I think this is supposed to be :
[sin x cos y + cos x sin y] / [ cos x cos y- sin x sin y] = [cot x + cot y]/ [cot x cot y - 1]
Work on the left side dividing every term by sin x sin y
[ cot y + cot x] / [ cot x cot y - 1] and this equals the right side
6sinθ−7=0 this has no solution........6 is as large as the first term can be....so
6 - 7 can never equal 0
14 = [ 2(h + 3) ] / 5 multiply both sides by 5
70 = 2 (h + 3) divide both sides by 2
35 = h + 3 subtract 3 from both sides
32 = h
\(sin\left(cos^{-1}\left(\frac{x^2}{4-x^2}\right)\right)\)
We are looking for the sin of an angle (theta) with a cosine of x^2 / ( 4 - x^2).....so we have...
sin (theta) = √ [ 1 - cos^2 (theta) ] = [ 1 - [ (x^2)/(4-x^2)]^2 ] = √ [x^4 - 8x^2 + 16 - x^4] / (4 - x^2) =
√ {16 - 8x^2) / (4- x^2) = √ [ 4 (4 - 2x^2) ] / (4 - x^2) = 2√(4 - 2x^2) / (4 - x^2)
L = ( 1 + 1/3) W = (4/3)W
And the area = L * W ....so....
4800 = 4(/3)W * W
4800 = (4/3)W^2 multiply both sides by (3/4)
3600 = W^2 take the positive square root
60 in = W and L = (4/3)* 60 = 80 in
Your Medal of Honor should be arriving shortly by mail, hectictar...!!!.....LOL!!!!
We have :
9! / [ (number of repeats of letter A) ! * (number of repeats of letter B) ! ] =
9! / ( 5! * 4!) = 126 identifiable words
2 + 3/4 = 11/4
1 + 1/4 = 5/4 .... so we have
(11/4) * (1/2) * (5/4) = 55 / 32
Thanks, Guest....!!!!
Just hit the "2nd" key......you will see asin, acos, atan, etc.
