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 I: Each correct answer will receive 2 credits.

*11. Which graph represents the function log _{2} x = y?
position?
*

**Answer: 1) (see image) **

The log

_{2}x function is the inverse of the exponential function 2

^{x}, a reflection across the line y = x.

When x = 1 then y = 0 because when 0 is the exponent of 2 then the result is 1.

*12. A circle is drawn to represent a pizza with a 12 inch diameter. The circle is cut into eight congruent pieces. What is the length of the outer edge of any one piece of this circle?
1) 3π/4
2) π
3) 3π/2
4) 3π
*

**Answer: 3) 3π/2 **

The circumference of the pizza is 12π. One eighth of that is 12π/8, which simplifies to 3π/2.

*13. What is the solution set for the equation √(5x + 29) = x + 3?
1) {4}
2) {-5}
3) {4, 5}
4) {-5, 4}
*

**Answer: 1) {4} **

You can work this out (see below) or you can check if 4, -5 or 5 work.

√(5x + 29) ?= x + 3

√(5(4) + 29) ?= (4) + 3

√(49) ?= 7

True, x = 4 is a solution. Eliminate Choice (2).

√(5x + 29) ?= x + 3

√(5(5) + 29) ?= (5) + 3

√(54) ?= 8

False, x = 5 is NOT a solution. Eliminate Choice (3).

√(5x + 29) ?= x + 3

√(5(-5) + 29) ?= (-5) + 3

√(4) ?= -2

False, x = 2 is NOT a solution. Eliminate Choice (4).

Choice (1) is the solution.

Alternatively, you could solve the equation:

5x + 29 = (x + 3)

^{2}

5x + 29 = x

^{2}+ 6x + 9

x

^{2}+ x - 20 = 0

(x + 5)(x - 4) = 0

x + 5 = 0 or x - 4 = 0

x = -5 or x = 4

Next check for extraneous values. As shown above x = -5 is NOT a solution. Discard it. Then x = 4 is the only solution.

*14. When factored completely, x ^{3} + 3x^{2} - 4x - 12 equals
*

1) (x + 2)(x - 2)(x - 3)

2) (x + 2)(x - 2)(x + 3)

3) (x^{2} - 4)(x + 3)

4) (x^{2} - 4)(x - 3)

**Answer: 2) (x + 2)(x - 2)(x + 3) **

It says "factored completely" and (x

^{2}- 4) is not factored completely. It can be broken down further. Eliminate Choices (3) and (4).

If you factor the polynomial by grouping:

^{3}+ 3x

^{2}- 4x - 12

= x

^{3}- 4x + 3x

^{2}- 12

= x(x

^{2}- 4) + 3(x - 4)

= (x

^{2}- 4)(x + 3)

= (x + 2)(x - 2)(x + 3)

*15. What is the middle term in the expansion of (x/2 - 2y) ^{6}?
1) 20x^{3}y^{3}
2) -15/4 x^{4}y^{2}
3) -20x^{3}y^{3}
4) 15/4 x^{4}y^{2}
*

**Answer: 3) -20x ^{3}y^{3} **

The expansion has 7 terms, starting with the x

^{6}y

^{0}term and ending with the x

^{0}y

^{6}term. The middle term would be the fourth one and it would be in the form x

^{3}y

^{3}, so eliminate Choices (2) and (4).

The coefficient of the fourth term would be _{6}C_{3} * (1/2)^{3} * (-2)^{3} = (20)(1/8)(-1/8) = -20.

More to come. Comments and questions welcome.

More Regents problems.

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