SO confused on my math homework right now.. Anyone good at math
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SO confused on my math homework right now.. Anyone good at math

[From: ] [author: ] [Date: 12-10-17] [Hit: ]
. that means y = some constant, call it K, divided by the cube root of x Here, y = 12 = K / 2 (the cube root of x= 8) so K is 24 since 24 / 2 = 12. Now,......
27) If y varies inversely as the cube root of x, and y = 12 when x = 8, find y when x = 1.
28) If y varies jointly as x and the cube root of z, and y = 120 when x = 3 and z = 8, find y when x = 4 and z = 27.
29) The graph of f(x)=1/x^2-c has a vertical asymptote at x = 3. Find c.

I'd appreciate it if you would also explain the steps in addition to the answer so I can know how to do this for future reference. Thank you!

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27)" If y varies inversely as the cube root of x...."
that means y = some constant, call it K, divided by the cube root of x
Here, y = 12 = K / 2 (the cube root of x= 8) so K is 24 since 24 / 2 = 12.
Now, when x = 1, the cube root of 1 is also 1, so y = 24/1 = 24.

28)" If y varies jointly as x and the cube root of z...."
That means y = K times x times the cube root of z. Again, we need to find K, the constant of variation.
Here, 120 = K*x*2 = K*3*2 = K*6 so K = 20.
Next, now that we know that K= 20, we can substitute:
y = 20*4*3 = 240

The key to both of these is to understand what it means to vary inversely, directly, jointly, then to find the constant of variation.


29) The graph of f(x)=1/x^2-c has a vertical asymptote at x = 3. Find c

The vertical asymptote is the place where the denomiantor is zero. So here, x^2 - c = 0 when x = 3. So c = 9.

It's good that you want to understand how this is done, not just get an answer. So I hope I've explained how this is done; if not, maybe what I've said together with what others may say may help you to understand the concepts behind the answers.

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27) y=k/x^(1/3)
12 = k/2
k = 24
y = 24/1 = 24

28) y = kxz^(1/3)
120 = k*3*2
k =20
y=20*4*3 = 240

29) When x = 3 denominator = 0
9-c = 0
c = 9
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