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3 years ago

62/8 With a remainder

Mathematics

2 answers:

patriot [66]3 years ago

8 0

Answer:

7 and 6/8

Step-by-step explanation:

scoundrel [369]3 years ago

5 0

Answer:

Step 1: 8 times what equals the closest number to 62?

Answer: 8*7=56

Step 2: what do you have left oveR?

answer: 6

Step 3: put it together.

ANSWER: 7 with a remander of 6 (7.6)

Step-by-step explanation:

7.6

sorry if it took to long i wanted to show you how to do it so next time you know:)

Brainliest is appreciated:)

Have a great day mate!

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

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