(12)(33x)(92-12x) (12)(33x)(92)-(12)(33x)(12x) 36432x-4752x2 if this restatement of the grouping in the original problem is correct.
3x+21x+9x=33x. Break it apart. The factor of x is x and the factors of 33 are 3 and 11. Therefore, the factors of 3x+21x+9x are 3, 11, and x.
2x + 33x - 54 35x - 54 We can not find x with the information given. You need to equate somethingto your equation. If your equation equals 0 then... 0=35x-54 35x=54 x=1.543 IF your equation equals x then... x=35x-54 -34x=-54 x=1.588
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33x
5x squared plus 33x plus 18 = (5x + 3)(x + 6) x = -6, -3/5
(12)(33x)(92-12x) (12)(33x)(92)-(12)(33x)(12x) 36432x-4752x2 if this restatement of the grouping in the original problem is correct.
33x + 9 = 18 ----> 33x = 9 ----> x = 3/11
3x2 + 33x + 54 =3(x2 + 11x + 18) = 3(x + 9)(x + 2)
3x+21x+9x=33x. Break it apart. The factor of x is x and the factors of 33 are 3 and 11. Therefore, the factors of 3x+21x+9x are 3, 11, and x.
9x + 2 + 24x + c = 33x + 2 + c If = 0 then c = -33x - 2 if = 1 then c = -33x - 1 if = 4 then c = -33x + 2 in general, c = -33x -2 + n2 where n is any integer.
21x + 12x -2 = 0 33x -2 = 0 33x = 2 x = 2/33 x = 0.06
25x+1=33x+7 25x=33x+6 -8x=6 x=-8/6 x=-4/3
I'm assuming that your solving for X. 21x + 12x - 2 = 0 33x - 2 = 0 33x = 2 x = 2/33 x =.06
33X Around the Sun - 2005 is rated/received certificates of: UK:15
7(x^2)+33x-10 This is the way I would approach this problem. Find the factors of the first and last number: Factors of 7: 1,7 Factors of -10: -10,-5,-2,-1,1,2,5,10 Set up the two quantities: (x +/- __)(7x +/- __ ) Figure out which factors of -10 when multiplied with the factors of 7 then added together will equal +33: 5(7)+(-2)=33 Therefore: (x+5)(7x-2) Double check answer by foiling: (x+5)(7x-2)=7(x^2)-2x+35x-10 7(x^2)+33x-10