## help_people one year ago Samuel bought a cement mixer for \$54,205. The value of the cement mixer depreciated at a constant rate per year. The table below shows the value of the cement mixer after the first and second years: Year 1 2 Value (in dollars) 47,158.35 41,027.76 Which function best represents the value of the cement mixer after t years? f(t) = 47,158.35(0.87)t f(t) = 54,205(0.13)t f(t) = 47,158.35(0.13)t f(t) = 54,205(0.87)t

1. Nnesha

1)post a question 2) tag :-)

2. help_people

|dw:1433768820354:dw|

3. Nnesha

depreciated meaning what ? any idea how to start ?

4. Nnesha

key words depreciated constant rate *per* year

5. help_people

multiply something but no idk where to start

6. Nnesha

depreciate =decrease

7. Nnesha

so $\huge\rm f(t)=P (1- \frac{ r }{ 100 })^t$ r=rate p =starting amount t= years substitute values :-) and then sovle

8. help_people

where are the vlues?

9. Nnesha

10. help_people

f(t)=54,205(1-r/100)^1?

11. Nnesha

alright great can you solve for r :-) and yeah when t=1 f(t) = 47158.35 :-)

12. Nnesha

$47158.35 =54,205(1-\frac{r}{100})^1$

13. help_people

you would diode 54,205 on the other side right

14. Nnesha

yep divide**

15. help_people

47185/54205=(1-6/100)^1

16. help_people

6 is r srry

17. Nnesha

yep right :-) solve for r

18. help_people

now I'm confused how to do that can u show me please?

19. Nnesha

47185/54205= ??

20. Nnesha

anything to the 1 power equal to same thing so (1-r/100)^1 = 1-r/100

21. help_people

-1?

22. Nnesha

nope 47185/54205= ?? divide this first

23. help_people

ok

24. help_people

0.87

25. Nnesha

r isn't .87

26. Nnesha

did you divide 47185/54205= ??

27. help_people

yes

28. Nnesha

show ur work how did you get .87 ?

29. help_people

on a calc ???

30. help_people

it was 0.870492

31. Nnesha

what did you put in to the cal?

32. help_people

what you exactly told me @Nnesha

33. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @help_people 47185/54205=(1-r/100)^1 $$\color{blue}{\text{End of Quote}}$$ this is right solve for r i asked a question how did you get .87 and what did you pout into the calculator ? bec i got difference answer for r

34. Nnesha

put*

35. help_people

i got .87 by dividing those two numbers like you told me i did not see r equals that @Nnesha

36. Nnesha

facepalm* i thought u r saying r=.87 >.^

37. Nnesha

wait we don't have to use that equation mhmm

38. amistre64

our rate of change from year 1 to year 2 is: $r=\frac{value_2-value_1}{value_1}$ since we want to know how much worth it has retained, we are interested in 1-r

39. amistre64

or rather 1 - |r| might be more precise

40. help_people

so i have to put in the values in r ?

41. amistre64

you are finding the percentage of difference, yes

42. help_people

ok

43. amistre64

for an estimate, what is (47-41)/47?

44. help_people

0.12766

45. amistre64

$\frac{47-41}{47}\implies \frac{47}{47}-\frac{41}{47}\implies 1-\frac{41}{47}$ if you want to see how it relates to Nneshas posts

46. help_people

round it would be 0.13

47. amistre64

48. amistre64

what is out value? in other words, how much do we retain as the years go on? which option do you choose now?

49. help_people

c

50. amistre64

.13 is NOT what we want ...

51. amistre64

what is 1 - .13?

52. amistre64

c tells us how much we have lost over the years. it doesnt really tell us outright what the value is for each year without further calculations.

53. help_people

then b?

54. amistre64

does b have .13 in it?

55. amistre64

do want to have a .13 in it?

56. amistre64

if we are losing 13% of our value each year, then our value is not 13% if i have 1 dollar, and i have to pay someone 13cents, how much value of that dollar do I have left over?

57. amistre64

another view of it is 47(.13) = 41 ?

58. help_people

i don't get it i thought it would be that is tme 13

59. help_people

@amistre64

60. amistre64

answer me this ... is 47(1.3) = 41 ? or even close to it?

61. amistre64

47(.13 ) = 41 that is

62. amistre64

if .13 is what we lose in value, then how much do we keep in value?

63. help_people

? I'm not understudying what ur asking?

64. amistre64

good luck then, because I really dont have any other way of asking it.

65. help_people

would we keep .87?

66. amistre64

yes

67. help_people

ok then i did understand srry

68. help_people

so based on that i find a to be the answer

69. amistre64

'a' is a trick we start with the original value of the equipment

70. amistre64

54(.87)^0 = 54 , for year 0 its brand new. 54(.87)^1 = 47 or there abouts 54(.87)^2 = 41 or there abouts

71. help_people

ok

72. help_people

but remember that equation you gave me earlier about r-value 1-value 2/value1 i used that and found the answer

73. help_people

so that does nt show the whole information?

74. amistre64

you found the percent of change from 47 to 41, its a difference calculation. 47 lost 13% to get to 41

75. amistre64

it shows a method of approaching the solution.

76. amistre64

41/47 = .87 is valid as well.

77. help_people

yeah so d? but i still am not getting why

78. help_people

wait yes i amm srry @amistre64

79. amistre64

good luck :)

80. Nnesha

P represents initial amount which is what in the question ?

81. help_people

thank yo so d is correct ? @amistre64

82. Nnesha

great explanation amis! you can also use this equation $47158.35 =54,205(1-\frac{r}{100})^1$ solve for r r=13 plug in $\large\rm f(t)=54,205(1-\frac{13}{100})^t$ remember p represent initial amount solve the parentheses gO_Od luck! :-)