MATH QUETION ASKED TO BE WORKED - correct me if im wrong
Nico-T said
14:52 03/12 2009
question 1
18x² + 4 - [6 (x² - 2 ) + 5 ]
18x² + 4 - [6x² - 12 + 5 ]- work from inside out to remove braket
18x² + 4 - 6x² +12 - 5 - the minus sing in front the bracket changes the signs in the bracket
18x² - 6x² + 4 +12 - 5 - then u group the common ones
12x² + 11- then u add and subtract common factors
question 2
(3x - 4) (2x + 5)
the first step is to times each factor in the first bracket by the two factors in the next braacket but do it one at a time. this will remove the brackets.
6x² + 15x - 8x - 20 - then u look if there is anything common and work them out
6x² + 7x - 20- the 15x and the 8x was common u we apply the addition rule to get the final ansa...but it can be simplified more using A,B,C, method.
18x² + 4 - [6 (x² - 2 ) + 5 ]
18x² + 4 - [6x² - 12 + 5 ]- work from inside out to remove braket
18x² + 4 - 6x² +12 - 5 - the minus sing in front the bracket changes the signs in the bracket
18x² - 6x² + 4 +12 - 5 - then u group the common ones
12x² + 11- then u add and subtract common factors
question 2(3x - 4) (2x + 5)
the first step is to times each factor in the first bracket by the two factors in the next braacket but do it one at a time. this will remove the brackets.
6x² + 15x - 8x - 20 - then u look if there is anything common and work them out
6x² + 7x - 20- the 15x and the 8x was common u we apply the addition rule to get the final ansa...but it can be simplified more using A,B,C, method.
-- Edited by Nico_T at 16:37, 2009-03-12
-- Edited by tajge at 16:02, 2009-03-12
refreashing!!!!