****OFFICIAL JUNE 2014 SAT MATH II THREAD****

<p>I multiplied out that ENTIRE thing. And my lovely TI-89 was able to do it, and got the big fat number I had above. It was 4, I’m pretty sure.</p>

<p>what do you guys think the curve is gonna be like?</p>

<p>How did you multiply?</p>

<p>The octohedron was 1/28, and not 1/16, wasn’t it :/</p>

<p>ahh no i hope not i got 1/16 @BassGuitar‌ </p>

<p>and srsly y’all google doc</p>

<p>It was 1/16, at least according to my brain. The combinations of 13 were 5/8,8/5,7/6,6/7, and multiply 8 by 8 by the counting rule and you get 64, 4/64=1/16.</p>

<p>And I just plugged in 2x4x6x8x12… as for SATManiac’s question.</p>

<p>I was thinking about it in the car ride home. I actually don’t think it’s 1/28:</p>

<p>5 and 8 = 13
6 and 7 = 13</p>

<p>1/2 to get of the 4 numbers above. 1/7 to get it’s match on the next roll. Multiply them, and oh poop it was 1/14…</p>

<p>pretty sure it was 1/16 :)</p>

<p>If it wasn’t 1/16, I’m going to be very, very upset.</p>

<p>It was 1/16. (1,2,3,4,5,6,7,8)*(1,2,3,4,5,6,7,8) = 64 different combinations. You can have [5,8] [6,7] [7,6] [8,5]. That’s 4/64 = 1/16.</p>

<p>@mrnephew What did you multiply up to?</p>

<p>That box problem really got on my nerves.</p>

<p>So did that g(x)(x-2)+r problem.</p>

<p>@mrnephew What did you multiply up to for that 2<em>4</em>6<em>8</em>12…</p>

<p>Yeah, wth was up with the g(x)(x-2)+r problem?</p>

<p>f(2)=g(2)(2-2)+r
f(2)=r
Its true for all values x, so this allows us to replace x with 2</p>

<p>Nope, I multiplied up to 98.</p>

<p>BassGuitar: I didn’t get that. I would have gone back to it, but I was just done at that point.</p>

<p>You can still draw a great circle connecting points A and B within the sphere. A and B will just have to lie on a different cross section, but that cross section will still have a diameter equal to the diameter of the sphere…</p>

<p>I don’t understand that. There’s no way you can connect both points with the line being equal to the diameter. </p>