₦airaland Forum

Make R the subject of the formula.

R = PR +PB/E

youngest85:
R = PR +PB/E

can you screenshot ur workings? NB: we have two R in the question

lytech1:
can you screenshot ur workings?
NB: we have two R in the question

Oga na off hand I do am o

youngest85:
Oga na off hand I do am o

put it into written bro lets learn frm you

R =1/(2P) × [(E² - 2Pb) ± E(E² - 4Pb)^½]

youngest85:
R = PR +PB/E

lol.. This is wrong

dejt4u:
R = ((1-2b) ± {√[(4E²/P)-4b+1]})/2

Boss!
Drop steps

dejt4u:
lol.. This is wrong

Oin

lytech1:
Make R the subject of the formula.

p = [RE/R+b]^2

take square root of both sides

₹p = ₹[RE/R+b]^2

₹p = RE/R+b

cross multiply
₹p(R+b) = RE

open bracket and simplify

₹p*R + ₹p*b = RE

COLLECT LIKE TERMS

₹p*b = RE - ₹p*R

₹p*b = R(E-₹p)

.divide both sides by E-₹p

R = ₹p*b/E-₹p........



Note that ₹ stands for square root and * for multiplication sign.....good luck.

Very wrong

NDSMELODY:
p = [RE/R+b]^2

take square root of both sides

₹p = ₹[RE/R+b]^2

₹p = RE/R+b

cross multiply
₹p(R+b) = RE

open bracket and simplify

₹p*R + ₹p*b = RE

COLLECT LIKE TERMS

₹p*b = RE - ₹p*R

₹p*b = R(E-₹p)

.divide both sides by E-₹p

R = ₹p*b/E-₹p........



Note that ₹ stands for square root and * for multiplication sign.....good luck.

dejt4u:
My answer is R =1/(2P) × [(E² - 2Pb) ± E(E² - 4Pb)^½]

show your workings from the begining till end sir

P= RE²
----
(R+b)²
P(R+b)²=RE²
(PR+Pb)²=RE²
PR²+Pb²=RE²
PR²+Pb²=E²
-------------
R
PR+Pb²=E²
PR=E² - Pb²
R=E² - Pb²
-----------
P

am lost

NDSMELODY:
show your workings from the begining till end sir

shown

dejt4u:
Very wrong

obviously it was a mistake since E only has the square

Here is my solution

exynos:
Boss!

Drop steps

dropped

lytech1:
Make R the subject of the formula.

Lol
Has it been solved?

Op, I think i have been able to answer your question.
It's more of a factorization game. Lol

R = E^2/P - 2b -b^2

Sorry, I cannot upload the proof. I'm using freebasics.

Mrshape:
Lol Has it been solved?

dejt4u got it correctly.

dprpikin:
dejt4u got it correctly.

Good

Equation of a mad man,even Chika obi can hardly solve this

Equation of a mad man,even Chika obi can hardly solve this

dejt4u:
dropped

Seen boss!

Check this out

I hope you understand the steps sha


Meanwhile, check my signature

This is the soln

Akechukwuka:
This is the soln

lol.. Wayo wayo.. Why will you let R² be 1 ??
This representation is not valid in mathematics

dejt4u:
lol.. Wayo wayo.. Why will you let R² be 1 ??
This representation is not valid in mathematics

Are u a mathematician? If yes, I pressasume u 've done laplace transform,,what happen when solving laplace transform eqn.? .
There 're some eqn dat need manipulation in order to arrive at wat u 're looking for, but the manipulation must tally with the problem given..
The 1 in quest can be a letter,but I chose 1.

Akechukwuka:
Are u a mathematician? If yes, I pressasume u 've done laplace transform,,what happen when solving laplace transform eqn.? .
There 're some eqn dat need manipulation in order to arrive at wat u 're looking for, but the manipulation must tally with the problem given..
The 1 in quest can be a letter,but I chose 1.

lol.. Effect of too much knowledge. You don't need to worry yourself about any manipulation. This is straight forward.

You chose to let R² be 1, can you justify why you chose 1. Why don't you choose 2, 3, 4 or any other number..

That your process is not valid at all unless you can justify yourself which I doubt you can

dejt4u:
lol.. Effect of too much knowledge. You don't need to worry yourself about any manipulation. This is straight forward.

You chose to let R² be 1, can you justify why you chose 1. Why don't you choose 2, 3, 4 or any other number..

That your process is not valid at all unless you can justify yourself which I doubt you can

��..I laff
U 're funny, how does dat eqn a striaght one,

The 1 in the soln of the eqn above is a constant..
I can decided to use k or any letter, but I chose d numb 1 as constant ,

Before further prove , and argument with u
What level 're u in maths?