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
Answer:
A, 1,2
Step-by-step explanation:
we translate it first into language of math
we use n for the number
n² = 3n - 2
then, do some algebra things
n² - 3n + 2 = 0
then factorize it
(n - 2)(n - 1) = 0
first n
n - 2 = 0
n = 2
2nd n
n - 1 = 0
n = 1
so n can be 1 or 2
Answer:
x=2/3
Step-by-step explanation:
(2x+1)/2 = 2 1/3/4
Multiply 4 on both sides:
2x+1=2 1/3
Rewrite:
2x+1=7/3
Subtract 1 on both sides:
2x=4/3
Divide 2 on both sides:
x=2/3
Thus x=2/3
Hope this helps!
Answer:
kbjkbhvucsorrymycomputerisbeingweirdforsomereason
Step-by-step explanation:
Im sorry, it is too blurry. I cannot read it.
