anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
So it looks like this? \[h = 48t-16t^2\]
anonymous
  • anonymous
yea
anonymous
  • anonymous
so now what?

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anonymous
  • anonymous
Graph it
anonymous
  • anonymous
cant graph that
anonymous
  • anonymous
Your h is 36
campbell_st
  • campbell_st
so you need to le the height = 0 and then solve 0 = 48t - 16t^2 for t this can be factored
anonymous
  • anonymous
Now try to put it into quadratic form like \[48t-16t^2-36=0\]
anonymous
  • anonymous
Then factor this and you should get 2 values.
anonymous
  • anonymous
4(8t-9)
anonymous
  • anonymous
No...use quadratic equation
anonymous
  • anonymous
t=3,0
anonymous
  • anonymous
yes
anonymous
  • anonymous
thats the amswer?
anonymous
  • anonymous
so the answer is 3 seconds
anonymous
  • anonymous
thx
anonymous
  • 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
  • anonymous
can u help with this
anonymous
  • anonymous
h(x) is this? \[4x^4\]
anonymous
  • anonymous
k
anonymous
  • anonymous
If it is product 3 will exceed all other price, it's increasing exponentially
anonymous
  • anonymous
so c
anonymous
  • anonymous
yes
anonymous
  • anonymous
wait
anonymous
  • anonymous
It's product 1, product 3 is only 4x
anonymous
  • anonymous
kk
anonymous
  • 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
  • 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
  • anonymous
that should seem clearer when i put it this way
anonymous
  • anonymous
Ok take the derivative of the function f(t)
anonymous
  • anonymous
I hope you know how to do that
anonymous
  • anonymous
can i just solve for t
anonymous
  • anonymous
no you can't solve for t because every value of t gives you something new, its a function
anonymous
  • anonymous
and you are finding the rate of change
anonymous
  • anonymous
This curve looks like this|dw:1438220902622:dw|
anonymous
  • anonymous
so you are solving for the tangent line like this|dw:1438221106783:dw|
anonymous
  • anonymous
omg
anonymous
  • anonymous
@saseal
anonymous
  • anonymous
yea?
anonymous
  • anonymous
this is hard
anonymous
  • 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
  • anonymous
so you are basically doing \[f'(x)=\frac{ f(5)-f(3) }{ 2 }\]
anonymous
  • anonymous
kk
anonymous
  • anonymous
how would i solve that
anonymous
  • anonymous
@saseal
anonymous
  • anonymous
f(5) is \[f(5)=-16(5)^2+48(5)+100\]
anonymous
  • anonymous
is this the answer
anonymous
  • anonymous
no this is just f(5)
anonymous
  • anonymous
remember you need to take f(5)-f(3) and divide by 5-3
anonymous
  • anonymous
so divide 5 and 3
anonymous
  • anonymous
umm no
anonymous
  • anonymous
guess ill show you the short way since i have to go soon its some calculus magic but it works
anonymous
  • anonymous
ok
anonymous
  • anonymous
It's -80
anonymous
  • anonymous
thx

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