The principle of drawing small signal equivalent circuit in analog circuit

First of all, a transistor amplifier circuit consists of two parts: (1) DC path (2) AC path DC path: the analysis of DC path is to determine the static working point. When analyzing DC path, the capacitor is regarded as an open circuit and the AC signal source is short circuited

How to draw such power and grounding symbols in Proteus? GND, VDD in the figure?
Trouble to say in detail, I understand a little bit of the problem, depressed

Select "default" in terminals mode, then edit and change the name to VCC and GND

This capacitance symbol, if you draw it in Proteus?
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This is an electrolytic capacitor. It's named cap-elec in Proteus. In capctors library, you just need to input "cap -"

It is known that the square of the quadratic equation x-px + q = 0 is x1, X2 and X1 + x2 = 7
One part of X1 plus one part of x2 = 3, find the value of P + Q. sorry, "X1 + x2 = 7" is wrong, it's "equal share of X1 + square of x2 = 7"

Junlang lieying team answers for you
X1^2+X2^2=(X1+X2)^2-2X1X2=p^2-2q=7,
1/X1+1/X2=(X1+X2)/(X1*X2)=3
∴3q=p,
9q^2-2q-7=0
(9q+7)(q-1)=0
Q = - 7 / 9 or 1,
P = - 7 / 3 or 3,
P + q = - 28 / 9 or 4
(q is negative, the equation must have real roots)

The line y = - x + T and X Λ 2 / 3 + y Λ 2 / 2 = 1 intersect two different points AB on the right side of Y axis, where p (1,0) satisfies 2 vector PN = vector PA + vector Pb, and it is proved that ∠ PNF is an acute angle

Where does f come from

Given x2-3x + 1 = 0, find the value X3 / X6 + X3 + 1 of the following formula

x²－3x＋1＝0
x＋1/x＝3
﹙x＋1/x﹚²＝3²
x²＋1/x²＝7
∵﹙x^6＋x³＋1﹚/x³
＝x³＋1/x³＋1
＝﹙x＋1/x﹚﹙x²－1＋1/x²﹚＋1
＝3×﹙7－1﹚＋1
＝18＋1
＝19
∴ x³/﹙x^6＋x³＋1﹚＝1/19.

Given the parallelepiped, prove: vector ab '+ vector AC + vector ad' = 2 vector AC '
Urgent! The process said in detail, please!

Let's not draw a graph. You can draw a graph by yourself. Pay attention to the corresponding points of the same name. Prove that (addition and subtraction of parallelepiped vectors): ∵ AC '= AC + AA' ∵ 2Ac '= 2Ac + 2AA' = AC + 2AA '+ AC ∵ AA' = ad '+ d'a' = ab '+ b'a' ∵ 2AA '= ad' + ab '+ d'a' + b'a 'and ∵ d'a' + b'a '= c'b' + c'd '=

(x + a) (x + b) = the square of X + (a + b) x + AB find the square of x-7x + 12 = the square of X + 3x-40=

x²-7x+12=(x-3)(x-4)
x²+3x-40=(x+8)(x-5)

A = {x | x is known

M ＜ - 6 or m ≥ 6

The principle of drawing small signal equivalent circuit in analog circuit