x^(2)+8x-20

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If f(x)=(x^(3)-8)/(x^(2)+x-20) , then

The function (x^(3)-8)/(x^(2)-x-20) is continuous over the domain :

The roots of the quadratic equation (x^(2) - 8)/(x^(2) + 20) = 1/2 are…

Factorise: (1)2x^(2)-x-6=0(2)a^(3)-0.216 (3) (x^(2)-3x)^(2)-8(x^(2)-3x)-20

x ^(2) + y^(2) + 8x -20y =28 What is the diameter of the circle given by the equation above ?

The domain of the function (x^(2) + 8x + 9)/(x^(2) - 9x + 20) is

(16x^(2)-20x+9)/(8x^(2)+12x+21)=(4x-5)/(2x+3)

Subtract 7x^(3)+3x^(2)+8x+16 from 9x^(3)+6x^(2)-6x+20