## Intermediate Algebra for College Students (7th Edition)

$9 \times 10^{-3}$
Write the numerator and the denominator in scientific notation to obtain: $=\dfrac{(7.2 \times 10^{-4}) \times (3 \times 10^{-3})}{2.4 \times 10^{-4}}$ Multiply the corresponding parts of the numerator to obtain: $=\dfrac{21.6 \times 10^{-4+(-3)}}{2.4 \times 10^{-4}} \\=\dfrac{21.6 \times 10^{-7}}{2.4 \times 10^{-4}} \\=\dfrac{2.16 \times 10^{-7+1}}{2.4 \times 10^{-4}} \\=\dfrac{2.16 \times 10^{-6}}{2.4 \times 10^{-4}}$ Divide the corresponding parts to obtain: $=\dfrac{2.16}{2.4} \times \dfrac{10^{-6}}{10^{-4}} \\=0.9 \times 10^{-6-(-4)} \\=0.9 \times 10^{-6+4} \\=0.9 \times 10^{-2} \\=9 \times 10^{-2-1} \\=9 \times 10^{-3}$