## College Algebra 7th Edition

(a) $28x^3y^7$ (b) $27yz^3$ (c) $32x$
(a) $(4x^{3}y^{2})(7y^{5})=4*7x^{3}y^{2+5}=28x^{3}y^{7}$ (b) $(9y^{-2}z^{2})(3y^{3}z)=9*3y^{-2+3}z^{2+1}=27yz^{3}$ (c) $(8x^{7}y^{2})(\displaystyle \frac{1}{2}x^{3}y)^{-2}=\frac{8x^{7}y^{2}}{(\frac{1}{2}x^{3}y)^{2}}=\frac{2^{2}8x^{7}y^{2}}{x^{3\cdot 2}y^{2}}=\frac{32x^{7}y^{2}}{x^{6}y^{2}}=32x^{7-6}y^{2-2}=32xy^0=32x$