\(\displaystyle \frac{1}{3}\) of the movies are action movies. \(\displaystyle \frac{3}{5}\) of \(\displaystyle \frac{2}{3}\) of the movies are comedies, or \(\displaystyle \frac{3}{5}*\frac{2}{3}\), or \(\displaystyle \frac{6}{15}\). Combining the comedies and the action movies (\(\displaystyle \frac{1}{3}\) or \(\displaystyle \frac{5}{15}\)), we get \(\displaystyle \frac{11}{15}\) of the movies being either action or comedy. Thus, \(\displaystyle \frac{4}{15}\) of the movies remain and all of them have to be historical.

Substitute the values of \(\displaystyle x\) and \(\displaystyle y\) into the given expression:

\(\displaystyle 2(\frac{1}{3})+3\left ( \frac{1}{2} \right )\)

\(\displaystyle =\frac{2}{3}+\frac{3}{2}\)

\(\displaystyle =\frac{4}{6}+\frac{9}{6}\)

\(\displaystyle =\frac{13}{6}\)

In order to find the sum of the first five terms, we will need to find the values of each of the first five terms using the equation given above. Essentially, we will let \(\displaystyle n\) range from \(\displaystyle 1\) to \(\displaystyle 5\) to determine each term.

\(\displaystyle a_{1}=\frac{\left ( -1 \right )^{1-1}}{1}=\frac{\left ( -1 \right )^{0}}{1}=\frac{1}{1}=1\)

Remember that anything to the power of zero is equal to 1. Therefore, \(\displaystyle -1\) raised to the zero power is also \(\displaystyle 1\).

\(\displaystyle a_{2}=\frac{\left ( -1 \right )^{2-1}}{2}=\frac{\left ( -1 \right )^{1}}{2}=-\frac{1}{2}\)

\(\displaystyle a_{3}=\frac{\left ( -1 \right )^{3-1}}{3}=\frac{\left ( -1 \right )^{2}}{3}=\frac{\left (-1 \right )\left ( -1 \right )}{3}=\frac{1}{3}\)

In general, when \(\displaystyle -1\) is raised to an even power, the result is \(\displaystyle +1\). Conversely, if \(\displaystyle -1\) is raised to an odd power, the result is \(\displaystyle -1\).

\(\displaystyle a_{4}=\frac{\left ( -1 \right )^{4-1}}{4}=\frac{\left ( -1 \right )^{3}}{4}=\frac{\left (-1 \right )\left ( -1 \right )\left ( -1 \right )}{4}=-\frac{1}{4}\)

\(\displaystyle a_{5}=\frac{\left ( -1 \right )^{5-1}}{5}=\frac{\left ( -1 \right )^{4}}{5}=\frac{1}{5}\)

Thus, the first five terms of the sequence are \(\displaystyle 1, -\frac{1}{2},\frac{1}{3},-\frac{1}{4}, and\, \frac{1}{5}\). We must now add these all together. Because we are adding fractions with unlike denominators, we need to find the smallest multiple that \(\displaystyle 2, 3, 4, and\, 5\) have in common. Because \(\displaystyle 4\) is a multiple of \(\displaystyle 2\), we really only need to worry about \(\displaystyle 3, 4, and \, 5\). If we were to list out the multiple of \(\displaystyle 3, 4, and \, 5\), e would see that the smallest one they have in common is \(\displaystyle 60\). Sometimes, the easiest way to find the least common multiple of several numbers is o multiply them together. The products of \(\displaystyle 3, 4, and \, 5\) is \(\displaystyle 60\).

We will now convert each fraction to an equivalent form with a denominator of \(\displaystyle 60\). For example, if we were to convert \(\displaystyle \frac{1}{2}\) to a fraction with a denominator of \(\displaystyle 60\), we would multiple both the numerator and denominator by \(\displaystyle 30\), as shown below:

\(\displaystyle \frac{1}{2}=\frac{\left ( 1 \right )\left ( 30 \right )}{\left ( 2 \right )\left ( 30 \right )}=\frac{30}{60}\)

\(\displaystyle 1+\frac{1}{2}+\frac{1}{3}+-\frac{1}{4}+\frac{1}{5}=\frac{60}{60}-\frac{30}{60}+\frac{20}{60}-\frac{15}{60}+\frac{12}{60}=\frac{60-30+20-15+12}{60}=\frac{47}{60}\)

The answer is \(\displaystyle \frac{47}{60}\).

\(\displaystyle x=\frac{7}{9}+\frac{3}{5}+\frac{2}{15}+\frac{7}{45}=\frac{35}{45}+\frac{27}{45}+\frac{6}{45}+\frac{7}{45}=\frac{75}{45}=\frac{5}{3}\)

Let us first get our value for the percentage of the first fraction. \(\displaystyle 20%\) of \(\displaystyle \frac{2}{7}\)  is found by multiplying \(\displaystyle \frac{2}{7}\) by \(\displaystyle \frac{2}{10}\) (or, simplified, \(\displaystyle \frac{1}{5}\)): \(\displaystyle \frac{2}{7}*\frac{1}{5}=\frac{2}{35}\)

Our addition is therefore \(\displaystyle \frac{2}{35}+\frac{1}{4}\). There are no common factors, so the least common denominator will be \(\displaystyle 35*4\) or \(\displaystyle 140\). Multiply the numerator and denominator of \(\displaystyle \frac{2}{35}\) by \(\displaystyle \frac{4}{4}\) and the numerator of \(\displaystyle \frac{1}{4}\) by \(\displaystyle \frac{35}{35}\)

This yields:

\(\displaystyle \frac{8}{140}+\frac{35}{140}=\frac{43}{140}\), which cannot be reduced.

Find the least common denominator to solve this problem

Multiply 27 with \(\displaystyle x\), and multiply \(\displaystyle 9x\) with 3 to obtain common denominators.

Convert the fractions.

\(\displaystyle \frac{2}{9x}+\frac{4}{27}=\frac{6}{27x}+\frac{4x}{27x}\)

Combine the terms as one fraction.

The answer is:  \(\displaystyle \frac{4x+6}{27x}\)

First we must find what fraction of hip-hop records are there in the store. We find a common denominator for the faction of pop records and rock records by multiplying \(\displaystyle \frac{1}{4}\) by \(\displaystyle \frac{7}{7}\) and \(\displaystyle \frac{3}{7}\) by \(\displaystyle \frac{4}{4}\) to get \(\displaystyle \frac{7}{28}\) records being pop records and \(\displaystyle \frac{12}{28}\) records being rock records. 

In order to get the faction of hip-hop records remaining in the store, we must write out:

\(\displaystyle \frac{7}{28}+\frac{12}{28}+\frac{x}{28}=\frac{28}{28}\) since the sum total of the fractions must add up to \(\displaystyle 1\). We solve to get \(\displaystyle x=9\).

This tells us that fraction of hip-hop records must be \(\displaystyle \frac{9}{28}\).

The question is asking how many records are stocked in the store, therefore we need a denominator of \(\displaystyle 364\). We can write out \(\displaystyle \frac{9}{28}=\frac{h}{364}\) and solve for \(\displaystyle h=117\).

Similarly, this question can be solved using deduction. If \(\displaystyle \frac{1}{4}\) of \(\displaystyle 364\) is \(\displaystyle 91\) and \(\displaystyle \frac{3}{7}\) of \(\displaystyle 364\) is \(\displaystyle 156\), knowing that we need the amount remaining, we can subtract \(\displaystyle 91+156\) from \(\displaystyle 364\) and get \(\displaystyle 117\).

Answer choice \(\displaystyle 91\) is wrong because it is finding the total amount of Pop records.

Answer choice \(\displaystyle 156\) is wrong because it is finding the total amount of rock records.

Answer choice \(\displaystyle 156\) is wrong because the factions were added incorrectly \(\displaystyle \frac{1}{4}+\frac{3}{7}\neq \frac{4}{11}\).

First, we must convert the song length from minutes and seconds to simply seconds.
\(\displaystyle \frac{2\, minutes}{x\, seconds}=\frac{1\, minute}{60\, seconds}\).  We add \(\displaystyle 26\) seconds to \(\displaystyle 120\) seconds and get \(\displaystyle 146\) seconds total song length. 

If the song is equal parts pre-chorus, chorus, and bridge that means \(\displaystyle \frac{1}{3}\) of the song is composed of pre-chorus, \(\displaystyle \frac{1}{3}\) of the song is composed of chorus, and \(\displaystyle \frac{1}{3}\) of the song is composed of bridges. Therefore, we divide \(\displaystyle 146\) by \(\displaystyle 3\) and get \(\displaystyle 48.67\) seconds of chorus.
Since a beat occurs every five seconds, we divide \(\displaystyle 48.67\) seconds by \(\displaystyle 5\) seconds to get the amount of beats in that time frame. \(\displaystyle \frac{48.67\, seconds}{5\, seconds}=9.73\, beats\). Since a beat can not be sectioned, the answer is rounded down to \(\displaystyle 9\).

Answer choice \(\displaystyle 29\) is incorrect because it is finding the total number of beats in the song.

Answer choice \(\displaystyle 10\) is incorrect because it rounds up to the nearest beat.

The rule for dividing fractions is invert the second fraction and multiply.  We keep the \(\displaystyle \frac{a}{b}\) as it is then change the division sign to a multiplication sign and invert \(\displaystyle \frac{c}{d}\) into \(\displaystyle \frac{d}{c}\).
\(\displaystyle \frac{a}{b}\div\frac{c}{d}\) becomes \(\displaystyle \frac{a}{b}\times \frac{d}{c}=\frac{ad}{bc}\).

Similarly, you can memorize, “top, bottom, bottom, top” meaning \(\displaystyle a\) is on top and it stays on top, \(\displaystyle b\) is labeled bottom and it stays on the bottom, \(\displaystyle c\) is labeled bottom and it moved to the bottom, then \(\displaystyle d\) is labeled top so it moves to the top.

To find the answer we must cross multiply. When we cross multiply we get:

\(\displaystyle x\times 15=9\times (x+4)\) or \(\displaystyle 15x=9x+36\)

We subtract \(\displaystyle 9x\) from \(\displaystyle 15x\)

\(\displaystyle 6x=36\)Then we solve for x by dividing by 6,

\(\displaystyle x=6\)