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:
Step-by-step explanation:
Multiplying Equation A by (1/3) and adding the result to Equation B will do the trick. Let's actually solve the problem!
Equation A: (5/3)x + 3y = 12
Equation B: 4x - 3y = 8
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(5/3 + 12/3)x = 15 Note how this has eliminated the variable
(17/3)x = 15 y.
x = (3/17)(15)
through my calculations x=8
Answer:
the answer is 2q + 11 because 2 and 9 are like terms so you can add them
