When multiplying radicals, just take the values inside the radicand and perfom the operation.

In this case, we have a perfect square so simplify that first.

Then, take that answer and multiply that with $\displaystyle \sqrt{3}$ to get the final answer.

$\displaystyle {\sqrt{9}}\cdot \sqrt{3}=3\cdot \sqrt{3}=3\sqrt{3}$.

When dividing radicals, check the denominator to make sure it can be simplified or that there is a radical present that needs to be fixed. Since there is a radical present, we need to eliminate that radical. To do this, we multiply both top and bottom by $\displaystyle \sqrt{3}$. The reason is because we want a whole number in the denominator and multiplying by itself will achieve that. By multiplying itself, it creates a square number which can be reduced to $\displaystyle 3$.

$\displaystyle \frac{\sqrt{5}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{15}}{\sqrt{9}}=\frac{\sqrt{15}}{3}$

With the denominator being $\displaystyle 3$, the numerator is $\displaystyle \sqrt{5\cdot 3}=\sqrt{15}$. Final answer is $\displaystyle \frac{\sqrt{15}}{3}$.

When dividing radicals, check the denominator to make sure it can be simplified or that there is a radical present that needs to be fixed.

Both $\displaystyle 9$ and $\displaystyle 4$ are perfect squares so they can be simplify.

Final answer is

 $\displaystyle \frac{\sqrt9}{\sqrt4}=\frac{\sqrt{3\cdot 3}}{\sqrt{2\cdot 2}}=\frac{3}{2}$.

When multiplying radicals, just take the values inside the radicand and perfom the operation.

Since there is a number outside of the radicand, multiply the outside numbers and then the radicand.

$\displaystyle 3{\sqrt{2}}\cdot 4\sqrt{3}=3\cdot 4\cdot \sqrt{2\cdot 3}=12\sqrt{6}$

When multiplying radicals, just take the values inside the radicand and perfom the operation.

Since there is a number outside of the radicand, multiply the outside numbers and then the radicand. 

$\displaystyle 3\sqrt{6}\cdot 6\sqrt{10}=3\cdot 6\cdot\sqrt{6\cdot10}=18\sqrt{60}$

Before we say that's the final answer, check the radicand to see that there are no square numbers that can be factored. A $\displaystyle 4$ can be factored and thats a perfect square. When I divide $\displaystyle 60$ with $\displaystyle 4$, I get $\displaystyle 15$ which doesn't have perfect square factors.

Therefore, our answer becomes

$\displaystyle 18{\sqrt{60}}=18\sqrt{4\cdot 15}=18\cdot2\sqrt{15}=36\sqrt{15}$. 

When multiplying radicals, just take the values inside the radicand and perfom the operation.

Since there is a number outside of the radicand, multiply the outside numbers and then the radicand. 

$\displaystyle 2\sqrt{15}\cdot \sqrt{15}=2\cdot\sqrt{15\cdot15}=2\cdot15=30$

Since we are dealing with exponents, lets break it down.

$\displaystyle (4{\sqrt{7}})^2=4^2\cdot ({\sqrt{7}})^2=16\cdot 7=112$

Remember to distribute.

When dividing radicals, check the denominator to make sure it can be simplified or that there is a radical present that needs to be fixed. Since there is a radical present, we need to eliminate that radical. However, there is a faster way to possibly elminate the denominator. Let's simplify the numerator.

$\displaystyle \sqrt{24}=\sqrt{8\cdot3}=\sqrt{4\cdot2\cdot3}=2\sqrt{2}\cdot\sqrt{3}$.

The reason I split it up is because I can cancel out the radicals and thus simplifying the question to give final answer of $\displaystyle 2\sqrt{3}$.

When dividing radicals, check the denominator to make sure it can be simplified or that there is a radical present that needs to be fixed. Since there is a radical present, we need to eliminate that radical. However, there is a faster way to possibly elminate the denominator. Let's simplify the numerator.

$\displaystyle \frac{2\sqrt{75}}{3\sqrt{175}}=\frac{2\cdot\sqrt{25\cdot3}}{3\cdot\sqrt{25\cdot7}}=\frac{2\cdot\sqrt{25}\cdot\sqrt{3}}{3\cdot\sqrt{25}\cdot\sqrt{7}}=\frac{2\sqrt{3}}{3\sqrt{7}}$

We still need to eliminate the radical so multiply top and bottom by $\displaystyle \sqrt{7}$.

$\displaystyle \frac{2\sqrt{3}\cdot\sqrt{7}}{3\sqrt{7}\cdot\sqrt{7}}=\frac{2\sqrt{21}}{3\cdot 7}=\frac{2\sqrt{21}}{21}$.

I would first multiply and put everything under the radical to then figure out what to simplify.

After mulitplying everything together, it should look like:

$\displaystyle \sqrt{36x^5}$.

Then, simplify.

$\displaystyle 36$ is a perfect square, and for every pair of x's, put one on the outside and cross out the pair.

We are left with an answer of:

$\displaystyle 6x^2\sqrt{x}$.