Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, June 2013)

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 2013

Part III: Each correct answer will receive 4 credits. Partial credit is available.


36. Solve the equation below algebraically, and express the result in simplest radical form:

13/x = 10 - x

Answer:

Multiply both sides by x and solve the quadratic equation.

13/x = 10 - x
13 = 10x - x²
x² - 10x + 13 = 0

x = -b/(2a) + SQRT(b² - 4ac)/(2a)
= -(-10)/(2) + SQRT((-10)² - 4(13))/(2)
= 5 + SQRT(100 - 52)/2
= 5 + SQRT(48)/2
= 5 + SQRT((16)(3))/2
= 5 + 4 * SQRT(3)/2
= 5 + 2 * SQRT(3)

37. A ranch in the Australian Outback is shaped like triangle ACE [No image given], with m∠A = 42, m∠E = 103, and AC = 15 miles. Find the area of the ranch, to the nearest square mile.

Answer:

The area of a triangle can be found using 1/2 a b sin C, but we need a second side.

Use the Law of Sines:

sin 103 / 15 = sin 42 / x
x * sin 103 = 15 * sin 42
x = 15 * sin 42 / sin 103 = 10.3

180 - 42 - 103 = 35

A = 1/2 a b sin C
= 1/2 (10.3) (15) sin 35
= 44.30... = 44

38. Ten teams competed in a cheerleading competition at a local high school. Their scores were 29, 28, 39, 37, 45, 40, 41, 38, 37, and 48.

How many scores are within one population standard deviation from the mean?

For these data, what is the interquartile range?

Answer:

Put the numbers into a List in your calculator and run One Variable Stats.

The Mean is 38.2. The standard deviation is 5.9. This means that the range for scores that are one standard deviation from the mean is 38.2 - 5.9 = 32.3 through 38.2 + 5.9 = 44.1. There are 6 scores within that range.

THe interquartile range is Q3 - Q1. From your calculator, you know that Q3 = 41 and Q1 = 37, so the IQR is 4.

End of Part III.

More to come. Comments and questions welcome.

More Regents problems.

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