₦airaland Forum

Mbahchiboy: cause from d ques.
x^5-32=0..
so (-2)^5-32=-64
and dat makes it invalid

You seem to be right sha. I did it by inspection. It's better if one leaves it as (x - 2) (x^4 + 2x^3 + 4x^2 + 8x + 16) then.

Mbahchiboy: @ortarico
how did u obtain those values of x after dividin d equ by x-2 to get a new one.
and i dont think x can b -2

By also factorizing the quotient: (x^4 + 2x^3 + 4x^2 + 8x + 16)
Why do you think it can't?

Mbahchiboy: @ortarico
i want it to b express d way i express sinhx and coshx

Wow, fine!
Though I only have the basic knowledge of its concept, can it be that way?

Mbahchiboy: to my generals plz help me out
(1)if
coshx=1+x^2/2!+x^4/4!....,
sinhx=x+x^3/3!+x^5/5!.....,
find tanhx?
(2)find x if x^5-32=0
[hint::x must have five root e.g a,b,c,d or e]

2. To your second question:
To find the factors of x^5 - 32 = 0, solve x^5 = 32
x^5 = 2^5
x = 2
Therefore, (x - 2) is a factor of (x^5 - 32)
Now simply divide (x^5 - 32) by (x - 2) using long division and you'll get:
(x^5 - 32) = (x - 2) (x^4 + 2x^3 + 4x^2 + 8x + 16)

But (x^5 - 32) should have five values which you denoted to be a, b, c, d and e
:. Further factorization gives their values to be:
(x - 2), (x + 2), (x - 2), (x + 2) and (x - 2)

Mbahchiboy: to my generals plz help me out
(1)if
coshx=1+x^2/2!+x^4/4!....,
sinhx=x+x^3/3!+x^5/5!.....,
find tanhx?
(2)find x if x^5-32=0
[hint::x must have five root e.g a,b,c,d or e]

Hmmmm. . .
1. Your first question demands to be proved using Euler's formula for identities and hyperbolic functions (e^x).
If coshx = 1 + x^2/2! + x^4/4!. . .
Sinhx = x + x^3/3! + x^5/5!. . .

Euler's identity (e^x) for coshx = (e^x + e^-x)/2
sinhx = (e^x - e^-x)/2
:. Tanhx would be = sinhx/coshx
=> (e^x - e^-x)/2 / (e^x + e^-x)/2
=> (e^x - e^-x)/2 * 2/(e^x + e^-x)
=> (e^x - e^-x)/(e^x + e^-x)

So tanhx = (e^x - e^-x)/(e^x + e^-x)

Calculusf(x):
...great master but i guess this is newton-raphson iterative method

The prof after the order of Sir Isaac Newton. Yes, you're right. It's otherwise called the nondeterministic polynomial , it's used for solving transcedential equations.

oladoya: yeaaaa how things now? Hem,,, hem,,, hem..'scratching head' ok this is ma acount number: +2348171862611; just flash it with yours
account name: whatsAPP. Just a friendly talk. Oga Ortarico where you dey
, ‘ , ‘ , ‘ , ‘ , ‘ , ‘ , ‘ ,
, ‘ , ‘ , ‘ , ‘ , ‘, ‘ , ‘ ,
,__,____,
/____,_/ \ .;’;';.
l__[]__l_ l ,,)(,,..
I am at the front of your house you can see the rain is much.

Eyaah e pele oo, econs wizzy mi. . .even if u com mai auz u nor go c me, i don buy job. But you sure sey na mai auz u tel mumsy sey you bin go? ¤winks¤

emmyeuler1: i have two solutions but i'll start wit d easier one. Let # denote "congruency" then since 3^x+4^x=5^x, 4^x#5^x(mod3) or by Fermat's Little theorem (4^x)^2#(5^x)^2#1(mod3), i.e 4^2x#5^2x#1(mod3) or 16^x#1(mod3) and 25^x#1(mod3) so by FLT, x must divide @(3)=3-1=2 Euler's phi function, i.e x|2 hence either x=1 or 2 but since 1 does not satisfy the eqn, x=2,

Even nondeterministic polynomial (NP) method can be used.
x = x1 - f(x1)/f'(x1)

3^x + 4^x = 5^x
3^x + 4^x - 5^x = 0
f(x) = 3^x + 4^x - 5^x
from polynomial:
f(x1) = 3^x + 4^x - 5^x
let x1 = 2
f(2) = 3^2 + 4^2 - 5^2
f(2) = 9 + 16 - 25
f(2) = 0
:. x = 2 is a factor

f'(x) = 3^xIn3 + 4^xIn4 - 5^xIn5
f'(2) = 9In3 + 16In4 - 25In5
f'(2) = 9.89 + 22.18 - 40.24
f'(2) = -8.17

i.e x1 = 2, f(2) = 0 and f'(2) = -8.17. . . .substituting into the formula above:
x = 2 - 0/-8.17
x = 2 + 0
x = 2

Boladearo: hmmmm, No 1 question is from further maths project 1, i asked my teacher that can a question combine both A.P and G.P together and he said yes, unfortunately 4 he gave it to me alone as an assignment, if i could still remember my ans was 1024, can't recall sha

It's 105.

Adokwu.Ondoma:
questions 1&2 are straight forward, for question 3, set x=y-7/2, generally, always take d mean of d first two brackets, equate to zero, then u solve 2 get d constant 2 b added 2 d new variable y.

Hmmmm. . .Although solved, thank you.

Mr Calculus: okeiii...datz 9z to hear..*holds lips together*
wheres my chief ortarico

Present sir!

Me loves this one: 'In this life, there's nothing more, the trick is never to give up'.

This one alof una dey stress the primacy of ideas so. Una no tell me sey una be idealist like Plato, Leibniz, Descartes and Berkeley in metaphysics. Well you guys are on point sha, ideas exist in our minds.

@farano:
Huh?! BTW aw is Omosive??

She's very fine, thanks for asking after her.

mathefaro: My brother, [x+1]log5 != log5x + log5 and
[x²-1]log3 != log3.x²-log3

it should be noted that
(a + b)log 5 = alog5 + blog5
==> log5^a + log5^b
and not
log5.a + log5.b
I hope you get my point.
Weldone generals.

You rightly identified his error sir.

2nioshine: just a contribution
log₃15 is d same as log15/log3
as log_yx =logx/logy
d rules f logartm bro..sti observn..nice wuk @ all

Exactly boss, that's why I said no calculator or tables. Thanks!

rhydex 247: integral xarcsinx dx.
solution
arcsinx implies sin@=x. drawing a right angle triangle with angle @. opp=x, hyp=1 and adj=sqrt(1-x^2). recalling from our SOHCAHTOA. Cos@=adj/hyp= sqrt(1-x^2). @= arcsinx. d@/dx= 1/ sqrt(1-x^2). dx=sqrt(1-x^2)d@. recall dat sqrt(1-x^2)=cos@. now substitute. we av
integral @sin@cos@d@
let u=@. du/d@=1 and v=sin@cos@. dv= sin^2@/2
using integration by part
uv- integral vdu
@sin^2@/2 - integral sin^2@/2 d@
@sin^2@/2 - 1/2 integral sin^2@d@.
@sin^2@/2 - 1/2 integral 1(1-cos2@)/2 d@
@sin^2@/2 - 1/4 [(integral d@ - integral cos2@d@)]
@sin^2@/2 - 1/4 (@-sin2@/2)
@sin^2@/2 - 1/4@+sin2@/8. now recall that sin2@=2sin@cos@
@sin^2@/2 -@/4 + 2sin@cos@/8
@sin^2@/2 - @/4 +sin@cos@/4
substituting we av
x^2arcsinx/2 - arcsinx/4 + xsqrt(1-x^2)/4
1/4[(2x^2-1)arcsinx+xsqrt(1-x^2)}+C

Alias prof series, welcome sir!

yemmytcm: i lyk you jere,you knw book brova

Hmmmmm. . . .na you sabi, you no like me b4 ah bin help you settle the matter ni?

Calculusf(x):
Integral e^(x^2)

Weird! Use Jacobian of the transformation rule and pls don't ask me cause I dunno it either!

honey: he didnt state that in the question, but if calculator and tables are restricted, then we use factorial method..... log3.x²-log5.x-log15= (x+1)(log3.x-log15), then x =-1 or x=(log15/log3)= 2.465....... my former solution still stands

Nah, you made a little error bro. The values of X are: -1 and log₃15. . . . .pls, cross check again.

The supplementary and direct entry admission lists, should be out by next week, you can mark my words.

@farano:
..someone is jealous here

Lol. . .cause he's your most legitimate hubby, everybody knows.

Calculusf(x):
Pls,explain ur question bro@lavosier.

What is his question prof? Can't find it too!

Calculusf(x):
...my oga at the top...u will still integrate the square root of (1-x^2)...

ok my prof, but I just saw that you've done the correction can't find it anymore oo

honey: changing the question to logarithimic form....we have [x+1]log5=[x²-1]log3, expanding the bracket and moving the LHS to the RHS,.....log3.x²-log5x-log15.... seeing that this is a quadratic equation, we use formula method.... x= (log5+_([log5]²+4.[log3*log15])^1/2)/2log3....x= (0.6990+1.6532)/0.9542 or (0.6990-1.6532)/0.9542....... x= 2.4651 or -1

You tried bro, x is equal to -1 as its first value but not 2.4651 as second! And mind you, it should be done without using tables or calculator.

honey: Electrical/Electronics Engineering

okies

honey: the question can be translated on a right angled triangled, since tan^-1[x/y]=y/x, we can see that the right angled triangle has an angle of (y/x)^o, opp=x, adj=y and hyp= -\|(x²+y²)....... the question can be restructured as tan[y/x]- x/y =0, differentiating implicitly, we have -1/y-sec²(y/x) + dy/dx(1/x.sec²(y/x)+x/y²), going back to our right angled triangled triangled triangle, we will see that sec²(y/x)= (x²+y²)/y², replacing it in the equation.......dy/dx(2x²+y²)/xy²= (2x²+y²)/x²y, cross and multiplying yields. dy/dx=(xy²)/(x²y)...dividing,....dy/dx= y/x

Unilag mate, what course are you about studying bruv?

Boladearo: Oga ortarico, na my boss o, make the whole ause no

Lol. . . .and make the house know sey you be my master now.

Odehfamily: Abeg please help me out again.
2/root 3(2/root 3-root 12/6). Solve wifout using cal.or table.

hmmmmm. . . .
2/_/3 (2/_/3 - _/12/6)
First of all, simplify the surds in the bracket by taking 6_/3 as its LCM:
2/_/3 (12 - 6/6_/3)
2/_/3 (6/6_/3)
clear bracket; 12/6_/9 = 12/18 = 2/3

Boladearo: @odehfamily
salem:sunday = 200k:50k = 100K:25K
Therefore for every 1naira that salem gets sunday is given 25k i.e 1/4naira
Sunday:shaka= N4:N2 = 2:1
i.e for every N1 sunday gets shaka get N1/2
salem:sunday: shaka= 1:1/4;1/2*1/4 = 1:1/4 :1/8 multiply tru by 8 (LCM)
salem ; sundea : shaka = 8 : 2 : 1
skaka's share = 1/11 *1100 = 100
My oga's at the tower, am i right.

On point boss!

Richiez: Nice job bro!

Thank you sir.

Calculusf(x):
...you have done a great job my brother...but where you put dv=arcsinx...v will be integration of arcsinx...which makes the question complex

Ok, i've corrected myself by making dv = x, u = arcsinx
Got the answer to be: (x^2)arcsinx + |_/1-x^2)|^1/2 + c

lavosier: 3/5of pool takes 8hrs.1pool shud tke 8/(3/5)=40/3hrs..so 40/3hrs - 8hrs=5hrs 20mins..
Op m rite sha.

Bravo, you killed it boss!