Without further context or information about the relationship between x and y, it is impossible to determine the value of y when x is 3. The value of y could be anything depending on the equation or function that defines their relationship. In mathematics, x and y are often related through equations or functions, and without that information, we cannot determine the specific value of y when x is 3.
y=2x-3 y=-x-3
It depends on the relationship between x and y.
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let y= xsqrt(x) -1 y= x^(3/2) -1 ---- since xsqrt(x) is the same as x^(3/2) y' = (3/2) x^(3/2-1) y' = (3/2) x^(1/2) y'' = (3/2) (1/2) x^(1/2-1) y'' =(3/2)(1/2) x^(-1/2) y'' = 3/4x^(1/2) y'' = 3 / 4sqrt(x)
the Y in the sentence X plus Y equals 5 is Y=3 or 2 if the X is 2 or 3
x - y = -3 -y = -x - 3 y = x + 3 y-int. = 3 x-int. = -3
Two integers (X & Y). X+Y=-3, X-Y=-11. x=-11+y --> (x)+y=-3 --> (-11+y)+y=-3 --> y=4 x=-11+4=-7 Hope that helps! Two integers (X & Y). X+Y=-3, X-Y=-11. x=-11+y --> (x)+y=-3 --> (-11+y)+y=-3 --> y=4 x=-11+4=-7 Hope that helps!
There are an infinite number of them. Here are a few: x= -2, y= -5/3 x= -1, y= -4/3 x= 0, y= -1 x= 1, y= -2/3 x= 2, y= -1/3 x= 3, y= 0 x= 12, y= 3
To find the inverse you switch the x and the y and then solve for y. x=2 radical( y + 3) radical(y + 3) = x/2 y+3= (x/2)² y = (x/2)² -3 So the answer is y = (x/2)² -3
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there are 4 possible answers. X= 1 , Y=2 Y=1 , X=2 X=0 , Y-3 Y=0 , X=3
1545(x-y)=3(x²-y²) (1545÷3)(x-y)=(x²-y²) 515(x-y)=(x²-y²) 515=(x²-y²)÷(x-y)
The problem to solve is: xy+x+3y+3 Multiply y and x Multiply the y and x Multiply y and x The y just gets copied along. The x just gets copied along. The answer is yx yx x*y evaluates to yx x*y+x evaluates to yx+x Multiply y and 3 Multiply y and 1 The y just gets copied along. The answer is y y 3*y evaluates to 3y The answer is yx+x+3y x*y+x+3*y evaluates to yx+x+3y The answer is yx+x+3y+3 x*y+x+3*y+3 evaluates to yx+x+3y+3 ---- The final answer isyx+x+3y+3----
y^2 X y^3 = y^(2 + 3) = y^5 You can only do this if the coefficient 'y' is the same for both terms. Remember y^2 = y X y y^3 = y X y X y Hence y^2 X y^3 = y X y X y X y X y = y^5 Similarly for division/subtraction y^3 / y^2 = y^(3 - 2 ) = y^1 = y The power of '1' is trivial and not normally shown. NB You CANNOT do z^2 X y^3 by adding the indices. z^2 X y^3 is (z^2)*(y^3)
X = -Y + 3 add Y to each side Y + X = 3 subtract X from each side Y = -X + 3
x+y=3 y=3-2x so x+3-2x=3 so x=0 0+y=3 so y=3