December SAT: Math

<p>yeah everyone.. its 1:4.. and im 99% positive that its D and E... i know too bad cuz its needs to be 99.999999%</p>

<p>LOL</p>

<p>I definitely think we all have different forms, since I'm 95% sure of BC.</p>

<p>I got DE and I was really sure about it. I redid the problem a few times to make sure.</p>

<p>BTW, don't want to make a new thread- </p>

<p>Do you guys think it's worth it getting the detailed score report for $10?</p>

<p>i'm 99% sure of DE :)</p>

<p>Are you saying you got DE even after redoing the problem which GlueEater posted on page 49? If you are, then please explain it to me.</p>

<p>I guess if your score was way lower than you expected, then the detailed score report might be useful.</p>

<p>yeah dude.. glueeater has the problem wrong.... its AX = 2XF.. trust me.. i know thats correct</p>

<p>It was DE. If you marked everything and the space apart they were, the 66% mark was between DE. That means there was twice as much distance one way, when compared to the other.</p>

<p>A B C D E F</p>

<p>So, if you draw out a numberline, A=0, B=.2 C=.4 D=.6 E=.8 F=1. If x is at .66 between D and E then it is .33 away from F and .66 away from A. And jsmith is correct.</p>

<p>Exactly! Riyia's got it right!</p>

<p>Someone posted this a while ago and I'm still wondering about it.</p>

<p>if (r^2)st = 200 and rst = 40 ... what value CANT rs be?</p>

<p>a. 5
b. 10
c. 20
d. 30
e. 40</p>

<p>I don't know if integers or not were specified, but if rs=30, then t would have to be 4/3. Does anyone know how to solve this, in terms of if you had integers vs if you could use any #?</p>

<p>I think the difference of opinion here is whether they said AX = 2FX or FX = 2AX. I felt pretty confident in my answer, but I could've read it it too quickly ^^;</p>

<p>I don't remember that problem, the RST one I mean. </p>

<p>I will 100% guarantee you that it said the distance between a and x is twice the distance between f and x.</p>

<p>RahoulVA,</p>

<p>If you could use any number, then all those values are valid.</p>

<p>Mmm...I'm still not entirely convinced (though AX = 2 XF would be the more logical way to present the question), but in this case I'll pray for a -1 Math curve ^^</p>

<p>Yeah I was thinking that you could eventually use any #(even if you get a long decimal). But if it did say integers, would 30 be the answer? Without knowing the answer, how would you do that problem quick enough? If RST=40, then what can't RS equal? Youcan say 30 without even using the (r^2)st=200 known, so either my answer is wrong or they extra information to throw you off.</p>

<p>What was the last gridin?</p>

<p>HMMMM....</p>

<p>Maybe s or t couldn't be a number with repeating decimals? (lol)</p>

<p>if (r^2)st = 200 and rst = 40 ... what value CANT rs be?</p>

<p>a. 5
b. 10
c. 20
d. 30
e. 40</p>

<p>Is it asking you to get certain variables like r by itself so that you get r=100*Square root of 2/Square root of (ST)? That's getting a little complicated, do you then plug in the S and the T into the rst thing? I'm thinking too many things, what's the quick solution to this problem?</p>

<p>If you find r by itself, you'll get 5. However, after that there's no way to find s and t specifically.</p>

<p>I think they said that r,s,and t had to be integers... so rs can't be 30 because 40 divided by 30 isn't an integer.</p>

<p>if (r^2)st = 200 and rst = 40 ... what value CANT rs be?</p>

<p>a. 5
b. 10
c. 20
d. 30
e. 40</p>

<p>look! it said all numbers were positive integers.</p>

<p>the question itself proves that r=5, so st should be 8!
possible values for s are= 1,2,4,and 8.</p>

<p>5x1= 5
5x2=10
5x4=20
5x8=40</p>

<p>therefore, d. 30 can't be the answer!</p>

<p>Souuuu desu ka</p>