2. [0\" Points] DETAILS PREVIOUS ANSWERS LARAPCALC1II] 4.1.006.M|. Beginning Iwith 25 grams of a radioactive element whose half-life is 35 years, the mass y (in grams) remaining after tvears i5 given by y 25(1)\"?5 t :n o E ' ' ' How much of the initial mass remains after 150 years? {Round your answer to three decimal places.) 1.275 x g weeaHem? @ \f.[Cll1 Points] _ETAILS _REVIOUS ANSWERS LARAPCALCH] 4.1..020 The sales Si (in billions of dollars] for Starbucks from 2009 through 2014 can be modeled by the exponential function so] = 3.710.112] where t is the time in years, with t= 9 corresponding to 2009.' [Round your answers to two decimal places.) (a) Use the model to estimate the sales in 20L6 in billions of dollars. $ 22.5 K billion {b} Use the model to estimate the sales in 2021 in billions of dollars. 3*, billion Neee Help? @ Submit Answer 7. [01'1 Points] DETAILS PREVIOUS ANSWERS LARAPCALCW 4.1.024.M|. With an annual rate of ination of 1.5% over the next 10 years, the approximate cost C of goods or services during any year in the decade is given by cm = m1.015}', o s r s 10 where t is time (in years) and P is the present cost. The price of an oil change for a car is presently $23.45. Estimate the price 10 years from now. {Round your answer to two decimal places.) 3; 27.43 x Need Help? _' @- \fstep 1 Question 7: P = 10, 000 Question 15: Solve the equation For X e - 1 / x = e 1/8 r =9010 or 0.09 t - 40 X = [- 8 n = IDNE Question 14. A - 5000 -, 0.07t Step 2: A = Pert ( 9) t = 1 year Step 3 : solve. A = 5000 e 0.07 (1) A = Pert A : 5342 . 54 = 10,000- (9%% ) ( 40 ) ( b ) t = 10 years = 10,000 - (0 . 09 ) 40 ). A : 5000 e 0.07 ( 10 ) A = 10,000 e 3. 6 A = 5000 e 0.70 A = 1006 8. 76 A = 365 982. 34 ( 6 ) + : 50 years Question 8: rate: 5.5 %% A = 5000 e 0-07 (50) (a annually = 14) %% A - 5000 e 3.50 ( b) semiannually = 15) to A = 145577. 26 (C) quarterly = 15)% ( a) monthly = 15 010 Question 11: Find the tangent line y = ( e 3x - 4 ) 2 , (0,9) ax ( (e 3x - 4 )2 ) 1 xco y = 1- 18 Question 12: dx dy exy +x ? - y 2 - 14 eyx . ax ( yx ) + 2 x + 0 + 0 c y x y ' - ax [X) + 2 x = exy, y . 1 + 2x men = ye + 2X ley + 2x Question 13 . Find the second derivative f ( x ) = (7 +ox ) e- 5 x f' ( x ) = - ( 30 * + 29 ) - 5 x f" = ( 150 * + 115 ) e - 5xQuestion : y= 25 ( 1/2) $/35 170,000 = p( 12 + 0.03 12 ( 30 ) y= 25 ( 1/2) 150 / 85 12 170, 000 * P ( 12+ 0.03 \\ 360 y : 1 1. 28 grams 12 Question 2: 5 (+ ) = 3. 71 ( 1.112) t 170,000 : P (2.40) (a). = 3.71 ( 1.112) 14 P: 49105 . 69 = (3.71 ) (5. 47 ) ( when += 40 170, 000 = P ( 12 + 0.03 12( 40 ) : $ 20. 29 billion 12 170, 600 = P ( 3 . 32 ) (b) 3.71 ( 1.112) 21 P : 51204. 82 = (3. 71) ( 9. 20) 010 acet (D whent t = 50 1534 . 48 billion 170, 000 = p ( 12 + 0.03 12150) WE . POOL 12 Question s: C(E ) = P (1. 015 ), Ost = 10 170 , 060 = P ( 4.47 ) CHU = ( 23. 45 ) (1 . 015 ) 10 P = 38031. 32 $ 27.21 t P 1 165048. 54 Question 5: A = P ( 1 + 1 ) nt 10 $ 125925. 93 (U r= 30 or 0.03 t : 1 1= 12 20 93 404 . 59 170, 000 = P ( 1 + 0. 08 ) 12 ( 1) 30 49105- 69 12 40 51204. 82 170 , 000 = P ( 12 + 0.03 ) 12 50 12 38031 . 32 170, 000 = P ( 1,03 ) P : 145048. 54 Question 4: 180, 000 1= 40 or 0.04 n= 345 ( 1 ) r = 31. or 0.03 t: 10 n= 12 (1 180 , 000 : P ( 1+ 0.04 ) 345 ( 1) 845 170,000 = P ( 1 + 0.03 ) 12 ( 10 ) [P : 17 294 2 . 4781 12 ( 11 ) 140, 000 = P ( 1 + 0.04 1345 (10) 170, 000 : P ( 12 + 0. 08 ) 120 12 P = 120640. 2527 170, 000 = P MNORD (1. 35 ) (111) 140, 600 - P ( 1 + 0.04 7 345 ( 20) Bus P = 125925. 93 ( P = 80 882. 75875 ( ) 170, 000 = P ( 12 + 0.03) 12 ( 20) ( IV ) 140,000 = P / 1+ 0.04 345 (30) 365 170, 000 : P (12+0.03 ) 240 P = 54 2 18 . 52282 12 ( V) 180, 000 # 1 ( 1 + 0.04 ) shes ( 40 ) 170, 000 = P (1.82 ) 865 t P P = 93406.59 P : 34 3 44. 55924 (VI) 180.000 = P ( 1+0.04 3605 ( 80) $17 2942 . 48 345 10 $ 120640. 25 1 P = 24363. 02056 20 $ 80 8 82 . 76 30 $ 542418. 52 40 $ 34344.56 56 1 $ 24363. 02