The equation for solving for the slope of a line is \(\displaystyle m=\frac{y2-y1}{x2-x1}\). 

Thus, if \(\displaystyle (x1,y1)=(4,12)\) and \(\displaystyle (x2,y2)=(8,20)\), then:

\(\displaystyle m=\frac{20-12}{8-4}=\frac{8}{4}=2\)

For the sake of simplicity, let us assume that the lengths of \(\displaystyle \overarc{NPO}\) and \(\displaystyle \overarc{N O}\) are 5 and 3; this reasoning depends only on their ratio and not their actual length. The circumference of the circle is the sum of the lengths, which is 8, so \(\displaystyle \overarc{NPO}\) and \(\displaystyle \overarc{N O}\) comprise \(\displaystyle \frac{5}{8}\) and \(\displaystyle \frac{3}{8}\) of the circle, respectively. Therefore,

\(\displaystyle m \overarc{NPO} = \frac{5}{8} \cdot 360 ^{\circ } = 225 ^{\circ }\); and

\(\displaystyle m \overarc{N O} = \frac{3}{8} \cdot 360 ^{\circ } = 135 ^{\circ }\).

If two tangents are drawn to a circle, the measure of the angle they form is half the difference of the measures of the arcs they intercept, so

\(\displaystyle m \angle NTO = \frac{1}{2}(m \overarc{NPO} - m \overarc{N O} ) = \frac{1}{2}(225 ^{\circ }- 135^{\circ } ) = \frac{1}{2}(90^{\circ } ) = 45^{\circ }\)

This is greater than \(\displaystyle 30 ^{\circ }\).

(a) The length of the line segment connecting \(\displaystyle (0,0)\) and \(\displaystyle (a,b)\) is 

\(\displaystyle \sqrt{(a -0)^{2}+(b -0)^{2} } = \sqrt{a^{2}+b^{2} }\).

(b) The length of the line segment connecting \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) is 

\(\displaystyle \sqrt{(a -0)^{2}+(0-b)^{2} } = \sqrt{a^{2}+\left ( - b \right ) ^{2} }= \sqrt{a^{2}+b^{2} }\).

The segments have equal length.

A regular hexagon with sidelength \(\displaystyle s = 1\) can be seen as a composite of six equilateral triangles, each with sidelength \(\displaystyle s = 1\). Since area is in direct proportion to the square of the sidelength, the area of the equilateral triangle with sidelength \(\displaystyle s = 2\) is equal to that of four of those triangles. This makes the hexagon greater in area, and it makes (a) the greater quantity.

The sides of a regular polygon are congruent, so in each case, multiply the sidelength by the number of sides to get the perimeter.

(a) Since one foot equals twelve inches, \(\displaystyle 5 \times 12 = 60\) inches.

(b) Multiply: \(\displaystyle 6 \times 10 = 60\) inches

The two polygons have the same perimeter.

The angles of a hexagon measure a total of \(\displaystyle 180 (6-2) = 720\). From the information, we know that:

\(\displaystyle (x - 5) + x + (x+5) + (y-10) + y + (y+10) = 720\)

\(\displaystyle x+ x + x + y+ y + y - 5+5 -10 +10= 720\)

\(\displaystyle 3x + 3y = 720\)

\(\displaystyle 3 (x + y) = 720\)

\(\displaystyle 3 (x + y) \div 3 = 720 \div 3\)

\(\displaystyle x + y = 240\)

The quantities are equal.

The angles of a hexagon measure a total of \(\displaystyle 180 (6-2) = 720\).  From the information, we know that:

\(\displaystyle (x-10)+x+(x+5)+ (y-10)+ y+ (y + 20)= 720\)

\(\displaystyle x + x + x + y + y + y -10+5 -10 + 20= 720\)

\(\displaystyle 3x + 3y +5= 720\)

\(\displaystyle 3x + 3y +5-5= 720-5\)

\(\displaystyle 3x + 3y = 715\)

\(\displaystyle 3\left ( x + y \right )= 715\)

\(\displaystyle 3\left ( x + y \right ) \div 3= 715\div 3\)

\(\displaystyle x + y = 238 \frac{1}{3} < 240\)

This makes (b) greater.

The sum of the measures of a hexagon is \(\displaystyle 180 ^{\circ } \times (6-2)= 720 ^{\circ }\) . Therefore,

\(\displaystyle A + B + 150 + 150 + 150 + 150 = 720\)

\(\displaystyle A + B +600= 720\)

\(\displaystyle A + B = 120\)

The sum of the measures of an octagon is \(\displaystyle 180 ^{\circ } \times (8-2)= 1,080 ^{\circ }\). Therefore,

\(\displaystyle C + D + 150 + 150 + 150 + 150+ 150 + 150 = 1,080\)

\(\displaystyle C + D + 900 = 1,080\)

\(\displaystyle C + D = 180\)

\(\displaystyle C + D > A + B\), so (B) is greater.

The sum of the measures of the angles of a pentagon is \(\displaystyle 180^{\circ } \times (5-2) =540^{\circ }\). Therefore, 

\(\displaystyle y + y + 110 + 110 + 110 = 540\)

\(\displaystyle 2y +330 = 540\)

\(\displaystyle 2y = 210\)

\(\displaystyle y = 105\)

The sum of the measures of a hexagon is \(\displaystyle 180^{\circ } \times (6-2) =720 ^{\circ }\) . Therefore,

\(\displaystyle z + z + 130 + 130 + 130 + 130 = 720\)

\(\displaystyle 2z + 520 = 720\)

\(\displaystyle 2z= 200\)

\(\displaystyle z = 100\)

\(\displaystyle y> z\), so (A) is greater.

By the Pythagorean Theorem, the hypotenuse of the right triangle is 

\(\displaystyle \sqrt{14^{2}+48^{2}} = \sqrt{196+2,304} = \sqrt{2,500} = 50\) inches, making its perimeter

\(\displaystyle 14 + 48 + 50 =112\) inches.

The regular hexagon, which has six sides of equal length, has the same perimeter, so each side measures

\(\displaystyle 112 \div 6 = 18 \frac{2}{3}\) inches.