To solve the equation \( \frac{1}{5} \times \frac{3}{4} - 31 = -6 \), let's first simplify the left side of the equation.
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Calculate \( \frac{1}{5} \times \frac{3}{4} \): \[ \frac{1 \times 3}{5 \times 4} = \frac{3}{20} \]
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Now substitute this back into the equation: \[ \frac{3}{20} - 31 = -6 \]
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Add \( 31 \) to both sides to isolate \( \frac{3}{20} \): \[ \frac{3}{20} = -6 + 31 \] \[ \frac{3}{20} = 25 \]
Next, let's check the correctness of the simplified form.
The left side \( \frac{3}{20} \) is not equal to \( 25 \). So the equation does not make sense. However, it appears we need to find which of the given options satisfies the result of some arithmetic that might relate back to this equation.
Let's look at the numbers given:
- 25
- 125
- 5
- 625
None of them are solutions to \( \frac{3}{20} \).
Given the options, if they are linked to an equation we need to solve or reverse, it’s unclear without extra context. However, we can analyze if one of these might result from further operations.
If we take a guess based on calculations involving raising or multiplying significant values, \( 625 \) stands out as it is \( 25^2 \). Let’s keep that in mind.
If you're looking for the most reasonable context for consistently getting results on checks (dependent on previous questions), then:
Since the equation was \(\frac{3}{20} - 31 = -6\): the adjustments don’t lead to logical values as expected.
Thus if there are evaluations required, then it might require complex arrangements.
If I misunderstood the reading, please specify more context, or how you would like to approach filling in the empty spaces.
From the earlier outlined details, revisiting might help:
In conclusion: Given the equation doesn't tie back to finds, \( -31 + 25 \) would consistently land at lesser than recognizable values so it points toward 25 to fit checks more closely.