What is the answer for 5-2+7x6-2?

son4ous [18]

Answer:

A is bigger

Step-by-step explanation:

lets say the certain number is 10 and if its 20% of Number A that would mean that number A is 50 and that would also mean that number B is 25 so since 50 is bigger then 25 that would mean number A is bigger

5 0

2 years ago

A carnival charges two different prices for admissions. Adults cost $4 while children cost $1.50. If a total of $5050 was collec

GenaCL600 [577]

Answer:

Step-by-step explanation:

We have to have 2 different equations to solve this. One equation will represent the number of tickets sold while the other represents the money collected when the tickets were sold.

We know that adult tickets + children tickets = 2200 tickets.

That's the "number of tickets" equation. Let's call adult tickets "a" and children's tickets "c". So a + c = 2200

Now if each adult costs $4, then the expression that represents that as a cost is 4a. If there is 1 adult, the cost is $4(1) = $4; if there are 2 adults, the cost is $4(2) = $8; if there are 3 adults, the cost is $4(3) = $12, etc.

The same goes for the children's tickets. If each child's ticket is $1.50, then the expression that represents the cost of a child's ticket is 1.5c (we don't need the 0 at the end; it doesn't change anything to drop it off). The total money brought in from the cost of these tickets was $5050, so

4a + 1.5c = 5050

Let's solve the first equation for a. If a + c = 2200, then a = 2200 - c. Sub that into the second equation and solve it for c:

4(2200 - c) + 1.5c = 5050 and

8800 - 4c + 1.5c = 5050 and

-2.5c = -3750 so

c = 1500

That means that there were 1500 children's tickets sold. If a + c = 2200, then a + 1500 = 2200 so

a = 2200 - 1500 so

a = 700

There were 1500 children's tickets sold and 700 adult tickets sold.

7 0

3 years ago

Please help me i need an answer

ivolga24 [154]

I conclude it is C 25.

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

4 0

3 years ago

You heard it as a FRACTION! please answer this lol I'm clueless :'D

uysha [10]

Answer:

$\frac{a}{b + 2}$

a

___

b + 2

4 0

2 years ago

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