3.

Question

The distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 6 inches with 80 pounds attached, how far will it stretch with 55 pounds attached? Round to the nearest tenth of an inch.

Answer 1

Given that the distance a spring stretches varies directly with the weight attached, we can use the formula for direct variation:

\[ d = k \cdot w \]

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 when \( w = 80 \) pounds, the spring stretches \( d = 6 \) inches. We can solve for \( k \):

\[ 6 = k \cdot 80 \]

Dividing both sides by 80:

\[ k = \frac{6}{80} = \frac{3}{40} \]

Now that we have \( k \), we can find out how far the spring will stretch with \( w = 55 \) pounds:

\[ d = k \cdot w = \frac{3}{40} \cdot 55 \]

Calculating this:

\[ d = \frac{3 \cdot 55}{40} = \frac{165}{40} \]

Now, perform the division:

\[ \frac{165}{40} = 4.125 \]

Rounding to the nearest tenth of an inch, we get:

\[ d \approx 4.1 \text{ inches} \]

Thus, the spring will stretch approximately 4.1 inches with 55 pounds attached.