help_people
  • help_people
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Nnesha
  • Nnesha
1)post a question 2) tag :-)
help_people
  • help_people
|dw:1433768820354:dw|
Nnesha
  • Nnesha
depreciated meaning what ? any idea how to start ?

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Nnesha
  • Nnesha
key words depreciated constant rate *per* year
help_people
  • help_people
multiply something but no idk where to start
Nnesha
  • Nnesha
depreciate =decrease
Nnesha
  • 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
  • help_people
where are the vlues?
Nnesha
  • Nnesha
read the question :-)
help_people
  • help_people
f(t)=54,205(1-r/100)^1?
Nnesha
  • Nnesha
alright great can you solve for r :-) and yeah when t=1 f(t) = 47158.35 :-)
Nnesha
  • Nnesha
\[47158.35 =54,205(1-\frac{r}{100})^1\]
help_people
  • help_people
you would diode 54,205 on the other side right
Nnesha
  • Nnesha
yep divide**
help_people
  • help_people
47185/54205=(1-6/100)^1
help_people
  • help_people
6 is r srry
Nnesha
  • Nnesha
yep right :-) solve for r
help_people
  • help_people
now I'm confused how to do that can u show me please?
Nnesha
  • Nnesha
47185/54205= ??
Nnesha
  • Nnesha
anything to the 1 power equal to same thing so (1-r/100)^1 = 1-r/100
help_people
  • help_people
-1?
Nnesha
  • Nnesha
nope 47185/54205= ?? divide this first
help_people
  • help_people
ok
help_people
  • help_people
0.87
Nnesha
  • Nnesha
r isn't .87
Nnesha
  • Nnesha
did you divide 47185/54205= ??
help_people
  • help_people
yes
Nnesha
  • Nnesha
show ur work how did you get .87 ?
help_people
  • help_people
on a calc ???
help_people
  • help_people
it was 0.870492
Nnesha
  • Nnesha
what did you put in to the cal?
help_people
  • help_people
what you exactly told me @Nnesha
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
  • Nnesha
put*
help_people
  • help_people
i got .87 by dividing those two numbers like you told me i did not see r equals that @Nnesha
Nnesha
  • Nnesha
facepalm* i thought u r saying r=.87 >.^
Nnesha
  • Nnesha
wait we don't have to use that equation mhmm
amistre64
  • 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
  • amistre64
or rather 1 - |r| might be more precise
help_people
  • help_people
so i have to put in the values in r ?
amistre64
  • amistre64
you are finding the percentage of difference, yes
help_people
  • help_people
ok
amistre64
  • amistre64
for an estimate, what is (47-41)/47?
help_people
  • help_people
0.12766
amistre64
  • 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
  • help_people
round it would be 0.13
amistre64
  • amistre64
then we lose about .13 a year we retain about 1-.13
amistre64
  • 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
  • help_people
c
amistre64
  • amistre64
.13 is NOT what we want ...
amistre64
  • amistre64
what is 1 - .13?
amistre64
  • 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
  • help_people
then b?
amistre64
  • amistre64
does b have .13 in it?
amistre64
  • amistre64
do want to have a .13 in it?
amistre64
  • 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
  • amistre64
another view of it is 47(.13) = 41 ?
help_people
  • help_people
i don't get it i thought it would be that is tme 13
help_people
  • help_people
@amistre64
amistre64
  • amistre64
answer me this ... is 47(1.3) = 41 ? or even close to it?
amistre64
  • amistre64
47(.13 ) = 41 that is
amistre64
  • amistre64
if .13 is what we lose in value, then how much do we keep in value?
help_people
  • help_people
? I'm not understudying what ur asking?
amistre64
  • amistre64
good luck then, because I really dont have any other way of asking it.
help_people
  • help_people
would we keep .87?
amistre64
  • amistre64
yes
help_people
  • help_people
ok then i did understand srry
help_people
  • help_people
so based on that i find a to be the answer
amistre64
  • amistre64
'a' is a trick we start with the original value of the equipment
amistre64
  • 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
  • help_people
ok
help_people
  • 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
  • help_people
so that does nt show the whole information?
amistre64
  • amistre64
you found the percent of change from 47 to 41, its a difference calculation. 47 lost 13% to get to 41
amistre64
  • amistre64
it shows a method of approaching the solution.
amistre64
  • amistre64
41/47 = .87 is valid as well.
help_people
  • help_people
yeah so d? but i still am not getting why
help_people
  • help_people
wait yes i amm srry @amistre64
amistre64
  • amistre64
good luck :)
Nnesha
  • Nnesha
P represents initial amount which is what in the question ?
help_people
  • help_people
thank yo so d is correct ? @amistre64
Nnesha
  • 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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