" (i) "ab(x+y)+bx(x+y)

The ratio of bases of two triangles is x:y and that of their areas is a:b. Then the ratio of their corresponding altitudes will be (a) ax:by (b) (a)/(x):(b)/(y) (c) ay:bx( d) (x)/(a):(x)/(y)

(bx ) /(a) + ( ay ) /( b) = a ^(2) + b ^(2) , x + y = 2ab

If x^(y)=y^(x), then ((x)/(y))^(x/y) is equal to a.x^(y/x) b.x^((x)/(y)-1) c.1d.x^((x)/(y))

Solve for x and y:x^(2)-y^(2)=xy-ab,(x+y)(ax+by)=2ab(a+b)

Plot y = abs x, y = abs (x-2) and y = (x +2)

Solve :a(x+y)+b(x-y)=a^(2)-ab+b^(2)a(x+y)-b(x-y)=a^(2)+ab+b^(2)

The solution of the differential equation, dy/dx=(x-y)^(2) , when y(1)=1, is (A) log_(e) abs((2-y)/(2-x))=2(y-1) (B) -log_(e)abs((1+x-y)/(1-x+y))=x+y-2 (C) log_(e) abs((2-x)/(2-y))=x-y (D) -log_(e) abs((1-x+y)/(1+x-y))=2(x-1)