The algebraic expression for \(\displaystyle x\) being Jamie's youngest brother's age is:

\(\displaystyle 3x+x+(x+2)=27\)

\(\displaystyle 5x+2=27\)

\(\displaystyle 5x=25\)

\(\displaystyle x=5\)

Jamie's youngest brother is five, the next oldest brother is seven, and Jamie is 15.

Given the graph of record sales, to find the fraction of records that were sold in 2004 to 2010 first identify the record sales in 2004 and the record sales in 2010.

Examining the graph,

Record sales in 2004:14 million

Record sales in 2010: 13 million

From here, to find the fraction of records sold during this time period, use the following formula.

\(\displaystyle \text{Fraction of Sales}=\frac{\text{Beginning Year Record Sales}}{\text{Ending Year Record Sales}}\)

\(\displaystyle \text{Fraction}=\frac{14}{13}\)

Fred has $100 in quarters and nickels initially.  We are also told that he has 260 quarters.  This is worth $65.  Thus Fred initially has $35 in nickels or 700 nickels.

Fred now exchanges some of his nickels for the dimes of a friend.  He ends up with 650 coins.  We know that Fred started with 960 coins (700 nickels + 260 quarters).  He ends up with 650 coins.  The number of quarters remains unchanged, meaning he now has 390 nickels and dimes.  These must have the same value as the initial 700 nickels, though, since he didn't lose any money.

Now we can finally set up our solution:

\(\displaystyle 700 * 0.05= 35 = x*0.05 + (390-x)*0.1 = .05x +39 -.1x = 39 - .05x\)

\(\displaystyle 35 = 39-.05x \Rightarrow .05x = 4 \Rightarrow x = 80\)

Thus Fred has 80 nickels and 310 dimes.

\(\displaystyle \line(1,0){250}\)

An alternative solution step is to notice that turning nickels into dimes always occurs in exactly one way: 2 nickels to 1 dime.  Every time you do this conversion, you will lose exactly one coin.  We then notice that the number of coins drops from 960 to 650, or drops by 310 coins.  We thus need to get rid of 310 coins.  Since we're only allowed to change nickels into dimes (and lose 1 coin each time), we simply do this 310 times to reach the requisite number of coin losses.  We are left with the proper number of coins with the proper value immediately.  Since every replacement replaced 2 nickels, we also lost \(\displaystyle 310\cdot 2=620\) nickels.  Our final number of nickels is thus \(\displaystyle 700-620=80\) nickels.

In order to find the volume of the figure, we will first need to find the volume of both cylinders.

Recall how to find the volume of the cylinder:

\(\displaystyle \text{Volume of Cylinder}=\pi\times\text{radius}^2\times\text{height}\)

Now, use the given radius and height to find the volume of the larger cylinder.

\(\displaystyle \text{Volume of Larger Cylinder}=\pi\times 8^2 \times 10=640\pi\)

Next, use the given radius and height to find the volume of the smaller cylinder.

\(\displaystyle \text{Volume of Smaller Cylinder}=\pi\times 2^2 \times 10=40\pi\)

Subtract the volume of the smaller cylinder from the volume of the larger one to find the volume of the figure.

\(\displaystyle \text{Volume of Figure}=640\pi-40\pi=600\pi=1884.96\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

Since both fractions in the expression have a common denominator of \(\displaystyle x^{2}\), we can combine like terms into a single numerator over the denominator:

\(\displaystyle \frac{7x-18}{x^{2}}+\frac{6x-14}{x^{2}}\)

\(\displaystyle =\frac{(7x-18)+(6x-14)}{x^{2}}\)

\(\displaystyle =\frac{13x-32}{x^{2}}\)

This problem requires knowledge of the compound interest formula, 

\(\displaystyle A=P(1+\frac{r}{n})^{nt}\)

Where \(\displaystyle A\) is the amount of money owed, \(\displaystyle P\) is the sum borrowed, \(\displaystyle r\) is the yearly interest rate, \(\displaystyle n\) is the amount of times the interest is compounded per year, and \(\displaystyle t\) is the number of years.  

We know that the student borrowed $200,000 compounded annually at a 6% interest rate, therefore by plugging in those numbers we find that after she graduates in 4 years she will owe $252,495.

\(\displaystyle \\A=\$200,000(1+\frac{0.06}{1})^{4} \\A=\$252,495\)

This problem requires knowledge of the compound interest formula, 

\(\displaystyle A=P(1+\frac{r}{n})^{nt}\)

Where \(\displaystyle A\) is the amount of money owed, \(\displaystyle P\) is the sum borrowed, \(\displaystyle r\) is the yearly interest rate, \(\displaystyle n\) is the amount of times the interest is compounded per year, and \(\displaystyle t\) is the number of years.  

We know that the student borrowed $200,000 compounded quarterly at a 6% interest rate, therefore by plugging in those numbers we find what she will owe after she graduates in 4 years.

\(\displaystyle \\P=\$200,00 \\r=6\%=0.06 \\n=4 \\t=4\)

\(\displaystyle \\A=\$200,000(1+\frac{0.06}{4})^{4\cdot 4} \\A=\$253,797\)

The area of \(\displaystyle A\) is \(\displaystyle n\), the area of \(\displaystyle B\) is \(\displaystyle 5.0625n\). Therefore, you can fit 5.06 \(\displaystyle A's\) in \(\displaystyle B\).

Alvin runs at a rate of seven miles an hour for twelve minutes, or \(\displaystyle \frac{12}{60} = \frac{1}{5}\) hours. The distance he runs is equal to his rate multiplied by his time, so, setting\(\displaystyle r = 7 , t = \frac{1}{5}\) in this formula:

\(\displaystyle d = rt\)

\(\displaystyle d = 7 \cdot \frac{1}{5} = \frac{7}{5}\) miles.

One mile comprises 5,280 feet, so this is equal to 

\(\displaystyle \frac{7}{5} \cdot 5,280 = 7,392\) feet

Since each side of the track measures 264 feet, this means that Alvin runs 

\(\displaystyle 7,392 \div 264 = 28\) sidelengths.

\(\displaystyle 28 \div 6 = 4 \textrm{ R }4\),

which means that Alvin runs around the track four complete times, plus four more sides of the track. Alvin stops when he is at Point E.

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