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Readme [11.4K]
3 years ago
13

Prove that: –7 is not the arithmetic square root of 49

Mathematics
1 answer:
frez [133]3 years ago
6 0

Answer:

–7 is not the arithmetic square root of 49 only in case we are taking it for natural number.

Step-by-step explanation:

  • The square root of x is that number which on squaring gives the result as x.
  • For example, the square root of 4 is +2 or -2. Because on squaring +2 or -2 we get the number 4.
  • Similarly for the case of 49, the square root might be both +7 or -7.
  • But if we are taking it as a natural number then -7 is not valid. Like for length, quantity or other sort of things -7  is not the arithmetic square root of 49.
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(a)
the list would be
-20, -12, -8, -1, 1, 5, 10
the other intergers would have to be greater than 1 in order to be on the correct side and cause 1 to be in the middle. They could be anything but 5 and 10 because these were already chosen.

(b)
the list would be like before
-3 was not included so one of the chosen would be -3 and the other would have to be less than -3 but not -8, -12, or -20 beacuse those were already chosen
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Step-by-step explanation:

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Step-by-step explanation:

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