\(\displaystyle \sqrt{3}(3\sqrt{2}+\sqrt{3})\)

Order of operations, first distributing the \(\displaystyle \sqrt{3}\) to all terms inside the parentheses. 

\(\displaystyle (1\cdot 3\sqrt{3}\cdot \sqrt{2})+(\sqrt{3}\cdot\sqrt{3})\)

\(\displaystyle 3\sqrt{3\cdot 2}+\sqrt{3\cdot 3}\rightarrow3\sqrt{6}+\sqrt{9}\rightarrow3\sqrt{6}+3\)

The final answer is \(\displaystyle 3\sqrt{6}+3\).

To square a number is to multiply that number by itself. Because 6 x 6 = 36 AND -6 x -6 = 36, both 6 and -6 are square roots of 36.

Multiplication of square roots is easy! You just have to multiply their contents by each other. Just don't forget to put the result "under" a square root! Therefore:

\(\displaystyle \sqrt{4}*\sqrt{12}*\sqrt{5}\)

becomes

\(\displaystyle \sqrt{4*12*5}=\sqrt{240}\)

Now, you need to simplify this:

\(\displaystyle \sqrt{240} = \sqrt{2*2*2*2*3*5}\)

You can "pull out" two \(\displaystyle 2\)s.  (Note, that it would be even easier to do this problem if you factor immediately instead of finding out that \(\displaystyle 4*12*5 = 240\).)

After pulling out the \(\displaystyle 2\)s, you get:

\(\displaystyle \sqrt{2*2*2*2*3*5} = 2*2(\sqrt{3*5}) = 4\sqrt{15}\)

\(\displaystyle x\sqrt{45}+x\sqrt{72}=\sqrt{18}\)

Notice how all of the quantities in square roots are divisible by 9

\(\displaystyle x\sqrt{9\times 5}+x\sqrt{9\times 8}=\sqrt{9\times 2}\)

\(\displaystyle x\sqrt{9}\sqrt{5}+x\sqrt{9}\sqrt{4\times 2}=\sqrt{9}\sqrt{2}\)

\(\displaystyle 3x\sqrt{5}+3x\sqrt{4}\sqrt{2}=3\sqrt{2}\)

\(\displaystyle 3x\sqrt{5}+6x\sqrt{2}=3\sqrt{2}\)

\(\displaystyle x(3\sqrt{5}+6\sqrt{2})=3\sqrt{2}\)

\(\displaystyle x=\frac{3\sqrt{2}}{3\sqrt{5}+6\sqrt{2}}\)

Simplifying, this becomes

\(\displaystyle x=\frac{\sqrt{2}}{\sqrt{5}+2\sqrt{2}}\)

1. 2² = 4. Also, following the rules of exponents, 4¹ = 1. 
2. One can therefore say that m = 1 and n = 2.
3. The question asks to solve for m/n. Since m = 1 and n = 2, m/n = 1/2.

\(\displaystyle \sqrt{320}=\sqrt{2\times 160}=\sqrt{2\times 2\times 80}=2\sqrt{80}=2\sqrt{2\times 40}=2\sqrt{2\times 2\times 20}=2\times 2\times \sqrt{20}=4\sqrt{2\times 10}=4\sqrt{2\times 2\times 5}=4\times 2\times \sqrt{5}=8\sqrt{5}\)

\(\displaystyle \sqrt{\frac{16}{121}}\)

Take the square root of both the top and bottom terms.

\(\displaystyle \frac{\sqrt{16}}{\sqrt{121}}\)

Simplify. 

\(\displaystyle \frac{4}{11}\)

The easiest way to narrow down a square root from a list is to look at the last number on the squared number – in this case 4 – and compare it to the last number of the answer. 

70 * 70 will equal XXX0

71 * 71 will equal XXX1

72 * 72  will equal XXX4

73 * 73 will equal XXX9

74 * 74  will equal XXX(1)6

Therefore 72 is the answer.  Check by multiplying it out.

\(\displaystyle 5(5^{x})=5^{10}\)

\(\displaystyle 5^{1}(5^{x})=5^{10}\)

\(\displaystyle 5^{x+1}=5^{10}\)

\(\displaystyle x=9\)

In order for the original statement to be true, the \(\displaystyle xy\) and \(\displaystyle y^{2}\) terms must be either both odd or both even. Looking at each of the statements individually,

I. States that \(\displaystyle 3y\) is odd, but only odd values multiplied by 3 are odd. If \(\displaystyle y\) was an even number, the result would be even. But \(\displaystyle y\) can be either odd or even, depending on what \(\displaystyle x\) equals. Thus this statement COULD be true but does not HAVE to be true.

II. States that \(\displaystyle \frac{y}{2}\) is an integer, and since only even numbers are cleanly divided by 2 (odd values result in a fraction) this ensures that \(\displaystyle y\) is even. However, \(\displaystyle y\) can also be odd, so this is a statement that COULD be true but does not HAVE to be true.

III. For exponents, only the base value determines whether it is even or odd - it does not indicate the value of y at all. Only even numbers raised to any power are even, thus, this ensures that \(\displaystyle x\) is even. But \(\displaystyle x\) can be odd as well, so this statement COULD be true but does not HAVE to be true.

An example of two integers that will work violate conditions II and III is \(\displaystyle x=1\) and \(\displaystyle y=3\).

\(\displaystyle (1)\cdot (3)+(3)^{2}=3+9=12\), and even number.

\(\displaystyle \frac{3}{2}\) is not an integer.

\(\displaystyle 1^{3}=1\) is not even.

Furthermore, any combination of 2 even integers will make the original statement true, and violate the Statement I.