answer:First, we need to calculate the total number of students in the school. Since there are 67 classrooms with 66 students in each, we multiply: 67 classrooms * 66 students/classroom = 4422 students Now, we need to determine how many buses are needed to accommodate 4422 students if each bus has 6 seats. We divide the total number of students by the number of seats per bus: 4422 students / 6 seats/bus = 737 buses However, since we can't have a fraction of a bus, we need to round up to the nearest whole number to ensure that there is enough space for all students. Therefore, the school would need boxed{738} buses for the field trip.

answer:Let's denote the number of books Janine read this month as ( x ). Since each book has 10 pages, the total number of pages Janine read last month for the 5 books is ( 5 times 10 = 50 ) pages. Given that Janine read a total of 150 pages in two months, we can calculate the number of pages she read this month by subtracting the pages she read last month from the total pages read in two months: ( 150 text{ pages (total)} - 50 text{ pages (last month)} = 100 text{ pages (this month)} ). Now, since each book has 10 pages, we can find the number of books Janine read this month by dividing the number of pages she read this month by the number of pages per book: ( x = frac{100 text{ pages}}{10 text{ pages/book}} = 10 text{ books} ). So, Janine read 10 books this month. The ratio of the number of books Janine read this month to the number of books she read last month is: ( frac{x}{5} = frac{10}{5} = 2 ). Therefore, the ratio is boxed{2:1} .

answer:To solve this problem, we need to determine the value of x, which represents the number of granola bars each child received. Let's break down the information given: Number of granola bars made: 200 Number of granola bars eaten by Monroe and her husband: 80 Number of children: 6 We can set up the equation as follows: Number of granola bars made - Number of granola bars eaten by Monroe and her husband - (Number of children * Number of granola bars each child received) = 0 200 - 80 - (6 * x) = 0 Let's simplify and solve for x: 120 - 6x = 0 To isolate x, we divide both sides of the equation by -6: (120 - 6x) / -6 = 0 / -6 20 - x = 0 To isolate x, we subtract 20 from both sides of the equation: 20 - x - 20 = 0 - 20 -x = -20 Finally, we multiply both sides of the equation by -1 to solve for x: x = 20 The value of x is 20. 20 The answer is: 20

answer:(I) When a=0, -x+2 > 0, x < 2; When aneq 0, f(x)=(ax-1)(x-2)=0, x_{1}= frac{1}{a}, x_{2}=2, When a < 0, frac{1}{a} < x < 2; When 0 < a < frac{1}{2}, frac{1}{a} > 2, x > frac{1}{a} or x < 2; When a= frac{1}{2}, frac{1}{a}=2, xneq2; When a > frac{1}{2}, frac{1}{a} < 2, x > 2 or x < frac{1}{a}. Summarizing the above, when a < 0, the solution set is left( frac{1}{a},2right); When a=0, the solution set is left(-infty,2right); When 0 < a < frac{1}{2}, the solution set is left(-infty,2right)cupleft( frac{1}{a},+inftyright); When a= frac{1}{2}, the solution set is left(-infty,2right)cupleft(2,+inftyright); When a > frac{1}{2}, the solution set is left(-infty, frac{1}{a}right)cupleft(2,+inftyright). (II) f(|x|)=ax^{2}-(2a+1)|x|+2, ygeqslant f(|x|) therefore x+ygeqslant a{|x|}^{2}-(2a+1)|x|+2+x therefore x+ygeqslant begin{cases}ax^{2}-(2a+1)x+2+x, & (xgeqslant 0) ax^{2}+(2a+1)x+2+x, & (x < 0)end{cases} That is, x+ygeqslant begin{cases}a{(x-1)}^{2}-a+2, & (xgeqslant 0) a{(x+ frac{a+1}{a})}^{2}-left(a+ frac{1}{a}right), & (x < 0)end{cases} When xgeqslant 0, x+ygeqslant -a+2, When x < 0, x+ygeqslant -a- frac{1}{a}, Since a < 0, therefore 2 > - frac{1}{a} therefore boxed{{(x+y)}_{min}=-a- frac{1}{a}} Equality holds when y=f(-x), x=-1- frac{1}{a}. Therefore, the minimum value of x+y is boxed{-a- frac{1}{a}}.