Explanation Formula used: Quadratic formula : “The solutions of the general quadratic equation a x 2 + b x + c = 0 ( a ≠ 0 ) is x = − b ± b 2 − 4 a c 2 a ”. Calculation: The quadratic equation is D 0 2 − D D 0 − 0.25 D 2 = 0 . If D = 3.625 then the given equation becomes, D 0 2 − ( 3.625 ) D 0 − 0.25 ( 3.625 ) 2 = 0 D 0 2 − 3.625 D 0 − 0.25 ( 13 .140625 ) = 0 D 0 2 − 3.625 D 0 − 3 .28515625 = 0 Compare the equation D 0 2 − 3.625 D 0 − 3 .28515625 = 0 with the general quadratic equation a x 2 + b x + c = 0 . Then a = 1 , b = − 3.625 and c = − 3.28515625 . Substitute a = 1 , b = − 3

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

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ISBN: 9780135902912

Author: Allyn J. Washington

Publisher: PEARSON

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Chapter 7.3, Problem 45E

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The outside diameter D0 by using the equation D02−DD0−0.25D2=0 if D=3.625 cm.

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Chapter 7.3, Problem 44E

Chapter 7.3, Problem 46E

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Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

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The outside diameter D 0 by using the equation D 0 2 − D D 0 − 0.25 D 2 = 0 if D = 3.625 cm .

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The outside diameter D0 by using the equation D02−DD0−0.25D2=0 if D=3.625 cm.

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Chapter 7 Solutions