You are given that n= 2^2 * 5 Write 40n as the product of its prime factors.
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$\displaystyle 40=2^3\cdot 5$. Then $\displaystyle 40n=2^3\cdot 5\cdot 2^2\cdot5=2^5\cdot 5^2$
Originally Posted by GAdams You are given that n= 2^2 * 5 Write 40n as the product of its prime factors. $\displaystyle n = 2^2 \cdot 5$ $\displaystyle \Rightarrow 40n = 40 \cdot 2^2 \cdot 5$ $\displaystyle = 8 \cdot 5 \cdot 2^2 \cdot 5$ $\displaystyle = 2^3 \cdot 5 \cdot 2^2 \cdot 5$ $\displaystyle = 2^5 \cdot 5^2$ EDIT: Why am i always the second one to post today?
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