Going out on a limb here and guessing that the function is
Please correct me if this isn't the case.
Recall that
which converges for
.
It follows that
A.) To find the maximum height, we can take the derivative of h(t). This will give us the rate at which the horse jumps (velocity) at time t.
h'(t) = -32t + 16
When the horse reaches its maximum height, its position on h(t) will be at the top of the parabola. The slope at this point will be zero because the line tangent to the peak of a parabola is a horizontal line. By setting h'(t) equal to 0, we can find the critical numbers which will be the maximum and minimum t values.
-32t + 16 = 0
-32t = -16
t = 0.5 seconds
b.) To find out if the horse can clear a fence that is 3.5 feet tall, we can plug 0.5 in for t in h(t) and solve for the maximum height.
h(0.5) = -16(0.5)^2 + 16(-0.5) = 4 feet
If 4 is the maximum height the horse can jump, then yes, it can clear a 3.5 foot tall fence.
c.) We know that the horse is in the air whenever h(t) is greater than 0.
-16t^2 + 16t = 0
-16t(t-1)=0
t = 0 and 1
So if the horse is on the ground at t = 0 and t = 1, then we know it was in the air for 1 second.
The answer to this question is C.2 I think
Answer:
h=3
Step-by-step explanation:
I divided -3.6 on both sides to get h by itself, and now I have h=-3.6/-10.8. combine like terms (-3.6/-10.8) and I got 3. My final answer would be h=3
The closer the points are/more uniform they are to the residual line, the better the fit. The more skewed and scattered they are, the model is a worse fit.
