Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 823: 117


The required equation is, $1682=16a+4b+c$.

Work Step by Step

It is provided that: x in $y=a{{x}^{2}}+bx+c$ is 4; The value of y is 1682. Substitute the value of x in the given equation: $\begin{align} & 1682=a{{\left( 4 \right)}^{2}}+4b+c \\ & 1682=16a+4b+c \\ \end{align}$ Thus, the solution is $1682=16a+4b+c$.
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