Q: How do you do (3x 4)2?

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3x=(33+42) 3x=75 x=75/3 x=25

(42/3x)*12=24 42/3X=2 42=6X x=7 if it'isn't 3x but x^3 then 21=X^3 21^(1/3)=2.76

y=-2/3x with any y-intercept except -2.

3x + 3x + 3x = 3* (3x) = 9x

to get these to cancel multiply the bottom line by -2 to give 3x-2y=42 -4x+2y=-52 cancel the Y's and you get -x = -10 so x=10 substitute back into the original equation to get 2(10) - y = 26 so rearrange -y = 26- 20 -y = 6 y = -6

Related questions

3x+42=6x+12 42=6x-3x+12 42=3x+12 42-12=3x 30=3x 10=x

3x + 12 = 42 3x = 30 x = 10

3x=(33+42) 3x=75 x=75/3 x=25

If 15 + 3x = 42, then x = 9

42

3x-7+4 = 42 3x-7 = 38 3x = 45 x=15

(42/3x)*12=24 42/3X=2 42=6X x=7 if it'isn't 3x but x^3 then 21=X^3 21^(1/3)=2.76

(27x3 - 64) = [(3x)3 - 43] = (3x - 4)[(3x)2 + (3x)(4) + 42] = (3x - 4)(9x2 + 12x + 16)

39

Algebraically. X = integer. X + (X + 1) + (X + 2) = - 123 3X + 3 = - 123 3X = - 126 X = - 42 X + 1 = - 41 X + 2 = - 40 - 42 + - 42 + - 40 = -123 =======

3x + 9 = 42 3x = 33 x = 11

3x+2y = 0 -9x+y = 42 => y = 42+9x Substitute the bottom rearranged equation into the top equation: 3x+2(42+9x) = 0 3x+82+18x = 0 21x = -84 Divide both sides by 21 to find the value of x: x = -4 Substitute the value of x into the original equations to find the value of y: So: x = -4 and y = 6