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

Typo. It should be 1/1 = 1.

elmajor:
Anyone with a step by step solution to the problem below?

There is a substance which can be diluted to a number of solutions. The concentration of the substance is 35%, which can be diluted to 12% and 3% concentration.
If one part of the 35% concentration is added to eleven parts of distilled water, it will give a solution of 3% concentration and one part of 12% concentration added to three parts of distilled water will also give a solution of 3% concentration.
The problem is how to get the solution of 12% concentration from the 35% concentration. So, how many parts of distilled water will be added to one part of the 35% concentration to give the solution of 12% concentration?

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.

hey , greetings .ma madam is back, where u go hide na ?

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

So u sabi yam b4 ? ...hmm its well sha.. weldon boss

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.

my prof, seen my msg ?

Karmanaut:

Typo.
It should be 1/1 = 1.

Bisi and fibie ages add up to 29. Seven yrs ago bisi was twice as old as fibie. Find their present ages. Pls smeone shud solve dis

Profmaojo:
Bisi and fibie ages add up to 29. Seven yrs ago bisi was twice as old as fibie. Find their present ages. Pls smeone shud solve dis

b + f =29...1
(b-7) = 2(f-7)
-b+2f=+7...2

3f = 36
f=12
b = 29-12 =17

Bisi is 17; Fibie is 12.

Tanks

MathsChic:
b + f =29...1 (b-7) = 2(f-7) -b+2f=+7...2
3f = 36 f=12 b = 29-12 =17
Bisi is 17; Fibie is 12.

Profmaojo:
Tanks

You welcome.

agentofchange1:
my prof, seen my msg ?

Yes, boss. There's no series solution.

Try this pls.
A Software engineer has the capability of thinking 100 lines of code in 5 mins and can type 100 lines of codes in 10 mins. He takes a break of 5 mins after every 10 mins. How many lines of codes will he complete typing after an hour?

Karmanaut:

Yes, boss.
There's no series solution.

I actually disagree. I think there could be a series solution at an ordinary point like x=1, but certainly not at the singularity x = 0.

Arithmetic:
Try this pls.
A Software engineer has the capability of thinking 100 lines of code in 5 mins and can type 100 lines of codes in 10 mins. He takes a break of 5 mins after every 10 mins. How many lines of codes will he complete typing after an hour?

250 lines

agentofchange1:
Given the D. E, xy'-y-x-1 =0 , state the conditions for the series solution to exist.

hence or otherwise solve the DE..

See below attachments for the solution. Typically would have loved to type them out here with html, but it'll be cumbersome. Don't know why this site is like that

But a solution can't exist at x=0. I've explained all the details. Please bear with my almost illegible handwriting

Solving normally...
xy'-y-x-1=0
y'-y/x =1+1/x
(x^-1y)'=1/x + 1/x²
(x^-1y) = Inx - 1/x
y=xInx - 1

The power series solution is simply an approximation of the solution curve at x=1, which should give a good solution at that point. Should be desire a solution at any other point, we can do x = x_o. We can't get a solution at x=0 because of the Inx curve. And the first order DE showed it in the coefficients of y' and y, since 1/x is not analytic at x=0.

I'll also post if i can see how the curve will look like.
DEs solved with series solutions are just taylor series approximations of actual curves, either approximating it totally or partially. At least, as far as i can see.

(refer to picture)...with the series solution approximating the curve at x =1 with increasing values of n.

Plz can u xplain vector 2 me plz @richlez

MathsChic:

I actually disagree. I think there could be a series solution at an ordinary point like x=1, but certainly not at the singularity x = 0.

I tried Frobenius method but didn't make much headway.
Solving linearly the solution is x*ln(x) - 1
Also, y(x) =Ax + ×*ln(x) - 1 satisfies the equation, that's the general solution.
setting y(x) = f(x)
f(x) is not differentiable at 0.

Karmanaut:

I tried Frobenius method but didn't make much headway.
Solving linearly the solution is x*ln(x) - 1
Also, y(x) =Ax + ×*ln(x) - 1 satisfies the equation, that's the general solution.
setting y(x) = f(x)
f(x) is not differentiable at 0.

Yeah, agree. Not differentiable at x=0, but differentiable at x=1

masperano:

250 lines

I need soln pls.

Solve

Kindly solve with explicit solutions.

x + y =5
X^y + Y^x = 17

Find x and y....
Clear solutions please.

Lastly.....
Solve for x:

x^4 + x^5 = 12x^x

Please clear solutions.

Shit! Bless God I got C6 in Waec

Madmathecian:
Solve

x-5y=4
2x+y=7

MathsChic:


x-5y=4
2x+y=7

x-5y=4......i
2x+y=7.....ii
multiply i by 2 nnd ii by 1
2*: x-5y=4
1*: 2x+y=7
thus, it becomes;
2x-10y=8.....iii
2x- y=7.....iv
subtracting iv from iii
it becomes
-9y=1
y= -¹/9
substute for y in ...ii
thus;
2x+(-¹/9)=7
2x-¹/9=7
2x=7+¹/9
2x= 63/9
18x=63
x=63/18
x=7/2

MathsChic:

hello.
Integrate (sinx)^2

Goodyshoes:
Kindly solve with explicit solutions.

x + y =5
X^y + Y^x = 17

Find x and y....
Clear solutions please.

x=2, y=3
Or
x=3, y=2

shaboti:
hello. Integrate (sinx)^2

1/2 (x-sin(x)*cos(x))+ C

Goodyshoes:
Lastly..... Solve for x:
x^4 + x^5 = 12x^x
Please clear solutions.

As you can see clearly from the graph the function has roots at x=2 and x=3.

Madmathecian:
Solve

3x+2y-z = 4
2x-3y+3z=1
x+2y-2z=2

shaboti:
hello.
Integrate (sinx)^2

Karmanaut:

1/2 (x-sin(x)*cos(x))+ C

That's the solution above, but ama try expatiate.
∫(sinx)²dx

We know that cos(2x) = 1 - 2(sinx)²
So

∫(1 - cos(2x))/2dx
∫(1/2 - cos(2x)/2)dx

=(1/2)x - sin(2x)/4 + c
Since ∫(cosx)dx = sinx

Therefore
=(1/2)(x - sin(2x)/2) + c

Sin(2x) itself is 2sinxcosx

Therefore
= (1/2)(x - sinx.cosx)

Madmathecian:
Solve

The A part you can solve as a quadratic equation/inequality.
You should get x>= 1/2(√13 -1) and x <= 1/2(-1-√13)

For the B part:
A function is negative on intervals (read the intervals on the x-axis), where the graph line lies below the x- axis .
As you can see from the graph below, the interval below the x-axis is approximately from -2.25 to 1.25.
You can write it in a different way, by denoting “ negative values” by f(x) < 0: (-2.25, 1.25)
You can verify that plugging any value of x in that range gives you a negative number.

Cheers.