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[tex] \frac{( {2a}^{3} )( {10a}^{5} )}{ {4a}^{ - 1} } [/tex]negative and zero exponents

Answer :

[tex]\begin{gathered} \frac{( {2a}^{3} )( {10a}^{5} )}{ {4a}^{ - 1} }=\frac{(2\cdot10)(a^3a^5)}{4}a \\ =\frac{20a^{3+5}a}{4} \\ =\frac{20}{4}a^8a \\ =5a^{8+1} \\ =5a^9 \end{gathered}[/tex]

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