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Q: How many solutions in y2?

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y2 = 169 Square root both sides: y = 13

y2 + y2 = 2y2

X2 + Y2 = 2X = sqrt(2 - Y2)Y can not be greater than 2 or a complex number would be yielded from under the radical. So, - infinity to 2, or something like that for Y's value.Same procedure for X.

y2

No, because there is more than one solution: y2 = x2 y = ±(x2)1/2 y = ±x Because there are multiple solutions for a single value of x, this does not qualify as a function.

To graph y2 = -3x, first solve for y. Doing this results in two solutions: y = √(-3x) and y = -√(-3x) Put the first solution into y1 and the second solution into y2. The two solutions together should form a sideways parabola.

The GCF is y2.

If: 2x+y = 5 and x2-y2 = 3 Then the solutions work out as: (2, 1) and ( 14/3, -13/3)

y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2

The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.

That there are no whole number solutions to the equation: xn + yn = zn when n > 2. If n = 2 this is: x2 + y2 = z2 is known as Pythagoras' Theorem, and has many whole number solutions, eg 32 + 42 = 52, 52 + 122 = 132.

4x-y2 = 2

y2 + 4y -12 = 0 y2 + 6y - 2y -12 = 0 y(y + 6) -2(y +6) = 0 (y -2) x (y + 6) = 0 Hence (y-2) is zero which gives y = 2 or (y + 6) is zero which gives y = -6 The roots or solutions to the equation y2 + 4y -12 = 0 are y = 2 and y = -6

0

It is: y2

We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.

3y2 y x y = y2 3 x y2 = 3y2

y to the power of 5 ---- y2 x y3 = y2+3 = y5

The solutions work out as: x = 52/11, y = 101/11 and x = -2, y = -11

yea me too dude. Mahleko :(

5x2 - 2y2 = -20 7x2 - y2 = 152 Eq1 - 2*Eq2: -9x2 = -324 so that x2 = 36 Substituting the value of x2 in Eq1: 2y2 = 200 so that y2 = 100 The four solutions are (-6, -10), (-6, 10), (6, -10) and (6,10)

88 + 5y - y2 66 - 3y + y2 Subtract: 22 + 8y -2y2

The expression Y-2 is 1/Y2. The reciprocal of 1/Y2 is Y2.

y2=-x2-8x+6

There is no expansion for x2 + y2