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

<p>@purplepigeon</p>

<p>maybe you can end the debate. did the problem concern surfaces or solids?</p>

<p>if solids, the answer is a point and a sphere</p>

<p>if surfaces, the answer is a circle and a point</p>

<p>@jamesjunkers</p>

<p>as you saw today, there was no calculus on the exam. if you want to improve next year, you’ll have to improve your algebra skills</p>

<p>@tetratic: Nope. To envelop means to surround. You can completely surround a smaller sphere with a larger sphere. Shak, please confirm for me if this was your intended message.</p>

<p>Hey what was the answer to the sin A = 1/2, with side A being 12 and side B being 6 where it asked for sin B?</p>

<p>I got 37.5… Is that right?</p>

<p>@jeffisaboss: the scores are available on June 20th</p>

<p>It was surfaces!! And what’s you guys get for number one lol the one with distinct roots. 2 right</p>

<p>Cone problem was 2. Radius was supposed to be kept constant, not any sort of the ratio stuff.</p>

<p>Alyssa </p>

<p>i think sin(B) was 1/4 or .25</p>

<p>"Hey what was the answer to the sin A = 1/2, with side A being 12 and side B being 6 where it asked for sin B?</p>

<p>I got 37.5… Is that right? "</p>

<p>That one was 1/4.</p>

<p>The problem isn’t even whether the question referred to 3-dimensional shapes or borders, lol. When we get the results back we will find that out. The problem is your misinterpretation of what I and others have said and your argumentative, annoyed response to my attempt at asking you if you thought the question was aimed at borders and not solid shapes. In any case, I don’t want to discuss with you anymore.</p>

<p>@alyssa193</p>

<p>37.5 is outside of the domain of sin… you had to use the law of cosines.</p>

<p>sin A / 12 = sin B/6</p>

<p>(1/2)/12 = sin B /6</p>

<p>sin B = 1/4</p>

<p>@globalwolf</p>

<p>alright, i’m done with the debate. even if you’re right (which you aren’t) I still got an 800</p>

<p>"@jamesjunkers</p>

<p>as you saw today, there was no calculus on the exam. if you want to improve next year, you’ll have to improve your algebra skills"</p>

<p>The area formed by the intersection of the two lines where they asked you to find the slope of the second line was easily done with integrals.</p>

<p>Disagree with cone answer being 2, problem specifically stated similar, new cone cannot be similar if the height is radius is kept constant.</p>

<p>Was number 1 both -2 and 2 or just 2? Having a mini heart attack right now…</p>

<p>“alright, i’m done with the debate. even if you’re right (which you aren’t) I still got an 800” </p>

<p>Both of you need to calm down</p>

<p>@tetratic:</p>

<p>nevermind, I got that :D. what was the problem to which the answer was 37.5??</p>

<p>oh i remember that, i did it with integrals lol</p>

<p>@Nuggers, both -2 and 2 were roots.</p>

<p>how was the cone not 2? my friend and i both got that</p>

<p>what was the letter answer for the hill question? tetraticcccccc</p>

<p>alyssa193,</p>

<p>someone was confused- the answer to the law of sines problem was 1/4, not 37.5</p>

<p>Does anyone remember the answer to the question where they gave you an imaginary number (a + bi) with s and r, and you had to solve for r?</p>

<p>@beforethedawn i completely guessed and got 6. PLEASE ANSWER MY QUESTION! :(</p>