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1 year ago

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2 answers:

Alika [10]1 year ago

7 0

X is side to side
Y is down to up

GenaCL600 [577]1 year ago

3 0

Answer:

Top one is the y-axis. the bottom line is the x-axis. The origin is 0. the first point is c the second is b the third is a

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Buchtal, a manufacturer of ceramic tiles, reports on average 2.3 job-related accidents per year. Accident categories include tri

Marianna [84]

Answereippcb.jrc.ec.europa.eu

Step-by-step explanation:

this I the wed go on it and you will get your answer

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3 years ago

What is the simplest form of 8/40?

ehidna [41]

1/5 is the Simples Form

5 0

3 years ago

Read 2 more answers

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .

4 0

3 years ago

Which value, when placed in the box, would result in a system of equations with infinitely many solutions?

Fantom [35]

Answer:

<h2>If we placed the number 10 in the box, we obtain a system of equations with infinitely many solutions.</h2>

Step-by-step explanation:

The given system is

$y=2x-5\\2y-4x=a$

<em>It's important to know that a system with infinitely many solutions, it's a system that has the same equation</em>, that is, both equation represent the same line, or as some textbooks say, one line is on the other one, so they have inifinitely common solutions.

Having said that, the first thing we should do here is reorder the system

$y=2x-5\\2y=4x+a$

This way, you can compare better both equations. If you look closer, observe that the second equation is double, that is, it can be obtained by multiplying a factor of 2 to the first one, that is

$y=2x-5\\2y=4x-10$

So, by multiplying such factor, we obtaine the second equation. Observe that $a$ must be equal to 10, that way the system would have infinitely solutions.

Therefore, the answer is 10.

4 0

3 years ago

When do you carry the high number in vertical multiplication

tangare [24]

When it is greater than 9.

Example

15
x 5
——-

5x5 is 25, so you would leave 5 and carry the 2

2
15
x. 5
——-
5

3 0

3 years ago