To save on typing, let e^(i.theta) = z.

Then, using the binomial expansion,

$$\sin^{5}\theta=\\ \left[(z-1/z)/2\imath\right]^{5}=(z^{5}-5z^{3}+10z-10/z+5/z^{3}-1/z^{5})/32\imath$$

so,

$$16\sin^{5}\theta=\left[(z^{5}-1/z^{5})-5(z^{3}-1/z^{3})+10(z-1/z)\right]/2\imath \\ =\sin5\theta-5sin3\theta + 10sin\theta$$

The diameter of a circle is 1. What is the area of the circle ?

$$\small{\text{Diameter $\,d = 1 $}}$$   

$$\small{\text{Radius $\,r= \frac{d}{2} = \frac{1}{2} $}}$$

$$\small{\text{Area $\,A=\pi\cdot r^2 $}}\\
\small{\text{$A=\pi\cdot \frac{1}{2}\cdot \frac{1}{2}$}}\\
\small{\text{$A=\frac{\pi}{4}$}}\\
\small{\text{$A=\frac{3.14159265359}{4}$}}\\
\small{\text{$A=0.78539816340$}}
$}}$$

The formula for the area of a circle is $${\mathtt{A}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{2}}}$$ (r = radius)

The diameter of a circle is twice the radius. 

So $${\mathtt{A}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{D}}\right)}^{{\mathtt{2}}}$$

$${\mathtt{A}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}^{{\mathtt{2}}}$$

$${\mathtt{A}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{0.5}}}^{{\mathtt{2}}}$$

$${\mathtt{A}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{0.25}}$$

$${\mathtt{A}} = {\mathtt{0.785}}$$ square units (3 d.c.p.)

yes.

For addition, you can imagine it like this. count the number of circles;

o o (2) + o o (2) = o o o o(4)    or for a different sum: o o o o o o (6) + o o o o (4) = o o o o o o o o o o (10)

When multiplying, we are making x amout of groups of y.

so 2 groups of 2. (2)+(2).

It may be easier to understand with larger numbers.

Take 3 times 4, or 3 groups of 4 units.

(3)+(3)+(3)+(3), which as you know is 12.

Inversely (4)+(4)+(4), or 3 groups of 4, also equals 12.

I hope this helped 

$$\sqrt{121}=11$$

$$11\times11=121$$

negative subtract a positive?

Let's try a few concrete examples:

-7 - +2 = -9

-7 - +10 = -17

-7 - +7 = -14

So we can generalize: the answer is negative.

➥ ➥

Melody wrote:

> You made a little mistake at the end Badinage :/

Oops, so I did. I appreciate your correction, Melody. 

What is the 50th digit of pi

$$\small{\text{$\pi=$}}$$

$$\\\small{\text{$
3.$}}\\
\small{\text{$1415926535 $}}\\
\small{\text{$8979323846 $}}\\
\small{\text{$2643383279 $}}\\
\small{\text{$5028841971 $}}\\
\small{\text{$69399375\textcolor[rgb]{1,0,0}{1}0$}}$$

The 50th digit of $$\small{\text{$\pi$}}$$ is 1

(tan x)^2 − tan x − 42 = 0

We substitute:  $$\small{\text{$z=\tan{(x)}$}}$$

So we have: $$\small{\text{$z^2 -z - 42 = 0$}}$$

$$\small{\text{$
z_{1,2}=\frac{ 1\pm \sqrt{1-4\cdot(-42)} }{2\cdot 1}
=\frac{ 1\pm \sqrt{1+168 } }{2}
=\frac{ 1\pm \sqrt{169 } }{2}
=\frac{ 1\pm 13 }{2}
$}}\\\\
\small{\text{
$\begin{array}{l|l}
\hline
\\
z_1=\frac{ 1 + 13 }{2} \quad & \quad z_2=\frac{ 1 - 13 }{2}\\\\
z_1=\frac{ 14 }{2} \quad & \quad z_2=-\frac{ 12 }{2}\\\\
z_1=7 \quad & \quad z_2=-6\\\\
\hline
\\
\tan{(x_1)}=z_1=7 \quad & \quad \tan{(x_2)}=z_2=-6 \\\\
x_1 = \arctan(7) \quad & \quad x_2 = \arctan(-6)\\\\
\boxed{x_1 = 81.8698976458\ensurement{^{\circ}} \pm k\cdot 180\ensurement{^{\circ}}} \quad & \quad \boxed{ x_2 = -80.5376777920\ensurement{^{\circ}}\pm k\cdot 180\ensurement{^{\circ}}}
\end{array}
$}}$$  

k= 0,1,2, ...

10 and 1/3 divided by 2 and 1/4

$$\small{\text{
$
\dfrac{ 10 \frac{1}{3} } { 2\frac{1}{4} }
=\dfrac{ \frac{3\cdot10+1}{3} } { \frac{4\cdot 2+ 1}{4} }
=\dfrac{ \frac{31}{3} } { \frac{9}{4} }
= \frac{31}{3} \cdot \frac{4}{9}
= \frac{31\cdot 4}{3\cdot 9}
= \frac{124}{27}
= 4\frac{16}{27}
$
}}$$

Thanks Nauseated    

Here, have some cake :)

Chris and all the regulars made it for me, it is delicious.  :))

Happy Birthday anniversary, Melody!

Here’s something better than a Scooby Snack. . . .

$$\\v=18x^3-2x+45x^2-5\\\\$$

$${factor}{\left({\mathtt{18}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{45}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}\right)} = \left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right){\mathtt{\,\times\,}}\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)$$

The dimensions of this box are      2x+5,  3x-1 and 3x+1

b) do you mean x=2

$${\mathtt{18}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{45}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{315}}$$

$${\mathtt{315}} = \left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right){\mathtt{\,\times\,}}\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right) \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{551}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{27}}\right)}{{\mathtt{12}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{551}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{27}}\right)}{{\mathtt{12}}}}\\
{\mathtt{x}} = {\mathtt{2}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{9}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.956\: \!115\: \!765\: \!718\: \!968\: \!9}}{i}\right)\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{9}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.956\: \!115\: \!765\: \!715\: \!987\: \!5}}{i}\\
{\mathtt{x}} = {\mathtt{2}}\\
\end{array} \right\}$$

The only real solution is x=2

The dimensions are    9, 5 and 7   cm

x=-b/(2a) = -6/2 = -3

y=(-3)^2+6*-3+8 = 9-18+8 =-1

vertex(-3,-1)

2√75 x √72

$$\\2\sqrt{25*3} \times \sqrt{9*8}\\\\
=2*5\sqrt{3} \times 3\sqrt{8}\\\\
=10\sqrt{3} \times 3\sqrt{4*2}\\\\
=10\sqrt{3} \times 3*2\sqrt{2}\\\\
=10*3*2\sqrt{3} \times \sqrt{2}\\\\
=60\sqrt{3*2} \\\\
=60\sqrt{6} \\\\$$

/////////////////////////////

////////////////////////////////////

2 (square75)* (square72)

2*5(square 3)*6( square 2)

60*(square 6)//

thatz the answer///

There's an Echo in here 

                              an Echo in here

                                                           Echo in here

                                                                        in here

                                                                                                  here

                                                                                                                                         here

$$\\1175=1000(1+r)^2\\\\
1.175=(1+r)^2\\\\
\sqrt{1.175}=1+r\\\\
r=\sqrt{1.175}-1$$

$${\sqrt{{\mathtt{1.175}}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0.083\: \!974\: \!169\: \!433\: \!94}}$$

That is almost 8.4% pa

do you mean

$$18-\sqrt[3]{27}^3 = 18-27 = -9$$

arc tan is the same as inverse tan so...

atan( )

inverse cos may display as acos

inverse cos and arc cos are the same thing :)

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{6}} \Rightarrow {\mathtt{x}} = {\mathtt{2}}$$

$${\mathtt{2}}{v} = {\mathtt{5}}{v}{\mathtt{\,-\,}}{\mathtt{6}} \Rightarrow {\mathtt{v}} = {\mathtt{2}}$$

$${\mathtt{v}} = {\mathtt{6}}{v}$$

$${\mathtt{v}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{v}}{\mathtt{\,\times\,}}{\mathtt{3}} \Rightarrow {\mathtt{v}} = {\mathtt{2}}{\mathtt{\,\times\,}}\left(\underset{{\tiny{\text{Error: Unknown Identifier}}}}{{\mathtt{v}}}\right){\mathtt{\,\times\,}}{\mathtt{3}}$$

You are right, the calculator won't do that.