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3 years ago

A square has a perimeter given by the expression 16x+32y. Write an expression for the length of one side of the square.

2 answers:

8 0

A square has 4 equal sides which make up the perimeter.

Therefore the length of one side = (16x + 32y) / 4 = 4x + 8y Answer

AnnZ [28]3 years ago

6 0

A square has 4 sides with equal length.
The perimeter is the sum of the 4 equal lengths.
The perimeter is also 4 times the length of one side.
Each side is 1/4 of the perimeter.

P = 16x + 32y

s = P/4 = (16x + 32y)/4 = 4x + 8y

Answer: 4x + 8y

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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

Now see if you can solve these word problems about time and distance.

ladessa [460]

Answer:

3.5 hours

Step-by-step explanation:

Ediburgh -------------------------------350 miles-------------------------London

Let the distance traveled by train from Edin to London be "x"

hence the distance traveled by train from London to Edin would be "350-x"

The rate of Edin to London train is 60
The rate of London to Edin train is 40

Let the time they meet be"t"

Now, we know distance formula to be D = RT

Where

D is distance

R is rate

and

T is time

Train from Edin to London:

x = 60t

Train from London to Edin:

350-x = 40t

Solving for x:

350 - 40t = x

We put this into first equation:

x = 60t

350 - 40t = 60t

350 = 100t

t = 350/100

t = 3.5

They meet after 3.5 hours

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3 years ago

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olya-2409 [2.1K]

Answer:

can u convert this to miles?

Step-by-step explanation:

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3 years ago

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Andrej [43]

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Step-by-step explanation:

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3 years ago

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9514 1404 393

Answer:

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2 years ago