Question
The distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 11 inches with 75 pounds attached, how far will it stretch with 65 pounds attached? Round to the nearest tenth of an inch.
9.5 in.
10.9 in.
11.2 in.
12.7 in.
9.5 in.
10.9 in.
11.2 in.
12.7 in.
Answers
GPT-4o mini
The relationship between the distance a spring stretches and the weight attached to it can be expressed as:
\[
d = k \cdot w
\]
where \( d \) is the distance the spring stretches, \( w \) is the weight attached, and \( k \) is the constant of proportionality.
From the problem, we know that with a weight of 75 pounds, the spring stretches 11 inches. We can first find the value of \( k \):
\[
11 = k \cdot 75
\]
To solve for \( k \):
\[
k = \frac{11}{75}
\]
Now, we can find the distance the spring stretches with a weight of 65 pounds:
\[
d = k \cdot 65 = \frac{11}{75} \cdot 65
\]
Calculating that:
\[
d = \frac{11 \cdot 65}{75}
\]
Calculating \( 11 \cdot 65 = 715 \):
\[
d = \frac{715}{75}
\]
Now, performing the division:
\[
d \approx 9.4667
\]
Rounding to the nearest tenth:
\[
d \approx 9.5
\]
Thus, the spring will stretch approximately **9.5 inches** with 65 pounds attached.
\[
d = k \cdot w
\]
where \( d \) is the distance the spring stretches, \( w \) is the weight attached, and \( k \) is the constant of proportionality.
From the problem, we know that with a weight of 75 pounds, the spring stretches 11 inches. We can first find the value of \( k \):
\[
11 = k \cdot 75
\]
To solve for \( k \):
\[
k = \frac{11}{75}
\]
Now, we can find the distance the spring stretches with a weight of 65 pounds:
\[
d = k \cdot 65 = \frac{11}{75} \cdot 65
\]
Calculating that:
\[
d = \frac{11 \cdot 65}{75}
\]
Calculating \( 11 \cdot 65 = 715 \):
\[
d = \frac{715}{75}
\]
Now, performing the division:
\[
d \approx 9.4667
\]
Rounding to the nearest tenth:
\[
d \approx 9.5
\]
Thus, the spring will stretch approximately **9.5 inches** with 65 pounds attached.