expressions. Regard all variables as non-zero. \( \begin{array}{ll}\text { b) } 2 a^{4} \times 5 a^{3} & \text { c) } 7 x^{4} \times 2 x^{5} \times x^{3} \\ \text { e) } \frac{a^{9}}{a^{5}} & \text { f) } \frac{p^{8} q^{5}}{p q^{3}} \\ \text { h) } 4 m^{2} n^{3} \times 5 m n^{5} & \text { i) } \frac{m^{11} n^{3}}{n^{2} m^{6}} \\ \text { k) }\left(m^{6}\right)^{4} & \text { 1) }\left(a^{2} b^{5}\right)^{3} \\ \text { n) }\left(3 k^{4}\right)^{2} & \\ \text { p) } \frac{36 a^{9} b^{3} c^{2}}{12 a^{6} b^{3}} \\ \text { r) }\left(2 a^{3} b^{4}\right)^{0} \times\left(2 a^{5} b^{3}\right)^{2} & 2 k^{2} m \times 8 k^{9} m^{2} \\ \text { v) } 40+3 \times(-2)^{0}+6(8-3)^{0} \\ \text { x) } \frac{18 x^{6} \times y^{3}}{9 x^{4} y}\end{array} \)
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When multiplying powers with the same base, simply add their exponents! For instance, in \( 2 a^{4} \times 5 a^{3} \), you should multiply the coefficients (2 and 5) to get 10, and then combine the \( a \) terms to get \( a^{4+3} = a^{7} \). Overall, you’ll end up with \( 10 a^{7} \). For division, remember to subtract the exponents of the same base. In \( \frac{a^{9}}{a^{5}} \), you subtract 5 from 9, giving you \( a^{9-5} = a^{4} \). So, you’d just have \( a^{4} \) as your final answer. Easy-peasy!
