Question
The weight of an object on Earth varies directly with the weight of the same object on the moon. A 300-pound object weighs only 48 pounds on the moon. What is the constant of variation?
Answers
GPT-4o mini
To find the constant of variation, we can use the relationship established by the direct variation formula:
\[
W_m = k \cdot W_e
\]
where:
- \( W_m \) is the weight on the moon,
- \( W_e \) is the weight on Earth,
- \( k \) is the constant of variation.
From the problem, we know that:
- \( W_e = 300 \) pounds (weight on Earth),
- \( W_m = 48 \) pounds (weight on moon).
We can rearrange the formula to solve for \( k \):
\[
k = \frac{W_m}{W_e}
\]
Now, substituting the known values:
\[
k = \frac{48}{300}
\]
Calculating this gives:
\[
k = \frac{48 \div 12}{300 \div 12} = \frac{4}{25}
\]
Thus, the constant of variation \( k \) is:
\[
\boxed{\frac{4}{25}}
\]
\[
W_m = k \cdot W_e
\]
where:
- \( W_m \) is the weight on the moon,
- \( W_e \) is the weight on Earth,
- \( k \) is the constant of variation.
From the problem, we know that:
- \( W_e = 300 \) pounds (weight on Earth),
- \( W_m = 48 \) pounds (weight on moon).
We can rearrange the formula to solve for \( k \):
\[
k = \frac{W_m}{W_e}
\]
Now, substituting the known values:
\[
k = \frac{48}{300}
\]
Calculating this gives:
\[
k = \frac{48 \div 12}{300 \div 12} = \frac{4}{25}
\]
Thus, the constant of variation \( k \) is:
\[
\boxed{\frac{4}{25}}
\]