ANSWERED!

LOCK IF YOU WANT TO

 

 

 

 

Paying 500k for an explanation for the following:

 

 

The problem and the answer: (I think I know it but I just want to clarify)

Simplified ->

 

Defined for all real numbers except (a≠0), (b≠0), (c≠0).

My question is, why exactly are those the solutions?

 

Post below please.

Edited October 27, 201510 yr by shai

21 & 28 can be / 7

 

giving you 3/4

 

not too sure on the b variable..maybe a typo? rip my failure at math skillz

 

'a' stays on numerator: think of it as 'a^3 - a^2 =  just a^1 or a

 

the c cancels out

Edited October 27, 201510 yr by TheObserver

not really sure what you're trying to ask here... from what you've said i'm guessing that it cant be 0 for any of them because a, b and c are on the bottom of the fraction in the original equation and you cannot divide by 0

 

can you maybe post the full quesiton?

21 & 28 can be / 7

 

giving you 3/4

 

not too sure on the b variable..maybe a typo? rip my failure at math skillz

 

'a' stays on numerator: think of it as 'a^3 - a^2 =  just a^1 or a

 

the c cancels out

I was referring to this: "Defined for all real numbers except (a≠0), (b≠0), (c≠0)."

I was referring to this: "Defined for all real numbers except (a≠0), (b≠0), (c≠0)."

oops * walks away *

Observer almost has it.

 

To simplify, you have to do to the top the same as you do to the bottom

 

Divide by 7A^2BC^6, leaves you with the equation, except it should be 3a/4b^3 if my calcs are correct

 

not really sure what you're trying to ask here... from what you've said i'm guessing that it cant be 0 for any of them because a, b and c are on the bottom of the fraction in the original equation and you cannot divide by 0

 

can you maybe post the full quesiton?

You actually answered it, I was thinking that it was "a, b and c are on the bottom of the fraction in the original equation and you cannot divide by 0"  But apparently my teacher made a typo so I thought otherwise. Thanks, pm me your ign if you wan't the money.

Who doesn't like a bit of math eh ;)

Observer almost has it.

 

To simplify, you have to do to the top the same as you do to the bottom

 

Divide by 7A^2BC^6, leaves you with the equation, except it should be 3a/4b^3 if my calcs are correct

You're right, I did make a typo on the 4b^3, thanks. But you missed the actual question at the bottom, like Observer lol.

You're right, I did make a typo on the 4b^3, thanks. But you missed the actual question at the bottom, like Observer lol.

 

Haha yeah i noticed after :P

You should read rules of Exponentiation.

 

Example:

 

a^x * a^y = a^(x+y)

 

or

 

a^x / a^y = a^(x-y)

 

21 and 28 can be divided by 7, gives 3 and 4.

a^3 / a^2 = a^(3-2) = a^1 = a

b^1 / b^4 = b^(1-4) = b^-3 = 1/b^3

c^6 / c^6 = 1

 

not too sure on the b variable..maybe a typo? rip my failure at math skillz

 

'a' stays on numerator: think of it as 'a^3 - a^2 =  just a^1 or a

'a^3 - a^2 =  just a^1 or a ??? 

 

Come on man...

You should read rules of Exponentiation.

 

Example:

 

a^x * a^y = a^(x+y)

 

or

 

a^x / a^y = a^(x-y)

 

21 and 28 can be divided by 7, gives 3 and 4.

a^3 / a^2 = a^(3-2) = a^1 = a

b^1 / b^4 = b^(1-4) = b^-3 = 1/b^3

c^6 / c^6 = 1

 

'a^3 - a^2 =  just a^1 or a ??? 

 

Come on man...

 

Woody fell into the same hole as me and observer :P

 

Woody fell into the same hole as me and observer

 

what hole?

what hole?

 

we all answered the wrong question, he had already simplified it (although got the b power wrong)

 

He was asking why a/b/c cannot be 0, and its because it'll make the bottom line of the equation 0, and you can't divide by 0.

 

Apaec got it spot on.

we all answered the wrong question, he had already simplified it (although got the b power wrong)

 

He was asking why a/b/c cannot be 0, and its because it'll make the bottom line of the equation 0, and you can't divide by 0.

 

Apaec got it spot on.

Oh damn it.

 

All I saw was the equation and I thought he needed some help with simplifying it.   on me aswell