Answer :
To express 140 as a product of prime numbers in index form, we need to break it down into its prime factors.
First, we find the prime factors of 140:
- 140 is divisible by 2, a prime number. Dividing 140 by 2, we get 70, so 140 = 2 × 70.
- 70 is also divisible by 2. Dividing 70 by 2, we get 35, so 70 = 2 × 35. Hence, 140 = 2 × 2 × 35, or [tex]\(2^2 \times 35\)[/tex].
- 35 is not divisible by 2, so we move on to the next smallest prime number, which is 3. 35 is not divisible by 3, so we move on to the next prime number, which is 5. Dividing 35 by 5, we get 7, so 35 = 5 × 7.
Since 7 is a prime number, we have factored 140 down to its prime components completely.
Combining all the prime factors, we get:
[tex]\[ 140 = 2^2 \times 5^1 \times 7^1 \][/tex]
So, 140 written as a product of prime numbers in index form is:
[tex]\[ 2^2 \times 5^1 \times 7^1 \][/tex]
First, we find the prime factors of 140:
- 140 is divisible by 2, a prime number. Dividing 140 by 2, we get 70, so 140 = 2 × 70.
- 70 is also divisible by 2. Dividing 70 by 2, we get 35, so 70 = 2 × 35. Hence, 140 = 2 × 2 × 35, or [tex]\(2^2 \times 35\)[/tex].
- 35 is not divisible by 2, so we move on to the next smallest prime number, which is 3. 35 is not divisible by 3, so we move on to the next prime number, which is 5. Dividing 35 by 5, we get 7, so 35 = 5 × 7.
Since 7 is a prime number, we have factored 140 down to its prime components completely.
Combining all the prime factors, we get:
[tex]\[ 140 = 2^2 \times 5^1 \times 7^1 \][/tex]
So, 140 written as a product of prime numbers in index form is:
[tex]\[ 2^2 \times 5^1 \times 7^1 \][/tex]
