Find the sum of 0.48 and 0.6

Hello there,

First you need to add together the numbers.
4+5+9 = 18
Divide the total by this,
180 / 18 = 10
Multiply each number by 10.
40:50:90.

Hope This Helps You!
Good Luck Studying :)

4 0

3 years ago

Read 2 more answers

Can someone plz help me on this plz I beg u plz

JulsSmile [24]

<h3>Answer: 25w+200 > 750</h3>

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Explanation:

He starts off with 200 cards. Then he adds on 25w more cards for each week (w). Overall, he'll have 200+25w cards

We can think of it like this:

After 1 week, he adds on 25*1 = 25 cards
After 2 weeks, he adds on 25*2 = 50 cards total
After 3 weeks, he adds on 25*3 = 75 cards total
After 4 weeks, he adds on 25*4 = 100 cards total, and so on.
After w weeks, he adds on 25w cards total

So that's another way to see where the 25w comes from.

The expression 200+25w is the same as 25w+200. This is because we can add two numbers in any order.

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Since he wants to know when he'll have more than 750 cards, this means we set 25w+200 greater than 750.

That's how we get to the answer of 25w+200 > 750

Notice how there isn't a line under the inequality sign. We aren't using the "greater than or equal to" symbol here. We want to know when the cards gets over 750, but we don't want to know when it's equal to 750.

5 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

3 years ago