answer:Let's denote the number of footballs as x, basketballs as y, and volleyballs as z. According to the problem, we have: x + y + z = 20 (1); And the total value equation: 60x + 30y + 10z = 330, which simplifies to: 6x + 3y + z = 33 (2), Subtracting equation (1) from equation (2) gives: 5x + 2y = 13, Since x, y, and z are positive integers, We find that x = 1, y = 4, Thus, we can deduce that z = 15. Therefore, there are 15 volleyballs. Hence, the correct answer is boxed{C}.

answer:1. Identify numbers divisible by 3: We need to find the positive whole numbers less than 100 that are divisible by 3. These numbers form an arithmetic sequence: (3, 6, 9, ldots, 99). The general term of this sequence can be written as: [ a_n = 3n ] where (a_n < 100). To find the largest (n): [ 3n < 100 implies n < frac{100}{3} implies n < 33.33 ] Since (n) must be an integer, the largest (n) is 33. Therefore, there are 33 numbers less than 100 that are divisible by 3.2. Identify numbers divisible by both 3 and 2 (i.e., divisible by 6): We need to find the positive whole numbers less than 100 that are divisible by 6. These numbers also form an arithmetic sequence: (6, 12, 18, ldots, 96). The general term of this sequence can be written as: [ a_n = 6n ] where (a_n < 100). To find the largest (n): [ 6n < 100 implies n < frac{100}{6} implies n < 16.67 ] Since (n) must be an integer, the largest (n) is 16. Therefore, there are 16 numbers less than 100 that are divisible by both 3 and 6 (i.e., divisible by 6).3. Calculate numbers divisible by 3 but not by 2: The numbers that are divisible by 3 but not by 2 are those that are divisible by 3 but not by 6. We subtract the count of numbers divisible by 6 from the count of numbers divisible by 3: [ 33 - 16 = 17 ]Conclusion:The number of positive whole numbers less than 100 that are divisible by 3 but not by 2 is (boxed{17}).

answer:Solution:(1) When xin[200,300], let the profit of the project be S, thenS=200x-left( frac{1}{2}x^{2}-200x+80000 right)=-frac{1}{2}x^{2}+400x-80000=-frac{1}{2}(x-400)^{2};When xin[200,300], S<0, indicating that the project will not be profitable;When x=300, S reaches its maximum value of -5000. Therefore, the state needs to provide a minimum monthly subsidy of boxed{5000} yuan to prevent the project from making a loss.(2) According to the problem, the average processing cost per ton of carbon dioxide is:frac{y}{x}=begin{cases} frac{1}{3}x^{2}-80x+5040, & xin[120,144) frac{1}{2}x+frac{80000}{x}-200, & xin[144,500)end{cases},Then:- When xin[120,144), frac{y}{x}=frac{1}{3}x^{2}-80x+5040=frac{1}{3}(x-120)^{2}+240; thus, when x=120, frac{y}{x} reaches its minimum value of 240;- When xin[144,500), frac{y}{x}=frac{1}{2}x+frac{80000}{x}-200geqslant 2sqrt{frac{1}{2}xcdotfrac{80000}{x}}-200=200; this minimum value is achieved if and only if frac{1}{2}x=frac{80000}{x}, i.e., x=400;Since 200<240, the lowest average processing cost per ton is achieved when the monthly processing volume is boxed{400} tons.

answer:To solve this problem, we need to understand the definitions of the coefficient and degree of a monomial. - The coefficient of a monomial is the numerical part of the monomial. In the given monomial 5a^{2}b, the numerical part is 5. Therefore, the coefficient is 5.- The degree of a monomial is the sum of the exponents of all its variables. In 5a^{2}b, the exponent of a is 2 and the exponent of b is 1 (since any variable without an explicit exponent has an exponent of 1). Thus, the degree is the sum of these exponents: 2 + 1 = 3.Therefore, the coefficient of the monomial 5a^{2}b is boxed{5}, and the degree is boxed{3}.

answer:From 2|overrightarrow{a}|=3|overrightarrow{b}|, we have 4overrightarrow{a}^{2}=9overrightarrow{b}^{2}, which implies that overrightarrow{b}^{2}=frac{4}{9}overrightarrow{a}^{2}.Given that overrightarrow{a}cdot(overrightarrow{a}-2overrightarrow{b})=overrightarrow{b}^{2}, we have overrightarrow{a}^{2}-2overrightarrow{a}cdotoverrightarrow{b}=overrightarrow{b}^{2}=frac{4}{9}overrightarrow{a}^{2}.This simplifies to overrightarrow{a}cdotoverrightarrow{b}=frac{5}{18}overrightarrow{a}^{2}.Now, we can find the cosine of the angle between overrightarrow{a} and overrightarrow{b} as follows:cos = frac{overrightarrow{a}cdotoverrightarrow{b}}{|overrightarrow{a}||overrightarrow{b}|} = frac{frac{5}{18}overrightarrow{a}^{2}}{|overrightarrow{a}|cdotfrac{2}{3}|overrightarrow{a}|} = frac{5}{12}.So, the answer is boxed{frac{5}{12}}.To solve this problem, we expanded the given equation overrightarrow{a}cdot(overrightarrow{a}-2overrightarrow{b})=overrightarrow{b}^{2} to get an expression for overrightarrow{a}cdotoverrightarrow{b} in terms of overrightarrow{a}^{2}, and then used the formula for the cosine of the angle between two vectors to find the answer. This problem tests understanding of vector dot products and angle calculations, and is of moderate difficulty.

answer: Solution:# Part 1: Finding the function expression between y and xGiven the linear relationship between the selling price x and the weekly sales quantity y, we can express it as y = kx + b. From the table, we have two points that can help us determine k and b:1. When x = 60, y = 80, which gives us the equation 60k + b = 80.2. When x = 70, y = 60, which gives us the equation 70k + b = 60.Subtracting the first equation from the second, we get:[70k + b - (60k + b) = 60 - 80][10k = -20][k = -2]Substituting k = -2 into the first equation:[60(-2) + b = 80][-120 + b = 80][b = 200]Thus, the function expression between y and x is y = -2x + 200. Therefore, we have:[boxed{y = -2x + 200}]# Part 2: Determining the selling price for maximum weekly profitLet the weekly profit be w yuan. The profit for selling each kilogram of fruit is the selling price minus the cost price, which is (x - 40) yuan. The total profit is this margin multiplied by the number of kilograms sold, which gives us:[w = (x - 40)(-2x + 200)][= -2x^2 + 280x - 8000][= -2(x^2 - 140x + 4000)][= -2(x - 70)^2 + 1800]Since the coefficient of x^2 is negative, the parabola opens downwards, indicating that the maximum profit occurs at the vertex of the parabola, which is at x = 70. At this point, the maximum profit is 1800 yuan. Therefore, we have:[boxed{text{Selling price} = 70 text{ yuan/kg, Maximum profit} = 1800 text{ yuan}}]# Part 3: Determining the range of values for mGiven the new cost price increases by m yuan per kilogram, the profit function becomes:[w = (x - 40 - m)(-2x + 200)][= -2x^2 + 2(m + 140)x - 200(m + 40)]The axis of symmetry for this quadratic function is at:[x = frac{-b}{2a} = frac{2(m + 140)}{2(-2)} = frac{m + 140}{2}]For the profit to decrease as the selling price increases beyond 76 yuan per kilogram, the axis of symmetry must be less than or equal to 76, which gives us:[frac{m + 140}{2} leq 76][m + 140 leq 152][m leq 12]Since m > 0, the range of values for m is 0 < m leq 12. Therefore, we have:[boxed{0 < m leq 12}]