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
Step-by-step explanation:
let the number be x
and according to question,
six + 2 times a number = 2x + 6
26 + 6 times a number = 6x + 26
→ 2x + 6 = 6x + 26
→ -6x + 2x = 26 - 6
→ -4x = 20
→ x = 20/4 = 5
→ x = 5
therefore, the number is 5.
hope this answer helps you dear...take care and may u have a great day ahead!
42.105263157895 I hope this is the what u wanted
Step-by-step explanation:
