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u have got to b kidding me...

Βόμβες Γωγούση. Nuμβεrs nevεr ενd.

Ξανθή δεχόσασταν ηγήτωρ θα όλο παύω ΓΓΚ ξοδεύουν Γωγώς Υόρκη ξεκίνησε.

9+10=dumb question.

The opposite sides of a rectangle have equal length. Since the width is 12cm, that means the two width sides added together give you a length of 24 cm.

Since the total length(perimeter) is 50cm, we subtract 24cm from 50cm to get 26 cm remaining for the last two sides. Remember these sides are equal length, so we can divide 26 by 2 to get the length of one side, that is 13cm.

Now area of a rectangle is width x height, so 12cm x 13cm = 156 cm squared.

seven 

$${\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}} = {\mathtt{19}}$$

I've given you some points anyway!

.

$$\\\\$Let's hope Steve has high stamina\\\\
\\So 40m every two minutes (2\times30=60 minutes) so\;40m\times30= a\;hour = 1200\;meters\\
\\$1200 meters\times\;24\;for\;a\;day\;=\;28000\;meters$\\
\\$28000\times\;365\;for\;a\;year\;=\;10512000\;meters$\\
\\10512000m\;is\;a\;year,\;a\;century\;is\;100\;years\;\\
\\10512000m\times\;100=1051200000m\\
\\\\Convert\;1051200000m\;into\;centermeters\;=\;105120000000 cm\\\\
Your\;answer\;is\;105120000000 cm$$

You beat me to it!

x - y = 8                  (1)

x + y = 11               (2)

Add (1) and (2)

2x = 19

x = 19/2 = 9.5

Put this back into (2)

9.5 + y = 11

y = 11 - 9.5

y = 1.5

.

20 metres in 1 minute = 20*60 metres in one hour = 20*60*24 metres in one day = 20*60*24*365 metres in one year = 20*60*24*365*100 metres in one century = 20*60*24*365*100*100 centimetres in one century

$${\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{365}}{\mathtt{\,\times\,}}{\mathtt{100}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{105\,120\,000\,000}}$$

.

x can be 10 and y can be 2: 10 - 2 = 8 

x can be 6 and y can be 5: 6 + 5 = 11 

We could mix it up also and do the question x - y = 8 + x + y = 11. 

Which it would go like this: 

x can be 10 and y can be 2: 10 - 2 = 8

Now 8 is three away from 11. 

x can be 2 and y can be 1: 2 + 1 = 3 

8 + 2 + 1 = 11 

Now lets finish our problem by simplifying it all: 

8 + 3 = 11 

I Hope That Helped :) 

@Alan, don't you get the same answer?

40m/2min

20m/1min

2*30=60

40*30=1200

1*60=60

20*60=1200

EDIT: It appears i have confused myself on the first bit i meant for an hour not minute I'll fix that.

Thanks Alan.

There are many ways to solve this answer, here's one way I would solve the problem:

-6 * 10 + 6 * 3 = -42

I hope that helped :) 

Exactly.

Here's your cookie :

The square root of a positive number n, is a number a which square is equal to n.

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{n}}$$

Let f = 1 + 2x + 3x^2 + 4x^5 + ...

First consider y = x + x^2 + x^3 + x^4 + x^5 + ...

y = x(1 + x + x^2 + x^3 + x^4 + ...) or y = x(1 + y)  (as long as y converges, which it does if |x| < 1)

so y(1 - x) = x

y = x/(1 - x)

Now from the original series expression for y we have dy/dx = 1 + 2x + 3x^2 + 4x^3 + ... or in other words

f = dy/dx.  Since y also equals x/(1 - x) we have dy/dx = 1/(1-x) + x/(1-x)^2 or dy/dx = (1-x+x)/(1-x)^2 or dy/dx = 1/(1-x)^2, so, f = 1/(1-x)^2 or

1 + 2x + 3x^2 + 4x^5 + ... = 1/(1 - x)^2 

.

what is -5/8?

$$-5/8 = -0.625$$

Hi MG

Maybe I just misunderstood what you meant - you have to be very precise in mathematics :)

Colis means package in French.

A package can be any size :)

V_cone =(1/3)*pi * r^2 *h  =   (1/3)*pi*(4)^2 * 3 =  about 50.27cm^3

This has probably been done to death but too bad.

The area of the triangle is

 $$\\3\times \frac{1}{2}\times r^2*sin \frac{2\pi}{3}\\\\
=\frac{3r^2}{2}sin\frac{\pi}{3}\\\\
=\frac{3r^2}{2}\times\frac{\sqrt{3}}{2}\\\\
=\frac{3\sqrt{3}r^2}{4}\\\\$$

 $$\\Area\\
=4\times\frac{1}{2}\times r^2sin\theta\\\\
=2r^2sin\theta\\\\
so\\\\
2r^2sin\theta=\frac{3\sqrt3*r^2}{4}\\\\
sin\theta=\frac{3\sqrt3}{8}\\\\
If\;opp=3\sqrt3\;\;and\;\;hyp=8\;\;then\;\;adj=\sqrt{64-27}=\sqrt{37}\\\\
so \;\;cos\theta=\frac{\sqrt{37}}{8}\\\\
-----\\
x^2=2r^2-2r^2cos\theta\\
x^2=2r^2-2r^2\times\frac{\sqrt{37}}{8}\\
x^2=\frac{8r^2}{4}-\frac{r^2\sqrt{37}}{4}\\
x^2=\frac{r^2}{4}\left(8-\sqrt{37}\right)\\\\
x=\frac{r\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies$$

$$\\xy=\frac{3\sqrt3r^2}{4}\\\\
y=\frac{3\sqrt3r^2}{4}\div x\\\\
y=\frac{3\sqrt3r^2}{4}\times \frac{1}{x} \\\\
y=\frac{3\sqrt3r^2}{4}\times \frac{2}{r\sqrt{8-\sqrt{37}}} \\\\
y=\frac{3\sqrt3r}{2}\times \frac{1}{\sqrt{8-\sqrt{37}}} \\\\
y= \frac{3r\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\$$

Is this answer the same as the others?