2 halves or 1 whole.
x/y = 5/(y^2) and y/x = 5/(x^2). So x/y + y/x = 5/(y^2) + 5/(x^2), which equals 5x^2/(x^2 y^2) + 5y^2/(x^2 y^2) equals 5(x^2 + y^2)/25 equals (x^2 + y^2)/5. x^2 = 25/y^2, so you get (25/y^2 + y^4/y^2)/5 equals ((25 + y^4)/y^2)/25, which shows that your math teacher is on crack. Seriously, I'm not sure that's true.
y=x+2/4 (x+2 is all over four)
y=x+2/4 (x+2 is all over four)
(-2,3) reflected over y = x is (3,-2) (400,-2) reflected over y = x is (-2,400) All you do is switch the ordered pair.
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 you mean y=(x+2)/x, the range is all reals except y=1 If you mean y=x+2/x, the range is (-inf, -2sqrt(2)] U [2sqrt(2),+inf)
Y = (X+4) / 2 is a straight line parallel to Y = 0.5 * X
(9) · (x) · (2/3) · (y) = 6 x y
I understand you want [(2/x) + (1/y)] as a single fraction. The common denominator is (xy): 2/x = (2y)/(xy); and 1/y = x/(xy), so the answer is (2y + x) / (xy)
When transposed and made the subject of the given equation: x = 2/y-3
No, it is not.
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