All categories

2 years ago

40 POINTS

Mathematics

1 answer:

LUCKY_DIMON [66]2 years ago

8 0

Answer:

2.5/5 acre's in 8/8 day's

Step-by-step explanation:

You might be interested in

The O'Brien family is ordering food for a family reunion. The family orders 16 pounds of potato salad. If 20 people attend the r

sukhopar [10]

Answer:

Each person would receive maximum 0.8 pounds of potato.

Step-by-step explanation:

Given:

Total Amount of Potato = 16 pounds.

Number of people attend reunion = 20

We need to find the largest amount of potato each person would receive.

Solution:

largest amount of potato each person would receive can be calculated by dividing Total Amount of Potato with Number of people attend the reunion.

framing in equation form we get;

largest amount of potato each person would receive = $\frac{16}{20} =0.8\ pounds$

Hence Each person would receive maximum 0.8 pounds of potato.

5 0

3 years ago

Just go down the row and say yes no yes no in order!you don't have to type the whole thing out

tangare [24]

Answer:

1) yes

2) no

3) yes

4)no

5)no

Step-by-step explanation:

8 0

2 years ago

Of every 300 magic cakes Farah makes, 90 are chocolate, what percentage of chocolate cakes Farah makes?

navik [9.2K]

Answer:

Total magic cakes =300

chocolate cakes =90

so if you want the percentage of chocolate you will divide the number of chocolate cakes over the total number of chocolates

% of chocolate cakes = 90/300×100%

=30%

5 0

2 years ago

How was y'alls day? :)

VikaD [51]

Not the best, teachers are so annoying lol. Hope you are doing well though, don’t give up❤️

8 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

2 years ago