Class of 2014 From Nepal

International Students
1521

<p>nice question by Gurung and nice reply by sarbaraj.</p>

<p>but are we asked such questions in SAT as posted by sarbaraj? derivatives are not asked in SAT, are they?</p>

<p>in TOEFL or even in SAT, if we are asked : whether you agree or disagree with this statement, can we agree as well as disagree in the same point. For example, if the question is: What do you prefer: studying foreign language in foreign country or in own country? Can we answer: ‘I prefer basic course of foreign language in own country and higher level courses in foreign country’?</p>

<p>And best of luck for your SAT guys!!</p>

1522

<p>@sarbaraj: haha. But, even if one wanted to do that question there would be no way he’d be able to type the solution here.</p>

1523

<p>@ccprofile</p>

<p>“if you have guts” was halka offending
tesaile 1st order ODEs sodhdya</p>

<p>@unitedacademy
No dude. You’ll learn this kinda stuff in second/third semester of your college. “Differential Equations” bhanne course maa… hehe</p>

1524

<p>@sarbaraj: Yep. True.
@unitedacdemy: As long as you develop your answer properly, it won’t matter. As a matter of fact most prep. books would also suggest you to pick a side and then answer rather than provide arguments for both sides.</p>

1525

<p>FUNNY guys, i was about to ask sarbaraj’s question to my brother. </p>

<p>@ccprofile:</p>

<p>is ur’s SAT also on Oct?</p>

1526

<p>@unitedacademy: Yeah,I have my Subject tests on Oct. And that question was from Grade 12 Science-Maths and the solution is quite simple but I can’t type it out.
So, you’ll be having your SAT I in oct. 10, right?</p>

1527

<p>@unitedacademy</p>

<p>LOL…</p>

<p>sorry for freaking you out tho</p>

1528

<p>Since, everyone here is in the mood for posing a maths question, I thought i might ask one too. :smiley: Here’s the question:
Given a triangle ABC as described below: Side AB = Side AC. Draw a line from C to side AB. call that line CD. Now draw a line from B to side AC. Call that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. The question–find what angle EDC is. (Do not do this trigonometrically ; do it geometrically to get an exact answer).</p>

1529

<p>yeah.</p>

<p>2nd order chai
d^2 y/dx^2 + Py^n = f(x) </p>

<p>form maa hunchha</p>

1530

<p>Okay. Haven’t done those yet and will most probably be doing it in the second semester of college.</p>

1531

<p>BTW, if anyone finds the solution to that problem in post#1528 then please do PM me. I tried solving it a few months ago but didn’t get the answer.</p>

1532

<p>hmmm… I got four equations… XD</p>

<p>x + y = 130
y + z = 140
w + x = 150
w + z = 160</p>

<p>x = reqd answer</p>

1533

<p>And what are w, y and z?</p>

1534

<p>other angles… but sadly, this system of equations can’t be solved with only this much info.</p>

<p>x = EDC
w = EDA
z = AED
y = DEB</p>

1535

<p>yo, I could come up with only this answer and this proves that the question is fallacious (SAT word),</p>

<p>since AC=AB and their respective opposite angles equal,</p>

<p>Line segments BE and CD would divide the sides AC and AB into ratio of the divided anles, <ABC and < ACB.</p>

<p>6:2 for AC and 7:1 for AB.</p>

<p>Now, EDA+ EDC = 150.</p>

<p>In triangle ADC, it seems as if the line DE divides AC into 6:2 parts. HENCE, its angle must also do the same.</p>

<p>150/8 = 18.75*6= 112.5</p>

<p>THEREFORE, <EDC = 112.5.</p>

<p>using similar relationship in triangle ABE, we’ll find that <DEB = 122.5,</p>

<p>BY now, u should have realised that this cannot happen.</p>

1536

<p>one question!!!</p>

<p>As it burns does a candle wax or wane???</p>

1537

<p>SO I DUNNO if this theory is correct: If any angle in a triangle is divided by a line segment, the line segment divides the opposite side of the corresponding angle into the same ratio as the angle was earlier divided.</p>

<p>I found another method,</p>

<p>I assumed anlge <AED to be 30 degree since that would make triangle ADE and traingle BOD(Where O is the intersection of BE and CD) SIMILAR</p>

<p>with this we can easily conclude that <CDE= 20 degree.</p>

<p>But I doubt we can make assumptions as such.</p>

1538

<p>@Gurung</p>

<p>The answer can be found using the cosine rule, as far as I know. Tara geometrically garr bhaneko chha…</p>

1539

<p>Yep but i could think of only all these…ehe</p>

1540

<p>I will solve that question for u if the question is correct. I finally am back to maths…wow. I will start it soon. </p>

<p>@puspakamal
the candle wane (decrease in size)</p>