Sunday, August 25, 2019

August 2019 Algebra 1 Regents, Part 2

The following are some of the multiple questions from the August 2019 New York State Common Core Algebra I Regents exam.

August 2019 Algebra I, Part II

Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.

25. If g(x) = -4x2 - 3x + 2, determine g(2).

g(2) = -4(-2)2 - 3(-2) + 2
= -4(4) - (-6) + 2
= -16 + 6 + 2 = -8

One point off for one computational error, such as (-2)2 = -4, or similar.

26. A student is in the process of solving an equation. The original equation and the first step are shown below.

Original: 3a + 6 = 2 - 5a + 7
Step one: 3a + 6 = 2 + 7 - 5a
Which property did the student use for the first step? Explain why this property is correct.

The student used the Commutative Property to switch the order of the last to terms, which allows the student to Combine Like Terms in the next step.

27. On the set of axes below, graph the line whose equation is 2y = -3x - 2.
This linear equation contains the point (2,k). State the value of k

Divide the equation by 2 to get it in slope-intercept form. You will see that the y-intercept is -1 and the slope is -3/2. You could use the graphing calculator at this point, or you can plot the point (0, 11) and then make you next point down 3 spaces and 2 to the right, and then repeat. When you get to the edge of the graph, go back the other way: 3 spaces up and 2 to the left, and repeat. And the arrows.

You can plainly see that the point (2, k) is (2, -4), so k = -4.

If you graphed this incorrectly, but your answer for k matches your graph, you will get credit for that portion.
If you used the equation 2y = -3(2) - 2 to find the y-coordinate, you will probably get the credit.

28. The formula a = (vf - vi) / t is used to calculate acceleration as the change in velocity over the period of time.
Solve the formula for the final velocity, vf, in terms of initial velocity, vi, acceleration, a, and time, t.

This is a real-world problem: calculating a final velocity when given an initial velocity, acceleration and a period of time.
Isolate the variable by moving everything else to the other side of the equation.

a = (vf - vi) / t
at =vf - vi
at + vi = vf

29. Solve 3/5 x + 1/3 < 4/5 x - 1/3 for x.

You can eliminate the fractions, if you want, by multiplying the entire equation by 15. Or you can simplify it a little first.

3/5 x + 1/3 < 4/5 x - 1/3
2/3 < 1/5 x
(15)(2/3) < (1/5 x)(15)
10 < 3x
10/3 < x

30. Is the product of two irrational numbers always irrational? Justify your answer.

No, sometimes the answer is rational. SQRT(2) * SQRT(2) = 2, which is rational.

31. Solve 6x2 - 42 = 0 for the exact values of x.

Exact values of x means that you do NOT estimate irrational numbers. Do not round anything.

6x2 - 42 = 0
6x2 = 42
x2 = 7
x = +SQRT(7)

If you go any further than that, you will lose a point, so STOP.

If you used the Quadratic Formula, you would have gotten something like + SQRT(1008) / 12. This is OKAY because that is an Exact value, which can be simplified into 12 * SQRT(7) / 12, but you wouldn't be required to do this. However, if you did attempt to simplify it, you would have to get it right.

32. Graph the function:

h(x) = { 2x - 3, x < 0
x2 - 4x - 5, 0 < x < 5

ONLY Graph up to x = 5 and no farther. No arrow. The left side, however, does get an arrow. The left portion ends with an OPEN circle. The parabola starts with a CLOSED circle. You could end it with a closed circle as well, if you want, or just a point.

End of Part II

How did you do?

Questions, comments and corrections welcome.

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