Question

Math help please would be greatly appreciated

Answer 1

E. 4.3

Explanation:

We have the equation S = -16t^2 + 37t + 211

Given S = 78, then

78 = -16t^2 + 37t + 211

-16t^2 + 37t + 211 - 78 = 0

-16t^2 + 37t + 133 = 0

Using quadratic equation ax^2 + bx + c = 0

x = [-b ± √(b^2 - 4ac)] / (2a)

t = [-37 ± √(37^2 - 4(-16)(133)] / 2(-16)

t = [-37 ± √(1369 - (-8512)] / (-32)

t = [-37 ± √(9881)] / (-32)

a. t = [-37 + √(9881)] / (-32)

t = (-37 + 99.403) / (-32)

t = -1.95

b. t = [-37 - √(9881)] / (-32)

t = (-37 - 99.403) / (-32) = 4.26

Since t can't be a negative number, we have t = 4.26 or 4.3

Please double check my calculation. Hope this helps.

Answer 2

4.3

Explanation:

78= -16t²+37t+211

0= -16t²+37t+133

Using the quadratic formula,

(-37±√(37²-4*-16*133))/(2*-16)

(-37±√9881)/(-32)

(-37-√9881)/ -32 = 4.2626= 4.3

While -1.95 is a solution to the quadratic formula, a negative value doesn't make sense in this context.