<p>wouldnt the series be like x+x^2+x^4+x^6…im really confusing myself here. I know about the coefficients but I didnt think there were any for this type of series.</p>
<p>ohok, i get it when i actually take the derivatives. I guess I was getting a little lazy lol.</p>
<p>Can anyone look at 2008 BC FRQ Form B Question 2(b)?</p>
<p>It makes no sense to me… Where did the number for dg/dx come from?</p>
<p>Strategy question: how many raw points necessary for a 5? For a 4? Are you toast if you just totally don’t know one of the free responses?</p>
<p>For a 5, you need above a 60 to be in the high 4’s / low 5’s depending on the curve. Above a 70, you’re definitely in the 5’s</p>
<p>for the official practice exam…how do you do 2 and 3 on the mc?</p>
<p>That was a confusing problem. I got flustered with what exactly they were looking for. I think you take the derivitave of g, because they are asking for the rate of change of gasoline use, which is a function g. So, you take the derivitave of g with respect to x, which is what you are solving for. Since they give you t instead of x, you have to use the top equation to solve for x by plugging in t=2. I think this is how you do it. Not positive, though.</p>
<p>Do you have the link for that?</p>
<p>just copy and paste what the guy wrote into google search.</p>
<p>How many raw points are there total? Sorry if this is a stupid question lol.</p>
<p>Well everyone, goodnight, and good luck.</p>
<p>108 raw points total</p>
<p>108 points in total. You need like a mid 60 to get a 5, depending on the curve.</p>
<p>r(t) is velocity so you just integrate that to get x as a function of time, and plug this x(t) into the g(x) to get g as a function of time</p>
<p>^ Oh… thanks. :)</p>
<p>Does z substitution or trig substitution ever come up? I haven’t seen anything involving those in the questions that I’ve looked at.</p>
<p>z substitution? Is that the same as u substitution? </p>
<p>Anyways, trig substitution may come up as a MC question.</p>
<p>Is Z substitution the same as “v” substitution like integral of (x)(2x+2)^1/2 dx where v = 2x+2 so x= (v-2)/2 and 2dx=dv so it becomes (1/2) the integral of(v-2)/2 *v^1/2 dv so it becomes 1/4 integral of [v^3/2 - 2v^1/2]dv = (1/4)[(2/5)v^5/2 - 2(2/3)v^3/2] + c. Sorry for writing that whole thing out. lol, once i wrote the initial thing, I had to figure it out for myself. I know that I could have distributed the x into the square root, but I just made it up on the top of my head but you would use v if like the outside factor was (2x+5) or something like that. My calculus teacher tested us on it so I assume that it’s fair game tomorrow.</p>
<p>I’m doing the free response problems from the past years, but what’s different about form b? Is that for people who took it late?</p>
<p>No, you substitute z = tan (x/2) in order to simplify trig expressions inside radicals</p>
<p>It can probably be circumvented by way of trig identities or by way of complex numbers unless the expression is really a mess.</p>