sqrt(32))^(x)-:2^(y+1)=1;16^(4-x/2)-8^(y)=0

Solve the equation for x and y(32)^(x)-:2^(y+1)=1,16^(4-(x)/(2))-8^(y)=0

Solve the equation for x\ a n d\ y (32)^x-:2^(y+1)=1,\ 16^(4-x/2)-8^y=0

Solve for x and y, if : (sqrt(32))^(x)+2^(y+1)=1 and 8^(y)-16^(4-(x)/(2))=0

Solve for x and y if : (sqrt(32))^(x)÷2^(y+1)= 1 and 8^(y)-16^(4-(x)/(2))=0

Solve for x and y , if : (sqrt( 27) )^(x) ÷ 3^(y+ 4) = 1 and 8^(4-(x)/(3)) - 16^y = 0 .

Find the eccentricity, length of latus rectum, centre, foci, vertices and the equatioin to the dircctriccs of the hyperola. (i) 4x^(2)-5y^(2)-16x+10y+31=0 (ii) 5x^(2)-4y^(2)+20x+8y-4=0 (iii) 4(y+3)^(2)-9(x-2)^(2)=1 (iv) 4x^(2)-9y^(2)-8x-32=0

The value of f(x,y)=((x^(3)y)^((1)/(4))-(y^(3)x)^((1)/(4)))/(sqrt(y)-sqrt(x))+(1+sqrt(xy))/((xy)^((1)/(4)))xx((1+2sqrt((y)/(x))+(y)/(x))^(4) when x=9 and y=0.04

If sqrt(1-x^(4))+sqrt(1-y^(4))=k(x^(2)-y^(2)), prove that (dy)/(dx)=(x sqrt(1-y^(4)))/(y sqrt(1-x^(4)))

The equation of the parabola with focus (0,0) and directrix x+y=4 is x^(2)+y^(2)-2xy+8x+8y-16=0x^(2)+y^(2)+8x+8y-16=0x^(2)+y^(2)-2xy+8x+8y=0x^(2)-y^(2)+8x+8y-16=0