<p>What do you think my score would be with -2 MC and -2 grid ins?</p>
<p>@nico</p>
<p>Correct: 50 - (2/4) = 49.5 → 50
Range is 710 to 740, depending on difficulty of this March 2012 math section.</p>
<p>Can anyone remember what letter choice was correct for the question where there was a table and it was like F(c) G(c) or something like that. I think the answer was 0</p>
<p>was the bar graph for the avg of the kids 3.2 a multiple choice</p>
<p>@mybman
Was it a chart with F(c) and G(c) and the problem asked where they were equal?</p>
<p>@Lanayru
Yes that was it. I just can’t remember if I accidentally filled in 2 instead of zero bc that was the output that both produced.</p>
<p>does anyone remember the exact question for the 1440 answer?</p>
<p>did you guys get 4/15 for the question about the circle cut into thirds and fifths and you eat one of each section?</p>
<p>The exact question was:
(2^2x+1)(3^4)(5^x) = (2^z-5)(3^y)(5^3y)</p>
<p>x = 12
y = 4
z = 30</p>
<p>I got all of this done then preceded to put 12*430 into my calculator giving me the wrong answer. SO ANGRY.</p>
<p>yup 4/15 for the cake q</p>
<p>@mybman
I remember that the value for c was -1. Something like this?</p>
<p>x | f(x) | g(x)
-1 | 2 | 2
0 | 1 | -1
1 | # | #
rest have different numbers</p>
<p>@caruso
try doing problems two different ways. use your calculator then double check by hand. :(</p>
<p>I know, I was running out of time. I rushed it and I was so satisfied that I “got it”. I just left out 1 multiplication sign. I normally would have double checked with my calculator again, but I barely finished that section because that one question took so long. Normally I could just say, whatever, it is one problem, but it is in math and that one question could cost me 30 or so points.</p>
<p>Also, I omitted the c= one, but I heard it was 0. It was a question that looking back on a should have answered, so I am 1 wrong 1 omitted, but it basically 2 omitted because the wrong one was a fill-in.</p>
<p>Yo okay do any of you remember this question.</p>
<p>There was a triangle with a vertex at (6,4), the base was along the x axis, and another vertex as the origin. It looked like:</p>
<p>/l
_</p>
<p>Anyway, it said what value for the base will make the area of the triangle between 5 and 9. What did you guys get for this one? It was a grid-in.</p>
<p>I think I said 4, but I don’t completely remember. The area = .5bh. The height is 4, if you make the base 4 the area equals 8.</p>
<p>No… because if you make the base 4, then the height would be 8/3. Think about it, if you decrease the size of the base, then you also decrease the height as well.</p>
<p>@lan</p>
<p>The value was 0, I checked it multiple times during the test.</p>
<p>@monkey</p>
<p>Greater than 2.5 but less than 4.5 would work</p>
<p>PS: When you calculate area of the triangle problems, the height stays constant.</p>
<p>In a right triangle, the height is the long leg.
In an acute triangle the height is the top point straight down to the base.
In an OBTUSE triangle, it is the highest point going straight down to where it is perpendicular to the base.</p>
<p>You have to look at it one way the whole time or else the base and height locations change.</p>
<p>This is because the line going towards (6,4) starts at the origin. Thus, the slope is 2/3</p>
<p>Drac, how would 4.5 not work?</p>
<p>If the slope of the line is (2/3), then when you plug in 4.5 for x, the y would be 3.</p>
<p>4.5 * 3 / 2 is 6.75</p>
<p>You might want to draw out the triangle to help you visualize this. I was just pointing this out because all of my and my friends’ (always 800 on math) answers matched the ones posted on here except for this particular question. I think you all misinterpreted the question and assumed that the height would remain a constant 4, but the height changes as the base changes to keep the triangles similar.</p>