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2 years ago

The table shows a proportional relationship between the mass, in kilograms

Mathematics

1 answer:

OlgaM077 [116]2 years ago

5 0

Answer:

C) Dog: 40 kg, Medicine: 1.6 mL

D) Dog: 35 kg, Medicine: 1.4 mL

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

Let

x ---> the mass, in kilograms of a dog

y ---> milliliters of flea medicine a veterinarian prescribes

Find the value of the constant of proportionality k for the given data

$k=\frac{y}{x}$

For x=50 kg, y=2 mL ----> $k=\frac{2}{50}=0.04\ mL/kg$

For x=10 kg, y=0.4 mL ----> $k=\frac{0.4}{10}=0.04\ mL/kg$

For x=5 kg, y=0.2 mL ----> $k=\frac{0.2}{5}=0.04\ mL/kg$

The linear direct equation is equal to

$y=0.04x$

Verify each choice

A) Dog: 15 kg, Medicine: 0.9 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{0.9}{15}= 0.06\ mL/kg$

$0.06 \neq 0.04$

therefore

These numbers could not be used as the missing values in the table

B) Dog: 25 kg, Medicine: 0.8 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{0.8}{25}= 0.032\ mL/kg$

$0.032 \neq 0.04$

therefore

These numbers could not be used as the missing values in the table

C) Dog: 40 kg, Medicine: 1.6 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{1.6}{40}= 0.04\ mL/kg$

$0.04 = 0.04$

therefore

These numbers could be used as the missing values in the table

D) Dog: 35 kg, Medicine: 1.4 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{1.4}{35}= 0.04\ mL/kg$

$0.04 \neq 0.04$

therefore

These numbers could be used as the missing values in the table

E) Dog: 20 kg, Medicine: 1.2 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{1.2}{20}= 0.06\ mL/kg$

$0.06 \neq 0.04$

therefore

These numbers could not be used as the missing values in the table

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