Which expression is equivalent to $\frac{\sqrt{10}}{\sqrt[4]{8}}$ ?
A. $\frac{\sqrt[4]{200}}{2}$
B. $\frac{\sqrt[4]{20}}{2}$
C. $\frac{2 \sqrt{5}}{5}$
D. $\frac{100}{8}$
Answer (2)
The expression 4 8 10 simplifies to 2 4 200 , making option A the correct choice. The other options do not match this expression. This shows effective manipulation and simplification of radical expressions.
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Rewrite the expression using exponents: 4 8 10 = 8 4 1 1 0 2 1 .
Simplify the expression by expressing with a common root (fourth root): 8 4 1 1 0 2 1 = 4 8 4 100 .
Further simplification leads to: 4 8 4 100 = 2 4 200 .
The equivalent expression is: 2 4 200 .
Explanation
Understanding the Problem We are asked to find an expression equivalent to 4 8 10 . Let's analyze the given expression and the options to find the equivalent one.
Simplifying the Expression First, let's rewrite the given expression using exponents: 4 8 10 = 8 4 1 1 0 2 1 Now, we can rewrite 8 as 2 3 , so we have: ( 2 3 ) 4 1 1 0 2 1 = 2 4 3 1 0 2 1 To compare this with the given options, let's try to express everything with a common root, such as the fourth root. We can rewrite 1 0 2 1 as ( 1 0 2 ) 4 1 = 10 0 4 1 . So the expression becomes: 2 4 3 10 0 4 1 = 4 2 3 4 100 = 4 8 4 100 = 4 8 100 = 4 2 25 = 4 2 × 8 25 × 8 = 4 16 200 = 4 16 4 200 = 2 4 200
Comparing with Options Now let's compare our simplified expression with the given options: Option 1: 2 4 200 . This matches our simplified expression. Option 2: 2 4 20 . This does not match. Option 3: 5 2 5 . This does not match. Option 4: 8 100 . This is a simple fraction, equal to 12.5, and does not match.
Final Answer Therefore, the expression equivalent to 4 8 10 is 2 4 200 .
Examples
Understanding how to simplify and manipulate radical expressions is useful in various fields, such as engineering and physics, where you often encounter complex formulas involving roots and exponents. For example, when calculating the impedance of an AC circuit, you might need to simplify expressions involving square roots and fourth roots to find the overall resistance. This skill also helps in computer graphics when dealing with transformations and scaling of images, where you need to efficiently compute and simplify expressions involving radicals.
