2 answers:
Answer:
5/4k^2
Step-by-step explanation:
P=5\dfrac{k}{6}\times \dfrac{3}{2k^3}.
We will be using the following property of exponents:
\dfrac{a^x}{a^y}=a^{x-y}.
We have
P\\\\\\=5\dfrac{k}{6}\times\dfrac{3}{2k^3}\\\\\\=\dfrac{5}{6}\times\dfrac{3}{2}k^{1-3}\\\\\\=\dfrac{5}{4}k^{-2}=\dfrac{5}{4k^2}.
Thus, the required product is \dfrac{5}{4k^2}.
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Answer: 5/4k^2
Step-by-step explanation:
Just multiply the fractions then simplify by a common factor of 3, then cancel out the common variable of k
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