If you don't already have the pattern memorized, use FOIL. It's best to write out the parentheses twice (as below) to avoid mistakes:

$\displaystyle (3n+5)^2$

$\displaystyle (3n+5)(3n+5)$

$\displaystyle 9n^2+15n+15n+25$

$\displaystyle 9n^2+30n+25$

x² = 36 -> it is important to remember that this leads to two answers. 

x = 6 or x = -6. 

  If x = 6: A = B.

  If x = -6: A < B.

Thus the relationship cannot be determined from the information given.

We can use Heron's formula to find the area of the triangle. We can let a = 6, b = 10, and c = 12.

In order to find s, we need to find one half of the perimeter. The perimeter is the sum of the lengths of the sides of the triangle.

Perimeter = a + b + c = 6 + 10 + 12 = 28

In order to find s, we must multiply the perimeter by one-half, which would give us (1/2)(28), or 14.

Now that we have a, b, c, and s, we can calculate the area using Heron's formula. 

$\displaystyle \sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\displaystyle \sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\displaystyle \sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$\displaystyle -4x^{2}y^4$

Use the distributive property for radicals. 

$\displaystyle \sqrt{5}(2\sqrt{3}+\sqrt{12})$

Multiply all terms by $\displaystyle \sqrt{5}$.

$\displaystyle (\sqrt{5}*2\sqrt{3})+(\sqrt{5}*\sqrt{12})$

Combine terms under radicals.

$\displaystyle 2\sqrt{3*5}+\sqrt{12*5}$

$\displaystyle 2\sqrt{15}+\sqrt{60}$

Look for perfect square factors under each radical. $\displaystyle 60$ has a perfect square of $\displaystyle 4$. The $\displaystyle 4$ can be factored out.

$\displaystyle 2\sqrt{15}+\sqrt{4*15}$

$\displaystyle 2\sqrt{15}+2\sqrt{15}$

Since both radicals are the same, we can add them.

$\displaystyle 4\sqrt{15}$

$\displaystyle \sqrt{(45)(9)+(27)(36)}$

When simplifying a square root, consider the factors of each of its component parts:

$\displaystyle \sqrt{(3^2)\times5\times(3^{2})+(3^{3})(2^2)(3^2)}$

Combine like terms:

$\displaystyle \sqrt{5(3^4)+(2^2)(3^5)}$

Remove the common factor, $\displaystyle 3^4$:

$\displaystyle \sqrt{(3^4)\times(5+3\times(2^2))}$

Pull the $\displaystyle \sqrt{3^4}$ outside of the equation as $\displaystyle 3^2$:

$\displaystyle (3^2)\sqrt{5+12}=9\sqrt{17}$                       

$\displaystyle \sqrt{(16)(8)+(32)(20)}$

First, break down the components of the square root:

$\displaystyle \sqrt{(2^{4})(2^{3})+(2^{5})(2^{2})\times 5}$

Combine like terms. Remember, when multiplying exponents, add them together:

$\displaystyle \sqrt{(2^{7})+(2^{7})\times5}$

Factor out the common factor of $\displaystyle 2^7$:

$\displaystyle \sqrt{(2^{7})(1+5)}$

$\displaystyle \sqrt{(2^7)\times6}$

Factor the $\displaystyle 6$:

$\displaystyle \sqrt{(2^7)\times2\times3}$

Combine the factored $\displaystyle 2$ with the $\displaystyle 2^7$:

$\displaystyle \sqrt{(2^{8})\times3}$

Now, you can pull $\displaystyle \sqrt{2^8}$ out from underneath the square root sign as $\displaystyle 2^4$:

$\displaystyle 2^4\sqrt{3}$

$\displaystyle \sqrt{(27)(45)(125)}$

First, break down the component parts of the square root:

$\displaystyle \sqrt{(3^{3})(5\times 3^{2})(5^{3})}$

Combine like terms in a way that will let you pull some of them out from underneath the square root symbol:

$\displaystyle \sqrt{(5^{4})(3^4)(3)}$

Pull out the terms with even exponents and simplify:

$\displaystyle (5^{2})(3^{2})\sqrt{3}=225\sqrt{3}$

The complex conjugate of a complex number $\displaystyle a+bi$ is $\displaystyle a - bi$, so $\displaystyle \frac{3}{2} + \frac{1}{4} i$ has $\displaystyle \frac{3}{2} - \frac{1}{4} i$ as its complex conjugate. Subtract the latter from the former:

$\displaystyle \left (\frac{3}{2} + \frac{1}{4} i \right ) - \left ( \frac{3}{2} - \frac{1}{4} i \right )$

$\displaystyle = \frac{3}{2} + \frac{1}{4} i - \frac{3}{2}+ \frac{1}{4} i$

$\displaystyle = \frac{3}{2}- \frac{3}{2} + \frac{1}{4} i + \frac{1}{4} i$

$\displaystyle = \frac{1}{2} i$

The complex conjugate of a complex number $\displaystyle a - bi$ is $\displaystyle a+ bi$. Therefore, the complex conjugate of $\displaystyle 17 - i \sqrt{17}$ is $\displaystyle 17 + i \sqrt{17}$; subtract the latter from the former by subtracting real parts and subtracting imaginary parts, as follows:

$\displaystyle (17 - i \sqrt{17})- (17 + i \sqrt{17})$

$\displaystyle = 17 - i \sqrt{17} - 17 - i \sqrt{17}$

$\displaystyle = 17 - 17 - i \sqrt{17} - i \sqrt{17}$

$\displaystyle = - 2 i \sqrt{17}$