The measures of the angles of a quadrilateral have sum \(\displaystyle 360^{\circ }\). If \(\displaystyle x\) is the measure of the unknown angle, then:

\(\displaystyle x + 100 + 105 + 110 = 360\)

\(\displaystyle x + 315= 360\)

\(\displaystyle x + 315-315= 360-315\)

\(\displaystyle x = 45\)

The measure of the fourth angle is \(\displaystyle 45^{\circ }\).

A quadrilateral has four angles totalling \(\displaystyle 360^{\circ}\). So, first add up the three angles given. The sum is \(\displaystyle 219^{\circ}\). Then, subtract that from 360. This gives you the missing angle, which is \(\displaystyle 141^{\circ}\).

The sum of the measures of the angles of a quadrilateral is \(\displaystyle 360^{\circ }\); the sum of the measures if a pentagon is \(\displaystyle 540^{\circ }\). Therefore, 

\(\displaystyle x + 80 + 80 + 80 = 360\)

\(\displaystyle x + 240 = 360\)

\(\displaystyle x = 120\)

and 

\(\displaystyle y + 100+ 100+ 100+ 100 = 540\)

\(\displaystyle y + 4 00 = 540\)

\(\displaystyle y = 140\)

\(\displaystyle y > x\), so (B) is greater.

The measures of the angles of a quadrilateral have sum \(\displaystyle 360^{\circ }\). If \(\displaystyle x\) is the measure of the unknown angle, then:

\(\displaystyle x + 100 + 105 + 110 = 360\)

\(\displaystyle x + 315= 360\)

\(\displaystyle x + 315-315= 360-315\)

\(\displaystyle x = 45\)

The measure of the fourth angle is \(\displaystyle 45^{\circ }\).

(a) One meter is equal to 100 centimeters; a square with this sidelength has perimeter \(\displaystyle 100 \times 4 = 400\) centimeters.

(b) A regular pentagon has five congruent sides; since the sidelength is 75 centimeters, the perimeter is \(\displaystyle 75 \times 5 = 375\) centimeters.

This makes (a) greater.

Let \(\displaystyle s\) be the sidelength of Square 1. Then the length of a diagonal of this square - which is \(\displaystyle \sqrt{2}\) times this sidelength, or \(\displaystyle s \sqrt{2}\) by the \(\displaystyle 45^{\circ }-45^{\circ }-90^{\circ }\) Theorem - is the same as the diameter of this circle, which, in turn, is equal to the sidelength of Square 2. 

Since the perimeter of a square is four times its sidelength, Square 1 has perimeter \(\displaystyle 4s\); Square 2 has perimeter \(\displaystyle 4 s \sqrt{2}\), which is \(\displaystyle \sqrt{2}\) times the perimeter of Square 1. \(\displaystyle \sqrt{2} < 2\), making the perimeter of Square 2 less than twice than the perimeter of Square 1.

The perimeters of the squares are 

\(\displaystyle 4 \times 1 = 4\) feet

\(\displaystyle 4 \times 2 = 8\) feet

\(\displaystyle 4 \times 3 = 12\) feet

\(\displaystyle 4 \times 4 = 16\) feet

\(\displaystyle 4 \times 5 = 20\) feet

The mean of the perimeters is their sum divided by five;

\(\displaystyle (4+8+12+16+20) \div 5 = 60 \div 5 = 12\) feet.

The median of the perimeters is the value in the middle when they are arranged in ascending order; this can be seen to also be 12 feet.

The quantities are equal.

First find the perimeters of the squares:

\(\displaystyle 4 \times 100 = 400\) centimeters (one meter being 100 centimeters)

\(\displaystyle 4 \times 100 = 400\) centimeters

\(\displaystyle 4 \times 120 = 480\) centimeters

\(\displaystyle 4 \times 140 = 560\) centimeters

The mean of the perimeters is their sum divided by four:

\(\displaystyle (400+400+480+560) \div 4 = 1,840 \div 4 = 460\) feet.

The median of the perimeters is the mean of the two values in the middle, assuming the values are in numerical order:

\(\displaystyle (400 + 480) \div 2 = 880 \div 2 = 440\)

The mean, (A), is greater.

The length of one side of a square is the square root of its area. The polynomial representing the area of the square can be recognized as a perfect square trinomial:

\(\displaystyle 36t ^{2} + 84t + 49\)

\(\displaystyle =( 6t )^{2} + 2\cdot 6t \cdot 7 + 7^{2}\)

\(\displaystyle = (6t+7)^{2}\)

Therefore, the square root of the area is

\(\displaystyle \sqrt{36t ^{2} + 84t + 49} = \sqrt{ (6t+7)^{2}} = |6t+7|\),

which is the length of one side.

The perimeter of the square is four times this length, or

\(\displaystyle 4 |6t+7| = |24t+28|\).

Let the length of one side of the first square be \(\displaystyle s_{1}\). Then the length of one side of the second-smallest square is \(\displaystyle s_{2}= s_{1}+ 2\), and the perimeters of the squares are

\(\displaystyle P_{1}= 4s_{1}\)

and 

\(\displaystyle P_{2} = 4 s_{2}= 4 (s_{1}+ 2) = 4s_{1} + 8 = P_{1}+ 8\)

This makes the common difference of the perimeters 8 units.

The perimeters of the squares being in arithmetic progression, the perimeter of the \(\displaystyle n\)th-smallest square is

\(\displaystyle P_{n} = P_{1} + (n-1)d\)

Since \(\displaystyle d = 8\), this becomes 

\(\displaystyle P_{n} = P_{1} + 8 (n-1)\)

The perimeter of the third-smallest square is 

\(\displaystyle P_{3} = P_{1} + 8 (3-1) = P_{1}+ 8 \cdot 2 = P_{1} + 16\)

The perimeter of the largest, or sixth-smallest, square is

\(\displaystyle P_{6} = P_{1} + 8 (6-1) = P_{1}+ 8 \cdot 5= P_{1} + 40\)

The length of one side of this square is one fourth of this, or

\(\displaystyle s_{6} = \frac{ P_{6} }{4}= \frac{ P_{1} + 40}{4} = \frac{1}{4} P_{1} + 10\)

Therefore, we are comparing \(\displaystyle P_{1} + 16\) and \(\displaystyle \frac{1}{4} P_{1} + 10\).

Since a perimeter must be positive,

\(\displaystyle \frac{1}{4} P_{1} < P_{1}\);

also, \(\displaystyle 10 < 16\).

Therefore, regardless of the value of \(\displaystyle P_{1}\),

\(\displaystyle P_{1} + 16 > \frac{1}{4} P_{1} + 10\),

and 

\(\displaystyle P_{3}> s_{6}\),

making (a) the greater quantity.