<p>The test seemed similar to other tests in terms of difficulty. I would say that 5 wrong (raw score 43.75) would be enough for an 800. 6 wrong is more iffy.</p>
<p>For the area of the triangle, the height was 2 because the max value of 2sin(2x) is 2, and the base was pi/2 because the period is pi as opposed to 2pi on a normal sine graph. Thus the area was bh/2 = [(pi/2)*2]/2 = pi/2.</p>
<p>For the question with the weird function with sine, the answer was June, because that was the first month where the function was above 32 degrees. Wait actually on second thought it was probably May, because somewhere in the month of May the function probably hit 32. Although we were only told that the variable t represents the first day of the t’th month of the year, so I’m not sure if the function was even defined for non-integer values of t… If someone else remembers the question more exactly then they can comment on this as well. Regardless I put down June.</p>
<p>For the problem with 8% annual interest compounded monthly, the answer was 105 months. Each month we get a .667% increase in the investment, so we want to solve 1.00667^t = 2 --> t = ~105.</p>
<p>For the vector one, we can translate the vectors nicely and the answer comes out to 11.0</p>
<p>For the last one with the triangle, we bash it out with law of sines and get an answer of 6.88.</p>
<p>I’m not sure about the May/June one; can anyone else who remembers the question better add to this?</p>
<p>Also, isn’t discussing the test not allowed as per what we signed?</p>
<p>I don’t remember the details of the sin graph question but I got pi. I might have missed something though (I might have overlooked the fact that it was sin2x, not sinx).</p>
<p>I got May for the Sine function question although I’m not confident at all about this answer.</p>
<p>I got 25 for the 355 students/360 chairs question for the same reason.</p>
<p>I got “at least one x intercept” (last value was -13 or something).</p>
<p>(0,2) for the parabola symmetric about x=4.</p>
<p>Can’t remember the months question.</p>
<p>I got -t^2 -1 as well.</p>
<p>As I said before, I got sqrt(16-x^2)/4</p>
<p>Can’t remember the answer to the vector one.</p>
<p>I screwed up on the pi thing too. The 32 degrees thing threw me for a loop because I couldn’t even get close with my calculations. I prob was messing it up somehow. </p>
<p>Vector one was 11 or something like that.
I’m fairly confident in all my answers; pretty sure May was right. Can someone confirm the last question was 11ish too? The diagram made it look bigger than the other side, which was 10 or something like that.</p>
<p>Yeah, I wasted soooo much time on that and ended up with glorified guesswork. Quite a few people got May though, so I still have hope! Annoyed about getting the sin one wrong if the answer is pi/2 and not pi. Typical of me to overlook such a simple detail.</p>
<p>@omnipotent24 11.09 for the last one? You just found the length of the bottom side of the triangle (that was in fact 11.09). However, the question asked for the distance of the ship to the bottom right vertex at the 2nd point where the ship was 6 units away from the top vertex. So we had to create an isosceles triangle and find its base, then subtract that from 11.09 to get the answer. 6.88 was surely the answer.</p>
<p>Also, for the May/June one, sure, it might’ve been 5.3, but the function was only defined for integer values of t, right? I may be wrong though.</p>
<p>So assuming I was wrong and the correct answer was May (this is probably the case) I would have gotten 49/50, which should be an 800. Yay :)</p>
<p>@happiface. I remember the exact numbers for the last question. The two sides given were 6 and 10, the angle was 32. Using a triangle solver found online, answer is 5.66 </p>
<p>Sigh… I’ll just post the entire solution here.</p>
<p>We got a triangle ABC with AB = 6, BC = 10.6, angle C = 32 degrees. We chose a point D on AC such that BD = 6. We want the length of DC. By Law of Sines, (sin 32)/6 = (sin A)/10.6. Solving with calc gives A = 69.42. Now note that since AB = BD, ABD is isosceles, so angle BDA = angle A = 69.42. Thus angle ABD = 180 - 2*69.42 = 41.16. By Law of Sines, (sin 41.16)/AD = (sin 69.42)/6. Solving gives AD = 4.22. Now return to the original triangle. angle ABC = 180 - 32 - 69.42 = 78.58. Thus (sin 78.58)/AC = (sin 32)/6. Solving gives AC = 11.10. Thus we have DC = AC - AD = 11.10 - 4.22 = 6.88, which is our answer.</p>
<p>How can you double your money in 10-11 months with 8% annual interest? Pretty sure it’s 105, but I may be wrong. @happiface ah I see my mistake. Didn’t read the problem correctly. Hm 49/50 I guess. </p>
<p>No need to be condescending here. I just had a problem with loading the triangle solver. The only reason our answers are different is that I used 10 as the side length instead of 10.6. Can somebody else confirm - 10 or 10.6? </p>