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.


10.9 in.


11.2 in.


9.5 in.


12.7 in.

Answer 1

The relationship between the weight attached to the spring and the distance it stretches is a direct variation. We can represent this relationship mathematically 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 information given:

\[
d = 11 \text{ inches} \quad \text{when} \quad w = 75 \text{ pounds}
\]

We can find \( k \):

\[
11 = k \cdot 75
\]

To find \( k \), we solve for it:

\[
k = \frac{11}{75}
\]

Now we want to find out how far the spring stretches with 65 pounds attached:

\[
d = k \cdot 65
\]

Substituting the value of \( k \):

\[
d = \frac{11}{75} \cdot 65
\]

Calculating that gives:

\[
d = \frac{11 \cdot 65}{75}
\]

Calculating the numerator:

\[
11 \cdot 65 = 715
\]

Now, dividing:

\[
d = \frac{715}{75} \approx 9.5333
\]

Rounding to the nearest tenth gives:

\[
d \approx 9.5 \text{ inches}
\]

So the answer is:

**9.5 in.**