Add the following addition, subtraction, multiplication and division or brackets 9 8 7 6 4 3 2 1 = 10 9 8 7 6 5 4 3 2 1 = 1000 9 8 7 6 5 4 3 2 1 = 1994

Add the following addition, subtraction, multiplication and division or brackets 9 8 7 6 4 3 2 1 = 10 9 8 7 6 5 4 3 2 1 = 1000 9 8 7 6 5 4 3 2 1 = 1994

9－8－7－6－5－4＋32－1=10
987+6+5+4-3+2-1=1000
（987+6+5-4+3）*2=1994
You made me pick up the pen I didn't use for a long time

The specification of an electrical appliance is "120V, 4A", which is connected in a 220 V circuit. How many ohm resistor should be connected in series in the circuit,
Only in this way can electrical appliances work normally

25 Ohm, series partial voltage, so the voltage on the resistance should be 220 v-120 v = 100 V, so the resistance should be 100 / 4 = 25 Ohm

The midpoint of a chord of parabola y & # 178; = - 12x is m (- 2, - 3). The equation of the secondary chord and the chord length are obtained
The linear equation has been solved. Y = 2x + 1. The chord length is 75

Your answer is correct, but the value of chord length needs to be simplified!
First question:
From the parabolic equation y ^ 2 = - 12x, we can see that the parabola is symmetric about the x-axis, but the point m (- 2, - 3) is not on the x-axis,
The equation of chord is y + 3 = K (x + 2), that is, y = KX + 2k-3
Simultaneous: y = KX + 2k-3, y ^ 2 = - 12x, eliminate y, get: (KX + 2k-3) ^ 2 = - 12x,
∴k^2x^2＋2（2k－3）kx＋（2k－3）^2＋12x＝0,
∴k^2x^2＋2［（2k－3）k＋6］x＋（2k－3）^2＝0.
The two ends of the string are on the straight line y = KX + 2k-3,
Let the coordinates of the two ends of the string be (m, KM + 2k-3) and (n, kn + 2k-3)
Obviously, m and N are two parts of the equation K ^ 2x ^ 2 + 2 [(2k-3) K + 6] x + (2k-3) ^ 2 = 0,
By Weida's theorem, M + n = - 2 [(2k-3) K + 6] / K ^ 2
From the midpoint coordinate formula, there are: (M + n) / 2 = - 2, ｜ m + n = - 4, ｜ - 2 [(2k-3) K + 6] / K ^ 2 = - 4,
∴（2k－3）k＋6＝2k^2,∴2k^2－3k＋6＝2k^2,∴k＝2,∴2k－3＝1.
The equation of string satisfying the condition is y = 2x + 1
Second question:
According to Weida's theorem, Mn = (2k-3) ^ 2 / K ^ 2 = [(2 × 2-3) / 2] ^ 2 = 1 / 4
The chord length
＝√［（m－n）^2＋（km－kn）^2］＝√［（1＋k^2）（m－n）^2］
＝√［（1＋4）（m－n）^2］＝√｛5［（m＋n）^2－4mn］｝
＝√｛5×［（－4）^2－4×（1/4）］｝＝√［5×（16－1）］＝5√3.

10 ohm and ohm and ohm and ohm and 40 ohm can make up which of the following group of resistances? Ohm two resistances
(A) 25 Ohm, 8 ohm (b) 8 ohm, 50 ohm (c) 30 ohm, 8 ohm (d) 50 ohm, 30 ohm

Series connection: r = R1 + R2 = 10 Ω + 40 Ω = 50 Ω
Parallel: 1 / r = 1 / R1 + 1 / r2
R=R1R2/（R1+R2）=10Ω×40Ω÷（10Ω+40Ω）=8Ω
So choose B

∫x^3/(9+x^2)dx=?
The next step is
∫xdx-9/2∫d（9+x^2)/(9+x^2)
How did you get out?

∫x^3/(9+x^2)dx=∫[x-9x/(9+x^2)]dx=∫xdx-9/2∫1/(9+x^2)dx^2=∫xdx-9/2∫1/(9+x^2)d(9+x^2)=1/2x^2-9/2ln(9+x^2)+c

A 25 watt bulb turns on for many hours
How can we figure it out? The bulb in our stairwell is ordinary lighting, not energy-saving lamp!

Hello
One kilowatt hour is equal to one thousand watt hours
The physical meaning of 1 kilowatt hour is the electric energy consumed by an electric appliance with power of 1 kilowatt working for 1 hour
Your light is 25 watts
1000 divided by 25 is the number of hours
In other words, a 25 watt light bulb consumes one degree of electricity in 40 hours
It can also be said that 40 25 watt light bulbs consume 1000 watt hours per hour, which is one degree

1.x^3+x^3y^3+y^3=12
x+xy+y=0
It's a system of equations

Solution; X + XY + y = 0, x + y = - XY
x^3+x^3y^3+y^3=12
The original formula can be changed to (x + y) (x ^ 2-xy + y ^ 2) + x ^ 3Y ^ 3 = 12
(x+y)[(x+y)^2-3xy]+x^3y^3=12
Substitute x + y = - XY to get - XY (x ^ 2Y ^ 2-3xy) + x ^ 3Y ^ 3 = 12
-x^3y^3+3x^2y^2+x^3y^3=12
3x^2y^2=12
Xy = 2 or xy = - 2
x+y=-2,xy=2
Or x + y = 2, xy = - 2
What's left are two equations. You can solve them by yourself, because there should be four sets of roots

An electric iron is marked with the word "220 V 880 W". If it is connected to a 110 V power supply, what is the current passing through it and its actual power?

The current is 0.5 × 880 / 220 = 2A
Actual power = 2A × 110V = 220W

[2 - (5 / 4 * 4 / 7 + 1.25 * 3 / 7)] / 0.375 = how to calculate will be easier

The original formula = [2-5 / 4 * (4 / 7 + 3 / 7)] / 0.375 = (2-1.25) / 0.375 is equal to 0.75/0.375 = 2

When a motor works normally, the voltage at both ends of the coil is 380V, the coil resistance is 2 Ω, and the current in the coil is 10A______ J. What is the heat produced______ J．

The electric energy consumed by the motor in normal operation for one second is w = uit = 380 × 10 × 1J = 3800j. The heat generated is q = i2rt = 102 × 2 × 1 = 200J, so the answer is: 3800; 200