**Official June 2013 SAT II Math 2 Thread**

<p>it was 1/4</p>

<p>yippee :D</p>

<p>what other questions did you guys have trouble with? i feel like we’ve covered a lot of the tougher ones, which is good.</p>

<p>I can’t believe the answer to the cone problem isn’t 2 .__.</p>

<p>I’m currently looking at 2 wrong at most (48).</p>

<p>Cone problem was 2?</p>

<p>No it was 3.2. Various people worked out the problem already.</p>

<p>Some question about how much a coin is worth after a certain year, remember getting C/D for it but don’t remember the numbers.</p>

<p>Also another one involving exponents and converting from hours to minutes, anyone remember that?</p>

<p>Same as cos was cos(2pi-x) right?</p>

<p>So far, I have 2 wrong 2 omit, which is a raw score of 46, which isn’t bad. Do you guys think the curve will be moved to 42 or 43, or do you think they’ll keep it at 44?</p>

<p>midpoint question was (-4,-6)
intersecting spheres was sphere and point</p>

<p>Pretty sure 27 is not right on that</p>

<p>Yes #27 isn’t that. For cos(2pi-x) I graphed and they were on top of eachother…</p>

<p>1913 for coin one</p>

<p>i plugged in cos(pi/4-x) and it works for 27.</p>

<p>What was the “some function problem” where you got 1.08? Or is that the same as 25</p>

<p>I redid the cliff problem and got 41. I made a calculator error and recorded sin 30º as tan 30º. Here is how I solved it:</p>

<p>Draw right triangle with base 30 + x and height h. 30º angle on the outside, 45º angle towards the center. tan 30º = h/(30+x) = 1/sqrt(3), tan 45º = h/x = 1</p>

<p>h = x = (1/sqrt(3))(30+x) = 17.321 + 0.577x</p>

<p>0.423x = 17.321</p>

<p>x = 40.95 = h ~=~ 41</p>

<p>Nvm. i think it’s the same.</p>

<p>i also got 1913 for the coin question.</p>

<p>how did you guys go about doing the reflection problem? the question was “which points remain the same when reflected over 2x+1?” or something like that x)</p>

<p>Was 1913 like 77 or 78 years after the starting point? I don’t remember the exact year I had…</p>

<p>@purplepigeon the answer was on the line, you just had to plug in to find a point on the line</p>

<p>Yup. If the point is on the line then it wouldn’t change.</p>