Given that m and N are real numbers, if the solution set of inequality (2m-n) x + 3m-4n ＜ 0 is x ＞ 49, find the solution set of inequality (m-4n) x + 2m-3n ＞ 0

Given that m and N are real numbers, if the solution set of inequality (2m-n) x + 3m-4n ＜ 0 is x ＞ 49, find the solution set of inequality (m-4n) x + 2m-3n ＞ 0

The solution set of ∵ inequality (2m-n) x + 3m-4n < 0 is x > 49, ∵ solution inequality (2m-n) x + 3m-4n < 0 is: x > 4N − 3m2m − n, and 2m-n < 0, ∵ 4N − 3m2m − n = 49, i.e. n = 78m, 2m-78m < 0, solution: m < 0, n < 0, ∵ (m-4n) x + 2m-3n > 0, ∵ (m-72m) x ＞ - 2m + 218m, - 52mx > 58m, X ＞ - 14, i.e. the solution set of inequality (m-4n) x + 2m-3n > 0 is x ＞ - 14

A kindergarten has several toys for children. If each child has three toys, then there are 59 toys left. If each child has five toys, then there are few toys left for the last child? How many children are there?

Suppose there are x children, the total number of toys should be (3x + 59), so the inequality group can be listed as 3x + 59 ＜ 5x3x + 59 ＞ 5 (x − 1). Solve the inequality (1) to get x ＞ 592, solve the inequality (2) to get x ＜ 32, so the solution set of the inequality group is 592 ＜ x ＜ 32, and because x is a positive integer, x = 30 or 31; at this time, 3x + 59 = 149 or 152; answer: there are 30 children, 149 toys or 31 children playing There are 152 pieces

It is not difficult to solve the inequality of degree one variable
Given that the solution of half of equation (X-5) - A + 1 = x is suitable for inequality - 1 / 2 x ≤ 0, find the value of A
By the way, what do you mean by worried heavy

Because: - 1 / 2x ≤ 0, so x ≤ 0
Because (X-5) / 2-A + 1 = X
So x = - 3-2a ≤ 0
a≥-1.5

1. Let two unequal positive numbers a and B satisfy a ^ 3-B ^ 3 = a ^ 2-B ^ 2, then the value range of a + B______
2. Compare log3 4____ log6 7
3. If a, B, C, D are positive numbers and satisfy a + B + C + D = 4, and M is the largest of a + B + C, a + B + D, a + C + D, B + C + D, then the minimum value of M is________
4. When n > = 3, n belongs to N, prove: 2 ^ n > = 2 (n + 1)
5. Given that real numbers a, B, C satisfy a > b > C, and have a + B + C = 1, a ^ 2 + B ^ 2 + C ^ 2 = 1, verification: 1

1、a^3-b^3=a^2-b^2,()(a-b)=()(a-b)
a+b=a^2+ab+b^2=(a+b)^2-ab>(a+b)^2-1/4*(a+b)^2=3/4*(a+b)^2
So: 0