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.
<span>c. 100x^8 because the square root or 10k is 100 and 64 is 8
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Answer:
B. 90°
Step-by-step explanation:
Quadrilateral inscribed in the circle with center O is a cyclic quadrilateral.
Opposite angles of a cyclic quadrilateral are supplementary.
Therefore,
(b + d) + (32° + 58°) = 180°
b + d + 90° = 180°
b + d = 180° - 90°
b + d = 90°
You simply switch the odds, the answer would be b:a
Answer:
the last one? i think.
Step-by-step explanation:
