Final-Post Test Review There are two examples for most problems. Formulas that you can have on your note card are listed with problems, the first time it appears. Use the PI button, not 3.14. On the test, WRITE OUT YOUR SET-UP! If I can give partial credit, I will, but only if I can see that you set it up correctly.

* * * – – 1a. Find the exact value of tan 150° IV III II * – 1b. Find the exact value of cos 330° I IV III II * –

* s = r Ө The arc is 59.6 m. The arc is 24.8 m. 2a. The radius of a circle is 12.64 m. What is the length of an arc of the circle intercepted by a central angle of radians? (1 dec. pl.) * s = r Ө The arc is 59.6 m. 2b. The radius of a circle is 9.85 m. What is the length of an arc of the circle intercepted by a central angle of radians? (1 dec. pl.) The arc is 24.8 m.

* Side a is 20.7 inches. Side a is 40.7 inches. 3a. In the right triangle ABC, the right angle is C. If A = 24° and c = 51 in., find the length of side “a”. (1 dec. pl.) * A How do the sides relate to the angle? 24° c = 51 b B Side a is 20.7 inches. C a = ? 3b. In the right triangle ABC, the right angle is C. If A = 71° and c = 43 in., find the length of side “a”. (1 dec. pl.) A 71° c = 43 b B Side a is 40.7 inches. C a = ?

4a. The angle of elevation from a point 15 4a. The angle of elevation from a point 15.3 feet from the base of the tower to the top of the tower is 65° . What is the height of the tower? (1 dec. pl.) How do the sides relate to the angle? = ? 65° The tower is 32.8 feet tall = 15.3 4b. The angle of elevation from a point 27.6 feet from the base of the tower to the top of the tower is 78° . What is the height of the tower? (1 dec. pl.) How do the sides relate to the angle? = ? 78° The tower is 129.8 feet tall = 27.6

5a. A triangle has three angles measuring 65°, 55°, and 60° If the side across from the 65° is 30 feet, how long is the side across from 55°? (1 dec. pl.) * Is this a right triangle? 65° x What do we have? 55° 60° Which Law? It is 27.1 feet. 30 5b. A triangle has three angles measuring 38°, 105°, and 37° If the side across from the 38° is 27 feet, how long is the side across from 105°? (1 dec. pl.) Is this a right triangle? 38° x What do we have? 105° 37° Which Law? It is 42.4 feet. 27

* A C B It is 3.7 miles. A C B It is 14.9 miles. 6a. The cities, A, B, and C create a triangle. If it is 2 miles to A from B, 3 miles to A from C and the angle at A is 95°, how far is it from B to C? (1 dec. pl.) * A Is this a right triangle? 2 95° 3 What do we have? C B SAS a It is 3.7 miles. Which Law? 6b. The cities, A, B, and C create a triangle. If it is 12 miles to A from B, 20 miles to A from C and the angle at A is 48°, how far is it from B to C? (1 dec. pl.) B C A Is this a right triangle? 12 48° 20 What do we have? SAS a It is 14.9 miles. Which Law?

7a. Suppose sec Ө = , in quadrant III 7a. Suppose sec Ө = , in quadrant III. Draw the triangle in the correct quadrant and compute exact values for sin Ө. * * – – 7b. Suppose tan Ө = , in quadrant IV. Draw the triangle in the correct quadrant and compute exact values for sin Ө. – –

* 8a Given one root, use Synthetic Division to solve for all roots: -2 1 2 -6 -12 -2 12 1 -6 a b c This leaves x2 – 6 = 0. There are two ways to find the next roots.

8b Given one root, use Synthetic Division to solve for all roots: 3 1 -3 -2 6 3 -6 1 -2 a b c This leaves x2 – 2 = 0. There are two ways to find the next roots.

* y = a*sin b * (x – h) + k 2 y = 3 3 4 2π “0” Max “0” Min “0” Amplitude……….._______ Vertical Translation_______ Period Frequency…_______ Period Length……._______ Phase shift……….._______ 5 y = 3 3 4 1 Beginning ¼ ½ ¾ End 1 Period 2π “0” Max “0” Min “0”

11. f(x) = x2 – 2 g(x) = 5x a. Find f(g(–4)) b. Find f(g(5)) First: g(–4) = 5*–4 g(–4) = –20 g(5) = 25 Then: f(–20) = (–20)2 – 2 Then: f(25) = 252 – 2 f(–20) = 400 – 2 f(25) = 625 – 2 f(–20) = 398 f(25) = 623

12. Find the inverse: a. f(x) = 2x – 19 y = 2x – 19 x = 2y – 19 x + 19= 2y

13a. Simplify: (4a-5)2 * 4a3 13b. Simplify: (5a-4)3 * 3a2 or or

* 52x – 1 = 304 Change of Base Rule: If: bx = a Then: 14a. Use logs to solve: (4 dec. plcs) 52x – 1 = 304 * Change of Base Rule: If: bx = a Then: 2x – 1 = 3.55… 2x = 4.55… x = 2.2761

73x + 2 = 12830 Change of Base Rule: If: bx = a Then: 14b. Use logs to solve (4 dec. places): 73x + 2 = 12830 Change of Base Rule: If: bx = a Then: 3x + 1 = 4.86… 3x = 3.86… x = 1.2871

(Check that the answer ensures the domain remains positive.) 15a) Solve: log 7 (x + 10) – log 7 6 = log 7 x * 7 7 (Check that the answer ensures the domain remains positive.)

(Check that the answer ensures the domain remains positive.) 15b) Solve: log 2 (x + 9) – log 2 x = log 2 4 2 2 (Check that the answer ensures the domain remains positive.)

16a) Suppose $7200 is invested at 5 16a) Suppose $7200 is invested at 5.3% compounded continuously for 10 years. Using: how much will there be? (The formula will be on the test.) A = 12,232.31 After 10 years, there will be $12,232.31.

16b) Suppose $10,578 is invested at 2 16b) Suppose $10,578 is invested at 2.3% compounded continuously for 15 years. Using: how much will there be? A = 14,936.03 After 15 years, there will be $ 14,936.03.

17a. If there are 32 g of Carbon 14, which has a half life of 5730 years, how much will there be in 10,000 years, using (One decimal place. The formula will be on the test.) A t = 9.545357966… After 10,000 years, there will be 9.5 g.

17b. If there are 253 g of Carbon 14, which has a half life of 5730 years, how much will there be in 14,800 years, using (One decimal place.) A t = 42.2270787… After 10,000 years, there will be 42.2 g.

Formulas you may have on your Card: No examples or pictures! Trig functions, Laws of Sines and Cosines. Arc length. 30-60-90 triangle gives 1, 2, √3. Quadratic Formula, Pythagorean Theorem. Standard Trig Graphing Equation, what symbols mean, & period frequency formula. Laws of Exponents (p170 & p175 red equations only…no numbers!) & Properties of Logs (p197). Change of Base Rule: COBR. You may also use your 1-20 sheet.