To express the diameter range of colloidal particles in cm is to convert 1nm ~ 100nm into cm. What is the result?

To express the diameter range of colloidal particles in cm is to convert 1nm ~ 100nm into cm. What is the result?

It is: 10 ^ - 7 ~ 10 ^ - 5 cm
1 nm ==10^-9 m
1 m==100 cm

Conversion between 1m and 1cm and 1mm 1nm and 1um
Nm (micron), um (nanometer), each between the need to have

1nm = 0.001 μ M = 0.000001 mm = 0.0000001cm = 0.00000001 decimeter = 0.000000001 meter

Write the polynomial x ^ 2-6xy + 12xy ^ 2-8y ^ 3 + 1 as the sum of two integers, so that one of them does not contain the letter X: the original formula = () + (); if it is written as a cubic term
Difference from another integer: original = () - ()

The original formula = (X & # 178; - 6xy + 12xy & # 178;) + (- 8y & # 179; + 1)
The original formula = (12xy & # 178; - 8y & # 179;) - (- X & # 178; + 6xy-1)

It is known that the square of M & # 178; - M-7 of the function y = (M & # 178; - 5m-6) x is an inverse proportional function,
And when x < 0, y decreases with the increase of X, what is the value of M?

m²-m-7=-1
We get (M-3) (M + 2) = 0
There are m = 3 or M = - 2
From M & # 178; - 5m-6 = (M-6) (M + 1), when m = 3, y = - 12 / X; when m = - 2, y = 8 / X
When x < 0, y decreases with the increase of X, then M = - 2

Let f (x) = x * m and f (2) < f (3), where m is a part of the equation x & sup2; + X-6 = 0

x²＋x－6=0;x1=2,x2=-3
The power function f (x) = x * m, and f (2) < f (3), M > 0
m=2

Given that the image of power function y = f (x) passes through point (2, 2 under the root), then f (9) =?
It is known that f (x) = {- X - |, (x)

Hello!
(1) ∵ power function y = f (x) image passing through point (2, √ 2),
Let y = f (x) = x ^ A, √ 2 = 2 ^ a
∴a=1/2
∴y=f(x)=x^1/2,
∴f(9)=9^1/2=3

Given that a, B and C are rational numbers, and the fifth power of AB multiplied by the fifth power of C is greater than 0, ACC, then ()
A.a0,b0,b>0,c

The fifth power of AB multiplied by the fifth power of C > 0 is equivalent to ABC > 0
So there are two or no negative numbers in ABC
According to ACC, C is negative
A is a positive number
Choose B

If the increasing function y = f (x) defined on R has f (x + y) = f (x) + F (y) for any x, y ∈ R, then (1) find f (0); & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (2) prove that f (x) is an odd function; (3) if f (K · 3x) + F (3x-9x-2) < 0 is constant for any x ∈ R, find the value range of real number K

(1) In F (x + y) = f (x) + F (y), let x = y = 0, f (0) = f (0) + F (0), then f (0) = 0, (2) let y = - x, then f (x-x) = f (x) + F (- x), and f (0) = 0, then 0 = f (x) + F (- x), then f (x) is proved to be an odd function; (3) because f (x) on R is

If the increasing function y = f (x) defined on R has f (x + y) = f (x) + F (y) for any x, y ∈ R, then (1) find f (0);
The answer is (1) in F (x + y) = f (x) + F (y),
Let x = y = 0, f (0) = f (0) + F (0),
Then f (0) = 0, but I don't know why f (0) = f (0) + F (0) can become f (0) = 0

f（0）=f（0）+f（0）
f（0）=2f（0）
transposition
-f（0）=0
f（0）=0

The function f (x) defined on R satisfies that f (x + y) = f (x) + F (y) is an odd function for any x, y ∈ R

It is proved that: let x = y = 0, substitute f (x + y) = f (x) + F (y), then f (0 + 0) = f (0) + F (0), that is, f (0) = 0. Let y = - x, substitute f (x + y) = f (x) + F (y), then f (x-x) = f (x) + F (- x), and f (0) = 0, then 0 = f (x) + F (- x). That is, f (- x) = - f (x) holds for any x ∈ R, so f (x) is an odd function