Juan graphed a line with a slope of −2/3 .

Mathematics

2 answers:

elena55 [62]3 years ago

5 0

Answer: Down 2, right 3

Step-by-step explanation:

jolli1 [7]3 years ago

4 0

Remember it’s rise over run. When the rise is negative you will be going down, therefore down 2. When run is positive you move to the right. Therefore, over (to the right) 3.

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denis-greek [22]

Got it, that answer is 1 5/12

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3 years ago

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A space is totally disconnected if its connected spaces are one-point-sets.Show that a finite Hausdorff space is totally disconn

marysya [2.9K]

Step-by-step explanation:

If X is a finite Hausdorff space then every two points of X can be separated by open neighborhoods. Say the points of X are $x_1, x_2, ..., x_n$ . So there are disjoint open neighborhoods $U_{12}$ and $U_2$ , of $x_1$ and $x_2$ respectively (that's the definition of Hausdorff space). There are also open disjoint neighborhoods $U_{13}$ and $U_3$ of $x_1$ and $x_3$ respectively, and disjoint open neighborhoods $U_{14}$ and $U_4$ of $x_1$ and $x_4$ , and so on, all the way to disjoint open neighborhoods $U_{1n}$ , and $U_n$ of $x_1$ and $x_n$ respectively. So $U=U_2 \cup U_3 \cup ... \cup U_n$ has every element of $X$ in it, except for $x_1$ . Since $U$ is union of open sets, it is open, and so $U^c$ , which is the singleton $\{ x_1\}$ , is closed. Therefore every singleton is closed.

Now, remember finite union of closed sets is closed, so $\{ x_2\} \cup \{ x_3\} \cup ... \cup \{ x_n\}$ is closed, and so its complemented, which is $\{ x_1\}$ is open. Therefore every singleton is also open.

That means any two points of $X$ belong to different connected components (since we can express X as the union of the open sets $\{ x_1\} \cup \{ x_2,...,x_n\}$ , so that $x_1$ is in a different connected component than $x_2,...,x_n$ , and same could be done with any $x_i$ ), and so each point is in its own connected component. And so the space is totally disconnected.