I am going to post a few questions that I found very

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Manager

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I am going to post a few questions that I found very [#permalink]

14 Nov 2006, 07:25

I am going to post a few questions that I found very interesting (and confusing). They come from a VERY imporant source... Read into that what you will.

Sorry I do not have the actual answers, I'm going off memory.

Answer is this problem simplfied, NOT solved
[(1/3)^2-(1/3)(1/4)]*[(1/3)^2+(1/3)(1/4)]

What is the y intercept:
y=x^2+2^x

Any help solving them would be great.

I'll post more if I can remember more.

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14 Nov 2006, 07:41

Answer is this problem simplfied, NOT solved
[(1/3)^2-(1/3)(1/4)]*[(1/3)^2+(1/3)(1/4)]

i didn't understand to simplify it?

It will be (1/3)^4 - (1/12)^2

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14 Nov 2006, 07:43

for this one What is the y intercept:
y=x^2+2^x
y intercept when x = 0 so we have y=0^2 + 2^0
y= 0 + 1 = 1

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14 Nov 2006, 08:19

1 .. (1/3)^4 -(1/12)^2
2.. y = 0+1 =1 thus y intercept is 1

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14 Nov 2006, 08:40

Thanks, question 2 was answered (I got it wrong). I don't really understand how you come up with your answers. Can anyone explain?

None of your answers to question 1 were possible answers on the test. The answer came under 1 fraction.

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14 Nov 2006, 09:29

If I am reading the problem correctly, it's just a form of (a-b)(a+b), which makes it (a^2-b^2). So 1/3^2 = 1/9, 1/3*1/4 = 1/12, 1/12^2 = 1/144. Then, 1/9-1/144 = 5/48 when simplified. Am I close??
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14 Nov 2006, 09:36

Yeah that's as far as I could get also. Somehow this breaks down even more to ONE fraction.

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14 Nov 2006, 10:17

for the first i got this
1/81 - 1/144 = 7/1296

for second any number in power 0 will be one it's a rule. and 0 in any power will be zero.

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14 Nov 2006, 10:32

1) 7/1296 just order of operations pretty straightforward

{4/36 - 3/36 } * {4/36 + 3/36}

2) 2 to the 0 power is 1, just like anything to the 0 power is, not 0

therefore Y = 1 when X = 0

[#permalink] 14 Nov 2006, 10:32

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I am going to post a few questions that I found very

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I am going to post a few questions that I found very

I am going to post a few questions that I found very

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