Score: 4 of 8 pts TA
23.1.59
A ball is thrown upward and outward from a height of 5 feet. The height of the ball, f(x), in feet, can be mo
f(x) = -0.2x² +2.1x+5
where x is the ball's horizontal distance, in feet from where it was thrown. Use this model to solve parts (
a. What is the maximum height of the ball and how far from where it was thrown does this occur?
The maximum height is feet, which occurs feet from the point of release
(Round to the nearest tenth as needed.)

  10.5 ft high

  5.3 ft horizontally

Step-by-step explanation:

The equation can be written in vertex form to answer these questions.

  f(x) = -0.2(x² -10.5x) +5

  f(x) = -0.2(x² -10.5x +5.25²) +5 +0.2(5.25²)

  f(x) = -0.2(x -5.25)² +10.5125

The vertex of the travel path is (5.25, 10.5125).

The maximum height is 10.5 feet, which occurs 5.3 feet (horizontally) from the point of release.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2