Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

End-of-Course Assessment - Page 968: 40

Answer

$1.8$ hrs, or (H)

Work Step by Step

Here, we are looking for the value of $x$ where $C(x) = 1.3$. So, we set the equation for $C(x)$ equal to $1.3$ and solve for $x$: $C(x) = \frac{10}{2x^2+1} = 1.3$ Multiply both sides by $2x^2+1$: $10 = 2.6x^2+1.3$ Isolate the $x^2$ term: $2.6x^2 = 8.7, x^2 = 3.35$ Solve for $x$ by taking the square root of 3.35, only worrying about the positive answer, because in this case, time after injection would never be negative. $x = \sqrt{3.35} = 1.83$
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