Help me with this plz

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

F(x) = 2(x+3)2 -7 a= h= k=

vlabodo [156]

Answer:

Step-by-step explanation:

In vertex form, the equation is

y = a(x-h)^2 + k

So just read off the values!

6 0

3 years ago

A jet climbs at a steady rate at an angle of inclination of 0.5 degree during game off what will its he height be after a 2.2 km

Aneli [31]

A right triangle will be formed with 2.2 km as the hypotenuse, 0.5 degrees as the angle, and an unknown height (see attachment).

Recall the trigonmetric ratios sine, cosine, and tangent that can be used for right triangles.
Sinθ = opposite side/hypotenuse
Cosθ = adjacent side/hypotenuse
tanθ = opposite side / adjacent side
θ (theta) refers to the angle (not the right angle). In this case, θ is the given angle of 0.5°.

In this problem, you will use sine because there is a side opposite to the angle (the unknown side h) and the hypotenuse, and sine relates those two sides.

Let the variable h represent the unknown opposite side, the height.
sin(0.5°) = h / (2.2km)
h = (2.2km)sin(0.5°)
h = 0.019 km

Hope I helped! If you have questions on any of the steps, please comment below.