Question

Plz help me!!!

Answer 1

To solve this we are going to use the exponential function: $f(t)=a(1(+/-)b)^t$
where
$f(t)$ is the final amount after $t$ years
$a$ is the initial amount
$b$ is the decay  or grow rate rate in decimal form
$t$ is the time in years

Expression A 
$f(t)=624(0.95)^{4t}$
Since the base (0.95) is less than one, we have a decay rate here.
Now to find the rate $b$, we are going to use the formula: $b=|1-base|$*100%
$b=|1-0.95|$*100%
$b=0.05$*100%
$b=$5%
We can conclude that expression A decays at a rate of 5% every three months.

Now, to find the initial value of the function, we are going to evaluate the function at $t=0$
$f(t)=624(0.95)^{4t}$
$f(0)=624(0.95)^{0t}$
$f(0)=624(0.95)^{0}$
$f(0)=624(1)$
$f(0)=624$
We can conclude that the initial value of expression A is 624.

Expression B
$f(t)=725(1.12)^{3t}$
Since the base (1.12) is greater than 1, we have a growth rate here.
To find the rate, we are going to use the same equation as before:
$b=|1-base|$*100%
$b=|1-1.12|$*100
$b=|-0.12|$*100%
$b=0.12$*100%
$b=$12%
We can conclude that expression B grows at a rate of 12% every 4 months.

Just like before, to find the initial value of the expression, we are going to evaluate it at $t=0$
$f(t)=725(1.12)^{3t}$
$f(0)=725(1.12)^{0t}$
$f(0)=725(1.12)^{0}$
$f(0)=725(1)$
$f(0)=725$
The initial value of expression B is 725.

We can conclude that you should select the statements:
- Expression A decays at a rate of 5% every three months, while expression B grows at a rate of 12% every fourth months. 

- Expression A has an initial value of 624, while expression B has an initial value of 725.