Q: Can you solve3 x - y 1?

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4

y = sin(x+y) cos( x + y )[(1 + y')] = y' cos(x + y ) + y'cos(x + y ) = y' y'-y'cos( x+ y) = cos( x + y ) y'[1-cos(x+y)]= cos(x+y) y'= [cos(x+y)]/ [1-cos(x+y)]

y = square root of x y = x(1/2) y' = (1/2)x(1/2 - 1) y' = (1/2)x(-1/2) y' = (1/2)(square root of x)

X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.

(x + y) / (x-1 + y-1) = (x + y) divided by ( 1/x + 1/y)Multiply numerator and denominator by 'xy' :xy (x + y) / (y + x)Then the whole thing falls apart, and we're left with:xy

Related questions

(y * x) - y = y * (x - 1)

x+y=1 x=1-y or y=1-x or if you fill that in 1-y+y=1 1 = 1 or 1-x+x = 1 1 = 1

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If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2

If [ y = x + 2 ], then x is not -1 when y = 5.If [ y = x + 2 ],then when x = -1, y = 1,and when y = 5, x = 3.

The assertion in the question is false. The result of a multiplication depends on the values. Given two numbers X and Y, if X < 0 and if Y < 0 then X*Y is greater than either; if X > 1 and if Y > 1 then X*Y is greater than either; if X < 0 and if Y > 1 then X*Y is smaller than either; if X > 1 and if Y < 0 then X*Y is smaller than either; if 0 < X < 1 and if 0 < Y < 1 then X*Y is smaller than either; If X < 0 and if 0 < Y < 1 then X*Y is greater than X but smaller than Y; If 0 < X < 1 and if Y > 1 then X*Y is greater than X but smaller than Y; If 0 < X < 1 and if Y < 0 then X*Y is smaller than X but greater than Y; If X > 1 and if 0 < Y < 1 then X*Y is smaller than X but greater than Y.

y = sin(x+y) cos( x + y )[(1 + y')] = y' cos(x + y ) + y'cos(x + y ) = y' y'-y'cos( x+ y) = cos( x + y ) y'[1-cos(x+y)]= cos(x+y) y'= [cos(x+y)]/ [1-cos(x+y)]

y = square root of x y = x(1/2) y' = (1/2)x(1/2 - 1) y' = (1/2)x(-1/2) y' = (1/2)(square root of x)

In the equations Y=X-1 and Y=-X+1, the solution is (1,0)

0

X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.

(x-y) + (xy - 1) = (x - 1)(y + 1)