<p>precocious-then did you have the fourth of a circle question that equaled pi/4?</p>
<p>Wait guys for this pi / 4 one. Did you have a circle inscribed in a square with half of a side being 2? So the area would be 4 pi. And pi would be the area of the shaded area. It gave that 2 was half of a side of the square. Are we talking about the same question?</p>
<p>2 was the length of the whole side</p>
<p>What did everyone get for the pressure and volume one?
It asked for an inverse relationship.</p>
<p>I put pv=500, because P1V1=P2V2, and pv=k, is Boyle’s gas law lol… and because if it equal to a constant, their values will be in an inverse relationship.</p>
<p>I didn’t have that question, it must be experimental…</p>
<p>kcajgnaw: yes, I did - and I’m pretty sure I had the extra CR section because I only had three math sections… and like a bajillion CR at the beginning…</p>
<p>I did have a pressure depth one, with the pressure increasing by 7 each time or something. But thats not what youre talking about, is it?</p>
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<p>that question was pretty poorly worded. usually when they indicate the length of a side they put a bracket next to it, but they didn’t on this problem. i was kinda confused too but then i assumed it was the whole side. they really should toss this one out.</p>
<p>Population of school: [(110-2x)/100)]P??? what was that question like?</p>
<p>ok, i’ve done it over and over, and the answer to the pyramid question has got to be (sqrt3)m/2. not only is it a 30-60-90 triangle, but if you use pythagorean theorum, you get:</p>
<p>(m^2)-(m/2)^2=h^2
m^2-[(m^2)/4]=h^2
multiply by 4/4 so you can subtract:
4m^2/4 - m^2/4=h^2
3m^2/4=h^2
square root it and you get (sqrt3)m/2</p>
<p>i think people forgot to square the denominator for m/2. it is definitely (m/2)^2, NOT (m^2)/2. am i wrong here, because it seems pretty clear to me. by all means, correct me if im missing something.</p>
<p>I got that it was 45-45-90 triangle</p>
<p>no, because you are dividing the triangle in two with the height, so the sides are m/2, m, and h. 30-60-90. look at the diagram on the previous page.</p>
<p>sorry, thinking of something else.</p>
<p>but I got m/sqrt2 for that question</p>
<p>Plug in. If M = 2 then the two diagnols from a vertex of the base are equal (45-45-90). So then they would each be square root of 2. now you know the base and you need height. You know that e = m or 2 in this case so…
square root of 2)^2 + H^2 = 2^2
2 + H^2 = 4
H = square root of 2. Only A fits.
I cant believe I missed the pi / 4 one. I ALWAYS MAKE STUPID MISTAKES LIKE THIS></p>
<p>its not m/sqrt2. it has to be (sqrt3)m/2. i explained why in the other post, and im pretty sure im right.</p>
<p>you start with this</p>
<p>(m^2)-(sqrt(.5)<em>m)^2=h^2 since the lower leg is sqrt(.5^2 + .5^2)</em>m, not m/2</p>
<p>(m^2)-(.5<em>m^2)=h^2
h = sqrt((m^2)-(.5</em>m^2))
h = m / sqrt(2)</p>
<p>
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<p>Lux no… listen you know the base and you know the hypotneus (the distance from one vertex to the top is hypotenuse). So all you need is that base or the diagnol from one vertex to center which happens to be m / square root of 2. Obviously the height would also have to m / square root of 2 in order to fit the pythag. theorem. Remember figure was NOT drawn to scale.</p>
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<p>sry, but m is the hypotenuse. h and 0.5m(root2) are the legs.</p>
<p>i know that, hence the m^2-(m/2)^2=h^2</p>
<p>Your mistake is that you say its m/2 squared in the first step. Draw it. how can m/2^2 + m/2^2 = m^2? That makes no sense. It MUST be</p>