Simplify /54.

Note: You will find the option to add a cubic root symbol in the Algebra (X)
keyboard.
(1 point)

11 answers

To simplify \(\sqrt[3]{54}\), we first factor 54 into its prime factors:

\[ 54 = 2 \times 27 = 2 \times 3^3 \]

Now we can express the cube root:

\[ \sqrt[3]{54} = \sqrt[3]{2 \times 3^3} \]

Using the property of cube roots, we can separate the factors:

\[ \sqrt[3]{54} = \sqrt[3]{2} \times \sqrt[3]{3^3} \]

Since the cube root of \(3^3\) is simply 3, we have:

\[ \sqrt[3]{54} = \sqrt[3]{2} \times 3 \]

Thus, the simplified form of \(\sqrt[3]{54}\) is:

\[ 3\sqrt[3]{2} \]

So, \(\sqrt[3]{54}\) simplifies to \(\boxed{3\sqrt[3]{2}}\).

What is the irrational square root V63 rewritten as the product of an integer and another irrational square root, in simplest form? (1 point)

To simplify \(\sqrt{63}\) as the product of an integer and another square root, we first factor 63 into its prime factors:

\[ 63 = 9 \times 7 = 3^2 \times 7 \]

Now we can express the square root:

\[ \sqrt{63} = \sqrt{9 \times 7} \]

Using the property of square roots, we can separate the factors:

\[ \sqrt{63} = \sqrt{9} \times \sqrt{7} \]

Since \(\sqrt{9} = 3\), we have:

\[ \sqrt{63} = 3\sqrt{7} \]

Thus, \(\sqrt{63}\) rewritten as the product of an integer and another square root in simplest form is:

\[ \boxed{3\sqrt{7}} \]

What is the irrational cube root 3/81 rewritten as a product of an integer and another irrational cube root?
Note: You will find the option to add a cubic root symbol in the Algebra (X)
keyboard.
(1 point)

To simplify \(\sqrt[3]{\frac{3}{81}}\) as the product of an integer and another irrational cube root, we start by simplifying the expression:

First, notice that \(81\) can be expressed as \(3^4\):

\[ \frac{3}{81} = \frac{3}{3^4} = \frac{1}{3^3} = 3^{-3} \]

Now, we can find the cube root:

\[ \sqrt[3]{\frac{3}{81}} = \sqrt[3]{3^{-3}} = \sqrt[3]{\frac{1}{27}} = \sqrt[3]{\frac{1}{3^3}} = \sqrt[3]{1} \div \sqrt[3]{27} = 1 \div 3 = \frac{1}{3} = \frac{1}{3} \cdot \sqrt[3]{1} \]

However, let's rewrite \( \frac{3}{81} = \frac{1}{27} = \frac{1}{3^3} \) appropriately in the original format:

\[ \sqrt[3]{\frac{3}{81}} = \frac{1}{3} \cdot \sqrt[3]{3} \]

Thus, the final answer for \(\sqrt[3]{\frac{3}{81}}\) rewritten as a product of an integer and another irrational cube root in simplest form is:

\[ \boxed{\frac{1}{3} \sqrt[3]{3}} \]

Find the square root of 169: 169. (1 point)

—13
• 84.5
• 13, -13
• 13

The square root of 169 is calculated as follows:

\[ \sqrt{169} = 13 \]

However, since the square root of a number can have both a positive and negative value, \(\sqrt{169}\) is typically expressed as:

\[ \pm 13 \]

But among the options given, the values we see are:

  • —13
  • 84.5
  • 13, -13
  • 13

Thus, the correct answer is:

\[ \boxed{13, -13} \]

Find the cube root of -729: 3-729 (1 point)
09
• -9
09, - 9
• 243

To find the cube root of \(-729\), we compute:

\[ \sqrt[3]{-729} \]

We know that \(729 = 9^3\), so:

\[ \sqrt[3]{-729} = \sqrt[3]{-(9^3)} = -9 \]

Therefore, the cube root of \(-729\) is:

\[ \boxed{-9} \]

Among the options provided, the correct choice is \(-9\).

Find the cube root of -8: 3-8 (1 point)

To find the cube root of \(-8\), we compute:

\[ \sqrt[3]{-8} \]

Since \(-8\) can be expressed as \(-2^3\), we have:

\[ \sqrt[3]{-8} = \sqrt[3]{-2^3} = -2 \]

Thus, the cube root of \(-8\) is:

\[ \boxed{-2} \]