For an AC circuit with inductance, capacitance and resistance, when the AC frequency decreases, does the impedance increase or decrease? How to analyze? I hope you can give me your advice The key point is, when the frequency decreases, is the increase of capacitive reactance greater or the decrease of inductive reactance greater?

For an AC circuit with inductance, capacitance and resistance, when the AC frequency decreases, does the impedance increase or decrease? How to analyze? I hope you can give me your advice The key point is, when the frequency decreases, is the increase of capacitive reactance greater or the decrease of inductive reactance greater?

You are talking about the problem of electric resonance. The point circuit composed of inductance, capacitance and resistance has a central frequency. That is to say, when the AC frequency is at a certain value, it is the smallest. When the AC frequency deviates from this frequency, the impedance increases rapidly. The higher the frequency is, the smaller the capacitance reactance of the capacitor is, and the lower the frequency is, the smaller the inductance reactance is

If the complex impedance of the sinusoidal AC circuit is Z = 40 + F30 ohm, what is the resistance of the circuit?
Speed, result
The option is 30.40.50.70

The complex impedance represents the property of blocking AC current. It is a complex number. The real part is the resistance R and the imaginary part is the reactance X. therefore, the DC resistance of the circuit is 40 Ω according to the complex impedance expression

Find the tangent equation and normal equation at the point M0 on the curve y = f (x) = 1 / X & # 178;, M0 (1,1)
y=-2x+3 y=(1/2)x+1/2

A:
f(x)=1/x²
Derivation:
f'(x)=-2/x³
Point m (1,1) on f (x)
When x = 1, f '(1) = - 2
Tangent slope k = f '(1) = - 2
Normal slope k = - 1 / F '(1) = 1 / 2
So:
The tangent is Y-1 = - 2 (x-1), y = - 2x + 3
The normal is Y-1 = (1 / 2) (x-1), y = (1 / 2) x + 1 / 2

7x-12% = 5.8

7x-120%=5.8
7X-1.2=5.8
7X=7
X=1

When k is an integer, the equation (K + 1) sin ^ 2x-4cosx + 3k-5 = 0 has a real solution

（k+1）sin^2x-4cosx+3k-5=0
（k+1）(1-cos^2x)-4cosx+3k-5=0
Expand the formula,
Simplify
k=(2+cosx)/(2-cosx)
It is easy to know that K cannot be less than or equal to 0 or greater than or equal to 4 (because cosx ranges from - 1 to 1)
So K is 1,2,3

The product of a number and its reciprocal is 4. What is the quarter of the reciprocal of this number?

There's no doubt that this question is wrong. It's obviously added. How did it become a product?

2x+4y=10 x+2y=5

2x+4y=10①
x+2y=5②
① - 2 × 2, 0 = 0
So the equations have the same solution, that is, the original equations have only one independent equation, x + 2Y = 5
So there are countless solutions

Let f (x) satisfy f (x) * f (x + 2) = 13 if f (1) = 2, then f (99)=

f(1)=2 f(1)*f(3)=13
f(3)=13/2 f(3)*f(5)=13
f(5)=2
……
……
……
f(4n+1)=2 n=0,1,2,3……
f(4n-1)=13/2 n=1,2,3,……
Mathematical induction
So f(99)=f(4*25-1)=13/2

If x ≠ 1 or X ≠ 2, then x ^ 2-3x + 2 ≠ 0 is true or false

Proposition p: if x ≠ 1, or X ≠ 2, then x ^ 2-3x + 2 ≠ 0 is not p: if x ≠ 1 and X ≠ 2, then x ^ 2-3x + 2 = 0 is false. So proposition p: if x ≠ 1, or X ≠ 2, then x ^ 2-3x + 2 ≠ 0 is true

It is known that the image of quadratic function passes through the point a (- 1,0), and the symmetry axis of B (0, - 3) is x = 1
Find the analytic expression of the quadratic function and write out the coordinates of vertex P

Can you do a function?
Let y = a (x-1) & sup2; + C according to x = 1
If we take a (- 1,0), B (0, - 3) into it, we can calculate a and C
This is the method,