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GarryVolchara [31]
3 years ago
13

During the 2010 baseball​ season, the number of wins for three teams was three consecutive integers. Of these three​ teams, the

first team had the most wins. The last team had the least wins. The total number of wins by these three teams was 252. How many wins did each team have in the 2010​ season?
Mathematics
1 answer:
attashe74 [19]3 years ago
3 0

Answer: 83,84 and 85

Step-by-step explanation:

Let the number of win for the teams be x, x+1 and x+2 respectively. From the question, we are told that the total number of wins by these three teams was 252 for the 2010​ season. Therefore,

x+x+1+x+2 = 252

3x+3 = 252

3x = 252 - 3

3x = 249

x = 249/3

x = 83

This means that the team with least wins has 83 while the second team has 84 and the team with most wins has 85.

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<u>Answer:</u>

<u>(625 • (x4)) -  28y4</u>

<u></u>

<u>  54x4 -  28y4 3.1      Factoring:  625x4-256y4  </u>

<u> </u>

<u>Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B) </u>

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<u>         A2 - B2Note :  AB = BA is the commutative property of multiplication. </u>

<u> </u>

<u>Note :  - AB + AB equals zero and is therefore eliminated from the expression. </u>

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<u>Check :  625  is the square of  25  </u>

<u>Check : 256 is the square of 16 </u>

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<u>Factorization is :       (25x2 + 16y2)  •  (25x2 - 16y2)  3.2      Factoring:  25x2 - 16y2  </u>

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<u>Check :  25  is the square of  5  </u>

<u>Check : 16 is the square of 4 </u>

<u>Check :  x2  is the square of  x1  </u>

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<u>Check :  y2  is the square of  y1  </u>

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3 years ago
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