Since the square of any real number is nonnegative, every point on the graph of the quadratic equation y = (x − 2)² in the xy-plane has a nonnegative y-coordinate. Thus, y ≥ 0 for every point on the graph. Therefore, the equation y = (x − 2)² has a graph for which y is always greater than or equal to −1.

Choices A, B, and D are incorrect because the graph of each of these equations in the xy-plane has a y-intercept at (0, −2). Therefore, each of these equations contains at least one point where y is less than −1.

Since the slope of the first line is 2, an equation of this line can be written in the form y = 2x + c, where c is the y-intercept of the line. Since the line contains the point (1, 8), one can substitute 1 for x and 8 for y in y = 2x + c, which gives 8 = 2(1) + c, or c = 6. Thus, an equation of the first line is y = 2x + 6. The slope of the second line is equal to \(\frac{1 − 2}{2 − 1}\) or −1.Thus, an equation of the second line can be written in the form y = −x + d,where d is the y-intercept of the line. Substituting 2 for x and 1 for y gives 1 = −2 + d, or d = 3. Thus, an equation of the second line is y = −x + 3.

Since a is the x-coordinate and b is the y-coordinate of the intersection point of the two lines, one can substitute a for x and b for y in the two equations,giving the system b = 2a + 6 and b = –a + 3. Thus, a can be found by solving the equation 2a + 6 = −a + 3, which gives a = −1. Finally, substituting −1 for a into the equation b = –a + 3 gives b = −(−1) + 3, or b = 4. Therefore, the value of a + b is 3.

Alternatively, since the second line passes through the points (1, 2) and (2, 1), an equation for the second line is x + y = 3. Thus, the intersection point of the first line and the second line, (a, b) lies on the line with equation x + y = 3. It follows that a + b = 3.

Choices A and C are incorrect and may result from finding the value of only a or b, but not calculating the value of a + b. Choice D is incorrect and may result from a computation error in finding equations of the two lines or in solving the resulting system of equations.

The relationship between n and A is given by the equation nA = 360. Since n is the number of sides of a polygon, n must be a positive integer, and so nA = 360 can be rewritten as A = \(\frac{360}{n}\) . If the value of A is greater than 50, it follows that \(\frac{360}{n}\) > 50 is a true statement. Thus, 50n < 360, or n < \(\frac{360}{50}\) = 7.2. Since n must be an integer, the greatest possible value of n is 7.

Choices A and B are incorrect. These are possible values for n, the number of sides of a regular polygon, if A > 50, but neither is the greatest possible value of n. Choice D is incorrect. If A < 50, then n = 8 is the least possible value of n, the number of sides of a regular polygon. However, the question asks for the greatest possible value of n if A > 50, which is n = 7.

Since the numerator and denominator of \(\frac{x^{a^{2}}}{x^{b^{2}}}\) have a common base, it follows by the laws of exponents that this expression can be rewritten as \(x^{a^{2}-b^{2}}\). Thus, the equation \(\frac{x^{a^{2}}}{x^{b^{2}}}\)= 16 can be rewritten as \(x^{a^{2}-b^{2}}\)= x¹⁶. Because the equivalent expressions have the common base x, and x >  1, it follows that the exponents of the two expressions must also be equivalent. Hence, the equation a² − b² = 16 must be true. The left-hand side of this new equation is a difference of squares, and so it can be factored: (a + b)(a − b) = 16. It is given that (a + b) = 2; substituting 2 for the factor (a + b) gives 2(a  − b) = 16. Finally, dividing both sides of 2(a − b) = 16 by 2 gives a − b = 8.

Choices B, C, and D are incorrect and may result from errors in applying the laws of exponents or errors in solving the equation a² − b² = 16.

Since lines ℓ and k are parallel, the lines have the same slope. Line ℓ passes through the points (−5, 0) and (0, 2), so its slope is \(\frac{0 − 2}{−5 − 0}\), which is ²⁄₅. The slope of line k must also be ²⁄₅. Since line k has slope ²⁄₅ and passes through the points (0, −4) and (p, 0), it follows that \(\frac{−4 − 0}{0 − p}\) = ²⁄₅, or ⁴⁄_p = ²⁄₅.Multiplying each side of ⁴⁄_p = ²⁄₅ by 5p gives 20 = 2p, and therefore, p = 10.

Choices A, B, and C are incorrect and may result from conceptual or calculation errors.

If a polynomial expression is in the form (x)² + 2(x)(y) + (y)², then it is equivalent to (x + y)². Because 9a⁴ + 12a² b² + 4b⁴ = (3a² )² + 2(3a² )(2b² ) + (2b² )² , it can be rewritten as (3a² + 2b² )².

Choice B is incorrect. The expression (3a + 2b)⁴ is equivalent to the product (3a + 2b)(3a + 2b)(3a + 2b)(3a + 2b). This product will contain the term 4(3a)³ (2b) = 216a³ b. However, the given polynomial, 9a⁴ + 12a² b² + 4b⁴, does not contain the term 216a³ b. Therefore, 9a⁴ + 12a² b² + 4b⁴ ≠ (3a + 2b)⁴. Choice C is incorrect. The expression (9a² + 4b² )² is equivalent to the product (9a² + 4b² )(9a² + 4b²). This product will contain the term (9a²) (9a²) = 81a⁴. However, the given polynomial, 9a⁴ + 12a² b² + 4b⁴ , does not contain the term 81a⁴. Therefore, 9a⁴ + 12a²b² + 4b⁴ ≠ (9a² + 4b²)².Choice D is incorrect. The expression (9a + 4b)⁴ is equivalent to the product(9a + 4b)(9a + 4b)(9a + 4b) (9a + 4b). This product will contain the term (9a)(9a)(9a)(9a) = 6,561a⁴. However, the given polynomial, 9a⁴ + 12a²b² + 4b⁴ , does not contain the term 6,561a⁴. Therefore, 9a⁴ + 12a² b² + 4b⁴ ≠ (9a + 4b)⁴.

The price of the job, in dollars, is calculated using the expression 60 + 12nh, where 60 is a fixed price and 12nh depends on the number of landscapers, n, working the job and the number of hours, h, the job takes those n landscapers. Since nh is the total number of hours of work done when n landscapers work h hours, the cost of the job increases by $12 for each hour a landscaper works. Therefore, of the choices given, the best interpretation of the number 12 is that the company charges $12 per hour for each landscaper.

Choice B is incorrect because the number of landscapers that will work each job is represented by n in the equation, not by the number 12. Choice C is incorrect because the price of the job increases by 12n dollars each hour, which will not be equal to 12 dollars unless n = 1. Choice D is incorrect because the total number of hours each landscaper works is equal to h. The number of hours each landscaper works in a day is not provided.

Multiplying each side of x + y = 0 by 2 gives 2x + 2y = 0.Then, adding the corresponding sides of 2x + 2y = 0 and 3x − 2y = 10 gives 5x = 10. Dividing each side of 5x = 10 by 5 gives x = 2. Finally, substituting 2 for x in x + y = 0 gives 2 + y = 0, or y = −2. Therefore, the solution to the given system of equations is (2, −2).

Alternatively, the equation x + y = 0 can be rewritten as x = −y, and substituting x for −y in 3x − 2y = 10 gives 5x = 10, or x = 2. The value of y can then be found in the same way as before.

Choices A, C, and D are incorrect because when the given values of x and y are substituted into x + y = 0 and 3x − 2y = 10, either one or both of the equations are not true. These answers may result from sign errors or other computational errors.

Subtracting 6 from each side of 5x + 6 = 10 yields 5x = 4.Dividing both sides of 5x = 4 by 5 yields x =⁴⁄₅. The value of x can now be substituted into the expression 10x + 3, giving 10 (⁴⁄₅) + 3 = 11.

Alternatively, the expression 10x + 3 can be rewritten as 2(5x + 6) − 9, and 10 can be substituted for 5x + 6, giving 2(10) − 9 = 11.

Choices A, B, and D are incorrect. Each of these choices leads to 5x + 6 ≠ 10, contradicting the given equation, 5x + 6 = 10. For example, choice A is incorrect because if the value of 10x + 3 were 4, then it would follow that x = 0.1, and the value of 5x + 6 would be 6.5, not 10.