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anonymous
 5 years ago
ok, i got a really bad grade. so how do you solve 2msquared + 2msquared?
anonymous
 5 years ago
ok, i got a really bad grade. so how do you solve 2msquared + 2msquared?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Define 'solve'. I don't see and equals sign :( .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And so you mean: \[(2m)^2 + (2m)^2 \text{or } 2m^2 + 2m^2 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes? They are different. \[(2m)^2 = 4m^2 \not= 2m^2 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, i need to simlify it to where i get an answer like 4m squared or something like that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right, but which one are you simplifying is what Newton is asking Is your equation: a) \(2m^2 + 2m^2\) b) \((2m)^2 + (2m)^2\) c) none of the above.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And what do you suppose the answer is? If I told you that \(m^2 = apple\), then you'd have \[2m^2 + 2m^2 = 2apple + 2apple = ?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank God for your patience  I'm out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04apple, so do you also add the exponets together?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hehe, no worries Newton. Go have a pint. =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don't add the exponents.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02(apple) + 2(apple) = 4(apple) \(\ne 4apple^2\) If you have 2 apples and I give you 2 more, you don't get square apples.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But now since apple is \(m^2\) you have \(4(apple) = 4m^2\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh!! :) ok, but what if it was m to the 4 + m to the 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then you cannot combine them.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But you can factor out a cubed m.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i thought it would be m to the 7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[m^4 + m^3 = (m*m*m*m) + (m*m*m)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Imagine m = 2. \[2^4 + 2^3 = 16 + 8 = 24\] \[2^7 = 128\] So no, \(m^4+m^3\ne m^7\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, 2 t the 4 power is 16, 2 to the 3 power is 8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But. \[2^4 + 2^3 = 24\]\[2^3(2 + 1) = 8(3) = 24\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's the same either way. I just switched the order.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Point is 16 + 8 is not 128

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, thaks anyways. i gotta go :(
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