Question

1. evaluate ( 27x^{0}y^{-2} ) for ( x = 4 ) and ( y = 3 ). solve

Explanation:

Step1: Recall the zero - exponent rule

Any non - zero number raised to the power of 0 is 1, i.e., \(a^{0}=1\) for \(a
eq0\). Also, recall the negative - exponent rule \(a^{-n}=\frac{1}{a^{n}}\) for \(a
eq0\).

First, substitute \(x = 4\) and \(y=3\) into the expression \(27x^{0}y^{-2}\).

For the term \(x^{0}\), since \(x = 4
eq0\), by the zero - exponent rule, \(x^{0}=4^{0}=1\).

For the term \(y^{-2}\), using the negative - exponent rule, \(y^{-2}=\frac{1}{y^{2}}\). When \(y = 3\), \(y^{2}=3^{2}=9\), so \(y^{-2}=\frac{1}{9}\).

Step2: Substitute the values into the expression

The original expression is \(27x^{0}y^{-2}\). Substitute \(x^{0}=1\) and \(y^{-2}=\frac{1}{9}\) into it:

\(27\times x^{0}\times y^{-2}=27\times1\times\frac{1}{9}\)

Step3: Simplify the expression

\(27\times\frac{1}{9}=\frac{27}{9} = 3\)

Answer:

\(3\)