To get the final price, you use \(\displaystyle x\) as the variable for the original price. So,

\(\displaystyle 0.8x = \$ $88\), and divide by 0.8 on both sides of the equations, and you get

\(\displaystyle \$110\) for the original price of the shirt.

Look at the work rates as "tanks per minute", not "minutes per tank".

The two pipes can fill the tank up at \(\displaystyle \frac{1}{45}\) tanks per minute and  \(\displaystyle \frac{1}{40}\) tanks per minute. 

Let \(\displaystyle t\) be the time it took, in minutes, to fill the tank up. Then, since rate multiplied by time is equal to work, then two pipes filled up \(\displaystyle \frac{1}{45} t\) and \(\displaystyle \frac{1}{40} t\) tanks; together, they filled up \(\displaystyle \frac{1}{45} t + \frac{1}{40} t\) tank - one tank. This sets up the equation to be solved:

\(\displaystyle \frac{1}{45} t + \frac{1}{40} t = 1\)

\(\displaystyle 360 \cdot \left ( \frac{1}{45} t + \frac{1}{40} t \right ) = 360 \cdot 1\)

\(\displaystyle \frac{360}{45} t + \frac{360}{40} t = 360\)

\(\displaystyle 8 t +9 t = 360\)

\(\displaystyle 17 t = 360\)

\(\displaystyle t = 360 \div 17 \approx 21.2\) minutes

Look at the work rates as "tanks per minute", not "minutes per tank".

The inlet pipe can fill the tank at a rate of \(\displaystyle \frac{1}{25}\) tank per minute. The drain is emptying the tank at a rate of \(\displaystyle \frac{1}{55}\) tank per minute.

Let \(\displaystyle t\) be the time it took, in minutes, to fill the tank up. Then, since rate multiplied by time is equal to work, then the inlet pipe let in \(\displaystyle \frac{1}{25} t\) tank of water, but the drain let out \(\displaystyle \frac{1}{55} t\) tank. Since they were working against each other, the work of the latter is subtracted from that of the former, and we can set up and solve the equation:

\(\displaystyle \frac{1}{25} t -\frac{1}{55} t = 1\)

\(\displaystyle 275 \cdot \left ( \frac{1}{25} t -\frac{1}{55} t \right ) = 275 \cdot 1\)

\(\displaystyle \frac{275 }{25} t -\frac{275 }{55} t = 275\)

\(\displaystyle 11 t -5 t = 275\)

\(\displaystyle 6t = 275\)

\(\displaystyle t = 275 \div 6 \approx 45.8\) minutes

The total of the measures of the angles of a pentagon is

\(\displaystyle 180 \cdot (5-2) = 540^{\circ }\).

One angle measures \(\displaystyle t ^{\circ }\); since the other four angles have the same measure as one another, we let \(\displaystyle X^{\circ }\) be their common measure. Then we set up the equation and solve for \(\displaystyle X\):

\(\displaystyle 4X + t = 540\)

\(\displaystyle 4X + t - t = 540- t\)

\(\displaystyle 4X= 540- t\)

\(\displaystyle 4X\div 4 = \left ( 540- t \right )\div 4\)

\(\displaystyle X =135 - \frac{1}{4} t\)

Let Julie's score be \(\displaystyle x\). Then John's score was \(\displaystyle x + 16\). Since the sum of their scores was 80,

\(\displaystyle x + x + 16 = 80\)

\(\displaystyle 2x+16 = 80\)

\(\displaystyle 2x+16- 16 = 80 - 16\)

\(\displaystyle 2x = 64\)

\(\displaystyle 2x \div 2 = 64 \div 2\)

\(\displaystyle x = 32\)

Julie scored 32, and John scored 16 higher, or 48.

Let David's score be \(\displaystyle y\). Then Dana's score was \(\displaystyle y + 10\). Since the sum of their scores was 60,

\(\displaystyle y + y + 10 = 60\)

\(\displaystyle 2y + 10 = 60\)

\(\displaystyle 2y + 10 - 10 = 60 - 10\)

\(\displaystyle 2y = 50\)

\(\displaystyle 2y\div 2 = 50 \div 2\)

\(\displaystyle y = 25\)

David's score was 25, and Dana's score was 10 higher, or 35.

In descending order of score, the students were John, Dana, Julie, David.

Since the mean of the four numbers \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\), and \(\displaystyle d\) is 125,

\(\displaystyle \frac{a + b + c + d}{4} = 125\)

\(\displaystyle \frac{a + b + c + d}{4} \cdot 4 = 125\cdot 4\)

\(\displaystyle a + b + c + d = 500\)

Similarly, 

\(\displaystyle \frac{ c + d+e+f}{4} = 150\)

\(\displaystyle \frac{ c + d+e+f}{4}\cdot 4 = 150\cdot 4\)

\(\displaystyle c + d+e+f = 600\)

Add the two sums:

\(\displaystyle a + b + c + d + c + d + e + f= 500 + 600\)

\(\displaystyle a + b + c + d + e + f+ c + d= 1,100\)

The mean of \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\), \(\displaystyle d\), \(\displaystyle e\), and \(\displaystyle f\) is \(\displaystyle M\), so 

\(\displaystyle \frac{a + b + c + d + e + f }{6}= M\)

\(\displaystyle \frac{a + b + c + d + e + f }{6}\cdot 6= M \cdot 6\)

\(\displaystyle a + b + c + d + e + f= 6M\)

So:

\(\displaystyle a + b + c + d + e + f+ c + d= 1,100\)

\(\displaystyle 6M+ c + d= 1,100\)

\(\displaystyle 6M+ c + d - 6M= 1,100- 6M\)

\(\displaystyle c + d = 1,100- 6M\)

\(\displaystyle \sqrt{x + 5}\) is the square root of \(\displaystyle x + 5\), which in turn can be written as "the sum of a number and five"; \(\displaystyle \sqrt{x + 5}\) can subsequently be written as "the square root of the sum of a number and five". Since \(\displaystyle \sqrt{x + 5} = x + 6\),  \(\displaystyle \sqrt{x + 5}\) is six greater than \(\displaystyle x\), the number, so the equation can be stated as "The square root of the sum of a number and five is six greater than the number."

The expression \(\displaystyle 5x\) can be written as "the product of five and a number".

The expression \(\displaystyle x -7\) can be written as " the difference of the number and seven"; the expression \(\displaystyle 3 (x-7)\) can be written as "three times the difference of the number and seven". 

Therefore, the equation

\(\displaystyle 5x = 3 (x - 7)\)

can be written as 

"The product of five and a number is equal to three times the difference of the number and seven".

If we let \(\displaystyle Y\) be Yancy's xcore, then, since Zack outscored Yancy by 22, Zack scored \(\displaystyle Y + 22\). Velma outscored Zack by 12, so Velma scored \(\displaystyle Y + 22 + 12 = Y + 34\).

The sum of the four scores is the team score, or 391. Since Wendell scored 98, we can add the four expressions to get 391, then solve for \(\displaystyle Y\):

\(\displaystyle Y + (Y+22)+ (Y + 34) + 98 = 391\)

\(\displaystyle 3Y +154 = 391\)

\(\displaystyle 3Y +154- 154 = 391 - 154\)

\(\displaystyle 3Y = 237\)

\(\displaystyle 3Y\div 3 = 237 \div 3\)

\(\displaystyle Y = 79\)

Yancy scored 79; Zack scored 22 more than Yancy, or 101; Velma scored 12 more than Zack, or 113. Wendell scored 98, so the correct ordering, greatest to least, is: Velma, Zack, Wendell, Yancy.

Call the fifth number \(\displaystyle N\). If the average of the five numbers is \(\displaystyle t\), then \(\displaystyle t\) is the sum of the numbers divided by five:

\(\displaystyle \left (67+ 83+ 72+ 80 +N \right ) \div 5 = t\)

Solve for \(\displaystyle N\) in the equation:

\(\displaystyle \left (302+N \right ) \div 5 = t\)

\(\displaystyle \left (302+N \right ) \div 5 \times 5 = t \times 5\)

\(\displaystyle 302+N = 5 t\)

\(\displaystyle 302+N-302 = 5 t -302\)

\(\displaystyle N= 5 t -302\)

The fifth number is \(\displaystyle 5 t -302\).