Question

Determine the LCM of 56f^3g^2 and 70fg^3.

a) LCM = 140f^3g^3
b) LCM = 560fg^3
c) LCM = 280f^3g^3
d) LCM = 70f^3g^3

Answer 1

The LCM of
`56f^3g^2 and 70fg^3 is 280f^3g^3`.

Step-by-step explanation:

To find the LCM of two or more numbers, we first find the highest power of each prime factor in each number and then multiply them together. Let's apply this method to find the LCM of 56f³g²and 70fg³.

First, we find the prime factors of both numbers:

56 = 2³ 14

70 = 2 5 * 14

f and g are variables, so we don't need to find their prime factors. Now, let's find the highest power of each prime factor in both numbers:

For f: Both numbers have f raised to the power of 3, so we don't need to do anything here.

For g: Both numbers have g raised to the power of 2, so we don't need to do anything here either.

For 2: The highest power of 2 in the first number is 3, and the highest power of 2 in the second number is also 3 (since it's a factor of both 14 and 5). The highest power of 2 in both numbers is therefore 3.

For 14: Both numbers have a factor of 14, so we don't need to do anything here.

Now that we have found the highest power of each prime factor in both numbers, we can multiply them together to get the LCM:

LCM =
`(2^3)(14)^2(f^3)(g^2) = (8)(16)(f^3)(g^2) = 280f^3g^3`.

Answer 2

It represents the smallest multiple that both 56f³g²and 70fgv³will divide into without leaving a remainder.Therefore the correct option is c) LCM = 280f³g³

Step-by-step explanation:

To find the least common multiple (LCM) of 56f³g² and 70fg³, start by breaking down each term into its prime factors:

56 = 2³ * 7

70 = 2 * 5 * 7

f³g² and fg³ remain as they are.

Now, identify the highest power of each prime factor that appears in either expression. For 2, the highest power is 3 in 56 and 1 in 70. For 5 and 7, they appear in 70 but not in 56. For f, the highest power is 3, and for g, the highest power is 3 as well.

Multiplying these highest powers together gives the LCM:

2³* 5 * 7 * f³* g³ = 280f³g³.

This means that 280f³g³ is the smallest multiple that both 56f³g² and 70fg³will divide evenly into without leaving a remainder.

Therefore the correct option is c) LCM = 280f³g³