To multiply the two expressions \( \frac{4}{9}k^5 \) and \( \frac{3k^3}{8} \), we follow these steps:
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Multiply the numerators: \[ 4 \times 3k^5 \times k^3 = 12k^{5+3} = 12k^8 \]
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Multiply the denominators: \[ 9 \times 8 = 72 \]
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Combine them: \[ \frac{12k^8}{72} \]
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Simplify the fraction: We can simplify \( \frac{12}{72} \) which becomes \( \frac{1}{6} \): \[ \frac{12k^8}{72} = \frac{12}{72}k^8 = \frac{1}{6}k^8 \]
Now, because you stated \( k = 0 \), substituting \( k = 0 \) into the expression gives: \[ \frac{1}{6} \times 0^8 = \frac{1}{6} \times 0 = 0 \]
Thus, the product in simplest form is: \[ \text{Product} = 0 \]