<p>The K-f(x) one what I did was just drew a horizontal line above the graph. Then i shaded in the area from the line to the graph. At each x value, g(x) would be equal to that difference of k and f(x) (k-f(x)). So it was the answer where the graph increases but levels out, moving towards a horizontal asymptote.</p>
<h1>49 is 8/11. It includes all real numbers within that range. It would take an infinite amount of time to account for every real number.</h1>
<p>lets just say im going to commit suicide right.........now!</p>
<p>my hopes for a triple 800 have now been completely. and utterly. dashed.
After analyzing the curve before test, I thought I would only answer 1-44 to 45, making sure they were all right, and just ignore the last couple, I could still get 800. Well, I was tired so I only got to 42 (my third test of the day). I had a couple mins left so I made a couple educated guesses. Of the 4 I guessed on, I got 4 wrong. And, I made 3 silly mistakes. That brings me down to a 40-38, so I'll be happy with a 750 at this point. God da mn it.
At least I probably got 800s on my other 2 tests (lit & ush)</p>
<p>I was lucky.. I was not only one who took a SAT test at my test centre..:p MY proctor gave me extra time and even asked me at the end if I want to finish or want to review the answers for a few minutes lol</p>
<p>OK here's my explanation, although it never came in use on the test :P.</p>
<p>say x is between -2 and 2. and we want x to be greater than or equal to 0. You can't say it's 2/4 plus the zero to make it 3/4. it is essentially 2/4. you don't add the 0 in because it is only one point.</p>
<p>it's not even out of twelve. it's out of 11. look at the +-2 again. you know it's out of 4. not 5. if i'm wrong please correct me, i didn't think it all the way through ....</p>
<p>you are wrong</p>
<p>i put 8/11 for it</p>
<p>i did all the questions except for the last one. i dont even remember what the questions is, just that i didnt noe where to start...did anyone else do it?</p>
<p>The same solutions one was indeed the odd one, which was like</p>
<p>x^5-3x^3+x=0</p>
<p>^does anyone remember the other answer choices for this question or what it said?? All I remember thinking when I read it was "try and find the even equation" so I didn't put that.</p>
<p>Oh yah, also for the one which used statistics, was the answer 5!/(3!2!)+5!(4!)=15? I think I missed that because I used a permutation instead of a combination =(</p>
<p>^Sorry, that's a 5!/(4!).</p>
<p>o the one about choosing 3-4 people groups out of 5 people?</p>
<p>5C3+5C4 = 15</p>
<p>Explanation:</p>
<p>the # of unordered 3 people group that can be formed by choosing from 5 people is 5C3 = 10
choosing 4 people is 5C4 = 5</p>
<p>(basicly what you said =p)</p>
<p>the probability one is 8/11</p>
<p>lemme attempt to recall the quetion.</p>
<p>P(A given B) =A/B (think everyone got this part, basicly all # in the set thats > 1 / all numbers in the set)</p>
<p>first yes there are infinite amount of numbers! </p>
<p>the solution is A = 4/3 - 1 = 1/3
B = 4/3- 7/8= 32/24 - 21/24 = 11/24
(1/3)/(11/24) = 8/11</p>
<p>this ratio can be easily represented on a number line
mark 7/8 ~4/3 area as B and 1~4/3 area as A =p</p>
<p>got 15 too.. but it actually the only thing I understand of that whole permutation and combination crap</p>
<p>And I got x^5-3x^3+x=0 too, but that's was also a guess...
Could anyone explain the question?</p>
<p>hey for the remainder of 147 q.. the answer was E right?cant remember the answer..</p>
<p>n for the no. of ppl u can chose from 5.. either 3 or four.. well teh ans. is 25.
5C4 + 5C3</p>
<p>??????
5C4+5C3= 15</p>
<p>and not 25</p>
<p>erm >.> 5C3 = 5! / (3!2!) = 5X4X3X2 / 3X2X2 (omited all the 1s) =5X2 = 10
5C4 = 5!/(4!1!) = 5X4X3X2/4X3X2 = 5
10 + 5 = 15
<3 sry but the answer remains 15</p>
<p>yah dude you made the same mistake as I did; you used a permutation instead of a combination. Permutations only when order matters. Gah I wish I had been paying more attention when I took the test! I realized it after I was already on the Chem SAT II. Can anyone explain the x^5-3x^3+x problem to me tho? I'm not sure I get that, although it does work when I plug it into my calculator. Anyone mind solving it algebraically?</p>
<p>sry skipped that question and didnt have the time to go back to it, so dun remember it at all .. what was the question? was it the roots that had to be equal?</p>
<p>at most how many wrongs can ensure an 800 for math 2c?</p>