The average of five natural numbers is 28. If the five numbers are arranged from small to large, the average of the first three numbers is 19, and the average of the last three numbers is 42, What's the number in the middle?

The average of five natural numbers is 28. If the five numbers are arranged from small to large, the average of the first three numbers is 19, and the average of the last three numbers is 42, What's the number in the middle?

The average of five natural numbers is 28
The sum of five natural numbers is 28 * 5 = 140
If the five numbers are arranged from small to large, then the average of the first three numbers is 19
Then the total number of the first three numbers is 19 * 3 = 57
The total of the last two numbers is 140-57 = 83
The average of the last three numbers is 42
The total number of the last three numbers is 42 * 3 = 126
The number in the middle is
Total number of last 3 - total number of last 2
126-83=43

There are five natural numbers with an average of 28. If the five numbers are arranged from small to large, the average of the first three decimals is 19, and the average of the last three large numbers is 42
What's the number in the middle?
three million one hundred and forty-five thousand six hundred and fourteen

43
Let these five natural numbers be a, B, C, D and e respectively
a+b+c+d+e=140 （1）
a+b+c=57 （2）
c+d+e=126 （3）
（1）-（2）
d+e=83 (4)
(3)-(4)
c=43

The first question is 0.45:1.5, the second is, two fifths: 0.8, the third is two ninths: one third, the fourth is, 0.25:3

1、
0.45:1.5=3：10
2、
Two fifths: 0.8 = 1:2
3、
Two in nine: one in three = 2:3
4、
0.25:3=1：12

1. How does (- x) ^ 3 · (- x ^ 4) calculate the general process of 2, a = 4 / √ 5-1

1、（﹣x）^3·﹙﹣x^4）
=-（x）^3·﹙﹣x^4）
=x^3·x^4
=x^(3+4)
=x^7
two
a=4／(√5－1)?
=4(√5+1)／[(√5－1)(√5+1)]
=4(√5+1)／[(√5)²－1²)]
=4(√5+1)／[5-1]
=√5+1

It is known that x satisfies the inequality: 4-2 (x-4) < 3 (x-1), simplifying | 6-2x | - | 2x-1|

4-2（x-4)＜3（x-1)
4-2x+8

Is there a rational number whose square is minus 4 / 9? - (2 / 3) is it?

No
The squares of rational numbers are nonnegative
2 / 3 squared is 4 / 9

It is known that the solution of the equation ax-3 = 3x-1 about X is x = 1, and the value of a is obtained

Substituting x = 1 into the original equation
a-3=3-1
a-3=2
a=5
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Calculate 1.1 * 11 * 1.1-1.1 * 1.1-1.1 with simple method

1.1*11*1.1-1.1*1.1-1.1
=1.1*1.1*（11-1）-1.1
=1.1*1.1*10-1.1
=1.1*11-1.1
=1.1*（11-1）
=1.1*10
=11

If y = ax ^ 2-2a ^ 2x + 1 in the interval [- 1 / 2,1 / 2], f (x) > 0 is constant, find the value range of A

Y = ax ^ 2-2a ^ 2x + 1 corresponds to x = a
(1) When a = 0, f (x) = 1 > 0 holds
(2) When a > 0, there are two cases
When 0 ＜ a ＜ 1 / 2
If f (x) min = f (a) = a ＾ 3-2a ＾ 3 + 1 > 0, then 0 ＾ a ＾ 1 / 2
When a ≥ 1 / 2
If f (x) min = f (1 / 2) = A / 4-A ＾ 2 + 1 ＞ 0, the solution is 1 / 2 ≤ a ＜ (1 ＋ 65) / 8
From (2) we know that 0 ＜ a ＜ (1 ＋ 65) / 8
(3) When a ＜ 0, f (x) min = f (1 / 2) = A / 4-A ＾ 2 + 1 ＞ 0, the solution is (1 - √ 65) / 8 ＜ a ＜ 0
In conclusion: (1 - √ 65) / 8 ＜ a ＜ (1 ＋ 65) / 8
If the calculation is not wrong, it should be this result, considering the method of taking the maximum value of quadratic function in a given interval and the problem of constancy

The polynomial (X & # 178; + MX + n) (X & # 178; - 3x + 4) has no X & # 179; term and X & # 178; term at the back. Try to find the value of M, n

(x²+mx+n)(x²-3x+4)=x⁴+(m-3)x³+(n+4-3m)x²+(4m-2n)x+4n
If the expansion does not contain X & # 179; term and X & # 178; term, then the coefficients of X & # 179; term and X & # 178; term are equal to 0
m-3=0
n+4-3m=0
Solution
m=3 n=5