Given the complex number Z1 = 1 – 2I, Z2 = 3 + 4I, I is an imaginary unit. ① if the point corresponding to complex number Z1 + bz2 is in the fourth quadrant, find the value range of real number a. ② if z = (z1-z2) / (z1 + Z2), find the conjugate complex number of Z

Given the complex number Z1 = 1 – 2I, Z2 = 3 + 4I, I is an imaginary unit. ① if the point corresponding to complex number Z1 + bz2 is in the fourth quadrant, find the value range of real number a. ② if z = (z1-z2) / (z1 + Z2), find the conjugate complex number of Z

【1】
z1＋bz2
=(1－2i)＋b(3＋4i)
=(1＋3b)＋(－2＋4b)i
If the complex number is in the fourth quadrant, then:
1＋3b>0、－2＋4b

How to restore the calculator that is set to rounding accidentally?
The calculator is set to round
I don't know how it was set up like this
Who knows how to restore
My calculator is a kind of scientific calculator, k.l.t brand, which is a little different from ordinary Casio,
There is no reset key
I tried 2 nd f.0, but I still couldn't
What else can I do

Let's be specific about the model of your calculator?
Generally speaking, it should be the problem of setting decimal places
If you have a scientific calculator,
You can press "2ndf" → "." 0 "first to see if it can be solved
【…… 】
I went back to have a good study. I found that there are many kinds of calculators in China,
The usage of various calculators is different
I have a look. I think I can try it from the following:
The general idea of setting decimal places is (I was wrong last time, sorry)
"2ndf" → "tab" → "(the number key of the decimal point you want; the key to cancel setting the decimal point is". ")"
(the "tab" key is generally set at the "f ←→ e" key or "=" key; or other keys...)
Of course, you can also look at the manual, which may contain some introduction
If it doesn't work, you can take the battery out for a while and then replace it

If the value of polynomial 2A ^ 2 + 3b-7 is 4, try to find the value of 6A ^ 2 + 9b-7

2A ^ 2 + 3b-7 = 4, so 2A ^ 2 + 3B = 11, so 6A ^ 2 + 9b-7 = 3 (2a ^ 2 + 3b) - 7 = 33-7 = 26

The function f (x) is defined at point x = x0 and is continuous ()
A. Sufficient and unnecessary condition B. necessary and insufficient condition C. sufficient and necessary condition D. neither sufficient nor necessary condition

From the definition that f (x) is continuous at point x = x0, we can see that f (x) is continuous at point x = x0 {function f (x) has a definition at point x = x0; otherwise, it does not hold. Therefore, it is a necessary and insufficient condition, so we choose: B

1 of a (a + 1) + (a + 1) (a + 2) + (a + 2005) (a + 2006)

1 of a (a + 1) + (a + 1) (a + 2) + (a + 2005) (a + 2006) 1 / 2
=[1/a-1/(a+1)]+[1/(a+1)-1/(a+2)]+[1/(a+2)-1/(a+3)]+.[1/(a+2005)-1/(a+2006)]
=1/a-1/(a+2006)
It is suggested that 1 / A-1 / (a + 1) should be divided into 1 / A-1 / (a + 1)

The square of X is more than 2x

Wrong. For example, when x is 1, is the square of 1 less than 2 times 1

6 / 35 = 26 / 45 / 25

6 / 35 = 26 / 45 / 25
x×35/6=26/45×(25/15)
x=26/45×(5/3)×6/35
x=52/315

There is a column of numbers, A1, A2, A3., an. Starting from the second number, each number is equal to the difference between 1 and the reciprocal of the number in front of it. If A1 = 2, then A2008 is ()

a1=2
a2=1/2
a3=-1
a4=2
loop
a2008=2

(10 + 2x) (76-4x) = 1080

760-40x＋152x－8x²＝1080
8x²－112x＋320＝0
x²－14x＋40＝0
（x－10）（x－4）＝0
∴x1＝4 2＝10

Let us know that the optimal solution of linear programming problem is (x1, X2, x3) = (- 5,0, - 1);
2. Write and find the optimal solution of the dual problem
min z=2x1-x2+2x3
s. T - > - X1 + x2 + X3 = 4; - X1 + x2-kx3 ≤ 6; x1 ≤ 0, X2 ≥ 0.x3 unconstrained
It's mainly about finding the value of K,

K = 1, the optimal solution of the dual problem is: (0, - 2)
The dual problem is as follows
max Z=4w1+6w2
s.t. -w1-w2 >= 2
w1+w2