A soccer ball was kicked in the air and follows the path h(x)=−2x2+1x+6, where x is the time in seconds and h is the height of the soccer ball. At what time will the soccer ball hit the ground?

At the time x = 2 seconds

Step-by-step explanation:

The height of the soccer ball is described by the function h(x), so if we want to know when the soccer ball will hit the ground, we just need to find the value of x that gives us the height zero, that is, h(x) = 0:

h(x) = −2*x^2 + 1*x + 6

0 = −2*x^2 + x + 6

2*x^2 - x - 6 = 0

Using Bhaskara's formula, we have:

Delta = b^2 - 4ac = 1 + 48 = 49

sqrt(Delta) = 7

x = (-b + sqrt(Delta)) / 2a = (1 + 7) / 4 = 2 seconds

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

555.99.

step-by-step explanation:

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