3e dR = kR dt dR = kR Obtains dt and states equ 1 and for equ 2 ln(R) = kt + c grade u sub t = 100, R = 150 : ln(150) = 100k + c equ1 Obtains dR = kR sub t = 200, R = 90 : dt ln(90) = 200k + c equ2 and calculates c subtracting : and for grade r k ln(90) ln(150) = 100k – so k = -0.005108 sub in equ 1 c = 5.5214 Obtains sub in the equ: ln(R) = kt + c t 490 days ≈ ln(R) = –0.005108t + 5.5214 with all correct mathematical if R = 20 : ln(20) = –0.005108t + 5.5214 statements for so t ≈ 494 days grade t.
3e dR = kR dt ln(R) = kt + c sub t = 100, R = 150 : ln(150) = 100k + c equ1 sub t = 200, R = 90 : ln(90) = 200k + c equ2 subtracting : ln(90) – ln(150) = 100k so k = -0.005108 sub in equ 1 c = 5.5214 sub in the equ: ln(R) = kt + c ln(R) = –0.005108t + 5.5214 if R = 20 : ln(20) = –0.005108t + 5.5214 so t ≈ 494 days Obtains dR = kR dt and states equ 1 and equ 2 for grade u Obtains dR = kR dt and calculates c and k for grade r Obtains t ≈ 490 days with all correct mathematical statements for grade t.
4b_practice_examination__number_4_schedule.docx