<p>What did you guys get for the question they asked about the resultant point after reflecting a point over the xy-plane?</p>
<p>"if ab is divisible by c", it was the only one in which neither b nor c was divisible by a. I think I put C. Don't remember answer though, but I'm sure I had correct answer. </p>
<p>And the recursive one was 2*previous + 1</p>
<p>-7, -13, -25 was the series</p>
<p>jabbermouth - I got most of those answers too. i got the period of that graph to be 2(pi^2) though, not 2/(pi^2), and I don't remember that 112 question either.</p>
<p>I wasn't really sure how to do the vector question? I knew what the x,y,z plane looked like because my math teacher just mentioned it briefly, but i wasn't even sure which was x, which was y, and which was z . . . .</p>
<p>for the vector problem i wasn't sure, i subtracted the angles and used law of cosines to find the third side (that's adding the vectors)</p>
<p>justatest - I did something similar. I moved vector B so that it's tail was at the head of vector A. I found out what that angle was between A and B. And since you were given the magnitudes of vector A and B, i just used the law of cosines to find the other side.</p>
<p>And I think someone else posted earlier about that question with the two sides and the included angle and you had to find the area. The short formula for that would be .5absin(C), in other words, .5(one side)(other side)(sin of the included angle)</p>
<p>for the triangle one with 15, 12, angle 25...isn't the formula 1/2 AB sin C</p>
<p>i forget who asked.</p>
<p>and how can 3 planes intersect 2 form a point? i couldn't figure that out...i only put a line. for the a divisible, b divisible...blah blah i put the one that started with....(6, 3, 8) i'm pretty sure</p>
<p>oh just realized galgrl said the 1/2 ab sin C thing</p>
<p>Wouldn't you just make the x,y coordinates negative if you reflect over the x-y axis?</p>
<p>edit: and it's not 1/2sidesidesinangle, because a&b have to be the height and the base.</p>
<p>i agree with jabbermouth on all of those except im sure the answer for the period one was 2pi^2, he just put divided. i got 9,000,000 and 112 and such also. yeah the answer was 38 for that one question, i used law of cosines.</p>
<p>I've heard that the recent curves have been that you can get 8 wrong (no omit) and get an 800...that you can have a score of 40 essentially...</p>
<p>So far I omitted 3 and got one wrong (I divided the period by pi, didn't multiply it dammit)...800?</p>
<p>Ohnoes, draw a picture of it, and think about it...when you reflect it over the axis formed by the x and y plane, all you do is reverse the z coordinate. In a normal graph, when you reflect over the x axis, do you reverse the x score? Nope ;)</p>
<p>dicked that test in the eye</p>
<p>yeah for the three planes intersecting
i know its a point and a line</p>
<p>but is it also another plane?</p>
<p>Piano, no it cannot be another plane...they're distinctive planes first off, and if the first two intersect and only form a line, then they all can't converge and form a plane.</p>
<p>ok i think i only put line and point</p>
<p>hopefully :(</p>
<p>ohnoes - yes the formula is .5absinC or .5(one side)(other side)(sin of the included angle). I think you're getting the more common formula for the area of a triangle (.5 x base x height) mixed up. For that one you actually need a base and a real height, for this you don't. You can derive this formula from the law of sines. So, yes .5absinC or .5(one side)(other side)(sin of the included angle) is definitely the equation.</p>
<p>Oh and I also got that 9 x 10^6 thing.</p>
<p>I wasn't sure about the point line question, because I thought it said that when the two plans are side by side all the points collectively could form which of the following (or something to that effect).</p>
<p>I've heard that they have a fairly generous curve for Math IIC, but I'm not sure if it's if you get 8 wrong you still get an 800 . . . .</p>
<p>Yea I was wrong, the curve is slightly less...you need a raw of 43 for an 800, so 6 wrong.</p>
<p>i omitted two.....including that area of the triangle that required you to know the formula when given SAS. i have seen that formula before but i really never use it.....ever. lol</p>
<p>last question was A i think . . . someone posted it earlier. You needed to know your trig identities, I think the answer was. ((arcsin (K-1))/2)</p>
<p>6 wrong for an 800? If that's right, that's still pretty good</p>
<p>so is the last question(arcsin..) E?</p>
<p>And does anyone know the answer for the x,y,z reflecting problem?</p>
<p>I changed the signs of x and y only. (-1,-2,-3) something like that.</p>
<p>**and should I take another IC Test if got 780 on it on January.
cuz although that's not a bad score, I seriously want to get 800.
But I'm really afraid that I'm gonna mess up again.
It will look so much better if i got 800 on IC cuz so many people get just 1 or 2 wrong (just like I did..).</p>
<p>I just posted that I'm pretty sure the last one was A? but I'm not positive.</p>
<p>The x, y, z problem was 1, 2, 3 you only change the z coordinate.</p>
<p>Yea, x,y,z was 1,2,3, explanation in an earlier post.</p>