A punter kicks a football upward with an initial speed of 48 feet per second. After how many seconds does the ball hit the ground?
Use the formula h = rt − 16t2 where h represents height in feet and r represents the initial speed.

- anonymous

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- anonymous

So it looks like this?
\[h = 48t-16t^2\]

- anonymous

yea

- anonymous

so now what?

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## More answers

- anonymous

Graph it

- anonymous

cant graph that

- anonymous

Your h is 36

- campbell_st

so you need to le the height = 0
and then solve
0 = 48t - 16t^2 for t
this can be factored

- anonymous

Now try to put it into quadratic form like
\[48t-16t^2-36=0\]

- anonymous

Then factor this and you should get 2 values.

- anonymous

4(8t-9)

- anonymous

No...use quadratic equation

- anonymous

t=3,0

- anonymous

yes

- anonymous

thats the amswer?

- anonymous

so the answer is 3 seconds

- anonymous

thx

- anonymous

A company launches 4 new products. The market price, in dollars, of the 4 products after a different number of years, x, is shown in the following table:
Product Function Year 1
(dollars) Year 2
(dollars) Year 3
(dollars)
Product 1 f(x) = x3 1 8 27
Product 2 g(x) = x2 + 11 12 15 20
Product 3 h(x) = 4x 4 16 64
Product 4 j(x) = 4x + 8 12 16 20
Based on the data in the table, for which product does the price eventually exceed all others?
Product 1
Product 2
Product 3
Product 4

- anonymous

can u help with this

- anonymous

h(x) is this?
\[4x^4\]

- anonymous

k

- anonymous

If it is product 3 will exceed all other price, it's increasing exponentially

- anonymous

so c

- anonymous

yes

- anonymous

wait

- anonymous

It's product 1, product 3 is only 4x

- anonymous

kk

- anonymous

A ball is thrown upward from the top of a building. The function below shows the height of the ball in relation to sea level, f(t), in feet, at different times, t, in seconds:
f(t) = −16t2 + 48t + 100
The average rate of change of f(t) from t = 3 seconds to t = 5 seconds is _____feet per second.

- anonymous

\[\left[\begin{matrix}Product & Function & Year 1 & Year 2 & Year 3\\ 1 & f(x)=x^3 & 1 &8 &27 \\ 2 & g(x) = x^2+11 & 12 &15&20 \\ 3 & h(x) = 4x & 4&16& 64 \\4 &j(x)=4x+8 &12 &16 &20\end{matrix}\right]\]

- anonymous

that should seem clearer when i put it this way

- anonymous

Ok take the derivative of the function f(t)

- anonymous

I hope you know how to do that

- anonymous

can i just solve for t

- anonymous

no you can't solve for t because every value of t gives you something new, its a function

- anonymous

and you are finding the rate of change

- anonymous

This curve looks like this|dw:1438220902622:dw|

- anonymous

so you are solving for the tangent line like this|dw:1438221106783:dw|

- anonymous

omg

- anonymous

@saseal

- anonymous

yea?

- anonymous

this is hard

- anonymous

Ok...I'll show you the formula first, you try it then I show to the other way
\[f'(x)= \frac{ f(x+h)-f(x) }{ h }\]
h is 5-3=2

- anonymous

so you are basically doing
\[f'(x)=\frac{ f(5)-f(3) }{ 2 }\]

- anonymous

kk

- anonymous

how would i solve that

- anonymous

@saseal

- anonymous

f(5) is
\[f(5)=-16(5)^2+48(5)+100\]

- anonymous

is this the answer

- anonymous

no this is just f(5)

- anonymous

remember you need to take f(5)-f(3) and divide by 5-3

- anonymous

so divide 5 and 3

- anonymous

umm no

- anonymous

guess ill show you the short way since i have to go soon its some calculus magic but it works

- anonymous

ok

- anonymous

It's -80

- anonymous

thx

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