To find g(f (x)) using the values of the functions f (x) and g(x) as given in the tables, think of g(f (x)) as applying the function g to f (x). So, first find f (x) for x = −1; f (−1) = 2. Then, find g(x) for x = 2 : g(2) = 3.

Since AC is the diameter of the circle, and D is the midpoint of AC, D is the center of the circle. Thus the distance from D to any point on the circle is equal to the radius, r. Since B is a point on the circle, the distance BD is equal to r. Since A is a point on the circle, the distance AD is also equal to r. Further, since AB is congruent to BC, the segment AB is perpendicular to the segment AD, thus the measure of angle ADB is 90◦. In the triangle ABD, two sides are equal to r, thus the triangle is isosceles with one of the angles equal to 90◦. The other two angles are then equal in measure and add up to 90◦. Therefore, angle ABD has a measure of 45◦

When square ABCD is divided as shown, each of the small triangles has legs of length 2 and 4. They also happen to be right triangles, because each shares one angle with the corner of the large square, which are right angles. To solve for the area of the shaded region, you can use the Pythagorean Theorem to calculate the length of the side of the shaded square. Alternatively, you can find the area of one triangle, quadruple that value, and subtract it from the area of square ABCD, which is 6², or 36 square inches.
The area of a triangle is equal to ¹⁄₂(base)(height). In a right triangle, either leg will work as the value for the base or the height, so the area of one triangle is A = ¹⁄₂(2)(4) = 4. The area of the 4 triangles together, then, is 4 × 4, or 16. Thus the area of the shaded region is 36 − 16, or 20.

When AB is 5 inches long, BC is 8 inches long, and AC is 3 inches long, ABC has a perimeter of 5 + 8 + 3 = 16. To find the perimeter of XYZ when its longest side is 20, you can set up proportions to find the lengths of the sides, since the sides of similar triangles are in proportion to each other. In this case, you can take the proportion of the longest sides, 20:8, and apply it to the perimeter p:

\(\frac{20}{8}=\frac{p}{16}\)

P = \(\frac{20 \times 16}{8}=40\)

Given that sin A = ³⁄₅,you might want to draw a picture like the one below to illustrate the situation.

In this case, the ratio of the side opposite angle A to the hypotenuse is ³⁄₅. To find tan A, the length of the other leg is needed. This can be obtained using the Pythagorean Theorem, or by knowing that this is a special 3–4–5 right triangle. According to the Pythagorean Theorem (in a right triangle with sides a, b, and c, where c is the hypotenuse, c² = a² + b²), 5² = 3² + x², where x is the length of the unknown leg. Solve for x:
x² = 5² − 3²
x² = 25 − 9
x² = 16
x = 4
Once the length of the third leg is obtained, tan A = \(\frac{opposite}{adjacent}\) = ³⁄₄.

The percent of students who pass both exams is (85%)(70%), or 0.85 × 0.70, which is 0.595. In a sample of 100, 100 × 0.595, or 59.5 (roughly 60) students would be expected to pass both exams.