Answer :
To determine the correct equation that relates the weight [tex]\( w \)[/tex] of a dog to the dosage [tex]\( d \)[/tex] of a medicine, we need to understand that if [tex]\( d \)[/tex] varies directly with [tex]\( w \)[/tex], then:
[tex]\[ d = k \cdot w \][/tex]
where [tex]\( k \)[/tex] is the constant of proportionality.
Given that a 30-pound dog receives [tex]\(\frac{3}{5}\)[/tex] milligram of the medicine, we can substitute these values into our direct variation equation to find [tex]\( k \)[/tex].
[tex]\[ \frac{3}{5} = k \cdot 30 \][/tex]
To solve for [tex]\( k \)[/tex], we divide both sides by 30:
[tex]\[ k = \frac{3/5}{30} \][/tex]
Simplify the expression on the right-hand side:
[tex]\[ k = \frac{3}{5 \cdot 30} \][/tex]
[tex]\[ k = \frac{3}{150} \][/tex]
[tex]\[ k = \frac{1}{50} \][/tex]
Now that we have the constant of proportionality, [tex]\( k = \frac{1}{50} \)[/tex], we can write the equation relating the weight [tex]\( w \)[/tex] and the dosage [tex]\( d \)[/tex]:
[tex]\[ d = \frac{1}{50} w \][/tex]
Thus, the correct equation is:
[tex]\[ d = \frac{1}{50} w \][/tex]
So the answer is:
[tex]\[ d = \frac{1}{50} w \][/tex]
[tex]\[ d = k \cdot w \][/tex]
where [tex]\( k \)[/tex] is the constant of proportionality.
Given that a 30-pound dog receives [tex]\(\frac{3}{5}\)[/tex] milligram of the medicine, we can substitute these values into our direct variation equation to find [tex]\( k \)[/tex].
[tex]\[ \frac{3}{5} = k \cdot 30 \][/tex]
To solve for [tex]\( k \)[/tex], we divide both sides by 30:
[tex]\[ k = \frac{3/5}{30} \][/tex]
Simplify the expression on the right-hand side:
[tex]\[ k = \frac{3}{5 \cdot 30} \][/tex]
[tex]\[ k = \frac{3}{150} \][/tex]
[tex]\[ k = \frac{1}{50} \][/tex]
Now that we have the constant of proportionality, [tex]\( k = \frac{1}{50} \)[/tex], we can write the equation relating the weight [tex]\( w \)[/tex] and the dosage [tex]\( d \)[/tex]:
[tex]\[ d = \frac{1}{50} w \][/tex]
Thus, the correct equation is:
[tex]\[ d = \frac{1}{50} w \][/tex]
So the answer is:
[tex]\[ d = \frac{1}{50} w \][/tex]