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Q: What is the solution set for the equation y2 plus 14y equals 0?

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It is not possible to solve a single linear equation in two variables.

2y-7 = 14y+29 2y = 14y + 36 -12y = 36 -y = 3 y = -3

x = -2 and y = 0

It equals 3 more than (14 times y)

If: 14y = 98 Then: y = 7

6x+14y+8x-18y = 14x-4y

The center of the circle is at (9, 7) on the Cartesian plane

If: 6+17y = 15+14y then 17y-14y = 15-6 so 3y = 9 and then y = 3

The equation 7 x + 14 y + 10 z = 98001 has infinite solutions. For any x and y we can have a z which satisfies this equation. For example, if we choose x=1 and y=1 then z=9798 satisfies. To have a specific solution there must be three equations for three unknowns (x,y,z), which can be solved simultaneously.

Without an equality sign and not knowing the plus or minus values of some of the given terms it can't be considered to be an equation.

14y - 2x = 28 14y = 2x + 28 y = x/7 + 2 Therefore, the slope is 1/7.

Equation: x^2 + y^2 +10x -14y +10 = 0 Completing the squares: (x+5)^2 + (y-7) -25-49 +10 = 0 So: (x+5)^2 + (y-7) = 64 Therfore the circle's centre is at (-5, 7) and its radius is 8

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