5-3[x-7(2x-6)]
please show process step-by-step
thanks in advance
2007-08-27 14:23:28 · 10 answers · asked by ailee 2 in Science & Mathematics ➔ Mathematics
Work from the inside out
5-3[x-7(2x-6)] =
5-3[x-7*2x-7*(-6)] =
5-3[x -14x + 42] =
5-3[-13x + 42] =
5 -3*(-13x) - 3*(42) =
5 + 39x -126 =
39x - 121
2007-08-27 14:29:15 · answer #1 · answered by Steve A 7 · 0⤊ 0⤋
This is not an equation. It is only an expression
I will try to simplify it.
5-3[x-7(2x-6)]
Open the single bracket
5-3[x-14x+42]
Open the square bracket
5-3x+42x-126
Add all like terms
39x-121...Ans.
========
2007-08-27 14:53:46 · answer #2 · answered by Joymash 6 · 0⤊ 0⤋
remember to work first on the innermost parenthesis
5-3[x-7(2x-6)]
5 - 3[x -14x + 42]
5 - 3x + 42x - 126
39x - 121
2007-08-27 14:27:57 · answer #3 · answered by peachi517 2 · 0⤊ 0⤋
5 - 3[x - 7(2x - 6)]
= 5 - 3[x - 14x + 42]
= 5 - 3x + 42x - 126
= 39x - 121
2007-08-27 14:36:01 · answer #4 · answered by tancy2411 4 · 0⤊ 0⤋
5-3[x-7(2x-6)]
okay: here you first need to distribute everything..
5-3[x-14x+42] - distribute 7
5-3[-13x+42] - subtract the x's
5+39x-126 - distribute 3
39x - 121 -- subtract 5 - ans
2007-08-27 14:31:03 · answer #5 · answered by toffer 3 · 0⤊ 0⤋
5- 3[x-7(2x-6)]
5- 3x + 21(2x-6)
5 - 3x + 42x - 126
39x-121
2007-08-27 14:26:38 · answer #6 · answered by way2hot2becool 3 · 0⤊ 0⤋
5 - 3[x-14x+42]
5 - 3x+42x-126
39x-121
2007-08-27 14:39:06 · answer #7 · answered by Anonymous · 0⤊ 0⤋
first remove the internal bracket you get
5-3(x-14x+42)then 5+39x-126=39x-121 is the answer.
2007-08-27 14:30:14 · answer #8 · answered by venkataramanan t 2 · 0⤊ 0⤋
(27x^3)-512=0 (27x^3)=512 x^3=512/27 x=(512/27)^one million/3 x= 8/3 this equation has all 3 zeros occurring on the comparable ingredient on the graph because of the fact the graph curves because it passes contained in direction of the x axis.
2016-10-09 08:48:36 · answer #9 · answered by lorts 4 · 0⤊ 0⤋
use the distributive property
2[x-7(2x-6)]
(2x-14)(2x-6)
4x^2-12x-28x+84
4x^2-40x+84
2007-08-27 14:33:47 · answer #10 · answered by flame_yoshi05 2 · 0⤊ 0⤋