If you've started pre-calculus, then you know that the derivative of h(t)
is zero where h(t) is maximum.
The derivative is h'(t) = -32 t + 96 .
At the maximum ... h'(t) = 0
32 t = 96 sec
t = 3 sec .
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If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.
In that case, the question GIVES you the maximum height.
Just write it in place of h(t), then solve the quadratic equation
and find out what 't' must be at that height.
150 ft = -16 t² + 96 t + 6
Subtract 150ft from each side: -16t² + 96t - 144 = 0 .
Before you attack that, you can divide each side by -16,
making it a lot easier to handle:
t² - 6t + 9 = 0
I'm sure you can run with that equation now and solve it.
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.
(Funny how the two widely different methods lead to the same answer.)
The answer is from AL2006
3c/5-1/2=-31/2 Add 1/2 both sides
+1/2 +1/2
3c/5=-30/2 Then simplify the fraction
(5)3c/5=-15(5) Multiply 5 both sides
3c=75 Finally, divide by 3 both sides
c=25
So the answer to your question is E.
Yes. depending on where you buy it and how much the ball costs
Aneesha is traveling at a rate of 4400 feet per second, which means that Morris is traveling a a rate of 4397 feet per second
Answer:
C. -5.3
Step-by-step explanation:
Simply add the negatives straight across to arrive at your answer.
I am joyous to assist you anytime.
