November 2009 SAT Math (US only)

Test Preparation SAT Preparation
321

<p>Yes crazy…(but I skipped a number writing them down…)
The folded paper one was 2 holes in each corner, and then 4 vertical lines just left and right of the folded line.</p>

322

<p>Or maybe it is what crazy said. Can’t really remember. -.-</p>

323

<p>the CB mixes all this stuff just to ***** with us. i say who cares we’ll find out in 3 weeks</p>

324

<p>@sahermit, it was a dot at each of the 4 corners, nd 4 dots at the center</p>

325

<p>My picture’s based from memory, so it could be incorrect. I think early in this thread someone drew a picture of it if you wanna go hunt for it.</p>

326

<p>are you sure the third choice was not
(a^2)(b^2)+(c^2)=-2abc
i’m pretty sure it was +(c^2), which would make it true.</p>

327

<p>the prime is not 995 you can break that down by 5.the answer was 997</p>

328

<p>2200orbust: alright thanks, thats what I put but I noticed that earlier in this thread someone had posted a different drawing for that question</p>

329

<p>You’re right, Perry!
Thank you for making me feel slightly more happy again. :p</p>

330

<p>Well if all numbers are positive integers (that’s what the question said right?), then the left side would be positive. The -2 would make the right side negative, so then they wouldn’t be equal.</p>

331

<p>It wasn’t that the number had to be prime, the sum of its digits needed to be prime. It was 995 because the sum of its digits is 23, which is prime.</p>

332

<p>For clarity, the paper folding question:

<a href=“http://i903.■■■■■■■■■■■■■■■/albums/ac232/Joshtch/PaperFolding.jpg[/IMG]”>http://i903.■■■■■■■■■■■■■■■/albums/ac232/Joshtch/PaperFolding.jpg

</a></p>

333

<p>yea i got that =D</p>

334

<p>@rawk
sorry bud you read the question wrong.
they asked for the sum to be prime not the number itself. 9+9+5=23
9+9+7 was 25</p>

335

<p>combine youre right
****tt -1</p>

336

<p>Another masterpiece by Josh!
Is there a mailing list or something I can join so I receive information about any future pieces of magnificent art?</p>

337

<p>If it really was (a^2)(b^2)+(c^2) = -2abc it would still fail when a=2 b=3 c=6 :</p>

<p>(a^2)(b^2)+(c^2) = 4*9+36 = 72
-2abc = -2(2)(3)(6) = -72
72 != -72</p>

<p>However, if the original equation was ab+c=0, then it would work for a=2 b=3 c=-6 :
(a^2)(b^2)+(c^2) = 4*9+36 = 72
-2abc = -2(2)(3)(-6) = 72
72 = 72</p>

<p>But it depends on what the original equation was. I thought it was ab-c=0, but I’m not so sure anymore. :/</p>

<p>And haha shapiro. :P</p>

338

<p>didnt someone get 5 for a gridin? i believe it was about a circle in a square or vice versa?</p>

339

<p>joshtch</p>

<p>the formula stated in the question was a*b+c=0. Therefore if you picked a to be 2, b to be 3, than c had to be -6, and thus the third answer would have also worked</p>

340

<p>Josh, you are really starting to **** me off. Stop playing with my emotions like this.</p>