∫ 1 / (x ^ 2-4x + 3) DX, indefinite integral,

Finding indefinite integral √ 4x ^ 2-1 DX

Let X be equal to 1 / 2 sect

Can 35 square three-phase four wire copper wire load about 120 kilowatts?
One of the central air conditioning is 80-90 kilowatts

If the cable length is less than 30 meters and the cable has good heat dissipation, it can be used in emergency

There is a large circular design drawing on the wall, where o is the center of the circle and a and B are on the circumference, as shown in the figure. Now I want to measure the distance between a and B, but the wall is very high,
Ladder, can't measure directly. If you are given a bamboo pole and a tape measure, can you measure the distance between ab bright spots? Tell me about your method

I didn't see your picture, but I suppose your picture is roughly the same as mine. A and B are in the upper part of the circle, and their positions are higher. 1. Stick the bamboo pole to the wall, let it pass through point a and center O, and mark a'2 at the intersection of the bamboo pole and the lower part of the circle. 2. Stick the bamboo pole to the wall, let it pass through point B and center O, and mark a'2 at the intersection of the bamboo pole and the lower part of the circle

What is the physical meaning of electric power?

Power -- a physical quantity indicating the speed of work done by an object
In quantity, it is equal to the work done by the current through the consumer in one second!
For reference only! Ask again if you have any questions!

Square of 15 = 225, square of 25 = 62

25 squared is 625

Physical circuit problem: R1 = 8 Ω, R2 = 20 Ω, R3 = 30 Ω, connect to the two ends of 20V power supply, and calculate the voltage on each resistor
In the circuit diagram, a voltage is connected in series with R1, and R2 and R3 (one up and one down) are connected in parallel with R1

If R2 and R3 are connected in parallel, the resistance is r = 20 * 30 / (20 + 30) = 12 Ω
If R1 = 8 Ω, then
I1=U/(R1+R)=20/(8+12)=1A
U1=1*8=8V
Then U2 = U3 = u-u1 = 20-8 = 12V
I2=U2/R2=12/20=0.6A,I3=U3/R3=12/30=0.4A

(2011 Zhongshan) as follows: 1 234 56789 10 11 12 13 14 15 16 17 18 19 20 22 24 25
1. The first number in line n
2. The last number in line n
3. How many numbers are there in line n
4. The sum of the numbers in line n
The first line is 1, the second line is 234, the third line is 56789, the fourth line is 10 11 12 13 14 15 16, and the fifth line is 17 18 19 20 21 22 24 25

12345678910 11 12 13 14 15 1617 18 19 20 21 22 23 24 25…… Obviously, the last number in each row is a complete square number, and its arithmetic square root is the number of rows. Therefore: the first number in the nth row = (n-1) ^ 2 + 1 = n ^ 2 - 2n + 2 the last number in the nth row = n ^ 2 how many numbers in the nth row = n ^ 2

The more lights are connected in the home circuit, the smaller the A: total current, the smaller the B: total resistance and the C: total voltage of the circuit

B: The smaller the total resistance is
Because the connected lamps are connected in parallel, and the more they are connected, the larger the cross-sectional area of the conductor is, so the smaller the resistance is

8 / 9 times [3 / 5 - (9 / 16-2 / 5)]

8/9x[3/5-(9/16-2/5)]
=8/9x[3/5+2/5-9/16]
=8/9x7/16
=7/18

∫ 1 / (x ^ 2-4x + 3) DX, indefinite integral,