thanks

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- May 23rd 2006, 02:56 PMJohnSenaPlease - Need help tonight on a couple more calc problems
thanks

- May 23rd 2006, 03:12 PMThePerfectHackerQuote:

Originally Posted by**JohnSena**

$\displaystyle 5(x^3+3)=5x^3+15$ - May 23rd 2006, 03:15 PMJohnSena
thanks for the quick answer

- May 23rd 2006, 03:16 PMThePerfectHackerQuote:

Originally Posted by**JohnSena**

$\displaystyle A(t)=100(.83)^t$

For the second part you need to find,

$\displaystyle 100(.83)^3\approx 57.18$ - May 23rd 2006, 07:47 PMCaptainBlackQuote:

Originally Posted by**JohnSena**

$\displaystyle

f(t)=-16t^2+96t+6=0

$

This will have two roots, one positive and the other negative, the

negative root is unphysical but the positive one is what you want.

The roots are:

$\displaystyle

-0.06186, 6.06186

$

So the ball is in the air $\displaystyle \approx 6.06$seconds.

b) The maximum height is achieved when $\displaystyle f(x)$ is

a maximum. As this is a quadratic we know this occurs midway

between the roots, so it occurs at $\displaystyle (-0.06186+6.06186)/2=3$,

so the maximum height is:

$\displaystyle

f(3)=16\ 3^2+96\ 3+6=438$ feet.

c) Already answered at $\displaystyle t=3$seconds.

RonL