HELP I'M SO ANGRY PLEASE HELP
determine the values of X for which g(x) is defined

Any real number x as long as and

In other words, anything but 0 or -2/3 is valid.

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Explanation:

Set the denominator equal to zero and solve for x

2(3x^2 + 2x) = 0

3x^2 + 2x = 0

x(3x + 2) = 0

x = 0 or 3x+2 = 0 .... zero product property

x = 0 or 3x = -2

x = 0 or x = -2/3

If either x = 0 or x = -2/3, then the denominator 2(3x^2 + 2x)  will be zero. But recall that we cannot have zero in the denominator. Dividing by zero is not allowed. The expression is undefined when we divide by zero.

Therefore, we must exclude x = 0 and x = -2/3 from the domain. Any other real number is valid as an x input.

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.

idk

step-by-step explanation: