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ni +na -2 using these relations derive I= where t=o 2n, n2 te relations n. mi = m2 = hiz , m12 ,m12=mi+m2 (1-nit) (1- n2t) (1-nnt) The Solution hiz (1-nnt) (1-nit) (1- nzt) ni2 ni (1-nit) + na(1-nzt) how did we get this ? 1- nız T (1-nit) (1-n2t) n. -n,nat + n2 - n, na T and this? 1- ni2 T |- n, T- nat +nin2 niz - n, T- nat +ni nzt' n.-n,nat + n2 - n, na T niz[1-(n,+na)T] = ni+n - 2ninit-ningt and this? -n2 ni2 T+ 2n, nz nizT? ni2-nihizł - nzknT =nith2-2minzT-nim2t -nanizt to niz = ni + n2 - 2ninz T %3D ni +na -n2 %3D 2n, n2

ni +na -2 using these relations derive I= where t=o 2n, n2 te relations n. mi = m2 = hiz , m12 ,m12=mi+m2 (1-nit) (1- n2t) (1-nnt) The Solution hiz (1-nnt) (1-nit) (1- nzt) ni2 ni (1-nit) + na(1-nzt) how did we get this ? 1- nız T (1-nit) (1-n2t) n. -n,nat + n2 - n, na T and this? 1- ni2 T |- n, T- nat +nin2 niz - n, T- nat +ni nzt' n.-n,nat + n2 - n, na T niz[1-(n,+na)T] = ni+n - 2ninit-ningt and this? -n2 ni2 T+ 2n, nz nizT? ni2-nihizł - nzknT =nith2-2minzT-nim2t -nanizt to niz = ni + n2 - 2ninz T %3D ni +na -n2 %3D 2n, n2

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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i have a question and its solution but i mark some steps i want you to explain from where we get this result by using any rule . i mark three steps with ( red and purple and green ) . thank you

 

ni th2 -n2 using these relations derive I = where t=0 %3D 2n, n2 the relations n. hiz m2 = , mi2 = ,miz=mi+mg (1-nit) (1-nzt) (1-nnt) The Solution his n. %3D (1-nnt) (1-nit) (1- nzt) niz n, (1-nit) + na(1- nit) how did we get this ? 1- nı2 T (1-nit) (1- nzt) ni2 ni -n,nat + nz - n, naT and this? %3D 1- nizt |- n, T - n2T + n, na T [ - n. T- nat +ni nz T? niz ł n. -n,nat + n2 - n, na niz [1-(n, +na)T] -| ni+n2 - 2niniT -ningt and this? - n2 n12 T+ 2n,nz nut' ni2 = ni +na - 2ninz T 2 ni na T = i+na -ni2 ni +ni -n2 2n, na
Transcribed Image Text:ni th2 -n2 using these relations derive I = where t=0 %3D 2n, n2 the relations n. hiz m2 = , mi2 = ,miz=mi+mg (1-nit) (1-nzt) (1-nnt) The Solution his n. %3D (1-nnt) (1-nit) (1- nzt) niz n, (1-nit) + na(1- nit) how did we get this ? 1- nı2 T (1-nit) (1- nzt) ni2 ni -n,nat + nz - n, naT and this? %3D 1- nizt |- n, T - n2T + n, na T [ - n. T- nat +ni nz T? niz ł n. -n,nat + n2 - n, na niz [1-(n, +na)T] -| ni+n2 - 2niniT -ningt and this? - n2 n12 T+ 2n,nz nut' ni2 = ni +na - 2ninz T 2 ni na T = i+na -ni2 ni +ni -n2 2n, na
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