Question

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

Answer 1

Define 'solve'. I don't see and equals sign :( .

Answer 2

And so you mean:

\[(2m)^2 + (2m)^2 \text{or } 2m^2 + 2m^2 \]

Answer 3

yes

Answer 4

Yes? They are different.
\[(2m)^2 = 4m^2 \not= 2m^2 \]

Answer 5

no, i need to simlify it to where i get an answer like 4m squared or something like that

Answer 6

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.

Answer 7

a

Answer 8

And what do you suppose the answer is?

If I told you that \(m^2 = apple\), then you'd have

\[2m^2 + 2m^2 = 2apple + 2apple = ?\]

Answer 9

Thank God for your patience - I'm out.

Answer 10

4apple, so do you also add the exponets together?

Answer 11

Hehe, no worries Newton. Go have a pint. =)

Answer 12

No.

Answer 13

Don't add the exponents.

Answer 14

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.

Answer 15

ok

Answer 16

But now since apple is \(m^2\) you have \(4(apple) = 4m^2\)

Answer 17

Oh!! :) ok, but what if it was m to the 4 + m to the 3

Answer 18

Then you cannot combine them.

Answer 19

But you can factor out a cubed m.

Answer 20

i thought it would be m to the 7

Answer 21

\[m^4 + m^3 = (m*m*m*m) + (m*m*m)\]

Answer 22

ok

Answer 23

Imagine m = 2.

\[2^4 + 2^3 = 16 + 8 = 24\]
\[2^7 = 128\]
So no, \(m^4+m^3\ne m^7\)

Answer 24

no, 2 t the 4 power is 16, 2 to the 3 power is 8

Answer 25

But.
\[2^4 + 2^3 = 24\]\[2^3(2 + 1) = 8(3) = 24\]

Answer 26

It's the same either way. I just switched the order.

Answer 27

Point is 16 + 8 is not 128

Answer 28

well, thaks anyways. i gotta go :(