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°.

Answer:

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 a_k+1 = 2a_k - 1, where a₁ = 3.

Answer:

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.

Answer:

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.

Answer:

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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