<p>somedude.
all i did was just plug and chug. i put in random numbers for x, solved for y, put it in the other equation w/ a and b, solved for a and b...</p>
<p>i think i got a = 2 and b = 4/rad3 (or maybe it was rad3/4? dont remember)
i'm pretty sure i got it right =/</p>
<p>darn, I think i got it wrong then. I just divided both sides by 16, and got </p>
<p>x^2 over 4 + y^2 over 16/3 = 1</p>
<p>but it was asking for a and b, not a^2 and b^2 right? If thats the case they fooled me.</p>
<p>yeah, I think it was asking for a and b, not their squares.
So you just took the square root of 4 and 16/3 = 2 & 4/root3</p>
<p>Cmmon Hersh</p>
<p>Anybody ready for Hrevnak's mini test tomorrow?</p>
<p>OMG I forgot about that...Im gonna fail...I didnt study at all</p>
<p>anyone get the answer for what is the lowest divisible term of (x^3)(y^5)(z^7)...</p>
<p>x3y6z9</p>
<p>Choice B.</p>
<p>When I went to take this test a crippled kid opened the door for me.</p>
<p>Nice kid.</p>
<p>I love that kid.</p>
<p>umm for the question on top I put x^3y^3z^3, it made sense so idk, when u plug in x=2,y=3,z=5 then u do 2^3(3^5)(5^7) u get 151875000, then u plug it in for the answer 2^3(3^3)(5^3) and u get 27000, if u divide it, they are divisible and it asked for the smallest one that is divisible. </p>
<p>Hope that is right.</p>
<p>For the question that asked about the smallest possible number that is a perfect cube and divisible by (x^3)(y^5)(z^7), the answer would be (x^3)(y^6)(z^9), because 3, 6, and 9 are all divisible by 3, which means it's a perfect cube (think about it...2^3 and 2^6 and (2^3)(3^3) are all examples of perfect cubes), plus 3, 6, and 9 are larger than 3, 5, and 7, which is necessary for it to be divisible.</p>
<p>For ellipse question,I actually did the same way as somedud did. you are supposed to divided it by 16 and take the square root out of it. doesn't that it? and what is all about the mini test????????!?!?!?!?!</p>
<p>For all those that took the Math II in May, which prep book seemed like it did best to prepare you? D will take it in June and has Barrons and PR. If she had to review only one book, which one would it be? How about the CB's SAT II prep book?</p>
<p>Thanks!</p>
<p>I have Kaplan's and Barron's book. As we all probably know, Barron's book is much much harder than the actual test. I'd be lucky to get about 25 outta 50 right on the Barron's practice test, but maybe thats just me. If you can score well on the Barron book though, you are way more than set to do well. </p>
<p>Kaplan's book was alright. About the same level as the May test, however it has quite a few typos I've noticed. </p>
<p>However, if you know your math, you shouldnt really need anybook. Just know what kind of questions to expect, do a quick review, and you should be set for at least 700.</p>
<p>Kaplan has hard geometry related problems, from my own experience, but maybe that's just me. Otherwise, Barron's is great to review from, I could get around 700 on average, 680-740 ish, on the practice tests from that book. And on the May SAT I am pretty sure I got over 780 easy. You should use barrons until the last few days, and I mean really use it, and then the last few days take the practice exams from the collegeboard book so you can feel good about yourself.</p>
<p>Thanks! Heard that Barrons is tough, but can you do it on your own? Would a Princeton Review tutor be worthwhile if a lot is riding on this Math II score? </p>
<p>How is the PR book for Math II? Is it the second choice to Barrons, or is it Kaplan that is the second choice?</p>
<p>Thanks again!</p>
<p>i would get both the PR and the barrons. I did just the barrons, was prepared for the hard problems, but was unprepared for the easier ones. With PR and barrons, youll be prepared for both.</p>
<p>The question about the polar coordinates! Ugh...</p>
<p>^ That one was easy. If I'm thinking of the right one.</p>