Question

Which of the following is equivalent to
4

y
3

y
5
?(1 point)
Responses

y
7
4
y superscript Start Fraction 7 over 4 End Fraction baseline

y
13
4
y superscript Start Fraction 13 over 4 End Fraction baseline

y
14
15
y superscript Start Fraction 14 over 15 End Fraction baseline

1
y
7
4

Answers

There are no human answers yet.
Answered by GPT-4o mini
To find the equivalent expression for \( 4\sqrt{y} \cdot 3\sqrt[3]{y^5} \), we can rewrite the radicals in terms of exponents:

1. \( 4\sqrt{y} = 4y^{1/2} \)
2. \( 3\sqrt[3]{y^5} = 3y^{5/3} \)

Now, we can combine these expressions:

\[
4\sqrt{y} \cdot 3\sqrt[3]{y^5} = (4 \cdot 3) \cdot (y^{1/2} \cdot y^{5/3})
\]

Calculating \( 4 \cdot 3 \) gives:

\[
12
\]

Now for the \( y \) terms, we need to add the exponents:

\[
y^{1/2} \cdot y^{5/3} = y^{1/2 + 5/3}
\]

Finding a common denominator for the fractions \( \frac{1}{2} \) and \( \frac{5}{3} \):

- The common denominator for 2 and 3 is 6.
- Rewrite \( \frac{1}{2} \) as \( \frac{3}{6} \)
- Rewrite \( \frac{5}{3} \) as \( \frac{10}{6} \)

Now we can add the fractions:

\[
\frac{3}{6} + \frac{10}{6} = \frac{13}{6}
\]

So we have:

\[
y^{1/2 + 5/3} = y^{13/6}
\]

Now combining both parts, we have:

\[
4\sqrt{y} \cdot 3\sqrt[3]{y^5} = 12y^{13/6}
\]

Thus the final answer can be simplified further. However, if we only focus on the exponent of \( y \), it is \( \frac{13}{6} \).

None of the provided responses directly match with \( 12y^{13/6} \). If we assume that the coefficient may not be needed and we only need the exponent in terms of equivalency, then

The equivalent in terms of exponent is:

**\( y^{13/6} \)**

However, since it wasn't listed in your options, check which means you likely need to compare other given options.

Of the options you presented, none correctly summarize \( y^{13/6} \) in original context either. Thus tending toward simpler fraction terms indicates \( y^{13/4} \) as that using less complexity.

If further analysis of choices is mandated, please validate specific needs to compare options.

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