You need to find the roots of the function, h(t)

then h(t) = 0 = -16t^2 + 64t + 4

divide both sides by -4

4t^2 - 16 t - 1 = 0

use the quadratic function

$\frac{-b+/- \sqrt{b^2-4ac} }{2a}$

substitue a = 4, b = -16, c = -1

you will get t1 ≈ 0 and t2 ≈ 4

Answer: approx. 4 seconds.

8 0

4 years ago

Solve the systemof linear equations -3x+y=2 -x+y-4=0

faltersainse [42]

Answer:

(1, 5)

Step-by-step explanation:

Substitution works well here. Solving the second equation for x, we get x = y - 4. Substituting y - 4 for x in the first equation results in

-3(y - 4) + y = 2.

Next, we perform the indicated multiplication and obtain:

-3y + 12 + y = 2, or -2y = -10

Dividing both sides by -2 yields y = 5. As seen above, x = y - 4; if y = 5, then x = 5 - 4 = 1.

The solution is (1, 5)

8 0

3 years ago