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In 3 - 1 lm 3 = = ln 3 why ? ; What happen to In 3-

In 3 - 1 lm 3 = = ln 3 why ? ; What happen to In 3-

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In 3 - 1 lm 3 why?; What happen to br 3- - 1/2 ln 3 = // In 3 why ? ; In 3
Transcribed Image Text:In 3 - 1 lm 3 why?; What happen to br 3- - 1/2 ln 3 = // In 3 why ? ; In 3
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I dont understand. they only think I want to know how to get from ln3 - (1/2)ln3 = (1/2)ln3. you are making me go backwards and I dont think that is the path I should go. This is the rest of the problem. I just need help at the end

In 3 - 1 lm 3 why?; What happen to br 3- - 1/2 ln 3 = // In 3 why ? ; In 3
Transcribed Image Text:In 3 - 1 lm 3 why?; What happen to br 3- - 1/2 ln 3 = // In 3 why ? ; In 3
10:05 PM Sun Mar 5 < x 8.5 Partial Fraction Decomposition 9 of 10 Po #43) : = 8.4 Trig Substitution (3 Cases) : 4/53 ~77/6 .TT/6 So suo do 4 440 dx → Case 3; √ a² + x² ; let x = a tano; a seco √x² +16 x = 4 tan o dx = 4 su ²o de 4 su ²o do √(4 tamo)² + 16 77/6 suo do 4 S suco do = In /sico +tomo | 7/6 In | seco + tamo || 10 = ln (3) - In (3) ² → mo negative coordinatio therefor digo 14 = ln (sec 7 + ton 1) - ln (sec 0 - tamo) 2 = ln (+/+*++) - L (1) = In ( 32/32 ) → przo. In ( ^^ ) = ln (x) - In (^3) = In (3) - In (√3) = ln 3 - 1 lm 3 =£en 3← why? Hw6 8.4 • → ln (x) = b ln(x) Bounds upper → x = 4√3 ; x = 4 tan o 4 tan o tan- T O 4 √3 4 4√3 6 = 4 (√3)= ak اسامه V = tan o (1,0) 53, 1/2) suo= Lodo lower → x=0; x=' = NET |- Hw6 8.4 tano = sino = loso ; x = 4 tam o 1/2 0=4 tano tan' (0) = 0 0 = tano o=e حم الله 2 √3 √3 2 √3 ; }=1\ ; i = 0 □ Hw5 8.3 22% ●●●
Transcribed Image Text:10:05 PM Sun Mar 5 < x 8.5 Partial Fraction Decomposition 9 of 10 Po #43) : = 8.4 Trig Substitution (3 Cases) : 4/53 ~77/6 .TT/6 So suo do 4 440 dx → Case 3; √ a² + x² ; let x = a tano; a seco √x² +16 x = 4 tan o dx = 4 su ²o de 4 su ²o do √(4 tamo)² + 16 77/6 suo do 4 S suco do = In /sico +tomo | 7/6 In | seco + tamo || 10 = ln (3) - In (3) ² → mo negative coordinatio therefor digo 14 = ln (sec 7 + ton 1) - ln (sec 0 - tamo) 2 = ln (+/+*++) - L (1) = In ( 32/32 ) → przo. In ( ^^ ) = ln (x) - In (^3) = In (3) - In (√3) = ln 3 - 1 lm 3 =£en 3← why? Hw6 8.4 • → ln (x) = b ln(x) Bounds upper → x = 4√3 ; x = 4 tan o 4 tan o tan- T O 4 √3 4 4√3 6 = 4 (√3)= ak اسامه V = tan o (1,0) 53, 1/2) suo= Lodo lower → x=0; x=' = NET |- Hw6 8.4 tano = sino = loso ; x = 4 tam o 1/2 0=4 tano tan' (0) = 0 0 = tano o=e حم الله 2 √3 √3 2 √3 ; }=1\ ; i = 0 □ Hw5 8.3 22% ●●●
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