The perimeter of a polygon is the sum of the lengths of its sides. Label each of the sides with a value of \(3\) and add.

We are looking for an odd integer between \(4\) and \(11\). D must be one of the integers \(5, 7, 9\). Since D is also between \(7\) and \(18\), the only integer that meets both conditions is \(9\).

Gary and Paul have a total of \(20 + 18 = 38\) operas put together. Since 4 of them are in common, subtract \(4\) from \(38\), giving \(34\) unique operas.

Solve for \(G\) by multiplying both sides by 6: \(G = 18 \times 6 = 108\). Substitute \(G = 108\) into the equation \(\frac{1}{3}G = \frac{1}{3} \times 108 = 36\).

The sailboat can float within a circular area with radius equal to the length of the string, 1 meter. The area of the circle is given by \(A = \pi r^2\), where \(r = 1\). The area is \(A = \pi\) square meters.

Let \(x\) represent the number of children. The number of adults is \(3x\), so the total number of people is \(x + 3x = 4x\). The total must be a multiple of 4.

A triangle can be made by cutting diagonally. A square can be made by cutting parallel to one side. A smaller rectangle can be made by cutting parallel to any side of the original rectangle.

The number of students in March = \(40\), April = \(60\), May = \(80\), June = \(80\). The average is \(\frac{40 + 60 + 80 + 80}{4} = 65\).

Substitute \(n = 6\) and \(r = 2\) into the equation: \(6 \clover 2 = (6 – 1) – \frac{6}{2} = 5 – 3 = 2\).

Here, you are given n = Q and r = 2. Use the equation given in the definition to set Q$2 equal to 3 and solve for Q: (Q-1) – 2 = 3. First, eliminate the denominator by multiplying both sides by 2:2 (Q- 1) – Q=6. Then, distribute the 2 through the parentheses: 2Q-2 – Q= 6. Third, isolate the Q: Q=8, (B).

Begin with $6 + cover price – $8 and simplify it. This can be simplified to cover price – $2, which means $2 below the cover price. (B) is correct.

Note here that each cube = 20 boxes. February has three cubes less than January, hence 3(20) = 60 boxes less.

In March, 5 cubes were sold and in February, 3 cubes were sold. Thus, in March, the number of boxes sold was 5/3 = 1⅔ times the number of boxes sold in February.

Let x = the number of ties for Team A; keep in mind that x is an integer here. Thus, Team A had 5x losses. Adding the losses and ties (there were no wins), the number of games the team played was x + 5x = 6x. Thus, the correct answer choice must be a multiple of 6 (because x is an integer). Only (C), 18, is a multiple of 6.

Systematically count the different rectangles in the figure. There are 5 basic rectangles. Then, there are 5 rectangles made up of 2 basic rectangles. Next, there is 1 rectangle made up of 3 basic rectangles, and 1 rectangle made up of 4 basic rectangles. In total, there are 5 + 5 + 1 + 1 = 12 different rectangles in the figure.

We are looking for the fraction that is NOT less than ¼, that is, a fraction that is greater than or equal to ¼. (D), 19/70, is correct because it simplifies to a number that is greater than ¼.

Begin by labeling each side 1. Using the answer choices, count the lengths of the paths: (A) = 4, (B) = 4, (C) = 4, (D) = 5, and (E) = 4. (D), A – B – E – G – H, is the longest path with a length of 5.

No long calculation is needed here. In order for a product of numbers to equal 0, at least one of the numbers must equal zero. Since 5 ⅓ is not 0, the other factor, 5 – x, must equal 0. So 5 – x = 0 and x = 5.

Since 1.08 has 2 places after the decimal point, write each answer choice with 2 places after the decimal point. (A) and (B) are more than 1.00 away from 1.08. (C), 1.90, is more than 0.80 away from 1.08, (D), 1.10, is 0.02 away from 1.08, and (E), 1.00, is 0.08 away from 1.08.

Write out the given inequality: X > 25. Next multiply both sides by (or divide both sides by) 5. We now have X/5 > 5, so the answer is greater than 5 (C).