The q-points of direct coupled multistage amplifier affect each other, () it can only amplify DC signal, () right or wrong

The q-points of direct coupled multistage amplifier affect each other, () it can only amplify DC signal, () right or wrong

The q-points at all levels of a directly coupled multistage amplifier interact with each other. It can only amplify the DC signal

Q-points at all levels of a directly coupled multistage amplifier interact with each other, and it can only amplify DC signals. Is this sentence correct? Why? Resistance capacitance coupling?

The interaction of q-points at all levels in direct coupled multistage amplifier
It can only amplify DC signals -- no, it can amplify both AC and DC signals
Resistance capacitance coupling: Q points at all levels do not affect each other, only amplify the AC signal

Calculation by polar coordinates
∫∫sin√x²+y² dxdy,
Where d = {(x, y) / Π & sup2; ≤ X & sup2; + Y & sup2; ≤ 4 Π & sup2;}

Let x = RCOs (T), y = rsin (T), 0 ≤ t ≤ 2 π, R1 ≤ R ≤ R2, DXDY = rdtdr,
∫sin(R)RdR=∫Rd(-cos(R))=-Rcos(R)+∫cos(R)dR

How to calculate polar coordinate method

R and theta

Judge the true or false of the converse proposition "if a ≥ 0, then x2 + X - a = 0 has real root"

The original proposition: "if a ≥ 0, then x2 + x-a = 0 has a real root". Its converse proposition: "if x2 + x-a = 0 has no real root, then a < 0". The judgment is as follows: ∵ x2 + x-a = 0 has no real root, ∵ Δ= 1 + 4A < 0, ∵ a < - 14 < 0, ∵ proposition "if x2 + x-a = 0 has no real root, then a < 0" is true

How to calculate the symmetry axis of quadratic function
For example, if there are any two points a (x1, Y1) and B (X2, Y2) on a quadratic function, the axis of symmetry is equal to x = (x1 + x2) / 2. Why, a and B may be on the same side of the axis of symmetry=

No, if Y1 = Y2
The formula of axis of symmetry is - 2A / b

If 2x ^ - 2 + (x-3) ^ 0 is meaningful, what conditions should x satisfy?

The formula is meaningful
∴x≠0
x-3≠0
∴x≠0,3

Answer: given that the square of equation (1-A) x + 2x-3 = 2 is a unary linear equation about X, then a = ()

Since it is a linear equation of one variable about X, the square of (1-A) x does not exist, so a = 1

(x + 2) ² = 16, solve the equation,

Solution
X + 2 = 4 or - 4
x1=2
x2=-6
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Simplifying [6 (X-2) & # 178; + 12 (x + 2) (X-2) - 3 (x-1) & # 178; (X-2)] / [3 (2-x)] step by step

Solution
[6(x-2)²+12(x+2)(x-2)-3(x-1)²(x-2)]÷[3(2-x)]
=3(x-2)[2(x-2)+4(x+2)-(x-1)²]÷[-3(x-2)]
=-[2(x-2)+4(x+2)-(x-1)²]
=-(2x-4+4x+8-x²+2x-1)
=-(8x-x²+3)
=x²-8x-3