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2 years ago

Mr.Russo s,changed 200 euros (European money) for 270 dollars.How many dollars would be get for 300 euros? Hellllllllp me

2 answers:

4 0

He would get 405 US dollars. Because 200 euros can be simplified to 2 = 270 divide 270 by 2 you get 135. Then multiply 135 by 3

Travka [436]2 years ago

3 0

Mr russo has 200 euros so another 100 euro

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Last Saturday the movie Plex theater sold 1456 tickets. The theater sold six times as many regular price tickets as it sold disc

Zepler [3.9K]

NO!!!!!!! 1450/7 is correct because 6x+x=1456 7x=1456 x=208

7 0

3 years ago

Frank works at a convenience store. He earns 7.50 per hour when he works during the day. He earns 12.50 per hour when he works a

fredd [130]

Answer:

7.50x + 12.50y = 300

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Find the measure of angle A

blagie [28]

Answer:

43°

Step-by-step explanation:

(x + 53)° + (x + 74)° + 73° = 180°

(x + 53 + x + 74)° = 180° - 73°

(2x + 127)° = 107°

2x + 127 = 107

2x = 107 - 127

2x = - 20

x = - 20/2

x = - 10

$m\angle A = (x +53)\degree \\\\m\angle A = (-10 +53)\degree \\\\m\angle A = 43\degree \\\\$

4 0

3 years ago

Solve the system of linear equations using the elimination method: 2x + 2y = 16 -2x +3y = 14

horrorfan [7]

Answer:

Step-by-step explanation:

soulution:

given, 2x +2x = 16 - 2x + 3y = 14

2x + 2x = -2y +3y = 14 -16

4x = y = -2

y = 4x = -2

y = x = 4/-2

x = y = -2 ans

4 0

3 years ago

Prove each statement that is true and ﬁnd a counterexample for each statement that is false. Assume all sets are subsets of a un

tekilochka [14]

Answer with Step-by-step explanation:

We are given that A, B and C are subsets of universal set U.

We have to prove that

$A\cap (B-C)=(A\cap B)-(A\cap C)$

Proof:

Let x $\in A\cap (B-C)$

Then $x\in A$ and $x\in(B-C)$

When $x\in ( B-C)$ then $x\in B$ but $x\notin C$

Therefore, $x\in( A\cap B)$ but $x\notin (A\cap C)$

Hence, it is true.

Conversely , Let $x\in(A\cap B)$ but $x\notin(A\cap C)$

Then $x\in A$ and $x\in B$

When $x\notin ( A\cap C)$ then $x\notin C$

Therefor, $x\in A\cap (B-C)$

Hence, the statement is true.

5 0

3 years ago