You can solve <span>9/x-4/y=8 for x or for y but NOT at the same time.
Solving for x: </span><span>9/x-4/y=8 Mult all 3 terms by xy to eliminate the fractions.
9(xy)/x - 4xy/y = 8xy => 9y - 4x = 8xy, or 9y = 4x + 8xy = 4x(1+2y)
then 9y = x [ 4(1+2y) ]
9y
therefore x = ------------
4(1+2y)
Solve for y using a similar approach.
</span>
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
Step-by-step explanation:
formula is
v = 4/3 * pi * r^3
r = 8.8
answer is 2854.54
There is only one solution to the equation. A quadratic equation has 2 solutions a positive and a negative, but since we are looking for length, the answer can only be postive. Area= side^2
400= x^2
sqrt(400)=x
20 inches = x
The length of one side is 20 inches
check: 20*20 = 400, which is the area of the square painting.
