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Let the side of the square be 4 inits.

DF = tan(20º) * 4  = 1.455880937

FC = 4 - DF = 2.544119063

BE = tan(25º) * 4 = 1.865230633

EC = 4 - BE = 2.134769367

tan(∠FEC) = FC / EC          ∠FEC = 50º

∠X = 180º - (65º + 50º) =  65º    

I think I get your solution, but is it possible to get the possible placement. I just need help with the placement. 

There are 780 permutations that sum up to 17.

Therefore, the probability is: 780 / 6^5 =65 / 648

a₁  =  1          a₄  =  1            a₇  =  1         a₁₀  =  1

a₂  =  2          a₅  =  ½           a₈  =  2         a₁₁  =  ½

a₃  =  2          a₆ =   ½           a₉  =  2           a_{12  =}  ½

If this pattern holds, if you divide the index (the subscript) by 3 and get a remainder of 1, the value will be 1.

2011  =  670 x 3 + 1, so  a₂₀₁₁  =  1

432 square units

5/1296

As follows:

Here's a more detailed solution that agrees with Geno, that it isn't possible to do it within the specified cost:

-2 - 2i  in  r·cis(theta) form:

r  =  sqrt( (-2)² + (-2)² )  =  sqrt(8)

theta  =  tan^-1( -2/-2 )  =  5pi/4

x³  =  sqrt(8) · cis ( 5pi/4 )     --->     x  =  [ sqrt(8) · cis ( 5pi/4 )  ]^1/3

To find the principal cube root, take the cube root of  sqrt(8)  =  (8^1/2)^1/3  =  (8^1/3)^1/2  =  sqrt(2)

and divide 5pi/4 by 3  =  5p/12.

So, the principal cube root is:          sqrt(2)·cis( 5pi/12 )

Adding 2p/3, the second root is:     sqrt(2)·cis( 13pi/12 )

and another 2p/3, the third root is:  sqrt(2)·cis( 21pi/12 )

So, I don't get any of the answers ... 

If these towns are in a straight East-West line, then, from West to East are:

   A  -- 30 miles  -- B  -- 85 miles --  C  -- 100 miles --  D

If you must connect each of these by a separate calbe to a power station (at $1000 per mile),

--  if you place the power station at B or C or anywhere between B and C, the total cost will be $300,000.

--  if you place the power station either west of B or east of C, the total cost will be higher.

So, if I am reading the question correctly, I can't solve the problem.

I am interested to see anyone else's solution.

a=1;b=1;c=1;d=1;e=1;p=0; cycle:n= a*10000+b*1000+c*100+d*10+e;if(a!=3 and a!=4 and b!=3 and b!=4 and c!=3 and c!=4 and d!=3 and d!=4 and e!=3 and e!=4 and a!=b and a!=c and a!=d and a!=e and b!=c and b!=d and b!=e and c!=d and c!=e and d!=e, goto loop, goto next); loop:printn,", ",;p=p+1; next:e++;if(e<10, goto cycle, 0);e=1;d++;if(d<10, goto cycle, 0);e=1;d=1;c++;if(c<10, goto cycle,0);c=1;d=1;e=1;b++;if(b<10, goto cycle,0);e=1;d=1;c=1;b=1;a++;if(a<10, goto cycle,0);print"Total = ",p

OUTPUT =2520 such numbers. This computer code exlcudes ALL 3's and 4's completely. They do not appear anywhere in the 2,520 numbers, which are all different digits such as: 12567 , 12568 , 12569 , 12576 , 12578 , 12579......etc.

Note: You can generate all these numbers using permutations of: 7 nPr 5 =2,520 permutations, which is what the computer code does.

The area of a parallelogram is the product of its base and its height.

DC = 27 = base.

EG = 16 = height.

Area  =  27 x 16

No, you aren't. Sorry for not responding earlier or putting it in the question itself.

To strike six o'clock, there are 5 intervals.

So, under the assumption that each strike is instantaneous, it will take 30 seconds / 5 intervals, or 6 seconds per interval.

To strike twelve o'clock, there are 11 intervals.

A 6 seconds per interval, this will take 11 x 6 = 66 seconds.

There are patterns that can be used to make a 5 x 5 magic square (not only are the sums of all the rows and all the columns the same, but also the sums of the two diagonals).

Using one of these patterns, I constructed a magic square that used the first 25 numbers, omitting 169.

Then, using the same pattern, I constructed a magic square that used the last 25 numbers, omitting 144.

So, there are at least two answers to your question: either 169 or 144.

Are you allowed to repeat digits such as: 11226 ??

in 2 1/2 hours the ship travels    2.5  x 15 = 37.5 miles

the x component is   37.5 cos 214^o = -31 miles

the y component is   37.5 sin 214  =   ~ - 21 miles

if 0^o  is north :

31 miles south    and 21 miles west        ( 214 degrees would be southwest )