The height of an object seconds after it is launched is modeled by the function h(t)=-4t^2+16t+180. When will the object hit the ground?

9 seconds after the object is launched

Step-by-step explanation:

The function models the height of the object as a function of the time after the object is launched. Therefore h(t) will be zero, and you will solve for t like a normal quadratic equation.

0 = -4t² + 16t + 180

First, you can factor out the -4 from all the terms.

0 = -4(t² - 4t - 45)

Next, factor what is inside the parentheses

0 = -4 (t + 5) (t - 9)

Then you set each of the factors individually to zero and solve

t + 5 = 0 and t - 9 =0

From this we get that time is:

t = -5 and t = 9

Now that time cannot be negative, we figure out that the time is 9 seconds, therefore the object will hit the ground after 9 seconds.

hence, the required result is

step-by-step explanation:

suppose we have two matrices a and b

we will assume the two matrices

we have to compute b-a which means subtracting corresponding elements of a from b.

-

yx = 36

step-by-step explanation:

cos f = opposite side/ hypotenuse

cos f = yw / yx

we know cos f = 4/9 and yw =16

4/9 = 16/yx

multiply each side by yx

4/9 yx = 16/ yx * yx

4/9 yx = 16

multiply each side by 9/4

9/4 * 4/9 yx = 16 *9/4

yx = 36