Algebra and Trigonometry: Structure and Method, Book 2

2000th Edition

ISBN: 9780395977255

Author: MCDOUGAL LITTEL

Publisher: McDougal Littell

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Question

Chapter 7.2, Problem 15P

To determine

Tofind:The two speeds of the hydrofoil.

Expert Solution & Answer

Answer to Problem 15P

Thehydrofoil has the speed of 45 km/h while going forward and the speed of 30 km/h while returning.

Explanation of Solution

Given:

The roundtrip of 144 km is made in 4 h.

The average speed of returning is 15 km/h less than the average speed of going out.

Calculation:

Consider the average speed of the hydrofoil is x . The total trip is of 144 km means each way is 72 km. The total time taken is 4 h.

4=72x+72x−154=72(x−15)+72xx(x−15)4=72x−1080+72xx2−15x4x2−204x+1080=0x2−51x+270=0

The solution of the given equation is obtained by using the quadratic formula.

For a quadratic equation in the form of ax2+bx+c=0 , the solution of this equation is given by

x1,2=−b±b2−4ac2a

Comparing the given equation with the standard quadratic equation, the coefficients are:

a=1b=−51c=270

Substituting these in the formula for the roots.

x1=−(−51)±(−51)2−4×1×(270)2×1=51±2601−10802=51±392x=6x=45

The value of x as 6 is not valid because it means that the hydrofoil is pushed back by the winds. Therefore the speed of hydrofoil while going is 45 km/h and while returning 45-15 = 30 km/h.

Chapter 7.2, Problem 14P

Chapter 7.2, Problem 1ST

Chapter 7 Solutions

Algebra and Trigonometry: Structure and Method, Book 2

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Tofind: The two speeds of the hydrofoil.

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Tofind:The two speeds of the hydrofoil.

Answer to Problem 15P

Explanation of Solution

Chapter 7 Solutions

Additional Math Textbook Solutions