₦airaland Forum

elmajor:
1(35%)l +11(DW)l =12(3%)l..........1
1(12%)l + 3(DW)l = 4(3%)l..........2
We want to know how many litres of distilled water (X) can be added to 1 litre of a substance with 35% concentration to produce Y litre(s) of a solution with 3% concentration. This leads to the third equation, which is given as follows.
1(35%)l + X(DW)l = Y(3%)l..........3
We're looking for X litre(s) of Distilled Water that will produce Y litre(s) of a solution with 12% concentration.
So, multiply eq. 2 by 3.
3[1(12%)l + 3(DW)l = 4(3%)l]
3(12%)l + 9(DW)l = 12(3%)l..........4
Equate eq.1 to eq.4
1(35%)l +11(DW)l = 3(12%)l + 9(DW)l
1(35%)l +11(DW)l - 9(DW)l = 3(12%)l
Therefore,
1(35%)l + 2(DW)l = 3(12%)l
So, 2 litres of Distilled Water needed to 1 litre of the substance with 35% concentration give 3 litres of a solution with 12% concentration.

jackpot:
0/1=0 or 1/1=1?

Sirneij:
Divide ( y^a/y^b )^a+b by ( y^a+b/y^{a-b)[sup]a^2/b} or solve it using this image

https://scontent-ams3-1.xx.fbcdn.net/hphotos-xap1/v/t1.0-9/11140269_657765617659139_3050089635620568439_n.jpg?efg=eyJpIjoiYiJ9&oh=7957b2270599fe05565559b347a83df7&oe=56B570E7

MathsChic:
Straight forward

Karmanaut:
Easy.
Set tan(x) as sin(x)/cos(x)
So you'll rewrite the question as:
lim^x->π/2 sinx^cosx/cosx^cosx
Then you substitute π/2 for x.
Since cosx^cosx resolves to 0^0 which is mathematically undefined (0 or 1, depending on who you ask) substitute y for cosx.
So you have 0/y^y
Then using the exponential function the denominator becomes e^{lny^y}
which is e^y*lny
Since y = cos π/2 = 0
we have:
e[sup]0[sup]
which is 1.
So we have 0/1
= 0.
Done.

Karmanaut:
There's a way to solve it without L'Hopital's rule.

Karmanaut:
The answer is 1.
But I want to see how you solved it.

Karmanaut:
This is elementary.

Defining the Wronskian as [img]http://tutorial.math.lamar.edu/Classes/DE/Wronskian_files/eq0036M.gif[/img]

The differential equation is:
t²y" +2t³y' - t²y = 0

First we divide all through by the coefficient of the second derivative (y" ) which leaves us with:

y" +2ty' - y = 0

Applying the Wronskian defined above we get:
W = c*e^{-∫2t dt}
W = c* e^{-t^2}
(Because the integral (anti derivative) of 2t = t² + C)
W = c* e^{-t^2}
where c is an arbitrary nonzero constant.
That's all.
Cheers.

Vixxie:
Thanks

Vixxie:
Looks pretty straightforward to me.



First, we should note: if, for example, 200ml of 30% conc is added to 200ml of water, it gives 400ml of x% conc (more diluted)
200/400 = x/30; x=15% conc

1 part of 35% conc added to 11 parts of water gives 3% conc.
, i.e. 1 part can be any volume, but we can assume 1 ml for now.
1/(x+1) = 3/35
x = 32
for 11 parts of water; we can have
3/(3 + (32/11).11) = 3/35
where 32/11 = ~3, which makes sense for 1 part of 3ml of solution and 11 parts of 3ml of water

1 part of 12% conc added to 3 parts of water gives 3% conc.
1:4= 3:12
where 4 = 3 parts of water + 1 part of solution

how to get 12% conc from 35% conc?
1 part of 35% conc, how many parts of water to get 12%?

1/(x+1) = 12/35

x = (35-12)/12 = 23/12 = ~2 parts of water

Ans: ~2 parts of water.

Anyways, I love maths and this was a good one to do!

masperano:
It is like the double integral doesn't converge.

ayokunlei:
thanks. Pls Help with the other part (b) pr that one
of them misses it?

Umartins1:
Boss, I'm not good at probability ooo but let me try. I'm sure agentofchange1 will correct me.

pr that P hits= 2/9

Pr that P misses= 1- 2/9 = 7/9


Pr that Q misses= 6/7

Pr that Q hits= 1-6/7 = 1/7

(a) pr that only one of them hits

= (pr that P hits + pr that Q misses) * (pr that P missed + pr that Q hits)







****************************


Please, let me wait a little and see if the bosses will confirm if I'm right so far.

MaxGraviton:
Thanks a lot cheif.

But one question cheif, Why does it have to be U=2sin@

Why can't it just be U=signatu @ or U=cos @ .
Coz from the look of things up there in the solution, those Trig notatations were overcrowding the solution. . . SO does it have to be like that?

MaxGraviton:
Thanks a lot cheif. Expecting...

MaxGraviton:
Thanks a lot. But pls I'm finding it dificult to understand it. Do you mind sending a shot of the solution?

Karmanaut:
Far from it. I'm just an internet troll.

Karmanaut:
Geez, I forgot.
No light so I have to type the answer.
2)2xy" + 3y' -y= 0
Recurrence relation is an = an-1/(2n+2r+1)(n+r)
The roots of the indicial equation are -1/2 and 0.
The first series is y1(x) = a0( 1 + x/3 + x²/630+....)
which is the same as the series of sinh√(2x) * 1/√(2x)
Therefore y1(×) = (a0*sinh√(2x))/ √(2×)
The second series is
y2= b0(1+x/30 + ) which is the same as cosh(2×)/√×
y2= (b0 * cosh√(2×)/√×)

You can rewrite it in terms of e.
By using the trigonometric identities:
sinh(x) = 1/2(e^x -e^-x)
cosh (x) = 1/2(e^x +e[/sup]-×[/sup])
Define P1 and P2 as the linear combination of the two solutions with P1= a0/2√2 + b0/√2 and P2 = b0/√2 - a0/2
y(x) = (P1 * e^√(2×))/√× - (P2 * e^-√(2×)) /√(2×)
Which simplifies to
e^-√(2×) (2P1 e^2√(2×) + P2 * √2) / 2√×

MaxGraviton:
Senior men and Women in the house. Pls I need help with this. . . Thanks in Advance.

Karmanaut:
Same answer.
Will do the rest later, perhaps this night.

ayokunlei:
thanks alot

Karmanaut:
Same answer.
Will do the rest later, perhaps this night.

ayokunlei:
Please help solve : integral of 1/ (1+tanx)dx

cc: umartins1, agentofchange1, mathefaro, laplacian, Karmanut and others... Thanks

Karmanaut:
I'll assume that number 1 question is:
4xy" + 2y' + y = 0.
Attached is the solution for question 1.
Took 3 Leaves (6 pages to solve)

ayokunlei:
Please help solve : integral of 1/ (1+tanx)dx

ok let t=tanx, then use partial fraction

Ans:0.5[ln(1+t) +arctan(t) -0.5ln(1+t^2)]+k

hope you get
cc: umartins1, agentofchange1, mathefaro, laplacian, Karmanut and others... Thanks