Wednesday, October 27, 2021

Algebra 2 Problems of the Day (Algebra 2 Regents, June 2012)

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 2012

Part II: Each correct answer will receive 2 credits. Partial credit is possible.

32. Find, to the nearest tenth, the radian measure of 216°.


The rate of conversion is 180° = &pi radians, 216°, which is about 1/6th more that 180, should be just a little bit under 4. (Figure about 3.14 plus another .52 or so. This is meant to measure a reasonable answer, not to be an accurate result.)

216 * π / 180 = 3.769911... = 3.8 radians.

33. Find the third term in the recursive sequence ak+1 = 2ak - 1, where a1 = 3.


Start with a1 and calculate the next two terms:

a1 = 3

a2 = 2(a1) - 1 = 2(3) - 1 = 5

a3 = 2(a2) - 1 = 2(5) - 1 = 9

34. The two sides and included angle of a parallelogram are 18, 22, and 60°. Find its exact area in simplest form.


The area of a parallelogram can be found using the formula A=a·b·sin(θ). (Use half of that for a triangle.)

A=a·b·sin(θ) = 18 * 22 * sin(60) = 396 * √(3) / 2 = 198 √(3)

35. Write an equation for the graph of the trigonometric function shown below.


The graph starts at 0, so it's sin, not cos. The amplitude is 3, and it starts in the negative direction instead of positive. And the period is π instead of 2π, meaning that it has a frequency of 2.

So the equation for this graph is y = -3 sin (2x).

Leave any part of that out, including the y =, and you'll lose one point. Make two mistakes and you will earn no points, unless you state both the amplitude = 3 and the frequency is 2 (which is worth a point).

End of Part II

More to come. Comments and questions welcome.

More Regents problems.

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