Aball is thrown upward and outward from a height of 8 feet. the height of the ball, f(x), in feet, can be modeled by 1(x) -0.1x2+14x+8 where x is the ball's horizontal distance, in feet, from where it was thrown use this model to solve parts (a) through (c). a. what is the maximum height of the ball and how far from where it was thrown does this occur? feet from the point of release. feet, which occurs the maximum height is (round to the nearest tenth as needed)

Step-by-step explanation:

A graphing calculator can show the answers to these questions.

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For quadratic ax²+bx+c, the x-value of the vertex is ...

  x = -b/(2a)

For the given expression, the vertex is ...

  x = -14/(2(-0.1)) = 70

The distance from the point of release to the point of maximum height is 70 feet.

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The height of the ball at that point is ...

  f(70) = (-0.1·70 +14)70 +8 = 7·70 +8 = 498 . . . . feet

The maximum height is 498 feet.

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

b. 1.25 hr

step-by-step explanation:

if you have a constant speed of 60 mph, then you have a ratio between 60 mi. to 1 hr. dividing both sides by 60 we get the ratio

1 mi. to

then mult. both sides by 75 we get the ratio

75 mi. to

simplifying we get

75 mi. to

or 1.25 hr.