The measurement of EF is EF=5.9
So what you would do is divide 37 minutes by 5 to get 7.4 minutes
answer: 7.4 minutes
Answer:
C and D I think??? Could be wrong. But definitely D is one of them.
Answer:
True, see proof below.
Step-by-step explanation:
Remember two theorems about continuity:
- If f is differentiable at the point p, then f is continuous at p. This also applies to intervals instead of points.
- (Bolzano) If f is continuous in an interval [a,b] and there exists x,y∈[a,b] such that f(x)<0<f(y), then there exists some c∈[a,b] such that f(c)=0.
If f is differentiable in [0,4], then f is continuous in [0,4] (by 1). Now, f(0)=-1<0 and f(4)=3>0. Thus, we have the inequality f(0)<0<f(4). By Bolzano's theorem, there exists some c∈[0,4] such that f(c)=0.
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
