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3 years ago

Which of the following values of x is a solution to the equation negative 2 x plus 8 equals 14?

Mathematics

1 answer:

andreev551 [17]3 years ago

5 0

Answer:

x= -3

Step-by-step explanation:

-2 x 3 =6 so 6+8=14

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Advocard [28]

Answer:

0759, 1019

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the volume of a sphere with the diameter of 4

givi [52]

$d=4 \Rightarrow r=2\\V=\dfrac{4}{3}\pi r^3\\
V=\dfrac{4}{3}\pi \cdot2^3\\
V=\dfrac{4}{3}\pi \cdot8\\
V=\dfrac{32}{3} \pi
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3 years ago

two students club plan to share a bus on a trip to the capital transportation at lunch for day cost $12 per student one club has

ozzi

Answer:

$516

Step-by-step explanation:

first you multiply 12 and 25 and you get 300

then you multiply 12 and 18 and you get 216

Then finally you add 216 and 300 together to get $516

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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

The eagles had a record of 16 wins and 12 loses. What was the ratio of wins to losses?

nata0808 [166]

Answer:

16/12 or 4/3

Math

6 0

2 years ago