\(\displaystyle a \blacklozenge b = \left | 2a- 2b + 5 \right |\)

\(\displaystyle (-4) \blacklozenge (-7) = \left | 2 (-4)- 2(-7) + 5 \right |\)

\(\displaystyle = \left |-8- (-14) + 5 \right |\)

\(\displaystyle = \left |-8+14 + 5 \right |\)

\(\displaystyle = \left |11 \right |\)

\(\displaystyle = 11\)

As is clear from the graph, in the interval between \(\displaystyle -2\) (\(\displaystyle -2\) included) to \(\displaystyle 1\), the \(\displaystyle f(x)\) is constant at \(\displaystyle 1\) and then from \(\displaystyle x=1\) (\(\displaystyle 1\) not included) to \(\displaystyle x=3\) (\(\displaystyle 3\) not included), the \(\displaystyle f(x)\) is a decreasing function.

The formula for inverse variation is as follows: \(\displaystyle y=\frac{k}{x}\)

Use the x and y values from the first part of the sentence to find k. 

\(\displaystyle 14=\frac{k}{3}\)

\(\displaystyle k=42\)

Then use that k value and the given x value to find y.

\(\displaystyle y=\frac{42}{28}\)

\(\displaystyle y=1\frac{1}{2}\)

A line can be named after any two points it passes through. The line \(\displaystyle \overleftrightarrow{CF}\) is indicated in green below.

The line does not pass through \(\displaystyle D\), so \(\displaystyle D\) cannot be part of the name of the line. Specifically, \(\displaystyle \overleftrightarrow{DF}\) is not a valid name.

The break-even point is where the costs equal the revenues.

Let \(\displaystyle x\) = # of frames sold

Costs:  \(\displaystyle C(x) = 10x +350\)

Revenues:  \(\displaystyle R(x) = 35x\)

Thus, \(\displaystyle 10x + 350 = 35x\)

So 14 picture frames must be sold each month to break-even.

Mrs. Smith's 8th grade class had a quiz last week. The graph below depicts the number of questions students got incorrect on their quiz and their corresponding quiz grade. In other words, the graph in this particular question is a dot plot and the question asks to find a correlation if one exists.

Recall that a correlation is a trend seen in the data. Graphically, trends can be either:

I. Positive 

II. Negative

III. Constant

IV. No trend

For a trend to be positive the x and y variable both increase. A trend is negative when the y variable (dependent variable) decreases as the x variable (independent variable) increases. A constant trend occurs when the y variable stays the same as the x variable increases. No trend exists when the data appears to be scattered with no association between the x and y variables.

Examining the graph given it is seen that the x variable is the number of questions missed and the y variable is the overall score on the quiz. It is seen that as the number of questions missed increases, the overall score on the quiz decreases. This describes  a negative trend.

In other words, the graph depicts a negative correlation.

To solve this problem, we need to know the conversions between yards to feet, and feet to inches.  Write their correct conversions.

\(\displaystyle 1 \textup{ yard}= 3\textup{ feet}\)

\(\displaystyle 1\textup{ foot}= 12 \textup{ inches}\)

Convert three yards to feet.

\(\displaystyle 3 \textup{ yards}\left(\frac{3 \textup{ feet}}{1 \textup{ yard}}\right)= 9 \textup{ feet}\)

Convert nine feet to inches.

\(\displaystyle 9 \textup{ feet}\left(\frac{12 \textup{ inches}}{1 \textup{ foot}}\right)= 108\textup{ inches}\)

Convert \(\displaystyle 75 ^{\circ } F\) to the Celsius scale by setting \(\displaystyle F = 75\) in the conversion formula and solving for \(\displaystyle C\):

\(\displaystyle 1.8 C + 32 = F\)

\(\displaystyle 1.8 C + 32 = 75\)

\(\displaystyle 1.8 C + 32 - 32 = 75 - 32\)

\(\displaystyle 1.8 C = 43\)

\(\displaystyle 1.8 C \div 1.8 = 43 \div 1.8\)

\(\displaystyle C \approx 23.9\)

The question is therefore asking for the number of days that the temperature topped \(\displaystyle 23.9^{\circ } C\). Examine the graph below:

The high temperature was greater than \(\displaystyle 23.9^{\circ } C\) on Tuesday, Friday, and Saturday - three different days.

\(\displaystyle a \blacktriangledown b= \frac{a+1}{\left | a\right |+ \left | b\right |}\), or, equivalently,

\(\displaystyle a \blacktriangledown b=\left ( a+1 \right ) \div ( | a |+ | b | )\)

\(\displaystyle \frac{4}{5} \blacktriangledown \left (-\frac{4}{5} \right )=\left ( \frac{4}{5}+1 \right ) \div \left (\left| \frac{4}{5} \right |+ \left |- \frac{4}{5}\right | \right )\)

\(\displaystyle = \frac{9}{5} \div \left ( \frac{4}{5} +\frac{4}{5} \right )\)

\(\displaystyle = \frac{9}{5} \div \frac{8}{5}\)

\(\displaystyle = \frac{9}{5} \times \frac{5}{8}\)

\(\displaystyle = \frac{9}{8}\)

\(\displaystyle =1 \frac{1}{8}\)

Looking at the second statement, isolate x on one side with all other constants and variables on the other side.

\(\displaystyle x + 2 = w - 8\)

\(\displaystyle x + 2 -2= w - 8-2\)

\(\displaystyle x = w - 10\)

Looking at the third statement, isolate z on one side with all other constants and variables on the other side. 

\(\displaystyle x- 9 = z + 1\)

\(\displaystyle x- 9 -1= z + 1-1\)

\(\displaystyle x -10= z\)

Looking at the first statement, isolate y on one side with all other constants and variables on the other side. 

\(\displaystyle z + 5 = y + 3\)

\(\displaystyle z + 5 -3 = y + 3 -3\)

\(\displaystyle z +2= y\)

From here, use these equivalencies so solve for y.

Substituting twice:

\(\displaystyle y = z+ 2\)

\(\displaystyle = (x-10 )+ 2\)

\(\displaystyle = x-8\)

\(\displaystyle = (w-10)-8\)

\(\displaystyle = w - 18\)