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# 3.9 Modeling using variation (Page 7/14)
Page 7 / 14
$f\left(x\right)={x}^{5}+4{x}^{4}+4{x}^{3}$
$f\left(x\right)={x}^{3}-4{x}^{2}+x-4$
4 with multiplicity 1
For the following exercises, based on the given graph, determine the zeros of the function and note multiplicity.
$\frac{1}{2}\text{\hspace{0.17em}}$ with multiplicity 1, 3 with multiplicity 3
Use the Intermediate Value Theorem to show that at least one zero lies between 2 and 3 for the function $\text{\hspace{0.17em}}f\left(x\right)={x}^{3}-5x+1$
## Dividing Polynomials
For the following exercises, use long division to find the quotient and remainder.
$\frac{{x}^{3}-2{x}^{2}+4x+4}{x-2}$
$\text{\hspace{0.17em}}{x}^{2}+4\text{\hspace{0.17em}}$ with remainder 12
$\frac{3{x}^{4}-4{x}^{2}+4x+8}{x+1}$
For the following exercises, use synthetic division to find the quotient. If the divisor is a factor, then write the factored form.
$\frac{{x}^{3}-2{x}^{2}+5x-1}{x+3}$
${x}^{2}-5x+20-\frac{61}{x+3}$
$\frac{{x}^{3}+4x+10}{x-3}$
$\frac{2{x}^{3}+6{x}^{2}-11x-12}{x+4}$
$2{x}^{2}-2x-3$ , so factored form is $\left(x+4\right)\left(2{x}^{2}-2x-3\right)$
$\frac{3{x}^{4}+3{x}^{3}+2x+2}{x+1}$
## Zeros of Polynomial Functions
For the following exercises, use the Rational Zero Theorem to help you solve the polynomial equation.
$2{x}^{3}-3{x}^{2}-18x-8=0$
$3{x}^{3}+11{x}^{2}+8x-4=0$
$2{x}^{4}-17{x}^{3}+46{x}^{2}-43x+12=0$
$4{x}^{4}+8{x}^{3}+19{x}^{2}+32x+12=0$
For the following exercises, use Descartes’ Rule of Signs to find the possible number of positive and negative solutions.
${x}^{3}-3{x}^{2}-2x+4=0$
0 or 2 positive, 1 negative
$2{x}^{4}-{x}^{3}+4{x}^{2}-5x+1=0$
## Rational Functions
For the following rational functions, find the intercepts and the vertical and horizontal asymptotes, and then use them to sketch a graph.
$f\left(x\right)=\frac{x+2}{x-5}$
Intercepts $\left(–2,0\right)\text{and}\left(0,-\frac{2}{5}\right)$ , Asymptotes $\text{\hspace{0.17em}}x=5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=1.$
$f\left(x\right)=\frac{{x}^{2}+1}{{x}^{2}-4}$
$f\left(x\right)=\frac{3{x}^{2}-27}{{x}^{2}+x-2}$
Intercepts (3, 0), (-3, 0), and $\text{\hspace{0.17em}}\left(0,\frac{27}{2}\right)\text{\hspace{0.17em}}$ , Asymptotes
$f\left(x\right)=\frac{x+2}{{x}^{2}-9}$
For the following exercises, find the slant asymptote.
$f\left(x\right)=\frac{{x}^{2}-1}{x+2}$
$f\left(x\right)=\frac{2{x}^{3}-{x}^{2}+4}{{x}^{2}+1}$
For the following exercises, find the inverse of the function with the domain given.
$f\left(x\right)={\left(x-2\right)}^{2},\text{\hspace{0.17em}}x\ge 2$
${f}^{-1}\left(x\right)=\sqrt{x}+2$
$f\left(x\right)={\left(x+4\right)}^{2}-3,\text{\hspace{0.17em}}x\ge -4$
$f\left(x\right)={x}^{2}+6x-2,\text{\hspace{0.17em}}x\ge -3$
${f}^{-1}\left(x\right)=\sqrt{x+11}-3$
$f\left(x\right)=2{x}^{3}-3$
$f\left(x\right)=\sqrt{4x+5}-3$
${f}^{-1}\left(x\right)=\frac{{\left(x+3\right)}^{2}-5}{4},\text{\hspace{0.17em}}x\ge -3$
$f\left(x\right)=\frac{x-3}{2x+1}$
## Modeling Using Variation
For the following exercises, find the unknown value.
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly as the square of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If when find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x=4.$
$y=64$
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies inversely as the square root of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ If when find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x=4.$
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies jointly as the cube of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and as $\text{\hspace{0.17em}}z.\text{\hspace{0.17em}}$ If when $\text{\hspace{0.17em}}x=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}z=2,\text{\hspace{0.17em}}$ $y=6,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}z=3.$
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies jointly as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and the square of $\text{\hspace{0.17em}}z\text{\hspace{0.17em}}$ and inversely as the cube of $\text{\hspace{0.17em}}w.\text{\hspace{0.17em}}$ If when $\text{\hspace{0.17em}}x=3,\text{\hspace{0.17em}}$ $z=4,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}w=2,\text{\hspace{0.17em}}$ $y=48,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x=4,\text{\hspace{0.17em}}$ $z=5,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}w=3.$
For the following exercises, solve the application problem.
The weight of an object above the surface of the earth varies inversely with the distance from the center of the earth. If a person weighs 150 pounds when he is on the surface of the earth (3,960 miles from center), find the weight of the person if he is 20 miles above the surface.
148.5 pounds
The volume $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ of an ideal gas varies directly with the temperature $\text{\hspace{0.17em}}T\text{\hspace{0.17em}}$ and inversely with the pressure P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K.
## Chapter test
Perform the indicated operation or solve the equation.
$\left(3-4i\right)\left(4+2i\right)$
$20-10i$
$\frac{1-4i}{3+4i}$
${x}^{2}-4x+13=0$
Give the degree and leading coefficient of the following polynomial function.
$f\left(x\right)={x}^{3}\left(3-6{x}^{2}-2{x}^{2}\right)$
Determine the end behavior of the polynomial function.
$f\left(x\right)=8{x}^{3}-3{x}^{2}+2x-4$
$As\text{\hspace{0.17em}}x\to -\infty ,\text{\hspace{0.17em}}f\left(x\right)\to -\infty ,\text{\hspace{0.17em}}as\text{\hspace{0.17em}}x\to \infty ,\text{\hspace{0.17em}}f\left(x\right)\to \infty$
$f\left(x\right)=-2{x}^{2}\left(4-3x-5{x}^{2}\right)$
Write the quadratic function in standard form. Determine the vertex and axes intercepts and graph the function.
$f\left(x\right)={x}^{2}+2x-8$
$f\left(x\right)={\left(x+1\right)}^{2}-9$ , vertex $\text{\hspace{0.17em}}\left(-1,-9\right)$ , intercepts $\text{\hspace{0.17em}}\left(2,0\right);\left(-4,0\right);\text{\hspace{0.17em}}\left(0,-8\right)$
Given information about the graph of a quadratic function, find its equation.
Vertex $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ and point on graph $\text{\hspace{0.17em}}\left(4,12\right).$
Solve the following application problem.
A rectangular field is to be enclosed by fencing. In addition to the enclosing fence, another fence is to divide the field into two parts, running parallel to two sides. If 1,200 feet of fencing is available, find the maximum area that can be enclosed.
60,000 square feet
Find all zeros of the following polynomial functions, noting multiplicities.
$f\left(x\right)={\left(x-3\right)}^{3}\left(3x-1\right){\left(x-1\right)}^{2}$
$f\left(x\right)=2{x}^{6}-6{x}^{5}+18{x}^{4}$
0 with multiplicity 4, 3 with multiplicity 2
Based on the graph, determine the zeros of the function and multiplicities.
Use long division to find the quotient.
$\frac{2{x}^{3}+3x-4}{x+2}$
$2{x}^{2}-4x+11-\frac{26}{x+2}$
Use synthetic division to find the quotient. If the divisor is a factor, write the factored form.
$\frac{{x}^{4}+3{x}^{2}-4}{x-2}$
$\frac{2{x}^{3}+5{x}^{2}-7x-12}{x+3}$
$2{x}^{2}-x-4$ . So factored form is $\text{\hspace{0.17em}}\left(x+3\right)\left(2{x}^{2}-x-4\right)$
Use the Rational Zero Theorem to help you find the zeros of the polynomial functions.
$f\left(x\right)=2{x}^{3}+5{x}^{2}-6x-9$
$f\left(x\right)=4{x}^{4}+8{x}^{3}+21{x}^{2}+17x+4$
$-\frac{1}{2}\text{\hspace{0.17em}}$ (has multiplicity 2), $\text{\hspace{0.17em}}\frac{-1±i\sqrt{15}}{2}\text{\hspace{0.17em}}$
$f\left(x\right)=4{x}^{4}+16{x}^{3}+13{x}^{2}-15x-18$
$f\left(x\right)={x}^{5}+6{x}^{4}+13{x}^{3}+14{x}^{2}+12x+8$
$\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ (has multiplicity 3), $\text{\hspace{0.17em}}±i$
Given the following information about a polynomial function, find the function.
It has a double zero at $\text{\hspace{0.17em}}x=3\text{\hspace{0.17em}}$ and zeroes at $\text{\hspace{0.17em}}x=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=-2\text{\hspace{0.17em}}$ . It’s y -intercept is $\text{\hspace{0.17em}}\left(0,12\right).\text{\hspace{0.17em}}$
It has a zero of multiplicity 3 at $\text{\hspace{0.17em}}x=\frac{1}{2}\text{\hspace{0.17em}}$ and another zero at $\text{\hspace{0.17em}}x=-3\text{\hspace{0.17em}}$ . It contains the point $\text{\hspace{0.17em}}\left(1,8\right).$
$f\left(x\right)=2{\left(2x-1\right)}^{3}\left(x+3\right)$
Use Descartes’ Rule of Signs to determine the possible number of positive and negative solutions.
$8{x}^{3}-21{x}^{2}+6=0$
For the following rational functions, find the intercepts and horizontal and vertical asymptotes, and sketch a graph.
$f\left(x\right)=\frac{x+4}{{x}^{2}-2x-3}$
Intercepts $\text{\hspace{0.17em}}\left(-4,0\right),\text{\hspace{0.17em}}\left(0,-\frac{4}{3}\right)\text{\hspace{0.17em}}$ , Asymptotes .
$f\left(x\right)=\frac{{x}^{2}+2x-3}{{x}^{2}-4}$
Find the slant asymptote of the rational function.
$f\left(x\right)=\frac{{x}^{2}+3x-3}{x-1}$
$y=x+4$
Find the inverse of the function.
$f\left(x\right)=\sqrt{x-2}+4$
$f\left(x\right)=3{x}^{3}-4$
${f}^{-1}\left(x\right)=\sqrt[3]{\frac{x+4}{3}}$
$f\left(x\right)=\frac{2x+3}{3x-1}$
Find the unknown value.
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies inversely as the square of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and when $\text{\hspace{0.17em}}x=3,\text{\hspace{0.17em}}$ $y=2.\text{\hspace{0.17em}}$ Find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x=1.$
$y=18$
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies jointly with $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and the cube root of $\text{\hspace{0.17em}}z.\text{\hspace{0.17em}}$ If when $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}z=27,\text{\hspace{0.17em}}$ $y=12,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x=5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}z=8.$
Solve the following application problem.
The distance a body falls varies directly as the square of the time it falls. If an object falls 64 feet in 2 seconds, how long will it take to fall 256 feet?
4 seconds
how can are find the domain and range of a relations
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
i want to sure my answer of the exercise
what is the diameter of(x-2)²+(y-3)²=25
how to solve the Identity ?
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim
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# Difference between revisions of "2017 AMC 8 Problems/Problem 22"
## Problem 22
In the right triangle $ABC$, $AC=12$, $BC=5$, and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle? $[asy] draw((0,0)--(12,0)--(12,5)--(0,0)); draw(arc((8.67,0),(12,0),(5.33,0))); label("A", (0,0), W); label("C", (12,0), E); label("B", (12,5), NE); label("12", (6, 0), S); label("5", (12, 2.5), E);[/asy]$
$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}$
## Solution
We can reflect triangle $ABC$ on line $AC.$ This forms the triangle $AB'C$ and a circle out of the semicircle. Let us call the center of the circle $O.$
We can see that Circle $O$ is the incircle of $AB'C.$ We can use the formula for finding the radius of the incircle to solve this problem. The area of $AB'C$ is $12\times5 = 60.$ The semiperimeter is $5+13 = 18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$
## Solution
We immediately see that $AB=13$, and we label the center of the semicircle $O$. Drawing radius $OD$ with length $x$ such that $OD$ is tangent to $AB$, we immediately see that $ODB\cong OCB$ because of HL congruence, so $BD=5$ and $DA=8$. By similar triangles $ODA$ and $BCA$, we see that $\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{D}}$.
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## Comparing fractions - 1 whole
Activity type: Interactive Activity
## Comparing fractions - 1 whole
Course
Mathematics
Year 5
Section
Fractions
Outcome
Comparing fractions - 1 whole
Activity Type
Interactive Activity
Activity ID
28408
## Testimonials
What a brilliant site you have!!! I love it, especially as it saves me hours and hours of hard work. Others who haven't found your site yet don't know what they are missing!
## New Zealand – National Standards
• ##### 5.NA – Number and algebra
• 5.NA.1 – Apply additive and simple multiplicative strategies and knowledge of symmetry to:
• 5.NA.1.b – find fractions of sets, shapes, and quantities
• ##### 6.NA – Number and algebra
• 6.NA.1 – Apply additive and simple multiplicative strategies flexibly to:
• 6.NA.1.b – Find fractions of sets, shapes, and quantities
## Australia – Australian Curriculum
• ##### Number and Algebra
• Fractions and decimals
• ACMNA077 – Investigate equivalent fractions used in contexts
## United Kingdom – National Curriculum
• ##### Year 3 programme of study
• KS2.Y3.N.F – Number - fractions
• Pupils should be taught to:
• KS2.Y3.N.F.3 – Recognise and use fractions as numbers: unit fractions and non-unit fractions with small denominators
## United States – Common Core State Standards
• ##### 3.NF – Number & Operations—Fractions¹ (Grade 3 expectations in this domain are limited to fractions with denominators 2, 3, 4, 6, 8.)
• Mathematics
• 3.NF.3 – Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
• 3.NF.3.c – Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Examples: Express 3 in the form 3 = 3/1; recognize that 6/1 = 6; locate 4/4 and 1 at the same point of a number line diagram.
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# Quadrilaterals II by Monica Yuskaitis.
## Presentation on theme: "Quadrilaterals II by Monica Yuskaitis."— Presentation transcript:
Definition Perimeter – (P) The distance around a figure. 5 9 in. 9 5 5 in. 5 in. + 9 28 in. 9 in. Copyright © 2000 by Monica Yuskaitis
Find the perimeter 9 cm. 9 9 cm. 9 cm. 9 9 + 9 36 cm. 9 cm.
A shortcut for the square is 4 x side. Copyright © 2000 by Monica Yuskaitis
Find the perimeter 7 cm. 9 9 cm. 9 cm. 7 9 + 7 36 cm. 7 cm. Copyright © 2000 by Monica Yuskaitis
Find the perimeter 10 ft. 10 10 ft. 10 ft. 6 10 + 10 36 ft. 6 ft. Copyright © 2000 by Monica Yuskaitis
Find the perimeter 18 m. 12 12 m. 15 m. 11 15 + 18 11 m. 56 m. Copyright © 2000 by Monica Yuskaitis
Area – (A) the measure, in square units, of the surface of a figure.
Definition Area – (A) the measure, in square units, of the surface of a figure. The area of this rectangle is 8 squares. Copyright © 2000 by Monica Yuskaitis
How to find the area of a square or rectangle.
Step 1 – Multiply the length times the width. (A = l x w) 5 x 3 3 cm. 15 5 cm. Copyright © 2000 by Monica Yuskaitis
How to find the area of a square or rectangle.
Step 2 – Indicate that the answer is in square centimeters by writing the exponent 2 after the unit of measure. 5 x 3 3 cm. 15 cm.2 5 cm. Copyright © 2000 by Monica Yuskaitis
How to find the area of a square or rectangle.
Step 3 – Read the answer: 15 square centimeters. 5 x 3 3 cm. 15 cm.2 5 cm. Copyright © 2000 by Monica Yuskaitis
What is the area? 5 x 6 5 ft. 30 ft.2 6 ft. You say: 30 square feet. Copyright © 2000 by Monica Yuskaitis
What is the area? 2 2 in. x 6 12 in.2 6 in. You say: 12 square inches. Copyright © 2000 by Monica Yuskaitis
What is the area? 7 4 m. x 4 28 m.2 7 m. You say: 28 square meters. Copyright © 2000 by Monica Yuskaitis
How to find the area of a parallelogram
Step 1 – Multiply the base times the height. (A = b x h) height 90º base Copyright © 2000 by Monica Yuskaitis
How to find the area of a parallelogram
Step 1 – Multiply the base times the height. (A = b x h) 2 13 height 17 x 13 in. 90º 9 1 + 1 3 base 17 in. 221 Copyright © 2000 by Monica Yuskaitis
How to find the area of a parallelogram
Step 2 – Indicate that the answer is in square inches by writing the exponent 2 after the unit of measure. 13 x 17 height 9 1 13 in. 90º 1 3 + 221 in. 2 base 17 in. Copyright © 2000 by Monica Yuskaitis
How to check the area of a parallelogram
height 2 cm. 90º base 4 cm, 4 x 2 = 8 cm.2 First find the area. Copyright © 2000 by Monica Yuskaitis
How to check the area of a parallelogram
4 x 2 = 8 cm.2 height 2 cm. 90º base 4 cm, Cut off the piece at the dotted line. Copyright © 2000 by Monica Yuskaitis
How to check the area of a parallelogram
4 x 2 = 8 cm.2 height 2 cm. 90º base 4 cm, Cut off the piece at the dotted line. Copyright © 2000 by Monica Yuskaitis
How to check the area of a parallelogram
4 x 2 = 8 cm.2 height 2 cm. 90º base 4 cm, Move this piece to the other side. Copyright © 2000 by Monica Yuskaitis
How to check the area of a parallelogram
4 x 2 = 8 cm.2 height 2 cm. 90º base 4 cm, Move this piece to the other side. Copyright © 2000 by Monica Yuskaitis
How to check the area of a parallelogram
4 x 2 = 8 cm.2 height 2 cm. 90º base 4 cm, Now you have a rectangle 4 x 2 cm. Copyright © 2000 by Monica Yuskaitis
How to check the area of a parallelogram
4 x 2 = 8 cm.2 width 2 cm. length 4 cm, What is the area of this rectangle? Copyright © 2000 by Monica Yuskaitis
How to check the area of a parallelogram
4 x 2 = 8 cm.2 width 2 cm. length 4 cm, That’s right! 8 cm2 The same as the parallelogram. Copyright © 2000 by Monica Yuskaitis
What is the area of this parallelogram?
5 height 5 cm. x 7 35 cm.2 base 7 cm. You say: 35 square centimeters Copyright © 2000 by Monica Yuskaitis
What is the area of this parallelogram?
height 5 ft. 5 x 6 30 ft.2 base 6 ft. You say: 30 square feet. Copyright © 2000 by Monica Yuskaitis
What is the area of this parallelogram?
height 10 m. 10 x 5 50 m.2 base 5 m. You say: 50 square meters. Copyright © 2000 by Monica Yuskaitis
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## NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.2
Chapter 13 Limits and Derivatives Exercise 13.2 NCERT Solutions for Class 11 Maths will help you in solving your doubts as these Class 11 Maths NCERT Solutions are prepared by Studyrankers experienced subject matter experts. These NCERT Solutions are very important for the purpose of examinations as it will help you knowing the basic concepts of the chapter.
1. Find the derivative of x 2 – 2 at x = 10
Let y = x 2 – 2
dy/dx = 2x dy/dx at x = 10 is equal to 20
2. Find the derivative of x 2 – 2 at x = 10
Find the derivative of 99x at x = 100 (By first principle)
Let f (x) = 99x. Derivative of f (x) at x = 100 is
Now, f (x) = 99x
f (100 + h)= 99(100 + h)
f (100)= 99 × 100
∴ f (100 + h) – f (100) = 99 (100 + h) – 99 × 100
= 99 [100 + h – 100] = 99 × h
3. Find the derivative of x at x = 1
Derivative of f (x) = x at x = 1
4. Find the derivative of the following functions from first principle:
(i) x 3 – 27 (ii) (x – 1)(x – 2)
(iii) 1/x2 (iv) (x + 1)/(x – 1)
5. For the function
f(x) = x100/100 + x99/99 + ……..+ x2/2 + x + 1
prove that f’(1) = 100 f’(0)
We know that d/dx (xn) = nxn-1
∴ For f(x) = x100/100 + x99/99 + ……+ x2/2 + x+ 1
f'(x) = 100x99/100 + 99. x98/99 + …..+ 2x/2 + 1
= x99 + x98 + ………….+ x + 1
Now, f’(x) = 1 + 1+ …….. to 100 term = 100
f'(0) = 1
∴ f'(1) = 100 × 1 = 100f’(0)
hence, f’(1) = 100f’(0)
6. Find the derivative of x n + axn – 1 + a 2n – 2 + ....+ a n – 1x + a n for some fixed real number a?
Let f (x) = xn + axn – 1 + a 2 x n – 2+ ........ + a n – 1x + a n
Now, d/dx xn = nxn-1 , d/dx xn-1 = (n – 1)xn-2.
etc and d/dx [ag(x)] = ag’(x), d/dx an = 0
f’(x) = nxn-1 + (n – 1)axn – 2 + (n – 2)a2xn-3 + ………..+ an-1
7. For some constants a and b, find the derivative of:
(i) (x – a)(x – b)
(ii) For some constant a and b find the derivation of (ax2 + b) 2 .
(iii) (x – a)/( x – b)
(i) Let f (x) = (x – a)(x – b)
Using product rule, we have
df(x)/dx = (x – a) d(x – b)/dx + (x-b) d(x-a)/dx
= (x – a)[d(x)/dx - d(b)/dx] + (x – b)[d(x)/dx - d(a)/dx]
= (x – a)[1 – 0] + (x – b)[1 – 0]
(ii) f(x) = (ax2 + b)2 = a2x+ 2abx2 + b2
now, d/dx x4 = 4x3 and
d/dx x2 = 2x, d/dx b2 = 0
f’(x) = a2. 4x3 + 2ab.2x + 0
= 4a2x3 + 4abx
(iii) Let f(x) = (x- a)/(x – b)
Using quotient rule, we have
= ((x – b)(d(x-a))/dx – (x – a) (d(x – b))/dx)/(x – b)2
8. Find the derivative of (xn – an)/(x – a) for some constant ‘a’.
We know d/dx (u/v) = (u’v – uv’)/v2
d/dx ((xn – an)/(x – a))
= ([d/dx (xn – an)](x – a)-(xn – an)d/dx (x – a))/(x – a)2
= (nxn-1(x – a)- (xn – an). 1)/(x – a)2
= (nxn – n.xn – 1 a – xn + an)/(x – a)2
= ((n – 1 )xn – naxn-1 + an)/(x – a)2
9. Find the derivative of
(i) 2x – 3/4
(ii) (5x3 + 3x – 1)(x – 1)
(iii) x-3(5 + 3x)
(iv) x5(3 – 6x-9)
(v) x-4(3 – 4x-5)
(vi) 2/(x + 1) - x2/(3x – 1)
(i) Let f(x) = 2x – 3/4
f'(x) = 2 (d/dx) x = 1, (d/dx)(3/4) = 0
(ii) let f(x) = (5x3 + 3x – 1)(x – 1)
(uv)’ = u’v + uv’
f'[(5x3 + 3x – 1)(x – 1)]
= [d/dx (5x3 + 3x – 1)](x – 1) + (5x3 + 3x – 1) d/dx (x – 1)
= (15x2 + 3)(x – 1) + (15x3 + 3x – 1).1
= x(15x2 + 3) – (15x2 + 3) + (5x3 + 3x – 1))
= 20x3 – 15x2 + 6x - 4
(iii) let f(x) = x-3 (5 + 3x) = 5x-3 + 3x-2
f’(x) = 5(-3)x-4 + 3(-2)x-3
= (-15/x4) – (6/x3) = (-(6x + 15))/x4 = (-3(2x + 5))/x4
(iv) Let f(x) = x5(3 – 6x-9) = 3x5 – 6x-4
f'(x) = (d(3x5 – 6x-4))/dx
= (d(3x5)/dx) – (d(6x-4)/dx)
= 3.(dx5/dx) – 6.(dx-4/dx)
= (3)(5)x4 – (6)(-4)x-5
= 15x4 + 24x-5
(v) let f(x) = x-4(3 – 4x-5)
f'(x) = x-4 . (d(3 – 4x-5)/dx) + (3 – 4x-5)(dx-4/dx)
= x-4[(d(3)/dx) – (d(4x-5)/dx)] + (3 – 4x-5)(-4)x-5
= x-4[0 – (4)(-5)x-6] + (3 – 4x-5)(-4)x-5
= x-4[20x-6]-x-5(12 – 16x-5)
= 20x-10 – 12x-5 + 16x-10 = 36x10 – 12x-5
10. Find the derivative of cos x from first principle.
f (x) = cos x By first Principle,
11. Find the derivative of the following functions:
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3 cot x + 5 cosec x
(vi) 5 sin x – 6 cos x + 7
(vii) 2tanx – 7 sec x
(i) Let f(x) = sinx cosx
f’(x) = u’v + uv’
= (d/dx sin x) cosx + sinx d/dx (cos x)
= cos x. cos x + sin x(-sin x)
cos2x – sin2x = cos2x
(ii)
(iii) Let f(x) = 5secx + 4cos x
f’(x) = (d(5sec x + 4cos x))/dx
= (d(5sec x)/dx) + (d(4cosx)/dx)
= 5.(d(sec x)/dx) + 4.(d(cos x)/dx)
= 5sec x tan x – 4 sin x
(iv) Let f(x) = cosec x
f'(x) = (d(cosec x))/dx = (d. 1/sinx)/dx
= (sinx (d(1)/dx) – (1.d(sin x)/dx))/(sin x)2
= (0 – cos x)/(sin 2 x) = (-cosx/sin x). (1/sin x)
= - cosec x cot x
(v) Now,
d/dx (3 cot x) = 3. d/dx (cot x) = -3 cosec2 x
Also, 5.d/dx (cosec x) = - 5cosec x cot x
d/dx (3cot x + 5 cosec x)
= -3 cosec2x – 5cosec x cot x
= -cosec x (3cosec x + 5 cot x)
(vi) Let f(x) = 5 sin x – 6 cos x + 7
f'(x) = (d(5 sin x – 6 cos x + 7))/dx
= (d(5 sin x))/dx - (d(6 cos x))/dx + (d(7))/dx
= 5cos x – 6(- sin x) + 0
= 5 cos x + 6 sin x
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# Subtracting Exponents
To understand subtraction of exponents, let's start from a simple example.
Example. Suppose, we want to find (2^7)/(2^4).
We already learned about positive integer exponets, so we can write, that 2^7=2*2*2*2*2*2*2 and 2^4=2*2*2*2.
So, (2^7)/(2^4)=(2*2*2*color(red)(2*2*2*2))/(color(red)(2*2*2*2))=2*2*2=2^3.
Let's see what have we done. We counted number of 2's in 2^7, then counted number of 2's in 2^4. Since we divided, we canceled common terms. Note, that 7-4=3.
It appears, that this rule works not only for positive integer exponents, it works for any exponent.
Rule for subtracting exponents: huge color(purple)((a^m)/(a^n)=a^(m-n)).
Word of caution. It doesn't work, when bases are not equal.
For example, (4^5)/(3^2)=(4*4*4*4*4)/(3*3) which is neither 4^3 nor 3^3.
Word of caution. Above rule doesn't work for addition and subtraction.
For example, 2^7-2^4!=2^3, because 2^7-2^4=128-16=112 and 2^3=8. Clearly, 112!=8.
Let's go through a couple of examples.
Example 2. Find (2^3)/(2^(-5)).
It doesn't matter, that exponent is negative.
Just proceed as always: (2^3)/(2^(-5))=(2^(3-(-5)))=2^(3+5)=2^8.
Even when exponents are fractional, we use the same rule!
Example 3. Find (3^(1/4))/(3^(2/3)).
(3^(1/4))/(3^(2/3))=3^(1/4-2/3)=3^(-5/12)=1/3^(5/12).
We can handle radicals, also, because radicals can be rewritten with the help of exponent.
Example 4. Rewrite, using positive exponent: root(8)(3)/root(7)(1/3^2).
First we rewrite numbers, using exponents and then apply the rule:
root(8)(3)/root(7)(1/3^2)=(3^(1/8))/root(7)(3^(-2))=(3^(1/8))/(3^(-2/7))=3^(1/8-(-2/7))=3^(23/56).
Finally, we can see now, why a^0=1.
Indeed, a^0=a^(n-n)=(a^n)/(a^n)=1.
Now, it is time to exercise.
Exercise 1. Find (3^5)/(3^2).
Answer: 3^3=27.
Exercise 2. Can we use rule for adding exponents to find (5^5)/(3^5)?
Answer: No, bases are not equal.
Exercise 3. Find (4^(5/3))/(4^(2/3)).
Answer: 4.
Exercise 4. Find (3^2)/(3^(-1/5)).
Answer: 3^(2+1/5)=3^(11/5)=root(5)(3^11).
Exercise 5. Find root(7)(1/27)/root(8)(9).
Answer: root(7)(3^(-3))/root(8)(3^2)=1/3^(19/28).
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Graphing Trig Functions
Presentation on theme: "Graphing Trig Functions"— Presentation transcript:
Graphing Trig Functions
Graphing Trig Functions SAMM Sections 2.7 and 2.8 Amplitude and Period - Notes
Graphing Terminology Amplitude—The maximum or minimum vertical distance between the graph and the x-axis. Amplitude is always positive because it is a distance. Period--The number of degrees or radians we must graph before it begins to repeat.
Amplitude Amplitude Period
This number controls the period.
Trig Functions Trig functions are written in the form of y = AsinBx or y = AcosBx y= 5 sin 2x This number controls the period. Amplitude
Graphing Sine and Cosine
To find the period length —
To find the angles to use for graphing –
Find the sum of the beginning and end angles and divide by two. Repeat step 2 with the beginning and middle angles. Do the same with the middle and end angles.
Quick graph of the Cosine function is Max-Zero-Min-Zero-Max
Quick graph of the Sine function is Zero-Max-Zero-Min-Zero Quick graph of the Cosine function is Max-Zero-Min-Zero-Max Sketch a sine or cosine curve Based on the points graphed.
The amplitude is ? The period length is?. Example #1 y= 2sin4x
Answer is 2. Remember, amplitude is always a positive number . The period length is?.
Graph y = 2sin4x
The amplitude is ? The period length is?. Example #2 y= -3cos2x
Answer is 3. Remember, amplitude is always a positive number . The negative means the graph will be flipped when graphed – How does this happen? The period length is?.
Graph y = -3cos2x
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Live truth instead of professing it
# What is monotonic and bounded sequence?
## What is monotonic and bounded sequence?
In this section, we will be talking about monotonic and bounded sequences. We will learn that monotonic sequences are sequences which constantly increase or constantly decrease. We also learn that a sequence is bounded above if the sequence has a maximum value, and is bounded below if the sequence has a minimum value.
What is meant by monotonic sequence?
Monotone Sequences. Definition : We say that a sequence (xn) is increasing if xn ≤ xn+1 for all n and strictly increasing if xn < xn+1 for all n. Similarly, we define decreasing and strictly decreasing sequences. Sequences which are either increasing or decreasing are called monotone.
### How do you know if a sequence is monotonic?
If a sequence is monotonic, it means that it’s always increasing or always decreasing. If a sequence is sometimes increasing and sometimes decreasing and therefore doesn’t have a consistent direction, it means that the sequence is not monotonic.
How do you tell if a sequence is increasing or decreasing or monotonic?
Definition 6.16. Monotonic Sequence.
1. If an
2. If an≤an+1 a n ≤ a n + 1 for all n, then the sequence is non-decreasing .
3. If an>an+1 a n > a n + 1 for all n, then the sequence is decreasing or strictly decreasing .
## What is bounded sequence?
A sequence is bounded if it is bounded above and below, that is to say, if there is a number, k, less than or equal to all the terms of sequence and another number, K’, greater than or equal to all the terms of the sequence. Therefore, all the terms in the sequence are between k and K’.
What is bounded and unbounded sequence?
A sequence an is bounded below if there exists a real number M such that. M≤an. for all positive integers n. A sequence an is a bounded sequence if it is bounded above and bounded below. If a sequence is not bounded, it is an unbounded sequence.
### What is monotonic sequence theorem?
In real analysis, the monotone convergence theorem states that if a sequence increases and is bounded above by a supremum, it will converge to the supremum; similarly, if a sequence decreases and is bounded below by an infimum, it will converge to the infimum.
What is unbounded sequence?
If a sequence is not bounded, it is an unbounded sequence. For example, the sequence 1/n is bounded above because 1/n≤1 for all positive integers n. It is also bounded below because 1/n≥0 for all positive integers n.
## What is the monotonic sequence theorem?
Monotone Sequence Theorem: (sn) is increasing and bounded above, then (sn) converges. Intuitively: If (sn) is increasing and has a ceiling, then there’s no way it cannot converge.
How can I prove that this sequence is monotonic?
– show the differences in consecutive terms have the same sign (or is zero) – show derivative of associated function has same sign (or is zero) – Given the sequence is always strictly positive or always strictly negative, show the absolute ratio between consecutive terms is always weakly greater than or always weakly less than 1
### How to show that this sequence is monotonic?
– Prove that an < 2 for all n ∈ N. – Prove that {an} is an increasing sequence. – Prove that limn→∞an = 2.
What is a monotone sequence?
If {an} is increasing or decreasing, then it is called a monotone sequence. The sequence is called strictly increasing (resp. strictly decreasing) if an < an + 1 for all n ∈ N (resp. an > an + 1 for all n ∈ N. It is easy to show by induction that if {an} is an increasing sequence, then an ≤ am whenever n ≤ m.
## What does monotonic mean in calculus?
In calculus and analysis. In calculus, a function defined on a subset of the real numbers with real values is called monotonic if and only if it is either entirely non-increasing, or entirely non-decreasing. That is, as per Fig. 1, a function that increases monotonically does not exclusively have to increase, it simply must not decrease.
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# Exponents and Roots
## Remember how we defined multiplication. It is addition done a given
number of times.
Similarly, the process of exponent is doing multiplication a given
number of times.
2 x 5 = 2 + 2 + 2 + 2 + 2 = 10
25 = 2 x 2 x 2 x 2 x 2 = 32
----------------------5 times
## The exponent function also called as the power function is defined
as a (Base)Power
In general terms it is defined as m n, where m and n are any real
numbers. Real numbers include all the numbers we have studied so far
natural numbers, whole numbers, integers, fractional numbers,
decimals, rational numbers and irrational numbers. All these numbers
combined are called real numbers
A few examples of exponents are
26 = 64
33 = 27
52 = 25
122 = 144
(-1)5 = -1
(2.2)2 = 4.84
and so on.
The idea of exponents is used across mathematics in varied number of
ways from calculating areas and volumes, to finding out bank
interests, to even finding the coordinates of a spaceship.
The GMAT poses questions on exponents both directly and indirectly.
An exponent has 2 parts the Base and the Power. Well first study
how powers impact the exponent and then explore about the bases.
The Power:
This is the number written in small size over a given number or an
expression. For instance in (x + 5)2, the expression (x + 5) is raised to
the power of 2.
## The power of 2 is called as square, and the power of 3 is called as
cube.
For any number raised to the power 0, the result is 1, i.e. n 0 =
1, where n is any real number.
So, 42 is called as square of 4 (or 4 squared), and 5 3 is called as the
cube of 5 (or 5 cubed).
If a number is the square of an integer, then it is called a perfect
square.
For eg: 81 is a perfect square (of 9), but 88 is not!
Similarly, if a number is the cube of an integer, then it is called a
perfect cube.
For eg: 125 is a perfect cube (of 5), but 111 is not!
The major perfect squares and perfect cubes are highlighted later in
the chapter.
n2 is always positive, regardless of the fact whether n is positive or
negative.
Because any number positive or negative multiplied by itself even
number of times is always positive, i.e.
Xn is always positive if n is an even number!
Take a few examples and check. If X is -121 and n is 4, the result will
be
-121 x -121 x -121 x -121
This will always be positive because the ve sign is being multiplied
even number of times, and we know that -ve x ve is +ve
In simpler words, EVEN powers EAT the sign!
There are some standard (perfect) squares and cubes that you should
be familiar with, ideally remember them.
02
12
22
32
=
=
=
=
0
1
4
9
42 = 16
52 = 25
62 = 36
72 = 49
82 = 64
92 = 81
102 = 100
112 = 121
122 = 144
132 = 169
152 = 225
202 = 400
252 = 625
502 = 2500
1002 = 10000
03 = 0
13 = 1
23 = 8
33 = 27
43 = 64
53 = 125
63 = 216
103 = 1000
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
28 = 256
29 = 512
210 = 1024
The Roots:
We have dealt with positive powers, with negative powers and even
with 0 as a power. Now, well deal with fractional, or even decimal,
powers.
X1/n or
## n X is called as the nth root of X
Remember roots are the same as powers, but roots are fractional
powers with the numerator as 1.
When you are looking for the n th root of X, you are basically looking for
a number that when multiplied by itself n number of times will give
you X as a result
Remember, when finding powers you were looking for the opposite, i.e.
you were trying to find X
Eg: 22 = 4, but 41/2 = 2
So, if you multiply 2 with itself 2 times you get 4 as a result. Therefore,
2 is the 2nd roots of 4.
Like powers, in roots too there are special names given to the 2 nd and
the 3rd roots.
The 2nd root is called the Square Root, and the 3rd root is called the
Cube Root.
If no number is mentioned on the root symbol, like this by default it is a square root.
Some standard square and cube roots are:
2 1 = 1
2 4 = 2
2 9 = 3
2 16 = 4
2 25 = 5
2 36 = 6
2 49 = 7
2 64 = 8
2 81 = 9
2 100 = 100
3 1 = 1
3 8 = 2
3 27 = 3
3 64 = 4
3 125 = 5
Also, for approximate calculation purposes,
2 2 = 1.4
n , then
2 3 = 1.7
If you notice carefully, this is the exact opposite of the standard
squares and cubes that we wrote earlier.
To find root (remember by default it means the square root), you just
need to break the number down.
Think of the process of finding root as escaping from a prison. Numbers
are trapped under the prison of the root symbol
125 , 3 56
To escape from this root-like prison, the numbers must form pairs.. If
its a square root, the numbers need to form pairs of 2, if its a cube
root, the numbers need to form pairs of 3, and so on.
So lets try this 125
125 can be broken into its factors as 25 x 5 or 5 x 5 x 5
=
5 x 5 x 5
<---->
1 pair of two 5s formed
## Now, since 1 pair of 2 numbers can be formed, one 5 is able to escape
the prison, and we get
=5
Lets try, 3 56
56 can be factorized as 7 x 8 or 7 x 2 x 2 x 2
= 3 7 x 2 x 2 x 2
<-------->
1 pair of three 2s formed
= 2 3 7
Rationalization:
Roots can be applied to fractional numbers as well.
For eg:
3 =
## Sometimes, its easier to simplify roots in the denominator by bringing
them up to the numerator.
For eg:
1 - 3
---------1 + 3
## can be simplified like this
Step 1: Multiply both num and den by a number with opposite sign as
the den
1 - 3
1 - 3
---------x
---------1 + 3
1 - 3
Step 2: Apply basic algebraic theorems* and simplify
(1 -
3 )2
---------(12 - 3
12 +
=
4 + 2 3
-------------2
3 )2 + 2x1x 3
---------------------------1 -3
2 ( 2 + 3
----------------- =
1 + 3 + 2
----------------------2
-2 -
-2
## * The 3 basic algebraic theorems are :
1) (a + b)2 = a2 + b2 + 2ab
2) (a b)2 = a2 + b2 2ab
3) (a2 b2) = (a-b) (a+b)
Lets try one more:
15
-------6 1
=
15
---------6 1
15 6
------------( 6 2 - 12
6 + 1
---------6 + 1
=
15 6
------------- =
61
## Notice that we havent use negative powers so far. The process of
exponent changes with negative powers. Lets take an example and
see what happens
2-3
---------
(2)
= 1/8
## Observe that 2-3 is positive.
Lets try it with a different number.
10-4
1
1
------ =
------ = 0.0001
104
10000
Again a positive number is the result.
=
## Infact no matter how bigger a negative power you put on a positive
number the result will always be positive, i.e.
Pn = +ve number, where P is any positive number and n is any real
number.
Similarly,
1
------ = 112 = 121
11-2
The key point to remember here is that when you need to change the
sign of powers move them from denominator to numerator (or vice
versa)
Negative roots go pretty much the same way as negative powers do.
However, remember like we saw before that the roots can be
negative, but the base (the number inside the root prison) HAS to be
positive if its a square root or any other even power root.
Negative Roots
Negative roots are a bit different. While 8- = 1/ (8) = 1/
But there is one major exception for ve roots!
## One thing you must remember is 8 ,
any negative number is NOT DEFINED.
or
1 , or square root of
For eg:
4 125 and
## 153 are not defined
12
The reason is very simple. The square (or even power) of any number
can never be negative. Therefore, the square (or the even root) can
never be negative as well!
However, odd roots of negative numbers are defined.
For eg:
And
1 = - 1
128 = -2
3
7
Thats because when you multiply -1 three times you get -1, and if you
multiply -2 seven times you get -128.
The common reason to the above observations can be defined like this
(-1)n = +1, if n is even
(-1)n = -1, if n is odd
This simple looking observation is mighty powerful and GMAT Loves
it!
Q) If n is any integer, then which is bigger: (-1)6n + 1 or (-1)82n ?
Ans: The only thing that matters here is whether the power is an odd
number or an even number
The 2 possible cases here are:
Case 1: n 0
Then, 6n + 1 is an odd number for any n (try it!), i.e. (-1)6n + 1 = -1
And, 82n is an even number for any n, i.e. (-1)82n = +1
Case 2: n = 0
Then, 6n + 1 = 1 which is an odd number, i.e. (-1)6n + 1 = -1
And, 82n = 0 which is an even number, i.e. (-1)82n = +1
Either way, (-1)82n > (-1)6n + 1
## Extent of powers and roots:
A very important factor to consider when dealing with powers is to
notice the extent of powers.
For eg:
2 x 145 is different than (2 x 14)5
x3 + y3 is different than (x + y)3
(3/8)2 is way different than 3/82
A lot of candidates often get confused in this. The key to crack such
cases is to observe the brackets, see what all terms they contain and
whether the power is inside or outside the bracket.
So, 2 x 145 = 2 x 14 x 14 x 14 x 14 x 14
But, (2 x 14)5 = 285
Similarly, 3 / 8
Fractional Roots:
Fractional roots or even decimal roots are a combination of powers
and roots. The numerator tells the power and the denominator tells the
root.
So, 8 can be rewritten as (82)
3 8 x 8 = 3 2 x 2 x 2 x 2 x 2 x 2
=2x2=4
<---------> <-------->
## Alternatively, you could have also done (8)2 = 22 = 4
Lets try one more example with (144)
144 can be broken into its factors as -> 24 x 32
(144) = (24 x 32) = ( 24) x (32) = (2)4 x x 33/2 = (2)3 x
27
27 = 8
Base:
Now, lets try playing with the base of an exponent. Remember the
whole thing is (Base)power
A few standard base values first.
(0)n = 0, i.e. 0 to the power anything (except 1) is 0
00 = 1
Also, 1n = 1, i.e. 1 to the power anything (including 0) is 1
And, -1n = 1, if n is even
-1n = -1, if n is odd
Which means that,
(negative number)n = +ve , if n is even, and
(negative number)n = -ve , if n is odd
The GMAT LOVES to play with these 3 standard values and form
questions around them.
For eg: if x = x2, then x must be either 0 or 1
And, if x = xx, then x must be either -1 or 1
You can break down the base too.
For eg: 562 = (7 x 8)2 = 72 x 82 = 49 x 64 = 3136
The base can be fractional too. So you can have something like this
()2 = (42/52) = 16/25
Or the base can be decimal like this
(0.5)2 = 0.25, (0.15)3 = 0.003375 or (1.3)2 = 1.69
Roots also work with decimal bases like this
0.25 = 0.5
If you have trouble finding roots in decimals, you can convert the
decimal value into fractions and solve it then
0.0009 =
9
10000
= 3/100 = 0.03
## Notice that, proper fractions or a decimals between 0 and 1 decrease
as the power increases. Normally, that is not the case.
This happens because of the decimal value. Try
Q) Which is bigger:
(i) (0.2) or (0.2)3
(ii) 2 or 23
Q) Solve:
i) (625)4/7
ii) (-72)2/5
iii) (2.88)3/2
Operations on Exponents
Operations on Exponents can happen ONLY WHEN THE BASE IS
SAME!!!
na x nb = n(a + b)
That makes sense because
23 x 22 = 8 x 4 = 32 = 25 = 2(3 + 2)
When divided, Powers get Subtracted!:
na nb = n(a - b)
That makes sense because
33 32 = 27 9 = 3 = 31 = 3(3 - 2)
Power inside a Power gets multiplied!:
(na)b = nab
That makes sense because
(52)3 = 253 = 25 x 25 x 25 = 15625 = 56 = 52x3
Make sure you dont confuse this with
Power on top of a power stays!:
ab
(n)
eg:
22
(2)
= 24 = 2 x 2 x 2 x 2 = 16
Be careful
1) You remember that these operations can be applied ONLY when
the base is Same.
2) You understand and appreciate the difference between each of
the operations illustrated above.
In our experience, GMAT LOVES posing questions on this concept. A
question exploiting this concept and its tricky nature is very likely to
appear in the test.
## Thats about performing operations on powers. However, operations on
roots are a bit tricky.
First thing ->
a b
a b
For eg:
72+24 72+ 24
7224 72 24
However, a b = a b , and
a b = a b
For eg: 125 5
125 5 = 5
= 625=25
5 x 5 = 5 x 5 = 25
125 5 = 25 = 5
125 5 = 5 5 5 = 5
Therefore, you can split up roots only in case of multiplication and
division and NOT in addition and subtraction. These techniques can be
used to solve roots that may otherwise seem difficult to solve at first.
For eg:
45 x 20 = 900 = 30
Solving Exponents:
1) You can take out the common term, provided that the base is
same.
Eg: 174 +175 = 174 (1 + 17) = 18 x 174
2) Sometimes if the base is not same, you can make it same
132 + 262 = 132 + (13x2)2 = 132 + (132 x 22) = 132 (1 + 4) = 5 x
132
Or, 83 x 442 = (23)3 x (2 x 11)2 = 29 x 22 x 112 = 211 x 112
3) The square (or even power) Eats the sign!
And so they disguise the real number beneath them.
Eg: if x2 = 9
Then x can be either +3 or -3
However, if x3 = 27
Then x has to be +3
You can use this in an equation too, like this
x2 5 = 21
x2 = 26
x = + 26 or -
26
Make sure you always know that x2 gives 2 solutions and NOT 1
solution
This concept would DEFINITELY be tested in the GMAT
The only exception is x 2 = 0, this has only 1 solution which is x =
0
However, odd powers show the real sign of the base!
Q) Solve for x:
i) x4 = 82
ii) x3 = 120 112
Q) Solve: 237 x 325 42 - 466
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Fractions: sub-module 1 of 4 of math tutorials
Fractions are numbers that are not whole; they represent parts of a whole number. You have likely encountered several examples of fractions in your day-to-day life. For example, a recipe may require 3 quarters of a cup of flour. If a pizza is divided into eight equal slices, this is a fraction: each slice is one-eighth of the whole pizza. This module will help you understand fractions and review key concepts so that you will feel comfortable working with fractions.
Top Tips
• Don’t be afraid to work with fractions! You are going to encounter fractions in math and in everyday life, so get comfortable working with them.
• To add and subtract fractions, you need a common denominator first. Once you have that, you can add or subtract the numerators and place them over the common denominator.
• To multiply fractions, multiply the numerators and denominators individually. Multiply the numerators of the fractions together and then multiply the denominators together to get the result.
• To divide fractions, multiply the first fraction by the reciprocal of the second fraction. To get the reciprocal, flip the second fraction so that the numerator becomes the denominator and vice versa. Multiply that by the first fraction.
• Always convert mixed numbers to improper fractions before multiplying or dividing fractions. Instead of a whole number with a fraction (e.g. 5 ⅔), you want a fraction where the numerator is bigger than the denominator (e.g. 17/3). You can also do this for addition and subtraction if you prefer this method.
• Always remember to simplify your fractions to the lowest terms in your final answer. If the numerator and the denominator have a common factor other than 1 (i.e. they can be divided by the same number), simplify them.
• Any whole number can be written as a fraction by placing it over 1. If you have a whole number and a fraction to work with, this will allow you to work with two fractions instead.
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# What is 51/296 as a decimal?
## Solution and how to convert 51 / 296 into a decimal
51 / 296 = 0.172
51/296 or 0.172 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Both are used to handle numbers less than one or between whole numbers, known as integers. The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. So let’s dive into how and why you can convert 51/296 into a decimal.
## 51/296 is 51 divided by 296
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. Fractions have two parts: Numerators and Denominators. This creates an equation. To solve the equation, we must divide the numerator (51) by the denominator (296). Here's how our equation is set up:
### Numerator: 51
• Numerators are the portion of total parts, showed at the top of the fraction. 51 is one of the largest two-digit numbers you'll have to convert. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Let's take a look below the vinculum at 296.
### Denominator: 296
• Denominators represent the total parts, located at the bottom of the fraction. 296 is one of the largest two-digit numbers to deal with. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Have no fear, large two-digit denominators are all bark no bite. So grab a pen and pencil. Let's convert 51/296 by hand.
## Converting 51/296 to 0.172
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 296 \enclose{longdiv}{ 51 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 296 \enclose{longdiv}{ 51.0 }$$
Uh oh. 296 cannot be divided into 51. So that means we must add a decimal point and extend our equation with a zero. Now 296 will be able to divide into 510.
### Step 3: Solve for how many whole groups you can divide 296 into 510
$$\require{enclose} 00.1 \\ 296 \enclose{longdiv}{ 51.0 }$$
Now that we've extended the equation, we can divide 296 into 510 and return our first potential solution! Multiply by the left of our equation (296) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.1 \\ 296 \enclose{longdiv}{ 51.0 } \\ \underline{ 296 \phantom{00} } \\ 214 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If there is a remainder, extend 296 again and pull down the zero
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. This is also true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Here are examples of when we should use each.
### When you should convert 51/296 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.172 to 51/296 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 51/296 = 0.172 what would it be as a percentage?
• What is 1 + 51/296 in decimal form?
• What is 1 - 51/296 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.172 + 1/2?
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# Choosing The Right Statement About The Graph Of X
When it comes to understanding graphs, it is important to be able to read and interpret the data accurately. Knowing what is true and what is false about a graph can help you make better decisions about how to act in your life. In this article, we will discuss how to choose the correct statement about the graph of x.
## Interpreting the Graph
The graph of x is a visual representation of the relationship between two variables. The graph is typically composed of two axes, the x-axis and the y-axis. The x-axis is the independent variable and the y-axis is the dependent variable. The graph may also contain labels, lines, and points. When looking at the graph, you should be able to identify which variables are being plotted and the relationship between them.
### Analyzing the Data
In order to choose the right statement about the graph of x, you must first analyze the data. To do this, you must look at the labels, lines, and points on the graph and determine what relationship they represent. You may also need to look at the range of the x-axis and y-axis to determine the relationship between the two variables. Once you have analyzed the data, you should be able to determine which statement about the graph of x is true and which statement is false.
### Making the Right Choice
Once you have determined which statements are true and which statements are false, you must then make the right choice. You must choose the statement that accurately describes the relationship between the two variables on the graph. In some cases, you may need to make an educated guess based on the information you have gathered from your analysis. Remember, the goal is to choose the statement that is most accurate and that best describes the relationship between the two variables on the graph.
## Conclusion
Choosing the right statement about the graph of x is an important skill to have when it comes to understanding graphs. By analyzing the data, you can determine which statement is true and which statement is false. This will help you make better decisions in your life and help you understand graphs more clearly. So the next time you look at a graph of x, remember to analyze the data and choose the statement that is most accurate and best describes the relationship between the two variables.
## A Restaurant Chain's Owners Are Trying To Decide
Pho owners consider sale of restaurant chain from www.cityam.com The restaurant industry is booming, with …
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# Breaking down Equations one step - Yr9AMM
```Breaking down Equations – 0ne step
When there is an unknown letter, we would want to solve it or basically get a
value for this letter. Another word used to call this letter is a pronumeral.
Hence, you may see such statements in textbooks which says, ‘solve the
unknown’ or ‘solve for the pronumeral’.
The letter or pronumeral would indicate that we would like to find the value of
something that we do not know.
An example for this maybe, 2 dvds cost \$21.00, so how much does one dvd
cost?
This can be represented as 2y= 21.50, y would represent the cost of a dvd, so 1
dvd would cost 21.50 divided 2, which would be \$10.50.
In solving equations, remember, the opposite operations
Opposite operation
+
x
÷
minus, subtract, take away
times, multiply,product
divide, over, quotient
-, minus
+, plus
÷, divide
x, times
In solving equations, what we are doing when using opposite operations is
simply to balance the equation so that we do not change the equation, what we
do to one side of the equal sign, we need to do the same to the other.
One step Equation
2y = 10, meaning 2 x y
Opposite operation
÷ or written as __
2 x y = 10
2
2
y=5
t + 7 = 12
t + 7 -7 = 12 -7
t= 5
opposite of +7 is -7
-
r =4
opposite of divide 3 is X3, so
3
3 X r = 4 X 3 , so r = 12
3
Solve for the pronumeral:
Opposite operation:
3t = 15
t = __________________
2g = 4
g = ___________________
r=6
2
r = ___________________
j – 5 = 11
j = ___________________
y = 10
3
y = ___________________
g+5=9
g = ___________________
```
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# How to Graph Inequalities
Co-authored by wikiHow Staff
Updated: March 29, 2019
You can graph a linear or quadratic inequality similarly to how you would graph an equation. The difference is that, since an inequality shows a set of values greater than or less than, your graph will show more than just a dot on a number line or a line on a coordinate plane. By using algebra and evaluating the inequality sign, you can determine which values are included in the solution of an inequality.
### Method 1 of 3: Graphing a Linear Inequality on a Number Line
1. 1
Solve for the variable. To solve the inequality isolate the variable using the same algebraic methods you would use to solve an equation.[1] Remember that when you multiply or divide by a negative number, you need to flip the inequality sign.
• For example, if you are solving the inequality ${\displaystyle 3y+9>12}$, isolate the variable by subtracting 9 from each side of the inequality, then dividing by 3:
${\displaystyle 3y+9>12}$
${\displaystyle 3y+9-9>12-9}$
${\displaystyle 3y>3}$
${\displaystyle {\frac {3y}{3}}>{\frac {3}{3}}}$
${\displaystyle y>1}$
• Your inequality should only have one variable. If your inequality has two variables, it is more appropriate to graph it on a coordinate plane using another method.
2. 2
Draw a number line. Include the relative value on your number line (the value you found the variable to be less than, greater than, or equal to). Make the number line as long or short as required.
• For example, if you found that ${\displaystyle y>1}$, make sure to include a point for 1 on the number line.
3. 3
Draw a circle indicating the relative value. If the value is less than (${\displaystyle <}$) or greater than (${\displaystyle >}$) this number, the circle should be open, since the solution does not include the value. If the value is less than or equal to (${\displaystyle \leq }$), or greater than or equal to (${\displaystyle \geq }$), the circle should be filled in, since the solution includes the value.[2]
• For example, if ${\displaystyle y>1}$, you would draw a circle at 1 on the number line. You would not fill in the circle, since 1 is not included in the solution.
4. 4
Draw an arrow indicating the included values. If the variable is greater than the relative value, your arrow should point to the right, since the solution includes values greater than that number. If the variable is less than the relative value, your arrow should point to the left, since the solution includes values less than that number.[3]
• For example, for the solution ${\displaystyle y>1}$, you would draw an arrow pointing to the right, since the solution includes values greater than 1.
### Method 2 of 3: Graphing a Linear Inequality on a Coordinate Plane
1. 1
Solve for . You want to find the equation of the line, so to do this you need to isolate the ${\displaystyle y}$ variable on the left side of the equation using algebra.[4] The right side of the equation should have the ${\displaystyle x}$ variable, and likely, a constant.
• For example, for the inequality ${\displaystyle 3y+9>9x}$, you would isolate the y variable by subtracting 9 from both sides, then dividing by 3:
${\displaystyle 3y+9>9x}$
${\displaystyle 3y+9-9>9x-9}$
${\displaystyle 3y>9x-9}$
${\displaystyle {\frac {3y}{3}}>{\frac {9x-9}{3}}}$
${\displaystyle y>3x-3}$
2. 2
Graph the line on a coordinate plane. To do this, turn the inequality into an equation, and graph as you would any equation of a line.[5] Plot the y-intercept, then use the slope to graph other points on the line.
• For example, if the inequality is ${\displaystyle y>3x-3}$, you would graph the line ${\displaystyle y=3x-3}$. The y-intercept (the point where the line crosses the y axis) is -3, and the slope is 3, or ${\displaystyle {\frac {3}{1}}}$. So, you would draw a point at ${\displaystyle (0,-3)}$. The point above the y-intercept is ${\displaystyle (1,0)}$. The point below the y-intercept is ${\displaystyle (-1,-6)}$.
3. 3
Draw the line. If the inequality is less than (${\displaystyle <}$) or greater than (${\displaystyle >}$), the line should be dashed, since the solution does not include values equal to the line. If the value is less than or equal to (${\displaystyle \leq }$), or greater than or equal to (${\displaystyle \geq }$), the line should be solid, since the solution includes values equal to the line.[6]
• For example, since the inequality is ${\displaystyle y>3x-3}$, the line should be dashed, since the values do not include points on the line.
4. 4
Shade in the appropriate area. If the inequality shows ${\displaystyle y>mx+b}$ you should shade in the area above the line. If the inequality shows ${\displaystyle y, you should shade the area below the line.[7]
• For example, for the inequality ${\displaystyle y>3x-3}$ you would shade above the line.
### Method 3 of 3: Graphing a Quadratic Inequality on a Coordinate Plane
1. 1
Determine whether you have a quadratic inequality. A quadratic inequality takes the form of ${\displaystyle ax^{2}+bx+c}$.[8] Sometimes there may not be an ${\displaystyle x}$ term or a constant, but there should always be an ${\displaystyle x^{2}}$ term on one side of the inequality, and an isolated ${\displaystyle y}$ variable on the other side.
• For example, you might need to graph the inequality ${\displaystyle y.
2. 2
Graph the line on a coordinate plane. To do this, turn the inequality into an equation, and graph the line as you normally would. Since you have a quadratic equation, the line will be a parabola.[9]
• For example, for the inequality ${\displaystyle y, you would graph the line ${\displaystyle y=x^{2}-10x+16}$. The vertex is at point ${\displaystyle (5,-9)}$, and the parabola crosses the x-axis at the points ${\displaystyle (2,0)}$ and ${\displaystyle (8,0)}$.
3. 3
Draw the parabola. Draw the parabola with a dashed line if the inequality is less than (${\displaystyle <}$) or greater than (${\displaystyle >}$). If the value is less than or equal to (${\displaystyle \leq }$), or greater than or equal to (${\displaystyle \geq }$), you should draw the parabola with a solid line, since the solution includes values equal to the line.
• For example, for the inequality ${\displaystyle y, you would draw the parabola with a dashed line.
4. 4
Find some test points. In order to determine which area to shade, you need to pick points from inside the parabola, and from outside the parabola.
• For example, the graph of the inequality ${\displaystyle y shows that the point ${\displaystyle (0,0)}$ is outside the parabola. This would be a good point to use to test the solution.
5. 5
Shade the appropriate area. To determine which area to shade, plug the values of ${\displaystyle x}$ and ${\displaystyle y}$ from your test points into the original inequality. Whichever point produces a true inequality indicates which area of the graph should be shaded in.[10]
• For example, plugging the values of ${\displaystyle x}$ and ${\displaystyle y}$ of the point ${\displaystyle (0,0)}$ into the original inequality, you get:
${\displaystyle y
${\displaystyle 0<0^{2}-0x+16}$
${\displaystyle 0<16}$
Since this is true, you would shade the area of the graph where the point ${\displaystyle (0,0)}$ is found. In this case, this is outside of the parabola, not inside of it.
## Community Q&A
Search
• Question
How do I graph the inequality y is less than or equal to X + 6?
Since you do not have an x-squared term, you are working with a linear inequality. Turn the inequality into the equation y = x + 6. Plot the line using this equation. Since it is in the form of y = mx + b, you can tell that the y-intercept is at the point (0, 6), and that the slope is 1, or 1/1. Since it is less than OR equal to, the line should be solid. Since y is less than the line, you would shade the area below the line.
• Question
How do I graph x - y is greater than 5 as a linear inequality?
First, substitute "greater than" with "equals," which gives you x - y = 5. Then rearrange the equation to get y = x - 5. Input some values for x. If you do 1, 5, 10, you get y-values -4, 0, 5. Plot these 3 points (1,-4), (5,0) and (10,5). Then pick a point on your graph (not on the line) and put this into your starting equation. Take the point (4,2) for example. You have 4 - 2 > 5, and 2 > 5 is false. This means that the graph area on the same side of the line as point (4,2) is not in the region x - y > 5. Therefore, any point on the other side of the line should be in this region. To check, use point (8,2). 8 - 2 > 5, 6 > 5, which is true. You can then draw arrows off your line indicating the region represented by x - y > 5.
• Question
How do I graph y=5x+1?
The Y-axis goes up and down; the X-axis goes left and right. Plug in x values: In the equation above, if x is 0, what is y? Plugging 0 in for x and doing the calculation gives a 1 for y. The first point, then, is (0,1). (When x is 0, y is 1.) Next, plug in 1 for x. (5 times 1) + 1 = 6. So the next point is (1,6). (When x is 1, y is 6.) Now, plug in 2 for x. (5 times 2) + 1 = 11. The next point is (2,11). Plot the points on the graph and draw a line through all of them.
• Question
How do I graph inequality equation -4x + 2 is less than or equal to 10?
Inequalities should be graphed on a number line. Solve the equation algebraically, and plot it on the line with a dot on the answer. The dot should be open, with an arrow pointing toward the left.
• Question
How do I graph a no solution?
Donagan
A graph of "no solution" would typically be two graph lines which never intersect.
• Question
How do I graph y is less than a number with no x, and how do I graph x is greater than a number with no y?
Donagan
Let's use two examples: The graph of y < 10 is all of the area below the horizontal line y = 10. The graph of x > 10 is all of the area to the right of the vertical line x = 10.
200 characters left
## Tips
• Always simplify the inequalities before graphing them.
Thanks!
• If you get really stuck, you can input the inequality into a graphing calculator and try to work backwards.
Thanks!
## Video.By using this service, some information may be shared with YouTube.
Co-Authored By:
wikiHow Staff Editor
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Together, they cited information from 10 references.
|
# Cylindrical to Cartesian coordinates – Formulas and Examples
Cylindrical coordinates have the form (r, θ, z), where r is the distance in the xy plane, θ is the angle formed with respect to the x-axis, and z is the vertical component in the z-axis. Similar to polar coordinates, we can relate cylindrical coordinates to Cartesian coordinates by using a right triangle and trigonometry. We use cosine to find the x component and sine to find the y component. The component in z stays the same.
Here, we will look at the formulas that we can apply to transform from cylindrical to Cartesian coordinates. Then, we will use these formulas to solve some practice exercises.
##### TRIGONOMETRY
Relevant for
Learning to transform from cylindrical to Cartesian coordinates.
See examples
##### TRIGONOMETRY
Relevant for
Learning to transform from cylindrical to Cartesian coordinates.
See examples
## How to transform from cylindrical coordinates to Cartesian coordinates?
Three-dimensional Cartesian coordinates are represented in the form . An alternate three-dimensional coordinate system is the cylindrical coordinate system. The cylindrical coordinate system is a three-dimensional extension of the polar coordinate system.
The polar coordinates are extended in the third dimension similar to how we extended the Cartesian coordinates, that is, we simply add z as the third dimension.
Then, the cylindrical coordinates have the form , where, r is the distance from the origin to the position of the point in the xy plane, θ is the angle with respect to the x-axis, and z is the coordinate in the z-axis.
The transformations for x and y are the same as those used in polar coordinates. To find the x component, we use the cosine function, and to find the y component, we use the sine function.
Also, the z component of the cylindrical coordinates is equal to the z component of the Cartesian coordinates.
## Cylindrical to Cartesian coordinates – Examples with answers
The formulas for the transformation of cylindrical coordinates to Cartesian coordinates are used to solve the following examples. Each example has its respective solution, but it is recommended that you try to solve the problems yourself before looking at the answer.
### EXAMPLE 1
We have the point (3, 30°, 6) in cylindrical coordinates. What is its equivalence in Cartesian coordinates?
We start with the values . Using these values and the formulas seen above, we have to find the values of x and y. Therefore, the value of x is:
The value of y is:
We know that the z component remains the same, so the coordinates of the point are (2.6, 1.5, 6).
### EXAMPLE 2
What are the Cartesian coordinates of the point (5, 45°, -4)?
We use the values to find the different components of the Cartesian coordinates. We start with the value of x:
Now, we find the value of y:
The z component is the same, so the Cartesian coordinates of the point are (3.54, 3.54, -4).
### EXAMPLE 3
We have a point with cylindrical coordinates (6, 120°, 7). What are the Cartesian coordinates of this point?
We find the Cartesian coordinates using the values along with the formulas seen above. Therefore, the value of x is:
The value of y is:
The z component is the same, so the coordinates of the point are (-3, 5.2, 7).
### EXAMPLE 4
We have the point (12, 90°, 8) in cylindrical coordinates. What is its equivalence in Cartesian coordinates?
We have the values . We find the value of x as follows:
We find the value of y as follows:
We know that the z component remains the same, so the coordinates of the point are (0, 12, 8).
## Cylindrical to Cartesian coordinates – Practice problems
Use the formulas seen above to solve the following problems and transform the cylindrical coordinates to Cartesian coordinates. Select an answer and check it to see if you got the correct answer.
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×
Back to all chapters
# Discrete Probability
How often will a die come up "4"? How likely is it to rain tomorrow? Probability is one of the most powerful frameworks for modeling the world around us.
# Probability - By Outcomes
If two six-sided dice are rolled, the probability that they both show the same number can be expressed as $$\frac{a}{b}$$ where $$a$$ and $$b$$ are coprime positive integers. What is the value of $$a + b$$?
A fair 6-sided die is rolled twice. The probability that the second roll is strictly less than the first roll can be written as $$\frac{a}{b}$$, where $$a$$ and $$b$$ are positive, coprime integers. What is the value of $$a+b$$?
Details and assumptions
The roll of a dice refers to the value on the top face of the dice.
Two players each flip a fair coin. The probability that they get the same result can be expressed as $$\frac{a}{b}$$ where $$a$$ and $$b$$ are coprime positive integers. What is the value of $$a + b$$?
Lily has two bags containing balls. The first bag contains $$10$$ balls, with each ball labelled by a distinct number from $$1$$ through $$10$$. The second bag contains one ball labelled $$1$$, two balls labelled $$2$$, etc, up to ten balls labelled with $$10$$. Suppose Lily draws one ball from each bag uniformly at random and let $$\frac{a}{b}$$ be the probability that the two balls have the same value, where $$a$$ and $$b$$ are coprime positive integers. What is the value of $$a+b$$?
If a 20-sided fair die with sides distinctly numbered 1 through 20 is rolled, the probability that the answer is a perfect square can be expressed as $$\frac{a}{b}$$ where $$a$$ and $$b$$ are coprime positive integers. What is the value of $$a + b$$?
×
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# How do you factor 6x^3+34x^2-12x?
Oct 9, 2015
$2 x \left(3 x - 1\right) \left(1 x + 6\right)$
#### Explanation:
Given
$\textcolor{w h i t e}{\text{XXX}} 6 {x}^{3} + 34 {x}^{2} - 12 x$
First extract the obvious common factor of $\left(2 x\right)$ from each term:
$\textcolor{w h i t e}{\text{XXX}} \left(2 x\right) \left(3 {x}^{2} + 17 x - 6\right)$
In the hopes of finding integer coefficient factors of $\left(3 {x}^{2} + 17 x - 6\right)$
we consider the integer factors of $3$ and of $\left(- 6\right)$
looking for a combination that will give us a sum of products $= 17$
There are only a few possibilities. The diagram below might help understand the process:
The "working" combination is $\left(3 x - 1\right) \left(1 x + 6\right)$
Which gives us the complete factorization:
$\textcolor{w h i t e}{\text{XXX}} 6 {x}^{3} + 34 {x}^{2} - 12 x$
$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} = \left(2 x\right) \left(3 x - 1\right) \left(1 x + 6\right)$
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4 months ago
# CLASS_11_MATHS_SOLUTIONS_NCERT
## Class XI Chapter
Class XI Chapter 11 – Conic Section Maths ______________________________________________________________________________ Miscellaneous Exercise Question 1: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. Solution 1: The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x-axis. This can be diagrammatically represented as The equation of the parabola is of the form y 2 = 4ax (as it is opening to the right). Since the parabola passes through point A (10, 5), 10 2 = 4a(5) ⇒ 100 = 20a ⇒ a = 100 20 = 5 Therefore, the focus of the parabola is (a, 0) = (5, 0), which is the mid-point of the diameter. Hence, the focus of the reflector is at the mid-point of the diameter. Question 2: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. Solution 2: The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x-axis. This can be diagrammatically represented as Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
Class XI Chapter 11 – Conic Section Maths ______________________________________________________________________________ The equation of the parabola is of the form y 2 = 4ax (as it is opening to the right). Since the parabola passes through point A (5, 10), 10 2 = 4a(5) ⇒ 100 = 20a ⇒ a = 100 20 = 5 Therefore, the focus of the parabola is (a, 0) = (5, 0), which is the mid-point of the diameter. Hence, the focus of the reflector is at the mid-point of the diameter. Question 3: An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? Solution 3: The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the positive y-axis. This can be diagrammatically represented as The equation of the parabola is of the form x 2 = 4ay (as it is opening upwards). Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
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+0
# )2PI, 0) solve it
0
87
1
sin2x-sqrt(3)*cosx=o
Guest Apr 18, 2017
Sort:
#1
0
Solve for x:
sin(2 x) - sqrt(3) cos(x) = 0
Expand trigonometric functions:
2 cos(x) sin(x) - sqrt(3) cos(x) = 0
Collecting terms, 2 cos(x) sin(x) - sqrt(3) cos(x) = (2 sin(x) - sqrt(3)) cos(x):
cos(x) (2 sin(x) - sqrt(3)) = 0
Split into two equations:
cos(x) = 0 or 2 sin(x) - sqrt(3) = 0
Take the inverse cosine of both sides:
x = π/2 + π n_1 for n_1 element Z
or 2 sin(x) - sqrt(3) = 0
x = π/2 + π n_1 for n_1 element Z
or 2 sin(x) = sqrt(3)
Divide both sides by 2:
x = π/2 + π n_1 for n_1 element Z
or sin(x) = sqrt(3)/2
Take the inverse sine of both sides:
x = π/2 + π n_1 for n_1 element Z
or x = π - sin^(-1)(sqrt(3)/2) + 2 π n_2 for n_2 element Z or x = sin^(-1)(sqrt(3)/2) + 2 π n_3 for n_3 element Z
sin(2 x) - sqrt(3) cos(x) ⇒ sin(2 (π/2 + π n_1)) - sqrt(3) cos(π/2 + π n_1) = 0:
So this solution is correct
sin(2 x) - sqrt(3) cos(x) ⇒ sin(2 (329 π - sin^(-1)(sqrt(3)/2))) - sqrt(3) cos(329 π - sin^(-1)(sqrt(3)/2)) = 0:
So this solution is incorrect
sin(2 x) - sqrt(3) cos(x) ⇒ sin(2 (184 π + sin^(-1)(sqrt(3)/2))) - sqrt(3) cos(184 π + sin^(-1)(sqrt(3)/2)) = 0:
So this solution is incorrect
The solution is:
Answer: | x = π/2 + π n_1 for n_1 element Z or x = π - sin^(-1)(sqrt(3)/2) + 2 π n_2 for n_2 element Z or x = sin^(-1)(sqrt(3)/2) + 2 π n_3 for n_3 element Z
Guest Apr 18, 2017
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Use adaptive quiz-based learning to study this topic faster and more effectively.
# Parametric and implicit functions
## Differentiation of parametric functions
The tangent at the point $(x_0,y_0) = \big(x(t_0),y(t_0)\big)$ to the parametric curve $\big(x(t),y(t)\big)$ is parallel to the vector $$\big(x'(t_0),y'(t_0)\big).$$
If $x'(t_0)\neq 0$, the parametric derivative or slope at $t_0$ is $$\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx}= \frac{y'(t_0)}{x'(t_0)}.$$ It is the slope of the tangent to the curve at the point $\big(x(t_0),y(t_0)\big)$. If $x'(t_0) = 0$ and $y'(t_0) \ne 0$, the tangent is vertical.
For instance, the parametric derivative for the circle at $t_0 = \pi/4$ is $$\frac{dy}{dx} = \frac{\sin'(t_0)}{\cos'(t_0)} = \frac{\cos(t_0)}{-\sin(t_0)}= - \cot(t_0) = -1.$$
Parametric equation of a circle
## Differentiation of implicit functions
We consider a function defined implicitly by the relation $$F\big(y(x),x\big) = 0$$ around a point $x_0$. This means that the value of the function at $x_0$ is $y_0$ with $F(y_0,x_0) = 0$.
The derivative of an implicit function is the slope of the tangent to the curve at $x_0$. It is obtained by differentiating the implicit equation directly.
For instance, the slope of the tangent to the circle $y^2+x^2=1$ is $$\displaystyle y'(x) = -x/y.$$ We get this by using the chain rule to differentiate the equation as a function of $x$: $$2yy' + 2x=0.$$
Functions of $x$ are differentiated in the usual way. $y$ differentiates to $y'$.
Derivative of the implicit representation of a circle $y^2+x^2=1$
## Implicit functions and partial derivative
The partial derivative is useful for the differentiation of implicit functions.
The partial derivative in $y$ of the two-variable function $F(y,x)$ is the derivative of the one-variable function $y\mapsto F(y,x)$. Here $x$ is frozen and seen as a parameter. The partial derivative is denoted $\displaystyle \frac{\partial F}{\partial y}$ or $F'_y$.
We define the partial derivative $F'_x$ of $F$ with respect to $x$ similarly.
The derivative of an implicit function $$F\big(y(x),x\big) = 0$$ is the slope of the tangent to the curve at $x_0$. It is given by the formula $$\frac{dy}{dx}(x_0) = \frac{-F'_x\big(y_0,x_0\big)}{F'_y\big(y_0,x_0\big)},\qquad F(y_0,x_0) = 0.$$
By the chain rule, we get $$0 = \frac{d}{dx} F\big(y(x),x\big) = y'(x) F'_y\big(y(x),x\big) + F'_x\big(y(x),x\big).$$
For instance, the slope of the tangent to the circle $y^2+x^2=1$ is $\displaystyle y'(x) = -x/y.$ We get this by differentiating the equation or computing the partial derivatives $$2yy' + 2x=0,\qquad F'_x = 2x,\quad F'_y = 2y.$$
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Definition 1:
Event X and event Y are independent
if P(X ? Y) = P(X) • P(Y). This implies that knowing the
outcome of X does not affect the probability of Y. This is also shown in
theorem 5.1 from source 1 where:
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If P(X) > 0 and P(X) > 0 and P(X) and P(Y) are independent,
then:
·
P(X|Y) = = = P(X)
Therefore: P(X|Y) = P(X)
·
P(Y|X) = = = P(Y)
Therefore: P(Y|X) = P(Y)
Venn Diagram illustrating two independent events X and Y:
Proof:
Show: If X and Y are independent
then P(X|Y) = P(X).
Let X and Y be events which are independent.
Using the definition of independence, we can see that: P(X ? Y) = P(X) • P(Y).
We
can use the definition of Conditional
Probability which demonstrate that: P(X|Y) = where P(Y)?0
According
to source 2: “Conditional
probability measures how an event effects another event happening, in the case
above, the conditional probability of when event X occurs given the information
we gathered from event Y that has occurred before.”
Since P(Y) ?
0, we can rearrange the equation P(X ? Y) = P(X) • P(Y) to: P(X) = = P(X|Y)
=> P(X) = P(X|Y)
?
Proof:
Show: P(X) = P(X|Y) => P(Y) = P(Y|X)
Suppose P(X)
= P(X|Y) Then by using definition of Conditional Probability:
P(X) =
Now if P(X) ? 0 and P(X ? Y) = P(Y ? X) this
means that:
= P(Y)
Hence P(Y) = P(Y|X)
?
Example 1:
Suppose that
event X and event Y are mutually exclusive events (disjointed), where P(X) ? 0
and P(Y) ? 0.
To work P(Y|X), we use
formula for conditional probability:
P(Y|X) =
In this case there is no
intersection as events X and Y are mutually exclusive therefore: P(X ? Y) = 0
Hence P(Y|X) = = = 0
Looking at this example we
can see that X and Y are dependent events as P(Y|X) ? P(Y). This demonstrates
that mutual exclusion doesn’t necessarily mean events are independent.
Example 2:
Let P(X) = 0.6, P(Y) =
0.2, P(X U Y) = 0.68
To work out P(X ? Y) we
use the Inclusion Exclusion Principle
which states that:
P(X ? Y) = P(X) + P(Y) – P(X U Y)
Subbing in the information
provided above: P(X ? Y) = 0.6 + 0.2 – 0.68 = 0.12
To show events X and Y are
independent they must satisfy P(X ? Y) = P(X) • P(Y), here P(X ? Y) = 0.6 • 0.2 = 0.12
Since both parts give the
same answer this shows that events X and Y are independent.
Mutual Independence
Suppose we have 3 events
X, Y and Z and they are pairwise independent meaning:
·
P(X ? Y) = P(X) • P(Y) Therefore X and Y are independent.
·
P(X ? Z) = P(X) • P(Z) Therefore X and Z are independent.
·
P(Y ? Z) = P(Y) • P(Z) Therefore Y and Z are independent.
For events X, Y and Z to
be mutually independent i.e. P(X ? Y ? Z) = P(X) • P(Y) • P(Z) they must follow a strict rule where; “each
event is independent of each intersection of other events and if they are not
the events become mutually dependent.” Source 7
Definition
2:
Events E1, E2,
…. , En are mutually independent if for 2 ?
p ? n and 1 ? x1 < x2 <…..< xp ? n there is: P( Ex1 ? Ex2 ? …… ? Exp ) = P(Ex1) • P(Ex2) • P(Ex3) • …….. • P(Exp) Example 3: Suppose we toss a fair coin twice which produces 3 events: Event X: first outcome is tails (T). Event Y: second outcome is tails (T). Event Z: both outcomes are the same (TT or HH). Sample space: { TH, HT, TT, HH } Therefore: P(X) = 1/2 P(Y) = 1/2 P(Z) = 1/2 So: P(X ? Y) = P(X) • P(Y) = 1/4 Hence events X and Y are independent. P(X ? Z) = P(X) • P(Z) = 1/4 Hence events X and Z are independent. P(Y ? Z) = P(Y) • P(Z) = 1/4 Hence events Y and Z are independent. However: P(X ? Y ? Z) = P(X) • P(Y) • P(Z) = 1/8 ? 1/4 Therefore events X, Y and Z are not mutually independent. Example 3 is based on one from source 9.
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# PART I. THE REAL NUMBERS
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1 PART I. THE REAL NUMBERS This material assumes that you are already familiar with the real number system and the representation of the real numbers as points on the real line. I.1. THE NATURAL NUMBERS AND INDUCTION Let N denote the set of natural numbers (positive integers). Axiom: If S is a nonempty subset of N, then S has a least element. That is, there is an element m S such that m n for all n S. Note: A set which has the property that each non-empty subset has a least element is said to be well-ordered. Thus, the axiom tells us that the natural numbers are well-ordered. Mathematical Induction. Let S be a subset of N. If S has the following properties: 1. 1 S, and. k S implies k +1 S, then S = N. Proof: Suppose S N. Let T = N S. Then T. Let m be the least element in T. Then m 1 / T. Therefore, m 1 S which implies that (m 1) + 1 = m S, a contradiction. Corollary: Let S be a subset of N such that 1. m S.. If k m S, then k +1 S. Then, S = {n N : n m}. Example Prove that n 1 = n 1 for all n N. SOLUTION Let S be the set of integers for which the statement is true. Since 0 =1= 1 1, 1 S. Assume that the positive integer k S. Then k 1 + k = ( k 1) + k = k 1+ k = k 1= k+1 1. Thus, k +1 S. 1
2 We have shown that 1 S and that k S implies k +1 S. It follows that S contains all the positive integers. Exercises Prove that n = n(n +1) for all n N.. Prove that n n(n + 1)(n +1) = 6 3. Let r be a real number r 1. Prove that for all n N 4. Prove that 1 + n 3 n for all n N. 5. Prove that for all n N. 1+r + r + r r n = 1 rn+1 1 r n n for all n N. 6. Prove that (1 1 )(1 13 ) (1 1n ) = n +1 for all n. n 7. True or False: If S is a non-empty subset of N, then there exists an element m S such that m k for all k S. I.. ORDERED FIELDS Let R denote the set of real numbers. The set R, together with the operations of addition (+) and multiplication ( ), satisfies the following axioms: Addition: A1. For all x, y R, x+ y R (addition is a closed operation). A. For all x, y R, x+ y = y + x (addition is commutative) A3. For all x, y, z R, x+(y + z) =(x + y)+z (addition is associative). A4. There is a unique number 0 such that x+0 = 0+x for all x R. (0 is the additive identity.) A5. For each x R, there is a unique number x R such that x +( x) =0. ( x is the additive inverse of x.) Multiplication: M1. For all x, y R, x y R (multiplication is a closed operation).
3 M. For all x, y R, x y = y x (multiplication is commutative) M3. For all x, y, z R, x (y z) =(x y) (multiplication is associative). M4. There is a unique number 1 such that x 1=1 x for all x R. (1 is the multiplicative identity.) M5. For each x R, x 0, there is a unique number 1/x = x 1 R such that x (1/x) =1. (1/x is the multiplicative inverse of x.) Distributive Law: D. For all x, y, z R, x (y + z) =x y + x z. A non-empty set S together with two operations, addition and multiplication which satisfies A1-A5, M1-M5, and D is called a field. The set of real numbers with ordinary addition and multiplication is an example of a field. The set of rational numbers Q, together with ordinary addition and multiplication, is also a field, a sub-field of R. The set of complex numbers C is another example of a field. Order: There is a subset P of R that has the following properties: a If x, y P, then x + y P. b If x, y P, then x y P. c For each x R exactly one of the following holds: x P, x =0, x P. The set P is the set of positive numbers. Let x, y R. Then x<y (read x is less than y ) if y x P. x<y is equivalent to y>x (read y is greater than x ). P = {x R : x>0}. x y means either x<y or x = y; y x means either y>x or y = x. The relation < has the following properties: O1. For all x, y R, exactly one of the following holds: x < y, x = y, x > y. (Trichotomy Law) O. For all x, y, z R, if x < y and y < z, then x < z. O3. For all x, y, z R, if x<y, then x + z<y+ z. O4. For all x, y, z R, if x<y and z>0, then x z<y z. {R, +,, <} is an ordered field. Any mathematical system {S, +,, <} satisfying these 15 axioms is an ordered field. In particular, the set of rational numbers Q, together with ordinary addition, multiplication and less than, is an ordered field, a subfield of R. 3
4 THEOREM 1. Let x, y R. If x y + ɛ for every positive number ɛ, then x y. Proof: Suppose that x>y and choose ɛ = x y x<y+ ɛ = y + x y a contradiction. Therefore, x y.. Then = x + y < x + x = x, Definition 1. Let x R. The absolute value of x, denoted x, is given by { x, if x 0, x = x, if x<0. The properties of absolute value are: for any x, y R. (1) x 0, () xy = x y, (3) x + y x + y. Exercises True False. Justify your answer by citing a theorem, giving a proof, or giving a counterexample. (a) If x, y, z R and x < y, then xz < yz. (b) If x, y R and x < y + ɛ for every positive number ɛ, then x < y. (c) If x, y R, then x y x + y. (d) If x, y R, then x y x y.. Prove: If x y <ɛ for every ɛ>0, then x = y. 3. Suppose that x 1,x,..., x n are real numbers. Prove that x 1 + x + + x n x 1 + x + + x n. I.3. THE COMPLETENESS AXIOM R and Q are each ordered fields. What distinguishes R from Q is the completeness axiom. As you know, Q is a proper subset of R; i.e., there are real numbers which are not rational numbers. Such numbers are called irrational numbers. THEOREM. is not a rational number. In general, if p is a prime number, then p is not a rational number. 4
5 Proof: Suppose = p/q where p, q N. Without loss of generality, assume that p, q have no integral factors > 1. Now p =q,so p is even. p even implies p must be even, so p =k for some k N. Consequently, q =k and so q is even. Thus p and q have the common factor, a contradiction. Other examples of irrational numbers are m where m is any rational number which is not a 3 perfect square, m where m is any rational number which is not a perfect cube, etc. Also, the numbers π and e are irrational. Definition. Let S be a subset of R. A number u R is an upper bound of S if s u for all s S. An element w R is a lower bound of S if w s for all s S. If an upper bound u for S is an element of S, then u is called the maximum (or largest element) of S. Similarly, if a lower bound w for S is an element of S, then w is called the minimum (or smallest element) of S. Examples: Give some examples to illustrate upper bounds, lower bounds, maximum and minimum elements. Definition 3. A set S R is said to be bounded above if S has an upper bound; S is bounded below if it has a lower bound. A subset S of R is bounded if it has both an upper bound and a lower bound. Definition 4. Let S R be a set that is bounded above. A number u R is called the supremum (least upper bound) of S, denoted by sup S, if it satisfies the conditions 1. s u for all s S.. If v is an upper bound for S, then u v. THEOREM 3. Let S R be bounded above, and let u = sup S. Then, given any positive number ɛ, there is an element s ɛ S such that u ɛ<s ɛ u. Proof: Suppose there exists an ɛ > 0 such that the interval (u ɛ, u] contains no points of S. Then s u ɛ for all s S, which implies that u ɛ is an upper bound for S which is less than u, a contradiction. Definition 5. :LetS R be a set that is bounded below. A number u R is called the infimum (greatest lower bound ) of S and is denoted by inf S if it satisfies the conditions 1. u s for all s S.. If v is a lower bound for S, then v u. The Completeness Axiom Axiom Every nonempty subset S of R that is bounded above has a least upper bound. That is, if S is bounded above, then sup S exists and is a real number. 5
6 The set of real numbers R is a complete, ordered, field. The set of rational numbers Q, although an ordered field, is not complete. For example, the set T = {r Q: r< } is bounded above, but T does not have a rational least upper bound. The Archimedean Property THEOREM 4. (The Archimedean Property) The set N of natural numbers is unbounded above. Proof: Suppose N is bounded above. Let m = sup N. By Theorem 3 there exists a positive integer k such that m 1 <k m. But then k + 1 is a positive integer and k +1>m, a contradiction. THEOREM 5. The following are equivalent: (a) The Archimedean Property. (b) For each z R, there exists an n N such that n>z. (c) For each x>0 and for each y R, there exists an n N such that nx > y. (d) For each x>0, there exists an n N such that 0 < 1/n<x. Proof: (a) (b). Suppose there exists a real number r such n r for all n N. Then N is bounded above by r, contradicting (a). (b) (c). Let z = y/x. Then, by (b), there exists n N such that n>y/x which implies nx > y. (c) (d). By (c) there exists n N such that n > 1/x which implies 1/n < x. Since n>0, 1/n > 0. Thus, 0 < 1/n < x. (d) (a). Suppose N is bounded above. Let m = sup N. Then n m for all n N which implies 1/m 1/n for all n N contradicting (d). THEOREM 6. There exists a real number x such that x =. In general, if p is a prime number, then there exists a real number y such that y = p. There is a technical proof which could be given here, but this result is an easy consequence of the Intermediate-Value Theorem in Part III. The Density of the Rational Numbers and the Irrational Numbers Lemma: Let y be a positive number. Then there exists an m N such that m 1 y<m. Proof: Let K = {n N : n>y}. By Theorem 5 (b), K is not empty. By the well-ordering axiom, K has a least element m. It follows that m 1 y<m. THEOREM 7. If x and y are real numbers, x<y, then there exists a rational number r such that x<r<y. 6
7 Proof: Assume first that x > 0. There exists a positive integer n such that n > 1/(y x), which implies nx +1<ny. By the lemma, there exists a positive integer m such that m 1 nx < m. Therefore, m nx +1<ny. We now have Take r = m/n. nx < m < ny which implies x< m n <y. For x 0, choose a positive integer k such that x + k>0 and apply the result above to find a rational number q such that x + k < q < y + k. Then r = q k satisfies x < r < y. THEOREM 8. If x and y are real numbers, x<y, then there exists an irrational number z such that x<z<y. Exercises True False. Justify your answer by citing a theorem, giving a proof, or giving a counterexample. (a) If a non-empty subset of R has a infimum, then it is bounded. (b) Every non-empty bounded subset of R has a maximum and a minimum. (c) If v is an upper bound for S u < v, then u is not an upper bound for S. (d) If w = inf S and z < w, then z is a lower bound for S. (e) Every nonempty subset of N has a minimum. (f) Every nonempty subset of N has a maximum.. True False. Justify your answer by citing a theorem, giving a proof, or giving a counterexample. (a) If x and y are irrational, then xy is irrational. (b) Between any two distinct rational numbers, there is an irrational number. (c) Between any two distinct irrational numbers, there is a rational number. (d) The rational and irrational numbers alternate. 3. Let S R be non-empty and bounded above and let u = sup S. Prove that u S if and only if u = max S. 4. (a) Let S R be non-empty and bounded above and let u = sup S. Prove that u is unique. (b) Prove that if each of m and n is a maximum of S, then m = n. 5. Let S R and suppose that v = inf S. Prove that for any positive number ɛ, there is an element s ɛ S such that v s ɛ <v+ ɛ. 6. Prove that if x and y are real numbers with x<y, then there are infinitely rational numbers in the interval [x, y]. 7
8 I.4. TOPOLOGY OF THE REALS Definition 6. Let S R. The set S c = {x R : x/ S} is called the complement of S. Definition 7. Let x R and let ɛ > 0. An ɛ-neighborhood of x (often shortened to neighborhood of x ) is the set The number ɛ is called the radius of N(x, ɛ). N(x, ɛ) ={y R : y x <ɛ}. Note that an ɛ-neighborhood of a point x is the open interval (x ɛ, x + ɛ) centered at x with radius ɛ. Definition 8. Let x R and let ɛ > 0. A deleted ɛ-neighborhood of x (often shortened to deleted neighborhood of x ) is the set N (x, ɛ) ={y R :0< y x <ɛ}. A deleted ɛ-neighborhood of x is an ɛ-neighborhood of x with the point x removed; N (x, ɛ)=(x ɛ, x) (x, x + ɛ). Definition 9. Let S R. A point x S is an interior point of S if there exists a neighborhood N of x such that N S. The set of all interior points of S is denoted by int S and is called the interior of S. Examples Make up some examples to illustrate interior point and interior of S. Definition 10. Let S R. A point z R is a boundary point of S if N S and N S c for every neighborhood N of z. The set of all boundary points of S is denoted by bd S and is called the boundary of S. Examples Make up some examples to illustrate boundary point and boundary of S. Open Sets and Closed Sets Definition 11. Let S R. S is open if every point of S is an interior point. That is, S is open if and only if S =ints. S is closed if and only if S c is open. Examples: A neighborhood N of a point x is an open set; an open interval (a, b) is an open set; R is an open set. A closed interval [a, b] is a closed set. THEOREM 9. Let S R. S is closed if and only if bd S S. Proof: Suppose S is closed. Let x S c. Since S c is open, there is a neighborhood N of x such that N S =. Therefore x is not a boundary point of S. Therefore bd S S. Now suppose bd S S. Let x S c. Then x / S and x / bd S. Therefore there is a neighborhood N of x such that N S =. This implies that S c is open and S is closed. 8
9 THEOREM 10. (a) The union of any collection of open sets is open. (b) The intersection of any finite collection of open sets is open. Proof: (a) Let G be a collection of open sets and let x G G G. Then x G for some G G. Since G is open, there is a neighborhood N of x such that N G. Since N G, N G G G. Therefore G G G is open. (b) Let G 1,G,...,G n be a (finite) collection of open sets, and let x G i. Since x G 1 and G 1 is open, there is an ɛ 1 -neighborhood N 1 of x such that N 1 G 1 ; since x G and G is open, there is an ɛ -neighborhood N of x such that N G ;...; since x G n and G n is open, there is an ɛ n -neighborhood N n of x such that N n G n. Let ɛ = min {ɛ 1,ɛ,...,ɛ n } and let N = N(x, ɛ). Now, N N i for i =1,,...,n which implies N G i for i =1,,...,n, and so N G i. Note: The restriction finite in( Theorem 10(b) is necessary. For example, the intersection of the (infinite) collection of open sets 1 n, 1+ 1 ), n =1,, 3,... is the closed interval [0, 1]. n COROLLARY (a) The intersection of any collection of closed sets is closed. (b) The union of any finite collection of closed sets is closed. The Corollary follows directly from the Theorem by means of De Morgan s Laws : Let {S α },α A be a collection of sets. Then 1. ( α A S α ) c = α A S c α. ( α A S α ) c = α A S c α Accumulation Points Definition 1. Let S R. A point x R is an accumulation point of S if every deleted neighborhood N of x contains a point of S. The set of accumulation points of S is denoted by S.If x S and x is not an accumulation point of S, then x is an isolated point of S. Examples Make up some examples to illustrate accumulation point and isolated point. Note: An accumulation point of S may or may not be a point of S. Definition 13. Let S R. The closure of S, denoted by S, is the set S = S S. THEOREM 11. Let S R. Then (a) S is closed if and only if S contains all its accumulation points. 9
10 (b) S is a closed set. Proof: (a) Suppose S is closed. Let x be an accumulation point of S and suppose x/ S. Then x S c, an open set. Therefore, there is a neighborhood N of x such that N S c. Now, N S = which implies x is not an accumulation point of S, a contradiction. Now suppose that S contains all its accumulation points. Let x S c. Then x/ S and x is not an accumulation point of S. Therefore, there is a neighborhood N of x such that N S = so N S c. This implies that S c is open and S is closed. (b) Let y be an accumulation point of S and let N be a neighborhood of y. Then N contains a point x S, x y. Therefore x is in S or x is an accumulation point of S. Suppose x is an accumulation point of S. Since N is an open set, there is a neighborhood N x of x such that N x N, and since x is an accumulation point of S, N x contains a point z of S, z x (we also choose N x such that z y). Thus, each neighborhood of N of y contains a point of S. Therefore y is an accumulation point of S so y S. Since S contains all its accumulation points, it is closed. Exercises True False. Justify your answer by citing a theorem, giving a proof, or giving a counterexample. (a) int S bd S =. (b) bd S S. (c) S is closed if and only if S =bds. (d) If x S, then x int S or x bd S. (e) bd S =bds c. (f) bd S S c.. True False. Justify your answer by citing a theorem, giving a proof, or giving a counterexample. (a) A neighborhood is an open set. (b) The union of any collection of open sets is open. (c) The union of any collection of closed sets is closed. (d) The intersection of any collection of open sets is open. (e) The intersection of any collection of closed sets is closed. (f) R is neither open nor closed. 3. Classify each set as open, closed, neither, or both. (a) N (b) Q { } 1 (c) n : n N 10
11 (d) n=1 ( 0, 1 ) n (e) {x : x > 0} (f) {x : x 3} 4. Let S be a bounded infinite set and let u = sup S. (a) Prove that if u/ S, then u S. (b) True or false: if u S, then u/ S? 5. Prove that if x is an accumulation point of S, then every neighborhood of x contains infinitely many points of S. I.5. COMPACT SETS Definition 14. Let S R. A collection G of open sets such that S G G G is called an open cover of S. A subcollection F of G which also covers S is called a subcover of S. Examples Make up some examples to illustrate open cover and subcover. Definition 15. A set S R is compact if and only if every open cover G of S contains a finite subcover. That is, S is compact if for every open cover G of S there is a finite collection of open sets G 1, G,... G n in G such that S n k=1 G k. Examples The interval (0, 1] is not compact; the open intervals (1/n, 1 + 1/n), n= 1,, 3,... form an open cover with no finite subcover. The interval [0, ) is not compact; the open intervals ( 1/n, n), n =1,, 3,... form an open cover with no finite subcover. THEOREM 1. If S R is non-empty, closed and bounded, then S has a maximum and a minimum. Proof: Since S is bounded, S has a least upper bound m and a greatest lower bound k. Since m is the least upper bound for S, given any ɛ>0, m ɛ is not an upper bound for S. Ifm/ S, then there exists an x S such that m ɛ < x < m which implies that m is an accumulation point of S. Since S is closed, m S and m = max S. A similar argument holds for k. THEOREM 13. (Heine-Borel Theorem) A subset S of R is compact if and only if it is closed and bounded. Proof: Suppose S is compact. Let G be the collection of open intervals I n =( n, n), n=1,,... Then G is an open cover of S. Since S is compact, G contains a finite subcover I n1,i n,..., I nk. Let m = max n i. Then S I m, and for all x S, x m. Therefore S is bounded. To show that S is closed, we must show that S contains all its accumulation points. Suppose that p is an accumulation point of S and suppose p / S. For each positive integer n, let G n =[p 1/n, p +1/n] c. Since the complement of a closed interval is an open set, G n is an open 11
12 for all n. Since p/ S, S G n. That is, the sets G n, n =1,, 3... form an open cover of S. Since S is compact, this open cover has a finite subcover, G n1,g n,..., G nk. Let m = max n i. Then the neighborhood N of p of radius 1/m contains no points of S contradicting the assumption that p is an accumulation point of S. We omit the proof that S closed and bounded implies that S is compact. THEOREM 14. (Bolzano-Weierstrass Theorem) If S R is a bounded infinite set, then S has at least one accumulation point. Proof: Let S be a bounded infinite set and suppose that S has no accumulation points. Then S is closed (vacuously), and S is compact. For each x S, let N x be a neighborhood of x such that S N x = {x}. The set of neighborhoods N x, x S is an open cover of S. Since S is compact, this open cover has a finite subcover N x1, N x,..., N xk. But S [N x1 N x... N xk ]= {x 1,x,...,x k } which implies that S is finite, a contradiction. Exercises True False. Justify your answer by citing a theorem, giving a proof, or giving a counterexample. (a) Every finite set is compact. (b) No infinite set is compact. (c) If a set is compact, then it has a maximum and a minimum. (d) If a set has a maximum and a minimum, then it is compact. (e) If S R is compact, then there is at least one point x R such that x is an accumulation point of S. (f) If S R is compact and x is an accumulation point of S, then x S.. Show that each of the following subsets S of R is not compact by giving an open cover of S that has no finite subcover. (a) S =[0, 1) (b) S = N (c) S = {1/n : n N} 3. Prove that the intersection of any collection of compact sets is compact. 4. Prove that if S R is compact and T is a closed subset of S, then T is compact. 1
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|
# Recursion Part 3
## Fibonacci Numbers
The Fibonacci numbers are a sequence of numbers starting with {1, 1}. Each subsequent number is then the sum of the two previous. We can find Fibonacci numbers with a loop, but we can also use recursion.
A recursive definition of the Fibonacci sequence is as follows:
$fib(1) = 1$
$fib(2) = 1$
$fib(x) = fib(x - 1) + fib(x - 2)$
We can use this definition to get the following method:
public static int fib(number) {
if(number < 3) {
return 1;
} else {
return fib(number - 1) + fib(number - 2);
}
}
This method is also $O(2^n)$. As you can see every call to the method expands into two more, so the number of calls explodes exponentially.
This is a case where the most natural recursive solution is not very efficient. Just because this particular algorithm is $O(2^n)$ doesn't mean that this problem is hard to solve. Remember that Big-O complexities are for algorithms, not problems. Some problems don't have efficient solutions, but this one happens to. There is a $O(n)$ solution, which we can write with a loop like this:
int fib(int number) {
int a = 1;
int b = 1;
int n = 1;
for(int i = 2; i < number; i++) {
n = a + b;
a = b;
b = n;
}
return n;
}
This Fibonacci method runs far faster. This algorithm is $O(n)$, which you can see from the fact that there is one loop which runs up to the number given.
## Faster Recursive Fibonacci Numbers
In this case, the slow solution is easiest implemented with recursion and the fast solution is easiest implemented with iteration. But we can write fast recursive solutions as well. We could also write the slow algorithm with a loop (but why would we want to?).
There are at least two fast recursive solutions to this problem. The simplest is to employ a technique called memoization which is basically to cache solutions to the problem in an array. When we call the recursive method, it first checks if the array holds the answer. If so, it returns it directly. If not, then it computes it from scratch (saving the result in the array for later). This reduces the repetitious work the original recursive solution does.
The Fib_Memoized.java code can be seen here:
// recursive fibonacci sequence w/ memoization
public static long recursive_fib(int number) {
// if we've already calculated this, use the saved result
if (results[number] != 0) {
return results[number];
}
// the base case of 0 or 1
if (number < 2) {
results[number] = 1;
return 1;
} else {
// figure it out, then save it in the array
long answer = recursive_fib(number - 1) + recursive_fib(number - 2);
}
}
This program runs far faster because it only computes each unique Fibonacci number one time. Memoization is nice because you can use it as an optimization to an existing solution without needing to rewrite the algorithm. Like the iterative solution, this runs in $O(n)$
## Using Recursion as a Stack
Because method calls cause a push onto the stack, and method returns cause a pop from the stack, we can use recursion to write any algorithm using a stack using recursion. For example, the depth first search algorithm was given as follows:
1. Set current to the start position.
2. While current is not the end position:
1. Mark current as visited.
2. If we haven't gone left, push left.
3. If we haven't gone right, push right.
4. If we haven't gone up, push up.
5. If we haven't gone down, push down.
6. If the stack is empty, there is no path!
7. Set current to pop().
We can write this using recursion by replacing the pushes with recursive calls. DFS_Recursive.java demonstrates a recursive depth first search.
## Designing Recursive Functions
• Break big problems into smaller ones.
• Make sure there is a base case.
• Make sure the general case gets us closer to the base case.
|
### Simplifying Square Roots (Review)
Let's review the steps involved in simplifying square roots:
1. Factor the number inside the square root sign.
2. If a factor appears twice, cross out both and write the factor one time to the left of the square root sign. If the factor appears three times, cross out two of the factors and write the factor outside the sign, and leave the third factor inside the sign.
Note: If a factor appears 4, 6, 8, etc. times, this counts as 2, 3, and 4 pairs, respectively.
3. Multiply the numbers outside the sign. Multiply the numbers left inside the sign.
4. Check: The outside number squared times the inside number should equal the original number inside the square root.
To simplify the square root of a fraction, simplify the numerator and simplify the denominator.
Example 1: Simplify
1. =
2. = 2×2
3. 2×2 = 4
4. Check: 42(3) = 48
Thus, = 4.
Example 2: Simplify .
First, reduce the fraction to lowest terms: =
=
Numerator:
1. =
2. = 2
3. 2 = 2
4. Check: 22(3) = 12
Denominator:
1. =
2. = 5
3. 5 = 5
4. Check: 52(7) = 175
Thus, = .
### Rationalizing the Denominator
In addition to simplifying the numerator and the denominator in a fraction, it is mathematical convention to rationalize the denominator--that is, to write the fraction as an equivalent expression with no roots in the denominator.
To rationalize a denominator, multiply the fraction by a "clever" form of 1--that is, by a fraction whose numerator and denominator are both equal to the square root in the denominator. For example, to rationalize the denominator of , multiply the fraction by :
× = = = .
Thus, = .
Often, the fraction can be reduced:
Rationalize the denominator of :
× = = = = 3.
Thus, = 3.
|
# Finding the Size of a Tree
## Introduction
A common operation when working with trees is finding the size of a tree, which refers to counting the total number of nodes in the tree. One of the most common and intuitive ways to find the size of a tree is through a recursive approach. This method involves traversing the tree and counting the nodes using a recursive function.
In this guide, we will explore methods to determine the size of a binary tree.
## What is the Size of a Tree?
The size of a tree is a straightforward concept—it represents the number of nodes in the tree. In a binary tree, it’s the total count of nodes, including both internal nodes (those with child nodes) and leaf nodes (those without children).
## Finding the Size of a Tree Example
### Consider a Binary Tree : Let us take an example.
The tree depicted above has a size of 7.
• To determine the tree’s size, we calculate it by adding the sizes of its left and right subtrees and then incrementing the result by 1.
• This process involves invoking a recursive function for both the left and right subtrees of the tree. If a subtree is absent, the function returns 0.
## Above Example Analysis 📊
The above example for finding the size of a tree is depicted as below :
• Size of node 13 = Size(11) + Size(15) + 1
• Size of node 13 = (Size(9) + Size(12) + 1) + (Size(14) + Size(17) + 1) + 1
• Size of node 13 = (Size(9) + Size(12) + 1) + (Size(14) + Size(17) + 1) + 1
• Size of node 13 = (1 + 1 + 1) + (1 + 1 + 1) + 1
• Size of node 13 = 7
### How to Calculate the Size of a Tree in Data Structure
A straightforward approach:
• Begin at the root.
• The size is determined by adding 1 (for the root) to the sizes of the left sub-tree and the right sub-tree.
• Resolve the sizes of the left and right sub-trees through recursive calculations.
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### Methods for Finding the Size of a Tree
There are mainly two approaches for finding the size of a tree :
### Recursive Approach
One of the most common and intuitive ways to find the size of a tree is through a recursive approach. This method involves traversing the tree and counting the nodes using a recursive function. Here’s a simple example in Python:
### Code :
```class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
def treeSize(root):
if root is None:
return 0
else:
return 1 + treeSize(root.left) + treeSize(root.right)
```
#### Iterative Approach
Another way to find the size of a tree is through an iterative approach, often utilizing a level order traversal (Breadth-First Search or BFS). The idea here is to start at the root node and traverse the tree while keeping track of the count. Below is an example of an iterative implementation in Python:
### Code :
```def treeSizeIterative(root):
if root is None:
return 0
size = 0
queue = []
queue.append(root)
while queue:
node = queue.pop(0)
size += 1
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return size
```
### Code :
```class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Recursive approach to find the size of a tree
def treeSizeRecursive(root):
if root is None:
return 0
else:
size = 1 + treeSizeRecursive(root.left) + treeSizeRecursive(root.right)
return "Size of the tree (recursive): " + str(size)
# Iterative approach to find the size of a tree
def treeSizeIterative(root):
if root is None:
return 0
size = 0
queue = []
queue.append(root)
while queue:
node = queue.pop(0)
size += 1
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return "Size of the tree (iterative): " + str(size)
# Example usage (creating a binary tree)
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
# Using the recursive approach to find the size of the tree
tree_size_recursive = treeSizeRecursive(root)
print(tree_size_recursive)
# Using the iterative approach to find the size of the tree
tree_size_iterative = treeSizeIterative(root)
print(tree_size_iterative)
```
### Output :
```Size of the tree (recursive): 7
Size of the tree (iterative): 7
```
### Explanation :
The code computes the size of a binary tree using both recursive and iterative methods. It finds the total number of nodes in the tree. The recursive approach uses a function that counts nodes by recursively traversing the tree. The iterative approach employs a queue for a level-order traversal to count the nodes iteratively. Both methods yield the same result of 7 for the given example tree.
### Applications
Following are the applications of finding the size of a tree :
### To Wrap it up:
In conclusion, Calculating the size of a tree is a fundamental operation when working with trees. Depending on your preferences and specific use cases, you can choose between a recursive or an iterative approach to find the size of a tree. Both methods are effective, and you can select the one that best fits your requirements.
Question 1.
What does “size of a tree” mean?
The size of a tree refers to the total number of nodes in the tree, including both internal (with children) and leaf nodes (without children).
Question 2.
Can I use the size of a tree to determine its height?
Yes, you can calculate the height of a tree using the size. The height is the longest path from the root to a leaf node, which can be found by finding the log base 2 of the size (in the case of a balanced binary tree).
Question 3.
How do I adapt the size-finding methods to n-ary trees or other tree structures?
The basic principles for finding the size of a binary tree can be adapted to n-ary trees by modifying the traversal and counting logic.
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1–2 WWLChen : Introduction to Complex Analysis Note the special case a =1and b =0. Addition / Subtraction - Combine like terms (i.e. The term “complex analysis” refers to the calculus of complex-valued functions f(z) depending on a single complex variable z. The horizontal axis representing the real axis, the vertical representing the imaginary axis. z= a+ ib a= Re(z) b= Im(z) = argz r = jz j= p a2 + b2 Figure 1: The complex number z= a+ ib. Complex Numbers and the Complex Exponential 1. Introduction to the introduction: Why study complex numbers? 1What is a complex number? Introduction to COMPLEX NUMBERS 1 BUSHRA KANWAL Imaginary Numbers Consider x2 = … Figure 1: Complex numbers can be displayed on the complex plane. (Note: and both can be 0.) z = x+ iy real part imaginary part. Complex numbers of the form x 0 0 x are scalar matrices and are called Introduction to Complex Numbers: YouTube Workbook 6 Contents 6 Polar exponential form 41 6.1 Video 21: Polar exponential form of a complex number 41 6.2 Revision Video 22: Intro to complex numbers + basic operations 43 6.3 Revision Video 23: Complex numbers and calculations 44 6.4 Video 24: Powers of complex numbers via polar forms 45 For instance, d3y dt3 +6 d2y dt2 +5 dy dt = 0 DEFINITION 5.1.1 A complex number is a matrix of the form x −y y x , where x and y are real numbers. Let i2 = −1. Well, complex numbers are the best way to solve polynomial equations, and that’s what we sometimes need for solving certain kinds of differential equations. Since complex numbers are composed from two real numbers, it is appropriate to think of them graph-ically in a plane. Complex Number – any number that can be written in the form + , where and are real numbers. COMPLEX NUMBERS 5.1 Constructing the complex numbers One way of introducing the field C of complex numbers is via the arithmetic of 2×2 matrices. View complex numbers 1.pdf from BUSINESS E 1875 at Riphah International University Islamabad Main Campus. Complex numbers are often denoted by z. A complex number z1 = a + bi may be displayed as an ordered pair: (a,b), with the “real axis” the usual x-axis and the “imaginary axis” the usual y-axis. The union of the set of all imaginary numbers and the set of all real numbers is the set of complex numbers. ∴ i = −1. Introduction to Complex Numbers. Introduction to Complex Numbers Adding, Subtracting, Multiplying And Dividing Complex Numbers SPI 3103.2.1 Describe any number in the complex number system. 3 + 4i is a complex number. Suppose that z = x+iy, where x,y ∈ R. 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introduction to complex numbers pdf 2021
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# Combining two differential equations
• Mar 24th 2013, 01:05 AM
alexander9408
Combining two differential equations
Two quantities, x and y, vary with time t. State the differential equation for each of the following.
(a) The rate of change of y with respect to t varies inversely as x.
(b) The rate of change of x with respect to t varies inversely as y. Combining the differential equation of (a) and (b), form a differential equation involving only x and y, solve the differential equation expressing y in terms of x.
How do I combine the two differential equations into one?
• Mar 24th 2013, 01:09 AM
Prove It
Re: Combining two differential equations
Quote:
Originally Posted by alexander9408
Two quantities, x and y, vary with time t. State the differential equation for each of the following.
(a) The rate of change of y with respect to t varies inversely as x.
(b) The rate of change of x with respect to t varies inversely as y. Combining the differential equation of (a) and (b), form a differential equation involving only x and y, solve the differential equation expressing y in terms of x.
How do I combine the two differential equations into one?
Do you understand what's meant by inverse proportion? If "a" is inversely proportional to "b", then $\displaystyle a = \frac{k}{b}$, where "k" is a constant.
• Mar 24th 2013, 02:08 AM
alexander9408
Re: Combining two differential equations
Yes, my problem was combining the two differential equations dy/dt = k/x and dx/dt = h/y (k and h are constants) into one differential equation and expressing y in terms of x.
• Mar 24th 2013, 03:53 AM
Prove It
Re: Combining two differential equations
Remember $\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dx}$, so
\displaystyle \begin{align*} \frac{\frac{dy}{dt}}{\frac{dx}{dt}} &= \frac{\frac{k}{x}}{\frac{h}{y}} \\ \frac{dy}{dx} &= \frac{k\,y}{h\,x} \end{align*}
This is a separable first order DE.
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# Prove that the curves
Question:
Prove that the curves $x=y^{2}$ and $x y=k$ cut at right angles if $8 k^{2}=1$. [Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.]
Solution:
The equations of the given curves are given as $x=y^{2}$ and $x y=k$.
Putting $x=y^{2}$ in $x y=k$, we get:
$y^{3}=k \Rightarrow y=k^{\frac{1}{3}}$
$\therefore x=k^{\frac{2}{3}}$
Thus, the point of intersection of the given curves is $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$.
Differentiating $x=y^{2}$ with respect to $x$, we have:
$1=2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 y}$
Therefore, the slope of the tangent to the curve $x=y^{2}$ at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ is $\left.\frac{d y}{d x}\right]\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)=\frac{1}{2 k^{\frac{1}{3}}}$.
On differentiating $x y=k$ with respect to $x$, we have:
$x \frac{d y}{d x}+y=0 \Rightarrow \frac{d y}{d x}=\frac{-y}{x}$
$\therefore$ Slope of the tangent to the curve $x y=k$ at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ is given by,
$\left.\left.\frac{d y}{d x}\right]\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)=\frac{-y}{x}\right]\left(k^{\frac{2}{3}, k^{\frac{1}{3}}}\right)=-\frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}}=\frac{-1}{k^{\frac{1}{3}}}$
We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ are
perpendicular to each other.
This implies that we should have the product of the tangents as − 1.
Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ is $-1 .$
i.e. $\left(\frac{1}{2 k^{\frac{1}{3}}}\right)\left(\frac{-1}{k^{\frac{1}{3}}}\right)=-1$
$\Rightarrow 2 k^{\frac{2}{3}}=1$
$\Rightarrow\left(2 k^{\frac{2}{3}}\right)^{3}=(1)^{3}$
$\Rightarrow 8 k^{2}=1$
Hence, the given two curves cut at right angels if $8 k^{2}=1$.
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## NCERT Solution Class X Mathematics Coordinate Geometry Question 3 (Ex 7.1)
Question 3:
Determine if the points (1, 5), (2, 3) and (− 2, − 11) are collinear.
Let the points (1, 5), (2, 3), and (−2, −11) be representing the vertices A, B, and C of the given triangle respectively.
Let A = (1,5), B = (2,3), C = (−2, −11)
Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.
## NCERT Solution Class X Mathematics Coordinate Geometry Question 2 (Ex 7.1)
Question 2:
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Distance between points (0,0) and (36,15)
Yes, we can find the distance between the given towns A and B.
Assume town A at origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
And hence, as calculated above, the distance between town A and B will be 39 km.
## NCERT Chapter Notes Class X Science Light – Reflection and Refraction 7
Sign Convention for Refraction by spherical lens
Similar to that of spherical mirror, only the difference is that all the measurement are made from optical centre ‘O’
Few Tips to Remember Sign Convention for Spherical Lens
Lens : f – u – v
Concave : –ve –ve –ve (Real)
+ve (virtual)
Convex : +ve, –ve +ve,
h – is always +ve
n´ – –ve for Real and
+ve for virtual & Errect.
Power of Lens
The degree of convergence or divergence of light ray achieved by a lens is known as power of a lens.
It is defined as the reciprocal of its focal length Represented by P.
SI unit of power of a lens is “dioptre” denoted by ‘D’
I dioptre or ID → It is the power of lens whose focal length is 1m
Power convex lens or converging lens is always positive
Power of concave lens or diverging lens is always negative
If any optical instrument have many lens, then net power will be
## NCERT Chapter Notes Class X Science Light – Reflection and Refraction 6
Spherical Lens
A transparent material bound by two surface, of which one or both surfaces are spherical, forms a lens.
Convex Lens
A lens may have two spherical surfaces, bulging outwards, is called double convex lens (or simply convex lens.
It is also known as converging lens because it converges the light.
Concave Lens
A lens bounded by two spherical surfaces, curved inwards is known as double concave lens (or simply concave lens)
It is also known as diverging lens because it diverges the light.
Few Basic Terms Related to Spherical Lens.
1. Centre of curvature : A lens, either a convex lens or a concave lens is a combination of two spherical surfaces. Each of these surfaces form a part of sphere. The centre of these two spheres are called centre of curvature represented by C1 and C2.
2. Principal axis : Imaginary straight line passing through the two centres of curvature
3. Optical Centre : The central point of lens is its optical centre (O). A ray of light, when passes through ‘O’ it remains undeviated i.e. it goes straight.
4. Aperture : The effective diameter of the circular outline of a spherical lens.
5. Focus of lens : Beam of light parallel is principal axis, after refraction from
1. Convex lens, converge to the point on principal axis, denoted by F, known as Principal focus
2. Concave lens, appear to diverge from a point on the principal axis, known as principal focus.
The distance OF2 and OF1 is called as focal length
Tips for drawing Ray diagram
(a) After refraction, a ray parallel to principal axis will pass through F.
(b) A ray passes through F, after refraction will emerge parallel to principal axis.
(c) A ray passes through optical centre ‘O’, passes without any deviation.
Image formation by concave lens
## NCERT Chapter Notes Class X Science Light – Reflection and Refraction 5
When a incident ray of light AO passes from a rarer medium (air) to a denser medium (glass) at point. O on interface KL, it will bends towards the normal. At point O1, on interface NM the light ray entered from denser medium (glass) to rarer medium (air) here the light ray will bend away from normal OO1is a refracted ray OB is an emergent ray. If the incident ray is extended to C, we will observe that emergent ray O1B is parallel to incident ray. The ray will slightly displaced laterally after refraction.
Note : When a ray of light is incident normally to the interface of two media it will go straight, without any deviation.
Laws of refraction of light
1. The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
2. The ratio of sine of angle of incidence to the sine of angle of refraction is a constant i.e.
for given colour and pair of media, this law is also known as Snells Law
Constant n is the refractive index for a given pair of medium. It is the refractive index of the second medium with respect to first medium.
Refractive Index
The refractive index of glass with respect to air is given by ratio of speed of light in air to the speed of light in glass.
C → Speed of light in vacuum = 3•108 m/s speed of light in air is marginally less, compared to that in vacuum.
Refractive index of air with respect to glass is given by
The absolute refractive index of a medium is simply called refractive index
Refractive index of water (nw) = 1.33
Refractive index of glass (ng) = 1.52
## NCERT Chapter Notes Class X Science Light – Reflection and Refraction 4
Magnification
It is expressed as the ratio of the height of the image to height of the object
Few tips to remember sign convention for Spherical mirror
Refraction of Light
Refraction of Light : Happens in Transparent medium when a light travels from one medium to another, refraction takes place.
A ray of light bends as it moves from one medium to another
Refraction is due to change in the speed of light as it enters from one transparent medium to another.
Speed of light decreases as the beam of light travel from rarer medium to the denser medium.
Some Commonly observed phenomenon due to Refraction
1. The stone at the bottom of water tub appears to be raised.
2. A fish kept in aquarium appears to be bigger than its actual size.
3. A pencil partially immersed in water appears to be displaced at the interface of air and water.
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# Completing the square
### Completing the square
#### Lessons
perfect squares:
• ${\left( {x + a} \right)^2} = {x^2} + 2ax + {a^2}$
• ${\left( {x - a} \right)^2} = {x^2} - 2ax + {a^2}$
• completing the square: adding a constant to a quadratic expression to make it a perfect square
• 1.
What is “COMPLETING THE SQUARE”?
a)
Review: expanding a perfect square.
b)
How to convert a polynomial into a perfect square
c)
How to complete the square
• 2.
Recognizing a Polynomial that Can Be Written as a Perfect Square
Convert the following expressions into perfect squares, if possible:
a)
${x^2} + 6x + {3^2}$ =
${x^2} - 6x + {\left( { - 3} \right)^2}$ =
b)
${x^2} + 20x + 100$ =
${x^2} - 20x + 100$ =
${x^2} - 20x - 100$ =
• 3.
Completing the Square
Add a constant to each quadratic expression to make it a perfect square.
a)
${x^2} + 10x + \;$_____ =
b)
${x^2} - 2x + \;$_____ =
c)
$2{x^2} + 12x + \;$_____ =
d)
$- 3{x^2} + 60x + \;$_____ =
e)
$\frac{2}{5}{x^2} - 8x + \;$_____ =
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On social media, the ‘What is the height of the table riddle’ is trending. You will be “forced” to recall your basic math courses as a result of the puzzle.
Most people in the country are currently self-isolating inside their homes due to the ongoing coronavirus outbreak. Keeping cool and focused during these tough moments might be difficult. Some people, on the other hand, have taken to solving puzzles in their free time to keep themselves occupied and enhance their minds. One such puzzle that is now trendy on social media and Whatsapp is the ‘What is the height of the table problem.’ Here is the solution to the riddle.
The puzzle seen above is currently trending on the internet. To solve the riddle, you must determine the table’s true height. A cat sits at the table’s edge in the first picture, while a tortoise stands beneath the table, directly beneath the cat. From the cat’s head to the tortoise’s shell, the total height is 170cm. The tortoise is now on top of the table, with the cat sitting beneath it in the second image. From the tortoise’s shell to the cat’s head, the total height is 130cm this time.
You must calculate the height of the table using these clues, excluding the heights of the cat and tortoise. While basic arithmetic abilities are required for this puzzle, it is not a simple math problem, and you will need to think creatively to solve it. In your free time, try to solve this puzzle. There is no time limit, so you have as much time as you like to consider the response.
Table + Cat + Tortoise = 170cm, according to the first photograph (Equation 1)
Table – Cat + Tortoise =130cm, according to the second image (Equation 2)
Table 1 + Table 2 is obtained by adding these two Equations, as the heights of the Cat and Tortoise cancel each other out.
170cm + 130cm = 300cm = Table 1 + Table 2
As a result, the height of one table is 300cm divided by two equals 150cm.
Do you like this? Share inspiration with your friends!
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# Difference between revisions of "2014 AMC 10A Problems/Problem 6"
The following problem is from both the 2014 AMC 12A #4 and 2014 AMC 10A #6, so both problems redirect to this page.
## Problem
Suppose that $a$ cows give $b$ gallons of milk in $c$ days. At this rate, how many gallons of milk will $d$ cows give in $e$ days?
$\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}$
## Solution 1
We need to multiply $b$ by $\frac{d}{a}$ for the new cows and $\frac{e}{c}$ for the new time, so the answer is $b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}$, or $\boxed{\textbf{(A)} \frac{bde}{ac}}$.
## Solution 2
We plug in $a=2$, $b=3$, $c=4$, $d=5$, and $e=6$. Hence the question becomes "2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?"
If 2 cows give 3 gallons of milk in 4 days, then 2 cows give $\frac{3}{4}$ gallons of milk in 1 day, so 1 cow gives $\frac{3}{4\cdot2}$ gallons in 1 day. This means that 5 cows give $\frac{5\cdot3}{4\cdot2}$ gallons of milk in 1 day. Finally, we see that 5 cows give $\frac{5\cdot3\cdot6}{4\cdot2}$ gallons of milk in 6 days. Substituting our values for the variables, this becomes $\frac{dbe}{ac}$, which is $\boxed{\textbf{(A)}\ \frac{bde}{ac}}$.
## Solution 3
We see that the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is $\dfrac{ac}{b}$.
Let $g$ be the answer to the question. We have $\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}$
## Solution 4
The problem specifics "rate," so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days $$\implies\text{rate}=\dfrac{b}{ac}$$
Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days $$\implies\boxed{\textbf{(A)} \dfrac{bde}{ac}}$$
## Solution 5
If $a$ cows give $b$ gallons of milk in $c$ days, that means that one cow will give $\frac{b}{a}$ gallons of milk in $c$ days. Also, we want to find the number of gallons of milk will $d$ cows give in $e$ days, so in $\frac{e}{c}$ days $d$ cows give $\frac{bd}{a}$ gallons of milk. Multiplying with the formula $d=rt$, we get $\boxed{(A)\frac{bde}{ac}}$
## Solution 6
In the right formula, plugging in $d=a$ and $e=c$ should simplify to $b$, as if it doesn't, we'd essentially be saying "$a$ cows give $b$ gallons of milk in $c$ days, but $a$ cows don't give $b$ gallons of milk in $c$ days." The only one of the answer choices that simplifies like his is $\boxed{\textbf{(A)}}$
## Video Solution
~savannahsolver
2014 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 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 10 Problems and Solutions
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# RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions
The RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions is based on Algebraic Identities. Students will get to learn about these expressions in detail. In this article, we will explain about using Algebraic Identities in the problems. Our experts prepare the questions in the PDF we attached below with easy solutions.
As we know, algebra is the part of Maths, to represent the quantities and number the several symbols and letters will be used in the form of formulas and equations. It is compulsory for the students to learn all the formulas and should have the knowledge when to use the particular formula in the questions.
Learn about RD Sharma Class 9 Chapter 4- Algebraic Identities
## Download RD Sharma Chapter 9 4 Class 9 Maths Exercise 4.1 Solutions PDF
Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.1
## Important Definition RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions
RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions is based on the following formulas-
Look down for some examples based on the formulas of algebraic identities mentioned above-
Ques)- (3p – 1/p)²
Solution- (3p – 1/p)²
= Algebraic Identity- (a-b)² = a² – 2ab + b²
= (3p – 1/p)² = (3p)² – (2 x 3p x 1/p) + (1/p)²
= 9p² – 6 + 1/p²
Ques)- (3x + y) (3x – y)
Solution- (3x + y) (3x – y)
= Algebric Identity- (a + b) (a – b) = a² – b²
= (3x + y) (3x – y) = 3x² – y²
= 3x² – y²
Ques)- 175 x 175 +2 x 175 x 25 + 25 x 25
Solution- 175 x 175 +2 x 175 x 25 + 25 x 25
= Algebraic Identity- a² + 2ab + b² = (a+b)²
= (175)² + 2 (175) (25) + (25)²
= (175 + 25)²
= (200)²
= 40000
### Benefits of RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions
Go to the following points to know the Benefits of the RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions-
1. The solutions we have prepared in the above PDF is based on the NCERT Syllabus with detailed information.
2. With the help of RD Sharma, students will get to learn tips and tricks to solve the questions by applying algebraic identities.
3. The answers to the questions are presented organized stepwise in an easy way.
Ques 1- How many algebraic identities are there?
Ans)- There are majorly four algebraic identities. The algebraic identities are used by using basic formulas mentioned below-
Ques 2- What is algebraic identity?
Ans)- An algebraic identity is an identity that exists for any values of its variables.
Ques 3- Are all identity equations?
Ans)- An equation satisfied for all variable values determining the expressions is termed as an identity. So, the identity is a unique case of an equation.
In other words, all identities can be equations, but not all equations will be identities.
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## Project Euler & HackerRank Problem 18 Solution
##### Maximum path sum I
by {BetaProjects} | MAY 17, 2009 | Project Euler & HackerRank
### Project Euler Problem 18 Statement
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
``` 3
7 5
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
```
Find the maximum total from top to bottom of the triangle below:
``` 75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
… data continues
```
NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one–hundred rows; it cannot be solved by brute force, and requires a clever method!
### Solution
To solve this problem and problem 67, which is much larger, start the search from the bottom to the top, adding the maximums along the way. This will “bubble” the maximum path total to the top of the triangle.
Let’s follow this technique, step–by–step, with the 4 row triangle example above to show how this works.
#### A step–by–step look at this algorithm
```Initial array After the first
iteration
3 3
7 5 7 5
2 4 6 10 13 15
8 5 9 3```
1. Starting at the bottom, take the first pair, 8 and 5, pick the maximum and replace the 2 in the previous row with their sum 8+2=10.
2. Take the next pair, 5 and 9, pick the maximum and replace the 4 in the previous row with their sum 9+4=13.
3. Finally, take the last pair, 9 and 3, pick the maximum and replace the 6 in the previous row with their sum 9+6=15.
If you think about it, we simply solved the maximum path from the last row to the second to the last row by considering each sub–problem for the following three triangles.
``` 2 4 6
8 5 5 9 9 3```
Keep that in mind and let’s do it again with our new array.
1. Take the larger of 10 and 13 and add it to 7 making 13+7=20.
2. Take the larger of 13 and 15 and add it to 5 making 15+5=20.
Now our array looks like:
``` 3
20 20```
At last we take the larger of 20 and 20 (yes, I know they’re the same) and add it to 3 making 20+2=23.
And our array looks like:
`23`
Which is the maximum total path in the triangle.
And if you follow this logic then you just witnessed dynamic programming in action—truly short and simple.
#### HackerRank version
HackerRank Project Euler 18 varies the number of rows in the triangle from 1 ≤ N ≤ 15 and runs 10 test cases. No changes required except to read from std input instead of a file.
### Last Word
• The data file, pe18.txt, is available on repl.it page by clicking the files icon in the top left corner.
• The first line of the program reads the data file into a two–dimensional array named `table`.
|
## What is Arithmetic Sequence
### Arithmetic Sequences Definition
A list of numbers with a clear pattern is known as an arithmetic sequence. We can tell if a series of numbers is arithmetic by taking any number and subtracting it by the number before it. In an arithmetic sequence, the difference between two numbers stays the same.
The letter $$d$$ stands for the common difference, the constant difference between the numbers in an arithmetic sequence. We have an increasing arithmetic sequence if $$d$$ is positive. We have a decreasing arithmetic sequence if $$d$$ is negative.
### Arithmetic Sequences Formula
We can use the following formula to find different terms for the arithmetic sequence:
$$a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)$$, where:
• $$a_1$$: The first term
• $$a_n$$: The $$n$$th term
• $$n$$: The term poistion
• $$d$$: Common difference
### How to figure out the terms in an arithmetic sequence?
To find any term in the arithmetic sequence, we need to know a common difference, the position of the term we want to find, and a term in the sequence.
### Example
Consider the arithmetic sequence: $$3, \ 7, \ 11, \ 15, \ 19, \ ...$$, Find $$a_{28} \ = \ ?$$
Solution
Before using the arithmetic sequence formula, we need to know the first term, the common difference, and the position of the term we want to find:
• The first term: $$a_1 \ = \ 3$$
• common difference: $$d \ = \ 4$$
• Term's position: $$n \ = \ 28$$
Now, we put these numbers into the formula and find the answer:
$$a_n \ = \ a_1 \ + \ d \ (n \ –\ 1) \ ⇒ \ a_{28} \ = \ 3 \ + \ 4(28 \ - \ 1) \ ⇒ \ a_{28} \ = \ 111$$
### Summing an Arithmetic Series
To summing the terms in an arithmetic sequence, use this formula:
$$\sum_{k \ = \ 0}^{n \ - \ 1} \ (a \ + \ kd) \ = \ \frac{n}{2} (2a \ + \ (n \ - \ 1)d)$$
### Example
Find the sum of the first $$12$$ terms for the sequence: $$1, \ 6, \ 11, \ 16, \ 21, \ ...$$
Solution
• The first term: $$a_1 \ = \ 1$$
• common difference: $$d \ = \ 5$$
• $$n \ = \ 12$$
So:
$$\sum_{k \ = \ 0}^{n \ - \ 1} \ (a \ + \ kd) \ = \ \frac{n}{2} (2a \ + \ (n \ - \ 1)d) \ ⇒$$
$$\sum_{k \ = \ 0}^{11} \ (1 \ + \ 5k) \ = \ \frac{12}{2} (2 \ + \ (11)5) \ = \ 6 \times (57) \ = \ 342$$
### Exercises for Arithmetic Sequences
1) Find the explicit formula: $$5, \ 14, \ 23, \ 32, \ 41, \ ...$$
2) Find the explicit formula: $$-7, \ -2, \ 3, \ 8, \ 13, \ ...$$
3) Find the explicit formula and $$a_{12}$$: $$-22, \ -16, \ -10, \ -4, \ 2, \ ...$$
4) Find the explicit formula and $$a_{21}$$: $$a_{30} \ = \ 72, \ d \ = \ -5$$
5) Find the first $$10$$ terms: $$a_{15} \ = \ 50, \ d \ = \ 4.5$$
6) Find the first $$7$$ terms: $$a_{41} \ = \ 178, \ d \ = \ 4$$
7) Find $$a_{30}$$: $$45, \ , \ 41.8, \ 38.6, \ 35.4, \ ...$$
8) Find $$a_{27}$$: $$a_{41} \ = \ 167, \ d \ = \ 6.4$$
9) Find the explicit formula and the sum of the first five terms: $$a_1 \ = \ 10, \ d \ = \ 3$$
10) Find the explicit formula and the sum of the first $$16$$ terms: $$-55, \ -48, \ -41, \ -34, \ -27, \ ...$$
1) Find the explicit formula: $$5, \ 14, \ 23, \ 32, \ 41, \ ...$$
$$\color{red}{a_1 \ = \ 5, \ d \ = \ 14 \ - \ 5 \ = \ 9, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ 5 \ + \ 9 \ (n \ – \ 1)}$$
2) Find the explicit formula: $$-7, \ -2, \ 3, \ 8, \ 13, \ ...$$
$$\color{red}{a_1 \ = \ -7, \ d \ = \ -2 \ - \ (-7) \ = \ 5, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ -7 \ + \ 5 \ (n \ – \ 1)}$$
3) Find the explicit formula and $$a_{12}$$: $$-22, \ -16, \ -10, \ -4, \ 2, \ ...$$
$$\color{red}{a_1 \ = \ -22, \ d \ = \ -16 \ - \ (-22) \ = \ 6, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ -22 \ + \ 6 \ (n \ – \ 1) \ ⇒ \ a_{12} \ = \ -22 \ + \ 6 \ (12 \ – \ 1) \ = \ 44}$$
4) Find the explicit formula and $$a_{21}$$: $$a_{30} \ = \ 72, \ d \ = \ -5$$
$$\color{red}{a_{30} \ = \ 72, \ d \ = \ -5, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_{30} \ = \ a_1 \ + \ (-5) \ (30 \ – \ 1) \ ⇒ \ a_1 \ = \ a_{30} \ + \ 5(29) \ = \ 72 \ + \ 145 \ = \ 217}$$
$$\color{red}{⇒ \ a_n \ = \ 217 \ - \ 5 \ (n \ – \ 1) \ ⇒ \ a_{21} \ = \ 217 \ - \ 5(20) \ = \ 117}$$
5) Find the first $$10$$ terms: $$a_{15} \ = \ 50, \ d \ = \ 4.5$$
$$\color{red}{a_{15} \ = \ 50, \ d \ = \ 4.5, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_{15} \ = \ a_1 \ + \ 4.5 \ (15 \ – \ 1) \ ⇒ \ a_1 \ = \ 50 \ - \ 4.5(14) \ = \ -13}$$
$$\color{red}{⇒ -13, \ -8.5, \ -4, \ 0.5, \ 5, \ 9.5, \ 14, \ 18.5, \ 23, \ 27.5, \ ...}$$
6) Find the first $$7$$ terms: $$a_{41} \ = \ 178, \ d \ = \ 4$$
$$\color{red}{a_{41} \ = \ 178, \ d \ = \ 4, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_{41} \ = \ a_1 \ + \ 4 \ (41 \ – \ 1) \ ⇒ \ a_1 \ = \ 178 \ - \ 4(40) \ = \ 18}$$
$$\color{red}{⇒ 18, \ 22, \ 26, \ 30, \ 34, \ 38, \ 42, \ ...}$$
7) Find $$a_{30}$$: $$45, \ , \ 41.8, \ 38.6, \ 35.4, \ ...$$
$$\color{red}{a_1 \ = \ 45, \ d \ = \ 41.8 \ - \ 45 \ = \ -3.2, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ 45 \ - \ 3.2 \ (n \ – \ 1) \ ⇒ \ a_{30} \ = \ 45 \ - \ 3.2 \ (30 \ – \ 1) \ = \ -47.8}$$
8) Find $$a_{27}$$: $$a_{41} \ = \ 167, \ d \ = \ 6.4$$
$$\color{red}{a_{41} \ = \ 167, \ d \ = \ 6.4, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_{41} \ = \ a_1 \ + \ 6.4 \ (41 \ – \ 1) \ ⇒ \ a_1 \ = \ 167 \ - \ 6.4(40) \ = \ -98}$$
$$\color{red}{⇒ \ a_n \ = \ -98 \ + \ 6.4 \ (n \ – \ 1) \ ⇒ \ a_{27} \ = \ -98 \ + \ 6.4(26) \ = \ 68.4}$$
9) Find the explicit formula and the sum of the first five terms: $$a_1 \ = \ 10, \ d \ = \ 3$$
$$\color{red}{a_1 \ = \ 10, \ d \ = \ 3, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ 10 \ + \ 3 \ (n \ – \ 1)}$$
$$\color{red}{\sum_{k \ = \ 0}^{n \ - \ 1} \ (a \ + \ kd) \ = \ \frac{n}{2} \ (2a \ + \ (n \ - \ 1)d)}$$
$$\color{red}{⇒ \ \sum_{k \ = \ 0}^{4} \ (10 \ + \ 3k) \ = \ \frac{5}{2} \ (20 \ + \ (4)3) \ = \ \frac{5}{2} \times (32) \ = \ 80}$$
10) Find the explicit formula and the sum of the first $$16$$ terms: $$-55, \ -48, \ -41, \ -34, \ -27, \ ...$$
$$\color{red}{a_1 \ = \ -55, \ d \ = \ -48 \ - \ (-55) \ = \ 7, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ -55 \ + \ 7 \ (n \ – \ 1)}$$
$$\color{red}{\sum_{k \ = \ 0}^{n \ - \ 1} \ (a \ + \ kd) \ = \ \frac{n}{2} \ (2a \ + \ (n \ - \ 1)d)}$$
$$\color{red}{⇒ \ \sum_{k \ = \ 0}^{15} \ (-55 \ + \ 7k) \ = \ \frac{16}{2} \ (-110 \ + \ (15)7) \ = \ 8 \times (-5) \ = \ -40}$$
## Arithmetic Sequences Practice Quiz
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|
How to Factor Algebraic Equations
In mathematics, factoring is the act of finding the numbers or expressions that multiply together to make a given number or equation. Factoring is a useful skill to learn for the purpose of solving basic algebra problems; the ability to competently factor becomes almost essential when dealing with quadratic equations and other forms of polynomials. Factoring can be used to simplify algebraic expressions to make solving simpler. Factoring can even give you the ability to eliminate certain possible answers much more quickly than you would be able to by solving manually.[1]
Method 1
Method 1 of 3:
Factoring Numbers and Basic Algebraic Expressions
1. 1
Understand the definition of factoring when applied to single numbers. Factoring is conceptually simple, but, in practice, can prove to be challenging when applied to complex equations. Because of this, it's easiest to approach the concept of factoring by starting with simple numbers, then move on to simple equations before finally proceeding to more advanced applications. A given number's factors are the numbers that multiply to give that number. For example, the factors of 12 are 1, 12, 2, 6, 3 and 4, because 1 × 12, 2 × 6, and 3 × 4 all equal 12.[2]
• Another way to think of this is that a given number's factors are the numbers by which it is evenly divisible.
• Can you find all the factors of the number 60? We use the number 60 for a wide variety of purposes (minutes in an hour, seconds in a minute, etc.) because it's evenly divisible by a fairly wide range of numbers.
• The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
2. 2
Understand that variable expressions can also be factored. Just as lone numbers can be factored, so too can variables with numeric coefficients be factored. To do this, simply find the factors of the variable's coefficient. Knowing how to factor variables is useful for simplifying algebraic equations that the variables are a part of.[3]
• For example, the variable 12x can be written as a product of the factors of 12 and x. We can write 12x as 3(4x), 2(6x), etc., using whichever factors of 12 are best for our purposes.
• We can even go as far as to factor 12x multiple times. In other words, we don't have to stop with 3(4x) or 2(6x) - we can factor 4x and 6x to give 3(2(2x) and 2(3(2x), respectively. Obviously, these two expressions are equal.
3. 3
Apply the distributive property of multiplication to factor algebraic equations. Using your knowledge of how to factor both lone numbers and variables with coefficients, you can simplify simple algebraic equations by finding factors that the numbers and variables in an algebraic equation have in common. Usually, to make the equation as simple as possible, we try to search for the greatest common factor. This simplification process is possible because of the distributive property of multiplication, which states that for any numbers a, b, and c, a(b + c) = ab + ac.[4]
• Let's try an example problem. To factor the algebraic equation 12 x + 6, first, let's try to find the greatest common factor of 12x and 6. 6 is the biggest number that divides evenly into both 12x and 6, so we can simplify the equation to 6(2x + 1).
• This process also applies to equations with negatives and fractions. x/2 + 4, for instance, can be simplified to 1/2(x + 8), and -7x + -21 can be factored to -7(x + 3).
Method 2
Method 2 of 3:
1. 1
Ensure the equation is in quadratic form (ax2 + bx + c = 0). Quadratic equations are of the form ax2 + bx + c = 0, where a, b, and c are numeric constants and a does not equal 0 (note that a can equal 1 or -1). If you have an equation containing one variable (x) that has one or more terms of x to the second power, you can usually shift the terms in the equation around using basic algebraic operations to get 0 on one side of equals sign and ax2, etc. on the other side.[5]
• For example, let's consider the algebraic equation. 5x2 + 7x - 9 = 4x2 + x - 18 can be simplified to x2 + 6x + 9 = 0, which is in the quadratic form.
• Equations with greater powers of x, like x3, x4, etc. can't be quadratic equations. They are cubic equations, quartic equations, and so on, unless the equation can be simplified to eliminate these terms of x above the power of 2.
2. 2
In quadratic equations where a = 1, factor to (x+d )(x+e), where d × e = c and d + e = b. If your quadratic equation it is in the form x2 + bx + c = 0 (in other words, if the coefficient of the x2 term = 1), it's possible (but not guaranteed) that a relatively simple shortcut can be used to factor the equation. Find two numbers that both multiply to make c and add to make b. Once you find these two numbers d and e, place them in the following expression: (x+d)(x+e). These two terms, when multiplied together, produce your quadratic equation - in other words, they are your quadratic equation's factors.
• For example, let's consider the quadratic equation x2 + 5x + 6 = 0. 3 and 2 multiply together to make 6 and also add up to make 5, so we can simplify this equation to (x + 3)(x + 2).
• Slight variations on this basic shortcut exist for slight variations in the equation itself:
• If the quadratic equation is in the form x2-bx+c, your answer is in this form: (x - _)(x - _).
• If it is in the form x2+bx+c, your answer looks like this: (x + _)(x + _).
• If it is in the form x2-bx-c, you answer is in the form (x + _)(x - _).
• Note: the numbers in the blanks can be fractions or decimals. For example, the equation x2 + (21/2)x + 5 = 0 factors to (x + 10)(x + 1/2).
3. 3
If possible, factor by inspection. Believe it or not, for uncomplicated quadratic equations, one of the accepted means of factoring is simply to examine the problem, then just consider possible answers until you find the right one. This is also known as factoring by inspection. If the equation is in the form ax2+bx+c and a>1, your factored answer will be in the form (dx +/- _)(ex +/- _), where d and e are nonzero numerical constants that multiply to make a. Either d or e (or both) can be the number 1, though this is not necessarily so. If both are 1, you've essentially used the shortcut described above.[6]
• Let's consider an example problem. 3x2 - 8x + 4 at first seems intimidating. However, once we realize that 3 only has two factors (3 and 1), it becomes easier, because we know that our answer must be in the form (3x +/- _)(x +/- _). In this case, adding a -2 to both blank spaces gives the correct answer. -2 × 3x = -6x and -2 × x = -2x. -6x and -2x add to -8x. -2 × -2 = 4, so we can see that the factored terms in parentheses multiply to become the original equation.
4. 4
Solve by completing the square. In some cases, quadratic equations can be quickly and easily factored by using a special algebraic identity. Any quadratic equation of the form x2 + 2xh + h2 = (x + h)2. So, if, in your equation, your b value is twice the square root of your c value, your equation can be factored to (x + (sqrt(c)))2.[7]
• For example, the equation x2 + 6x + 9 fits this form. 32 is 9 and 3 × 2 is 6. So, we know that the factored form of this equation is (x + 3)(x + 3), or (x + 3)2.
5. 5
Use factors to solve quadratic equations. Regardless of how you factor your quadratic expression, once it is factored, you can find possible answers for the value of x by setting each factor equal to zero and solving. Since you're looking for values of x that cause your equation to equal zero, a value of x that makes either of your factors equal zero is a possible answer for your quadratic equation.[8]
• Let's return to the equation x2 + 5x + 6 = 0. This equation factored to (x + 3)(x + 2) = 0. If either of the factors equals 0, the entire equation equals 0, so our possible answers for x are the numbers that make (x + 3) and (x + 2) equal 0. These numbers are -3 and -2, respectively.
6. 6
Check your answers - some of them may be extraneous! When you've found your possible answers for x, plug them back in to your original equation to see if they are valid. Sometimes, the answers you find don't cause the original equation to equal zero when plugged back in. We call these answers extraneous and disregard them.[9]
• Let's plug -2 and -3 into x2 + 5x + 6 = 0. First, -2:
• (-2)2 + 5(-2) + 6 = 0
• 4 + -10 + 6 = 0
• 0 = 0. This is correct, so -2 is a valid answer.
• Now, let's try -3:
• (-3)2 + 5(-3) + 6 = 0
• 9 + -15 + 6 = 0
• 0 = 0. This is also correct, so -3 is also a valid answer.
Method 3
Method 3 of 3:
Factoring Other Forms of Equations
1. 1
If the equation is in the form a2-b2, factor it to (a+b)(a-b). Equations with two variables factor differently than basic quadratics. For any equation a2-b2 where a and b do not equal 0, the equation factors to (a+b)(a-b).
• For example, the equation 9x2 - 4y2 = (3x + 2y)(3x - 2y).
2. 2
If the equation is in the form a2+2ab+b2, factor it to (a+b)2. Note that, If the trinomial is in the form a2-2ab+b2, the factored form is slightly different: (a-b)2.
• The equation 4x2 + 8xy + 4y2 can be re-written as 4x2 + (2 × 2 × 2)xy + 4y2. We can now see that it's in the correct form, so we can say with confidence that our equation factors to (2x + 2y)2
3. 3
If the equation is in the form a3-b3, factor it to (a-b)(a2+ab+b2). Finally, it bears mentioning that cubics and even higher-order equations can be factored, though the factoring process quickly becomes prohibitively complicated.
• For instance, 8x3 - 27y3 factors to (2x - 3y)(4x2 + ((2x)(3y)) + 9y2)
Community Q&A
Search
• Question
How do I factor simple addition?
Find a common factor for the two numbers. For example, for 6 + 8, 6 and 8 share a factor of two. You can then rewrite it as 2 (3 + 4).
• Question
How would I factor -24x+4x^2?
Donagan
Both terms have 4x as a factor. Therefore, -24x + 4x² = 4x(-6 + x) = 4x(x - 6).
• Question
Can you demonstrate an easier problem? I have problems like 42r - 18 to factor.
Find a common factor of 42r and 18, e.g. 6. This number will go on the outside of the bracket so 6(...). Then divide the original number(s) you had by 6. We end up with 7r-3. This will go on the inside of the bracket to make the final answer: 6(7r-3). You can check your answer by expanding the brackets again: if the answer matches what you started with, then the answer is correct!
200 characters left
Tips
• If you have a trinomial in the form x2+bx+ (b/2)2, the factored form is (x+(b/2))2. (You may have this situation while completing the square.)
⧼thumbs_response⧽
• a2-b2 is factorable, a2+b2 isn't factorable.
⧼thumbs_response⧽
• Remember how to factor constants- it might help.
⧼thumbs_response⧽
Things You'll Need
• Paper
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• Math Book (if necessary)
Co-authored by:
This article was co-authored by David Jia. David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. This article has been viewed 642,266 times.
Co-authors: 44
Updated: September 30, 2022
Views: 642,266
Categories: Algebra
Article SummaryX
To factor a basic algebraic equation, start by looking for the largest factor that all the numbers in the equation have in common. For instance, if your equation is 6x + 2 = 0, the largest common factor that can be divided evenly into both terms on the left side of the equation is 2. Divide each term by the largest common factor, then rewrite the expression in the form a(b + c), where a is the largest common factor. After factoring, our example equation would become 2(3x + 1) = 0. If the equation contains variables that are common factors in multiple terms, you can factor those out as well. For instance, in the equation 4x² – 2x = 0, each term contains the common factors 2 and x. To factor the left side of the equation, divide each term by those factors to get 2x(2x – 1) = 0. If you want to learn how to factor quadratic equations, keep reading the article!
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# Perimeter and Area of an Equilateral Triangle – Formulas and Examples
The perimeter of an equilateral triangle is the length of the triangle’s outline. On the other hand, the area is a measure of the space occupied by the triangle. We can find the perimeter of an equilateral triangle by adding the lengths of its three sides, and we can find its area by multiplying the length of its base by its height and dividing by 2.
In this article, we will learn all about the perimeter and area of an equilateral triangle. We will explore its formulas and apply them to solve some practice problems.
##### GEOMETRY
Relevant for
Learning about the perimeter and area of an equilateral triangle.
See examples
##### GEOMETRY
Relevant for
Learning about the perimeter and area of an equilateral triangle.
See examples
## How to find the perimeter of an equilateral triangle?
To calculate the perimeter of an equilateral triangle, we have to add the lengths of its three sides. Recalling that an equilateral triangle has all its sides of the same length, we only have to multiply the length of one of the sides by 3:
where, a is the length of one side of the triangle.
This means that we only need to know the length of one of the sides of the equilateral triangle to calculate its perimeter.
## How to find the area of an equilateral triangle?
To calculate the area of any triangle, we can multiply its base by its height and divide by 2. In the case of equilateral triangles, we can use the following formula to calculate their area:
where, a is the length of one of the sides of the equilateral triangle.
### Proof of the formula for the area of an equilateral triangle
To prove the formula for the area of an equilateral triangle, we are going to use the following diagram, where we draw a bisector perpendicular to the base with height h:
We recall that the area of any triangle can be calculated with the following formula:
$latex \text{Area}= \frac{1}{2} \times \text{base} \times \text{height}$
Here, the base is equal to “a” and the height is equal to “h“.
Using the Pythagorean theorem to calculate the height, we have:
$latex {{a}^2}={{h}^2}+{{( \frac{a}{2})}^2}$
⇒ $latex {{h}^2}={{a}^2}- \frac{{{a}^2}}{4}$
⇒ $latex {{h}^2}=\frac{3{{a}^2}}{4}$
⇒ $latex h=\frac{\sqrt{3}~a}{2}$
Now that we have an expression for h, we can use it in the formula for the area of a triangle:
$latex \text{Area}= \frac{1}{2} \times \text{base} \times \text{height}$
$latex A=\frac{1}{2}\times a \times \frac{\sqrt{3}~a}{2}$
⇒ $latex A=\frac{\sqrt{3}~{{a}^2}}{4}$
## Perimeter and area of an equilateral triangle – Examples with answers
The following examples are solved using the perimeter and area formulas of an equilateral triangle. Try to solve the problems yourself before looking at the answer.
### EXAMPLE 1
Find the perimeter of an equilateral triangle that has sides with a length of 5 inches.
We use the formula for the perimeter with the value $latex a=5$. Therefore, we have:
$latex p=3a$
$latex p=3(5)$
$latex p=15$
The perimeter of the equilateral triangle is equal to 15 inches.
### EXAMPLE 2
What is the area of an equilateral triangle that has sides with a length of 10 feet?
We use the formula for the area with the length a=10:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex A= \frac{ \sqrt{3}}{4}({{10}^2})$
$latex A= \frac{ \sqrt{3}}{4}(100)$
$latex A=43.3$
The area of the equilateral triangle is equal to 43.3 ft².
### EXAMPLE 3
Find the perimeter of an equilateral triangle that has sides with a length of 8 yards.
Using the value $latex a=8$ in the formula for the perimeter, we have:
$latex p=3a$
$latex p=3(8)$
$latex p=24$
The perimeter of the equilateral triangle is equal to 24 yd.
### EXAMPLE 4
Find the area of an equilateral triangle that has sides with a length of 14 inches.
Applying the formula for the area with the given length, we have:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex A= \frac{ \sqrt{3}}{4}({{14}^2})$
$latex A= \frac{ \sqrt{3}}{4}(196)$
$latex A=84.87$
The area of the equilateral triangle is equal to 84.87 in².
### EXAMPLE 5
What is the perimeter of an equilateral triangle that has sides with a length of 15 inches?
Applying the formula for the perimeter with the value $latex a=15$:
$latex p=3a$
$latex p=3(15)$
$latex p=45$
The perimeter of the triangle is equal to 45 inches.
### EXAMPLE 6
What is the area of an equilateral triangle that has sides with a length of 15 feet?
We use the length $latex a=15$ in the formula for the area:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex A= \frac{ \sqrt{3}}{4}({{15}^2})$
$latex A= \frac{ \sqrt{3}}{4}(225)$
$latex A=97.43$
The area of the equilateral triangle is equal to 97.43 ft².
### EXAMPLE 7
Find the length of the sides of an equilateral triangle that has a perimeter of 39 ft.
In this example, we know the perimeter of the triangle, and we have to find the length of one side. Therefore, we use the perimeter formula and solve for a:
$latex p=3a$
$latex 39=3a$
$latex a=13$
The sides of the triangle have a length of 13 ft.
### EXAMPLE 8
Find the length of the sides of an equilateral triangle that has an area of 35.07 ft².
In this case, we know the area, and we need to find the length of the sides. Therefore, we use the formula for the area and solve for a:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex 35.07= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex 35.07=0.433{{a}^2}$
$latex {{a}^2}=81$
$latex a=9$
The sides of the triangle have a length of 9 ft.
### EXAMPLE 9
Find the length of the sides of an equilateral triangle that has a perimeter of 102 in.
We are going to use the formula for the perimeter with the value $latex p=102$ and solve for a:
$latex p=3a$
$latex 102=3a$
$latex a=34$
The length of one side of the triangle is 34 in.
### EXAMPLE 10
Find the length of the sides of an equilateral triangle that has an area of 73.18 ft².
We use the formula for the area with the given value and solve for a:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex 73.18= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex 73.18=0.433{{a}^2}$
$latex {{a}^2}=169$
$latex a=13$
The length of the sides of the triangle is 13 ft.
## Perimeter and area of an equilateral triangle – Practice problems
Solve the following problems by applying everything you have learned about the perimeter and area of equilateral triangles. Click “Check” to make sure that you got the correct answer.
Choose an answer
Choose an answer
Choose an answer
Choose an answer
Choose an answer
Choose an answer
## See also
Interested in learning more about perimeters and areas of geometric figures? Take a look at these pages:
LEARN MORE
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# Problems on Law of Inequality
Here we will solve various types of problems on law of inequality.
(i) If m + 6 > 15 then m - 6 > 3
(ii) If 4k > - 24 then - k > 6.
Solution:
(i) m + 6 > 15
⟹ m + 6 - 12 > 15 - 12, [Subtracting 12 from both sides]
⟹ m – 6 > 3
Therefore the sentence is true.
(ii) 4k > - 24
⟹ $$\frac{4k}{-4}$$ < $$\frac{-24}{-4}$$, [Dividing both sides by -4]
⟹ -k < 6
Therefore the sentence is false.
2. If 3z + 4 < 16 and z ∈ N then find z.
Solution:
3z + 4 < 16
⟹ 3z < 16 - 4, [Using the Rule of transferring a positive term]
⟹ 3z < 12
⟹ $$\frac{3z}{3}$$ < $$\frac{12}{3}$$, [using the Rule of division by a positive number]
⟹ z < 4
According to the given question z is natural number.
Therefore, z = 1, 2, and 3.
3. If (m – 1)(6 – m) > 0 and m ∈ N then find m.
Solution:
We know that xy > 0 then x > 0, y > 0 or x < 0, y < 0
Therefore, m – 1 > 0 and 6 – m > 0 ....................... (1)
or, m – 1 < 0 and 6 – m < 0....................... (2)
From (1) we get, m – 1 > 0 ⟹ m > 1,
and 6 – m > 0 ⟹ 6 > m
Therefore form (1), m > 1 as well as m < 6
From (2) we get, m – 1 <0 ⟹ m < 1
and 6 – m < 0 ⟹ 6 < m
Therefore form (2), m < 1 as well as m > 6
This is not possible because is m is less than 1, it cannot be greater than 6.
Thus (1) is possible and it gives 1 < m < 6, i.e., m lies between 1 and 6.
But according to the given question m is natural number. So, m = 2, 3 , 4 and 5.
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# Which rectangle A, B, C or D fits into the big square without turning?
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Hint: We all have played Puzzle games, and that is the base for this question, as you can see that parts are to be fixed in the big main square board. In this case we also have to give a look at the length and breadth of the rectangles, whether it fits the length and breadth of the rectangles in the main square board. Also, they have given a restriction that parts cannot be rotated, we have to fix it as it is.
In this problem our aim is to find which of the given rectangles fits into the bid square without rotating it. We first let us take option (a) A.
That is,
When we see the length and breadth of this rectangle A, its matches a place in the big square picture.
Thus, without rotating the rectangle A, we can fit it into the shaded part of the big square, as its length and breadth looks equal.
Let us take the next option (b) B. That is,
Even this rectangle B can also look as equal as the shaded portion of the big square. That is,
But the breadth of rectangle B is smaller than the length of the rectangle part in the big square. Thus, we cannot fit rectangle B into it.
Let us see the next option (c) C. That is,
It looks the same as the shaded portion of the big square.
But the length of the rectangle C doesn’t fit to the shaded rectangle part of the big square.
Let’s see the last option (d) D. that is,
It can be fixed in the shaded portion of the big square if it is rotated.
But the condition is we are not supposed to rotate the rectangle given. Thus, rectangle D cannot be fitted into it.
Hence, only rectangle A can be fitted into the big main square.
Note: In puzzle games, we cannot rotate the parts of it, if we rotate it and fix, it may or may not be fitted into it, but the picture doesn’t look good as it may be tilted and doesn’t look like the original picture. That is the concept they have used here. Thus we have to adhere to the condition given in the question.
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# Recursion. Understand how the Fibonacci series is generated Recursive Algorithms Write simple recursive algorithms Analyze simple recursive algorithms.
## Presentation on theme: "Recursion. Understand how the Fibonacci series is generated Recursive Algorithms Write simple recursive algorithms Analyze simple recursive algorithms."— Presentation transcript:
Recursion
Understand how the Fibonacci series is generated Recursive Algorithms Write simple recursive algorithms Analyze simple recursive algorithms Understand the drawbacks of recursion Name other recursive algorithms and data structures John Edgar2
What happens if you put a pair of rabbits in a field? More rabbits! Assume that rabbits take one month to reach maturity and that Each pair of rabbits produces another pair of rabbits one month after mating. John Edgar3
How many pairs of rabbits are there after 5 months? Month 1: start – 1 Month 2: the rabbits are now mature and can mate – 1 Month 3: – the first pair give birth to two babies – 2 Month 4: the first pair give birth to 2 babies, the pair born in month 3 are now mature – 3 Month 5: the 3 pairs from month 4, and 2 new pairs – 5 John Edgar4
AAfter 5 months there are 5 pairs of rabbits ii.e. the number of pairs at 4 months (3) plus the number of pairs at 3 months (2) WWhy? WWhile there are 3 pairs of bunnies in month 4 only 2 of them are able to mate tthe ones alive in month 3 TThis series of numbers is called the Fibonacci series 5 month: pairs: 1 1 2 2 3 3 4 4 5 5 6 6 1 1 1 1 2 2 3 3 5 5 8 8
TThe n th number in the Fibonacci series, fib(n), is: 00 if n = 0, and 1 if n = 1 ffib(n – 1) + fib(n – 2) for any n > 1 ee.g. what is fib(23) EEasy if we only knew fib(22) and fib(21) TThe answer is fib(22) + fib(21) WWhat happens if we actually write a function to calculate Fibonacci numbers like this? John Edgar6
Let's write a function just like the formula fib(n) = 0 if n = 0, 1 if n = 1, otherwise fib(n) = fib(n – 1) + fib(n – 2) John Edgar7 int fib(int n){ if(n == 0 || n == 1){ return n; }else{ return fib(n-1) + fib(n-2); } int fib(int n){ if(n == 0 || n == 1){ return n; }else{ return fib(n-1) + fib(n-2); } The function calls itself C++
The Fibonacci function is recursive A recursive function calls itself Each call to a recursive method results in a separate call to the method, with its own input Recursive functions are just like other functions The invocation is pushed onto the call stack And removed from the call stack when the end of a method or a return statement is reached Execution returns to the previous method call John Edgar8
int fib(int n){ if(n == 0 || n == 1) return n; else return fib(n-1) + fib(n-2); } John Edgar9 fib(5) fib(4) fib(3) fib(2) fib(1) fib(0) fib(2) fib(1) fib(0) fib(1) fib(2) fib(1) fib(0) fib(1) 1111100011 21 3 2 5
When a function is called it is pushed onto the call stack This applies to each invocation of that function When a recursive call is made execution switches to that method call The call stack records the line number of the previous method where the call was made from Once a method call execution finishes, returns to the previous invocation John Edgar10
January 2010Greg Mori11
Recursive functions do not use loops to repeat instructions But use recursive calls, in if statements Recursive functions consist of two or more cases, there must be at least one Base case, and one Recursive case John Edgar12
The base case is a smaller problem with a simpler solution This problem’s solution must not be recursive ▪ Otherwise the function may never terminate There can be more than one base case John Edgar13
The recursive case is the same problem with smaller input The recursive case must include a recursive function call There can be more than one recursive case John Edgar14
Define the problem in terms of a smaller problem of the same type The recursive part e.g. return fib(n-1) + fib(n-2); And the base case where the solution can be easily calculated This solution should not be recursive e.g. if (n == 0 || n == 1) return n; John Edgar15
How can the problem be defined in terms of smaller problems of the same type? By how much does each recursive call reduce the problem size? By 1, by half, …? What are the base cases that can be solved without recursion? Will a base case be reached as the problem size is reduced? John Edgar16
January 2010Greg Mori17
Linear Search Binary Search Assume sorted array John Edgar18
John Edgar19 C++ int linSearch(int *arr, int n, int x){ for (int i=0; i < n; i++){ if(x == arr[i]){ return i; } } //for return -1; //target not found } int linSearch(int *arr, int n, int x){ for (int i=0; i < n; i++){ if(x == arr[i]){ return i; } } //for return -1; //target not found } The algorithm searches the array one element at a time using a for loop
Base cases Target is found at first position in array The end of the array is reached Recursive case Target not found at first position ▪ Search again, discarding the first element of the array John Edgar20
John Edgar21 int linSearch(int *arr, int n, int x){ return recLinSearch(arr,n,0,x); } int recLinSearch(int *arr, int n, int i, int x){ if (i >= n){ return -1; } else if (x == arr[i]){ return i; } else return recLinSearch(arr, n, i + 1, x); } int linSearch(int *arr, int n, int x){ return recLinSearch(arr,n,0,x); } int recLinSearch(int *arr, int n, int i, int x){ if (i >= n){ return -1; } else if (x == arr[i]){ return i; } else return recLinSearch(arr, n, i + 1, x); } C++
Of course, if it’s a sorted array we wouldn’t do linear search John Edgar22
Each sub-problem searches a subarray Differs only in the upper and lower array indices that define the subarray Each sub-problem is smaller than the last one In the case of binary search, half the size There are two base cases When the target item is found and When the problem space consists of one item ▪ Make sure that this last item is checked John Edgar23
John Edgar24 C++ int binSearch(int *arr, int lower, int upper, int x){ int mid = (lower + upper) / 2; if (lower > upper){ return - 1; //base case } else if(arr[mid] == x){ return mid; //second base case } else if(arr[mid] < x){ return binSearch(arr, mid + 1, upper, x); } else { //arr[mid] > target return binSearch(arr, lower, mid - 1, x); } int binSearch(int *arr, int lower, int upper, int x){ int mid = (lower + upper) / 2; if (lower > upper){ return - 1; //base case } else if(arr[mid] == x){ return mid; //second base case } else if(arr[mid] < x){ return binSearch(arr, mid + 1, upper, x); } else { //arr[mid] > target return binSearch(arr, lower, mid - 1, x); }
January 2010Greg Mori25
Merge Sort Quicksort John Edgar26
January 2010Greg Mori27
What’s the easiest list to sort? A list of 1 number John Edgar28
Let’s say I have 2 sorted lists of numbers How can I merge them into 1 sorted list? John Edgar29 1 1 3 3 5 5 12 22 23 42 99 output 1 1 12 22 23 3 3 5 5 42 99 List 1List 2
If I have a list of n numbers, how should I sort them? I know two things How to sort a list of 1 number How to merge 2 sorted lists of numbers into 1 sorted list Smells like recursion John Edgar30
John Edgar31 mergeSort (array) if (array is length 1) // base case, one element return the array else arr1 = mergeSort(first half of array) arr2 = mergeSort(second half of array) return merge(arr1,arr2) mergeSort (array) if (array is length 1) // base case, one element return the array else arr1 = mergeSort(first half of array) arr2 = mergeSort(second half of array) return merge(arr1,arr2)
What is the time complexity of a merge? John Edgar32 1 1 3 3 5 5 12 22 23 42 99 output 1 1 12 22 23 3 3 5 5 42 99 List 1List 2
How many recursive steps are there? How large are the merges at each recursive step? Merge takes O(n) time for n elements John Edgar33
John Edgar34 2341338107191145 Sort entire array 2341338107191145 Sort halves 2341338107191145 Sort quarters 2341338107191145 Sort eighths 2341338107191145 Sorted quarters 2333418107111945 Sorted halves 0711192333414581 Sorted entire array
John Edgar35 2341338107191145 Sort entire array 2341338107191145 Sort halves 2341338107191145 Sort quarters 2341338107191145 Sort eighths How many recursive steps are there? How large are the merges at each recursive step? Merge takes O(n) time for n elements
John Edgar36 2341338107191145 Sort entire array 2341338107191145 Sort halves 2341338107191145 Sort quarters 2341338107191145 Sort eighths How many recursive steps are there? O(log n) steps: split array in half each time How large are the merges at each recursive step? In total, merge n elements each step Time complexity is O(n log n)
Mergesort Best case: O(n(log 2 n)) Average case: O(n(log 2 n)) Worst case: O(n(log 2 n)) John Edgar37
Quicksort is a more efficient sorting algorithm than either selection or insertion sort It sorts an array by repeatedly partitioning it We will go over the basic idea of quicksort and an example of it See text / on-line resources for details John Edgar39
Partitioning is the process of dividing an array into sections (partitions), based on some criteria "Big" and "small" values Negative and positive numbers Names that begin with a-m, names that begin with n-z Darker and lighter pixels Quicksort uses repeated partitioning to sort an array John Edgar40
John Edgar41 Partition this array into small and big values using a partitioning algorithm 31 12 07 23 93 02 11 18
John Edgar42 Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot 3112072393021118 smalls < 18 bigs > 18 pivot 18
John Edgar43 31120723930211 18 arr[low ] is greater than the pivot and should be on the right, we need to swap it with something We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high Partition this array into small and big values using a partitioning algorithm
John Edgar44 31120723930211 18 arr[low ] (31) is greater than the pivot and should be on the right, we need to swap it with something arr[high] (11) is less than the pivot so swap with arr[low ] Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high
John Edgar45 31120723930211183111 Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high
John Edgar46 120723930218 repeat this process until: 31230211 Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high 1207
John Edgar47 12 07 93 18 repeat this process until: 31 23 02 11 high and low are the same Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high
John Edgar48 repeat this process until: high and low are the same We'd like the pivot value to be in the centre of the array, so we will swap it with the first item greater than it Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high 12 07 93 18 31 23 02 11 93 18
John Edgar49 smallsbigs pivot Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high 12 07 93 18 31 23 02 11
John Edgar50 Use the same algorithm to partition this array into small and big values 00 08 07 01 06 02 05 09 bigs! pivot 00 08 07 01 06 02 05 09 smalls
John Edgar51 Or this one: 09 08 07 06 05 04 02 01 bigs pivot 01 08 07 06 05 04 02 09 smalls
The quicksort algorithm works by repeatedly partitioning an array Each time a subarray is partitioned there is A sequence of small values, A sequence of big values, and A pivot value which is in the correct position Partition the small values, and the big values Repeat the process until each subarray being partitioned consists of just one element John Edgar52
How long does quicksort take to run? Let's consider the best and the worst case These differ because the partitioning algorithm may not always do a good job Let's look at the best case first Each time a subarray is partitioned the pivot is the exact midpoint of the slice (or as close as it can get) ▪ So it is divided in half What is the running time? John Edgar53
John Edgar54 08 01 02 07 03 06 04 05 bigs pivot 04 01 02 03 05 06 08 07 smalls First partition
John Edgar55 big1 pivot1 02 01 04 05 06 08 sm1 04 01 02 03 05 06 08 07 Second partition 07 03 pivot1pivot2 big2sm2
John Edgar56 pivot1 02 03 04 05 06 07 08 Third partition 02 01 03 04 05 06 07 08 pivot1done 01
Each subarray is divided exactly in half in each set of partitions Each time a series of subarrays are partitioned around n comparisons are made The process ends once all the subarrays left to be partitioned are of size 1 How many times does n have to be divided in half before the result is 1? log 2 (n) times Quicksort performs around n * log 2 (n) operations in the best case John Edgar57
John Edgar58 09 08 07 06 05 04 02 01 bigs pivot 01 08 07 06 05 04 02 09 smalls First partition
John Edgar59 bigs pivot 01 08 07 06 05 04 02 09 smalls 01 08 07 06 05 04 02 09 Second partition
John Edgar60 bigs pivot 01 02 07 06 05 04 08 09 Third partition 01 08 07 06 05 04 02 09
John Edgar61 pivot 01 02 07 06 05 04 08 09 smalls Fourth partition 01 02 07 06 05 04 08 09
John Edgar62 bigs pivot 01 02 04 06 05 07 08 09 Fifth partition 01 02 07 06 05 04 08 09
John Edgar63 pivot 01 02 04 06 05 07 08 09 smalls Sixth partition 01 02 04 06 05 07 08 09
John Edgar64 pivot 01 02 04 05 06 07 08 09 Seventh(!) partition 01 02 04 06 05 07 08 09
Every partition step results in just one partition on one side of the pivot The array has to be partitioned n times, not log 2 (n) times So in the worst case quicksort performs around n 2 operations The worst case usually occurs when the array is nearly sorted (in either direction) John Edgar65
With a large array we would have to be very, very unlucky to get the worst case Unless there was some reason for the array to already be partially sorted ▪ In which case first randomize the position of the array elements! The average case is much more like the best case than the worst case John Edgar66
January 2010Greg Mori67
Recursive algorithms have more overhead than similar iterative algorithms Because of the repeated method calls This may cause a stack overflow when the call stack gets full It is often useful to derive a solution using recursion and implement it iteratively Sometimes this can be quite challenging! John Edgar68
Some recursive algorithms are inherently inefficient e.g. the recursive Fibonacci algorithm which repeats the same calculation again and again Look at the number of times fib(2) is called Such algorithms should be implemented iteratively Even if the solution was determined using recursion John Edgar69
It is useful to trace through the sequence of recursive calls This can be done using a recursion tree Recursion trees can be used to determine the running time of algorithms Annotate the tree to indicate how much work is performed at each level of the tree And then determine how many levels of the tree there are John Edgar70
January 2010Greg Mori71
Recursion is similar to induction Recursion solves a problem by Specifying a solution for the base case and Using a recursive case to derive solutions of any size from solutions to smaller problems Induction proves a property by Proving it is true for a base case and Proving that it is true for some number, n, if it is true for all numbers less than n John Edgar72
Prove, using induction that the algorithm returns the values fact(0) = 0! =1 fact(n) = n! = n * (n – 1) * … * 1 if n > 0 John Edgar73 int fact (int x){ if (x == 0){ return 1; } else return n * fact(n – 1); } int fact (int x){ if (x == 0){ return 1; } else return n * fact(n – 1); } C++
Basis: Show that the property is true for n = 0, i.e. that fact(0) returns 1 This is true by definition as fact(0) is the base case of the algorithm and returns 1 Establish that the property is true for an arbitrary k implies that it is also true for k + 1 Inductive hypothesis: Assume that the property is true for n = k, that is assume that fact(k) = k * ( k – 1) * ( k – 2) * … * 2 * 1 John Edgar74
IInductive conclusion: Show that the property is true for n = k + 1, i.e., that fact(k + 1) returns ((k + 1) * k * (k – 1) * (k – 2) * … * 2 * 1 BBy definition of the function: fact(k + 1) returns ((k + 1) * fact(k) – the recursive case AAnd by the inductive hypothesis: fact(k) returns kk * (k – 1) * (k – 2) * … * 2 * 1 TTherefore fact(k + 1) must return ((k + 1) * k * (k – 1) * (k – 2) * … * 2 * 1 WWhich completes the inductive proof John Edgar75
Recursive sum Towers of Hanoi – see text Eight Queens problem – see text Sorting Mergesort Quicksort John Edgar76
Linked Lists are recursive data structures They are defined in terms of themselves There are recursive solutions to many list methods List traversal can be performed recursively Recursion allows elegant solutions of problems that are hard to implement iteratively ▪ Such as printing a list backwards John Edgar77
January 2010Greg Mori78
Recursion as a problem-solving tool Identify base case where solution is simple Formulate other cases in terms of smaller case(s) Recursion is not always a good implementation strategy Solve the same problem many times Function call overhead Recursion and induction Induction proves properties in a form similar to how recursion solves problems John Edgar79
Carrano Ch. 2, 5 John Edgar80
Download ppt "Recursion. Understand how the Fibonacci series is generated Recursive Algorithms Write simple recursive algorithms Analyze simple recursive algorithms."
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# Search by Topic
#### Resources tagged with Addition & subtraction similar to Tricky Track:
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### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Roll These Dice
##### Stage: 2 Challenge Level:
Roll two red dice and a green dice. Add the two numbers on the red dice and take away the number on the green. What are all the different possibilities that could come up?
### The Twelve Pointed Star Game
##### Stage: 2 Challenge Level:
Have a go at this game which involves throwing two dice and adding their totals. Where should you place your counters to be more likely to win?
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Connect Three
##### Stage: 3 and 4 Challenge Level:
Can you be the first to complete a row of three?
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Got it Article
##### Stage: 2 and 3
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
##### Stage: 1 and 2 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Cayley
##### Stage: 3 Challenge Level:
The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Cubes Within Cubes
##### Stage: 2 and 3 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Asteroid Blast
##### Stage: 2 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### More Carroll Diagrams
##### Stage: 2 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Two and Two
##### Stage: 2 and 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Number Daisy
##### Stage: 3 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### Totality
##### Stage: 1 and 2 Challenge Level:
This is an adding game for two players.
##### Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Cycling Squares
##### Stage: 2 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Tug Harder!
##### Stage: 2 Challenge Level:
In this game, you can add, subtract, multiply or divide the numbers on the dice. Which will you do so that you get to the end of the number line first?
### Reach 100
##### Stage: 2 and 3 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### On Target
##### Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### Hubble, Bubble
##### Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Super Value Shapes
##### Stage: 2 Challenge Level:
If each of these three shapes has a value, can you find the totals of the combinations? Perhaps you can use the shapes to make the given totals?
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### How Old?
##### Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Sitting Round the Party Tables
##### Stage: 1 and 2 Challenge Level:
Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.
|
## Poissonizing the Multinomial
### Poissonizing the Multinomial¶
Bernoulli trials come out in one of two ways. But many trials come out in multiple different ways, all of which we might want to track. A die can land six different ways. A jury member can have one of several different identities. In general, an individual might belong to one of several classes.
The multinomial distribution is an extension of the binomial to the case where there are more than two possible outcomes of each trial. Let's look at it first in an example, and then we will define it in general.
A box contains 2 blue tickets, 5 green tickets, and 3 red tickets. Fifteen draws are made at random with replacement. To find the chance that there are 4 blue, 9 green, and 2 red tickets drawn, we could start by writing all possible sequences of 4 B's, 9 G's, and 2 R's.
Each sequence would have chance $0.2^4 0.5^9 0.3^2$, so all we need to complete the probability calculation is the number of sequences we could write.
• There are $\binom{15}{4}$ ways of choosing places to write the B's.
• For each of these ways, there are $\binom{11}{9}$ ways of choosing 9 of the remaining 11 places to write the G's.
• The remaining 2 places get filled with R's.
So
\begin{align*} P(\text{4 blue, 9 green, 2 red}) &= \binom{15}{4} \cdot \binom{11}{9} 0.2^4 0.5^9 0.3^2 \\ \\ &= \frac{15!}{4!11!} \cdot \frac{11!}{9!2!} 0.2^4 0.5^9 0.3^2 \\ \\ &= \frac{15!}{4!9!2!} 0.2^4 0.5^9 0.3^2 \end{align*}
Notice how this simply extends the binomial probability formula by including a third category in exactly the same way.
Analogously or by induction, you can extend the formula to any finite number of categories or classes.
### Multinomial Distribution¶
Fix a positive integer $n$. Suppose we are running $n$ i.i.d. trials where each trial can result in one of $k$ classes. For each $i = 1, 2, \ldots, k$, let the chance of getting Class $i$ on a single trial be $p_i$, so that $\sum_{i=1}^k p_i = 1$.
For each $i = 1, 2, \ldots , k$, let $N_i$ be the number of trials that result in Class $i$, so that $N_1 + N_2 + \ldots + N_k = n$.
Then the joint distribution of $N_1, N_2, \ldots , N_k$ is given by
$$P(N_1 = n_1, N_2 = n_2, \ldots , N_k = n_k) = \frac{n!}{n_1!n_2! \ldots n_k!}p_1^{n_1}p_2^{n_2} \cdots p_k^{n_k}$$
where $n_i \ge 0$ for $1 \le i \le k$ and $\sum_{i=1}^k n_i = n$.
When there just two classes then $k = 2$ and the formula reduces to the familiar binomial formula, written as the joint distribution of the number of successes and the number of failures:
$$P(N_1 = n_1, N_2 = n_2) = \frac{n!}{n_1!n_2!} p_1^{n_1}p_2^{n_2} ~~ \text{where } p_1+p_2=1 \text{ and } n_1+n_2=n$$
Notice that the marginal distribution of each $N_i$ is binomial $(n, p_i)$. You don't have to sum the joint distributions to work this out. $N_i$ is the number of Class $i$ individuals in the sample; each sampled individual is in Class $i$ with probability $p_i$; and there are $n$ independent draws. That's the binomial setting.
### Poissonization¶
If you replace the fixed number $n$ of trials by a Poisson $(\mu)$ random number of trials, then the multinomial gets Poissonized as follows:
• For each $i = 1, 2, \ldots , k$, the distribution of $N_i$ is Poisson $(\mu p_i)$.
• The counts $N_1, N_2, \ldots , N_k$ in the $k$ different categories are mutually independent.
We won't go through the proof, which is a straightforward extension of the proof in the case $k=2$ given in an earlier section. Rather, we will look at why the result matters.
When the number of trials is fixed, $N_1, N_2, \ldots , N_k$ are all dependent on each other in complicated ways. But when you let the sample size be a Poisson random variable, then the independence of the counts $N_1, N_2, \ldots , N_k$ lets you quickly calculate the chance of any particular configuration of classes in the sample.
For example, if in your population the distribution of classes is as follows:
• Class 1: 20%
• Class 2: 30%
• Class 3: 50%
and you draw $N$ independent times where $N$ has the Poisson $(20)$ distribution, then the chance that you will get at least 3 individuals in each class is about 71.27%.
(1 - stats.poisson.cdf(2, 4))*(1-stats.poisson.cdf(2, 6))*(1-stats.poisson.cdf(2, 10))
0.71270362753222383
The number of factors in the answer is equal to the number of classes, unlike the inclusion-exclusion formula in which the amount of work increases much more with each additional class, as you have seen in exercises.
This helps data scientists tackle questions like, "How many times must I sample so that my chance of seeing at least one individual of each class exceeds a given threshold?" The answer depends on the distribution of classes in the population, of course, but allowing the sample size be a Poisson random variable can make calculations much more tractable. For applications, see for example the Abstract and References of this paper.
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Tips and Tricks: Height & Distance
# Height and Distance Tips and Tricks for Government Exams
## Height Definition
The measurement of an object in the vertical direction known as Height.
## Distance Definition
The measurement of an object from a particular point in the horizontal direction Known as Distance .
### How to solve word problems that involve angle of elevation or depression Shortcuts and Easy Tips & Tricks to solve questions
• Step 1: Draw a sketch of the situation given.
• Step 2: Mark in the given angle of elevation or depression and other information.
• Step 3: Use trigonometry to find the required missing length.
## Height and Distance Tips and Tricks & Shortcuts
There are two primary ways to express the angular relationship between the height of an object and an observer's line of sight:
• Angle of Elevation: This refers to the angle formed between the horizontal line and the line of sight from the observer's eye to an object situated above the observer.
• Angle of Depression: When the object is positioned below the observer's level, the angle between the horizontal line and the observer's line of sight to the object is termed the angle of depression.
### Type 1: Find the distance/height/base/length when angle is given
Example: A boy is flying a kite in the evening. The thread of the kite was 120 m long and the angle of elevation with the boy’s eyes was 30°. Find the height of the kite?
(a) 30 m
(b) 60 m
(c) 40 m
(d) 55 m
Ans:
(b)
Sin 30° = Perpendicular/Hypotenuse = AC/AB
1/2 = ℎ/120
h = 60 m
### Type 2: Find the angle when distance/height/base/length is given
Example: Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 m long?
(a) 45°
(b) 60°
(c) 30°
(d) 90°
Ans:
(b)
Let AB be the pole and CB be the shadow
Given that AB = 18 m and CB = 6√3
Let the angle of elevation, ACB = θ
From the right ΔABC, Tan θ = Perpendicular/Base =
Therefore, θ = tan–1(√3) = 60°
The document Height and Distance Tips and Tricks for Government Exams is a part of the SSC CGL Course Quantitative Aptitude for SSC CGL.
All you need of SSC CGL at this link: SSC CGL
## Quantitative Aptitude for SSC CGL
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# Numerical Analysis. TOPIC Interpolation TOPIC Interpolation.
## Presentation on theme: "Numerical Analysis. TOPIC Interpolation TOPIC Interpolation."— Presentation transcript:
Numerical Analysis
TOPIC Interpolation TOPIC Interpolation
Interpolation is the process of estimating the value of function for any intermediate value of the variable with the help of its given set of values. Let us assume that the function y=f(x) is known for certain values of x say a for x 0,x 1,x 2,………x n. As f(x 0 ),f(x 1 ),……..f(x n ). The process of finding the value of f(x) corresponding to x=x i. Where x 0 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/8/2271909/slides/slide_3.jpg", "name": "Interpolation is the process of estimating the value of function for any intermediate value of the variable with the help of its given set of values.", "description": "Let us assume that the function y=f(x) is known for certain values of x say a for x 0,x 1,x 2,………x n. As f(x 0 ),f(x 1 ),……..f(x n ). The process of finding the value of f(x) corresponding to x=x i. Where x 0
1.Finite Difference Operators 2.Newton’s Forward Difference Interpolation Formula 3.Newton’s Backward Difference Interpolation Formula 4.Lagrange’s Interpolation Formula
rth forward difference kth backward difference
Thus Similarly
Shift operator, E
The inverse operator E -1 is defined as Similarly,
Average Operator,
Differential Operator, D
Important Results
Newton’s Forward Difference Interpolation Formula
This is known as Newton’s forward difference formula for interpolation, which gives the value of f (x 0 + ph) in terms of f (x 0 ) and its leading differences.
This formula is also known as Newton-Gregory forward difference interpolation formula. Here p=(x-x 0 )/h. An alternate expression is
Here,
The required cubic polynomial.
Let y = f (x) be a function which takes on values f (x n ), f (x n -h), f (x n -2h), …, f (x 0 ) corresponding to equispaced values x n, x n -h, x n -2h, …, x 0. Suppose, we wish to evaluate the function f (x) at (x n + ph),
Binomial expansion yields,
This formula is known as Newton’s backward interpolation formula. This formula is also known as Newton’s-Gregory backward difference interpolation formula.
Example:- For the following table of values, estimate f (7.5).
Difference Table
In this problem,
-3
and
Newton’s interpolation formula gives Therefore,
Let y = f (x) be a function which takes the values, y 0, y 1,…y n corresponding to x 0, x 1, …x n. Since there are (n + 1) values of y corresponding to (n + 1) values of x, we can represent the function f (x) by a polynomial of degree n. DERIVATION:-
or in the form
Here, the coefficients a k are so chosen as to satisfy this equation by the (n + 1) pairs (x i, y i ). Thus we get Therefore,
and
The Lagrange’s formula for interpolation
We can easily observe that, and Thus introducing Kronecker delta notation
Example:- Find Lagrange’s interpolation polynomial fitting the points y(1) = -3, y(3) = 0, y(4) = 30, y(6) = 132. Hence find y(5).
Using Lagrange’s interpolation formula, we have
On simplification, we get which is required Lagrange’s interpolation polynomial. Now, y(5) = 75.
Solution:- Using Lagrange’s formula,
Therefore
1. Evaluate : (x.log x) 2. Evaluate : (sin2x.cos4x) 3. Estimate the missing term: x01234 Y139_81
4.Derive newton forward formula for interpolation. 5. Derive newton backward formula for interpolation.
1.Evaluate (x 2 +sinx), the interval of difference being . 2. Find the lowest degree polynomial which satisfies the following table: X012345 F(x)038152435
3.The population of a town is as follow: YEAR196119711981199120012011 POPULATION202435404251
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# Solving problems involving surface area, volume and capacity
## Worked Example 17.15: Calculating the height of a prism
The volume of the prism is $$4\ 032\text{ cm}^3$$. The area of the shaded region is $$224 \text{ cm}^2$$. Calculate the value of the length, $$x$$.
### Use the general formula for volume to find height of prism.
$\text{Volume of prism} = \text{area of base} \times h$
We know the volume of the prism, and we are given the area of the base of the prism, so we can solve for the height, $$x$$:
\begin{align} 4\ 032 &= 224 \times x \\ \frac{4\ 032}{224} &= x \\ 18 &= x \end{align}
The height of the prism $$x = 18 \text{ cm}$$.
## Worked Example 17.16: Calculating the surface area of a rectangular prism
Dintle wants to paint a wooden crate. If she uses $$\text{0,25} \text{ ml}$$ of paint per square centimetre, calculate how much paint she will need to paint the crate. Assume that Dintle will paint every outside surface of the crate. Express your answer in litres.
### Find the surface area of the crate.
\begin{align} \text{Surface area of crate} &= 2(\text{area of sides}) + 2(\text{area of ends}) + 2(\text{area of top/bottom}) \\ &= 2(85 \times 32) + 2(48 \times 32) + 2(85 \times 48) \\ &= 2(2\ 720) + 2(1\ 536) + 2(4\ 080) \\ &= 5\ 440 + 3\ 072 + 8\ 160 \\ &= 16\ 672 \end{align}
The surface area of the crate is $$16\ 672 \text{ cm}^2$$.
### Calculate the volume of paint needed.
Dintle will use $$\text{0,25} \text{ ml}$$ of paint per square centimetre.
So, we can calculate:
\begin{align} \text{Volume of paint} &= \text{0,25} \times \text{16 672} \\ &= 4\ 168 \end{align}
Volume of paint needed is $$\text{4 168} \text{ ml}$$.
\begin{align} \text{4 168} \text{ ml} &= \frac{\text{4 168}}{\text{1 000}} \\ &= \text{4,2} \text{ litres} \end{align}
Dintle will need $$\text{4,2} \text{ litres}$$ of paint.
## Worked Example 17.17: Calculating the volume of a complex solid
Calculate the volume of the solid. All measurements are in centimetres.
### Find the volume of the cuboid.
\begin{align} \text{Volume of cuboid} &= l \times b \times h \\ &= 8 \times 15 \times 11 \\ &= 1\ 320 \end{align}
### Calculate the volume of the triangular prism.
\begin{align} \text{Volume of triangular prism} &= \frac{1}{2}(b \times h) \\ &= \frac{1}{2}(8 \times 6) \\ &= 24 \end{align}
\begin{align} \text{Volume of solid} &= \text{Volume of cuboid} + \text{Volume of the triangular prism} \\ &= 1\ 320 + 24 \\ &= 1\ 344 \text{ cm}^3 \end{align}
|
# AP Statistics Curriculum 2007 Estim L Mean
## General Advance-Placement (AP) Statistics Curriculum - Estimating a Population Mean: Large Samples
### Point Estimation of a Population Mean
For any process, the population mean may be estimated by a (large) sample average. That is the smaple average $\overline{X_n}={1\over n}\sum_{i=1}^n{X_i}$, constructed from a random sample of the procees {$X_1, X_2, X_3, \cdots , X_n$}, is an unbiased estimate of the population mean μ, if it exists! Note that the sample average may be susseptible to outliers.
### Interval Estimation of a Population Mean
For large samples, interval estimation of the population means (or Confidence intervals) are constructed as follows. Choose a confidence level (1 − α)100%, where α is small (e.g., 0.1, 0.05, 0.025, 0.01, 0.001, etc.). Then a (1 − α)100% confidence interval for μ will be
Failed to parse (unknown function\lapha): CI(\lapha): \overline{x} \pm z_{\alpha\over 2} E,
• The margine of error E is defined as
$E = \begin{cases}{\sigma\over\sqrt{n}},& \texttt{for-known}-\sigma,\\ {SE},& \texttt{for-unknown}-\sigma.\end{cases}$
• The Standard Error of the estimate $\overline {x}$ is defined by Failed to parse (syntax error): SE(\overline {x}) = {1\over \sqrt{n} \sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}
• $z_{\alpha\over 2}$ is the critical value for a Standard Normal distribution at ${\alpha\over 2}$.
### Example
Market researchers use the number of sentences per advertisement as a measure of readability for magazine advertisements. A random sample of the number of sentences found in 30 magazine advertisements is listed. Use this sample to find point estimate for the population mean μ.
16 9 14 11 17 12 99 18 13 12 5 9 17 6 11 17 18 20 6 14 7 11 12 5 18 6 4 13 11 12
A confidence interval estimate of μ is a range of values used to estimate a population parameter (interval estimates are normally used more than point estimates because it is very unlikely that the sample mean would match exactly with the population mean) The interval estimate uses a margin of error about the point estimate. For example if you have a point estimate of 12. 25 with a margin of error of 1.75, then the interval estimate would be (10.5 to 14).
Before you find an interval estimate, you should first determine how confident you want to be that your interval estimate contains the population mean.
80% confidence (0.80), α = 0.1, z = 1.28
90% confidence (0.90), α = 0.05, z = 1.645
95% confidence (0.95), α = 0.025, z = 1.96
99% confidence (0.99), α = 0.005, z = 2.575
#### Known Variance
Suppose that we know the variance for the number of sentences per advertisement example above is known to be 256 (so the population standard deviation is σ = 16).
• For α = 0.1, the 90%CI(μ) is constructed by:
$\overline{x}\pm 1.28{16\over \sqrt{30}}=14.77 \pm 1.28{16\over \sqrt{30}}=[11.03;18.51]$
• For α = 0.05, the 90%CI(μ) is constructed by:
$\overline{x}\pm 1.645{16\over \sqrt{30}}=14.77 \pm 1.645{16\over \sqrt{30}}=[9.96;19.57]$
• For α = 0.005, the 90%CI(μ) is constructed by:
$\overline{x}\pm 2.575{16\over \sqrt{30}}=14.77 \pm 2.575{16\over \sqrt{30}}=[7.24;22.29]$
Notice the increase of the CI's (directly related to the decrease of α) reflecting our choice for higher confidence.
#### Unknown variance
Suppose that we do not know the variance for the number of sentences per advertisement but use the sample variance 273 as an estimate (so the sample standard deviation is σ = 16.54).
• For α = 0.1, the 90%CI(μ) is constructed by:
$\overline{x}\pm 1.28{16\over \sqrt{30}}=14.77 \pm 1.28{16.54\over \sqrt{30}}=[10.90;18.63]$
• For α = 0.05, the 90%CI(μ) is constructed by:
$\overline{x}\pm 1.645{16\over \sqrt{30}}=14.77 \pm 1.645{16.54\over \sqrt{30}}=[9.80;19.73]$
• For α = 0.005, the 90%CI(μ) is constructed by:
$\overline{x}\pm 2.575{16\over \sqrt{30}}=14.77 \pm 2.575{16.54\over \sqrt{30}}=[6.99;22.54]$
Notice the increase of the CI's (directly related to the decrease of α) reflecting our choice for higher confidence.
|
Solving Simple Trigonometric Equations
# Solving Simple Trigonometric Equations
• PRACTICE (online exercises and printable worksheets)
A trigonometric equation is an equation that uses at least one variable inside a trigonometric function.
Here are some examples of trigonometric equations. Follow the links to see the solutions.
• Example 1: a single trigonometric function with input $\,x\,$, no other appearance of the variable:
$\sqrt 2\sin x - 1 = 0$
• Example 2: a single trigonometric function with input $\,kx\,$, no other appearance of the variable:
$2\cos(3x) = 1$
$2\sin^2 x - \sin x - 1 = 0$
• Example 4: use graphical methods to approximate solutions to trigonometric equations:
find all solutions of the equation ‘$\,2\cos x = -3\sin x\,$’ in the interval $\,[0,2\pi]\,$
• Example 5: use WolframAlpha to solve trigonometric equations:
$\tan(5x-3)\sec(x^2 - 7) = 4$
Note that an equation like ‘$\,5 - x\sin 3 = 7\,$’ is not a trigonometric equation,
because it does not have a variable inside a trigonometric function.
Here, $\,\sin 3\,$ is just a constant. This is a linear equation in one variable with unique solution $\displaystyle\,x = \frac{-2}{\sin 3}\,$.
## Example 1:
### (a single trigonometric function with input $\,x\,$, no other appearance of the variable)
The equation ‘$\,\sqrt 2\sin x - 1 = 0\,$’ is a trigonometric equation because of the ‘$\,x\,$’ inside the sine function.
Note that there is no other appearance of $\,x\,$ in this equation.
In cases such as this, we start by isolating the trigonometric function containing the variable.
(Recall that isolate means to get all by itself on one side of the equation.)
$\sqrt 2\sin x - 1 = 0$ original equation
$\displaystyle\sin x = \frac{1}{\sqrt 2}$ Isolate the trigonometric function with variable input:
add $\,1\,$ to both sides (using the addition property of equality);
divide both sides by $\,\sqrt 2\,$ (using the multiplication property of equality).
• Is the equation ever true? Can an output from the sine function ever equal $\displaystyle\,\frac{1}{\sqrt 2}\,$?
The range of the sine function is $\,[-1,1]\,$.
Since $\,\sqrt 2 \approx 1.4 > 1\,$, we have $\,\frac{1}{\sqrt 2} \approx 0.7 < 1 \,$.
So, the number $\,\frac{1}{\sqrt 2}\,$ is in the range of the sine function.
Thus, there is at least one value for which the equation is true.
• How many solutions does the equation have?
A quick sketch (below) show that there are infinitely many solutions.
This is typical of trigonometric equations—they often have infinitely many solutions,
because of the periodic nature of trigonometric functions.
reporting the solutions in degrees:
$x = 45^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$, or
$x = 135^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$
$\displaystyle x = \frac{\pi}{4} + 2\pi k\,$ for all integers $\,k\,$, or
$\displaystyle x = \frac{3\pi}{4} + 2\pi k\,$ for all integers $\,k\,$
Precisely what angles have sine equal to $\,\frac{1}{\sqrt 2}\,$?
From the special triangle below, we see that $\,\sin 45^\circ = \frac{1}{\sqrt 2}\,$;
so, $\,45^\circ\,$ (or $\,\displaystyle\frac{\pi}{4}\,$ radians) is one of the solutions of the equation.
Recall that the sine function gives the $y$-values of points on the unit circle.
From the unit circle above, we can now get all the angles with $y$-value equal to $\,\frac{1}{\sqrt 2}\,$, as follows:
Recall that, when laying off angles:
counterclockwise is the positive direction
and
clockwise is the negative direction
Also recall that the integers are: $$\ldots, -3,-2,-1,0,1,2,3,\ldots$$ Start at the terminal point for $\,45^\circ\,$, in the first quadrant:
Rotate counterclockwise (up) to get additional solutions: $45^\circ+ 360^\circ\,$ $45^\circ+ 2\cdot360^\circ\,$ $45^\circ+ 3\cdot360^\circ\,$, and so on Rotate clockwise (down) to get additional solutions: $45^\circ - 360^\circ\,$ $45^\circ - 2\cdot360^\circ\,$ $45^\circ - 3\cdot360^\circ\,$, and so on
See the pattern? Together, these give all the solutions in the first quadrant: $$x = 45^\circ + {360k\,}^\circ\ \ \text{for all integers } k$$ Next, start at the terminal point for $\,180^\circ - 45^\circ = 135^\circ\,$, in the second quadrant:
Rotate counterclockwise (down) to get additional solutions: $135^\circ+ 360^\circ\,$ $135^\circ+ 2\cdot360^\circ\,$ $135^\circ+ 3\cdot360^\circ\,$, and so on Rotate clockwise (up) to get additional solutions: $135^\circ - 360^\circ\,$ $135^\circ - 2\cdot360^\circ\,$ $135^\circ - 3\cdot360^\circ\,$, and so on
See the same pattern? Together, these give all the solutions in the second quadrant: $$x = 135^\circ + {360k\,}^\circ\ \ \text{for all integers } k$$
## Example 2:
### (a single trigonometric function with input $\,kx\,$, no other appearance of the variable)
The equation ‘$\,2\cos(3x) = 1\,$’ is a trigonometric equation because of the ‘$\,x\,$’ inside the cosine function.
Note that there is no other appearance of $\,x\,$ in this equation.
Again, we start by isolating the trigonometric function containing the variable.
$\,2\cos(3x) = 1\,$ original equation
$\displaystyle\cos(3x) = \frac{1}{2}$ Isolate the trigonometric function with variable input.
The cosine function takes on the value $\,\frac 12\,$ infinitely many times,
so again there are infinitely many solutions.
to get $\,x\,$ in degrees:
$3x = 60^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$, or
$3x = -60^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$
to get $\,x\,$ in radians:
$\displaystyle 3x = \frac{\pi}{3} + 2\pi k\,$ for all integers $\,k\,$, or
$\displaystyle 3x = -\frac{\pi}{3} + 2\pi k\,$ for all integers $\,k\,$
For the moment, ignore the ‘$\,3x\,$’ inside the cosine function. (We'll use it in a minute.)
Ask: What angles have cosine equal to $\displaystyle\,\frac{1}{2}\,$?
From the special triangle below, we see that $\displaystyle\,\cos 60^\circ = \frac{1}{2}\,$.
Recall that the cosine function gives the $x$-values of points on the unit circle.
As in Example 1, get all the angles with $x$-value equal to $\,\frac{1}{2}\,$: $$60^\circ + {360k\,}^\circ \ \ \text{ for all integers k } \quad \text{(in quadrant I)} \tag{*}$$ $$-60^\circ + {360k\,}^\circ \ \ \text{ for all integers k } \quad \text{(in quadrant IV)} \tag{**}$$ These are all possible angles with cosine equal to $\,\frac 12\,$.
Now, we'll use the ‘$\,3x\,$’.
Note that if $\,x\,$ has units of degrees, then $\,3x\,$ also has units of degrees.
The angle inside the cosine function (which in this example is $\,3x\,$)
must equal one of the angles in (*) or (**), in order to have cosine equal to $\,\frac 12\,$.
Thus, for all integers $\,k\,$, and for $\,x\,$ in degrees: $$\begin{gather} 3x = 60^\circ + {360k\,}^\circ\cr \text{or}\cr 3x = -60^\circ + {360k\,}^\circ \end{gather}$$
degree measure:
$x = 20^\circ + {120k\,}^\circ\,$ for all integers $\,k\,$, or
$x = -20^\circ + {120k\,}^\circ\,$ for all integers $\,k\,$
$\displaystyle x = \frac{\pi}{9} + \frac{2\pi k}{3}\,$ for all integers $\,k\,$, or
$\displaystyle x = -\frac{\pi}{9} + \frac{2\pi k}{3}\,$ for all integers $\,k\,$
Solve for $\,x\,$.
Done!
## Example 4:
### (use graphical methods to approximate solutions to trigonometric equations)
Graphical methods for solving equations (and inequalities) were studied thoroughly in earlier sections:
Even if you're working with a simple equation (as in Examples 1 and 2),
graphical methods are great for giving yourself confidence in your answers and/or catching mistakes.
Graphical methods typically provide approximate solutions, not exact solutions.
The equation ‘$\,2\cos x = -3\sin x\,$’ is a trigonometric equation because there is at least one variable inside a trigonomeric function.
In this equation, there are two trigonometric functions with variable inputs;
isolating either one would still cause a variable expression (not a constant) to appear on the other side of the equation.
Thus, the technique illustrated in Example 1 and Example 2 doesn't work.
Both sides of the equation ‘$\,2\cos x = -3\sin x\,$’ are easy to graph—it is well-suited to graphical methods.
Graph the left-hand side; graph the right-hand side; approximate the intersection point(s) in a desired interval.
$2\cos x = -3\sin x$ original equation; find solutions in the interval $\,[0,2\pi]\,$ $x\approx 2.6\,$ or $\,x\approx 5.7\,$ Graph, on the interval $\,[0,2\pi]\,$: the left-hand side of the equation (in red): cosine curve, vertical stretch by a factor of $\,2\,$ the right-hand side of the equation (in black): sine curve, vertical stretch by a factor of $\,3\,$, reflect about the $x$-axis It is clear there are exactly two solutions in the interval $\,[0,2\pi]\,$; these correspond to the intersection points of the two graphs: $\,x_1\,$ is a bit less than $\,\pi\,$: $\color{red}{2\cos x_1} \approx -3\sin x_1$ $\,x_2\,$ is a bit less than $\,2\pi\,$: $\color{red}{2\cos x_2} \approx -3\sin x_2$ A graphing calculator or WolframAlpha (see Example 5) can be used to get decimal approximations for $\,x_1\,$ and $\,x_2\,$.
## Example 5:
### (use WolframAlpha to solve trigonometric equations)
WolframAlpha (opens in a new window) is a powerful, reliable, and free resource for mathematical information.
WolframAlpha understands conventional math syntax and abbreviations.
You can also type in a reasonable guess, and WolframAlpha will try to make sense of it.
For example, if you type in ‘sine of x’ and press Enter, the ‘Input:’ area will display ‘$\,\sin(x)\,$’.
This lets you know that WolframAlpha has understood you correctly.
Always check the ‘Input:’ area, to make sure that you and WolframAlpha are on the same page!
Try solving each of the examples from this page (or the exercises) at WolframAlpha.
WolframAlpha is a good resource for checking answers, gaining additional insight, or when you just need a quick reliable answer.
You can cut-and-paste each of the following. Have fun!
tan(5x-3)sec(x^2 - 7) = 4
sin(x) = 1/sqrt(2)
2cos(3x) = 1
2sin^2 x - sin x - 1 = 0
2cos(x) = -3sin(x), 0 <= x <= 2pi
(WolframAlpha gives exact solutions; click ‘Approximate forms’ if desired)
Master the ideas from this section
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# Basics of Data Interpretation for Bihar State Exams
Updated : February 11th, 2021
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Dear Aspirants,
In this post, we will discuss the basics of Data Interpretation to help you all understand this topic better.
Data Interpretation is a major part of Quantitative Aptitude section for any exam. Now, first and foremost, What does Data Interpretation exactly mean?
Interpretation is the process of making sense of numerical data that has been collected, analysed, and presented. Interpreting data is an important critical thinking skill that helps you comprehend textbooks, graphs and tables
## Basics of Data Interpretation
Majority of the questions asked in the Data Interpretation Section are based on the following topics of the Arithmetic Section -
1. Ratios
2. Averages
3. Percentages
If the basics of these topics are clear, attempting DI in the exams becomes comparatively easy.
Now, let us go through the types of DI graphs/charts that you may encounter in the exams -
1. Pie Charts
2. Line Charts
3. Bar Graphs
4. Tabular Charts
5. Mixed Graphs
We shall now have a look at the types of questions that are asked under these Data Interpretation Graphs -
Consider the following data presented in the bar graph -
Percentage of Students who like different sports in two different years is provided in the following graph. Total number of Students is 1000 for both the years.
Year 2011 and Year 2012
Now the following types of questions may be asked from this data -
1. Sum or Difference based -
These are the most basic questions that may be asked in a DI set. For instance,
What was the sum of the total number of students who like Badminton and Cricket in both the years?
Now for such questions, first find the number of students who like the two sports in the two years -
2011 - Badminton = (12/100) * 1000 = 120
Cricket = (45/100) * 1000 = 450
Total = 570
2012 - Badminton = (20/100) * 1000 = 200
Cricket = (37/100) * 1000 = 370
Total = 570
Sum = 570 + 570 = 1140
2. Averages based Questions -
Average based questions are very commonly asked in the Data Interpretation sets. For instance,
What is the average number of students who like badminton, cricket and football in 2011?
Total students who like badminton, cricket and football in 2011 = (12 + 45 + 22) = 79% of 1000
Required average = 790/3
3. Ratio based question -
Another arithmetic operation based question that may be asked is Ratio based.
Now, these questions may be asked directly or in combination with the above. For instance,
What is the ratio of the students who like football and tennis in 2011 and those who like volleyball and squash in 2012?
Students who like football and tennis in 2011= (22 + 4) = 26% of 1000
Students who like volleyball and squash in 2012 = (10 + 5) = 15% of 1000
Remember for such questions, you do not need to do the entire calculation, because such numbers will eventually cancel out while calculating the ratios.
Required ratio = (26% of 1000) : (15% of 1000) = 26 : 15
4. Percentage based question -
These are yet other arithmetic problems that are usually asked in DI questions.
These problems again may be asked individually or in combination with the sum or difference based problems. For instance,
The students who like badminton and squash in 2011 is what per cent of the students who like football and swimming in 2011?
Students who like badminton and squash in 2011 = (12 + 2) = 14% of 1000
Students who like football and swimming in 2011 = (22+7) = 29% of 1000
Here again, do not calculate the entire value.
Required % = (14% of 1000) / (29% of 1000) * 100 = 1400/29%
Same data may be presented in the form of other graphs as well, however, the approach to attempt the questions would remain the same. You may find numbers in place of percentages or vice - versa, so do read the question carefully before proceeding.
Line Graph -
Year 2011 and Year 2012
Tabular Chart -
Pie Chart -
One more variety of question that may be asked in pie charts is the angle based. For instance,
What is the central angle corresponding to football and volleyball together for 2012?
Angle = (20 + 10)% * 360 = (30/100) * 360 = 108
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A Basic Linear Algebra Principle
2. Implication of size on existence/uniqueness
By combining the numerical implications of the existence and the uniqueness on the size, we have
Ax = b has a unique solution for any b
⇒ number of rows of A = number of columns of A (A is a square matrix)
⇔ number of equations = number of variables
Conversely, assume A is an n by n matrix. Then we have (if you find the general argument too difficult, try 3 by 3 matrix first)
Ax = b has solutions for any b
⇒ All rows are pivot
⇒ Number of pivot rows is n (because there are n rows)
⇒ Number of pivot columns is n (by this equality)
⇒ All columns are pivot (because there are n columns)
⇒ The solution of a consistent system Ax = b is unique
In other words, if the number of equations is equal to the number of variables, then always existence implies uniqueness. By similar argument, we can also prove that uniqueness implies always existence.
In summary, we have the following basic principle of linear algebra.
For a square matrix A, the following are equivalent
Always Existence + Uniqueness
Ax = b has a unique solution for any b
Always Existence
Ax = b has solutions for any b
Uniqueness
The solution of a consistent system Ax = b is unique
Our discussion also tells us when the above happens from computational viewpoint.
For a square matrix A, the following are equivalent
• Ax = b has a unique solution for any b
• All rows of A are pivot
• All columns of A are pivot
• A can be row operated to become I
For the claim that A can be row operated to become I, please check out more details in this exercise.
Example The system
x1 - x2 + 2x3 = 1 3x1 + x2 - 2x3 = 3 2x1 - x2 + 2x3 = 2
has x1 = 1, x2 = x3 = 0 as an obvious solution. It is also easy to see that x1 = 1, x2 = 2, x3 = 1 is another solution. Therefore the system has many solutions. Since the system
x1 - x2 + 2x3 = b1 3x1 + x2 - 2x3 = b2 2x1 - x2 + 2x3 = b3
has the same coefficient matrix, by the basic principle, it does not always have solutions.
The significance of the basic principle is the following: We may consider (always) existence and the uniqueness as two complementary aspects of systems of linear equations. In general, there is no relation between the two aspects. However, in case the size is right (square coefficient matrix, or number of variables = number of equations), the two aspects are equivalent.
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# Perimeter of a Pentagon
## Perimeter of a Pentagon Lesson
### The Perimeter Formulas
There are two formulas for finding perimeter of a pentagon. For a regular pentagon where all sides are the same length, the formula is given as:
P = 5s
Where P is the perimeter and s is the side length.
For an irregular pentagon where the sides are not all the same length, the formula is given as:
P = a + b + c + d + e
Where ae are the lengths of each side.
INTRODUCING
### Perimeter of a Pentagon Example Problems
Let's go through a couple of example problems together to practice finding the perimeter of a pentagon.
#### Example Problem 1
Find the perimeter of a regular pentagon with a side length of 15.
Solution:
1. Since we know this is a regular pentagon, we can plug the side length 15 into the regular pentagon formula.
2. P = 5s
3. P = 5(15) = 75
4. The perimeter is 75.
#### Example Problem 2
In irregular pentagon has side lengths a = 2.36, b = 4.01, c = 3.12, d = 3.22, and e = 4.41. What is the perimeter?
Solution:
1. Let's plug the side lengths into the irregular pentagon formula.
2. P = a + b + c + d + e
3. P = 2.36 + 4.01 + 3.12 + 3.22 + 4.41 = 17.12
4. The perimeter is 17.12.
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Chapter 9.5, Problem 50E
### Mathematical Applications for the ...
12th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781337625340
Chapter
Section
### Mathematical Applications for the ...
12th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781337625340
Textbook Problem
# Endangered species population It is determined that a wildlife refuge can support a group of up to 120 of a certain endangered species. If 75 are introduced onto the refuge and their population after t years is given by p ( t ) = 75 ( 1 + 4 t t 2 + 16 ) find the rate of population growth after t years. Find the rate after each of the first 7 years.
To determine
To calculate: The rate of population growth of a certain endangered species after t years and after each of the first 7 years, if the population after t years is given as p(t)=75(1+4tt2+16) when 75 endangered species introduced onto the refuge.
Explanation
Given Information:
The expression is p(t)=75(1+4tt2+16)=(75+300tt2+16)
Formula Used:
The quotient rule for the derivative of the two function, ddx(fg)=g⋅dfdx−f⋅dgdxg2.
The sum and difference rule of derivate of functions, ddx[u(x)±v(x)]=ddxu(x)±ddxv(x).
The simple power rule of derivative ddx(xn)=nxn−1.
Calculation:
Consider the provided population growth of a certain endangered species after t years is p(t)=75(1+4tt2+16).
Differentiate the provided population growth function,
dp(t)dt=ddt[75(1+4tt2+16)]=ddt[75+300tt2+16]
Use the sum and difference rule of derivative, ddx[u(x)±v(x)]=ddxu(x)±ddxv(x).
dp(t)dt=ddt[75(1+4tt2+16)]=ddt(75)+ddt(300tt2+16)=ddt(300tt2+16)
Use the quotient rule for the derivative of the two function f(x) and g(x) is, ddx(fg)=g⋅dfdx−f⋅dgdxg2.
dp(t)dt=(t2+16)ddt(300t)−(300t)ddt(t2+16)(t2+16)2
Use the simple power rule of derivative ddx(xn)=nxn−1.
dp(t)dt=(t2+16)(300)−(300t)(2t+0)(t2+16)2=(300t2+4800)−600t2(t2+16)2=−300t2+4800(t2+16)2
Therefore, the rate of population growth after t years is p'(t)=−300t2+4800(t2+16)2.
Now, the rate of change population growth for each year is given as,
Substitute 1 for t in the rate function p'(t)=−300t2+4800(t2+16)2.
p′(1)=(−300(1)2+4800((1)2+16)2)=(−300(1)+4800(17)2)≈15
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# Probability Week 4 GT00303. Probability A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will.
## Presentation on theme: "Probability Week 4 GT00303. Probability A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will."— Presentation transcript:
Probability Week 4 GT00303
Probability A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur. 4-2
Ways of Assigning Probability There are three ways to assign a probability to an outcome: 1.CLASSICAL APPROACH Based on the assumption that the outcomes of an experiment are equally likely. 2.RELATIVE FREQUENCY APPROACH The probability of an event happening is the fraction of the time similar events happened in the past. 3.SUBJECTIVE APPROACH The likelihood (probability) of a particular event happening that is assigned by an individual based on subjective judgment. 4-3
(1) Classical Approach Consider an experiment of rolling a six-sided die. What is the probability of the event “an even number of spots appear face up”? The possible outcomes are: Probability (Even Number) = 3/6 = 0.5 4-4
P(10) = 3/36P(6) = 5/36 P(2) = 1/36 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 Consider an experiment of rolling 2 six-sided dice and observing the total. What is the probability of the event (a) “the total is 2”; (b) “the total is 6”; (c) “the total is 10”? The possible outcomes are: 4-5
Bits & Bytes Computer Shop tracks the number of desktop computer systems it sells over a month (30 days): Desktops Sold# of Days 01 12 210 312 45 (2) Relative Frequency Approach From this past record/ historical data, we can construct the probabilities of an event (i.e. the number of desktop sold on a given day). 4-6
“There is a 40% chance Bits & Bytes will sell 3 desktops on any given day” Desktops Sold# of DaysDesktops Sold 011/30 =.03 122/30 =.07 21010/30 =.33 31212/30 =.40 455/30 =.17 ∑ = 1.00 4-7
(3) Subjective Approach If there is little or no past experience or information on which to base a probability, it may be arrived at subjectively. For example: Estimating the likelihood the New England Patriots will play in the Super Bowl next year. Estimating the likelihood you will be married before the age of 30. Estimating the likelihood the U.S. budget deficit will be reduced by half in the next 10 years. 4-8
Complement of Event Intersection of Events Union of Events Mutually Exclusive Events A A’A’ AB AB AB Basic Relationships of Probability 4-9
The complement of event A is defined to be the event consisting of all sample points that are “not in A”. Complement of A is denoted by A ’ The Venn diagram below illustrates the concept of a complement. P(A) + P(A ’ ) = 1 OR P(A) = 1 - P(A ’ ) (1) Complement of An Event A A’A’ 4-10
Illustration: An automatic Shaw machine fills plastic bags with a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size of the beans and other vegetables, a package might be underweight or overweight. Use the complement rule to show the probability of a satisfactory bag is 0.90. P(B) = 1 – P(B ’ ) = 1 – P(A or C) = 1 – [P(A) + P(C)] = 1 – [0.025 + 0.075] = 1 - 0.10 = 0.90 4-11
The intersection of events A and B is the set of all sample points that are in both A and B. The joint probability of A and B is the probability of the intersection of A and B, i.e. P(A and B) OR (2) Intersection of Two Events AB 4-12
A = tosses where first toss is 1 ={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)} B = tosses where the second toss is 5 = {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)} The intersection of events A and B contains those points where the first toss is 1 and the second toss is 5. The intersection is {(1,5)}. Illustration: 4-13
The union of two events A and B, is the event containing all sample points that are in A or B or both: Union of A and B is denoted: A or B. P (A or B) OR (3) Union of Two Events AB 4-14
A = tosses where first toss is 1 ={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)} B = tosses where the second toss is 5 = {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)} The union of events A and B contains those points where the first toss is 1 or the second toss is 5 or both. The union is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,5), (3,5), (4,5), (5,5), (6,5)}. Illustration: 4-15
Events are mutually exclusive if the occurrence of any one event means that none of the others can occur at the same time. Hence, their joint probability is 0. AB (4) Mutually Exclusive Events 4-16
Why are some mutual fund managers more successful than others? One possible factor is where the manager earned his or her MBA. The following table compares mutual fund performance against the ranking of the school where the fund manager earned their MBA: Mutual fund outperforms the market Mutual fund doesn’t outperform the market Top 20 MBA program 0.110.29 Not top 20 MBA program 0.060.54 E.g. This is the probability that a mutual fund outperforms AND the manager was in a top- 20 MBA program; it’s a joint probability. Illustration: 4-17
Alternatively, we could introduce shorthand notation to represent the events: A = Fund manager graduated from a top-20 MBA program A ’ = Fund manager did not graduate from a top-20 MBA program B = Fund outperforms the market B ’ = Fund does not outperform the market BB’ A 0.110.29 A’ 0.060.54 B and B ’ are mutually exclusive events. 4-18
BB’ A 0.110.29 A’ 0.060.54 4-19 What’s the probability that a fund outperforms the market or the manager graduated from a top-20 MBA program? ✔✔ ✔ OR Return
Types of Probability JOINT PROBABILITY A probability that measures the likelihood two or more events will happen concurrently. 4-20 MARGINAL PROBABILITY A probability that measures the likelihood that a specific event will happen. CONDITIONAL PROBABILITY The probability of a particular event occurring, given that another event has occurred.
4-21 Illustration: BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 What is the probability a fund manager isn’t from a top school? What is the probability a fund outperforms the market? Conditional probability is not directly observable from contingency table.
4-22 BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 What’s the probability that a fund will outperform the market given that the manager graduated from a top-20 MBA program?
Thus, there is a 27.5% chance that that a fund will outperform the market given that the manager graduated from a top-20 MBA program. BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 4-23
One of the objectives of calculating conditional probability is to determine whether two events are related. In particular, we would like to know whether they are independent, that is, if the probability of one event is not affected by the occurrence of the other event. Two events A and B are said to be independent if: P(A|B) = P(A) or P(B|A) = P(B) Independence of Events 4-24
4-25 Illustration: BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 Are the two events A and B independent? Since P(A|B) ≠ P(A), A and B are not independent events.
Since P(B|A) ≠ P(B), A and B are not independent events. Stated another way, A and B are dependent. That is, the probability of one event (B) depends on the occurrence of the other event (A). 4-26
Probability Rules 4-27 (1) Complement Rule
4-28 (2) Multiplication Rule OR This applies when events A and B are dependent. What if they are independent?
4-29 If events A and B are independent:
A graduate statistics course has 7 male and 3 female students. The professor wants to select two students at random to help her conduct a research project. What is the probability that the two students chosen are female? Illustration: Let A = The first student is female B = The second student is female P(A) = 3/10 = 0.3 P(B|A) = 2/9 = 0.22 4-30 Return
4-31 What is the probability that the two students chosen are female? Since events A and B are dependent: There is a 6.7% chance that the professor will choose 2 female students from her grad class of 10.
The professor who teaches the course is suffering from the flu and will be unavailable for two classes. The professor’s replacement will teach the next two classes. His style is to select one student at random and pick on him or her to answer questions during that class. What is the probability that the two students chosen are female? Illustration: Let A = The first student is female B = The second student is female P(A) = 3/10 = 0.3 P(B) = 3/10 = 0.3 4-32
4-33 What is the probability that the two students chosen are female? Since events A and B are independent (the student selected in the 1 st class can still be chosen in the 2 nd class):
Illustration: Let A = The first shirt is white B = The second shirt is white P(A) = 9/12 = 0.75 P(B|A) = 8/11 = 0.727 A golfer has 12 golf shirts in his closet. Suppose 9 of these shirts are white and the others blue. He gets dressed in the dark, so he just grabs a shirt and puts it on. He plays golf two days in a row and does not do laundry. What is the likelihood both shirts selected are white? 4-34
4-35 What is the probability that both shirts selected are white? Since events A and B are dependent: There is a 54.5% chance that the golfer will choose 2 white shirts from his closet.
ABAB = +– If A and B are mutually exclusive, then this term goes to zero (3) Addition Rule Confirm with Slide 19! (Click here)here 4-36
4-37 If events A and B are mutually exclusive: AB
In a large city, two newspapers are published, the Sun and the Post. The circulation departments report that 22% of the city’s households have a subscription to the Sun and 35% subscribe to the Post. A survey reveals that 6% of all households subscribe to both newspapers. What is the probability of selecting a household at random that subscribes to the Sun or the Post or both? Illustration: Let A = Subscription to the Sun B = Subscription to the Post P(A) = 0.22 P(B) = 0.35 4-38
4-39 What is the probability of selecting a household at random that subscribes to the Sun or the Post or both? There is a 51% probability that a randomly selected household subscribes to one or the other or both papers
Illustration: An automatic Shaw machine fills plastic bags with a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size of the beans and other vegetables, a package might be underweight or overweight. What is the probability that a particular package will be either underweight or overweight? 4-40
4-41 BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 Contingency Tables Contingency Tables should be used when marginal and joint probabilities are given. Conditional probability is not needed!
This is P(F|F), the probability of selecting a second female student, given that a female was already chosen first First selectionSecond selection P(F) = 3/10 P( M) = 7/10 P(F|M) = 3/9 P(F|F) = 2/9 P( M|M) = 6/9 P( M|F) = 7/9 This is P(F), the probability of selecting the first female student Probability Trees When marginal and conditional probabilities are given, it is best to use Probability Trees. 4-42 See Slide 30!Slide 30
At the ends of the “branches”, we calculate joint probabilities as the product of the individual probabilities on the preceding branches. First selectionSecond selection P(F) = 3/10 P( M) = 7/10 P(F|M) = 3/9 P(F|F) = 2/9 P( M|M) = 6/9 P( M|F) = 7/9 P(F F)=(3/10)(2/9) P(F M)=(3/10)(7/9) P(M F)=(7/10)(3/9) P(M M)=(7/10)(6/9) Joint probabilities 4-43
3/9 + 6/9 = 9/9 = 1 2/9 + 7/9 = 9/9 = 1 3/10 + 7/10 = 10/10 = 1 The probabilities associated with any set of branches from one “node” must add up to 1.00… First selectionSecond selection P(F) = 3/10 P( M) = 7/10 P(F|M) = 3/9 P(F|F) = 2/9 P( M|M) = 6/9 P( M|F) = 7/9 Handy way to check your work ! 4-44
Suppose we have our grad class of 10 students again, but make the student sampling independent, that is “with replacement” – a student could be picked first and picked again in the second round. F MFMF M FMFM P(F) = 3/10 P( M) = 7/10 P(F|M) = 3/10 P(F|F) = 3/10 P( M|M) =7/10 P( M|F) = 7/10 P(F F)=(3/10)(3/10) P(F M)=(3/10)(7/10) P(M F)=(7/10)(3/10) P(M M)=(7/10)(7/10) Illustration: 4-45
Law school grads must pass a bar exam. Suppose pass rate for first-time test takers is 72%. They can re-write if they fail and 88% pass their second attempt. What is the probability that a randomly grad passes the bar? P(Pass) = 0.72 P(Fail and Pass) =( 0.28)(0.88) =0.2464 P(Fail and Fail) = (0.28)(0.12) = 0.0336 First exam P(Pass) = 0.72 P( Fail) = 0.28 Second exam P(Pass|Fail) = 0.88 P( Fail|Fail) = 0.12 Illustration: 4-46
What is the probability that a randomly grad passes the bar? P(Pass) = P(Pass 1 st ) + P(Fail 1 st and Pass 2 nd ) = 0.7200 + 0.2464 = 0.9664 P(Pass) = 0.72 P(Fail and Pass) = (0.28)(0.88)=0.2464 P(Fail and Fail) = (0.28)(0.12) = 0.0336 First exam P(Pass) = 0.72 P( Fail) = 0.28 Second exam P(Pass|Fail) = 0.88 P( Fail|Fail) =.12 There is a 96.64% chance they will pass the bar 4-47
Illustration: 4-48
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# Difference between revisions of "2017 AMC 10B Problems/Problem 24"
## Problem 24
The vertices of an equilateral triangle lie on the hyperbola $xy=1$, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$
## Diagram
$[asy] size(15cm); Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 2.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return 1/x; } draw(graph(f,-8,-0.125)); draw(graph(f,0.125,8)); [/asy]$
## Solution 1
Without loss of generality, let the centroid of $\triangle ABC$ be $I = (-1,-1)$. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, $A = (1,1)$, so $AI = BI = CI = 2\sqrt{2}$, so since $\triangle AIB$ is isosceles and $\angle AIB = 120^{\circ}$, then by Law of Cosines, $AB = 2\sqrt{6}$. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to $\frac {s}{\sqrt{3}}$. Therefore, the area of the triangle is $\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}$, so the square of the area of the triangle is $\boxed{\textbf{(C) } 108}$.
## Solution 2
Without loss of generality, let the centroid of $\triangle ABC$ be $G = (-1,-1)$. Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let $A = (1,1)$. Then, point $B$ must be the reflection of $C$ across the line $y=x$, so let $B = \left(a,\frac{1}{a}\right)$ and $C=\left(\frac{1}{a},a\right)$, where $a <-1$. Because $G$ is the centroid, the average of the $x$-coordinates of the vertices of the triangle is $-1$. So we know that $a + 1/a+ 1 = -3$. Multiplying by $a$ and solving gives us $a=-2-\sqrt{3}$. So $B=(-2-\sqrt{3},-2+\sqrt{3})$ and $C=(-2+\sqrt{3},-2-\sqrt{3})$. So $BC=2\sqrt{6}$, and finding the square of the area gives us $\boxed{\textbf{(C) } 108}$.
## Solution 3
Without loss of generality, let the centroid of $\triangle ABC$ be $G = (1, 1)$ and let point $A$ be $(-1, -1)$. It is known that the centroid is equidistant from the three vertices of $\triangle ABC$. Because we have the coordinates of both $A$ and $G$, we know that the distance from $G$ to any vertice of $\triangle ABC$ is $\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}$. Therefore, $AG=BG=CG=2\sqrt{2}$. It follows that from $\triangle ABG$, where $AG=BG=2\sqrt{2}$ and $\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}$, $[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$ using the formula for the area of a triangle with sine $\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)$. Because $\triangle ACG$ and $\triangle BCG$ are congruent to $\triangle ABG$, they also have an area of $2\sqrt{3}$. Therefore, $[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}$. Squaring that gives us the answer of $\boxed{\textbf{(C) }108}$.
## Solution 4
Without loss of generality, let the centroid of $\triangle ABC$ be $G = (1, 1)$. Assuming we don't know one vertex is $(-1, -1)$ we let the vertices be $A\left(x_1, \frac{1}{x_1}\right), B\left(x_2, \frac{1}{x_2}\right), C\left(x_3, \frac{1}{x_3}\right).$
Since the centroid coordinates are the average of the vertex coordinates, we have that $\frac{x_1+x_2+x_3}{3}=1$ and $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1.$
We also know that the centroid is the orthocenter in an equilateral triangle, so $CG \perp AB.$ Examining slopes, we simplify the equation to $x_1x_2x_3 = -1$. From the equation $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1,$ we get that $x_1x_2+x_1x_3+x_2x_3 = -3$. These equations are starting to resemble Vieta's:
$x_1+x_2+x_3=3$
$x_1x_2+x_1x_3+x_2x_3 = -3$
$x_1x_2x_3=-1$
$x_1,x_2,x_3$ are the roots of the equation $x^3 - 3x^2 - 3x + 1 = 0$. This factors as $(x+1)(x^2-4x+1)=0 \implies x = -1, 2 \pm \sqrt3,$ for the points $(-1, -1), (2+\sqrt3, 2-\sqrt3), (2-\sqrt3, 2+\sqrt3)$. The side length is clearly $\sqrt{24}$, so the square of the area is $\boxed{108}.$
$\sim\textbf{Leonard\_my\_dude}\sim$
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Random Variables and their PDFs Main Concepts | Demonstration | Activity | Teaching Tips | Data Collection & Analysis | Practice Questions | Milestone | Fathom Tutorial
Milestone
This milestone develops insight into random variables. Thanks to Chris Olsen for creating these.
1. Warm-up
Suppose that random variables D= the number of pips after rolling one fair die and C = number of heads after flipping one fair coin.
a) Calculate the mean and the standard deviation for D and for C.
b) Now define a new random variable D + C. Calculate its mean and standard deviation by finding the pdf for D + C.
c) Now compare your results to the mean and SDs you get for using the "rules" for adding random variables. How do these compare?
2. Racing on the Island of Nog
On the Island of Nog there are three species: horses, three-toed sloths, and humans. We classify horses as one-toed animals. We randomly select a create from the island. If we define the random variable T = number of toes on one foot, then the probability distribution of T is
t 1 3 5 P(T=t) 0.4 0.1 0.5
a) From the table, calculate the mean, variance, and SD of T.
The annual Nog Spring Picnic features a two-legged race, where creatures are paired together in the following manner: The left hind leg of a randomly selected creature is bound to the right hind leg of another randomly selected creature. We define a new variable S to be the sum of toes for the bound pair, X = number of toes of the first creature and Y = number of toes of the second creature.
b) Now construct a table for S. there are nine ways to make teams.
c) What is the probability that a team will have more than five toes?
d) What is the probability that a team will have more than 8 toes, given that one member of the team is human?
e) Find the mean and variance for S.
f) Because of the differences between the species, the committee suggests that some teams be given a "head start", depending on the number of toes of the bound legs. The number of meters of head start is determined by the formula: H = 3 + 7S. Calculate the mean and the variance for H.
Write up your solutions and name them ms7yourlastname and drop it in the Digital Dropbox.
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# What is a fifth of 50
## This is math 2, textbook
Percentage calculation C1 78 Ella discovers a T-shirt while selling in a clothing store. Her mother says: “Very good. It's 50 percent off. ”Not sure what this means, Ella asks. The mother explains: “One percent is a hundredth. 50 percent therefore equals half. The T-shirt previously cost € 18, now it only costs half as much, i.e. €. Ella represents the meaning of one percent in mathematical notation: 1% š 1 ___ 100 = 0.01. She recognizes that you can represent percentages both in fractions and as a decimal number. Ella makes a table to show her math teacher to check. Percentage fraction Decimal number 1 percent 1% 1 ___ 100 0.01 1 hundredth 5 percent 5% 5 ___ 100 = 1 __ 20 0.05 5 hundredths = 1 twentieth 10 percent 10% 10 ___ 100 = 1 __ 10 10 hundredths = 1 tenth 20 percent 20% 20 ___ 100 = 1 _ 5 0.20 20 hundredths = 1 fifth 25 percent 25% 25 ___ 100 = 1 _ 4 25 hundredths = 1 quarter 50 percent 50% 50 ___ 100 = 1 _ 2 50 hundredths = 1 Half 75 percent 75% 75 ___ 100 = 3 _ 4 75 hundredths = 3 quarters 100 percent 100 ___ 100 = 1 1.00 100 hundredths = 1 whole 200 percent 200% 200 ___ 100 = 2 2.00 200 hundredths = 2 whole ( the double) Note Often the percentage value W is also referred to as percentage A. The base value G corresponds to the whole or 100% and the percentage value W corresponds to p% of the base value. interactive preliminary exercise 7by8a4 AH p. 23 One percent corresponds to the hundredth part of a whole. It can be represented in both fractional and decimal notation. 1% š 1 ___ 100 = 0.01 (1 hundredth) One hundred percent is a whole. 100% š 1 Three terms are particularly important for the percentage calculation: the base value G, the percentage value W and the percentage p%. Percent Example Of the 28 children in class 2 B, 14 like math, that's 50%. Basic value G Percentage W Percentage p% 1 Basic terms For testing purposes only - property of the publisher öbv
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# How to Find Volume and Surface Area of Cubes? (+FREE Worksheet!)
Here you can learn how to find the volume and surface area of cubes using cubes volume and surface area formulas.
## Step-by-step guide to finding the Volume and Surface Area of Cubes
• A cube is a three-dimensional solid object bounded by six square sides.
• Volume is the measure of the amount of space inside of a solid figure, like a cube, ball, cylinder, or pyramid.
• Volume of a cube $$=$$ (one side)$$^3$$
• surface area of cube $$=6×$$ (one side)$$^2$$
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### Cubes – Example 1:
Find the volume and surface area of this cube.
Solution:
Use volume formula: volume $$=$$ (one side)$$^3$$
Then: volume $$=$$ (one side)$$^3=(4)^3=64$$ $$cm^3$$
Use surface area formula: surface area of cube: $$=6$$ (one side)$$^2$$
Then: surface area of cube $$=6$$ (one side)$$^2=6(4)^2=6(16)=96$$ $$cm^2$$
### Cubes – Example 2:
Find the volume and surface area of this cube.
Solution:
Use volume formula: volume $$=$$ (one side)$$^3$$
Then: volume $$=$$ (one side)$$^3=(2)^3=8$$ $$cm^3$$
Use surface area formula: surface area of cube: $$=6$$ (one side)$$^2$$
Then: surface area of cube $$=6$$ (one side)$$^2=6(2)^2=6(4)=24$$ $$cm^2$$
## Exercises for Finding Volume and Surface Area of Cubes
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### Find the volume of each cube.
1. $$\color{blue}{8 \ ft^3}$$
2. $$\color{blue}{125 \ m^3}$$
3. $$\color{blue}{27 \ in^3}$$
4. $$\color{blue}{216 \ miles^3}$$
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# Cool Math Equations That Mean Something latest 2023
You are searching about Cool Math Equations That Mean Something, today we will share with you article about Cool Math Equations That Mean Something was compiled and edited by our team from many sources on the internet. Hope this article on the topic Cool Math Equations That Mean Something is useful to you.
Page Contents
## Elimination Method To Solve System Of Linear Equations
The method of elimination is most often used by students to solve a system of linear equations. Moreover, this method is easy to understand and involves the addition and subtraction of polynomials. Students should be able to add and subtract polynomials involving two or three variables.
In the method of elimination, the coefficients of the same variable are identical, then the two equations are subtracted to eliminate this variable. The resulting equation involves only one variable and can be easily simplified. For example; Consider that there are two equations in the system of linear equations with variables “x” and “y” as shown below:
2x – 5a = 11
3x + 2a = 7
To solve the above equation by the method of elimination, we need to make the coefficients of one of the variables (either “x” or “y”) the same by multiplying the equation with some numbers, and these numbers can be obtained by finding the least common multiple of the coefficients. Consider that we want the coefficients of “x” to be the same in both equations. For this we need to find the least common multiple of “2” and “3” which is “6”.
To get “6” as the coefficient of the “x” variable in the equations, we need to multiply the first equation by “3” and the second equation by “2”, as shown below:
(2x – 5a = 11) * 3
(3x + 2y = 7) * 2
The new set of equations after multiplication is obtained as shown below:
6x – 15y = 33
6x + 4a = 14
Now we have the same coefficient of variable “x” in both equations. Once a variable has obtained the same coefficient, subtract one equation from the other. We will subtract the second equation from the first as shown below:
(6x – 15y = 33) – (6x + 4y = 14)
In the next step, combine similar terms:
6x – 6x – 15y – 4y = 33 – 14
– 19a = 19
y = – 1
So far we have solved the equations for one variable. To find the value of the other variable “x”, we will substitute the value of “y” into one of the equations given in the question.
Substitute the value of “y = – 1” in the equation 2x – 5y = 11 to find the value of “x” as shown in the next step:
2x – 5 (- 1) = 11
2x + 5 = 11
2x = 11 – 5
2x = 6
x = 3
Therefore, we have solved the two equations to find the value of the variables and our solution is x = 3 and y = – 1. You can take the same approach to solve the system of linear equations by eliminating one of the variables.
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## Cool Math Don T Look Back latest 2023
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## Cool Math Cool Math Cool Math Games latest 2023
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# The coin flip conundrum
When the Wright brothers had to decide who would be the first to fly their new airplane off a sand dune, they flipped a coin. That was fair: we all know there’s an equal chance of getting heads and tails. But what if they had a more complicated contest?
What if they flipped coins repeatedly, so that Orville would win as soon as two heads showed up in a row on his coin, and Wilbur would win as soon as heads was immediately followed by tails on his? Would each brother still have had an equal chance to be the first in flight? At first, it may seem they’d still have the same chance of winning. There are four combinations for two consecutive flips.
And if you do flip a coin just twice, there’s an equal chance of each one — 25%. So your intuition might tell you that in any string of coin flips, each combination would have the same shot at appearing first. Unfortunately, you’d be wrong. Wilbur actually has a big advantage in this contest. Imagine our sequence of coin flips as a sort of board game, where every flip determines which path we take. The goal is to get from start to finish. The heads/tails board looks like this.
It has two identical steps, each with a 50/50 chance of staying in place or moving forward. Option 1: If we stay in place by getting tails, we waste one flip. Since we’re back in the same place, on average we must flip x more times to advance one step. Together with that first flip, this gives an average of x + 1 total flips to advance. Option 2: If we get heads and move forward, then we have taken exactly one total flip to advance one step. We can now combine option 1 and option 2 with their probabilities to get this expression.
Solving that for x gives us an average of two moves to advance one step. Since each step is identical, we can multiply by two and arrive at four flips to advance two steps. For heads/heads, the picture isn’t as simple. This time, let y be the average number of flips to move from start to finish. There are two options for the first move, each with 50/50 odds. Option 1 is the same as before, getting tails sends us back to the start, giving an average of y+1 total flips to finish. In Option 2, there are two equally likely cases for the next flip.
With heads we’d be done after two flips. But tails would return us to the start. Since we’d return after two flips, we’d then need an average of y+2 flips in total to finish. So our full expression will be this. And solving this equation gives us six flips. So the math calculates that it takes an average of six flips to get heads/heads, and an average of four to get heads/tails.
And, in fact, that’s what you’d see if you tested it for yourself enough times. Of course, the Wright brothers didn’t need to work all this out; they only flipped the coin once, and Wilbur won. But it didn’t matter: Wilbur’s flight failed, and Orville made aviation history, instead. Tough luck, Wilbu
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# math median
please tell me you find the median in a group of numbers :)
1. 👍
2. 👎
3. 👁
1. place the data set of numbers in order from lowest to highest.
so for instance you are given:
5, 8 , 9 ,15, 20, 11
so then you place them in order:
5, 8 , 9, 11, 15, 20
to find the median which number falls in the middle.
since here there are six numbers so you get the two sets of numbers that fall in the middle add them and divide them by two.
so here would be:
9 +11 = 20 divided by 2 = 15
if your set of numbers are only 5 instead of 6 as my example. then what ever numbers falls in the middle is your middle number. but if you have two in the middle remember to add first then divide by them.
1. 👍
2. 👎
## Similar Questions
1. ### Math
Lesson 8: Mean, Median, Mode, and Range Math 7 A Unit 2: Decimals and Integers 1: FIND THE MEAN OF THE DATA SET 9,4,7,3,10,9 A: 6 B: 7 C: 8 D: 9 2: FIND THE MEDIAN OF THE DATA SET: 9,6,7,3,10,9 A: 6 B: 7 C: 8 D: 9 3: FIND THE MODE
2. ### ALGEBRA
Find the mean, median, and mode of the data set. Round to the nearest tenth. 15, 13, 9, 9, 7, 1, 11, 10, 13, 1, 13 (1 point) • mean = 9.3, median = 8, mode = 13 • mean = 8.5, median = 10, mode = 13 • mean = 9.3, median = 10,
3. ### Math
Here is the question:Consider a data set of 15 distinct measurements with mean A and median B. (a) If the highest number were increased, what would be the effect on the median and mean? Explain. The mean would remain the same
4. ### Math
Six whole numbers have A median of 10 A mode of 11 A range of 4 Work out a possible set of a six numbers. Write the numbers in order. Thank you.
1. ### Algebra
The table shows the number of hours that a group of friends spent in their first week training to run a marathon. In the second week, they each add five hours to their training times. What are the mean, median, mode, and range of
2. ### AP Statistics (math)
Find a set of numbers that will satisfy the following conditions: - the median of a set of 20 numbers is 24. - the range is 42. - to the nearest whole number the mean is 24. - no more than three numbers are the same.
3. ### Math
Pleas check my answers Directions: Calculate the mean, median, mode & range for the following data sets. If a data set does not contain a mode, write 'no mode' for mode. 1) 61, 98, 60, 60, 57, 81, 93 Mean Median Mode Range 2) 72,
4. ### algebra 1 question pls help!!!
1. Find the mean, median, and mode of the data set: 15, 16, 21, 23, 25, 25, 25, 39 a.mean = 23.6, median = 25, mode = 24 b.mean = 24, median = 23.6, mode = 25 c.mean = 24.6, median = 24, mode = 25 d.mean = 23.6, median = 24, mode
1. ### Algebra
3. the table shows the number of hours that a group of students spend studying for the SAT during their first week of preparation. the students each add 4 hours to their study times in the second week. what were the mean, median,
2. ### Statistics
One item on a questionnaire asks, "How many siblings (brothers and sisters) did you have when you were a child?" A researcher computes the mean, the median, and the mode for a set of n = 50 responses to this question. Which of the
3. ### Math
A back-to-back stem-and-leaf plot showing the points scored be each player on two different basketball teams is shown below. Team 1 | | Team 2 6 1| 0| 3 7 8 4 2| 1|4 6 8 9 9 6 3| 2|5 7 Key: 4|0|2 means 4 points for Team 1 2points
4. ### math
Please help me to find the first and third quartiles, Q1 and Q3, of the following set of numbers.13, 7, 15, 13, 13, 12, 9, 11, 19, 9 Qrrange the numbers in ascending order, like this: 7, 9, 9, 11, 12, 13, 13, 13, 15, 19 Now divide
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# What is a statistic?
Exploring the idea of a statistic by simulating dice rolls in R
Author
Max Rohde
Published
December 8, 2020
Code
library(tidyverse)
## Introduction
When we collect data from a data-generating process, we can calculate values from that data. These values are called statistics.
Common example include:
• mean and median (measures of center)
• variance and IQR (measures of spread)
• order statistics, such as the minimum and the maximum
We can even create arbitrary statistics that appear to have little use, such as adding only the first and third elements of the data and dividing by 17.
## Simulating statistics of dice rolls
As a simple data-generating process, let’s consider rolling 5 dice. Each time we roll, we obtain 5 numbers, each from 1 to 6. We will call each one of these vectors of 5 numbers,
$(x_1, x_2, x_3, x_4, x_5)$
a sample. We then will compute statistics from these samples. The main question we seek to answer is: how are the statistics distributed? When I calculate the mean of 5 dice, what will the most likely result be? We can ask this question about any statistic.
We’ll write a function to roll n dice called roll().
Code
# A function to roll n dice
roll <- function(n){
sample(x = 1:6, size=n, replace=TRUE)
}
Then we’ll use purrr:map() to generate 100,000 rolls of 5 dice.
Code
# Roll 5 dice 100,000 times
data <- map(1:1e5, ~roll(5))
Here’s an example of running the function.
Code
# Look at first three rolls
data[1:3]
[[1]]
[1] 3 1 4 2 5
[[2]]
[1] 5 3 3 1 1
[[3]]
[1] 5 3 3 3 2
For each of these rolls, we can calculate the value of a statistic.
We’ll calculate the following statistics:
• median
• mean
• minimum
• maximum
• second order statistic $$X_{(2)}$$
• range
Code
# Returns the nth order statistic of the sample
order_stat <- function(x, n){
x <- sort(x)
return(x[n])
}
# Generate various statistics for each roll
medians <- map_dbl(data, ~median(.x))
means <- map_dbl(data, ~mean(.x))
minimums <- map_dbl(data, ~min(.x))
maximums <- map_dbl(data, ~max(.x))
second_order_stat <- map_dbl(data, ~order_stat(x=.x, n=2))
ranges <- maximums - minimums
Code
# Create a data frame from our computed statistics
df <- tibble(medians, means, minimums, maximums, second_order_stat, ranges)
# Pivot the data into long format for plotting
df <- pivot_longer(df, cols = everything())
Now using the data from our simulation, we can plot the sampling distribution of the each of the statistics.
Code
df$name <- recode(df$name,
medians = "Median",
means = "Mean",
minimums = "Minimum",
maximums = "Maximum",
second_order_stat = "2nd Order Statistic",
ranges = "Range")
df$name <- as.factor(df$name)
df$name <- fct_relevel(df$name,
c("Minimum",
"2nd Order Statistic",
"Maximum",
"Range",
"Mean",
"Median"))
df %>%
ggplot(aes(x = value)) +
geom_bar(aes(y = ..prop..),
width = 0.2, fill = "gray", color = "black") +
scale_x_continuous(breaks = 0:6) +
facet_wrap(~name, scales = "free_x") +
labs(x = "Value",
y = "Estimated Probability",
title = "Distribution of various statistics for 100,000 rolls of 5 dice",
caption = "Monte Carlo estimate with 100,000 simulations") +
ggthemes::theme_solarized() +
theme(text = element_text(size = 12, family = "Source Sans Pro"))
A few things to note:
• Because of averaging, the mean can take on more possible values than the other statistics. Qe can see it taking on the characteristic bell shape of the normal distribution due to the central limit theorem.
• The median is always a whole number because we are rolling an odd number of dice.
• Some of these distributions are tedious to work out analytically, and with more complicated data-generating processes there may be no closed form solutions.
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# Difference between revisions of "2019 AMC 8 Problems/Problem 17"
## Problem
What is the value of the product
$$\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?$$
$\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50$
## Solution 1(Telescoping)
We rewrite: $$\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}$$
The middle terms cancel, leaving us with
$$\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}$$
## Solution 2
If you calculate the first few values of the equation, all of the values tend to $\frac{1}{2}$, but are not equal to it. The answer closest to $\frac{1}{2}$ but not equal to it is $\boxed{\textbf{(B)}\frac{50}{99}}$.
## Solution 3
Rewriting the numerator and the denominator, we get $\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}$. We can simplify by canceling 99! on both sides, leaving us with: $\frac{100 \cdot 98!}{2 \cdot 99!}$ We rewrite $99!$ as $99 \cdot 98!$ and cancel $98!$, which gets $\boxed{(B)\frac{50}{99}}$.
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# Square Roots Of Rational Numbers Worksheet
A Rational Phone numbers Worksheet may help your son or daughter be more knowledgeable about the concepts powering this rate of integers. In this worksheet, individuals will be able to fix 12 distinct difficulties relevant to rational expression. They will likely discover ways to multiply two or more phone numbers, class them in pairs, and determine their goods. They will also training simplifying rational expression. As soon as they have learned these methods, this worksheet might be a valuable tool for advancing their reports. Square Roots Of Rational Numbers Worksheet.
## Reasonable Numbers can be a rate of integers
The two main varieties of figures: rational and irrational. Reasonable amounts are described as complete amounts, in contrast to irrational amounts tend not to perform repeatedly, and get an infinite variety of digits. Irrational amounts are low-zero, non-terminating decimals, and sq . beginnings which are not ideal squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To outline a rational variety, you must understand what a logical variety is. An integer is a total quantity, along with a realistic amount can be a percentage of two integers. The proportion of two integers is the number at the top divided by the number at the base. If two integers are two and five, this would be an integer, for example. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They are often made into a small fraction
A realistic quantity has a denominator and numerator that are not absolutely no. Consequently they may be expressed being a small percentage. Together with their integer numerators and denominators, realistic figures can furthermore have a negative importance. The adverse worth should be positioned on the left of and its absolute worth is its length from zero. To simplify this instance, we are going to state that .0333333 can be a fraction that can be written as being a 1/3.
Together with bad integers, a reasonable variety can be created in to a portion. For example, /18,572 is a rational number, although -1/ is just not. Any small percentage made up of integers is realistic, provided that the denominator is not going to include a and may be written for an integer. Also, a decimal that ends in a level can be another logical amount.
## They make sense
Even with their name, logical amounts don’t make a lot feeling. In math, these are one organizations using a unique duration around the quantity series. This means that whenever we count some thing, we could get the shape by its proportion to its authentic quantity. This retains real even if you will find limitless reasonable amounts involving two certain phone numbers. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To get the length of a pearl, as an example, we could count up its size. A single pearl weighs in at 10 kilos, that is a realistic number. Additionally, a pound’s excess weight equals 15 kilograms. As a result, we must be able to divide a lb by twenty, without the need of concern yourself with the length of a single pearl.
## They could be depicted being a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal quantity might be created being a several of two integers, so 4x 5 is equal to 8. The same difficulty necessitates the repeated fraction 2/1, and each side ought to be split by 99 to obtain the right solution. But how do you make your conversion? Here are several good examples.
A rational variety will also be designed in various forms, including fractions as well as a decimal. A good way to symbolize a reasonable quantity inside a decimal would be to break down it into its fractional equal. There are 3 ways to break down a reasonable amount, and each one of these techniques yields its decimal equal. One of these techniques is always to break down it into its fractional equivalent, and that’s what’s known as a terminating decimal.
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# How Is 1730 the Sum of Consecutive Squares?
Contents
### Today’s Puzzle:
Write all the numbers from 1 to 12 in both the first column and the top row so that those numbers are the factors of the given clues.
### Factors of 1730:
• 1730 is a composite number.
• Prime factorization: 1730 = 2 × 5 × 173.
• 1730 has no exponents greater than 1 in its prime factorization, so √1730 cannot be simplified.
• The exponents in the prime factorization are 1, 1, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8. Therefore 1730 has exactly 8 factors.
• The factors of 1730 are outlined with their factor pair partners in the graphic below.
### More about the number 1730:
1730 is the sum of two squares in two different ways:
41² + 7² = 1730, and
37² + 19² = 1730.
1730 is the hypotenuse of FOUR Pythagorean triples:
520 1650 1730 which is 10 times (52-165-173)
574 1632 1730 calculated from 2(41)( 7), 41² – 7², 41² + 7²,
1008 1406 1730 calculated from 37² – 19², 2(37)(19), 37² + 19², and
1038 1384 1730 which is 346 times (3-4-5).
Finally, OEIS.org informs us that 1730 is the sum of consecutive squares in two different ways. What are those two ways? I figured it out. Can you?
Here’s a hint: It is the sum of three consecutive squares as well as twelve consecutive squares. That means √(1730/3) rounded is included in one sum and √(1730/12) rounded is included in the other. The solution can be found in the comments. Have fun finding them yourself though!
## One thought on “How Is 1730 the Sum of Consecutive Squares?”
1. ivasallay
23² + 24² + 25² = 1730, and
6² + 7² + 8² + 9² + 10² + 11² + 12² + 13² + 14² + 15² + 16² + 17² = 1730.
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## Find the roots of basic equations containing roots
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Last week Sherri bought 324 square yards of sod to grass a square play area for the children of her day care. Now she has to fence the area to keep the children safe but does not know how many yards of fencing she needs to buy. All Sherri can figure out is that all the sides are the same length because the grassy area is a square.
How can she determine how many yards of fencing to buy?
In this concept, you will learn to solve equations involving radicals.
When you solve an equation you are trying to find the value for the variable that will make the equality statement true. The steps applied to solving an equation are inverse operations. To solve an equation involving radicals, inverse operations are used to solve for the variable.
A radical involving the square root of a number can be evaluated by determining the square root of the number under the radical sign. If the radicand is a perfect square then its square will be the number which multiplied by itself twice will give the value of the radicand. For example the square root of 81 can be denoted by . What number times itself twice gives 81?
Taking the square root of a number is the inverse operation of squaring and vice versa.
Let’s look at an equation involving radicals.
The variable ‘’ is squared and its value is 121. To solve this equation the value of needs to be determined. The inverse operation of squaring is taking the square root. Remember, whatever operation is applied to one side of the equation must also be applied to the other side.
First, take the square root of both sides of the equation.
Next, verify the answer by substituting the value for ‘’ into the original equation.
Then, perform any indicated operations.
The value of 11 made the equality statement true – both sides of the equation are the same.
Let’s look at one more.
Solve the following equation involving radicals:
Notice the left side of the equation is a radical. The inverse operation of taking the square root is squaring.
Remember, the square of the square root of anything is the anything. In other words
First, square both sides of equation.
Next, perform any indicated operations and simplify the equation.
Then, subtract 2 from both sides of the equation to solve the equation for ‘’.
Next, verify the answer by substituting the value for ‘’ into the original equation.
Then, perform any indicated operations and simplify the equation.
The value of 34 made the equation true.
### Examples
#### Example 1
Earlier, you were given a problem about Sherri and the fence for the grassy square. She needs to figure out the perimeter of the square.
First, write an equation for the area of the grassy square.
Such that is the area and is the length of any side of the square.
Next, fill in the value for the area.
Next, solve the equation for ‘’ by taking the square root of both sides of the radical equation.
The length of each side of the square is 18 yds.
Next, write an equation for finding the perimeter of the square. The perimeter is the distance all around the grassy area.
Such that ‘’ is the perimeter and ‘’ is the length of one side of the square.
Next, fill in the value for ‘’ and perform the indicated operation.
Sherri needs to buy 72 yards of fencing.
#### Example 2
Solve the following equation involving radicals to the nearest tenth.
First, the opposite operation of squaring is taking the square root. Take the square root of both sides of the equation.
Notice that the number 26 is not a perfect square. Use the TI calculator to find the square root of 26.
Next, on the calculator press 2nd enter.
Then, round the value shown on the screen to one place after the decimal.
#### Example 3
Solve the following radical equation for the variable ‘’.
First, take the square root of both sides of the equation.
Next, simplify both sides of the equation.
#### Example 4
Solve the following radical equation for the variable ‘’.
First, apply the inverse operation of taking the square root to both sides of the equation.
Next, perform any indicated operations and simplify the equation.
Then, add eight to both sides of the equation to solve for the variable ‘’.
Next, verify the answer by substituting the value for ‘’ into the original equation.
Then, perform any indicated operations and simplify the equation.
#### Example 5
Solve the following equation for the variable.
First, isolate the variable by subtracting 4 from both sides of the equation.
Next, simplify both sides of the equation.
Next, perform the inverse operation of cubing – taking the cube root, on both sides of the equation.
Then, perform the indicated operations.
### Review
Solve each equation involving radical expressions.
1.
2.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Base When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent.
Cubed The cube of a number is the number multiplied by itself three times. For example, "two-cubed" = $2^3 = 2 \times 2 \times 2 = 8$.
Exponent Exponents are used to describe the number of times that a term is multiplied by itself.
Extraneous Solution An extraneous solution is a solution of a simplified version of an original equation that, when checked in the original equation, is not actually a solution.
Fractional Power A fractional power is an exponent in fraction form. A fractional exponent of $\frac{1}{2}$ is the same as the square root of a number. A fractional exponent of $\frac{1}{3}$ is the same as the cube root of a number.
Perfect Square A perfect square is a number whose square root is an integer.
Quadratic Equation A quadratic equation is an equation that can be written in the form $=ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real constants and $a\ne 0$.
Quadratic Formula The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Squared Squared is the word used to refer to the exponent 2. For example, $5^2$ could be read as "5 squared". When a number is squared, the number is multiplied by itself.
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## Lesson Tutor : Algebra Lesson 12 : Multiplying Positives and Negatives
/ Lesson Tutor : Algebra Lesson 12 : Multiplying Positives and Negatives
Algebra Lesson 12: Multiplying Positives and Negatives
by Elaine Ernst Schneider
Before starting: Review/ complete Algebra Lesson 11
As we learned last lesson: Additive Inverses are opposites. Two numbers are opposites if their sum equals zero. For example, -8 and 8 are additive inverses because their sums total zero. This makes them opposites. You can do the same thing with variables. For example, -(-x) = x.
When this principle is used in multiplication, these rules of thumb emerge:
1. A negative times a negative gives a positive answer.
2. A negative times a positive is negative.
3. A positive times a positive is a positive result.
Examples
(-6)(8) = -48
(-3)(-12) = 36
(12)(11) = 132
Now, what happens when there are three numbers to multiply? Simply work in sequence, following the rules you have learned.
Example: (-5) (-4) (-20)
(20) (-20) [multiply –5 times –4 to get positive 20]
-400 [multiply positive 20 from last step by –20 to get –400]
And what about exponents? Just write them out and follow the rules.
Example: -4 = (-4) (-4) (-4) (-4) (-4)
(16) (-4) (-4) (-4)
-64 (-4) (-4)
256 (-4)
-1024
Exercises:
1. (10) (-8)
2. (-5) (-26)
3. -5³
4. (-2) (-3)²
5. (-1.5) (4)
6. (-5) (22) (-2)
7. (-3) (-5) (-4)
8. (-1) (-3)
9. (-1)
10. (-2)² (-3)
1. –80
2. 130
3. –125
4. –18
5. –6
6. 220
7. –60
8. –81
9. –1 *
10. 324
*Note: Did you learn something about exponents? If the exponent is even, the answer is positive. If the exponent is odd, the answer is negative.
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# Math: Middle School: Grades 6, 7 and 8 Quiz - Measurement - Inches and Feet (Questions)
This Math quiz is called 'Measurement - Inches and Feet' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.
It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us
This quiz, Measurement - Inches and Feet, challenges you to find the equivalent. A horse’s height is often defined in terms of hands, i.e., 14 hands. In the U.S., most measurements begin with the use of a ruler. A ruler is divided in inches. The standard sized ruler has twelve equal sized inches on it. The full twelve inches is equal to one foot.
The foot measurement often confuses people because they believe that a foot on a ruler is the same as the foot they put into a shoe. For some people that might be true but for most it is not.
A foot in measurement is always equal to twelve (12) inches. Just as a foot is equal to 12 inches, the 14 hands mentioned above would equal 56 inches.
If you were given a measurement of 48 inches, you could turn that number into a more simplified measurement of feet. As there are 12 inches in a foot, you would divide 48 by 12 or 48 ÷ 12 = 4. This tells us that 48 inches is the same measurement as 4 feet.
In turn, if you are given a measurement of 5 feet, you could convert that back into inches. For this we would multiply 5 by 12 (the number of inches in one foot) so, 5 x 12 = 60. There are 60 inches in 5 feet.
If you are given the measurement of 19 inches and wanted to convert that to feet you would divide 19 by 12 as 19 ÷ 12 = 1 with a remainder of 7. This means that 19 inches equals 1 foot 7 inches.
1. 24 feet = ____ inches
[ ] 36 inches [ ] 48 inches [ ] 288 inches [ ] 312 inches
2. 49 inches = ____ feet
[ ] 3 feet 11 inches [ ] 4 feet 1 inch [ ] 4 feet 9 inches [ ] 4 feet 10 inches
3. 68 feet = _____ inches
[ ] 816 inches [ ] 136 inches [ ] 272 inches [ ] 748 inches
4. 132 inches = _____ feet
[ ] 9 feet [ ] 11 feet [ ] 12 feet [ ] 13 feet
5. 864 inches = _____ feet
[ ] 68 feet [ ] 70 feet [ ] 72 feet [ ] 74 feet
6. 28 feet = ____ inches
[ ] 56 inches [ ] 206 inches [ ] 306 inches [ ] 336 inches
7. 107 inches = _____ feet
[ ] 8 feet 11 inches [ ] 8 feet 7 inches [ ] 7 feet 9 inches [ ] 7 feet 11 inches
8. 75 feet = ____ inches
[ ] 875 inches [ ] 900 inches [ ] 975 inches [ ] 1025 inches
9. 41 inches = _____ feet
[ ] 2 feet 1 inch [ ] 2 feet 5 inches [ ] 2 feet 11 inches [ ] 3 feet 5 inches
10. 156 inches = ____ feet
[ ] 10 feet 10 inches [ ] 11 feet 6 inches [ ] 13 feet [ ] 14 feet
Math: Middle School: Grades 6, 7 and 8 Quiz - Measurement - Inches and Feet (Answers)
1. 24 feet = ____ inches
[ ] 36 inches [ ] 48 inches [x] 288 inches [ ] 312 inches
To convert feet into inches we multiply the number by 12 or 24 x 12 = 288 inches. Answer (c) is the correct answer
2. 49 inches = ____ feet
[ ] 3 feet 11 inches [x] 4 feet 1 inch [ ] 4 feet 9 inches [ ] 4 feet 10 inches
To convert inches into feet we must divide 49 inches by 12. 49 ÷ 12 = 4 with a remainder of 1 giving us 4 feet 1 inch - Answer (b) is the correct answer
3. 68 feet = _____ inches
[x] 816 inches [ ] 136 inches [ ] 272 inches [ ] 748 inches
To convert feet into inches we multiply the number by 12 or 68 x 12 = 816 inches. Answer (a) is the correct answer
4. 132 inches = _____ feet
[ ] 9 feet [x] 11 feet [ ] 12 feet [ ] 13 feet
To convert inches into feet we must divide 132 inches by 12. 132 ÷ 12 = 11 feet - Answer (b) is the correct answer
5. 864 inches = _____ feet
[ ] 68 feet [ ] 70 feet [x] 72 feet [ ] 74 feet
To convert inches into feet we must divide 864 inches by 12. 864 ÷ 12 = 72 feet - Answer (c) is the correct answer
6. 28 feet = ____ inches
[ ] 56 inches [ ] 206 inches [ ] 306 inches [x] 336 inches
To convert feet into inches we multiply the number by 12 or 28 x 12 = 336 inches. Answer (d) is the correct answer
7. 107 inches = _____ feet
[x] 8 feet 11 inches [ ] 8 feet 7 inches [ ] 7 feet 9 inches [ ] 7 feet 11 inches
To convert inches into feet we must divide 107 inches by 12. 107 ÷ 12 = 8 with a remainder of 11 giving us 8 feet 11 inches - Answer (a) is the correct answer
8. 75 feet = ____ inches
[ ] 875 inches [x] 900 inches [ ] 975 inches [ ] 1025 inches
To convert feet into inches we multiply the number by 12 or 75 x 12 = 900 inches. Answer (b) is the correct answer
9. 41 inches = _____ feet
[ ] 2 feet 1 inch [ ] 2 feet 5 inches [ ] 2 feet 11 inches [x] 3 feet 5 inches
To convert inches into feet we must divide 49 inches by 12. 41 ÷ 12 = 3 with a remainder of 5 giving us 3 feet 5 inches - Answer (d) is the correct answer
10. 156 inches = ____ feet
[ ] 10 feet 10 inches [ ] 11 feet 6 inches [x] 13 feet [ ] 14 feet
To convert inches into feet we must divide 156 inches by 12. 156 ÷ 12 = 13 feet - Answer (c) is the correct answer
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$A\left( 1,1 \right)$, $B\left( -2,3 \right)$ are two points. If a point $P$ forms a triangle of area $2$ square units with $A,B$, then find the locus of $P$.
Last updated date: 29th Feb 2024
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Hint: We will assume the point $P$ as $\left( h,k \right)$. From the given data we will calculate the area of the triangle formed by the point $P$ with $A\left( 1,1 \right)$, $B\left( -2,3 \right)$ by using the formula $A=\dfrac{1}{2}\left| {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right|$. Substituting the values of ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{y}_{1}},{{y}_{2}},{{y}_{3}}$ from the points $P\left( h,k \right),A\left( 1,1 \right),B\left( -2,3 \right)$. In the problem they have mentioned the area of the triangle as $2$ square units, so we will equate the calculated area and given area. Now simplifying the obtained equation, we will get the equations of the point $P$. To write the locus equation we need to convert the equation in terms of $x,y$.
Given that, $A\left( 1,1 \right)$, $B\left( -2,3 \right)$ are two points.
Let the point $P$ is at $\left( h,k \right)$.
If a triangle is formed with these points, then
$\left( {{x}_{1}},{{y}_{1}} \right)=\left( h,k \right)$,
$\left( {{x}_{2}},{{y}_{2}} \right)=\left( 1,1 \right)$,
$\left( {{x}_{3}},{{y}_{3}} \right)=\left( -2,3 \right)$.
Now the area of the triangle is given by
$A=\dfrac{1}{2}\left| {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right|$
Substituting the values, we have, then we will get
\begin{align} & A=\dfrac{1}{2}\left| h\left( 1-3 \right)+1\left( 3-k \right)-2\left( k-1 \right) \right| \\ & \Rightarrow A=\dfrac{1}{2}\left| -2h+3-k-2k+2 \right| \\ & \Rightarrow A=\dfrac{1}{2}\left| -2h-3k+5 \right| \\ \end{align}
In the problem we have given that the area of the triangle as $2$square units.
\begin{align} & \therefore A=2 \\ & \Rightarrow \dfrac{1}{2}\left| -2h-3k+5 \right|=2 \\ & \Rightarrow \left| -2h-3k+5 \right|=4 \\ \end{align}
To remove the modulus in the above equation we will take the sign $\pm$ to the opposite side of the modulus sign.
$\therefore -2h-3k+5=\pm 4$
Equating $-2h-3k+5$ to $+4$ and $-4$ individually, first we will equate to $+4$, then we will get
\begin{align} & -2h-3k+5=4 \\ & \Rightarrow -2h-3k=4-5 \\ & \Rightarrow -2h-3k=-1 \\ & \Rightarrow 2h+3k-1=0 \\ \end{align}
$2h+3k-1=0...\left( \text{i} \right)$
Now equating $-2h-3k+5$ to $-4$, then we will get
\begin{align} & -2h-3k+5=-4 \\ & \Rightarrow 2h+3k-9=0 \\ \end{align}
$2h+3k-9=0...\left( \text{ii} \right)$
From equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ we can write the locus of the point $P$ as $2x+3y-1=0$ and $2x+3y-9=0$.
Note: Students may use the formula of the triangle as $A=\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|$ and they don’t consider the modulus sign by taking it as determinant. Now students only get a single value for the calculated area which gives us a single equation for the locus of the point. Which is not a correct procedure. So, students should consider the modulus sign while calculating the area in any method. One more point is the locus equation must be in terms of $x,y$, so don’t forget to convert the obtained equations of locus into $x,y$ terms.
|
# How kids can quickly learn multiplication tables?
Multiplication is one of four basic rules of mathematics (addition, subtraction, multiplication, and division). A multiplication table is used to get the product of two numbers. If we learn the basic concept of multiplication, it should be very interesting to solve any multiplication.
In this post, we learn about the definition, notation, rules, and how to find the multiplication table.
## What is a Multiplication table?
In mathematics, a multiplication table is a mathematical table used to define a multiplication operation for an algebraic operation. It is also known as a times table. In simple algebra, multiplication is the process of calculating the results when a number is taken times. The result of the multiplication is called product and each number is known as the factors of the product.
Multiplication sign also known as times sign or dimension sign is the symbol x used in mathematics to denote the multiplication operation and its resulting product.
We can also use the multiplication chart. A multiplication chart is a table that shows the product of two numbers.
A 12×12 multiplication table is given below.
We can use the multiplication chart up to infinite numbers but here we made a multiplication chart for few tables in order to understand the main concept of the multiplication chart.
## Rules of multiplication table
Some basic rules are very important in the multiplication table.
1. Any number multiplied by zero is zero e.g., 2×0 = 0, 5×0 = 0, 100×0 = 0.
2. Any number multiplied by one stays the same or give the result itself e.g., 2×1 = 2, 5×1 = 5, 100×1 = 100.
3. When a number is multiplied by two, we are doubling that number e.g., 2×2 = 4, 5×2 = 10, 100×2 = 200.
4. When a number is multiplied by ten, we simply write zero at the end of that number e.g., 2×10 = 20, 5×10 = 50.
5. When a number is multiplied by a hundred, we simply write two zeros at the end of that number, and in the case of a thousand we simply write three zeros at the end of that number.
## Tips for students
Many educators believe it is necessary to memorize the table up to 9×9. You should learn l, 2, 5, and 10 times tables first. These tables are pretty easy to learn for the students as these tables are simple as compared to others. After this, you should learn 3, 4, 6, 7, 8, 9 times tables. If you want to learn and practice the multiplication tables, use an online multiplication table whichwill help you learn and practice the times tables easily.
## How to learn times tables?
If you want to work out the times table for 2, start with 2 and then add 2 in each step. The result obtained in every step is a multiple of 2 and is known as multiplication fact.
• 2 x 1 = 2
• 2 x 2 = 4
• 2 x 3 = 6
• 2 x 4 = 8
• 2 x 5 = 10
• 2 x 6 = 12
• 2 x 7 = 14
• 2 x 8 = 16
• 2 x 9 = 18
• 2 x 10 = 20
If you want to work out the times table for 5, start with 5 and then add 5 in each step. The result obtained in every step is a multiple of 5.
If you want to work out the times table for 10, start with10and then add 10 in each step. The result obtained in every step is a multiple of 10.
## Benefits of learning Multiplication table
Learning basic multiplication tables will make it easier to learn more challenging ones e.g. if kids know their 2 times table, they will be able to work out their 4 times table by doubling the results.
• 2 x 1 = 2 4 x 1 = 4
• 2 x 2 = 4 4 x 1 = 8
• 2 x 3 = 6 4 x 1 = 12
• 2 x 4 = 8 4 x 1 = 16
• 2 x 5 = 10 4 x 1 = 20
• 2 x 6 = 12 4 x 1 = 24
• 2 x 7 = 14 4 x 1 = 28
• 2 x 8 = 16 4 x 1 = 32
• 2 x 9 = 18 4 x 1 = 36
• 2 x 10 = 20 4 x 1 = 40
When kids successfully recall their time tables they will grow in confidence, which will not only tackle more challenging math problems but will help to keep them motivated and engaged across other objects too.
Recalling times tables improves memory skills, which is a transferrable skill that will help kids throughout school and into adult life.
Kids will find it easier to solve math problems and to do mental arithmetic if they have already memorized their times tables.
Multiplication is used throughout adulthood, whether it is working out price reductions, doubling recipes, or splitting bills.
## Summary
In this post, we have learned the definition of multiplication tables, rules, and some basic concepts. Once you grab the knowledge of this topic you will be master in it.
|
# how many solutions has equation |4x-4|=12? ( one or two? )
jess1999 | Student
| 4x - 4 | = 12
Since there's an absolute value sign, change this equation to
4x - 4 = -12 and 4x -4 =12 now add 4 on both sides on both equation
4x = -8 and 4x = 16 now divide by 4 on both sides and on both equation
By dividing, your equation should look like
x = -2 and x = 4
So there are two solutions
neela | Student
To find the number of solutions : |4x-4| = 12.
When 4x-4 >= 0, |4x-4| = 12 implies 4x- 4. = 12.
4x-4+4 = 12+4 = 16.
4x = 16
4x/4 = 16/4 = 4.
x = 4.
When 4x-4 < = 0, | 4x-4| = 12 implies 4x-12 = -12.
4x-4 +4 = -12 +4 = -8
4x = -8.
Therefore 4x/4 = -8/2 = -2.
x= -2.
Thus there are two solution for x: x= 4 and x = -2.
giorgiana1976 | Student
By definition, the absolute value means:
|p| = a>0
We'll have to solve 2 cases:
1) 4x-4 = 12
4x = 12 + 4
4x = 16
We'll divide by 4:
x = 4
2) 4x-4 = -12
We'll add 4 both sides, to isolate x to the left side:
4x = -12 + 4
4x = -8
We'll divide by 4:
x = -2
The equation has 2 solutions : {-2 ; 4}.
|
# Combined Transformations
Lesson
A transformation is a change, so when we transform a shape, we change it in some way. There are three kinds of transformations: reflections, rotations and transformations. Let's recap these first.
## Reflection (Flip)
We see reflections all the time- in mirrors, in pools of water and so on. A flip is a reflection over a line or axis. We can see in the picture below that the blue object has been reflected over the vertical axis to create the green image. Notice how they are exactly the same distance from the $y$y-axis?
## Rotation (Turn)
A shape is rotated around a centre point in a circular motion. It does not have to be turned in a full circle, otherwise it would be back and the same point. We commonly see $90^\circ$90° turns (also known as quarter turns), $180^\circ$180° turns (half turns) and $270^\circ$270° turns (three-quarter turns). The triangle below has been rotated anticlockwise $90^\circ$90°.
## Translation (Slide)
The whole shape moves the same distance in the same direction, without being rotated or flipped. In the picture below, we can see the diamond has been translated (slid) to the right.
After any of those transformations (rotations, reflections and translations), the shape still has the same size, area, angles and line lengths. However, a shape may be transformed in more than one way. Let's look through some examples now.
#### Examples
##### Question 1
What two transformations would be needed to get from Flag $A$A to Flag $B$B?
1. Rotation and translation
A
Two translations
B
Two reflections
C
Reflection and translation
D
Rotation and translation
A
Two translations
B
Two reflections
C
Reflection and translation
D
##### Question 2
When the original image is rotated $90^\circ$90° clockwise about point $O$O and then translated $3$3 units up, what is the new image?
1. $J$J
A
$H$H
B
$N$N
C
$K$K
D
$J$J
A
$H$H
B
$N$N
C
$K$K
D
##### Question 3
A shape is translated, then rotated about its centre.
The same result can always be obtained by a rotation about its centre, followed by a translation.
True or false?
1. True
A
False
B
True
A
False
B
### Outcomes
#### GM4-8
Use the invariant properties of figures and objects under transformations (refl ection, rotation, translation, or enlargement)
|
# How do I solve for m and n? mx-y=23 nx+y=12
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
mx - y = 23.............(1)
nx + y = 12............(2)
Let us add (1) and (2):
==> nx + mx = 35
==> (n+m) x = 35
==> n + m = 35/ x
==> n = 35/x - m
But : mx - y = 23
==> m = (y+ 23)/x
==> n = 35/x - (y+23)/ x
=> m = 35/x - n
But : nx + y = 12
==> n = 12-y)/ x
==> m = (35/x) - (12-y)/x
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
We have to find m and n using the equations :
mx - y = 23...(1)
nx + y = 12...(2)
Now we see that there are 4 terms m, n, x and y in the two equations given. So we can only write 2 of them in terms of the others.
Using (1), we can write, m in terms of x and y as
mx - y = 23
=> m = (23 + y) / x
Using (2), we can write n in terms of x and y as
nx + y = 12
=> n = (12 - y) / x
Therefore m and n in terms of x and y are m=(23 + y)/x and n=(12 - y)/x
givingiswinning | Student, Grade 10 | (Level 1) Valedictorian
Posted on
mx - y = 23
mx = y + 23
`m = (y + 23)/x`
The second one :
nx + y= 12
nx = 12 - y
` n = (12 - y)/x `
atyourservice | Student, Grade 11 | (Level 3) Valedictorian
Posted on
mx-y=23
the first step is to add y
mx = 23 + y
divide by x because were are trying to solve for m
`x = 23/x + y/x ` this can also be written as `x = (23 + y) / x`
nx+y=12
subtract y
nx = 12 - y
divide by x
`n = 12/x - y/x` which can also be written as `n=(12-y)/x`
Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian
Posted on
mx-y=23
nx+y=12
To solve for a variable in terms of another variable means to isolate the variable you want from the rest.
Lets deal with the first one :
mx - y = 23
mx = y + 23
m = (y + 23)/x
Now the second one :
nx + y= 12
nx = 12 - y
n = (12 - y)/x
neela | High School Teacher | (Level 3) Valedictorian
Posted on
To solve for m and n in equations:
mx-y=23...(1).
nx+y=12...(2).
We can treat both the graphs as independent, as there are no relationship given between the graphs at (1) and (2).
Both equations represent different straight lines.
From the first equation, mx-y = 23,
mx = 23+y
m = (23+y)/x is the solution for m.
From the 2nd equation nx+y = 12. we get:
nx = 12-y
n = (12-y)/x.
Therefore the solution for n and m are m = (23+y)/x and n = (12-y)/x.
|
Home Polynomial Functions Introduction Investigating Polynomial Functions Polynomial Division Remainder & Factor Theorems Polynomial Equations Polynomial Inequalities Rational Functions Review&Test
UNIT 4 : POLYNOMIAL AND RATIONAL FUNCTIONS
LESSON 1: POLYNOMIAL FUNCTIONS INTRODUCTION
Example 1:
f(x) = x3 4x2 + x + 6 is a cubic polynomial function because the largest exponent of the variable is 3. Make a table of values or use
a graphing calculator to draw its graph [see below].
Note: The function has 3 zeros (x intercepts), x = -1, 2, 3 and two
turning points A and B. Turning point A is a local maximum point as
the function changes from increasing to decreasing at A. Turning point B
is a local minimum point as the function changes from decreasing to increasing
at B. The function has a local maximum value of y = 6.1 at point A and a
local minimum value of y = -0.9 at point B.
Example 2:
Use your graphing calculator to graph and study the cubic functions below.
a) f(x) = x3 x2 + 5x 3 b) f(x) = (x 3)3 + 1
c) f(x) = x3 + 3x2 + 3x + 2
Note: From the examples above we can make the following observations.
• A cubic function can have 1, 2 or 3 zeros (x-intercepts)
• It can have 0 or 2 turning points.
• The coefficient of x3 is called the leading coefficient (k). In (a) the leading coefficient is negative(k = -1) and the function rises to the left and falls to the right.
• The leading coefficient(k) determines the end behaviours. If k > 0, the function falls to the left and rises to the right (ex. 1, 2 b,c)
If k < 0, the function rises to the left and falls to the right (ex. 2a). This characteristic is true for all polynomial functions of odd degree (1, 3, 5,
).
Example 3:
Use your graphing calculator to graph and study the quartic functions below.
a) f(x) = x4 + x3 5x2 3x b) y = x4 + x3 2x2 3x
c) y = -0.5x4 x3 + 2x2 5 d) y = x4 + x3 2x2 3x + 3
Note: From the examples above we can make the following observations.
• A quaric function can have 0,1, 2, 3 or 4 zeros (x-intercepts)
• It can have 1 or 3 turning points.
• The coefficient of x4 is called the leading coefficient (k). In (c) the leading coefficient is negative(k = -0.5) and the function falls to the left and right.
In (a,b,d) the leading coefficient is positive and the function rises to the left and right
• The leading coefficient(k) determines the end behaviours. If k > 0, the function rises to the left and right (ex. 3 a,b,d)
If k < 0, the function falls to the left and right (ex. 3c). This characteristic is true for all polynomial functions of even degree (2, 4, 6,
).
Polynomial Functions in Factored Form f(x) = k(x a)(x b)(x c) . . . etc.
# Linear Functions in factored Form:f(x) = k(x a)
# Cubic Functions in factored Form:f(x) = k(x a)(x b)(x c)
x y -2 6 0 -6
# Quartic Functions in factored Form:f(x) = k(x a)(x b)(x c)(x d)
x y -1 16 2 16
|
Sectors, Areas, and Arcs
# Sectors, Areas, and Arcs | The Complete SAT Course - Class 10 PDF Download
### Introduction
• As you may remember from geometry, the area A of a circle having a radius of length r is given:
A = πr2
• The circumference C (that is, the length around the outside) of that same circle is given by:
C = 2πr
• These are the formulas give us the area and arc-length (that is, the length of the "arc", or curved line) for the entire circle. But sometimes we need to work with just a portion of a circle's revolution, or with many revolutions of the circle. What formulas do we use then?
• If we start with a circle with a marked radius line, and turn the circle a bit, the area marked off looks something like a wedge of pie or a slice of pizza; this is called a "sector" of the circle, and the sector looks like the green portion of this picture:
• The angle marked off by the original and final locations of the radius line (that is, the angle at the center of the pie / pizza) is the "subtended" angle of the sector. This angle can also be referred to as the "central" angle of the sector. In the picture above, the central angle is labelled as "θ" (which is pronounced as "THAY-tuh").
Affiliate
• What is the area A of the sector subtended by the marked central angle θ? What is the length s of the arc, being the portion of the circumference subtended by this angle?
• To determine these values, let's first take a closer look at the area and circumference formulas. The area and circumference are for the entire circle, one full revolution of the radius line. The subtended angle for "one full revolution" is 2π. So the formulas for the area and circumference of the whole circle can be restated as:
C = (2π)r
• What is the point of splitting the angle value of "once around" the circle? I did this in order to highlight how the angle for the whole circle (being 2π) fits into the formulas for the whole circle. This then allows us to see exactly how and where the subtended angle θ of a sector will fit into the sector formulas. Now we can replace the "once around" angle (that is, the 2π) for an entire circle with the measure of a sector's subtended angle θ, and this will give us the formulas for the area and arc length of that sector:
s = (θ) r
Note: If you are working with angles measured in degrees, instead of in radians, then you'll need to include an extra conversion factor:
Confession: A big part of the reason that I've explained the relationship between the circle formulas and the sector formulas is that I could never keep track of the sector-area and arc-length formulas; I was always forgetting them or messing them up. But I could always remember the formulas for the area and circumference of an entire circle. So I learned (the hard way) that, by keeping the above relationship in mind, noting where the angles go in the whole-circle formulas, it is possible always to keep things straight.
### Example
1. Given a circle with radius r = 8 units and a sector with subtended angle measuring 45°, find the area of the sector and the length of the arc.
They've given me the radius and the central angle, so I can just plug straight into the formulas, and simplify to get my answers. For convenience, I'll first convert "45°" to the corresponding radian value of π/4. Then I'll do my plug-n-chug:
area A = 8π square units, arc-length s = 2π units
• Notice how I put "units" on my answers. If they'd stated a specific unit for the radius, like "centimeters" or "miles" or whatever, then I could have been more specific in my answer. As it was, I had to be generic.
• Many times, if the question doesn't state a unit, or just says "units", then you can probably get away without putting "units" on your answer. However, this often leads to the bad habit of ignoring units entirely, and then — surprise! — the instructor counts off on the test because you didn't include any units. It's probably better to err on the side of caution, and always put some unit, even if it's just "units", on your answers.
2. Given a sector with radius r = 3 cm and a corresponding arc length of 5π radians, find the area of the sector.
• For this exercise, they've given me the radius and the arc length. To find the area of the sector, I need the measure of the central angle, which they did not give me. However, the formula for the arc length includes the central angle. So I can plug the radius and the arc length into the arc-length formula, and solve for the measure of the subtended angle. Once I've got that, I can plug-n-chug to find the sector area.
• So the central angle for this sector measures 5/3 π. Then the area of the sector is:
• And this value is the numerical portion of my answer. Since this value stands for "area", which is a square dimension, I'll want to remember to put "squared" on the units they gave me for the radius.
A = 15π/2 cm2
• Sometimes, an exercise will give you information, but, like the above, it might not seem like it's the information that you actually need. Don't be afraid to fiddle with the values and the formulas; try to see if you can figure out a back door in to a solution, or some other manipulation that'll give you want you need. It's okay not to know, right at the beginning, how you're going to reach the end. (And, if they give you, or ask for, the diameter, remember that the radius is half of the diameter, and the diameter is twice the radius.)
3. A circle's sector has an area of 108 cm2, and the sector intercepts an arc with length 12 cm. Find the diameter of the circle.
• They've asked me for the diameter. The formulas I've learned use the radius. But I can find the radius, and then double it to get the diameter, so that's not a problem. However, they've asked me for a length, given the arc length and the area, each of which uses the radius and the subtended angle. And I have neither of those values. So what do I do?
• When I can't think of anything else to do, I plug whatever they've given me into whatever formulas might relate, and I hope something drops out of it that I can use. So:
s = θr = 12
• I can substitute from the second line above into the first line above (after some rearrangement), and see if the result helps me at all:
• I found the value for the radius! I don't have the value for the central angle, but they didn't ask for that, and it turns out that I didn't need it anyway. They asked me for the diameter, which is twice the radius, so my answer (including the units!) is:
d = 36 cm
The document Sectors, Areas, and Arcs | The Complete SAT Course - Class 10 is a part of the Class 10 Course The Complete SAT Course.
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## The Complete SAT Course
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### Posts Tagged ‘Math10’
Using elimination to solve a system:
*note you are trying to get zero pairs.
Example:
Step 1: to get a zero pair you need a negative and a positive number or variable. so with that said if you take a look at the example, what do you think is the easiest coefficient to multiply into any of these two equations to get a zero pair. remember there is more than one type of coefficient that can be multiplied into an equation to get a zero pair. I chose to multiply the second equation by -3.
Step 2: start solving for y by plugging in the x value into the first equation.
How to use substitution while solving a system.
Example:
*note we are trying to find the values of x and y.
Step 1: choose the equation with the with a simple variable, which is usually the one that has a coefficient of 1 in front of it. In this case since there is a 1 in front of both the y and x, you can choose whichever one you want. I am going to choose the y.
Step 2: we have to get the y by itself, so we have to move the 2x to the other side making it negative.
Step 3: plug in the y = -2x + 7 into the other equation and let algebra take over.
Step 4: once you have your answer, plug it into one of your equations. It doesn’t really matter which one you plug it into, but remember to work smart and choose which one is the easier choice. Once you have plugged it in, let algebra take over again and solve for y.
This week I learned the equation, slope y- intercept form. Which is this formula y=mx+b. Your main goal here is to isolate the y, below I have an example of how to get an equation into slope y. Which really means, to isolate the y.
Ex.
Step 1: move #’s to the other side to isolate the y.
Step 2: use algebra to get the y by itself.
This week I learned how to find the slope of parallel and perpendicular lines. What this means is that we can use the slope of a line to tell us whether two lines are parallel or perpendicular. Two lines are parallel if they have the same slope, and never cross. Two lines are perpendicular if the products of their slope is -1.
Wonky Initials:
Initials:
How to solve an equation using an output:
When solving an equation using an output you have to look at it backwards.
For example take a look at this equation.
You’re still trying to get x by itself and in order to do this you have to make a zero pair by adding 4 to each side. After you have done that you use BEDMAS which tells you to divide both sides by 7 which gives you x = 6.
How to find x and y intercepts in an equation:
Let’s say we have the following equation: y = 4x + 7
keep in mind that the x intercept – (x,0) y always = 0 and that the y intercept – (y,0) x always = 0.
Also remember that you’ll have to solve for x and y separately which will give you the coordinates for x and y on a graph.
Solving for x:
Solving for y:
Taking too long?
| Open in new tab
How to use a table of values to show a relation:
An example would be: The cost of sour keys is related to the weight. The first step would be to figure out what the independent and dependent variables are. A good way to look at this is to think about which one depends on the other. Does the cost determine the weight of the sour keys, or does the weight of the sour keys determine the cost of the sour keys. I guess we can all say that when we buy sour keys at a store the amount (weight) of the sour keys always determines how much (cost) of the sour keys will be. The cost is the dependent variable and the weight is the independent variable, because the cost depends on the weight of the sour keys.
When we put this into a table this is how it will look:
Since we are finding out the cost of something we have to use some common sense, which in this case is that money can’t be negative numbers so we have to start at 0 and make our way up.
The equation for this relation would be:
The to figure out this equation would be to find out what the independent values which is the y column, goes up by.
This week I didn’t understand how to get the answer for the following question because I didn’t know the steps that had to be taken in order to solve this question. After asking for help I understood what to do to answer this question correctly. Below are the steps I took in order to solve this question.
```Factoring simple trinomials when they have something in common:
```
|
eigen values
# eigen values - Eigenvalues and Eigenvectors More Direction...
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1 Eigenvalues and Eigenvectors, More Direction Fields and Systems of ODEs First let us speak a bit about eigenvalues. Defn. An eigenvalue λ of an nxn matrix A means a scalar (perhaps a complex number) such that A v = λ v has a solution v which is not the 0 vector. We call such a v an eigenvector of A corresponding to the eigenvalue λ . Note that A v = λ v if and only if 0 = A v - λ v = (A- λ I) v , where I is the nxn identity matrix. Moreover, (A- λ I) v =0 has a non-0 solution v if and only if det(A- λ I)=0. This gives: Theorem. The scalar λ is an eigenvalue of the nxn matrix A if and only if det(A- λ I)=0. Eigenvalue Example. Find the eigenvalues of the matrix By the preceding theorem, we need to solve det(A- λ I)=0. That is 21 . 12 A ⎛⎞ = ⎜⎟ ⎝⎠ 1 0 det 0 1 2 1 det 0. λ ⎩⎭ −− ==
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2 (2+ λ ) 2 -1=0 4+4 λ + λ 2 -1=0 3+4 λ + λ 2 =0 (3+ λ )(1+ λ )=0 λ =-1,-3. These are the eigenvalues of A. You can use Gaussian elimination to find the corresponding eigenvectors. (A+3I) v = 0 is solved as follows: 21 3 0 1 1 3. 12 0 3 1 1 AI ⎛⎞ += + = ⎜⎟ ⎝⎠ To find v , do the Gauss thing: 11 0 1 1 0 . 0 0 0 0 Here we replaced row 2 by row 2 – row 1. Now a solution v of (A+3I) v =0 means 1 2 0. v v such that v v v =+ = So v 1 =-v 2 . Take v 2 =-1 (the free unknown coefficient) and then v 1 =1. An eigenvector of A for the eigenvalue -3 is Any scalar multiple of v is also an eigenvector. 1 . 1 v =
3 Similarly we find a solution w of (A+I) w =0.
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## This note was uploaded on 10/19/2009 for the course MATH MATH 20D taught by Professor Staff during the Spring '09 term at UCSD.
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eigen values - Eigenvalues and Eigenvectors More Direction...
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# dx/dy in three different ways
Up a level : Differentiation, derivatives
Previous page : Implicit differentiation
Next page : The circle and implicit derivatives
Say you have a function
$y = {x^2},\quad x \geqslant 0$
The derivative of this is then
$y' = \frac{{dy}}{{dx}} = 2x$
Now, what is dx/dy? The derivative dy/dx gives us how fast y is changing as we change x, and dx/dy gives us how fast x is changing as we change y. We can find this in several ways.
Using the inverse function
The most natural way might be to make x the subject and then take the derivative. We thus switch the roles of x and y. We get
$x = \sqrt y$
and then
$\frac{{dx}}{{dy}} = \frac{1}{{2\sqrt y }}$
Here we have to remember that x is now a function of our variable y.
Using the reciprocal
If dy/dx=m, could it be that dx/dy=1/m? Could it be that simple?
Normally we see dy/dx as one symbol the derivative of y with respect to x. But could the parts dy and dx take on roles as separate entities? Let us say we have some function, and that the derivative at a particular point is m, and if we draw a tangent line at that point, then the tangent line would have the slope m at every point.
We may then choose to let dy and dx represent any two values such that dy/dx=m. If so then we may start to manipulate them as any variables and thus do the following
$\frac{{dy}}{{dx}} = m,\quad dy = mdx,\quad 1 = m\frac{{dx}}{{dy}},\quad \frac{1}{m} = \frac{{dx}}{{dy}}$
We can also show this by
$\begin{gathered} \frac{1}{{\;\frac{{dy}}{{dx}}\;}} = \frac{1}{{\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}}}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{\frac{{\Delta y}}{{\Delta x}}}} \hfill \\ \quad \quad = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta x}}{{\Delta y}} = \mathop {\lim }\limits_{\Delta y \to 0} \frac{{\Delta x}}{{\Delta y}} = \frac{{dx}}{{dy}} \hfill \\ \end{gathered}$
The last step can be done since, if the derivative exists, both Δx and Δy will go toward 0 simultaneously.
This would now give us, for our function, y=x2 that
$\frac{{dx}}{{dy}} = \frac{1}{{2x}}$
but since
$x = \sqrt y$
we get
$\frac{{dx}}{{dy}} = \frac{1}{{2\sqrt y }}$
Using implicit differentiation
We may also take the derivative of y=x2 with respect to x to get
$1 = 2x \cdot x' = 2x\frac{{dx}}{{dy}}$
or
$\frac{{dx}}{{dy}} = \frac{1}{{2x}} = \frac{1}{{2\sqrt y }}$
Another example
Let
$y = {x^3}$
• The derivative of the inverse function
$x = {y^{1/3}}$
So
$\frac{{dx}}{{dy}} = \frac{1}{3}{y^{\frac{1}{3} - 1}} = \frac{1}{3}{y^{ - \frac{2}{3}}} = \frac{1}{{3{y^{2/3}}}}$
• The reciprocal
The derivative dy/dx is y´=3x2, so
$\frac{{dx}}{{dy}} = \frac{1}{{3{x^2}}}$
But
$x = {y^{1/3}}$
so
${x^2} = {y^{2/3}}$
and thus
$\frac{{dx}}{{dy}} = \frac{1}{{3{y^{2/3}}}}$
• The implicit differential
The derivative with respect to y of
$y = {x^3}$
Gives us
$1 = 3{x^2}\frac{{dx}}{{dy}}$
So
$1 = 3{x^2}\frac{{dx}}{{dy}} = \frac{1}{{3{x^2}}}$
And then we do as above.
Up a level : Differentiation, derivatives
Previous page : Implicit differentiation
Next page : The circle and implicit derivativesLast modified: Dec 27, 2023 @ 21:30
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Simple Explanation Of Rabin-Karp Pattern Matching Algorithm
In this post, I will try to explain Rabin-Karp algorithm & its worst time complexity. Rabin-Karp algorithm is based on hash matching.
Let’s take an example.
`Text: aadabcPattern: abc`
Let’s create a table as below & represent each letter with some arbitrary number. We will use it to calculate hash.
`a b c d e1 2 3 4 5`
So how do we calculate the hash value? Our hash function is simple. We will add up the number values as per the table.
`Pattern hash: a + b + c = 1 + 2 + 3 = 6`
For comparing the text, we will take first 3 characters as our pattern length is 3. Then we will shift one letter in each step & recalculate hash.
`Text hash: t[1] + t[2] + t[3] = a + a + d = 1 + 1 + 4 = 6`
So hash matched. We need to compare each character from 1 to 3 position with the characters in the pattern.
`t: aadabcp: abc`
t[1] = p[1], match length 1, increment both t & p.
`t: aadabcp: abc`
t[2] ≠ p[2], matching failed. So t[1] to t[3] is not matching with the pattern. Now we will match pattern against text characters from t[2] to t[4].
`Text hash: t[2] + t[3] + t[4] = a + d + a = 1 + 4 + 1 = 6`
Patten hash & text hash matched again. We need to compare each text character from 2 to 4 position with the characters in the pattern.
`t: aadabcp: abc`
t[2] = p[1], match length 1, increment both t & p.
`t: aadabcp: abc`
t[3] ≠ p[2], matching failed. So t[2] to t[4] is not matching with the pattern. We need to match pattern against t[3] to t[5].
`Text hash: t[3] + t[4] + t[5] = d + a + b = 4 + 1 + 2 = 7`
Hash matching failed. We will move forward & match pattern against t[4] to t[6].
`Text hash: t[4] + t[5] + t[6] = a + b + c = 1 + 2 + 3 = 6`
So hash matched. We need to compare each character from 4 to 6 position with the characters in the pattern.
`t: aadabcp: abc`
t[4] = p[1], match length 1, increment both t & p.
`t: aadabcp: abc`
t[5] = p[2], match length 2, increment both t & p.
`t: aadabcp: abc`
t[6] = p[3], match length 3, so we found our match.
Now let’s discuss few drawbacks of above example. When we are shifting one position in the text, we are recalculating the entire hash.
If we have pattern length 20, we will add up 20 values in each text hash calculation step. If we have pattern length 10, we will add 10 values in each text hash calculation step. So it’s not constant & time complexity of hash function depends on pattern length m.
If we use normal hash function, worst time complexity will be O(mn). That’s why we need to use concept of rolling hash function. It uses sliding window technique.
When we shift one position, we will remove first position value from previous hash & add current position value to that. So whatever the value of m is, we can recalculate text hash with only two operations in each step.
That is constant & that will bring down time complexity to O(n).
Here is how we can do that.
```Text hash: t[1] + t[2] + t[3] = a + a + d
= 1 + 1 + 4
= 6```
To get hash for t[2] to t[4], we can do following.
`t[1] + t[2] + t[3] = 66 - t[1] = 55 + t[4] = 6`
So, t[2] + t[3] + t[4] = 6
There is one more thing. Whenever there is a hash match, we need to match pattern with that portion of text & verify if characters are same. In the example above, we saw that hash value is same but characters are not.
These false positives will increase time complexity. And if you are using a bad hash function, then you might get false positives in every step. You might need to compare half of the pattern characters on an average before you find a mismatch.
So time complexity on that case becomes (m / 2) * n or O(mn). That’s why Rabin-Karp algorithm has a worst time complexity of O(mn).
We can do some improvement on the hash function that can reduce time complexity to O(m+n) in most of the cases. I will try to write it in some future post.
But I hope you have got a clear understanding of this string searching algorithm & why its worst time complexity is O(mn).
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# How do you solve abs(x+1)+ abs(x-1)<=2?
May 30, 2015
With moduli it is useful to split into cases at the point that the sign of the enclosed expression changes. $\left(x + 1\right)$ changes sign at $x = - 1$ and $\left(x - 1\right)$ changes sign at $x = 1$....
Case (a) : $x < - 1$
$x + 1 < 0$ so $\left\mid x + 1 \right\mid = - \left(x + 1\right)$
$x - 1 < 0$ so $\left\mid x - 1 \right\mid = - \left(x - 1\right)$
$2 \ge \left\mid x + 1 \right\mid + \left\mid x - 1 \right\mid = - \left(x + 1\right) + - \left(x - 1\right)$
$= - x - 1 - x + 1$
$= - 2 x$
Dividing both ends by $- 2$ and reversing the inequality we get:
$x \ge - 1$
Notice that we have to reverse the inequality, because we are dividing by a negative number.
So in Case (a) we have $x < - 1$ and $x \ge - 1$
These conditions cannot be satisfied at the same time, so Case (a) yields no solutions.
Case (b) : $- 1 \le x \le 1$
$x + 1 \ge 0$ so $\left\mid x + 1 \right\mid = \left(x + 1\right)$
$x - 1 \le 0$ so $\left\mid x - 1 \right\mid = - \left(x - 1\right)$
$\left\mid x + 1 \right\mid + \left\mid x - 1 \right\mid = \left(x + 1\right) + - \left(x - 1\right) = 1 + 1 = 2 \le 2$
So the target inequality is satisfied for all $x$ in Case (b).
That is $- 1 \le x \le 1$.
Case (c) : $x > 1$
$x + 1 > 0$ so $\left\mid x + 1 \right\mid = \left(x + 1\right)$
$x - 1 > 0$ so $\left\mid x - 1 \right\mid = \left(x - 1\right)$
$\left\mid x + 1 \right\mid + \left\mid x - 1 \right\mid = \left(x + 1\right) + \left(x - 1\right) = 2 x > 2$
So in Case (c) the inequality is never satisfied.
So the solution is $- 1 \le x \le 1$
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Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 2 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among Class 10 students for Math Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.
#### Question 1:
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
${x}^{2}+7x+12$
${x}^{2}+7x+12=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x+3x+12=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+4\right)+3\left(x+4\right)=0$
Sum of zeroes =
Product of zeroes =
#### Question 2:
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
${x}^{2}-2x-8$
${x}^{2}-2x-8=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-4x+2x-8=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-4\right)+2\left(x-4\right)=0$
Sum of zeroes = 4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes = (4)(3)=121=constant termcoefficient of x2
#### Question 3:
Find the zeros of the quadratic polynomial (x2 + 3x − 10) and verify the relation between its zeros and coefficients.
#### Question 4:
Find the zeros of the quadratic polynomial 4x2 − 4x − 3 and verify the relation between the zeros and the coefficients.
#### Question 5:
Find the zeros of the quadratic polynomial 5x2 − 4 − 8x and verify the relationship between the zeros and the coefficients of the given polynomial.
#### Question 6:
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$2\sqrt{3}{x}^{2}-5x+\sqrt{3}$
$2\sqrt{3}{x}^{2}-5x+\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒2\sqrt{3}{x}^{2}-2x-3x+\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒2x\left(\sqrt{3}x-1\right)-\sqrt{3}\left(\sqrt{3}x-1\right)=0$
Sum of zeroes = 4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes = (4)(3)=121=constant termcoefficient of x2
#### Question 7:
Find the zeros of the quadratic polynomial 2x2 − 11x + 15 and verify the relation between the zeros and the coefficients.
#### Question 8:
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$4{x}^{2}-4x+1$
Sum of zeroes = 4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes = (4)(3)=121=constant termcoefficient of x2
#### Question 9:
Find the zeros of the quadratic polynomial (x2 − 5) and verify the relation between the zeros and the coefficients.
#### Question 10:
Find the zeros of the quadratic polynomial (8 x2 − 4) and verify the relation between the zeros and the coefficients.
#### Question 11:
Find the zeros of the quadratic polynomial (5y2 + 10y) and verify the relation between the zeros and the coefficients.
Hence, the relation has been verified.
#### Question 12:
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$3{x}^{2}-x-4$
$3{x}^{2}-x-4=0\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-4x+3x-4=0\phantom{\rule{0ex}{0ex}}⇒x\left(3x-4\right)+1\left(3x-4\right)=0$
Sum of zeroes = 4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes = (4)(3)=121=constant termcoefficient of x2
#### Question 13:
Find the quadratic polynomial whose zeros are 2 and −6. Verify the relation between the coefficients and the zeros of the polynomial.
#### Question 14:
Find the quadratic polynomial whose zeros are $\frac{2}{3}$ and $\frac{-1}{4}$. Verify the relation between the coefficients and the zeros of the polynomial.
#### Question 15:
Find the quadratic polynomial, sum of whose zeros is 8 and their product is 12. Hence, find the zeros of the polynomial.
#### Question 16:
Find the quadratic polynomial, the sum of whose zeros is 0 and their product is −1. Hence, find the zeros of the polynomial.
#### Question 17:
Find the quadratic polynomial, the sum of whose zeros is $\left(\frac{5}{2}\right)$ and their product is 1. Hence, find the zeros of the polynomial.
#### Question 18:
Find the quadratic polynomial, the sum of whose root is $\sqrt{2}$ and their product is $\frac{1}{3}$
We can find the quadratic equation if we know the sum of the roots and product of the roots by using the formula
x2 − (Sum of the roots)x + Product of roots = 0
$⇒{x}^{2}-\sqrt{2}x+\frac{1}{3}=0\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-3\sqrt{2}x+1=0$
#### Question 19:
If $x=\frac{2}{3}$ and $x=-3$ are the roots of the quadratic equation $a{x}^{2}+7x+b=0$ then find the values of a and b.
Given: $a{x}^{2}+7x+b=0$
Since, $x=\frac{2}{3}$ is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
Since, $x=-3$ is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
From (1) and (2), we get
#### Question 20:
If $\left(x+a\right)$ is a factor of the polynomial $2{x}^{2}+2ax+5x+10$, find the value of a.
Given: $\left(x+a\right)$ is a factor of $2{x}^{2}+2ax+5x+10$
So, we have
$x+a=0\phantom{\rule{0ex}{0ex}}⇒x=-a$
Now, It will satisfy the above polynomial.
Therefore, we will get
$2{\left(-a\right)}^{2}+2a\left(-a\right)+5\left(-a\right)+10=0\phantom{\rule{0ex}{0ex}}⇒2{a}^{2}-2{a}^{2}-5a+10=0\phantom{\rule{0ex}{0ex}}⇒-5a=-10\phantom{\rule{0ex}{0ex}}⇒a=2$
#### Question 21:
One zero of the polynomial $3{x}^{3}+16{x}^{2}+15x-18$ is $\frac{2}{3}$. Find the other zeroes of the polynomial.
Given: $x=\frac{2}{3}$ is one of the zero of $3{x}^{3}+16{x}^{2}+15x-18$
Now, we have
$x=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒x-\frac{2}{3}=0$
Now, we divide $3{x}^{3}+16{x}^{2}+15x-18$ by $x-\frac{2}{3}$ to find the quotient
So, the quotient is $3{x}^{2}+18x+27$
Now,
$3{x}^{2}+18x+27=0\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}+9x+9x+27=0\phantom{\rule{0ex}{0ex}}⇒3x\left(x+3\right)+9\left(x+3\right)=0$
#### Question 1:
Verity that 3, −2, 1 are the zeros of the cubic polynomial p(x) = x3 − 2x2 − 5x + 6 and verify the relation between its zeros and coefficients.
#### Question 2:
Verify that 5, −2 and $\frac{1}{3}$ are the zeros of the cubic polynomial p(x) = 3x3 − 10x2 − 27x + 10 and verify the relation between its zeros and coefficients.
#### Question 3:
Find a cubic polynomial whose zeroes are 2, −3 and 4
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
${x}^{3}-\left(a+b+c\right){x}^{2}+\left(ab+bc+ca\right)x-abc$ ...(1)
Let
Substituting the values in (1), we get
${x}^{3}-\left(2-3+4\right){x}^{2}+\left(-6-12+8\right)x-\left(-24\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}-3{x}^{2}-10x+24$
#### Question 4:
Find a cubic polynomial whose zeroes are $\frac{1}{2}$, 1 and −3.
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
${x}^{3}-\left(a+b+c\right){x}^{2}+\left(ab+bc+ca\right)x-abc$ ...(1)
Let
Substituting the values in (1), we get
${x}^{3}-\left(\frac{1}{2}+1-3\right){x}^{2}+\left(\frac{1}{2}-3-\frac{3}{2}\right)x-\left(\frac{-3}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}-\left(\frac{-3}{2}\right){x}^{2}-4x+\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{3}+3{x}^{2}-8x+3$
#### Question 5:
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes are 5, −2 and −24 respectively.
We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as
x3 −(Sum of the zeroes)x2 + (sum of the product of the zeroes taking two at a time)x − Product of zeroes
Therefore, the required polynomial is
${x}^{3}-5{x}^{2}-2x+24$
#### Question 6:
Find the quotient and the remainder when:
$f\left(x\right)={x}^{3}-3{x}^{2}+5x-3$ is divided by $g\left(x\right)={x}^{2}-2$
Quotient $q\left(x\right)=x-3$
Remainder $r\left(x\right)=7x-9$
#### Question 7:
Find the quotient and remainder when:
$f\left(x\right)={x}^{4}-3{x}^{2}+4x+5$ is divided by $g\left(x\right)={x}^{2}+1-x$
Quotient $q\left(x\right)={x}^{2}+x-3$
Remainder $r\left(x\right)=8$
#### Question 8:
Find the quotient and remainder when
$f\left(x\right)={x}^{4}-5x+6$ is divided by $g\left(x\right)=2-{x}^{2}$
We can write
and
Quotient $q\left(x\right)=-{x}^{2}-2$
Remainder $r\left(x\right)=-5x+10$
#### Question 9:
By actual division, show that ${x}^{2}-3$ is a factor of $2{x}^{4}+3{x}^{3}-2{x}^{2}-9x-12$
Let $f\left(x\right)=2{x}^{4}+3{x}^{3}-2{x}^{2}-9x-12$ and $g\left(x\right)={x}^{2}-3$
Quotient $q\left(x\right)=2{x}^{2}+3x+4$
Remainder $r\left(x\right)=0$
Since, the remainder is 0.
Hence, ${x}^{2}-3$ is a factor of $2{x}^{4}+3{x}^{3}-2{x}^{2}-9x-12$
#### Question 10:
On dividing, $3{x}^{3}+{x}^{2}+2x+5$ by a polynomial g(x), the quotient and remainder are $3x-5$ and $9x+10$ respectively. Find g(x)
By using division rule, we have
Divided = Quotient × Divisor + Remainder
$\therefore 3{x}^{3}+{x}^{2}+2x+5=\left(3x-5\right)g\left(x\right)+9x+10\phantom{\rule{0ex}{0ex}}⇒3{x}^{3}+{x}^{2}+2x+5-9x-10=\left(3x-5\right)g\left(x\right)\phantom{\rule{0ex}{0ex}}⇒3{x}^{3}+{x}^{2}-7x-5=\left(3x-5\right)g\left(x\right)\phantom{\rule{0ex}{0ex}}⇒g\left(x\right)=\frac{3{x}^{3}+{x}^{2}-7x-5}{3x-5}$
$\therefore g\left(x\right)={x}^{2}+2x+1$
#### Question 11:
Verify division algorithm for the polynomials $f\left(x\right)=8+20x+{x}^{2}-6{x}^{3}$ and $g\left(x\right)=2+5x-3{x}^{2}$
We can write and
Quotient = $2x+3$
Remainder = $x+2$
By using division rule, we have
Divided = Quotient × Divisor + Remainder
#### Question 12:
It is given that −1 is one of the zeros of the polynomial x3 + 2x2 − 11x − 12. Find all the given zeros of the given polynomial.
#### Question 13:
If 1 and −2 are two zeros of the polynomial (x3 − 4x2 − 7x + 10), find its third zero.
#### Question 14:
If 3 and −3 are two zeros of the polynomial (x4 + x3 − 11x2 − 9x + 18), find all the zeros of the given polynomial.
#### Question 15:
If 2 and −2 are two zeros of the polynomial (x4 + x3 − 34x2 − 4x + 120), find all the zeros of given polynomial.
#### Question 16:
Find all the zeros of (x4 + x3 − 23x2 − 3x + 60), if it is given that two of its zeros are $\sqrt{3}$ and $-\sqrt{3}$.
#### Question 17:
Find all the zeros of (2x4 − 3x3 − 5x2 + 9x − 3), it being given that two of its zeros are $\sqrt{3}$ and $-\sqrt{3}$.
#### Question 18:
Obtain all other zeros of (x4 + 4x3 − 2x2 − 20x − 15) if two of its zeros are $\sqrt{5}$ and $-\sqrt{5}$.
#### Question 19:
Find all the zeros of the polynomial (2x4 − 11x3 + 7x2 + 13x), it being given that two if its zeros are $3+\sqrt{2}$ and $3-\sqrt{2}$.
#### Question 1:
If one zero of the polynomial ${x}^{2}-4x+1$ is $2+\sqrt{3}$ . Write the other zero. [CBSE 2010]
Let the other zeroes of ${x}^{2}-4x+1$ be a.
By using the relationship between the zeroes of the quadratic ploynomial.
We have, Sum of zeroes =
$\therefore 2+\sqrt{3}+a=\frac{-\left(-4\right)}{1}\phantom{\rule{0ex}{0ex}}⇒a=2-\sqrt{3}$
Hence, the other zeroes of ${x}^{2}-4x+1$ is $2-\sqrt{3}$.
#### Question 2:
Find the zeros of the polynomial x2 + x − p(p + 1). [CBSE 2011]
$f\left(x\right)={x}^{2}+x-p\left(p+1\right)$
By adding and subtracting px, we get
$f\left(x\right)={x}^{2}+px+x-px-p\left(p+1\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+\left(p+1\right)x-px-p\left(p+1\right)\phantom{\rule{0ex}{0ex}}=x\left[x+\left(p+1\right)\right]-p\left[x+\left(p+1\right)\right]$
So, the zeros of f(x) are −(p + 1) and p.
#### Question 3:
Find the zeros of the polynomial x2 − 3x − m(m + 3). [CBSE 2011]
$f\left(x\right)={x}^{2}-3x-m\left(m+3\right)$
By adding and subtracting mx, we get
$f\left(x\right)={x}^{2}-mx-3x+mx-m\left(m+3\right)\phantom{\rule{0ex}{0ex}}=x\left[x-\left(m+3\right)\right]+m\left[x-\left(m+3\right)\right]\phantom{\rule{0ex}{0ex}}=\left[x-\left(m+3\right)\right]\left(x+m\right)$
So, the zeros of f(x) are −m and m + 3.
#### Question 4:
If α, β are the zeros of a polynomial such that α + β = 6 and αβ = 4 the write the polynomial. [CBSE 2010]
If the zeroes of the quadratic polynomial are α and β then the quadratic polynomial can be found as
x2 − (α + β)x + αβ .....(1)
Substituting the values in (1), we get
x2 − 6x + 4
#### Question 5:
If one zero of the quadratic polynomial kx2 + 3x + k is 2 then find the value of k.
Given: x = 2 is one zero of the quadratic polynomial kx2 + 3x + k
Therefore, It will satisfy the above polynomial.
Now, we have
$k{\left(2\right)}^{2}+3\left(2\right)+k=0\phantom{\rule{0ex}{0ex}}⇒4k+6+k=0\phantom{\rule{0ex}{0ex}}⇒5k+6=0\phantom{\rule{0ex}{0ex}}⇒k=-\frac{6}{5}$
#### Question 6:
If 3 is a zero of the polynomial 2x2 + x + k, find the value of k. [CBSE 2010]
Given: x = 3 is one zero of the polynomial 2x2 + x + k
Therefore, It will satisfy the above polynomial.
Now, we have
$2{\left(3\right)}^{2}+3+k=0\phantom{\rule{0ex}{0ex}}⇒21+k=0\phantom{\rule{0ex}{0ex}}⇒k=-21$
#### Question 7:
If −4 is a zero of the quadratic polynomial x2x − (2k + 2) then find the value of k.
Given: x = −4 is one zero of the polynomial x2x −(2k + 2)
Therefore, It will satisfy the above polynomial.
Now, we have
${\left(-4\right)}^{2}-\left(-4\right)-\left(2k+2\right)=0\phantom{\rule{0ex}{0ex}}⇒16+4-2k-2=0\phantom{\rule{0ex}{0ex}}⇒-2k=-18\phantom{\rule{0ex}{0ex}}⇒k=9$
#### Question 8:
If 1 is a zero of the polynomial ax2 − 3(a − 1) x − 1, then find the value of a.
Given: x = 1 is one zero of the polynomial ax2 − 3(a − 1) x − 1
Therefore, It will satisfy the above polynomial.
Now, we have
$a{\left(1\right)}^{2}-3\left(a-1\right)1-1=0\phantom{\rule{0ex}{0ex}}⇒a-3a+3-1=0\phantom{\rule{0ex}{0ex}}⇒-2a=-2\phantom{\rule{0ex}{0ex}}⇒a=1$
#### Question 9:
If −2 is a zero of the polynomial 3x2 + 4x + 2k then find the value of k.
Given: x = −2 is one zero of the polynomial 3x2 + 4x + 2k
Therefore, It will satisfy the above polynomial.
Now, we have
$3{\left(-2\right)}^{2}+4\left(-2\right)+2k=0\phantom{\rule{0ex}{0ex}}⇒12-8+2k=0\phantom{\rule{0ex}{0ex}}⇒k=-2$
#### Question 10:
Write the zeros of the polynomial x2 −− 6
$f\left(x\right)={x}^{2}-x-6\phantom{\rule{0ex}{0ex}}={x}^{2}-3x+2x-6\phantom{\rule{0ex}{0ex}}=x\left(x-3\right)+2\left(x-3\right)$
So, the zeros of f(x) are 3 and −2.
#### Question 11:
If the sum of the zeros of the quadratic polynomial kx2 − 3x + 5 is 1, write the value of k.
By using the relationship between the zeros of the quadratic ploynomial.
We have
Sum of zeroes =
$⇒1=\frac{-\left(-3\right)}{k}\phantom{\rule{0ex}{0ex}}⇒k=3$
#### Question 12:
If the product of the zeros of the quadratic polynomial x2 − 4x + k is 3 then write the value of k.
By using the relationship between the zeros of the quadratic ploynomial.
We have
Product of zeroes =
$⇒3=\frac{k}{1}\phantom{\rule{0ex}{0ex}}⇒k=3$
#### Question 13:
If (x + a) is a factor of (2x2 + 2ax + 5x + 10), find the value of a. [CBSE 2010]
Given: (x + a) is a factor of 2x2 + 2ax + 5x + 10
We have
$x+a=0\phantom{\rule{0ex}{0ex}}⇒x=-a$
Since, (x + a) is a factor of 2x2 + 2ax + 5x + 10
Hence, It will satisfy the above polynomial
$\therefore 2{\left(-a\right)}^{2}+2a\left(-a\right)+5\left(-a\right)+10=0\phantom{\rule{0ex}{0ex}}⇒-5a+10=0\phantom{\rule{0ex}{0ex}}⇒a=2$
#### Question 14:
If (a − b), a and (a + b) are zeros of the polynomial 2x3 − 6x2 + 5x − 7, write the value of a.
By using the relationship between the zeroes of the cubic ploynomial.
We have
Sum of zeroes =
$⇒a-b+a+a+b=\frac{-\left(-6\right)}{2}\phantom{\rule{0ex}{0ex}}⇒3a=3\phantom{\rule{0ex}{0ex}}⇒a=1$
#### Question 15:
If x3 + x2ax + b is divisible by (x2x), write the values of a and b.
Equating x2x to 0 to find the zeros, we will get
Since, x3 + x2ax + b is divisible by x2x.
Hence, the zeros of x2x will satisfy x3 + x2ax + b
$\therefore {\left(0\right)}^{3}+{0}^{2}-a\left(0\right)+b=0\phantom{\rule{0ex}{0ex}}⇒b=0\phantom{\rule{0ex}{0ex}}$
and
#### Question 16:
If α and β are the zeroes of a polynomial 2x2 + 7x + 5, write the value of α + β + αβ. [CBSE 2010]
By using the relationship between the zeros of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
#### Question 17:
State division algorithm for polynomials.
“If f(x) and g(x) are two polynomials such that degree of f(x) is greater than degree of g(x) where g(x) ≠ 0, then there exists unique polynomials q(x) and r(x) such that
f(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).
#### Question 18:
The sum of the zero and the product of zero of a quadratic polynomial are $\frac{-1}{2}$ and −3 respectively, write the polynomial.
We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula
x2 − (Sum of the zeros)x + Product of zeros
$⇒{x}^{2}-\left(-\frac{1}{2}\right)x+\left(-3\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{2}x-3\phantom{\rule{0ex}{0ex}}$
Hence, the required polynomial is ${x}^{2}+\frac{1}{2}x-3$.
x22x+13=03x232x+1=0
#### Question 19:
Write the zeros of the quadratic polynomial f(x) = 6x2 − 3
To find the zeros of the quadratic polynomial we will equate f(x) to 0
$\therefore f\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}-3=0\phantom{\rule{0ex}{0ex}}⇒3\left(2{x}^{2}-1\right)=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-1=0$
$⇒2{x}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒x=±\frac{1}{\sqrt{2}}$
Hence, the zeros of the quadratic polynomial f(x) = 6x2 − 3 are $\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$.
#### Question 20:
Find the zeros of the quadratic polynomial $f\left(x\right)=4\sqrt{3}{x}^{2}+5x-2\sqrt{3}$
To find the zeros of the quadratic polynomial we will equate f(x) to 0
$\therefore f\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒4\sqrt{3}{x}^{2}+5x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒4\sqrt{3}{x}^{2}+8x-3x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒4x\left(\sqrt{3}x+2\right)-\sqrt{3}\left(\sqrt{3}x+2\right)=0$
Hence, the zeros of the quadratic polynomial $f\left(x\right)=4\sqrt{3}{x}^{2}+5x-2\sqrt{3}$ are .
#### Question 21:
If α and β are the zeroes of a polynomial f(x) = x2 − 5x + k, such that αβ = 1, find the value of k.
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Solving αβ = 1 and α + β = 5, we will get
α = 3 and β = 2
Substituting these values in $\alpha \beta =\frac{k}{1}$, we will get
k = 6
#### Question 22:
If α and β are the zeroes of a polynomial f(x) = 6x2 + x − 2, find the value of $\left(\frac{\alpha }{\beta }+\frac{\beta }{\alpha }\right)$
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
#### Question 23:
If α and β are the zeroes of a polynomial f(x) = 5x2 − 7x +1, find the value of $\left(\frac{1}{\alpha }+\frac{1}{\beta }\right)$
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
#### Question 24:
If α and β are the zeroes of a polynomial f(x) = x2 + x − 2, find the value of $\left(\frac{1}{\alpha }-\frac{1}{\beta }\right)$
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
$\because {\left(\frac{1}{\alpha }-\frac{1}{\beta }\right)}^{2}=\frac{9}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\alpha }-\frac{1}{\beta }=±\frac{3}{2}$
#### Question 25:
If the zeros of the polynomial f(x) = x3 − 3x2 + x + 1 are (a − b), a and (a + b), Find a and b.
By using the relationship between the zeroes of the cubic ploynomial.
We have, Sum of zeroes =
$\therefore a-b+a+a+b=\frac{-\left(-3\right)}{1}\phantom{\rule{0ex}{0ex}}⇒3a=3\phantom{\rule{0ex}{0ex}}⇒a=1$
Now, Product of zeros =
#### Question 1:
Which of the following is a polynomial?
(a) ${x}^{2}-5x+4\sqrt{x}+3$
(b) ${x}^{3/2}-x+{x}^{1/2}+1$
(c) $\sqrt{x}+\frac{1}{\sqrt{x}}$
(d) $\sqrt{2}{x}^{2}-3\sqrt{3}x+\sqrt{6}$
(d) is the correct option.
A polynomial in x of degree n is an expression of the form p(x) =ao +a1x+a2x2 +...+an xn, where an $\ne$0.
#### Question 2:
Which of the following is not a polynomial?
(a) $\sqrt{3}{x}^{2}-2\sqrt{3}x+5$
(b) $9{x}^{2}-4x+\sqrt{2}$
(c) $\frac{3}{2}{x}^{3}+6{x}^{2}-\frac{1}{\sqrt{2}}x-8$
(d) $x+\frac{3}{x}$
It is because in the second term, the degree of x is −1 and an expression with a negative degree is not a polynomial.
#### Question 3:
The zeros of the polynomial x2 − 2x − 3 are
(a) −3, 1
(b) −3, −1
(c) 3, −1
(d) 3, 1
#### Question 4:
The zeros of the polynomial ${x}^{2}-\sqrt{2}x-12$ are
(a) $\sqrt{2},-\sqrt{2}$
(b)
(c)
(d)
#### Question 5:
The zeros of the polynomial $4{x}^{2}+5\sqrt{2}x-3$ are
(a) $-3\sqrt{2},\sqrt{2}$
(b) $-3\sqrt{2},\frac{\sqrt{2}}{2}$
(c) $\frac{-3\sqrt{2}}{2},\frac{\sqrt{2}}{4}$
(d) none of these
#### Question 6:
The zeros of the polynomial ${x}^{2}+\frac{1}{6}x-2$ are
(a) −3, 4
(b) $\frac{-3}{2},\frac{4}{3}$
(c) $\frac{-4}{3},\frac{3}{2}$
(d) none of these
#### Question 7:
The zeros of the polynomial $7{x}^{2}-\frac{11}{3}x-\frac{2}{3}$ are
(a) $\frac{2}{3},\frac{-1}{7}$
(b) $\frac{2}{7},\frac{-1}{3}$
(c) $\frac{-2}{3},\frac{1}{7}$
(d) none of these
#### Question 8:
The sum and product of the zeros of a quadratic polynomial are 3 and −10 respectively. The quadratic polynomial is
(a) x2 − 3x + 10
(b) x2 + 3x −10
(c) x2 − 3x −10
(d) x2 + 3x + 10
#### Question 9:
A quadratic polynomial whose zeros are 5 and −3, is
(a) x2 + 2x − 15
(b) x2 − 2x + 15
(c) x2 − 2x − 15
(d) none of these
#### Question 10:
A quadratic polynomial whose zeros are $\frac{3}{5}$ and $\frac{-1}{2}$, is
(a) 10x2 + x + 3
(b) 10x2 + x − 3
(c) 10x2x + 3
(d) 10x2 x – 3
Multiply by 10, we get
$10{x}^{2}-x-3$
#### Question 11:
The zeros of the quadratic polynomial x2 + 88x + 125 are
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal
#### Question 12:
If α and β are the zero of x2 + 5x + 8, then the value of (α + β) is
(a) 5
(b) −5
(c) 8
(d) −8
#### Question 13:
If α and β are the zero of 2x2 + 5x − 8, then the value of (αβ) is
(a) $\frac{-5}{2}$
(b) $\frac{5}{2}$
(c) $\frac{-9}{2}$
(d) $\frac{9}{2}$
#### Question 14:
If one zero of the quadratic polynomial kx2 + 3x + k is 2, then the value of k is
(a) $\frac{5}{6}$
(b) $\frac{-5}{6}$
(c) $\frac{6}{5}$
(d) $\frac{-6}{5}$
#### Question 15:
If one zero of the quadratic polynomial (k − 1) x2 + kx + 1 is −4, then the value of k is
(a) $\frac{-5}{4}$
(b) $\frac{5}{4}$
(c) $\frac{-4}{3}$
(d) $\frac{4}{3}$
#### Question 16:
If −2 and 3 are the zeros of the quadratic polynomial x2 + (a + 1) x + b, then
(a) a = −2, b = 6
(b) a = 2, b = −6
(c) a = −2, b = −6
(d) a = 2, b = 6
#### Question 17:
If one zero of 3x2 + 8x + k be the reciprocal of the other, then k = ?
(a) 3
(b) −3
(c) $\frac{1}{3}$
(d) $\frac{-1}{3}$
#### Question 18:
If the sum of the zeros of the quadratic polynomial kx2 + 2x + 3k is equal to the product of its zeros, then k = ?
(a) $\frac{1}{3}$
(b) $\frac{-1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{-2}{3}$
#### Question 19:
If α, β are the zeros of the polynomial x2 + 6x + 2, then $\left(\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}\right)=?$
(a) 3
(b) −3
(c) 12
(d) −12
#### Question 20:
If α, β, γ are the zeros of the polynomial x3 − 6x2x + 30, then (αβ + βγ + γα) = ?
(a) −1
(b) 1
(c) −5
(d) 30
#### Question 21:
If α, β, γ are the zeros of the polynomial 2x3x2 − 13x + 6, then αβγ = ?
(a) −3
(b) 3
(c) $\frac{-1}{2}$
(d) $\frac{-13}{2}$
#### Question 22:
If α, β, γ be the zeros of the polynomial p(x) such that (α + β + γ) = 3, (αβ + βγ + γα) = −
10 and αβγ = −24, then p(x) = ?
(a) x3 + 3x2 − 10x + 24
(b) x3 + 3x2 + 10x −24
(c) x3 − 3x2 −10x + 24
(d) None of these
#### Question 23:
If two of the zeros of the cubic polynomial ax3 + bx2 + cx + d is 0, then the third zeros is
(a) $\frac{\mathit{-}\mathit{b}}{\mathit{a}}$
(b) $\frac{b}{a}$
(c) $\frac{c}{a}$
(d) $\frac{-d}{a}$
#### Question 24:
If one of the zeros of the cubic polynomial ax3 + bx2 + cx + d is 0, then the product of the other two zeros is
(a) $\frac{-c}{a}$
(b) $\frac{c}{a}$
(c) 0
(d) $\frac{-b}{a}$
#### Question 25:
If one of the zeros of the cubic polynomial x3 + ax2 + bx + c is −1, then the product of the other two zeros is
(a) ab − 1
(b) b a − 1
(c) 1 − a + b
(d) 1 + a b
#### Question 26:
If α, β be the zero of the polynomial 2x2 + 5x + k such that α2 + β2 + αβ = $\frac{21}{4}$, then k = ?
(a) 3
(b) −3
(c) −2
(d) 2
#### Question 27:
On dividing a polynomial p(x) by a non-zero polynomial q(x), let g(x) be the quotient and r(x) be the remainder, than p(x) = q(x)⋅g(x) + r(x), where
(a) r(x) = 0 always
(b) deg r (x) <deg g(x) always
(c) either r(x) = 0 or deg r(x) <deg g(x)
(d) r(x) = g(x)
#### Question 28:
Which of the following is a true statement?
(a) x2 + 5x − 3 is a linear polynomial.
(b) x2 + 4x − 1 is a binomial.
(c) x + 1 is a monomial.
(d) 5x3 is a monomial.
#### Question 1:
Zeros of p(x) = x2 − 2x − 3 are
(a) 1, −3
(b) 3, −1
(c) −3, −1
(d) 1, 3
(b) 3,-1
Here, ${\mathrm{p}\left(\mathrm{x}\right)=x}^{2}-2x-3\phantom{\rule{0ex}{0ex}}$
#### Question 2:
If α, β, γ are the zeros of the polynomial x3 − 6x2x + 30, then the value of (αβ + βγ + γα) is
(a) −1
(b) 1
(c) −5
(d) 30
(a) −1
Here,
Comparing the given polynomial with , we get:
#### Question 3:
If α, β are the zeroes of kx2 − 2x + 3k such that α + β = αβ, then k = ?
(a) $\frac{1}{3}$
(b) $\frac{-1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{-2}{3}$
(c) $\frac{2}{3}$
Here, $\mathrm{p}\left(x\right)={x}^{2}-2x+3k$
Comparing the given polynomial with $a{x}^{2}+bx+c$, we get:
It is given that
are the roots of the polynomial.
Also, =$\frac{c}{a}$
#### Question 4:
It is given that the difference between the zeroes of 4x2 − 8kx + 9 is 4 and k > 0. Then, k = ?
(a) $\frac{1}{2}$
(b) $\frac{3}{2}$
(c) $\frac{5}{2}$
(d) $\frac{7}{2}$
(c) $\frac{5}{2}$
Let the zeroes of the polynomial be .
Here,
p
Comparing the given polynomial with $a{x}^{2}+bx+c$, we get:
a = 4, b = −8k and c = 9
Now, sum of the roots$=-\frac{b}{a}$
#### Question 5:
Find the zeros of the polynomial x2 + 2x − 195.
Here, p
#### Question 6:
If one zero of the polynomial (a2 + 9)x2 + 13x + 6a is the reciprocal of the other, find the value of a.
#### Question 7:
Find a quadratic polynomial whose zeros are 2 and −5.
It is given that the two roots of the polynomial are 2 and −5.
Let
Now, sum of the zeroes, $\mathrm{\alpha }+\mathrm{\beta }$ = 2 + (5) = 3
Product of the zeroes, $\mathrm{\alpha \beta }$ = 2$×$5 = 10
∴ Required polynomial = ${x}^{2}-\left(\mathrm{\alpha }+\mathrm{\beta }\right)x+\mathrm{\alpha \beta }$
$={x}^{2}—\left(-3\right)x+\left(-10\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+3x-10$
#### Question 8:
If the zeroes of the polynomial x3 − 3x2 + x + 1 are (ab), a and (a + b), find the values of a and b.
The given polynomial and its roots are .
#### Question 9:
Verify that 2 is a zero of the polynomial x3 + 4x2 − 3x − 18.
Let p$\left(\mathrm{x}\right)={x}^{3}+4{x}^{2}-3x-18$
#### Question 10:
Find the quadratic polynomial, the sum of whose zeroes is −5 and their product is 6.
Given:
Sum of the zeroes = −5
Product of the zeroes = 6
∴ Required polynomial =
$={x}^{2}-\left(-5\right)x+6\phantom{\rule{0ex}{0ex}}={x}^{2}+5x+6$
#### Question 11:
Find a cubic polynomial whose zeros are 3, 5 and −2.
#### Question 12:
Using remainder theorem, find the remainder when p(x) = x3 + 3x2 − 5x + 4 is divided by (x − 2).
#### Question 13:
Show that (x + 2) is a factor of f(x) = x3 + 4x2 + x − 6.
#### Question 14:
If α, β, γ are the zeroes of the polynomial p(x) = 6x3 + 3x2 − 5x + 1, find the value of $\left(\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}+\frac{1}{\mathrm{\gamma }}\right)$
Comparing the polynomial with ${x}^{3}-{x}^{2}\left(\alpha +\beta +\gamma \right)+x\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)-\alpha \beta \gamma$, we get:
#### Question 15:
If α, β are the zeros of the polynomial f(x) = x2 − 5x + k such that α − β = 1, find the value of k.
#### Question 16:
Show that the polynomial f(x) = x4 + 4x2 + 6 has no zeroes.
#### Question 17:
If one zero of the polynomial p(x) = x3 − 6x2 + 11x − 6 is 3, find the other two zeroes.
#### Question 18:
If two zeroes of the polynomial p(x) = 2x4 − 3x3 − 3x2 + 6x − 2 are $\sqrt{2}$ and $-\sqrt{2}$, find its other two zeroes.
#### Question 19:
Find the quotient when p(x) = 3x4 + 5x3 − 7x2 + 2x + 2 is divided by (x2 + 3x + 1).
#### Question 20:
Use remainder theorem to find the value of k, it being given that when x3 + 2x2 + kx + 3 is divided by (x − 3), then the remainder is 21.
|
Chapter 5 Orthogonality
```5.1
Orthogonality
Definitions
A set of vectors is called an orthogonal set if all pairs of
distinct vectors in the set are orthogonal.
An orthonormal set is an orthogonal set of unit vectors.
An orthogonal (orthonormal) basis for a subspace W of
n
R is a basis for W that is an orthogonal (orthonormal) set.
An orthogonal matrix is a square matrix whose columns
form an orthonormal set.
Examples
1) Is the following set of vectors orthogonal? orthonormal?
a) 1
2
3 2 1
, 4 , 1
1 2
b) { e1 , e 2 ,..., e n }
2) Find an orthogonal basis and an orthonormal basis
n
for the subspace W of R
x
W y : x y 2 z 0
z
Theorems
All vectors in an orthogonal set are linearly independent.
Let {v1, v2,…, vk } be an orthogonal basis for a subspace
n
W of R and w be any vector in W. Then the unique
scalars c1 ,c2 , …, ck such that w = c1v1 + c2v2 + …+ ckvk
are given by
w vi
ci
for i 1,..., k
vi vi
Proof: To find ci we take the dot product with vi
w vi = (c1v1 + c2v2 + …+ ckvk ) vi
Examples
3) The orthogonal basis for the subspace W in previous
example is 1 1
1 , 1
0 1
Pick a vector in W and express it in terms of the vectors
in the basis.
4) Is the following matrix orthogonal?
3
A
1
2
2
4
1
1
1
2
0
B 1
0
0
0
1
1
0
0
cos
C
sin
sin
cos
If it is orthogonal, find its inverse and its transpose.
Theorems on Orthogonal Matrix
The following statements are equivalent for a matrix A :
1) A is orthogonal
-1
T
2) A = A
n
3) ||Av|| = ||v|| for every v in R
n
4) Av1∙ Av2 = v1∙ v2 for every v1 ,v2 in R
Let A be an orthogonal matrix. Then
1) its rows form an orthonormal set.
-1
2) A is also orthogonal.
3) |det(A)| = 1
4) |λ| = 1 where λ is an eigenvalue of A
5) If A and B are orthogonal matrices, then so is AB
5.2
Orthogonal Complements
and Orthogonal Projections
Orthogonal Complements
Recall: A normal vector n to a plane is orthogonal to
every vector in that plane. If the plane passes through
the origin, then it is a subspace W of R3 .
Also, span(n) is also a subspace of R3
Note that every vector in span(n) is orthogonal to
every vector in subspace W . Then span(n) is called
orthogonal complement of W.
Definition:
A vector v is said to be orthogonal to a subspace W of
n
R if it is orthogonal to all vectors in W.
The set of all vectors that are orthogonal to W is called
the orthogonal complement of W, denoted W ┴ . That is
W perp
W
{v R
n
: v w 0 w W}
http://www.math.tamu.edu/~yvorobet/MATH304-2011C/Lect3-02web.pdf
Example
3
1) Find the orthogonal complements for W of R .
1
a) W span 2
3
b) W plane with direction
c) (subspace
x
W y : x y 2 z 0
z
spanned
by) vectors
1
0
1 and 1
0
1
Theorems
n
Let W be a subspace of R .
n
┴
1) W is a subspace of R .
2) (W ┴)┴ = W
3) W ∩ W ┴ = {0}
4) If W = span(w1,w2,…,wk), then v is in W ┴ iff v∙wi = 0
for all i =1,…,k.
Let A be an m x n matrix. Then
(row(A))┴ = null(A)
and
(col(A))┴ = null(AT)
Proof?
Example
2) Use previous theorem to find the orthogonal complements
3
for W of R .
1
0
a) W plane w ith direction (subspace spann ed by) vectors 1 and 1
0
1
b)
3
2
W subspace spanned by vectors 0 ,
1
4
1
2
2 an d
0
1
3
2
6
2
5
Orthogonal Projections
u v
w 1 proj v u
v 2
u
w2
w1
u v
v
v v
v
w 2 = perp v u u - w 1
v
Let u and v be nonzero vectors.
w1 is called the vector component of u along v
(or projection of u onto v), and is denoted by projvu
w2 is called the vector component of u orthogonal to v
Orthogonal Projections
n
Let W be a subspace of R with an orthogonal basis
{u1, u2,…, uk }, the orthogonal projection of v onto W is
defined as:
projW v = proju v + proju v + … + proju v
1
2
k
The component of v orthogonal to W is the vector
perpW v = v – projw v
n
n
Let W be a subspace of R and v be any vector in R .
Then there are unique vectors w1 in W and w2 in W ┴
such that v = w1 + w2 .
Examples
3) Find the orthogonal projection of v = [ 1, -1, 2 ] onto W and
the component of v orthogonal to W.
1
a) W span 2
3
1
-1
b) W subspace spanned by 1 and
1
0
1
x
c) W y : x y 2 z 0
z
5.3
The Gram-Schmidt Process
And the QR Factorization
The Gram-Schmidt Process
Goal: To construct an orthogonal (orthonormal) basis for
n
any subspace of R .
each vector vi in the basis one at a time by finding the
component of vi orthogonal to W = span(x1, x2,…, xi-1 ).
Let {x1, x2,…, xk } be a basis for a subspace W. Then
choose the following vectors:
v1 = x1,
v2 = x2 – projv x2
1
v3 = x3 – projv x3 – projv x3
1
2
… and so on
Then {v1, v2,…, vk } is orthogonal basis for W .
We can normalize each vector in the basis to form an
orthonormal basis.
Examples
1) Use the following basis to find an orthonormal basis for R
3 1
, ,
1 2
3
2) Find an orthogonal basis for R that contains the vector
1
1
1
2 ,
1
, 0
1
1
1
2
The QR Factorization
If A is an m x n matrix with linearly independent columns,
then A can be factored as A = QR where R is an invertible
upper triangular matrix and Q is an m x n orthogonal
n
matrix. In fact columns of Q form orthonormal basis for R
which can be constructed from columns of A by using
Gram-Schmidt process.
Note: Since Q is orthogonal, Q-1 = QT and we have R = QT A
Examples
3) Find a QR factorization for the following matrices.
3
A
1
1
2
1
A 2
1
-1
1
-1
- 1
0
1
5.4
Orthogonal Diagonalization
of Symmetric Matrices
Example
1) Diagonalize the matrix.
3
A
2
2
6
Recall:
A square matrix A is symmetric if AT = A.
A square matrix A is diagonalizable if there exists a
matrix P and a diagonal matrix D such that P-1AP = D.
Orthogonal Diagonalization
Definition:
A square matrix A is orthogonally diagonalizable if there
exists an orthogonal matrix Q and a diagonal matrix D
such that Q-1AQ = D.
Note that Q-1 = QT
Theorems
1. If A is orthogonally diagonalizable, then A is symmetric.
2. If A is a real symmetric matrix, then the eigenvalues of A
are real.
3. If A is a symmetric matrix, then any two eigenvectors
corresponding to distinct eigenvalues of A are orthogonal.
A square matrix A is orthogonally diagonalizable
if and only if it is symmetric.
Example
2) Orthogonally diagonalize the matrix
0
A 1
1
1
0
1
1
1
0
and write A in terms of matrices Q and D.
Theorem
If A is orthogonally diagonalizable, and QTAQ = D
then A can written as
A 1 q1 q1 2 q 2 q 2 ... n q n q n
T
T
T
where qi is the orthonormal column of Q, and λi is
the corresponding eigenvalue.
This fact will help us construct the matrix A given
eigenvalues and orthogonal eigenvectors.
Example
3) Find a 2 x 2 matrix that has eigenvalues 2 and 7, with
corresponding eigenvectors
v1
1
2
v
2
1
2
5.5
Applications
A quadratic form in x and y :
a
T
2
2
ax by cxy x 1
2 c
c
x
b
1
2
A quadratic form in x,y and z:
a
2
2
2
ax by cz dxy exz fyz x T 12 d
12 e
where x is the variable (column) matrix.
1
2
d
b
1
2
f
e
1
f
x
2
c
1
2
A quadratic form in n variables is a function
n
f : R R of the form:
f (x) x Ax
T
where A is a symmetric n x n matrix and x is in R
A is called the matrix associated with f.
z f ( x , y ) x y 8 xy
2
2
z f ( x, y ) 2 x 5 y
2
2
n
The Principal Axes Theorem
Every quadratic form can be diagonalized. In fact,
if A is a symmetric n x n matrix and if Q is an
orthogonal matrix so that QTAQ = D then the change
of variable x = Qy transforms the quadratic form into
x A x y D y 1 y1 2 y 2 ... n y n
T
T
2
2
2
Example: Find a change of variable that transforms the
Quadratic into one with no cross-product terms.
z f ( x , y ) x y 8 xy
2
2
z f ( x, y ) 2 x 5 y
2
2
```
|
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# What percentage of the square's area is shaded?
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29 Nov 2017, 03:52
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Difficulty:
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Question Stats:
38% (02:47) correct 62% (02:51) wrong based on 148 sessions
### HideShow timer Statistics
What percentage of the square's area is grey-shaded, with AB = 4 and DE = 1? (see picture)
A: 30%
B: 37,5%
C: 40%
D: 45%
E: 50%
*found it on brilliant.org and I guess it's a tough gmat-style questions
Attachments
GeoProblemTriangle.jpg [ 47.75 KiB | Viewed 6757 times ]
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### Show Tags
29 Nov 2017, 09:34
5
1
septwibowo
Why the two triangles are indeed similar is shown graphically below.
Angles a, b, and c are identical, making the two triangles similar. Hence, the base to height ratio is similar and we can set up a proportion to find the height:
$$\dfrac {4-h}{4}=\dfrac {h}{1}$$
h = 0.8
So, height of the shaded triangle is 3.2 and since we have the base (4) and the height (3.2) we can calculate the area.
$$3.2 = \dfrac{16}{5}$$
$$\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}$$
Percentage of the square's area:
$$\dfrac {\dfrac {32}{5}}{16} = \dfrac{2}{5} \rightarrow \dfrac{2}{5}\times 100 = 40\%$$
Attachments
GeoProblemTriangleSol1.jpg [ 55.06 KiB | Viewed 6531 times ]
GeoProblemTriangleSol2.jpg [ 62.96 KiB | Viewed 6546 times ]
##### General Discussion
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### Show Tags
29 Nov 2017, 04:05
4
Option C
The two triangles are proportionate. Hence the height of the shaded triangle is 4/(4+1)*4 =3.2. Hence area of the shaded region =1/2*4*3.2=6.4. The ratio is 6.4/(4*4)*100=40%.
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### Show Tags
29 Nov 2017, 07:31
Sasindran wrote:
Option C
The two triangles are proportionate. Hence the height of the shaded triangle is 4/(4+1)*4 =3.2. Hence area of the shaded region =1/2*4*3.2=6.4. The ratio is 6.4/(4*4)*100=40%.
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Dear Sasindran , would u please elaborate more why these triangles are proportionate?
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### Show Tags
29 Nov 2017, 07:57
3
septwibowo wrote:
Sasindran wrote:
Option C
The two triangles are proportionate. Hence the height of the shaded triangle is 4/(4+1)*4 =3.2. Hence area of the shaded region =1/2*4*3.2=6.4. The ratio is 6.4/(4*4)*100=40%.
Posted from my mobile device
Dear Sasindran , would u please elaborate more why these triangles are proportionate?
Sure.
We all know that AB and CD are parallel lines. When two lines intersect between parallel lines the triangles so formed by them are always proportional to each other. Also the distance of intersection from either of the lines are also proportional to the length of the parallel lines.
For our case lets consider the intersecting point as O. Here we have AB=4 and DE=1. Hence by the above theorem we have perpendicular Distance between O and line AB is proportional to the perpendicular distance between O and the line DE.
i.e., AB/DE = per dist between O and AB/per dist between O and DE
4/1=height of triangle ABO/height of triagle DEO
also we know that height of triangle AEO+height of triagle DEO=4 (divided at O in the ratio 4:1)
Hence height of triangle AEO=4/5=3.2
height of triangle DEO=1/5=0.8
Hence area of triangle AEO=1/2*4*3.2=6.4
And you know the remaining septwibowo
Thanks
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29 Nov 2017, 08:11
1
septwibowo
Although may find 1000s of resources I am providing one:
https://www.varsitytutors.com/hotmath/h ... rtionality
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29 Nov 2017, 23:09
Sasindran wrote:
septwibowo wrote:
Sasindran wrote:
Option C
The two triangles are proportionate. Hence the height of the shaded triangle is 4/(4+1)*4 =3.2. Hence area of the shaded region =1/2*4*3.2=6.4. The ratio is 6.4/(4*4)*100=40%.
Posted from my mobile device
Dear Sasindran , would u please elaborate more why these triangles are proportionate?
Sure.
We all know that AB and CD are parallel lines. When two lines intersect between parallel lines the triangles so formed by them are always proportional to each other. Also the distance of intersection from either of the lines are also proportional to the length of the parallel lines.
For our case lets consider the intersecting point as O. Here we have AB=4 and DE=1. Hence by the above theorem we have perpendicular Distance between O and line AB is proportional to the perpendicular distance between O and the line DE.
i.e., AB/DE = per dist between O and AB/per dist between O and DE
4/1=height of triangle ABO/height of triagle DEO
also we know that height of triangle AEO+height of triagle DEO=4 (divided at O in the ratio 4:1)
Hence height of triangle AEO=4/5=3.2
height of triangle DEO=1/5=0.8
Hence area of triangle AEO=1/2*4*3.2=6.4
And you know the remaining septwibowo
Thanks
Thank you very much Sasindran for your generosity! Kudo for you!
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Joined: 27 Dec 2016
Posts: 227
Concentration: Marketing, Social Entrepreneurship
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### Show Tags
29 Nov 2017, 23:10
drexxie wrote:
septwibowo
Why the two triangles are indeed similar is shown graphically below.
Angles a, b, and c are identical, making the two triangles similar. Hence, the base to height ratio is similar and we can set up a proportion to find the height:
$$\dfrac {4-h}{4}=\dfrac {h}{1}$$
h = 0.8
So, height of the shaded triangle is 3.2 and since we have the base (4) and the height (3.2) we can calculate the area.
$$3.2 = \dfrac{16}{5}$$
$$\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}$$
Percentage of the square's area:
$$\dfrac {\dfrac {32}{5}}{16} = \dfrac{2}{5} \rightarrow \dfrac{2}{5}\times 100 = 40\%$$
Thank you drexxie ! I forgot this theorem, my bad!
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### Show Tags
04 Jan 2018, 01:17
well, this question is hardly a gmat question. The source is also unreliable. However, this question is really a tough question.
Intern
Joined: 31 Dec 2017
Posts: 4
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GPA: 3.05
### Show Tags
25 Mar 2018, 07:45
drexxie wrote:
Why the two triangles are indeed similar is shown graphically below.
Angles a, b, and c are identical, making the two triangles similar. Hence, the base to height ratio is similar and we can set up a proportion to find the height:
$$\dfrac {4-h}{4}=\dfrac {h}{1}$$
h = 0.8
So, height of the shaded triangle is 3.2 and since we have the base (4) and the height (3.2) we can calculate the area.
$$3.2 = \dfrac{16}{5}$$
$$\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}$$
Percentage of the square's area:
$$\dfrac {\dfrac {32}{5}}{16} = \dfrac{2}{5} \rightarrow \dfrac{2}{5}\times 100 = 40\%$$
This is real compact and accurate answer. Thanks mate. Kudo for you!
Re: What percentage of the square's area is shaded? [#permalink] 25 Mar 2018, 07:45
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# Mathbook: Introduction to Sets
Quite a while ago, an endeavoring individual tried to start an open-source repository of mathematical information called Mathbook. I contributed an article, which I’ll put down in two parts on this blog.
It seems that the project has died, although the website is still available. While this is a bit of a shame, I would like to give some of my own little lessons here. The creator’s idea behind Mathbook was to focus on giving people an understanding of why we do math in a certain way. This is missing from mathematical curriculum today, but it is vital to understand that when math was developed, decisions were made for specific reasons. Moving forward, I’ll occasionally add a new post here to that effect. The people in my life don’t always understand the math I learned, so this is part of my effort in showing it.
## Introduction to Sets
We will learn the basics of how sets are used in mathematics. It is important to understand basic arithmetic before diving in, but nothing else.
### Understanding Sets
In any field of mathematics, it is important to be able to deal with objects and structures. At the lowest level of mathematical objects and structures are sets. Most simply, a set is a collection of objects. We can think of the set of all flowers in Hawaii, or the set of whole numbers between 10 and 37. Typically, we use curly brackets (braces) to denote a set, such as $\{1, 2, 3\}$. If we are using the same set many times in a row, or talking about a set that cannot easily be written down, we can use some other symbol. Throughout this tutorial we will let $S$ be the set $\{1, 2, 3\}$, and $H$ be the set of all flowers in Hawaii.
There are certain rules and terms used with sets that allow mathematicians to be consistent when using and talking about sets. For example, we want to know what to call the objects in our sets in general, and how we can write sets.
##### Definition
Element: Each object or member of a set is called an element of the set. Each element can only occur once in a set. $\{1,2,1,3\}$ is not a valid set since the $1$ occurs twice. In addition, the order of elements in a set does not matter. $S = \{1,2,3\} = \{3,1,2\}$.
To say an element $s$ is in the set $S$, we write $s \in S$.
It is also natural to discuss how many elements are in a set.
##### Definition
Cardinality: The number of elements in a set is called the cardinality of the set. The cardinality is often denoted by putting vertical bars around the set. For example, since $S$ has three elements, we write $|S| = 3.$
For this tutorial, we will only be looking at sets with finite cardinality; this means we will always be able to list and count every element in the set. Future tutorials may explore larger sets, which becomes an even more powerful (and fun!) mathematical tool.
Often we want to look at some of the elements in a set, but not all of them. For example, we might want the elements of $H$ which are red flowers. This is a very common pattern in mathematics: given an object or structure, how can we look at smaller objects that have a similar structure?
##### Definition
Subset: If every element in some set $A$ is in a set $B$, we say that $A$ is a subset of $B$, and we write $A\subset B$. For example, the set $\{1,3\}$ is a subset of $S$.
When doing mathematics, it is good practice to look at the simplest example of any object you are interested in exploring. When it comes to sets, it becomes natural to ask “What if my set has no elements?”
##### Definition
Empty Set: The empty set is defined to be the set which has no elements. The most common notation is $\emptyset$, but you may see $\{\}$, especially in older math texts. This second notation emphasizes it is a set with no elements.
Almost everything you see and do has sets hiding in the background. They are a universal way of communicating mathematical ideas, and are thus very important to understand.
In a future post, I’ll give the second half of this post: Understanding Functions. Once you have established your structure (the set), and have explored some basic ideas (subsets, the empty set), it is important to discuss how you can have two sets interact. The simplest way to have two sets interact is via a function.
|
# 2009 AMC 8 Problems/Problem 1
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie 3 apples, keeping 4 apples for herself. How many apples did Bridget buy?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 14$
## Solution
If we set up an equation, we find out $x=(3+4)\cdot 2$ because 3 apples were left after giving half, then four away. We can simplify the equations to $x=7\cdot 2=14.$ The answer is $\text{(E) } 14.$
~John0412
## Solution 2
You can work backwards and add 3 apples that she gave to Cassie to the 4 she currently has, which results in 7, and then multiply by 2 since she gave half the apples to Ann, resulting in $\qquad\textbf{(E)}\ 14$. ~Anabel.disher
~savannahsolver
|
# What Is Segment Angle?
## What are the types of angle?
Different Types of AnglesZero Angles.Acute Angles.Right Angles.Obtuse Angles.Straight Angles.Reflex Angles.Complete Angle..
## What is a segment in math?
In geometry, a line segment is a part of a line that is bounded by two distinct end points, and contains every point on the line between its endpoints. A closed line segment includes both endpoints, while an open line segment excludes both endpoints; a half-open line segment includes exactly one of the endpoints.
## How do you find an angle in radians?
The radian measure of a central angle θ of a circle is defined as the ratio of the length of the arc the angle subtends, s, divided by the radius of the circle, r. Note that when s = r, we get θ expressed as one radian.
## What is minor sector?
A sector is a region bounded by two radii of a circle and the intercepted arc of the circle. … A sector with a central angle less than 180° is called a minor sector. A sector with a central angle greater than 180° is called a major sector.
## What is the angle in a major segment?
We know that angle subtended by an arc of a circle at its center is twice the angle subtended by it at any point of the alternate segment of circle. Since AB is minor arc and angle ACB is the angle formed by it in alternate segment. ∴,2∠ACB=∠AOB.
## What is a basic angle?
The reference angle is the positive acute angle that can represent an angle of any measure. … The reference angle is always the smallest angle that you can make from the terminal side of an angle (ie where the angle ends) with the x-axis. A reference angle always uses the x-axis as its frame of reference.
## What is an inscribed angle of a circle?
In geometry, an inscribed angle is the angle formed in the interior of a circle when two secant lines intersect on the circle. It can also be defined as the angle subtended at a point on the circle by two given points on the circle.
## How do you solve an inscribed angle?
Inscribed Angle Theorem: The measure of an inscribed angle is half the measure of the intercepted arc. That is, m∠ABC=12m∠AOC. This leads to the corollary that in a circle any two inscribed angles with the same intercepted arcs are congruent.
## What is minor segment?
A segment is a region bounded by a chord of a circle and the intercepted arc of the circle. A segment with an intercepted arc less than a semicircle is called a minor segment. A sector with an intercepted arc greater than a semi-circle is called a major segment.
## What is the formula of segment?
Area of a Segment of a Circle FormulaFormula To Calculate Area of a Segment of a CircleArea of a Segment in RadiansA = (½) × r2 (θ – Sin θ)Area of a Segment in DegreesA = (½) × r 2 × [(π/180) θ – sin θ]
## How do you find the angle of a segment of a circle?
Circular Segment Equation and CalculatorRadius: R = h + d = h / 2 + c2 / ( 8h )Arc Length: s = arcsin ( c / ( h + c2 / 4h ) ) ( h + c2 / 4h )Chord Length: θ given in radians. c = 2 R sin ( θ / 2 )Angle: θ = 2 arctan ( c / ( 2 d ) )Sagitta: θ given in radians. h = R ( 1 – cos ( θ / 2) )Area:
## What are the 7 types of angles?
Types of Angles – Acute, Right, Obtuse, Straight and Reflex…Acute angle.Right angle.Obtuse angle.Straight angle.Reflex angle.
## How do you find an inscribed angle?
An inscribed angle is an angle whose vertex is on a circle and whose sides contain chords of a circle. This is different than the central angle, whose vertex is at the center of a circle. If you recall, the measure of the central angle is congruent to the measure of the minor arc.
## Can a central angle be 180 degrees?
There are two types of central angles. A convex central angle, which is a central angle that measures less than 180 degrees and a reflex central angle, which is a central angle that measures more than 180 degrees and less than 360 degrees. These are both part of a complete circle.
## What is a segment circle?
More formally, a circular segment is a region of two-dimensional space that is bounded by an arc (of less than 180°) of a circle and by the chord connecting the endpoints of the arc.
## What is the perimeter of a segment?
The perimeter of segment is the sum of the length of the arc and the chord.
## What is the definition of an angle?
In Euclidean geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. Angles formed by two rays lie in the plane that contains the rays.
## What is acute angle?
Acute angles measure less than 90 degrees. Right angles measure 90 degrees. Obtuse angles measure more than 90 degrees.
|
# 6.3: Boxes and Boxes Extras
Difficulty Level: At Grade Created by: CK-12
Extras for Experts - Boxes and Boxes – Interpret pan balances to determine values of variables
Solutions
\begin{align*}1. \quad t = 2 \ \text{pounds}; \ u = 4 \ \text{pounds}\!\\ {\;} \quad \ \text{From} \ D, 3u = 12, \ \text{so} \ u = 4 \ \text{pounds}.\!\\ {\;} \quad \ \text{From} \ C, 2t = 4, \ \text{so} \ t = 2 \ \text{pounds}.\end{align*}
\begin{align*}2. \quad v = 3 \ \text{pounds}; w = 1 \ \text{pounds}\!\\ {\;} \quad \ \text{From} \ H, v < 5 \ \text{or} \ 1, 2, 3 \ \text{or} \ 4 \ \text{pounds}.\!\\ {\;} \quad \ \text{From} \ G, v = 3w, \ \text{so} \ v \ \text{must be a multiple of} \ 3.\!\\ {\;} \quad \ \text{Then} \ 3 = 3w \ \text{and} \ w \ \text{is} \ 1 \ \text{pound}.\end{align*}
\begin{align*}3. \quad y = 4, 8, \ \text{or} \ 12 \ \text{pounds}; z = 1, 2, \ \text{or} \ 3 \ \text{pounds}\!\\ {\;} \quad \ \text{From} \ J, y < 16 \ \text{or} \ 1, 2, 3, ..., 15 \ \text{pounds}.\!\\ {\;} \quad \ \text{From} \ K, y = 4z, \ \text{so} \ y = \ \text{must be a multiple of} \ 4.\!\\ {\;} \quad \ \text{The possible multiples of} \ 4 \ \text{that are less than} \ 16 \ \text{are} \ 4, 8, \ \text{and} \ 12.\!\\ {\;} \quad \ \text{If} \ y = 4, 8, \ \text{or} \ 12, \ \text{then} \ z = 1, 2, \ \text{or} \ 3 \ \text{pounds}.\end{align*}
\begin{align*}4. \quad r = 1, 2 \ \text{or} \ 3 \ \text{pounds}; s = 10 \ \text{pounds}\!\\ {\;} \quad \ \text{From} \ B, 2s = 20, \ \text{so} \ s = 10 \ \text{pounds}.\!\\ {\;} \quad \ \text{From} \ A, 3r < 10, \ \text{or} \ 1, 2, 3, ..., 9 \ \text{pounds. So} \ r = 1, 2, \ \text{or} \ 3 \ \text{pounds}.\end{align*}
All weights are whole numbers of pounds.
What could be the weights? Tell how you figured it out.
All weights are whole numbers of pounds.
What could be the weights? Tell how you figured it out.
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# RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions
RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions is concerned with Heron’s formula. The students will study “how to find the area of a triangle using heron’s formula?” with the help of several solved problems. The experts have used the step-by-step problem-solving methods of the examples mentioned in this article.
Students can freely download RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions PDF and grind their skills. The PDF consists of varieties of questions based on Heron’s Formula. The questions were explained stepwise, which is easily understandable to the students. By practicing through PDF, learners will get to know the several types of questions that can be formed. Practicing with these problems help students to score well in the exam.
## Download RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions PDF
Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1
## Important Definitions RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions
We can calculate any triangle if we know the length of all three (3) sides of that triangle by using Heron’s Formula, which has been known for approximately 2000 years. It is known as “Heron’s Formula” after Hero of Alexandria.
### Heron’s Formula
Area of a triangle= √s (s-a) (s-b) (s-c)
Semi Perimeter= s=(a+b+c)2, where, a, b, and c are the sides of the triangle.
### Examples of Heron’s Formula RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions
Ques- In a triangle PQR, PQ = 15cm, QR = 13cm, and PR = 14cm. Find the area of a triangle PQR and hence its altitude on PR.
Solution-
Let the sides of the given triangle be PQ = p, QR = q, PR = r, respectively.
Here,
= p = 15 cm
= q = 13 cm
= r = 14 cm
From Heron’s Formula;
Area of a triangle= √s (s-p) (s-q) (s-r)
Semi Perimeter= s=(p+ q+ r)2
Where, p, q, and r are sides of a triangle.
= s= (15 + 13 + 14)/ 2
= s= 21
= Area= √21 (21-15) (21-13) (21-14)
= √21 (6 x 7 x 8)
= √7056
= 84
= Area = 84 cm2
= Let, QT is a perpendicular on PR
Now, area of triangle = ½ x Base x Height
= ½ × QT × PR = 84
= QT = 12cm
= The altitude is 12 cm (Hence Proved).
## Frequently Asked Questions (FAQs) of RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions
Ques- What is S in a Triangle of Hero’s Formula?
Ans- The other is Heron’s formula, which gives the area in terms of the three sides of a triangle, specifically, as the square root of the product s(s – a)(s – b)(s – c), where, ‘s’ is the semi perimeter of a triangle. So, s = (a + b + c)/2.
Ques- What is the semi perimeter of a triangle?
Ans- A semi perimeter of a triangle is equivalent to the perimeter of its medial triangle.
Ques- Who gave Heron’s formula?
Ans- Hero of Alexandria, a great mathematician who derived the formula to calculate the area of the triangle using the length of all three (3) sides.
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# Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 24
$${y^2} - y = \frac{{{x^3}}}{3} + 5x + 110$$
#### Work Step by Step
\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{x^2} + 5}}{{2y - 1}};\,\,\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 11 \cr & {\text{Separating variables leads to}} \cr & \left( {2y - 1} \right)dy = \left( {{x^2} + 5} \right)dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\left( {2y - 1} \right)dy} = \int {\left( {{x^2} + 5} \right)} dx \cr & {\text{integrate by using the power rule}} \cr & {y^2} - y = \frac{{{x^3}}}{3} + 5x + C\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 0 \right) = 1 \cr & \,y\left( 0 \right) = 11{\text{ implies that }}y = 11{\text{ when }}x = 0 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & {\left( {11} \right)^2} - 11 = \frac{{{{\left( 0 \right)}^3}}}{3} + 5\left( 0 \right) + C \cr & C = 110 \cr & {\text{substitute }}C = 110{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & {y^2} - y = \frac{{{x^3}}}{3} + 5x + 110 \cr}
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# If 5 tan θ − 4 = 0, then the value of 5 sin θ−4 cos θ5 sin θ+4 cos θ is
Question:
If $5 \tan \theta-4=0$, then the value of $\frac{5 \sin \theta-4 \cos \theta}{5 \sin \theta+4 \cos \theta}$ is
(a) $\frac{5}{3}$
(b) $\frac{5}{6}$
(c) 0
(d) $\frac{1}{6}$
Solution:
Given that $5 \tan \theta-4=0 .$ We have to find the value of the following expression
$\frac{5 \sin \theta-4 \cos \theta}{5 \sin \theta+4 \cos \theta}$
Since $5 \tan \theta-4=0 \Rightarrow \tan \theta=\frac{4}{5}$
We know that: $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$\Rightarrow$ Base $=5$
$\Rightarrow$ Perpendicular $=4$
$\Rightarrow$ Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$
$\Rightarrow$ Hypotenuse $=\sqrt{16+25}$
$\Rightarrow$ Hypotenuse $=\sqrt{41}$
Since $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ and $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
Now we find
$\frac{5 \sin \theta-4 \cos \theta}{5 \sin \theta+4 \cos \theta}$
$=\frac{5 \times \frac{4}{\sqrt{41}}-4 \times \frac{5}{\sqrt{41}}}{5 \times \frac{4}{\sqrt{41}}+4 \times \frac{5}{\sqrt{41}}}$
$=\frac{\frac{20}{\sqrt{41}}-\frac{20}{\sqrt{41}}}{\frac{20}{\sqrt{41}}+\frac{20}{\sqrt{41}}}$
$=0$
Hence the correct option is $(c)$
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# Integers
```Section 7.3 – Ratio and Proportion
Definition. A ratio is an ordered pair of numbers, denoted as either a : b or
a
and read as the ratio of a to b.
b
Generally, ratios relate values in one of three types: part-to-part, part-to-whole, or whole-to-part.
For example, suppose a + b = c, we may write the following ratios as:
a
b
part-to-part
a : b or b : a In the other notation,
or .
b
a
a
b
part-to-whole
a : c or b : c In the other notation,
or .
c
c
c
c
whole-to-part
c : a or c : b In the other notation,
or .
a
b
Examples.
1.
In a school with 420 students and 35 teachers:
(a) What is the ratio of students to teachers?
(b) What is the ratio of teachers to students?
(c) Which type of a ratio did you use for each of the above?
2.
A school had 180 boys and 240 girls attending.
(a) What is the ratio of boys to girls?
(b) What is the ratio of girls to boys?
(c) What is the ratio of students to boys?
(d) What is the ratio of girls to students?
(e) Which type of ratio was used for each of the above?
3.
A recipe calls for 2 parts sugar, 5 parts flour, and 3 parts milk.
(a) What is the ratio of sugar to flour to milk?
(b) What is the ratio of sugar to the whole recipe?
Note. The ratio a : b is equivalent to the ratio an : bn,. We can use this relationship to solve some types of problems that
involve ratios.
Example. The ratio of boys to girls in a class is 3:4. The class has 35 students.
How many boys and girls are in the class?
Solution: The ratio 3:4 is equivalent to the ratio 3n : 4n where 3n represents the number of boys and 4n
represents the number of girls.
3n + 4n = 35
7n = 35
n=5
The number of boys is 3n = 3 ∙ 5 = 15.
The number of girls is 4n = 4 ∙ 5 = 20.
-1-
The class had 15 boys and 20 girls.
Definition. A proportion is a statement that gives the equality of two ratios, denoted as either a : b :: c : d or
a c
.
b d
The a and d are called the extremes and the b and c are called the means.
We use two methods for solving problems involving proportions. We will call the first method Equivalent Ratios. The
procedure is the same as finding equivalent fractions. The second method is called Cross-Product Algorithm (note the
comment on this method on p. 480 in the book—the discussion gives several reasons why the method is not good for
teaching proportional reasoning).
Equivalent Ratios
The method is the similar to finding equivalent fractions. Write the ratios in the fractional form and then change the
fractions with a common denominator.
Examples.
1. The dog to cat ratio for Mathville is two to three. If there are 18 cats, how many dogs are there?
Solution:
We use the ratio
dogs
and let N represent the number of dogs.
cats
2 N
3 18
26 N
3 6 18
12 N
18 18
N = 12, since the numerators must be equal.
There are twelve dogs in Mathville.
2.
Kim was assessed property taxes of \$1420 on a house valued at \$62,000. Approximately, what would
the assessment be on a \$140,000 house?
Solution:
We use the ratio of taxes to value and let N represent taxes on a \$140,000 house.
1420
N
62,000 140,000
1420 140
62 N
Note that 62,000 ∙ 140 = 62 ∙ 1,000 ∙ 140 = 62 ∙ 140,000.
62,000 140 62 140,000
62N = 1420 ∙ 140, since the numerators are equal when the denominators are equal.
N
1420 140 198800
3206.45
62
62
The \$140,000 house would have assessed property taxes of \$3206.45.
3.
Pat used three gallons of gas to drive seventy miles. If Pat used eight gallons of gas, approximately
how many miles could Pat drive?
Solution:
We use the ratio of gallons to miles. (Note this is opposite of the usual ratio of miles per gallon.)
3
8
70 N
3 N
70 8
70 N 70 N
-2-
3N = 560
2
N = 186 .
3
Pat would drive approximately 187 miles.
How could we simplify this process? What is method that requires fewer steps?
Examine problem 3 above: does the solution give a method that would require fewer steps?
Why does this method work?
Cross-Product Algorithm
The second method for solving problems is often called Cross-Multiplication. The procedure is essentially the same as
the shortcut used to show fractions are equivalent or to compare two fractions where we did not write the common
denominator. The justification follows from the equivalent ratios method.
Cross-Multiplication. a : b :: c : d if and only if ad = bc.
As your grandparents or great-grandparents would have said,
"the product of the means is equal to the product of the extremes."
a c
In the other notation, if and only if ad = bc.
b d
Examples.
4.
If a six-foot man standing near a flag pole casts a shadow four and one-half feet on a sunny day, how tall is a
flag pole that casts a shadow of eighteen and one-third feet?
5.
A store has two sizes of peanut butter, an economy size of 41.2 ounces for \$3.60 and a regular size of
25.5 ounces for \$2.30. Which size offers the better deal?
6.
ASU has 5200 students and 390 faculty members; whereas, BSU has 17,800 students and 1480 faculty
members. Assuming the institutions have similar missions, which college should be able to offer the
students the most individual attention?
Property of Proportions
3 9
7 21
3 7
9 21
,
,
and
.
7 21
3 9
9 21
3 7
Note that all four represent the same proportional relationships between the values 3, 7, 9, and 21.
Consider the four proportions:
Property of Proportions.
a c
b d
a b
c d
if and only if
if and only if
if and only if
b d
a c
c d
a b
where a, b, c, and d are nonzero.
This property can often simplify the steps in solving a proportion.
320 960
.
731
N
960 3
N
3
, we may rewrite the above as
.
Since
320 1
731 1
Hence, N = 3 ∙ 731 = 2193.
By using the Property of Proportions, a person could do this problem mentally.
-3-
Example. Solve
Problems and Exercises
1.
The distribution of final grades in a mathematics class showed 4 A's, 6 B's, 12 C's, 8 D's, and 2 F's. What is the ratio
of: A's to the class, A's to F's, C's to A's, the class to C's, and B's to D's?
2.
The pitch of a roof is the ratio of the rise to the half-span. What is the pitch of a roof with a rise of six feet and a span of
eighteen feet?
3.
What is the ratio of: a nickel to a dime? a dozen to a gross? 20 minutes to 45 minutes? 1 foot 6 inches to 4 yards?
4.
What is the ratio of 325 miles to 5 hours? Interpret this ratio.
5.
Kim saved \$25 in 9 weeks. At that rate, how long will it take Kim to save \$175?
6.
A picture two and one-half inches wide and three and one-half inches high is to be enlarged so that the height will be
seven inches. How wide will it be?
7.
Solve each proportion.
3 n
(a)
8 56
(b)
14 n
18 27
(c)
10 5
d 16
(d)
9 3
c 53
(e)
16 96
7
x
(f)
4 15
13 y
-4-
```
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### Featured GED Math Problem Video
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# Reducing Fractions Worksheets
One method in reducing fractions is to find a common factor that can be divided evenly into both the numerator (top number) and the denominator (bottom number).
A quick way to determine a common factor is to consider the divisibility rules for 2, 3, 5, and 10.
**A number is divisible by 2 IF the last digit of the number is an even number. That is, if the last digit is a 0, 2, 4, 6, or 8.
Positive example: 346 is divisible by 2 because the last digit 6 is an even number.
Negative example: 467 is not divisible by 2 because the last digit 7 is NOT an even number.
**A number is divisible by 3 if the sum of the digits is divisible by 3.
Positive example: 246 is divisible by 3 because 2+4+6= 12 and 12 is divisible by 3.
Negative example: 211 is NOT divisible by 3 because 2+1+1= 4 and 4 is not divisible by 3.
**A number is divisible by 5 if the last digit is a “0” or “5”.
Positive example: 12,435 is divisible by 5 because the last digit is a “5”.
Negative example: 11,231 is NOT divisible by 5 because the last digit is a 1, not a “0” or “5”.
** A number is divisible by 10 if the last digit is a “0”.
Positive example: 12, 340 is divisible by 10 because the last digit is a “0”.
Negative example: 986 is NOT divisible by 10 because the last digit is a 6, not a “0”.
Remember, when you reduce fractions….. you use a form of 1. That means that both the numerator and denominator must be divisible by the SAME number.
Example:
4/6 can be divided by 2/2 and reduced to 2/3.
Reducing Fractions Worksheet 1
Reducing Fractions Worksheet 1- Answer Key
Reducing Fractions
### 3 comments to Reducing Fractions Worksheets
• instructor
You may wonder if there is any divisibility rule for 7. I am sorry to say there is no easy way to determine if a number is divisible by 7. If it is a large number, grab a calculator and check it out.
• Kit
I can’t seem to download some items (specifically “Comparing Fractions”) even though I am logged in as a member. It just goes back to the registration page, then says “You are already registered at this level.
Is there another (maybe paid) membership level?
Thank you for your help! LOVE the site – keep up the good work!
• Instructor
Try it now. It should be fixed. Let me know if it still is not working.
|
Perpendicular at any point on a line
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PERPENDICULAR AT ANY POINT ON A LINE Figure 18-4 shows a line AB with point C between A and B. A perpendicular to AB is erected at C as follows: Figure 18-3.-Bisecting a line geometrically. Figure 18-4.-Erecting a perpendicular at a point. 1. Using any convenient point above the line (such as 0) as a center, draw a circle with radius OC. This circle cuts AB at C and at D. 2. Draw line DO and extend it to intersect the circle at E. 3. Draw line EC. This line is perpendicular to AB at C. BISECTING AN ANGLE Let angle AOB in figure 18-5 be an angle which is to be bisected. Using 0 as a center and any convenient radius, draw an arc intersecting OA and a second arc intersecting OB. Label these intersections C and D. Using C and D as centers, and any convenient radius, draw two arcs intersecting halfway between lines OA and OB. A line from 0 through the intersection of these two arcs is the bisector of angle AOB. Figure 18-5.-Bisecting an angle. SPECIAL ANGLES Several special angles may be constructed by geometric methods, so that an instrument for measuring angles is not necessary in these special cases. Figure 18-4 illustrates a method of constructing a right angle, DCE, by inscribing a right triangle in a semicircle. But an alternate method is needed for those situations in which drawing circles is inconvenient. The method described herein makes use of a right triangle having its sides in the ratio of 3 to 4 to 5. -It is often used in laying out the foundations of buildings. The procedure is as follows: 1. A string is stretched as shown in figure 18-6, forming line AC. The length of AC is 3 feet. 2. A second string is stretched, crossing line AC at A, directly above the point intended as the corner of the foundation. Point D on this line is 4 feet from A. 3. Attach a third string, 5 feet long, at C and D. When AC and AD are spread so that line CD is taut, angle DAC is a right angle. A 60° angle is constructed as shown in figure 18-7. With AB as a radius and A and B as centers, draw arcs intersecting at C. When A and B are connected to C by straight lines, all three angles of triangle ABC are 60° angles. The special angles already discussed are used in constructing 45° and 30° angles. A 90° angle is bisected to form two 45° angles, and a 60° angle is bisected to form two 30° angles.
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# Ordinary Differential Equations/Separable 2
First-Order Differential Equations
# Separable Variables: Real-World Examples
This page gives some examples of where simple separable variable DEs are found in the world around us.
## Acceleration, velocity, and position
The classic real world example of differential equations is the relationship between acceleration, velocity, and position.
$a(t)=v'(t)=x''(t) \,$
So if you're given an equation for acceleration, you can figure out both velocity and position.
### Example 1 - Constant Acceleration
Lets say that accelearation is a constant, g (the acceleration due to gravity, or about 10 m s-2. The initial velocity at t=0 is v0. The initial position is x0. Solve for v and x.
First, you need to solve for v. I.w.r.t.x.
$v'=a=g \,$
$\int v' dv=\int g dt \,$
$v=gt+C \,$
Now plug in to find C.
$v_0=g \times 0+C \,$
$C=v_0 \,$
$v(t)=gt+v_0 \,$
Now we solve for x.
$x'=v=gt+v_0 \,$
$\int x' dt=\int (gt+v_0)dt$
$x=\frac{1}{2}gt^2+v_0t+C$
Again, now we solve for C.
$x_0=\frac{1}{2}g \times 0^2+v_0 \times 0+C$
$x_0=C \,$
$x(t)=-\frac{1}{2}gt^2+v_0t+x_0$
Anyone who has studied physics will recognize this as the basic equation for position for an object undergoing a constant force in one dimension.
### Example 2 - Resistive Medium
Lets say we're traveling through a medium that resists movement. In this medium, $a=-v^2$. Solve for v and x, given that the initial velocity was 10 m/s and the initial displacement was 0 m.
$a=v'=-v^2 \,$
$\frac{dv}{dt}=-v^2$
$\int \frac{1}{v^2}dv=\int -1 dt$
$-\frac{1}{v}=-t+C$
$v=\frac{1}{t+C}$
Notice that as t increases, velocity decreases. This is what you'd expect if the medium was resisting your movement and slowing you down over time. Now substitute in the initial velocity at t=0
$10=\frac{1}{0+C}$
$C=\frac{1}{10}$
$v=\frac{1}{t+0.1}$
Now to solve for x.
$v=x'=\frac{1}{t+0.1}$
$\frac{dx}{dt}=\frac{1}{t+0.1}$
$\frac{dx}{1}=\frac{dt}{t+0.1}$
$\int 1 dx=\int \frac{1}{t+0.1} dt$
$x=ln(t+0.1)+D \,$
The position increases throughout, but it increases ever more slowly as time goes on. This is again what you'd expect for a medium resisting motion. Since velocity is never less than 0, we never stop going forward, but we go exponentially less far over time. Putting in our boundary conditions,
$0=ln(0+0.1)+D \,$
$D=-ln(0.1) \,$
$x=ln(t+0.1)-ln(0.1)=ln(10t+1) \,$
## Exponential Growth and Decay
One of the most common differential equations in science is
$y'=ky \,$.
The solution to this is
$y=Ce^{kt} \,$.
If k is positive, this is called exponential growth. If k is negative, its exponential decay. Both are used in science, for very different reasons.
### Population Growth
Lets say we have a group of animals in the wild. We want to know how many animals there will be in t years. We know how many there are now. We also know the birth rate and death rate. Can we solve this problem?
Of course we can. First, we need to figure out the rate of growth. If the birth rate is B, and the death rate is D, the total rate of change is (B-D). Since this is the rate, we need to multiply it by the current population to get the population growth. The final equation looks like
$\Delta P=(B-D)P \,$
where P is the population. That looks like the equation for exponential growth, doesn't it? As a matter of fact, change the 'delta' to a differential, and it is. The growth factor is (B-D).
#### Example 3
In a certain population of rabbits, the birth rate is 10%. The death rate is 15%. The initial population is 100. How many rabbits are there after 10 years? Will we always have rabbits?
From our solution to the exponential equation:
$P=Ce^{(B-D)t} \,$
$100=Ce^{(B-D)\times 0}=Ce^0 \,$
$C=100 \,$
$P=100e^{-0.05t} \,$
$P(10)=100e^{-0.5}\approx 61$
Unfortunately, we will not always have rabbits. Since the growth rate is negative, they will eventually go extinct. (Note: we will never actually hit 0, but in real life you can't have less than 1 rabbit. If we were measuring a continuous property instead of a discrete one, we would always have something, it would just get very small).
Another situation for exponential growth is radioactive isotopes. If you have a sample of radioactive material, individual atoms will randomly decay or not decay. While you can't know exactly how many atoms decay and when, you do know the average rate of decay. Every λ years, half of the atoms left will decay. This period of time, λ, is called a half life. The activity of the sample (how many decays per second) is called A. Mathematically, this looks like
$\frac{dA}{dt}=\lambda A$
These problems look just like the problem above. Just like rabbits, we will eventually run out of radioactive atoms as well.
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Pvillage.org
# What is the division property of multiplication?
## What is the division property of multiplication?
The multiplication property of equality and the division property of equality are similar. Multiplying or dividing the same number to both sides of an equation keeps both sides equal.
Is commutative property of multiplication?
Commutative property only applies to multiplication and addition. However, subtraction and division are not commutative.
### What is an example of the distributive property of multiplication?
The distributive property of multiplication over addition is used when we multiply a value by the sum of two or more numbers. For example, let us solve the expression: 5(5 + 9). This expression can be solved by multiplying 5 by both the addends. So, 5(5) + 5(9) = 25 + 45 = 70.
What does unity mean in a multiplication equation?
Unity, or one, is also an identity element when applied to numerical multiplication equations as any real number multiplied by unity remains unchanged (e.g., a x 1 = a and 1 x a = a).
## Why is unity called the multiplicative identity element?
Unity, or one, is also an identity element when applied to numerical multiplication equations as any real number multiplied by unity remains unchanged (e.g., a x 1 = a and 1 x a = a). It is because of this unique characteristic of unity that is called the multiplicative identity.
Which is an example of unity in math?
For example, in the addition of real numbers, zero (0) is an identity element as any number added to zero remains unchanged (e.g, a + 0 = a and 0 + a = a). Unity, or one, is also an identity element when applied to numerical multiplication equations as any real number multiplied by unity remains unchanged (e.g., a x 1 = a and 1 x a = a).
### What are the three main properties of multiplication?
Explore the commutative, associative, and identity properties of multiplication. In this article, we’ll learn the three main properties of multiplication. Here’s a quick summary of these properties:
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POWER SET OF A SET
The set of all subsets of A is said to be the power set of the set A.
The power set of A is denoted by P(A).
Example 1 :
Let A = {1, 2, 3}. Find the power set of A.
Solution :
We know that the power set is the set of all subsets.
Here, the given set A contains 3 elements.
Then, the number of subsets = 23 = 8.
Therefore,
P(A) = {{1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}, { }}
Example 2 :
Let A = {a, b, c, d, e} find the cardinality of power set of A.
Solution :
The formula for cardinality of power set of A is given below.
n[P(A)] = 2n
Here 'n' stands for the number of elements contained by the given set A.
The given set A contains five elements. So n = 5.
Then, we have
n[P(A)] = 25
n[P(A)] = 32
The cardinality of the power set of A is 32.
Subset of a Set
A set X is a subset of set Y if every element of X is also an element of Y.
In symbol we write
x ⊆ y
Read ⊆ as "X is a subset of Y" or "X is contained in Y"
Read as "X is a not subset of Y" or "X is not contained in Y"
Proper Subset
A set X is said to be a proper subset of set Y if X ⊆ Y and X ≠ Y.
In symbol, we write X ⊂ Y.
Read X ⊂ Y as "X is proper subset of Y".
The figure given below illustrates this.
Super Set
A set X is said to be a proper subset of set Y if X ⊆ Y and X ≠ Y.
In symbol, we write X ⊂ Y
Here,
Y is called super set of X
Formula to Find Number of Subsets
If A is the given set and it contains n number of elements, we can use the following formula to find the number of subsets.
Number of subsets = 2n
Formula to find the number of proper subsets :
Number of proper subsets = 2- 1
Cardinality of Power Set
We already know that the set of all subsets of A is said to be the power set of the set A and it is denoted by P(A).
If A contains n number of elements, then the formula for cardinality of power set of A is
n[P(A)] = 2n
Note :
Cardinality of power set of A and the number of subsets of A are same.
Null Set is a Subset or Proper Subset
Null set is a proper subset for any set which contains at least one element.
For example, let us consider the set A = {1}.
It has two subsets. They are { } and {1}.
Here null set is proper subset of A. Because null set is not equal to A.
If Null Set is a Super Set
If null set is a super set, then it has only one subset. That is { }.
More clearly, null set is the only subset to itself. But it is not a proper subset.
Because, { } = { }.
Therefore, A set which contains only one subset is called null set.
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Solve the equation: x = 4(cube root x)
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First, I will demonstrate how cubing a cube root or squaring a square root results in them canceling each other out. But, to do this, you need to cube the entire expression. So cubing the entire expression will result in: x cubed = 64x We have 64 because 4x4x4 = 64. The next step is having the equation equal zero, so subtract 64x from both sides: x cubed - 64x = 0 Next, we should recognize that 64 is a perfect square, meaning its square root is a perfect integer, 8. Once we know that squaring 64 gives us 8, we should make sure that everything else can be squared as well because we can't apply one function to one part of the expression; it needs to be applied to the whole expression. BUT, there's nothing else that can be squared, because the x is cubed. So we need to create the square root by factoring out ONE of the THREE x's. So: x(x squared - 64) = 0 NOW, we can find the square root of (x squared - 64), which is: (x-8)(x+8) The reason why we have a positive and negative 8 is because if we just had them positive, we would have 64, because 8x8=64, but in the previous step we found that it was (x squared MINUS 64), so that means two things multiplied resulted in a negative 64. They can't both be negative, because negative multiplied by negative results in a positive number. Hence, we have one positive and one negative. So what we have now is: 0 = x(x-8)(x+8) Finally, we can see that our answer is: x = 0, 8, or -8, because if we input all of those for x, we will end up with 0 = 0 We should prove that by inputting each of them to double check that we are correct.
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First, I will outline what they have in common: cell membrane, nucleus, mitochondria, and vacuoles. I will then explain the function of each organelle. Next, their differences: 1. Plant cells have a cell wall, but animals cells do not 2. Plant cells have chloroplasts, but animal cells do not 3. Plant cells usually have one or more large vacuole(s), while animal cells have smaller vacuoles I will also explain the function of each, and why animals don't have or need them, ex. plants have chloroplasts because they enable them to photosynthesize to make food, and animals do not photosynthesize because the amount of energy that photosynthesis would give animals is nowhere near enough what they need to be alive. Plants can survive on the minimal energy from the sun because they are still.
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Simplify the following algebraic expression: 3(x + 7) + 2(-x + 4) + 5x
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Well, first I would begin by introducing what the distributive property is. This will explain why the 3 and 2 get multiplied by what's in their adjacent parentheses. Next, I will introduce the concept of 'like terms,' which will help explain why we can combine 5x, -2x, and 3x. 3(x + 7) + 2(-x + 4) + 5x ---> (original) = 3x + 21 - 2x + 8 + 5x ----> (expanded, distributive property) = (3x - 2x + 5x) + (21 + 8) ----> (combine like terms) = 6x + 29 ----> (final answer)
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# Checking Factoring The checking of factoring can be done with the calculator. Graph the following expressions: 1.x 2 + 5x – 6 2.(x – 3)(x – 2) 3.(x.
## Presentation on theme: "Checking Factoring The checking of factoring can be done with the calculator. Graph the following expressions: 1.x 2 + 5x – 6 2.(x – 3)(x – 2) 3.(x."— Presentation transcript:
Checking Factoring The checking of factoring can be done with the calculator. Graph the following expressions: 1.x 2 + 5x – 6 2.(x – 3)(x – 2) 3.(x + 6)(x – 1) What do you notice? Are they the same graph? Discuss what you can conclude from the graphs.
Roots What is the value of 2 2 ? What is x? However: x = In this case 2 is considered the principal root or the nonnegative root, when there is more than one real root. Finding the square root of a number and squaring a number are inverse operations. WHY? Should be 2, right?
Roots What is the value of The values found are known as the nth roots and are also principal roots. The following is the format for a radical expression. index radical sign radicand
Roots of Summary of the real nth roots. Real nth Roots of b,, or – nb > 0b < 0b = 0 even one positive root one negative root no real roots one real root, 0 odd one positive root no negative roots no positive roots one negative root
Practice with Roots Simplify the following. 1. 2. 3. 4. 5. has to be absolute value to identify principal root Estimated between 5 and 6 because 5 2 = 25 and 6 2 = 36.
Radical Expressions Radical “like expressions” have the same index and same radicand. Product and Quotient Properties:
Simplifying Radical Expressions 1.The index, n, has to be as small as possible 2.radicand NO factors, nth roots 3.radicand NO fractions 4.NO radical expressions in denominator For example:
More examples: Simplifying Radical Expressions
More examples: Simplifying Radical Expressions
Radical Expressions Conjugates ± same terms Multiply the following:
Radical Expressions Deduction about conjugates: Product of conjugates is always a rational number. For example:
Radical Expressions Simplify the following
Radical Expressions In-Class work Rationalize the denominator:
Download ppt "Checking Factoring The checking of factoring can be done with the calculator. Graph the following expressions: 1.x 2 + 5x – 6 2.(x – 3)(x – 2) 3.(x."
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# 3.5 Addition of velocities (Page 4/12)
Page 4 / 12
## Calculating relative velocity: an airline passenger drops a coin
An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?
Strategy
Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes.
Solution for (a)
Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation:
${{v}_{y}}^{2}={{v}_{0y}}^{2}-2g\left(y-{y}_{0}\right)\text{.}$
Substituting known values into the equation, we get
${{v}_{y}}^{2}={0}^{2}-2\left(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(-1\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{m}-0 m\right)=\text{29}\text{.}4\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}{\text{/s}}^{2}$
yielding
${v}_{y}=-5\text{.}\text{42 m/s.}$
We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.
Solution for (b)
Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is ${v}_{y}=-5.42\phantom{\rule{0.25em}{0ex}}\text{m/s}$ , the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and . The x - and y -components of velocity can be combined to find the magnitude of the final velocity:
$v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}\text{.}$
Thus,
$v=\sqrt{\left(\text{260 m/s}{\right)}^{2}+\left(-5\text{.}\text{42 m/s}{\right)}^{2}}$
yielding
$v=\text{260}\text{.}\text{06 m/s.}$
The direction is given by:
$\theta ={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)={\text{tan}}^{-1}\left(-5\text{.}\text{42}/\text{260}\right)$
so that
$\theta ={\text{tan}}^{-1}\left(-0\text{.}\text{0208}\right)=-1\text{.}\text{19º}\text{.}$
Discussion
In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v in part (b) is not ; rather, it is . The velocity’s magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See [link] .) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path.
write an expression for a plane progressive wave moving from left to right along x axis and having amplitude 0.02m, frequency of 650Hz and speed if 680ms-¹
how does a model differ from a theory
what is vector quantity
Vector quality have both direction and magnitude, such as Force, displacement, acceleration and etc.
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what's electromagnetic induction
electromagnetic induction is a process in which conductor is put in a particular position and magnetic field keeps varying.
Lukman
wow great
Salaudeen
what is mutual induction?
je
mutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.
Johnson
how to undergo polarization
show that a particle moving under the influence of an attractive force mu/y³ towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v²k² and distance uk²/√u-vk as origin
show that a particle moving under the influence of an attractive force mu/y^3 towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v^2k^2 and distance uk^2/√u-k as origin
No idea.... Are you even sure this question exist?
Mavis
I can't even understand the question
yes it was an assignment question "^"represent raise to power pls
Gabriel
Gabriel
An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º .
no ideas
Augstine
if u at an angle of 12 degrees their period will be same so as their velocity, that means they both move simultaneously since both both hovers at same length meaning they have the same length
Modern cars are made of materials that make them collapsible upon collision. Explain using physics concept (Force and impulse), how these car designs help with the safety of passengers.
calculate the force due to surface tension required to support a column liquid in a capillary tube 5mm. If the capillary tube is dipped into a beaker of water
find the time required for a train Half a Kilometre long to cross a bridge almost kilometre long racing at 100km/h
method of polarization
Ajayi
What is atomic number?
The number of protons in the nucleus of an atom
Deborah
type of thermodynamics
oxygen gas contained in a ccylinder of volume has a temp of 300k and pressure 2.5×10Nm
why the satellite does not drop to the earth explain
what is a matter
Yinka
what is matter
Yinka
what is matter
Yinka
what is a matter
Yinka
I want the nuclear physics conversation
Mohamed
because space is a vacuum and anything outside the earth 🌎 can not come back without an act of force applied to it to leave the vacuum and fall down to the earth with a maximum force length of 30kcm per second
Clara
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# ICSE Revision Notes for Co-ordinate Geometry Class 10 Maths
Chapter Name Co-ordinate Geometry Topics Covered Reflection of FiguresSection FormulaSlope of a LineSlope-Intercept Form of Straight LinesPoint-Slope and Two-Point Form of Straight Lines Related Study
### Reflection of Figures
Let us consider the following pictures.
For each picture, let us draw vertical lines exactly at the middle as shown below.
After drawing lines, we can observe that the left half of the pictures is exactly the same as the right half of the pictures. These pictures are known as symmetrical pictures. The line through which the figure is divided is called line of symmetry.
Here, the dotted lines of these pictures are lines of symmetry. If we consider only one-half of these images and place a mirror instead of the dotted line, then we will get a mirror reflection of the image, which will be the missing half of the original image. Thus, we can also say that the two halves obtained by dividing the figure through the line of symmetry are mirror images of each other. For example, the left portion of each image is the mirror image of the right portion of the image and vice-versa.
We come across different objects in our day to day life where symmetry is shown by mirror reflection. For example, if we look at the following picture, we can see the reflected image of the buildings in the water. The water surface acts as a mirror or as a line of symmetry. We can observe the symmetry of the objects by their reflection, though the image is not very clear.
We can also take the example of Rangoli patterns.
There are many lines of symmetry for these types of patterns as shown below. We can observe reflection pattern in the patterns along their lines of symmetry acting as mirror lines.
We can see that in each figure, one half of the pattern is the reflection of the other half.
Let us consider the given figure of ΔPQR on a grid paper, where AB is a mirror line.
Let us draw the image of ΔPQR with reference to the mirror line AB.
Here, ΔPQR and ΔP′Q′R′ are symmetrical with reference to the mirror line AB.
Now, we can say that ΔP′Q′R′ is the mirror image of ΔPQR with reference to the mirror line AB. If we fold the grid paper along the mirror line, then we will observe that the two triangles overlap. It can also be observed that:
The lengths of the sides of ΔP′Q′R′ are equal to the corresponding sides of ΔPQR. • The angles of ΔP′Q′R′ are equal to the corresponding angles of ΔPQR.
Every portion of ΔP′Q′R′ is at the same distance from the mirror line as that of the corresponding portions of ΔPQR.
Now, let us extend the concept of mirror further to study about image of a point. If we place a point in front of a mirror, then what is the nature of the image formed?
Let XY be a mirror. Let A be a point (object) placed in front of it. We obtain its image A′ as shown below:
Can we notice anything in the above figure?
We can notice that:
1. The distance of the image (A′) behind the mirror is same as the distance of the object (A) from it i.e., PA = PA′
2. The mirror line XY is perpendicular to the line joining the object and the image i.e., XY ⊥ AA′
Here, XY (the mirror line) is called the axis of reflection or mediator. What would happen if the point A lies on XY?
If the point A lies on XY, then its image will be this point itself. In such case, A is called an invariant point with respect to mirror line XY.
Reflection of a point in the lines x = a and y = a.
x = a is a line parallel to the y-axis and at a distance of a units from it. If we have to find the reflection of point A(−2, 4) from the line x = 2 we follow the below given steps:
Step 1: Line PQ represents x = 2 which is a straight line parallel to the y-axis and at a distance of 2 units from it.
Step 2: Mark a point A(−2, 4) on the same graph.
Step 3: From the point A, draw a straight line perpendicular to PQ. Mark a point A' behind the straight line PQ at the same distance as A(−2, 4) is before it. A'(6, 4) is the required reflection of the point A(−2, 4) in the line x = 2.
Similarly, we have y = a which is parallel to x-axis and is at a distance of a units from it. Suppose we have to find the reflection of point B(2, 2) from the line y + 2 = 0 we follow the below given steps:
Step 1: Line PQ represents y = − 2 which is a straight line parallel to the x-axis and at a distance of 2 units from it.
Step 2: Mark a point B(2, 2) on the same graph.
Step 3: From the point B, draw a straight line perpendicular to PQ. Mark a point B' below this straight line PQ at the same distance as B(2, 2) is above it. B'(2, −6) is the required reflection of the point B(2, 2) in the line y + 2 = 0.
In order to understand these concepts better, let us look at some examples.
Example 1: Draw the mirror reflection of the following figure where AB is the mirror line.
The mirror reflection of the given figure with respect to mirror line AB can be drawn as
Example 2: The given figure shows a wall with end points A and B. Sanjana is standing at position P. She has to come to position Q after touching the wall. Find the shortest path for Sanjana to come from P to Q.
Let us imagine that wall AB acts as a mirror. Then P′ is the position of the image of point P. The object and its mirror image are at the same distance from the mirror. Therefore, points P and P′ are at the same distance from the wall.
The shortest distance between two points is the straight line joining the two points.
Therefore, the shortest distance between P′ and Q is P′Q. Let us join the points P′ and Q by a straight line which passes through the wall at point R.
Now, P′Q = P′R + RQ
But PR is the mirror reflection of P`R.
Or we can say that PR = P′R
Therefore, P′Q = PR + RQ
The path from P to R and then from R to Q is the shortest path which should be followed by Sanjana.
Example 3: The quadrilateral ABCD whose vertices are A (−3, −2), B (2, −3), C (3, 1), and D (−2, 4) is on a co-ordinate plane. Draw its reflection in origin.
(1) The reflection of the points A (−3, −2), B (2, −3), C (3, 1), and (D (−2, 4) in the origin are:
A’(3, 2), B’(-2, 3), C’(-3, -1), and D’(2, -4)
By joining A’B’, B’C’, C’D’, and A’D’, we obtain the quadrilateral A'B'C'D', which is the reflection of the given quadrilateral ABCD in the origin as shown below.
Example 4: The line AB joining the points A (−3, −4) and B (2, 1) is on co-ordinate plane. Draw its reflection
(i) A'B' about x-axis
(ii) A" B" about y-axis
(i) The refection of points A (−3, −4) and B (2, 1) about x-axis are A' (−3, 4) and B' (2, − 1). By joining A'B' , we obtain the reflection of the given line AB as shown below.
(ii) The reflection of points A (−3, −4) and B (2, 1) about y-axis are A" (3, −4) and (− 2, 1).
By joining A"B", we obtain the reflection of the given line AB as shown below.
### Section Formula
Bageecha Singh’s garden is rectangular in shape and its length and breadth are 10 m and 20 m respectively. Two lamp posts in the garden are placed at the ends of a diagonal of the garden. Bageecha Singh wants to place one more lamp post between the two lamp posts that will divide the line segment joining the two lamp posts in the ratio 3:5.
Now, can you help him to find the position of the new lamp post? External division of a line segment:
Observe the figure given below.
Here, AB is a line segment and P is a point outside the line segment AB such that A – B – P (or P – A – B). So, it can be said that the point P divides the line segment AB externally in the ratio AB : BP. Point P is known as the point of external division.
Let the coordinates of points A and B be (x1y1) and (x2y2) respectively. Also, let the point P divide the line segment AB externally in the ratio then the coordinates of point P are given by the following formula.
Coordinates of P = (mx2 – nx1)/(m – n), (my2 – ny1)/(m – n)
This formula is known as the section formula for external division.
Centroid of a Triangle: If A(x1, y1) B(x2, y2) and C(x3,y3) are the vertices of a ∆ABC, then the centroid of ∆ABC is
G (x, y) = G (x1 + x2 + x3)/3, (y1 + y2 + y3)/3
Let us solve some examples based on the section formula.
Example 1: Find the coordinates of the point which divides the line segment joining the points (2, 3) and (–1, 7) internally in the ratio 1:2.
Let (2, 3) and (–1, 7) be denoted by A and B respectively.
Let C be the point that divides the line segment AB internally in the ratio 1:2.
Using section formula, we obtain
Coordinates of C = {1(-1) + 2(2)/(1 + 2), 1(7) + 2(3)/(1 + 2)}
= {(-1 + 4)/3, (7 + 6)/3}
= (3/3, 13/3)
= (1, 13/3)
Thus, (1, 13/3) are the required coordinates of the point.
Thus, are the required coordinates of the point.
Example 2: Find the coordinates of the point which divides the line segment joining the points (4, –5) and (6, 2) externally in the ratio 3:2.
Let (4, –5) and (6, 2) be denoted by A and B respectively.
Let P be the point that divides the line segment AB externally in the ratio 3:2.
Using section formula, we obtain
Coordinates of P
= {3(6) - 2(4)/(3 – 2), 3(2) – 2(-5)/(3 – 2)}
= (18 – 8)/1, (6 + 10)/1)
= (10, 16)
Thus, (10, 16) are the required coordinates of the point.
Example 3: Find the coordinates of the points of trisection of a line segment joining the points (−2, 1) and (4, –3).
Let (–2, 1) and (4, –3) be denoted by A and B respectively.
Let C and D be the points of trisection. This means that C divides the line segment AB in the ratio 1 : 2 and D divides the line segment AB in the ratio 2 : 1.
Using section formula, we have
Coordinates of C = {1(4) + 2(-2)/(1 + 2)/(1 + 2), 1(-3) + 2(1)/(1 + 2)
= (4 - 4)/3, (-3 + 2)/3
= (0, -1/3)
and Coordinates of D = {2(4) + 1(-2)}/(1 + 2), 2(-3) + 1(1)/(1 + 2)}
= (8 - 2)/3, (-6 + 1)/3
= (6/3, -5/3)
= (2, -5/3)
Thus, (0, -1/3) and (2, -5/3) are the points of trisection of a line segment joining the points (–2, 1) and (4, –3).
Example 4: The mid-point of a portion of a line that lies in the first quadrant is (3, 2). Find the points at which the line intersects the axes.
The line has been shown in the following graph:
Let A and B be the points of intersection with y and x-axes respectively.
Let the coordinates of A and B be (0, b) and (a, 0).
Here, C is the mid-point of A and B.
∴ (a + 0)/2, (0 + b)/2 = (3, 2)
(a/2, b/2) = (3, 2)
= (3, 2)
On equating the x and y-coordinates on both sides, we obtain
a/2 = 3 and b/2 = 2
a = 6 and b = 4
Thus, the coordinates of A and B are (0, 4) and (6, 0) respectively.
### Slope of a Line
Have you ever wondered why it is difficult to climb a mountain while it is easy to walk down a straight road?
In such cases, we generally use the term ‘slope’ and say that the slope of the mountain is steep.
But do we actually know what slope is and how it is calculated?
Here, we will study about the slopes of straight lines. To understand what we mean by slope, let us first understand what we mean by inclination of a line.
Consider a straight line l, as shown in the figure.
Observe that the line l makes an angle θ with the positive direction of x-axis when measured in the anticlockwise direction. We say that this angle θ is the inclination of the line l.
The angle which a straight line makes with the positive direction of x-axis measured in the anticlockwise direction is called the inclination (or angle of inclination) of the line.
Now, from this definition, we can observe the following points:
1. Inclination of a line parallel to y-axis or the y-axis itself is 90°.
2. Inclination of a line parallel to x-axis or the x-axis itself is 0°.
Now that we have understood what we mean by inclination, let us now understand the meaning of the slope of a line.
In the above figure, we have seen that the inclination of line l is θ. In this case, we say that tan θ is the slope of line l
If θ is the inclination of a line l with the positive direction of x-axis, then tan θ is called the slope or gradient of line l. The slope of a line is denoted by m.
For example, the slope of the line which makes an inclination of 45° with the positive direction of x-axis is given by m = tan 45° = 1
Note:
1. Since tan θ is not defined for θ = 90°, we say that the slope of a vertical line is not defined. We also conclude that the slope of y-axis is not defined.
2. The slope of x-axis is 0.
Now, if we have a line which passes through two given points, then can we find the slope of that line?
Yes, we can find the slope of that line using the formula given below.
If P(x1y1) and Q(x2y2) are two points on a non-vertical line whose inclination is θ, then the slope of line is given by m = (y2 – y1)/(x2 – x1)
Let us prove this formula.
We have two points P(x1y1) and Q(x2y2) on a line whose inclination is θ as shown in the following figure.
Let us draw perpendiculars from P and Q to X-axis which meet X-axis at A and B respectively.
Also, let us draw PC ⊥ QB.
∴ PC || AB
It can be seen that PQ is transversal with respect to X-axis and PC such that PC || X axis.
Now,
∠QMB = θ (Given)
∠QPC = ∠QMB (Corresponding angles)
∴ ∠QPC = θ
Also, we have
OA = x1 and OB = x2 ∴ AB = x2 – x1
PA = y1 and OB = y2 ∴ QC = y2 – y1
Since AB = PC
∴ PC = x2 – x1
In right-angled triangle ΔPQC, we have
∠QPC = θ
tan θ = (Side opposite to angle θ)/(Side adjacent to angle θ)
⇒ tan θ = QC/PC
⇒ tan θ = (y2 – y1)/(x2 – x1)
Slope of the PQ = Slope of the line l = tan θ
Slope of PQ = Slope of line l = (y2 – y1)/(x2 – x1)
Hence proved.
Using this formula, we can find the slope of any line passing through two distinct points.
For example, the slope of the line passing through the points (3, −7) and (5, 1) is (1 - (-7))/(5 - 3) = 8/2 = 4
Now, we know that if there are two lines in a coordinate plane, then they will be either parallel or perpendicular. In either of the two cases, a relation between the slopes of the two lines is exhibited. The relation is explained as follows:
Two non-vertical lines l1 and l2 are parallel, if and only if their slopes are equal. In other words, if m1 and m2 are the slopes of lines l1 and l2 respectively, then the lines l1 and l2 are parallel to each other, if m1m2
Two non-vertical lines l1 and l2 are perpendicular to each other, if and only if their slopes are negative reciprocals of each other. In other words, if m1 and m2 are the slopes of lines l1 and l2 respectively, then the lines l1 and l2 are perpendicular to each other, if m1m2 = −1.
Now, if we have three points A, B, and C, then we can conclude the following statement:
Three points A, B, and C will lie on a line i.e., they will be collinear, if and only if the slope of AB is the same as the slope of BC.
Let us now look at some examples to understand the concept of slope better.
Example 1: A line l1 passes through points (5, −3) and (4, −6). Another line, l2, passes through points (8, 1) and (2, 3). Are lines l1 and l2 perpendicular, parallel or neither of the two?
We will first find the slopes of the two lines.
We know that if a line passes through points (x1y1) and (x2y2), then the slope of that line is given by m = (y2 – y1)/(x2 – x1)
Thus,
Slope of line l1 is given by
m1 = {(-6) – (-3)}/(4 – 5) = -3/-1 = 3
Slope of line l2 is given by
m2 = (3 – 1)/(2 – 8) = 2/-6 = -1/3
Here, we can observe that m1 m2 = −1. Hence, lines l1 and l2 are perpendicular to each other.
Example 2: The line passing through points (0, 2) and (8, 4) is parallel to the line passing through points (4, 8/5) and (2, p). Find the value of p
We know that two lines are parallel if and only if their slopes are equal. The slope of a line passing through points (x1, y1) and (x2, y2) is given by m = (y2 – y1)/(x2 – x1)
Therefore,
Slope of the line passing through points (0, 2) and (8, 4) is given by m1 = (4 – 2)/(8 – 0) = 2/8 = 1/4
Slope of the line passing through points (4, 8/5) and (2, p) is given by
m2 = (p – 8/5)/(2 – 4) = (5p – 8)/(-2 × 5) = (-5p + 8)/10
Since the two lines are parallel,
m1 = m2
⇒ 1/4 = (- 5p + 8)/10
⇒ 10 = -20p + 32
⇒ 5 = - 10p + 16
⇒ 10p = 16 – 5 = 11
⇒ p = 11/10
Thus, the value of p is 11/10.
Example 3: The given graph shows the temperature of water, which was kept on fire for some time, at different intervals of time.
What will be the temperature of water at 8 p.m. if it was kept in the same conditions from 2 p.m. to 9 p.m.?
Since line AB passes through points A (2:00 p.m., 60°C) and B (5:00 p.m., 45°C), its slope is (45 – 60)/(5 – 2) = (-15)/3 = -5
Let y be the temperature of water at 8:00 p.m. Accordingly, on the basis of the given graph, line AB must pass through point C (8:00 p.m., y).
∴ Slope of AB = Slope of BC
⇒ - 5 = (y – 45)/(8 – 5)
⇒ -5 = (y – 45)/3
⇒ -15 = y – 45
⇒ y = - 15 + 45
⇒ y = 30
Thus, the temperature of water will be 30°C at 8:00 p.m.
### Slope-Intercept Form of Straight Lines
Slope-intercept Form
If a line with slope m makes y-intercept as c, then the equation of the line is given by y = mx + c.
In other words, we can say that point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c.
If a line with slope m makes x-intercept as d, then the equation of the line is given by y = m(x d).
A general equation Ax + By + C = 0 can be written in slope-intercept form as follows:
y = -(A/B)x – C/B , if B ≠ 0, where m = -(A/B) and C = -(C/B)
x = -(C/A), if B = 0, which is a vertical line whose slope is undefined and whose x-intercept is –(C/A) .
Solved Examples
Example 1: The equation of a line is given by 12x + 8y − 9 = 0. Find the angle made by this line with the positive direction of the x-axis.
The equation of the line is given by
12x + 8y − 9 = 0
⇒ 8y = 9 − 12
⇒ y = 9/8 – 12/8.x
⇒ y = 9/8 – 3/2.x
Comparing this equation with the general form y = mx + c, we obtain the slope of the line as
m = -(3/2) = tan θ
Thus, the angle made by the line with the positive direction of the x-axis is tan-1(-3/2).
Example 2: Find the equation of the line that makes x-intercept as 5 and is perpendicular to the line 16x + 4y = 5.
It is given that the line is perpendicular to the line 16x + 4y = 5.
The slope of this line can be calculated as
4y = −16x + 5
⇒ y = - 4x + 5/4
Thus, the slope of this line is −4. Therefore, the slope of the required line is 1/4. Also, it is given that the line makes x-intercept as 5.
By using the slope-intercept form, we get the required equation of the line as
y = 1/4 (x – 5)
⇒ 4y = x – 5
⇒ x – 4y – 5 = 0
### Point-Slope and Two-Point Form of Straight Lines
Point-Slope Form
The equation of a non-vertical line with slope m and passing through the point (x1, y1) is given by (y y1) = m(x x1).
In other words, the point (x, y) lies on the line with slope m through the fixed point (x1, y1) if and only if its coordinates satisfy the equation (y y1) = m(x x1).
The point-slope form of the equation is used when the information about the slope of the line and a point through which it passes is given.
Two-Point Form
The equation of a non-vertical line passing through two given points (x1, y1) and (x2, y2) is given by (y – y1)/(x – x1) = (y2 – y1)/(x2 – x1).
Solved Examples
Example 1: Find the equation of the line that passes through the points and (−5, 2).
Using the two-point form of the equation of line, we know that the equation of the line passing through the points (x1y1) and (x2y2) is given by (y – y1)/(x – x1) = (y2 – y1)/(x2 – x1
Thus, the required equation of the line passing through the points (1/3, 1/2) and (−5, 2) is given by
Example 2: Find the equation of the line that passes through the point (8, 5) and makes an inclination of 210° with the x-axis.
It is given that the line makes an inclination of 210° with the x-axis. Therefore, the slope of the line is given by
m = tan 210° = tan (180° + 30°) = tan 30° = 1/√3
Using point-slope form of equation of line, we know that the equation of the line with slope that passes through the point (x1y1) is given by
(− y1) = m(− x1)
Thus, the required equation of the line is given by
y – 5 = 1/√3 = (x – 8)
⇒ √3(y – 5) = (x – 8)
⇒ √3y – x - 5√3 + 8 = 0
Example 3: Find the equation of the line that passes through the intersection of lines 2x + y + 6 = 0 and x y + 9 = 0 and is perpendicular to the line that passes through points (−6, 3) and (4, 5).
It is given that the required line passes through the intersection of lines 2x + y + 6 = 0 and x y + 9 = 0.
The points of intersection of lines 2x + y + 6 = 0 and x y + 9 = 0 can be found by adding the two equations. Hence,
3x + 15 = 0
x = −5
∴ y = 4
Thus, the required equation passes through the point (−5, 4).
It is also given that the required line is perpendicular to the line that passes through points (−6, 3) and (4, 5).
We know that the slope of a line that passes through points (x1y1) and (x2y2) is given by m = (y2 – y1)/(x2 – x1).
Thus, the slope of the line that passes through points (−6, 3) and (4, 5) is given by = (5 – 3)/(4 + 6) = 2/10 = 1/5
We also know that if two lines are perpendicular to each other, then their slopes are negative reciprocals of each other.
Thus, the slope of the required line is −5.
Thus, we are required to find the equation of the line that passes through the point (−5, 4) having slope −5.
Using point-slope form, the equation of the required line is given by
− 4 = (−5)(+ 5)
y − 4 = −5x – 25
⇒ 5x + y + 21 = 0
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Food for thought
## Solution
We denote the product of the first $20$ natural numbers by $20!$ and call this $20$ factorial.
1. What is the highest power of $5$ which is a divisor of $20$ factorial?
We have $20! = 20 \times 19 \times 18 \times \dotsm \times 3 \times 2 \times 1$.
Since $5$ is a prime number, we can just add the highest power of $5$ dividing each of the numbers $1$, $2$, $3$, …, $20$. This is $0$ for each number not divisible by $5$. There are only four numbers in this range ($5$, $10$, $15$, $20$) that are divisible by $5^1$, and none of them is divisible by $5^2$. So the highest power of $5$ dividing $20!$ is $5^4$.
Just how many factors does $20!$ have altogether?
The first thing we need to do is find the prime factorisation of $20!$. As $20!=20 \times 19 \times 18 \times \dotsm \times 3 \times 2 \times 1$, we can can do this by finding the prime factorisation of each of the numbers $2$, $3$, …, $20$ (ignoring $1$ since it won’t contribute any primes) and multiplying them together.
Number Prime factorisation
$2$ $2$
$3$ $3$
$4$ $2^2$
$5$ $5$
$6$ $2 \times 3$
$7$ $7$
$8$ $2^3$
$9$ $3^2$
$10$ $2 \times 5$
$11$ $11$
$12$ $2^2 \times 3$
$13$ $13$
$14$ $2 \times 7$
$15$ $3 \times 5$
$16$ $2^4$
$17$ $17$
$18$ $2 \times 3^2$
$19$ $19$
$20$ $2^2 \times 5$
Multiplying the prime factorisations in the right-hand column together and simplifying, we get $20!=2^{18} \times 3^8 \times 5^4 \times 7^2 \times 11 \times 13 \times 17 \times 19.$
Now we can use this to calculate the number of divisors of $20!$. Each divisor will have a unique prime factorisation, which must be ‘contained’ within the prime factorisation of $20!$. Let $m$ be a divisor of $20!$. Then there are nineteen possible values for the highest power of $2$ dividing $m$ ($0$, $1$, …, $18$). Similarly, there are nine possible values for the highest power of $3$ dividing $m$ ($0$, $1$, …, $8$). Continuing in this way for all the prime factors of $20!$, we can calculate that there are $19 \times 9 \times 5 \times 3 \times 2 \times 2 \times 2 \times 2 = 41040$ divisors of $20!$.
1. Show that the highest power of $p$ that divides $500!$, where $p$ is a prime number and $p^t<500 < p^{t+1}$, is $\lfloor 500/p\rfloor+\lfloor 500/p^2\rfloor+\dotsb+\lfloor 500/p^t\rfloor,$ where $\lfloor x\rfloor$ (the floor of $x$) means to round $x$ down to the nearest integer. (For example, $\lfloor 3\rfloor=3$, $\lfloor 4.7\rfloor=4$, $\lfloor -2.7\rfloor=-3$, and so on.)
We can see that $\lfloor 500/p\rfloor$ is the number of multiples of $p$ that are less than or equal to $500$. For example, if $p$ goes into $500$ “seven and a bit” times, this means that $p$, $2p$, …, $7p$ are less than $500$ but $8p$ is greater than $500$.
Similarly, $\lfloor 500/p^2\rfloor$ is the number of multiples of $p^2$ that are less than or equal to $500$, and so on.
Now, we can work out the highest power of $p$ that divides $500!$ by considering the number of multiples of $p$, $p^2$, …, $p^t$ less than $500$.
We need to count all the multiples of $p$. We need to count the multiples of $p^2$ twice, since they contribute $2$ to the exponent, but we have already counted them once as they are also multiples of $p$ so we need to count them just once more. Similarly, we need to count the multiples of $p^3$ three times in total, but we have already counted them twice (once in the multiples of $p$ and once in the multiples of $p^2$), so we need to count them just once more. And so on for the remaining powers. So the highest power of $p$ that divides $500!$ has exponent $\left\lfloor \frac{500}{p}\right\rfloor+\left\lfloor \frac{500}{p^2}\right\rfloor+ \dotsb +\left\lfloor \frac{500}{p^t}\right\rfloor$
Note that we had to assume that $p$ was prime. What would have gone wrong if it had not been?
1. How many factors does $n!$ have?
We can generalise the above result beyond the case of $500!$. The highest power of $p$ that divides $n!$, where $p^t \leq n < p^{t+1}$, is equal to $\left\lfloor \frac{n}{p}\right\rfloor + \left\lfloor \frac{n}{p^2}\right\rfloor + \dotsb + \left\lfloor \frac{n}{p^t}\right\rfloor,$ by exactly the same reasoning as in (b).
We can use this information to find the prime factorisation of $n!$. Let $P$ be the largest prime with $P\leq n$. Also, for any number $m$, let $t_m$ be the integer such that $m^{t_m} \leq n< m^{t_m+1}$.
Using the information from part (b), we can then calculate the prime factorisation of $n!$: we have \begin{align*} n! &= 2^{\left(\lfloor n/2\rfloor+\lfloor n/2^2\rfloor + \dotsb + \lfloor n/2^{t_2}\rfloor\right)} \times 3^{\left(\lfloor n/3\rfloor+\lfloor n/3^2\rfloor + \dotsb + \lfloor n/3^{t_3}\rfloor\right)} \times \dotsm\\ &\qquad\times P^{\left(\lfloor n/P\rfloor+\lfloor n/P^2\rfloor + \dotsb + \lfloor n/P^{t_P}\rfloor\right)}. \end{align*}
Then we can use the same reasoning as at the end of part (a) to calculate the number of factors of $n!$: we get
\begin{align*} &\left(1 + \left\lfloor \frac{n}{2}\right\rfloor+\left\lfloor \frac{n}{2^2}\right\rfloor+ \dotsb + \left\lfloor \frac{n}{2^{t_2}}\right\rfloor\right) \times \left(1 + \left\lfloor \frac{n}{3}\right\rfloor + \left\lfloor \frac{n}{3^2}\right\rfloor + \dotsb + \left\lfloor \frac{n}{3^{t_3}}\right\rfloor\right) \times \dotsm\\ &\qquad\times \left(1 + \left\lfloor \frac{n}{P}\right\rfloor + \left\lfloor \frac{n}{P^2}\right\rfloor + \dotsb +\left\lfloor \frac{n}{P^{t_P}}\right\rfloor\right). \end{align*}
We can check that this expression works for the example of $20!$:
\begin{align*} &\left(1 + \left\lfloor \frac{20}{2}\right\rfloor+\left\lfloor \frac{20}{4}\right\rfloor+\left\lfloor \frac{20}{8}\right\rfloor+\left\lfloor \frac{20}{16}\right\rfloor \right) \times \left(1 + \left\lfloor \frac{20}{3}\right\rfloor + \left\lfloor \frac{20}{9}\right\rfloor\right) \times \\ &\qquad \left(1 + \left\lfloor \frac{20}{5}\right\rfloor\right) \times \left(1 + \left\lfloor \frac{20}{7}\right\rfloor\right) \times \left(1 + \left\lfloor \frac{20}{11}\right\rfloor\right) \times \\ &\qquad \left(1 + \left\lfloor \frac{20}{13}\right\rfloor\right) \times \left(1 + \left\lfloor \frac{20}{17}\right\rfloor\right) \times \left(1 + \left\lfloor \frac{20}{19}\right\rfloor\right)\\ &\quad=(10+5+2+1+1) \times (6+2+1) \times (4+1) \times (2+1) \times\\ &\qquad (1+1) \times (1+1) \times (1+1) \times (1+1)\\ &\quad=19 \times 9 \times 5 \times 3 \times 2 \times 2 \times 2 \times 2\\ &\quad= 41040. \end{align*}
This is the same answer as in part (a), suggesting that this formula works!
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# Polynomial Exponents Lessons
The previous lesson explained how to simplify exponents of a single term inside parentheses, like the problem below.
(x3y4)5
This lesson covers how to simplify exponents on parentheses that contain a polynomial (more than one term), like the problem below.
(x3 + y4)2
Because the two terms inside parentheses are not being multiplied or divided, the exponent outside the parentheses can not just be "distributed in". Instead, a 1 must be multiplied by the entire polynomial the number of times indicated by the exponent. In this problem the exponent is 2, so it is multiplied two times:
1(x3 + y4)(x3 + y4)
Use the FOIL Method to simplify the multiplication above, then combine like terms.
x6 + x3y4 + x3y4 + y8
x6 + 2x3y4 + y8
## Multiplication and Division
Examine the problem below.
(x3y4)5
Recall that multiplication is implied when there is no sign between a variable or set of parentheses and a number, another variable, or another set of parentheses. Therefore in this problem, the x3 and y4 are being multiplied.
In the next problem the x2 and x are being multiplied. The difference is that a * is present which explicitly indicates multiplication. We will solve this problem, then return to the first problem on the page.
(x2 * x)3
Because there is no addition or subtraction inside the parentheses, the exponent can be just "distributed" in and simplified:
(x2*3 * x3)
x6 * x3
x9
Notice that this gives the same result as if we had simplified the inside of the parentheses first, as we have done below.
(x2 * x)3
(x3)3
x3*3
x9
So why are there two different methods of solving this problem? The first method, where the exponent was distributed in can be applied to the first problem on this page, whereas the second method cannot.
We will now apply the "distribute in" method to the first problem presented on this page.
(x3y4)5
(x3*5y4*5)
x15y20
This method will also work when the terms are being divided, like the problem below:
(x2 / x)3
Again, the exponent is just "distributed" in:
(x2*3 / x3)
(x6 / x3)
x3
## Fractions
Fractions are really just a division problem which is shown in a special form. Since we can just "distribute" in the exponents for an ordinary division problem, we can do the same for a fraction. Look over the example below:
We can just distribute in the 3, as in the other problems.
As you can see, once the 3 was distributed, the parentheses could be removed. Then the 23 was simplified.
## Exponents of Polynomials (Parentheses) Resources
Practice Problems / WorksheetPractice all of the methods you learned in this lesson. Next Lesson: Order of OperationsLearn how to use the Order of Operations to simplify expressions containing more than one operation.
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