Ask question

Ask question

All categories

3 years ago

7

The population of master town was 23000 in 2012. Assume that master town's population increases at a rate of 2% per year Write a

n equation to model the population of master town based on the number of years since 2012.

1 answer:

LiRa [457]3 years ago

6 0

This is an example of exponential growth...

F=Ir^t

P(y)=23000(1.02)^(y-2012)

You might be interested in

What is the product of the polynomials below?

alisha [4.7K]

The answer of the question is A

8 0

3 years ago

What is <br> A= 3000(1+.036/12)^72<br> Algebra 2 compound interest

mihalych1998 [28]

First we need to simplify $\frac{.036}{12}$ to 0.003.

We get 3000(1+0.003) $^{72}$

Now we simplify 1+0.0031+0.003 to 1.0031.003.

3000×1.00 $3^{72}$

Simplify {1.003}^{72}1.003 to 1.2407011.240701.

3000×1.240701

Then we simplify to <u>3722.103387.</u>

<u>Have an amazing day, thanks for using Brainly!</u>

4 0

3 years ago

What addition double fact can help you find 4+3? Explain how you know

cestrela7 [59]

3+4 because they are both the same answer

6 0

3 years ago

hassan and eva rode their bikes from the same starting point. Hassan rode his bike 5/6 of a mile, while eve rode hers 4/7 of a m

NARA [144]

Answer: B

Step-by-step explanation:

8 0

2 years ago

Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

Take an element $y\in Y$ , then $f\circ h(y)=y$ . Now, write $x=h(y)$ and notice that $x\in X$ . Also, with this we have that $f(x)=y$ .

So, for every element $y\in Y$ we have found that an element $x\in X$ (recall that $x=h(y)$ ) such that $y=f(x)$ , which is equivalent to the fact that $f$ is surjective. Therefore, the prove is complete.

3 0

3 years ago