The function \(\displaystyle x\ \blacklozenge \ y\) is defined by \(\displaystyle (x^{3}-y)^{2}-x\). This means that for whatever value is in the space of the \(\displaystyle x\) before the symbol \(\displaystyle \blacklozenge\), in our case 2, is inserted into any \(\displaystyle x\) in the defined function:

\(\displaystyle \dpi{100} (2^{3}-y)^{2}-2\)

For the value that follows \(\displaystyle \blacklozenge\), the \(\displaystyle y\) value of 1 in our case, is inserted into any \(\displaystyle y\) variable in the defined function.

\(\displaystyle \dpi{100} (2^{3}-1)^{2}-2\)

Then simplify:

\(\displaystyle (8-1)^{2}-2\)

\(\displaystyle 7^{2}-2\)

\(\displaystyle 49-2\)

\(\displaystyle 47\)

\(\displaystyle 47\) is our correct answer after all the simplification.

The word problem can be simplified into two steps. First, subtracting the 15% discount:

\(\displaystyle New Price = (Original Price - (Original Price \cdot 0.15))\)

Second, adding the sales tax to the result from the first step:

\(\displaystyle Final Price = (New Price + (New Price \cdot 0.07))\)

The percentage is represented as a decimal by dividing the percentage by 100.

So, for the numbers provided:

New Price = \(\displaystyle (\$80.00-(\$80.00\cdot 0.15))=\$68.00\)

Final Price = \(\displaystyle \$68.00+(\$68.00\cdot 0.07)=\$72.76\)

Plug in the values for the variables and use PEMDAS to dictate the order of the operations. PEMDAS stands for Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction. Using this order of operations we get

 \(\displaystyle (x-2) = 0\) so the first part of the expression is \(\displaystyle 0\).

Next we deal with the exponents,

\(\displaystyle -3^2\)\(\displaystyle =9\).

Then, we deal with the multiplication, 

\(\displaystyle 2\cdot -3 = -6.\) 

Finally, we plug these simplified values back into the equation and solve,

\(\displaystyle 9 - (-6) = 15\).

Since \(\displaystyle 153\) tickets were sold in advance, the group earned a revenue of \(\displaystyle 153*\$ 5= \$ 765\). In order to reach their goal of \(\displaystyle \$ 2,000\), they would still need to earn an extra \(\displaystyle (\$ 2000-\$ 765)= \$ 1235.\)

Since each ticket costs \(\displaystyle \$ 7\) at the door, the theatre would need to sell \(\displaystyle \$ 1235/7\)tickets, or \(\displaystyle 177\) tickets (rounded to the next whole number).

Note: the answer is NOT \(\displaystyle 176\) tickets, because if \(\displaystyle 176\) tickets are sold, the revenue would be \(\displaystyle (176*7) + (153*5) = \$ 1997\) , which does not reach the goal of \(\displaystyle \$ 2,000\).

Since \(\displaystyle x=-4\) this value can be substituted for \(\displaystyle x\) in the function.

\(\displaystyle \frac{x^2-1}{x+1}}\)

\(\displaystyle =\frac{(-4)^2-1}{(-4)+1}\)

\(\displaystyle =\frac{16-1}{-3}\)

Here, the denominator becomes \(\displaystyle -3\), while the numerator becomes \(\displaystyle 15\).

\(\displaystyle =\frac{15}{-3}\)

Hence, 

\(\displaystyle \frac{15}{-3} = -5\).

When substituting \(\displaystyle (-2)\) for \(\displaystyle x\) in the provided equation, we get

\(\displaystyle 4(-2)^2 +2(-2) -4\),

which can be simplified to

\(\displaystyle 16-4-4\), or \(\displaystyle 8\).

May corresponds to the month Abraham collected the most stamps, adding 24 to his collection. April, July, October, and November correspond to 18, 6, 21, 12, respectively.

Alex’s bonus of 

\(\displaystyle 2700 = \frac{s^3}{10} \rightarrow 27000 = s^3\rightarrow s = 30.\)

 \(\displaystyle |-7| = 7\). Thus

 \(\displaystyle 2x + 7 = 7\rightarrow 2x = 0 \rightarrow x = 0\)

To evaluate, simply plug in \(\displaystyle 2\) for \(\displaystyle x\). Thus:

\(\displaystyle 2^3+\frac{2}{2}-4*2^2+2!=8+1-16+2=-5\)