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.
Find the average rate of change of h over the interval
[3, 6].

During the last three seconds of the objects' time in the air, it is
falling at an average rate of feet per second.

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

for this case we must apply distributive property to the following expression:

the distributive property states that:

then, when applying the property we have to multiply the -2 for each term within the parenthesis:

42.4 6th grade students prefer hip hop

step-by-step explanation:

40 percent of 106 = 42.4