Question

1. \\(\frac{3x cdot 3y^3}{2x^{-2}y^2}\\)\
2. \\(\frac{}{4x}\\)\
3. \\(\frac{2y^0}{(x^4y^4)^4 cdot 2x}\\)\
4. \\(\frac{(2}{2^n}\\)\
5. \\(\frac{(2n^{-3})^2}{0^4}\\)\

\\(\frac{2mn cdot nm^3}{}\\)\

1. \\(\frac{2}{}\\)

Explanation:

Let's solve problem (11) first: $\frac{3x \cdot 3y^3}{2x^{-2}y^2}$

Step 1: Simplify the numerator and use exponent rules

First, multiply the coefficients and use the rule $a^m \cdot a^n = a^{m + n}$ for the variables. The numerator is $3x \cdot 3y^3 = 9xy^3$. The denominator is $2x^{-2}y^2$.

Step 2: Use the rule $\frac{a^m}{a^n}=a^{m - n}$ for $x$ and $y$

For $x$: $\frac{x}{x^{-2}} = x^{1 - (-2)} = x^{3}$. For $y$: $\frac{y^3}{y^2} = y^{3 - 2} = y^{1}=y$. The coefficient is $\frac{9}{2}$.

Step 3: Combine the results

Multiply the coefficient, $x$ term, and $y$ term together: $\frac{9}{2}x^{3}y$

Now problem (13): $\frac{2y^0}{(x^4y^4)^4 \cdot 2x}$

Step 1: Simplify $y^0$ and the power of a product

Recall that $a^0 = 1$ (for $a
eq0$), so $y^0 = 1$. Then, $(x^4y^4)^4 = x^{4\times4}y^{4\times4}=x^{16}y^{16}$ by the rule $(ab)^n = a^n b^n$ and $(a^m)^n = a^{mn}$.

Step 2: Simplify the denominator

The denominator is $x^{16}y^{16} \cdot 2x = 2x^{16 + 1}y^{16}=2x^{17}y^{16}$ (using $a^m \cdot a^n = a^{m + n}$). The numerator is $2\times1 = 2$.

Step 3: Simplify the fraction

$\frac{2}{2x^{17}y^{16}}=\frac{1}{x^{17}y^{16}}$ (cancel the 2s)

Problem (15): $\frac{(2n^{-3})^2}{2m^0n^4 \cdot mn^3}$

Step 1: Simplify the numerator and $m^0$

First, $(2n^{-3})^2 = 2^2n^{-3\times2}=4n^{-6}$ (using $(ab)^n = a^n b^n$ and $(a^m)^n = a^{mn}$). Also, $m^0 = 1$.

Step 2: Simplify the denominator

The denominator is $2m^0n^4 \cdot mn^3 = 2\times1\times m^{1}n^{4 + 3}=2mn^7$ (using $a^m \cdot a^n = a^{m + n}$).

Step 3: Simplify the fraction

$\frac{4n^{-6}}{2mn^7}=\frac{4}{2}\cdot\frac{n^{-6}}{mn^7}=2\cdot\frac{1}{mn^{6 + 7}}=\frac{2}{mn^{13}}$ (using $\frac{a^m}{a^n}=a^{m - n}$ and $a^{-n}=\frac{1}{a^n}$)

Answer:

s:
(11) $\boldsymbol{\frac{9}{2}x^{3}y}$
(13) $\boldsymbol{\frac{1}{x^{17}y^{16}}}$
(15) $\boldsymbol{\frac{2}{mn^{13}}}$