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
Answer:
$24.35
Step-by-step explanation:
sale price⇒ total price-discount price
s.p=$24.60-25/100 ∴25%⇒25/100
s.p⇒$2460/100-25/100 ∴24.60⇒2460/100
s.p⇒$24.35
⇒$24.35
Answer:
937.5
Step-by-step explanation:
$18.75 - 1 hour
$x. - 50hours
cross multiply the 18.75 with 50.
I have seen you ask this question twice before, so I will give you the answer that HugoD gave you:
In mathematics, "of" is translated as a multiplication.
So, you have :
.
First you need to find the volume of the cylinder. Then find the volume of all 4 rubber balls (spheres). Subtract the cylinder's volume by the spheres' volumes.
Volume of cylinder: πr²h
π2.5²(20) = 392.7 cm³
Volume of sphere: 4/3πr³
4/3π2.5³ = 65.45 cm³
392.7 - 65.45 = 327.25
The amount of space in the container unoccupied by the rubber balls is 327 cm³.
