<p>I think the answer for this one was (1,pi/2), because it asked for POLAR coords, when you convert (1,pi/2) to rectangular coords you get (0,1) which was the only one present on the cos theta graph.
If it asked for rectangular coords, you’d be right.</p>
<p>Can someone tell me the exact method to do the last question? It was the one about percentiles. I had never seen that sort of question before.</p>
<p>Questions for the October test takers:
What was the October test curve like? 44-45? Was this test easier/harder than that one?</p>
<p>r=costheta, (r,theta) the cosine of one does not equal pi/2.</p>
<p>r=costheta, (r,theta) the cosine of one does not equal pi/2.</p>
<p>To put into perspective how hard october was 
I have 800 Math 1, and so far no mistakes in this test (maybe 1 i cant remember which one i circled). I finished this test in 40 minutes. In comparison, in october I rand out of time, omitted 3, got around 6 wrong.</p>
<p>I cancelled my score though so dunno how the curve was for october</p>
<p>r=costheta, (r,theta) the cosine of one does not equal pi/2.</p>
<p>In regards to the (x+y)^2 mentioned, your issue is that it asked for x^2 + y^2. So the answer is 104.
And yes, it was II only.</p>
<p>Is 3 omit 3 wrong good enough for an 800 depending on the curve or am I definitely below?</p>
<p>Convert (1,pi/2) to rect coords:
x=1<em>cos(pi/2)=0
y=1</em>sin(pi/2)=1</p>
<p>And (0,1) is contained in the cos(theta) graph.</p>
<p>For anyone struggling with the last problem, it was simply a matter of knowing some statistics distributions. The empirical rule of statistics dictates that on normally distributed data (such as test scores)…± 1 StrDev = 68%, ± 2 StrDev = 95%, ± 3 StrDev = 99.7%.</p>
<p>That being said, the problem asks for the percentile with which the score (84) would be in in relation to data with a mean of 74 and a StrDev of 2. Because 84 is 5 StrDev away from the mean, it can only be over the 99th percentile.</p>
<p>the coordinates were polar though… so r=cos(theta), polar are coordinates are (r, theta)
theta was pi/3, therefore r is .5, so the coordinates are (.5,pi/3)</p>
<p>wast the x= 24/104 q
like the x-y=8, xy= 20
so (x-y)^2=64
x^2-2xy+y^2=64
so wouldnt x^2+y^2=24?</p>
<p>r=costheta, (r,theta) the cosine of one does not equal pi/2.</p>
<p>x-y=8, xy=20
y=20/x
x-(20/x)=8
(x^2-20)/x=8
x^2-8x-20=0
(x+2)(x-10)=0
x=-2,10
Since you subbed in, x=-2, y=10 (which one is which does not matter)
x^2+y^2=(-2)^2+10^2=4+100=104</p>
<p>That’s how I did it at least.</p>
<p>how’d u get from</p>
<p>"x^2-2xy+y^2=64
to
so wouldnt x^2+y^2=24? "</p>
<p>I just picked the values 10 and 2 which fit the conditions and subbed in and got 104</p>
<p>Sorry. I did not realize that hitting the refresh button would make that same post come up over and over again. I got 104 for the system one. Referring to Whatisthis1:</p>
<p>x^2-40+y^2=64… x^2+y^2=104. Just a dumb math error. I hate when stuff like that happens to me.</p>
<p>oh well this test was extremely easy compared to when I took it back in June, I finished within 30 min and got to go back and catch all my errors. I expect the curve to be harsh on this one :///</p>
<p>With a harsh curve, would missing three and omitting none still be an 800?</p>
<p>I don’t get it, afair the question was smth like this:</p>
<p>Which of the following polar coordinates describes a point on a cos theta graph.
I chose the answer (1,pi/2), because it’s the point (0,1) in rectangular coordinates, which belongs to the cos theta graph because cos(0)=1.</p>
<p>If I am wrong, please point out my mistake.</p>
<p>they’re talking about a polar graph not a rectangular/cartesian graph, I guess you thought about it too much</p>