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Find the constant of variation for the relation and use it to write an equation for the statement. Then solve the equation.

If [tex]$y$[/tex] varies directly as [tex]$x$[/tex] and [tex]$z$[/tex], and [tex]$y=4$[/tex] when [tex]$x=6$[/tex] and [tex]$z=1$[/tex], find [tex]$y$[/tex] when [tex]$x=7$[/tex] and [tex]$z=4$[/tex].

A. [tex] y = \frac{8}{3} x z \ ; \ y(7,4) = \frac{224}{3} [/tex]
B. [tex] y = \frac{3}{8} x z \ ; \ y(7,4) = \frac{21}{2} [/tex]
C. [tex] y = \frac{3}{2} x z \ ; \ y(7,4) = 42 [/tex]
D. [tex] y = \frac{2}{3} x z \ ; \ y(7,4) = \frac{56}{3} [/tex]



Answer :

GinnyAnswer
To solve the problem, we need to find the constant of variation for the relation and use it to write an equation, then determine the value of [tex]\( y \)[/tex] when [tex]\( x = 7 \)[/tex] and [tex]\( z = 4 \)[/tex].

1. Understand the relationship:

The problem states that [tex]\( y \)[/tex] varies directly as [tex]\( x \)[/tex] and [tex]\( z \)[/tex]. This means we can express [tex]\( y \)[/tex] as:
[tex]\[ y = kxz \][/tex]
where [tex]\( k \)[/tex] is the constant of variation.

2. Given values to find [tex]\( k \)[/tex]:

We are given:
[tex]\[ y = 4 \quad \text{when} \quad x = 6 \quad \text{and} \quad z = 1 \][/tex]

Substitute these values into the equation to find [tex]\( k \)[/tex]:
[tex]\[ 4 = k \cdot 6 \cdot 1 \][/tex]

Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{4}{6} = \frac{2}{3} \][/tex]

3. Write the equation using [tex]\( k \)[/tex]:

Now that we have [tex]\( k \)[/tex], the equation becomes:
[tex]\[ y = \frac{2}{3} x z \][/tex]

4. Solve for [tex]\( y \)[/tex] when [tex]\( x = 7 \)[/tex] and [tex]\( z = 4 \)[/tex]:

Substitute [tex]\( x = 7 \)[/tex] and [tex]\( z = 4 \)[/tex] into the equation:
[tex]\[ y = \frac{2}{3} \cdot 7 \cdot 4 \][/tex]

Simplify the expression:
[tex]\[ y = \frac{2}{3} \cdot 28 = \frac{56}{3} \][/tex]

Therefore, the correct constant of variation is [tex]\(\frac{2}{3}\)[/tex] and the value of [tex]\( y \)[/tex] when [tex]\( x = 7 \)[/tex] and [tex]\( z = 4 \)[/tex] is [tex]\(\frac{56}{3}\)[/tex].

So the correct answer is:
[tex]\[ \text{d. } y = \frac{2}{3} x z; \quad y(7, 4) = \frac{56}{3} \][/tex]

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