anonymous
  • anonymous
ok, i got a really bad grade. so how do you solve 2msquared + 2msquared?
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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 :(

Looking for something else?

Not the answer you are looking for? Search for more explanations.