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6d6d6

S=65; N=10; D=12;sumfor(k, 0, ((S - N)/D), ((-1)^k * (N nCr k) * (S - D*k - 1) nCr (N - 1)) / D^N

139259657 / 12^10 = 3.6%

E Q U A L S

5 nPr 4 = 360 permutations.
Each 4-letter  permutation begins with one of the 6 letters this many times: 360 / 6=60 times, and that many times with the letter L, and ends with the remaining 5 letters this many times:60 / 5 = 12 permutations ending in the letter Q. 

Therefore, there are: 360 - 60  - 60=240 permutations with no L at the beginning and no Q at the end.

Are the rest of them correct? Thank you!

Idk it can be really annoying just ingore them and maybe they will stop

Thank you so much for all your help CPhill! I greatly appreciate your help and time! 😊

y = k * x * z

20  =  (k) * (4)  * ( 3)

20 =  12k

k =  20/12 =  5/3

y =  (5/3)  (2) ( 3)

y =  10

16 / N  =  .20

N  = 16/.20  =   80  questions =  100%

4.

tan  [ arcsin (-1/2)  ]

________________

cos [ arctan (1)] 

Let  arcsin  (-1/2)  = A      and  arctan (1)  =  B  .....so we  have

tan (A)

______

cos (B)

If arcsin  (-1/2)  = A....rhis  angle  is a Q4  angle....x will  be  positive here

Then   x =  sqrt  [ 4 - 1 ] =   sqrt (3)

So tan (A) =   -1/ sqrt (3)

If  arctan (1)   =  B.......this is a Q1  angle

r=  sqrt  (1 + 1 )   =  sqrt (2)

So  cos B=   1/ sqrt (2)

So we have

(-1/sqrt (3) )  

__________   =

( 1/sqrt (2))

-sqrt (2)  /sqrt (3)   =

- sqrt (6)  / 3

Okay, so the two asymptotes are -(3+√3)/3 & -(3-√3)/3

So, add both asymptotes.

\(-\frac{3+\sqrt{3}}{3}+\frac{3-\sqrt{3}}{3}\)\(=\frac{-\left(3+\sqrt{3}\right)+3-\sqrt{3}}{3}\)

Expand:

\(=\frac{-2\sqrt{3}}{3}\) \(=-\frac{2\sqrt{3}}{3}\)

The answer is \(=-\frac{2\sqrt{3}}{3} \)

Here is the graph:

3.   a  = 8      c  =11   C  = 33

sin A  /a = sin C/ c

sin A / 8   = sin 33 / 11

sin A  =  (8/11) sin 33

arcsin  [ (8/11) * sin 33 ] = A  = 23.3° 

B =  180  -33  -  23.3   =  123.7°

b/ sin B =  c /sin C

b /  sin 123.7 = 11/sin 33

b =  11 sin 123.7 / sin 33  =   16.8

1) Product  of a complex  number  and a  real  number  is just a scalar of the complex number

I do not understand the algebraic answers above. I mean i do not see the logic that Spring has used

\(\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}} \qquad=\;\;?\\ \text{First, since the square root is in the question, the answer must be positive.}\\ let\\ y=\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}} \\ then\\ y=\sqrt{12-y}\\ y^2=12-y\\ y^2+y-12=0\\ (y-3)(y+4)=0\\ y=3\;\;or \;\;-4\\ \text{y cannot be negative}\\ y=3 \)

Thank you so much for all your help CPhill! 

What do these ones end with

10!

11!

12!

and all the way to the end?

So what do you need to consider.?

There are 20 cars in my building's parking lot. All of the cars are red or white. Also, all the cars are either 2-door or 4-door. 15 of them are red, 18 of them are 4-door, and 3 of them are 2-door and white. How many of the cars are 4-door and red?

Unless I am missing something the question does not make sense.

18 are  four-door which means that 2 are two-door.

So 3 cannot be 2-door and white.

( 98 * k)^(1/3)  =

( 7^2 * 2 * k)^(1/3)

We need k  to be   7 * 2^2  =  28

Check  :    (98 * 28)^(1/3)  =  (2744)^(1/3)   =   14

What is the smallest positive integer such that (98*k)^(1/3) is an integer

98 = 2 * 7^2

k =2^2  x  7 = 28

The hypotenuse  is  EG  =   8

FG  = 6

So....  EF      =    sqrt  (8^2 - 6^2)  = sqrt  (64  - 36) =  sqrt  ( 28)  =  2sqrt (7)

sin E  =   FG / EG =   6/8   =  3/4

cot G  =   FG / EF  =    6/ (2sqrt (7)  =   3sqrt (7)  / 7

sec G =   EG / FG  =  8 / 6  =  4/3

E

2sqrt (7)           8

F             6                         G

GOOD JOB, Spring   !!!!

Get lost Kavvy,

Go do your own homework.

You have had quite enough homework done for you for one day.

y=  4 ( x - - 2)  + 3

y= 4 ( x + 2) + 3

y=  4x + 8 + 3

y= 4x + 11

Yo find "a"  just set the equations equal

5x^2  + 6  =  11x       rearramge as

5x^2   - 11x  +  6  =   0      actor

(5x   - 6) ( x - 1)  = 0

The  smallest x = "a" will occur when

x - 1  = 0

x  = 1  =  "a"

See the graph here :    https://www.desmos.com/calculator/ryznv6php2