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GR8677 #75
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Alternate Solutions |
wittensdog
2009-07-26 14:02:16 | In terms of simplifying the radical expression, there's a method I found somewhat useful. If you write out 18 in terms of prime factors, you get 2*3*3. It is then easy to write the square of 18 in terms of prime factors, since you just double the number of each factor, and you have 2*2*3*3*3*3. If you call one of the 3's a 2, which is a somewhat reasonable approximation, then you have 3 3's and 3 2's, resulting in (2*2*2)*(3*3*3). Clearly, taking the cube root of that results in 2*3 = 6. It's not super exact, but certainly close enough to get the right answer. |  |
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Comments |
wittensdog
2009-07-26 14:02:16 | In terms of simplifying the radical expression, there's a method I found somewhat useful. If you write out 18 in terms of prime factors, you get 2*3*3. It is then easy to write the square of 18 in terms of prime factors, since you just double the number of each factor, and you have 2*2*3*3*3*3. If you call one of the 3's a 2, which is a somewhat reasonable approximation, then you have 3 3's and 3 2's, resulting in (2*2*2)*(3*3*3). Clearly, taking the cube root of that results in 2*3 = 6. It's not super exact, but certainly close enough to get the right answer.
wittensdog
2009-10-07 17:55:00 |
To convince yourself that you shouldn't be worried about the error, even without going and finding the exact values to compare, notice that you're introducing an error on the order of (3/2) ^ 1/3, by flipping a 3 to a 2 underneath the radical.
Now, just by looking at the expression,
(1.5) ^ (1/3)
or the cube root of 1.5, you can tell that it's certainly more than one and less than 1.5, probably closer to the one side (the actual result is something around 1.145, but pretend you don't know that). Now, for sure, this at least tells us we're not introducing anything like a factor of two in error, and looking at the answer choices and how much they're spaced out, we know it's definitely not 3 or 18, since going from a value of 6 to one of those would mean at least a factor of two of error. So you can definitely feel safe about 7.
I know when I type it all out it seems sort of long, but I think being able to approximate like this, and also being aware of how good your approximation is, is something that would be good to be able to do quickly without thinking too much. I mean seriously, how else are you going to calculate something like 18^(2/3) ?? Anyone who thinks they're going to approach it by doing out 18^2 to find 324, and then trying to take the cube root of that, is not going to get far on the test. Even ball-parking it as 400 and trying to take the cube root isn't exactly a walk in the park.
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| | herrphysik
2006-10-30 00:32:29 | No need to solve for a proportionality factor, just use the form ^2=(s_1/s_2)^3) . Same thing. Also, the exact result I get is  , so I suppose it would actually be  and not  due to the gravitational force being less at the higher orbit?
tin2019
2007-10-29 10:52:42 |
The exact result is 6.868
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