answer:Mark gives the cashier 7.00 and his total purchase is 2.05. The change he should receive is 7.00 - 2.05 = 4.95. The cashier has only 1 quarter (25 cents) and 1 dime (10 cents) to give Mark, which totals 25 + 10 = 35 cents. The rest of the change must be given in nickels. Mark receives 8 nickels, and since each nickel is worth 5 cents, the total value of the nickels is 8 * 5 = 40 cents. Now, let's add the value of the quarter, the dime, and the nickels to find out how much change Mark gets in coins: 35 + 40 = 75 cents. The change Mark should receive is 4.95, and he gets 75 cents in coins, so the remaining amount must have been given in dollar bills: 4.95 - 0.75 = 4.20. Since Mark gets 4.20 in dollar bills, and we know that dollar bills can only be given in whole dollars, the cost of the cheese must be such that it rounds the total to a number that allows for whole dollar bills to be given as change. Let's assume the cost of the loaf of bread is B and the cost of the cheese is C. We know that B + C = 2.05. If Mark gets 4.20 in dollar bills, the cheese must cost 0.80 (since 0.80 + 0.20 = 1.00, and 4.20 is 0.20 more than 4.00, which is the next whole dollar amount below 4.20). So, if the cheese costs 0.80, the cost of the loaf of bread would be B = 2.05 - 0.80 = 1.25. Therefore, the loaf of bread costs boxed{1.25} .

answer:Our goal is to divide the factors of 10! into three groups such that the products of the factors in each group are as close together as possible, and x < y < z. Begin by writing 10! = 10 cdot 9 cdot 8 cdot 7 cdot 6 cdot 5 cdot 4 cdot 3 cdot 2. Given 30^3 < 10! < 40^3, we estimate the cube root of 10! to be between 30 and 40. We aim to group the factors to form integers close to this range. - Group 7 and 3 to form x = 21. - Group 8, 5, and 2 to form y = 80. - Group 10, 9, 6, and 4 to form z = 2160. Thus, the assignment (x, y, z) = (21, 80, 2160) satisfies x cdot y cdot z = 10!. The smallest possible value of z - x is: z - x = 2160 - 21 = 2139 Therefore, the minimum value of z - x is boxed{2139}.

answer:Since the derivative of the function f(x) is f'(x) = 2 + sin x, we can set f(x) = 2x - cos x + c, Since f(0) = -1, we have -1 + c = -1, thus c = 0. Therefore, f(x) = 2x - cos x. Since the sequence {a_n} is an arithmetic sequence with a common difference of frac{pi}{4}, we have a_n = a_1 + (n-1) times frac{pi}{4}, Since f(a_2) + f(a_3) + f(a_4) = 3pi, we get 2(a_2 + a_3 + a_4) - (cos a_2 + cos a_3 + cos a_4) = 3pi, thus 6a_2 + frac{3pi}{2} - cos a_2 - cos(a_2 + frac{pi}{4}) - cos(a_2 + frac{pi}{2}) = 3pi, thus 6a_2 - cos(a_2 + frac{pi}{4}) = frac{3pi}{2}. Let g(x) = 6x - cos(x + frac{pi}{4}) - frac{3pi}{2}, then g'(x) = 6 + sin(x + frac{pi}{4}) is monotonically increasing on mathbb{R}, and g(frac{pi}{4}) = 0. Therefore, a_2 = frac{pi}{4}. Thus, frac{a_{2016}}{a_{2}} = frac{frac{pi}{4} + 2014 times frac{pi}{4}}{frac{pi}{4}} = 2015. Hence, the correct choice is: boxed{text{B}}. The problem examines the general formula of an arithmetic sequence and its properties, the use of derivatives to study the monotonicity of functions, and tests reasoning and computational skills, making it a medium-level question.

answer:According to the definition of mapping and condition ② f(a) neq f(b), we can deduce: - The value of f(a) can be chosen in 3 ways, - The value of f(b) can be chosen in 2 ways (since f(a) neq f(b)), - The values of f(c), f(d), and f(e) can each be chosen in 3 ways, Therefore, the number of functions satisfying condition ② is 3 times 2 times 3 times 3 times 3 = 162. Among these, the functions that do not satisfy condition ①, which means the range of function f only includes two elements out of {1, 2, 3}, are counted as 3 times 2 times 2 times 2 times 2 = 48. Thus, the number of functions that simultaneously satisfy conditions ① and ② is 162 - 48 = 114. Therefore, the answer is boxed{114}.