Use the square of a sum pattern, substituting \(\displaystyle 3.7x\) for \(\displaystyle A\) and \(\displaystyle 1.4 y\) for \(\displaystyle B\) in the pattern:

\(\displaystyle (A+ B)^{2}= A^{2}+ 2AB + B^{2}\)

\(\displaystyle \left (3.7x + 1.4 y \right )^{2}= (3.7x)^{2}+ 2 (3.7x)(1.4 y) + (1.4 y) ^{2}\)

\(\displaystyle = 13.69x^{2}+ 10.36x y +1.96 y ^{2}\)

Simplify the base of 9 and 27 in order to have a common base.

(3^x)(9)=27²

= (3^x)(3²)=(3³)²

=(3^x+2)=3⁶

Therefore:

x+2=6

x=4

The terms of \(\displaystyle 4x^{4}+ 36 x^{2}\) have \(\displaystyle 4x^{2}\) as their greatest common factor, so

\(\displaystyle 4x^{4}+ 36 x^{2} = 4x^{2} (x^{2}+9)\)

\(\displaystyle x^{2}+ 9\) is a prime polynomial. 

Of the five choices, only \(\displaystyle x^{2}+ 9\) is a factor.

The easiest way to approach this problem is to break everything into exponents. \(\displaystyle 8^{3}\) is equal to \(\displaystyle 2^{9}\) and 27 is equal to \(\displaystyle 3^{3}\). Therefore, the expression can be broken down into \(\displaystyle \frac{2^{9} \times 3}{2^{6}\times 3^{3}}\). When you cancel out all the terms, you get \(\displaystyle \frac{2^{3}}{3^{2}}\), which equals \(\displaystyle \frac{8}{9}\).

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

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:

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

To find an equivalency we must rationalize the denominator.

To rationalize the denominator multiply the numerator and denominator by the denominator.

\(\displaystyle \frac{12}{\sqrt{18}}*\frac{\sqrt{18}}{\sqrt{18}}=\) 

\(\displaystyle \frac{12*\sqrt{18}}{18}=\)

Factor out 6,

 \(\displaystyle \frac{2*\sqrt{18}}{3}=\)

Extract perfect square 9 from the square root of 18.

\(\displaystyle \sqrt{18}=3*\sqrt{2}\)

\(\displaystyle \frac{2*3*\sqrt{2}}{3}=\) 

\(\displaystyle 2*\sqrt{2}\)

A complex number is a combination of a real and imaginary number. To subtract complex numbers, subtract each element separately.

In equation \(\displaystyle a\), \(\displaystyle 3\) is the real component and \(\displaystyle 1\) is the imaginary component (designated by \(\displaystyle i\)). In equation \(\displaystyle b\), \(\displaystyle 4\) is the real component and \(\displaystyle -2\) is the imaginary component. Solving for \(\displaystyle b - a\),

\(\displaystyle b-a = \left ( 4-2i\right ) - \left ( 3 + i\right ) = \left ( 4-3\right ) + i\left ( -2-1\right ) = 1 - 3i\)

When you have an exponent on the outside of parentheses while another is on the inside of the parentheses, such as in \(\displaystyle (3^{6})^{2}\), multiply the exponents together to get the answer: \(\displaystyle 3^{12}\).

This is different than when you have two numbers with the same base multiplied together, such as in \(\displaystyle x^{2} \cdot x^{3} = x^{5}\). In that case, you add the exponents together.