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:
Divide both sides by 9.
Step-by-step explanation:
To find the second step of the equation, we actually have to solve parts of the equation until we get to step 2.
Step 1: Add 23 to both sides.
Step 2: Divide both sides by 9.
Therefore, the second step is divide both sides by 9.
Have a lovely rest of your day/night, and good luck with your assignments! ♡
Answer: y=x+60
So y=mx+b is slope intercept form
y=x+60
m=1/1 because we go equally up as we go right so just put x
b=60 because we start at 60
Hope this helps
Answer:
Step-by-step explanation:
The student currently has $50 and plans to save $15 every month.
Let x represent the number of months that the student will save enough money to buy the microscope.
Let y represent the amount that the student saves after x months.
The function that represents the amount y (in dollars) of money that the student saves after x months will be
y = 50 + 15x
The 50 remains constant because she has already saved it