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.
Ok
in a 30-60-90 right triangle, if the leg oposite the 30 degree angle is x then the hypontuse is 2x
if the leg oposite the 30 degree angle is 12 inches then the hypontuse is 12*2 or 24 inches
I assume you're looking for n.
so, take 2n/10-3n/10=2/10. Making everything have the same denominator is always easier.
This way, you have -n/10=2/10.
n=-2.
Answer:
x = -1 ; y = 1
Step-by-step explanation:
x = -1
y = 1
Plug in the corresponding numbers to the corresponding variables:
y < -x + 1
(1) < -(-1) + 1
1 < 1 + 1
1 < 2 (True)
y > x
1 > -1 (True).
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