Melody

You are right. Not everyone uses this site to cheat from.

I am sure some people learn a great deal here.

If I did not believe that then I would not be here. 

Next time make sure you delete all the meaningless $ signs.  So that your question presents properly.

\(\text{Let F(x) and G(x) be polynomials such that }\\ f(x)=F(x)(x+3)(x-4)(x-8)\\ g(x)=G(x)(x+3)(x-4)(x-8)(x+5)(x-2)\\ \frac{g(x)}{f(x)}=\frac{G(x)(x+3)(x-4)(x-8)(x+5)(x-2)}{F(x)(x+3)(x-4)(x-8)}\\ \frac{g(x)}{f(x)}=\frac{G(x)(x+5)(x-2)}{F(x)}\\ \)

g(x) is only divisable by f(x) if F(x) is a factor of G(x)(x+5)(x-2)

for instance in the following example g(x) is not divisable by f(x)

\(g(x)=7(x+5)(x-2)(x+3)(x-4)(x-8)\\ f(x)=(x^8+x^5-4x)(x+3)(x-4)(x-8)\\\)

Thanks

I just added this to our list of useful calculators. 

One way to enter    6C5 is     nCr(6,5)

If you want an old question temporarily opened you can privately message a moderator and request it. 

It seems to me that this is going to be the same as rolling a dice 5 times and finding the probability that 4 is rolled any 2 times in a row.

call a 4 a success and any other a failure then   for any individual roll   P(S)=1/6     P(F)=5/6     

44444  Any 5 fours        1 way                                                   (1/6)^5

4444  any 4 fours           5 ways                                                 5*(1/6)^4*(5/6)

444      an 3 fours          5C3 ways = 10 ways                           10*(1/6)^3*(5/6)^2

except   4*4*4     no but all other 3 fours ok.    -1 way                (1/6)^3*(5/6)^2

No fours 1 way    (5/6)^5                                                               

P(4*4**)=P(4**4*)=P(4***4)=P(*4*4*)=P(*4**4)=P(**4*4) = (1/6)^2*(5/6)^3

\(\text {P(rolling 2 fours together in 5 rolls ) }\\ =\text{1-P(no fours)-P(1 four)-P(2 fours but not together) - P(3 fours but none together) } \\=1-(\frac{5}{6})^5- 5(\frac{1}{6})^1(\frac{5}{6})^4 - 6(\frac{1}{6})^2(\frac{5}{6})^3 - (\frac{1}{6})^3(\frac{5}{6})^2\)

1-(5/6)^5-5*(1/6)^1*(5/6)^4-6*(1/6)^2*(5/6)^3-(1/6)^3*(5/6)^2 = 0.0965792181069959   \(=\frac{751}{7776}\)

2 fours in the right places

44***

*44**

**44*

***44

P(2 fours in the right places) = 4*(1/6)^2(5/6)^3

OR

P(rolling 2 fours together)

=P(5 fours)+P(4 fours)+ P(3 fours) - P(4*4*4) +P(2 fours in the right places)

(1/6)^5+5*(1/6)^4*(5/6)^1+10*(1/6)^3*(5/6)^2-(1/6)^3*(5/6)^2+4*(1/6)^2(5/6)^3 = 0.0965792181069959    \(=\frac{751}{7776}\)

That is excellent.

I did it 2 different ways and got the same answer both times :)

\(y=(\sqrt2)^{(\frac{7\pi i}{4})}\\ y=2^{(\frac{7\pi }{8}i)}\\ log y=log 2^{(\frac{7\pi }{8}i)}\\ ln y=(\frac{7\pi }{8}i) ln 2\\ e^{ln y}=e^{(\frac{7ln 2\pi }{8}i) }\\ y=e^{(\frac{7(ln 2)\pi }{8})i }\\ y=cos(\frac{7(ln 2)\pi }{8})+isin(\frac{7(ln 2)\pi }{8})\)

You had better check that very carefully.

Find the sum of all positive rational numbers that are less than 5, and that have a denominator of 30 when written in lowest terms.

A question for you my friend.

Which of these 5  will be included in the sum?  There might be more than 1 correct answer

   \(1) \frac{1}{30}\\ 2) \frac{5}{30}\\ 3) \frac{30}{40}\\ 4) \frac{30}{30}\\ 5) 3\frac{7}{30}\)

Answer this for me and then someone (maybe me) will help you further.  (If you need more help)

Please no one answer over me.  I am trying to facilitate learning and interaction.

I'm just playing.  Like EP said, what are the choices?

If   

\(y=(log_3(m))^{-1}\\ then\\ \frac{1}{y}=log_3(m)\\ 3^{\frac{1}{y}}=m\\ m^y=3\\ \)

I'd be interested in seeing this one done too   

Oblique asymptotes are a mystery to me.