Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

*Algebra 2/Trigonometry Regents, June 2011*

Part I: Each correct answer will receive 2 credits.

**6.** What is the value of x in the equation log_{5} x = 4?

1) 1.16

2) 20

3) 625

4) 1,024

**Answer: 3) 625**

If log base 5 of x is 4 then 5 to the power of 4 is x.

5^{4} = 625, which is choice (3).

Choice (1) is the approximate answer to 4^{x} = 5.

Choice (2) is just 4 times 5.

Choice (4) is 4^{5}, if you reversed the expression that you wanted.

**7.** The expression ^{4}√(16x^{2}y^{7} is equivalent to

1) 2x^{1/2}y^{7/4}

2) 2x^{8}y^{28}

3) 4x^{1/2}y^{7/4}

4) 4x^{8}y^{28}

**Answer: 1) 2x ^{1/2}y^{7/4} **

The fourth root of 16 is 2 because 2

^{4}= 16. Eliminate Choices (3) and (4).

The 4th root is the same as taking the 1/4 power, so divide the exponents under the radical by 4, and you get 1/2 and 7/4. This is Choice (1).

DON'T multiply them.

**8.** Which equation is represented by the graph below?

1) y = 5

2) y = 0.5

3) y = 5

4) y = 0.5

1) y = 5

^{x}2) y = 0.5

^{x}3) y = 5

^{-x}4) y = 0.5

^{-x}

**Answer: 2) y = 0.5 ^{x} **

Choices (1) and (4) would both be increasing functions, not decreasing, so eliminate them.

(0.5)^{x} = (1/2)^{x} and (5)^{-x} = (1/5)^{x}

If you look at x = -1, y = 2, and (1/2)^{(-1)} = 2. The answer is Choice (1).

**9.** What is the fifteenth term of the geometric sequence −√(5), √(10), −2√(5), . . . ?

1) −128√(5)

2) 128√(10)

3) −16384√(5)

4) 16384√(10)

**Answer: 1) −128√(5) **

The common ratio in the sequence is √(10)/-√(5) = -√(2). First, you can see that the 15th term will be negative, so eliminate Choices (2) and (4).

The 50th term would be -√(5)*(-√(2))^{14}, which is -√(5)*(2)^{7}, which is -128√(5).

**10.** In △ABC, a = 15, b = 14, and c = 13, as shown in the diagram below.
What is the m∠C, to the nearest degree?

1) 53

2) 59

3) 67

4) 127

1) 53

2) 59

3) 67

4) 127

**Answer: 1) 53 **

The angle across from the smallest side is the smallest angle. The triangle is 13-14-15, which is almost equilateral, so the angles are close to 60, but angle C must be smaller than 60. Eliminate Choices (3) and (4).

The question is does the difference in sides reduce the angle all the way down to 53 degrees? My hunch before working out this problem is that 53 is way more likely than 59.

Since we don't know any of the angles, we aren't going to use the Law of Sines. Instead, we'll use the Law of Cosines

cos(C) = (a^{2} + b^{2} − c^{2}) / (2ab)

cos(C) = (15^{2} + 14^{2} − 13^{2}) / (2(14)(15))

cos(C) = 3/5

C = cos^{-1} 3/5 = 53.1 degrees

Looks like my hunch was good, but I'm glad I worked it out anyway.

More to come. Comments and questions welcome.

More Regents problems.

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