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 I: Each correct answer will receive 2 credits.


26. A right triangle contains a 38° angle whose adjacent side measures 10 centimeters. What is the length of the hypotenuse, to the nearest hundredth of a centimeter?

1) 7.88
2) 12.69
3) 12.80
4) 16.24

Answer: 2) 12.69

You have the adjacent and you want to find the hypotenuse, so you need to use cosine. Since you are looking for the longest side of the triangle, you know that you can eliminate Choice (1) which is smaller than 10.

cos 38 = 10/x
x = 10 / cos 38 = 12.69....

I'm surprised that one of the choices isn't the answer you would get if your calculator was in radians mode. If you 10.47..., that is why.

27. Which ordered pair is in the solution set of the system of inequalities shown in the graph below?

1) (-2,-1)
2) (-2,2)
3) (-2,-4)
4) (2,-2)

Answer: 2) (-2,2)

The solution set is all the points in the area that is double-shaded (criss-cross) along with any points on a SOLID line border of that region of the graph. The dashed line points are NOT part of the solution, but a boundary marker.

Choice (1) is the intersection of the two lines, but one of those lines is broken so it is NOT a solution to BOTH inequalities.

Choice (2) is in the double-shaded region. This is the answer.

Choice (3) is in the unshaded region. It doesn't satisfy either ineqaulity.

Choice (4) is only in a region that is only shaded once. You might think it was a solution if you mixed up the x and y coordinates.

28. A garden is in the shape of an isosceles trapezoid and a semicircle, as shown in the diagram below. A fence will be put around the perimeter of the entire garden.
Which expression represents the length of fencing, in meters, that will be needed?

1) 22 + 6π
2) 22 + 12π
3) 15 + 6π
4) 15 + 12π

Answer: 1) 22 + 6π

You want the perimeter and cirucumference, not the area. Note that the either side of the isosceles trapezoid is not labeled but it has a length of 7.

The three sides of the trapezoid add up to 22, so eliminate Choices (3) and (4).

The circumference of a semicircle is 1/2 C = 1/2 π d = 1/2 (12)π = 6π. So Choice (1) is the answer.

29. Which expression represents 36x² − 100y⁶ factored completely?

1) 2(9x + 25y³)(9x − 25y³)
2) 4(3x + 5y³)(3x − 5y³)
3) (6x + 10y³)(6x − 10y³)
4) (18x + 50y³)(18x − 50y³)

Answer: 2) 4(3x + 5y³)(3x − 5y³)

The given expression is a Difference of Two Perfect Squares, but always be careful when they say "factor completely" because that usually means that there will be more than one step involved. In this case, notice that both terms are multiples of 4, which can be factored first.

So do the following:

36x² − 100y⁶

(4)(9x² − 25y⁶)

(4)(3x + 5y³)(3x - 5y³)

Choice (3) is not factored completely. You can factor (2) from EACH of the two binomials, and (2)(2) = 4.

Choice (4) is if you divided the coefficients by two instead of taking square roots. Note that Choice (1) is NOT the same as Choice (4) because both binomials were divided by 2.

30. What is the quotient of x / (x + 4) divided by 2x / (x² - 16)?

1) 2 / (x - 4)
2) 2x² / (x - 4)
3) 2x² / (x² - 16)
4) (x - 4) / 2

Answer: 4) (x - 4) / 2

To divide rational expressions, multiply by the reciprocal of the second fraction. (You know: "Keep, Change, Change".)

x / (x + 4) ÷ 2x / (x² - 16)

x / (x + 4) * (x² - 16) / 2x

x / (x + 4) * (x - 4)(x + 4) / 2x

x / (x + 4) * (x - 4)(x + 4) / 2x

1 / 1 * (x - 4) / 2

(x - 4) / 2

More to come. Comments and questions welcome.

More Regents problems.

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