The average rate of change of h over the interval is -48 feet per second.
Given,
The height of an object off the ground, h (in feet) t seconds after it is launched into the air is given by
h(t) = −16t2 + 96t, 0 ≤ t ≤ 6.
We need to find the average rate of change of h over the interval [3, 6].
<h3>How do we find the average rate of change of a function over an interval?</h3>
If we have an interval [a, b] and a function f(x).
The rate of change is given by:
= f(b) - f(a) / b - a
We have,
h(t) = −16t² + 96t and [3, 6].
h(6) = -16 x 6² + 96 x 6 = -576 + 576 = 0
h(3) = -16 x 3² + 96 x 3 = -144 + 288 = 144
The average rate of change of h is:
= h(6) - h(3) / 6 - 3
= 0 - 144 / 3
= -144 / 3
= -48
Thus the average rate of change of h over the interval is -48 feet per second.
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It's 360 feet (The person above)
A, C, and D are all equal to that
Answer:
½ = x
Step-by-step explanation:
As discussed in one of my You-Tube videos on Exponential Rules, aᵐ\ⁿ, any base to the half power is the SAME AS taking the exponential denominator of that power, and setting that to the square root of the base, leaving the exponential numerator on the outside:
25¹\² = √25
25 = a [Base]
1 = m [Exponential Numerator]
2 = n [Exponential Denominator]
As you can see, it is unnecessary to put the two because by definition, any vacant root is automatically assumed as the square root, and 1, which is the exponential numerator, is on the outside, which again is unnecessary because 5¹ is 5.
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Answer:
Step-by-step explanation:
