<p>ya… if i remember it correctly, it was 0.2 mile for that person to go back.
and i think the difference was .2 between both people’s houses in distance from school
so i said it was same for both of them…</p>
<p>by the way, what’s the difference between R test form and Q test form?
Are questions just in different order or are even answer choices in different order?</p>
<p>If what I think is correct and applicable, and assuming the same difficulty as past tests, the curve should be about 70. Remember, for 2003, the curve was 63 or 64? out of 108 for a 5.</p>
<p>I dont know the differences between R and Q test forms. Most likely order of test questions.</p>
<p>thanks man
i asked some people who are pretty good at math, and they objectively thought it wasn’t as hard as 2007 or 2008 tests that they took from online collegeboard.
so i’m guessing the curve will be tougher…
i really hope that doesn’t happen</p>
<p>To anyone who made a “stupid mistake” on the bike problem:
I didn’t get part d because I couldn’t figure out how to find the value of the constant added. ON A DEFINITE INTEGRAL.
I got a 91 on the 2008 exam that we took in class, too. looks like that didn’t happen this time.
I think I just had a really bad testing day, but it definitely seemed harder than last year’s so I hope the curve is nicer. [Well, the non-calc MC was a lot easier, but I usually only get a few wrong on that anyways…why do you hate me, Collegeboard?]</p>
<p>I actually did pretty well. I got full credit (all 9 pts) on the first three FRQs. Got part b) of the area/volume problem wrong (they tricked me hard with that one; it was so easy…). I think I got 5.d wrong and #6 was just completely off lol.</p>
<p>You didn’t even HAVE to square it.
It was just the integral of A(x) from 0 to 2.</p>
<p>I can’t believe I over-confused myself on that one.
I multiplied the given cross-sectional area and the area of the shaded region inside the integral. I’m guessing no credit at all for that (well atleast for the limits lol).</p>
<p>Wait AxeBack I multiplied the cross section by the integral too.</p>
<p>That should be right because it says R is the base of the solid. If you just took the integral of A(x) from 0 to 2 that doesn’t take into account the base of the solid.</p>
<p>It doesn’t matter what the base is. They give you CROSS-SECTIONAL area (which is already dependent on the base), not depth or something. So if you integrate A(x) from 0 to 2, you’ll get the volume.</p>