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Baelor · 07-24-2008, 01:18 AM

Okay, here's the problem. You have y twice, in two places that don't work together (I was looking at this last night).

To say that *x* can be expressed as y^2 implies that *x*=y^2, but also equals y^3/2. That doesn't work because the example provided doesn't satisfy both of those equalities. How about this?

If x can be expressed as y^2, then *x*=y^3/2. This should make the question more reasonable.

*m*=4
y^3/2=4 for some y
y=2 by solving.

So m=y^2=4, which means *4m*=*16*

16=4^2, so *4m* = *16* = 4^3/2 = 64/2 = 32

Is that correct?

07-24-2008, 01:18 AM	#3
Baelor Member Join Date: May 2008 Location: Northwest Posts: 993	Okay, here's the problem. You have y twice, in two places that don't work together (I was looking at this last night). To say that x can be expressed as y^2 implies that x=y^2, but also equals y^3/2. That doesn't work because the example provided doesn't satisfy both of those equalities. How about this? If x can be expressed as y^2, then x=y^3/2. This should make the question more reasonable. m=4 y^3/2=4 for some y y=2 by solving. So m=y^2=4, which means 4m=16 16=4^2, so 4m = 16 = 4^3/2 = 64/2 = 32 Is that correct? Last edited by Baelor; 07-24-2008 at 01:25 AM.