#### Answer

$y=k(50t - \frac{t^2}{2})+C$

#### Work Step by Step

$$\frac{dy}{dt} = k(50-t)$$
$$\int dy = \int k(50-t^1) dt$$
$$y=k(50t - \frac{t^{1+1}}{1+1})+C$$
$$y=k(50t - \frac{t^2}{2})+C$$

Published by
Brooks Cole

ISBN 10:
1-28505-709-0

ISBN 13:
978-1-28505-709-5

$y=k(50t - \frac{t^2}{2})+C$

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