The total electricity consumption is 3887220 V, 10 (60) a, 300 R / kWh, 50 Hz, and the power consumption of sub meters is 911220 V, 3 (6) a, 2000 R / kWh, 50 Hz My family shares a master meter with another family. The other family has a sub meter, but we don't have one. We always subtract their sub meter from the master meter, which is our electricity bill. I feel that their household electricity is more than ours, and our electricity bill is more than theirs. The master meter shares 3887 degrees. This is the meter of Power Supply Bureau, 220 V, 10 (60) a, 300 R / kWh, 50 Hz, 3 (6) a, 2000r / kWh, 50. They live at 911 degrees. I want to know how much electricity we have. Can we use the formula to express it?

The "R" in 300r / kWh and 2000r / kWh refers to the "rotation" of the disc type electric meter (inductive mechanical type). This kind of meter takes a long time, and the error (electric power term is meter loss) is large. It is unreasonable to calculate the electricity charge for a long time. How much electricity do you use = 3887-911-line loss meter loss in addition to the meter code provided by you
The total electricity consumption of the meter is 3887 degrees, 220 V, 10 (60) a, 300 R / kWh, 50 Hz, and that of the sub meter is 911 degrees, 220 V, 3 (6) a, 2000 R / kWh, 50 Hz
My family shares a master meter with another family. The other family has a sub meter, but we don't have one. We always subtract their sub meter from the master meter, which is our electricity bill. I feel that their household electricity is more than ours, and our electricity bill is more than theirs. The master meter shares 3887 degrees. This is the meter of Power Supply Bureau, 220 V, 10 (60) a, 300 R / kWh, 50 Hz, 3 (6) a, 2000r / kWh, 50. They live at 911 degrees. I want to know how much electricity we use. Can we use the formula? Our two families live in the house almost at the same time (more than two years). Our two families have a fluorescent lamp (they have more than our family), a 17 inch TV (this may use the same), an electric cooker (they are cooking two meals, Our family has more power, electric pressure cooker (I don't know if they have it or not), an electric water heater (we use it for about 30 minutes every day, I don't know if they have it), and so many electrical appliances in our family, I don't know if they have a water dispenser (I don't know if it's used to boil water, but I guess it often has) and other electrical appliances. I don't know how our two families are so different in electricity. It's almost twice as much. Although I know their electricity bill may not be more than that, I still don't say anything. We share a toilet, The light bulb in the toilet uses their home electricity. I don't know how many watts the light bulb is. I guess it's 15 watts. It should be because the light is very dark. In fact, we don't use their home electricity for more than 20 minutes a day. Their people everywhere publicize that we use their home electricity, which makes me very angry. I want to know the real power consumption of our two families, If you know the answer,
How much power can this meter provide? 5 (60) a, 220 V, 50 Hz, 1200 imp / kWh
5 A is long-term use, 60 A is short-term use, such a large power for a long time will heat up!
Let the sum of the first n terms of the sequence {BN} be Sn, and BN = 2-2sn; let the sequence {an} be an arithmetic sequence, and A5 = 14, a7 = 20, if CN = an · BN [n = 1,2,3 Let tn be the sum of the first n terms of the sequence {CN}, find TN and write out the solution of each step.)
(1) : 2Sn = 2-BN (1) 2sn-1 = 2-bn-1 (2) (1) - (2): 2bn = - BN + bn-13bn = bn-1bn / bn-1 = 1 / 3, n ≥ 2. When n = 1, B1 = 2 / 3, so BN is equal ratio, the first term? Common ratio? The general term formula is BN = 2 / 3? (2): the second question is too long, I will listen to it for you
(1):2Sn=2-bn (1)
2Sn-1=2-(bn-1) (2)
(1)-(2):2bn=-bn+bn-1
3bn=bn-1
Let q = BN / bn-1 = 1 / 3 (n ≥ 2)
When n = 1, B1 = 2 / 3
... unfold
(1):2Sn=2-bn (1)
2Sn-1=2-(bn-1) (2)
(1)-(2):2bn=-bn+bn-1
3bn=bn-1
Let q = BN / bn-1 = 1 / 3 (n ≥ 2)
When n = 1, B1 = 2 / 3
So BN is an equal ratio sequence of the first term 2 / 3 and the common ratio 1 / 3.
The general term formula is BN = B1. Q ^ (n-1) (2): find the general term formula of an: an = (3n-1)
Express CN. Cn=b1.q^(n-1)(3n-1)
Tn=C1+C2+…… +Cn
?Tn=?×2+1/32×5+… ＋1/3?(3n-1)①
?×?Tn=1/32×2＋1/33×5＋… ＋1/3?(3n-4)＋1/3?×?(3n-1)②
②-①:?Tn=…… (this is an equal ratio, which is easy to find.)
Finally, TN is simplified as: TN = 7 / 2 - (11 / 2 + 3n) / 3?
When TN ＜ 7 / 2, put it away
Mathematical simplification of senior one (1 + sin α + 2Sin α cos α) / (1 + sin α + cos α)
The title should be (1 + sin α + cos α + 2Sin α cos α) / (1 + sin α + cos α) = [(sin α + cos α) + (sin α) ^ 2 + (COS α) ^ 2 + 2Sin α cos α] / (1 + sin α + cos α) = [(sin α + cos α) + (sin α + cos α) ^ 2] / (1 + sin α + cos α) = [(1 + sin α + cos α) (sin α + cos α)] /
There seems to be something wrong with the title?! ~
The known set a = {x | (4x ^ 2-16x + 21) / (x ^ 2-2x + 1) > = 5}, B = {x | x ^ 2-ax + 9}
First, (x ^ 2-2x + 1) is always greater than zero, so it can be multiplied to the right of the inequality, and then it can be simplified to get - 8
What is the condition for the equal sign of this absolute trigonometric inequality to hold?
AB is less than 0
Follow up:
Let the sum of the first n terms of the sequence BN be Sn, and BN = 2-2sn. The sequence an is an arithmetic sequence, and A5 = 10, a7 = 14. (1) find the general formula of the sequence an and {BN}
(2) CN = 1 / 2anbn, TN is the sum of the first n terms of the sequence CN, find TN
Bn=2-Sn,Bn-B(n-1)=Bn Bn=B(n-1)/2 =B1/2^(n-1)=1/3*(1/2)^(n-2)
An=2n
Cn=1/2AnBn=n/3*(1/2)^(n-2)=8n/3*(1/2)^(n+1)
Tn=C1+...+Cn 2Cn-C(n-1)=8/3*(1/2)^n
Tn=2Tn-Tn=2C1+2C2+...+2Cn-C1-...-Cn=2C1+(2C2-C1)+(2C3-C2)+...+(2Cn-C(n-1))-Cn
=2C1-Cn+8/3*[(1/2)^2+(1/2)^3+...+(1/2)^n]=8/3-(2+n)/[3*2^(n-2)]
a∨n＝2n,b∨n＝2/3∧n
T ∨ n to use dislocation subtraction, first write the equation of t ∨ n, and then write 1 / 3T ∨ n, two simultaneous subtraction, it is OK. I can't write here. Why don't you give me an address and I'll take photos of the process and send them to you.
Bn=2-Sn，Bn+1=2-Sn+1 ,Bn+1-Bn=-Bn+1
2Bn+1 = Bn ,B1=2/3 ,Bn=2/3 * (1/2)^(n-1)
d=(14-10)/2=2 ,An=2*n
(1)
a(5)=a(1)+4d=10
a(7)=a(1)+6d=14
a(1)=2
D=2
a(n)=2+(n-1)2=2n
b(n)=2-2s(n)
s(n)=1-b(n)/2
s(n-1)=1-b(n-1)/2
b(n)=s(n)-s(n-1)=[b(n-1)-b(n)]/2
3b(n)=b(n-1)
Q... unfold
(1)
a(5)=a(1)+4d=10
a(7)=a(1)+6d=14
a(1)=2
D=2
a(n)=2+(n-1)2=2n
b(n)=2-2s(n)
s(n)=1-b(n)/2
s(n-1)=1-b(n-1)/2
b(n)=s(n)-s(n-1)=[b(n-1)-b(n)]/2
3b(n)=b(n-1)
q=b(n)/b(n-1)=1/3
b(1)=2-2s(1)=2-2b(1)
b(1)=2/3
b(n)=b(1)q^(n-1)=2/3*(1/3)^(n-1)=2(1/3)^n
(2)
c(n)=1/2a(n)b(n)
=1/2*2n*2(1/3)^n
=2n/3^n
t(n)=2/3+4/9+6/27+...+2(n-2)/3^(n-2)+2(n-1)/3^(n-1)+2n/3^n
3t(n)=2+4/3+6/9+...+2(n-2)/3^(n-3)+2(n-1)/3^(n-2)+2(n-1)/3^(n-1)
3t(n)-t(n)=2t(n)=2+2/3+2/9+2/27+...+2/3^(n-3)+2/3^(n-2)+2/3^(n-1)-2n/3^n
t(n)=1+1/3+1/9+1/27+...+1/3^(n-3)+1/3^(n-2)+1/3^(n-1)-2n/3^n
=1*[1-(1/3)^n]/(1-1/3)-2n/3^n
=3/2*(1-1/3^n)-2n/3^n
=^ 2n / 3
It is proved that Tan α· sin α Tan α − sin α = Tan α + sin α Tan α· sin α
In order to make Tan α· sin α Tan α − sin α = Tan α + sin α Tan α· sin α tenable, we only need (Tan α· sin α) 2 = (Tan α + sin α) (Tan α - sin α) tenable, ∵ tan2 α - sin2 α = sin2 α Cos2 α − sin2 α = (sin2 α) (1cos2 α − 1) = sin2 α· 1 − Cos2 α cos
Has set a = {x | x square + 2x (a + 1) + a square - 1 = 0}, B = {x | x square + 4x = 0}, a ∩ B = a, find the value range of real number a, the teacher said to discuss!
B={0,-4}
Anb = a, then a is a subset of B
1) If a = Φ, then Δ = 4 (a + 1) ^ 2-4 (a ^ 2-1)

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