Expressions
### Learning Outcomes
• Define and simplify rational expressions
• Identify the domain of a rational expression
Rational expressions are fractions that have a polynomial in the numerator, denominator, or both. Although rational expressions can seem complicated because they contain variables, they can be simplified using the techniques used to simplify expressions such as $\frac{4x^3}{12x^2}$ combined with techniques for factoring polynomials. There are a couple ways to get yourself into trouble when working with rational expressions, equations, and functions. One of them is dividing by zero, and the other is trying to divide across addition or subtraction.
## Determine the Domain of a Rational Expression
One sure way you can break math is to divide by zero. Consider the following rational expression evaluated at $x = 2$:
Evaluate $\frac{x}{x-2}$ for $x=2$
Substitute $x=2$
$\begin{array}{l}\frac{2}{2-2}=\frac{2}{0}\end{array}$
This means that for the expression $\frac{x}{x-2}$, $x$ cannot be $2$ because it will result in an undefined ratio. In general, finding values for a variable that will not result in division by zero is called finding the domain. Finding the domain of a rational expression or function will help you not break math.
### Domain of a rational expression or equation
The domain of a rational expression or equation is a collection of the values for the variable that will not result in an undefined mathematical operation such as division by zero. For a = any real number, we can notate the domain in the following way:
$x$ is all real numbers where $x\neq{a}$
The reason you cannot divide any number c by zero $\left( \frac{c}{0}\,\,=\,\,? \right)$ is that you would have to find a number that when you multiply it by $0$ you would get back $c \left( ?\,\,\cdot \,\,0\,\,=\,\,c \right)$. There are no numbers that can do this, so we say “division by zero is undefined”. When simplifying rational expressions, you need to pay attention to what values of the variable(s) in the expression would make the denominator equal zero. These values cannot be included in the domain, so they are called excluded values. Discard them right at the start before you go any further.
(Note that although the denominator cannot be equal to $0$, the numerator can—this is why you only look for excluded values in the denominator of a rational expression.)
For rational expressions, the domain will exclude values where the the denominator is $0$. The following example illustrates finding the domain of an expression. Note that this is exactly the same algebra used to find the domain of a function.
### Example
Identify the domain of the expression. $\frac{x+7}{{{x}^{2}}+8x-9}$
## Simplify Rational Expressions
Before we dive in to simplifying rational expressions, let us review the difference between a factor, a term, and an expression. This will hopefully help you avoid another way to break math when you are simplifying rational expressions.
Factors are the building blocks of multiplication. They are the numbers that you can multiply together to produce another number: $2$ and $10$ are factors of $20$, as are $4, 5, 1, 20$.
Terms are single numbers, or variables and numbers connected by multiplication. $-4, 6x$ and $x^2$ are all terms.
Expressions are groups of terms connected by addition and subtraction. $2x^2-5$ is an expression.
This distinction is important when you are required to divide. Let us use an example to show why this is important.
Simplify: $\dfrac{2x^2}{12x}$
The numerator and denominator of this fraction consist of factors. To simplify it, we can divide without being impeded by addition or subtraction.
$\begin{array}{cc}\dfrac{2x^2}{12x}\\=\dfrac{2\cdot{x}\cdot{x}}{2\cdot3\cdot2\cdot{x}}\\=\dfrac{\cancel{2}\cdot{\cancel{x}}\cdot{x}}{\cancel{2}\cdot3\cdot2\cdot{\cancel{x}}}\end{array}$
We can do this because $\frac{2}{2}=1\text{ and }\frac{x}{x}=1$, so our expression simplifies to $\dfrac{x}{6}$.
Compare that to the expression $\dfrac{2x^2+x}{12-2x}$. Notice the denominator and numerator consist of two terms connected by addition and subtraction. We have to tip-toe around the addition and subtraction. When asked to simplify, it is tempting to want to cancel out like terms as we did when we just had factors. But you cannot do that, it will break math!
Be careful not to break math when working with rational expressions.
In the examples that follow, the numerator and the denominator are polynomials with more than one term, and we will show you how to properly simplify them by factoring. This turns expressions connected by addition and subtraction into terms connected by multiplication.
### Example
Simplify and state the domain for the expression. $\frac{x+3}{{{x}^{2}}+12x+27}$
### Example
Simplify and state the domain for the expression. $\frac{x^{2}+10x+24}{x^{3}-x^{2}-20x}$
We will show one last example of simplifying a rational expression. See if you can recognize the special product in the numerator.
### Example
Simplify $\frac{{x}^{2}-9}{{x}^{2}+4x+3}$ and state the domain.
In the following video, we present additional examples of simplifying and finding the domain of a rational expression.
### Steps for Simplifying a Rational Expression
To simplify a rational expression, follow these steps:
• Determine the domain. The excluded values are those values for the variable that result in the expression having a denominator of $0$.
• Factor the numerator and denominator.
• Cancel out common factors in the numerator and denominator and simplify.
## Summary
An additional consideration for rational expressions is to determine what values are excluded from the domain. Since division by $0$ is undefined, any values of the variable that result in a denominator of $0$ must be excluded from the domain. Excluded values must be identified in the original equation, not from its factored form.Rational expressions are fractions containing polynomials. They can be simplified much like numeric fractions. To simplify a rational expression, first determine common factors of the numerator and denominator then cancel them out. |
Miscellaneous
Chapter 8 Class 11 Sequences and Series
Serial order wise
### Transcript
Misc 4 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. Let the numbers in G.P. be a, ar, and ar2. It is given that Sum of these number is 56 a + ar + ar2 = 56 ar2 = 56 a ar Also, When 1, 7, 21 subtracting from these number respectively, the new numbers are in AP So, (a 1) , (ar 7) , (ar2 21) are in AP Common difference is same (ar 7) (a 1) = (ar2 21) (ar 7) ar a 6 = ar2 ar 14 ar2 ar ar + a 14 + 6 = 0 ar2 2ar + a 8 = 0 From (1) putting ar2 = 56 a ar (56 a ar) 2ar + a 8 = 0 56 a ar 2ar + a 8 = 0 -a + a ar 2ar + 56 8 = 0 0 3ar + 48 = 0 -3ar = -48 ar = ( 48)/( 3) ar = 16 a = 16/ Putting a = (16 )/ in (1) a + ar + ar2 = 56 a(1 + r + r2) = 56 16/ (1 + r + r2) = 56 (16 (1+ + 2))/ = 56 (16 (1+ + 2))/ = 56 16(1 + r + r2) = 56r 16 + 16r + 16r2 56r = 0 16r2 + 16r 56r + 16 = 0 8(2r2 5r + 2) = 0 (2r2 5r + 2) = 0/8 2r2 5r + 2 = 0 2r2 4r r + 2 = 0 2r(r 2) 1(r 2) = 0 (2r 1)(r 2) = 0 2r 1 = 0 or r 2 = 0 2r = 1 or r = 2 r = 1/2 or r = 2 Now, finding numbers Thus the numbers are 8, 16, 32 for r = 1/2 & a =32 & 32, 16, 8 for r = 2 & a = 8 |
# Question Video: Solving Word Problems Involving Addition of Numbers up to 99 Mathematics • 1st Grade
Hannah is 52 years old and Charlotte is 37. What is the sum of their ages?
01:58
### Video Transcript
Hannah is 52 years old and Charlotte is 37. What is the sum of their ages?
First, we need to highlight the key numbers and words in the question. Hannah is 52 and Charlotte is 37. The question asks us to find the sum of the ages.
In math, the word sum means the total of two or more numbers. If we’re finding the total of two or more numbers, do we use plus or minus? Sum is another way of saying add the numbers together.
Now we can write our equation: 52 plus 37. Let’s use column addition and a place value chart to help us calculate the answer. 52 has five tens and two ones, plus 37, 37 has three tens and seven ones.
We always start in the ones column. Here we have to add two plus seven, which makes nine. Next, we move over into the tens column and calculate five plus three, which is eight.
52 plus 37 equals 89. Hannah is 52 and Charlotte is 37. The sum of their ages is 89. When we recognized that sum means add together, we calculated our answer using column addition. |
# By Donna Greenwood
Given line segments j, k, m.If these are the medians of a triangle, construct the triangle
Click here for a sketch to manipulate. You can manipulate the medians, j, k and m to see how it impacts the constructed triangle ABC. This is the “backwards” version of creating a triangle of medians for any given triangle, so that’s the approach I took for the construction.
# Steps to a Triangle Construction
1. Replicate one of the medians (m in this sketch) by drawing a parallel line to segment m through an arbitrary point (C). Construct a circle with radius m at C. One of the intersections of the line with the circle and point C determine the endpoints of m.
2. Create a triangle out of the given segments. At the endpoints of the new segment m, create two circles of radius j and k respectively. One of the intersections of the two circles is the third vertex of the triangle, so pick one.
3. Now we have a triangle whose leg lengths are j, k and m. From now on, references to j, k and m are to those segments, rather than the givens. Let one of the segments of the triangle be one of the medians of the new triangle (again, m in this sketch). The vertex is at point C. Note that if we could use measurement in constructions (which we can’t, of course), then we could simply measure 2/3 of the way along this median to find out where the medians intersect. Instead, construct the centroid of this triangle of medians.
4. Create a parallel at L to segment k in the sketch and replicate the segment (BL in this sketch). The endpoint is at point B in the “new” triangle.
5. By the definition of a median, segment CL is 1/2 the length of the side of the triangle. To get vertex A, construct the ray CL and a circle at L with radius CL. The intersection is the third vertex of the triangle, point A.
6. Connect up the points, take measurements and it works!
# Observations and Final Notes
Why does this work? There are a couple of observations made and used in the construction that haven’t been proven. The first: Is point L, the centroid of the triangle of medians, really the midpoint of side AC of DABC? This was the foundation of the construction. Here’s a sketch:
Point L was constructed as the centroid of the triangle of medians, DGEC.LB was constructed as the line parallel to segment k (FG in this sketch). We let m be a median of the constructed, giving point G as the midpoint of side AB of the constructed triangle. So, by triangle mid-segment theorem, DALB is similar to DAFG, and the similarity ratio is 1/2, so AF = 1/2 AL, or AF = FL. Since L is the centroid of DGEC, FL (=AF) = 1/3 FC, and LC = 2/3 of FC, and L is the midpoint of AC.
The second observation/question is: why does creating parallel line segments “work”? See the sketch:
The answer is basically the same as for the first question. Constructing the parallels creates similar triangles. In this sketch, look at DGEC and DOHC. The measure of angle EGC = the measure of angle HOC. The other angles aren’t marked, but you can visually follow the sets of similar triangles to see the congruent angles. |
34 + 15 + 6 - 1 = 54.
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Addition and subtraction sums include the plus or minus sign. If you are worried about your addition and subtraction - don't be! With practice, they are very easy operations to carry out.
In this quiz you'll get some practice with dealing with really easy numbers. TIP: With numbers less than 100, add the tens first and then add the units. For example:
• 12 + 16 + 22 = (10 + 10 + 20) + (2 + 6 + 2) = 40 + 10 = 50
1.
23 + 6 - 4 = ?
24
23
26
25
23 + 6 - 4 = (20) + (3 + 6 - 4) = 20 + 5 = 25
2.
12 + 12 + 12 = ?
35
34
36
32
12 + 12 + 12 = (10 + 10 + 10) + (2 + 2 + 2) = 30 + 6 = 36. Obviously, if you know that 12 × 3 = 36, you won't follow this procedure. Always BE ON THE LOOKOUT for short cuts
3.
8 + 14 - 8 + 6 = ?
14
16
18
20
8 + 14 - 8 + 6 = (14 + 6) = 20. Cross out identical pairs of numbers that differ only in sign, e.g. +8 - 8 = 0. Don't waste your time adding them to your final calculations
4.
0 + 2 + 12 + 0 = ?
24
14
10
16
0 + 2 + 12 + 0 = (10) + (2 + 2) = 10 + 4 = 14. You could do this sum just by looking at it
5.
12 + 6 - 2 = ?
16
15
14
17
TIP: With numbers less than 100, add the tens first and then add the units. 12 + 6 - 2 = (10) + (2 + 6 - 2) = 10 + 6 = 16
6.
17 - 8 = ?
7
8
9
6
17 - 8 = (10) + (7 - 8) = 10 - 1 = 9
7.
46 - 20 + 12 = ?
38
37
36
35
46 - 20 + 12 = (40 + 10 - 20) + (6 + 2) = 30 + 8 = 38
8.
34 + 15 + 6 - 1 = ?
44
55
54
56
34 + 15 + 6 - 1 = (30 + 10) + (4 + 5 + 6 - 1) = 40 + 14 = 54
9.
10 + 23 + 5 - 1 = ?
36
35
37
38
10 + 23 + 5 - 1 = (10 + 20) + (3 + 5 - 1) = 30 + 7 = 37
10.
2 + 4 + 6 + 7 + 1 = ?
20
19
18
21
2 + 4 + 6 + 7 + 1 = (4 + 6) + (7 + 2 + 1) = 20. Try to group your numbers in lots of ten
Author: Frank Evans |
# 2018 AMC 10A Problems/Problem 8
## Problem
Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$
## Solution 1
Let $x$ be the number of 5-cent coins that Joe has. Therefore, he must have $(x+3)$ 10-cent coins and $(23-(x+3)-x)$ 25-cent coins. Since the total value of his collection is 320 cents, we can write $5x + 10(x+3) + 25(23-(x+3)-x) = 320 \Rightarrow 5x + 10x + 30 + 500 - 50x = 320 \Rightarrow 35x = 210 \Rightarrow x = 6$ Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is $8-6 = \boxed{\textbf{(C) } 2}$
~Nivek
## Solution 2
Let n be the number of 5 cent coins Joe has, d be the number of 10 cent coins, and q the number of 25 cent coins. We are solving for q - n.
We know that the value of the coins add up to 320 cents. Thus, we have 5n + 10d + 25q = 320. Let this be (1).
We know that there are 23 coins. Thus, we have n + d + q = 23. Let this be (2).
We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes. Thus, we have d - 3 = n.
Plugging d-3 into the other two equations for n, (1) becomes 2d + q - 3 = 23 and (2) becomes 15d + 25q - 15 = 320. (1) then becomes 2d + q = 26, and (2) then becomes 15d + 25q = 335.
Multiplying (1) by 25, we have 50d + 25q = 650 (or 25^2 + 25). Subtracting (2) from (1) gives us 35d = 315, which means d = 9.
Plugging d into d - 3 = n, n = 6.
Plugging d and q into the (2) we had at the beginning of this problem, q = 8.
Thus, the answer is 8 - 6 = $\boxed{\textbf{(C) } 2}$.
## Solution 3
Let the number of 5-cent coins be x, the number of 10-cent coins be x+3, and the number of quarters be y.
Set up the following two equations with the information given in the problem: $$5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290$$ $$x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20$$
From there, you multiply the second equation by 25 to get $$50x+25y=500$$
You subtract the first equation from the multiplied second equation to get $$35x=210 \Rightarrow x=6$$ You can plug that value into one of the equations to get $$y=8$$ So, the answer is $8-6=\boxed{\textbf{(C) } 2}$.
- mutinykids
~savannahsolver
~ pi_is_3.14 |
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Found in: Page 335
Physics For Scientists & Engineers
Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271
Question: A car accelerates uniformly from rest and reaches a speed of 22.0m/s in 9.00s. Assuming the diameter of a tire is 58.0cm, (a) find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?
a. The number of revolutions the tire makes during the car accelerates uniformly, assuming that no slipping occurs is $N=54.4 rev$.
b. The final angular speed of a tire in revolutions per second is $n=477 rev/s$.
See the step by step solution
Step 1: Defining the final angular speed formula
Hence, the final angular speed formula is, $\omega =v/R$, where $\omega$ is the angular speed, v is the velocity, R is the radius.
Step 2: Calculating the final angular speed to determine the number of revolutions
(a)
Consider the given values
$v=22.0 m/s,\phantom{\rule{0ex}{0ex}}t=9.00 s,$
D = 58.0 cm
Hence, the final angular speed can be found by using, $\omega =\frac{v}{R}$.
Substitute.$v=22.0 m/s,R=0.58 m$
$\begin{array}{rcl}\omega & =& \frac{v}{R}\\ & =& \frac{22.0}{0.58/2}\\ & =& 75.9 \text{rad}/\text{s}\end{array}$
Thus, to find the number of revolutions, use the formula, $N=\frac{\varphi }{2\pi }$.
Solve for $\varphi$.
$\begin{array}{rcl}\varphi & =& \frac{\omega +{\omega }_{0}}{2}×t\\ & =& \frac{75.9+0.00}{2}×9.00\\ & =& 341 \text{rad}\end{array}$
Substitute $\varphi =341$ in the number of revolutions formula.
$\begin{array}{rcl}N& =& \frac{\varphi }{2\pi }\\ & =& \frac{341}{2\pi }\\ & =& 54.4\end{array}$
Therefore, the number of revolutions the tire makes during the acceleration of the car is $54.4$.
Step 3: Calculating the final angular speed in revolution per second
(b)
Thus, the formula to find the final angular speed in revolution per second is
$n=2\pi \omega$
Substitute $\omega =75.9 rad/s$.
$\begin{array}{rcl}n& =& 2\pi \omega \\ & =& 6.28×75.9 \text{rad}/\text{s}\\ & =& 477 \text{rev}/\text{s}\end{array}$
Hence, the final angular speed in revolution per second is $n=477 \text{rev}/\text{s}$. |
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# Read How to Convert Units of Measurement
WHAT IS CONVERTING UNITS OF MEASUREMENT?
You can convert between units to solve problems involving measurement.
To better understand converting units of measurement…
WHAT IS CONVERTING UNITS OF MEASUREMENT?. You can convert between units to solve problems involving measurement. To better understand converting units of measurement…
## LET’S BREAK IT DOWN!
### The Imperial System of ft and in.: Iguanas
Converting units of length can help you determine if your iguana will fit inside the tank when he is fully-grown. Suppose April has a tank that is 36 inches long. She wants to know if the iguana will fit in the tank when he grows to 6 feet long. We can figure this out by finding how many inches are in 6 feet. Each foot is 12 inches. So, in 6 feet there are 12 + 12 + 12 + 12 + 12 + 12 = 72 inches. Repeated addition is multiplication, so we could also multiply 6 × 12 = 72 inches. We get the same answer. The tank is 36 inches long and the iguana will be 72 inches long, so the tank will not be big enough. Now you try. Suppose Marcos gets a spiney-tailed iguana that only grows to about 36 inches. Will a tank that is 4 feet long be big enough for his iguana?
The Imperial System of ft and in.: Iguanas Converting units of length can help you determine if your iguana will fit inside the tank when he is fully-grown. Suppose April has a tank that is 36 inches long. She wants to know if the iguana will fit in the tank when he grows to 6 feet long. We can figure this out by finding how many inches are in 6 feet. Each foot is 12 inches. So, in 6 feet there are 12 + 12 + 12 + 12 + 12 + 12 = 72 inches. Repeated addition is multiplication, so we could also multiply 6 × 12 = 72 inches. We get the same answer. The tank is 36 inches long and the iguana will be 72 inches long, so the tank will not be big enough. Now you try. Suppose Marcos gets a spiney-tailed iguana that only grows to about 36 inches. Will a tank that is 4 feet long be big enough for his iguana?
### Imperial Weights: Bridges!
Converting units of weight can help you decide how much weight a bridge can hold. Suppose Marcos made a bridge from blocks that can hold a rock that weighs 32 ounces. He wants to know how many pounds the bridge can hold. There are 16 ounces in 1 pound. We can break 32 into groups of 16. 16 + 16 = 32. There are 2 groups of 16 in 32. Each group is 1 pound, so 32 ounces is the same as 2 pounds. Making equal groups of 16 is the same as dividing. We could divide 32 by 16. We get the same answer, 2. His bridge can hold 2 pounds. Now you try. Suppose Marcos builds another bridge that can hold a bag of flour that weighs 80 ounces. How many pounds can the bridge hold?
Imperial Weights: Bridges! Converting units of weight can help you decide how much weight a bridge can hold. Suppose Marcos made a bridge from blocks that can hold a rock that weighs 32 ounces. He wants to know how many pounds the bridge can hold. There are 16 ounces in 1 pound. We can break 32 into groups of 16. 16 + 16 = 32. There are 2 groups of 16 in 32. Each group is 1 pound, so 32 ounces is the same as 2 pounds. Making equal groups of 16 is the same as dividing. We could divide 32 by 16. We get the same answer, 2. His bridge can hold 2 pounds. Now you try. Suppose Marcos builds another bridge that can hold a bag of flour that weighs 80 ounces. How many pounds can the bridge hold?
### Metric Distance of cm, m, and km: Bike Rides
Converting metric units can help you adjust your bike seat to the right height. Suppose the manual for April’s bike says that the height of her bike seat should be 1.24 meters. She measured the height of her seat and found that it is 112 centimeters high. There are 100 centimeters in 1 meter. We can multiply 1.24 meters by 100 to find how many centimeters high the seat should be. When we multiply by 10, we can just move the decimal point one place to the right. Multiplying by 100 is the same as multiplying by 10 twice. So, we can move the decimal point 2 places to the right. 1.24 × 100 = 124. This means that the height of her seat should be 124 centimeters. April’s seat is 112 centimeters high, so she should raise the seat. Now you try. Suppose the manual for Marcos’s bike says that his seat should be 1.37 meters high. He measured the height and found that it is 147 centimeters. Does he need to adjust the seat up, down, or not at all?
Metric Distance of cm, m, and km: Bike Rides Converting metric units can help you adjust your bike seat to the right height. Suppose the manual for April’s bike says that the height of her bike seat should be 1.24 meters. She measured the height of her seat and found that it is 112 centimeters high. There are 100 centimeters in 1 meter. We can multiply 1.24 meters by 100 to find how many centimeters high the seat should be. When we multiply by 10, we can just move the decimal point one place to the right. Multiplying by 100 is the same as multiplying by 10 twice. So, we can move the decimal point 2 places to the right. 1.24 × 100 = 124. This means that the height of her seat should be 124 centimeters. April’s seat is 112 centimeters high, so she should raise the seat. Now you try. Suppose the manual for Marcos’s bike says that his seat should be 1.37 meters high. He measured the height and found that it is 147 centimeters. Does he need to adjust the seat up, down, or not at all?
### Pie/Division
Converting metric units of mass can help you make an apple pie. Let’s say you need 500 grams of sugar for apple pie filling. You can weigh the sugar on a scale, but the scale only measures in kilograms. We can find how many kilograms of sugar you need for the pie so you can use the scale. There are 1,000 grams in 1 kilogram. We need to convert grams to kilograms, so we are converting from a smaller unit to a larger unit. We can divide. When we divide by powers of 10, we can move the decimal point to the left. Dividing by 1,000 is like dividing by 10 three times. So, we can move the decimal point 3 places to the left. 500 divided by 1,000 is 0.5. You need 0.5 kilogram of sugar. Now you try. Suppose you need 700 grams of flour for the pie crust. How many kilograms of flour do you need?
Pie/Division Converting metric units of mass can help you make an apple pie. Let’s say you need 500 grams of sugar for apple pie filling. You can weigh the sugar on a scale, but the scale only measures in kilograms. We can find how many kilograms of sugar you need for the pie so you can use the scale. There are 1,000 grams in 1 kilogram. We need to convert grams to kilograms, so we are converting from a smaller unit to a larger unit. We can divide. When we divide by powers of 10, we can move the decimal point to the left. Dividing by 1,000 is like dividing by 10 three times. So, we can move the decimal point 3 places to the left. 500 divided by 1,000 is 0.5. You need 0.5 kilogram of sugar. Now you try. Suppose you need 700 grams of flour for the pie crust. How many kilograms of flour do you need?
## CONVERT UNITS OF MEASUREMENT VOCABULARY
Unit
A quantity used as the standard to tell how much makes up ‘1’ of the measurement. For example, units of length include inches, feet, centimeters, and meters.
Unit conversion
The process of changing from one unit of measurement to another.
US Customary system
A system of measurements used to measure length, weight, capacity, and temperature.
Metric system
A system of measurement that uses meters to measure length and distance, liters to measure capacity (volume), and grams to measure weight (mass).
Convert (measurement)
To change from one unit to another.
Length
A measurement of the distance between two endpoints.
A measurement of how heavy an object is.
## CONVERT UNITS OF MEASUREMENT DISCUSSION QUESTIONS
### How do you convert 7 feet to inches? How many inches are in 7 feet?
Convert 7 feet to inches by multiplying 7 by 12. There are 84 inches in 7 feet.
### How did you know to multiply when converting from feet to inches?
I converted from larger units to smaller units.
### How do you convert 80 ounces to pounds? Explain how you know.
Divide 80 by 16 because I start with a smaller unit and convert to a larger unit. There are 16 ounces in a pound.
### Which is longer, 1.2 meters or 121 centimeters? How do you know?
121 centimeters is longer; 1.2 meters is 120 centimeters. 121 is greater than 120.
### How can you remember to use the number 1,000 to convert grams to kilograms and kilograms to grams?
‘Kilo’ means 1,000. So, a kilogram is 1,000 grams. To convert kilograms to grams, multiply by 1,000. To convert grams to kilograms, divide by 1,000.
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## Algebra basics (ASL)
### Course: Algebra basics (ASL)>Unit 7
Lesson 6: Factoring quadratics 2 (ASL)
Learn how to factor quadratic expressions as the product of two linear binomials. For example, 2x²+7x+3=(2x+1)(x+3).
#### What you need to know before taking this lesson
The grouping method can be used to factor polynomials with 4 terms by taking out common factors multiple times. If this is new to you, you'll want to check out our Intro to factoring by grouping article.
We also recommend that you review our article on factoring quadratics with a leading coefficient of 1 before proceeding.
#### What you will learn in this lesson
In this article, we will use grouping to factor quadratics with a leading coefficient other than 1, like 2, x, squared, plus, 7, x, plus, 3.
## Example 1: Factoring $2x^2+7x+3$2, x, squared, plus, 7, x, plus, 3
Since the leading coefficient of left parenthesis, start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 7, end color #e07d10, x, start color #aa87ff, plus, 3, end color #aa87ff, right parenthesis is start color #11accd, 2, end color #11accd, we cannot use the sum-product method to factor the quadratic expression.
Instead, to factor start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 7, end color #e07d10, x, start color #aa87ff, plus, 3, end color #aa87ff, we need to find two integers with a product of start color #11accd, 2, end color #11accd, dot, start color #aa87ff, 3, end color #aa87ff, equals, 6 (the leading coefficient times the constant term) and a sum of start color #e07d10, 7, end color #e07d10 (the x-coefficient).
Since start color #01a995, 1, end color #01a995, dot, start color #01a995, 6, end color #01a995, equals, 6 and start color #01a995, 1, end color #01a995, plus, start color #01a995, 6, end color #01a995, equals, 7, the two numbers are start color #01a995, 1, end color #01a995 and start color #01a995, 6, end color #01a995.
These two numbers tell us how to break up the x-term in the original expression. So we can express our polynomial as 2, x, squared, plus, 7, x, plus, 3, equals, 2, x, squared, plus, start color #01a995, 1, end color #01a995, x, plus, start color #01a995, 6, end color #01a995, x, plus, 3.
We can now use grouping to factor the polynomial:
\begin{aligned}&\phantom{=}~~2x^2+1x+6x+3\\\\ &=({2x^2+1x}){+(6x+3)}&&\small{\gray{\text{Group terms}}}\\ \\ &=x({2x+1})+3({2x+1})&&\small{\gray{\text{Factor out GCFs}}}\\ \\ &=x(\maroonD{2x+1})+3(\maroonD{2x+1})&&\small{\gray{\text{Common factor!}}}\\\\ &=(\maroonD{2x+1})(x+3)&&\small{\gray{\text{Factor out } 2x+1}} \end{aligned}
The factored form is left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis.
We can check our work by showing that the factors multiply back to 2, x, squared, plus, 7, x, plus, 3.
### Summary
In general, we can use the following steps to factor a quadratic of the form start color #11accd, a, end color #11accd, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #aa87ff, c, end color #aa87ff:
1. Start by finding two numbers that multiply to start color #11accd, a, end color #11accd, start color #aa87ff, c, end color #aa87ff and add to start color #e07d10, b, end color #e07d10.
2. Use these numbers to split up the x-term.
3. Use grouping to factor the quadratic expression.
1) Factor 3, x, squared, plus, 10, x, plus, 8.
2) Factor 4, x, squared, plus, 16, x, plus, 15.
## Example 2: Factoring $6x^2-5x-4$6, x, squared, minus, 5, x, minus, 4
To factor start color #11accd, 6, end color #11accd, x, squared, start color #e07d10, minus, 5, end color #e07d10, x, start color #aa87ff, minus, 4, end color #aa87ff, we need to find two integers with a product of start color #11accd, 6, end color #11accd, dot, left parenthesis, start color #aa87ff, minus, 4, end color #aa87ff, right parenthesis, equals, minus, 24 and a sum of start color #e07d10, minus, 5, end color #e07d10.
Since start color #01a995, 3, end color #01a995, dot, left parenthesis, start color #01a995, minus, 8, end color #01a995, right parenthesis, equals, minus, 24 and start color #01a995, 3, end color #01a995, plus, left parenthesis, start color #01a995, minus, 8, end color #01a995, right parenthesis, equals, minus, 5, the numbers are start color #01a995, 3, end color #01a995 and start color #01a995, minus, 8, end color #01a995.
We can now write the term minus, 5, x as the sum of start color #01a995, 3, end color #01a995, x and start color #01a995, minus, 8, end color #01a995, x and use grouping to factor the polynomial:
\begin{aligned}&&&\phantom{=}~6x^2+\tealD{3}x\tealD{-8}x-4\\\\ \small{\blueD{(1)}}&&&=({6x^2+3x}){+(-8x-4)}&&\small{\gray{\text{Group terms}}}\\ \\ \small{\blueD{(2)}}&&&=3x({2x+1})+(-4)({2x+1})&&\small{\gray{\text{Factor out GCFs}}}\\ \\ \small{\blueD{(3)}}&&&=3x({2x+1})-4({2x+1})&&\small{\gray{\text{Simplify}}}\\ \\ \small{\blueD{(4)}}&&&=3x(\maroonD{2x+1})-4(\maroonD{2x+1})&&\small{\gray{\text{Common factor!}}}\\\\ \small{\blueD{(5)}}&&&=(\maroonD{2x+1})(3x-4)&&\small{\gray{\text{Factor out } 2x+1}}\\ \end{aligned}
The factored form is left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, 3, x, minus, 4, right parenthesis.
We can check our work by showing that the factors multiply back to 6, x, squared, minus, 5, x, minus, 4.
Take note: In step start color #11accd, left parenthesis, 1, right parenthesis, end color #11accd above, notice that because the third term is negative, a "+" was inserted between the groupings to keep the expression equivalent to the original. Also, in step start color #11accd, left parenthesis, 2, right parenthesis, end color #11accd, we needed to factor out a negative GCF from the second grouping to reveal a common factor of 2, x, plus, 1. Be careful with your signs!
3) Factor 2, x, squared, minus, 3, x, minus, 9.
4) Factor 3, x, squared, minus, 2, x, minus, 5.
5) Factor 6, x, squared, minus, 13, x, plus, 6.
## When is this method useful?
Well, clearly, the method is useful to factor quadratics of the form a, x, squared, plus, b, x, plus, c, even when a, does not equal, 1.
However, it's not always possible to factor a quadratic expression of this form using our method.
For example, let's take the expression start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 2, end color #e07d10, x, start color #aa87ff, plus, 1, end color #aa87ff. To factor it, we need to find two integers with a product of start color #11accd, 2, end color #11accd, dot, start color #aa87ff, 1, end color #aa87ff, equals, 2 and a sum of start color #e07d10, 2, end color #e07d10. Try as you might, you will not find two such integers.
Therefore, our method doesn't work for start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 2, end color #e07d10, x, start color #aa87ff, plus, 1, end color #aa87ff, and for a bunch of other quadratic expressions.
It's useful to remember, however, that if this method doesn't work, it means the expression cannot be factored as left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis where A, B, C, and D are integers.
## Why is this method working?
Let's take a deep dive into why this method is at all successful. We will have to use a bunch of letters here, but please bear with us!
Suppose the general quadratic expression a, x, squared, plus, b, x, plus, c can be factored as left parenthesis, start color #11accd, A, end color #11accd, x, plus, start color #e07d10, B, end color #e07d10, right parenthesis, left parenthesis, start color #1fab54, C, end color #1fab54, x, plus, start color #aa87ff, D, end color #aa87ff, right parenthesis with integers A, B, C, and D.
When we expand the parentheses, we obtain the quadratic expression left parenthesis, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, right parenthesis, x, squared, plus, left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, x, plus, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff.
Since this expression is equivalent to a, x, squared, plus, b, x, plus, c, the corresponding coefficients in the two expressions must be equal! This gives us the following relationship between all the unknown letters:
left parenthesis, start underbrace, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, end underbrace, start subscript, a, end subscript, right parenthesis, x, squared, plus, left parenthesis, start underbrace, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, end underbrace, start subscript, b, end subscript, right parenthesis, x, plus, left parenthesis, start underbrace, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff, end underbrace, start subscript, c, end subscript, right parenthesis
Now, let's define m, equals, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54 and n, equals, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff.
left parenthesis, start underbrace, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, end underbrace, start subscript, a, end subscript, right parenthesis, x, squared, plus, left parenthesis, start underbrace, start overbrace, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, end overbrace, start superscript, m, end superscript, plus, start overbrace, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, end overbrace, start superscript, n, end superscript, end underbrace, start subscript, b, end subscript, right parenthesis, x, plus, left parenthesis, start underbrace, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff, end underbrace, start subscript, c, end subscript, right parenthesis
According to this definition...
m, plus, n, equals, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, equals, b
and
m, dot, n, equals, left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, right parenthesis, left parenthesis, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, equals, left parenthesis, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, right parenthesis, left parenthesis, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff, right parenthesis, equals, a, dot, c
And so start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54 and start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff are the two integers we are always looking for when we use this factorization method!
The next step in the method after finding m and n is to split the x-coefficient left parenthesis, b, right parenthesis according to m and n and factor using grouping.
Indeed, if we split the x-term left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, x into left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, right parenthesis, x, plus, left parenthesis, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, x, we will be able to use grouping to factor our expression back into left parenthesis, start color #11accd, A, end color #11accd, x, plus, start color #e07d10, B, end color #e07d10, right parenthesis, left parenthesis, start color #1fab54, C, end color #1fab54, x, plus, start color #aa87ff, D, end color #aa87ff, right parenthesis.
In conclusion, in this section we...
• started with the general expanded expression a, x, squared, plus, b, x, plus, c and its general factorization left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis,
• were able to find two numbers, m and n, such that m, n, equals, a, c and m, plus, n, equals, b left parenthesiswe did so by defining m, equals, B, C and n, equals, A, D, right parenthesis,
• split the x-term b, x into m, x, plus, n, x, and were able to factor the expanded expression back into left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis.
This process shows why, if an expression can indeed be factored as left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis, our method will ensure that we find this factorization.
Thanks for pulling through! |
Posted on
## Limits vs Values
Look at the function f(x) in orange below:
We are going to answer 4 questions about this graph. They are all related to each other, but different questions. Seeing the difference will help us sort out the difference between a function value and a limit.
Q1: Find f(1)
Q2: Find limx→1 f(x)
Q3: Find limx→1+ f(x)
Q4: Find limx→1 f(x)
OK….. Try to answer these questions with what you know… Then continue reading to see the answers and explanations!
Posted on
## Basic Concept of a Limit
Here is a brief example on the concept of a limit:
Look at the function f(x) in orange below:
Question 1. Find f(3)
Explanation of question 1: Find the value of the function when you plug in 3. What is the height of the function at the exact moment when x=3?
Answer 1: The function is undefined at x=3. There is a hole when x=3.
So, f(3) is undefined.
Question 2: Find limx→3f(x)
Explanation of question 2: We are being asked to find what the function is doing around (but not at) 3. What is happening to the path of the function on either side of 3?
In order to find limx→3f(x), we must confirm that limx→3+ f(x) and limx→3– f(x) both exist and are equal to each other.
So, let’s find limx→3+ f(x). What is happening to the function values as you approach x=3 from the right-hand side? Literally run your finger along as if x=4, then x=3.5, then x=3.1. What value is the function getting closer to?
The function is approaching a height of 4.
Let’s find limx→3– f(x). What is happening to the function values as you approach x=3 from the left-hand side? Literally run your finger along as if x=1, then x=2, then x=2.9. What value is the function getting closer to?
The function is also approaching a height of 4.
So:
limx→3+ f(x) = 4
limx→3– f(x) = 4
Since, the left-handed limit at 3 and right-handed limit at 3 exist and are equal, this gives:
limx→3f(x) = 4.
So, to summarize, here are 4 different things we found. They are related, but not necessarily the same:
f(3) is undefined
limx→3+ f(x) = 4
limx→3– f(x) = 4
limx→3f(x) = 4
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## One-sided Limit Example
Q: Find the one-sided limit (if it exists):
limx→-1– (x+1)/(x4-1)
A: So we need to find the limit of this function (x+1)/(x4-1) as x approaches -1 from the left. Remember, from the left means as x gets closer and closer to -1, but is still smaller.
The concept: What is happening to this function as x = -2, x = -1.5, x = -1.1, x = -1.0001, etc…
We test first and plug -1 into the function: (-1+1)/((-1)4-1) = 0/0
Whenever you get 0/0, that is your clue that maybe you need to do “more work” before just plugging in or jumping to conclusions.
So, let’s try “more work” — usually that means simplifying. I see that the denominator can factor. We have:
(x+1) / (x4-1) = (x+1) / [(x2-1)(x2+1)]
Let’s keep factoring the denominator:
(x+1) / [(x-1)(x+1)(x2+1)]
Now, it appears there is a “removable hole” in the function. This means, we can remove this hole by reducing the matching term in the numerator with the matching term in the denominator:
(x+1) / [(x-1)(x+1)(x2+1)]
= 1 / [(x-1)(x2+1)]
Notice that hole exists when x = -1 (and it was removable! This is good news for us since we are concerned with the nature of the function as x approaches -1)
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## Finding the limit example
Q: Find the limit (as x approaches 3) of (x3 – 27) / (3-x)
A: The first thing to do when finding a basic limit is try plugging in the number in question (3).
So, plug in 3 to get:
(33 – 27) / (3-3) = 0/0 <– if you get 0/0 or infinity/infinity that means there is more work to be done. However, if you had just got a number like 4 or something, that would’ve been your answer!
OK, we got 0/0 so that means more work. More work could mean many things (apply different rules, factor and cancel, simplify, etc). In this case, it appears we can factor, so we try that:
(x3 – 27) / (3-x) = (x – 3)(x2 + 3x + 9)/ (3 – x)
Now, here comes some tricky insight. I notice that the (x – 3) on top is very similar to the (3 – x) on the bottom. I am going to factor a “-1” out of the (x – 3) that is on top.
Notice: -1(3 – x) = (x – 3)
So, the numerator becomes:
-1(x – 3)(x2 + 3x + 9)/ (3 – x)
Now, the (x – 3) term cancels from the top and bottom to leave:
-1(x2 + 3x + 9)
So, we are trying to find the limit (as x approaches 3) of -1(x2 + 3x + 9). We have “removed the hole” — the factor (3 – x) was a “removable hole” that was causing calculation problems. In the simplified version, we can plug in the value 3 to now calculate where that hole was occuring:
Lim (x –> 3) of -1(x2 + 3x + 9) = -1(32 + 3(3) + 9) = -1(9 + 9 + 9) = -27 |
# Rekenrek Activities
In the realm of early mathematics education, fostering number sense and understanding place value are foundational skills. However, these concepts can often be abstract and challenging for young learners. One effective tool that many educators utilize to make these concepts more tangible and accessible is the rekenrek. The rekenrek, also known as the arithmetic rack, was designed by Adrian Treffers, a mathematics curriculum researcher at the Freudenthal Institute in Holland, to support the natural development of number sense in children. Smaller versions consist of two rows of 10 beads, while larger versions have ten rows of ten beads. Each row has five red beads and five white beads. This structure helps students visualize numbers in terms of fives and tens, laying a solid foundation for understanding place value. Using 5 and 10 as anchors for counting, adding and subtracting is obviously more efficient than one-by-one counting. The rekenrek provides learners with the visual models they need to discover number relationships and develop a variety of addition and subtraction strategies, including doubles plus or minus one, making tens, and compensation, thereby leading to automaticity of basic facts.
## Benefits of Using Rekenreks
➡️Concrete Representation: Rekenreks offer a physical representation of numbers, making abstract mathematical concepts more tangible for young learners. Students can physically move beads to manipulate numbers, aiding in conceptual understanding.
➡️Developing Number Sense: By using rekenreks, students develop a deep understanding of number relationships and operations. They can visually see how numbers combine and decompose, facilitating mental math strategies and fluency.
➡️Place Value Understanding: The organization of beads in groups of five and ten helps students grasp the concept of place value. They learn to recognize that ten ones make a group of ten, laying the groundwork for understanding two-digit numbers and beyond.
## Possible Rekenrek Activities
💡Meet the Rekenrek: Begin by asking children what they notice about the rekenrek. Then introduce the ‘start position’(all beads over to the far right) and have them practice sliding beads in groups rather than one by one “Put your beads in start position. Now, without touching the beads, count the first three beads in your mind. On the count of three, slide all three beads at once across the string. One… two…three!” Repeat with other numbers.
💡Show me 0-10: Say a number, or hold up a numeral card (0-10). Ask students to show the given number by moving the beads with one push.
💡Show me 11-20: As above but ask students to show the given number of beads using only two pushes.
💡Quick Images to Build Subitizing Skills: Push some beads across and display them briefly before covering them with a piece of cloth or card. Ask, “How many beads did you see? How do you know?” Asking children to draw or write what they saw on a dry erase board ensures that everyone is actively involved and serves as a quick assessment. If using a 100 bead rack gradually add rows until you are displaying quick images to 100. This can be extended by asking students to show the number that is one more/one less/ten more/ten less than/double the number flashed.
💡Building Missing Addends: Ask a student to be your partner. Tell the class that you and your partner are going to build the number 6 as a team. You will move beads on the top row and your partner will move beads on the bottom row. “I am going to slide 4 beads to the left on the top row. Now in one move, you slide beads on the bottom row to build the number 6.” Pair students up and have them turn over the top card in a stack of numeral cards and work with their partner to build that number in as many different ways as possible. Begin with cards 1-10, later increase to 1-20. An arithmetic rack with ten rows can be used with students who are ready to represent numbers larger than 20.
💡Finding Different Ways to Make a Given Number: Initially use only the top row of beads. Cover the bottom row with a folded sheet of card or piece of fabric. Begin by sliding the red beads to the left and the white beads to the right on the top row of the rekenrek. Choose a number to build. “Let’s see how many ways we can build 6 by sliding beads from each side to the middle. What if I slide 4 red beads from the left and 2 white beads from the right. Does that make 6 beads? Can you think of another way to make 6? Record the different ways 6 can be built. This activity should be repeated many times using different numbers from 1-10. Once children are confident using the top row, combinations can be found using both the top and bottom rows. Children can record the different ways they find to build the given number.
Click on the blue text below to download prompt cards and flashcards to use during mental math sessions. Print on cardstock and place on a ring for teacher reference or place in a math center for children to use when working with a partner.
Interested in using rekenreks in your class but don't have the budget to buy a class set? See these instructions on how to make a class set for just a fraction of the price of store bought models.
🟢Show 5: How many different ways can you show 5 on your rekenrek? (repeat with different numbers)
🟢Doubles: How many different doubles facts can you show on your rekenrek? Record.
🟢Near Doubles: How many different ‘near doubles’ can you show? Record.
🟢Turn Around Facts: Show an addition fact. What would the turn around fact look like? Repeat.
Number Stories: Have children use individual arithmetic racks as a tool to solve various types of addition and subtraction number stories. This may be used as a journal or oral activity, with the focus on children explaining their strategy for solving the problem. Be sure to include open-ended problems that have more than one solution that children can model on the arithmetic rack, such as the following:
➡️There were 8 children on a bunk bed. Some were on the top bunk and some were on the bottom bunk. How many children were on the top bunk? How many children were on the bottom bunk? Show as many different solutions as you can.
➡️There were 12 passengers on a double-decker bus. Some passengers were on the top deck and some were on the bottom deck. How many passengers were on the top deck? How many passengers were on the bottom deck? Show as many different solutions as you can.
The below two work samples were completed by a Kindergarten student who represented her work very clearly on the unlined pages in her Math Journal using pictures, numbers and words. To begin with some Kindergarten students may find it difficult to represent their work on paper. Having recording paper available for children who choose to use it is one way to scaffold early attempts at drawing.
Incorporating rekenreks into kindergarten and first-grade classrooms is an effective way to build number sense and place value understanding. Through hands-on exploration and manipulation of beads, students develop a deep conceptual understanding of mathematical concepts. From counting and cardinality to addition, subtraction, and place value, rekenreks offer a versatile tool for engaging young learners and laying a solid foundation for future mathematical success. By integrating these activities into the curriculum, educators can create meaningful mathematical experiences that empower students to become confident and proficient mathematicians.
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# Number Lines
Created by: Team Maths - Examples.com, Last Updated: April 27, 2024
## Number Lines
Mathematical journey with our definitive guide to number lines, an essential tool for educators and students. This resource demystifies number lines, illustrating their importance in understanding the fundamentals of mathematics. Through practical examples and straightforward explanations, we aim to enhance numerical literacy, providing learners with the skills needed to tackle a wide range of mathematical challenges. Whether used for simple addition or complex algebra, this guide ensures number lines are accessible and engaging for all.
## What are Number Lines?
Number lines are fundamental mathematical tools that represent numbers as points along a straight line. Each point corresponds to a number, with equal intervals indicating equal value differences. They serve as a visual aid to understand concepts such as addition, subtraction, and other operations, making abstract ideas tangible. By incorporating number lines into lessons, teachers can significantly improve studentsβ comprehension of numerical relationships and mathematical reasoning.
## What is the Best Example of a Number Line?
The best example of a number line is its use in visualizing addition and subtraction. For instance, to add 2 + 3, start at the point labeled 2, move three units to the right, and arrive at 5. Conversely, to subtract 3 from 5, start at 5 and move three units to the left, landing on 2. This visual method simplifies these operations, making it easier for students to grasp the concept of movement along the number line, thereby enhancing their mathematical understanding and skills.
## How to Draw a Number Line?
Mastering how to draw a number line is pivotal for both educators and students, facilitating a deeper understanding of mathematical concepts. This guide outlines the simple steps to create a number line: start with a straight line, mark a point for zero, and evenly space increments both to the right for positive numbers and to the left for negative numbers. This visual tool is indispensable in teaching numerical order, addition, subtraction, and introducing the concept of integers, making math more accessible and engaging.
Examples:
1. Creating a Basic Number Line:
• Draw a horizontal line and evenly space marks along it. Label the center as 0, marks to the right with positive numbers, and to the left with negative numbers.
• On your number line, decide on the interval (e.g., 1s, 2s, 5s) and mark accordingly. This helps in understanding scale and distance between numbers.
3. Including Fractions:
• For a more advanced number line, between any two whole numbers, add marks for fractions or decimals to illustrate these concepts visually.
4. Incorporating Colored Segments:
• Use different colors to highlight sections of the number line, such as positive and negative areas, to enhance comprehension of number relationships.
5. Interactive Number Line:
• Create a number line with movable markers or labels to allow students to physically engage with addition, subtraction, and number placement exercises.
## Negative and Positive Number Line
A negative and positive number line visually represents both negative and positive integers on either side of zero, serving as a fundamental tool in mathematics for illustrating the concept of opposites. It aids in teaching the addition and subtraction of negative numbers, understanding absolute values, and exploring the relationships between numbers. This dual-sided approach not only simplifies complex concepts but also enriches students’ numerical fluency and analytical skills, making it an essential element in math education.
Examples:
1. Understanding Zeroβs Role:
• The point where the negative side transitions to the positive, zero serves as a central reference, emphasizing its importance as neither positive nor negative.
2. Comparing Temperatures:
• Use a number line to compare temperatures below and above freezing, illustrating how negative numbers represent colder temperatures than zero.
3. Elevation Above and Below Sea Level:
• Depict elevations using a number line, where positive numbers indicate above sea level and negative numbers represent below sea level, making abstract concepts tangible.
4. Financial Balances:
• A number line can demonstrate financial concepts, such as debts (negative values) versus savings (positive values), providing real-world applications.
5. Time Zones:
• Illustrate time differences from a reference point (e.g., GMT) using a number line, with eastward zones as positive and westward as negative.
## Parts of a Number Line
Understanding the parts of a number line is crucial for students to navigate and apply mathematical concepts effectively. A number line consists of a horizontal line marked with numbers at equal intervals, a zero point that divides the line into positive and negative sides, and often includes tick marks or points to represent specific numbers or values. By dissecting a number line into its component parts, educators can guide students through various mathematical operations and principles, enhancing their grasp of linear measurements, operations, and number properties.
Examples:
1. Zero Point:
• The central point of a number line, dividing positive and negative values, crucial for understanding the concept of polarity in numbers.
2. Intervals:
• Regularly spaced marks that indicate the distance between numbers, essential for measuring and calculating differences or sums.
3. Positive Side:
• The right side of zero, representing positive numbers, crucial for understanding growth, addition, and positive changes.
4. Negative Side:
• The left side of zero, showcasing negative numbers, vital for depicting reductions, subtraction, and debts.
5. Tick Marks:
• Small lines used to represent individual numbers, helping users precisely identify and locate specific values on the number line.
Each component plays a significant role in utilizing number lines as educational tools, facilitating a comprehensive understanding of numerical concepts and their applications.
## Number Line with Decimals
Dive into the precision of mathematics with our guide on number lines with decimals. This tool is pivotal for teachers and students to understand fractions and decimals visually. It breaks down complex concepts, making them easier to grasp by illustrating the values between whole numbers. By placing decimals accurately on the number line, learners can compare and perform operations with decimals, enhancing their mathematical fluency and confidence in handling real-world problems.
Examples:
1. Placing 0.5 on the Number Line:
• Locate 0.5 between 0 and 1, showing how decimals represent values less than a whole. This illustrates the concept of halves, foundational for understanding fractions and decimals.
2. Comparing 0.75 and 0.25:
• By marking 0.75 and 0.25 on the number line, students visually understand that 0.75 is closer to 1, emphasizing the idea of quarter values and their comparison.
3. Adding Decimals: 0.3 + 0.6:
• Start at 0.3, move 0.6 units to the right, and land on 0.9. This example teaches adding decimals, providing a clear visual representation of the sum.
4. Subtracting Decimals: 0.8 – 0.3:
• From the point 0.8, move 0.3 units to the left to reach 0.5. Demonstrates subtraction of decimals, reinforcing the concept through visual aids.
5. Finding Midpoints: Between 0.2 and 0.4:
• Identifying the midpoint at 0.3 showcases how number lines can determine values exactly between two decimals, fostering a deeper understanding of number relationships.
## Inequalities on a Number Line
Explore the realm of inequalities through our insightful guide on representing them on a number line. Ideal for educators aiming to demystify algebraic concepts, this resource uses number lines to visually express inequalities, making abstract concepts tangible. Students learn to plot and interpret inequalities, gaining a solid foundation in understanding mathematical relationships. This visual approach not only clarifies the nature of inequalities but also prepares learners for more complex algebraic reasoning and problem-solving.
Examples:
1. Plotting x > 2:
• Draw an open circle at 2 and shade the line to the right, indicating all values greater than 2. This visual helps students grasp the concept of greater-than inequalities.
2. Representing x β€ -1:
• Place a closed circle at -1 and shade to the left, showing x is less than or equal to -1. It visually demonstrates how to include the endpoint in the solution set.
3. Graphing x < 3:
• An open circle at 3 with shading to the left illustrates values less than 3, teaching students about less-than inequalities and their graphical representations.
4. Illustrating x β₯ 0.5:
• With a closed circle at 0.5 and shading to the right, students see that x is any value greater than or equal to 0.5, emphasizing the equality component of inequalities.
5. Solving Compound Inequalities: 1 < x β€ 4:
• Plot an open circle at 1 and a closed circle at 4, shading between them. This teaches the concept of compound inequalities, showing values that x can take within a range.
These detailed explanations and examples for number lines with decimals and inequalities on a number line are crafted to enhance the learning experience, providing teachers and students with effective tools to master mathematical concepts.
## Graphing Inequalities on a Number Line
Graphing inequalities on a number line offers a visual representation to solve and understand inequalities. This method highlights the range of values that satisfy the inequality, using open or closed circles to indicate whether endpoints are included. By incorporating this technique, educators can enhance students’ comprehension of mathematical concepts, providing a clear and intuitive way to grasp inequalities.
Examples:
1. Less Than Five (x < 5):
• Use an open circle at 5 and shade left to represent all numbers less than 5, illustrating values not reaching 5.
2. Greater Than or Equal to Three (x β₯ 3):
• Place a closed circle at 3 and shade right, indicating inclusivity of 3 and all numbers greater than 3.
3. Not Equal to Two (x β 2):
• An open circle at 2 with no shading demonstrates that all numbers are valid except 2, emphasizing exclusion.
4. Between One and Four (1 < x < 4):
• Open circles at 1 and 4 with shading between them show values strictly between 1 and 4, excluding endpoints.
5. At Most Six (x β€ 6):
• A closed circle at 6 shaded leftward indicates all numbers up to and including 6 are solutions, showing inclusivity.
## The Importance of Number Lines
Number lines are invaluable in education, offering a straightforward method for visualizing and understanding numerical concepts. They aid in the comprehension of basic operations, inequalities, and absolute values, fostering a deep understanding of mathematical principles. By integrating number lines into teaching, educators can simplify complex ideas, making mathematics more accessible and less intimidating for students.
Examples:
1. Understanding Absolute Value:
• Demonstrating distance from zero on a number line helps clarify the concept of absolute value, making it tangible.
2. Simplifying Fraction Comparison:
• Placing fractions on a number line allows for easy comparison, enhancing understanding of their relative sizes.
3. Visualizing Integer Operations:
• Using number lines to add or subtract integers provides a clear visual path of movement, aiding in operation comprehension.
4. Decimals and Place Value:
• Number lines illustrate how decimals fit between integers, clarifying the concept of place value and decimal positions.
5. Concept of Negative Numbers:
• Introducing negative numbers on a number line helps students grasp the idea of values less than zero, expanding their number sense.
## Different Types of Numbers on a Number Line
Number lines can represent a variety of number types, including integers, fractions, decimals, and even irrational numbers. This versatility makes number lines a powerful tool for teaching a broad spectrum of mathematical concepts. Understanding how different types of numbers are positioned on a number line enhances students’ numerical fluency and mathematical reasoning.
Examples:
1. Integers:
• Marked at regular intervals, integers on a number line show the progression of whole numbers, including negatives.
2. Fractions:
• Fractions are positioned between integers, illustrating their values in relation to whole numbers and each other.
3. Decimals:
• Decimals appear between integers and fractions, demonstrating their precise value and helping compare decimal quantities.
4. Irrational Numbers:
• While not marked exactly, irrational numbers’ approximate locations can be shown, bridging gaps between rational numbers.
5. Real Numbers:
• A number line represents the continuum of real numbers, encompassing all rational and irrational numbers within its span.
## Number Lines as Mathematical Tools
Number lines are essential mathematical tools that facilitate the understanding of numerical relationships and operations. They serve as the foundation for developing algebraic thinking, enabling students to visualize and solve equations, understand functions, and explore the properties of numbers. By using number lines, educators can create a dynamic learning environment that encourages exploration and enhances students’ mathematical intuition.
Examples:
1. Solving Equations:
• Demonstrating solutions to simple equations on a number line helps students understand the concept of equality and balance.
2. Exploring Functions:
• Graphing functions on a number line can introduce the idea of change and rate, foundational for algebraic concepts.
3. Understanding Sequences:
• Visualizing arithmetic and geometric sequences on a number line aids in recognizing patterns and relationships between terms.
4. Coordinate Geometry:
• Number lines lay the groundwork for understanding coordinates, crucial for graphing and spatial reasoning in geometry.
5. Probability and Statistics:
• They can represent probability distributions and statistical data, providing a basis for understanding randomness and variation.
Number lines stand as a fundamental tool in the realm of mathematics, bridging the gap between abstract concepts and visual understanding. By facilitating a deeper comprehension of numerical relationships, operations, and inequalities, they empower educators to deliver lessons that enhance students’ analytical skills and confidence in math. Incorporating number lines into educational practices promises to enrich the learning experience and foster a lifelong appreciation for mathematics.
Text prompt |
## ← Subtract Rational Expressions Practice 2 - Visualizing Algebra
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Showing Revision 5 created 05/24/2016 by Udacity Robot.
1. To subtract the second rational expression we really want to add the opposite.
2. We want to distribute our subtraction sign to each term in the numerator which
3. gives us a negative 1. Now we factor each denominator and try to find the lowest
4. common denominator between the two fractions. The lowest common denominator is x
5. minus 1 times x minus 1 times x plus 1. We know this since these two
6. denominators share a common factor of x minus 1. The other factor for this
7. denominator is x minus 1. And the other factor for this denominator is x plus 1.
8. We multiply each fraction by the missing factors of the lowest common
9. denominator. For the first fraction, we distribute 6x to the quantity x plus 1.
10. We'll get 6x squared plus 6x. For this second fraction, we'll have negative 1
11. times the quantity x minus 1. We really just changed these two signs. So we'll
12. have negative x plus 1. Now that we have two fractions with common denominators
13. We can add the numerators together. We just add the like terms. When we add the
14. like terms of 6x and negative x together we'll get 6x squared plus 5x plus 1.
15. Now remember, we have a new numerator here. And let's see if we can factor it.
16. We want factors of positive 6 that's sum to 5. Well yes that would be 3 and 2.
17. This would be the factored form if we use factoring by grouping. But notice that
18. this doesn't help us out. None of these factors appears in our denominator. We
19. can't simply any more. I final answer would be 6 x squared plus 5 x plus 1.
20. Divided by the factors of this denominator. |
We have seen that it is possible to find the probability of compound events, where we have the occurrence of more than one simple event in a sequence. When working with more than one event, you have to be concerned as to whether the first event affects the second event.
When determining if events are independent, you are determining if the events are affecting one another.
Independent Events
Two events are said to be independent if the result of the second event is not affected by the result of the first event. The probability of one event does not change the probability of the other event.
If A and B are independent events, the probability of both events occurring is the product of the probabilities of the individual events.
If A and B are independent events, P(A∩B) = P(A and B) = P(A) • P(B). (referred to as the "Probability Multiplication Rule")
Notice the connection between "AND" and "multiplication".
What is the probability of tossing a head on a penny and then choosing an ace from a standard deck of cards? These are independent events as the second event is not affected by the first. The probability of BOTH of these events is found by the Multiplication Rule. The events are independent. P(head then ace) = P(head) • P(ace) = 1/2 • 4/52 = 2/52 = 1/26.
A drawer contains 3 red paper clips, 4 green paper clips, and 5 blue paper clips. One paper clip is taken from the drawer and then replaced. Another paper clip is taken from the drawer. What is the probability that the first paper clip is red and the second paper clip is blue? Because the first paper clip is replaced, the sample space of 12 paper clips does not change from the first event to the second event. The events are independent. P(red then blue) = P(red) • P(blue) = 3/12 • 5/12 = 15/144 = 5/48.
When you toss a coin, the probability of getting a head is 1 out of 2 or ½.
If you toss the coin again, the probability of getting a head is still 1 out of 2 or ½. If you toss a coin 10 times and get a head each time, you may think that your luck of tossing a tail is increasing since it has not yet appeared. This is not the case. These events are independent events and do not affect one another. The probability of tossing a tail is 1 out of 2 or ½ regardless of how many heads were tossed previously.
Dependent Events
If the result of one event IS affected by the result of another event,
the events are said to be dependent.
If A and B are dependent events, the probability of both events occurring
is the product of the probability of the first event and the probability of the second event
once the first event has occurred.
If A and B are dependent events, and A occurs first, P(A and B) = P(A) • P(B, once A has occurred) ... and is written as ... P(A∩B) = P(A and B) = P(A) • P(B | A)
The notation P(B | A) is called a "conditional probability"
the probability of event B given that event A has occurred".
A bag contains 3 green marbles and 2 red marbles. A marble is drawn, not replaced, and then a second marble is drawn. What is the probability of drawing a green marble followed by drawing a red marble? By not replacing the marble after the first draw, the probability of the second draw is affected. The sample space of the second draw has changed, leaving only 4 marbles. The events are dependent. P(green then red) = P(green) • P(red given green occurred) = 3/5 • 2/4 = 6/20 = 3/10.
A drawer contains 3 red paper clips, 4 green paper clips, and 5 blue paper clips. One paper clip is taken from the drawer and is NOT replaced. Another paper clip is taken from the drawer. What is the probability that the first paper clip is red and the second paper clip is blue? Because the first paper clip is NOT replaced, the sample space of the second event is changed. The sample space of the first event is 12 paper clips, but the sample space of the second event is now 11 paper clips. The events are dependent. P(red then blue) = P(red) • P(blue) = 3/12 • 5/11 = 15/132 = 5/44.
Sampling with, and without, replacement:
When working with the probability of two (or more) events occurring, it is important to determine if finding the probability of one of the events has an effect on any of the other events.
Consider the following example:
What is the probability of drawing a red marble,
then drawing a blue marble from this jar?
The probability of drawing a red marble = 2/5.
The probability of drawing a blue marble = 1/5.
BUT...
• The 1/5 probability of drawing a blue marble assumes all 5 marbles are in the jar.
• What happens if you draw the first marble and do NOT put that marble back in the jar before drawing the second marble? If the marble is not "replaced", the probability of the second drawing changes, since there are less marbles in the jar.
The probability of drawing a red marble = 2/5.
The probability of drawing a blue marble is now = 1/4.
Let's compare the two different answers:
With Replacement: Without Replacement:
The probability of drawing a red marble = 2/5. Put the marble back in the jar. The probability of drawing a blue marble = 1/5. (of the 5 in the jar) Answer: 2/5 • 1/5 = 2/25
The probability of drawing a red marble = 2/5. Do not put marble back in jar. The probability of drawing a blue marble = 1/4. (of the 4 left in jar) Answer: 2/5 • 1/4 = 2/20 = 1/10
In relation to probability, the word "replacement" most often refers to situations where something can be "removed" (drawn, chosen, etc.) from the sample set, and then replaced (or not replaced).
• "
With replacement": Choosing a ball, a card, a marble, or other object, and then replacing the item back into the sample space each time an event occurs.
Example:
Choosing a card from a deck and then putting the card back into the deck before drawing another card.
• "Without replacement": Choosing a ball, a card, a marble, or other object, and then NOT replacing the item back into the sample space before choosing another object.
Example:
Choosing a card from a deck and not replacing it to the deck before drawing another card.
The sample space for the second card draw has now been changed to one less card.
Be on the lookout for the word "replacement" as a clue.
With Replacement: the events are independent. Probabilities do NOT affect one another. Without Replacement: the events are dependent. Probabilities DO affect one another. |
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# Exponential Terms Raised to an Exponent
## Multiply to raise exponents to other exponents
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Power Properties of Exponents
There are 1,000 bacteria present in a culture. When the culture is treated with an antibiotic, the bacteria count is halved every 4 hours. How many bacteria remain 24 hours later?
### Watch This
Watch the second part of this video, starting around 3:30.
### Guidance
The last set of properties to explore are the power properties. Let’s investigate what happens when a power is raised to another power.
#### Investigation: Power of a Power Property
1. Rewrite $(2^3)^5$ as $2^3$ five times.
$(2^3)^5 = 2^3 \cdot 2^3 \cdot 2^3 \cdot 2^3 \cdot 2^3$
2. Expand each $2^3$ . How many 2’s are there?
$(2^3)^5 = \underbrace{2 \cdot 2 \cdot 2}_{2^3} \cdot \underbrace{2 \cdot 2 \cdot 2}_{2^3} \cdot \underbrace{2 \cdot 2 \cdot 2}_{2^3} \cdot \underbrace{2 \cdot 2 \cdot 2}_{2^3} \cdot \underbrace{2 \cdot 2 \cdot 2}_{2^3} = 2^{15}$
3. What is the product of the powers?
$3 \cdot 5 = 15$
4. Fill in the blank. $(a^m)^n = a^{-^\cdot-}$
$(a^m)^n = a^{mn}$
The other two exponent properties are a form of the distributive property.
Power of a Product Property: $(ab)^m = a^m b^m$
Power of a Quotient Property: $\left( \frac{a}{b} \right)^m = \frac{a^m}{b^m}$
#### Example A
Simplify the following.
(a) $(3^4)^2$
(b) $(x^2 y)^5$
Solution: Use the new properties from above.
(a) $(3^4)^2 = 3^{4 \cdot 2} = 3^8 = 6561$
(b) $(x^2 y)^5 = x^{2 \cdot 5} y^5 = x^{10} y^5$
#### Example B
Simplify $\left( \frac{3a^{-6}}{2^2 a^2} \right)^4$ without negative exponents.
Solution: This example uses the Negative Exponent Property from the previous concept. Distribute the $4^{th}$ power first and then move the negative power of $a$ from the numerator to the denominator.
$\left( \frac{3a^{-6}}{2^2 a^2} \right)^4 = \frac{3^4 a^{-6 \cdot 4}}{2^{2 \cdot 4} a^{2 \cdot 4}} = \frac{81a^{-24}}{2^8 a^8} = \frac{81}{256a^{8+24}} = \frac{81}{256a^{32}}$
#### Example C
Simplify $\frac{4x^{-3} y^4 z^6}{12x^2 y} \div \left( \frac{5xy^{-1}}{15x^3 z^{-2}} \right)^2$ without negative exponents.
Solution: This example is definitely as complicated as these types of problems get. Here, all the properties of exponents will be used. Remember that dividing by a fraction is the same as multiplying by its reciprocal.
$\frac{4x^{-3} y^4 z^6}{12x^2 y} \div \left( \frac{5xy^{-1}}{15x^3 z^{-2}} \right)^2 &= \frac{4x^{-3} y^4 z^6}{12x^2 y} \cdot \frac{225x^6 z^{-4}}{25x^2 y^{-2}}\\&= \frac{y^3 z^6}{3x^5} \cdot \frac{9x^4 y^2}{z^4}\\&= \frac{3x^4 y^5 z^6}{x^5 z^4}\\&= \frac{3y^5 z^2}{x}$
Intro Problem Revisit To find the number of bacteria remaining, we use the exponential expression $1000 (\frac{1}{2})^n$ where n is the number of four-hour periods.
There are 6 four-hour periods in 24 hours, so we set n equal to 6 and solve.
$1000 (\frac{1}{2})^6$
Applying the Power of a Quotient Property, we get:
$1000 (\frac{1^6}{2^6}) = \frac {1000 \cdot 1}{2^6} = \frac {1000}{64} = 15.625$
Therefore, there are 15.625 bacteria remaining after 24 hours.
### Guided Practice
Simplify the following expressions without negative exponents.
1. $\left( \frac{5a^3}{b^4} \right)^7$
2. $(2x^5)^{-3} (3x^9)^2$
3. $\frac{(5x^2 y^{-1})^3}{10y^6} \cdot \left( \frac{16x^8 y^5}{4x^7} \right)^{-1}$
1. Distribute the 7 to every power within the parenthesis.
$\left( \frac{5a^3}{b^4} \right)^7 = \frac{5^7 a^{21}}{b^{28}} = \frac{78,125a^{21}}{b^{28}}$
2. Distribute the -3 and 2 to their respective parenthesis and then use the properties of negative exponents, quotient and product properties to simplify.
$(2x^5)^{-3} (3x^9)^2 = 2^{-3} x^{-15} 3^2 x^{18} = \frac{9x^3}{8}$
3. Distribute the exponents that are outside the parenthesis and use the other properties of exponents to simplify. Anytime a fraction is raised to the -1 power, it is equal to the reciprocal of that fraction to the first power.
$\frac{\left(5x^2 y^{-1}\right)^3}{10y^6} \cdot \left( \frac{16x^8 y^5}{4x^7} \right)^{-1} &= \frac{5^3 x^{-6} y^{-3}}{10y^6} \cdot \frac{4x^7}{16x^8 y^5}\\&= \frac{500xy^{-3}}{160x^8 y^{11}}\\&= \frac{25}{8x^7 y^{14}}$
### Vocabulary
Power of Power Property
$(a^m)^n = a^{mn}$
Power of a Product Property
$(ab)^m = a^m b^m$
Power of a Quotient Property
$\left( \frac{a}{b} \right)^m = \frac{a^m}{b^m}$
### Practice
Simplify the following expressions without negative exponents.
1. $(2^5)^3$
2. $(3x)^4$
3. $\left( \frac{4}{5} \right)^2$
4. $(6x^3)^3$
5. $\left( \frac{2a^3}{b^5} \right)^7$
6. $(4x^8)^{-2}$
7. $\left( \frac{1}{7^2 h^9} \right)^{-1}$
8. $\left( \frac{2x^4 y^2}{5x^{-3} y^5} \right)^3$
9. $\left( \frac{9m^5 n^{-7}}{27 m^6 n^5} \right)^{-4}$
10. $\frac{(4x)^2 (5y)^{-3}}{(2x^3 y^5)^2}$
11. $(5r^6)^4 \left( \frac{1}{3} r^{-2} \right)^5$
12. $(4t^{-1} s)^3 (2^{-1} ts^{-2})^{-3}$
13. $\frac{6a^2 b^4}{18a^{-3} b^4} \cdot \left( \frac{8b^{12}}{40a^{-8} b^5} \right)^2$
14. $\frac{2(x^4 y^4)^0}{2^4 x^3 y^5 z} \div \frac{8z^{10}}{32x^{-2} y^5}$
15. $\frac{5g^6}{15g^0 h^{-1}} \cdot \left( \frac{h}{9g^{15} j^7} \right)^{-3}$
16. Challenge $\frac{a^7 b^{10}}{4a^{-5} b^{-2}} \cdot \left[ \frac{(6ab^{12})^2}{12a^9 b^{-3}} \right]^2 \div (3a^5 b^{-4})^3$
17. Rewrite $4^3$ as a power of 2.
18. Rewrite $9^2$ as a power of 3.
19. Solve the equation for $x$ . $3^2 \cdot 3^x = 3^8$
20. Solve the equation for $x$ . $(2^x)^4 = 4^8$
### Vocabulary Language: English
Power of a Power Property
Power of a Power Property
The power of a power property states that $(a^m)^n = a^{mn}$.
Power of a Quotient Property
Power of a Quotient Property
The power of a quotient property states that $\left( \frac{a}{b} \right)^m = \frac{a^m}{b^m}$. |
Dividing a Quantity in Three given Ratios
Rules of dividing a quantity in three given ratios is explained below along with the different types of examples.
If a quantity K is divided into three parts in the ratio X : Y : Z, then
First part = X/(X + Y + Z) × K,
Second part = Y/(X + Y + Z) × K,
Third part = Z/(X + Y + Z) × K.
For example, suppose, we have to divide $1200 among X, Y, Z in the ratio 2 : 3 : 7. This means that if X gets 2 portions, then Y will get 3 portions and Z will get 7 portions. Thus, total portions = 2 + 3 + 7 = 12. So, we have to divide$ 1200 into 12 portions and then distribute the portions among X, Y, Z according to their share.
Thus, X will get 2/12 of $1200, i.e., 2/12 × 1200 =$ 200
Y will get 3/12 of $1200, i.e., 3/12 × 1200 =$ 300
Z will get 7/12 of $1200, i.e., 7/12 × 1200 =$ 700
Solved examples:
1. If $135 is divided among three boys in the ratio 2 : 3 : 4, find the share of each boy. Solution: The sum of the terms of the ratio = 2 + 3 + 4 = 9 Share of first boy = 2/9 × 135 =$ 30.
Share of second boy = 3/9 × 315 = $45. Share of first boy = 4/9 × 315 =$ 60.
Thus, the required shares are $30,$ 45 and \$ 60 respectively.
2. Divide 99 into three parts in the ratio 2 : 4 : 5.
Solution:
Since, 2 + 4 + 5 = 11.
Therefore, first part = 2/11 × 99 = 18.
Second part = 4/11 × 99 = 36.
And, third part = 5/11 × 99 = 45.
3. 420 articles are divided among A, B and C, such that A gets three-times of B and B gets five-times of C. Find the number of articles received by B.
Solution:
Let the number of articles C gets = 1
The number of article that B gets = five times of C = 5 × 1 = 5.
And, the number of articles that A gets = three times of B = 3 × 5 = 15.
Therefore, A : B : C = 15 : 5 : 1
And, A + B + C = 15 + 5 + 1 = 21
The number of articles received by B = 5/21 × 420 = 100
The above examples on dividing a quantity in three given ratios will help us to solve different types of problems on ratios.
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# 6.2 - Volumes. Definition: Right Cylinder Let B 1 and B 2 be two congruent bases. A cylinder is the points on the line segments perpendicular to the bases.
## Presentation on theme: "6.2 - Volumes. Definition: Right Cylinder Let B 1 and B 2 be two congruent bases. A cylinder is the points on the line segments perpendicular to the bases."— Presentation transcript:
6.2 - Volumes
Definition: Right Cylinder Let B 1 and B 2 be two congruent bases. A cylinder is the points on the line segments perpendicular to the bases that join B 1 and B 2. B1B1 B2B2 Alternate Definition A cylinder is a surface whose cross section in every plane parallel to a given plane are the same.
Volumes of Solids P x is perpendicular to the x axis. A(x) is a function that represents the cross-sectional area where a ≤ x ≤ b.
Volumes of Solids Establish n subintervals on [a, b] and let Δx. Choose the midpoint to be x i. Then a cylinder is formed with volume V(S i ) = A(x i ) Δx
Estimate of the volume of the figure using n = 7 subintervals. The volume can be approximated by
Definition of Volume Let S be a solid that lies between x = a and x = b. If the cross- sectional area of S in the plane P x, through x and perpendicular to the x-axis, is A(x), where A is a continuous function, then the volume of S is V = Area × Width
Volumes of Solids of Revolution: Example Let R be the region bounded by y = 4 – x 2 and y = 0. Find the volume of the solids obtained by revolving R about the y-axis. 2-D Graph What is the function that represents the radius? 3-D Graph What is the function that represents the area of the disk?
It will be easier to integrate along the y-axis since it is perpendicular to the disks. Thus, the radius, x, is the distance from the y-axis to the function or The area of the disk is given by The volume of revolution is given by
Volumes of Solids of Revolution: Example Let R be the region bounded by the graphs of y = x 2, x = 0, and y = 1. Compute the volume of the solid formed by revolving R about the x- axis. 2-D Graph What is the function that represents the radius? 3-D Graph What is the function that represents the area of the washer?
Now we need to determine the area function for the washer.
The volume can now be found by
Volumes of Solids of Revolution: Example Let R be the region bounded by the graphs of y = x 2, x = 0, and y = 1. Compute the volume of the solid formed by revolving R about the line y = 2. 2-D Graph What is the function that represents the radius? 3-D Graph What is the function that represents the area of the washer?
Now we need to determine the area function for the washer. The volume is given by
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# What is 5/12 + 1/4?
5 12
+
1 4
## Step 1
Our denominators (bottom numbers) don't match, so we can't add yet.
Since 4 is evenly divided by 12, we can multiply just one term to get a common denominator.
Multiply 1 by 3, and get 3, then we multiply 4 by 3 and get 12.
The problem now has new fractions to add:
5 12
+
3 12
## Step 2
Since our denominators match, we can add the numerators.
5 + 3 = 8
The sum we get is
8 12
## Step 3
The last step is to reduce the fraction if we can.
To find out, we try dividing it by 2...
Are both the numerator and the denominator evenly divisible by 2? Yes! So we reduce it:
8 12
÷ 2 =
4 6
Now, try the same number again.
Are both the numerator and the denominator evenly divisible by 2? Yes! So we reduce it:
4 6
÷ 2 =
2 3
Now, try the same number again.
Nope. Try the next prime number, 3...
No good. 3 is larger than 2. So we're done reducing.
Congratulations! Here's your final answer to 5/12 + 1/4
5 12
+
1 4
=
2 3
© 2014 Randy Tayler |
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# Probability
### Example 1: using the probability rules
An experiment consists of rolling a single die. Find the probability that the number appearing uppermost is:
a) a $$5$$ or a $$6$$?
b) a number less than $$5$$?
c) an $$8$$?
d) an even number or a number less than $$4$$?
Solution:
An experiment consists of rolling a single die. The probability distribution will be:
$$x$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$P(x)$$ $$\displaystyle \frac{1}{6}$$ $$\displaystyle \frac{1}{6}$$ $$\displaystyle \frac{1}{6}$$ $$\displaystyle \frac{1}{6}$$ $$\displaystyle \frac{1}{6}$$ $$\displaystyle \frac{1}{6}$$
a) \begin{eqnarray*}
&& P(x = 5 \mbox{ or } x = 6) \\
&=& P(x = 5) + P(x = 6) \\
&=& \frac{1}{6} + \frac{1}{6} \\
&=& \frac{1}{3}
\end{eqnarray*}
b) \begin{eqnarray*}
&& P(x < 5) \\
&=& P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) \\
&=& \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\
&=& \frac{2}{3}
\end{eqnarray*}
c) $P(x=8)=0$
d) \begin{eqnarray*}
&&P(x \mbox{ is even, or } x < 4) \\
&=& P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) \\ |
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# Y Mx + C in Geometry: An Explanation for Students
If you're taking a geometry class, you've probably seen the equation y = mx + c on a chalkboard or whiteboard at some point. But what does this equation actually mean? In this blog post, we'll break down the meaning of each term in the equation so that you can better understand how to use it.
Y: The Y coordinate is the point where the line intersects with the Y axis.
M: M is the slope of the line. Slope is defined as the ratio of the rise (the vertical distance between two points on a line) to the run (the horizontal distance between two points on a line). You can calculate slope using the following formula:
### Slope = (y2 - y1)/(x2 - x1)
C: C is the y-intercept, which is the point where the line intersects with the y-axis.
Now that we know what each term in the equation means, let's take a look at how to use it. Suppose we have a line with the following coordinates: (2,4), (4,8), and (6,12). We can plug these coordinates into the equation as follows:
### 12 = m(6) + c
From here, we can solve for m and c. First, we'll solve for c. We can do this by plugging in one of our coordinate pairs and solving for c. Let's use (2,4). We know that 4 = m(2) + c, so we can rearrange the equation to solve for c like this:
### c = 4 - 2m
Now we'll plug our value for c back into one of our equations and solve for m. We'll use 8 = m(4) + c. We know that c = 4 - 2m, so we can substitute that into our equation and solve for m like this:
### m = 3
Now that we know both m and c, we can plug them back into our original equation and write it out in standard form like this: y = 3x + 4. And there you have it! That's all there is to understanding and using the y = mx + c equation in geometry.
## FAQ
### What is the Y MX C equation called?
The Y MX C equation is also known as the slope-intercept form of a linear equation.
### What does MX mean in geometry?
M is the slope of a line and X is the x-coordinate of a point on that line.
### What is the slope of Y MX C?
The slope of the Y MX C equation is 3.
### What is slope and intercept in y MX C?
The slope is the number that is multiplied by the x-coordinate, and the intercept is the number that is added to the product of the slope and the x-coordinate. In the equation y = 3x + 4, 3 is the slope and 4 is the intercept. |
Master the 7 pillars of school success
Common Core Standard 6.NS.4
What is the Distributive Property?
When do you use the Distributive Property?
What type of problems use the Distributive Property
Distributive Property
Transcript
Today we are going to talk about the distributive property. Here’s our example: Two times five X minus two Y times three. That’s ten X minus four Y plus 6. Now let’s go over the rules for the distributive property. The rules for the distributive property are very simple. We are going to multiply each term by the led number or led coefficient in front of our parenthesis. So we are multiplying. That’s one of the biggest mistakes I see students commit. They add instead of multiplying. Now let’s go back and break it into slow mo. We are going to take the two and multiply it by each term. Two times five X which equals 10 and I bring down my X and two times negative two Y so I have a negative and the two times two Y is four Y and I bring down the variable. Then I take the two times the positive three I take two bring down my positive, two times three gives me positive six. So there is the distributive property. Now let’s look at one that has a variable in it. We are going to distribute the three X into a trinomial, because we have three terms. So we are going to take three X times two X squared. So we take the two times the three, which gives us six times an X squared which gives us an X cubed because we are going to add the exponents. Now we take the three X minus three Y three times three is X just gives us an XY Now the final term I’m going to bring down my addition Three X times positive eight is twenty four there are no other variables so I will tack on my X That is the distributive property
# Disributive Property Definition
Multiply the led coefficient (2), by the numbers inside the parentheses one at a time.
The numbers inside the parentheses cannot be added together because they are not like terms
Same rule applies with exponents, multiple the led coefficient, which in this example is 3x, by the numbers inside the parentheses one at a time.
What is the distributive propery?
A formal definition of the Distributive Property would sound something like this," The distributive property states that multiplying a sum by a number gives the same result as multiplying each number, and then adding the products together."
In other words you can add the numbers in the parenthesis, and then multiply, or multiply the number outside the parenthesis, and then add.
In either case you get the same result. This example is called the Distributive property over addition.
For example 4(2+3)
4(2+3)= 4 * 5 = 20 (add then multiply)
or 4(2+3)= 4*2 + 4*3 = 8 + 12 = 20 (multiply then add)
The Distributive property over subtraction works the same way.
You can subtract the numbers inside the parenthesis and then add, or multiply the number outside the parenthesis and then subtract.
For example: 4(5-2) = 4(3) = 12 (Subtract then multiply
or 4*5 - 4*2 = 20-8 =12 (Multiply then subtract)
The Distributive property is written as:
a(b + c) = ab + ac
When using the Distributive Property follow these rules.
1. Multiply each term inside the parenthesis by the lead term, which is the number outside the parentheses
2. Do not combine unlike terms inside the parenthesis, and the multiply this term by the lead term.
3. Do not multiply the lead term, and then add the lead term to the next term or terms.
For example,it would be incorrect to remove the parenthesis and then add.
For example: 4(5 +4x)= 4*5 + 4+4x
20+8x=28x this is incorrect.
According to the distributive property you must multiply the lead terms by the terms inside the parenthesis.
Correct:
4(5+5x) = 4*5 + 4*5x
20+20x = 40x
## Algebra distributive property
Use the Distributive Property to multiply in Algebra
## How to use the distributive property to simplify algebraic equations
4(x + 3) - 2x
2y - 4(3y + 5)
5 - 2(6 - 3y)
4b + 3(7 - 2a) + 5
x - 3(x - 6) + 5(3x + 2)
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# Circumference
## Pi times the diameter or the distance around the circle.
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Circumference
What if you wanted to find the "length" of the crust for an entire pizza? A typical large pizza has a diameter of 14 inches and is cut into 8 or 10 pieces. Think of the crust as the circumference of the pizza. Find the circumference. After completing this Concept, you'll be able to solve this problem.
### Guidance
Circumference is the distance around a circle. The circumference can also be called the perimeter of a circle. However, we use the term circumference for circles because they are round. The term perimeter is reserved for figures with straight sides. In order to find the formula for the circumference of a circle, we first need to determine the ratio between the circumference and diameter of a circle.
##### Investigation: Finding \begin{align*}\pi\end{align*} (pi)
Tools Needed: paper, pencil, compass, ruler, string, and scissors
1. Draw three circles with radii of 2 in, 3 in, and 4 in. Label the centers of each \begin{align*}A, B\end{align*}, and \begin{align*}C\end{align*}.
2. Draw in the diameters and determine their lengths. Are all the diameter lengths the same in \begin{align*}\bigodot A\end{align*}? \begin{align*}\bigodot B\end{align*}? \begin{align*}\bigodot C\end{align*}?
3. Take the string and outline each circle with it. The string represents the circumference of the circle. Cut the string so that it perfectly outlines the circle. Then, lay it out straight and measure, in inches. Round your answer to the nearest \begin{align*}\frac{1}{8}\end{align*}-inch. Repeat this for the other two circles.
4. Find \begin{align*}\frac{circumference}{diameter}\end{align*} for each circle. Record your answers to the nearest thousandth. What do you notice?
From this investigation, you should see that \begin{align*}\frac{circumference}{diameter}\end{align*} approaches 3.14159... The bigger the diameter, the closer the ratio was to this number. We call this number \begin{align*}\pi\end{align*}, the Greek letter “pi.” It is an irrational number because the decimal never repeats itself. Pi has been calculated out to the millionth place and there is still no pattern in the sequence of numbers. When finding the circumference and area of circles, we must use \begin{align*}\pi\end{align*}. \begin{align*}\pi\end{align*}, or pi is the ratio of the circumference of a circle to its diameter. It is approximately equal to 3.14159265358979323846... To see more digits of \begin{align*}\pi\end{align*}, go to http://www.eveandersson.com/pi/digits/.
From this Investigation, we found that \begin{align*}\frac{circumference}{diameter}=\pi\end{align*}. In other words, \begin{align*}C= \pi d\end{align*}. We can also say \begin{align*}C=2 \pi r\end{align*} because \begin{align*}d=2r\end{align*}.
#### Example A
Find the circumference of a circle with a radius of 7 cm.
Plug the radius into the formula.
\begin{align*}C=2 \pi (7)=14 \pi \approx 44 \ cm\end{align*}
Depending on the directions in a given problem, you can either leave the answer in terms of \begin{align*}\pi\end{align*} or multiply it out and get an approximation. Make sure you read the directions.
#### Example B
The circumference of a circle is \begin{align*}64 \pi\end{align*}. Find the diameter.
Again, you can plug in what you know into the circumference formula and solve for \begin{align*}d\end{align*}.
\begin{align*}64 \pi= \pi d=14 \pi\end{align*}
#### Example C
A circle is inscribed in a square with 10 in. sides. What is the circumference of the circle? Leave your answer in terms of \begin{align*}\pi\end{align*}.
From the picture, we can see that the diameter of the circle is equal to the length of a side. Use the circumference formula.
\begin{align*}C=10 \pi \ in.\end{align*}
Watch this video for help with the Examples above.
#### Concept Problem Revisited
The entire length of the crust, or the circumference of the pizza is \begin{align*}14 \pi \approx 44 \ in\end{align*}.
### Vocabulary
A circle is the set of all points that are the same distance away from a specific point, called the center. A radius is the distance from the center to the outer rim of the circle. A chord is a line segment whose endpoints are on a circle. A diameter is a chord that passes through the center of the circle. The length of a diameter is two times the length of a radius. Circumference is the distance around a circle. \begin{align*}\pi\end{align*}, or “pi” is the ratio of the circumference of a circle to its diameter.
### Guided Practice
1. Find the perimeter of the square. Is it more or less than the circumference of the circle? Why?
2. The tires on a compact car are 18 inches in diameter. How far does the car travel after the tires turn once? How far does the car travel after 2500 rotations of the tires?
3. Find the radius of circle with circumference 88 in.
1. The perimeter is \begin{align*}P=4(10)=40 \ in\end{align*}. In order to compare the perimeter with the circumference we should change the circumference into a decimal.
\begin{align*}C=10 \pi \approx 31.42 \ in\end{align*}. This is less than the perimeter of the square, which makes sense because the circle is smaller than the square.
2. One turn of the tire is the circumference. This would be \begin{align*}C=18 \pi \approx 56.55 \ in\end{align*}. 2500 rotations would be \begin{align*}2500 \cdot 56.55in \ \approx 141,375 \ in\end{align*}, 11,781 ft, or 2.23 miles.
3. Use the formula for circumference and solve for the radius.
\begin{align*}C&=2 \pi r\\ 88 &=2 \pi r\\ \frac{44}{\pi}&=r\\ r&\approx 14 \ in\end{align*}
### Practice
Fill in the following table. Leave all answers in terms of \begin{align*}\pi\end{align*}.
1. 15
2. 4
3. 6
4. \begin{align*}84 \pi\end{align*}
5. 9
6. \begin{align*}25 \pi\end{align*}
7. \begin{align*}2 \pi\end{align*}
8. 36
1. Find the radius of circle with circumference 88 in.
2. Find the circumference of a circle with \begin{align*}d=\frac{20}{\pi} cm\end{align*}.
Square \begin{align*}PQSR\end{align*} is inscribed in \begin{align*}\bigodot T\end{align*}. \begin{align*}RS=8 \sqrt{2}\end{align*}.
1. Find the length of the diameter of \begin{align*}\bigodot T\end{align*}.
2. How does the diameter relate to \begin{align*}PQSR\end{align*}?
3. Find the perimeter of \begin{align*}PQSR\end{align*}.
4. Find the circumference of \begin{align*}\bigodot T\end{align*}.
1. A truck has tires with a 26 in diameter.
1. How far does the truck travel every time a tire turns exactly once?
2. How many times will the tire turn after the truck travels 1 mile? (1 mile = 5280 feet)
2. Jay is decorating a cake for a friend’s birthday. They want to put gumdrops around the edge of the cake which has a 12 in diameter. Each gumdrop is has a diameter of 1.25 cm. To the nearest gumdrop, how many will they need?
3. Bob wants to put new weather stripping around a semicircular window above his door. The base of the window (diameter) is 36 inches. How much weather stripping does he need?
4. Each car on a Ferris wheel travels 942.5 ft during the 10 rotations of each ride. How high is each car at the highest point of each rotation?
### Vocabulary Language: English Spanish
chord
A line segment whose endpoints are on a circle.
Circumference
The circumference of a circle is the measure of the distance around the outside edge of a circle.
diameter
A chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.
pi
(or $\pi$) The ratio of the circumference of a circle to its diameter.
Congruent
Congruent figures are identical in size, shape and measure.
The radius of a circle is the distance from the center of the circle to the edge of the circle.
Regular Polygon
A regular polygon is a polygon with all sides the same length and all angles the same measure. |
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# Divisible subsets
Armaan Nougai
### Problem statement
Given a multiset with n integers, find such a non-empty subset of it that the sum of the subset’s elements is divisible by N.
If many valid subsets are there, then print any.
Pre-requisites:
• The pigeonhole principle
• Subarrays
### Problem explanation
Let’s look at the following example.
• N=4
• array = [4, 6, 10, 3]
Condition = the sum of the subset must be divisible by N (that is 4 in this example )
Some of the valid subsets are:
• {4}
• {6, 10}
### Solution
We’ll prove that there will always exist a subarray of the array whose sum is divisible by N.
Let’s proceed with an example.
• N = 5
• array = [1, 2, 3, 5, 7]
We’ll first create a cumulative sum array, cumArray = [1, 3, 6, 11, 18]
Here, cumArray [i] represents the sum of subarray (0, i). We’ll change this cumArray to cumArray%N.
cumArray = [1, 3, 1, 1, 3]
Now, if any cumArray[ i ] == 0, this will mean that:
• The sum of subarray (0, i)%N==0
• Thus subarray found.
Else:
• cumArray has N values.
• Since we changed cumArray to cumArray%N, this means values in cumArray can only be in the range [0, N-1]. Which means there could be at max N different values in cumArray. But since we already checked there exists no 0 in cumArray, thus cumArray can now only have N-1 different value.
• Now, According to the pigeonhole principle, at least 1 value in cumArray will occur twice.
• This means there always will exist at least one (a,b) pair such that:
• cumArray[ a]%N = cumArray[ b]%N
• Therefore, the sum of subarray (a, b) is divisible by N.
• Thus the subarray is found.
Hence, we proved that in either case, a subarray will exist with sum divisible by N.
Now, as we know it, we now do not need it to generate the whole cumArray first and then find the required subarray. We could simply find the required subarray whilst calculating the cumArray.
If x is the sum until ith index, then we just need to check if either x=0 or x occurred in cumArray previously. If yes, then we got the required subarray otherwise add x to cumArray and check for i+1th index.
### Solution
#### Input
• The first line of the input contains an integer T denoting the number of test cases.
• The first line of each test consists of a single integer N which is the size of the multiset.
• The second line of each test contains N space-separated integers which is the multiset’s elements.
Note. Press enter/ return key(↵) after each line, in order to input the next line.
### Output
• Print -1 if the required subset doesn’t exist.
• Otherwise print two lines, first the size of the subset, then the space-separated indices of the subset.
### Example
• 1
• 5
• 2 4 2 9 3
#### Output
• 3
• 2 3 4
t = int(input())
# t - test cases
for _ in range(t):
N = int(input())
# N - size of array
array = list(map(int,input().split()))
# array - array of N numbers.
sumx = 0
sumxIndex = {}
for i in range(0,N):
sumx = (sumx + array[i])%N
if (sumx==0) :
# sum of subarray [0,i] is divisible by N.
left = 0
right = i
break
if (sumx in sumxIndex.keys()):
# sum of subarray [prev index of sumx in sumIndex, i] is divisible by N
left = sumxIndex[sumx]+1
right = i
break
sumxIndex[sumx] = i
print(right-left+1)
for index in range(left,right+1):
print(index+1,end=" ")
print()
Enter the input below
Python Code
### Code explanation
• Line 1: Take the number of test cases( t) as input.
• Line 3: Start a for loop for each test case.
• Line 5: Take the number of elements(N) as input.
• Line 7: Store N space-separated integers in a list (array) as input.
• Line 10: sumx will store the sum until the current index.
• Line 11: sumxIndex will store the sumx with respective indices.
• Example:
• N = 5
• Array = [1,3,4,10,5]
• At index 0 -> sumx=1%5=1 -> sumxIndex = { 1:0 }
• At index 1 -> sumx=3%5=3 -> sumxIndex = { 1:0 , 3:4 }
• At index 2 -> sumx=7%5=2 -> sumxIndex = { 1:0 , 3:4 , 2:2 }
• Line 13: Start for loop from [0-N].
• Line 14: Since the sum until the ith element = the sum until i-1th element + array[i]. Thus, sumx = ( sumx + array[i] )%N.
• Line 16: Check if sumx=0.
• Line 15: If yes, then subarray [0, i] will be the required subarray. Thus, putting left = 0, right = i and breaking the loop.
• Line 22: Check if sumx already occurred previously in sumxIndex.
• Line 24: If yes, then left = previous index at which sumx occured in sumxIndex + 1. , right = i and breaks the loop.
• Line 31: Print the size of found subarray i.e. right-left+1.
• Line 33: Now, for printing the indices of the subarray starting for loop for i from range = [left,right].
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# Precalculus : Rational Functions
## Example Questions
1 2 3 4 6 Next →
### Example Question #8 : Rational Equations And Partial Fractions
Solve the inequality:
Possible Answers:
1 < z < 31
z < 1
No solution
z < 1 or z > 31
Correct answer:
z < 1 or z > 31
Explanation:
First, find the LCD, which is 6(z-1). Then, multiply the entire equation by this:
This simplifies to:
Now, make a number line with points at 1 and 31, and vertical lines in between zones:
Let's test three points, one in each zone:
f(0):
f(10):
f(35):
Therefore, the inequality is true for values of z such that z < 1 or z > 31.
### Example Question #9 : Rational Equations And Partial Fractions
Decompose into partial fractions.
Possible Answers:
Correct answer:
Explanation:
Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.
Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y.
Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD.
Looking back to our original fraction, note that the values p = 1 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.
First, let p=1.
Second, let p=-1
Now, substitute these values in for X and Y.
Therefore, .
### Example Question #10 : Rational Equations And Partial Fractions
Decompose into partial fractions.
Possible Answers:
Correct answer:
Explanation:
Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.
Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y.
Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD.
Looking back to our original fraction, note that the values b = 0 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.
First, let b=0.
Second, let b=-1
Now, substitute these values in for X and Y.
Therefore, .
1 2 3 4 6 Next → |
# Triangle Inequality Theorem - Explanation
## What is Triangle Inequality Theorem ?
If I want to give you a perfect definition for Triangle Inequality then I can say : -
The sum of the lengths of any two sides of a triangle is always greater than the length of the third side of that triangle.
It follows from the fact that a straight line is the shortest path between two points. The inequality is strict if the triangle is non-degenerate (meaning it has a non-zero area).
So in other words we can say that : It is not possible to construct a triangle from three line segments if any of them is longer than the sum of the other two. This is known as The Converse of the Triangle Inequality theorem .
So suppose we have three sides lengths as 6 m, 4 m and 3 m then can we draw a triangle with this side ? The answer will be YES we can.
Suppose side a = 3 m
length of side b = 4 m
Length of side c = 6 m
if side a + side b > side c then only we can draw the triangle or
side b + side c > side a or
side a + side c > side b
So from the above example we can find that 4 m + 3 m > 6 m
But look if we try to take 4 m + 6 m $\geq$ 3 m .
This inequality is particularly useful and shows up frequently on Intermediate level geometry problems. It also provides the basis for the definition of a metric spaces and analysis.
## Problem using Triangle Inequality :
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
• 43
• 44
• 45
• 46
### Key Concepts
Triangle Inequality
Inequality
Geometry
AMC - 2006 - 10 B - Problem 10
Secrets in Inequalities.
## Try with Hints
This can be a very good example to show Triangle Inequality
Let ' x ' be the length of the first side of the given triangle. So the length of the second side will be 3 x and that of the third side be 15 . Now apply triangle inequality and try to find the possible values of the sides.
If we apply Triangle Inequality here then the expression will be like
$3 x < x + 15$
$2 x < 15$
$x < \frac {15}{2}$
x < 7.5
Now do the rest of the problem ...........
I am sure you have already got the answer but let me show the rest of the steps for this sum
If x < 7.5 then
The largest integer satisfying this inequality is 7.
So the largest perimeter is 7 + 3.7 +15 = 7 + 21 + 15 = 43.
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# KSEEB Solutions for Class 10 Maths Chapter 10 Quadratic Equations Ex 10.2
Students can Download Class 10 Maths Chapter 10 Quadratic Equations Ex 10.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
## Karnataka State Syllabus Class 10 Maths Chapter 10 Quadratic Equations Ex 10.2
Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) $$\sqrt{2}$$x2 + 7x + 5$$\sqrt{2}$$ = 0
(iv) 2x2 – x + $$\frac{1}{8}$$ = 0
(v) 100 x2 – 20x + 1 = 0
i) x2 – 3x – 10 = 0
x2 – 5x + 2x – 10 = 0
x (x – 5) + 2 (x – 5) = 0
(x – 5) (x + 2) = 0
x – 5 = 0 (or) x + 2 = 0
x = 5 (or) x = – 2
∴ x = 5, x = – 2
∴ 5 & – 2 are the roots of the equation x2 – 3x – 10 = 0
ii) 2x2 + x – 6 = 0
2x2 + 4x – 3x – 6 = 0
2x (x + 2) – 3 (x + 2) = 0
(x + 2) (2x – 3) = 0
x + 2 = 0 (or) 2x – 3 = 0
x = $$\frac{3}{2}$$
x = – 2 (or) 2x = 3
x = $$\frac{3}{2}$$
x = – 2, $$\frac{3}{2}$$
∴ – 2 & $$\frac{3}{2}$$ are the roots of the equation 2x2 + x – 6 = 0
iii) $$\sqrt{2}$$ + 7x + 5$$\sqrt{2}$$ = 0
(iv) 2x2 – x + $$\frac{1}{8}$$ = 0
16x2 – 8x + 1 = 0
16x2 – 4x – 4x + 1 = 0
4x(4x – 1) – 1(4x- 1) = 0
(4x – 1) (4x – 1) = 0
(4x – 1)2 = 0
If 4x – 1 = 0, then 4x = 1
∴ x = $$\frac{1}{4}$$
∴ Two roots are $$\frac{1}{4}, \frac{1}{4}$$
v) 100x2 – 20x + 1 =0
100x2 – 10x – 10x + 1 =0
10x(10x – 1) – 1(10x – 1) = 0
(10x – 1)(10x – 1) = 0
10x – 1 = 0 (or) 10x – 1 = 0
10x = 1 or 10x = 1
x = $$\frac{1}{10}$$ (or) x = $$\frac{1}{10}$$
x = $$\frac{1}{10}$$, $$\frac{1}{10}$$
∴ $$\frac{1}{10}$$ & $$\frac{1}{10}$$ are the roots of the equation of 100x2 – 20x + 1 =0
Question 2.
Solve the problems given in Example 1.
i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Let the number of marbles John had be ‘x’ & Jivanti had be 45 – x
John lost 5 marbles = (x – 5) & Jivanti lost 5 marbles = 45 – x – 5 = 40 – x.
their product = (x – 5) (40 – x) = 124
40x – 200 – x2 + 5x = 124
– x2 + 45x – 200 – 124 = 0
– x2 + 45x – 324 = 0
Multiply by ‘_’
x2 – 45x + 324 = 0
x2 – 36x – 9x + 324 = 0
x (x – 36) – 9 (x – 36) = 0
(x – 36) (x – 9) = 0
x – 36 = 0 or x – 9 = 0
x = 36 (or) x = 9
x = 36, 9
∴ Therefore John and Jivanti had 36 and 9 (or) 9 and 36 marbles.
ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.
Let the number of toys produced on that day be ‘x’
The cost of production of each toy = 55 – x
The total cost of Production that day
= x (55 – x)
∴ x (55 – x) = 750
55x – x2 = 750
x2 – 55x + 750 = 0
x2 – 30x – 25x + 750 = 0
x (x – 30) – 25 (x – 30) = 0
(x – 30) (x – 25) = 0
x – 30 = 0 or x – 25 = 0
x = 30 (or) x = 25
The number of toys produced on that day has 25 (or) 30.
Question 3.
Find two numbers whose sum is 27 and the product is 182.
In those numbers, let one of the numbers be ‘x’.
Another number is (27 – x).
Their product is 182.
∴ x (27 – x) = 182
27x – x2 = 182
-x2 + 27x – 182 = 0
x2 – 27x + 182 = 0
x2 – 14x – 13x + 182 = 0
x(x – 14) – 13(x – 14) = 0
(x – 14) (x – 13) = 0
If x – 14 = 0, then x = 14
If x – 13 = 0, then x = 13
∴ x = 14 OR x = 13
∴ The Numbers are 14 and 13.
Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Let two consecutive Positive integers be x and x + 1
Sum of their squares = 365
x2 + (x + 1)2 = 365
x2 + x2 + 2x + 1 = 365
2x2 + 2x + 1 – 365 = 0
2x2 + 2x – 364 = 0
x2 + x – 182 = 0
x2 + 14x – 13x – 182 = 0
x(x + 14) – 13 (x + 14) = 0 (x + 14) (x – 13) = 0
x + 14 = 0 (or) x – 13 = 0
x = – 14 or x = 13
x= 13
Two consecutive numbers are x, x + 1
∴ 13, 14
Question 5.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let the base of right angle triangle be ‘x’ and altitude be x – 7, hypotenuse = 13 cm
(hypotenuse)2 = (base)2 + (Alitutude)2 [By Pythagoras theorem]
(13)2 = x2 + (x – 7)2
169 = x2 + x2 – 14x + 49
2x2 – 14x +49 – 169 = 0
2x2 – 14x – 120 = 0 Divide by 2
x2 – 7x – 60 = 0
x2 – 12x + 5x – 60 = 0
x(x – 12) + 5 (x – 12) = 0
(x – 12)(x + 5) = 0
x – 12 = 0 (or) x + 5 = 0
x = 12 or x = – 5
x = 12
∴ base of Right angle triangle
= x = 12 cm
∴ Altitude of Right angle triangle
= x – 7 = 12 – 7 = 5 cm
∴ Other two sides are 12 cm & 5 cm.
Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
Let the number of pottery articles be ‘x’.
Cost of production of each (in rupees) is = (2x + 3)
∴ x(2x + 3) = 90
2x2 + 3x = 90
2x2 + 3x – 90 = 0
2x2 – 12x + 15x – 90 = 0
2x(x – 6) + 15 (x – 6) = 0
(x – 6) (2x + 15) = 0
If x – 6 = 0, then x = 6
If 2x + 15 = 0, then 2x = -15
∴ x = $$\frac{-15}{2}$$
∴ Number of pottery articles produced is 6.
Cost of production of 6 pottery articles is Rs. 90
Cost of production of 1 pottery article ……….. ?
$$\frac{1 \times 90}{6}$$ = Rs. 15
OR Cost of Production= 2x + 3
= 2 × 6 + 3
= 12 + 3
= Rs. 15. |
## Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.8
Exercise 3.8
1. The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
The numerator of the fraction is 4 less the denominator. Thus, we have
x = y – 4
⇒ x – y = -4
If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is 8 times the numerator. Thus, we have
y + 1 = 8(x-2)
⇒ y + 1 = 8x – 16
⇒ 8x – y = 1 + 16
⇒ 8x – y = 17
So, we have two equations
x – y = -4
8x – y = 17
Here, x and y are unknowns. We have to solve the above equations for x and y.
Subtracting the second equation from the first equation, we get
(x-y) – (8x – y) = -4 -17
⇒ x – y - 8x + y = -21
⇒ - 7x = -21
⇒ 7x = 21
⇒ x = 21/7
⇒ x = 3
Substituting the value of x in the first equation , we have
3 – y = -4
⇒ y = 3 + 4
⇒ y =7
Hence, the fraction is 3/7.
2. A fraction becomes 9/11 if 2 is added to both numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
If 2 is added to both numerator and the denominator, the fraction becomes 9/11 . Thus, we have
x+2/y+2 = 9/11
⇒ 11(x+2) = 9 (y+2)
⇒ 11x + 22 = 9(y+2)
⇒ 11x + 22 = 9y + 18
⇒ 11x – 9y = 18 – 22
⇒ 11x – 9y + 4 = 0
If 3 is added to both numerator and the denominator, the fraction becomes 5/6 . Thus , we have
x + 3/y + 3 = 5/6
⇒ 6(x+3) = 5 (y+3)
⇒ 6x + 18 = 5y + 15
⇒ 6x – 5y = 15 - 18
⇒ 6x – 5y + 3 = 0
So, we have two equations
11x -9y + 4 =0
6x -5y + 3 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
⇒ x = 7, y = 9
Hence, the fraction is 7/9
3. A fraction becomes 1/3 if 1 is subtracted from both its numerator and denominator. It 1 is added to both the numerator and denominator, it becomes 1/2. Find the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y .
If 1 is substracted from both numerator and the denominator, the fraction becomes 1/3.
x-1/y -1 = 1/3
⇒ 3(x-1) = y-1
⇒ 3x – 3 = y -1
⇒ 3x – y – 2 = 0
If 1 is added to both numerator and the denominator, the fraction becomes 1/2 .
x+1/y +1 = 1/2
⇒ 2(x+1) = y+1
⇒ 2x + 2 = y + 1
⇒ 2x – y + 1 = 0
So, we have two equations
3x – y – 2 = 0
2x – y + 1 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y.
By using cross – multiplication, we have
Hence, the fraction is 3/7 .
4. If we add 1 to the numerator and subtract 1 from the denominator, a fraction becomes 1. It also becomes 1/2 if we only add 1 to the denominator. What is the fraction ?
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y .
If 1 is added to the numerator and 1 is subtracted from the denominator, the fraction becomes 1. Thus, we have
x+1/y -1 = 1
⇒x + 1 = y – 1
⇒ x + 1 – y + 1 = 0
⇒ x – y + 2 = 0
If 1 is added to the denominator , the fraction becomes 1/2 . Thus, we have
x/y+1 = ½
⇒ 2x = y + 1
⇒ 2x – y – 1 = 0
So, we have two equations
x – y + 2 = 0
2x – y – 1 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
5. If the numerator of a fraction is multiplied by 2 and the denominator is reduced by 5 the fraction becomes 6/5. And, if the denominator is doubled and the numerator is increased by 8, the fraction becomes 2/5. find the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
If the numerator is multiplied by 2 and the denominator is reduced by 5, the fraction becomes 6/5 . Thus, we have
2x/y–5 = 6 / 5
⇒ 10x = 6 (y-5)
⇒ 10x = 6y – 30
⇒ 10x – 6y + 30 = 0
⇒ 2 (5x – 3y + 15) = 0
⇒ 5x – 3y + 15 = 0
If the denominator is doubled and the numerator is increased by 8, the fraction becomes 2/5 . Thus, we have
x+8/2y = 2/5
⇒ 5(x+8) = 4y
⇒ 5x + 40 = 4y
⇒ 5x – 4y + 40 = 0
So, we have two equations
5x – 3y + 15 = 0
5x – 4y + 40 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
6. When 3 is added to the denominator and 2 is subtracted from the numerator a fraction becomes 1/4. And when 6 is added to numerator and the denominator is multiplied by 3, it becomes 2/3. Find the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y .
If 3 is added to the denominator and 2 is subtracted from the numerator , the fraction becomes 1/4.
Thus , we have
x-2/y+3 = 1/4
⇒ 4(x-2) = y+3
⇒ 4x – 8 = y + 3
4x – y – 11 = 0
If 6 is added to the numerator and the denominator is multiplied by 3, the fraction becomes 2/3 . Thus, we have
x+6/3y = 2/3
⇒ 3(x+6) = 6y
⇒ 3x + 18 = 6y
⇒ 3x – 6y + 18 = 0
⇒ 3(x - 2y + 6) = 0
⇒ x – 2y + 6 = 0
So, we have two equations
4x – y – 11 = 0
x – 2y + 6 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
7. The sum of a numerator and denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to 1/3. Find the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
The sum of the numerator and the denominator of the fraction is 18. Thus, we have
x + y = 18
⇒ x + y – 18 = 0
If the denominator is increased by 2, the fraction reduces by 2, the fraction reduces to 1/3 . Thus, we have
x/y+2 = 1/3
⇒ 3x = y + 2
⇒ 3x – y – 2 = 0
So, we have two equations
x + y – 18 = 0
3x – y – 2 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
8. If 2 is added to the numerator of a fraction, it reduces to 1/2 and if 1 is subtracted from the denominator, it reduces to 1/3. Find the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
If 2 is added to the numerator of the fraction, it reduces to 1/2 . Thus, we have
x+2/y = 1/2
⇒ 2(x+2) = y
⇒ 2x + 4 = y
⇒ 2x – y + 4 = 0
If 1 is subtracted from the denominator, the fraction reduces to 1/3 . Thus, we have
x/y – 1 = 1/3
⇒ 3x = y – 1
⇒ 3x – y + 1 = 0
So, we have two equations
2x –y + 4 = 0
3x – y + 1 = 0
Here x and y are unknowns . We have to solve the above equations for x and y .
By using cross – multiplicatiion , we have
9. The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
The sum of the numerator and denominator of the fraction is 4 more than twice the numerator. Thus, we have
x+y = 2x + 4
⇒ 2x + 4 – x – y = 0
⇒ x – y + 4 = 0
If the numerator and denominator are increased by 3, they are in the ratio 2:3. Thus, we have
x + 3 : y + 3 = 2 : 3
⇒ x + 3 / y + 3 = 2/3
⇒ 3(x+3) = 2(y+3)
⇒ 3x + 9 = 2y + 6
⇒ 3x – 2y + 3 = 0
So, we have two equations
x – y + 4 = 0
3x – 2y + 3 = 0
Here x and y are unknowns. We have to solve the above equations for x and y. By using cross-multiplication, we have
10. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
The sum of the numerator and denominator of the fraction is 3 less than twice the denominator.
Thus, we have
x+y = 2y – 3
⇒ x+y-2y+3 = 0
⇒ x–y+3= 0
If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Thus, we have
x-1 = 1/2 (y-1)
⇒ x-1/y-1 =1/2
⇒ 2(x-1) = y-1
⇒ 2x-2=y-1
⇒ 2x-y-1=0
So, we have two equations
x-y+3=0
2x-y-5=0
Here x and y are unknowns . We have to solve the above equations for x and y.
By using cross – multiplication, we have
11. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y The sum of the numerator and denominator of the fraction is 12. Thus, we have
x+y=12
⇒ x+y-12=0
If the denominator is increased by 3, the fraction becomes 1/2. Thus, we have
x/y+3 = 1/2
⇒ 2x = y + 3 = 0
⇒ 2x – y – 3 = 0
So, we have two equations
x + y – 12 = 0
2x – y – 3 = 0
Here x and y unknowns . We have to solve the above equations for x and y. By using cross – multiplication, we have |
# 346. Area & Volume conversion
This is a quick post on how I teach metric unit conversion for area and volume. All you need is a big whiteboard and coloured board pens.
Start by stressing that all diagrams are not to scale/accurate.
Two colours
1. Draw a square on the board
2. Pen colour 1: Label it as 1cm
3. What is the area? Show the calculation
4. What is 1cm in mm?
5. Pen colour 2: Label it as 10mm
6. What is the area in mm? Show the calculation
7. What is the scale factor between the sides? the area? why?
Three colours
1. Draw a square on the board
2. Pen colour 3: Label it as 1m
3. What is the area? Show the calculation
4. What is 1m in cm?
5. Pen colour 2: Label it as 100cm
6. What is the area in cm? Show the calculation
7. What is the scale factor between the sides? the area? why?
8. Repeat in pen colour 1 for mm
Four colours
Well not actually four colours – pens 2,3 & 4 only. Repeat the process for kilometres to metres and centimetres.
Volume – same process, just three dimensions
Why all the colours?
By coding each unit of measurement with a colour students can see the progression of the calculations and the links between area/volume and scale factor. After all, an okay mathematician can reproduce memorised facts, but a great mathematician doesn’t need to memorise – they understand where the calculations came from.
# 345. Practical percentage skills
It’s perfectly obvious that fluency in the use of multiplication tables directly impacts students ability to divide. This grows into confidence with algebra and reverse operations. Students are able to see the links between the concepts. Our understanding of the importance of such skills is part of the success of programmes such as TTRockstars and Numeracy Ninjas.
Why is it then that so many textbooks, websites and resource banks keep the manipulation of percentages as separate skills sets? Percentage increase / Percentage change / Reverse percentages. We know that when concepts overlap, fluency increases when these links are pursued. So that’s what I set out to do.
I have a bright Year 8 class and started working on percentages with them. It didn’t take much to have them confident using equivalent decimal multipliers to find percentages of amounts. Using a multiplier for increase/decrease was a walk in the park. Then finding percentage change came up. Over the years I’ve seen a lot of students get very confused with half remembered methods:
“Which do I take away?”
“What number do I divide by?”
“Is this calculation the right way around?”
I tend to teach new value divided by old value and interpret the answer. It got me thinking – why am I teaching them this? They can increase by a percentage using a multiplier, why can’t they rearrange their working to find the actual percentage? Same goes for reversing a percentage.
After a good discussion, I used this worksheet to recap and develop their skills:
Percentages Linking concepts questions |
# 2011 AMC 12A Problems/Problem 16
## Problem
Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 2880 \qquad \textbf{(C)}\ 3120 \qquad \textbf{(D)}\ 3250 \qquad \textbf{(E)}\ 3750$
## Solution
We can do some casework when working our way around the pentagon from $A$ to $E$. At each stage, there will be a makeshift diagram.
1.) For $A$, we can choose any of the 6 colors.
A : 6
2.) For $B$, we can either have the same color as $A$, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and $D$ will be affected by both $A$ and $B$.
A : 6
B:1 B:5
3.) For $C$, we cannot have the same color as $A$. Also, we can have the same color as $B$ ($E$ will be affected), or any of the other 4 colors. Because $C$ can't be the same as $A$, it can't be the same as $B$ if $B$ is the same as $A$, so it can be any of the 5 other colors.
A : 6
B:1 B:5
C:5 C:4 C:1
4.) $D$ is affected by $A$ and $B$. If they are the same, then $D$ can be any of the other 5 colors. If they are different, then $D$ can be any of the (6-2)=4 colors.
A : 6
B:1 B:5
C:5 C:4 C:1
D:5 D:4 D:4
5.) $E$ is affected by $B$ and $C$. If they are the same, then $E$ can be any of the other 5 colors. If they are different, then $E$ can be any of the (6-2)=4 colors.
A : 6
B:1 B:5
C:5 C:4 C:1
D:5 D:4 D:4
E:4 E:4 E:5
6.) Now, we can multiply these three paths and add them: $(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)=600+1920+600=3120$
7.) Our answer is $C$!
## Solution 2
Right off the bat, we can analyze three things:
1.) There can only be two of the same color on the pentagon.
2.) Any pair of the same color can only be next to each other on the pentagon.
3.) There can only be two different pairs of same colors on the pentagon at once.
Now that we know this, we can solve the problem by using three cases: no same color pairs, one same color pair, and two same color pairs.
1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count $6!=720$ cases. No rotation is necessary because all permutations are accounted for.
2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex.
We get $6\times5\times4\times3=360$.
However, there are 5 different locations the pair could be at. Therefore we get $360\times5=1800$ possibilities for two pairs.
3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.
We get $6\times5\times4=120$.
Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get $120\times5=600$.
5.) If we add all of three cases together, we get $720+1800+600=3120$. The answer is $C$.
## See also
2011 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Shifts of Square Root Functions
## Translate square root functions vertically and horizontally
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Shifts of Square Root Functions
### Shifts of Square Root Functions
We will now look at how graphs are shifted up and down in the Cartesian plane.
Graph the functions \begin{align*}y=\sqrt{x}, y=\sqrt{x} + 2\end{align*} and \begin{align*}y=\sqrt{x} - 2\end{align*}.
When we add a constant to the right-hand side of the equation, the graph keeps the same shape, but shifts up for a positive constant or down for a negative one.
#### Graphing Multiple Functions
Graph the functions \begin{align*}y=\sqrt{x}, y=\sqrt{x - 2},\end{align*} and \begin{align*}y = \sqrt{x + 2}\end{align*}.
When we add a constant to the argument of the function (the part under the radical sign), the function shifts to the left for a positive constant and to the right for a negative constant.
Now let’s see how to combine all of the above types of transformations.
#### Combining Transformations
Graph the function \begin{align*}y = 2\sqrt{3x - 1} + 2\end{align*}.
We can think of this function as a combination of shifts and stretches of the basic square root function \begin{align*}y = \sqrt{x}\end{align*}. We know that the graph of that function looks like this:
If we multiply the argument by 3 to obtain \begin{align*}y = \sqrt{3x}\end{align*}, this stretches the curve vertically because the value of \begin{align*}y\end{align*} increases faster by a factor of \begin{align*}\sqrt{3}\end{align*}.
Next, when we subtract 1 from the argument to obtain \begin{align*}y = \sqrt{3x - 1}\end{align*} this shifts the entire graph to the left by one unit.
Multiplying the function by a factor of 2 to obtain \begin{align*}y = 2 \sqrt{3x - 1}\end{align*} stretches the curve vertically again, because \begin{align*}y\end{align*} increases faster by a factor of 2.
Finally we add 2 to the function to obtain \begin{align*}y = 2 \sqrt{3x - 1} + 2\end{align*}. This shifts the entire function vertically by 2 units.
Each step of this process is shown in the graph below. The purple line shows the final result.
Now we know how to graph square root functions without making a table of values. If we know what the basic function looks like, we can use shifts and stretches to transform the function and get to the desired result.
### Example
#### Example 1
Graph the function \begin{align*}y = -\sqrt{x +3} -5\end{align*}.
We can think of this function as a combination of shifts and stretches of the basic square root function \begin{align*}y = \sqrt{x}\end{align*}. We know that the graph of that function looks like this:
Next, when we add 3 to the argument to obtain \begin{align*}y = \sqrt{x +3}\end{align*} this shifts the entire graph to the right by 3 units.
Multiplying the function by -1 to obtain \begin{align*}y = - \sqrt{x +3}\end{align*} which reflects the function across the \begin{align*}x\end{align*}-axis.
Finally we subtract 5 from the function to obtain \begin{align*}y = - \sqrt{x +3}-5\end{align*}. This shifts the entire function down vertically by 5 units.
### Review
Graph the following functions.
1. \begin{align*}y = \sqrt{2x - 1}\end{align*}
2. \begin{align*}y = \sqrt{x - 100}\end{align*}
3. \begin{align*}y = \sqrt{4x + 4}\end{align*}
4. \begin{align*}y = \sqrt{5 - x}\end{align*}
5. \begin{align*}y = 2\sqrt{x} + 5\end{align*}
6. \begin{align*}y = 3 - \sqrt{x}\end{align*}
7. \begin{align*}y = 4 + 2 \sqrt{x}\end{align*}
8. \begin{align*}y = 2 \sqrt{2x + 3} + 1\end{align*}
9. \begin{align*}y = 4 + \sqrt{2 - x}\end{align*}
10. \begin{align*}y = \sqrt{x + 1} - \sqrt{4x - 5}\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
square root function
A square root function is a function with the parent function $y=\sqrt{x}$.
Transformations
Transformations are used to change the graph of a parent function into the graph of a more complex function. |
# Section 5.6 Solving Quadratic Equations by Factoring
Solving Equations by Factoring Definition of Quadratic Equations Zero-Factor Property Strategy for Solving Quadratics Standard Form Quadratic Equation Quadratic equations can be written in the form ax2 + bx + c = 0 where a, b, and c are real numbers with a 0.
Standard form for a quadratic equation is in descending order equal to zero. BACK Examples of Quadratic Equations Standard Form 2 p 81 18 p 2 x 9 x 18
2 y 25 2 p 18 p 81 0 2 x 9 x 18 0 2
y 25 0 BACK Zero-Factor Property If a and b are real numbers and if ab =0, then a = 0 or b = 0. BACK Solve the equation (x + 2)(2x - 1)=0 By the zero factor property we know...
Since the product is equal to zero then one of the factors must be zero. ( x 2) 0 x 2 1 x { 2, } 2 (2 x
1) 0 OR 2 x 1 1 x 2x 1 2 2 2 BACK
Solve the equation. Check your answers. ( x 5)( x 2) 0 x 5 0 OR x 2 0 x 5 x 2 Solution Set x { 2, 5}
BACK Solve each equation. Check your answers. x(5 x 3) 0 x 0 x 0 OR Solution Set 3 x { , 0} 5
5 x 3 0 5 x 3 5 3 x 5 5 3 x 5 BACK
Solving a Quadratic Equation by Factoring Step 1 Write the equation in standard form. Step 2 Factor completely. Step 3 Use the zero-factor property. Set each factor with a variable equal to zero. Step 4 Solve each equation produced in step 3. BACK
2 Solve. x 9 x 18 2 x 9 x 18 0 ( x 6)( x 3) 0 x {6, 3}
BACK Solve. 2 x 7 x 0 x ( x 7) 0 x {0, 7} BACK Number Of Solutions
The degree of a polynomial is equal to the number of solutions. 3 2 x 2 x 3x Three solutions!!! Example
x (x + 1)(x 3) = 0 Set each of the three factors equal to 0. x=0 x+1=0 x = -1 x3=0 x=3 Solve the resulting equations.
Write the solution set. x = {0, -1, 3} BACK Solve the following equations. 1. x2 25 = 0 x 5 x 5 0 5, 5 2. x2 + 7x 8 = 0 x 8 x 1 0 1, 8 3. x2 12x + 36 = 0 x
4. c2 8c = 0 6 x 6 0 c(c 8) 0 0,8 6 1. Get a value of zero on one side of the equation.
2. Factor the polynomial if possible. 3. Apply the zero product property by setting each factor equal to zero. 4. Solve for the variable. Solving Equations by Factoring Definition of Quadratic Equations Zero-Factor Property Strategy for Solving Quadratics
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# What is the domain and range of y = sqrt(x-10) + 5?
Aug 3, 2015
Domain: $\left[10 , + \infty\right)$
Range: $\left[5 , + \infty\right)$
#### Explanation:
The only restriction you have will depend on sqrt(x-10. Since the square root of a number will produce a real value only if that number if positive, you need $x$ to satisfy the condition
$\sqrt{x - 10} \ge 0$
which is equivalent to having
$x - 10 \ge 0 \implies x \ge 10$
This means that any value of $x$ that is smaller than $10$ will be excluded from the function's domain.
As a result, the domain will be $\left[10 , + \infty\right)$.
The range of the function will depend on the minimum value of the square root. Since $x$ cannot be smaller than $10$, f(10 will be the starting point of the function's range.
$f \left(10\right) = \sqrt{10 - 10} + 5 = 5$
For any $x > 10$, $f \left(x\right) > 5$ because $\sqrt{x - 10} > 0$.
Therefore, the range of the function is $\left[5 , + \infty\right)$
graph{sqrt(x-10) + 5 [-3.53, 24.95, -3.17, 11.07]}
SIDE NOTE Move the focus of the graph 5 points up and 10 points to the right of the origin to see function. |
Unitary Method with Percentages
Lesson
So far we're feeling pretty confident about finding a certain percentage of an amount, but is there a way to find the total amount after being given a percentage of it? For instance, let's say $40%$40% of a dog's weight is $10$10kg, how heavy is the dog? We can use what is called the unitary method here, which means finding out what one unit of something is first. In our case we're going to find what one percent is first. Let's see how this works in the following diagram:
Can you see how there are three stages in this method?
1. Start with the amount that we know and its related percentage
2. Convert both numbers (percentage and amount) to what $1%$1% would be
3. Multiply by $100$100 to get $100%$100%, which is the whole
Sometimes questions will involve starting amounts over $100%$100%.
For example, say we knew that a bank account was worth $\$770$$770 after 10%10% interest was paid. To find the original 100%100% we would first need to figure out what the starting percentage is. If \770$$770 is the amount after interest then it equals $100%+10%=110%$100%+10%=110%.
Then to find the total amount we would follow Step $2$2 above and divide everything by $110$110 to get $1%$1%, and then finally multiply by $100$100 to get the whole amount.
So the original amount in the bank account would be:
$\frac{770}{110}\times100$770110×100 $=$= $7\times100$7×100 $=$= $\$700700
Wow, so the unitary method also works for amounts more than $100%$100%!
Worked Examples
Question 1
$12%$12% of a quantity is $24$24.
1. What is $1%$1% of the quantity?
2. Hence find the total quantity.
Question 2
Find the number if $10%$10% of the number is $12$12.
Question 3
$6%$6% of a number is $36$36.
1. Find $1%$1% of the number.
2. Hence find $36%$36% of the number.
Question 4
After $10%$10% GST was added, the price of an iPod was $£238$£238. What was its original price before GST? |
## Georgia Learning Connections
*TITLE:
Arc Length and Area of a Sector
*Annotation
This lesson is about teaching students to use proportions and GSP to see that the ratio of the arc length to the circumference is equal to the ratio of the area of the sector to area of the circle. The lesson will begin with a review of how to find the area and circumference of a circle and a review of the concept of central angles. The overall objective is for students to realize that the ratio of the arc length to the circumference is equal to the ration of the area of the sector to the area of the circle, regardless of the measure of the central angle
## *Primary Learning Outcome:
The primary learning outcome is for students to know that the ratio of the arc length to the circumference is equal to the ratio of the area of the sector to the area of the circle
For students to achieve the primary outcome, they must first be able to perform a number of simpler skills. These would include setting up a proportion, finding the circumference of a circle, finding the area of a circle, and identifying a central angle. Students will also need to be familiar with GSP so they can draw the required sketches and use the measure and compute features of GSP.
*Assessed QCC:
28
*Total Duration:
Procedures:
## Step One
Review the following concepts, demonstrating examples as necessary:
Circumference of a circle
Area of a circle
Central angle of a circle
Proportions
Estimated Time:
Ideally, this step would only take five to ten minutes.
###### Step Two
Instruct students to use GSP to create a sketch of a circle. There are no conditions on the radius. Students may draw circles of various sizes. Hide the point on the perimeter of the circle that is left from doing this.
###### Step Three
Students should select two points on the perimeter of the circle. Put a third point in between these two points Use GSP to create a sector from these two points.
###### Step Four
Draw two segments connecting the points on the perimeter to the center of the circle. Do this by selecting the point at the center of the circle and one of the two new points. From the construct menu, select line segment. Repeat this process for the other segment on the perimeter of the segment (make sure that the only item selected is the center point and the second point on the perimeter). See the image below for an example of such a sketch created in GSP.
Step Five
Find the length of the sector and the circumference of the circle using GSP.
Step Six
Find the area of the sector and the area of the circle using GSP.
Step Seven
Use the calculate feature of GSP to find the ratio of the arc length to the circumference.
Step Eight
Use the calculate feature of GSP to find the ratio of the area of the sector to the area of the circle.
The image below shows how these images would appear in GSP along with the image of the circle.
Step Nine
Click on either of the two points ( the points at the ends of the segment, not the point in the middle) placed on the perimeter of the circle. Drag the point around the circle. As the point is dragged around the circle, observe how the ratios changed. Make notes using paper and pencil about observations made. Make a general conjecture about how the three ratios are related.
Click here to explore this topic in GSP prior to attempting this lesson.
*Assessment:
Students will be assessed based on the observations they submit. Teachers should look for evidence of recognition that the two ratios will always be equal regardless of the measure of the central angle. Hopefully, students will observe that each ratio is equal to the ratio of the measure of the central angle to 360.
Extension:
As an extension, students could be asked to create congruent sectors. They could be asked how many congruent sectors can be created. They could also be asked to determine the measure of the central angle that would give the largest and/or smallest ratio of the area of the sector to the area of the circle.
Remediation:
Students that struggle with this concept could break the concepts into fewer pieces. For example, the students could first find only the arc length and circumference. They could set up a proportion comparing the ratio of the arc length to the circumference to the ratio of the measure of the central angle to 360. Once the students are able to recognize that these two are equal, then they could move forward to work with the area of the sector.
Accommodation:
Students with exceptional needs may be given a peer helper to answer questions they may encounter. The instructor could also set up the sketch, measurements, and calculations in GSP ahead of time for some students. This would still provide the students the opportunity to work with GSP. Another possible would be to provide the vision impaired students with a magnifying program so they can still use GSP.
Modification:
Possible modifications could include offering a peer helper for assistance. Assessment could also be modified so that these students were only expected to notice that the ratios were equal for a particular example rather than make a conjecture about all examples.
Title: Intermath Investigations for Geometry
URL:
//www.intermath-uga.gatech.edu/topics/geometry/circles/r06.htm
Annotation:
This web site offers suggestions for additional investigations. |
# Elementary Number Theory MCQ Quiz - Objective Question with Answer for Elementary Number Theory - Download Free PDF
Last updated on May 13, 2024
## Latest Elementary Number Theory MCQ Objective Questions
#### Elementary Number Theory Question 1:
Use Euclid's division lemma to find out the HCF of 155 and 1385.
1. 15
2. 11
3. 5
4. More than one of the above
5. None of the above
Option 3 : 5
#### Elementary Number Theory Question 1 Detailed Solution
Concept:
Euclid's division lemma: Given positive integers a and b there exist whole numbers q and r satisfying,
a = bq + r, where, 0 ≤ r ≤ b
Calculation:
Given integers are 155 and 1385. Clearly, 1385 > 155. Applying Euclid's division lemma to 155 and 1385, we get
1385 = 155 × 8 + 145
Since the remainder 145 ≠ 0. So, we apply the division lemma to the divisor 155 and remainder 145 to get
155 = 145 × 1 + 10
Now, we apply the division lemma to the new divisor 145 and the new remainder 10 to get
145 = 10 × 14 + 5
We now consider the new divisor 10 and the new remainder 5 and apply the division lemma to get
10 = 5 × 2 + 0
The remainder at this stage is zero. So, the divisor at this stage or the remainder at the previous stage i.e. 5 is the HCF of 155 and 1385.
Hence, the HCF of 155 and 1385 is 5.
#### Elementary Number Theory Question 2:
What is the GCD of 4598 and 3211 ?
1. 11
2. 40
3. 19
4. More than one of the above
5. None of the above
Option 3 : 19
#### Elementary Number Theory Question 2 Detailed Solution
Concept:
Euclid division algorithm: If a and b are any integers and b > 0then there exists unique integers q and r such that a = bq + r where 0 ≤ r ≤ b.
Greatest common divisor: Let a and b be any two positive integers with atleast one of them different from zero then the greatest common divisor (GCD) of a and b is the positive integer d satisfying the following two conditions.
1. d | a and d | b
2. If c | a and c | b the c ≤ b then c ≤ d. Further c | d.
Calculations:
To find : gcd(4598, 3211) = ?
By using Euclid´s division algorithm, we have
$$4598 = 3211 \times 1 + 1387$$
$$3211= 1387 \times 2 + 437$$
⇒ $$1387 = 437 \times 3 + 76$$
⇒ $$437 = 76 \times 5 + 57$$
⇒ $$76= 57 \times 1 + 19$$
⇒ $$57 = 19 \times 3 + 0$$
∴ gcd(4598, 3211) = 19
Hence, the correct answer is option 3)
#### Elementary Number Theory Question 3:
Which of the following relations defines the Euclid's Division Lemma:
1. a = bq + r, 0 ≤ r < b
2. a = br + q, 0 ≤ r ≤ b
3. a = bq + r, 0 < r ≤ b
4. More than one of the above
5. None of the above
Option 1 : a = bq + r, 0 ≤ r < b
#### Elementary Number Theory Question 3 Detailed Solution
Explanation:
Euclid's division lemma:
If a and b are two positive integers, then there exist unique positive integers q and r such that;
a = bq + r, where 0 ≤ r < b
If b ∣ a then r = 0 otherwise r must satisfy stronger inequality 0 < r < b.
#### Elementary Number Theory Question 4:
Euclid's division algorithm can be applied to:
1. All integers
2. All integers except zero
3. Only positive integers
4. More than one of the above
5. None of the above
Option 3 : Only positive integers
#### Elementary Number Theory Question 4 Detailed Solution
Concept:
Euclid's division lemma: If a and b are two positive integers, then there exists unique positive integers q and r such that
$$a=bq+r$$ where $$0\leq{r}\ <\ {b}$$ .
If b ∣ a then r = 0 otherwise r must satisfy stronger inequality 0 < r < b.
Explanation:
From the above discussion, we can conclude that Euclide division lemma can be applied in all positive integers
Hence, option 3 is correct.
• Euclid’s division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers.
• The HCF of two positive integers a and b is the largest positive integer d that divides both a and b.
#### Elementary Number Theory Question 5:
If the HCF of 65 and 117 is expressible in the form 65m - 117, then the value of m is
1. 4
2. 2
3. 1
4. More than one of the above
5. None of the above
Option 2 : 2
#### Elementary Number Theory Question 5 Detailed Solution
Concept:
Euclid's division lemma: Given positive integers a and b there exist whole numbers q and r satisfying,
a = bq + r, where, 0 ≤ r ≤ b
Calculation:
Using Euclid's division lemma,
117 = 65 × 1 + 52
⇒ 65 = 52 × 1 + 13
⇒ 52 = 13 × 4 + 0
∴ HCF(65, 117) = 13
According to question,
65m - 117 = 13
⇒ 65m = 117 + 13 = 130
⇒ m = 2
## Top Elementary Number Theory MCQ Objective Questions
#### Elementary Number Theory Question 6
If the HCF of 65 and 117 is expressible in the form 65m - 117, then the value of m is
1. 4
2. 2
3. 1
4. 3
Option 2 : 2
#### Elementary Number Theory Question 6 Detailed Solution
Concept:
Euclid's division lemma: Given positive integers a and b there exist whole numbers q and r satisfying,
a = bq + r, where, 0 ≤ r ≤ b
Calculation:
Using Euclid's division lemma,
117 = 65 × 1 + 52
⇒ 65 = 52 × 1 + 13
⇒ 52 = 13 × 4 + 0
∴ HCF(65, 117) = 13
According to question,
65m - 117 = 13
⇒ 65m = 117 + 13 = 130
⇒ m = 2
#### Elementary Number Theory Question 7
for any two positive integers a and b, there exist unique integers q and r. If b = 5, then which is not the value of r.
1. 2
2. 3
3. 4
4. 5
Option 4 : 5
#### Elementary Number Theory Question 7 Detailed Solution
Concept:
Euclid's division lemma: If a and b are two positive integers, then there exist unique positive integers q and r such that
a = bq + r, where 0 ≤ r < b
If b ∣ a then r = 0 otherwise r must satisfy stronger inequality 0 < r < b.
Calculation:
As discussed above, according to Euclid division lemma,
0 ≤ r < b
Here, b = 5
Therefore, r should be less than 5
Hence, can not take the value 5.
• Euclid’s division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers.
• The HCF of two positive integers a and b is the largest positive integer d that divides both a and b.
#### Elementary Number Theory Question 8
Find the general solution of $$x\ \equiv 5 \ mod(25)$$ and $$x\ \equiv 32 \ mod(23)$$
1. 800 + 55k for k ∈ Z.
2. 55 - 800k for k ∈ Z.
3. 55 + 575k for k ∈ Z
4. 800 - 55k for k ∈ Z.
Option 3 : 55 + 575k for k ∈ Z
#### Elementary Number Theory Question 8 Detailed Solution
Concept:
Chinese remainder theorem(CRT):
Let a and b be two relatively prime (coprime) numbers such that 0 ≤ r < a and 0 ≤ s < b. There exists a unique number N such that 0 ≤ N < ab and
$$r\ \equiv N \ mod(a)$$
$$s\ \equiv N \ mod(b)$$
i.e. N has remainder r on dividing by a and remainder s on dividing by b.
Calculation:
Given simultaneous congruences are
$$x\ \equiv 5 \ mod(25)$$ ----(1)
$$x\ \equiv 32 \ mod(23)$$ ----(2)
That means we have to find x such that, when it is divided by 25, gives remainder 5 and when divided by 23, gives a remainder 32.
⇒ x = 25a + 5
Where, a = 0, 1, 2, ...
⇒ x = 5, 30, 55, 80, ...
x = 23b + 32
⇒ x = 32, 55, 87, ....
x will satisfy both equations simultaneously.
x = 55
∵ LCM of 25 & 23 is 575
Hence, the general solution of x will be 55 + 575k
(3) is correct
#### Elementary Number Theory Question 9
If a is a whole number and p is a prime number then according to Fermat's theorem:
1. 2p - 1 - 2 is divisible by p
2. 2p - 1 is divisible by p
3. 2p - 2 is not divisible by p
4. 2p - 2 is divisible by p
Option 4 : 2p - 2 is divisible by p
#### Elementary Number Theory Question 9 Detailed Solution
Concept:
Fermat’s little theorem: It states that if p is a prime number, then for any integer a, the number a p – a is an integer multiple of p.
• Here p is a prime number ap ≡ a (mod p).
• If 2p ≡ 2 (mod p) i,e2p - 2 is divisible by p.
#### Elementary Number Theory Question 10
Which of the following relations defines the Euclid's Division Lemma:
1. a = bq + r, 0 ≤ r < b
2. a = br + q, 0 ≤ r ≤ b
3. a = bq + r, 0 < r ≤ b
4. a = qr + b, 0 ≤ r < b
Option 1 : a = bq + r, 0 ≤ r < b
#### Elementary Number Theory Question 10 Detailed Solution
Explanation:
Euclid's division lemma:
If a and b are two positive integers, then there exist unique positive integers q and r such that;
a = bq + r, where 0 ≤ r < b
If b ∣ a then r = 0 otherwise r must satisfy stronger inequality 0 < r < b.
#### Elementary Number Theory Question 11
Apply Chinese Remainder theorem to solve x = 3 (mod9), x = 7 (mod13). the common solution is
1. x = 107 (mod 117)
2. x = 103 (mod 117)
3. x = 111 (mod 117)
4. x = 105 (mod 117)
Option 3 : x = 111 (mod 117)
#### Elementary Number Theory Question 11 Detailed Solution
Given:
X = 3 (mod 9), X = 7 (mod 13)
Concept:
Chinese Remainder Theorem:
If m1, m2, .., mk are pairwise relatively prime positive integers, and if a1, a2, .., ak are any integers, then
The simultaneous congruences
x ≡ a1 (mod m1),
x ≡ a2 (mod m2)
x ≡ ak (mod mk)
have a solution, and the solution is unique modulo M, where
M = m1m2⋅⋅⋅mk .
Now, the solution x is given by
X = M1X1a1 + M2X2a2+......... + MkXkak
Where,
$$M_i = \frac{M}{m_i}$$, I = 1, 2, 3, ....K
and
MiX ≡ 1 mod(mi)
Calculation:
We have given,
x = 3 (mod9),
x = 7 (mod13)
m1 = 9 and m2 = 13 which are relatively co-prime.
Also, a1 = 3 and a2 = 7, which are integer.
Hence the condition of the Chinese Remainder Theorem satisfied.
So, according to the theorem, there will be the unique solution of x for modulo
M = 9 × 13 = 117
Now, using the relation,
$$M_i = \frac{M}{m_i}$$
we will get
M1 = 13 and M2 = 9
To calculate X1 and X2 we can use the relation,
MiX ≡ 1 (mod mi)
⇒ 13X1 ≡ 1 (mod 9)
⇒ 4X1 ≡ 1 (mod 13)
Multiplying both sides with 7, we will get
28X1 ≡ 7 (mod 9)
⇒ X1 ≡ 7 (mod 9)
⇒ X1 = 7
Again, using the above relation
9X2 ≡ 1 mod(13)
⇒ 27X2 ≡ 3 mod(13)
⇒ X2 ≡ 3 mod(13)
⇒ X2 = 3
Hence the unique solution of x will be
X = M1X1a1 + M2X2a2
⇒ X = 13 × 7 × 3 + 9 × 3 × 7 = 138
⇒ X ≡ 462 mod(117)
⇒ X ≡ 111 mod(117)
#### Elementary Number Theory Question 12
What is the remainder when 220 + 330 + 440 + 550 + 117 is divided by 7?
1. 2450
2. 36
3. 0
4. 3
Option 4 : 3
#### Elementary Number Theory Question 12 Detailed Solution
Concept:
Fermat´s theorem:
• If p is a prime and p doesn´t divide a, then a p - 1 ≡ 1(mod p).
• If a ≡ b (mod n) and c ≡ d (mod n) then a + c ≡ b + d (mod n) where a, b, c, d are any integers.
Calculation:
Here 7 is a prime And gcd(2, 7) = gcd(3, 7) = gcd(4, 7) = gcd(5, 7) = gcd(11, 7) = 1
So,by using Fermat´s theorem,
Consider, 27 - 1 ≡ 1(mod 7) ⇒ 26 ≡ 1(mod 7) ⇒ ( 26)3 ≡ 1(mod 7)
⇒ 218 ≡ 1(mod 7)
multiply throughout by 22,we get
⇒ 218 . 22 ≡ 22 (mod 7) ⇒ 220 ≡ 4(mod 7) .....(1)
Next consider, 37 - 1 ≡ 1(mod 7) ⇒ 36 ≡ 1(mod 7) ⇒ ( 36)5 ≡ 1(mod 7)
⇒ 330 ≡ 1(mod 7) ......(2)
Next consider, 47 - 1 ≡ 1(mod 7) ⇒46 ≡ 1(mod 7) ⇒ ( 46)6 ≡ 1(mod 7)
⇒ 436 ≡ 1(mod 7)
multiply throughout by 44,we get
⇒ 436. 44 ≡ 44 (mod 7) ⇒ 440 ≡ 256 (mod 7) ⇒ 440 ≡ 4(mod 7) ......(3)
Next consider, 57 - 1 ≡ 1(mod 7) ⇒56 ≡ 1(mod 7) ⇒ ( 56)8 ≡ 1(mod 7)
⇒ 548 ≡ 1(mod 7)
multiply throughout by 52,we get
⇒ 548 . 52 ≡ 52 (mod 7) ⇒ 550 ≡ 25 (mod 7) ⇒ 550 ≡ 4 (mod 7) ....(4)
Finally consider, 117 - 1 ≡ 1(mod 7) ⇒116 ≡ 1(mod 7)
multiply throughout by 11,we get
⇒116 . 11 ≡ 11 (mod 7) ⇒ 117 ≡ 11(mod 7) ⇒ 11 ≡ 4(mod 7) ......(5)
Now adding (1), (2), (3), (4) and (5) we get,
220 + 330 + 440 + 550 + 117 ≡ (4 + 1 + 4 + 4 + 4 )(mod 7) | using (6)
⇒ 220 + 330 + 440 + 550 + 117 ≡ 17 (mod 7) ⇒ 220 + 330 + 440 + 550 + 117 ≡ 3 (mod 7)
Hence, the correct answer is option 4)
#### Elementary Number Theory Question 13
Let n >1 be fixed and a, b, c, d be arbitrary integers. If a ≡ b (mod n) and c ≡ d (mod n), then:
1. a + c $$\not\equiv$$ b + d (mod n) and ac $$\not\equiv$$ bd (mod n)
2. a + c $$\not≡$$ b + d (mod n) but ac ≡ bd (mod n)
3. a + c ≡ b + d (mod n) and ac ≡ bd (mod n)
4. a + c ≡ b + d (mod n) but ac $$\not≡$$ bd (mod n)
Option 3 : a + c ≡ b + d (mod n) and ac ≡ bd (mod n)
#### Elementary Number Theory Question 13 Detailed Solution
Concept :
For any fixed integer n > 1 and for any arbitrary integers a, b, c and d.
If b (mod n) and c d (mod n) then
• a + c b + d (mod n) and
• ac bd (mod n)
Proof:
Firstly,
Given , a ≡ b (mod n) and c ≡ d (mod n)
⇒ n | (a - b) and n | (c - d)
⇒ a - b = hn and c - d = kn ; for some integers h and k ---(1)
Now consider, a - b + c - d = nh + nk
⇒ (a + c) - (b + d) = n (h + k) ( ∵ h + k is an integer )
⇒ n | [ (a + c) - (b + d) ]
⇒ a + c ≡ b + d (mod n)
Next,
From (1) we have,
a = b + hn and c = d + kn
Consider, ac = (b + hn) × (d + kn)
⇒ ac = bd + bkn + hnd + hknn
⇒ ac - bd = n (bk + hd + hkn)
⇒ n | ac - bd ( ∵ (bk + hd + hkn) is an integer )
⇒ ac ≡ bd (mod n)
Hence, the correct answer is option 3)
#### Elementary Number Theory Question 14
By applying Euclid's division lemma, the cube of any positive integer is of the form:
1. 9 m, 9 m + 2 or 9 m + 7
2. 9 m, 9 m + 1 or 9 m + 8
3. 9 m, 9 m + 4 or 9 m + 5
4. 9 m, 9 m + 3 or 9 m + 6
Option 2 : 9 m, 9 m + 1 or 9 m + 8
#### Elementary Number Theory Question 14 Detailed Solution
Concept:
Euclid's division lemma: If a and b are two positive integers, then there exists unique positive integers q and r such that $$a=bq+r$$ where $$0\leq{r}\leq{b}$$ .
Formula used :
• (xy)n = xyn
• (a + b)= a3 + 3a2b + 3ab2 + b3
CalculationLet a be any positive integer and b=3.
Therefore we can write,$$a=3q+r,q\geq0,0\leq{r}<3$$.
Now ,we have to solve the problem in 3 cases:
Case 1: When r = 0,
we get a = 3q
Now by taking cube on both the sides we get:
a3 = (3q)3
a3 =27q3 ; a = 9m where m = 3q3
⇒a3 = 9(3q3
Case 2 :When r = 1,
We get, a = 3q+1
Now by taking cube on both the sides we get:
a3 = (3q+1)3
we know that,(a + b)3 = a+ 3a2b + 3ab+ b3
here, a = 3q and b= 1,therefore our equation becomes:
$$a^3=(3q)^3+3\times(3q)^2\times1+3\times3q\times1^2$$
$$a^3=27q^3+3\times3^2q^2+9q+1$$
$$a^3=27q^3+27q^2+9q+1$$
From the first three terms, take out 9 as common, we get:
$$a^3=9(3q^3+3q^2+q)+1$$
$$a^3=9m+1,$$ where m = $$3q^3+q^2+q$$
Case 3: When r = 2
We get, a = 3q+2
Now by taking cube on both the sides we get:
a3 = (3q+2)3
$$a^3=(3q)^3+3\times(3q)^2\times2+3\times3q\times2^2$$
$$a^3=27q^3+3\times3^2q^2\times2+9q\times4+8$$
$$a^3=27q^3+54q^2+36q+8$$
From the first three terms, take out 9 as common, we get:
$$a^3=9(3q^3+6q^2+4q)+8$$
$$a^3=9m+8,$$ where $$m =3q^3+6q^2+4q$$
Therefore ,we can say that a cube of any integer is of the form 9m,9m+1 or 9m+2.
Hence, the correct answer is option 2).
#### Elementary Number Theory Question 15
For any positive integers 'a' and 3, there exist unique integers 'q' and 'r' such that a = 3q + r, where r must satisfy:
1. 1 < r < 3
2. 0 < r < 3
3. 0 ≤ r < 3
4. 0 < r ≤ 3
Option 3 : 0 ≤ r < 3
#### Elementary Number Theory Question 15 Detailed Solution
Concept:
Euclid's division lemma: It states that if 'a' and 'b' are positive integers then there exists 'q' and 'r' such that
a = bq + r where, 0 r < b
Calculation:
Here in the given problem, we have 'a' and 'b = 3' as any positive integers, then by applying the above lemma there exists 'q' and 'r' such that
a = 3q +r where,
0 ≤ r < 3 ( As b = 3)
Hence, the correct answer is 0 ≤ r < 3. |
# Into Math Grade 4 Module 9 Lesson 1 Answer Key Apply the Area Formula to Rectangles
We included HMH Into Math Grade 4 Answer Key PDF Module 9 Lesson 1 Apply the Area Formula to Rectangles to make students experts in learning maths.
## HMH Into Math Grade 4 Module 9 Lesson 1 Answer Key Apply the Area Formula to Rectangles
I Can find the area of a rectangle by using the formula for area.
Making banners is a very old craft. Banners are usually made of cloth and can have a symbol, logo, slogan, or any other message. How can you find the area of each banner?
The area can be calculated by using the formula,
Area = Length x width.
for the first banner
Area = 5 x 8 = 40 square feet
for the second banner,
Area = 4 x 10 = 40 square feet.
Turn and Talk What do you notice about the perimeter and area of each banner?
Perimeter gives the total circumference of the rectangle whereas, Area gives the total number of units that fit in the given figure or banner.
Build Understanding
1. Sports fans use banners to show support. How can you find the area of the banner?
A. What does it mean to find the area of a flat surface?
__________________________________
Area is a measure of how much space there is on a flat surface.
To find the area, think about the measure of the sides of the rectangle.
B. How many unit squares make up the length and width of the rectangle?
__________________________________
28 unit squares make up the length and width of the rectangle
C. Count the number of unit squares to find the area. __________________________________
28 unit squares required to find the area.
D. What is the relationship between the length and width and the area? __________________________________
the relationship between the length and width and the areais
Area = Length X Width.
E. How can you describe a mathematical rule or formula that you would use to find the area (A) of the rectangle? Write the formula. Represent the length with l and the width with w.
_________________________________
Area = Length X Width.
Connect to Vocabulary
You can also write the formula for the area (A) of a rectangle as A = b × h. Think of the base (b) of a rectangle as the measure of any side. The height (h) of the rectangle is the measure of a side perpendicular to the base.
F. How can you use the area formula to find the area of the rectangle?
A = l × w
A = __ × ___
A = ___
A = l × w
A = 7 x 4
A = 28 square units.
G. The rectangle has an area of ___ square units.
The rectangle has an area of 28 square units
Step It Out
2. The Hawks’ fans are making a giant banner for a competition. The banner has the dimensions shown. What is the area of the banner?
A. Write the formula that you can use to solve the problem.
Area = Length x Width
B. Write the measure of the length, l, and the width, w, of the banner.
From the figure,
Length = 5 meters
Width = 7 meters.
C. Find the area of the banner.
From formula,
Area = Length x width
Area = 5 x 7 = 35 square meters.
D. The area of the banner The area is ___ square meters.
The area of the banner The area is 35 square meters.
Area = Length x width
Area = 5 x 7
Area = 35 square meters.
Turn and Talk How would the area of the banner change if the banner was turned so that the width is 7 meters and the length is 5 meters?
there will be no change in the area of the rectangle. Because, 5 x 7 = 7 x 5 = 35 square meters only.
Check Understanding Math Board
Question 1.
Mr. Nielsen is going to buy sod for a portion of his yard, but first, he must know the area of the surface that he wants to cover. What is the area?
___________________
From the figure,
Area = Length x width
Area = 18 x 25
Area = 450 square feet.
Therefore, Area of the figure is 450 square feet.
Find the area.
Question 2.
From the figure,
Area = Length x width
Area = 9 x 6
Area = 54 square m.
Therefore, Area of the figure is 54 square m.
Question 3.
From the figure,
Area = Length x width
Area = 10 x 10
Area = 100 square ft.
Therefore, Area of the figure is 100 square ft.
Question 4.
Jason has a sticker on a folder. What is the area of the sticker?
From the figure,
Area = Length x width
Area = 7 x 3
Area = 21 square cm.
Therefore, Area of the sticker on folder is 21 square cm.
Question 5.
Lisa is setting square tiles on a backsplash. What is the area of each tile?
From the figure,
Area = Length x width
Area = 5 x 5
Area = 25 square in
Therefore, Area of the each tile is 25 square in.
Question 6.
Reason Mr. Wells knows that the area of the square floor in a toolshed is 36 square feet. What is the length of one of the sides? How do you know?
We know,
Area of square = side x side
36 = side x side
side = square root of 36 = 6
Therefore, length of one of the sides = 6 ft.
Use Structure Find its area.
Question 7.
From the figure,
Area = Length x width
Area = 14 x 6
Area =84 square cm
Therefore, Area of the figure is 84 square cm.
Question 8.
From the figure,
Area = Length x width
Area = 3 x 11
Area = 33 square in.
Therefore, Area of the figure is 33 square in.
Question 9.
From the figure,
Area = Length x width
Area = 8 x 8
Area = 64 square cm.
Therefore, Area of the figure is 64 square cm.
I’m in a Learning Mindset!
What parts of solving problems using the area formula am I comfortable with?
_________________________________
_________________________________
Scroll to Top
Scroll to Top |
## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.15
10th Maths Exercise 3.15 Solutions Question 1.
Graph the following quadratic equations and state their nature of solutions,
(i) x2 – 9x + 20 = 0
Solution:
Step 1:
Points to be plotted : (-4, 72), (-3, 56), (-2, 42), (-1, 30), (0, 20), (1, 12), (2, 6), (3, 2), (4, 0)
Step 2:
The point of intersection of the curve with x axis is (4, 0)
Step 3:
The roots are real & unequal
∴ Solution {4, 5}
(ii) x2 – 4x + 4 = 0
Step 1: Points to be plotted : (-4, 36), (-3, 25), (-2, 16), (-1, 9), (0, 4), (1, 1), (2, 0), (3, 1), (4, 4)
Step 2: The point of intersection of the curve with x axis is (2, 0)
Step 3:
Since there is only one point of intersection with x axis, the quadratic equation x2 – 4x + 4 = 0 has real and equal roots.
∴ Solution{2, 2}
(iii) x2 + x + 7 = 0
Let y = x2 + x + 7
Step 1:
Step 2:
Points to be plotted: (-4, 19), (-3, 13), (-2, 9), (-1, 7), (0, 7), (1, 9), (2, 13), (3, 19), (4, 27)
Step 3:
Draw the parabola and mark the co-ordinates of the parabola which intersect with the x-axis.
Step 4:
The roots of the equation are the points of intersection of the parabola with the x axis. Here the parabola does not intersect the x axis at any point.
So, we conclude that there is no real roots for the given quadratic equation,
(iv) x2 – 9 = 0
Let y = x2 – 9
Step 1:
Step 2:
The points to be plotted: (-4, 7), (-3, 0), (-2, -5), (-1, -8), (0, -9), (1,-8), (2, -5), (3, 0), (4, 7)
Step 3:
Draw the parabola and mark the co-ordinates of the parabola which intersect the x-axis.
Step 4:
The roots of the equation are the co-ordinates of the intersecting points (-3, 0) and (3, 0) of the parabola with the x-axis which are -3 and 3 respectively.
Step 5:
Since there are two points of intersection with the x axis, the quadratic equation has real and unequal roots.
∴ Solution{-3, 3}
(v) x2 – 6x + 9 = 0
Let y = x2 – 6x + 9
Step 1:
Step 2:
Points to be plotted: (-4, 49), (-3, 36), (-2, 25), (-1, 16), (0, 9), (1, 4), (2, 1), (3, 0), (4, 1)
Step 3:
Draw the parabola and mark the co-ordinates of the intersecting points.
Step 4:
The point of intersection of the parabola with x axis is (3, 0)
Since there is only one point of intersection with the x-axis, the quadratic equation has real and equal roots. .
∴ Solution (3, 3)
(vi) (2x – 3)(x + 2) = 0
2x2 – 3x + 4x – 6 = 0
2x2 + 1x – 6 = 0
Let y = 2x2 + x – 6 = 0
Step 1:
Step 2:
The points to be plotted: (-4, 22), (-3, 9), (-2, 0), (-1, -5), (0, -6), (1, -3), (2, 4), (3, 15), (4, 30)
Step 3:
Draw the parabola and mark the co-ordinates of the intersecting point of the parabola with the x-axis.
Step 4:
The points of intersection of the parabola with the x-axis are (-2, 0) and (1.5, 0).
Since the parabola intersects the x-axis at two points, the, equation has real and unequal roots.
∴ Solution {-2, 1.5}
Question 2.
Draw the graph of y = x2 – 4 and hence solve x2 – x – 12 = 0
Solution:
Point of intersection (-3, 5), (4, 12) solution of x2 – x – 12 = 0 is -3, 4
10th Maths Graph 3.15 Answers Question 3.
Draw the graph of y = x2 + x and hence solve x2 + 1 = 0.
Solution:
Draw the parabola by the plotting the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12), (4, 20), (5, 30)
To solve: x2 + 1 = 0, subtract x2 + 1 = 0 from y = x2 + x.
x2 + 1 = 0 from y = x2 + x
Plotting the points (-2, -3), (0, -1), (2, 1) we get a straight line. This line does not intersect the parabola. Therefore there is no real roots for the equation x2 + 1 = 0.
10th Maths Guide Graph Question 4.
Draw the graph of y = x2 + 3x + 2 and use it to solve x2 + 2x + 1 = 0.
Solution:
Draw the parabola by plotting the point (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20), (4, 30).
To solve x2 + 2x + 1 = 0, subtract x2 + 2x + 1 = 0 from y = x2 + 3x + 2
Draw the straight line by plotting the points (-2, -1), (0, 1), (2, 3)
The straight line touches the parabola at the point (-1,0)
Therefore the x coordinate -1 is the only solution of the given equation
10th Graph Exercise 3.15 Solutions Question 5.
Draw the graph of y = x2 + 3x – 4 and hence use it to solve x2 + 3x – 4 = 0. y = x2 + 3x – 4
Solution:
Draw the parabola using the points (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14), (4, 24).
To solve: x2 + 3x – 4 = 0 subtract x2 + 3x – 4 = 0 from y = x2 + 3x – 4 ,
The points of intersection of the parabola with the x axis are the points (-4, 0) and (1, 0), whose x – co-ordinates (-4, 1) is the solution, set for the equation x2 + 3x – 4 = 0.
10th Maths Graph Answers Question 6.
Draw the graph of y = x2 – 5x – 6 and hence solve x2 – 5x – 14 = 0.
Solution:
Draw the parabola using the points (-5, 44), (-4, 30), (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10)
To solve the equation x2 – 5x – 14 = 0, subtract x2 – 5x – 14 = 0 from y = x2 – 5x – 6.
The co-ordinates of the points of intersection of the line and the parabola forms the solution set for the equation x2 – 5x – 14 = 0.
∴ Solution {-2, 7}
10th Class Maths Graph Pdf Question 7.
Draw the graph of y = 2x2 – 3x – 5 and hence solve 2x2 – 4x – 6 = 0. y = 2x2 – 3x – 5
Solution:
Draw the parabola using the points (-4, 39), (-3, 22), (-2, 9), (-1, 0), (0, -5), (1, -6), (2, -3), (3, 4), (4, 15).
To solve 2x2 – 4x – 6 = 0, subtract it from y = 2x2 – 3x – 5
Draw a straight line using the points (-2, -1), (0, 1), (2, 3). The points of intersection of the parabola and the straight line forms the roots of the equation.
The x-coordinates of the points of intersection forms the solution set.
∴ Solution {-1, 3}
10th Maths Graph Question 8.
Draw the graph of y = (x – 1)(x + 3) and hence solve x2 – x – 6 = 0.
Solution:
y = (x – 1)(x + 3) = x2 – x + 3x – 3 = 0
y = x2 + 2x – 3
Draw the parabola using the points (-4, 5), (-3, 0), (-2, -3), (-1,-4), (0, -3), (1, 0), (2, 5), (3, 12), (4, 21)
To solve the equation x2 – x – 6 = 0, subtract x2 – x – 6 = 0 from y = x2 – 2x – 3.
Plotting the points (-2, -3), (-1, 0), (0, 3), (2, 9), we get a straight line.
The points of intersection of the parabola with the straight line gives the roots of the equation. The co¬ordinates of the points of intersection forms the solution set.
∴ Solution {-2, 3} |
# The Labouchére system
2016-07-22 09:12:54 -
Along with the Martingale system, the Labouchére system is perhaps the best known roulette systems. Labouchére is a simple roulette strategy based on crossing out numbers in a number sequence. That is why a pen and paper will come in handy.
If you thought that the system originated from France, well, you'll be surprised. The system is attributed to the British aristocrat and politician Henry Du Pré Labouchére, who transformed his specific interest in roulette systems into the discovery of a new one.
## The basic principle of the system
Even though the Labouchére system might seem complicated at first, that's not really the case. In fact it is very simple and first-grade mathematics are all you'll need. First, we need an arbitrary numerical sequence where each subsequent number must be 1 greater than the previous one.
The Labouchére system will then be summing up the first and last number in the sequence for each bet. The sum of these two numbers will determine the bet. If you lose, place this sum on the end of the sequence. On the other hand, if you win then cross out the first and last number in the sequence. We recommend using the Labouchére system when betting on one possibility out of two (even / uneven, red / black); however, some players also use it for 1/3 bets.
## Example of a game using the Labouchére system
To really see that the system is not difficult, let's have a look at a simulated game with the Labouchére system.
• Select an arbitrary numerical sequence, for instance 1, 2, 3, 4, 5, 6, and bet on black. Your first bet is 1 + 6 = 7 €.
• What happens if you lose this spin? Well, you lost 7 €. Write down this number at the end of the numerical sequence. So, the sequence now is: 1, 2, 3, 4, 5, 6, 7.
• Your next bet will be 8 € (1+7) on black.
• This spin, you win. You won 16 €. Do not forget to cross out the first and last number of the sequence. The sequence now is: 2, 3, 4, 5, 6.
• Let's proceed to the next bet. Based on the sequence, this should again be 8 € (2+6).
• Assume you win again. This means you got an additional 16 €. So, once again cross out the first and last number in the sequence. There's only three numbers left: 3, 4 and 5.
• The next bet will once again be 8 € (3+5). Proceed until you completely cross out the whole sequence.
As you can see, it is not difficult to learn the system by heart fairly quickly. It is also suitable for less experienced players.
## The many advantages of the Labouchére system
The Labouchére system is popular among players especially because it is less financially demanding than for instance Martingale, and it is also more "generous" to players. If you take the time and calculate your probabilities, you'll see that a 33 % success rate is sufficient to ensure you are not losing money. In practice, winning one out of three bets is enough to have an even budget.
## Evaluation
The Labouchére strategy is highly popular among players thanks to its many positives. Together with its simplicity, quick application and low financial risk, the system is an excellent tool to improve your play. We highly recommend giving it a try.
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# Ohio - Grade 1 - Math - Measurement and Data - Interpreting Data - 1.MD.4
### Description
Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another.
• State - Ohio
• Standard ID - 1.MD.4
• Subjects - Math Common Core
### Keywords
• Math
• Measurement and Data
## More Ohio Topics
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem. See Table 1, page 95.
Apply properties of operations as strategies to add and subtract. For example, if 8 + 3 = 11 is known, then 3 + 8 = 11 is also known (Commutative Property of Addition); to add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12 (Associative Property of Addition). Students need not use formal terms for these properties.
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 - 8 by finding the number that makes 10 when added to 8.
1.OA.5 Relate counting to addition and subtraction, e.g., by counting on 2 to add 2.
1.OA.6 Add and subtract within 20, demonstrating fluency with various strategies for addition and subtraction within 10. Strategies may include counting on; making ten, e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14; decomposing a number leading to a ten, e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9; using the relationship between addition and subtraction, e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4; and creating equivalent but easier or known sums, e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13.
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6; 7 = 8 – 1; 5 + 2 = 2 + 5; 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations: 8 + __ = 11; 5 = __ - 3; 6 + 6 = __ .
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: 10 can be thought of as a bundle of ten ones — called a “ten;” the numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones; and the numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
## Here is the skill that Ohio requires you to master
• State Test Ohio's State Tests
• State Standards Ohio’s New Learning Standards
• Subject Math
• Topic Name Interpreting Data
• Standard ID 1.MD.4
• Description
Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another.
### Education Galaxy’s Ohio State Assessment preparation program
Education Galaxy’s Ohio State Assessment preparation program provides online assessment and practice for students in Grades K-6 to help build mastery towards Ohio’s New Learning Standards. Our unique online program is easy to use and enjoyable for both teachers and students. Students work on their Study Plans practicing important concepts while teachers pull formative assessment reports to identify the strengths and weaknesses of their classroom and individual students. |
# PSEB 8th Class Maths Solutions Chapter 7 ਘਣ ਅਤੇ ਘਣਮੂਲ Ex 7.2
Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 7 ਘਣ ਅਤੇ ਘਣਮੂਲ Ex 7.2 Textbook Exercise Questions and Answers.
## PSEB Solutions for Class 8 Maths Chapter 7 ਘਣ ਅਤੇ ਘਣਮੂਲ Exercise 7.2
1. ਅਭਾਜ ਗੁਣਨਖੰਡ ਵਿਧੀ ਦੁਆਰਾ ਹੇਠਾਂ ਲਿਖੀਆਂ ਵਿਚੋਂ ਹਰੇਕ ਸੰਖਿਆ ਦਾ ਘਣਮੂਲ ਪਤਾ ਕਰੋ :
ਪ੍ਰਸ਼ਨ (i).
64
ਹੱਲ:
64
∴ 64 = $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$
∴ $$\sqrt[3]{64}$$ = 2 × 2 = 4.
ਪ੍ਰਸ਼ਨ (ii).
512
ਹੱਲ:
512
∴ 512 = $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$
∴ $$\sqrt[3]{512}$$ = 2 × 2 × 2 = 8.
ਪ੍ਰਸ਼ਨ (iii).
10648
ਹੱਲ:
10648
∴ 10648 = $$\underline{2 \times 2 \times 2}$$ × $$\underline{11 \times 11 \times 11}$$
∴ $$\sqrt[3]{10648 }$$ = 2 × 11
= 22
ਪ੍ਰਸ਼ਨ (iv).
27000
ਹੱਲ:
27000
∴ 27000 = $$\underline{2 \times 2 \times 2}$$ × $$\underline{3 \times 3 \times 3}$$ × $$\underline{5 \times 5 \times 5}$$
∵ $$\sqrt[3]{27000 }$$ = 2 × 3 × 5
= 30
ਪ੍ਰਸ਼ਨ (v).
15625
ਹੱਲ:
15625
∴ 15625 = $$\underline{5 \times 5 \times 5}$$ × $$\underline{5 \times 5 \times 5}$$
∴ $$\sqrt[3]{15625 }$$ = 5 × 5
= 25
ਪ੍ਰਸ਼ਨ (vi).
13824
ਹੱਲ:
13824
∴ 13824 = $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$ × $$\underline{3 \times 3 \times 3}$$
∴ $$\sqrt[3]{13824 }$$ = 2 × 2 × 2 × 3
= 24
ਪ੍ਰਸ਼ਨ (vii).
110592
ਹੱਲ:
110592
∴ 110592 = $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$ × $$\underline{3 \times 3 \times 3}$$
∴ $$\sqrt[3]{110592 }$$ = 2 × 2 × 2 × 2 × 3
= 48
ਪ੍ਰਸ਼ਨ (viii).
46656
ਹੱਲ:
46656
∴ 46656 = $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$ × $$\underline{3 \times 3 \times 3}$$ × $$\underline{3 \times 3 \times 3}$$
∴ $$\sqrt[3]{46656 }$$ = 2 × 2 × 3 × 3
= 36
ਪ੍ਰਸ਼ਨ (ix).
175616
ਹੱਲ:
175616
∴ 175616 = $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$ × $$\underline{2 \times 2 \times 2}$$ × $$\underline{7 \times 7 \times 7}$$
∴ $$\sqrt[3]{175616 }$$ = 2 × 2 × 2 × 7
= 56
ਪ੍ਰਸ਼ਨ (x).
91125.
ਹੱਲ:
91125
∴ 91125 = $$\underline{3 \times 3 \times 3}$$ × $$\underline{3 \times 3 \times 3}$$ × $$\underline{5 \times 5 \times 5}$$
∴ $$\sqrt[3]{91125 }$$ = 3 × 3 × 5
= 45
ਪ੍ਰਸ਼ਨ 2.
ਦੱਸੋ ਸੱਚ ਹੈ ਜਾਂ ਝੂਠ :
(i) ਕਿਸੀ ਵੀ ਟਾਂਕ ਸੰਖਿਆ ਦਾ ਘਣ ਜਿਸਤ ਹੁੰਦਾ ਹੈ ।
(ii) ਇਕ ਪੂਰਨ ਘਣ ਦੋ ਸਿਫ਼ਰਾਂ ‘ਤੇ ਖ਼ਤਮ ਨਹੀਂ ਹੁੰਦਾ ਹੈ ?
(iii) ਜੇ ਕਿਸੇ ਸੰਖਿਆ ਦਾ ਵਰਗ 5 ’ਤੇ ਖ਼ਤਮ ਹੁੰਦਾ ਹੈ, ਤਾਂ ਉਸਦਾ ਘਣ 25 ’ਤੇ ਖ਼ਤਮ ਹੁੰਦਾ ਹੈ ।
(iv) ਇਸ ਤਰ੍ਹਾਂ ਦਾ ਕੋਈ ਪੂਰਨ ਘਣ ਨਹੀਂ ਹੈ ਜੋ 8 ‘ ਤੇ ਖ਼ਤਮ ਹੁੰਦਾ ਹੈ ।
(v) ਦੋ ਅੰਕਾਂ ਦੀ ਸੰਖਿਆ ਦਾ ਘਣ ਤਿੰਨ ਅੰਕਾਂ ਵਾਲੀ ਸੰਖਿਆ ਹੋ ਸਕਦੀ ਹੈ ।
(vi) ਦੋ ਅੰਕਾਂ ਦੀ ਸੰਖਿਆ ਦੇ ਘਣ ਵਿਚ ਸੱਤ ਜਾਂ ਜ਼ਿਆਦਾ | ਅੰਕ ਹੋ ਸਕਦੇ ਹਨ ।
(vii) ਇਕ ਅੰਕ ਵਾਲੀ ਸੰਖਿਆ ਦਾ ਘਣ ਇਕ ਅੰਕ ਵਾਲੀ ਸੰਖਿਆ ਹੋ ਸਕਦੀ ਹੈ ।
ਉੱਤਰ :
(i) ਝੂਠ
(ii) ਸੱਚ
(iii) ਝੂਠ
(iv) ਸੱਚ
(v) ਝੂਠ
(vi) ਝੂਠ
(vii) ਸੱਚ ।
ਪ੍ਰਸ਼ਨ 3.
ਤੁਹਾਨੂੰ ਇਹ ਦੱਸਿਆ ਜਾਂਦਾ ਹੈ ਕਿ 1331 ਇਕ ਪੂਰਨ ਘਣ ਹੈ | ਕੀ ਬਿਨ੍ਹਾਂ ਗੁਣਨਖੰਡ ਕੀਤੇ ਤੁਸੀਂ ਇਹ ਅਨੁਮਾਨ ਲਗਾ ਸਕਦੇ ਹੋ ਕਿ ਇਸਦਾ ਘਣਮੂਲ ਕੀ ਹੈ ? ਇਸੇ ਤਰ੍ਹਾਂ 4913, 12167 ਅਤੇ 32768 ਦੇ ਘਣਮੂਲਾਂ ਦੇ ਅਨੁਮਾਨ ਲਗਾਉ ।
ਹੱਲ:
ਅਸੀਂ ਜਾਣਦੇ ਹਾਂ ਕਿ 0, 1, 4, 5, 6 ਅਤੇ 9 ਉੱਤੇ ਖ਼ਤਮ ਹੋਣ ਵਾਲੀਆਂ ਸਿਖਿਆਵਾਂ ਦੇ ਘਣ ਵੀ 0, 1, 4, 5 ਅਤੇ 9 ਉੱਤੇ ਖ਼ਤਮ ਹੁੰਦੇ ਹਨ ।
ਕਿਉਂਕਿ 1331 ਦਾ ਇਕਾਈ ਅੰਕ 1 ਹੈ । ਕਿਉਂਕਿ
1331 = 11 × 11 × 11 ਇਸ ਲਈ ਇਸਦਾ ਮੁਲ = 11
4913 = 17 × 17 × 17
∴ 4913 ਦਾ ਘਣਮੂਲ = 17
12167 = 23 × 23 × 23
12167 ਦਾ ਘਣਮੂਲ = 23
32768 = 32 × 32 × 32
32768 ਦਾ ਘਣਮੂਲ = 32. |
# Real Numbers Exercise 1.2
Real Numbers Exercise 1.2
Page No: 11
1. Express each number as product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
(i) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
(i) 26 = 2 × 13
91 =7 × 13
HCF = 13
LCM =2 × 7 × 13 =182
Product of two numbers, 26 × 91 = 2366
Product of HCF and LCM, 13 × 182 = 2366
Hence, product of two numbers = product of HCF × LCM
(ii) 510 = 2 × 3 × 5 × 17
92 =2 × 2 × 23
HCF = 2
LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of two numbers, 510 × 92 = 46920
Product of HCF and LCM, 2 × 23460 = 46920
Hence, product of two numbers = product of HCF × LCM
(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
HCF = 2 × 3 = 6
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024
Product of two numbers, 336 × 54 =18144
Product of HCF and LCM, 6 × 3024 = 18144
Hence, product of two numbers = product of HCF × LCM.
3. Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
(i) 12 = 2 × 2 × 3
15 =3 × 5
21 =3 × 7
HCF = 3
LCM = 2 × 2 × 3 × 5 × 7 = 420
(ii) 17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 1 × 17 × 19 × 23 = 11339
(iii) 8 =1 × 2 × 2 × 2
9 =1 × 3 × 3
25 =1 × 5 × 5
HCF =1
LCM = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
We have the formula that,
Product of LCM and HCF = product of number
LCM × 9 = 306 × 657
Divide both side by 9 we get
LCM = (306 × 657)/9 = 22338
5. Check whether 6n can end with the digit 0 for any natural number n.
If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5.
So value 6n should be divisible by 2 and 5 both 6n is divisible by 2 but not divisible by 5 So it can not end with 0.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
7 × 11 × 13 + 13
Taking 13 common, we get
13 (7 x 11 +1 )
13(77 + 1 )
13 (78)
It is product of two numbers and both numbers are more than 1 so it is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 common, we get
5(7 × 6 × 4 × 3 × 2 × 1 +1)
5(1008 + 1)
5(1009)
It is product of two numbers and both numbers are more than 1 so it is a composite number.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
They will be meet again after LCM of both values at the starting point.
18 = 2 × 3 × 3
12 = 2 × 2 × 3
LCM = 2 × 2 × 3 × 3 = 36
Therefore, they will meet together at the starting point after 36 minutes. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 6.4: Compound Inequalities
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Write and graph compound inequalities on a number line.
• Solve a compound inequality with “and”.
• Solve a compound inequality with “or”.
• Solve compound inequalities using a graphing calculator (TI family).
• Solve real-world problems using compound inequalities
## Introduction
In this section, we will solve compound inequalities. In previous sections, we obtained solutions that gave the variable either as greater than or as less than a number. In this section we are looking for solutions where the variable can be in two or more intervals on the number line.
There are two types of compound inequalities:
1. Inequalities joined by the word “and”.
The solution is a set of values greater than a number and less than another number.
a<x<b\begin{align*}a
In this case we want values of the variable for which both inequalities are true.
2. Inequalities joined by the word “or”.
The solution is a set of values greater than a number or less than another number.
x<a\begin{align*}x or x>b\begin{align*}x>b\end{align*}
In this case, we want values for the variable in which at least one of the inequalities is true.
## Write and Graph Compound Inequalities on a Number Line
Example 1
Write the inequalities represented by the following number line graphs.
a)
b)
c)
d)
Solution
a) The solution graph shows that the solution is any value between -40 and 60, including -40 but not 60. Any value in the solution set satisfies both inequalities.
x40\begin{align*}x \geq -40\end{align*} and x<60\begin{align*} x<60\end{align*}
This is usually written as the following compound inequality.
40x<60\begin{align*}-40 \leq x <60 \end{align*}
b) The solution graph shows that the solution is any value greater than 1 (not including 1) or any value less than -2 (not including -2). You can see that there can be no values that can satisfy both these conditions at the same time. We write:
x>1\begin{align*}x >1\end{align*} or x<2\begin{align*}x <-2\end{align*}
c) The solution graph shows that the solution is any value greater than 4 (including 4) or any value less than -1 (including -1). We write:
x4\begin{align*}x \geq 4\end{align*} or x1\begin{align*} x \leq -1\end{align*}
d) The solution graph shows that the solution is any value less than 25 (not including 25) and any value greater than -25 (not including -25). Any value in the solution set satisfies both conditions.
x>25\begin{align*}x>-25\end{align*} and x<25\begin{align*}x <25\end{align*}
This is usually written as 25<x<25\begin{align*}-25.
Example 2
Graph the following compound inequalities on the number line.
a) 4x6\begin{align*}-4 \leq x \leq 6\end{align*}
b) x<0\begin{align*}x <0\end{align*} or x>2\begin{align*} x>2\end{align*}
c) x8\begin{align*}x \geq -8\end{align*} or x20\begin{align*}x \leq -20\end{align*}
d) 15<x85\begin{align*}-15
Solution
a) The solution is all numbers between -4 and 6 including both -4 and 6.
b) The solution is either numbers less than 0 or numbers greater than 2 not including 0 or 2.
c) The solution is either numbers greater than or equal to -8 or less than or equal to -20.
d) The solution is numbers between -15 and 85, not including -15 but including 85.
## Solve a compound Inequality With “and”
When we solve compound inequalities, we separate the inequalities and solve each of them separately. Then, we combine the solutions at the end.
Example 3
Solve the following compound inequalities and graph the solution set.
a) 2<4x511\begin{align*}-2<4x-5\leq 11\end{align*}
b) 3x5<x+95x+13\begin{align*}3x-5
Solution
a) First, we rewrite the compound inequality as two separate inequalities with and. Then solve each inequality separately.
2334<4x5<4x<xand4x5114x16x4\begin{align*}-2 & <4x-5 & & & & 4x-5 \leq 11\\ 3 & < 4x & & \text{and} & & 4x \leq 16\\ \frac{3}{4}&
Answer 34<x\begin{align*} \frac{3} {4} and x4\begin{align*} x\leq 4\end{align*}. This can be written as 34x4\begin{align*} \frac{3} {4}x\leq 4\end{align*}.
b) Rewrite the compound inequality as two separate inequalities by using and. Then solve each inequality separately.
3x52xx<x+9<14<7andx+95x+1344x1 x or x1\begin{align*}3x-5&
Answer x<7>\begin{align*} x<7> \end{align*} and x1\begin{align*} x\geq -1\end{align*}. This can be written as 1x<7\begin{align*}-1 \leq x <7\end{align*}.
## Solve a Compound Inequality With “or”
Consider the following example.
Example 4
Solve the following compound inequalities and graph the solution set.
a) 92x3\begin{align*} 9-2x \leq 3\end{align*} or 3x+106x\begin{align*}3x+10 \leq 6-x\end{align*}
b) x262x4\begin{align*} \frac{x-2} {6}\leq2x-4\end{align*} or x26>x+5\begin{align*}\frac{x-2} {6}>x+5\end{align*}
Solution
a) Solve each inequality separately.
92x2xx363or3x+106x4x4x1\begin{align*}9-2x & \leq 3 & & & & 3x+10 \leq 6-x\\ -2x & \leq -6 & & \text{or} & & 4x \leq -4\\ x & \geq 3 & & & & x \leq -1\end{align*}
Answer x3\begin{align*} x \geq 3\end{align*} or x1\begin{align*} x \leq -1\end{align*}
b) Solve each inequality separately.
x26x2x22222x46(2x4)12x2411xxorx26>x+5x2>6(x+5)x2>6x+3032>5x6.4>x\begin{align*}\frac{x-2} {6} & \leq 2x-4 & & & & \frac{x-2} {6} > x+5\\ x-2 & \leq 6(2x-4) & & & & x-2 > 6(x+5)\\ x-2 & \leq 12x-24 & & \text{or} & & x-2 > 6x+30\\ 22 & \leq 11x & & & & -32>5x\\ 2 & \leq x & & & & -6.4>x\end{align*}
Answer x2\begin{align*} x \geq 2\end{align*} or x<64\begin{align*} x <-64\end{align*}
## Solve Compound Inequalities Using a Graphing Calculator (TI-83/84 family)
This section explains how to solve simple and compound inequalities with a graphing calculator.
Example 5
Solve the following inequalities using the graphing calculator.
a) 5x+2(x3)2\begin{align*}5x+2(x- 3) \geq 2\end{align*}
b) 7x2<10x+1<9x+5\begin{align*}7x-2<10x+1<9x+5\end{align*}
c) 3x+210\begin{align*}3x+2 \leq 10\end{align*} or 3x+215\begin{align*} 3x+2 \geq 15\end{align*}
Solution
a) 5x+2(x3)2\begin{align*}5x+2(x-3) \geq 2\end{align*}
Step 1 Enter the inequality.
Press the [Y=] button.
Enter the inequality on the first line of the screen.
Y1=5x+2(x3)2\begin{align*} Y_1 = 5x+2(x-3) \geq 2\end{align*}
The \begin{align*} \geq \end{align*} symbol is entered by pressing [TEST] [2nd] [MATH] and choose option 4.
Press the [GRAPH] button.
Because the calculator translates a true statement with the number 1 and a false statement with the number 0, you will see a step function with the y\begin{align*}y-\end{align*}value jumping from 0 to 1. The solution set is the values of x\begin{align*}x\end{align*} for which the graph shows y=1\begin{align*}y = 1\end{align*}.
Note: You need to press the [WINDOW] key or the [ZOOM] key to adjust window to see full graph.
The solution is x87=1.42857\begin{align*}x \geq \frac{8}{7}=1.42857\ldots\end{align*}, which is why you can see the \begin{align*}y\end{align*} value changing from 0 to 1 at 1.14.
b) \begin{align*}7x-2 <10x+1<9x+5\end{align*}
This is a compound inequality \begin{align*}7x- 2 <10x + 1\end{align*} and \begin{align*}10x + 1 < 9x + 5 \end{align*}.
To enter a compound inequality:
Press the [Y=] button.
Enter the inequality as \begin{align*} Y_1 = (7x-2<10x+1)\end{align*} AND \begin{align*}(10x+1<9x+5)\end{align*}
To enter the [AND] symbol press [TEST], choose [LOGIC] on the top row and choose option 1.
The resulting graph looks as shown at the right.
The solution are the values of \begin{align*}x\end{align*} for which \begin{align*}y=1\end{align*}.
In this case \begin{align*}-1.
c) \begin{align*}3x+2 \leq 10\end{align*} or \begin{align*} 3x+2 \geq 15 \end{align*}
This is a compound inequality \begin{align*}3x + 2 \leq 10\end{align*} or \begin{align*}3x + 2 \geq 15\end{align*}
Press the [Y=] button.
Enter the inequality as \begin{align*} Y_1=(3x+2 \leq 10)\end{align*} OR \begin{align*}(3x+2 \geq 15) \end{align*} To enter the [OR] symbol press [TEST], choose [LOGIC] on the top row and choose option 2.
The resulting graph looks as shown at the right. The solution are the values of \begin{align*}x\end{align*} for which \begin{align*}y=1\end{align*}. In this case, \begin{align*}x \leq 2.7\end{align*} or \begin{align*}x \geq 4.3\end{align*}.
## Solve Real-World Problems Using Compound Inequalities
Many application problems require the use of compound inequalities to find the solution.
Example 6
The speed of a golf ball in the air is given by the formula \begin{align*}v =-32t+ 80\end{align*}, where \begin{align*}t\end{align*} is the time since the ball was hit. When is the ball traveling between 20 ft/sec and 30 ft/sec?
Solution
Step 1
We want to find the times when the ball is traveling between 20 ft/sec and 30 ft/sec.
Step 2
Set up the inequality \begin{align*}20 \leq v \leq 30\end{align*}
Step 3
Replace the velocity with the formula \begin{align*}v=-32t+80\end{align*}.
\begin{align*}20 \leq -32t +80 \leq 30\end{align*}
Separate the compound inequality and solve each separate inequality.
\begin{align*}&20 \leq -32t+80 & & & & -32t+80 \leq 30\\ &32t \leq 60 & & \text{and} & & 50 \leq 32t\\ &t \leq 1.875 & & & & 1.56 \leq t\end{align*}
Answer \begin{align*}1.56 \leq t \leq 1.875 \end{align*}
Step 4 To check plug in the minimum and maximum values of \begin{align*}t\end{align*} into the formula for the speed.
For \begin{align*} t =1.56, v=-32t+80=-32(1.56)+80=30 \ ft/sec\end{align*}
For \begin{align*}t=1.875, v=-32t+80= -32(1.875)+ 80 = 20 \ ft/sec\end{align*}
So the speed is between 20 and 30ft/sec. The answer checks out.
Example 7
William’s pick-up truck gets between 18 to 22 miles per gallon of gasoline. His gas tank can hold 15 gallons of gasoline. If he drives at an average speed of 40 miles per hour how much driving time does he get on a full tank of gas?
Solution
Step 1 We know
The truck gets between 18 and 22 miles/gallon
There are 15 gallons in the truck’s gas tank
William drives at an average of 40 miles/hour
Let \begin{align*}t =\end{align*} driving time
Step 2 We use dimensional analysis to get from time per tank to miles per gallon.
\begin{align*}\frac{t \ \bcancel{hours}}{1 \ \bcancel{tank}} \times \frac{1 \ \bcancel{tank}}{15 \ gallons} \times \frac{40 \ miles}{1 \ \bcancel{hours}} = \frac{40t}{45} \ \frac{miles}{gallon}\end{align*}
Step 3 Since the truck gets between 18 to 22 miles/gallon, we set up the compound inequality.
\begin{align*} 18 \leq \frac{40t} {15} \leq 22\end{align*}
Separate the compound inequality and solve each inequality separately.
\begin{align*}18 & \leq \frac{40t} {15} & & & & \frac{40t} {15}\leq 22\\ 270 & \leq 40t & & \text{and} & & 40t \leq 330\\ 6.75 & \leq t & & & & t \leq 8.25\end{align*}
Answer \begin{align*}6.75 \leq t \leq 8.25\end{align*}. Andrew can drive between 6.75 and 8.25 hours on a full tank of gas.
Step 4
For \begin{align*} t = 6.75\end{align*}, we get \begin{align*} \frac{40t} {15} \frac{40(6.75)} {15}=18 \ miles\end{align*} per gallon.
For \begin{align*} t = 8.25\end{align*}, we get \begin{align*} \frac{40t} {15} \frac{40(8.25)} {15}=18 \ miles\end{align*} per gallon.
## Lesson Summary
• Compound inequalities combine two or more inequalities with “and” or “or”.
• “And” combinations mean the only solutions for both inequalities will be solutions to the compound inequality.
• “Or” combinations mean solutions to either inequality will be solutions to the compound inequality.
## Review Questions
Write the compound inequalities represented by the following graphs.
Solve the following compound inequalities and graph the solution on a number line.
1. \begin{align*} -5 \leq x-4 \leq 13\end{align*}
2. \begin{align*} 1 \leq 3x+4 \leq 4\end{align*}
3. \begin{align*} -12 \leq 2-5x \leq 7\end{align*}
4. \begin{align*} \frac{3} {4} \leq 2x+9 \leq \frac{3} {2}\end{align*}
5. \begin{align*} -2\frac{2x-1} {3}<-1\end{align*}
6. \begin{align*} 4x-1 \geq 7\end{align*} or \begin{align*}\frac{9x} {2}<3\end{align*}
7. \begin{align*} 3-x<-4\end{align*} or \begin{align*}3-x>10\end{align*}
8. \begin{align*} \frac{2x+3} {4}<2\end{align*} or \begin{align*}-\frac{x} {5}+3\frac{2} {5}\end{align*}
9. \begin{align*} 2x-7 \leq -3\end{align*} or \begin{align*}2x-3>11\end{align*}
10. \begin{align*} 4x+3 \leq 9\end{align*} or \begin{align*}-5x+4 \leq -12\end{align*}
11. To get a grade of B in her Algebra class, Stacey must have an average grade greater than or equal to 80 and less than 90. She received the grades of 92, 78, 85 on her first three tests. Between which scores must her grade fall if she is to receive a grade of B for the class?
1. \begin{align*}-40 \leq x \leq 70\end{align*}
2. \begin{align*}x <-2 \end{align*} or \begin{align*}x \geq 5 \end{align*}
3. \begin{align*} -8< x < 0 \end{align*}
4. \begin{align*}x \leq -2\end{align*} or \begin{align*}x> 1.5 \end{align*}
5. \begin{align*}-1 \leq x \leq 17\end{align*}
6. \begin{align*}-\frac{4}{3} \leq x \leq -\frac{1}{3}\end{align*}
7. \begin{align*}-1 \leq x \leq \frac{14}{5} \end{align*}
8. \begin{align*}-\frac{33}{8} \leq x \leq -\frac{15}{4}\end{align*}
9. \begin{align*}-\frac{5}{2} \leq x <-1 \end{align*}
10. \begin{align*}x \geq 2\end{align*} or \begin{align*}x < \frac{2}{3}\end{align*}
11. \begin{align*}x>7 \end{align*} or \begin{align*} x<-7\end{align*}
12. \begin{align*}x< \frac{5}{2}\end{align*} or \begin{align*}x > 13 \end{align*}
13. \begin{align*}x \leq 2\end{align*} or \begin{align*}x> 7 \end{align*}
14. \begin{align*}x < \frac{3}{2}\end{align*} or \begin{align*}x \geq \frac{16}{5}\end{align*}
15. \begin{align*}65 \leq x < 105 \end{align*}
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# How do you divide (-3x^3-23x^2-4x+1)/(x-4) ?
How do you divide (-3x^3-23x^2-4x+1)/(x-4) ?
Answer 1
color(green)(=>"Quotient " -3x^2 - 35x - 144, " Remainder " -575/ (x-4)
#### Explanation:
$\frac{3 {x}^{3} - 23 {x}^{2} - 4 x + 1}{x - 4}$
$\textcolor{w h i t e}{a a a} - - - - - - - - - - - - -$
$4 \textcolor{w h i t e}{a a} | \textcolor{w h i t e}{a a} - 3 \textcolor{w h i t e}{a a a} - 23 \textcolor{w h i t e}{a a a a} - 4 \textcolor{w h i t e}{a a a a} + 1$
$\textcolor{w h i t e}{a a a} | \textcolor{w h i t e}{a a a a} \downarrow \textcolor{w h i t e}{a a} - 12 \textcolor{w h i t e}{a a} - 140 \textcolor{w h i t e}{a a} - 576$
$\textcolor{w h i t e}{a a a} | - - - - - - - - - - - - -$
$\textcolor{w h i t e}{a a a} | \textcolor{w h i t e}{a a} - 3 \textcolor{w h i t e}{a a a} - 35 \textcolor{w h i t e}{a a} - 144 \textcolor{w h i t e}{a a} - 575$
color(green)(=>"Quotient " -3x^2 - 35x - 144, " Remainder " -575/ (x-4)
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