Nairaland Mathematics Clinic - Education (146)

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Re: Nairaland Mathematics Clinic by dejt4u(m): 5:05am On Aug 21, 2014

[quote author=benbuks][/quote]
i⁵ = (i⁴ ) i = (1)i = i.
Kudos to you bro!

Re: Nairaland Mathematics Clinic by Kendzyma(m): 7:28am On Aug 21, 2014

dejt4u:
i⁵ = (i⁴ ) i = (1)i = -i.i
Kudos to you bro!

Re: Nairaland Mathematics Clinic by dejt4u(m): 7:55am On Aug 21, 2014

Kendzyma:

yea, tnx

Re: Nairaland Mathematics Clinic by kolamilan(m): 12:41am On Aug 22, 2014

Help Gurus At Home, pls i need a solution to these questions.1: Solve using simultaneous differential equation (D-1)x-Dy=2t+1 and (2D + 1)x + 2Dy=1

Re: Nairaland Mathematics Clinic by kolamilan(m): 12:44am On Aug 22, 2014

expand f(x)=x/2 , 0<x <2 pie. using fourier series.

Re: Nairaland Mathematics Clinic by kolamilan(m): 12:46am On Aug 22, 2014

investigate the stationary values of z=sqr(x)+xy+sqr (y)+5x-5x+3

Re: Nairaland Mathematics Clinic by kolamilan(m): 12:53am On Aug 22, 2014

show that the laplace transform of cosh at= a/sqr(s)-sqr(a)

question 2. d2y/dx2 +3dy/dx +2y= sin 2t

determine the laplace inverse of {4sqr(s) -5s + 6/(s + 1)(sqr(s) + 4)}. Thanks.

Re: Nairaland Mathematics Clinic by Soneh(m): 1:09pm On Aug 22, 2014

benbuks:

ok...you"ll have to go back to your basics of complex numbers

rationally , negative roots are invalid e.g ± √-1 , ±√-2 ,±√-3 e.t c.

solving x² + 1 =0 ..........(*)

produces complex roots , however mathematicians have worked severally on such irrational roots , which later discovered & agreed that solution to say (*) =±√-1

we now use ' i ' or ' j ' = ±√-1 ..........(**)

hence solution of (*) ,x = ± i

having believed that x=± i

=>i² =-1 .......( from (**) )

i³ = i(i²) =i(-1) =-i

i⁴ = (i² ) (i² ) = (-1)(-1)=1

i⁵ = (i⁴ ) i = (1)i = i

we could observe that powers of ' i ' in .

2 ,6 , 10 ,14 . . . =-1

3 ,7 , 11 ,15 . . . =-i

4 , 8 , 12 , 16 . . . = 1

5 , 9, 13 , 17 , . . . =i

& their multiples in that order , gives same result .
.

thanks

Re: Nairaland Mathematics Clinic by Arithmetic(m): 1:14pm On Aug 22, 2014

benbuks:

--------SOLUTION--------------
(a)

= 3(i¹⁰⁴ )i + 5(i⁷² )i -8(i² )²⁵ -2(i¹¹⁸ ) i

=>3i(1) +5i(1) - 8(-1) -2i(-1)

= 10i +8

(b)

by difference of two squares.
we have
[(3i-5)² ]² - [(2-i)² ]²

=>(9i² -30i+25)² -(4-4i+i²)²

=>(16-30i)² - (3-4i)²

=> (256-960i +900i² )-(9-24i+16i² )
=>-544-960i +7 +24i

hence we have ,

-537 -936i

@boss, (b), can be solved also by DE MOIVRE'S Theorem... My opinion.

Re: Nairaland Mathematics Clinic by Kendzyma(m): 10:33pm On Aug 22, 2014

Benbuks ,dejit4u,arithmetic,drniyi4u and oda math generals..abeg help with diz question...
Integrat squaroot(x² - 4)³

pls solv without using trig subtitution

Re: Nairaland Mathematics Clinic by Obinoscopy(m): 11:13am On Aug 23, 2014

Dear Maths Guru,

Cauchy Integral Test: Can someone discuss with examples?

Re: Nairaland Mathematics Clinic by Nobody: 2:25pm On Aug 23, 2014

thought I knew maths, bt mehn the stuff here is giving me a headache.
you guys should try and drop O" level problems, for guys like me to solve.

BTW, solve √90 + 3√3 + √3

cheap ryt wink

Re: Nairaland Mathematics Clinic by dejt4u(m): 2:56pm On Aug 23, 2014

Preboy: thought I knew maths, bt mehn the stuff here is giving me a headache.
you guys should try and drop O" level problems, for guys like me to solve.

BTW, solve √90 + 3√3 + √3

cheap ryt

yea..very cheap..
Answer is 3√10 + 4√3

Re: Nairaland Mathematics Clinic by Arithmetic(m): 3:17pm On Aug 23, 2014

@ kendzyma , u can only solve using trig. subs.
§¥(x^2-4)^3dx, ¥ rep. sq.root.
SOLUTION:
Let x=2sec@, dx=2tan@sec@d@.
§¥(4sec^2@-4)^3dx.
§{¥4(sec^2@-1)}^3dx.
Recall, 1+tan^2@=sec^2@.
§(2tan@)³.2tan@sec@d@.
16§tan⁴@sec@d@.
tan⁴@=(sec²@-1)².
16§(sec⁴@+1-2sec²@)sec@d@.
16[§(sec⁵@d@+§sec@d@-2§sec³@d@].
Taking one by one, we have;
§sec³@d@. Integrating by part,
u=sec@,v=tan@,du=tan@sec@d@.
=tan@sec@-§tan²@sec@d@.
§(sec²@-1)sec@d@=§sec³@d@-§sec@d@.
Let I=§sec³@d@.
I=tan@sec@-I+§sec@d@,
§sec³@d@= {tan@sec@+ln|sec@+tan@|}/2+c.
§sec@d@=ln|sec@+tan@|.
§sec⁵@d@=§sec³@.sec²@d@. Integrating by part,
§sec⁵@d@=sec³@tan@/4+3tan@sec@/8+3ln|sec@+tan@|.
Substituting, into the real integral and also x=2sec@.

x³¥(x²-4)/4-5x¥(x²-4)/2+6ln|{x+¥(x²-4)}/2|.

Re: Nairaland Mathematics Clinic by Nobody: 4:24pm On Aug 23, 2014

A = {factors of 50}
B = {odd numbers less than 10}

solve,
a) A U B
b) A Π B
c) n(A U B)
c) n (A Π B

this should be way cheaper

Re: Nairaland Mathematics Clinic by Nobody: 4:25pm On Aug 23, 2014

dejt4u:
yea..very cheap..
Answer is 3√10 + 4√3

correct √

Re: Nairaland Mathematics Clinic by Drniyi4u(m): 4:42pm On Aug 23, 2014

Preboy:
correct √

Really?? I think m lost. Are u trying to test d gurus, trying to educate yourself or trying to educate others??

Re: Nairaland Mathematics Clinic by Nobody: 4:52pm On Aug 23, 2014

Drniyi4u: Really?? I think m lost. Are u trying to test d gurus, trying to educate yourself or trying to educate others??

educate myself mostly

Re: Nairaland Mathematics Clinic by Drniyi4u(m): 5:39pm On Aug 23, 2014

Preboy:
educate myself mostly

okay

Re: Nairaland Mathematics Clinic by Kendzyma(m): 9:03pm On Aug 23, 2014

Arithmetic: @ kendzyma , u can only solve using trig. subs.
§¥(x^2-4)^3dx, ¥ rep. sq.root.
SOLUTION:
Let x=2sec@, dx=2tan@sec@d@.
§¥(4sec^2@-4)^3dx.
§{¥4(sec^2@-1)}^3dx.
Recall, 1+tan^2@=sec^2@.
§(2tan@)³.2tan@sec@d@.
16§tan⁴@sec@d@.
tan⁴@=(sec²@-1)².
16§(sec⁴@+1-2sec²@)sec@d@.
16[§(sec⁵@d@+§sec@d@-2§sec³@d@].
Taking one by one, we have;
§sec³@d@. Integrating by part,
u=sec@,v=tan@,du=tan@sec@d@.
=tan@sec@-§tan²@sec@d@.
§(sec²@-1)sec@d@=§sec³@d@-§sec@d@.
Let I=§sec³@d@.
I=tan@sec@-I+§sec@d@,
§sec³@d@= {tan@sec@+ln|sec@+tan@|}/2+c.
§sec@d@=ln|sec@+tan@|.
§sec⁵@d@=§sec³@.sec²@d@. Integrating by part,
§sec⁵@d@=sec³@tan@/4+3tan@sec@/8+3ln|sec@+tan@|.
Substituting, into the real integral and also x=2sec@.

x³¥(x²-4)/4-5x¥(x²-4)/2+6ln|{x+¥(x²-4)}/2|.

nice1 bro...guess dis d only Method of solving it.

Re: Nairaland Mathematics Clinic by akokoodide(m): 9:56pm On Aug 23, 2014

Arithmetic: @ kendzyma , u can only solve using trig. subs.
§¥(x^2-4)^3dx, ¥ rep. sq.root.
SOLUTION:
Let x=2sec@, dx=2tan@sec@d@.
§¥(4sec^2@-4)^3dx.
§{¥4(sec^2@-1)}^3dx.
Recall, 1+tan^2@=sec^2@.
§(2tan@)³.2tan@sec@d@.
16§tan⁴@sec@d@.
tan⁴@=(sec²@-1)².
16§(sec⁴@+1-2sec²@)sec@d@.
16[§(sec⁵@d@+§sec@d@-2§sec³@d@].
Taking one by one, we have;
§sec³@d@. Integrating by part,
u=sec@,v=tan@,du=tan@sec@d@.
=tan@sec@-§tan²@sec@d@.
§(sec²@-1)sec@d@=§sec³@d@-§sec@d@.
Let I=§sec³@d@.
I=tan@sec@-I+§sec@d@,
§sec³@d@= {tan@sec@+ln|sec@+tan@|}/2+c.
§sec@d@=ln|sec@+tan@|.
§sec⁵@d@=§sec³@.sec²@d@. Integrating by part,
§sec⁵@d@=sec³@tan@/4+3tan@sec@/8+3ln|sec@+tan@|.
Substituting, into the real integral and also x=2sec@.

x³¥(x²-4)/4-5x¥(x²-4)/2+6ln|{x+¥(x²-4)}/2|.

can anybody solve this question for me.
if a clock takes 4 seconds to strike 6.00 (with 6 equallyspaced chimes)
how long will it takes to srtike 12.00 (wiyh 12 equally spaced chimes)
PLS VERY VERY URGENT, THANKS

Re: Nairaland Mathematics Clinic by Nobody: 2:22pm On Aug 24, 2014

hmmm. wer do i start from?

Re: Nairaland Mathematics Clinic by Nobody: 2:23pm On Aug 24, 2014

Kendzyma: nice1 bro...guess dis d only Method of solving it.

yea u cn also use series

Re: Nairaland Mathematics Clinic by Nobody: 2:40pm On Aug 24, 2014

Preboy: A = {factors of 50}
B = {odd numbers less than 10}

solve,
a) A U B
b) A Π B
c) n(A U B)
c) n (A Π B

this should be way cheaper

A={ 1,2,5,10,25, 50}
B={1,3,5,7,9}

AuB={1,2,3,5,7,9,10,25,50}

AnB={1,5}

n(AnB)=9
n(AnB)=2

i think you should bring up something more serious/challenging that will help you & others know mathematics better rather than testing others .

hope u get bro ?

Re: Nairaland Mathematics Clinic by Nobody: 2:41pm On Aug 24, 2014

dejt4u:
i⁵ = (i⁴ ) i = (1)i = i.
Kudos to you bro!

thanks man

Re: Nairaland Mathematics Clinic by Nobody: 2:53pm On Aug 24, 2014

kolamilan: Help Gurus At Home, pls i need a solution to these questions.1: Solve using simultaneous differential equation

(D-1)x-Dy=2t+1................(^)
(2D + 1)x + 2Dy=1...........(^^)

use elimination (^) *2 & (^^) *-1

Re: Nairaland Mathematics Clinic by oyeludef(m): 5:09pm On Aug 24, 2014

Richiez: We diagnose and solve math problems here
This thread is the meeting point for nairaland math gurus...I dare anyone to ask a question in mathematics without me having an answer to them, LETS START

plenty people come from front,back,left,right n dey meet for center hw many dem go b for center

Re: Nairaland Mathematics Clinic by Obinoscopy(m): 11:20pm On Aug 24, 2014

Guys solve this:

If √2x^3 + 2x^2 + 3√2x - 5 = 0 has three roots namely a, b and c. Find:

1. a + b + c

2. ab + bc + ac

3. abc

Re: Nairaland Mathematics Clinic by Abuklaw(m): 9:10am On Aug 25, 2014

I am following.

Re: Nairaland Mathematics Clinic by akpos4uall(m): 12:10pm On Aug 25, 2014

Please help rephrase this question and get the solution if possible. My aim is to place two bulbs in a room such that the point in the room with the least brightness as a result of the position of the two bulbs is maximized assuming both bulbs are of the same capacity and the illumination at any point is proportional to the sum of the squares of the distance from the point to the two bulbs.
Let me put it this way: in rectangle ABCD with coordinates A(0,0), B(10,0), C(10,4), D(0,4), find the two points U(a,b) & V(c,d) inside the rectangle such that the point whose sum of squares of distance to U & V within the rectangle is the highest as a result of position of U & V will be minimized.
Analytical, numerical, programming solution etc all welcomed

Re: Nairaland Mathematics Clinic by Nobody: 12:44pm On Aug 25, 2014

Obinoscopy: Guys solve this:

If √2x^3 + 2x^2 + 3√2x - 5 = 0 has three roots namely a, b and c. Find:

1. a + b + c

2. ab + bc + ac

3. abc

_>>>>>>_____SOLUTION_________<<<<<<<<_

By roots of cubic equations of the form

px³ +qx² +rx + s = 0 , for p=/= 0

sum of roots (a +b+c) = -q/p

sum of product of roots (ab + ac + bc ) = r/ p

product of roots ( abc) = - s/a

thus we have

a+b+c = -2/√2 =-2√2/2 = -√2

ab+ac+bc= 3√2/√2 =3

abc =5/√2 = 5√2/ 2

that's it sir.

Re: Nairaland Mathematics Clinic by Nobody: 12:51pm On Aug 25, 2014

akokoodide:
can anybody solve this question for me.
if a clock takes 4 seconds to strike 6.00 (with 6 equallyspaced chimes)
how long will it takes to srtike 12.00 (wiyh 12 equally spaced chimes)
PLS VERY VERY URGENT, THANKS

v=d/t
v=6/4
v=3/2
when d=12
3/2=12/t

hence t= 8sec.

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