# Maths Help!

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• iokthemonkey 1 Sep 2010 23:41:33 4,662 posts
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 My kid is doing some maths and hasn't a clue how to do it (and given me and my wife left school 20+ years ago, neither do we.) Can somebody explain how you simplify expressions like: 2y^2 * 3y^0 I understand if the base number is the same, you simply add the exponents, so 3^2 * 3^4 = 3^6, but what do you do when the base values are different? Any help is appreciated.
• stephenb 1 Sep 2010 23:44:28 3,367 posts
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 At twenty to midnight?
• iokthemonkey 1 Sep 2010 23:46:06 4,662 posts
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 stephenb wrote: At twenty to midnight? I'm in the US, hence the reason I haven't a fucking clue how they do their crazy moon ''math'' here...
• mwtb 1 Sep 2010 23:47:06 2,381 posts
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 Do they not give kids books these days? Try this.
• Deleted user 1 September 2010 23:48:01
 y^0 = 1 should be your first hint.
• MrCarrot 1 Sep 2010 23:48:52 3,524 posts
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 y^0 is 1 So I think you end up with 6y^2
• Saucy 1 Sep 2010 23:49:57 2,739 posts
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 Anything to the power 0 equals 1. Edit: what they said.
• Deleted user 1 September 2010 23:50:15
 When multiplying, add the powers When dividing, subtract them. Your expression becomes 6y^2 if it was 2y^2 * 3y^2 it would be 6y^4 (as 2 +2, the top powers = 4)
• Rauha 1 Sep 2010 23:51:23 4,271 posts
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 I lol'ing for a reason But the right answer is: y^0 is 1 So I think you end up with 6y^2
• heyyo 1 Sep 2010 23:52:26 14,356 posts
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 3y^2 x 4y^2 = 12y^4
• Deleted user 1 September 2010 23:52:59
 In fact lets do some examples y^2 * y^3 = y^5 4y^2 * 5y^3 = 20y^5 y^3 / y^2 = y^1 or just y 10y^3 / 5y^2 = 2y^1 or 2y (10/5=2)
• Deleted user 1 September 2010 23:54:14
 richardtock wrote: (10/5=2) YOU'RE GOING TOO FAST
• Deleted user 1 September 2010 23:55:21
 I just wanted him to know where it came from. Obviously he knows 10/5 is 2 Also: Where do you live? Term hasn't started in most places yet!
• iokthemonkey 1 Sep 2010 23:58:37 4,662 posts
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 Okay, so what about if it's something like 3ab * 2b^2?
• iokthemonkey 1 Sep 2010 23:59:05 4,662 posts
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 Oh and to explain: I'm from the UK originally, I now live in the US, hence the odd maths and odder hour of posting...
• heyyo 2 Sep 2010 00:00:02 14,356 posts
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 iokthemonkey wrote: Okay, so what about if it's something like 3ab * 2b^2? = 6ab^3 You can only multiply like terms with like terms so basically there, you do 3 x 2 = 6 b x b^2 or ( b x b x b ) = b^3 there's no other a terms so you leave it = a
• Deleted user 2 September 2010 00:04:26
 *cough* Split it to: 3a^1*3b^1*2b^2 And go from there.
• dsmx 2 Sep 2010 00:05:03 8,179 posts
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 3ab*2b^2 = 3*a*b*2*b^2 =3*a*b*2*b*b = 6ab^3 for those that need a bit of explaining.
• Deleted user 2 September 2010 00:07:12
 If you're stuck, write long hand. So 3ab x 2b^2 becomes 3 x a x b x 2 x b^2 Collect the numbers 6 x a x b x b^2 b is the same as b^1 So b x b^2 could be written as b^1 x b^2 (we know how to do this. We collect the top numbers to make b^3) So now we have 6 x a x b^3 Get rid of those annoying x symbols and you get 6ab^3
• iokthemonkey 2 Sep 2010 00:07:56 4,662 posts
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 Right, I think I've got it - cheers for the help, fellas!
• Deleted user 2 September 2010 00:10:20
 I'm just glad that none of this has any practical application in any industry any more, including engineering (luckily for me).
• kentmonkey 2 Sep 2010 00:12:04 22,380 posts
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 It looks like a page of text where people have fallen asleep on their keyboards.
• dsmx 2 Sep 2010 00:15:41 8,179 posts
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 I find this kind of maths piss easy there's no complex rules to remember and there's no transform rules that need to be remembered so long as you know what the symbols actually mean your fine.
• Zomoniac 20 Feb 2011 21:03:54 9,472 posts
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 Right you lovely lot. I'm doing remedial maths for remedials and I'm such a remedial that I'm stuck. Can someone please tell me how to simplify this linear recurrence sequence to remove the dots so I can work out for any value without having to enter an endless string: xn = 10r^n-1 + 5r^n-2 + 5r^n-3 + 5r^n-4 ... + 5r + 5.
• chopsen 20 Feb 2011 22:15:18 19,970 posts
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 5r^n-1 + 5(r^n - 1)/(r-1) assuming the 10 really is supposed to be a 10, and not another 5. I think.
• Zomoniac 20 Feb 2011 22:27:32 9,472 posts
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 Thanks for trying, but it's not working. I have X1 = 10, and Xn+1 = 0.6Xn + 5. From that I've got X2 = 11, X3 = 11.6, X4 = 11.96 and X5 = 12.176. Using that I made the sequence in the above post which I've tested and it works (r given for 0.6), but I can't work out how to shorten it. The 10 was meant to be a 10.
• chopsen 20 Feb 2011 22:36:52 19,970 posts
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 I assumed you were talking about the sum of a geometric sequence, as I couldn't make any sense of it otherwise. The ... implies the indicies continue to reduce, until you get to 5r^n-infinity, and then you get to 5r^1 and then 5r^0, which doesn't make sense as that doesn't follow.
• Zomoniac 20 Feb 2011 22:42:22 9,472 posts
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 Sorry, I'm new to this so my terminology and whatnot is probably wrong. The ... was meant to denote "and so on until I run out of numbers". So if n were 50, then: X50 = 10r^n-1 (so 49) + 5r^48 + 5r^47 + 5r^46 ... +5r^3 + 5r^2 + 5r + 5.
• chopsen 20 Feb 2011 22:46:15 19,970 posts
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 a better way of writing would be to go in the other direction. so xn = 5r^0 + 5r^1 + 5r^2 ..... 5r^n-2 + 10r^n-1 is that what you're trying to say?
• Zomoniac 20 Feb 2011 22:49:27 9,472 posts
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 That would be exactly what I'm trying to say! Feel stupid now.
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