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madam [21]
3 years ago
12

If 2(a^2+b^2)=(a+b)^2 then, >a+b=0, >ab=0, >a=b, >2a=b

Mathematics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

2·(a^2 + b^2) = (a + b)^2

2·a^2 + 2·b^2 = a^2 + 2·a·b + b^2

a^2 + b^2 = 2·a·b

a^2 - 2·a·b + b^2  = 0

(a - b)^2 = 0

a = b


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Answer:

the average of this new list of numbers is 94

Step-by-step explanation:

Hello!

To answer this question we will assign a letter to each number for the first list and the second list of numbers, remembering that the last number of the first list is 80 and the last number of the second list is 96

for the first list

\frac{a+b+c+80}{4} =90

for the new list

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To solve this problem consider the following

1.X is the average value of the second list

2. We will assign a Y value to the sum of the numbers a, b, c.

a + b + c = Y to create two new equations

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for the second list

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4 0
3 years ago
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Answer:

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Step-by-step explanation:

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Answer:

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