Melody

Mmm

\((x^2+y^2-z^2)^2-4x^2y^2\\ [(x^2+y^2-z^2)-2xy][(x^2+y^2-z^2)+2xy]\\ [(x^2+y^2-2xy)-z^2][(x^2+y^2+2xy)-z^2]\\ [(x-y)^2-z^2][(x+y)^2-z^2]\\ [((x-y)-z)((x-y)+z)][((x+y)-z)((x+y)+z)]\\ [(x-y-z)(x-y+z)][(x+y-z)(x+y+z)]\\ (x-y-z)(x-y+z)(x+y-z)(x+y+z)\\ \)

Are the fruit baskets identical too?

Not that it makes much difference. You would just have to double the answer if they are not.

How many ways are there to assign each of 6 friends to either the chemistry class or the biology class if one of these six, Manoj, refuses to be in a class without any of his friends?

Yes there are many more than 8 possibilities.

I think I will assign the 5 first and then assign the sixth one later.

number of ways that there can be all 5 in one subject is   1*2 = 2    (because there are 2 subjects)

Manoj would have to be with the others so when he is added there are still only 2 ways

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number of ways that there can be 1 in one subject and 4 in the other in one subject is   5*2 = 10   

number of ways that there can be 2 in one subject and 3 in the other in one subject is   5C2*2 = 10*2=20

So that is 30 ways the 5 can be assigned if they are not all in the same class.

with each of these Manoj can be in either class so when I add him the number of possibilities double.

So that is 60 ways where they are not all in the same class.

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Add the other 2 and I think there are   62 ways that the children can be placed into subjects where Manoj is not left on his own.

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If you have been given an answer then please let me know what it is. If it is different from mine maybe we can work out why.

How many ordered pairs (x,y) satisfy BOTH conditions below?

 Condition I: x=1 or y=0 or y=2

 Condition II: x=0 or x=2 or x=1

Since all points where x=1 are included in both sets, there is an infinite number of points that satisfy both conditions.

ex  (1,-10)(1,-3)!1,0), 1,100) etc etc etc

The actual points that satisfy both contitions is all points where x=1 plus (0,0), (0,2),(2,0),(2,2)

Suppermanaccz,

It is good that you have stated that you do not understand.

BUT

You have not even said which answer you do not understand.  There are 2 different question!

You need to say WHAT question and also try to say where you get lost.

If you do not understand the first line then say so.

If you do understand the first line then tell GYanggg what you do not understand!

This is the question.

Find all numbers r for which the system of congruences

\(\begin{align*} x &\equiv r \pmod{6}, \\ x &\equiv 9 \pmod{20}, \\ x &\equiv 4 \pmod{45} \end{align*}\)

has a solution.

I get r=1

Lets see....  (All the pronumerals I will use are integers)

x/20 = k remainder 9 so    x=20k+9

x/45 = t remainder 4 so     x=45t+4

so

20k+9 = 45t+4

45t-20k = 9-4

45t - 20k = 5

9t - 4k = 1

one obvious solution is t=1 and k=2

9( 1 ) - 4 (2) =1

9(1+4n)-4(2+9n)=1      (because I have added 36n and then taken 36n away again)

so a general solution is   t=1+4n   and  k=2+9n

x=20(2+9n)+9 = 40+180n+9 = 49+180n

or

x=45(1+4n)+4 = 45+180n+4 = 49+180n      Good they are the same.

Now we have x is equivalent to r mod 6

So

(49+180n)/6 = Q remainder r

49+180n = 6Q+r

48+180n+1=6Q+r

6(8+30n)+1=6Q+r

r = 1

I think...

\(f^{-1}\left(\frac{13+2}{5}\right) = f^{-1}(3) =1\)

Yes you are right. That was very remiss of me. 

I'll hide my erroneous answer for now. While I think on it.

Thanks for pointing that out guest.

Please do not delete your questions.

Somebody might be trying to answer it already!

You can say you do not need an answer any more. That is fine.

But please do not delete the original question.