<p>Oh, I see. I wasn’t sure. Thanks</p>
<p>@basketball1 I got C i believe, which ever distance solved from point B to Point C was 2sqrt(13).</p>
<p>For the unit circle chord one, the diagram =
you know the radius is 1, because it is a unit circle. You have a triangle formed between two radius’s, with the distance between the two points from the radius on the edge of the circle being the chord. Also, the triangle is bisected, forming two identical angles.</p>
<p>I did it a long way, because
if you look at just one of the right triangles(formed by bisector), you know that sin theta = x/1, 1 being the radius of the circle, also the hypotenuse of the right triangle. So you know that half of the chord is sin theta, because sin theta = x(half of the chord)
so you multiply both sides,
you get 2 sin theta, = 2x(length of whole chord.)</p>
<p>is it possible to get an 800 if i got 43…?</p>
<p>Also, there was the cube question with each side being 7 and a triangle was formed from diagonals.
I got that each diagonal was a bit over 9 in length, multiplied by three, and ended up with 27 or 28 or something. Sound right?</p>
<p>Also, could someone explain what the (x^2)^3 thing was? I have no clue what’s being talked about.</p>
<p>
</p>
<p>The raw score cutoff is usually 43 or 44.</p>
<p>also, what I did for that infamous last question was this:
I figured out where the laser became pointed. I set up a right triangle with base 1188000 ft, and the height being x.
I set the angle to .05º and solved for x. I got 1087 I think. Then subtracted 50 to find how far that is from the center.
So I got 1037.</p>
<p>Or 987. I don’t remember. LOL</p>
<p>@rust, got that too.</p>
<p>Was the (x^2)^3 one like 2.4 or something?</p>
<p>@rustgust
well one diagonal was 7sqrt(2), that multiplied by 3, is 21sqrt(2) which is 29.7</p>
<p>I’m pretty sure there was a question where none of the choices corresponded to the question.</p>
<p>(a^3)^2 = 123.1</p>
<p>None of the answers fit so I chose the closest answer 2.2.</p>
<p>Hey was the 40 miles per hr/ 60 miles per hour simply avg is 50?
or was there some trick to it</p>
<p>@wang, yeah I noticed that. If you plug in 2.23 though, you get the answer. I think the SAT II rounded the decimal, but it threw us all off.</p>
<p>@ Member123 - i thought it was less than fifty, i plugged in a random distance and did rate times time equals distance for both, it ended up that it was 48 average…</p>
<p>@rockn @wangh haha I had a ti-89, just used solve! for the (a^3)^2=123.1</p>
<p>I give a hearty ‘wordup’ to wang.</p>
<p>wasnt it just his average mph speed? or did we have to incorporate distance into that?</p>
<p>@member123
it would be less than 50
because if its same distnace there and back
the speed there was 60
speed back was 40.</p>
<p>But think about it. Wouldnt you take longer to go the same distance at 40, than at 60?
so you’re spending more of your time at 40mph, thus, the average speed would be closer to 40 than 60.</p>
<p>Member-</p>
<p>It’s E, less than 50mph. The most obvious choice was 50mph, and by the last ten questions of the test, the obvious answers are almost always the ones the College Board use to throw test takers off. :/</p>
<p>@ member
the average speed is 48mph.
I set the distance from x to y as 240miles.
so he took 4 hours + 6 hours total.
d = rt. so total distance(480)/10 = 48mph</p>
<p>how is the f(3) - f(1) = 6</p>
<p>-_- I got 4</p>
<h1>whatisthis - oops. i spent way more time on that problem than i had to. :D</h1>