If `1960=2^(a)xx5^(b)xx7^(c )`, calculate the value of `2^(-a)xx7^(b)xx5^(-c )`

Answer

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) from the prime factorization of \( 1960 \), and then use these values to calculate \( 2^{-a} \times 7^{b} \times 5^{-c} \).

### Step-by-Step Solution:

1. **Factorize 1960**:
We start by factorizing \( 1960 \) into its prime factors.

- Divide by \( 5 \):
\[ 1960 \div 5 = 392 \]
- Divide \( 392 \) by \( 2 \):
\[ 392 \div 2 = 196 \]
- Divide \( 196 \) by \( 2 \):
\[ 196 \div 2 = 98 \]
- Divide \( 98 \) by \( 2 \):
\[ 98 \div 2 = 49 \]
- Divide \( 49 \) by \( 7 \):
\[ 49 \div 7 = 7 \]
- Divide \( 7 \) by \( 7 \):
\[ 7 \div 7 = 1 \]

So, the prime factorization of \( 1960 \) is:
\[ 1960 = 2^3 \times 5^1 \times 7^2 \]

2. **Identify values of \( a \), \( b \), and \( c \)**:
From the factorization, we can see:
- \( a = 3 \)
- \( b = 1 \)
- \( c = 2 \)

3. **Substitute values into the expression**:
We need to calculate \( 2^{-a} \times 7^{b} \times 5^{-c} \):
\[ 2^{-a} \times 7^{b} \times 5^{-c} = 2^{-3} \times 7^{1} \times 5^{-2} \]

4. **Calculate each part**:
- \( 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \)
- \( 7^{1} = 7 \)
- \( 5^{-2} = \frac{1}{5^2} = \frac{1}{25} \)

5. **Combine the results**:
Now, we combine these values:
\[ 2^{-3} \times 7^{1} \times 5^{-2} = \frac{1}{8} \times 7 \times \frac{1}{25} \]
This can be simplified as:
\[ = \frac{7}{8 \times 25} = \frac{7}{200} \]

### Final Answer:
Thus, the value of \( 2^{-a} \times 7^{b} \times 5^{-c} \) is:
\[ \frac{7}{200} \]