http://www.sgforums.com/forums/2297/posts en-US 60 <blockquote> <div class="quote_from">Originally posted by jiaxing2:</div> <div class="quote_body"> <p>the methods are the same....we have different normal vectors thus different answers...</p> <p>if (2 -2 -1) is correct then <span style="color: #333399;">1/9 ( -13i + 49j - 25k)&nbsp;should be correct...</span></p> </div> </blockquote> <p><br /> i agree with u. sry for the duplicate then.</p> Fri, 04 Jul 2008 08:56:12 +0800 www.sgforums.com:2297:321748:8216366 Uncertain http://www.sgforums.com/forums/2297/topics/321748 <p>the methods are the same....we have different normal vectors thus different answers...</p> <p>if (2 -2 -1) is correct then <span style="color: #333399;">1/9 ( -13i + 49j - 25k)&nbsp;should be correct...</span></p> Thu, 03 Jul 2008 22:45:49 +0800 www.sgforums.com:2297:321748:8215665 jiaxing2 http://www.sgforums.com/forums/2297/topics/321748 <p>Ok, i solve the question ... seems correct though.</p> <p>&nbsp;</p> <p>(<span style="color: #333399;"><strong>i</strong> + 3<strong>j</strong> - 4<strong>k</strong> ) x (11<strong>i</strong> + 7<strong>j</strong> + 8<strong>k)...u get 52,-52,-26</strong></span></p> <p>Thus the normal vector is (2i -2j -k)&nbsp; &lt;--- same as jiaxing2</p> <p>&nbsp;</p> <p>Line through OA = (1, 3, -4) +&nbsp;<span style= "color: #ff0000;">b</span> (2, 2, -1) ,&nbsp;<span style= "color: #ff0000;">b</span> is a constant</p> <p>Since (2, 2, -1) is perpendicular to plane <span style= "font-size: 12pt; font-family: Arial;"><span style= "color: #333399;"><span style="color: #000000;">π ,</span></span></span></p> <p><span style="font-size: 12pt; font-family: Arial;"><span style= "color: #333399;"><span style= "color: #000000;">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></span></span><span style="font-size: 12pt; font-family: Arial;"><span style="color: #333399;"><span style="color: #000000;">&nbsp;its equation becomes&nbsp;&nbsp;&nbsp;&nbsp; r. (2, 2, -1) = (-15, 10, 5) . (2, 2, -1)&nbsp; ---- <strong>[1]</strong></span></span></span></p> <p><span style="font-size: 12pt; font-family: Arial;"><span style= "color: #333399;"><span style="color: #000000;">Then sub <strong>r = (1+2b, 3+2b, -4-b)</strong> &lt;-- eqn of line OA&nbsp; into <strong>[1]</strong> to find OL<strong>,</strong></span></span></span></p> <p><span style="font-size: 12pt; font-family: Arial;"><span style= "color: #333399;"><strong><span style="color: #000000;">u will get</span> <span style="color: #000000;">b = -3 , OL = (-5, -3, -1)</span></strong></span></span></p> <p><span style="font-size: 12pt; font-family: Arial;"><span style= "color: #333399;"><strong><span style="color: #000000;">AM = 1/5 (AL)</span></strong></span></span></p> <p><span style="font-size: 12pt; font-family: Arial;"><span style= "color: #333399;"><strong><span style="color: #000000;">OM - OA = 1/5 (OL - OA)</span></strong></span></span></p> <p><span style="font-size: 12pt; font-family: Arial;"><span style= "color: #333399;"><span style="color: #800000;">OM = 1/5 OL&nbsp;+ 4/5OA =&nbsp; <strong><span style="text-decoration: underline;">1/5 (-1, 9, -17)</span></strong></span>&nbsp;</span></span></p> Thu, 03 Jul 2008 22:28:40 +0800 www.sgforums.com:2297:321748:8215628 Uncertain http://www.sgforums.com/forums/2297/topics/321748 <p>hello bryan..<img src="/images/emoticons/classic/icon_wink.gif" alt="icon_wink.gif" /></p> Thu, 03 Jul 2008 22:01:35 +0800 www.sgforums.com:2297:259804:8215547 QX179R http://www.sgforums.com/forums/2297/topics/259804 <p>hi everybody <img src= "/images/emoticons/kde-3.5.8/KMess-Cartoon/smile.png" alt= "smile.png" /></p> Thu, 03 Jul 2008 21:58:33 +0800 www.sgforums.com:2297:259804:8215535 bryanw http://www.sgforums.com/forums/2297/topics/259804 <p>help my fren do survey thx!</p> <p><a href= "http://www.surveymonkey.com/s.aspx?sm=W8p29bdaOgz3CuGzm8mthA_3d_3d" rel= "nofollow">http://www.surveymonkey.com/s.aspx?sm=W8p29bdaOgz3CuGzm8mthA_3d_3d</a></p> <p>sorri those ppl pm me i nv reply cos i lazy&nbsp;liao. anyway im fine thx!</p> Thu, 03 Jul 2008 20:19:59 +0800 www.sgforums.com:2297:322993:8215116 jiayingg http://www.sgforums.com/forums/2297/topics/322993 <p>Survey from&nbsp; <em><strong>Diamond Andry</strong></em></p> <p>&nbsp;</p> <p>hi, i need data for my project, kindly check my online survey form <a href= "http://www.surveymonkey.com/s.aspx?sm=9nc0m_2f1h40L639d8v33uwQ_3d_3d" rel= "nofollow">http://www.surveymonkey.com/s.aspx?sm=9nc0m_2f1h40L639d8v33uwQ_3d_3d</a></p> <p>thank you very much</p> Thu, 03 Jul 2008 18:42:05 +0800 www.sgforums.com:2297:322988:8214889 Darkness_hacker99 http://www.sgforums.com/forums/2297/topics/322988 <p>=.=" okie.</p> <p>&nbsp;</p> Thu, 03 Jul 2008 18:40:31 +0800 www.sgforums.com:2297:259804:8214885 Darkness_hacker99 http://www.sgforums.com/forums/2297/topics/259804 <p>hi, i need data for my project, kindly check my online survey form <a href= "http://www.surveymonkey.com/s.aspx?sm=9nc0m_2f1h40L639d8v33uwQ_3d_3d" rel= "nofollow">http://www.surveymonkey.com/s.aspx?sm=9nc0m_2f1h40L639d8v33uwQ_3d_3d</a></p> <p>thank you very much<a href= "http://www.surveymonkey.com/s.aspx?sm=9nc0m_2f1h40L639d8v33uwQ_3d_3d" rel="nofollow"></a></p> Thu, 03 Jul 2008 18:03:55 +0800 www.sgforums.com:2297:259804:8214792 Diamond andry http://www.sgforums.com/forums/2297/topics/259804 <p>Another easy method to solve for k+1 for the complex roots:</p> <p>You draw the Im-Real axis. You can also see that one of the roots is cube root of 1.5, which is a real number.</p> <p>Then you rotate the point around the origin by 120 degrees (360/3)</p> <p>one of the complex roots of k+1 would be:<br /> - 1.145 sin 30 + 1.145 cos 30 = -0.5725 + 0.9916i</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>I think both of the above methods would be easier than using and understanding theory of equations</p> Thu, 03 Jul 2008 15:25:26 +0800 www.sgforums.com:2297:322601:8214445 eagle http://www.sgforums.com/forums/2297/topics/322601 <blockquote> <div class="quote_from">Originally posted by jiaxing2:</div> <div class="quote_body"> <p>there should be 2 more roots which are complex numbers...</p> <p>can be solved by theory of eqn</p> <p>sum of the 3 roots = -3</p> <p>product of 3 roots = 0.5</p> <p>since there are no complex coefficients to the cubic eqn...then if complex roots exist, there exist in conjugate pairs...ie in the form of a + ib and a - ib where a and b are real numbers and i = root of ( -1 )</p> <p>let k be the real root.&nbsp; 2a + k = -3</p> <p>solving a = -&nbsp;[ 1 + 0.5(1.5)^1/3 ]</p> <p>similarly ( a^2 + b^2&nbsp;) * k =0.5 (product of 3 roots)</p> <p>go solve for b...and u get ur 2 complex roots</p> <p>for ur info only...unless this qn carries&nbsp;alot of&nbsp;marks...i dun think u are really required to find&nbsp;out what the complex roots are...&nbsp;</p> </div> </blockquote> <p>No need do so hard to solve the complex roots also</p> <p>Simplified version</p> <p>(k+1)^3 - (cube root[1.5])^3 = 0</p> <p>you can use the formula x^3 - y^3 = (x^2 + xy + y^2)(x-y)<br /> where x = k+1 and y = cube root of 1.5</p> <p>Then you can do the normal use equation to solve for quadratic equations method to determine the final complex roots, in which your discriminant should be -ve</p> Thu, 03 Jul 2008 15:20:46 +0800 www.sgforums.com:2297:322601:8214436 eagle http://www.sgforums.com/forums/2297/topics/322601 <p><a href= "http://www.surveymonkey.com/s.aspx?sm=zwPqYZahDA2Xh1Esj5_2bysw_3d_3d" rel= "nofollow">http://www.surveymonkey.com/s.aspx?sm=zwPqYZahDA2Xh1Esj5_2bysw_3d_3d</a></p> Thu, 03 Jul 2008 14:12:40 +0800 www.sgforums.com:2297:322927:8214322 Darkness_hacker99 http://www.sgforums.com/forums/2297/topics/322927 <p>hmmmm</p> <p>&nbsp;</p> <p>eat dead people...</p> Thu, 03 Jul 2008 00:51:41 +0800 www.sgforums.com:2297:322745:8213349 airgrinder http://www.sgforums.com/forums/2297/topics/322745 <p>geography right?</p> <p>to solve, study the green and blue revolution</p> Thu, 03 Jul 2008 00:40:29 +0800 www.sgforums.com:2297:322745:8213334 wishboy http://www.sgforums.com/forums/2297/topics/322745 <p>Im not sure if this is spam, but moderators can close it.</p> <p>It would be nice if people could do this survey for this stranger:</p> <p> http://www.surveymonkey.com/s.aspx?sm=zwPqYZahDA2Xh1Esj5_2bysw_3d_3d</p> <p>Thanks!</p> Wed, 02 Jul 2008 23:49:31 +0800 www.sgforums.com:2297:322927:8213278 LostAccount http://www.sgforums.com/forums/2297/topics/322927 <p>there should be 2 more roots which are complex numbers...</p> <p>can be solved by theory of eqn</p> <p>sum of the 3 roots = -3</p> <p>product of 3 roots = 0.5</p> <p>since there are no complex coefficients to the cubic eqn...then if complex roots exist, there exist in conjugate pairs...ie in the form of a + ib and a - ib where a and b are real numbers and i = root of ( -1 )</p> <p>let k be the real root.&nbsp; 2a + k = -3</p> <p>solving a = -&nbsp;[ 1 + 0.5(1.5)^1/3 ]</p> <p>similarly ( a^2 + b^2&nbsp;) * k =0.5 (product of 3 roots)</p> <p>go solve for b...and u get ur 2 complex roots</p> <p>for ur info only...unless this qn carries&nbsp;alot of&nbsp;marks...i dun think u are really required to find&nbsp;out what the complex roots are...&nbsp;</p> Wed, 02 Jul 2008 21:10:51 +0800 www.sgforums.com:2297:322601:8212865 jiaxing2 http://www.sgforums.com/forums/2297/topics/322601 <blockquote> <div class="quote_from">Originally posted by hiphop2009:</div> <div class="quote_body"> <p>hey guys~! long time no online here cause busy, but i will come here more often as i am a bit more free nw.</p> <p>&nbsp;</p> <p>ok, your suggestion is noted. :)</p> </div> </blockquote> <p>me busy :(</p> Wed, 02 Jul 2008 15:51:04 +0800 www.sgforums.com:2297:259804:8212308 eagle http://www.sgforums.com/forums/2297/topics/259804 <p>Whenever the question says "81.1% by mass", let the sample mass of compound be 100g. Then, sample mass of Ba in compound&nbsp;is 81.1g, and sample mass of oxygen in compound is 100 - 81.1 = 18.9g.</p> <p>&nbsp;</p> <p>Next, work out number of moles of each element in the compound, using the formula "No. of moles = Sample mass&nbsp;/ Molar mass". Simplify the mole ratio to simplest ratio (ie. divide all numerical values by the smallest value), and you've got your empirical formula.</p> <p>&nbsp;</p> Wed, 02 Jul 2008 08:37:12 +0800 www.sgforums.com:2297:322824:8211589 UltimaOnline http://www.sgforums.com/forums/2297/topics/322824 <p>When barium metal is burned in air, an oxide&nbsp;is formed with 81.1% by mass of barium. Calculate the formula of the compound. [Mr of barium=137, oxygen=16]</p> <p>The answer is simple (BaO2), but I'm looking for a proper working instead of guess &amp; check.</p> <p>I tried doing this but I failed:</p> <p>Let compound be Ba(x)O(y)</p> <p>137x / (137x + 16y) = 0.811</p> <p>...</p> Wed, 02 Jul 2008 00:50:05 +0800 www.sgforums.com:2297:322824:8211329 secretliker http://www.sgforums.com/forums/2297/topics/322824 <p>tricky bastards ah, these ppl.</p> <p>pay them million dollars to solve world crisis, then they cannot think up anything, then give this issue as assignment to kids, then if got good idea, steal. grr.</p> <p>&nbsp;</p> <p>anyway, lack of food = must grow more food lor.</p> Tue, 01 Jul 2008 21:39:48 +0800 www.sgforums.com:2297:322745:8210794 dumbdumb! http://www.sgforums.com/forums/2297/topics/322745 <p>It's how the food trickles down from the production side... Look at those who've the money and waste the precious food, versus those who go hungry...</p> Tue, 01 Jul 2008 21:39:29 +0800 www.sgforums.com:2297:322745:8210793 deathscythe99 http://www.sgforums.com/forums/2297/topics/322745 <p>all families are encouraged to plant bean sprout at home for food</p> Tue, 01 Jul 2008 21:25:09 +0800 www.sgforums.com:2297:322745:8210724 thoreldan http://www.sgforums.com/forums/2297/topics/322745 <p><img src="/images/emoticons/kde-3.5.8/plain/confused.png" alt= "confused.png" />haha... so many funny ans......</p> Tue, 01 Jul 2008 21:18:44 +0800 www.sgforums.com:2297:322745:8210702 XxPOoPOoxX http://www.sgforums.com/forums/2297/topics/322745 <p>plant more food</p> <p>clones animals for eating</p> Tue, 01 Jul 2008 20:19:37 +0800 www.sgforums.com:2297:322745:8210466 MyPillowTalks http://www.sgforums.com/forums/2297/topics/322745 <p>Try not to eat too much until u so fat like M.Aisuke aka the guy with unusal name. Try to give some food to the poor if possible.</p> Tue, 01 Jul 2008 20:08:02 +0800 www.sgforums.com:2297:322745:8210432 bus555 http://www.sgforums.com/forums/2297/topics/322745