The number of milligrams of a certain medicine a veterinarian gives to a dog varies directly with the weight of the dog. If the veterinarian gives a 30-pound dog [tex] \frac{3}{5} [/tex] milligram of the medicine, which equation relates the weight, [tex] w [/tex], and the dosage, [tex] d [/tex]?

A. [tex] d = \frac{1}{50} w [/tex]
B. [tex] d = \frac{3}{5} w [/tex]
C. [tex] d = 18 w [/tex]
D. [tex] d = 50 w [/tex]



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]