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
Okay first we add them:
0.89 + 1.95 = 2.84
Then we estimate them to the tenths place:
2.84 (estimated to) 2.80
Since the number in the ones place is less than 5, we estimate to 80.
Answer:
the new ones and the old ones that you didn’t delete
Step-by-step explanation:
For the first one its 7/2 and the second one is 7/6
and u cant covert them into a proper fraction you can only covert them into a improper fracion
I'm guessing you would need the formula for area of a circle. Find the whole dimensiosn of that circle and then subtract what you need to find the shaded part. That make sense?
