Melody

This question makes no sense to me.

Have you tried plotting A and C it on a grid and then trying to work it out from that?

2n^4−15/6n^4+17

\(\displaystyle \lim_{n\rightarrow \infty}\;\;\frac{2n^4−15}{6n^4+17}\\ \qquad \text{At an educated guess I would say 2/6 = 1/3}\\ \qquad \text{But I should be more mathematical about it}\\~\\ \text{divide through by n^4 top and bottom}\\~\\ \displaystyle \lim_{n\rightarrow \infty}\;\;\frac{2−\frac{15}{n^4}}{6+\frac{17}{n^4}}\\~\\ =\frac{2}{6}\\ =\frac{1}{3} \)

Please proof read your questons and get rid of meaningles dollar signs

ABCDEFG  that is 7 friends

Cut into 9 peices that is 8 bars.     

7+8=15

15C8=6435   ways to distribute.

If D gets 2 peices than you can take him and the 2 choc he is getting out  

then you are left with

ABCEFG   6 friends    divided into 7 parts that is 6 bars

6+6=12

12C6 = 924 possible ways

so that is    924/6435 =         28 / 195

I used the Stars and bars method to solve it. 

I could have make a logic error. You need to decide for yourelf whether to accept this answer.

Nice logic Imcool  

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

It says the deck is shown below.

I do not see it.

More words needed.

I thinhk you are totally confused and have seriously stuffed up this question.

I will discuss it though.

sin(arcsin(5/4))

arc sin cannot be greater than 1 so I will change the expression.......

sin(arcsin(4/5)) there is only one answer,    4/5

Lets look at your second part and see if that makes more sense

sin(pi/6) = 1/2  (not sqrt3/2)        and cos(pi/6) =sqrt3/2  (not 1/2)

I just draw the right angled triangle and also when necessary think about the sign    

You don't have a terminal point for the sin of an angle

You have a terminal point on the unit circle for an angle.

the terminal point for   pi/6 is   ((sqrt3)/2 , 0.5)