Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2011)

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, January 2011

Part I: Each correct answer will receive 2 credits.


11. The conjugate of 7 − 5i is

1) −7 − 5i
2) −7 + 5i
3) 7 − 5i
4) 7 + 5i

Answer: 4) 7 + 5i

To find the conjugate of a complex number, change the sign of the imaginary part.

So the conjuage of 7 - 5i is 7 + 5i, which is Choice (4).

Similarly, the conjugate of the binomial (7x - 5y) is (7x + 5y).

12. If sin^-1 (5/8) = A, then

1) sin A = 5/8
2) sin A = 8/5
3) cos A = 5/8
4) cos A = 8/5

Answer: 1) sin A = 5/8

Remember that sin(sin^-1(5/8) = 5/8. Therefore,

sin^-1 (5/8) = A

sin( sin^-1 (5/8) ) = sin(A)

5/8 = sin A

That is Choice (1).

13. How many distinct triangles can be formed if m∠A = 35, a = 10, and b = 13?

1) 1
2) 2
3) 3
4) 0

Answer: 2) 2

You are given two sides and an angle that is not included between those two sides. We learned in Geometry that there is no SSA Theorem because there are two possible triangles in this situation.

The exception to this rule is the Hypotenuse Leg Theorem. This triangle is a right triangle if and only if the sin 35 = 10/13, so check this.

sin 35 degrees = 0.573...

10/13 = 0.7692...

They are not the same, so the triangle is not right.

14. When 3/2 x² − 1/4 x - 4 is subtracted from 5/2 x² - 3/4 x + 1, the difference is

1) -x² + 1/2 x - 5
2) x² - 1/2 x + 5
3) -x² - x - 3
4) x² - x - 3

Answer: 2) x² - 1/2 x + 5

The "from" term goes first in the subtraction:

(5/2 x² - 3/4 x + 1) - (3/2 x² − 1/4 x - 4)

= 5/2 x² - 3/2 x² - 3/4 x - (-1/4 x) + 1 - (-4)

= 2/2 x² - 2/4 x + 5

= x² - 1/2 x + 5, which is Choice (2).

Note that you would've gotten choice (1) if you subtracted backwards.

15. The solution set of the inequality x² − 3x > 10 is

1) {x | −2 < x < 5}
2) {x | 0 < x < 3}
3) {x | x < -2 or x > 5}
4) {x | x < -5 or x > 2}

Answer: 3) {x | x < -2 or x > 5}

Factor the inequality.

x² − 3x > 10

x² - 3x - 10 > 0

(x - 5)(x + 2) > 0

For the product to be positive, both binomials must be positive or both must be negative.

They are both positive is x > 5. They are both negative if x < -2. This is Choice (3).

More to come. Comments and questions welcome.

More Regents problems.

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