EBK ALGEBRA 2
EBK ALGEBRA 2
11th Edition
ISBN: 9780544714304
Author: Larson
Publisher: HMH PUB
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Chapter 3.6, Problem 7MRPS
Solution

a.

To state: The inverse function if an object is launched upward from the ground level and reaches a maximum height of h feet The initial velocity v (in feet per second) of the object is given by the function v=8h .

The inverse function is h=v264 .

Given information:

The given function is v=8h .

Calculation:

Write the given function:

  v=8h .

Divide both sides of the equation by 8,

  v8=8h8v8=h

Square both sides of the equation,

  v82=h2v264=hh=v264

Therefore, the inverse function is h=v264 .

b.

To state: A problem that can be solved using an inverse function also show how to solve the problem.

The problem is “The time period of a pendulum, in seconds, can be calculated using the function T=1.11l , where l is the length of the pendulum in inches. Find the inverse function which gives the length in terms of the time period”.

Given information:

The given statement says to write a problem that can be solved using the inverse function also show how to solve the problem.

Explanation:

The time period of a pendulum, in seconds, can be calculated using the function T=1.11l , where l is the length of the pendulum in inches. Find the inverse function which gives the length in terms of the time period.

The function is T=1.11l .

Divide both sides of the function by 1.11,

  T1.11=1.11l1.11T1.11=l

Square both sides of the function,

  T1.112=l2T1.112=ll=T1.112

Therefore, the inverse function is l=T1.112 .

Chapter 3 Solutions

EBK ALGEBRA 2

Ch. 3.1 - Prob. 11GPCh. 3.1 - Prob. 12GPCh. 3.1 - Prob. 13GPCh. 3.1 - Prob. 14GPCh. 3.1 - Prob. 15GPCh. 3.1 - Prob. 16GPCh. 3.1 - Prob. 17GPCh. 3.1 - Prob. 18GPCh. 3.1 - Prob. 19GPCh. 3.1 - Prob. 1ECh. 3.1 - Prob. 2ECh. 3.1 - Prob. 3ECh. 3.1 - Prob. 4ECh. 3.1 - Prob. 5ECh. 3.1 - Prob. 6ECh. 3.1 - Prob. 7ECh. 3.1 - Prob. 8ECh. 3.1 - Prob. 9ECh. 3.1 - Prob. 10ECh. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.1 - Prob. 18ECh. 3.1 - Prob. 19ECh. 3.1 - Prob. 20ECh. 3.1 - Prob. 21ECh. 3.1 - Prob. 22ECh. 3.1 - Prob. 23ECh. 3.1 - Prob. 24ECh. 3.1 - Prob. 25ECh. 3.1 - Prob. 26ECh. 3.1 - Prob. 27ECh. 3.1 - Prob. 28ECh. 3.1 - Prob. 29ECh. 3.1 - Prob. 30ECh. 3.1 - Prob. 31ECh. 3.1 - Prob. 32ECh. 3.1 - Prob. 33ECh. 3.1 - Prob. 34ECh. 3.1 - Prob. 35ECh. 3.1 - Prob. 36ECh. 3.1 - Prob. 37ECh. 3.1 - Prob. 38ECh. 3.1 - Prob. 39ECh. 3.1 - Prob. 40ECh. 3.1 - Prob. 41ECh. 3.1 - Prob. 42ECh. 3.1 - Prob. 43ECh. 3.1 - Prob. 44ECh. 3.1 - 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