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dangina [55]
2 years ago
7

If x and y are positive numbers with x>y, show that a triangle with sides of lengths 2xy, x^2-y^2, and x^2+y^2 and is always

a right triangle?
Mathematics
1 answer:
Marianna [84]2 years ago
5 0

Answer: 2xy

Step-by-step explanation: We know that x%5E2+%2B+y%5E2 must be the longest side of the triangle (it cannot be 2xy, I will demonstrate why afterwards). By the Pythagorean theorem,

. Therefore this is a right triangle.

The only "assumption" we had to make was that x%5E2+%2B+y%5E2+%3E+2xy, but this is easy to prove. By the AM-GM inequality, %28x%5E2+%2B+y%5E2%29%2F2+%3E=+sqrt%28x%5E2y%5E2%29+=+xy. Multiplying both sides by 2, x%5E2+%2B+y%5E2+%3E=+2xy. Equality occurs only when x+=+y, but this cannot be true, so x%5E2+%2B+y%5E2 is strictly greater than 2xy.

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Write an equation to represent the N'th term of the sequence (2,-1.-4,-7,...)
mr_godi [17]

Answer: a(n) = 5 - 3n

the sequence has:

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.........

we can see that: a2 - a1 = a3 - a2 = -3

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Step-by-step explanation:

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Solve the equation 3/4x+3-2x = -1/4+1/2x+5
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Let's solve this equation step by step!
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First collect terms.
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Subtract (1/2)x from both sides.
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Subtract 3 from both sides.
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