kwakayekaa: Let Q>0 be the set of positive rational numbers. Let f : Q>0 ! R be a function satisfying
the following three conditions:
(i) for all x; y 2 Q>0, we have f(x)f(y)  f(xy);
(ii) for all x; y 2 Q>0, we have f(x + y)  f(x) + f(y);
(iii) there exists a rational number a > 1 such that f(a) = a.
Prove that f(x) = x for all x 2 Q>0.

kwakayekaa: Let Q>0 be the set of positive rational numbers. Let f : Q>0 ! R be a function satisfying
the following three conditions:
(i) for all x; y 2 Q>0, we have f(x)f(y)  f(xy);
(ii) for all x; y 2 Q>0, we have f(x + y)  f(x) + f(y);
(iii) there exists a rational number a > 1 such that f(a) = a.
Prove that f(x) = x for all x 2 Q>0.

Laplacian:
hey, wat symbol is between f(x)f(y) and f(xy)?
Wats d symbol between f(x + y) and f(x) + f(y)?
Write this in words f : Q>0 ! R

benbuks: Am working as a computer operator in a company,my manager general, jackpot ask me to creat a 5-digit password for one our agents, such that
1. the square root of the first is the second
2. The second is 5 more than the third
3.the third and the forth are consecutive even and their sum is 10
4. The sum of the whole digits is 30

if i dont creat the right password i might lose my job, u kw my madam, na no nonsense woman o abeg help me create this password...

benbuks: Am working as a computer operator in a company,my manager general, jackpot ask me to creat a 5-digit password for one our agents, such that
1. the square root of the first is the second
2. The second is 5 more than the third
3.the third and the forth are consecutive even and their sum is 10
4. The sum of the whole digits is 30

if i dont creat the right password i might lose my job, u kw my madam, na no nonsense woman o abeg help me create this password...

LogoDWhiz: 39468

Calculusf(x):
.hmmm...from 1.the square root of the first is the second...hope it's not the square root of the second is the first,if it's that.then,let the digits be abcde...from my modified (1).a=sqrt.b ...from 2 b=c+5 ...from 3.c+d=10(but don't forget here that the third(c) and fourth(d) are consecutives even numbers.then d=c+2...from 4.a+b+c+d+e=30...considering 3...c+d=10 and d=c+2.then c+c+2=10...2c=8.c=4...and d=4+2=6...from 2,b=c+5...b=4+5=9...from 1,a=sqrt.b=sqrt9...therefore a=3...so far,a=3,b=9,c=4 and d=6...don't forget from 4 that a+b+c+d+e=30...substitute,3+9+4+6+e=30...22+e=30...e=8...then the password is 39468

benbuks: ..wow thanks man, you just safed my job.

....*madam jackpot , a'v cracked it ma..,oya give me my allowance, else i 'll go on strike.*.....lolz.

Laplacian:
y re u alwaz swappin ur gender?

benbuks: ...sorry maybe it was an intentional mistake

Sir-Tunechi:
Help me out, cot@ + sec@=5 find cos@

benbuks: ..wow thanks man, you just safed my job.

....*madam jackpot , a'v cracked it ma..,oya give me my allowance, else i 'll go on strike.*.....lolz.

smurfy:

Here goes...

cot@ = cos@/sin@ and sec@ = 1/cos@

So, cot@ + sec@ = 5 becomes
cos@/sin@ + 1/cos@ = 5

Multiply through by sin@cos@, the LCM:

cos^2@ + sin@ = 5sin@cos@

cos^2@ = sin@(5cos@ - 1)

Square both sides to change the sine to cosine:

cos^4@ = sin^2@(5cos@ - 1)^2

cos^4@ = (1 - cos^2@)(25cos^2@ - 10cos@ + 1)

Open brackets to get:

26cos^4@ - 10cos^3@ - 24cos^2@ + 10cos@ - 1 = 0

The equation above is comparable to
26x^4 - 10x^3 - 24x^2 + 10x - 1 = 0

So, cos@ = -0.9865, 0.2089, 0.9699

@Generals

Is there a more novel way of tackling this question? I'd like to know, please.

aysuccess99:
oga mi sir, i think the ansa 2 dis qstn is 14* to the nearest degree i.e @ is 14 to the nearest degree...i too have tried solving it, bt when i got 2 a stage, i got lost....so our ogas in the thread shld pls help with dis qstn o.....twale 4 u guys o.pls o...help show d workings o...bt i hope dis my ans(@=14 is correct)

benbuks: Mourning a great hero today,4me, no solving math today..till tomorrow...

smurfy:

Here goes...

cot@ = cos@/sin@ and sec@ = 1/cos@

So, cot@ + sec@ = 5 becomes
cos@/sin@ + 1/cos@ = 5

Multiply through by sin@cos@, the LCM:

cos^2@ + sin@ = 5sin@cos@

cos^2@ = sin@(5cos@ - 1)

Square both sides to change the sine to cosine:

cos^4@ = sin^2@(5cos@ - 1)^2

cos^4@ = (1 - cos^2@)(25cos^2@ - 10cos@ + 1)

Open brackets to get:

26cos^4@ - 10cos^3@ - 24cos^2@ + 10cos@ - 1 = 0

The equation above is comparable to
26x^4 - 10x^3 - 24x^2 + 10x - 1 = 0

So, cos@ = -0.9865, 0.2089, 0.9699

@Generals

Is there a more novel way of tackling this question? I'd like to know, please.

Sir-Tunechi:

good, the ansa @d back of the buk is in fraction form and its 12/13

smurfy:

He's asking for the value of cos@, not @. I know the working is correct. I just want to know if there's another way that'll bypass that horrible looking equation above.

By the way, where have you been all these years? You finish quiz con disappear, se?

aysuccess99:
it's nt lyk dat oga mi, i went on strike like ASUU bt i've called off my own strike.....oga mi, i'm feeling u most especially in d games section...kudos 2 u o.....o ya...continue d wrkings na....u fit do am.

jackpot: lolz. I've got a better offer na. What would a N250.00k allowance do for you?

I am thinking about raising your retirement age from 65 to 80 years.

how about that?

smurfy:

Here goes...

cot@ = cos@/sin@ and sec@ = 1/cos@

So, cot@ + sec@ = 5 becomes
cos@/sin@ + 1/cos@ = 5

Multiply through by sin@cos@, the LCM:

cos^2@ + sin@ = 5sin@cos@

cos^2@ = sin@(5cos@ - 1)

Square both sides to change the sine to cosine:

cos^4@ = sin^2@(5cos@ - 1)^2

cos^4@ = (1 - cos^2@)(25cos^2@ - 10cos@ + 1)

Open brackets to get:

26cos^4@ - 10cos^3@ - 24cos^2@ + 10cos@ - 1 = 0

The equation above is comparable to
26x^4 - 10x^3 - 24x^2 + 10x - 1 = 0

So, cos@ = -0.9865, 0.2089, 0.9699

@Generals

Is there a more novel way of tackling this question? I'd like to know, please.

Calculusf(x):
.which method did you use to solve the polynomial master?

indoorlove: Log3^x+Logx^3=10/3. Guru in the house, help me with above question