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lions [1.4K]
3 years ago
7

Damarcus and Xavier went to the arcade to play their two favorite games. Damarcus played 4 rounds of game 1 and 6 rounds of game

2 for $18.00. Xavier played 5 rounds of each game for $20. What is is the cost of game 1?
Mathematics
2 answers:
nlexa [21]3 years ago
8 0

Answer:

I think the cost of game 1 is $8

Step-by-step explanation:

dezoksy [38]3 years ago
4 0
The cost of one game is 8
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Write rational number in number line -7 upon 2​
Leno4ka [110]

Answer:

el numero 1

Step-by-step explanation:

7 0
2 years ago
Kenneth signed up to receive Internet service for $13 per month plus a $30 start-up fee. Which equation could be used to find th
zimovet [89]

The equation 13x+30=134 can be used to find the number of months Kenneth can receive internet service.

Step-by-step explanation:

Given,

Start up fee to receive internet = $30

Per month charges = $13

Total amount = $134

Let,

x be the number of months.

y be the total cost.

y = 13x+30

Putting y=134

134=13x+30\\13x+30=134

The equation 13x+30=134 can be used to find the number of months Kenneth can receive internet service.

Keywords: equation, addition

Learn more about addition at:

  • brainly.com/question/1836777
  • brainly.com/question/1993757

#LearnwithBrainly

7 0
2 years ago
X/6 = 15/18<br><br> I NEED ASAP!!!
nadya68 [22]

X/6 = 15/18

Cross multiply:


18x = 15 times 6

18x = 90

Divide both sides by 18

X = 5


5/6 = 15/18

4 0
2 years ago
Read 2 more answers
Pls help me.........
Natali [406]
Wouldn’t it just be (0,0) ?
5 0
3 years ago
When adding two numbers, such as 123 and 423, care is taken to first line them up and then add like digits. How does expanding t
zavuch27 [327]
Think of 10^2, 10^1, and 10^0 as x^2, x^1, and x^0.
When you add polynomials, you can only combine like terms.
When you add expanded numbers, you can only combine like powers of 10.

123 + 423 =

= (100 + 20 + 3) + (400 + 20 + 3)

= (1 * 10^2 + 2 * 10^1 + 3 * 10^0) + (4 * 10^2 + 2 * 10^1 + 3 * 10^0)

= (1 * 10^2 + 4 * 10^2) + (2 * 10^1 + 2 * 10^1) + (3 * 10^0 + 3 * 10^0)

= 5 * 10^2 + 4 * 10^1 + 6 * 10^0

= 500 + 40 + 6

= 546
7 0
2 years ago
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