Complete Question:
The profile of a dam is modeled by the equation y = 24 (StartFraction x Over 10 EndFraction) squared, where x represents horizontal distance and y represents height, both in meters.
Point Z is at a horizontal distance of 15 meters from the left-most point of the dam. What is the height of the dam at point Z?
Answer:
The height is 36 meters
Step-by-step explanation:
Given
Required
The interpretation of the question is to determine the value of y when x = 15
Substitute 15 for x in:
<em>The height is 36 meters</em>
Step-by-step explanation:
If x=20 3x/5=12
Then if y=30 3y/5=18
It shows that A'B'C' smaller than ABC
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
d i believe so it might be d if not i’m sorry
Answer:
3. 9 - (n ÷ 5 ) 5. 8 x ( 5 x t ) (( i don't know the others ))
Step-by-step explanation:
