If you mean: 3x+2y = 6 and -4x+5y = 15 then it works out as x = 0 and y = 3
It depends very much on what the question is!
I assume you mean......3X - 2Y = 6- 2Y = - 3X + 6Y = (3/2)X - 3=============now, zero out variables in turnY = (3/2)0 - 3Y = - 3===Y intercept = (0, - 3)X intercept = (2, 0)
5 to the tenth power would be 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5, or 9,765,625.
5x5x5x5x5x5=15,625.
x + 2y6 cannot be simplified, nor evaluated, nor solved.
2(x - y)(x + y)(x2 - xy + y2)(x2 + xy + y2)
Given the linear equation 3x - 2y^6 = 0, the x and y intercepts are found by replacing the x and y with 0. This gives the intercepts of x and y where both = 0.
If you mean: 3x+2y = 6 and -4x+5y = 15 then it works out as x = 0 and y = 3
2y6 is an expression not an equation. This is an equation: 2y = 6 To solve that equation you divide each side of the equation by 2: 2y/2= 6/2 y = 3
It depends very much on what the question is!
If the slopes of the lines are negative reciprocals, then the lines are perpendicular.2x - 5y = -3 subtract 2x to both sides-5y = -2x - 3 divide both sides by -5y = (2/5)x +3/55x + 2y = 6 subtract 5x to both sides2y = -5x + 6 divide both sides by 2y = (-5/2)x + 3Yes, the lines are perpendicular.
I assume you mean......3X - 2Y = 6- 2Y = - 3X + 6Y = (3/2)X - 3=============now, zero out variables in turnY = (3/2)0 - 3Y = - 3===Y intercept = (0, - 3)X intercept = (2, 0)
5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 = 78,125
5 to the tenth power would be 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5, or 9,765,625.
5 x 5 x 5 x 5 x 5 = 3,125
5x5x5x5x5x5=15,625.