Subtract 4 from both sides
R/10 = 5 - 4
Simplify 5 - 4 to 1
R/10 = 1
Multiply both sides by 10
R = 1 × 10
Simplify 1 × 10 to 10
<u>R = 10</u>
Step-by-step explanation:
difference is 1.8
hope it is correct
Answer:
-35
Step-by-step explanation:
PEMDAS
Answer:
Step-by-step explanation:
3(x-3)+x(r+4)
3x-9+xr+4x
7x+9+xr
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
