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.
7229 feet = 1.369129 miles
561 seconds =0.155833 hours
She can run 1.369129 miles per 0.155833 hours
Answer:
F
Step-by-step explanation:
the square root of 51 = 7.141
7 square root 2 = 9.899
7 = 7
8 = 8
Answer:
x = 25
Step-by-step explanation:
4x+50 and 150 are vertical angles which means they are equal.
4x+50 =150
Subtract 50 from each side
4x+50-50 = 150-50
4x = 100
Divide each side by 4
4x/4 = 100/4
x = 25
