column 1
1. $2x^{5}y \cdot 3x^{2}y$
2. $(4x^{4}y)^{2} \cdot 2x^{3}y^{4}$
3. $\frac{36x^{9}y^{4}}{4x^{7}y^{3}}$
4. $\frac{(2xy^{5})^{3}}{2x^{3}y^{8}}$
5. $(-5x^{6}y^{2})^{2} - 12x^{12}y^{4}$
6. $\frac{6x^{10}y^{4}}{3x^{5}y^{7}}$
7. $6x^{-1}y^{5} \cdot 4x^{-4}y^{-2}$
8. $(3x^{-6}y^{2})^{3} \cdot 2x^{10}y^{-7}$
9. $\frac{(-2xy)^{2} \cdot 10x^{3}y^{11}}{8x^{10}y^{4}}$
10. $\frac{8x^{3} \cdot 12xy^{7}}{3x^{2}y^{4}} - 15x^{2}y^{3}$

Question

Accepted Answer

### Problem 1: $ 2x^5y \cdot 3x^2y $
# Explanation:
## Step 1: Multiply coefficients
Multiply the coefficients $ 2 $ and $ 3 $.
$ 2 \times 3 = 6 $
## Step 2: Multiply $ x $-terms
Use the rule $ a^m \cdot a^n = a^{m + n} $ for $ x^5 $ and $ x^2 $.
$ x^5 \cdot x^2 = x^{5 + 2} = x^7 $
## Step 3: Multiply $ y $-terms
Use the rule $ a^m \cdot a^n = a^{m + n} $ for $ y $ (which is $ y^1 $) and $ y $ (which is $ y^1 $).
$ y \cdot y = y^{1 + 1} = y^2 $
## Step 4: Combine results
Multiply the results from steps 1, 2, and 3.
$ 6 \times x^7 \times y^2 = 6x^7y^2 $
# Answer:
$ 6x^7y^2 $

### Problem 2: $ (4x^4y)^2 \cdot 2x^3y^4 $
# Explanation:
## Step 1: Simplify the power of a product
Use the rule $ (ab)^n = a^n b^n $ and $ (a^m)^n = a^{mn} $ for $ (4x^4y)^2 $.
$ (4x^4y)^2 = 4^2 \cdot (x^4)^2 \cdot y^2 = 16x^8y^2 $
## Step 2: Multiply coefficients
Multiply the coefficients $ 16 $ and $ 2 $.
$ 16 \times 2 = 32 $
## Step 3: Multiply $ x $-terms
Use the rule $ a^m \cdot a^n = a^{m + n} $ for $ x^8 $ and $ x^3 $.
$ x^8 \cdot x^3 = x^{8 + 3} = x^{11} $
## Step 4: Multiply $ y $-terms
Use the rule $ a^m \cdot a^n = a^{m + n} $ for $ y^2 $ and $ y^4 $.
$ y^2 \cdot y^4 = y^{2 + 4} = y^6 $
## Step 5: Combine results
Multiply the results from steps 2, 3, and 4.
$ 32 \times x^{11} \times y^6 = 32x^{11}y^6 $
# Answer:
$ 32x^{11}y^6 $

### Problem 3: $ \frac{36x^9y^4}{4x^7y^3} $
# Explanation:
## Step 1: Divide coefficients
Divide the coefficient $ 36 $ by $ 4 $.
$ \frac{36}{4} = 9 $
## Step 2: Divide $ x $-terms
Use the rule $ \frac{a^m}{a^n} = a^{m - n} $ for $ x^9 $ and $ x^7 $.
$ \frac{x^9}{x^7} = x^{9 - 7} = x^2 $
## Step 3: Divide $ y $-terms
Use the rule $ \frac{a^m}{a^n} = a^{m - n} $ for $ y^4 $ and $ y^3 $.
$ \frac{y^4}{y^3} = y^{4 - 3} = y $
## Step 4: Combine results
Multiply the results from steps 1, 2, and 3.
$ 9 \times x^2 \times y = 9x^2y $
# Answer:
$ 9x^2y $

### Problem 4: $ \frac{(2xy^5)^3}{2x^3y^8} $
# Explanation:
## Step 1: Simplify the power of a product
Use the rule $ (ab)^n = a^n b^n $ and $ (a^m)^n = a^{mn} $ for $ (2xy^5)^3 $.
$ (2xy^5)^3 = 2^3 \cdot x^3 \cdot (y^5)^3 = 8x^3y^{15} $
## Step 2: Divide coefficients
Divide the coefficient $ 8 $ by $ 2 $.
$ \frac{8}{2} = 4 $
## Step 3: Divide $ x $-terms
Use the rule $ \frac{a^m}{a^n} = a^{m - n} $ for $ x^3 $ and $ x^3 $.
$ \frac{x^3}{x^3} = x^{3 - 3} = x^0 = 1 $ (we can ignore the $ x $-term since it's $ 1 $)
## Step 4: Divide $ y $-terms
Use the rule $ \frac{a^m}{a^n} = a^{m - n} $ for $ y^{15} $ and $ y^8 $.
$ \frac{y^{15}}{y^8} = y^{15 - 8} = y^7 $
## Step 5: Combine results
Multiply the results from steps 2, 3, and 4.
$ 4 \times 1 \times y^7 = 4y^7 $
# Answer:
$ 4y^7 $

### Problem 5: $ (-5x^6y^2)^2 - 12x^{12}y^4 $
# Explanation:
## Step 1: Simplify the power of a product
Use the rule $ (ab)^n = a^n b^n $ and $ (a^m)^n = a^{mn} $ for $ (-5x^6y^2)^2 $.
$ (-5x^6y^2)^2 = (-5)^2 \cdot (x^6)^2 \cdot (y^2)^2 = 25x^{12}y^4 $
## Step 2: Subtract the terms
Subtract $ 12x^{12}y^4 $ from $ 25x^{12}y^4 $.
$ 25x^{12}y^4 - 12x^{12}y^4 = (25 - 12)x^{12}y^4 = 13x^{12}y^4 $
# Answer:
$ 13x^{12}y^4 $

### Problem 6: $ \frac{6x^{10}y^4}{3x^5y^7} $
# Explanation:
## Step 1: Divide coefficients
Divide the coefficient $ 6 $ by $ 3 $.
$ \frac{6}{3} = 2 $
## Step 2: Divide $ x $-terms
Use the rule $ \frac{a^m}{a^n} = a^{m - n} $ for $ x^{10} $ and $ x^5 $.
$ \frac{x^{10}}{x^5} = x^{10 - 5} = x^5 $
## Step 3: Divide $ y $-terms
Use the rule $ \frac{a^m}{a^n} = a^{m - n} $ for $ y^4 $ and $ y^7 $.
$ \frac{y^4}{y^7} = y^{4 - 7} = y^{-3} = \frac{1}{y^3} $ (we can write it with positive exponent or keep as $ y^{-3} $, but usually positive is preferred for simplicity here, but let's follow the rule)
Wait, actually, let's do it step by step:
$ \frac{y^4}{y^7} = y^{4 - 7} = y^{-3} $, but we can also express it as $ \frac{1}{y^3} $, but let's combine with the other terms.
## Step 4: Combine results
Multiply the results from steps 1, 2, and 3.
$ 2 \times x^5 \times y^{-3} = \frac{2x^5}{y^3} $ or $ 2x^5y^{-3} $. But let's check the exponents again. Wait, $ 4 - 7 = -3 $, so $ y^{-3} $ is correct. But maybe the problem expects positive exponents, so $ \frac{2x^5}{y^3} $.
Wait, let's re-express:
$ \frac{6x^{10}y^4}{3x^5y^7} = \frac{6}{3} \times \frac{x^{10}}{x^5} \times \frac{y^4}{y^7} = 2 \times x^{5} \times y^{-3} = \frac{2x^5}{y^3} $
# Answer:
$ \frac{2x^5}{y^3} $ (or $ 2x^5y^{-3} $)

### Problem 7: $ 6x^{-1}y^5 \cdot 4x^{-4}y^{-2} $
# Explanation:
## Step 1: Multiply coefficients
Multiply the coefficients $ 6 $ and $ 4 $.
$ 6 \times 4 = 24 $
## Step 2: Multiply $ x $-terms
Use the rule $ a^m \cdot a^n = a^{m + n} $ for $ x^{-1} $ and $ x^{-4} $.
$ x^{-1} \cdot x^{-4} = x^{-1 + (-4)} = x^{-5} = \frac{1}{x^5} $
## Step 3: Multiply $ y $-terms
Use the rule $ a^m \cdot a^n = a^{m + n} $ for $ y^5 $ and $ y^{-2} $.
$ y^5 \cdot y^{-2} = y^{5 + (-2)} = y^3 $
## Step 4: Combine results
Multiply the results from steps 1, 2, and 3.
$ 24 \times \frac{1}{x^5} \times y^3 = \frac{24y^3}{x^5} $ or $ 24x^{-5}y^3 $
# Answer:
$ \frac{24y^3}{x^5} $ (or $ 24x^{-5}y^3 $)

### Problem 8: $ (3x^{-6}y^2)^3 \cdot 2x^{10}y^{-7} $
# Explanation:
## Step 1: Simplify the power of a product
Use the rule $ (ab)^n = a^n b^n $ and $ (a^m)^n = a^{mn} $ for $ (3x^{-6}y^2)^3 $.
$ (3x^{-6}y^2)^3 = 3^3 \cdot (x^{-6})^3 \cdot (y^2)^3 = 27x^{-18}y^6 $
## Step 2: Multiply coefficients
Multiply the coefficients $ 27 $ and $ 2 $.
$ 27 \times 2 = 54 $
## Step 3: Multiply $ x $-terms
Use the rule $ a^m \cdot a^n = a^{m + n} $ for $ x^{-18} $ and $ x^{10} $.
$ x^{-18} \cdot x^{10} = x^{-18 + 10} = x^{-8} = \frac{1}{x^8} $
## Step 4: Multiply $ y $-terms
Use the rule $ a^m \cdot a^n = a^{m + n} $ for $ y^6 $ and $ y^{-7} $.
$ y^6 \cdot y^{-7} = y^{6 + (-7)} = y^{-1} = \frac{1}{y} $
## Step 5: Combine results
Multiply the results from steps 2, 3, and 4.
$ 54 \times \frac{1}{x^8} \times \frac{1}{y} = \frac{54}{x^8y} $ or $ 54x^{-8}y^{-1} $
# Answer:
$ \frac{54}{x^8y} $ (or $ 54x^{-8}y^{-1} $)

### Problem 9: $ \frac{(-2xy)^2 \cdot 10x^3y^{11}}{8x^{10}y^4} $
# Explanation:
## Step 1: Simplify $ (-2xy)^2 $
Use the rule $ (ab)^n = a^n b^n $ and $ (a^m)^n = a^{mn} $ for $ (-2xy)^2 $.
$ (-2xy)^2 = (-2)^2 \cdot x^2 \cdot y^2 = 4x^2y^2 $
## Step 2: Multiply numerator terms (coefficients and variables)
Multiply $ 4x^2y^2 $ and $ 10x^3y^{11} $.
First, multiply coefficients: $ 4 \times 10 = 40 $
Multiply $ x $-terms: $ x^2 \cdot x^3 = x^{2 + 3} = x^5 $
Multiply $ y $-terms: $ y^2 \cdot y^{11} = y^{2 + 11} = y^{13} $
So numerator becomes $ 40x^5y^{13} $
## Step 3: Divide by denominator
Now divide $ 40x^5y^{13} $ by $ 8x^{10}y^4 $
Divide coefficients: $ \frac{40}{8} = 5 $
Divide $ x $-terms: $ \frac{x^5}{x^{10}} = x^{5 - 10} = x^{-5} = \frac{1}{x^5} $
Divide $ y $-terms: $ \frac{y^{13}}{y^4} = y^{13 - 4} = y^9 $
## Step 4: Combine results
Multiply the results from step 3: $ 5 \times \frac{1}{x^5} \times y^9 = \frac{5y^9}{x^5} $ or $ 5x^{-5}y^9 $
# Answer:
$ \frac{5y^9}{x^5} $ (or $ 5x^{-5}y^9 $)

### Problem 10: $ \frac{8x^3 \cdot 12xy^2}{3x^2y^4} - 15x^2y^3 $
# Explanation:
## Step 1: Simplify the numerator of the fraction
Multiply $ 8x^3 $ and $ 12xy^2 $
Multiply coefficients: $ 8 \times 12 = 96 $
Multiply $ x $-terms: $ x^3 \cdot x = x^{3 + 1} = x^4 $
Multiply $ y $-terms: $ y^2 $ (since the first term has no $ y $ except in the second term, wait, $ 8x^3 $ has no $ y $, $ 12xy^2 $ has $ y^2 $, so $ y^0 \cdot y^2 = y^2 $
So numerator is $ 96x^4y^2 $
## Step 2: Divide the fraction
Divide $ 96x^4y^2 $ by $ 3x^2y^4 $
Divide coefficients: $ \frac{96}{3} = 32 $
Divide $ x $-terms: $ \frac{x^4}{x^2} = x^{4 - 2} = x^2 $
Divide $ y $-terms: $ \frac{y^2}{y^4} = y^{2 - 4} = y^{-2} = \frac{1}{y^2} $
So the fraction becomes $ 32x^2y^{-2} = \frac{32x^2}{y^2} $
## Step 3: Subtract the terms
Subtract $ 15x^2y^3 $ from $ \frac{32x^2}{y^2} $. Wait, this seems inconsistent. Wait, maybe I made a mistake. Wait, let's

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QUESTION IMAGE

Question

Explanation:

Problem 1: \( 2x^5y \cdot 3x^2y \)

Step 1: Multiply coefficients

Step 2: Multiply \( x \)-terms

Step 3: Multiply \( y \)-terms

Step 4: Combine results

Step 1: Simplify the power of a product

Step 2: Multiply coefficients

Step 3: Multiply \( x \)-terms

Step 4: Multiply \( y \)-terms

Step 5: Combine results

Step 1: Divide coefficients

Step 2: Divide \( x \)-terms

Step 3: Divide \( y \)-terms

Step 4: Combine results

Answer:

Problem 2: \( (4x^4y)^2 \cdot 2x^3y^4 \)