To solve the equations 2011x + 2010Y = 4024 2010x + 2011y = 4018

To solve the equations 2011x + 2010Y = 4024 2010x + 2011y = 4018

2011x+2010y=4024
2010x+2011y=4018
The sum of two formulas 4021x + 4021y = 8042
So x + y = 2
So y = 2-x, substituting (1)
We get 2011x + 2010 (2-x) = 4024
The solution is x = 4, substituting y = 2-x, the solution is y = 2-4 = - 2
The solution of the original equations is x = 4, y = - 2

If x / y = 2013, then x + Y / x-2012y=

Known X / y = 2013
We get x = 2013y
（x+y）/（x-2012y）
=（2013y+y)/(2013y-2012y)
=2014y/y
=2014

The resistance of electric lamp L is 24 ohm, and the allowed current is 0.2A. To connect it to the circuit with 1A current, a 6 ohm resistance should be connected in parallel
Why can't you add a resistor in series? Is it different if the resistance becomes larger and the current becomes smaller

Yes
But after series connection, the voltage will be divided and the brightness will be reduced

A regular number: 1,1,2,3,5,8,13,21,34 How many multiples of 5 are there in the top 100 of the string?

100 / 5 = 20

Close the switch and find that bulb a is brighter than bulb B. what is the correct judgment about current, voltage, resistance and consumed electric power?

If it is a series circuit, then the current is equal (the current relationship of the series circuit), the electric power is higher (the power of the light bulb is higher), the voltage is higher, and the resistance is higher (all these can be deduced according to the formula)
If it is in parallel, then the voltage is equal (the voltage relationship of the parallel circuit), the electric power a is larger (as above), the current a is larger, and the resistance B is larger

Given a & # 178; - A-1 = 0, find A3 + 2A & # 178; + 2007

Is a & # 178; + A-1 = 0
a²+a=1
Original formula = A & # 179; + A & # 178; + A & # 178; + 2007
=a(a²+a)+a²+2007
=a+a²+2007
=2008

The voltage at both ends of a section of conductor is 4V, and the current in the conductor is 0.4A. If the voltage increases to 6V, the current in the conductor and the resistance of the conductor are?

You should know that the resistance of the conductor will not change with the voltage at both ends: r = 4V / 0.4A = 10 ohm
When the voltage increases to 6V: I '= 6V / 10ohm = 0.6A, r = 10ohm

The known proposition p: equation x2m − 4 + y2m − 2 = 1 represents hyperbola with focus on Y axis; proposition q: the solution set of inequality x2-2x + m > 0 about X is R; if "P ∨ Q" is false proposition and "P ∨ Q" is true proposition, the value range of real number m is obtained

∫ proposition p: the equation x2m − 4 + y2m − 2 = 1 represents the hyperbola with focus on the y-axis ∫ when proposition p is true, m − 2 ＞ 0m − 4 ＜ 0, ∫ solution obtains the value range of real number m, 2 ＜ m ＜ 4 ∫ proposition q: the solution set of inequality x2-2x + m ＞ 0 about X is R ∫ when proposition q is true, △ = 4-4m ＜ 0 ∫ solution obtains the value range of real number m ＞ 1 ∫ if "P ∧ Q" is a false proposition, "P ∨ Q" is a false proposition If P is a true proposition and Q is a false proposition, then one is a true proposition. ① when p is a true proposition and Q is a false proposition, 2 ＜ m ＜ 4m ≤ 1 holds, and the solution is m ∈φ. ② when q is a true proposition and P is a false proposition, m ≤ 2, or, m ≥ 4m ＞ 1 holds, and the solution is 1 ＜ m ≤ 2 or m ≥ 4

How much is 1g ohm, m ohm, 1m ohm? Thank you!

1g ohm equals 1024m ohm, 1m ohm equals 10 ^ 6 ohm

Let f (x) = x Λ & #178; - 4 be the interval of increasing function

Is this it
f(x)=x²-4
solution
f(x)=x²-4
f‘(x)=2x
When x > 0, f '(x) > 0
F (x) is an increasing function
The increasing interval of F (x) is: (0, + ∞)