Question

Help please

Answer 1

Can you use a calculator for this?

Answer 2

I did

Answer 3

Did you guys learn how to do derivatives?

Answer 4

Calculus approach:

Take the derivative of that equation, you get

\[h'(t)=-32t+70\]
Find out where -32t + 70 = 0 , and that's where the maximum point is (cause the slope changes from positive to negative at that point).

Then you can plug t back into h(t) to get the height.

Answer 5

you can do this without calc. you are looking for the vertex
the first coordinate of the vertex of \(ax^2+bx+c\) is always \(-\frac{b}{2a}\) which in your case is \[-\frac{70}{2\times (-16))}\]

Answer 6

since you are going to use a calculator anyways, round that to \(2.19\) and plug it in, i.e. computer \[-16\times (2.19)^2+70\times (2.19)+2\]

Answer 7

let me know when you get the answer

Answer 8

85.22

Answer 9

i got \(78.56\)

Answer 10

you need to square the first one

Answer 11

\[-16\times (2.19)^2+70\times (2.19)+2\]

Answer 12

oh okay then with that I got the same

Answer 13

whew

Answer 14

Haha

Answer 15

So D =)

Answer 16

Thank you

Answer 17

oh yeah, D

Answer 18

\[\color\magenta\heartsuit\]