No because think about 12 and 6. the greatest common factor is 3 which isn't even
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:
93.4 cm²
Step-by-step explanation:
Area of the shaded region = area of the square - area of half of the circle
Area of the shaded region = s² + ½(πr²)
Where,
r = 6.2 cm
s = length of square = diameter of circle = 2*r = 2*6.2
s = 12.4 cm
Plug in the values
Area of the shaded region = 12.4² - ½(π × 6.2²)
= 153.76 - 60.381411
= 93.378589
≈ 93.4 cm² (nearest tenth)
Answer:
9
Step-by-step explanation:
