12y+2=146 12y=146-2 12y=144 144/12 12x12=144 y=12
(75y - 12x^2) * y = 75y^2 -12x^2y 75y - (12x^2 * y) = 75y -12x^2y
y = x - 4 × 2 + 12 y = x - 8 + 12 y = x - 20 That is the simplest form of the equation, as you can only solve completely by having only 1 variable to solve.
y to the 36 power
If you mean: y = -8 and y = -2x-12 then the solution is x = -2 and y = -8
12y+2=146 12y=146-2 12y=144 144/12 12x12=144 y=12
x+y=16; y=16-xx^2 + y^2=146; y=±√(146-x^2) So, 16-x=±√(146-x^2)(16-x)^2=146-x^2256-32x+x^2=146-x^22x^2-32x+110=0x^2-16x+55=0Quadratic formula:x=[16±√(256-220)]/2x=(16±6)/2x=8±3x=11 or x=5. Plug these values back into the original equations, and you will see that y also equals 11 or 5. So, the two pairs of numbers that satisfy the original equations are: x=11, y=5; and x=5, y=11
I assume you mean 2 times y times y times y. Since y is multiplied by itself three times, you have 2 times y3, also known as 2 times y cubed, which is written as 2y3.
It is the inequality 12 < y/2.
(75y - 12x^2) * y = 75y^2 -12x^2y 75y - (12x^2 * y) = 75y -12x^2y
y = x - 4 × 2 + 12 y = x - 8 + 12 y = x - 20 That is the simplest form of the equation, as you can only solve completely by having only 1 variable to solve.
The equation is y + 2 = 12.
y = 12. It's the transitive property. y = 12 = 3x.
Let the tens digit be x, and ones digit be y. So we have: x + y = 12 x = y + 2 Substitute y + 2 for x to the first equation: x + y = 12 (y + 2) + y = 12 y + y + 2 - 2 = 12 - 2 2y = 10 2y/2 = 10/2 y = 5 the ones digit y + 2 = 5 + 2 = 7 the tens digit Thus, the number is 75.
18
y to the 36 power
If you mean: y = -8 and y = -2x-12 then the solution is x = -2 and y = -8