Algebra Problems of the Day (Integrated Algebra Regents, August 2011)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones. The Integrated Algebra Regents covered most of the same material as the current Algebra Regents, with a few differences.

More Regents problems.

Integrated Algebra Regents, August 2011

Part III: Each correct answer will receive 3 credits. Partial credit is available.


34. On the set of axes below, graph and label the equations y = |x| and y = 3|x| for the interval −3 ≤ x ≤ 3.

Explain how changing the coefficient of the absolute value from 1 to 3 affects the graph.

Answer:

Only graph the domain you were given. Do not go beyond it. Do not put arrows on the graph. Label each line.

The coefficient of 3 makes the graph more narrower. It goes up faster.

35. A trapezoid is shown below.

Calculate the measure of angle x, to the nearest tenth of a degree.

Answer:

You don't know the height of the trapezoid or the triangle, but you can calculate the base of the triangle from the two bases of the right trapezoid. That will give you the side OPPOSITE the angle. You know the Hypotenuse, so you can use SINE to find the angle.

36 - 28 = 8.

Sin x = 8/12

x = sin^-1(8/12) = 41.81... = 41.8 degrees.

36. Express 16 √(21) / (2 √(7)) - 5 √(12) in simplest radical form.

Answer:

The square root of 21 divided by the square root of 7 is the same as the square root of (21 divided by 7). The square root of 12 can be simplified by factoring out the largest perfect square (which is 4).

16 √(21) / (2 √(7)) = 8 √(3)

5 √(12) = 5 √(4 * 3) = 5 √(4) √(3) = 5 * 2 √(3) = 10 √(3)

8 √(3) - 10 √(3) = -2 √(3)

End of Part III.

More to come. Comments and questions welcome.

More Regents problems.

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