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30 July, 06:00

We know the following about the numbers a, b and c:

(a + b) ^2 = 9,

(b + c) ^2 = 25, and

(a + c) ^2 = 81.

If a + b + c ≥ 1, determine the number of possible values for a + b + c.

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  1. 30 July, 08:41
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    Step-by-step explanation:

    (a + b) ² = 9

    (b + c) ² = 25

    (a + c) ² = 81

    Taking the square root:

    a + b = ±3

    b + c = ±5

    a + c = ±9

    By adding these three equations together and dividing both sides by 2, we get the value of a + b + c.

    Possible combinations for a + b + c such that the sum is greater than or equal to 1 are:

    a + b + c = (-3 + 5 + 9) / 2 = 11/2

    a + b + c = (3 - 5 + 9) / 2 = 7/2

    a + b + c = (3 + 5 + 9) / 2 = 17/2
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