For the second one I got stuck here

Y = ln(x) + ln(y)

Mrshape:
Can Anyone In This Forum kill This?

my answer is n = ([-(d+2a) + or - ((d+2a)^2 - 4d(2a-2s))^1/2]/2d) + 1

Deicide:
For the second one I got stuck here

Y = ln(x) + ln(y)

Oh!
I doubt if anyone can solve thus in this forum not more than 2 people.

It takes the special grace to have seen the process required to solve it

2^c(5), y(54)_6{2b}[~2] <[9]=9d
Simple.

Mrshape:

Oh!
I doubt if anyone can solve thus in this forum not more than 2 people.

It takes the special grace to have seen the process required to solve it

check my solution above

josh123:
Who mathematics help.

Ask the Chinese!

Amotolongbo:
n² = [2S + d-2a]/d
Or

n = [(2S + d-2a)/d] ^0.5

bravo...let me break ur heart

u actually want to use another approach but u miss some...I can see ur mistake

and this correct

n =+- [(2S + d-2a)/d] ^0.5

because n is a real number from must obey the both the - & + number line(abstract algebra or real analysis)
same as the almighty formula solution not to fail

Scott200:
my answer is n = ([-(d+2a) + or - ((d+2a)^2 - 4d(2a-2s))^1/2]/2d) + 1

What is the last +1 for?

Mrshape:

Oh!
I doubt if anyone can solve thus in this forum not more than 2 people.

It takes the special grace to have seen the process required to solve it

I would try sha just need to get some facts

Deicide:
I would try sha just need to get some facts

Alright.
I will be glad to see your solution.
The solution is not up to 12 steps

Mrshape:

What is the last +1 for?

I changed n- 1 to be C did some manipulation arranged in format of quadractic equation, used almighty formula then changed C back to n - 1, so to get n u have to add 1 to the other side of the equation

Either
n = [(d - 2a) + (4a^2 - 4ad + d^2 + 8ds)^(1/2) ] /2d
or
n= [(d - 2a) - (4a^2 -4ad + d^2 +8ds)^(1/2)] /2d
...chai

LamaNewton:
Either
n = [(d - 2a) + (4a^2 - 4ad + d^2 + 8ds)^(1/2) ] /2d
or
n= [(d - 2a) - (4a^2 -4ad + d^2 +8ds)^(1/2)] /2d
chai

nice attempt but I think it is faulty

Mrshape:
Can Anyone In This Forum kill This?

I also feel a closing bracket is missing. Op check the question

You no go see all dem foolish olosho on dis kain thread

Mrshape:
This is the correct question.
Lalasticaala this has already been solved you would have put the other one on FP
Everyone is afraid of it

https://www.nairaland.com/5794496/make-y-subject-formula

Check the other thread.

yek:
What can one do with this in real life?

Quite a lot.
1. In reinforced concrete design of structures, the derived equations from the rectangular-parabolic compressive stress block (assumed rectangular in the british code of practice, BS8110) used in determining the lever arm (distance from compression block to tensile force) is derived from equations such as this.
We deplore this equation in the design of reinforced concrete structures subjected to flexural stresses such as low/high rise buildings, retaining walls, foundations, etc

2. In structural dynamics or vibrational mechanics, for free damped, free undamped, forced damped or forced undamped systems the derived equations to understudy the response of a structure or structural system under the action of pulsating loads (forced vibration) requires the solution of a second order differential non-homogenous equation resulting to algebraic equations such as the one put up here

3. In the structural design of sheet piles walls under the action of lateral surcharge, lateral soil stresses, soil pore pressures, etc the determination of the minimum penetration depth of sheet piles or cofferdam from equilibrium equations for stability, narrows down to the resolution of polynomials such as this, and making the penetration depth 'd' the subject of the formula


There are several other citations I can't mention at this time. So these are very applicable. It is also used in the shell design of weapon systems such as the Russian Khinzal weapon by solving the bi-harmonic equations for thin plates and shells
The world generally evolves around applied mathematics.

..

eminemkayc:

[s]Quite a lot.
1. In reinforced concrete design of structures, the derived equations from the rectangular-parabolic compressive stress block (assumed rectangular in the british code of practice, BS8110) used in determining the lever arm (distance from compression block to tensile force) is derived from equations such as this.
We deplore this equation in the design of reinforced concrete structures subjected to flexurural stresses such as low/high rise buildings, retaining walls, foundations, etc[/s]

[s]2. In structural dynamics or vibrational mechanics, for free damped, free undamped, forced damped or forced undamped systems the derived equations to understudy the response of a structure or structural system under the action of pulsating loads (forced vibration) requires the solution of a second order differential non-homogenous equation resulting to algebraic equations such as the one put up here[/s]

[s]There are several other citations I can't mention at this time. So these are very applicable. It is also used in the shell design of weapon systems such as the Russian Khinzal weapon by solving the bi-harmonic equations for thin plates and shells
The world generally evolves around applied mathematics.[/s]

Solve the problem of Hunger plaguing Nigerians now!

Mrshape:

Alright.
I will be glad to see your solution.
The solution is not up to 12 steps

Lol what's the answer?

S=n/2(2a+(n-1)d)
Clear the fraction by multiplying both sides by 2
2S=n(2a+(n-1)d)
swap the d and the (n-1)
2s=n(2a+d(n-1)
remove the inner parentheses on the right:
2s=n(2a+dn-d)
remove the remaining parentheses on the right
2s=2an+dn^2-dn
get O on the left by subtracting 2s from both sides
O=2an+dn^2-dn-2s
swap sides
2an+dn^2-dn-2s=O
dn^2+2an-dn-2s=O
dn^2+n(2a-d)-2s=O
dn^2+(2a-d)n-2s=O
Using quadratic formula
n=-(2a-d)+-(2a-d)^2-4(d)(-2s)/2d
finally
n=-2a+d+-4a^2-4ad+d^2+8ds/2d

Mrshape:
Can Anyone In This Forum kill This?

This is arithmetic progression

yek:
What can one do with this in real life?

Interpolation
Extrapolation
Error modeling and analysis
Error correction in observed spatial data

mila1121:
QED

Where is n in the second line

n=[- 4a+2d+or-(16a^2-4d^2+8ds) ^1/2]/4d
Slept off on it

Check it out.

OlawaleBammie:

Guy wetin b dis

check am well,...... That's your answer

Na solvient union na

AnyaIgbo:
Hi everyone, I'm back again to this forum. Sorry, I'm just seeing this now.
Lalasticlala, any prize for it?
Let's start by simplifying
s = [n(2a+(n-1)d]/2
Expand the bracket => s = (2an+dn²—d)/2
Noticed, we have exponent of "n", hence we use Calculus/Differentiation. We use the "Quotient Rule" => ds/dn = [vdu/dn - udv/dn]/V² .
where ds/dn means differentiate "s" under "n".
"u" is the numerator, while "v" is the denominator.
ds/dn = [vdu/dn - udv/dn]/v², solving

ds/dn = [2(2a+2dn-d) - (2an+dn²—d)0]/2²
=> ds/dn = 2(2a+2dn—d)/4
simplifying, we have: s = 2a+2dn—d/2
Then, making "n" subject: 2s = 2a+2dn—d
=> 2dn = 2s—2a+d
Therefore,
n = (2s—2a+d)/2d.

cc: Look at the right answer and how I solved it. Prove me wrong, mrshape dejt4u matex97 pyrex23 stennyb, co...

N.B: We have a virtual lesson platform on Facebook for those interested in learning (General/Basic) Physics in an easy way.
Search the page on Facebook and join for free:
PHYSICS VIRTUAL LEARNING - PVL

Let the AP be 1,2
S=3
a=1
d=2
From your formula n=(6-2+2)/4=6/4=1.5
So your formula is wrong My dear

Can you do this