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Number of total cards in the deck = 52. Number of face cards in the deck = 12. Now, we need to find the number of ways such that the first card is always a face card. Now, first card is face card so number of cards left to be arranged = 52 - 1 = 51. We take the factorial of that so we can find out how many diffrent ways we can re-arange them, but them we have to multiply it by 12, beacus of the 12 face cards we can hoose from.

Im not 100 percent sure, but i think the answere should be 51!*12, pls tell whether it right or not.

The question asks that the two R's are adjacent and the two O's are adjacent. RR and OO don't have to be adjacent to each other. It could be xooxxrrx

I think you intepreted it as "The two R's and the two O's are adjacent". Then you would be correct.

j = 13 + b         and in 8 years   j +8 = 2(b+8)   (or  (j+8)/2 -8=b)

j = 13 +  (j+8)/2 - 8

1/2 j = 13 +4-8

j = 18        b = 5                  in 8 years    j = 26   b = 13

So we can express jacks curent age as x, and bao's current age as x-13. It says that 'In 8 years, he will be twice as old as Bao" and we can express that as x+8=2(x-13+8) which we can simplify to x+8=2x-10 from here x=18, so jack is curently 18.

Thanx, TLB     but I think you calc'd   oo  and rr being ANYWHERE in the sequence....not the probability of them being NEXT to each other??

The total number of ways to arrange TRAPDOOR is \(\frac{8!}{2!*2!}\). 8 choices for the first letter, 7 choices for the second etc. We divide by 2!*2! because there are repeating cases with the two R's and the two O's. 

Now onto the cases that the two R's are adjacent and the two O's are adjacent. 

Think of the two R's as a "block" or a "bundle". Think of the two O's the same way.

So now we find the number of ways to arrange 6 letters (The 4 distinct letters T, A, P, D & the 2 "bundles" of R and O). This is just 6!. We don't have to divide by anything because they are all distinct.

\(\frac{6!}{\frac{8!}{2!*2!}}\)= \(\frac{1}{14}\)

(k + 4)x^2 + (k + 1)x = -1

(k + 4)x^2 + (k + 1)x  +1= 0    must have discriminant =0

b^2-4ac = 0

(k+1)^2 - 4 ((k+4)(1)) = 0

k^2 + 2k+1 - 4k-16 =0

k^2-2k-15 = 0

(k-5)(k+3) = 0    k = 5 or -3

There are 5 ways for the boys to be seated in a block next to each other  AND there are 24 ways to arrange them inside of the blocks

24 x 5 = 120 

out of a possible   8 P r = 40320 possible seating arrangements

120/40320 = 1/336        (not sure of this answer)

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See TLB answer below !

Thanx , TLB for the clarification below !!!!

Correct , CTG....   when you ar inactive for a time period (I do not know what that period IS)  your account gets taken off of the board....If CPhil signs in again, it will reappear where it last was.....   Maybe he is on vacation or taking a break for a while.....   

No, it is not. 5/2 will give you James's age:18 / 7 = 2.5714.....etc., which doesn't make sense!

I am going to assume the    w + l + h = 2.04

    or is it the sum of ALL of the edges (all 12 of them)?

Here is a graph to look at to help make things clearer:

https://www.desmos.com/calculator/c5w2klg0xw

when you shift it  (by subtracting 8)    the value at  x = 12  .........becomes the value that WAS at x = 4

   12-8 = 4

For the didtance to be 4 times the distance from B as A   the segment must be divided into (4+1) = 5 segments

   so we want the point that is 1/5 of the way from A to B

x    from A   to B   is     2   to   - 6   =   -8 units       1/5 * -8    +  2  =    0.4

y    from A   to B   is     7 to    -3    =   -10 units      1/5 * -10  +  7    = 5     

x^2 + 5 x + 8 = 0         has roots (solutions)     - 5/2   +- sqrt (7)/2  i         (obtained by Quadratic Formula) 

                                                            so     a   + b^2    =   -5/2 + (7/4)  =  - 10/4  + 7/4  = - 3/4

when you subtract from the object 'x'   you will get the value that is usually at the smaller 'x'       so if you parabola has a value     of   10   at x= 4

    when you shift it  (by subtracting 8)    the value at  x = 12      will be the value that WAS at x = 4         so the parabola gets shifted to the right 

  I hope I explained it so you can understand......

(a) We use the vertical asymptote, x=3, to help us solve for c. We let x be 3. We need to find c so that the denominator, x+c, is 0, and the fraction is undefined.

\(x+c=0\\3+c=0\\c=-3.\) 

We let f(x) equal to y and plug in the value of c. We then have

\(y=\frac{ax+b}{x-3}\\y(x-3)=\frac{ax+b}{x-3}(x-3)\\y(x-3)=ax+b\\xy-3y=ax+b\\xy-ax=b+3y\\x(y-a)=b+3y\\x=\frac{3y+b}{-a+y}.\) 

Now we use the horizontal asymptote y=-4 to help us solve for a. Using similar logic as we solved for c to solve for a, we get

 \(-a+y=0\\-a-4=0\\a=-4.\)

We then plug the values of a and c into \(f(x)=\frac{ax+b}{x-3}\) to get \(f(x)=\frac{-4x+b}{x-3}\). We plug in the point (1, 0) into the equation to solve for b. \(f(x)=\frac{-4x+b}{x-3}\\0=\frac{-4(1)+b}{1-3}\\0=\frac{-4+b}{-2}\\0=-4+b\\b=4.\)

Therefore \(f(x) = \frac{ax+b}{x+c}\) is \(\boxed{f(x) = \frac{-4x+4}{x-3}.}\) 

Use similar logic to solve for part (b).

(b) - See Alan's answer here:  https://web2.0calc.com/members/alan/?answerpage=751

(a) - sum_(n=1)^26 (-1)^(n + 1) (105 - 4 n)^2 = 5304

Using difference of squares is the most efficient way. I don't know how to do part (b), but I can certainly explain part (a).

Yes, thank you.