QUESTION IMAGE
Question
algebra 1
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name
product and quotient rules exit ticket
simplify. your answer should contain only positive exponents.
- ( 3b^2 cdot 3b^2 cdot 2b )
- ( y^2 cdot 2x^3 y^3 )
- ( \frac{3x}{x^5} )
- ( \frac{2yx^3}{3x^2 y^3} )
Problem 1: \( 3b^2 \cdot 3b^2 \cdot 2b \)
Step 1: Multiply coefficients
Multiply the numerical coefficients: \( 3 \times 3 \times 2 = 18 \)
Step 2: Apply exponent rule for multiplication (\( a^m \cdot a^n = a^{m + n} \))
For the variable \( b \): \( b^2 \cdot b^2 \cdot b = b^{2 + 2 + 1} = b^5 \)
Step 3: Combine results
Multiply the coefficient and the variable term: \( 18 \times b^5 = 18b^5 \)
Step 1: Multiply coefficients
The coefficient of \( y^2 \) is 1, and the coefficient of \( 2x^3y^3 \) is 2. So, \( 1 \times 2 = 2 \)
Step 2: Apply exponent rule for multiplication on \( y \) terms
\( y^2 \cdot y^3 = y^{2 + 3} = y^5 \)
Step 3: The \( x \) term remains as \( x^3 \) (since there's only one \( x \) term)
Step 4: Combine all terms
Multiply the coefficient, \( x \) term, and \( y \) term: \( 2 \times x^3 \times y^5 = 2x^3y^5 \)
Step 1: Apply exponent rule for division (\( \frac{a^m}{a^n} = a^{m - n} \))
For the variable \( x \): \( \frac{x}{x^5} = x^{1 - 5} = x^{-4} \)
Step 2: Rewrite with positive exponent (\( a^{-n} = \frac{1}{a^n} \)), but we can also simplify the coefficient and the \( x \) term directly
Alternatively, we can think of \( \frac{3x}{x^5} = 3 \times \frac{x}{x^5} = 3 \times x^{1 - 5} = 3x^{-4} \), and then rewrite as \( \frac{3}{x^4} \) (since \( x^{-4} = \frac{1}{x^4} \))
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\( 18b^5 \)