Holt Mcdougal Larson Algebra 2: Student Edition 2012

1st Edition

ISBN: 9780547647159

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Expert Solution & Answer

Chapter 3.4, Problem 47PS

Solution

(a)

Find the inverse of the given model.

l−1=2w−6

Given information:

When calibrating a spring scale, you need to know how far the spring stretches for various weights. Hooke’s law states that the length a spring stretches is proportional to the weight attached to it. A model for one scale is l=0.5w+3 where l is the total length (in inches) of the stretched spring is and w is the weight (in pounds) of the object.

Concept used:

In order to find the inverse of a function y=fx, follow below steps:

Step 1: interchange x and y in the given function

Step 2: solve for y and that will give the required inverse function

Calculation:

Hooke’s law states that the length ( l ) a spring stretches is proportional to the weight ( w ) attached to it.

l=0.5w+3

The inverse of l=0.5w+3 is-

l=0.5w+3y=ly=0.5w+3swap w with yw=0.5y+3solve for yw=0.5y+30.5y=w−3y=w−30.5y=w0.5−30.5y=2w−6l−1=2w−6

Thus the inverse of the given function l=0.5w+3 is l−1=2w−6 .

(b)

If you place a weight on the scale and the spring stretches to a total length of 6.5 inches, how heavy is the weight?

7 pounds.

Given information:

When calibrating a spring scale, you need to know how far the spring stretches for various weights. Hooke’s law states that the length a spring stretches is proportional to the weight attached to it. A model for one scale is l=0.5w+3 where l is the total length (in inches) of the stretched spring is and w is the weight (in pounds) of the object.

Concept used:

In order to find the inverse of a function y=fx, follow below steps:

Step 1: interchange x and y in the given function

Step 2: solve for y and that will give the required inverse function

Calculation:

Hooke’s law states that the length ( l ) a spring stretches is proportional to the weight ( w ) attached to it.

l=0.5w+3

If we place a weight on the scale and the spring stretches to a total length ( l ) of 6.5 inches.

Then the weight is -

l=0.5w+3 6.5=0.5w+36.5−3=0.5w 3.5=0.5w w=3.50.5 w=7

So, the weight = 7 pounds

Chapter 3.4, Problem 46PS

Chapter 3.4, Problem 48PS

Chapter 3 Solutions

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Ch. 3.1 - Prob. 1GP Ch. 3.1 - Prob. 2GP Ch. 3.1 - Prob. 3GP Ch. 3.1 - Prob. 4GP Ch. 3.1 - Prob. 5GP Ch. 3.1 - Prob. 6GP Ch. 3.1 - Prob. 7GP Ch. 3.1 - Prob. 8GP Ch. 3.1 - Prob. 9GP Ch. 3.1 - Prob. 10GP

Ch. 3.1 - Prob. 11GP Ch. 3.1 - Prob. 12GP Ch. 3.1 - Prob. 13GP Ch. 3.1 - Prob. 14GP Ch. 3.1 - Prob. 15GP Ch. 3.1 - Prob. 16GP Ch. 3.1 - Prob. 17GP Ch. 3.1 - Prob. 18GP Ch. 3.1 - Prob. 19GP Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E Ch. 3.1 - Prob. 11E Ch. 3.1 - Prob. 12E Ch. 3.1 - Prob. 13E Ch. 3.1 - Prob. 14E Ch. 3.1 - Prob. 15E Ch. 3.1 - Prob. 16E Ch. 3.1 - Prob. 17E Ch. 3.1 - Prob. 18E Ch. 3.1 - Prob. 19E Ch. 3.1 - Prob. 20E Ch. 3.1 - Prob. 21E Ch. 3.1 - Prob. 22E Ch. 3.1 - Prob. 23E Ch. 3.1 - Prob. 24E Ch. 3.1 - Prob. 25E Ch. 3.1 - Prob. 26E Ch. 3.1 - Prob. 27E Ch. 3.1 - Prob. 28E Ch. 3.1 - Prob. 29E Ch. 3.1 - Prob. 30E Ch. 3.1 - Prob. 31E Ch. 3.1 - Prob. 32E Ch. 3.1 - Prob. 33E Ch. 3.1 - Prob. 34E Ch. 3.1 - Prob. 35E Ch. 3.1 - Prob. 36E Ch. 3.1 - Prob. 37E Ch. 3.1 - Prob. 38E Ch. 3.1 - Prob. 39E Ch. 3.1 - Prob. 40E Ch. 3.1 - Prob. 41E Ch. 3.1 - 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Solution Summary: The author explains how Hooke's law states that the length of a spring stretches is proportional to the weight attached to it.

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Chapter 3 Solutions