anonymous
  • anonymous
ok, i got a really bad grade. so how do you solve 2msquared + 2msquared?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Define 'solve'. I don't see and equals sign :( .
anonymous
  • anonymous
And so you mean: \[(2m)^2 + (2m)^2 \text{or } 2m^2 + 2m^2 \]
anonymous
  • anonymous
yes

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anonymous
  • anonymous
Yes? They are different. \[(2m)^2 = 4m^2 \not= 2m^2 \]
anonymous
  • anonymous
no, i need to simlify it to where i get an answer like 4m squared or something like that
anonymous
  • anonymous
Right, 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
  • anonymous
a
anonymous
  • anonymous
And 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
  • anonymous
Thank God for your patience - I'm out.
anonymous
  • anonymous
4apple, so do you also add the exponets together?
anonymous
  • anonymous
Hehe, no worries Newton. Go have a pint. =)
anonymous
  • anonymous
No.
anonymous
  • anonymous
Don't add the exponents.
anonymous
  • anonymous
2(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
  • anonymous
ok
anonymous
  • anonymous
But now since apple is \(m^2\) you have \(4(apple) = 4m^2\)
anonymous
  • anonymous
Oh!! :) ok, but what if it was m to the 4 + m to the 3
anonymous
  • anonymous
Then you cannot combine them.
anonymous
  • anonymous
But you can factor out a cubed m.
anonymous
  • anonymous
i thought it would be m to the 7
anonymous
  • anonymous
\[m^4 + m^3 = (m*m*m*m) + (m*m*m)\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
Imagine m = 2. \[2^4 + 2^3 = 16 + 8 = 24\] \[2^7 = 128\] So no, \(m^4+m^3\ne m^7\)
anonymous
  • anonymous
no, 2 t the 4 power is 16, 2 to the 3 power is 8
anonymous
  • anonymous
But. \[2^4 + 2^3 = 24\]\[2^3(2 + 1) = 8(3) = 24\]
anonymous
  • anonymous
It's the same either way. I just switched the order.
anonymous
  • anonymous
Point is 16 + 8 is not 128
anonymous
  • anonymous
well, thaks anyways. i gotta go :(

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