TRUEBOSS72
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PLEASE HELP ME QUICKLY ON THIS QUESTION!!!
Brainliest will be given to the correct answer
A famous fast food chain opened two branches in different parts of a city. Branch A made a profit of $50,000 in the first year, and the profit increased by 4% every year. Branch B made a profit of $35,000 in the first year, and the profit increased by 5.5% every year.

Which function can the fast food chain use to determine its total profit, P(x), after x years, and how much money will the chain have made in profit after 4 years?

a. P(x) = 5,000(10(1.055)x + 7(1.04)x); $97,341.65
b. P(x) = 5,000(10(1.055)x + 7(1.04)x); $106,576.25
c. P(x) = 5,000(10(1.04)x + 7(1.055)x); $101,851.79
d. P(x) = 5,000(10(1.04)x + 7(1.055)x); $73,017.93


Please explain

Respuesta :

sqdancefan
Selections C and D have the correct formula.

Do the arithmetic to find that selection C has the correct 4-year profit value.
mathmate
Let
x=number of years AFTER the first year (year 0).
A=amount earned during year 0.
r=increase in profit every year thereafter.

Then the profit in year x would be
p(x)=A(1+r)^x   [note the exponential operator ^]

If there are two stores, then
[tex]P(x)=p_1(x)+p_2(x)[/tex]
[tex]=A_1(1+r_1)^x+A_2(1+r_2)^x[/tex] ..............(1)
For store 1, 
A1=50,000, r1=0.04 (4%),
For store 2,
A2=35,000, r2=0.055
Now substitute numbers in (1)
[tex]=50,000(1+0.04)^x+35,000(1+0.055)^x[/tex]
Factor out the common factor 5000,
[tex]=P(x)=5,000(10(1.04)^x+7(1.055)^x)[/tex].................(1A)
which corresponds to equations c OR d.

Substituting x=4 yearsAFTER the first  year,  into (1A) :
[tex]P(4)=5,000(10(1.04)^4+7(1.055)^4)[/tex]
[tex]=58492.93+43358.86[/tex]
P(4)=$101,851.79






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