If the perimeter of a septagon (in which all sides are equal) is \(\displaystyle 5(x^{2}+1)+2x(x-4)\), then the length of one side will be one seventh of this expression. 

To find one seventh, the value must first be simplified and then divided by 7. 

\(\displaystyle 5(x^{2}+1)+2x(x-4)\)

\(\displaystyle 5x^{2}+5+2x^{2}-8x\)

\(\displaystyle 7x^{2}-8x+5\)

When this is divided by 7, the result is:

\(\displaystyle x^{2}-\frac{8x}{7}+\frac{5}{7}\)

This value is thus the length of one side of the septagon. 

The number of degrees in an internal angle of a regular polygon can be solved using the following equation where n equals the number of sides in the polygon:

\(\displaystyle \frac{180(n-2)}{n}=\frac{180(7-2)}{7}=\frac{900}{7}=128.6\)

The measure of an interior angle of a regular polygon can be determined using the following equation where n equals the number of sides:

\(\displaystyle int=\frac{180(n-2)}{n}=\frac{180(9-2)}{9}=\frac{1260}{9}=140\)

The answer can be more clearly seen by extending \(\displaystyle \overline{PA}\) to a ray \(\displaystyle \overrightarrow{PA}\):

Note that angles have been newly numbered.

\(\displaystyle \angle 2\) and \(\displaystyle \angle 4\) are exterior angles of a (five-sided) regular pentagon in relation to two parallel lines, so each has a measure of \(\displaystyle \left (360 \div 5 \right )^{\circ}= 72^{\circ}\). \(\displaystyle \angle 3\) is a corresponding angle to \(\displaystyle \angle 4\), so its measure is also \(\displaystyle 72^{\circ}\).

By angle addition, 

\(\displaystyle m \angle 1 = m\angle 2 + m\angle 3 = 72^{\circ}+ 72^{\circ} = 144^{\circ}\)

The answer can be more clearly obtained by extending the top of the two parallel lines as follows:

Note that two angles have been newly labeled.

\(\displaystyle \angle 2\) is an interior angle of a regular heptagon and therefore has measure

\(\displaystyle \frac{\left ( 7-2\right )180^{\circ }}{7}= 128\frac{4}{7}^{\circ }\)

By the Isosceles Triangle Theorem, since the two sides of the heptagon that help form the triangle are congruent, so are the two acute angles, and

\(\displaystyle m \angle 3 =\frac{1}{2}\left ( 180 -128\frac{4}{7} \right ) ^{\circ }= 25\frac{5}{7}^{\circ }\)

\(\displaystyle \angle 1\) is supplementary to \(\displaystyle \angle 3\), so 

\(\displaystyle m \angle 1 = 180^{\circ } - m \angle 3 = 180^{\circ } -25\frac{5}{7}^{\circ }= 154\frac{2}{7}^{\circ}\)

The answer can be more clearly seen by extending the lower right side of the heptagon to a ray, as shown:

Note that angles have been newly numbered.

\(\displaystyle \angle 2\) and \(\displaystyle \angle 4\) are exterior angles of a (seven-sided) regular heptagon, so each has a measure of \(\displaystyle \left (360 \div 7 \right )^{\circ}= 51\frac{3}{7}^{\circ}\). \(\displaystyle \angle 3\) is a corresponding angle to \(\displaystyle \angle 4\) in relation to two parallel lines, so its measure is also \(\displaystyle 51\frac{3}{7}^{\circ}\).

By angle addition, 

\(\displaystyle m \angle 1 = m\angle 2 + m\angle 3 = 51\frac{3}{7}^{\circ}^{\circ}+ 51\frac{3}{7}^{\circ}^{\circ} = 102\frac{6}{7}^{\circ}^{\circ}\)

The sum of the degree measures of the angles of Quadrilateral \(\displaystyle XEYZ\) is 360, so

\(\displaystyle m \angle XZY+ m \angle EXZ + m \angle E + m \angle EYZ = 360^{\circ}\)

Each interior angle of a regular pentagon measures 

\(\displaystyle \frac{(5-2)180^{\circ}}{5} = 108^{\circ}\),

which is therefore the measure of \(\displaystyle \angle E\).

It is also given that \(\displaystyle m \angle EXZ = 90^{\circ}\) and \(\displaystyle m \angle EYZ = 121^{\circ}\), so substitute and solve:

\(\displaystyle m \angle XZY+ m \angle EXZ + m \angle E + m \angle EYZ = 360^{\circ}\)

\(\displaystyle m \angle XZY+ 90^{\circ} +108^{\circ} +121^{\circ} = 360^{\circ}\)

\(\displaystyle m \angle XZY+ 319^{\circ} = 360^{\circ}\)

\(\displaystyle m \angle XZY=41 ^{\circ}\)

The sum of the degree measures of the angles of Quadrilateral \(\displaystyle TNYZ\) is 360, so

\(\displaystyle m \angle YZT + m \angle N+ m \angle T + m \angle ZYN = 360^{\circ}\).

Each interior angle of a regular pentagon measures 

\(\displaystyle \frac{(5-2)180^{\circ}}{5} = 108^{\circ}\),

which is therefore the measure of both \(\displaystyle \angle N\) and \(\displaystyle \angle T\).

\(\displaystyle \angle ZYN\) and \(\displaystyle \angle ZYE\) form a linear pair, making them supplementary. Since \(\displaystyle m \angle ZYE = 121^{\circ}\),

\(\displaystyle m \angle ZYN = 180^{\circ} - m \angle ZYE = 180^{\circ} -121^{\circ} = 59^{\circ}\).

Substitute and solve:

\(\displaystyle m \angle YZT + 108 ^{\circ} + 108^{\circ} + 59^{\circ} = 360^{\circ}\)

\(\displaystyle m \angle YZT + 275^{\circ} = 360^{\circ}\)

\(\displaystyle m \angle YZT = 85^{\circ}\)

The total of the degree measures of any eight-sided polygon is

\(\displaystyle \left (8-2 \right )180^{\circ} =1,080^{\circ}\).

In an arithmetic sequence, the terms are separated by a common difference, which we will call \(\displaystyle d\). Since the greatest of the degree measures is \(\displaystyle 166^{\circ }\), the measures of the angles are

\(\displaystyle 166-7d, 166-6d...166-d, 166\)

Their sum is 

\(\displaystyle \left (166-7d \right )+\left ( 166-6d \right ) +...+ \left (166-d \right )+166 = 1,080\)

\(\displaystyle 1,328 -28d = 1,080\)

\(\displaystyle 28d=248\)

\(\displaystyle d= 8\frac{6}{7}\)

The least of the angle measures is 

\(\displaystyle 166-7d = 166 - 7 \cdot 8\frac{6}{7} = 166 - 62 = 104\)

The correct choice is \(\displaystyle 104^{\circ}\).

The sum of the angles in a polygon can be found using the equation below, in which t is equal to the total sum of the angles, and n is equal to the number of sides. 

\(\displaystyle t=(n-2)(180)\)

Therefore, the equation for the sum of the angles in a 9 sided polygon would be:

\(\displaystyle t=(9-2)(180)\)

\(\displaystyle t=7\cdot 180\)

Therefore, \(\displaystyle \frac{1}{7}\) of the total sum of degrees in a 9 sided polygon would be equal to 180 degrees.