MATH SOLVE

9 months ago

Q:
# The amount of radioactive substance left at the end of x days can be represented by an exponential function. The initial amount is 100 grams, and the amount left after three days is 72.9 grams. Which function models the amount of radioactive substance left at the end of x number of days?a.f(x)=0.85^xb.f(x)=0.90^xc.f(x)=100(0.85)^xd. f(x)=100(0.90)^x STEPS PLEASE WITH THE ANSWER!! I WAS NEVER TAUGHT THIS!!!!

Accepted Solution

A:

The general formula for exponential growth/decay is:

[tex]$f(x) = (\text{initial amount})\cdot(\text{growth/decay rate r})^x[/tex]

so in this case we have:

[tex]f(x)=100\cdot r^x[/tex]

and correct answer is C or D. Now we have two ways to solve this:

1. Substitute x = 3 days and check wich answer is correct (where we get 72.9).

C)

[tex]f(3)=100(0.85)^3\approx\boxed{61.41}\qquad\qquad\text{(wrong answer)}[/tex]

D)

[tex]f(3)=100(0.90)^3=\boxed{72.9}\qquad\qquad\text{(correct answer)}[/tex]

Correct answer is D)

2. We know that [tex]f(3)=72.9[/tex] so we can find value of r.

[tex]f(3)=72.9\\\\100\cdot r^3=72.9\qquad|:100\\\\r^3=0.729\qquad|\sqrt[3]{(\ldots)}\\\\\boxed{r=0.9}[/tex]

Our function is:

[tex]f(x)=100(0.90)^x[/tex]

and again correct answer is D)

[tex]$f(x) = (\text{initial amount})\cdot(\text{growth/decay rate r})^x[/tex]

so in this case we have:

[tex]f(x)=100\cdot r^x[/tex]

and correct answer is C or D. Now we have two ways to solve this:

1. Substitute x = 3 days and check wich answer is correct (where we get 72.9).

C)

[tex]f(3)=100(0.85)^3\approx\boxed{61.41}\qquad\qquad\text{(wrong answer)}[/tex]

D)

[tex]f(3)=100(0.90)^3=\boxed{72.9}\qquad\qquad\text{(correct answer)}[/tex]

Correct answer is D)

2. We know that [tex]f(3)=72.9[/tex] so we can find value of r.

[tex]f(3)=72.9\\\\100\cdot r^3=72.9\qquad|:100\\\\r^3=0.729\qquad|\sqrt[3]{(\ldots)}\\\\\boxed{r=0.9}[/tex]

Our function is:

[tex]f(x)=100(0.90)^x[/tex]

and again correct answer is D)