Question

1. $6xy^{0}z^{-3} \cdot 2yx^{-2}z^{-8}$

Explanation:

Step1: Multiply coefficients

Multiply the numerical coefficients \(6\) and \(2\).
\(6\times2 = 12\)

Step2: Combine \(x\) terms

Use the rule \(a^m\cdot a^n=a^{m + n}\) for \(x\) terms: \(x^1\cdot x^{-2}=x^{1+( - 2)}=x^{-1}=\frac{1}{x}\)

Step3: Combine \(y\) terms

For \(y\) terms: \(y^0\cdot y^1=y^{0 + 1}=y^1 = y\) (since \(y^0 = 1\))

Step4: Combine \(z\) terms

For \(z\) terms: \(z^{-3}\cdot z^{-8}=z^{-3+( - 8)}=z^{-11}=\frac{1}{z^{11}}\)

Step5: Multiply all parts together

Multiply the results from steps 1 - 4: \(12\times\frac{1}{x}\times y\times\frac{1}{z^{11}}=\frac{12y}{x z^{11}}\)

Answer:

\(\frac{12y}{x z^{11}}\)