how can you use plug-in for #13 on page 607?

To get this one by plugging in, pick values for all the variables on the right side of the equals sign: t, v, and k. For example, say t = 2, v = 3, and k = 5. Use your calculator to see what that gives you for h:

    \begin{align*}h&=-16(2)^2+(3)(2)+5\\h&=-53\end{align*}

Now plug –53 in for h, 2 in for t, and 5 in for k in each answer choice. The one that gives you 3 (our original value for v) is the right answer.

Choice D does the trick:

    \begin{align*}v&=\dfrac{h-k}{t}+16t\\3&=\dfrac{-53-5}{2}+16(2)\\3&=-29+32\\3&=3\end{align*}

Note that I’m not saying you need to do the question this way, just that it CAN be done this way. On a more complicated algebraic manipulation question, you might be happy to have a method like this in your back pocket.

Because I know someone will ask, here’s the algebra:

    \begin{align*}h&=-16t^2+vt+k\\h+16t^2-k&=vt\\\dfrac{h-k+16t^2}{t}&=v\\\dfrac{h-k}{t}+\dfrac{16t^2}{t}&=v\\\dfrac{h-k}{t}+16t&=v\end{align*}

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