Find the value of (13xy + 6zx - 19yz), if x = 1, y = 2, and z = 1.

- 5

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- 6

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Solution

The correct option is B - 6
(13xy + 6zx - 19yz)
Substituting the value of x = 1, y = 2, and z = 1 in the above expression, we get
(13xy + 6zx - 19yz) = {13(1)(2) + 6(1)(1) - 19(2)(1)}
$= (13 \times 1 \times 2) + (6 \times 1 \times 1) - (19 \times 1 \times 2)$
= 26 + 6 - 38
= - 6

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