lesson 8 homework 2.8 answer key

Home > CC3 > Chapter Ch8 > Lesson 8.2.2

Lesson 8.1.1, lesson 8.1.2, lesson 8.1.3, lesson 8.2.1, lesson 8.2.2, lesson 8.2.3, lesson 8.2.4, lesson 8.3.1.

© 2022 CPM Educational Program. All rights reserved.

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Answer Key 8.2

• [latex]\dfrac{8x^2}{9}\cdot \dfrac{9}{2}\Rightarrow \dfrac{\cancel{2}\cdot 4\cdot x^2}{\cancel{9}}\cdot \dfrac{\cancel{9}}{\cancel{2}}\Rightarrow 4x^2[/latex]
• [latex]\dfrac{8x}{3}\div \dfrac{4x}{7}\Rightarrow \dfrac{8x}{3}\cdot \dfrac{7}{4x} \Rightarrow \dfrac{2\cdot \cancel{4}\cdot \cancel{x}}{3}\cdot \dfrac{7}{\cancel{4}\cdot \cancel{x}}\Rightarrow \dfrac{14}{3}[/latex]
• [latex]\dfrac{5x^2}{4}\cdot \dfrac{6}{5}\Rightarrow \dfrac{\cancel{5}\cdot x^2}{2\cdot \cancel{2}}\cdot \dfrac{\cancel{2}\cdot 3}{\cancel{5}}\Rightarrow \dfrac{3x^2}{2}[/latex]
• [latex]\dfrac{10p}{5}\div \dfrac{8}{10}\Rightarrow \dfrac{10p}{5}\cdot \dfrac{10}{8}\Rightarrow \dfrac{\cancel{2}\cdot 5\cdot p}{\cancel{5}}\cdot \dfrac{\cancel{2} \cdot \cancel{5}}{2\cdot \cancel{2} \cdot \cancel{2}}\Rightarrow \dfrac{5p}{2}[/latex]
• [latex]\dfrac{\cancel{(m-6)}}{7\cancel{(7m-5)}}\cdot \dfrac{5m\cancel{(7m-5)}}{\cancel{m-6}}\Rightarrow \dfrac{5m}{7}[/latex]
• [latex]\dfrac{7(n-2)}{10(n+3)}\div \dfrac{n-2}{(n+3)}\Rightarrow \dfrac{7\cancel{(n-2)}}{10\cancel{(n+3)}}\cdot \dfrac{\cancel{(n+3)}}{\cancel{n-2}}\Rightarrow \dfrac{7}{10}[/latex]
• [latex]\dfrac{7r}{7r(r+10)}\div \dfrac{r-6}{(r-6)^2}\Rightarrow \dfrac{7r}{7r(r+10)}\cdot \dfrac{(r-6)^2}{r-6}\Rightarrow \dfrac{\cancel{7r}}{\cancel{7r}(r+10)}\cdot \dfrac{(r-6)\cancel{(r-6)}}{\cancel{r-6}}\Rightarrow \dfrac{r-6}{r+10}[/latex]
• [latex]\dfrac{6x(x+4)}{(x-3)}\cdot \dfrac{(x-3)(x-6)}{6x(x-6)}\Rightarrow \dfrac{\cancel{6x}(x+4)}{\cancel{(x-3)}}\cdot \dfrac{\cancel{(x-3)}\cancel{(x-6)}}{\cancel{6x}\cancel{(x-6)}}\Rightarrow x+4[/latex]
• [latex]\dfrac{x-10}{35x+21}\div \dfrac{7}{35x+21}\Rightarrow \dfrac{x-10}{7\cancel{(5x+3)}}\cdot \dfrac{\cancel{7}\cancel{(5x+3)}}{\cancel{7}}\Rightarrow \dfrac{x-10}{7}[/latex]
• [latex]\dfrac{v-1}{4}\cdot \dfrac{4}{v^2-11v+10}\Rightarrow \dfrac{\cancel{v-1}}{\cancel{4}}\Rightarrow \dfrac{\cancel{4}}{\cancel{(v-1)}(v-10)}\Rightarrow \dfrac{1}{v-10}[/latex]
• [latex]\dfrac{x^2-6x-7}{x+5}\cdot \dfrac{x+5}{x-7}\Rightarrow \dfrac{\cancel{(x-7)}(x+1)}{\cancel{(x+5)}}\cdot \dfrac{\cancel{(x+5)}}{\cancel{(x-7)}}\Rightarrow x+1[/latex]
• [latex]\dfrac{1}{a-6}\cdot \dfrac{8a+80}{8}\Rightarrow \dfrac{1}{a-6}\cdot \dfrac{\cancel{8}(a+10)}{\cancel{8}}\Rightarrow \dfrac{a+10}{a-6}[/latex]
• [latex]\dfrac{4m+36}{m+9}\cdot \dfrac{m-5}{5m^2}\Rightarrow \dfrac{4\cancel{(m+9)}}{\cancel{m+9}}\cdot \dfrac{m-5}{5m^2}\Rightarrow \dfrac{4(m-5)}{5m^2}[/latex]
• [latex]\dfrac{2r}{r+6}\div \dfrac{2r}{74+42}\Rightarrow \dfrac{\cancel{2r}}{\cancel{r+6}}\cdot \dfrac{7\cancel{(r+6)}}{\cancel{2r}}\Rightarrow 7[/latex]
• [latex]\dfrac{n-7}{6n-12}\cdot \dfrac{12-6n}{n^2-13n+42}\Rightarrow \dfrac{\cancel{(n-7)}}{\cancel{6}(n-2)}\cdot \dfrac{\cancel{6}(2-n)}{(n-6)\cancel{(n-7)}}\Rightarrow \dfrac{-1\cancel{(n-2)}}{\cancel{(n-2)}(n-6)}\Rightarrow \dfrac{-1}{n-6}[/latex]
• [latex]\dfrac{x^2+11x+24}{6x^3+18x^2}\cdot \dfrac{6x^3+6x^2}{x^2+5x-24}\Rightarrow \dfrac{\cancel{(x+3)}\cancel{(x+8)}}{\cancel{6x^2}\cancel{(x+3)}}\cdot \dfrac{\cancel{6x^2}(x+1)}{\cancel{(x+8)}(x-3)}\Rightarrow \dfrac{x+1}{x-3}[/latex]
• [latex]\dfrac{27a+36}{9a+63}\div \dfrac{6a+8}{2}\Rightarrow \dfrac{\cancel{9}\cancel{(3a+4)}}{\cancel{9}(a+7)}\cdot \dfrac{\cancel{2}}{\cancel{2}\cancel{(3a+4)}}\Rightarrow[/latex] [latex]\dfrac{1}{a+7}[/latex]
• [latex]\dfrac{k-7}{k^2-k-12}\cdot \dfrac{7k^2-28k}{8k^2-56k}\Rightarrow \dfrac{\cancel{k-7}}{\cancel{(k-4)}(k+3)}\cdot \dfrac{7\cdot \cancel{k}\cancel{(k-4)}}{8\cdot \cancel{k}\cancel{(k-7)}}\Rightarrow \dfrac{7}{8(k+3)}[/latex]
• [latex]\dfrac{x^2-12x+32}{x^2-6x-16}\cdot \dfrac{7x^2+14x}{7x^2+21x}\Rightarrow \dfrac{\cancel{(x-8)}(x-4)}{\cancel{(x-8)}\cancel{(x+2)}}\cdot \dfrac{\cancel{7x}\cancel{(x+2)}}{\cancel{7x}(x+3)}\Rightarrow \dfrac{x-4}{x+3}[/latex]
• [latex]\dfrac{9x^3+54x^2}{x^2+5x-14}\cdot \dfrac{x^2+5x-14}{10x^2}\Rightarrow \dfrac{9\cancel{x^2}(x+6)}{10\cancel{x^2}}\Rightarrow \dfrac{9(x+6)}{10}[/latex]
• [latex](10m^2+100m)\cdot \dfrac{18m^3-36m^2}{20m^2-40m}\Rightarrow \cancel{10m}(m+10)\cdot \dfrac{\cancel{2}\cdot 9m^2\cancel{(m-2)}}{\cancel{2}\cdot \cancel{10m}\cancel{(m-2)}}\Rightarrow[/latex] [latex]9m^2(m+10)[/latex]
• [latex]\dfrac{n-7}{n^2-2n-35}\div \dfrac{9n+54}{10n+50}\Rightarrow \dfrac{\cancel{n-7}}{\cancel{(n-7)}\cancel{(n+5)}}\cdot \dfrac{10\cancel{(n+5)}}{9(n+6)}\Rightarrow \dfrac{10}{9(n+6)}[/latex]
• [latex]\\ \dfrac{x^2-1}{2x-4}\cdot \dfrac{x^2-4}{x^2-x-2}\div \dfrac{x^2+x-2}{3x-6}\Rightarrow[/latex] [latex]\dfrac{\cancel{(x-1)}\cancel{(x+1)}}{2\cancel{(x-2)}}\cdot \dfrac{\cancel{(x+2)}\cancel{(x-2)}}{\cancel{(x-2)}\cancel{(x+1)}}\cdot \dfrac{3\cancel{(x-2)}}{\cancel{(x+2)}\cancel{(x-1)}}\Rightarrow \dfrac{3}{2}[/latex]
• [latex]\dfrac{a^3+b^3}{a^2+3ab+2b^2}\cdot \dfrac{3a-6b}{3a^2-3ab+3b^2}\div \dfrac{a^2-4b^2}{a+2b}\Rightarrow[/latex] [latex]\dfrac{\cancel{(a+b)}\cancel{(a^2-ab+b^2)}}{(a+2b)\cancel{(a+b)}}\cdot \dfrac{\cancel{3}\cancel{(a-2b)}}{\cancel{3}\cancel{(a^2-ab+b^2)}}\cdot \dfrac{\cancel{a+2b}}{\cancel{(a-2b)}\cancel{(a+2b)}}\Rightarrow \dfrac{1}{a+2b}[/latex]

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Share This Book

Engage your students with effective distance learning resources. ACCESS RESOURCES>>

• Notifications 0

• Add Friend ($5)

As a registered member you can:

• View all solutions for free
• Request more in-depth explanations for free
• Ask our tutors any math-related question for free
• Email your homework to your parent or tutor for free
• Grade 5 HMH Go Math - Answer Keys

Explanation:

Erica is saving her money to buy a dining room set that costs $580. If she saves $29 each month, how many months will she need to save to have enough money to buy the set?

Yes, email page to my online tutor. ( if you didn't add a tutor yet, you can add one here )

Thank you for doing your homework!

Submit Your Question

8.1 Non-right Triangles: Law of Sines

α = 98 ∘ a = 34.6 β = 39 ∘ b = 22 γ = 43 ∘ c = 23.8 α = 98 ∘ a = 34.6 β = 39 ∘ b = 22 γ = 43 ∘ c = 23.8

β ≈ 5.7° , γ ≈ 94.3° , c ≈ 101.3 β ≈ 5.7° , γ ≈ 94.3° , c ≈ 101.3

about 8.2 8.2 square feet

8.2 Non-right Triangles: Law of Cosines

a ≈ 14.9 , a ≈ 14.9 , β ≈ 23.8° , β ≈ 23.8° , γ ≈ 126.2° . γ ≈ 126.2° .

α ≈ 27.7° , α ≈ 27.7° , β ≈ 40.5° , β ≈ 40.5° , γ ≈ 111.8° γ ≈ 111.8°

Area = 552 square feet

about 8.15 square feet

8.3 Polar Coordinates

( x , y ) = ( 1 2 , − 3 2 ) ( x , y ) = ( 1 2 , − 3 2 )

r = 3 r = 3

x 2 + y 2 = 2 y x 2 + y 2 = 2 y or, in the standard form for a circle, x 2 + ( y − 1 ) 2 = 1 x 2 + ( y − 1 ) 2 = 1

8.4 Polar Coordinates: Graphs

The equation fails the symmetry test with respect to the line θ = π 2 θ = π 2 and with respect to the pole. It passes the polar axis symmetry test.

Tests will reveal symmetry about the polar axis. The zero is ( 0 , π 2 ) , ( 0 , π 2 ) , and the maximum value is ( 3 , 0 ) . ( 3 , 0 ) .

The graph is a rose curve, n n even

Rose curve, n n odd

8.5 Polar Form of Complex Numbers

| z | = 50 = 5 2 | z | = 50 = 5 2

z = 3 ( cos ( π 2 ) + i sin ( π 2 ) ) z = 3 ( cos ( π 2 ) + i sin ( π 2 ) )

z = 2 ( cos ( π 6 ) + i sin ( π 6 ) ) z = 2 ( cos ( π 6 ) + i sin ( π 6 ) )

z = 2 3 − 2 i z = 2 3 − 2 i

z 1 z 2 = − 4 3 ; z 1 z 2 = − 3 2 + 3 2 i z 1 z 2 = − 4 3 ; z 1 z 2 = − 3 2 + 3 2 i

z 0 = 2 ( cos ( 30° ) + i sin ( 30° ) ) z 0 = 2 ( cos ( 30° ) + i sin ( 30° ) )

z 1 = 2 ( cos ( 120° ) + i sin ( 120° ) ) z 1 = 2 ( cos ( 120° ) + i sin ( 120° ) )

z 2 = 2 ( cos ( 210° ) + i sin ( 210° ) ) z 2 = 2 ( cos ( 210° ) + i sin ( 210° ) )

z 3 = 2 ( cos ( 300° ) + i sin ( 300° ) ) z 3 = 2 ( cos ( 300° ) + i sin ( 300° ) )

8.6 Parametric Equations

x ( t ) = t 3 − 2 t y ( t ) = t x ( t ) = t 3 − 2 t y ( t ) = t

y = 5 − 1 2 x − 3 y = 5 − 1 2 x − 3

y = ln x y = ln x

x 2 4 + y 2 9 = 1 x 2 4 + y 2 9 = 1

y = x 2 y = x 2

8.7 Parametric Equations: Graphs

The graph of the parametric equations is in red and the graph of the rectangular equation is drawn in blue dots on top of the parametric equations.

8.8 Vectors

3 u = 〈 15 , 12 〉 3 u = 〈 15 , 12 〉

u = 8 i − 11 j u = 8 i − 11 j

v = 34 cos ( 59° ) i + 34 sin ( 59° ) j v = 34 cos ( 59° ) i + 34 sin ( 59° ) j

Magnitude = 34 34

θ = tan − 1 ( 5 3 ) = 59.04° θ = tan − 1 ( 5 3 ) = 59.04°

8.1 Section Exercises

The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle.

When the known values are the side opposite the missing angle and another side and its opposite angle.

A triangle with two given sides and a non-included angle.

β = 72° , a ≈ 12.0 , b ≈ 19.9 β = 72° , a ≈ 12.0 , b ≈ 19.9

γ = 20° , b ≈ 4.5 , c ≈ 1.6 γ = 20° , b ≈ 4.5 , c ≈ 1.6

b ≈ 3.78 b ≈ 3.78

c ≈ 13.70 c ≈ 13.70

one triangle, α ≈ 50.3° , β ≈ 16.7° , a ≈ 26.7 α ≈ 50.3° , β ≈ 16.7° , a ≈ 26.7

two triangles, γ ≈ 54.3° , β ≈ 90.7° , b ≈ 20.9 γ ≈ 54.3° , β ≈ 90.7° , b ≈ 20.9 or γ ′ ≈ 125.7° , β ′ ≈ 19.3° , b ′ ≈ 6.9 γ ′ ≈ 125.7° , β ′ ≈ 19.3° , b ′ ≈ 6.9

two triangles, β ≈ 75.7° , γ ≈ 61.3° , b ≈ 9.9 β ≈ 75.7° , γ ≈ 61.3° , b ≈ 9.9 or β ′ ≈ 18.3° , γ ′ ≈ 118.7° , b ′ ≈ 3.2 β ′ ≈ 18.3° , γ ′ ≈ 118.7° , b ′ ≈ 3.2

two triangles, α ≈ 143.2° , β ≈ 26.8° , a ≈ 17.3 α ≈ 143.2° , β ≈ 26.8° , a ≈ 17.3 or α ′ ≈ 16.8° , β ′ ≈ 153.2° , a ′ ≈ 8.3 α ′ ≈ 16.8° , β ′ ≈ 153.2° , a ′ ≈ 8.3

no triangle possible

A ≈ 47.8° A ≈ 47.8° or A ′ ≈ 132.2° A ′ ≈ 132.2°

370.9 370.9

29.7° 29.7°

x = 76.9° or x = 103.1° x = 76.9° or x = 103.1°

110.6° 110.6°

A ≈ 39.4 , C ≈ 47.6 , B C ≈ 20.7 A ≈ 39.4 , C ≈ 47.6 , B C ≈ 20.7

430.2 430.2

A D ≈ 13.8 A D ≈ 13.8

A B ≈ 2.8 A B ≈ 2.8

L ≈ 49.7 , N ≈ 56.3 , L N ≈ 5.8 L ≈ 49.7 , N ≈ 56.3 , L N ≈ 5.8

The distance from the satellite to station A A is approximately 1716 miles. The satellite is approximately 1706 miles above the ground.

2.6 ft

5.6 km

371 ft

5936 ft

24.1 ft

19,056 ft 2

445,624 square miles

8.65 ft 2

8.2 Section Exercises

two sides and the angle opposite the missing side.

s s is the semi-perimeter, which is half the perimeter of the triangle.

The Law of Cosines must be used for any oblique (non-right) triangle.

not possible

B ≈ 45.9° , C ≈ 99.1° , a ≈ 6.4 B ≈ 45.9° , C ≈ 99.1° , a ≈ 6.4

A ≈ 20.6° , B ≈ 38.4° , c ≈ 51.1 A ≈ 20.6° , B ≈ 38.4° , c ≈ 51.1

A ≈ 37.8° , B ≈ 43.8 , C ≈ 98.4° A ≈ 37.8° , B ≈ 43.8 , C ≈ 98.4°

177.56 in 2

292.4 miles

8.3 Section Exercises

For polar coordinates, the point in the plane depends on the angle from the positive x- axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations.

Determine θ θ for the point, then move r r units from the pole to plot the point. If r r is negative, move r r units from the pole in the opposite direction but along the same angle. The point is a distance of r r away from the origin at an angle of θ θ from the polar axis.

The point ( − 3 , π 2 ) ( − 3 , π 2 ) has a positive angle but a negative radius and is plotted by moving to an angle of π 2 π 2 and then moving 3 units in the negative direction. This places the point 3 units down the negative y -axis. The point ( 3 , − π 2 ) ( 3 , − π 2 ) has a negative angle and a positive radius and is plotted by first moving to an angle of − π 2 − π 2 and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y -axis.

( − 5 , 0 ) ( − 5 , 0 )

( − 3 3 2 , − 3 2 ) ( − 3 3 2 , − 3 2 )

( 2 5 , 0.464 ) ( 2 5 , 0.464 )

( 34 , 5.253 ) ( 34 , 5.253 )

( 8 2 , π 4 ) ( 8 2 , π 4 )

r = 4 csc θ r = 4 csc θ

r = s i n θ 2 c o s 4 θ 3 r = s i n θ 2 c o s 4 θ 3

r = 3 cos θ r = 3 cos θ

r = 3 sin θ cos ( 2 θ ) r = 3 sin θ cos ( 2 θ )

r = 9 sin θ cos 2 θ r = 9 sin θ cos 2 θ

r = 1 9 cos θ sin θ r = 1 9 cos θ sin θ

x 2 + y 2 = 4 x x 2 + y 2 = 4 x or ( x − 2 ) 2 4 + y 2 4 = 1 ; ( x − 2 ) 2 4 + y 2 4 = 1 ; circle

3 y + x = 6 ; 3 y + x = 6 ; line

y = 3 ; y = 3 ; line

x y = 4 ; x y = 4 ; hyperbola

x 2 + y 2 = 4 ; x 2 + y 2 = 4 ; circle

x − 5 y = 3 ; x − 5 y = 3 ; line

( 3 , 3 π 4 ) ( 3 , 3 π 4 )

( 5 , π ) ( 5 , π )

r = 6 5 cos θ − sin θ r = 6 5 cos θ − sin θ

r = 2 sin θ r = 2 sin θ

r = 2 cos θ r = 2 cos θ

x 2 + y 2 = 16 x 2 + y 2 = 16

y = x y = x

x 2 + ( y + 5 ) 2 = 25 x 2 + ( y + 5 ) 2 = 25

( 1.618 , − 1.176 ) ( 1.618 , − 1.176 )

( 10.630 , 131.186° ) ( 10.630 , 131.186° )

( 2 , 3.14 ) o r ( 2 , π ) ( 2 , 3.14 ) o r ( 2 , π )

A vertical line with a a units left of the y -axis. 

A horizontal line with a a units below the x -axis.

8.4 Section Exercises

Symmetry with respect to the polar axis is similar to symmetry about the x x -axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the line θ = π 2 θ = π 2 is similar to symmetry about the y y -axis.

Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., then plot points at θ = 0 , π 2 , θ = 0 , π 2 , π π and  3 π 2 , 3 π 2 , and sketch the graph.

The shape of the polar graph is determined by whether or not it includes a sine, a cosine, and constants in the equation.

symmetric with respect to the polar axis

symmetric with respect to the polar axis, symmetric with respect to the line θ = π 2 , θ = π 2 , symmetric with respect to the pole

no symmetry

symmetric with respect to the pole

one-loop/dimpled limaçon

inner loop/two-loop limaçon

Archimedes’ spiral

They are both spirals, but not quite the same.

Both graphs are curves with 2 loops. The equation with a coefficient of θ θ has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to 4 π 4 π to get a better picture.

When the width of the domain is increased, more petals of the flower are visible.

The graphs are three-petal, rose curves. The larger the coefficient, the greater the curve’s distance from the pole.

The graphs are spirals. The smaller the coefficient, the tighter the spiral.

( 4 , π 3 ) , ( 4 , 5 π 3 ) ( 4 , π 3 ) , ( 4 , 5 π 3 )

( 3 2 , π 3 ) , ( 3 2 , 5 π 3 ) ( 3 2 , π 3 ) , ( 3 2 , 5 π 3 )

( 0 , π 2 ) , ( 0 , π ) , ( 0 , 3 π 2 ) , ( 0 , 2 π ) ( 0 , π 2 ) , ( 0 , π ) , ( 0 , 3 π 2 ) , ( 0 , 2 π )

( 8 4 2 , π 4 ) , ( 8 4 2 , 5 π 4 ) ( 8 4 2 , π 4 ) , ( 8 4 2 , 5 π 4 ) and at θ = 3 π 4 , θ = 3 π 4 , 7 π 4 7 π 4 since r r is squared

8.5 Section Exercises

a is the real part, b is the imaginary part, and i = − 1 i = − 1

Polar form converts the real and imaginary part of the complex number in polar form using x = r cos θ x = r cos θ and y = r sin θ . y = r sin θ .

z n = r n ( cos ( n θ ) + i sin ( n θ ) ) z n = r n ( cos ( n θ ) + i sin ( n θ ) ) It is used to simplify polar form when a number has been raised to a power.

14.45 14.45

4 5 cis ( 333.4° ) 4 5 cis ( 333.4° )

2 cis ( π 6 ) 2 cis ( π 6 )

7 3 2 + i 7 2 7 3 2 + i 7 2

− 2 3 − 2 i − 2 3 − 2 i

− 1.5 − i 3 3 2 − 1.5 − i 3 3 2

4 3 cis ( 198° ) 4 3 cis ( 198° )

3 4 cis ( 180° ) 3 4 cis ( 180° )

5 3 cis ( 17 π 24 ) 5 3 cis ( 17 π 24 )

7 cis ( 70° ) 7 cis ( 70° )

5 cis ( 80° ) 5 cis ( 80° )

5 cis ( π 3 ) 5 cis ( π 3 )

125 cis ( 135° ) 125 cis ( 135° )

9 cis ( 240° ) 9 cis ( 240° )

cis ( 3 π 4 ) cis ( 3 π 4 )

3 cis ( 80° ) , 3 cis ( 200° ) , 3 cis ( 320° ) 3 cis ( 80° ) , 3 cis ( 200° ) , 3 cis ( 320° )

2 4 3 cis ( 2 π 9 ) , 2 4 3 cis ( 8 π 9 ) , 2 4 3 cis ( 14 π 9 ) 2 4 3 cis ( 2 π 9 ) , 2 4 3 cis ( 8 π 9 ) , 2 4 3 cis ( 14 π 9 )

2 2 cis ( 7 π 8 ) , 2 2 cis ( 15 π 8 ) 2 2 cis ( 7 π 8 ) , 2 2 cis ( 15 π 8 )

3.61 e − 0.59 i 3.61 e − 0.59 i

− 2 + 3.46 i − 2 + 3.46 i

− 4.33 − 2.50 i − 4.33 − 2.50 i

8.6 Section Exercises

A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, x = f ( t ) x = f ( t ) and y = f ( t ) . y = f ( t ) .

Choose one equation to solve for t , t , substitute into the other equation and simplify.

Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions.

y = − 2 + 2 x y = − 2 + 2 x

y = 3 x − 1 2 y = 3 x − 1 2

x = 2 e 1 − y 5 x = 2 e 1 − y 5 or y = 1 − 5 l n ( x 2 ) y = 1 − 5 l n ( x 2 )

x = 4 log ( y − 3 2 ) x = 4 log ( y − 3 2 )

x = ( y 2 ) 3 − y 2 x = ( y 2 ) 3 − y 2

y = x 3 y = x 3

( x 4 ) 2 + ( y 5 ) 2 = 1 ( x 4 ) 2 + ( y 5 ) 2 = 1

y 2 = 1 − 1 2 x y 2 = 1 − 1 2 x

y = x 2 + 2 x + 1 y = x 2 + 2 x + 1

y = ( x + 1 2 ) 3 − 2 y = ( x + 1 2 ) 3 − 2

y = − 3 x + 14 y = − 3 x + 14

y = x + 3 y = x + 3

{ x ( t ) = t y ( t ) = 2 sin t + 1 { x ( t ) = t y ( t ) = 2 sin t + 1

{ x ( t ) = t + 2 t y ( t ) = t { x ( t ) = t + 2 t y ( t ) = t

{ x ( t ) = 4 cos t y ( t ) = 6 sin t ; { x ( t ) = 4 cos t y ( t ) = 6 sin t ; Ellipse

{ x ( t ) = 10 cos t y ( t ) = 10 sin t ; { x ( t ) = 10 cos t y ( t ) = 10 sin t ; Circle

{ x ( t ) = − 1 + 4 t y ( t ) = − 2 t { x ( t ) = − 1 + 4 t y ( t ) = − 2 t

{ x ( t ) = 4 + 2 t y ( t ) = 1 − 3 t { x ( t ) = 4 + 2 t y ( t ) = 1 − 3 t

yes, at t = 2 t = 2

answers may vary: { x ( t ) = t − 1 y ( t ) = t 2  and  { x ( t ) = t + 1 y ( t ) = ( t + 2 ) 2 { x ( t ) = t − 1 y ( t ) = t 2  and  { x ( t ) = t + 1 y ( t ) = ( t + 2 ) 2

answers may vary: , { x ( t ) = t y ( t ) = t 2 − 4 t + 4  and  { x ( t ) = t + 2 y ( t ) = t 2 { x ( t ) = t y ( t ) = t 2 − 4 t + 4  and  { x ( t ) = t + 2 y ( t ) = t 2

8.7 Section Exercises

plotting points with the orientation arrow and a graphing calculator

The arrows show the orientation, the direction of motion according to increasing values of t . t .

The parametric equations show the different vertical and horizontal motions over time.

There will be 100 back-and-forth motions.

Take the opposite of the x ( t ) x ( t ) equation.

The parabola opens up.

{ x ( t ) = 5 cos t y ( t ) = 5 sin t { x ( t ) = 5 cos t y ( t ) = 5 sin t

a = 4 , a = 4 , b = 3 , b = 3 , c = 6 , c = 6 , d = 1 d = 1

a = 4 , a = 4 , b = 2 , b = 2 , c = 3 , c = 3 , d = 3 d = 3

The y y -intercept changes.

y ( x ) = − 16 ( x 15 ) 2 + 20 ( x 15 ) y ( x ) = − 16 ( x 15 ) 2 + 20 ( x 15 )

{ x ( t ) = 64 t cos ( 52 ° ) y ( t ) = − 16 t 2 + 64 t sin ( 52 ° ) { x ( t ) = 64 t cos ( 52 ° ) y ( t ) = − 16 t 2 + 64 t sin ( 52 ° )

approximately 3.2 seconds

1.6 seconds

8.8 Section Exercises

lowercase, bold letter, usually u , v , w u , v , w

They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1.

The first number always represents the coefficient of the i , i , and the second represents the j . j .

〈 7 , − 5 〉 〈 7 , − 5 〉

7 i − 3 j 7 i − 3 j

− 6 i − 2 j − 6 i − 2 j

u + v = 〈 − 5 , 5 〉 , u − v = 〈 − 1 , 3 〉 , 2 u − 3 v = 〈 0 , 5 〉 u + v = 〈 − 5 , 5 〉 , u − v = 〈 − 1 , 3 〉 , 2 u − 3 v = 〈 0 , 5 〉

− 10 i – 4 j − 10 i – 4 j

− 2 29 29 i + 5 29 29 j − 2 29 29 i + 5 29 29 j

− 2 229 229 i + 15 229 229 j − 2 229 229 i + 15 229 229 j

− 7 2 10 i + 2 10 j − 7 2 10 i + 2 10 j

| v | = 7.810 , θ = 39.806 ° | v | = 7.810 , θ = 39.806 °

| v | = 7.211 , θ = 236.310° | v | = 7.211 , θ = 236.310°

〈 4 , 1 〉 〈 4 , 1 〉

v = − 7 i + 3 j v = − 7 i + 3 j

3 2 i + 3 2 j 3 2 i + 3 2 j

i − 3 j i − 3 j

x = 7.13 x = 7.13 pounds, y = 3.63 y = 3.63 pounds

x = 2.87 x = 2.87 pounds, y = 4.10 y = 4.10 pounds

4.635 miles, 17.764° N of E

17 miles. 10.318 miles

Distance: 2.868. Direction: 86.474° North of West, or 3.526° West of North

4.924°. 659 km/hr

( 0.081 , 8.602 ) ( 0.081 , 8.602 )

21.801°, relative to the car’s forward direction

parallel: 16.28, perpendicular: 47.28 pounds

19.35 pounds, 231.54° from the horizontal

5.1583 pounds, 75.8° from the horizontal

Review Exercises

Not possible

C = 120° , a = 23.1 , c = 34.1 C = 120° , a = 23.1 , c = 34.1

distance of the plane from point A : A : 2.2 km, elevation of the plane: 1.6 km

b = 71.0° , C = 55.0° , a = 12.8 b = 71.0° , C = 55.0° , a = 12.8

( 0 , 2 ) ( 0 , 2 )

( 9.8489 , 203.96° ) ( 9.8489 , 203.96° )

r = 8 r = 8

x 2 + y 2 = 7 x x 2 + y 2 = 7 x

y = − x y = − x

symmetric with respect to the line θ = π 2 θ = π 2

cis ( − π 3 ) cis ( − π 3 )

2.3 + 1.9 i 2.3 + 1.9 i

60 cis ( π 2 ) 60 cis ( π 2 )

3 cis ( 4 π 3 ) 3 cis ( 4 π 3 )

25 cis ( 3 π 2 ) 25 cis ( 3 π 2 )

5 cis ( 3 π 4 ) , 5 cis ( 7 π 4 ) 5 cis ( 3 π 4 ) , 5 cis ( 7 π 4 )

x 2 + 1 2 y = 1 x 2 + 1 2 y = 1

{ x ( t ) = − 2 + 6 t y ( t ) = 3 + 4 t { x ( t ) = − 2 + 6 t y ( t ) = 3 + 4 t

y = − 2 x 5 y = − 2 x 5

• { x ( t ) = ( 80 cos ( 40° ) ) t y ( t ) = − 16 t 2 + ( 80 sin ( 40° ) ) t + 4 { x ( t ) = ( 80 cos ( 40° ) ) t y ( t ) = − 16 t 2 + ( 80 sin ( 40° ) ) t + 4
• The ball is 14 feet high and 184 feet from where it was launched.
• 3.3 seconds

− 3 10 10 − 3 10 10 i − 10 10 − 10 10 j

Magnitude: 3 2 , 3 2 , Direction: 225° 225°

Practice Test

α = 67.1° , γ = 44.9° , a = 20.9 α = 67.1° , γ = 44.9° , a = 20.9

1712 miles 1712 miles

( 1 , 3 ) ( 1 , 3 )

y = − 3 y = − 3

− 5 2 + i 5 3 2 − 5 2 + i 5 3 2

4 cis ( 21° ) 4 cis ( 21° )

2 2 cis ( 18° ) , 2 2 cis ( 198° ) 2 2 cis ( 18° ) , 2 2 cis ( 198° )

y = 2 ( x − 1 ) 2 y = 2 ( x − 1 ) 2

−4 i − 15 j

2 13 13 i + 3 13 13 j 2 13 13 i + 3 13 13 j

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/precalculus/pages/1-introduction-to-functions

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Precalculus
• Publication date: Oct 23, 2014
• Location: Houston, Texas
• Book URL: https://openstax.org/books/precalculus/pages/1-introduction-to-functions
• Section URL: https://openstax.org/books/precalculus/pages/chapter-8

© Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

• Texas Go Math
• Big Ideas Math
• enVision Math
• EngageNY Math
• McGraw Hill My Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Texas Go Math Grade 5 Lesson 2.8 Answer Key Division

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 2.8 Answer Key Division.

Unlock the Problem

Sean and his crew operate a fishing charter company. They caught a blue marlin and amberjack. The weight of the blue marlin was 12 times as great as the weight of the amberjack. The combined weight of both fish was 1,014 pounds. How much did each fish weigh?

What do I need to find? I need to find the weight of each fish.

What information am I given?

I know that they caught a total of 1,014 pounds of fish and the weight of the blue marlin was 12 times as great as the weight of the amberjack.

I can use the strategy of Addition and Substitution and then divide. I can draw and use a strip diagram to write the division problem that helps me find the weight of each fish.

Try Another Problem

Jason, Murray, and Dana went fishing. Dana caught red snapper. Jason caught a tuna with a weight 3 times as great as the weight of the red snapper. Murray caught a sailfish with a weight 12 times as great as the weight of the red snapper. If the combined weight of the three fish was 208 pounds, how much did the tuna weigh?

What do I need to find? Answer: I need to find the weight of tuna

What information am I given? Answer: I know that Jason, Murray, and Dana went fishing. Dana caught red snapper. Jason caught a tuna with a weight 3 times as great as the weight of the red snapper. Murray caught a sailfish with a weight 12 times as great as the weight of the red snapper and the combined weight of the three fish was 208 pounds

What is my plan or strategy? Answer: I can use the strategy of Addition and Substitution and then divide

Solve It is given that Jason, Murray, and Dana went fishing. Dana caught red snapper. Jason caught a tuna with a weight 3 times as great as the weight of the red snapper. Murray caught a sailfish with a weight 12 times as great as the weight of the red snapper and the combined weight of the three fish was 208 pounds Now, Let the weight of the red snapper be: x Pounds So, The weight of the fish caught by Dana = x Pounds The weight of the tuna caught by Jason = 3x Pounds The weight of the saltfish caught by Murray = 12x Pounds Now, According to the combined information, x + 3x + 12x = 208 Pounds 16x = 208 x = \(\frac{208}{16}\) = 13 Pounds So, The weight of the tuna caught by Jason = 3x = 3 × 13 = 39 Pounds So, The tuna weighed 39 pounds.

How can you check if your answer is correct? Answer: Substitute the weight of the tuna in the total weight of the three fishes and check whether the answer is equal to the combined weight of the fishes caught by Dana and Murray

Math Talk Mathematical Processes

Explain how you could use another strategy to solve this problem. Answer: The other strategy to solve the above problem is: Step 1: Read the given information Step 2: Plan the strategy how to find the required quantity from the given information Step 3: Solve the problem and find the required quantity

Share and Show

Problem Solving

Daily Assessment Task

Fill in the bubble completely to show your answer.

Texas Test Prep

Texas Go Math Grade 5 Lesson 2.8 Homework and Practice Answer Key

Eddie’s Moving Company moved 3 boxes out of Mrs. Diaz’s house. The box labeled “Kitchen” was 4 times heavier than the box labeled “Bedroom.” The box labeled “Library” was 14 times heavier than the box marked “Bedroom.” If the combined weight of the boxes was 817 pounds, how much did the box labeled “Kitchen” weigh?

Question 1. What do I need to find? Answer: I need to find the weight of the box labeled “Kitchen”

Question 2. What information am I given? Answer: I know that Eddie’s Moving Company moved 3 boxes out of Mrs. Diaz’s house. The box labeled “Kitchen” was 4 times heavier than the box labeled “Bedroom.” The box labeled “Library” was 14 times heavier than the box marked “Bedroom” and the combined weight of the boxes was 817 pounds

Question 4. Solve. Let the weight of the box labeled “Bedroom” be: x Pounds So, x + 4x + 14x = 817 Pounds 19x = 817 So, 817 ÷ 19 = 43 So, I can multiply 4 by 43 to find the weight of the “Kitchen” box. So, The box labeled “Kitchen” weighed 172 pounds.

Question 5. An athlete ran 1,200 laps around a track over a two-month period. He ran 4 times the number of laps in the second month than he ran in the first month. How many laps did the athlete run the first month? Answer: It is given that An athlete ran 1,200 laps around a track over a two-month period. He ran 4 times the number of laps in the second month than he ran in the first month Now, Let the number of laps run by an athlete in the first month be: x laps So, The number of laps run by an athlete in the first month be: 4x laps Now, According to the given information, x + 4x = 1,200 5x = 1,200 x = \(\frac{1,200}{5}\) x = 240 laps Hence, from the above, We can conclude that The number of laps did the athlete run in the first month is: 240 laps

Lesson Check

Share this:

Leave a comment cancel reply.

You must be logged in to post a comment.

• Texas Go Math
• Big Ideas Math
• Engageny Math
• McGraw Hill My Math
• enVision Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Eureka Math Grade 8 Module 2 Lesson 13 Answer Key

Engage ny eureka math 8th grade module 2 lesson 13 answer key, eureka math grade 8 module 2 lesson 13 exploratory challenge answer key.

We want to prove that the angle sum of any triangle is 180°. To do so, we use some facts that we already know about geometry: → A straight angle is 180° in measure. → Corresponding angles of parallel lines are equal in measure (corr. ∠’s, \(\overline{A B}\) || \(\overline{C D}\). → Alternate interior angles of parallel lines are equal in measure (alt. ∠’s, \(\overline{A B}\) || \(\overline{C D}\).

a. Name the three interior angles of triangle ABC. Answer: ∠ABC, ∠BAC, ∠BCA

b. Name the straight angle. Answer: ∠BCD Our goal is to show that the measures of the three interior angles of triangle ABC are equal to the measures of the angles that make up the straight angle. We already know that a straight angle is 180° in measure. If we can show that the interior angles of the triangle are the same as the angles of the straight angle, then we will have proven that the sum of the measures of the interior angles of the triangle have a sum of 180°.

c. What kinds of angles are ∠ABC and ∠ECD? What does that mean about their measures? Answer: ∠ABC and ∠ECD are corresponding angles. Corresponding angles of parallel lines are equal in measure (corr. ∠’s, \(\overline{A B}\) || \(\overline{\boldsymbol{C E}}\)).

d. What kinds of angles are ∠BAC and ∠ECA? What does that mean about their measures? Answer: ∠BAC and ∠ECA are alternate interior angles. Alternate interior angles of parallel lines are equal in measure (alt. ∠’s, \(\overline{\boldsymbol{A B}}\) || \(\overline{\boldsymbol{C E}}\)).

d. We know that m∠BCD=m∠BCA+m∠ECA+m∠ECD=180°. Use substitution to show that the measures of the three interior angles of the triangle have a sum of 180°. Answer: m∠BCD=m∠BCA+m∠BAC+m∠ABC=180° (∠ sum of △)

a. Name the triangle in the figure. Answer: △BCF

b. Name a straight angle that will be useful in proving that the sum of the measures of the interior angles of the triangle is 180°. Answer: ∠GFE As before, our goal is to show that the sum of the measures of the interior angles of the triangle are equal to the measure of the straight angle. Use what you learned from Exploratory Challenge 1 to show that the measures of the interior angles of a triangle have a sum of 180°.

c. Write your proof below. Answer: The straight angle ∠GFE is comprised of ∠GFB, ∠BFC, and ∠EFC. Alternate interior angles of parallel lines are equal in measure (alt. ∠’s, \(\overline{A D}\) || \(\overline{C E}\)). For that reason, ∠BCF=∠EFC and ∠CBF=∠GFB. Since ∠GFE is a straight angle, it is equal to 180°. Then, ∠GFE=∠GFB+∠BFC+∠EFC=180°. By substitution, ∠GFE=∠CBF+∠BFC+∠BCF=180°. Therefore, the sum of the measures of the interior angles of a triangle is 180° (∠ sum of △).

Eureka Math Grade 8 Module 2 Lesson 13 Exit Ticket Answer Key

Eureka Math Grade 8 Module 2 Lesson 13 Problem Set Answer Key

Students practice presenting informal arguments about the sum of the angles of a triangle using the theorem to find the measures of missing angles.

Leave a Comment Cancel Reply

You must be logged in to post a comment.