[Fyer Werks]

Originally Posted by tonyC4L

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.

[tonyC4L]

Originally Posted by Fyer Werks

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.

[DJWildefire]

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.

[DJWildefire]

Originally Posted by tonyC4L

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.

[tonyC4L]

Originally Posted by DJWildefire

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

[DJWildefire]

Originally Posted by tonyC4L

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.

[musicfan9795]

Got an Algebra 2/Pre-Calc question:
1. What is the inverse of the function "f(x)=-3/2x - 3"

[fishguts182]

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.

[DJWildefire]

Originally Posted by musicfan9795

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)

Originally Posted by fishguts182

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.

[SpyKi]

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.