A cell phone costs $750 and loses 28% of its value each year. What is the value of the phone after 2 years? 1. $216.54
2. $388.80
3. $1228.80
4. $456.30

Initial costs 750 and loses .28 of value each year, so still worth 1-.28=.72 of value at start of next year. 750*(.72)^n is the value after n years since for n=0, new, it is 750*.72^0=750*1=750, and after 1 year it is worth 750*.72^1=750*.72 means it lost 28% value after one year. If you with to express it as an exponential exp(..) you need to convert the .72^n into exp(log(.72)*n)=exp(-.3285*n) so you get

750*exp(-.3285*n) as the value after n years.

388

Step-by-step explanation:

first year it loses 210

750×.28=210

second year phone value is 540

540×.28=151.2

third year phone value is 388

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!