Separated

factorial1: @benbuks... The 1st qt has been solved, by the way, the 2nd qt is nt cleared enough...is it [(1/(x+2))^1/2 - (1/x)^1/2]....u know wat I mean...as in the x+2 is 2geda or separated like this 1/x + 2...?

benbuks: ...bravo,...try d 2nd 1.....kinda long sha..

re write the second question bro.

Laplacian:
is it dat i didnt use d method u have in mind or d solution is completely wrong...if it is wrong, can u b kind enough 2 point out my error?...

...check rhydex's solution..

K, [(x^2 +1)^2 + [1+ (x^2 +1)^2]^1/2 ]^1/2 or should i snap it?

rhydex 247:
re write the second question bro.

factorial1: @benbuks... The 1st qt has been solved, by the way, the 2nd qt is nt cleared enough...is it [(1/(x+2))^1/2 - (1/x)^1/2]....u know wat I mean...as in the x+2 is 2geda or separated like this 1/x + 2...?

..here.

benbuks: K, [(x^2 +1)^2 + [1+ (x^2 +1)^2]^1/2 ]^1/2 or should i snap it?

..here.,differentiate...wrt. X.

the soln to dis questn long o

rhydex 247: the soln to dis questn long o

..na just 1page solution na....e no long reach general jacpot..enigma.....just try past tam..or snap n post..

smurfy:

@ = 74, ¥ = 19.

not correct sir....
plz recheck ur calculations

Mr Calculus: not correct sir....
plz recheck ur calculations

@ = 28, ¥ = 30

Approximate acute angle values only.

benbuks: ..here

limit equals 0.

jackpot: limit equals 0.

.on point .my darlin, kindly show workin na?..so dat others cn learn......long tym o...i didn't only miss u , bt i 'lack' u....die...,
(dont think otherwise o...me no b bad boy oo.)

^lolz. . .okay dear


so, here we go!

express that number inside the limit sign as that same over 1.

Now, to make things look nice, multiply numerator and denominator with the square root of x.

If we substitute now, you will get (1-1)/0=0/0

so, we can now apply L'Hospital's rule to get the limit as tends to zero of (2sqrt(x))/(sqrt(1+2x))

substitute 0 now to get 0/1=0

Kia kia

jackpot: ^lolz. . .okay dear


so, here we go!

express that number inside the limit sign as that same over 1.

Now, to make things look nice, multiply numerator and denominator with the square root of x.

If we substitute now, you will get (1-1)/0=0/0

so, we can now apply L'Hospital's rule to get the limit as tends to zero of (2sqrt(x))/(sqrt(1+2x))

substitute 0 now to get 0/1=0

Kia kia

....nice1,...

if 3+5 qual 8 prove that 1+1 equal 2

if ,m and n are open set than show that the intersection is also open

Humphrey77: if 3+5 qual 8 prove that 1+1 equal 2

hmm

[quote author=Humphrey77]if 3+5 qual 8 prove that 1+1 equal 2


solution.
this is real analysis.
here it goes.
well known that 1.a=a from the axiom ii.
from 1+1=2
1.a+1.a=1(a+a)=2a. hence 1+1=2 Q.E.D

[quote author=rhydex 247][/quote]

#GBAM

FIND THE SUM OF 2.99999999999999 UP TO INFINITY

Humphrey77: FIND THE SUM OF 2.99999999999999 UP TO INFINITY

Here goes...

2.999... = 2 + 0.999...
= 2 + (0.9 + 0.09 + 0.009 + ...)

Dealing with expression in brackets:

Sum to infinity = a/(1-r), where a = 0.9 and r = 0.1

Sum to infinity = 0.9/(1-0.1) = 1

So 2 + (0.999...) becomes 2 + 1 = 3

Compute the derivative of..
**] sqrt[ (x^2 +1)^2 +sqrt (1 + (x^2 +1)^2 ) ]
nb:"sqrt." Means square root [/quote]
SOLUTION BY A.K.A SERIES.
HERE IT GOES.
USING THE CHAIN RULE,
d/dx(sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)]=dsqrt(u)/du*du/dx , where u=(x^2+1)^2+sqrt(1+(x^2+1)^2)
and d/du(sqrt(u)= 1/2sqrt(u).
d/dx[(x^2+1)^2+sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
differentiatin the sum term by term
d/dx[(x^2+1)^2]+d/dx[sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
Using the chain rule, d/dx[(x^2+1)^2]= du^2/du*du/dx, where u=x^2+1 and d/du(u^2)=2u.
d/dx[sqrt(1+(x^2+1)^2)]+4x^3+4x/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
Using chain rule d/dx[sqrt(1+(x^2+1)^2)] d/du(sqrt(u))*du/dx, where u=1+(x^2+1)^2 and d/du(sqrt(u))=1/2sqrt(u).
4x^3+4x+[d/dx(1+(x^2+1)^2)/2sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
4x^3+4x+[d/dx(1)+d/dx(x^2+1)^2)]/2sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
4x^3+4x+[2(2x)(x^2+1)/2sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
4x^3+4x+[2x(x^2+1)/sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
that's all. All is well.

As punishment for missing most of Nairaland Maths Quiz Competition, General Richiez asked General Q to list all possible four-letter words that can be formed from the word MEDITERRANEAN such that the first letter is E and the last letter is R.

How many of such four-letter words are possible?

@factorial, there were nothing wrong with my answers. The difference was that I expressed the change in units. E.g. 8(pi)cm2 as my first answer is the same answer as 8(3.142)cm2 = 25.136

smurfy: As punishment for missing most of Nairaland Maths Quiz Competition, General Richiez asked General Q to list all possible four-letter words that can be formed from the word MEDITERRANEAN such that the first letter is E and the last letter is R.

How many of such four-letter words are possible?

They're listed below but here is how to calculate how many there are
E _ _ R We want to fill those two blanks.
Case 1. The two letters in the blanks are the same.
The only such possibilities are these 3: EAAR, EEER, ENNR.
Case 2. The two letters in the blanks are different. Then
we must choose them from these 8: M,E,D,I,T,R,A,N. We can
choose the letter for the first blank in 8 ways, and for
each of those ways we can choose the letter for the second
blank 7 ways. That's 8×7 or 56 ways.
So 3 ways from case 1 plus 56 ways from case 2, gives us
59 ways total.
Here are all 59 ways listed:
1. EAAR
2. EADR
3. EAER
4. EAIR
5. EAMR
6. EANR
7. EARR
8. EATR
9. EDAR
10. EDER
11. EDIR
12. EDMR
13. EDNR
14. EDRR
15. EDTR
16. EEAR
17. EEDR
18. EEER
19. EEIR
20. EEMR
21. EENR
22. EERR
23. EETR
24. EIAR
25. EIDR
26. EIER
27. EIMR
28. EINR
29. EIRR
30. EITR
31. EMAR
32. EMDR
33. EMER
34. EMIR
35. EMNR
36. EMRR
37. EMTR
38. ENAR
39. ENDR
40. ENER
41. ENIR
42. ENMR
43. ENNR
44. ENRR
45. ENTR
46. ERAR
47. ERDR
48. ERER
49. ERIR
50. ERMR
51. ERNR
52. ERTR
53. ETAR
54. ETDR
55. ETER
56. ETIR
57. ETMR
58. ETNR
59. ETRR

Prove the couple's formula: 1 + 1 = 1. . . . . . . . PS: your proof shouldn't be less than a page, Lolzzzzzzz

Ortarico:

They're listed below but here is how to calculate how many there are
E _ _ R We want to fill those two blanks.
Case 1. The two letters in the blanks are the same.
The only such possibilities are these 3: EAAR, EEER, ENNR.
Case 2. The two letters in the blanks are different. Then
we must choose them from these 8: M,E,D,I,T,R,A,N. We can
choose the letter for the first blank in 8 ways, and for
each of those ways we can choose the letter for the second
blank 7 ways. That's 8×7 or 56 ways.
So 3 ways from case 1 plus 56 ways from case 2, gives us
59 ways total.
Here are all 59 ways listed:
1. EAAR
2. EADR
3. EAER
4. EAIR
5. EAMR
6. EANR
7. EARR
8. EATR
9. EDAR
10. EDER
11. EDIR
12. EDMR
13. EDNR
14. EDRR
15. EDTR
16. EEAR
17. EEDR
18. EEER
19. EEIR
20. EEMR
21. EENR
22. EERR
23. EETR
24. EIAR
25. EIDR
26. EIER
27. EIMR
28. EINR
29. EIRR
30. EITR
31. EMAR
32. EMDR
33. EMER
34. EMIR
35. EMNR
36. EMRR
37. EMTR
38. ENAR
39. ENDR
40. ENER
41. ENIR
42. ENMR
43. ENNR
44. ENRR
45. ENTR
46. ERAR
47. ERDR
48. ERER
49. ERIR
50. ERMR
51. ERNR
52. ERTR
53. ETAR
54. ETDR
55. ETER
56. ETIR
57. ETMR
58. ETNR
59. ETRR

Wow! Outstandingly correct! Now we all know who General Q is, don't we?

Ortarico: Prove the couple's formula: 1 + 1 = 1. . . . . . . . PS: your proof shouldn't be less than a page, Lolzzzzzzz

for God's sake, mathematicians take it easy.. WTH is Couple's formulae that make 1+1=2.. Smh

smurfy: As punishment for missing most of Nairaland Maths Quiz Competition, General Richiez asked General Q to list all possible four-letter words that can be formed from the word MEDITERRANEAN such that the first letter is E and the last letter is R.

How many of such four-letter words are possible?

Hahahaha very funny smurfy, nice question though

Richiez:

Hahahaha very funny smurfy, nice question though

smurfy:

@ = 28, ¥ = 30

Approximate acute angle values only.

d ans to d ques are 16.78' & 31.1'.
plz i just need d workings

Mr Calculus: plz guys help me out wit dis......
if 3sin2@=5sin2¥....
find @ & ¥ if(tan@=2tan¥)

plz my generals i need ur help to dis o.@laplacian,rechiez,ben,fx,smurfy,ortarico.e.t.c
i dont know but d answers to d ques are 16.78' & 31.1'.
plz i want d workings not only d answers