Xiao Dong's mother received 380 yuan of bonus from her unit, of which 80 were 2 yuan, 5 yuan and 10 yuan. The number of 5 yuan and 10 yuan were equal. How many of these three kinds of RMB? Urgent! Need analysis

Xiao Dong's mother received 380 yuan of bonus from her unit, of which 80 were 2 yuan, 5 yuan and 10 yuan. The number of 5 yuan and 10 yuan were equal. How many of these three kinds of RMB? Urgent! Need analysis


5 yuan and 10 yuan are equal,
Let 5 yuan have X pieces, then 10 yuan also have X pieces,
∵ there are 80 tickets for 2 yuan, 5 yuan and 10 yuan,
There are 80-x-x = 80-2x pieces for 2 yuan,
Xiaodong's mother received a bonus of 380 yuan from her unit,
The sum of the three denominations of RMB is 380,
That is, 2 × (80-2x) + 5x + 10x = 380
160-4x+5x+10x=380
11x=220
x=20
There are 20 for 5 yuan, 20 for 10 yuan, and 80-20 × 2 = 40 for 2 yuan,
A: slightly



Xiaotian saved 118 pieces of RMB 2 yuan, 5 yuan and 10 yuan, totaling 500 yuan, of which 5 yuan and 10 yuan are equal in number?


The answer to the equation
Suppose 5 yuan and 10 yuan are x sheets, then 2 yuan is 118-2x sheets
Then (10 + 5) x + 2 (118-2x) = 500
11X=500-236 11X=264 X=24
Then the number of 2 yuan sheets is 118-48 = 70
So the answer is 2 yuan 70, 5 yuan 24, 10 yuan 24



Master Wang has 118 personal name coins of 2 yuan, 5 yuan and 10 yuan, totaling 500 yuan, of which 5 yuan and 10 yuan are equal. How many are there in each of the three kinds of RMB
We need to solve the equation


2 is x, 5, 10 is y
x + 2y = 118
2x + 5y + 10y = 500
Let's solve it



Uncle Li has 118 pieces of 2 yuan, 5 yuan and 10 yuan, totaling 500 yuan. The number of 5 yuan and 10 yuan is equal. How many pieces of each of the three kinds of RMB?


Let 5.10 yuan have X, 2 yuan have y long
10X+5X+2Y=500
2X+y=118
X=24 Y=70



On the existence theorem of zeros
If the image of the function y = f (x) on the interval [a, b] is an uninterrupted curve, how to understand the "uninterrupted" in it


Any x value between [a, b] has y value corresponding to it



380V three-phase four wire calculation formula P = root 3 * u * I * cos Φ, where is I current the sum of three-phase currents or one of them?
380V three-phase four wire calculation formula P = root 3 * u * I * cos Φ & nbsp; where I current is the sum of three-phase current or one phase current?


1 I = phase current = current of a live wire
The premise is three-phase balance
If the voltage is unbalanced, it depends on whether the voltage is unbalanced. Multiply the phase voltage and add it
4 if balanced, it is 220 (for example) times (63 + 42 + 58) = 35860 = 35.86kva (apparent power)
If you have any questions, please continue to ask!
Baidu maintenance electrician wholeheartedly answer for you!



1 / x-2y, 1 / y ^ 2-1


The common denominator of solution 1 / x-2y, 1 / y ^ 2-1 is (x-2y) (y ^ 2-1)
So 1 / x-2y = (y ^ 2-1) / (x-2y) (y ^ 2-1)
1/(y^2-1)=(x-2y)/(x-2y)(y^2-1)



Can 25W and 60W bulbs with rated voltage of 220 work normally when connected in series with 220 V power supply? Who is brighter


After series connection, the partial voltage of each bulb is less than 220 V, which obviously can not work normally. As for who is brighter, their resistance can be compared. When they are all connected to 220 V, the current passing through them is 25 / 220 A and 60 / 220 a respectively, and the latter is larger. According to r = u / I, the resistance of the former is larger. When they are in series connection, the current passing through them is the same, according to P = I ^ 2



The polynomial 12x ^ 2-10xy + 2Y ^ 2 + 11x-5y + M can be decomposed into the product of two first-order factors
If I do (3x-y) (4x-2y) + 11x-5y + m, I can't go on


It is known that the polynomial 12x & # 178; - 10xy + 2Y & # 178; + 11x - 5Y + M can be decomposed into the product of two first-order factors
Analysis: the original must be type, such as (AX + by + C) (DX + e + F)
The coefficients of X and y can be easily assigned by double cross multiplication,
Let the original formula = (2x - y + a) (6x - 2Y + b)
= 12x² - 10xy + 2y² + (6a + 2b)x + (-2a - b)y + ab
Compared with the original formula, 6a + 2B = 11, - 2A - B = - 5, ab = M
The solution of the first two equations is a = 1 / 2, B = 4
So m = (1 / 2) * 4 = 2



There is an electric energy meter marked with 3000revs / kW · h. now use this electric energy meter to measure the power of a certain electric appliance, turn off other electric appliances, and only let this electric appliance work. In one minute, the electric energy meter rotates for 15 revolutions. What is the power of this electric appliance? (write down the specific steps)


1. If an electric appliance turns 15 times per minute, it should turn 900 times per hour. (15 times 60 minutes). 2. The watt hour meter & nbsp; "3000revs / kW · H" & nbsp; means 1000 wh (watt hour) turns 3000 times, and 900 times per hour is (900 times 3000) 0.3 kW