<p>There goes Harvard. WH is gonna be like a 740 at best, so my SAT IIs will be 800, 780, and 730ish…dammit</p>
<p>say hi to an 800 most 93% chance</p>
<p>blanche, it was defff I and III… Instead of graphing, you should’ve gotten the inverse of all of them by swapping all the Xs and Ys. When you did so, both I and III look exactly the same, once you solve for Y.</p>
<p>what? rofl? what d’ya mean? So 6 omits is an 800?</p>
<p>I am always amazed by the rate with which the Math II threats proliferate.</p>
<p>yeaaaaa!!!</p>
<p>^ you might say that it has a growth rate of 25%, compounded every minute 
teehee</p>
<p>It can be an 800 if you get everything else correct. Cheer !!</p>
<p>hmm, so plugging the numbers in for the vector question, the answer turns out to be 11…but only 12.0 was a choice i believe, right?</p>
<p>^ nah I think 11 was an answer…that was one that I skipped due to time constraints.</p>
<p>Hey quick question, guys - do you remember one that was like, if b and c are real numbers, how many answers satisfy…x = …?</p>
<p>was this one “infinitely” many solutions?</p>
<p>it was 2. that type of quadratic equation has exactly 2 “real” roots.</p>
<p>i got answer = 2</p>
<p>on a similar note, you guys remember the one about the cubic equations? something like a(x+b)…?? It showed a graph and asked about the relationships between variables. I said a had to be positive, because the cubic graph wasn’t inverted, and that b=c, because two of the zeros were on the same x-axis point. thoughts?</p>
<p>I put 2 as well for the x=…+c question, after a bit of contemplation. I just took it as x must equal y.</p>
<p>And hookem, the graphing method works (I was trying to conserve time), but I must’ve typed it in wrong. I remembered choice 3 and now I see it is most definitely symmetric w/ respect to the line y=x.</p>
<p>^ yeah for the quadratic, just remember that every quadratic equation, when using all positive numbers, will give you exactly 2 real solutions. They could be integers or decimals, but it doesn’t matter</p>
<p>Yeah but weren’t b and c just “real numbers”? So they didn’t have to be positive…maybe I’m getting confused with another question (after all, I saw way too many variables today) but the point is, I put 2, so we’re all good. :D</p>
<p>Btw for your question on the cubic equation, I put the same thing as you I think…</p>
<p>I omitted 8 and probably missed 2 - is that over a 700??</p>
<p>The cubic had two roots, one of which had a multiplicity of two. The a value had to be positive.</p>
<p>^ So b=c, and c does not = d? That would yield two distinct roots and one multiplicity…</p>