All categories

3 years ago

Does each function describe exponential growth or decay?

Mathematics

1 answer:

Aleksandr [31]3 years ago

5 0

<h3>Answer: Only first two are exponential growth function and last three functions are exponential decay functions.</h3>

Step-by-step explanation: We need to describe exponential growth or decay for the given functions.

The standard exponential function equation is

$y=a(b)^x$ .

Where a is the initial value and b is the growth factor.

Note: If value of b > 1, it would be an exponential growth and if b < 1, it would be an exponential decay.

Let us check them one by one.

$y=(1.008^{1/12}})^{12t}$ => $y=(1.008^})^{1/12\times{12t} }$

=> $y= (1.008)^t$ .

Value of b is 1.008 > 1, therefor it's an exponential growth function.

y=250(1+0.004)^t, also have b>1 therefor it's an exponential growth function.

All other functions has b values less than 1, therefore only first two are exponential growth function and last three functions are exponential decay functions.

You might be interested in

Sean earns $8.50 per hour and made a total of $80.75 today. How many hours did he work?

Vesnalui [34]

Answer: 9.5 hours

In order to solve this, you must divide his total (80.75) by his hourly wage (8.50) When you do that, you get 9.5, so he worked 9.5 hours

Step-by-step explanation:

3 0

3 years ago

How do I solve this

solniwko [45]

There is your answer

3 0

4 years ago

The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .