Is it possible to have a regular polygon with measure of each exterior angle as 580? why?

No because a circle is only 360 degrees which means no angle can go beyond that without overlapping. If we use a square as an example, each interior and is 90 degrees which means that each exterior angle will be 360 - 90 which is 270 degrees. It is impossible for an exterior angle to go above 359 degrees because no angle can exceed 359 degrees as an exterior angle. A circle is he only shape that can reach and that is only because without drawing them in a circle has no angles. To find the measure of exterior angles you take the measure of the interior angle and subtract it from 360 degrees. Hope this makes sense but the overall answer in no. No polygon, regular or irregular in fact, can have an exterior angle that exceeds 360 degrees.

you can't solve x because it is 2 linear equation and you only have one equation

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.