11) $\frac{(2m^2)^{-1}}{m^2} =$
12) $\frac{2x^3}{(x^{-1})^3}$
13) $(a^{-3}b^{-3})^0$
14) $x^4y^3 cdot (2y^2)^0$
15) $ba^4 cdot (2ba^4)^{-3}$
16) $(2x^0y^2)^{-3} cdot 2yx^3$
17) $\frac{2k^3 cdot k^2}{k^{-3}}$
18) $\frac{(x^{-3})^4x^4}{2x^{-3}}$
19) $\frac{(2x)^{-4}}{x^{-1} cdot x}$
20) $\frac{(2x^3z^2)^3}{x^3y^4z^2 cdot x^{-4}z^3}$

Question

Accepted Answer

### Problem 11: $\boldsymbol{\frac{(2m^2)^{-1}}{m^2}}$
# Explanation:
## Step 1: Apply negative exponent rule
Recall that $(ab)^n = a^n b^n$ and $(a^m)^n = a^{mn}$, also $a^{-n}=\frac{1}{a^n}$. So, $(2m^2)^{-1}=2^{-1}(m^2)^{-1}=\frac{1}{2}m^{-2}$.
Now the expression becomes $\frac{\frac{1}{2}m^{-2}}{m^2}$.

## Step 2: Divide with same base (subtract exponents)
When dividing with the same base $a^m\div a^n = a^{m - n}$. Here, for the $m$ terms: $m^{-2}\div m^2=m^{-2 - 2}=m^{-4}$. And we still have the $\frac{1}{2}$ factor.
So, $\frac{1}{2}m^{-4}=\frac{1}{2m^4}$.

# Answer:
$\frac{1}{2m^4}$

### Problem 12: $\boldsymbol{\frac{2x^3}{(x^{-1})^3}}$
# Explanation:
## Step 1: Simplify the denominator
Using the power of a power rule $(a^m)^n=a^{mn}$, so $(x^{-1})^3 = x^{-1	imes3}=x^{-3}$.
The expression is now $\frac{2x^3}{x^{-3}}$.

## Step 2: Divide with same base (add exponents)
$a^m\div a^n=a^{m - n}$, so $x^3\div x^{-3}=x^{3-(-3)}=x^{6}$. Multiply by the coefficient 2.
So, $2x^{6}$.

# Answer:
$2x^6$

### Problem 13: $\boldsymbol{(a^{-3}b^{-3})^0}$
# Explanation:
## Step 1: Apply zero exponent rule
Any non - zero number (or non - zero expression) raised to the power of 0 is 1. Since $a^{-3}b^{-3}=\frac{1}{a^3b^3}$ (assuming $a
eq0$ and $b
eq0$), then $(a^{-3}b^{-3})^0 = 1$.

# Answer:
$1$

### Problem 14: $\boldsymbol{x^4y^3\cdot(2y^2)^0}$
# Explanation:
## Step 1: Simplify the term with zero exponent
Using the zero exponent rule, $(2y^2)^0 = 1$ (as long as $y
eq0$).
## Step 2: Multiply the remaining terms
So, $x^4y^3\cdot1=x^4y^3$.

# Answer:
$x^4y^3$

### Problem 15: $\boldsymbol{ba^4\cdot(2ba^4)^{-3}}$
# Explanation:
## Step 1: Apply negative exponent rule to the second term
Using $(ab)^n=a^n b^n$ and $(a^m)^n=a^{mn}$ and $a^{-n}=\frac{1}{a^n}$, we have $(2ba^4)^{-3}=2^{-3}b^{-3}(a^4)^{-3}=\frac{1}{8}b^{-3}a^{-12}$.
## Step 2: Multiply the two terms
Now, $ba^4\cdot\frac{1}{8}b^{-3}a^{-12}=\frac{1}{8}b^{1+( - 3)}a^{4+( - 12)}$.
## Step 3: Simplify the exponents
For $b$: $1-3=-2$, for $a$: $4 - 12=-8$. So the expression is $\frac{1}{8}b^{-2}a^{-8}=\frac{1}{8a^8b^2}$.

# Answer:
$\frac{1}{8a^8b^2}$

### Problem 16: $\boldsymbol{(2x^0y^2)^{-3}\cdot2yx^3}$
# Explanation:
## Step 1: Simplify $x^0$
Since $x^0 = 1$ (for $x
eq0$), the first term becomes $(2	imes1	imes y^2)^{-3}=(2y^2)^{-3}$.
## Step 2: Apply negative exponent rule to $(2y^2)^{-3}$
Using $(ab)^n=a^n b^n$ and $a^{-n}=\frac{1}{a^n}$, we get $2^{-3}(y^2)^{-3}=\frac{1}{8}y^{-6}$.
## Step 3: Multiply with the second term
Now, $\frac{1}{8}y^{-6}\cdot2yx^3$. Multiply the coefficients: $\frac{1}{8}	imes2=\frac{1}{4}$. Multiply the $y$ terms: $y^{-6}\cdot y=y^{-6 + 1}=y^{-5}$. And we have the $x^3$ term.
So, $\frac{1}{4}x^3y^{-5}=\frac{x^3}{4y^5}$.

# Answer:
$\frac{x^3}{4y^5}$

### Problem 17: $\boldsymbol{\frac{2k^3\cdot k^2}{k^{-3}}}$
# Explanation:
## Step 1: Multiply the numerator terms
Using $a^m\cdot a^n=a^{m + n}$, $k^3\cdot k^2=k^{3 + 2}=k^5$. So the numerator is $2k^5$.
## Step 2: Divide by $k^{-3}$
Using $a^m\div a^n=a^{m - n}$, $2k^5\div k^{-3}=2k^{5-(-3)}=2k^{8}$.

# Answer:
$2k^8$

### Problem 18: $\boldsymbol{\frac{(x^{-3})^4x^4}{2x^{-3}}}$
# Explanation:
## Step 1: Simplify $(x^{-3})^4$
Using $(a^m)^n=a^{mn}$, $(x^{-3})^4=x^{-12}$.
## Step 2: Multiply with $x^4$ in the numerator
Using $a^m\cdot a^n=a^{m + n}$, $x^{-12}\cdot x^4=x^{-12 + 4}=x^{-8}$. So the numerator is $x^{-8}$.
## Step 3: Divide by $2x^{-3}$
Using $\frac{a^m}{a^n}=a^{m - n}$, $\frac{x^{-8}}{2x^{-3}}=\frac{1}{2}x^{-8-(-3)}=\frac{1}{2}x^{-5}=\frac{1}{2x^5}$.

# Answer:
$\frac{1}{2x^5}$

### Problem 19: $\boldsymbol{\frac{(2x)^{-4}}{x^{-1}\cdot x}}$
# Explanation:
## Step 1: Simplify the denominator
Using $a^m\cdot a^n=a^{m + n}$, $x^{-1}\cdot x=x^{-1 + 1}=x^0 = 1$ (for $x
eq0$).
## Step 2: Simplify the numerator
Using $(ab)^n=a^n b^n$, $(2x)^{-4}=2^{-4}x^{-4}=\frac{1}{16}x^{-4}$.
## Step 3: Divide numerator by denominator
Since the denominator is 1, the expression is $\frac{1}{16}x^{-4}=\frac{1}{16x^4}$.

# Answer:
$\frac{1}{16x^4}$

### Problem 20: $\boldsymbol{\frac{(2x^3z^2)^3}{x^3y^4z^2\cdot x^{-4}z^3}}$
# Explanation:
## Step 1: Simplify the numerator
Using $(ab)^n=a^n b^n$ and $(a^m)^n=a^{mn}$, $(2x^3z^2)^3=2^3(x^3)^3(z^2)^3 = 8x^9z^6$.
## Step 2: Simplify the denominator
Using $a^m\cdot a^n=a^{m + n}$, for $x$ terms: $x^3\cdot x^{-4}=x^{3-4}=x^{-1}$. For $z$ terms: $z^2\cdot z^3=z^{2 + 3}=z^5$. The $y$ term remains $y^4$. So the denominator is $x^{-1}y^4z^5$.
## Step 3: Divide numerator by denominator
$\frac{8x^9z^6}{x^{-1}y^4z^5}$. For $x$ terms: $x^9\div x^{-1}=x^{9-(-1)}=x^{10}$. For $z$ terms: $z^6\div z^5=z$. And we have the coefficient 8 and the $y^{-4}$ term (since we divide by $y^4$). So the expression is $8x^{10}z y^{-4}=\frac{8x^{10}z}{y^4}$.

# Answer:
$\frac{8x^{10}z}{y^4}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 11: $\boldsymbol{\frac{(2m^2)^{-1}}{m^2}}$

Step 1: Apply negative exponent rule

Step 2: Divide with same base (subtract exponents)

Step 1: Simplify the denominator

Step 2: Divide with same base (add exponents)

Step 1: Apply zero exponent rule

Answer:

Problem 12: $\boldsymbol{\frac{2x^3}{(x^{-1})^3}}$