11b and 11c I am struggling with these problems thank you! aJAS,11. Oil is being pumped continuously from a Texas oil well at a rate proportional to the amount of oildydtleft in the well; that is,=ky, where y is the number of gallons of oil left in the well at any time t(in years). Initially there are 1,000,000 gallons of oil in the well, and 6 years later there are500,000 remaining. Assume that the well is no longer profitable to pump oil when there are fewerthan 50,000 gallons remaining.a. Write an equation for y in terms of t.Hobpost1½ dy ngktat = ky zy = eSTORMS & dy = f k dtу= сексLuly |kt + ctas os dwhen there are 600,000 gallons ofb. At what rate is the amount of oil in the well decreasing whoil remaining in the well?6kin t+ 1/12 =600,000 = 1,000,000 €ekt12/2/2 = e16 ²/² = 6k100⁰0001000000y = 1,000,000 e500,000 = 1,000,000 ek1,000,000In ²}/² = 14/1/20k = ln6 tloop. 09/06C.How long will the well be profitable?lui22 years3:30kt50,000 = 1,007,0966Int=5100Lu 1⁄2t:t = 6=eاف(nin the£.(n =t· In 1/1019e화тvr/w()e luz det37بدارt =t = 4.122*dydue = kydtdy k. 600,00022dtIF-

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11. Oil is being pumped continuously from a Texas oil well at a rate proportional to the amount of oil dt left in the well; that is, dy =ky, where y is the number of gallons of oil left in the well at any time t (in years). Initially there are 1,000,000 gallons of oil in the well, and 6 years later there are 500,000 remaining. Assume that the well is no longer profitable to pump oil when there are fewer than 50,000 gallons remaining. a. Write an equation for y in terms of t. ку ў y = e S dy = f k dt y = Cekt tos os d ArtiNi (uly 1 = kt + c b. At what rate is the amount of oil in the well decreasing when there are 600,000 gallons of oil remaining in the well? 6k e u 600,000 = 1,000,000 € - kt y = 1,000,000 e 10 ²/² = 6k 1000000 1000000 6k 500,000= 1,000,000 € (n = k= 3 (n = = 1 / + t こ 1000,000 1,000,000 5 7 t = In ²³² 6 c. How long will the well be profitable? in t t = 4.22 years) = 4 fir kt 3150,900 = 1,007,096 e 5 Init e 100 ~ 1/²/3 = (2+ (n 20 dy = ky -1 t = t = 6 (n = · 23 In 20 ✓ e lut dy k. dt IF 600,00 D 2

11. Oil is being pumped continuously from a Texas oil well at a rate proportional to the amount of oil dt left in the well; that is, dy =ky, where y is the number of gallons of oil left in the well at any time t (in years). Initially there are 1,000,000 gallons of oil in the well, and 6 years later there are 500,000 remaining. Assume that the well is no longer profitable to pump oil when there are fewer than 50,000 gallons remaining. a. Write an equation for y in terms of t. ку ў y = e S dy = f k dt y = Cekt tos os d ArtiNi (uly 1 = kt + c b. At what rate is the amount of oil in the well decreasing when there are 600,000 gallons of oil remaining in the well? 6k e u 600,000 = 1,000,000 € - kt y = 1,000,000 e 10 ²/² = 6k 1000000 1000000 6k 500,000= 1,000,000 € (n = k= 3 (n = = 1 / + t こ 1000,000 1,000,000 5 7 t = In ²³² 6 c. How long will the well be profitable? in t t = 4.22 years) = 4 fir kt 3150,900 = 1,007,096 e 5 Init e 100 ~ 1/²/3 = (2+ (n 20 dy = ky -1 t = t = 6 (n = · 23 In 20 ✓ e lut dy k. dt IF 600,00 D 2

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter6: Exponential And Logarithmic Functions

Section6.8: Fitting Exponential Models To Data

Problem 3TI: Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to...

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