Question

1. \frac{}{3a^{-3}b^{-4}}\
2. \frac{4x^{-2}y^{0}\cdot x^{4}y^{4}}{x^{-1}\cdot y}\
3. \frac{3vu^{2}}{3vu^{0}\cdot 3u}\
4. \frac{3x\cdot 3y^{3}}{2x^{-2}y^{2}}\
5. \frac{2y^{0}}{(x^{4}y^{4})^{4}\cdot 2x}\
6. \frac{(2n^{-3})^{2}}{2m^{0}n^{4}\cdot nm^{3}}

Explanation:

Let's solve problem 7: \(\frac{4x^{-2}y^{0} \cdot x^{4}y^{4}}{x^{-1} \cdot y}\)

Step 1: Simplify exponents for \(x\) and \(y\) in the numerator

Recall that \(y^0 = 1\) and when multiplying variables with exponents, we add the exponents (\(a^m \cdot a^n = a^{m + n}\)).
For \(x\) in the numerator: \(x^{-2} \cdot x^{4}=x^{-2 + 4}=x^{2}\)
For \(y\) in the numerator: \(y^{0} \cdot y^{4}=y^{0 + 4}=y^{4}\)
So the numerator becomes \(4 \cdot x^{2} \cdot y^{4}\) (since \(y^0 = 1\), it doesn't change the product)

Step 2: Simplify the denominator

The denominator is \(x^{-1} \cdot y\), which remains as is for now.

Step 3: Divide the numerator by the denominator

When dividing variables with exponents, we subtract the exponents (\(a^m \div a^n=a^{m - n}\))
For \(x\): \(x^{2}\div x^{-1}=x^{2-(-1)} = x^{2 + 1}=x^{3}\)
For \(y\): \(y^{4}\div y = y^{4 - 1}=y^{3}\)
The coefficient is \(4\) (since there is no coefficient in the denominator to divide with, we just keep it)

Putting it all together, we have \(4x^{3}y^{3}\)

Answer:

\(4x^{3}y^{3}\)