7 answers
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.
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:
\[ \sqrt[3]{48} = 2\sqrt[3]{6} \]
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.
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.
