Please help with this question!

D

Step-by-step explanation:

When the ball hits the ground then h = 0 , that is

- 5t² + 13t + 6 = 0 ( multiply through by - 1 )

5t² - 13t - 6 = 0

Consider the factors of the product of the coefficient of the h² term and the constant term which sum to give the coefficient of the h- term.

product = 5 × - 6 = - 30 and sum = - 13

The factors are - 15 and + 2

Use these factors to split the h- term

5t² - 15t + 2t - 6 = 0 ( factor the first/second and third/fourth terms )

5t(t - 3) + 2(t - 3) = 0 ← factor out (t - 3) from each term

(t - 3)(5t + 2) = 0

Equate each factor to zero and solve for t

t - 3 = 0 ⇒ t = 3

5t + 2 = 0 ⇒ 5t = - 2 ⇒ t = -

Now t > 0 , so t= 3

The ball hits the ground after 3 seconds

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°