A potato was launched in the air using a potato gun. The function for the situation was h(t) = -16t^2 + 100t + 10, where t was the time in seconds and h(t) was the height in feet. 1. What was the lowest time that applies to the actual situation? 2. What was the greatest time that applies to the actual situation? please help.

1. 0 seconds.

2. 6.35 seconds.

Step-by-step explanation:

There are two lowest times in the situation, and both occur when the potato is 0 feet from the ground.

0 = -16t^2 + 100t + 10

16t^2 - 100t - 10 = 0

8t^2 - 50t - 5 = 0

To solve for t, use the quadratic formula, where a = 8, b = -50, and c = -5.

[please ignore the A-hat; that is a typo]

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So, one value is...

[25 - sqrt(665)] / 8 = (25 - 25.78759392) / 8 = -0.7875939165 / 8 = -0.0984482396

BUT... the value is negative, and time cannot be negative. So, the LOWEST time that applied to the actual situation is 0 seconds.

The other value is...

[25 + sqrt(665)] / 8 = (25 + 25.78759392) /  8 = 50.78759392 / 8 = 6.34844924

Since the value is positive, it is valid. So, the GREATEST time that applies to the situation is about 6.35 seconds.

Hope this helps!

a rotation about point l

step-by-step explanation:

quizlet

hi there!

question 1 answer: p/q can be terminating when q = 5^m * 2^n

question 2 answer: -1 and -1

question 3 answer: -4

question 4 answer: 5/12

hope this : )