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xz_007 [3.2K]
3 years ago
13

What is the y-intercept of the equation 2x+3y=6?

Mathematics
1 answer:
Natali [406]3 years ago
4 0
(0,2) is the y-intercept
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Are these functions?
likoan [24]

Answer:

Relation 1: No

Relation 2: Yes

Step-by-step explanation: Domain cannot repeat(have more than one output)

6 0
4 years ago
(V-W)T=a. Solve for V
jasenka [17]
(V - W)T = a             Use the Distributive Property
VT - WT = a             Add WT to both sides
         VT = a + WT   Divide both sides by T
           V = \frac{a + WT}{T}   Cancel out the \frac{T}{T}
           V = \frac{a}{T} + W
4 0
4 years ago
Read 2 more answers
8.
diamong [38]

Answer:

Savings

Step-by-step explanation:

If you don't have a lot of money in general and an emergency occurs savings is the easiest way to go because you money money saved up for issues like these. But if u don't even have a savings account credit cards might be the way to go but it will damage your credit score pretty badly. So Savings is my final answer.

3 0
3 years ago
An item is regularly priced at $40. It is on sale for 30% off the regular price. What is the sale price?
morpeh [17]

40$ - 100% (full price)

x$ - 70% (100-30=70%)

40 * 70 = 100x

2800 = 100x

x = 28

Answer 28$

or

40$ = 100%

then 1% = 40/100 = 0.4$

so 70% will be 0.4 * 70 = 28$

8 0
3 years 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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