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Go online to this link and have all your three angles solved there, using their special calculator:

http://www.calculatorsoup.com/calculators/geometry-plane/triangle-theorems.php

Radius=Diameter / 2

Radius=8 / 2 =4

V=4/3.Pi.r^3

V=4/3.3.141592.4^3

V=268.08 Cubic units.

how do i write this in a more compact way? 0.00000000000000000000000001169

=1.169 x 10^-26

Thanks genos! I got the same orbital size you got, but my period was way off

2/3-1/2+(-5/7) = ?

2/3-1/2 (find commum denominators) -> 4/6-3/6 = 1/6 + (-5/7) (find the commum denominators again) -> 7/42 - 30/42 = -23/42

:)

using the law: c^2=a^2+b^2-2ab*cos(angle)

so, 11^2=6^2+7^2-(2*6*7)*cos x

121=36+49-84*cos x

120=cos x

cos x = -1/2

1,169*10^-29 :)

here, i used paint, sorry if you couldn't understand it ; EDIT: forgot to put the unit ^3

by solving it

The point-slope forms includes a known point (x₁, y₁) and a known slope m.

     y - y₁  =  m(x - x₁)

To me, it seems that one full wave occurs between days 2 and 12:  period = 10 days.

The maximum value seems to be 3.75 and the minimum -3.75:  size = 3.75

(x + 4) (x - 1)  =

x^2 + 4x -1x - 4  =

x^2 + 3x  - 4

I suggest you make sure the web2.0 calc is set to degrees and then type in tan(260) and then press equals.

I don't think that the brackets are needed either.

Reading the question more carefully, I see I have misinterpreted the angles.  The angle between incident and reflection is 45 degrees not 90 degrees, so we need to use the cosine rule:

a^2 = 100^2 + 120^2 - 2*100*120*cos(45°)   where a is the distance  between pocket and initial strike.

so a = sqrt(100^2 + 120^2 - 2*100*120*cos(45°))

a = 86 cm (to the nearest cm).

 x^2 + y^2 + 6y = 7      complete the square on y

x^2  + y^2 + 6y + 9   = 7 + 9        simplify

x^2 + ( y + 3)^2  = 16       which we can write as

(x - 0)^2 + (y + 3)^2   = 16

So.......the center is   (0, -3)    and the radius  is 4

[2/3-1/6]/[2/5-1/3]    =      get  a common denominator of 6 on top and 15 on bottom

[ 4/6 - 1/6]  /   [ 6/15 - 5/15]  =     simpiify

[3/6] / [ 1/15]    =    

[1/2] /  [1/15]   =      invert tthe denominator and multiply

1/2 *  15/1   =

15/2   =

7 + 1/2

[2/3-1/6]/[2/5-1/3]

[1/2] / [1/15] =

1/2 x 15 =15/2 =7 1/2

And, since H = 1/2gT^2 we can say the final position is H - H/9 = 8H/9 meters above the ground.

.

Since the height of the tower is given as H in this instance, the final position here is given by:

H - 1/2g(T/3)^2 meters above the ground (where g = 9.81 m/s^2)

With the angles of incidence and reflection both 45degrees, the ball travels along two sides of a right angled triangle.  The distance between pocket and initial strike is therefore that of the hypotenuse of the triangle.  

Distance = sqrt(100^2 + 120^2) cm.

Use this formula for your problem. H=1/2 g t^2, where H=Height, g=9.81 m/s^2 gravitational constant, t=time in seconds.

Sometimes "d" is used instead of "H", which stands for "distance".

Just enter 10 first. Then click on "x^y" key. Then enter 5 and change it to negative using +- key. Then click on = key to get your answer.

Look at the bottom left hand corner.It says "deg", "rad".

Solve for n:
n^2-9 n-10 = 0

The left hand side factors into a product with two terms:
(n-10) (n+1) = 0

Split into two equations:
n-10 = 0 or n+1 = 0

Add 10 to both sides:
n = 10 or n+1 = 0

Subtract 1 from both sides:
Answer: |  n = 10     or     n = -1