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??.

There's a sort of triangular region (with two curvy lines) enclosed by the two circles and the tangent.

The third circle lies within that.

It can also be computed this way:

[6 C 2]  *  [4 C 3]  *  4 (the remaining die can have any of 4 faces: 3, 4, 5, 6] ==240 permutations.

Hello Guest #2!

I have isolated y in all lines: I divide the line by the y-coefficient.

The x-coefficient is the slope of the line.

If two lines have the same slopes,

they are a good pair. Parallel course.

If two lines have the same, slope and negative reciprocal of the slope,

then they are a good pair. Perpendicular course

  !

9.9/9.9/-193 = ≈ 0.005  

Between 300 and 600 - 150 odd integers.

\( \displaystyle\sum_{n=1}^{150} \mbox{ } ( 2n + 299 ) = 67500\), as BB said.

For positive integers, x = 223 and y = + 3 / y = - 3 and x = 2007 and y = + 1 / - 1.

I believe I already have seen this question, idk.

Thanks Alan :)

Yes I just thought of somethin i missed out on.   I get 240 too.

I will edit my first post and talk about it there.

I get 17 points of intersection:

The computer is listing 240 such permutations, but don't know how it arrived at it!

Hi Guest,

Can you please explain how your got this answer.

I am noty seeing how that is possible.

I mean I do not see how a third circle can be externally tangential to both the other 2 as well as tangential to line l.

1^2 + 2^2 + 3^2 + ... +60^2    is divided by 15.

Im going to split this up into

\((1^2+2^2+ \dots +14^2)+(15^2\dots+29^2)+(30^2\dots+44^2)+(45^2\dots+56^2)+60^2 \mod(15)\\ =4(1^2+2^2+ \dots +14^2)+0 \mod(15)\\ =4(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2+14^2) \mod(15)\\ =4(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2+(-7)^2+(-6)^2+(-5)^2+(-4)^2+(-3)^2+(-2)^2+(-1)^2) \mod(15)\\ =8(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2) \mod(15)\\ =8(1+4 -6+1-5+6+4) \mod(15)\\ =8( 5) \mod(15)\\ =40 \mod(15)\\ =30+10 \mod(15)\\ =10 \mod(15)\)

You need to check for careless errors

LaTex

(1^2+2^2+ \dots +14^2)+(15^2\dots+29^2)+(30^2\dots+44^2)+(45^2\dots+56^2)+60^2 \mod(15)\\
=4(1^2+2^2+ \dots +14^2)+0 \mod(15)\\
=4(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2+14^2) \mod(15)\\
=4(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2+(-7)^2+(-6)^2+(-5)^2+(-4)^2+(-3)^2+(-2)^2+(-1)^2) \mod(15)\\
=8(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2) \mod(15)\\
=8(1+4 -6+1-5+6+4) \mod(15)\\
=8( 5) \mod(15)\\
=40 \mod(15)\\
=30+10 \mod(15)\\
=10 \mod(15)

Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly three of the dice show a 2? Express your answer as a common fraction.

1,1,2,2,2,*  any order

5C2*6 = 60    

  Edit:  I forgot to take into account that the extra number can be 3,4,5,6   so I have to myultiply 60 by 4 to get 240.

60/6^6 *4= 10*4/6^5 = 10*4/7776 = 5*4/3888 = 5/972

That is what I get.

-------------------------------------------

I will try and explain how I got 240 permutations

It is easier to think of all the diec being different colours.  So, for instance, a red 2 is different from a green 2.

The numbers that you need are  1,1,2,2,2,*   where star can be 3,4,5, or 6

First how many ways can I order 1,1,2,2,2.   

There are 5 positions, how many ways are there that I can choose 2 of them?

That would be  5C2 = 10 ways.   

Now I need to choose a number for the wild card.  There are 4 to choose from.

Now Now I need to choose a position for the wild number.  There are 6 to choose from.   

so there are    10*4*6 = 240 favourable permutations.

And the sample space is 6^6 = 46656   

So the prob is    240/46656 = 5/972

Thanks guest,

Here is a neat way to do it.

\(3*3^2*3^3*...*3^{43}*3^{44}*3^{45}\mod(11)\\ \text{Lets look for a pattern}\\ 3^1=3\\ 3^2=9\mod(11)=-2\mod(11)\\ 3^3=-2*3\mod(11)=-6\mod(11)=5\mod(11)\\ 3^4=5*3\mod(11)=4\mod(11)\\ 3^5=4*3\mod(11)=1\mod(11)\\ 3^6=1*3\mod(11)=3\mod(11)\\ \text{The pattern has started to repeat.} \)

there is 45 terms, 5*9 =45 so I will have to rasie the procuct to a power of 9

so I have  

(3*-2*5*4*1) ^9    mod(11)

=(-120)^9   mod(11)

=(-121+1)^9  mod(11)

=(1)^9  mod(11)

=1   mod(11)

LaTex

3*3^2*3^3*...*3^{43}*3^{44}*3^{45}\mod(11)\\
\text{Lets look for a pattern}\\
3^1=3\\
3^2=9\mod(11)=-2\mod(11)\\
3^3=-2*3\mod(11)=-6\mod(11)=5\mod(11)\\
3^4=5*3\mod(11)=4\mod(11)\\
3^5=4*3\mod(11)=1\mod(11)\\
3^6=1*3\mod(11)=3\mod(11)\\
\text{The pattern has started to repeat.}
 

Given that is a multiple of 23478

Something is missing in your question.

substitute eacgh point into the equation and then you wil have 2 equations with 2 unknowns b and c.  Solve simultaneously.

f(x^3)  will have a degree of 12 and  g(x^2) will have a degree of  10 

You can finish it

Is this a part of a bigger question?  It looks like it is.

9.9/9.9/-193 =1/-193     =  -1/193

 i don't now

 How many pairs are good?

Hello Guest!

a  y = 3x + 5           m = 3

b  2y = 4x + 5         m = 2

c  3y = x - 2            m = 1/3

d  2y = 3x - 2          m = 3/2

e  4y = x - 5            m = 1/4

No pairs are good.

  !

productfor(n, 1, 45, 3^n)  mod 11==1 - the remainder.

What is the longest possible integer length of the third side of the triangle?

Hello Guest!

11 + 19 = 30

30 - 1 = 29

The longest possible integer length of the third side of the triangle is 29.

  !

sumfor(n,1, 60, n^2)  mod 15 ==10 - the remainder