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:
-1
Step-by-step explanation:
-4(-1)-5=4-5=-1
The value of the variable x in the equation –x + 8 + 3x = x – 6 is - 14.
<h3>How to find variable in an equation?</h3>
The variables of an equation can be found as follows:
–x + 8 + 3x = x – 6
The variable in the equation is x.
Therefore,
–x + 8 + 3x = x – 6
combine like terms on the left side
-x + 3x + 8 = x - 6
2x + 8 = x - 6
subtract x form both sides of the equation
2x - x + 8 = - 6
x + 8 = -6
subtract 8 from both sides
x = -6 - 8
x = - 14
learn more on equation here: brainly.com/question/1347990
#SPJ4
561 percent because we will move the decimal 2 paces and we will put the percent so it will be 561
Answer:
c≥5
Step-by-step explanation:
hope that helps :)
