<p>^most of the SAT questions are like that… it’s only tricky b/c of the time constraint</p>
<p>That’s true Walton, maybe the ‘8’ in the diagram only counted the portion of the pail not including the lip? I wouldn’t know though didn’t look at it closely.</p>
<p>Oh and um, was the answer to the oatmeal cookie tin “E. Not enough information to determine the answer”?</p>
<p>I’m gonna be ****ed if there was some sort of gimmick to that -_-</p>
<p>out of all of these questions no one seemed to have trouble with #16 (the last question) on the 20 min math section??? It was the one with the right triangles and sides of 8. Anyone remember that one?</p>
<p>^ yes it was</p>
<p>and yea i remeber that q dont remember what i put</p>
<p>Cup explanation, i think this makes sense:</p>
<p>ok, imagine your cup with height 10. from the top to a certain point is 2, and the rest is 8. So you start out with 2+8. Then each additional cup adds 2, which makes the equation make sense.</p>
<p>8 doesnt refer to the height of the cup, it refers to the height minus 2.</p>
<p>I think that was experimental, the last question for my 20 min math was the 0 1 2 3 4 thing with the n’s.</p>
<p>Anyone know the answer to the oatmeal cookies?</p>
<p>oatmeal was E. can someone chekc my cup explanation?</p>
<p>i forget what i put for the 8 sided triangles, might have left blank??? That was extremely annoying and I didnt figure it out, but not sure if/what i guessed.</p>
<p>why wasn’t the answer 4 for the last question of a section</p>
<p>and 0 is an integer, but they asked for a positive integer. Therefore the answer was D)3. </p>
<p>i was flying through the questions and missed the POSITIVE part, so I included 0 in my answer as well. Greatttt.</p>
<p>Oh was it 3? I didn’t see whether it was positive integer or just integer, I hope it’s 3 =[</p>
<p>hmm…actually I don’t remember. Maybe it didn’t ask for a positive integer? ahh does anyone remember?</p>
<p>It was just integer.</p>
<p>sweet, then the answer is 4. 0 is in fact an integer.</p>
<p>For the absolute value of f(x), the line had a negative slop, and since y=f(x), wouldn’t that make the absolute value(also equal to y, which is negative?) negative, which meant that it was an UPSIDE DOWN “V” instead of a regular V? </p>
<p>I just feel so bad that I gave this question away, I don’t know the answer for sure, but I have a good feeling that it was the regular V. I had the regular V first, but then I thought about it being a negative slope so I changed it just knowing I would get it wrong but I didn’t know what to do, the negative slop looked like something to fool you.</p>
<p>its a regular V, sorry.</p>
<p>Can someone compile a list of answers?</p>
<p>It was the regular V.</p>
<p>The key is taking the absolute value of the y values.</p>
<p>Everything with a negative x value already had a positive y value. The absolute value of a positive is STILL positive.</p>
<p>Everything with a positive x value originally had a negative y value. The absolute value of the negative y values reflected that half of the line over the x axis.</p>
<p>Thus: Regular V.</p>
<p>But wasn’t it a negative absolute value? A negative absolute value will always make it negative.</p>
<p>I don’t think so…</p>
<p>I thought it was just a regular old absolute value.</p>
<p>Nevermind I got it wrong! The little things like this will keep me out of 700 I think. This and possibly a question with a parallelogram where I might have added the area of one of the triangles to the rectangle when I wasn’t supposed to, but maybe I didn’t. Still, this absolute value one doesn’t make me happy since we’ve been doing absolute values in Pre-Calculus class recently! That’s what we’re doing now, and I had a feeling that my way of thinking was a bit too advanced for a medium level question. If I get a 700, I’ll be both surprised and overjoyed considering some of these mistakes.</p>