A ball is thrown in air and it’s height, h(t) in feet, at any time, t in seconds, is represented by the equation h(t) =-t^2+7t. When is the ball higher than 10 feet off the ground?

The ball is higher than 10 feet off the ground when the time is between 2 to 5 seconds

Step-by-step explanation:

In the above question, we are given the equation:

h(t) =-t²+7t.

We are asked to find the time (t) when the ball will be higher than 10 feet.

h(t) = height = 10 feet.

So we have:

10 = -t² + 7t

t² - 7t + 10 = 0

We have a quadratic equation hence we factorise.

t² - 5t - 2t + 10 = 0

(t² - 5t) +(2t - 10)= 0

t(t - 5) + 2 (t - 5) = 0

(t - 2) (t - 5) = 0

t - 2 = 0, t = 2

t - 5 = 0, t = 5

t = 5 and t = 2

Therefore, the ball is higher than 10 feet off the ground when the time is between 2 to 5 seconds.

y=mx+b

step-by-step explanation:

m is slope, b is y-intercept

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

~