Q: Can you solve 6x2 33x 15?

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6x2=-7

It is: 6x2

solve -3x-3y=-15

6x2+13x-5 = (2x+5)(3x-1) when factored

6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y) 6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y) 6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y) 6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y)

Related questions

3/(x2+4x-1) = 6/(2x2-3x+5) Cross - multiply in order to eliminate the fractions: 6*(x2+4x-1) = 3*(2x2-3x+5) 6x2+24x-6 = 6x2-9x+15 6x2-6x2+24x+9x = 15+6 33x = 21 x = 21/33 => x = 7/11

x3 6x2-x-30

33X Around the Sun - 2005 is rated/received certificates of: UK:15

6x2=-7

5x squared plus 33x plus 18 = (5x + 3)(x + 6) x = -6, -3/5

-3(2x + 1)(4 + 3x) = -3[8x + 6x2 + 4 + 3x] = -3[6x2 + 11x + 4] = -18x2 - 33x - 12.

33x + 9 = 18 ----> 33x = 9 ----> x = 3/11

80

They are 6x1, 6x2, 6x3 and so on until 6x15.They are 6x1, 6x2, 6x3 and so on until 6x15.They are 6x1, 6x2, 6x3 and so on until 6x15.They are 6x1, 6x2, 6x3 and so on until 6x15.

(6 × 2) + 3 = 15

(12)(33x)(92-12x) (12)(33x)(92)-(12)(33x)(12x) 36432x-4752x2 if this restatement of the grouping in the original problem is correct.

6x2 - x - 15 = 0 since (6)(-15) = (-10)(9) and -10 + 9 = -1, then factor as: [(6x + 9)/3][6x - 10)/2] = 0 (2x + 3)(3x - 5) = 0 let each factor be zero: 2x + 3 = 0 or 3x - 5 = 0 solve for x: The solutions are: x = -3/2 and x = 5/3