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3 years ago

Not easy help me please

Mathematics

1 answer:

Yuki888 [10]3 years ago

3 0

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

In the number line attached in my attachment,

The red line represents point :

$\dfrac{10}{2}$

The blue line represents point :

$\dfrac{ - 9}{2}$

I hope it helps ~

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Use the area of the rectangle to find the area of the triangle.<br> 10 ft

77julia77 [94]

Answer: 15

Aka the first one

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3 years ago

In a movie theater, the number of seats in a row is 8 greater than the total number of rows. How many rows are in the movie thea

Harman [31]

Answer:

26 rows

Step-by-step explanation:

this is like a rectangle length×width situation.

seats per row = s

number of rows = r

s × r = 884

s = r + 8

so, we can use e.g. the second equation in the first :

(r + 8) × r = 884

r² + 8r = 884

r² + 8r - 884 = 0

the general solution to such a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = r

a = 1

b = 8

c = -884

r = (-8 ± sqrt(8² - 4×1×-884))/(2×1) =

= (-8 ± sqrt(64 + 3536))/2 = (-8 ± sqrt(3600))/2 =

= (-8 ± 60)/2 = -4 ± 30

r1 = -4 + 30 = 26

r2 = -4 - 30 = -34

a negative number did not make any sense for the number of rows, so r = 26 is our answer.

8 0

2 years ago

3√75<br> simplest rationalizing factor <br> PLEASE DO ANSWER ONLY IF YOU KNOW IT

rjkz [21]

Answer:

the square root of 75 is 8.6

and 8.6 times 3 is

the answer is 25.8

Hope This Helps!!!

4 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 $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .

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3 years ago

Which pair shows equivalent expressions?

Murrr4er [49]

Answer:

2(2/5x+2)=4x/5+4

Step-by-step explanation:

2(2/5x+2)=4x/5+4

4x/5 +4=4x/5 +4

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3 years ago