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:
Rounded to the nearest tenth they are -14.5 and 12.5
Step-by-step explanation:
The zeros of a function are the x-intercepts or roots where the function crosses the x-axis. To find them, graph the function x^2 +2x-180 and zoom in on the x-axis.
See attached picture.
Lets say the final price of the bike is SP plus 13% VAT. So:
1.13 * SP = 6356.25
SP = 5625
Now SP was after 10% discount on Original Price, say MP. So:
0.9 * MP = 5625
MP = 6250
Therefore, the percent of Marked Price above Cost Price is:
(6250 - 5000) * 100/5000 = 25%