Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

1st Edition

ISBN: 9780078682278

Author: McGraw-Hill, Berchie Holliday

Publisher: Glencoe/McGraw-Hill

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Question

Chapter 10.8, Problem 40E

To determine

Write the second degree system of equations.

Expert Solution

Answer to Problem 40E

x2+y2=2500x2+(y−30)2=1600(x−35)2+(y−18)2=169

Explanation of Solution

Given information:

Seismograph reading indicate that the epicenter of the earthquake is 50 km from the first station, 40 km from the second station 13 km from the third station. First station is at origin, the second at (0,30) and third station at (35,18)

Calculation:

Consider (x,y) be the epicenter of the map, First station is at origin (0,0) , 50 km from the epicenter,

The equation will be,

(x−0)2+(y−0)2=50x2+y2=502x2+y2=2500

This is equation of circle at (0,0) origin and radius 50 .

Now, second station is at (0,30) , 40 km from the epicenter

The equation will be,

(x−0)2+(y−30)2=40x2+(y−30)2=402x2+(y−30)2=1600

This is equation of circle at (0,30) and radius 40 .

Now, third station is at (35,18) , 13 km from the epicenter

The equation will be,

(x−35)2+(y−18)2=13(x−35)2+(y−18)2=169

This is equation of circle at (35,18) and radius 13 .

Hence, the systems of second degree equations are,

x2+y2=2500x2+(y−30)2=1600(x−35)2+(y−18)2=169

To determine

Graph the system of equations.

Expert Solution

Answer to Problem 40E

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition, Chapter 10.8, Problem 40E , additional homework tip 1

Explanation of Solution

Given information:

Seismograph reading indicates that the epicenter of the earthquake is 50 km from the first station, 40 km from the second station 13 km from the third station. First station is at origin, the second at (0,30) and third station at (35,18)

Calculation:

Graph the system of equations,

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition, Chapter 10.8, Problem 40E , additional homework tip 2

To determine

Find the location of epicenter by solving the equations.

Expert Solution

Answer to Problem 40E

(40,30)

Explanation of Solution

Given information:

Calculation:

Consider equation,

x2+y2=2500x2+(y−30)2=1600(x−35)2+(y−18)2=169

Now solve the equations,

x2+y2=2500x2=2500−y2put x2=2500−y2 in x2+(y−30)2=1600(2500−y2)+(y−30)2=16002500−y2+y2−60y+900=16002500−60y+900=160060y=1800y=30so x2=2500−y2 will bex2=2500−302x2=2500−900x2=1600x=±40

The solution from two equations is (40,30) and (−40,30)

put (40,30) in (x−35)2+(y−18)2=169(40−35)2+(30−18)2=16925+144=169169=169LHS=RHSsatisfy the equationput (−40,30) in (x−35)2+(y−18)2=169(−40−35)2+(30−18)2=169−752+122=1695625+144=169LHS=RHSDoes not satisfy,The epicenter is (40,30)

Chapter 10.8, Problem 39E

Chapter 10.8, Problem 41E

Chapter 10 Solutions

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

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