can you tell me what is the irrational square root 28 simplified

The square root of 28 can be simplified as follows:

\[
\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}
\]

So the simplified form of \(\sqrt{28}\) is \(2\sqrt{7}\). Since \(\sqrt{7}\) is an irrational number, \(2\sqrt{7}\) is also irrational.

rewrite the irrational cube root squareroot^3 48 as a product of an integer and another irrational cube root

To rewrite the expression \(\sqrt[3]{48}\) as a product of an integer and another cube root, we start by simplifying \(\sqrt[3]{48}\).

First, factor 48 into its prime factors:

\[
48 = 16 \times 3 = 2^4 \times 3
\]

Next, we can express \(48\) in terms of perfect cubes. The largest perfect cube that divides \(48\) is \(8\) (which is \(2^3\)). Therefore:

\[
48 = 8 \times 6
\]

Now, take the cube root:

\[
\sqrt[3]{48} = \sqrt[3]{8 \times 6} = \sqrt[3]{8} \times \sqrt[3]{6}
\]

Since \(\sqrt[3]{8} = 2\), we can express it as:

\[
\sqrt[3]{48} = 2 \times \sqrt[3]{6}
\]

So the expression \(\sqrt[3]{48}\) can be rewritten as a product of an integer and another irrational cube root as:

\[
\sqrt[3]{48} = 2\sqrt[3]{6}
\]

what is the irrational cube root squareroot^3 -24 rewritten as a product of an integer and another irrational cube root

To rewrite \(\sqrt[3]{-24}\) as a product of an integer and another irrational cube root, we start by factoring \(-24\).

First, note that \(24\) can be factored as:

\[
24 = 8 \times 3
\]

And since \(-24\) can be expressed as:

\[
-24 = -1 \times 24 = -1 \times (8 \times 3)
\]

Next, we take the cube root:

\[
\sqrt[3]{-24} = \sqrt[3]{-1 \times (8 \times 3)} = \sqrt[3]{-1} \times \sqrt[3]{8} \times \sqrt[3]{3}
\]

Since \(\sqrt[3]{-1} = -1\) and \(\sqrt[3]{8} = 2\), we have:

\[
\sqrt[3]{-24} = -1 \times 2 \times \sqrt[3]{3} = -2 \sqrt[3]{3}
\]

Thus, the expression \(\sqrt[3]{-24}\) can be rewritten as:

\[
\sqrt[3]{-24} = -2 \sqrt[3]{3}
\]

This is in the form of an integer and another irrational cube root.

what is the irrational cube root squareroot^3 40 rewritten as a product of an integer and another irrational cube root

To rewrite \(\sqrt[3]{40}\) as a product of an integer and another irrational cube root, we start by factoring \(40\).

First, we can express \(40\) as:

\[
40 = 8 \times 5
\]

Here, \(8\) is a perfect cube, which is \(2^3\). Now we can take the cube root:

\[
\sqrt[3]{40} = \sqrt[3]{8 \times 5} = \sqrt[3]{8} \times \sqrt[3]{5}
\]

Since \(\sqrt[3]{8} = 2\), we find:

\[
\sqrt[3]{40} = 2 \times \sqrt[3]{5}
\]

Therefore, the expression \(\sqrt[3]{40}\) can be rewritten as:

\[
\sqrt[3]{40} = 2 \sqrt[3]{5}
\]

This is in the form of an integer multiplied by another irrational cube root.