step!AiAs3s“ + 2s +5こ(s+2(s+4)s+6)StaSt6A3let s=-6StepaA,et Sニ-335°42s+5 =A,(sty) (SHG)Aスlet s=-4= AaAzニlor4-afraで4s )4(-97+S) |733(-6)2 t2(-6) +J 3(367+(-14)+s(4)(-2)3(-47° +2(-4)+5(-4+2) (-4t4)45=A1(4 4)(マ6)(Q) (4)(-2) (2)101(ht >-7(erケ-)Step 31313/8こメ。受 k= %ー4keat10122ウ-454k= -参=4Std多evte Prefinali Tasd 2as -5S+3のFの%3D(S+1) (S.2) (s+3)

Answered: Image /qna-images/answer/5f59b8c5-4fa7-4a9d-8c1b-667e356cf5b9.jpg

Answered: Prefinal Tad a O Fn =

Engineering

Electrical Engineering

Prefinal Tad a O Fn =

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

step!
Ai
As
3s“ + 2s +5
こ
(s+2(s+4)s+6)
Sta
St6
A3
let s=-6
Stepa
A,
et Sニ-3
35°42s+5 =A,
(sty) (SHG)
Aス
let s=-4
= Aa
Az
ニ
lor
4-afraで4s )4(-97+S) |73
3(-6)2 t2(-6) +J 3(367+(-14)+s
(4)(-2)
3(-47° +2(-4)+5
(-4+2) (-4t4)
45
=A1
(4 4)(マ6)
(Q) (4)
(-2) (2)
101
(ht >-7(erケ-)
Step 3
13
13/8
こメ。
受 k= %
ー4
keat
101
2
2ウ
-45
4
k= -参
=4
Std
多evt
e

Prefinali Tasd 2
as -5S+3
のFの
%3D
(S+1) (S.2) (s+3)

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