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Nimfa-mama [501]
3 years ago
15

What is the unit rate for 25 rulers in 5 groups

Mathematics
1 answer:
Artist 52 [7]3 years ago
5 0

Answer:

5 rulers/ group

Step-by-step explanation:

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The dimensions of a rectangle are shown. What is the area of the rectangle?
ivanzaharov [21]

Answer:

=(3x+8) (5x+2)

=15x2 + 46x + 16 units

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3 years ago
Solve the system using substitution.<br> 5x + 3y = -4<br> y - 2x = 6<br> ([?], [? ])
sp2606 [1]

Answer:

(-2,2)

Step-by-step explanation:

i tried sorry if it's wrong

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DESPERATE (please answer with picture of the graph) <br>WILL MARK BRAINLIST AND THANKS​
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4.8

Step-by-step explanation:

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you have three purple chips and two green chips in a bag. you draw one chip, and then without replacing it, you draw another chi
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1 year ago
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.

4 0
4 years ago
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