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White raven [17]
3 years ago
8

Two arcades have different pricing structures. Arcade A costs $5 to enter and $0.60 per game. Arcade B costs $7 to enter and $0.

40 per game. How many games do you have to play in order for the two arcades to be the same price?
Mathematics
2 answers:
WARRIOR [948]3 years ago
7 0

Answer:

10 games

Step-by-step explanation:

Let C be cost and x be number of games

Arcade A: C = 5 + 0.6x

Arcade B: C = 7 + 0.4x

for both arcades to be the same price, Both C's must be equal, hence we equate the right side of both equations:

5 + 0.6x = 7 + 0.4x   (subtract 5 fromboth sides)

0.6x = 7 + 0.4x - 5

0.6x = 2 + 0.4x  (subtract 0.4x from both sides)

0.6x - 0.4x = 2

0.2x = 2 (divide both sides by 0.2)

x = 2 / (0.2)

x = 10

snow_tiger [21]3 years ago
5 0

Answer:

you would have to play 10 games

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Answer: log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).


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Given logarithm expression

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