<p>I'm not sure what you mean by the Sk=k problem, but for the polar form question you form a right triangle with the x and y coordinates and use an inverse tangent to find the angle. The angle for that problem would be equal to arctan(y/x).</p>
<p>FxImpeccable, it helps to draw a diagram for that one.
If you solve for hypotenuse, you will get 1. (which the answer choices already gave you)</p>
<p>If you draw it you will find that it's in the fourth quadrant, which eliminates all the other answer choices except E, which I think was 5pi/3.</p>
<p>i omitted one and got 4-5 wrong...can i still get an 800???</p>
<p>With one omit and 4-5 wrong, we're talking 43-44 raw score, which is 790/800.</p>
<p>And for the x^2 coefficient, I got 5.64 x 2. Why is it 5.64 x 4?</p>
<p>what would a 15 omit score be?!?!?!?!?!?!?!?!?!?</p>
<p>hazhulkhen:
(2x)^2 =4*x^2
so multiply coefficient by 4</p>
<p>because when you plug 2x for x, you would get 5.64(2x)^2. So:</p>
<p>5.64(2x)^2
5.64(4(x^2))
(5.64*4)(x^2)
22.56(x^2)</p>
<p>Thanks for the explanation.</p>
<p>Now I have 3 wrong, and 1 omit. Hopefully I didn't botch any other problems.</p>
<p>i also have 3 definitely wrong and 1 omit. ughhh i really really hope i didn't mess anything else up.. i shouldn't have gotten any of those three wrong.. :(</p>
<p>expect<em>the</em>impossible, what letter was 22.56?</p>
<p>i thought it was hard.</p>
<p>I think it's B.</p>
<p>hmmm... do you remember what C was? i think i may have done x^3 by accident... but i'm pretty sure i put down B or C...</p>
<p>I didn't take it yesterday...
but wouldnt....
cosx^2=-root(1-sinx^2)
have no real values because the first part will always be positive and the second part will always be negative?</p>
<p>oh...nevermind....I wrote the question wrong. So should the answer be quad II/III?</p>
<p>i think it was cosx, not cosx^2</p>
<p>yes it was quad II/III</p>
<p>i'm getting so jittery reading this- i keep forgetting what answer i put down, then freaking out, and now all the problems are just sort of swimming around in my head...
i wish i could just know my score now.. the wait is going to drive me crazy</p>
<p>polar coordinate one was (1,5pi/3)</p>
<p>y = rSinA
x = rCosA</p>
<p>let r = 1, then just plug in the numbers, only tricky thing was to determine which quadrant it was in, and I believe it was in the 4th quadrant, therefore 5pi/3.</p>
<p>I forget what I did. I think I found the angle that would give rt(3)/2 and -.5, then converted it to radians. 1, 5pi/3 was what I had though.</p>