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Okay thanks!

1. P(7)  = 0    [there isn't a "7" on a single die ]

2. P(Red) = [Number of red marbles] / [ Total number of marbles ]  =   1 / 7

Okay Thanks!!!!

sin^2θ/1-cosθ=1+cosθ   .......working with the left hand side, we have

(1 - cos^2θ) / (1 - cosθ)  =

[( 1 + cosθ) (1 - cosθ) ] / ( 1 - cosθ)  =

1 + cosθ

y = rs + t    .......

y = (3)(4) + 1  =   12 + 1  = 13

thanks so much!!!!!

We have 18 marbles of which 6 are green.......the probability of drawing a green one is 6/18  = 1/3

the possibility of drawing a green marble is 6/12 = 1/2

N^2 - N - 90  = 0     factor

(N - 10) (N + 9)  = 0        setting each factor to 0, we have that N = 10 or N = -9

Nice presentation Alan       

I've often looked are ways of presenting polynomial division  :/

Here it is with synthetic division....

1       1   -1    1   -1   1   -4

               1    0    1   0    1

       --------------------------

        1     0    1    0   1   -3

The resulting polynomial   is   x^4 + x^2 + 1    with a remainder of  -3/(x - 1)

Well this is a simple one and in answering your question i hope you will answer mine back so my farther does not do it again. so it is pi X radius sqaured (3.14 times the radius sqaured)

Thanks Chris

$$\begin{array}{rlllll}
&3&^19&&\\
1&\not{4}&\not{A}&^16&_{12}\quad&\mbox{Take 12 from the 2nd column 12+6-9=9 etc}\\
&5&B&9&_{12}\quad&\mbox{12+9-11=10=A}\\
&&&&\quad&\mbox{12+3-5=10=A}\\\\\hline
&A&A&9&_{12}\\
\end{array}$$

Do you understand Mellie ?     I have done this like an ordinary subtraction  

Great explanation DarkBlaze347  

pi times radius sqared

$$\\\frac{(-6)^2}{(-3)}+\qquad\frac{-8}{(-1/2)}\qquad-5 \\\\
=\frac{36}{(-3)}+\qquad-8\div\frac{-1}{2}\qquad-5 \\\\
=\frac{36}{(-3)}+\qquad 8\div\frac{1}{2}\qquad-5 \qquad $A neg divided by a neg = a pos$\\\\
=-12+\qquad 8\times\frac{2}{1}\qquad-5 \\\\
=-12+\qquad 16\qquad-5 \\\\
=-17+16\\\\
=-1$$

.

Usual training distance = 5*2*(260 + 480) m 

Diagonal distance = √(260² + 480²) m

Heavy rain training distance = 5*(260 + 480 + √(260² + 480²)) m

Difference between these:

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{260}}{\mathtt{\,\small\textbf+\,}}{\mathtt{480}}\right){\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}\left({\mathtt{260}}{\mathtt{\,\small\textbf+\,}}{\mathtt{480}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{{\mathtt{260}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{480}}}^{{\mathtt{2}}}}}\right) = {\mathtt{970.531\: \!187\: \!208\: \!763\: \!874\: \!4}}$$

difference = 971 m to nearest metre

.

Area     A = a*b = 4 cm * 11 cm = 44 cm²

$${\frac{{\mathtt{3}}}{{\mathtt{7}}}} = {\mathtt{0.428\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

= $$0.\overline{428571}$$

$${\mathtt{A}} = {{\mathtt{r}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}$$

 Look here:

Thank Alan 

0.8571428571....