<p>My proctor screwed up and shortchanged us 3 minutes and didn't even give us warnings past the 20 minute one (she should have at the very least given us a 10 or 5 minute warning), which means that I ended up with 7 omits. I'm confident in all the questions I answered, but it's entirlely possible I missed a couple. I'm so ****ed right now. Is it stupid to retake with a 750+? I know I could've got an 800 if I hadn't been jipped.</p>
<p>i think i skipped about 15-16. no kidding. i'm contemplating whether its worth retaking, i could still not finish, you know....</p>
<p>calizoso, if you're confident, you SHOULD retake it. </p>
<p>jipped is an interesting word..</p>
<p>do u guys remember any hard questions? what did u get for #50?</p>
<p>I'm 99% sure I would've gotten an 800 if I had had those three minutes. What I was wondering though is will it even make a difference in college admissions to selective universities if I make a 790 (I think this is what I get with 7 omits) and retake to an 800?</p>
<p>motion detector?</p>
<p>Anyone know the answer to these questions?
A ball went up in the air, initial height was (0,1.5) Max was like (3 or 4, 34), they wanted to know when the ball hit the ground.
Was it 5.2 secs or 6.0 secs?
I guessed 5.2
The other was for some angle value between 0 and 2pi, which of the following were true?
A. Sin2theta>sintheta
B. Costheta<cost2theta c.="" tantheta="">(sintheta or costheta)
A. I
B. II
C. III
D. I and II
E. II and III</cost2theta></p>
<p>I put E and it was the angle between 0 and pi/2. I just looked at the graphs.</p>
<p>yeah i too filled in 5.2</p>
<p>since from 3s-4s the height dropped by 14m therefore in the next minute it will drop even further, but may be not to zero(this was some strange intuition that forced my logic in thid direction) so 5.2</p>
<p>btw, could anyone tell me what the last question, the 50th was</p>
<p>it was 4.9, the information was all there, aka you can set up:</p>
<p>ax^2+bx+c=f(x), where f(x) is the hight, and x is the time, and from the table of values, you can solve for a, b, and c</p>
<p>the quadratic equation was like -4.7x^2+somethingx+something=f(x), and it had a root at at x=4.9, where height=f(x) was obviously 0 since it asked for when it hit the ground.</p>
<p>the answer was 4.9</p>
<p>btw, pranjal i don't think your logic works b/c the projectile path was a parabola that does not "start out" at t=0 b/c it has a positive y intercept, so you can't use the symmetry logic</p>
<p>when x=4.9 the baall returns to the same ht that it was hit from i.e 1.5 meters, and it'll take more than 4.9s</p>
<p>what was the last ques..
it completely is out of my head</p>
<p>pranjal it was 4.9, which was the root of the quadratic that they pretty much told you to make</p>
<p>yeah just plug in the data in calc and find roots</p>
<p>what was the answer to the question: find the value of one of the roots of the quadratic that has two roots, one is two times the other.</p>
<p>last question i got C III only i think</p>
<p>what was the last question?</p>
<p>the one about the lines and planes. i forgot the question but i put c only III</p>
<p>oh i think i put down all but II i think...which is the one where the third plane is parallel to both other planes</p>
<p>It can't be 4.9. There are other coefficents for the T and ofcourse some constant C at the end.
So its something like this: y=-4.9t^2+bt+c. At t=4.9, you'll most likely get something greater than y=0 if C is positive. Which I believe it should be. This was just my instinct. Did anyone know how to solve it? The only way I knew to solve it was to find the equation. If the points that they gave me were not symmetric around the maximum, then the equation was difficult. So I didnt bother figuring it out. I thought it was 5.2. Can anyone confirm this?</p>
<p>What was the answer to the BC chord thing? I forgot how to do that..</p>