<span>Quadratic because you have x^2</span>
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:
1 / 216
Step-by-step explanation:
Remember, when you divide fractions you have to multiply by the reciprocal
1/6 ÷ 6/1 ÷ 6/1
1st step: 1/6 ÷ 6/1 = 1/6 · 1/6 = 1/36
2nd step: 1/36 ÷ 6/1 = 1/36 · 1/6 = 1 /216
Answer:
Negative 3 less-than y less-than-or-equal-to 3
Step-by-step explanation:
By looking at a coordinate plane, we can easily see the range
remember, a closed dot means includes , and an open dot means does not include
[closed dot]
(open dot)
the range is all possible y-values. so, all y values shown are
3 which it includes
to -3 which it excludes / does not include
so, all y values are greater than -3 and less than or equal to 3
Answer:
Quotient: 2
Remainder: 0
50 = 25 × 2 + 0
Step-by-step explanation:
