To solve the problem, we first determine the rates at which the hose and the sprinkler fill the pool.
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Determine the rate of the hose: The hose fills the kiddie pool in 8 minutes. The rate of the hose \( r_{\text{hose}} \) is: \[ r_{\text{hose}} = \frac{1 \text{ pool}}{8 \text{ minutes}} = \frac{1}{8} \text{ pools per minute} \]
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When using the lawn sprinkler and the hose together: It takes 5 minutes to fill the pool. Thus, the combined rate \( r_{\text{combined}} \) is: \[ r_{\text{combined}} = \frac{1 \text{ pool}}{5 \text{ minutes}} = \frac{1}{5} \text{ pools per minute} \]
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Let \( r \) be the rate of the lawn sprinkler: We can express the combined rate as the sum of the rates of the hose and the sprinkler: \[ r_{\text{hose}} + r = r_{\text{combined}} \] Substituting the known values: \[ \frac{1}{8} + r = \frac{1}{5} \]
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Rearranging the equation: To isolate \( r \), we first need to get a common denominator. The least common multiple of 8 and 5 is 40. We rewrite the fractions: \[ \frac{1}{8} = \frac{5}{40}, \quad \frac{1}{5} = \frac{8}{40} \] So our equation becomes: \[ \frac{5}{40} + r = \frac{8}{40} \]
Rearranging gives: \[ r = \frac{8}{40} - \frac{5}{40} = \frac{3}{40} \]
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Now, let's find the correct form of our equation: We can multiply the whole equation by 40 to eliminate the fractions. The original equation: \[ \frac{1}{8} + r = \frac{1}{5} \] becomes: \[ 5 + 40r = 8 \] Rearranging gives: \[ 40r = 8 - 5 = 3 \]
However, we can relate this back to the options given. The best choice aligned with the original equation setup is: \[ \frac{5}{8} + 5r = 1 \]
So the answer that fits this context and setup is: B. \( \frac{5}{8} + 5r = 1 \).
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