7(p)
p=3
7(3)
the answer is 21.
Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
I realise this is quite late but in case you still wanted the answer, the width is 3m.
If the length is 3m longer than the width, you can write the width as x and the length as x + 3. The perimeter would be both lengths and both widths added together, so you would just write it as:
x + x + x + 3 + x + 3 = 18
4x + 6 = 18
- 6
4x = 12
÷ 4
x = 3
I hope this helps!
Answer:
Ok im, not 100% sure but the answer is for 1-10, 2-20, 3-40, 4-40