3-7+11-15+19-23+27-13 To simplify the operation, I know the answer - 16, I don't know the process. This is the last question

3-7+11-15+19-23+27-13 To simplify the operation, I know the answer - 16, I don't know the process. This is the last question

(11 + 19) + (27 + 3) - (7 + 23) - 13-15 = - 2. The answer is - 2

The most difficult mathematical reasoning of civil servants
[example] 0,12,24,14, 120,16, ()
A.280 B.32 C.64 D.336
[example] 0,1,0,5,8,17,19, ()
A.21 B.37 C.102 D.106
[example] 67,54,46,35,29, ()
A.13 B.15 C.18 D.20

[example] 0,12,24,14120,16, () a.280 b.32 c.64 d.336 answer: D odd number is 0,24120,) 0 = 1 ^ 3-1, 24 = 3 ^ 3-3, 120 = 5 ^ 3-5 (336) = 7 ^ 3-7 [example] 0,1,0,5,8,17,19, () a.21 b.37 c.102 d.106 answer: D two now add 0 + 1 = 1 (5 ^ 0) 0 + 5

On the mathematical sequence inference of civil servants
-1,0,1,2,9,（） A.11 B.82 C.459 D 730
How do you calculate that

730
The third power of 9 + 1
This is also the case with the previous ones

Given the equation x ^ 2 + (4a + 1) x + 4A ^ 2-1 = 0 (1), if the equation has a constant real number solution, find the value range of real number a
Given the equation x ^ 2 + (4a + 1) x + 4A ^ 2-1 = 0 (1) if the equation has a constant real number solution, find the value range of real number a (2) if the equation has a constant non negative real number solution, find the value range of real number a

1) If there is a real solution, then the discriminant > = 0
That is, (4a + 1) ^ 2-4 (4a ^ 2-1) > = 0
8a+5>=0
a>=-5/8
2) If there are always nonnegative real roots, first of all, the discriminant > = 0 is a > = - 5 / 8
If there is no negative root, then two sum > = 0, two product > = 0, that is - (4a + 1) > = 0, 4A ^ 2-1 > = 0

Linear algebra, vector, in order to know that n-dimensional vectors A1, A2, A3 are linearly independent, prove that 3A1 + 2A2, A2-A3, 4a3-5a1 are linearly independent,

Prove: because A1, A2, A3 are linearly independent, so r (A1, A2, A3) = 3
From the known (3A1 + 2A2, a2-a3,4a3-5a1) = (A1, A2, A3) k
K=
3 0 -5
2 1 0
0 -1 4
= 22 ≠0
So K is an invertible matrix
So r (3A1 + 2A2, a2-a3,4a3-5a1) = R ((A1, A2, A3) k) = R (A1, A2, A3) = 3
So 3A1 + 2A2, A2-A3, 4a3-5a1 are linearly independent

The solution of 1.5x = 19 + X

1.5X＝19＋x
1.5x-x=19
0.5x=19
x=38

We know the equation 4 ^ X-2 ^ (x + 1) - B = 0 about X. if the equation has a solution, we can find the value range of B. when the equation has a solution, we can discuss the number of real roots and find the solution of the equation
The main idea is this kind of topic

Let t = 2 ^ x > 0
T ^ 2-2t-b = 0, the equation has positive roots
Therefore, there are: delta = 4 + 4B > = 0 > > b > = - 1
Because the sum of two is 2, there must be a positive root, so when b > = - 1, the original equation must have a solution
The product of two is - B, if - 1=

Add an integral to the square + 4 of the polynomial x to make it a complete square. It is to write two integers satisfying the above conditions. Please make the description easy to understand
By the way, what is an integral and what is a complete square formula?

x²+4+4*x=（x+2）²
x²+4-4*x=（x-2）²
monomial:
(single number)
The product of letters
Product of number and letter)
A polynomial is the sum of several monomials
Monomials and polynomials are called integers
The complete square formula means that if integral A and integral B satisfy a = B & # 178;, then a is called the complete square formula

It is known that f (x) = (a × 2 ^ x-1) / (2 ^ x + 1) a ∈ R is an odd function on R. for any k ∈ (0, + ∞) solution, f ^ - 1 (x) > log2 (1 + x) / K is an inequality
In f ^ - 1 (x) > log2 (1 + x) / K, 2 is the base; (1 + x) / K is the true number

Why is this question so tangled
F (x) = (a * 2 ^ x-1) / (2 ^ x + 1) is an odd function on R = = > F (0) = 0 = = > A = 1
F (x) = (2 ^ x-1) / (2 ^ x + 1) = = > f ^ - 1 (x) = log2 [(1 + x) / (1-x)] > log2 (1 + x) / K (1 > x > - 1) = = > (1 + x) / (1 + x) > (1 + x) / k = = > 1-x1-x > = 1 = = > no solution

VB for loop statement
Dim i,j,c As Integer
Dim a,b As String
c = 0
Adodc1.recordsource = "select * from inf where order time between #" & dtp1.value & "#" and # "& dtp2.value &" # "
Adodc1.Refresh
For j = 0 To Combo1.ListCount - 1
b = Combo1.List(j)
For i = 0 To Adodc1. Recordset.RecordCount - 1
a = Adodc1. Recordset.Fields (15).Value
If b = a Then
c = 1 + c
End If
Next i
Text8. Text = Combo1. List (J) + "total" + str (c) + "record"
Next j
There is a ComboBox control, which has the names of several provinces (not repeated), and the fields (fields (15)) in a table also have the names of provinces (repeated). Now I want to count the names of several provinces in the fields (fields (15)) in the table during this period, and the results are displayed in text8.text! For example, there are 8 records in Henan, 3 records in Inner Mongolia, 9 records in Henan

Use count, otherwise, if there are too many records, you will keep moving next and judging B = A. dim I as integer, n as long, s as stringtext8 = "adodc1. RecordSource =" select the name of field 15, count (*) from inf where order time between "&