Question

given the expression \\(\frac{(2^{-3})(2^{5})}{(2^{7})}\\), select the three equivalent numerical expressions. \\(\frac{1}{32}\\) \\(\frac{2^{5}}{2^{3 - 7}}\\) \\((2^{-3 + 5})(2^{-7})\\) 32 \\(\frac{2^{5}}{2^{10}}\\)

Explanation:

Step1: Simplify the original expression

First, use the rule of exponents for multiplication: \(a^m \cdot a^n = a^{m + n}\) for the numerator. So, \((2^{-3})(2^{5}) = 2^{-3 + 5}=2^{2}\). Then the original expression \(\frac{(2^{-3})(2^{5})}{2^{7}}\) becomes \(\frac{2^{2}}{2^{7}}\). Now, use the rule for division of exponents: \(a^m\div a^n=a^{m - n}\), so \(\frac{2^{2}}{2^{7}} = 2^{2-7}=2^{-5}\). And \(2^{-5}=\frac{1}{2^{5}}=\frac{1}{32}\).

Step2: Analyze each option

• Option \(\frac{1}{32}\): As we calculated, \(2^{-5}=\frac{1}{32}\), so this is equivalent.
• Option \(\frac{2^{5}}{2^{3 - 7}}\): Simplify the denominator: \(3 - 7=-4\), so the expression becomes \(\frac{2^{5}}{2^{-4}}\). Using the division rule, \(2^{5-(-4)} = 2^{9}\), which is not equivalent. Wait, maybe I made a mistake. Wait, no, let's re - check the option. Wait, the option is \(\frac{2^{5}}{2^{3 - 7}}\)? Wait, no, maybe it's a typo? Wait, no, let's re - evaluate the original expression again. Wait, the original expression is \(\frac{(2^{-3})(2^{5})}{2^{7}}\). Let's try another approach. \((2^{-3})(2^{5})=2^{-3 + 5}=2^{2}\), so the expression is \(\frac{2^{2}}{2^{7}}=2^{-5}\). Now, let's look at the option \((2^{-3 + 5})(2^{-7})\): First, \(2^{-3+5}=2^{2}\), then multiply by \(2^{-7}\): \(2^{2}\cdot2^{-7}=2^{2-7}=2^{-5}\), which is equivalent. And \(\frac{2^{5}}{2^{10}}\): Using the division rule, \(2^{5-10}=2^{-5}\), which is equivalent. Wait, let's re - check the options:

Wait, the options are: \(\frac{1}{32}\), \(\frac{2^{5}}{2^{3 - 7}}\) (wait, \(3 - 7=-4\), so \(\frac{2^{5}}{2^{-4}}=2^{5+4}=2^{9}\), no), \((2^{-3 + 5})(2^{-7})\), \(32\), \(\frac{2^{5}}{2^{10}}\).

Wait, let's recalculate \((2^{-3 + 5})(2^{-7})\): \(2^{-3+5}=2^{2}\), \(2^{2}\times2^{-7}=2^{2-7}=2^{-5}=\frac{1}{32}\), correct. \(\frac{2^{5}}{2^{10}}=2^{5 - 10}=2^{-5}=\frac{1}{32}\), correct. And \(\frac{1}{32}\) is also correct.

Wait, maybe I misread the second option. Let's re - check the second option: \(\frac{2^{5}}{2^{3 - 7}}\). \(3-7=-4\), so \(\frac{2^{5}}{2^{-4}}=2^{5-(-4)} = 2^{9}\), which is not equivalent. So the three equivalent expressions are \(\frac{1}{32}\), \((2^{-3 + 5})(2^{-7})\), and \(\frac{2^{5}}{2^{10}}\).

Answer:

The three equivalent numerical expressions are \(\frac{1}{32}\), \((2^{-3 + 5})(2^{-7})\), and \(\frac{2^{5}}{2^{10}}\) (i.e., the first, third, and fifth options).