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01:25 PM on 05/28/12 
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DJWildefire
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Moraga, CA
Male - 19 Years Old
I have a take home test for algebra on things that we've done in the class, this is easy work but we did it all so long ago that I can't for the life of me remember how to do it. I'd really appreciate it if someone could help me by tonight. (x stands for multiplication unless noted)

1. -3 x m x 4 x n x n


2. [ 3a(to the 4th power) t(to the 4th power)] to the 3rd power

3. [2/5] to the 3rd power

4. 2/23 = x/92

5. 10/a = 15/30

6. 2/9 divided by [- 3/27]


Sorry if the way I typed this doesn't make sense. I just don't remember where to start on these problems
Alright.
1. When you multiply together numbers/variables with the same base, you can simply add their exponents. n can be written as n^1 so
n x n is simply n^2. -3 x 4 = -12. I like to arrange expressions like these in terms of decreasing degree, so I'd write out the answer as -12(n^2)m. Parentheses wouldn't be needed when you write out the answer on paper because the superscript makes it apparent that m is not part of the exponent.
2. When you exponentiate an expression that already has been raised to a power, you multiply the exponents together. Thus, when you have [(3a^4)(t^4)]^3 both the a and t will be raised to the 12th power and the 3 is raised to the third power. Therefore, the simplification would be 27(a^12)t^12. However, though you didn't write it this way, if all of 3a was put to the fourth power (e.g. (3a)^4), the answer would be different so I'm hoping you didn't mistype.
3. When you exponentiate a number, you simply multiply the number by itself for as many times as the value of the exponent. So (2/5)^3 = 2/5 x 2/5 x 2/5 = 8/125. Alternatively you could think about this problem similarly to the previous 2. (2/5)^3 = 2^3/5^3 = 8/125.
4. This is just algebra. What do we need to do to get x by itself? Currently x is being divided by 92 so to get x by itself we must multiply both sides of the equation by 92. In numbers, 92 x 2/23 = x/92 x 92. Simplified, 8 = x.
5. This is essentially the same problem as 4 except that the variable is in the denominator now. So for this one, you could either "flip" both sides of the equation (e.g. a/10 = 30/15, then approach the problem the same way as 4) or cross-multiply (e.g. 10 x 30 = 15a and divide by 15 on both sides to get a by itself). Either way, a = 20.
6. When dividing by fractions, it's often easiest to just multiply by the reciprocal. So 2/9 x -27/3 = 2/9 x -9 = -2.
06:43 PM on 05/28/12 
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Fyer Werks
indiemo kid
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This is just a fun little algebraic/geometric problem that I found kind of interesting to do with the A series of paper. A4, A3, you know what I mean.

When you go from a sheet of A(n) paper to a sheet of A(n+1) paper two things happen: The area of the sheet is halved, but the ratio of length to width remains the same. There's only one ratio where this is possible. What is it?
08:07 PM on 05/28/12 
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xhmnimx
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Male - 22 Years Old
Alright.
1. When you multiply together numbers/variables with the same base, you can simply add their exponents. n can be written as n^1 so
n x n is simply n^2. -3 x 4 = -12. I like to arrange expressions like these in terms of decreasing degree, so I'd write out the answer as -12(n^2)m. Parentheses wouldn't be needed when you write out the answer on paper because the superscript makes it apparent that m is not part of the exponent.
2. When you exponentiate an expression that already has been raised to a power, you multiply the exponents together. Thus, when you have [(3a^4)(t^4)]^3 both the a and t will be raised to the 12th power and the 3 is raised to the third power. Therefore, the simplification would be 27(a^12)t^12. However, though you didn't write it this way, if all of 3a was put to the fourth power (e.g. (3a)^4), the answer would be different so I'm hoping you didn't mistype.
3. When you exponentiate a number, you simply multiply the number by itself for as many times as the value of the exponent. So (2/5)^3 = 2/5 x 2/5 x 2/5 = 8/125. Alternatively you could think about this problem similarly to the previous 2. (2/5)^3 = 2^3/5^3 = 8/125.
4. This is just algebra. What do we need to do to get x by itself? Currently x is being divided by 92 so to get x by itself we must multiply both sides of the equation by 92. In numbers, 92 x 2/23 = x/92 x 92. Simplified, 8 = x.
5. This is essentially the same problem as 4 except that the variable is in the denominator now. So for this one, you could either "flip" both sides of the equation (e.g. a/10 = 30/15, then approach the problem the same way as 4) or cross-multiply (e.g. 10 x 30 = 15a and divide by 15 on both sides to get a by itself). Either way, a = 20.
6. When dividing by fractions, it's often easiest to just multiply by the reciprocal. So 2/9 x -27/3 = 2/9 x -9 = -2.
Thanks a lot, I believe you helped me before. I appreciate it! I'll definitely be back in here later this week for some geometry, unfortunately. Hahah. Thanks again!
04:35 PM on 05/29/12 
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tonyC4L
tech-savvy at-risk youth
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Lake Forest, CA
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This is just a fun little algebraic/geometric problem that I found kind of interesting to do with the A series of paper. A4, A3, you know what I mean.

When you go from a sheet of A(n) paper to a sheet of A(n+1) paper two things happen: The area of the sheet is halved, but the ratio of length to width remains the same. There's only one ratio where this is possible. What is it?
Should be sqrt(2). Take a sheet with length 2a and width b, so if you halve the area it creates two sheets of length b and width a. If the ratio of length to width is to be the same, then 2a/b = b/a. Thus 2a^2 = b^2, and b^2/a^2 = 2, so b/a (which we've already identified as a ratio of length to width) is equal to sqrt(2).

That's an interesting fact though; I had no idea that's how the A paper sizes worked. They don't seem to be common in the states as far as I've seen (then again, why would they be? We already go against nice math with other measurement standards, seems fitting that we'd use arbitrarily sized letter and legal paper as well).
04:43 PM on 05/29/12 
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Fyer Werks
indiemo kid
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Should be sqrt(2). Take a sheet with length 2a and width b, so if you halve the area it creates two sheets of length b and width a. If the ratio of length to width is to be the same, then 2a/b = b/a. Thus 2a^2 = b^2, and b^2/a^2 = 2, so b/a (which we've already identified as a ratio of length to width) is equal to sqrt(2).

That's an interesting fact though; I had no idea that's how the A paper sizes worked. They don't seem to be common in the states as far as I've seen (then again, why would they be? We already go against nice math with other measurement standards, seems fitting that we'd use arbitrarily sized letter and legal paper as well).

Bang on, well done. :D The definition of an A0 piece of paper is one with an area of 1m^2 and a length to width ratio of √2:1. A1 is 500cm^2 and so on. It makes it much easier for manufacturers to scale up and down.
04:46 PM on 05/29/12 
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tonyC4L
tech-savvy at-risk youth
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Lake Forest, CA
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Bang on, well done. :D The definition of an A0 piece of paper is one with an area of 1m^2 and a length to width ratio of √2:1. A1 is 500cm^2 and so on. It makes it much easier for manufacturers to scale up and down.
Ahh yeah I imagine so, seems a great advantage of sizing them that way.
08:57 PM on 09/01/12 
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DJWildefire
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Moraga, CA
Male - 19 Years Old
Bump. Might need to start utilizing this thread soon because I'm taking multivariable calc at the community college. Dunno how hard that'll get.
09:08 PM on 09/04/12 
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DJWildefire
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Moraga, CA
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Ahh yeah I imagine so, seems a great advantage of sizing them that way.

Hey! Sorry to bother you, but I have a math problem I'm not sure I'm approaching correctly.

Here it is -- Given point P(1,2,1) and the line whose parametric equations are x=2+t, y=-2+2t, and z=t, find a vector whose beginning point is P and whose terminal point is on the given line so that the magnitude of the vector is the shortest distance to the line.

What I did is found a point (P1) on the line by plugging in 0 for t. then I found vector PP1 to be <1,-4,-1> and crossed that with the directional vector of the line, <1,2,1>. This gave me <-2, -2, 6> which can be reduced to <-1, -1, 3> which I'm hoping is the correct answer?

Thanks for any help you can give.
12:13 AM on 09/05/12 
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tonyC4L
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Hey! Sorry to bother you, but I have a math problem I'm not sure I'm approaching correctly.

Here it is -- Given point P(1,2,1) and the line whose parametric equations are x=2+t, y=-2+2t, and z=t, find a vector whose beginning point is P and whose terminal point is on the given line so that the magnitude of the vector is the shortest distance to the line.

What I did is found a point (P1) on the line by plugging in 0 for t. then I found vector PP1 to be <1,-4,-1> and crossed that with the directional vector of the line, <1,2,1>. This gave me <-2, -2, 6> which can be reduced to <-1, -1, 3> which I'm hoping is the correct answer?

Thanks for any help you can give.
Well let me first apologize for being a little drunk and probably not too helpful but here is this http://mathworld.wolfram.com/Point-L...mensional.html

I know you are a smart cookie though and just giving you a formula is not awesome so I'll try to explain my thought process. You need to find a vector that is perpendicular to the line and passes through P, because then that vectors magnitude when starting from P and ending at the line will be the shortest distance from the point to the line. Easier to see in the picture in that link. So like when you crossed PP1 with <1,2,1> that vector would be perpendicular to both of those other vectors but I think you want it to be in the same plane as PP1 not perpendicular to it

I'm rambling, I'll come back to this in the morning
12:58 AM on 09/05/12 
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DJWildefire
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Moraga, CA
Male - 19 Years Old
Well let me first apologize for being a little drunk and probably not too helpful but here is this http://mathworld.wolfram.com/Point-L...mensional.html

I know you are a smart cookie though and just giving you a formula is not awesome so I'll try to explain my thought process. You need to find a vector that is perpendicular to the line and passes through P, because then that vectors magnitude when starting from P and ending at the line will be the shortest distance from the point to the line. Easier to see in the picture in that link. So like when you crossed PP1 with <1,2,1> that vector would be perpendicular to both of those other vectors but I think you want it to be in the same plane as PP1 not perpendicular to it

I'm rambling, I'll come back to this in the morning

Thanks. Yeah what's screwing me up is having to leave it as a vector. If the question asked for the shortest distance, I wouldn't have a problem, but I can't seem to wrap my head around how to find the vector whose magnitude is the shortest distance.

If the dot product of two vectors equals 0 they are perpendicular though right? And <-1, -1, 3><1, 2, 1>=0. So I dunno.
04:20 PM on 09/05/12 
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musicfan9795
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Got an Algebra 2/Pre-Calc question:
1. What is the inverse of the function "f(x)=-3/2x - 3"
05:17 PM on 09/05/12 
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fishguts182
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got a simple question that I am not seeing when I am working on a proof for an algorithms class

I am not seeing how

n*2^(n+2) + 2= n*2^(n+1) + n*2^(n+1) + 2

I am now seeing how the exponent terms on the right hand side are being simplified to get the left hand side.
06:48 PM on 09/05/12 
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DJWildefire
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Moraga, CA
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Got an Algebra 2/Pre-Calc question:
1. What is the inverse of the function "f(x)=-3/2x - 3"
switch f(x) with x (i.e. x=-3/2f(x) - 3) and solve for f(x)

got a simple question that I am not seeing when I am working on a proof for an algorithms class

I am not seeing how

n*2^(n+2) + 2= n*2^(n+1) + n*2^(n+1) + 2

I am now seeing how the exponent terms on the right hand side are being simplified to get the left hand side.

You need to combine like terms and get rid of excess for this one.
Get rid of the 2's.
Then combine like terms so that n x 2^(n+2) = 2n x 2^(n+1).
Divide by 2n on both sides and think of 2 as 2^1.
12:20 PM on 10/14/12 
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SpyKi
This Charming Man
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Wales
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Let (X1; X2; :::; Xm) be a sequence of L.I. columns in Fnx1 and let A in Fnxn be a
nonsingular matrix. Prove that the sequence (AX1; AX2; :::; AXm) in Fnx1 is also L.I.



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