11 + 6k = 65 Subtract 11 from both sides: 6k = 65-11 6k = 54 Divide by 6 to get k: k = 54/6 k = 9
k=4
11 - 5k. If k is a number then substitute it in. Example if k were to equal 2, then 11 - 5*2 = 11-10 = 1. It depends on the value of k.
k = 2 7 + (5 x 2) = 17 (8 x 2) + 1 = 17
5 Σ (10)^k-1 K=1 I recently had to do this problem. Pain in the a** if you ask me but I solved it for you, so it's okay :D
2k + 5
-10
6 + 2K = 11 subtract 6 from each side - 6 + 6 + 2K = 11 - 6 2K = 5 divide the integers both sides by 2 K = 5/2 check 6 + 2(5/2) = 11 6 + 10/2 = 11 6 + 5 = 11 11 = 11 checks
No.
I'm assuming this is a combinations problem. nCk = n!/(k!(n-k)!) is the general equation. In your case n= 11 and k = 5 for 11!/(5!(11-5)!) = 11!/(5!*6!) = (11*10*9*8*7)/(5*4*3*2*1) = 462
Equation: x^2 +2kx +10x +k^2 +5 = 0 Using the discriminant: (2k +10)^2 -4*1*(k^2 +5) = 0 Solving the discriminant: k = -2
k = 5
11 + 6k = 65 Subtract 11 from both sides: 6k = 65-11 6k = 54 Divide by 6 to get k: k = 54/6 k = 9
72 - k = 31 Therefore, k = 72 - 31 k = 41
k=4
7.5 x 55 is the ONLY ammunition to be used in the K-31. The Swiss designation for it is 7,5mm Gewehrpatrone 11 (7,5mm GP11)
16 - 3k + 5 and k is 4, then 16 -12 + 5 = 9