The voltage of a section of resistor is 4V, and the current passing through it is 0.2A. With the last resistor, the current in the trunk increases by 0.4A. Ask r sum r?

The voltage of a section of resistor is 4V, and the current passing through it is 0.2A. With the last resistor, the current in the trunk increases by 0.4A. Ask r sum r?

R=U/I
The voltage is constant 4V
R = 4 / 0.2 = 20 ohm
Total current after parallel resistance = 0.2 + 0.4 = 0.6
R total = 4 / 0.6 = 20 / 3 = 6.6666666666

The curve represented by the equation x & # 178; + 2XY + Y & # 178; = 4 is
A. Two parallel lines
B. Two straight lines perpendicular to each other
C. One point
D. None of the above conclusions are correct

x²+2xy+y²=4
(x+y)^2=4
X + y = 2 or x + y = - 2
Two parallel lines

The known voltage is 380V, the current is 60A, the power is 100kW, and the distance is 100m

Ha ha, the condition is not enough, there is a problem. If you use three-phase or two-phase, the power of three-phase is 1.732 * 380 * 60 = 39490w, which is less than 100kW. The difference between two phases is further. What's your limit? The current can't be determined, it should be determined the power and voltage, the current should be deduced, and the conductor cross-sectional area should be selected according to the current carrying capacity

Are the following functions increasing or decreasing in the specified interval? F (x) = x & # 178; + 1, X ∈ (0, + ∞)
f(x)=x²+1,x∈(0,+∞) f(x)=﹣1/x,x∈﹙0,+∞﹚

They are all increasing functions

Is 600 watts a high-power electrical appliance? Schools are not allowed to use high-power electrical appliances. I don't know if this is considered. What's more, how many watts is a notebook

Whether 600W is a high-power electrical appliance should be determined according to the regulations of the school. If there is no specific stipulation on the specific wattage, I think 600W is not a high-power electrical appliance, and at least 1000W can be regarded as a high-power electrical appliance
Notebook is generally about 100 watts

Simple calculation of 99 times 64 plus 64
Fast

99*64+64
=64*（99+1）
=64*100
=6400

Why should the power supply voltage in the home be set at 220 V? What about the current?
Why does the current on the third floor increase with the increase of electrical appliances and power?

The voltage is 220 v. I can't change it. If you want to install a booster with high voltage
When the voltage is fixed, the greater the total power of the electrical appliances in the power grid, the greater the current, P = UI
U is the voltage and I is the current

(0.25)^(-0.5)+(1/27)^(-1/3)-625^0.25

(0.25)^(-0.5)+(1/27)^(-1/3)-625^0.25
=(0.5^2)^(-0.5)+[(1/3)^3]^(-1/3)-5^4^0.25
=0.5^[2*(-0.5)]+(1/3)^[3*(-1/3)]-5^(4*0.25)
=0.5^(-1)+(1/3)^(-1)-5
=2+3-5
=0

Physical movement and force
Chapter 7 motion and force, how to classify the things studied in this chapter! Learned, force, measurement of force, gravity, composition of two forces on the same straight line, balance of two forces, what is the relationship between the size of friction, and the relationship between motion and force

First of all: we have learned the basic concept of force, such as the size, unit, representation, schematic diagram and measurement method of force, so this must be clarified. Then we have learned the most common force gravity, so we also need to understand the three elements of gravity, and then we naturally need to learn the relationship between force and force

(2x-1) & # + 179; + 216 = 0 is a calculation problem,
①.(2x-1﹚³+216=0
②.1/4（2x+3）³=16

The steps are as follows: use computer 1 (2x-1) &# 179; + 216 = 0 (2x-1) &# 179; = - 216 (negative 216) (press the triple root of the computer, then press 216, and then press equal, the result is - 6) 2x-1 = - 6 (negative 6) 2X = - 5x = 5 / 2 (5 / 2) 2 1 / 4 (2x + 3) &# 179; = 16 (2x + 3) &# 17