CPhill

Nice, BJ....!!!

No triangle is presented to us......???

tan (Θ/2)  =  sin Θ/ (1 + cosΘ)

sin(Θ/2)/cos(Θ/2) =

√ [(1-cosΘ)/2] / √[(1 + cosΘ)/2] =

√[(1 - cos Θ)/(1 + cos Θ)] = multiply top and bottom by √[(1 + cosΘ)/(1 + cosΘ)] =

√[(1 - cosΘ)(1+cosΘ)] / [1 + cosΘ]  =

√[1 - cos^2 Θ] / (1 + cosΘ] =

√[sin^2  Θ] / (1 + cos Θ] =

sin Θ  / [1 + cos Θ]       

Notice, that since we are in the first quadrant, the tangent is positive everywhere except at 0 and pi/2....so the root wias positive.

That's pretty neat, Bertie....!!!!

I believe that this is correct.....the probability is given by...

C(20,1) (.15)(.85)^19  = about 13.7%

 (x^4)^5/(x^3)^3=  x^20 / x^9  = x^11

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 (a^2b^3/ab^2)....I'm going to assume this is  (a^2b^3)/(ab^2).... if so, we have

(a^2)/(a) * (b^3)/(b^2) =

a * b  =

ab

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(2/3)^-3×(a^3/b)^-2.......notice that we can use this"trick' to get rid of the negative exponents.....

(3/2)^3 * (b/a^3)^2    ....so we have...

(27/8) * (b^2 / a^6) =

(27b^2)/(8a^6)

4x - [6/y] = 15

4x - [6/(-2)]  = 15

4x - [- 3 ]  = 15    

4x + 3 = 15     subtract 3 from both sides

4x  = 12          divide both sides by 4

x = 3

Always happy to save a life.......or even bail somebody out on Trig homework..LOL!!!

Sorry, but

-5^2  = -1(5)^2 = -1*25 = -25

The exponent only applies to the 5, not to the "understood" -1 out front.

Note however.......

(-5)^2  = (-1*5)^2  = (-1)^2*(5)^2 = 1 * 25  = 25

The exponent applies to both the 5 and the "understood" -1 in front.