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

Answer 1

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.

Answer 2

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

Answer 3

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:

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

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

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.