If the value, to the nearest thousandth, of cos α is −0.385, which of the following could be true about α?

Solution
You must know something about the values of cosine functions at certain values of α to answer this question correctly.
Work through each of the answer choices, as follows:
Answer choice F: (\(\frac{2\pi }{3}\) ≤ α ≤ π) When α is \(\frac{2\pi }{3}\) , cos α = −^{1}⁄_{2}; when α is π, cos α = −1.
The value −0.385 is not between −^{1}⁄_{2} and −1, so eliminate answer choice F.
Answer choice G: ( ^{π}⁄_{2} ≤ α ≤ \(\frac{2\pi }{3}\) ) When α is ^{π}⁄_{2}, cos α = 0; when α\(\frac{2\pi }{3}\) , cos α = −^{1}⁄_{2}.
The value −0.385 is between 0 and −^{1}⁄_{2}, so answer choice G is correct.
If c is directly proportional to s^{2} and c = \(\frac{7}{16}\) when s = ^{1}⁄_{4}, what is the value of s when c = 175?

Solution
Since c is directly proportional to s^{2}, then c = as^{2}, for some constant a. Given that c = \(\frac{7}{16}\) when s = ^{1}⁄_{4}, you can solve for a, as follows:
\(\frac{7}{16}=a(\frac{1}{4})^{2}\)
\(\frac{7}{16}=(\frac{1}{16})a\)
a = 7
You now know that a = 7, and c = as^{2}. Substitute 175 for c and 7 for a and solve for s, as follows:
175 = 7s^{2}
25 = s^{2}
s = 5
In the figure below, all line segments are either horizontal or vertical, and the dimensions given are in feet. What is the perimeter, in feet, of the figure?

Solution
To solve this problem, find the missing side lengths. Because the lengths 2 and 4 make up the entire left vertical side of the figure, you can deduce that the entire right vertical side of the figure will be 2 + 4 = 6.
Likewise,because the entire bottom horizontal side of the figure is 8, you can deduce that the missing horizontal length will be 2, because 8 − 6 = 2.
Therefore, the perimeter of the figure is 6 + 6 + 8 + 8 = 28.
The trapezoid below is divided into 2 triangles and 1 rectangle. Lengths are given in centimeters. What is the combined area, in square centimeters, of the 2 shaded triangles?

Solution
You can solve this problem by first calculating the area of the trapezoid.
Use the formula A = ^{1}⁄_{2}(b_{1} + b_{2})h:
A = ^{1}⁄_{2}(4 + 8)3
A = ^{1}⁄_{2}(12)3
A = ^{1}⁄_{2}(36)
A = 18
Next, calculate the area of the rectangle: 3×4 = 12.
Because the shaded regions comprise the difference between the area of the trapezoid and the area of the rectangle, the area of the shaded regions is 18 − 12 = 6.
If x < y, then x − y is equivalent to which of the following?

Solution
One way to solve this problem is to pick numbers for x and y. You are given that x < y, so make x = 2 and y = 3.
Now, substitute those values into the absolute value given: x − y=2 − 3=− 1 = 1.
Now, substitute the values into each answer choice.
The answer choice that yields a value of 1 will be correct:
Answer choice F: 2 + 3 = 5; eliminate answer choice F.
Answer choice G: −(2 + 3) = −5; eliminate answer choice G.
Answer choice H: √2 − 3 ≠ 1; eliminate answer choice H.
Answer choice J: 2−3 = −1; eliminate answer choice J
Answer choice K: −(2 − 3) = −(−1) = 1.
Answer choice K is correct. For any values of x and y when x < y, x − y=−(x − y).
A classroom has 10 tables that will seat up to 4 students each. If 20 students are seated at tables, and NO tables are empty, what is the greatest possible number of tables that could be filled with students?

Solution
You are given that no table can be empty, which means that there must be at least one student seated at each table.
There are 20 students in all and 10 tables, so if one student is seated at each table, you have accounted for 10 students. Now simply fill tables with the remaining 10 students until no students remain:
Table 1 already has 1 student; add 3 more so that it is filled. You now have 7 more students to seat.
Table 2 already has 1 student; add 3 more so that it is filled. You now have 3 more students to seat.
Table 3 already has 1 student; add the 3 remaining students so that it is filled.
You have accounted for all of the students and have filled 3 tables.
You cannot fill more than 3 tables with students while leaving no tables empty.
A square and a regular pentagon have equal perimeters. If the pentagon has sides of length 12, what is the area of the square?

Solution
A regular pentagon (5sided figure) is a pentagon with equal sides and equal angles. If each side of the pentagon has a length of 12, its perimeter is 60. For the square to have a perimeter of 60, each side of the square must have a length of 15. If each side of the square has length of 15, its area is 225.
If 6−2x>9, which of the following is a possible value of x?

Solution
This problem tests your knowledge of absolute values and inequalities.
The inequality 6−2x>5 can be written as two separate inequalities: 6 − 2x > 9 or 6 − 2x < −9. Because the original inequality was a greater than, the word in between the two new inequalities must be “or.” This means that 6 − 2x > 9 = −2x > 3 = x <−^{3}⁄_{2}, or that 6 − 2x < −9 = −2x < −15 = x > \(\frac{15}{2}\).(Remember to reverse your inequality when multiplying or dividing by negative numbers.)
The only answer choice that fits one of the inequalities is −2, which is less than −^{3}⁄_{2}. There are no answer choices that are greater than \(\frac{15}{2}\).
What rational number is halfway between ^{1}⁄_{6} and ^{1}⁄_{2}?

Solution
A rational number is any number that can be written as a ratio; in other words, a fraction.
So simply determine the fraction that is exactly halfway between ^{1}⁄_{6}and ^{1}⁄_{2}. One way to do this is to find the average, as follows:
\(\frac{\left ( \frac{1}{2}+\frac{1}{6} \right )}{2}=\frac{2}{6}=\frac{1}{3}\)
Another way to solve this problem is to find the Lowest Common Denominator (LCD):
^{1}⁄_{2} = ^{3}⁄_{6}, so the LCD is 6.
The value that is halfway between ^{1}⁄_{6} and ^{3}⁄_{6} is ^{2}⁄_{6}, which is equivalent to ^{1}⁄_{3}.
The lengths of the sides of right triangle ABC are shown in the figure below. What is the cotangent of ∠B?

Solution
Cotangent is defined by he length of the adjacent side over the length of the opposite side. Therefore, the cotangent of ∠B is ^{x}⁄_{z}.