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.
Answer:
Twelve pence made a shilling
I hope this is the ans
Answer:
$570
Step-by-step explanation:
500 * 1.2 = $600 after 20% pay raise
600 * .95 = $570 after 5% pay cut
To answer this question you will use the formula for circumference of a circle to find how far around one revolution is.
C = pi x d
3.14 x 32
C = 100.48 feet
Multiply the distance around one time by 4.3 to get the distance traveled in one revolution and then multiply it by 3 for the 3 minutes.
100.48 x 4.3 x 3 = 1296.19 feet
This is approximate and is closest to answer choice D.
