10^(2y)=25," then "10^(-y)=?

If (2.5)^(x)=(0.025)^(y)=10^(3) , then the value of (1)/(x)-(1)/(y) is

If log_(10)x+log_(10)y=2, x-y=15 then :

If log_(10)x+log_(10)y=2, x-y=15 then :

(x/10+y)(x/10-y)=?

The equation to the locus of a point P for which the distance from P to (0,5) is double the distance from P to y -axis is 1) 3x^(2)+y^(2)+10y-25=0 2) 3x^(2)-y^(2)+10y+25=0 3) 3x^(2)-y^(2)+10y-25=0 4) 3x^(2)+y^(2)-10y-25=0

Form the differential equation by eliminating A,B from y=e^(5x) (Ax+B) is (A) (d^(2)y)/dx^(2) +10(dy /dx) +25y=0 (B) (d^(2)y) /dx^(2) -10(dy /dx)+25y=0 (C) (d^(2)y) /dx^(2) +10(dy /dx) -25y=0 (D) (d^(2)y) /dx^(2) -10(dy /dx) -25y=0

{:((10)/(x + y) - (4)/(x - y)= -2),((15)/(x + y) + (7)/(x - y) = 10):}

{:((10)/(x + y) - (4)/(x - y)= -2),((15)/(x + y) + (7)/(x - y) = 10):}