To subtract the expression \( \frac{1}{3}(z+ \frac{2}{5}) - 4\left( \frac{1}{2}z - 2\right) \), let's first distribute and simplify.
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Distributing \( \frac{1}{3} \): \[ \frac{1}{3}(z + \frac{2}{5}) = \frac{1}{3}z + \frac{1}{3} \cdot \frac{2}{5} = \frac{1}{3}z + \frac{2}{15} \]
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Distributing \( -4 \): \[ -4\left( \frac{1}{2}z - 2 \right) = -4 \cdot \frac{1}{2}z + 4 \cdot 2 = -2z + 8 \]
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Now combine the two results: \[ \frac{1}{3}z + \frac{2}{15} - 2z + 8 \]
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Combining like terms (for the \( z \) terms and the constant terms):
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For the \( z \) terms: \[ \frac{1}{3}z - 2z = \frac{1}{3}z - \frac{6}{3}z = \frac{1 - 6}{3}z = -\frac{5}{3}z \]
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For the constants: \[ 8 + \frac{2}{15} = \frac{120}{15} + \frac{2}{15} = \frac{122}{15} \]
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So, putting it all together, we have: \[ -\frac{5}{3}z + \frac{122}{15} \]
To convert \(-\frac{5}{3}z\) into its equivalent with a common denominator of 15: \[ -\frac{5}{3}z = -\frac{25}{15}z \]
Combining, we get: \[ -\frac{25}{15}z + \frac{122}{15} \]
Thus the final result is: \[ -\frac{25}{15}z + \frac{122}{15} \]
Now, let's evaluate the options given:
- Option 1: \(-1 \frac{2}{3} z + 8 \frac{2}{15}\) → Equivalent to \(-\frac{5}{3}z + \frac{122}{15}\), but it's not simplified correctly in terms of fractions.
- Option 2: \(-\frac{1}{3}z +4 \frac{1}{2} z\) → Not equivalent to our result.
- Option 3: \(-1 \frac{2}{3} - 7 \frac{12}{15}\) → Doesn't match.
- Option 4: \(- \frac{3}{15}z + 4\) → Not equivalent.
None of the options from 1 to 4 match the calculated result of \(-\frac{25}{15}z + \frac{122}{15}\). If you're looking for a specific rewording or adjustment, please clarify!
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