Choice A is correct. Subtracting 8 from both sides of the given equation yields
p+3 = 2. Subtracting 3 from both sides of this equation yields p=-1.

Choice B is incorrect and may result from conceptual or calculation errors.
Choice C is incorrect and may result from conceptual or calculation errors.
Choice D is incorrect and may result from conceptual or calculation errors.

Choice D is correct. An appropriate model should follow the trend of the data
points and should have data points both above and below the model. The
scatterplot shows that the data points have an increasing trend that is curved.
Therefore, an appropriate model should be an increasing curve with data points
both above and below the model. Of the given choices, only the model in choice D
is an increasing curve with data points both above and below the model.

Choice A is incorrect. Since the trend of the data points isn’t linear, a line isn’t the
most appropriate model for the data. Choice B is incorrect. Since the trend of the
data points is increasing and isn’t linear, a decreasing line isn’t the most
appropriate model for the data. Choice C is incorrect. All the data points are below
the model shown in this graph.

**Choice D** is correct. Adding $53$ to each side of the given equation yields $k^2 = 144$. Taking the square root of each side of this equation yields $k = \pm 12$. Therefore, the positive solution to the given equation is $12$.

*Choice A* is incorrect. This is the positive solution to the equation $k^2 – 53 = 20,683$, not $k^2 – 53 = 91$. *Choice B* is incorrect. This is the positive solution to the equation $k^2 – 53 = 5,131$, not $k^2 – 53 = 91$. *Choice C* is incorrect. This is the positive solution to the equation $k^2 – 53 = 1,391$, not $k^2 – 53 = 91$.

Choice D is correct. It’s given that during a portion of a flight, a small airplane’s
cruising speed varied between 150 miles per hour and 170 miles per hour. It’s
also given that s represents the cruising speed, in miles per hour, during this
portion of the flight. It follows that the airplane’s cruising speed, in miles per hour,
was at least 150, which means s$150, and was at most 170, which means
s#170. Therefore, the inequality that best represents this situation is
150 1 # #s 70.

Choice A is incorrect and may result from conceptual or calculation errors.
Choice B is incorrect and may result from conceptual or calculation errors.
Choice C is incorrect and may result from conceptual or calculation errors.

Choice A is correct. It’s given that the variable y represents the height, in meters,
of the object above the ground. The graph shows that the height of the object
was increasing from x=0 to x=2, and decreasing from x=2 to x=4. Therefore,
the height of the object was increasing for the entire interval of time from x=0 to
x=2.

Choice B is incorrect. The height of the object wasn’t increasing for this entire
interval of time, as it was decreasing from x=2 to x=4. Choice C is incorrect.
The height of the object was decreasing, not increasing, for this entire interval of
time. Choice D is incorrect. The height of the object was decreasing, not
increasing, for this entire interval of time.

The correct answer is $31$. It’s given that $1$ yard is equal to $36$ inches. Therefore, $1,116$ inches is equivalent to $(1,116\text{ inches})\left(\frac{1\text{ yard}}{36\text{ inches}}\right)$, or $31$ yards.

Choice D is correct. It’s given that when \( x = 0 \), \( f(x) = 30 \). Substituting 0 for \( x \) and 30 for \( f(x) \) in the given function yields \( 30 = 0 + b \), or \( 30 = b \). Therefore, the value of \( b \) is 30.

Choice A is incorrect and may result from conceptual or calculation errors.
Choice B is incorrect and may result from conceptual or calculation errors.
Choice C is incorrect and may result from conceptual or calculation errors.

The correct answer is 11. It’s given that the function \( f(x) = 14 + 4x \) represents the total cost, in dollars, of attending an arcade when \( x \) games are played. Substituting 58 for \( f(x) \) in the given equation yields \( 58 = 14 + 4x \). Subtracting 14 from each side of this equation yields \( 44 = 4x \). Dividing each side of this equation by 4 yields \( 11 = x \). Therefore, 11 games can be played for a total cost of $58.

Choice B is correct. The function P gives the estimated number of marine
mammals in a certain area, where t is the number of years since a study began.
Since the value of P^ h 0 is the value of P t^ h when t=0, it follows that P^ h 0 1 = ,800
means that the value of P t^ h is 1 8, 00 when t=0. Since t is the number of years
since the study began, it follows that t=0 is 0 years since the study began, or
when the study began. Therefore, the best interpretation of P^ h 0 1 = ,800 in this
context is the estimated number of marine mammals in the area was 1 8, 00 when
the study began.

Choice B is correct. It’s given that the shop’s inventory starts with 4,500 paper cups and that the manager estimates that 70 of these paper cups are used each day. Let \( x \) represent the number of days in which the estimated supply of paper cups will reach 1,700. The equation \( 4,500 – 70x = 1,700 \) represents this situation. Subtracting 4,500 from both sides of this equation yields \( -70x = -2,800 \). Dividing both sides of this equation by \( -70 \) yields \( x = 40 \). Therefore, based on this estimate, the supply of paper cups will reach 1,700 in 40 days.

Choice A is incorrect. After 20 days, the estimated supply of paper cups would be \( 4,500 – 70(20) \), or 3,100 cups, not 1,700 cups. Choice C is incorrect. After 60 days, the estimated supply of paper cups would be \( 4,500 – 70(60) \), or 300 cups, not 1,700 cups. Choice D is incorrect. After 80 days, the estimated supply of paper cups would be \( 4,500 – 70(80) \), or \( -1,100 \) cups, which isn’t possible.

Choice A is correct. In each choice, the values of \( x \) are 2, 4, and 6. Substituting the first value of \( x \), 2, for \( x \) in the given inequality yields \( y > 4(2) + 8 \), or \( y > 16 \). Therefore, when \( x = 2 \), the corresponding value of \( y \) must be greater than 16. Of the given choices, only choice A is a table where the value of \( y \) corresponding to \( x = 2 \) is greater than 16. To confirm that the other values of \( x \) in this table and their corresponding values of \( y \) are also solutions to the given inequality, the values of \( x \) and \( y \) in the table can be substituted for \( x \) and \( y \) in the given inequality. Substituting 4 for \( x \) and 30 for \( y \) in the given inequality yields \( 30 > 4(4) + 8 \), or \( 30 > 24 \), which is true. Substituting 6 for \( x \) and 41 for \( y \) in the given inequality yields \( 41 > 4(6) + 8 \), or \( 41 > 32 \), which is true. It follows that for choice A, all the values of \( x \) and their corresponding values of \( y \) are solutions to the given inequality.

Choice B is incorrect. Substituting 2 for \( x \) and 8 for \( y \) in the given inequality yields \( 8 > 4(2) + 8 \), or \( 8 > 16 \), which is false. Choice C is incorrect. Substituting 2 for \( x \) and 13 for \( y \) in the given inequality yields \( 13 > 4(2) + 8 \), or \( 13 > 16 \), which is false. Choice D is incorrect. Substituting 2 for \( x \) and 13 for \( y \) in the given inequality yields \( 13 > 4(2) + 8 \), or \( 13 > 16 \), which is false.

Choice B is correct. The expression \( (x^{2} + 11)^{2} \) can be written as \( (x^{2} + 11)(x^{2} + 11) \), which is equivalent to \( x^{2}(x^{2} + 11) + 11(x^{2} + 11) \). Distributing \( x^{2} \) and 11 to \( (x^{2} + 11) \) yields \( x^{4} + 11x^{2} + 11x^{2} + 121 \), or \( x^{4} + 22x^{2} + 121 \). The expression \( (x – 5)(x + 5) \) is equivalent to \( (x – 5)x + (x – 5)5 \). Distributing \( x \) and 5 to \( (x – 5) \) yields \( x^{2} – 5x + 5x – 25 \), or \( x^{2} – 25 \). Therefore, the expression \( (x^{2} + 11)^{2} + (x – 5)(x + 5) \) is equivalent to \( (x^{4} + 22x^{2} + 121) + (x^{2} – 25) \), or \( x^{4} + 22x^{2} + 121 + x^{2} – 25 \). Combining like terms in this expression yields \( x^{4} + 23x^{2} + 96 \).

Choice A is incorrect. Equivalent expressions must be equivalent for any value of \( x \). Substituting 0 for \( x \) in this expression yields \( -14 \), whereas substituting 0 for \( x \) in the given expression yields 96. Choice C is incorrect. Equivalent expressions must be equivalent for any value of \( x \). Substituting 0 for \( x \) in this expression yields 121, whereas substituting 0 for \( x \) in the given expression yields 96. Choice D is incorrect. Equivalent expressions must be equivalent for any value of \( x \). Substituting 0 for \( x \) in this expression yields 146, whereas substituting 0 for \( x \) in the given expression yields 96.

The correct answer is \( \frac{1}{2} \). The value of \( h(2) \) is the value of \( h(x) \) when \( x = 2 \). Substituting 2 for \( x \) in the given equation yields \( h(2) = \frac{8}{5(2) + 6} \), which is equivalent to \( h(2) = \frac{8}{16} \), or \( h(2) = \frac{1}{2} \). Therefore, the value of \( h(2) \) is \( \frac{1}{2} \). Note that 1/2 and .5 are examples of ways to enter a correct answer.

The correct answer is \( \frac{15}{2} \). The area, \( A \), of a triangle is given by the formula \( A = \frac{1}{2}bh \), where \( b \) is the length of the base of the triangle and \( h \) is the height of the triangle. In the right triangle shown, the length of the base of the triangle is 5 inches, and the height is 3 inches. It follows that \( b = 5 \) and \( h = 3 \). Substituting 5 for \( b \) and 3 for \( h \) in the formula \( A = \frac{1}{2}bh \) yields \( A = \frac{1}{2}(5)(3) \), which is equivalent to \( A = \frac{1}{2}(15) \), or \( A = \frac{15}{2} \). Therefore, the area of the triangle, in square inches, is \( \frac{15}{2} \). Note that 15/2 and 7.5 are examples of ways to enter a correct answer.

Choice B is correct. It’s given that the graph models the number of active projects a company was working on x months after the end of November 2012. Therefore, the value of x that corresponds to the end of November 2012 is 0. The point at which x=0 is the y-intercept of the graph. It follows that the y-intercept of the graph shown is the point ^ h 0 5, . Therefore, according to the model, the predicted number of active projects the company was working on at the end of November 2012 is 5.

Choice A is incorrect. This is the value of x that corresponds to the end of November 2012, not the predicted number of active projects the company was working on at the end of November 2012. Choice C is incorrect. This is the predicted number of active projects the company was working on 2 months after the end of November 2012. Choice D is incorrect. This is the predicted number of active projects the company was working on 4 months after the end of November 2012.

Choice C is correct. It’s given that the relationship between \( x \) and \( y \) is linear. An equation representing a linear relationship can be written in the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the \( y \)-coordinate of the \( y \)-intercept of the graph of the relationship in the \( xy \)-plane. It’s given that for every increase in the value of \( x \) by 1, the value of \( y \) increases by 8. The slope of a line can be expressed as the change in \( y \) over the change in \( x \). Thus, the slope, \( m \), of the line representing this relationship can be expressed as \( \frac{8}{1} \), or 8. Substituting 8 for \( m \) in the equation \( y = mx + b \) yields \( y = 8x + b \). It’s also given that when the value of \( x \) is 2, the value of \( y \) is 18. Substituting 2 for \( x \) and 18 for \( y \) in the equation \( y = 8x + b \) yields \( 18 = 8(2) + b \), or \( 18 = 16 + b \). Subtracting 16 from each side of this equation yields \( 2 = b \). Substituting 2 for \( b \) in the equation \( y = 8x + b \) yields \( y = 8x + 2 \). Therefore, the equation \( y = 8x + 2 \) represents this relationship.

Choice A is incorrect. This equation represents a relationship where for every increase in the value of \( x \) by 1, the value of \( y \) increases by 2, not 8, and when the value of \( x \) is 2, the value of \( y \) is 22, not 18. Choice B is incorrect. This equation represents a relationship where for every increase in the value of \( x \) by 1, the value of \( y \) increases by 2, not 8, and when the value of \( x \) is 2, the value of \( y \) is 12, not 18. Choice D is incorrect. This equation represents a relationship where for every increase in the value of \( x \) by 1, the value of \( y \) increases by 3, not 8, and when the value of \( x \) is 2, the value of \( y \) is 32, not 18.

Choice D is correct. If a data set contains an odd number of data values, the median is represented by the middle data value in the list when the data values are listed in ascending or descending order. Since the numbers of employees are given as ranges of values rather than specific values, it’s only possible to determine the range in which the median falls, rather than the exact median. Since there are 17 restaurants included in the data set and the numbers of employees are listed in ascending order, it follows that the median number of employees will be represented by the ninth restaurant in the list. Since the first 2 restaurants each have 2 to 7 employees, numbers of employees in the 2 to 7 range would be represented by the first and second restaurants in the list. The next 4 restaurants each have 8 to 13 employees. Therefore, numbers of employees in the 8 to 13 range will be represented by the third through sixth restaurants in the list. The next 2 restaurants each have 14 to 19 employees. Therefore, numbers of employees in the 14 to 19 range will be represented by the seventh and eighth
restaurants in the list. Since the next 7 restaurants each have 20 to 25 employees, numbers of employees in the 20 to 25 range will be represented by the ninth through fifteenth restaurants in the list. This means that the ninth restaurant in the list, which has the median number of employees for the restaurants in this town, has a number of employees in the 20 to 25 range. Of the given choices, the only number of employees in the 20 to 25 range is 21.

Choice A is incorrect and may result from conceptual or calculation errors.
Choice B is incorrect. This is the position of the median in the list, not the value of
the median. Choice C is incorrect and may result from conceptual or calculation
errors.

Choice D is correct. Adding 40 to both sides of the given equation yields \( w^{2} + 12w = 40 \). To complete the square, adding \( (\frac{12}{2})^{2} \), or \( 6^{2} \), to both sides of this equation yields \( w^{2} + 12w + 6^{2} = 40 + 6^{2} \), or \( (w + 6)^{2} = 76 \). Taking the square root of both sides of this equation yields \( w + 6 = \pm\sqrt{76} \), or \( w + 6 = \pm 2\sqrt{19} \). Subtracting 6 from both sides of this equation yields \( w = -6 \pm 2\sqrt{19} \). Therefore, the solutions to the given equation are \( -6 + 2\sqrt{19} \) and \( -6 – 2\sqrt{19} \). Of these two solutions, only \( -6 + 2\sqrt{19} \) is given as a choice.

Choice A is incorrect and may result from conceptual or calculation errors.
Choice B is incorrect and may result from conceptual or calculation errors.
Choice C is incorrect and may result from conceptual or calculation errors.

The correct answer is \( \frac{189}{5} \). A \( y \)-intercept of a graph in the \( xy \)-plane is a point where the graph intersects the \( y \)-axis, which is a point with an \( x \)-coordinate of 0. Substituting 0 for \( x \) in the given equation yields \( \frac{3(0)}{7} = -\frac{5y}{9} + 21 \), or \( 0 = -\frac{5y}{9} + 21 \). Subtracting 21 from both sides of this equation yields \( -21 = -\frac{5y}{9} \). Multiplying both sides of this equation by \( -9 \) yields \( 189 = 5y \). Dividing both sides of this equation by 5 yields \( \frac{189}{5} = y \). Therefore, the \( y \)-coordinate of the \( y \)-intercept of the graph of the given equation in the \( xy \)-plane is \( \frac{189}{5} \). Note that 189/5 and 37.8 are examples of ways to enter a correct answer.

The correct answer is \( -24 \). Since the graph passes through the point \( (0, -6) \), it follows that when the value of \( x \) is 0, the value of \( y \) is \( -6 \). Substituting 0 for \( x \) and \( -6 \) for \( y \) in the given equation yields \( -6 = 2(0)^{2} + b(0) + c \), or \( -6 = c \). Therefore, the value of \( c \) is \( -6 \). Substituting \( -6 \) for \( c \) in the given equation yields \( y = 2x^{2} + bx – 6 \). Since the graph passes through the point \( (-1, -8) \), it follows that when the value of \( x \) is \( -1 \), the value of \( y \) is \( -8 \). Substituting \( -1 \) for \( x \) and \( -8 \) for \( y \) in the equation \( y = 2x^{2} + bx – 6 \) yields \( -8 = 2(-1)^{2} + b(-1) – 6 \), or \( -8 = 2 – b – 6 \), which is equivalent to \( -8 = -4 – b \). Adding 4 to each side of this equation yields \( -4 = -b \). Dividing each side of this equation by \( -1 \) yields \( 4 = b \). Since the value of \( b \) is 4 and the value of \( c \) is \( -6 \), it follows that the value of \( bc \) is \( (4)(-6) \), or \( -24 \).

Alternate approach: The given equation represents a parabola in the \( xy \)-plane with a vertex at \( (-1, -8) \). Therefore, the given equation, \( y = 2x^{2} + bx + c \), which is written in standard form, can be written in vertex form, \( y = a(x – h)^{2} + k \), where \( (h, k) \) is the vertex of the parabola and \( a \) is the value of the coefficient on the \( x^{2} \) term when the equation is written in standard form. It follows that \( a = 2 \). Substituting 2 for \( a \), \( -1 \) for \( h \), and \( -8 \) for \( k \) in this equation yields \( y = 2(x – (-1))^{2} + (-8) \), or \( y = 2(x + 1)^{2} – 8 \). Squaring the binomial on the right-hand side of this equation yields \( y = 2(x^{2} + 2x + 1) – 8 \). Multiplying each term inside the parentheses on the right-hand side of this equation by 2 yields \( y = 2x^{2} + 4x + 2 – 8 \), which is equivalent to \( y = 2x^{2} + 4x – 6 \). From the given equation \( y = 2x^{2} + bx + c \), it follows that the value of \( b \) is 4 and the value of \( c \) is \( -6 \). Therefore, the value of \( bc \) is \( (4)(-6) \), or \( -24 \).

Choice D is correct. It’s given that in 2008 Zinah earned 14% more than in 2007. Let \( h \) represent the amount Zinah earned in 2007 and let \( j \) represent the amount that Zinah earned in 2008. This situation can be represented by the equation \( j = (1 + \frac{14}{100})h \), or \( j = 1.14h \). It’s also given that in 2009 Zinah earned 4% more than in 2008. Let \( k \) represent the amount Zinah earned in 2009. This situation can be represented by the equation \( k = (1 + \frac{4}{100})j \), or \( k = 1.04j \). Substituting \( 1.14h \) for \( j \) in the equation \( k = 1.04j \) yields \( k = (1.04)(1.14h) \), or \( k = 1.1856h \). If Zinah earned \( y \) times as much in 2009 as in 2007, then the value of \( y \) is 1.1856.

Choice C is correct. In the triangle shown, the measure of angle \( B \) is 30° and angle \( C \) is a right angle, which means that it has a measure of 90°. Since the sum of the angles in a triangle is equal to 180°, the measure of angle \( A \) is equal to \( 180^{\circ} – (30 + 90)^{\circ} \), or 60°. In a right triangle whose acute angles have measures 30° and 60°, the lengths of the legs can be represented by the expressions \( x \), \( x\sqrt{3} \), and \( 2x \), where \( x \) is the length of the leg opposite the angle with measure 30°, \( x\sqrt{3} \) is the length of the leg opposite the angle with measure 60°, and \( 2x \) is the length of the hypotenuse. In the triangle shown, the hypotenuse has a length of 54. It follows that \( 2x = 54 \), or \( x = 27 \). Therefore, the length of the leg opposite angle \( B \) is 27 and the length of the leg opposite angle \( A \) is \( 27\sqrt{3} \). The tangent of an acute angle in a right triangle is defined as the ratio of the length of the leg opposite the angle to the length of the leg adjacent to the angle. The length of the leg opposite angle \( A \) is \( 27\sqrt{3} \) and the length of the leg adjacent to angle \( A \) is 27. Therefore, the value of \( \tan A \) is \( \frac{27\sqrt{3}}{27} \), or \( \sqrt{3} \).

Choice A is incorrect and may result from conceptual or calculation errors. Choice B is incorrect. This is the value of \( \frac{1}{\tan A} \), not the value of \( \tan A \). Choice D is incorrect. This is the length of the leg opposite angle \( A \), not the value of \( \tan A \).