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(3h^(2)j^(0)k^(0))/(3hj^(-4)k^(-1))

$4$ ) $\frac{3 h^{2} j^{0} k^{0}}{3 h j^{-4} k^{-1}}$

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Multiplication with rational exponents

Full solution

Q.

4

)

\frac{3 h^{2} j^{0} k^{0}}{3 h j^{-4} k^{-1}}

Identify and Apply Exponent Properties: Simplify the expression by identifying and applying the properties of exponents. $\newline$ We have the expression $(3h^{2}j^{0}k^{0})/(3hj^{-4}k^{-1})$ . We know that any number raised to the power of $0$ is $1$ , so $j^{0}$ and $k^{0}$ both equal $1$ .
Apply Simplification: Apply the simplification from Step $1$ to the expression. $\newline$ The expression becomes $(3h^{2}\cdot 1\cdot 1)/(3h\cdot 1^{-4}\cdot k^{-1})$ , which simplifies to $(3h^{2})/(3h\cdot k^{-1})$ .
Simplify Coefficients and Terms: Simplify the coefficients and the $h$ terms. $\newline$ We can cancel out the $3$ in the numerator and the denominator, and apply the property $h^{a}/h^{b} = h^{(a-b)}$ to the $h$ terms. $\newline$ This gives us $h^{(2-1)}/k^{(-1)}$ , which simplifies to $h^{1}/k^{(-1)}$ .
Simplify Negative Exponents: Simplify the expression with negative exponents. $\newline$ We know that $a^{-n} = \frac{1}{a^n}$ , so we can rewrite $k^{-1}$ as $\frac{1}{k^1}$ . $\newline$ The expression now becomes $\frac{h}{k}.$
Check for Further Simplifications: Check for any further simplifications. The expression $\frac{h}{k}$ is already in its simplest form, so no further simplification is needed.

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