CPhill

28° 50 .....convert this to decimal form.....  

28.833°  x  pi / 180   = about 0.5032 rads

Wow, Melody....you made that one look easy....very nice........!!!!!!

Thanks, Melody......I "stole" that pic from Wikipedia.....LOL!!!

Try  7 and  -1

(1/2) sin(2θ) + (1/4)sin(4θ) =

(1/2)*2sinθcosθ + (1/4)sin(2θ + 2θ) =

sinθcosθ + (1/4)[ sin2θcos2θ +  sin2θcos2θ ] =

sinθcosθ  + (1/4) [ (2)[sin2θ][cos2θ] ] =

sinθcosθ  + (1/4) (2) [ [sin2θ][cos2θ] ] =

sinθcosθ + (1/2)[2sinθcosθ]* [2cos^2θ - 1] =

sinθcosθ + sinθcosθ * [ 2cos^2θ - 1 ] =

sinθcosθ + sinθcosθ *  2cos^2θ  - sinθcosθ =

 sinθcosθ *  2cos^2θ  =

2 sinθcosθ *  cos^2θ    

sin(2θ) * cos^2θ      and replacing θ  with pi * x .....we have

sin(2* pi *x ) * cos^2(pi * x)

And that's it, GL  ....!!!!

V = (1\3)*3.14*(3^2)*8 =

(1/3) * 3.14 * 9 * 8 =

(1/3) * 3.14 * 72 =

24 * 3.14 =

75.36 units³

2800 / 5  = 560   = 5.6 x 10²

I'm assuming this is

3^352^(log10^(3^(1/4))

3^(1/4) = 1.3160740129524925  ... so we have

3^352^log10^1.3160740129524925 =

3^352^1.3160740129524925 =

3^463.25805255927736

In terms of base 10 .... 3 = 0.4771212547196624

So, we have

(10^0.4771212547196624)^463.25805255927736  =

10^221.030263296069725261145127163264 =

10^221 * 10^0.030263296069725261145127163264 = 

10^221 * 1.072   {approximately } 

Or, in scientifiic notation

1.072 x  10^221  = 1072 followed by 218 zeroes..... (more or less)

Here's a picture of a  regular octahedron :

There are 6 vertices and the number of ways of pairing any two of them is C(6,2)  = 15

But notice that only 12 pairs of them form edges. {The top and bottom vertices aren't connected to each other and neither are the two opposite vertices on the "sides." }

So, the probability that any two form an edge is  12/15 = 4/5

We can use the sine here

sin 13 = 400 / d     where d is the distance on the hypotenuse it has traveled

Rearrange..and we have

d = 400/sin 13 = about 1800 ft (rounded)