A ball is thrown vertically upward with an initial velocity of 32 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s equals 32 t minus 16 t squared. (a) At what time t will the ball strike the ground? (b) For what time t is the ball more than 12 feet above the ground?

(a) Therefore the ball will strike the ground after s.

(b)Therefore the ball is more than 12 feet above the ground for 1 s.

Step-by-step explanation:

Given that,

a ball is thrown vertically upward with an initial velocity of 32 feet per second.

The distance of the ball from the ground is

where s is distance in feet and t is time in second.

(a)

When the ball strike the ground, the height of the ball will be zero.

That is S=0

Therefore the ball will strike the ground after 2 s.

(b)

At some time t₁ between the ball being thrown upward direction and the maximum height, the ball is at a height of 12 ft.

After attain the maximum height, the ball come down towards ground. So, the ball again attain a height 12 ft after t₂ second.

To find the value of t₁ and t₂, we have to solve the following equation

Therefore the ball is more than 12 feet above the ground for

                                                                                                      =1 s

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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