<p>Remember, no specific questions until 12:00 Eastern Time.</p>

<p>However, overall I thought the test was ok. I really didn't prep at all (2 practice tests in Kaplans) and I only omitted 5. On some of the "harder" ones I did indeed get an exact answer they wanted and the questions I omitted weren't really "hard" ones parsay, just really weird ones.</p>

<p>I thought it was ok. Specifics are ok now, 11:00. The hardest one i thought was the area of the triangle with a 30*angle and sides 4, 6</p>

<p>yeah, it was a bit tricky. What you do is drop an altitude from the corner of the 4 length side and unknown length side to the base of 6. That cuts the big triangle into two smaller ones. The left triangle has hypotenuse 4 and is a 30-60-90 rt triangle - so the "base" of that is 2 (which is the height of the big whole triangle). 1/2bh = .5(6)(2) = 6</p>

<p>I put 6, but now i think its worng. if you cut the bog trianle into 2, the angle wouldnt be 30 any more, right?</p>

<p>i didnt think it was too bad, i skipped 3, and guessed on 4.</p>

<p>What about the one with probablity of the 2 numbers being greater than 5?</p>

<p>ya that was one of the ones i skipped.</p>

<p>i had .90 as the probability of numbers greater than 5. I did a chart thingy for the sums of the beginning numbers (like 1,2,3,4,5) and you get 10 numbers less than or equal to five and there are 100 total sums so its 90/100 = .90</p>

<p>i used law of cosines to figure out that 4 and 6 one and got...3.23
anyone else get that? I skipped that probability one, what about the ferris wheel one at the end? I got 10.25 b/c it asked for interval, so i graphed the function and the line y=30 found the intersection points, and subtracted to get 10.25 is that what others got? someone told me it was 7.87 but i thought that was just the first interval where h=30 and past that h is less than 30 until like 18 something....</p>

<p>i got .9 for the probability one
the vector problem was 7.. i believe?
what was the problem with with f(100)? i think i guess 1.</p>

<p>i thought the area was 6 but oh well... the ferris wheel i got 7.87 seconds as well</p>

<p>i omitted the f(100) one. f(0) was 0, but f(10) was 1. I think every 10 it went up 1. for f(20) was 2, f(30) was 3, and f(40) was 0. That would put f(100) at 2 but i wasnt sure so i omitted.</p>

<p>WTF with #49 split-field thingy!!!!!!!!</p>

<p>the area definately wasnt 3.23.. if you used law of cosines that just gave u the length of the third side.
I dont remeber what i put for the area though.. i think i put 12, was that a choicE? I probably forgot to divide by 2.</p>

<p>f(100) = 1</p>

<p>Wait, what was the exact wording of the 30 degree angle problem?</p>

<p>dangit im forgetting all the problems, i think i forgot to solve for area of the triangle one</p>

<p>They gave you a side-angle-side triangle. side 1= 4 angle=30 side 2=6 and they asked for the area of the triangle.</p>

<p>how could it be 6.46? I dont thnink that was a choice.
Also, did anyone get the answer for the 4 by 4 pyramid question?</p>

<p>The answer to the ferris wheel problem is simple.</p>

<p>The question is how many seconds is the ferries wheel below 30 ft after it starts from the bottom.</p>

<p>The equation given is h(t) = 29 + 25 * cos(pi/10 * (t-3)).</p>

<p>Set h(t) = 30.</p>

<p>30 = 29 + 25 * cos(pi/10 * (t-3))</p>

<p>1/25 = cos(pi/10*(t-3))</p>

<p>arccos of 1/25 is apprx. 1.5307. divide by (pi/10) and you get 4.8726. Add 3. Answer is 7.87.</p>

<p>The pyramid question was easy. A 4 by 4 square base of a pyramid with height 8. Find the length of one triangular face from the center of the base to the top (paraphrasing).</p>

<p>Look at it from the side. Set up a right triangle with side 8 and side 2 (because the distance to the edge from the base is half of the total base). Sqrt(8^2 + 2^2) = 8.246</p>