A household electric energy meter is marked with 220 v. a household electric energy meter is marked with "220 V 10A 3000 R / kWh". When all the electrical appliances are working, the electric energy meter turns 45 r in one minute. Therefore, the total power of all the electrical appliances in the household can be estimated as W? If the total current of all the electrical appliances in the household is not more than the rated current of the electric energy meter, the electrical appliances below w can be added to the circuit Why multiply by 10? Why divide by 3000?

220 * 10 = 2200W = 2.2kW voltage multiplied by current equals power
45*60=2700r
2700 / 3000 = 0.9Kw to 3000rpm equals 1KW
2.2-0.9=1.3kw
The total power of the household electrical appliances is 0.9Kw
The circuit can also add 1.3KW or less electrical appliances
I can't type the formula, only the answer. The total power is 900W
It can also add 1300W electrical appliances
The name plate of the electric energy meter is marked with "220 V 5 (10) a, 50 Hz, 2500 R / kW. H",
With a watch, even if the dial turns 25 times in three minutes, what is the power of the TV?
(60/3)*25/2500 = 0.2(KW) = 200W
The name plate of an electric energy meter is marked with "220V 10A" and "3000r / kW &; H". Connect an air conditioner to the electric energy meter and measure the dial
The name plate of an electric energy meter is marked with "220V 10A" and "3000r / kW &; H". If an air conditioner is connected to the electric energy meter separately, the power of the air conditioner will be zero if the dial turns 450 revolutions in 10 minutes____________ After using this air conditioner, it can be connected to the air conditioner_________ A "220 V 60 W" incandescent lamp
450*60/10/3000=0.9kw
(220*10/1000-0.9)/0.06=21
What is the power of the air conditioner____ 0.9kw___ After using this air conditioner, it can be connected to the air conditioner____ 21___ A "220 V 60 W" incandescent lamp
The absolute value of inequality x + (x-m) > 1 is constant on R, and the range of M is obtained
X + | x-m | > 1 is constant on R,
Then f (x) = x + | x-m | - 1 > 0 is constant on R
When x = m, f (x) = x + x-m-1 = 2x-m-1 > = M-1,
Therefore, if M > 1, f (x) > 0 is constant on R
The range of M is: (1, + ∞)
m> 2x-1: how to find out
10. Let the first n terms of sequence {an} and Sn = 3 / 2n ^ 2-1 / 2n, the sequence {BN} be equal ratio sequence, and A1 = B1, B2 (a2-a1) = B1, find the sequence {an}, {BN}
The general term formulas of are?
For an, it can be found as follows
an = Sn - Sn-1
For BN
b1 = a1 = S1
According to the general term formula of an, A1 and A2 are obtained, so that B1, A1 and A2 are known. According to B2 (a2-a1) = B1, B2 is obtained, so that
The common ratio Q of {BN} = B2 / B1
The first term and the common ratio of {BN} are solved, BN = B1 ^ (Q-1)
【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2+θ)-sin(3π/2-θ)】
Find the value of F (π / 3)
f(θ)=【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2+θ)-sin(3π/2-θ)】
=(2cos²θ+sin²θ+cosθ-3)/(2+2cos²θ+cosθ)
=(cos²θ+cosθ-2)/(2cos²θ+cosθ+2)
f(π/3)=[cos²(π/3)+cos(π/3)-2]/[2cos²(π/3)+cos(π/3)+2]
=(1/4+1/2-2)/(2*1/4+1/2+2)
=(-5/4)/3
=-5/12
It is known that the set M = {Y / y = x ^ 2-4x + 3, X belongs to Z}, and the set n = {Y / y = - x ^ 2-2x, X belongs to Z}. There are two kinds of wrong solutions: the first one is that 2x ^ 2-2x + 3 = 0 is obtained from the simultaneous equations of y = x ^ 2-4x + 3 and y = - x ^ 2-2x, so the intersection is empty; the second one is that M = {Y / y is greater than or equal to - 1}, n = {Y / y is less than or equal to 1}, so the intersection is {Y / - 1 ≤ y ≤ 1}, 1} What is wrong with the above two methods? What is the correct solution?
In the second solution: X belongs to Z, so y belongs to Z. then M = {Y / y = - 1,0,3,8...} = {Y / 0, (n ^ 2) - 1, n belongs to positive integer} n = {Y / y = 1,0, - 3, - 8.} = {Y / 1,1 - (m ^ 2)}
If inequality 3 (x + 1) ≥ 5x-2, then | 2x-5|=______ .
3 (x + 1) ≥ 5x-2, remove the brackets, 3x + 3 ≥ 5x-2, all move to the left side of the equal sign, get, - 2x + 5 ≥ 0, ∵ 2x-5 ≤ 0. ∵ the absolute value of a non positive number is its opposite number, ∵ 2x-5 | = 5-2x
The sum of the first n terms of the sequence {an} is Sn = 2n & sup2;, {BN} is an equal ratio sequence, and A1 = B1, B2 (a2-a1) = B1. Let
Let CN = an / BN, find the first n terms and TN of sequence
The general term of {an}
an=Sn-S(n-1)=2n2-2(n-1)2=2(n+n-1)(n-n+1)=4n-2
So: A1 = 2, A2 = 6
Because A1 = B1, B2 (a2-a1) = B1
So B1 = 2, B2 = 1 / 2
Because {BN} is an equal ratio sequence, the common ratio q = B2 / B1 = 1 / 4
So the first term of {BN} is 2 and the common ratio is 1 / 4
bn=2×(1/4)^(n-1)
When an = 4n-2, BN = 2 × (1 / 4) ^ (n-1), we can get the following formula:
Cn=an／bn=（2n-1）×4^(n-1)
so
Tn= 1 + 3×4 + 5×4^2 +7×4^3 +…… +Formula (2n-1) × 4 ^ (n-1) → (1)
(1) The left and right sides of the equation are multiplied by 4 at the same time to obtain:
4Tn= 4 + 3×4^2 + 5×4^3 +…… +Formula (2n-3) × 4 ^ (n-1) + (2n-1) × 4 ^ n → (2)
Then compare the two formulas (1) and (2) as follows. Note that except for the first term of formula (1) and the last term of formula (2), the indexes of all terms in the middle are the same, that is, the first power of 4, the second power up to the n-1 power all correspond one by one
You will also find that (3-1) = (5-3) = (7-5) = (2n-1) - (2n-3) = 2
Tn= 1 + 3×4 + 5×4^2 +7×4^3 +…… +Formula (2n-1) × 4 ^ (n-1) → (1)
4Tn= 4 + 3×4^2 +5×4^3 +…… +Formula (2n-3) × 4 ^ (n-1) + (2n-1) × 4 ^ n → (2)
(1) If (2) is subtracted from (2), we can get:
(1-4)Tn=1+ 2×(4+4^2+4^3+…… +4 ^ (n-1)) - (2n-1) × 4 ^ n (this method is called dislocation subtraction)
That is - 3tn = 1 + 2 × (4 ^ n-4) / 3 - (2n-1) × 4 ^ n
Namely
Tn= (2n-1)×4^n + 2×(4-4^n) -1
__________ ________ ____
3 9 3
An = 2 (2n-1), A1 = B1 = 2, B2 / B1 = 1 / 4, BN = 2. (1 / 2) ^ 2 (n-1) = 2 ^ (3-2n)
CN = (2n-1). 2 ^ (2n-2) = 2n.2 ^ (2n-2) - 4 ^ (n-1) = 1 / 2. N.4 ^ n - 4 ^ (n-1). The difficulty lies in finding the sum of the first n terms of n.4 ^ n, en = 1.4 ^ 1 + 2.4 ^ 2 + 3.4 ^ 3 +... + (n-1). 4 ^ (n-1) + n.4 ^ n.... （1）
... unfold
An = 2 (2n-1), A1 = B1 = 2, B2 / B1 = 1 / 4, BN = 2. (1 / 2) ^ 2 (n-1) = 2 ^ (3-2n)
CN = (2n-1). 2 ^ (2n-2) = 2n.2 ^ (2n-2) - 4 ^ (n-1) = 1 / 2. N.4 ^ n - 4 ^ (n-1). The difficulty lies in finding the sum of the first n terms of n.4 ^ n, en = 1.4 ^ 1 + 2.4 ^ 2 + 3.4 ^ 3 +... + (n-1). 4 ^ (n-1) + n.4 ^ n.... （1）
4.En= 1.4^2+2.4^3+3.4^4+...+（n-1）.4^n+n.4^（n+1）。。。（2）
(1) - (2): - 3EN = 1.4 ^ 1 + 4 ^ 2 + 4 ^ 3 +... + 4 ^ n-n.4 ^ (n + 1)
Given tan2 θ = 3 / 4 (π / 2 ＜θ＜π), find the value of (2cos & # 178; × θ / 2 + sin θ - 1) / √ 2cos (θ + π / 4)
solution
tan2θ=(2tanθ)/(1-tan^2θ)=3/4
That is, 3 (1-tan ^ 2 θ) = 8tan θ
That is, 3tan ^ 2 θ + 8tan θ - 3 = 0
(3tanθ-1)(tanθ+3)=0
∵π/2

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