Posted by **Sammy** on Saturday, October 26, 2013 at 2:05pm.

Find the domain of:

6/(9-4x)

7/(x^2-8x)

(x^2+9x)/(x^3-13x^2+40x)

- Algebra Help -
**Kuai**, Saturday, October 26, 2013 at 2:39pm
6/(9-4x)

9-4x =0

9-4x +4x = 0+4x

9= 4x

x =9/4

Domain of x is 9/4

7/(x^2-8x)

x^2 -8x = 0

x(x-8) =0

x =0

x -8 =0

x -8+8 = 0

x =8

domain of x are 0, 8

(x^2+9x)/(x^3-13x^2+40x))

x^3-13x^2+40x =0

x (x^2-13x +40) =0

x(x-5)(x-8) =0

x =0

x = 5

x = 8

Domain of x are 0, 5, and 8

- Algebra Help -
**Steve**, Saturday, October 26, 2013 at 4:54pm
the special values are correct, but the domain is all real numbers **except** those values.

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