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Q: Can you solve 6x2 33x 15?

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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

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

x3 6x2-x-30

-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

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

80

(6 × 2) + 3 = 15

(2x + 3)(3x - 5)

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.

6x2 - x You don't have enough information to take this any further or solve it

It will be: 6x2 +8x -15

6x2

(x - 1)(x^2 - 5x + 2)

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.

9+6x2=21

33x

2x(3x+6) = 0 x = 0 or x = -2

solve -3x-3y=-15

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

Yes you have to factor it. (3x^2+5)(2x^2-3)

6x2 + 10x = 2x(3x + 5)

(2x^2 - 5)(4x^3 + 3)