Nairaland Mathematics Clinic - Education (90) - Nairaland
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| Re: Nairaland Mathematics Clinic by MrCalculus(m): 10:31pm On Oct 25, 2013 |
[quote author=jackpot] in the form a+ib z=-16+0i [s]sorry, Mathematicians dont write the imaginary part with j like Physicists and Engineers [/s] now, for the fourth root, i'll prefer working with the exponential form to avoid the impossible task of typing out square roots here. z=-16+0i=16e^i(pi)=w4 Using the nth root of a complex number formula w=nth root of |z| times e^i[(Argz+2k pi)/n] for k= 0, 1, 2 ,. . ., n-2, n-1 Thus, the fourth roots are simply w= 2e^i(pi/4), 2e^i(3 |
| Re: Nairaland Mathematics Clinic by MrCalculus(m): 10:50pm On Oct 25, 2013 |
hmm.......@jackpot.. first of all u change my ques from -16 to -16+oi....i followed that. then later to 16e^i(pi) and w^4......i was totally loss by dat move cause u ignored d -ve sign attach to d 16..... ALL D SAME DIS IS D ANSWER,BUT WAT I WANT IS D SOLUTION:: root2(1+j),,root2(-1+j),,root2(-1-j),,root2(1-j)..... PLZ WEN SOLVIN TRY TO USE SIMPLE TERMS CAUS ME NO TOO GUD 4 MATHS CAUS AM STILL A FRESHER..... TNX IN ADVANCE@ALL MY GENERALS PLZ HELP ME OUT WIT D SOLUTION. |
| Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 11:17pm On Oct 25, 2013 |
| Re: Nairaland Mathematics Clinic by jackpot(f): 11:17pm On Oct 25, 2013 |
Mr Calculus: & DAT MAKES UR ANSWER WRONG......it doesnt make my answer wrong, since I didn't answer your question  hope you got my drift. WEN I DID IT I GOTyou are totally, entirely, capitally and wholistically wrong since there are serious radicals in your denominator. You started with a question having only one radical, and your wrong answer had 2 radicals in the denominator!!! 1 Like |
| Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 11:19pm On Oct 25, 2013 |
| Re: Nairaland Mathematics Clinic by MrCalculus(m): 11:24pm On Oct 25, 2013 |
jackpot: it doesnt make my answer wrong, since I didn't answer your questionhahaha......na because of fail way i fail am 9 make u dy writ all those vocalbs?.... i dint espect to b right,dats y i posted it..... |
| Re: Nairaland Mathematics Clinic by Nobody: 11:35pm On Oct 25, 2013 |
How come no one has attempted the simultaneous equations I posted on page 88? Is my LEGIT question scary? |
| Re: Nairaland Mathematics Clinic by jackpot(f): 12:16am On Oct 26, 2013 |
Mr Calculus: hmm.......@jackpot..guy, e^i(pi)=-1. Sorry, your high-end username misled me into thinking that you can be able to fill in the gaps well, let me be a little bit explanatory. Let T=theta then the Euler formula is e^iT=cos T+ i sin T its easy to see that e^i(pi)=cos pi + i sin pi=-1+i0=-1 and so, -16=16(-1)=16e^i(pi). Hope you now get it? then, about that w^4, if w is a fourth root of a complex number z (i.e., w=z1/4), then, w^4=z. Simple elementary algebra. So, if you use the Euler number on my first answer, you will have 2ei(pi/4)= 2(cos pi/4 + i sin pi/4) =2[(root2)/2+i(root2)/2 ]=root2 + iroot2 = root2 (1+i) did you notice that it correspond with your first answer? I'll leave the remaining three for you to simplify in rectangular form as I just did. They should equally correspond with what you have there. Sorry, I'll apologize once again that your username misled me into assuming that you are good in Math. My bad!!!  |
| Re: Nairaland Mathematics Clinic by jackpot(f): 12:22am On Oct 26, 2013 |
Mr Calculus: hahaha......na because of fail way i fail am 9 make u dy writ all those vocalbs?....the question I answered is even 3 times tougher than what you asked. Your question cheap sef. 1/(cuberoot2)-1= [ (22/3)/2 ] - 1 |
| Re: Nairaland Mathematics Clinic by jackpot(f): 12:35am On Oct 26, 2013 |
smurfy: Solve the simultaneous equationstaking Log10 of both sides, we have x+yLog4=0. . . . . . . . . . . .1 xLog8-y=1. . . . .. . . . . . . .2 from eqn 1, x= -yLog4. Substitute this for x in Eqn 2 -(Log4)(Log8)y-y=1 thus, y= -1/(1+ Log8 Log4) x=-yLog4= Log4/(1 + Log8 Log 4) thats all! |
| Re: Nairaland Mathematics Clinic by Nobody: 1:01am On Oct 26, 2013 |
jackpot: taking Log10 of both sides, we have Excellent! The answers I have here are in log 2 format, but they are basically the same. Hats off...! |
| Re: Nairaland Mathematics Clinic by Ortarico(m): 3:33am On Oct 26, 2013 |
Alpha Maximus: ...Brilliant solution!!! You failed the remaining two sha! (B)= in 3 years I.e: 1/15 *75 =45/9 (c) =69 yrs old!! All right bro, thanks but can you help explain the 'C'? |
| Re: Nairaland Mathematics Clinic by Ortarico(m): 3:35am On Oct 26, 2013 |
Alpha Maximus: @ Ortarico: No, I'm an Engineering student....why did you think I did Social Sciences? Nothing perse just the way you masticate business maths. |
| Re: Nairaland Mathematics Clinic by MrCalculus(m): 4:56am On Oct 26, 2013 |
TNX alot@jackpot. abeg no mind my username oo.... but i no i will still uphold dat name wit a little time......tnx again |
| Re: Nairaland Mathematics Clinic by mickywonder: 5:55am On Oct 26, 2013 |
Pls guyz help me solve this.its easy and technical the sum of 3 7/8 and 1 1/3 is less than the difference between 3/8 and 1 2/3 by...........answer |
| Re: Nairaland Mathematics Clinic by Nobody: 7:23am On Oct 26, 2013 |
Edited |
| Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 7:28am On Oct 26, 2013 |
Ortarico:....Their present ages are 72,42 and 6...the sum of these ages is 120$$$the objective is to find how old Mr Coker was when the the sum of their ages was 9 less the 120, which is 111, thus we have: (72-x)+(42-x)+(6-x)=111 72-x+42-8+6-x=111 120-3x=111 -3x=111-120 -3x=-9 3x=9 X=3 This shows that 3 years ago, the sum of their ages was 111, and thus: 72-3=69..Mr Coker was 69 years old....still kudos to you on your 'A' solution 1 Like |
| Re: Nairaland Mathematics Clinic by mickywonder: 7:32am On Oct 26, 2013 |
Pls nawww Pls guyz help me solve this.its giving me headache the sum of 3 7/8 and 1 1/3 is less than the difference between 3/8 and 1 2/3 by...........answer i begg naww |
| Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 7:38am On Oct 26, 2013 |
| Re: Nairaland Mathematics Clinic by rhydex247(m): 7:47am On Oct 26, 2013 |
Here is my question. 1. Solve for x. If x+a^(x^a)=b. Where x=(x1,x2,x3). Note that ^ means cap. 2. Let W be a subspace of real space R^3. Give a geometrical describtion of W in terms of its dimension. |
| Re: Nairaland Mathematics Clinic by Nobody: 7:48am On Oct 26, 2013 |
Mr Calculus: & DAT MAKES UR ANSWER WRONG......ur answer seems wrong too ... [1/(cuberoot2)] -1 . =(1 - 21/3)/21/3 x 22/3 / 22/3 . =(22/3-2) / 2 . =(22/3 / 2) - 1 ... |
| Re: Nairaland Mathematics Clinic by Nobody: 9:29am On Oct 26, 2013 |
The first term of an A.P is 1, and the sum of n terms is n2. Find the A.P. |
| Re: Nairaland Mathematics Clinic by Nobody: 9:36am On Oct 26, 2013 |
Mr Calculus: WRONGThank You Sir! |
| Re: Nairaland Mathematics Clinic by Nobody: 9:38am On Oct 26, 2013 |
Swagalord18: The first term of an A.P is 1, and the sum of n terms is n2. Find the A.P.guy recheck ur question |
| Re: Nairaland Mathematics Clinic by Nobody: 9:49am On Oct 26, 2013 |
I-am-Winner:no be my question oo ....i c am 4 one txtbook .... . Do u think d question iz rong? |
| Re: Nairaland Mathematics Clinic by Idenyijoshua(m): 10:07am On Oct 26, 2013 |
[1/(cuberoot2)] -1 . =(1 - 21/3)/21/3 x 22/3 / 22/3 . =(22/3-2) / 2 . =(22/3 / 2) - 1 ...[/quote] Getting the same thing  |
| Re: Nairaland Mathematics Clinic by Nobody: 10:24am On Oct 26, 2013 |
Swagalord18: The first term of an A.P is 1, and the sum of n terms is n2. Find the A.P..1,3,5,7,9,... |
| Re: Nairaland Mathematics Clinic by Ortarico(m): 10:42am On Oct 26, 2013 |
Alpha Maximus: ....Their present ages are 72,42 and 6...the sum of these ages is 120$$$the objective is to find how old Mr Coker was when the the sum of their ages was 9 less the 120, which is 111, thus we have: Oops! I've seen my errors. Thanks. |
| Re: Nairaland Mathematics Clinic by Nobody: 10:45am On Oct 26, 2013 |
benbuks: .1,3,5,7,9,...lolz ...we all know that ....i need proof .....i've modified d question ...answer it now |
| Re: Nairaland Mathematics Clinic by Nobody: 10:49am On Oct 26, 2013 |
Modified |
| Re: Nairaland Mathematics Clinic by Nobody: 10:49am On Oct 26, 2013 |
prove that if the first term of an A.P is 1, and the sum of n terms is n2, the A.P is 1,3,5.....9 |
| Re: Nairaland Mathematics Clinic by Nobody: 11:04am On Oct 26, 2013 |
Swagalord18: prove that if the first term of an A.P is 1, and the sum of n terms is n2, the A.P is 1,3,5.....9them suppose give us clue na. Like the value of d, l or n.. Dats y i said rechek the question.. U don solve am?? |
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