<p>Julina and that would be choice D?</p>
<p>what’s the answer?
for some reason i remember getting a = 4, and b = 16/3 but i might be wrong</p>
<p>Yeah, the circle thing was D. But I thought I read in this thread that ppl thought the new center was (8, -1)? And I got a = 2, b = 4/root 3 for 49…</p>
<p>Edit: Sunnex3, the values you have are for a^2 and b^2.</p>
<p>because a and b are both originally squared, so it’s a=2 b=4/RAD(3)</p>
<p>well i just asked my friend who is a math genius (got 800 on the Math IIC) about #49 and he got 4 and 16/3 also
maybe someone could explain how they went about getting 2, 4/square root of 3?</p>
<p>well tell your “math genius” friend that it’s a^2 and b^2 not a and b</p>
<p>I had bot hthose answers for the ones you posted on the last page aznosamaboy.</p>
<p>4 and 16/3 are a^2 and b^2</p>
<p>so a and be would be the square root of that</p>
<p>Sunnex3, the problem was like this:</p>
<p>(4)(X^2) - (3)(Y^2) = 16, right?</p>
<p>Divide both sides by 16 –> (X^2)/4 - (Y^2)/(16/3) =1</p>
<p>In the standard eq for an ellipse, 4 and 16/3 would be a^2 and b^2, respectively. So a = root 4 = 2, and b = root 16/3 = 4/root3. </p>
<p>Hang on, was the equation that of an ellipse or hyperbola?</p>
<p>hey i’m not trying to insult anyone here, i was just saying that he is really smart. no point in being sarcastic and mean you know.</p>
<p>i’m just thinking, if it’s x^2/a^2 and y^2/b^2 and you’re finding a and b, wouldn’t it have to be x/a + y/b, becaus eyou can’t just take the square root of one part of the fraction right? and since it asks for x^2/a + y^2/b how can a and b be the roots of a^2 and b^2 when the tops are not the roots of x^2 and y^2?
or am i wrong.
i’m not the strongest at math -.-</p>
<p>Does anyone remember the question sin2x = a, what is 2sinx? I know the answer is a/cosx, but does anyone remember (a) (secx) as like Choice E or something? I think there might have been (a)(cscx) too, and I might have picked that in a hurry…</p>
<p>haha we know you’re not trying to</p>
<p>so what did everyone get for the # of intersections near #40 or so</p>
<p>Julina, yes it was E a*sec(x)</p>
<p>Sunnex3, well, you’re just trying to find the values of a and b, so using the standard equation of an ellipse (or hyperbola?), you can square root a^2 and b^2 without worrying about X and Y.</p>
<p>Number of intersections…was that the one where it was xy^5 - 2X = blah blah?</p>
<p>yea what did you get for that</p>
<p>yea what did you get for that</p>
<p>I think I set both equations = 0 and graphed them both. But ppl have been saying they got either 0 intersections or 2 intersections and I got neither. So I’m probably wrong. Sigh.</p>
<p>Does anyone remember the right side of the equation? I definitely remember the left side being x(y^5) - 2X…</p>
<p>Actually, I think it might be 0. I think I might have accidently reduced 2x/x to x instead of 2. - -;</p>
<p>did anyone get 9sec(x) for an answer? i forgot what the question asked though</p>