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thanks 

thx xd

Assuming the front and rear wheels are the same size (the DO make bikes with different wheel sizes)

circumference of wheel = pi x d = pi x .6m = 1.885 m

360 rotations x 1.885m =  678.58 m

Same lol I hate math lol 

Press the "2nd" button and then atan

I don't know what you said, but I am sure I can help with a better sentence.

ыоу шилл дие сомедаы фром а чоррибле цомпутер вирус

Things like this only take 20 seconds to learn, you got 20 seconds !!!!

They provide a visual understanding of where you need place the valance electrons so that the bonding atoms are satisfied. This is called the octet rule. It can be broken because of expections in some bonds between atoms or it can not meet the rule, and simply have less than 8 valence electrons. There is so much more to explain but im tired. The answer to your question is that they determine the placement of valence electrons, determine molecular/geomtric shapes and can tell you if have enough plutonium to make a gamma laser gun.

I know this, one dad is the grandpa, the son of him is the dad, the dad is the second dad and the son is the second son.

Yee?

If you can find the forum, then you can find the button

oh wait no ned lol

diu

They're 2 sons and one dad, the reason there's 2 "dads" is because each son has got their own dad? So they're only 3 people? :D 

It's 3.4 :) 

I'm confused

You have written on the pic  BC=5

but in the writing it says  |BC|=3,4       do you mean 3.4?

Anyway, what is BC?

http://imgur.com/a/agpUw
This is where im at.. Looks good? 

liliiilililililili

43 years old? 
400/12=33 years 
Ten years before you so I'll add 10 years :D 
=43 

" Figure 1 is showing a model of a tablet. The cover is foldable. This will create a triangle |BC|=3,4
The tablet is connected in the cover 0.5cm down on one side shown on the figure. The tablets glas-surface is 13.5 cm from top to bottom. 

a) Prove the angles in triangle ABC 

B) Prove the angle v, as the tablet connects with. 

I started by breaking 6x into 6(x^1/2)² and thinking of x^1/2 as k, so that h is then 6k², which leaves g as sin²h.

It is not a unique breakdown! For example we could also have k = (6x)^1/3, h = x^3, g = sin²x

or an infinite number of other possibilities!

.

Find the value of x that makes each statement true

sin (x/3 + 10) = cos x

1. x₁ = ?

\(\begin{array}{|rcll|} \hline \sin(\frac{x}{3}+10^{\circ}) &=& \cos(x) \quad & | \quad \cos(x) = \sin(90^{\circ}-x) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \sin(90^{\circ}-x) \\ \frac{x}{3}+10^{\circ} &=& 90^{\circ}-x \\ x+\frac{x}{3}+10^{\circ} &=& 90^{\circ} \\ x+\frac{x}{3} &=& 80^{\circ} \\ x\cdot (1+\frac{1}{3}) &=& 80^{\circ} \\ x\cdot (\frac{4}{3}) &=& 80^{\circ} \\ x &=& 80^{\circ}\cdot \frac{3}{4} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{60^{\circ}} \\ \hline \end{array}\)

2. x₂ = ?

\(\begin{array}{|rcll|} \hline \sin(\frac{x}{3}+10^{\circ}) &=& \cos(x) \quad & | \quad \cos(x) = \cos(-x) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \cos(-x) \quad & | \quad \cos(-x) = \sin(90^{\circ}-(-x)) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \sin(90^{\circ}-(-x)) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \sin(90^{\circ}+x) \\ \frac{x}{3}+10^{\circ} &=& 90^{\circ}+x \\ x-\frac{x}{3} &=& -80^{\circ} \\ x\cdot (1-\frac{1}{3}) &=& -80^{\circ} \\ x\cdot (\frac{2}{3}) &=& -80^{\circ} \\ x &=& -80^{\circ}\cdot \frac{3}{2} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-120^{\circ} } \\ \hline \end{array}\)

How'd you attain these posibilities?? Thanks heaps by the way.