To solve the expression \((-3/5) \times (5/16) \times (7/12) / (-3/16) \times (-5/7)\), we will apply the properties of operations, as well as simplifying step-by-step.
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We first rewrite the expression for clarity:
\[ \frac{(-3/5) \times (5/16) \times (7/12)}{(-3/16) \times (-5/7)} \]
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Let's simplify the numerator and the denominator separately.
Numerator: \[ (-3/5) \times (5/16) \times (7/12) \] We can multiply the fractions together:
\[ = \frac{-3 \times 5 \times 7}{5 \times 16 \times 12} \]
Notice that the \(5\) in the numerator cancels with the \(5\) in the denominator:
\[ = \frac{-3 \times 7}{16 \times 12} \]
Calculating that:
\[ = \frac{-21}{192} \]
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Denominator: \[ (-3/16) \times (-5/7) \] Here, since both terms are negative, their product will be positive:
\[ = \frac{3 \times 5}{16 \times 7} \]
Thus:
\[ = \frac{15}{112} \]
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Combining our results, we rewrite our original expression as:
\[ \frac{-21/192}{15/112} \]
Dividing by a fraction is the same as multiplying by its reciprocal:
\[ = \frac{-21}{192} \times \frac{112}{15} \]
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Now, we multiply the fractions:
\[ = \frac{-21 \times 112}{192 \times 15} \]
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Let's simplify the fraction. We will find common factors. First, we'll calculate each term:
\(-21\) can be factored into \(-3 \times 7\).
\(192\) can be factored (by its prime factorization) as \(2^6 \times 3\).
\(15\) is \(3 \times 5\).
\(112\) is \(2^4 \times 7\).
Substituting these factorizations, we get:
\[ = \frac{-3 \times 7 \times (2^4 \times 7)}{(2^6 \times 3) \times (3 \times 5)} \]
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Now cancel common factors:
- The \(3\) in the numerator and denominator.
- The \(7\) in the numerator and denominator.
After canceling the common factors, we simplify:
\[ = \frac{-1 \times 16}{64 \times 5} = \frac{-16}{320} \]
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Now we can simplify \(-16/320\):
\[ = \frac{-1}{20} \quad \text{ (by dividing both numerator and denominator by 16)} \]
Thus, the final solution is:
\[ \boxed{-\frac{1}{20}} \]
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