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. 
Maths Help!

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iokthemonkey 4,664 posts
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stephenb 2,677 posts
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iokthemonkey 4,664 posts
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Registered 6 years agostephenb 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 2,381 posts
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Registered 10 years agoDo they not give kids books these days?
Try this. 
y^0 = 1 should be your first hint. 
MrCarrot 3,516 posts
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Registered 9 years agoy^0 is 1
So I think you end up with 6y^2 
SaucyGeoff 2,480 posts
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Registered 4 years agoAnything to the power 0 equals 1.
Edit: what they said. 
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 4,271 posts
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Registered 4 years agoI lol'ing for a reason
But the right answer is:
y^0 is 1
So I think you end up with 6y^2 
heyyo 14,374 posts
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Registered 8 years ago3y^2 x 4y^2
= 12y^4 
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)

richardtock wrote:
(10/5=2)
YOU'RE GOING TOO FAST 
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 4,664 posts
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Registered 6 years agoOkay, so what about if it's something like
3ab * 2b^2?

iokthemonkey 4,664 posts
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Registered 6 years agoOh 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 14,374 posts
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Registered 8 years agoiokthemonkey 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 
*cough*
Split it to:
3a^1*3b^1*2b^2
And go from there.

dsmx 7,412 posts
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Registered 9 years ago3ab*2b^2
= 3*a*b*2*b^2
=3*a*b*2*b*b
= 6ab^3
for those that need a bit of explaining."If we hit that bullseye the rest of the dominoes will fall like a a house of cards, checkmate." Zapp Brannigan

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 4,664 posts
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Registered 6 years agoRight, I think I've got it  cheers for the help, fellas! 
I'm just glad that none of this has any practical application in any industry any more, including engineering (luckily for me). 
kentmonkey 20,076 posts
Seen 8 hours ago
Registered 9 years agoIt looks like a page of text where people have fallen asleep on their keyboards. 
dsmx 7,412 posts
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Registered 9 years agoI 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. "If we hit that bullseye the rest of the dominoes will fall like a a house of cards, checkmate." Zapp Brannigan

Zomoniac 7,402 posts
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Registered 10 years agoRight 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^n1 + 5r^n2 + 5r^n3 + 5r^n4 ... + 5r + 5. 
Chopsen 15,176 posts
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Registered 9 years ago5r^n1 + 5(r^n  1)/(r1)
assuming the 10 really is supposed to be a 10, and not another 5.
I think. 
Zomoniac 7,402 posts
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Registered 10 years agoThanks 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 15,176 posts
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Registered 9 years agoI 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^ninfinity, and then you get to 5r^1 and then 5r^0, which doesn't make sense as that doesn't follow. 
Zomoniac 7,402 posts
Seen 9 hours ago
Registered 10 years agoSorry, 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^n1 (so 49) + 5r^48 + 5r^47 + 5r^46 ... +5r^3 + 5r^2 + 5r + 5. 
Chopsen 15,176 posts
Seen 7 hours ago
Registered 9 years agoa better way of writing would be to go in the other direction.
so
xn = 5r^0 + 5r^1 + 5r^2 ..... 5r^n2 + 10r^n1
is that what you're trying to say? 
That would be exactly what I'm trying to say! Feel stupid now. 
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