(a) Answer Equation of circle: x 2 + y 2 = 1 Equation of line: y = − 1 7 x + 5 7 Explanation Given: To be eligible for a parking pass on a college campus. A student must live at least 1 mile from the campus center. First find the equation of circle. According to diagram, center of circle is ( 0 , 0 ) and radius is 1 mi. Equation of circle: x 2 + y 2 = 1 According to diagram, the passing points of Oak line are ( − 2 , 1 ) and ( 5 , 0 ) Equation of line: y − 0 = 1 − 0 − 2 − 5 ( x − 5 ) y = − 1 7 x + 5 7 (b) To determine To solve: The system of equations x 2 + y 2 = 1 and x + 7 y = 5 Answer ( − 0.6 , 0.8 ) and ( 0.8 , 0.6 ) Explanation Given: To be eligible for a parking pass on a college campus. A student must live at least 1 mile from the campus center. Equation of circle: x 2 + y 2 = 1 Equation of line: x + 7 y = 5 Substitute the value of x in circle equation. ( 5 − 7 y ) 2 + y 2 = 1 25 + 49 y 2 − 70 y + y 2 = 1 50 y 2 − 70 y + 24 = 0 y = 0.6 , 0.8 For y = 0.6 , x = 0.8 For y = 0.8 , x = − 0.6 Intersection points are ( − 0.6 , 0.8 ) and ( 0.8 , 0.6 ) . (c) To determine To find: The distance where parking not eligible. Answer 4 2 miles Explanation Given: To be eligible for a parking pass on a college campus. A student must live at least 1 mile from the campus center. Intersection points are ( − 0.6 , 0.8 ) and ( 0.8 , 0.6 ) Distance between: ( − 2 , 1 ) and ( − 0.6 , 0.8 ) , ( 0.8 , 0.6 ) and ( 5 , 0 ) Length of Oak Lane where not eligible for a parking pass, = ( − 2 + 0.6 ) 2 + ( 1 − 0.8 ) 2 + ( 5 − 0.8 ) 2 + ( 0.6 − 0 ) 2 = 1.96 + 0.04 + 17.64 + 0.36 = 2 + 18 = 4 2 miles

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

15th Edition

ISBN: 9781608408405

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Cengage Learning

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Question

Chapter 3.5, Problem 59E

(a)

To determine

To Find:The equation that represents the circle and Oak Lane.

(a)

Expert Solution

Answer to Problem 59E

Equation of circle: x2+y2=1

Equation of line: y=−17x+57

Explanation of Solution

Given:To be eligible for a parking pass on a college campus. A student must live at least 1 mile from the campus center.

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 3.5, Problem 59E , additional homework tip 1

First find the equation of circle.

According to diagram, center of circle is (0,0) and radius is 1 mi.

Equation of circle: x2+y2=1

According to diagram, the passing points of Oak line are (−2,1) and (5,0)

Equation of line:

y−0=1−0−2−5(x−5)y=−17x+57

(b)

To determine

To solve: The system of equations x2+y2=1 and x+7y=5

(b)

Expert Solution

Answer to Problem 59E

(−0.6,0.8) and (0.8,0.6)

Explanation of Solution

Given: To be eligible for a parking pass on a college campus. A student must live at least 1 mile from the campus center.

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 3.5, Problem 59E , additional homework tip 2

Equation of circle: x2+y2=1

Equation of line: x+7y=5

Substitute the value of x in circle equation.

(5−7y)2+y2=125+49y2−70y+y2=150y2−70y+24=0y=0.6,0.8

For y=0.6,x=0.8

For y=0.8,x=−0.6

Intersection points are (−0.6,0.8) and (0.8,0.6).

(c)

To determine

To find: The distance where parking not eligible.

(c)

Expert Solution

Answer to Problem 59E

42 miles

Explanation of Solution

Given: To be eligible for a parking pass on a college campus. A student must live at least 1 mile from the campus center.

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 3.5, Problem 59E , additional homework tip 3

Intersection points are (−0.6,0.8) and (0.8,0.6)

Distance between: (−2,1) and (−0.6,0.8) ,(0.8,0.6) and (5,0)

Length of Oak Lane where not eligible for a parking pass,

=(−2+0.6)2+(1−0.8)2+(5−0.8)2+(0.6−0)2=1.96+0.04+17.64+0.36=2+18=42 miles

Chapter 3.5, Problem 58E

Chapter 3.5, Problem 60E

Chapter 3 Solutions

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

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