<p>yeah sorry but (5-x)(5-x) doesn’t equal 25-x^2 you’re getting confused with (5-x)(5+x) which you can’t pull out of a radical</p>
<p>I put integral under lnx-k from 1 to the intersection plus integral under 5-x-k from the intersection to 5 and set it equal to the answer from part a divided by 2.</p>
<p>i know…but i messed up most of the area/volume frq so im expecting like 1 point</p>
<p>@Johnhughzy </p>
<p>I did part (a) before anything else, and just ended up making my own equations to solve…I found the line segment from 1 to 2 (for example, don’t remember the specifics), then integrated it. For the g(-2) one, I found the area of a half circle, and subtracted it from the area of the triangle made to the left of it. </p>
<p>Is that what we were supposed to do? It wasn’t specific about what it wanted, so does that still get credit?</p>
<p>syndekit, you could have used y’s too right? like the integral from 0 to k = the integral from k to whatever?</p>
<p>The test was not nearly as bad as I thought it was going to be. I am hoping for a 5. The calculator MC was incredibly easy; I really only guessed on 1 or 2 questions. The non-calc MC was the hardest part of the test, in my opinion. The free response was nice.</p>
<p>Now I’m eagerly waiting for my scores!</p>
<p>Which part do y’all think was easiest? hardest?</p>
<p>I did the integral of lnx from 1 to k and set it equal to the integral of lnx from k to the intersection plus the integral of 5-x from the intersection to 5. But I have no clue if that’s right or not.</p>
<p>Yeah I think so, didn’t think of that! Does my answer sound right?</p>
<p>I really thought the non-calc MC was the easiest, then the calc FRQ. The hardest were non-calc FRQ then calc MC, IMO.</p>
<p>****t… was that one to set up an intergral where a y=k splits in half I read what I wanted to read and assumed it was a vertical line that was splitting the integral in half… Grrrrrrrrr. And mada if I’m following you correctly then I believe you did that one right. If it asked for g(2) then you should have found the area under the line segment between 1 and 2 on the graph of f etc</p>
<p>I don’t think that would have worked since y=k is a horizontal line, not a vertical line.</p>
<p>YESSSSSSSSSSSSSSSS
That was my guess!!! :D:D:D:D:D:D:D</p>
<p>There was a question about linear increase of population</p>
<p>did anyone get</p>
<p>“dP/dt = 200” ???</p>
<p>Your answer actually does make sense when I think about it. Although I set mine up relative to a vertical line… so if it’s right that’s pure luck. =P</p>
<p>The g(-2) was the area of the circle plus the area of the triangle with the signs on both flipped because you were integrating backwards.</p>
<p>@Johnhughzy Thanks! </p>
<p>I did what another person here said for the k-split problem. integral from 0 to k of ((5-y)-e^y) = integral from k to 1.34 [the intersection point…I forget] of ((5-y)-e^y)</p>
<p>I’m forgetting the numbers and equations, but that’s basically what I did–in terms of y!</p>
<p>dP/dt = 200 was correct</p>
<p>pkim850 I chose that answer for the linear pop question too</p>
<p>also
I don’t remember the problem but
I picked [e^(3/2) , infinity] ----> range</p>
<p>for concave down… i have no idea what the problem was…
anyone else get that??</p>
<p>Okay I can say that everyone in my class read that line K question the same way I did. There are going to be tons of kids around the country who assumed it was a vertical line splitting up the intergral and therefore set it up that way do you think it will get any points?</p>