x^(2)-(x-2)^(2)=32

y^(2) -x^(2) = 32, y - x = 2

Solve for x ((2)/(5))^(2x-2) = (32)/(3125)

(ii) Solve the following equation 2^(2x)+2^(x+2)-32=0

x^(3)-7x^(2)+2x-3,2+(1)/(2)y-(3)/(2)y^(2)+4y^(3) are cubic polynomials.

Lim_(x rarr2)(x^(5)-32)/(x^(2)-4)

The equation of the image of the circle x^(2)+y^(2)+16x-24y+183=0 in the mirror 4x+7y+13=0 is x^(2)+y^(2)+32x-4y+235=0x^(2)+y^(2)+32x+4y-235=0x^(2)+y^(2)+32x-4y-235=0x^(2)+y^(2)+32x+4y-235=0x^(2)+y^(2)+32x+4y+235=0

underset(xrarr2)"lim"(x^(5)-32)/(x-2)= ?

Evaluate Lim_(xto 2) (x^(5)-32)/(x^(2)-4)

The value of k for which f(x)={{:(,(x^(2^(32))-2^(32)x+4^(16)-1)/((x-1)^(2)),x ne 1),(,k,x=1):} is continuous at x=1, is

The value of k for which f(x)={{:(,(x^(2^(32))-2^(32)x+4^(16)-1)/((x-1)^(2)),x ne 1),(,k,x=1):} is continuous at x=1, is