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:
60
Step-by-step explanation:
when you multiply 100 by .6 it equals 60
Answer:
0.206897
Step-by-step explanation:
it is 0.206897 because you have take the whole make uneven fraction an add it to 60 in 60/72 to get 348 divided equals 0.206897
Answer:
Step-by-step explanation:
%change=100(final-initial)/(initial)
For just the ticket
%change=100(375-300)/300=25%
For the ticket and baggage fee of $50
%change=100(375+50-300)/300=42%
I think number 6 it should be right I think
