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2y + 2xy' + 4y + 4xy' = 0 6y + 6xy' = 0 y' = -y/x
Imagine that it is a square of a binomial.It would have to be of the form (X+Y)^2, where Y is some real number.Using the FOIL method, expanding the square should get you...X^2+XY+YX+Y^2, which you can simplify to X^2+2YX+Y^2.So, X^2+50X+100 = X^2+2YX+Y^2Y^2 must equal 100, and 2Y must equal 50.But, there's no value for Y like that, is there?
Assume we want to graph -2yx = 8. Divide both sides by -2x, so we have: y = -4/x This equation gives two hyperbolas, one in Quad. II and another in Quad. IV. You can graph the function by substituting each x value for the expression to determine the full coordinates of the points. Finally, plot them on the graph and connect them with a line.
If you mean: 2y = x-4 then y = 0.5x-2 whereas the slope is 0.5 or a 1/2 and the y intercept is -2
76 plus 54 plus 92 plus 88 plus 76 plus 88 plus 75 plus 93 plus 92 plus 68 plus 88 plus 76 plus 76 plus 88 plus 80 plus 70 plus 88plus 72 equal 1,440