Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Section 4.3 - Systems of Linear Equations and Problem Solving - Exercise Set - Page 234: 56

Answer

2.1 inches of rain

Work Step by Step

We write the system of equations: $$ 2.47=(a)(4^2)+(4)(b)+c \\ .58=(a)(7^2)+(7)(b)+c \\ 1.07=(a)(8^2)+(8)(b)+c$$ We solve for a: $$(a)(4^2)+(4)(b)+c \\ a=\frac{2.47-4b-c}{16}$$ Thus, in the second and third equations: $$0.58=\frac{2.47-4b-c}{16}\cdot 7^2+7b+c \\ 1.07=\frac{2.47-4b-c}{16}\cdot 8^2+8b+c$$ So: $$0.58=-5.25b-2.0625c+7.56 $$ Substituting b: $$ c = 12.8 $$ Thus, plugging this into the equation for b: $$ b = -3.71$$ Finally: $$ a= .28$$ In September, which is the 9th month: $$ y = (.28)(9^2)-3.71(9)+ 12.83 \\ y = 2.1 \ inches $$
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