Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.4 - Quadratic Equations and Applications - 1.4 Exercises - Page 112: 117

Answer

$4:00~p.m.$.

Work Step by Step

$$L=-0.270t^2+3.59t+83.1$$ Substituting $L=93$ into the equation: $$93=-0.270t^2+3.59t+83.1$$ $$9300=-27t^2+359t+8310$$ $$-27t^2+359t-990=0$$ Using the quadratic formula: $$t=\frac{-359\pm\sqrt{359^2-4(-27)(-990)}}{2(-27)}=\frac{-359\pm\sqrt{21961}}{-54}$$ $$t_1=\frac{-359+\sqrt{21961}}{-54}=3.90$$ $$t_2=\frac{-359-\sqrt{21961}}{-54}=9.39$$ With the constraint $2\leq t\leq7$, take $t=3.9\approx4$. Then, the time is: $$time=0+4=4:00~p.m.$$
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