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3 years ago

Which expression is equivalent to 5! 5^4*3^2 5-4-3-2-1 5.4.3-2.1 5+4+3+2+1 5.1

1 answer:

6 0

The first one is like 5x5x5x5 which is 625 multiplied by 3x3 which is 9 and together the product will be 5,625.

The second one is 5-4 which is 1-3 which is -2 - 2 = -4 - 1 which is -5 so it’s not it because it’s the opposite of the number we need to find.

The third one is kind of messed up because there’s no such thing as 5.4.3 so sorry I can’t do that one.

The fourth one is 15 because you basically add the numbers up.

And the fifth is not 5. It’s 5.1 so that’s not it either.

So I’m sorry I can’t find the answer but at least I have you some background :/

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X = y - 3

mojhsa [17]

Hello,
{x=y-3
{x+3y=13;

{x=y-3
{y-3+3y=13;

{x=y-3
{4y=13+3;

{x=y-3
{4y=16;

{x=y-3
{y=16:4=4;

{x=4-3=1
{y=4;

{x=1
{y=4

the solution to the system of equations is (1,4)

Bye :-)

8 0

3 years ago

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

2 years ago

If a neutral atom gains an electron, what is its charge?

DENIUS [597]

C. Negative because since the neutral atom stands at 0 and positive ions are 1+ and electrons are negative if it gains the electron ( negative ) it will become negative.

7 0

2 years ago

C is equidistant from A and B. What is the value of x?<br> A. 6<br> B. 8<br> C. 12<br> D. 14

VLD [36.1K]

The correct answer that i came up with is D

6 0

2 years ago

What does x equal? Similarity

Mashcka [7]

Check the picture below.

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2 years ago

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