Algebra

[Factfinder]

Please help (and explain) to fill in the blanks:-

kc(squared)+ 5 - c(squared)k + ck = _ + _

[fo3nix]

Not quite sure what you're wanting here; this seems very simple.

kc(squared) and c(squared)k are the same thing. The order of the letters does not matter for this type of algebra. As the first occurance of this has a + sign, and the next one has a - sign, they cancel out to leave 5 + ck.

Thus the answer = 5 + ck.

[mibn2cweus]

kc(squared) + 5 - c(squared)k + ck = ? + ?

kckc + 5 - cck + ck = ? + ?

[Factfinder]

Thanks fo3nix, but I thought kc(squared) and c(squared)k were different so couldn't be added togetehr, subtracted etc

[mibn2cweus]

5 + 2 = 7 ?

[stevie21]

is kc� meant to say that you multiply k and c and then square the result?
(as opposed to c�, THEN multiplied by k)..

(kc)� is not equal to (c�)k

Incidentally, you can get the wee "squared" symbol by holding the "Alt" key and typing 0178 on the keypad (not the numbers above the "qwerty" part of the keyboard).

[Factfinder]

the question was asked as

kc� + 5 - c�k + ck = _ + _

(thanks for the tip on alt + 0178 Stevie 21!)

[mibn2cweus]

kc� + 5 - c�k + ck = _ + _
{(order of operation)}
{k(c�)} + 5 - {(c�)k} + ck = _ + _
. . . same as . . .
kcc + 5 - cck + ck = _ + _
. . . same as . . .
kcc - cck + 5 + ck = _ + _
. . . same as . . .
1 - 1 + 5 + ck = 5 + ck

[Factfinder]

thanks all

[fo3nix]

so I was right :P

[mibn2cweus]

I take it that was not a ? bn as U took the ? @ {:~)value, good choice!