# Algebra

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

17:10 Sun 30th Oct 2005

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 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. 17:52 Sun 30th Oct 2005 kc(squared) + 5 - c(squared)k + ck = ? + ?kckc + 5 - cck + ck = ? + ? 18:19 Sun 30th Oct 2005 Question Author Thanks fo3nix, but I thought kc(squared) and c(squared)k were different so couldn't be added togetehr, subtracted etc 18:31 Sun 30th Oct 2005 5 + 2 = 7 ? 18:43 Sun 30th Oct 2005 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). 19:36 Sun 30th Oct 2005 Question Author the question was asked as kc� + 5 - c�k + ck = _ + _ (thanks for the tip on alt + 0178 Stevie 21!) 19:56 Sun 30th Oct 2005 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 20:50 Sun 30th Oct 2005 Question Author thanks all 09:29 Mon 31st Oct 2005 so I was right :P 10:26 Mon 31st Oct 2005 I take it that was not a ? bn as U took the ? @ {:~)value, good choice! 13:03 Tue 01st Nov 2005

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