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It is incorrect

(a + b) / (1 - a*b)

(5/2  + 1/2) / (1  - (5/2*1/2))

(2.5  +  1/2) / (1  -  (1.25))

( 3 )  /  ( - 0.25) == - 12

We are looking for the smallest n such that n(n - 1)(n - 2)(n - 3) > 100.

Since 4*3*2*1 = 24 and 5*4*3*2 = 120, the smallest n that works is n = 5.

You say " every digit in the number is either a 2 or next to a 2 ". And yet you list 45, 810 !! So, how are they "cozy" given your statement in quotes ?? Make up your mind !.

x/(x - 5) = 4/(x - 4) + 12/(x^2 - 9x + 20)

Note that x^2 - 9x + 20 can be written as (x-5)(x-4), so multiply all terms by   (x-5)(x-4) to get

x(x-4) = 4(x-5) + 12, or x^2 -4x = 4x -20 + 12  or  x^2 - 8x +8 = 0.  

Use the quadratic equation to find the values of x.

Like so:

let r and s be the root of the quadratic x^2 + bx + c where b and c are constants. If (r-1)(s-1)=7, find b+c

the roots of  1x^2+bx+c     are when    

 x^2 + bx + c  = 0

\(x = {-b \pm \sqrt{b^2-4*1*c} \over 2*1}\\ x = {-b \pm \sqrt{b^2-4c} \over 2}\\ r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\ ~\\ r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\~\\ (r-1)(s-1)=7\\ rs-s-r+1=7\\ rs-(s+r)=6\\~\\ rs=\frac{b^2-(b^2-4c)}{4}=c\\~\\ s+r=-b\\~\\ rs-(s+r)=6\\ c--b=6\\ c+b=6 \)

You should check this answer.

LaTex

x = {-b \pm \sqrt{b^2-4*1*c} \over 2*1}\\
x = {-b \pm \sqrt{b^2-4c} \over 2}\\
r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\
~\\
r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\~\\
(r-1)(s-1)=7\\
rs-s-r+1=7\\
rs-(s+r)=6\\~\\
rs=\frac{b^2-(b^2-4c)}{4}=c\\~\\
s+r=-b\\~\\
rs-(s+r)=6\\
c--b=6\\
c+b=6

Converting to polar form, we get 1 + sqrt(3)*i = sqrt(2)(cos 60 + i sin 60).

Then (1 + sqrt(3)*i)^12 = sqrt(2)^12 (cos 60 + i sin 60)^12

= 64 (cos 720 + i sin 720)

= 64

The answer is 64.

Average speed = Total distance / Total time

2  x  [100 +  120  +  180] ==800 m - total distance walked back and forth

[1  +  1  +  1  +  180/100 + 1 + 100/180 ==4  +  1.8  +  5/9 ==6.35555....hours - total time spent walking

800 / 6.35555.... ==125.874 m/h - her average speed

Note: When the distance back and forth is the same, then:

Average speed ==Harmonic mean

Harmonic mean: 3 / (100^-1 + 120^-1 + 180^-1)==125.581 m/h - which is very close to the actual average speed. The reason they are not exactly the same, is because the time back and forth is not exactly the same because of the uphill and downhill walking makes a difference of 0.3555 of an hour due to the 180 m downhill + 100 m uphill.

You can divide into cases:

4-0-0: 3 ways
3-1-0: 3 x 2 = 6 ways
2-2-0: 3 x 2 = 6 ways
1-1-2: 3! = 6 ways

Total is then 3 + 6 + 6 + 6 = 21 ways.

Guest #3

You have the right idea, but how did you get n=23

23 C 4 = 8,855 !! All you need is that: nC4 >=100 days

So, n = 9, because: 9 C 4 = 126 unique combinations - so that 4 DISTINCT people can easily be picked out of 9 for at least 100 days without having the exact same group of 4 people reviewing a book.

Please give me the correct answer

tooo many wrong answers today

[5 + 3 - 1] C [5] = 21 ways

Source: https://en.wikipedia.org/wiki/Twelvefold_way

              [Bottom table, item # 7]

There are 10 ways to put 5 balls in 3 boxes. You can either have all the balls in one box, all but one ball in one box, or one ball in each box. Here are the combinations:

5 balls in one box, 0 in the other two

4 balls in one box, 1 in another box, 0 in the last box

4 balls in one box, 0 in another box, 1 in the last box

3 balls in one box, 2 in another box, 0 in the last box

3 balls in one box, 1 in another box, 1 in the last box

3 balls in one box, 0 in another box, 2 in the last box

2 balls in one box, 2 in another box, 1 in the last box

2 balls in one box, 1 in another box, 2 in the last box

1 ball in one box, 2 in another box, 2 in the last box

0 balls in one box, 3 in another box, 2 in the last box

This gives us a total of 5 + 4 + 4 + 3 + 3 + 3 + 2 + 2 + 1 = 27 ways.

For each day, we have 4 distinct members to pick from. Hence, the number of possible combinations of 4 members is given by n choose 4, or nC4. The smallest positive integer n such that nC4 is greater than or equal to 100 is n = 23. This means that with a club of at least 23 members, it's possible to pick 4 different members every day for 100 days without ever having the exact same group twice.

The sum of the squares of the roots is given by (a^2 + bc)/b^2