Let's know: 12's x power = 3, 12's y power = 2 to find: 8 and (1-x + y) of (1-2x)

Let's know: 12's x power = 3, 12's y power = 2 to find: 8 and (1-x + y) of (1-2x)

X=log12 3
Y=log12 2
It's OK to bring it in. Isn't it difficult

1. : 1.2 = 12, find x 2, 1.8: x = 2 and 2 / 3

14.4:1.2=12
x=14.4
1.8：x=8：3
So x = 1.8 * 3 / 8 = 0.675

3 to the power of x = 12 to the power of y = 8, find the value of 1 / X-1 / y

3^x=8
x=log3(8) 1/x=log8(3)
12^y=8
y=log12(8) 1/y=log8(12)
1/x-1/y=log8(3)-log8(12)=log8(1/4)=-2/3

Factorization (6 18:24:43)
1. The square of a and the cube of B-B
2. &# 160; the square of 9x + 16-24x

a^2b-b^3
=b(a^2-b^2)
=b(a+b)(a-b)
9x^2+16-24x
=(3X-4)^2

Given that point a (3x-6, 4Y + 15) and point B (5Y, x) are symmetric about X axis, then the value of X + y is ()
A. 0B. 9C. -6D. -12

∵ point a (3x-6, 4Y + 15), point B (5Y, x) are symmetric about the X axis, ∵ 3x-6 = 5Y; 4Y + 15 + x = 0; the solution is: x = - 3, y = - 3, ∵ x + y = - 6, so C

x+210=(x+360)*9/19

Multiply both sides of the equation by 19
19 (x + 210) = 9 (x + 360)
The result is 19x + 19 * 210 = 9x + 360 * 9
Subtract 10x from both sides at the same time
We get 10x + 19 * 210 = 360 * 9
Subtract 19 * 210 from both sides at the same time
We get 10x = 360 * 9-19 * 210
The reduction result is x = - 75
Finished

Determinant of linear algebraic matrix
Let a be a third-order matrix, and | a | = - 2, find | a | a ^ 2 A ^ t |
I didn't understand the first step of the process
=|A|^3 |A^2A^T|
=|A|^3 |A^2| |A^T|
=|A|^6
=64

Note that | a | is a number. Use the formula | Ka | = k ^ n | a |, where k = | a |, n = 3

Given the dry bulb humidity and wet bulb temperature, how to calculate the relative humidity?
As the title!
How to calculate it? By what?

First look at the temperature above the dry bulb thermometer, and then look at the temperature above the wet bulb thermometer. Then subtract the temperature of the wet bulb thermometer from the reading of the dry bulb thermometer, and then compare the number with the relative thermometer to get the relative humidity
Relative humidity meter download:

When x tends to 1, what is the limit of 2 / 2 of the square of x minus 1 / 1 of X?
When the limit of X tends to 1, what is the limit of 2 / (Square-1 of x) - 1 / (x-1)?

It is - 1 / 2. First, we divide it into (1-x) / (the square of x-1); because x tends to be 1, so x is not equal to 1, so we can reduce (x-1) and divide it into negative 1 / (x + 1); we can substitute 1 to get - 1 / 2
Encounter two fractions of the limit problem, generally first general

How to calculate (8-0.08) * 1.25

(8-0.08)*1.25=8*1.25-0.08*1.25=10-0.1=9.9