Key to "Let n(X⋅Y)=24,n(X⋅Z)=15, and n(Y⋅Z)=40. Find n(X⋅Y⋅Z)."

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

nicekikah

Answered question

2021-06-13

Let $n (X \cdot Y) = 24, n (X \cdot Z) = 15$ , and $n (Y \cdot Z) = 40$ . Find $n (X \cdot Y \cdot Z) .$

Answer & Explanation

Talisha

Skilled2021-06-14Added 93 answers

We know that: $n (X \cdot Y) = n (X) \cdot n (Y)$ , $n (X \cdot Z) = n (X) \cdot n (Z)$ , $n (Y \cdot Z) = n (Y) \cdot n (Z)$ ,
and
$x (X \cdot Y \cdot Z) = n (X) \cdot n (Y) \cdot n (Z)$
Therefore, $n (X) \cdot n (Y) = 24, n (X) \cdot n (Z) = 15$
and
$n (Y) \cdot n (Z) = 40$
Multiplying the three equations, we get: ${(n (X))}^{2} {(n (Y))}^{2} {(n (Z))}^{2} = 24 \cdot 15 \cdot 40 = 14400$
Thus, ${(n (X) \cdot n (Y) \cdot n (Z))}^{2} = 14400 \to n (X) \cdot n (Y) \cdot n (Z) = 120$
Finally, using (*), $n (X \cdot Y \cdot Z) = 120$

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