III. AB Review. The material in this section is a review of AB concepts Illegal to post on Internet

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1 III. AB Review The mteril in this section is review of AB concepts Illegl to post on Internet

2 R1: Bsic Differentition The derivtive of function is formul for the slope of the tngent line to the grph of tht function. Students re required to know how to tke derivtive of bsic functions, trig functions, logrithmic nd exponentils, nd inverse trig functions. The bsic rules: f x + h Definition: f ( x) = lim h 0 h Power Rule: d dx xn ( ) f ( x) ( ) = nx n 1 Product Rule: d dx f x Quotient Rule: d f ( x) = g( x) f ( x) f ( x) g ( x) 2 Chin Rule: d dx g( x) g( x) dx f g x d dx sin x ( ) = cos x d ( ) = sin x d ( ) = sec2 x d dx ln x ( ) = 1 x dx cos x d dx ex dx tn x d ( ) = e x ( ) = x ln dx x d dx csc x d dx sin 1 x ( ) g( x) = f ( x) g x ( ( )) g( x) = f g x ( ) + g x ( ) ( ) f x ( ) = csc x cot x d ( ) = sec x tn x d ( ) = csc2 x ( ) = 1 1 x 2 dx sec x d dx cos 1 x ( ) = 1 1 x 2 ( ) dx cot x d dx tn 1 x ( ) = Students lso must be ble to tke derivtives implicitly. The technique of logrithmic differentition is one tht should be mstered x 2 For ech problem, find f ( x). 1. f ( x) = x 4 3x 2 + ln x e π 2. f ( x) = 5x5 3x 3 + 7x e x 2 3. f ( x) = ( 8x 2 4x 1) 3 4. f ( x) = x sin x 5. f ( x) = 4x sin2x 6. f ( x) = ln( π x)e π x 7. f ( x) = sin 2 3x 8. f ( x) = tn x 3 9. f ( x) = x cos 1 ( e x ) Illegl to post on Internet

3 10. f ( x) = sin( cos2 2 x ) 11. lim sec x + h h 0 h ( ) sec x 12. lim h 0 1 x + h 1 h h Find dy dx 13. x 2 + 5xy + y = x y = e xy 15. sin2x + cos3y = cos2x sin3y 16. y = x x 17. y = ( 2x +1) sin x 18. y = ( sin x) ex Find d 2 y dx y = sec 4x 20. y = e 2 x y = ln( x + x 2 ) 22. y = x + y sin y = x y Illegl to post on Internet

4 R2: Liner Approximtions You typiclly hve function f nd n x-vlue x = c, nd wish to find the slope of the tngent line to f t c. Since the slope of the line is f ( x), the slope of the tngent line to f t c is f ( c). To find the eqution of the tngent line to the curve t point x 1, y 1 ( ) on f, use the point-slope formul: y y 1 = m( x x 1 ) where m = f ( x 1 ). To pproximte f ( c), where x is vlue close to x 1, we cn use the point-slope formul. This uses the concept of locl linerity the closer you get to point on curve, the more the curve looks like line. Students should lso be ble to find the slope of the tngent line to f 1, the inverse function to f t x = c. Find the eqution of the tngent line to f t x = c nd use it to pproximte f 1 2 on #1 nd #2 1. f ( x) = x 3 x 2 2x 2, c = 1 2. f ( x) = e x2 x, c = 1 3. f ( x) = 2sin x tn x, c = π 4 Use the tble below to find h ( 1) for 4. h( x) = 1 f ( x) f g f g h x ( ) = f ( x) + g( x) f ( x) π 1 6. h x ( ) = f g( x) ( ) 7. h x ( ) 8. h x ( ) = g f ( 2x) ( ) = f 1 ( x) Illegl to post on Internet

5 If y = f ( x), R3: Limits, Continuity, nd Differentibility f is defined t x = if f lim f x x ( ) exists if lim x ( ) exists. ( ) = lim f x x + f is continuous t x = if lim f x x f is differentible t x = if lim f f ( x). ( ) = f x ( ). ( x) = lim f x + ( x) = f ( ). If f is differentible, it must be continuous. If f is continuous, it does not hve to be differentible. 1. In the grph of f ( x) below, determine whether the function is defined t the given loctions, whether the limit exists, the function is continuous nd the function is differentible. ( ) b c d e f ( x) defined f x b c d e limit exists continuous differentible Find the vlue of k tht mkes the function continuous. 2. f x ( ) = kx, x < 2 x 2 x, x 2 3. f ( x) = 4x + k, x < π cos x + 2, x π Find the vlues of nd b tht mke the function differentible. 4. f x x 2 bx + 3, x 2 ( ) = x2 4x 2, x < 2 5. f ( x) = x + b, x < 1 x 3 + x + 2b, x Illegl to post on Internet

6 R4: Relted Rtes In relted rtes problems, ny quntities, given or sked to clculte, tht re chnging re derivtives with respect to time. If you re sked to find how volume is chnging, you re being sked for dv. Typiclly in dt relted rtes problems, you need to write the formul for quntity in terms of single vrible nd differentite tht eqution with respect to time, remembering tht you re ctully performing implicit differentition. If the right side of the formul contins severl vribles, typiclly you need to link them somehow. Constnts my be plugged in before the differentition process but vribles tht re chnging my only be plugged in fter the differentition process. 1. If sphericl snowbll melts so tht its surfce re decreses t rte of 1.5 in 2 min, find the rte t which the volume decreses when the dimeter is 6 inches. 2. A street light is mounted t the top of 20-ft tll pole. A boy 5 feet tll is wlking his dog, 2 feet tll t the rte of 4 feet/sec. How fst re the tips of their shdows seprting when they re 15 feet from the pole? 3. A crew of men is trying to move pedestl tht is to be centered t point P which is directly below pulley. They re using rope tht is 50 feet long nd wlk bckwrd t the rte of 2 ft/sec. How fst does the pedestl move t the instnt tht the men re 9 feet from P? 12 ft P Illegl to post on Internet

7 4. An inverted conicl tnk is being driven by truck to irrigte crops. Wter comes out the bottom t the rte of 30 gllons/min (4.01 ft 3 /min). At the sme time wter is being pumped into the top of the tnk. The tnk hs height 12 feet nd the dimeter t the top is 8 feet. If the wter level rises in the tnk t the rte of 1 ft/min, t the time the height of the wter in the tnk is 3 feet high, find the rte in which the wter is being pumped into the tnk. h r 5. Atlntic Boulevrd runs est-west. Powerline Rd. nd Lyons Rd., one mile prt, run north-south nd cross Atlntic. At certin time, two crs re stopped t red lights on these rods t Atlntic. The red lights chnge to green t the sme time. The northbound cr trvels t 20 mph on Lyons nd the southbound cr trvels t 30 mph on Powerline. How fst re the crs seprting one minute lter? Lyons Atlntic b Powerline 6. A tire swing is connected to tree brnch by 20-foot rope. The tire is pushed t the rte of 12 ft/sec. How fst is the swing rising fter 1 second nd how fst is θ chnging t the sme time? 20 ft 20 ft y x Illegl to post on Internet

8 R5: Function Anlysis You hve function f ( x). You wnt to find intervls where f x concve down. You lso wnt to find vlues of x where there is reltive minimum, reltive mximum, nd points of inflection. Criticl vlues vlues x = c where f ( c) = 0 or f ( c) is undefined. f ( x) is incresing for vlues of k such tht f ( k) > 0 f ( x) is decresing for vlues of k such tht f ( k) < 0. If f ( x) switches from incresing to decresing t c, there is reltive (locl) mximum t ( c, f ( c) ). If f ( x) switches from decresing to incresing t c, there is reltive (locl) minimum t ( c, f ( c) ). This is commonly clled the first derivtive test. f ( x) is concve up for vlues of k such tht f ( k) > 0. f ( x) is concve down for vlues of k such tht f ( k) < 0. If f ( x) switches concvity t c, there is point of inflection t ( c, f ( c) ). The second derivtive test is used to find reltive extrem s well: Find criticl vlues c where f ( c) = 0. If f ( c) > 0, f is concve up t c nd there is reltive minimum t x = c. If f c If f ( ) is incresing nd decresing, concve up nd ( ) < 0, f is concve down t c nd there is reltive mximum t x = c. ( c) = 0, the 2nd derivtive test is inconclusive. ( ex. f ( x) = x 4 t x = 0) [ ], occurring t point of reltive extrem or n endpoint. [ ] nd differentible on (, b), there is t lest ( c) = f ( b) f ( ) Absolute extrem lwys exists on n intervl, b The Men-Vlue Theorem sttes tht if f is continuous on, b one vlue c such tht f. In this review section, we ssume tht students hve no knowledge b of the Fundmentl Theorem of Clculus nd the ccumultion function. We look t function nlysis in terms of the derivtive only. The following grphs re f ( x). ) Find ll vlues of x on [ 2, 4] where f is incresing, decresing nd ttins reltive extrem. b) Find ll vlues of x on [ 2, 4] where f is concve up, concve down, nd inflection points Illegl to post on Internet

9 For the following functions f, find ) intervls where f is incresing nd decresing, b) the locl mximum nd minimum vlues of f nd c) intervls of concvity nd inflection points. 3. f ( x) = ( x 2 4) 3 4. f ( x) = 2sin x + cos 2 x [ 0, 2π ] 5. f ( x) = 1 ln( x 2 +1) 6. f ( x) = e 1 x 7. Suppose f x whether f g x ( ) is incresing nd concve down nd g( x) is decresing nd concve down, determine ( ( )) is incresing/decresing nd its concvity. Explin your resoning Illegl to post on Internet

10 Use the 2 nd derivtive test to determine whether f hs reltive extrem t its criticl points. 8. f ( x) = x( 6 x) f ( x) = cos2x + x [ 0, 2π ] 10. The mount of money in bnk ccount fter t 11. Find the bsolute mximum nd bsolute yers is given by P( t) = te 5 t2 8. Find minimum vlue of f ( x) = the mximum nd minimum mount of money in the ccount over 10-yer period. x x +1 on [0, 4]. 12. Determine ll numbers c which stisfy the 13. Suppose f is differentible function such tht conclusion of the Men-Vlue Theorem f ( 4) = 5. If for ll vlues of x, 3 f x for f ( x) = x x 16 on [ 0, 4]. then wht rnge of vlues cn f ( 10) hve? ( ) 2, Illegl to post on Internet

11 R6: Integrtion Techniques Integrtion is nti-differentition. If d dx f ( x) dx = g( x) then g( x) dx = f ( x) + C. While derivtive cn be tken for ny expression, not ll expressions cn be integrted. Just s differentition of complicted expressions involve u-substitution, substitution cn be used to integrte some expressions. It is importnt to dd on the constnt of integrtion when performing nti-differentition. Students need to know these integrtion formuls. u n du = un+1 n +1 + C du u = ln u + C e u = e u + C u = 1 ln u + C sinu du = cosu + C cosu du = sinu + C sec 2 u du = tnu + C cscu du = cotu + C secutnu = secu + C cscucotu du = cscu + C du u = sin 1 2 u 2 + C du = 1 u 2 + u 2 tn 1 + C du u u 2 2 = 1 u sec 1 + C 1. dx = 2. ( 5x2 + 2x 1) 3 dx = 3. x 3 dx = dx = 5. x 4x ( 2 1 ) 3 dx = ( x 3 x) 2 1 dx = 10 x 7. tn5x dx = 8. sin 3 x dx = 9. 5 x dx = 5 x Illegl to post on Internet

12 10. xe x2 sin( e ) x2 dx = 11. sec 3 ( 0.1x)tn( 0.1x ) dx = 12. e 4 x 1 dx = x 3 x +1+ tn 1 x dx = x x + 9 dx = 15. 4x x +1 dx = 2x x 2 dx = 3x Illegl to post on Internet

13 R7: The Definite Integrl nd the Accumultion Function Students should know tht the definite integrl b f ( x) dx represents the re between the x-xis nd f ( x) from x = to x = b. Although re is lwys positive number, the vlue of definite integrl cn be positive, negtive or zero. If M = b f ( x)dx Students should know the bsic rules of integrtion: b f ( x ) dx + f ( x) dx = f x c b c ( ) dx, then the sign of M is determined by the tble: f ( x ) dx = 0, f ( x) dx = f x b ( ) > 0 f ( x) < 0 f x < b + > b + b ( ) dx for continuous functions. The ccumultion function F x, s well s x ( ) = f ( t) represents the ccumulted re under the curve f strting t t = nd going out to the vlue of x. The derivtive of the ccumultion function is given by the formul: d dx x f ( t) dt = f ( x). 1. A function f ( x), mde up of lines nd qurter circle, is defined on [ 2, 4]. Let F x F x F ( ) ( x) x dt. Complete the tble. ( ) = f ( t) dt 1b. Determine intervls on [ 2, 4] where F increses/decreses, where F hs reltive mximum/intervl, nd wht those vlues re. 1c. Determine intervls on [ 2, 4] where F is concve up 1d. Sketch grph of F x ( ). nd concve down nd loctions of inflection points Illegl to post on Internet

14 2. It is erly My nd Phildelphi experiences brutl het wve tht lsts 4 dys. Mny people turn on their ir-conditioners nd find tht they don t work. They strt clling for service nd the service compnies cn t hndle the volume of requests nd hve to schedule ppointments mny dys lter. One of those compnies chrts r( t), the rte tht service requests come in to the compny, mesured in requests per dy, s shown in the figure to the right. Time t is mesured in dys strting when the het wve strts. The compny cn do 20 service clls dy nd lredy hd 10 people witing for service t t = 0. Requests per dy Time (dys) ) How mny people hve requested service by the end of the het wve? b) How mny people re witing for service t the end of the het wve? c) Is the number of people witing for service incresing, decresing, or stying the sme on the first dy of the het wve? Explin. d) Assuming tht no one clls for service fter the het wve ends, how mny more dys will it tke for ll service requests to be filled? Explin. e) On wht dy were the number of people witing for service the longest? How mny people re witing t tht time? Justify your nswer Illegl to post on Internet

15 R8: Riemnn Sums Riemnn sums re pproximtions for definite integrls, which we know represent res under curves. There re numerous rel-life models for res under curves so this is n importnt concept. Typiclly these types of problems show up when we re given dt points s opposed to lgebric functions. Given dt points: Assuming eqully spced x-vlues: x i +1 x i = b Left RS: S = b f ( x 0 ) + f ( x 1 ) + f x 2 [ ( ) f ( x n 1 )] Right RS: S = b f ( x 1 ) + f ( x 2 ) + f ( x 3 ) f ( x n ) [ ] Trpezoids: S = b ( 2 x + 2x + 2x x + 2x + x n 2 n 1 n ) If bses re not the sme (typicl in AP questions), you hve to compute the re of ech trpezoid: 1 2 x i +1 x i Midpoints (ssuming n is even) : S = 2b f x 1 x 0 x 1 x 2... x n 2 x n 1 x n ( ) f ( x 1 ) f ( x 2 ) f ( x n 2 ) f ( x n 1 ) f ( x n ) f x 0 [ ] ( ) f ( x i ) + f ( x i +1 ) ( ) + f ( x 3 ) + f ( x 5 ) +... f ( x n 1 ) 1. A freezer is set to 5 F. I open the freezer to get some ice cubes nd the temperture in the freezer chnges. The chnge c t Fhrenheit is given t selected vlues of t in the tble to the right. 40 ( ) in temperture in degrees. Approximte c( t ) dt with left, right nd midpoint Riemnn sums nd 0 explin their mening using pproprite units. t ( sec) c( t) F sec b. Use trpezoidl sum with 8 trpezoids to pproximte the temperture in the freezer fter 40 seconds. 2. Let F x x dt. Use the trpezoidl rule with four equl subdivisions to pproximte F e ( ) = t sin( t) 0 ( ). 3. Find the best pproximtion to Illegl to post on Internet

16 R9: The Fundmentl Theorem of Clculus Students should know tht the Fundmentl Theorem of Clculus sys tht b f ( x) dx F ( x) is n ntiderivtive of f ( x). This leds to the fct tht F ( b) = F ( ) + f ( x) dx b = F ( b) F ( ) where. Students should be prepred to use substitution methods to integrte nd be ble to chnge the limits of integrtion using this substitution. Also, they should know the 2 nd FTC: d dx x f ( t)dt = f ( x) nd d dx ( ) g x f ( t)dt = f x Finlly, they should know tht the verge vlue of function on the intervl [, b] is given by: Find the following definite integrls. Confirm by clcultor. ( ) g ( x). b f ( x)dx b ( 3x 1) 2 dx 2. 3 e 5 x 5 2 x dx x 2 x + 4 x dx 4. π 4 sin x cos 2 x dx x dx 6. 2sin x 1 dx π π Illegl to post on Internet

17 ( 4x + 3) 8 dx x dx 9. e 8 dx x ln x e 1 0 x 2 1+ x 6 dx 11. Let F x If F 0 ( ) be n ntiderivtive of sin xe cos x. 12. Let F ( x) be n ntiderivtive of x 3 + x 2 + x +1. ( ) = 2, find F ( π ) If F ( 4) = 5, find F ( 1). x 13. d e t2 t dt = 15. Let f x dx 14. d dx 1 cos x 2 sint dt = 0 g( x) ( ) = cos x nd g( x) = x 2. If h( x) = f t ( )dt, find h π ( ) Find the verge vlue of x cos x 2 on 0, π If f t ( ) = cost + 2t 4, wht is the verge vlue of [ ] s x pproches 0? f on 0, x Illegl to post on Internet

18 R10: Stright-Line Motion In AB Clculus, we first studied stright-line motion from the derivtive point of view. You usully re given position function x( t) nd tke its derivtive to find the velocity function v( t)nd tke the velocity s derivtive to get the ccelertion function ( t). Questions sking mximum velocity or minimum ccelertion cn then be nswered. Speed = v t ( ). A prticle speeds up when its velocity nd ccelertion hve the sme signs. Using integrls, we work bckwrds. Typicl questions involve knowing the velocity function, the position of the prticle t the strt of the problem, nd integrting to find the position function nd then finding when the prticle is frthest to the right or left. x( t) = v( t ) dt + C nd v( t) = ( t ) dt + C. Using these equtions is essentilly solving DEQ s with n initil condition, usully the position t time t = 0 or velocity t t = 0. Two other concepts tht come into ply re displcement nd distnce over some time intervl [ t 1,t 2 ]. Displcement: the difference in position over [ t 1,t 2 ]: x( t 2 ) x( t 1 ) = v( t) t 2 dt. Displcement cn be positive, negtive or zero. Distnce: how fr the prticle trveled over [ t 1,t 2 ]: v( t ) dt. Distnce is lwys positive. t 1 t 2 t 1 1. A mn leves his rft 100 feet from shore on n ocen bech. The surf moves his rft t the velocities shown in the tble t vrious times s well s grph. The velocity v( t) of the rft is mesured in feet per second where positive velocity mens moving wy from the bech.. At wht intervl of times is the ccelertion positive? t( sec) v( t) ft sec b. Give n pproximtion for ccelertion of the rft t t = 25. Use proper units. c. For wht vlues of t is the rft speeding up? Explin. d. Wht is n pproximtion to frthest the rft is from the bech? Explin. e. Wht ws the rft s verge speed over the 60-second period of time? Illegl to post on Internet

19 Given the velocity of prticle in ft/sec on the given time intervl, find the displcement nd distnce trveled s well s the times tht the prticle is speeding up. 2. v( t) = 12 4t [ 0, 5] 3. v( t) = cosπt 1 [ 2 0,1 ] Given the ccelertion of n object moving long the x-xis, its initil velocity nd initil position, find its new position in the given time intervl, its verge speed, nd the time intervls in which the object is speeding up. Set up using integrls without bsolute vlue nd evlute using the clcultor. 4. ( t) = 6t 3, v( 0) = 6, x( 0) = 5, [ 0,3] 5. ( t) = 1 2 t +1, v 0 ( ) = 1, x ( 0 ) = 4 [ 0,8] Illegl to post on Internet

20 R11: Are nd Volume Are problems usully involve finding the re of region under curve or the re between two curves between two vlues of x. Given two curves f x between f nd g on [, b] is given by A = the x-xis, these problems sometimes show up in terms of y: A = x=b ( ) nd g( x) with f ( x) g( x) on n intervl [, b], the re f ( x ) g ( x ) dx. While integrtion is usully done with respect to x= y=d m( y) n( y) dy. Volume problems usully involve finding the volume of solid when rotting curve bout line. Disks nd Wshers: The method I recommend is to estblish the outside rdius R, the distnce from the line of rottion to the outside curve, nd, if it exists, the inside rdius r, the distnce from the line of rottion to the inside curve. The formul when rotting these curves bout line on n intervl is given by: x=b ( ) V = π R ( x ) 2 r ( x ) 2 dx or V = π R( y) 2 r ( y ) 2 dy x= y=c y=d y=c ( ) A fvorite type of problem is creting solid with the region R being the bse of the solid. Cross sections perpendiculr to n xis re typiclly squres, equilterl tringles, right tringles, or semi-circles. Rther thn give formuls for this, it is suggested tht the figure be drwn, estblish its re in terms of x or y, nd integrte tht expression on the given intervl. 1. Find the re between y = 5x 2 nd y = 2x Find the re between e x,e 2 x nd x = Find the re between y = x 2 nd y = 16 x y2 4. Find the re between x = 5 nd y = x Illegl to post on Internet

21 5. Find the volume if the re between y = e x, x = 1, the x-xis, nd the y-xis is rotted bout:. the x-xis b. the line y = 1 c. the y-xis 3 6. Find the volume if the first-qudrnt region between y = x nd y = x is rotted bout: 4. the x-xis b. the y-xis c. the line y = Illegl to post on Internet

22 7. The ellipse 4x 2 + 9y 2 = 36 is the bse of solid. Cross sections perpendiculr to the x-xis re drwn using the geometric shpes below. Find the volume of the solid.. Squres b. Equilterl tringles c. Isosceles tringles d. semicircles (hypotenuse in bse) 8. The curve y = x 1+ sin x 3 ( ) from x = 0 to x = 9π 3 is drwn onto block of wood nd the rotted bout the x-xis. It is then put on lthe to crete 3- dimensionl shpe s shown on the right tht will be plced on the top of bnnister. Wht is the exct volume of the shpe? 9. Two towns re circulr with rdius of 4 miles. The popultion density of ech is 1000 people mile 2 t the center. Town A s popultion decreses linerly from the center to the edge of the circle where no one lives. Town B s popultion decreses linerly from the center to the edge of the town where no one lives, but long the rdius of the circle. Wht re ech town s popultion? Illegl to post on Internet

23 R12: Differentil Equtions nd Growth/Decy Problems A differentil eqution (DEQ) is in the form of dy dx = ( Algebric Expression). The gol of solving DEQ is to work bckwrds from the derivtive to the function; tht is to write n eqution in the form of y = f ( x) + C (since the technique involves integrtion, the generl solution will hve constnt of integrtion). If the vlue of the function t some vlue of x is known (n initil condition), the vlue of C cn be found. This is clled specific solution. In Clculus AB, the only types of DEQ s studied re clled seprble. Seprble DEQ s re those thn tht cn be in the form of f ( y) dy = g( x) dx. Once in tht form, both sides cn be integrted with the constnt of integrtion on only one side, usully the right. Word problems involving chnge with respect to time re usully models of DEQ s. A fvorite type is problem using the words: The rte of chnge of y is proportionl to some expression. The eqution tht describes this sttement is: dy dt = k expression respect to time. Differentil equtions go hnd in hnd with slope fields. Simply clculte the slopes using the given derivtive formul nd plot them on the given grph. Usully the slopes will be integer or simple frctionl vlues. ( ) whose solution is y = Ce kt. The rte of chnge is usully with 1. Consider the differentil eqution dy dx = 1 y x 2, x 0.. On the xes provided, sketch slope field for this DEQ. b. On the slope field, drw n pproximte solution to the DEQ given tht f 2 ( ) = 0. c. Find the specific solution y = f ( x) to the DEQ given tht f ( 2) = 0. d. Find lim f ( x) nd lim f ( x). x x Illegl to post on Internet

24 Solve the following differentil equtions with the given initil condition. 2. dy dx = y2 3. dy dx = yln y, y e ( ) = e 4. dy dx = x( 1+ y2 ), y( π ) = 1 5. A refrigertor door is left open nd t first, the cool ir escpes t rte proportionl to the current temperture. If the refrigertor is 34 F t 9 PM, the time the door is opened, nd five minutes lter it is 36 F, wht will the temperture be t 10 PM? 6. A supermrket hs mny jrs on its shelves. Every dy, people buy penut butter nd the shelves re restocked. Also, 10 jrs re tken off the shelf dily becuse they hve reched their expirtion dte. The continuous rte of chnge of penut butter jrs is given by dp = 0.05P 10. ) If there re 300 jrs initilly, dt wht will be the number of jrs of penut butter on the shelves in week? b) If there re 100 jrs initilly, when will the shelves be empty? Illegl to post on Internet

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