They are member of the infinite set of numbers of the form 3108*k where k is an integer.
To find the common multiple of 4, 5, and 9, we need to identify the least common multiple (LCM) of the numbers. First, we find the prime factors of each number: 4 = 2^2, 5 = 5^1, and 9 = 3^2. Next, we take the highest power of each prime factor that appears in any of the numbers: 2^2 * 3^2 * 5^1 = 4 * 9 * 5 = 180. Therefore, the common multiple of 4, 5, and 9 is 180.
Common multiples include any multiple of 168.
The LCM is 144. Any multiple of 144 will also be a multiple of 16 and 18.
Any member of the infinite set of numbers of the form 48*k where k is an integer.
-6k + 7k = k or 7k- 6k = k or Factor out 'k' k(-6 + 7) = k(1) = k
7 = seven k = k se-k 7k sek sick
26-7k=6k First, add 7k to get k on one side. 26-7k+7k=6k+7k 26=13k Divide both sides by 13 to get k by itself. 26/13=13k/13 2=k The answer is 2.
7(k + 5)(k - 4)
-6K+7K=1K or K
-6k + 7k = k
A polygon with 7k sides, where k is any positive integer.A polygon with 7k sides, where k is any positive integer.A polygon with 7k sides, where k is any positive integer.A polygon with 7k sides, where k is any positive integer.
To solve an equation, use the order of operations in reverse:7k-52 = 9(addition)7k = 61(division)k = (61/7) or ~8.71
3k-5=7k+73k-7k=7+5-4k=12k=-3
-6k+7k = 1k or normally k on its own.
It is: 7k
They are members of the infinite set of numbers of the form n*k where n is an integer and k is the least common multiple of 12 and 42.