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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Step-by-step Solution
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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