Let's go through each question:
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*Which of the following is an equivalent expression to \(5^5^{-13}\) with only positive exponents?
To find an equivalent expression, we use the property \(a^m \cdot a^n = a^{m+n}\). Thus: \[ 5^1 \cdot 5^{-13} = 5^{1 - 13} = 5^{-12} \] For positive exponents, we rewrite it as: \[ 5^{-12} = \frac{1}{5^{12}} \]
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Which property of exponents was used to generate the equivalent expression \(3^{14}\) from \(\frac{3^5}{3^{-9}}\)?
We use the property of exponents which states that: \[ \frac{a^m}{a^n} = a^{m-n} \] Therefore, applying that: \[ \frac{3^5}{3^{-9}} = 3^{5 - (-9)} = 3^{5 + 9} = 3^{14} \]
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Which of the following is an equivalent expression to \(\frac{15^0 \cdot 7^{-2}}{(-4)^{-3}}\) with only positive exponents?
We first simplify: \(15^0 = 1\), so we have: \[ \frac{1 \cdot 7^{-2}}{(-4)^{-3}} = \frac{7^{-2}}{(-4)^{-3}} = 7^{-2} \cdot (-4)^3 \] Now rewrite it with positive exponents. The equivalent expression becomes: \[ 7^{-2} \cdot (-4)^3 = \frac{(-4)^3}{7^2} \]
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Which of the following is the equivalent expression to \(\frac{(15^{-3} \cdot 4^{7})^0}{4^{-3}}\) that has been generated by applying the properties of integer exponents?
Since any expression raised to the power of 0 is equal to 1: \[ (15^{-3} \cdot 4^{7})^0 = 1 \] Therefore: \[ \frac{1}{4^{-3}} = 4^{3} \]
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Which of the following is an equivalent expression to \(\frac{(2^{-3} \cdot 9^{3})^{-4}}{2^{9} \cdot 9^{-10}}\) with only positive exponents?
First, simplify the numerator: \[ (2^{-3} \cdot 9^{3})^{-4} = 2^{12} \cdot 9^{-12} = \frac{2^{12}}{9^{12}} \] Our expression now looks like: \[ \frac{\frac{2^{12}}{9^{12}}}{2^{9} \cdot 9^{-10}} = \frac{2^{12}}{9^{12}} \cdot \frac{1}{2^{9} \cdot 9^{-10}} = \frac{2^{12}}{2^{9} \cdot 9^{12}} \cdot 9^{10} \] This ultimately equals: \[ \frac{2^{12 - 9}}{9^{12 - 10}} = \frac{2^{3}}{9^{2}} = \frac{8}{81} \]
Note that the exact answers to the multiple-choice questions would depend on the specific options available. However, these calculated forms with only positive exponents are general answers.