Ehrlich

a+(-b)=a-b

-a-b=0-a-b

What does M have to do with it? is "M

if thats right, then according to the law of cosines 

(AB)²=(CB)²+(CA)²-2(CB)*(CA)*(cos(M 2+13 ²-2*29*13*cos(41)=440.948976511858

(AB)²=440.948976511858

AB=(440.948976511858)^1/2=20.9987851199029607

therefore AB=20.9987851199029607

You forgot one ')', please correct it and ill try to explain ( there are 4 ('s and three )'s)

The answer is 3

Proof:

P=3*n₁+a₁

Q=3*n₂+a₂

R=3*n₃+a₃

a1,a2,a3=1 OR 2

P²+Q²+R²=(3*n₁+a₁)²+(3*n₂+a₂)²+(3*n₃+a₃)²=(3n₁)²+(3n₂)²+(3n₃)²+2*3*n₁*a₁+2*3*n₂*a₂₊2*3*n₃*a₃+a₁²+a₂²​+a₃²​

If a₁=a₂=a₃=1 OR 2 the formula is divisible by 3.

same applies if one of them is 1 and if 2 of them are 1.

meaning it will always be divisible. but something here doesnt make sense right? thats right, if one if them is 3 it works. meaning one of them has to be 3. we already know none of them can be 2, and 1 is not a prime, so P=3 (always).

Because you are bad.

Are you 100% sure its solveable?

i appreciate people that come up with their own questions, but it might be a very hard question, a question im afraid im not able to solve.

a prime is a number that is not divisble by any number.

2*r+2*q+2*p can sometimes be a prime, according to you.

are you sure 2*r+2*q+2*p can be a prime? think about it (its a hint)

beta=B

sec(B)=1/(cos)

tan(B)=sin(B)/cos(B)

tan^2(B)=5-4sec(B) ......multiply by cos^2(B)

sin^2(B)=5*cos^2(B)-4cos(B)

using the identity sin^2(X)+cos^2(X)=1

sin^2(B)=1-cos^2(B)=5cos^2(B)-4*cos^2(B)........add (cos^2(B)-1)

0=6*cos^2(B)-4cos^2(B)-1.......divide by 6

0=cos^2(B)-(2/3)*cos(B)-1/6=(cos(B)-(1/3))^2-1/18......add 1/18

1/18=(cos(B)-(1/3))^2

i think you can continue from here

are c, e and p constants? if not then there is no answer my friend

dipende da cosa polinomio si sta parlando. Io non credo che ci sia un modo per fare questo per ogni polinomiale. è possibile utilizzare la formula quadratica di farlo con un'equazione di secondo grado. Spero che aiutato!