EBK PRECALCULUS W/LIMITS

4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 7.4, Problem 60E

(a)

To determine

The partial fraction for the equation R=5000(4−3x)(11−7x)(7−4x); 0<x≤1

(a)

Expert Solution

Answer to Problem 60E

The partial fraction equation will be

R=2760(4−3x)(11−7x)(7−4x)=−2760(11−7x)+2760(7−4x)

Explanation of Solution

Given:

The given equation is R=5000(4−3x)(11−7x)(7−4x); 0<x≤1

Formula used:

Substitution method is used.

Calculation:

The Partial fraction of the equation is

2760(4−3x)(11−7x)(7−4x)=A(11−7x)+B(7−4x) 2760(4−3x) =A(7−4x)+B(11−7x) 11040−8280x=7A+11B−(4A+7B)x

Comparing both side

11040=7A+11B and 8284A+7B

Substituting both equations we get

A=−2760 and B=2760

So, the partial equation be

R=2760(4−3x)(11−7x)(7−4x)=−2760(11−7x)+2760(7−4x)

Conclusion:

The partial fraction equation will be

R=2760(4−3x)(11−7x)(7−4x)=−2760(11−7x)+2760(7−4x)

(b)

To determine

To calculate:The values of Ymax and Ymin .

(b)

Expert Solution

Answer to Problem 60E

The values are Y max=|2760(7−4x)|;Ymin=|2760(11−7x)|

Explanation of Solution

Given:

The given equation is R=5000(4−3x)(11−7x)(7−4x); 0<x≤1

Formula used:

Substitution method is used

Calculation:

The partial fraction in difference form is

R=2760(4−3x)(11−7x)(7−4x)=2760(7−4x)−2760(11−7x)

Hence,

Y max=|2760(7−4x)|Ymin=|2760(11−7x)|

Conclusion:

The values are Y max=|2760(7−4x)|;Ymin=|2760(11−7x)|

(c)

To determine

To sketch: The graph for the values of Ymax and Ymin

(c)

Expert Solution

Answer to Problem 60E

Y max=|2760(7−4x)|;Ymin=|2760(11−7x)|

Explanation of Solution

Given:

The given equation is R=5000(4−3x)(11−7x)(7−4x); 0<x≤1

Formula used:

Substitution method is used

Calculation:

The Graph of both Y max and Y min is shown below as,

Y max=|2760(7−4x)|;Ymin=|2760(11−7x)|

EBK PRECALCULUS W/LIMITS, Chapter 7.4, Problem 60E

Here, red line is Y max and blue line is Y min.

Conclusion:

The values of Ymax and Ymin are Y max=|2760(7−4x)|;Ymin=|2760(11−7x)| and it is plotted. The value gets increases continuously.

(d)

To determine

To calculate: The maximum and minimum frequencies for the relative load.

(d)

Expert Solution

Answer to Problem 60E

The maximum and minimum frequencies are Y max(0.5)=552 , Y min(0.5)=368

Explanation of Solution

Given:

The given equation is R=5000(4−3x)(11−7x)(7−4x); 0<x≤1

Formula used:

Substitution method is used.

Calculation:

Relative load is 0.5.

Now,

Y max(0.5)=2760(7−4×0.5)Y max(0.5)=552

Y min(0.5)=2760(11−7×0.5)Y min(0.5)=368

Conclusion:

The maximum and minimum frequencies are Y max(0.5)=552 , Y min(0.5)=368

Chapter 7.4, Problem 59E

Chapter 7.4, Problem 61E

Chapter 7 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 7.1 - Prob. 1E Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

Ch. 7.1 - Prob. 11E Ch. 7.1 - Prob. 12E Ch. 7.1 - Prob. 13E Ch. 7.1 - Prob. 14E Ch. 7.1 - Prob. 15E Ch. 7.1 - Prob. 16E Ch. 7.1 - Prob. 17E Ch. 7.1 - Prob. 18E Ch. 7.1 - Prob. 19E Ch. 7.1 - Prob. 20E Ch. 7.1 - Prob. 21E Ch. 7.1 - Prob. 22E Ch. 7.1 - Prob. 23E Ch. 7.1 - Prob. 24E Ch. 7.1 - Prob. 25E Ch. 7.1 - Prob. 26E Ch. 7.1 - Prob. 27E Ch. 7.1 - Prob. 28E Ch. 7.1 - Prob. 29E Ch. 7.1 - Prob. 30E Ch. 7.1 - Prob. 31E Ch. 7.1 - Prob. 32E Ch. 7.1 - Prob. 33E Ch. 7.1 - Prob. 34E Ch. 7.1 - Prob. 35E Ch. 7.1 - Prob. 36E Ch. 7.1 - Prob. 37E Ch. 7.1 - Prob. 38E Ch. 7.1 - Prob. 39E Ch. 7.1 - Prob. 40E Ch. 7.1 - Prob. 41E Ch. 7.1 - Prob. 42E Ch. 7.1 - Prob. 43E Ch. 7.1 - Prob. 44E Ch. 7.1 - Prob. 45E Ch. 7.1 - Prob. 46E Ch. 7.1 - Prob. 47E Ch. 7.1 - Prob. 48E Ch. 7.1 - Prob. 49E Ch. 7.1 - Prob. 50E Ch. 7.1 - Prob. 51E Ch. 7.1 - Prob. 52E Ch. 7.1 - Prob. 53E Ch. 7.1 - Prob. 54E Ch. 7.1 - Prob. 55E Ch. 7.1 - Prob. 56E Ch. 7.1 - Prob. 57E Ch. 7.1 - Prob. 58E Ch. 7.1 - Prob. 59E Ch. 7.1 - Prob. 60E Ch. 7.1 - 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