<p>It was 90 degrees. If you set it up so the unknown angles on the inner triangle were Z and Y, you have 140 + Z + Y =180, so Z+Y = 40. Taking these angles over, you have X + Z + Y + (angle given 1, dont remember it) + (angle given 2, dont remember it) = 180. As z + y = 40, you add it to the given angles, subtract it from 180, x=90.</p>
<p>
I got 90 degrees as well</p>
<p>@ready, I don’t think I had a question like that.</p>
<p>so far im looking at 2 wrong, i omitted, could that break 750?</p>
<p>sorry, i can’t remember the mean/media/mode one.</p>
<p>wasn’t the mode 70, median 75, avg …highest?
so mode <median< avg?</p>
<p>Does anyone remember if for the last quadratics question, the answer was choice E?</p>
<p>No. Median was 80, mean was 76, mode was 70.</p>
<p>I dont even remember the “90” question…</p>
<p>no, it was c- 12</p>
<p>^ read previous posts, it’s 12 but I skipped it.</p>
<p>ur gay metboyjon</p>
<p>what was the vertices question? apparently the answer was 1/5. I don’t remember this. Was this experimental?</p>
<p>“im still not sure about that…it definitely wouldnt be 4x3x2x1 because we were only solving for 2 spots…so if you had 4 choices for spot 1, then you could only have only 3 wires left for the second spot…again, 4x3=1”</p>
<p>Yes, that’s true, but you have to divide by two for duplicates. It said lowest possible to test all wires.</p>
<p>for the chart with scissors each representing 400, you just add all the scissor up multiply by 400 and divide it by the number of days, right?</p>
<p>remember the parallelogram? with one angle = 16 and the other like 85?
what was x?</p>
<p>I dont think the median was 80, it was between 70 and 80, the “11.5” term…the 11th term was 70, and the 12th term was 80, making the median 75</p>
<p>the one with the 100th term and the 101st term… it wasn’t the 101st and 102nd terms? I can’t believe this I think that is the only one I got wrong if that is the case. I put two and that would be right if it was 101 and 102</p>
<p>i didnt have “If, then” problem or vertices…</p>
<p>ready: 101</p>
<p>there were a bunch of shapes, some parallelagrams and a pentagon and a hexagon. only on with more than 5 sides, so 1/5</p>
<p>for the sphere one, some people got 6, but the question asked for the longest possible distance on the SURFACE of the sphere.</p>
<p>the longest possible distance would be 6 if you were talking about the diameter of the circle, but it was asking for the largest arc length, so</p>
<p>the radius was 3, and to find the farthest distance, it would be on the opposite side, which would be half of the circumference, so the answer would be 3pi</p>
<p>am I wrong on this?</p>