12b2c3
280a4b2
The LCM of 28 and 42 is 84. Since y2 and x2 are multiples of y and x respectively, they are automatically the LCM. That makes the answer 84x2y2
The LCM of 8 and 34 is 136, which is the multiple of the highest power of prime factors from both numbers (23 x 17 = 2 x 2 x 2 x 17 = 136).
The LCM of 4 and 5 is 20, which is the multiple of the highest power of prime factors from both numbers (22 x 5 = 2 x 2 x 5 = 20).
To find the least common multiple (LCM) of 2, 9, and 18, we first need to factorize each number into its prime factors. 2 = 2^1 9 = 3^2 18 = 2^1 * 3^2 Next, we take the highest power of each prime factor that appears in any of the numbers: 2^1 and 3^2. Multiplying these prime factors together, we get LCM(2, 9, 18) = 2^1 * 3^2 = 2 * 9 = 18. Therefore, the LCM of 2, 9, and 18 is 18.
Oh, what a happy little question! To find the Least Common Multiple (LCM) of 2b^2 and 6c^3, first, we break down the numbers into their prime factors. The LCM is the product of all the highest powers of all prime factors in both numbers, so the LCM of 2b^2 and 6c^3 is 6b^2c^3. Just like painting a beautiful landscape, simplifying math problems can be quite calming and enjoyable.
To find the least common multiple (LCM) of 2b^2 and 6c^3, we first need to factorize both terms. The prime factorization of 2b^2 is 2 * b * b, and the prime factorization of 6c^3 is 2 * 3 * c * c * c. To find the LCM, we take the highest power of each prime factor that appears in either term, which gives us 2 * 3 * b^2 * c^3 = 6b^2c^3. Therefore, the LCM of 2b^2 and 6c^3 is 6b^2c^3.
Assuming you mean 2b^-6, with the 2b not in parenthesis, it simplifies to 2/b6, or 2 over b to the sixth.
-2b + 7 -3b = 2 -2b +7 -3b +2b = 2 + 2b 7 -3b + 3b = 2 + 2b +3b 7-2 = 2-2 + 2b + 3b 5 = 5b 1 = b
The Least Common Multiple (LCM) for 2 6 7 is 42.
b3c2
The question is somewhat ambiguous: the answers are LCM[3T2, 5T] =15T2 or LCM[(3T)2, 5T] =45T2
To find the least common multiple (LCM) of two terms, we need to identify the highest power of each unique factor present in both terms. In this case, the LCM of a³b² and a²b⁵ would be a³b⁵, as it includes the highest power of both 'a' and 'b' present in either term. Therefore, the LCM of a³b² and a²b⁵ is a³b⁵.
8b + 11 - 3b = 2b + 2 5b + 11 = 2b + 2 5b - 2b = 2 - 11 3b = -9 b = -3
a = 2b - c a + c = 2b (a+c)/2 = b b = (a+c)/2
5 over 2 thats the answer
Make them additive for convenience.4s3 + 36s2factor out 4s24s2(s + 9)---------------4s2====LCM