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In the concept of plural, 3. If (x + y) + (Y-1) I = (2x + 3Y) + (2Y + 1) I, find the value of real number x, y I'm a rookie with poor grades and poor foundation, Do you have a more detailed idea of why I have to move like this? I'm really a rookie. Please prawn. I want everything, ha ha.
The real part is equal to the real part, and the imaginary part is equal to the imaginary part (that is, the one with I), so that the formula is
X + y = 2x + 3Y and Y-1 = 2Y + 1
Just solve these two equations
The solution is either?... x, y all move to the same side
The first equation is moved to the end, x + 2Y = 0; the second equation is y = - 2
Substituting y = - 2 into the first equation gives x = 4
The two lamps marked with "220 V 40 W" are connected in parallel to the home circuit, and the total power consumed by the two lamps is
The two lamps marked with "220 V 40 W" are connected in parallel to the home circuit, and the total power consumed by the two lamps is
40w*2=80w
(2a²-ab+1)-2(a-2ab+3)
(2a²-ab+1)-2(a-2ab+3)
=2a²-ab+1-2a+4ab-6
=2A square + 3ab-2a-5
When the rated voltage of 6V is connected in series with the resistance of 4 ohm, it will work normally. When the circuit current is 0.5A, the resistance of the electric appliance is 0______ The power supply voltage is______ V.
Because the rated voltage of the consumer is 6V and it works normally after being connected in series, the voltage occupied by the consumer is 6V and the current is 0.5A, which means that the resistance of the consumer is r = V / I = 12 Ω and the power supply voltage is u = I (R1 + R2) = 8V
I want to know the volume of this container
If it can be solved tonight, there will be additional points
S (bottom area) = π (circumference) × a × B / 4 (where a and B are the length of major axis and minor axis of ellipse respectively)
Volume of container = sh (height) = 3.2 × 3.14 × 1.3 × 1.1 / 4 = 3.59216
What is the power consumption of a 100 watt bulb in 10 hours
1 kwh = 1 kwh, so a 100W bulb consumes 1 kwh in 10 hours
Find the value of the algebraic formula: 5A & # 178; b-7ab & # 178; - 8A & # 178; B + 9ab & # 178;, where a = 2, B = - 1 / 4
The original formula = 5A & # 178; b-7ab & # 178; - 8A & # 178; B + 9ab & # 178;
=（5a²b-8a²b）+（9ab²-7ab²）
=-3a²b+2ab²
When a = 2, B = - 1 / 4,
Original formula = - 3 × 4 × (- 1 / 4) + 2 × 2 × (1 / 16)
=3+1/4
=Three and a quarter
There are two metal wires a and B of the same length, same quality and different materials. It is known that the density of a is larger than that of B, and the resistivity of a is smaller than that of B. then a and B are two wires
There are two metal wires a and B with the same length, the same mass and different materials. It is known that the density of a is larger than that of B, and the resistivity of a is smaller than that of B. compare the resistance of a and B? What is the relationship between mass, density and cross-sectional area?
Because the length and mass of AB are the same, and the density of a is larger, the volume of a is smaller under the condition of equal mass, and because the length is the same, the cross-sectional area of a is smaller
When the length is the same, the cross-sectional area of a is small, if it is the same material or the resistivity of a is large, then the resistance of a must be large
Unless there is a specific value to calculate
All the odd numbers are arranged in regular order. The first line is 1, the second line is 3, 5, the third line is 7, 9, 11, 2011 is which line
one
3 5
7 9 11
13 15 17 19
21 23 25 27 29
What line is 2011
2011 is the (2011 + 1) / 2 = 1006th odd number
The sum of odd numbers is n (n + 1) / 2. When n = 44, n (n + 1) / 2 = 990
1006-990 = 16
So 2011 is on line 45, number 16
How big is the current passing through the bulb when the pz220v100w bulb is connected to the 220V circuit? How big is the resistance of the bulb at this time? What is the resistance of the pz220v40w bulb?
To find out the current through a 220v100w bulb, you can use the formula: P = IV, I = P / v = 100 / 220 = 0.45 (a) to find out the resistance of the bulb. You can find out P = u ^ 2 / RR = u ^ 2 / P = 220x220 / 100 = 484 (Ω) according to the parameters of the bulb. Similarly, the current of a 220v40w bulb is: I = P / u = 40 / 220 = 0.18 (a
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