<p>hm…was D t >=3? can someone confirm/(hopefully say no)</p>
<p>@whatkneww, it was D.</p>
<p>I remember this part clearly… All of the options had t (something) 3x so you guys are safe.</p>
<p>thanks lanaryu!</p>
<p>Does anyone remember the what #50 (with the laser and the disk) was actually asking? It seems that the consensus here is 930-something feet, but during the test I was nearly fully confident in my answer of 1037 ft. If I remember correctly, it said an ERROR of .05 degrees – and I’ve checked it over and still am getting 1037.</p>
<p>If I’m wrong, well, that sucks. But can anyone explain what the question was asking?</p>
<p>do you think 43 will be 800…</p>
<p>imma cry i skipped 5 and got 2 wrong D:</p>
<p>@mooniistlye
apparently a 43 still got you a 800 on the june 2010 test(exact one we took)
hoping for the best!</p>
<p>@untitledmind72</p>
<p>To the best of my knowledge, this was question 50 (wording might be different but concept is here):</p>
<p>A laser is aimed towards the center of a satellite disk. The satellite disk is positioned in space 225 miles directly above the location of the laser. In aiming the laser, there was an error of 0.05 degrees. By how many feet will the laser miss the edge of the disk? (1 mile = 5280 feet)</p>
<p>Construct a triangle, use simple trig, and you’ll find that the laser “misses” the original intended target (the center of the disk) by 1037 feet. However, the question asks how much the laser misses the EDGE of the disk - so one has to subtract 50 feet, for a final answer of 987 feet.</p>
<p>I omitted 4 and got 2 wrong so far. This is NOT looking god for me. I would die if I got lower than an 800. I would have to retake it in OCtober</p>
<p>I also got 987 ft for #50.
Did anyone find a solution (i.e. not plugging really small numbers into calculator) that didn’t involve calculus for lim x^x? The only real solution i could find used l’hopital’s rule.</p>
<p>@ RandomHSer I graphed it</p>
<p>@RandomHSer</p>
<p>I used my TI-Nspire CX CAS. It can solve limits. :D</p>
<p>Pretty sure E was t>=3</p>
<p>@random
x^x was something my ap calc teacher made us memorize, so I’m guessing there’s no better way to do it.</p>
<p>What? Can’t you just substitute in different positive values for x in x^x? 1^1, .1^.1, and .0000001^.0000001 all yield very close to 1.</p>
<p>Its not like the limit is unsolvable… you could do something like this. L=lim x^x
ln(L)=xln(x), ln(L)=lim ln(x)/(1/x) and by l’hopital, ln(L)=lim(1/x)/(-1/x^2)=-x->0
and if ln(L)=0, then L=1.
Yes i did actually do that in the test</p>
<p>@aqua, yes you could substitute, but i was wondering if there was an easy way to actually solve it</p>
<p>…Substituting seems much easier though o_O; do 1^1 and you instantly know your answer.</p>
<p>You can’t really substitute in 1 if the limit is to zero.</p>
<p>Any number raised to the 0th power is 1…including 0. That’s how u do it.</p>
<p>RandomHSer , technically , L’hopital’s rule is only for 0/0 or infinity/infinity .
I didn’t do it with Calc because there is always the pressure of time and making a graph and plugging in x=10^(-8) gave 1</p>