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Mon 20/4/15

FTJ means "For the Juniors"

Thank you .. 

Oh well, better late than never!

That pic must be huge - it took forever to load :/

Here have some of my birthday cake. It is delicious  :))

Chris cooked it and some of the regulars help him :)

Yes Zectico, you can see which one you have answered.  

On the left hand side of your profile page there is a "search users posts" tab.

Also, in  the search box on the normal forum page.  you can type in

by:zectico               and all you posts will be thrown up :)

that was really late!

my legs are k*****g me waiting for your to accept my greeting!

$$(\sqrt{4d})^6=[(4d)^{1/2}]^6\;=(4d)^{6/2}=(4d)^3=4^3d^3=64d^3$$

Thumbs up from me too MG :)

It is not as good as calculus though Brittany :)

$$9\times3\sqrt{3}=27\sqrt3$$

f(x)=2(x+2)(x+1) 

this is a concave up parabola

roots are -2 and -1

axis of symm is    x=(-2-1)/2 = -3/2    this is the x value of the vertex

sub in x=-3/2 to find y value

f(-3/2)=2(1/2)(-1/2)=-1/2

Vertex (-3/2, -1/2)

mcm = most common multiple??

$$\\x^2-1+4x-3\\
x^2+4x-4\\$$

Greetings mistress Rosala :)

$$6^3\; mm^3$$

0.002kg = 2grams=2000mg

500/2000*100    percent

Thanks MG     (on behalf of Chris)

PART A

Consider the nonogon vertices being labeled 1 to 9 in a clockwise direction

Let one of the triangle vertices be at 1

These are the triangles that will include the centre

126

136

137

146

147

148

156

157

158

159

That is 10 triangles that will include the centre

P(centre is included)   =    $$\frac{10}{8C2}=\frac{10}{8C2}=\frac{10}{28}=\frac{5}{14}$$

hmmmmm....how about MathG.....MathGenuis

and i just want to say how r u?

Thanks Chris   

Zetico have you done calculus yet - were you expected to do it a differerent way?

I decided to do it independently of Chris.

I tried a variety of ordinary geometry methods but eventually I also resorted to calculus.

Our answers are the same but maybe our methods are a little different.  

Let the bottom corner is the point (0,0)

Each of these curves is a quadrant.  I am only interested in the quadrant centred at (0,0)

The equation is     

$$\\x^2+y^2=1\\
y^2=1-x^2\\
y=\sqrt{1-x^2}\\\\
When \;\;x=0.5\;\;y=\sqrt{1-0.25}=\sqrt{0.75}\\\\
$The top intersection point is $(0.5,\sqrt{0.75})\\\\
$The right most intersection point is (\sqrt{0.75},0.5)\\\\
0.25* Area=\int_{0.5}^{\sqrt{0.75}}\;(\sqrt{1-x^2}-0.5 )\;\;dx\\\\
Area=4*\int_{0.5}^{\sqrt{0.75}}\;(\sqrt{1-x^2}-0.5 )\;\;dx\\\\
Area\approx 0.315147\;units^2$$

I also used wolfram alpha to do the calculation but maybe i could do it without wolfram|alpha if I had to :/

It's amazing. 

How you solve these things...?!

O.o 

Same here!

I don't understand it. (Well just. A bit)

But every time you commented a new try I'd be staring.

That is some good logic

I just wanted to say hello.

It's 19................... :/

But I wanna be a MathGod... Pwease.

3+7+4=14

3+7=10

10+4=14