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Zepler [3.9K]
3 years ago
14

Solve this equation -98=7(n-7)

Mathematics
2 answers:
iris [78.8K]3 years ago
7 0
-98=7(n-7)\\\\ -98=7n-7\cdot 7 \\ \\-98=7n-49\\ \\7n=-98+49\\\\7n=-49\ \ /:7\\\\n=-7
Sergeeva-Olga [200]3 years ago
3 0
-98=7(n-7)

-98=7n-49

7n=-98+49

7n=-49

Therefore n=-7, as 7(-7)=-49.
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Masteriza [31]

Answer:80 degree angle

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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .
kogti [31]

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)

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

c is another arbitrary prime number. (3)

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:

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

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Then, we have the following system of equations:

x = a (2)

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y = c (4)

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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:

1 + 2\cdot a\cdot c \ge 4

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a\cdot c \ge \frac{3}{2}

But the lowest possible product made by two prime numbers is 2^{2} = 4. Hence, a\cdot c \ge 4.

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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