<p>I thought the question said that at most they only shared one root.</p>
<p>not sure if it was roots or values</p>
<p>If its roots then its only 3 or 4</p>
<p>michael2013, how did you get 99th percentile for number 3?</p>
<ol>
<li>Which cannot be the sum? ANS: 150 degrees</li>
</ol>
<p>I forgot what the problem was, can someone tell me what 23 was?</p>
<p>@dotdotdot
no idea, someone else said so</p>
<p>@kobe
question was sinx=1/2 cosy = 1/2
What is not a possible value for x+y</p>
<p>@michael2013, was that question towards the end or the beginning of the test?</p>
<p>@dotdotdot77 All I did was do inverse Normal to find where the cutoff for the 99th percentile was. If I had a z-score table, norm PDF probably woulda been the way to go 
(TI-89 btw)</p>
<p>I’m 100000% sure that the standard deviation question was over 99%. There were a ridiculous number of standard deviations, I think it was around 5. I general rule for the standard deviations of normally distributed things is that anything 3 standard deviations or more is greater than 99% of other things.</p>
<p>This is actually so nice, I’ve never gotten this many right when I talked about it after a test. So far I only have either 1 or 2 wrong, which is still pretty far from not being an 800 :)</p>
<p>Am I the only one that bombed this?</p>
<p>@WhatsanSAT
Same here! I think the words were “the least possible value.” I left it blank. What do you think the curve will be?</p>
<p>@JuniorMint</p>
<p>Don’t worry, I did too. I was getting 700+ on Barrons and only 1 question wrong on the Official CB test, but I already know of 5-6 questions I absolutely bombed today. Stupid mistakes galore, things that I just couldn’t figure out for some reason, etc.</p>
<p>Is applying for engineering reasonable with a low-700 Math II score? I mean, since it appears that a huge percent of applicants get 800…</p>
<p>Don’t talk to your senior member like that young man Rythen 
</p>
<p>No, but really, I am pretty certain now that I actually put z=0 and the x=0, y=0 was the thing I crossed out when my options were limited to 2 answers and I realized that it was defined for any point up and down the z axis. I can’t remember squat. Props to you guys remembering actual numerical values. Sorry for being glad that I didn’t have to draw polar flowers while solving imaginary modulus circular points while trying to remember trig half angle and power reduction formulas and projected dot product vector ****. Cheers to anybody else who was pleasantly surprised with how they did/how reasonable the test was…</p>
<p>By the way, for the question 2 on periods, didn’t it start with f(a-1), then F(a), then F(a+1)
and it said that the period was 3. I don’t know. Maybe, I’m remembering it wrong again…</p>
<p>@Rythen</p>
<p>It’s actually more direct after finding z score = (x-mean)/(standard deviation) to just go and plug it into normalcdf as upperbound with mu=0 sigma=1, so that you aren’t guessing and checking through indirect methods.</p>
<p>Get at me bro.</p>
<p>just warning u guys…we cant talk about the test</p>
<p>but guys what was the curve for recent tests?</p>
<p>Yay. After the fact, I realize it’d be much better to do the conic in polar form, doing BC hw meh.</p>
<p>@meatkabob oh, z=0 is the equation for the plane that is made up by the two lines x=0 and y=0. This plane divides the space in half, one side contains the point (0, 0, 5) and the other side contains the point (0,0,-5)</p>
<p>i really hope the curve is a 42 or a 43 is an 800</p>
<p>^Me too. I thought this was significantly more difficult than the CB practice tests I took beforehand. Nowhere near as difficult as Barron’s, but still a lot more so.</p>
<p>let (x,y,z) be the set of points equidistant from (0,0,5) and (0,0,-5)
then, squareroot( (x-0)^2 + (y-0)^2 + (z-5)^2 ) = squareroot( (x-0)^2 + (y-0)^2 + (z+5)^2 )
simply, solve the equation to get z=0.
it’s similar to a question in the last practice test in Barron’s (5th ed); i’m pretty thankful I finished the Barron’s book haha. (:</p>
<p>anyone remember the entire question of the “number of zeros of f(x)*g(x)” question? sigh…</p>