Yay for Math!

[Minister]

I thought I'd make use of the absolutely awesome minds of Helplife2.net and ask for some help on a couple of Math problems I'm struggling with.

Rearrange to make n the subject,

X = (1/R)*((4n^20/n^2 - 4))


and...

Find the values of x which satisfy the inequality,

(1/x-4) > (1/x-6)


I love you guys.

 

[xcellerate]

if i read the problem right...

n = (2^(8/9) * (r*x=4)^(1/18)) / 2

 

[Mellish]

:l. This has gave me an insight into the future of maths for me. ****. We are only doing trinomials (factorizing).

 

[theotherguy]

:l. This has gave me an insight into the future of maths for me. ****. We are only doing trinomials (factorizing).

Trust me, it gets much, much worse than this.

Questions on my concepts homework:

1.Show that
2^n=1+Sigma(2^i from i=0 to i=n-1)

By counting in two ways. (Hint: Partition 2^n in such a way that produces the sum on the right, with the empty set added on)

2. Prove that all integers have unique prime factorization.

Aka, integer N=p1*p2*p3*p4*...*pk, where p1, p2, etc. are primes
Assume N also equals q1*q2*q3*...*ql, where q1, q2, etc. are primes.

3. Find x, y such that
ax+by=GCD(a,b) exists for any a, b

Questions for my calculus homework:
1. y''+3y'+4y=e^2x+x^2, y(0)=0, y'(0)=1

Solve for y.

2. Prove that Sigma(a*r^k, k=0 to k=infinity), where a is a real number between -1 and 1, converges on 0.

 

[Minister]

Ya, I only do Engineering. We put use out of the numbers

 

[Raziaar]

Trust me, it gets much, much worse than this.

Questions on my concepts homework:

1.Show that
2^n=1+Sigma(2^i from i=0 to i=n-1)

By counting in two ways. (Hint: Partition 2^n in such a way that produces the sum on the right, with the empty set added on)

2. Prove that all integers have unique prime factorization.

Aka, integer N=p1*p2*p3*p4*...*pk, where p1, p2, etc. are primes
Assume N also equals q1*q2*q3*...*ql, where q1, q2, etc. are primes.

3. Find x, y such that
ax+by=GCD(a,b) exists for any a, b

Questions for my calculus homework:
1. y''+3y'+4y=e^2x+x^2, y(0)=0, y'(0)=1

Solve for y.

2. Prove that Sigma(a*r^k, k=0 to k=infinity), where a is a real number between -1 and 1, converges on 0.

Spoiler

 

[Minister]

I'm pretty sure xcellerate is joking, because I have absolutely no idea how he came up with that.

Theotherguy, you seem like a math kind of guy. Any clues on my questions? :E

 

[<RJMC>]

Rearrange to make n the subject,

X = (1/R)*((4n^20/n^2 - 4))

20 kicks the n(iggas) out of the() because is racist and then one of the n backstab the other n and give a supposed coupon for prostitutes and then X leave

(1/x-4) > (1/x-6)

the X's fuse together theyr magic rings and one transform into a gorilla and the other in a armor of ice so X turns into a ice armored gorilla to kick the asses of the others

 

[Pulse]

As someone who self-diagnosed himself with Dysgraphia, let me be the first to say:

WAT

 

[Druckles]

Maths makes me awesome.

 

[Dog--]

Maths is stupid, I'm so glad my career choice will not need math (well, it'll need math up to the number 7 )

 

[vanillacrazycake]

(1/x-4) > (1/x-6)

Multiply throughout by (x-4)^2. Since a squared quantity is never negative, there is no need to reverse the inequality.

(x-4) > (x-4)^2 * (1/x-6)

Bringing all terms to left hand side:

(x-4) - [(x-4)^2 * (1/x-6)] > 0

Since (x-4) is a common factor:

(x-4) [1- (x-4) * (1/x-6)] > 0

(x-4) [1 - (x-4/x-6)] > 0

(x-4) [ (x-6-x+4)/(x-6) ] > 0

(x-4) [-2/(x-6)] > 0

(-2x + 8)/(x-6) > 0

Multiply throughout by (x-6)^2:

(-2x + 8) (x-6) > 0

f(x) = (-2x + 8) (x-6)

Critical values: x=4 and x=6

The critical values give you three regions: x<4, 6>x>4, x>6.

Choose a value in each of these regions, and calculate the value of the expression f(x). Look for positive values, since f(x) > 0.

6>x>4 is the only region that gives a positive value for f(x), thus it is the solution to the inequality.

 

[Jintor]

And so's your mother.

Seriously though I did maths as a proper topic for the last time on Monday. Barring unfortunate Uni choices, of course.

 

[RakuraiTenjin]

Yay for meth?

owait

 

[Minister]

Thanks vanilla!

Christ I need some practice, I would never even think to start it that way :O

Help is much appreciated :E

 

[soulslicer]

Only Vanilla was correct. The rest of you are wrong as I see it for the 2nd question.

Oh and you have this

you can solve it in 15 seconds

 

[Jintor]

Goddamnit I ****ed up somewhere, ended up with x-6/x-4 > 0.

 

[Gemma]

wtf
WTF
wtfffff........
im lost