Evans Bolton
09/27/2024 · Elementary School

Properties of Exponents and Scientific Notation 11. Simplify. \( \begin{array}{ll}\frac{3 z^{5}}{9 y z} & \text { c. }\left(\frac{7 z^{5}}{y}\right)^{2} \\ \text { b. }\left(2 x \cdot x^{4}\right)^{3} & \text { d. } 3 x^{0}\end{array} \)

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Let's simplify each expression one by one. ### a. \(\frac{3 z^{5}}{9 y z}\) 1. Simplify the coefficients: \(\frac{3}{9} = \frac{1}{3}\). 2. Simplify the \(z\) terms: \(\frac{z^{5}}{z} = z^{5-1} = z^{4}\). 3. Combine the results: \[ \frac{3 z^{5}}{9 y z} = \frac{1}{3} \cdot \frac{z^{4}}{y} = \frac{z^{4}}{3y} \] ### b. \(\left(2 x \cdot x^{4}\right)^{3}\) 1. Combine the \(x\) terms: \(2 x \cdot x^{4} = 2 x^{1 + 4} = 2 x^{5}\). 2. Apply the exponent: \((2 x^{5})^{3} = 2^{3} \cdot (x^{5})^{3} = 8 x^{15}\). ### c. \(\left(\frac{7 z^{5}}{y}\right)^{2}\) 1. Apply the exponent to both the numerator and the denominator: \[ \left(\frac{7 z^{5}}{y}\right)^{2} = \frac{(7)^{2} (z^{5})^{2}}{(y)^{2}} = \frac{49 z^{10}}{y^{2}}. \] ### d. \(3 x^{0}\) 1. Recall that any non-zero number raised to the power of 0 is 1: \(x^{0} = 1\). 2. Therefore, \(3 x^{0} = 3 \cdot 1 = 3\). ### Summary of Simplified Results: - a. \(\frac{z^{4}}{3y}\) - b. \(8 x^{15}\) - c. \(\frac{49 z^{10}}{y^{2}}\) - d. \(3\)

Quick Answer

a. \(\frac{z^{4}}{3y}\) b. \(8 x^{15}\) c. \(\frac{49 z^{10}}{y^{2}}\) d. \(3\)
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