20x(7X+4)=(20X-7X)+4
2006-10-03 07:08:52 · 8 answers · asked by THE MOTHERFUCKING PRINCESS 2 in Science & Mathematics ➔ Mathematics
there all wrong
when you use pemdas you get
140x +80=13x+4
127x=76
x=1.67
2006-10-03 07:32:20 · answer #1 · answered by powder 1 · 0⤊ 1⤋
20X(7X+4)=(20X-7X)+4 simplifies to
140x^2 + 80X = 13X + 4. Subtracting (13x + 4) from both sides gets us
140x^2 +67X - 4 = 0. I don't see an easy way to factor this, use the Quadratic Equation.
Note that many of the earlier answers made the mistakes of having 20X * 7X = 140X, when it should actually = 140X^2; and 20X - 7X = 14X, when it should be 13X.
Of course, it would be clearer if we knew if your first x (next to the 20) is the same as the other X, or if you were trying to indicate multiplication ( * ).
Good luck!
2006-10-03 07:18:47 · answer #2 · answered by Anonymous · 0⤊ 0⤋
20 * ( 7x + 4 ) = ( 20x -7x ) + 4
140x + 80 = 13x + 4
127x = - 76
x = -76 / 127
that only i cud understand ;
vaise is question mein karna kya tha , x ki value find karni thi na
2006-10-03 07:17:26 · answer #3 · answered by pooja 2 · 0⤊ 0⤋
the earlier answer was wrong...
140x + 80 = 14x + 4
126x = -76
x = -0.603174
2006-10-03 07:24:31 · answer #4 · answered by Leonardo 1 · 0⤊ 0⤋
140X + 80 = 13X + 4
140X - 13X = 4 - 80
127X = -76
X = -76/127
X = -0.598
2006-10-03 07:16:30 · answer #5 · answered by Onil N 1 · 0⤊ 0⤋
as far as i can tel the asnwer is X = -76/127 but i'm not 100%
2006-10-03 07:23:20 · answer #6 · answered by Anonymous · 0⤊ 0⤋
140x^2+80x=14x+4
140x^2+66x-4=0
70x^2+33x-2=0
i forgot after that
2006-10-03 07:14:41 · answer #7 · answered by Anonymous · 0⤊ 0⤋
da funk?
2006-10-03 07:16:26 · answer #8 · answered by Paris, je t'aime 5 · 0⤊ 0⤋