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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

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

No.

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

72 - k = 31 Therefore, k = 72 - 31 k = 41

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

k = 5

79 + k = 84 84 - 79 = k k = 5

16 - 3k + 5 and k is 4, then 16 -12 + 5 = 9

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.

5 + 12*k where k is any integer.

Using the discriminant of b^2 -4ac = 0 the value of k works out as -2

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)

7(k + 5)(k - 4)

When factored it is: (3k+5)(k-1)

2k + 3 = 11(- 3 from both sides)2k = 8(divide both sides by 2)k = 4

First, factor out the k. So, k(4+2) = 30 or 6k = 30 Then divide by 6. k = 30/6 = 5 So k = 5.

Print "Type the upper limit (n) ?" Input n K = -1 WHILE K < = n K = K + 2 Sum = Sum + K WEND Print "The sum of all odd numbers up to "; n; "is "; Sum

K+

60k plus k is 61k

Equation: x^2 +2kx +10x +k^2 +5 = 0 Using the discriminant: (2k +10)^2 -4*1*(k^2 +5) = 0 Multiplying out the brackets: 4k^2 +40K +100 -4k^2 -20 = 0 Collecting like terms: 40k +80 = 0 => 40k = -80 => k = -80/40 Therefore the value of k = -2

k = 2 7 + (5 x 2) = 17 (8 x 2) + 1 = 17