You can change it at the bank for one jiao and two jiao and one yuan and two yuan and five yuan and ten yuan and one hundred yuan I'm talking about the fourth set of RMB The third set is OK

You can change it at the bank for one jiao and two jiao and one yuan and two yuan and five yuan and ten yuan and one hundred yuan I'm talking about the fourth set of RMB The third set is OK

If you know the people inside the bank, it's still possible, because money is not only the past but also the present. It's possible for people to benefit from money. It's very convenient for people inside the bank

The value of K is ()
A. - 1 or 2B. 2C. - 1D. 1 + 3

Let a (x1, Y1), B (X2, Y2). From y = KX − 2Y2 = 8x, we can get k2x2 - (4K + 8) x + 4 = 0, from △ = [(4K + 8)] 2-16k2 = 64K + 64 > 0, we can get k ＞ - 1; X 1 + x 2 = 4K + 8K 2. The abscissa of a and B midpoint is 2, ∩ 4K + 8K 2 = 4, and the solution is k = - 1 (rounding) or K = 2. Therefore, if the line y = kx-2 and parabola y 2 = 8x intersect at a and B, and the abscissa of AB midpoint is 2, the value of K is 2

If the length of an aluminum conductor is 300m and the sectional area s is 4mm2, how many ohms is the resistance of the conductor?
The resistivity of aluminum at 20 ℃ is p = 0.0283 * 10 minus 6 ohm

The resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area
300 * 0.0283/4 = 2.1225 ohm

Find ∫ DX / (X & # 178; - x + 1) ^ (3 / 2)

How many kilowatt hours of electric energy will be consumed when the rice cooker with rated power of 500 watts works normally for 30 minutes?
If the rice cooker with rated power of 500 watts works normally for 30 minutes, it will consume kilowatt hours of electric energy; if the electric lamp with rated power of 100 watts works normally, it will consume 1 kilowatt hours of electric energy; if the air conditioner with rated power of 100 watts works normally, 1 kilowatt hours of electric energy will be consumed

(1) W = Pt = 500W × 30min = 0.5KW × 0.5h = 0.25kwh; (2) according to w = Pt, we can deduce: T = w / P = 1kwh △ 100W = 1kwh △ 0.1kw = 10h; (3) according to w = Pt, we can deduce: P = w / T = 1kwh △ 1H = 1kW

High school mathematics is in a hurry
The size of A2 (square) - 2A and A-3 requires a process

The former > the latter
a^2-2a-(a-3)=(a-3/2)^2+3/4>0
so.

After connecting the two bulbs marked with "220 V & nbsp; & nbsp; 200 W" and "220 V & nbsp; & nbsp; 40 W" in series into the 380 V circuit, what will be burned is______ ．

The resistance of bulb 1 is: R1 = u2p = (220V) 2200W = 242 Ω, the resistance of bulb 2 is: R2 = u ′ 2p ′ = (220V) 240W = 1210 Ω, the two bulbs are in series, r = R1 + R2 = 242 Ω + 1210 Ω = 1452 Ω, the current in the circuit is: I = ur = 380v1452 Ω = 0.26A, the voltage at both ends of bulb 1 is: U1 = IR1 = 0.26A × 242 Ω

What is the product of 7 minus 1 / 4 divided by 5 / 8 and multiplying the difference by 5 / 6?

5.5

The electromotive force of power supply e = 42V, internal resistance R = 1 Ω, r = 20 Ω, M is DC motor, its armature resistance R / = 1 Ω, the motor works normally, and the reading of voltmeter is 21V
(1) Current in motor M
(2) The power that a motor converts into mechanical energy
(3) The input power of the motor
(4) The heating power of the circuit
(5) The total power of the power supply
(6) The output power of the power supply

Is r = 20 Ω in this question connected in series with the motor in the circuit, and the voltage measured by the voltmeter is the voltage at both ends of the motor, if so,
1, the current of the motor I = u / r = (42-21) / (20 + 1) = 1a. Note: the voltage of the normal operation of the motor is 21V, indicating that the voltage drop consumed by the internal resistance and resistance R of the power supply is 21V
2, the power of the motor is p = UI = 21 * 1 = 21W, the power consumed by the armature resistance of the motor is p = 1W, then the power converted into mechanical energy of the motor is p = 20W
3. The input power of the motor is p = 21W
4. Heating power P = I * I * r = 20W
5, P total = EI = 42 * 1 = 42W
6, P transfusion = P total - P internal = 42-1 = 41w, P internal = I * I * r = 1W

67 times 63 plus 32 times 63 plus 63

67X63+32x63+63=67X63+32x63+63x1=(67+32+1 )x63=100x63=6300