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

- help_people

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- schrodinger

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- Nnesha

1)post a question
2) tag
:-)

- help_people

|dw:1433768820354:dw|

- Nnesha

depreciated meaning what ?
any idea how to start ?

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## More answers

- Nnesha

key words
depreciated
constant rate *per* year

- help_people

multiply something but no idk where to start

- Nnesha

depreciate =decrease

- Nnesha

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

- help_people

where are the vlues?

- Nnesha

read the question :-)

- help_people

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

- Nnesha

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

- Nnesha

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

- help_people

you would diode 54,205 on the other side right

- Nnesha

yep divide**

- help_people

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

- help_people

6 is r srry

- Nnesha

yep right :-) solve for r

- help_people

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

- Nnesha

47185/54205= ??

- Nnesha

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

- help_people

-1?

- Nnesha

nope 47185/54205= ??
divide this first

- help_people

ok

- help_people

0.87

- Nnesha

r isn't .87

- Nnesha

did you divide 47185/54205= ??

- help_people

yes

- Nnesha

show ur work how did you get .87 ?

- help_people

on a calc ???

- help_people

it was 0.870492

- Nnesha

what did you put in to the cal?

- help_people

what you exactly told me @Nnesha

- 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

- Nnesha

put*

- help_people

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

- Nnesha

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

- Nnesha

wait we don't have to use that equation mhmm

- 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

- amistre64

or rather 1 - |r| might be more precise

- help_people

so i have to put in the values in r ?

- amistre64

you are finding the percentage of difference, yes

- help_people

ok

- amistre64

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

- help_people

0.12766

- 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

- help_people

round it would be 0.13

- amistre64

then we lose about .13 a year
we retain about 1-.13

- amistre64

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

- help_people

c

- amistre64

.13 is NOT what we want ...

- amistre64

what is 1 - .13?

- 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.

- help_people

then b?

- amistre64

does b have .13 in it?

- amistre64

do want to have a .13 in it?

- 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?

- amistre64

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

- help_people

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

- help_people

@amistre64

- amistre64

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

- amistre64

47(.13 ) = 41 that is

- amistre64

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

- help_people

? I'm not understudying what ur asking?

- amistre64

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

- help_people

would we keep .87?

- amistre64

yes

- help_people

ok then i did understand srry

- help_people

so based on that i find a to be the answer

- amistre64

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

- 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

- help_people

ok

- help_people

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

- help_people

so that does nt show the whole information?

- amistre64

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

- amistre64

it shows a method of approaching the solution.

- amistre64

41/47 = .87 is valid as well.

- help_people

yeah so d? but i still am not getting why

- help_people

wait yes i amm srry @amistre64

- amistre64

good luck :)

- Nnesha

P represents initial amount which is what in the question ?

- help_people

thank yo so d is correct ? @amistre64

- 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! :-)

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