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ExtremeBDS [4]
3 years ago
9

Solve 6 1 - (2 4/16 +3 2/3)Do the operation in parentheses first Show all your steps.​

Mathematics
1 answer:
amid [387]3 years ago
6 0

Answer:

55\frac{1}{12}

Step-by-step explanation:

61 - (2 4/16 +3 2/3) = 61 - (5 +( 4/16 + 2/3))

                              = 61 - (5 +( 6/24 + 16/24))

                             = 61 - (5 + 22/24)

                             = (61 - 5) - 22/24

                             = 56 - 11/12

                             = 55 1/12

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Answer:

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

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Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if (x+y)^{2} = x^{2} + 2\cdot x\cdot y  + y^{2}. From statement, we must fulfill the following identity:

a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}

By Associative and Commutative properties, we can reorganize the expression as follows:

a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2} (1)

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x = a (2)

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By (2) and (4) in (3), we have the following expression:

(b^{2} - 1) = 2\cdot a \cdot c

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b = \sqrt{1 + 2\cdot a\cdot c}

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, a, b, c > 1. If a, b and c are prime numbers, then  2\cdot a\cdot c must be an even composite number, which means that a and c can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, b must be a natural number, which means that:

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The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Example: a = 2, c = 2

b = \sqrt{1 + 2\cdot (2)\cdot (2)}

b = 3

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