<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…
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>