The value of an antique plate after x years can be modeled by f(x) = 18(1.05)x. Which graph can be used to approximate the number of years it will take for the plate’s value to be $30?


The value of an antique plate after x years can be modeled by fx 18105x Which graph can be used to approximate the number of years it will take for the plates v class=
The value of an antique plate after x years can be modeled by fx 18105x Which graph can be used to approximate the number of years it will take for the plates v class=
The value of an antique plate after x years can be modeled by fx 18105x Which graph can be used to approximate the number of years it will take for the plates v class=

Respuesta :

we have that

[tex] f(x)=18*(1.05^{x} ) [/tex]

using a graph tool

see the attached figure N 1

for f(x)=$30

the number of years x is equal

x=10.47 years

the answer in the attached figure N 2

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Answer:

Option A                  

Step-by-step explanation:

Given : The value of an antique plate after x years can be modeled by

[tex]f(x) = 18(1.05)^x[/tex]    

To find : Which graph can be used to approximate the number of years it will take for the plate’s value to be $30.

Solution: First we find the value of x at y=30

[tex]f(x) = 18(1.05)^x[/tex]    

[tex]30 = 18(1.05)^x[/tex]    

[tex]\frac{30}{18}=(1.05)^x[/tex]    

[tex]1.6666=(1.05)^x[/tex]    

Taking log both side and apply property  [tex]logx^a=alogx[/tex]  

 [tex]log(1.6666)=xlog(1.05)[/tex]  

 [tex]0.221848749616=x0.0211892990699[/tex]  

[tex]\frac{0.221848749616}{0.0211892990699}=x[/tex]    

[tex]x=10.4698484308[/tex]    

Therefore, from the given graph where the value (x,y)=(10.47,30) lie and the curve from (0,18) to (10.47,30) is the solution of the answer.

Hence, option A is correct (also correct graph is attached)


                       

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