function

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function

by magical cook » Mon Dec 17, 2007 3:04 pm
if f is the function defined for all k by f(k)=k^5/16, what is f(2k) in terms of f(k)?

a. 1/8 f(k)
b. 5/8 f(k)
c. 2 f(k)
d. 10 f(k)
e. 32 f(k)

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by gmatguy16 » Mon Dec 17, 2007 3:11 pm
imo e f(2k)= (2k) ^ 5 / 16 = 32 k ^5/16 = 32 f(k)

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by camitava » Mon Dec 17, 2007 8:33 pm
Agree with gmayguy16. IMO E too.
Correct me If I am wrong


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Amitava

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by magical cook » Mon Dec 17, 2007 9:00 pm
that's the answer - thank you!

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by andes1 » Tue Dec 18, 2007 6:42 pm
gmatguy16 wrote:imo e f(2k)= (2k) ^ 5 / 16 = 32 k ^5/16 = 32 f(k)
i was lost

i tried this:
(2k)^(5/16)

then
{(2^5)(k^5)}^(1/16)
next:
[32k^5]^(1/16)

so i did it until here.... was i doing something wrong?
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by camitava » Tue Dec 18, 2007 9:34 pm
andes1,
U r getting wrong man! The Qs is saying - f(2k)= (2k) ^ 5 / 16 means f(2k)= ((2k) ^ 5) / 16, I think now u can get the idea where u misunderstood the Qs!
Correct me If I am wrong


Regards,

Amitava

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by andes1 » Wed Dec 19, 2007 4:59 pm
camitava wrote:andes1,
U r getting wrong man! The Qs is saying - f(2k)= (2k) ^ 5 / 16 means f(2k)= ((2k) ^ 5) / 16, I think now u can get the idea where u misunderstood the Qs!
haha thanks a lot... was a silly error from my part.
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