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Kruka [31]
3 years ago
13

Brianna went to a carnival. She played five games and rode six rides. Justin went to the same carnival and played seven games an

d rode eight rides. If Brianna paid $11.75 and Justin paid $16.15, how much did one game cost to play?
$0.50

$1.45

$1.25

$1.50
Urgent!
Mathematics
2 answers:
erastovalidia [21]3 years ago
6 0

Answer:

  $1.45

Step-by-step explanation:

To answer the question, we would like to have an equation that has the cost of a game as its only variable. That is, we would like to eliminate the cost of a ride from the system of equations we must write.

Let g and r represents the costs of a game and a ride, respectively. Then the expenses of the two carnival-goers can be described by ...

  5g +6r = 11.75

  7g +8r = 16.15

We note that the ratio of coefficients in the variable (r) that we want to eliminate is 3:4. So we can subtract 4 times the first equation from 3 times the second to eliminate that variable.

  3(7g +8r) -4(5g +6r) = 3(16.15) -4(11.75)

  g = 1.45 . . . . . . simplify

The cost to play one game was $1.45.

andreyandreev [35.5K]3 years ago
3 0

Answer:

$1.45

Step-by-step explanation:

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Answer:

y = -5/14x + 12

Step-by-step explanation:

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6 0
2 years ago
______<br> What is 0.142857 as a fraction
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3 years ago
What value of Y will satisfy the given equation y divided by 2y — 15 = 7/5​
guapka [62]

Answer:

\huge\boxed{\sf y = \frac{41}{5}  \ OR \ y = 8\frac{1}{5} }

Step-by-step explanation:

\sf 2y-15 = \frac{7}{5} \\\\Adding \ 15 \ to \ both \ sides\\\\2y = \frac{7}{5} + 15\\\\2y = \frac{7+75}{5} \\\\2y = \frac{82}{5} \\\\Dividing \ both \ sides \ bu 2\\\\y = \frac{82}{5*2} \\\\y = \frac{41}{5}

Hope this helped!

<h2>~AnonymousHelper1807</h2>
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3 years ago
The first term in a sequence is 3<br> The term-to-term rule is + 8<br> What is the 5th term?
galben [10]
2nd term: 3+8
= 11
3rd term: 11+8
= 19
4th term: 19+8
= 27
5th term: 27+8
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OR
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2 years ago
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Vanyuwa [196]
<h3>What is the greatest common factor of 15x³y² and 20x⁴y⁴? ​</h3>

​<em>First we have to find the factors of 15x³y²</em>

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<em>​Then we have to find the factors of 20x⁴y⁴</em>

<em>​20x⁴y⁴ = 2²x¹y² * 5x³y²</em>

<em>Now we have to find the common factors to both numbers.​</em>

<em>The common factors are </em><em>5x³y²</em>

Answer : GCF = (15x³y²,20x⁴y⁴) = 5x³y²​

​Hope this helps!​​​​

​​\textit{\textbf{Spymore}}​​​​​​

6 0
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