34.
Given:  A = {18, 6, -3, -12}
Determine all elements of set A that are in the solution of the inequality $\frac{2}{3}x+3 < -2x - 7$

There are two good approaches:
a) Plug in each value
b) Solve the inequality for x

a)
18?     $\frac{2}{3}(18)+3 < -2(18) - 7$  …  12 + 3 < -36 – 7 …. 15 <  -43 … false
6?     $\frac{2}{3}(6)+3 < -2(6) - 7$ …   4 + 3   < -12 – 7  ….. 7 < -19  … false
-3?    $\frac{2}{3}(-3)+3 < -2(-3) - 7$  …  -2 + 3 < 6 – 7 ….  1 < -1 …. false
-12?   $\frac{2}{3}(-12)+3 < -2(-12) - 7$  …  -8 + 3 < 24 – 7 ….. -5 < 17 …. true
Therefore only -12 is in A and is a solution to the given inequality

b)
$\frac{2}{3}x+3 < -2x - 7$ (start)
$2x + 9 < -6x - 21$ (multiply each term by 3 to get rid of denominators*)
$8x < -30$
$x < \frac{-30}{8} \rightarrow x < -3.75$
and we quickly conclude that only -12 meets that condition.

* Getting rid of denominators is not necessary, but it can make it easier to manipulate the numbers and expressions

35. Graph and label the following equations on the set of axes below

$y = |x|$
$y = |\frac{1}{2}x|$

Explain how decreasing the coeffecient of x affects the graph of the equation $y = |x|$.

Hmm. the first is a V-shaped graph. Start at (0,0) and graph y = x in the first quadrant, and graph y = -x in the second.

For the second, start at (0,0) and do the same thing, but with slopes one-half and negative one-half (a flattened V, kind of like a seagull)

Your calculators can make the pictures. If you push the MATH button, and move one across on the top, to NUM, “abs(” is the first item on the list. You graph y = abs(x). Done.

Decreasing the coefficient flattens the graph. It slows how quickly it rises. It decreases the slope on the right side…  Decrease it to 0 and you get a horizontal line. Decrease it past 0, for example to -2, and it makes the graph sink into the 3rd and 4th quadrants. Neat that the corner (the vertex) stays at the origin.

I wonder how we will grade this.

36. (see photo)

I assume (and am not certain) that we will look for the following:

• A straight line, drawn with a straight-edge (a ruler is good) that passes near most of the points.
• The y-intercept should fall between 1 and 3 (just above 2 looks very good).
• The line should extend to at least the 18th month, and probably the 20th month. It should cross the 18th month line somewhere between 12 and 16 thousand dollars.
• You should write something like “the prediction line I have drawn shows that they will have not reached their goal of \$20,000 profit for their 18th month of business”
• I would probably mark (18,20) on the graph and label it “goal” but they did not ask for that.
• Notice, the table with numbers at the top? Perfectly safe to completely ignore.

This part was worth 9 points, 3 points each. I do not know how they will penalize errors. From past experience, they force most part-credit down to 1 point (few are awarded 2), but I do not know that that will be the case here.

25 Comments leave one →
June 20, 2010 am30 9:49 am 9:49 am

how many points would you get off if you got the explanation for 35 was wrong

• June 20, 2010 am30 10:58 am 10:58 am

one or two. I haven’t seen the rubric.

June 20, 2010 pm30 1:23 pm 1:23 pm

I saw the conversion chart on jmap.org

June 21, 2010 pm30 9:44 pm 9:44 pm

Check your knowledge of abs val….I don’t think the graph ever goes into the 3rd or 4th quadrants!!!!

• June 21, 2010 pm30 10:25 pm 10:25 pm

If we have the coefficient outside the absolute value… (but we don’t. That’s strange. It should be outside)

So in the problem, as written, you are correct.

In a more normal place, y = -2|x| would be what I meant… and exists at the origin and in the 3rd and 4th Quadrants.

June 20, 2010 am30 10:35 am 10:35 am

“It should cross the 18th month line somewhere between 12 and 16 thousand dollars.”

Hrm, I entered the points into my calculator and calculated the line of best fit. According to that, the line is approximately y=0.5717x+2.3258. When you go into table, it says the 18th point is (18, 12.616). If the margin of error is 2, then shouldn’t the range of acceptable answers be more like 10.5-14.5 thousand dollars?

• June 20, 2010 am30 11:33 am 11:33 am

I’ll trust you are right for y = .57x + 2.3. That would make your calculation better than my estimate. However….

1. we’ll accept anything that’s not crazy.
2. “margin of error” here is not well-defined. I doubt that “2” does it. The regression should put some +/- on the coefficient and the constant, and we could go from there, but I don’t know how to find those.

June 20, 2010 am30 11:52 am 11:52 am

Oh, alright. I was just nervous because my line showed (18, 11.5). :P Thanks for the reply.

3. June 20, 2010 pm30 12:49 pm 12:49 pm

If I got two wrong on the mutliple choice and got the rest of the exam correct what would be my score. Would a curve apply (negative or positive)?

June 20, 2010 pm30 1:20 pm 1:20 pm

95. Good job :)

June 20, 2010 pm30 1:22 pm 1:22 pm

What if I got 10 wrong in the mutiple choice?
Will that majorly hurt my grade?????

June 20, 2010 pm30 6:37 pm 6:37 pm

Unfortunately it can go from a 100 to about 86

6. June 21, 2010 pm30 6:59 pm 6:59 pm

is 3 multiple choice wrong and partial credit on a constructive response a good grade??? I really want a 90 or above!!!!! that’s and 89.6 without the curve…..is the conversion chart online yet?? by the way thanks for this, it really brought my stress level down!!!!

7. June 21, 2010 pm30 7:06 pm 7:06 pm

and for number 35…can’t you just say that it makes it wider??

8. June 21, 2010 pm30 7:07 pm 7:07 pm

For #36, a best fit line can be any line that reasonably goes through 2 points…they can’t mark that wrong…also should your explanation apply to YOUR best fit line or the calculator’s??

9. June 21, 2010 pm30 7:21 pm 7:21 pm

What happens if i get 4 multiple choice wrong and then like 2/3 or 1/3 points on one constructive response???

June 21, 2010 pm30 7:32 pm 7:32 pm

That’s either a 93 or a 94.

10. June 21, 2010 pm30 9:20 pm 9:20 pm

i dont think that u should take off if we didnt write that the negatives change it, because that is too complicated for a regents.

11. suffolk teacher permalink
June 22, 2010 pm30 2:35 pm 2:35 pm

graded this questiion it was a nightmare the state doesnt give enough room for error. if the kids graphed two lines it was a zero
we had alot of zeroes

12. suffolk teacher permalink
June 22, 2010 pm30 2:38 pm 2:38 pm

yes the state was looking for the answer that if you lower the cooefficient the graph gets wider. saying a less slope or decreased slope is incorrect unless you showed the V and stated that the slope changed on both sides, bring the grpah closer to the x axis. Horrible question for an algebra 1 exam.

August 17, 2010 pm31 2:22 pm 2:22 pm

Question 34:
printed the June 2010 regents from NYS regents website to study for Aug exam.
Repeated q 34 5x with wrong answer of 18.

Question written as as 2/3x+3<2x-7 instead of what is posted here 2/3X+3<-2x-7.

Could there have been a misprint on the original exam in June?

Where did you get the question.

August 17, 2010 pm31 2:47 pm 2:47 pm

I typed those up very quickly – that might be a typo. I am far from my desk and cannot check. Sorry.

August 17, 2010 pm31 3:11 pm 3:11 pm

there was deffinetly a misprint because i saw the same exam andcompard anwsers and they are the same both ways so thecorrect anwser is infact 18….

August 17, 2010 pm31 3:14 pm 3:14 pm

can I please have a step by step instruction on how to do #35 please I find it very confusing????