POSSIBLE POLYGONS Determine if it is possible for a regular polygon to have an interior angle with the given angle measure. Explain your reasoning a. 165 b. 171 c. 75 d. 40​

yes for all 4.

Step-by-step explanation:

I am not sure I see the whole information.

but if the question truly is simply, whether a polygon can have an internal angle of 165 or 171 or 75 or 40 degrees, then the answer is a resounding yes to all 4.

the simplest polygon is a triangle.

the sum of all angles in a triangle is always 180 degrees.

so, any angle smaller than 180 degrees is a possible internal angle in a triangle (and therefore in polygons in general), because the difference to 180 can then be split into 2 angles (e.g. half/half).

so, if one angle is 165 degrees, then the other two angles have to have 15 degrees together (e.g. 10 and 5, or 7 and 8, or 2 and 13, or ...). they will be narrow angles, but they are still angles.

similar for 171 degrees. now the other two angles have to have only 9 degrees together. like 5 and 4.

for 75 degrees the other two angles must have 180-75 = 105 degrees together. like 20 and 85.

for 40 degrees the other two angles must have 180-40 = 140 degrees together. like 90 and 50.

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.