The length of a rectangle is 6 ft more than its width. The perimeter of the rectangle is 68 ft. What are the dimensions of the rectangle?

Question

Let w represent the width of the rectangle.
Enter the expression for length in the box.The length of a rectangle is 6 ft more than its width. The perimeter of the rectangle is 68 ft. What are the dimensions of the rectangle?
Let w represent the width of the rectangle.

l = __

Answer 1

To express the length \( l \) of the rectangle in terms of the width \( w \), we can use the information given in the problem. Since the length is 6 ft more than the width, we can write:

\[ l = w + 6 \]

Now we have the expression for the length of the rectangle.

Next, we can use the perimeter formula for a rectangle, which is given by:

\[ \text{Perimeter} = 2(l + w) \]

Since the perimeter is 68 ft, we write:

\[ 2(l + w) = 68 \]

Now, we can substitute the expression for \( l \) into the perimeter equation:

\[ 2((w + 6) + w) = 68 \]

This simplifies to:

\[ 2(2w + 6) = 68 \]

Dividing both sides by 2 gives us:

\[ 2w + 6 = 34 \]

Next, we subtract 6 from both sides:

\[ 2w = 28 \]

Now divide by 2:

\[ w = 14 \]

Now we can find the length using the expression we found earlier:

\[ l = w + 6 = 14 + 6 = 20 \]

Thus, the dimensions of the rectangle are:

• Width \( w = 14 \) ft
• Length \( l = 20 \) ft.

To summarize:

• The length \( l \) of the rectangle in terms of width \( w \) is: \[ l = w + 6 \]
• The dimensions are:
  • Width = 14 ft
  • Length = 20 ft

Answer 2

At noon, Doug drove west from a gas station to a motel. He reached the motel at 3 p.m. Also leaving at noon, Ashley drove east from the same gas station to her sister’s house. She arrived at 2:30 p.m. The driving distance from the motel to Ashley’s sister’s house is 341 mi. Ashley and Doug both drove at the same speed, r. At what speed did Doug and Ashley both drive?

Answer 3

Let \( r \) be the speed in miles per hour that both Doug and Ashley drove.

For Doug:

• He drove from the gas station to the motel from noon to 3 p.m., which is a total of 3 hours.
• The distance Doug traveled can be expressed as: \[ d_{Doug} = r \times 3 \]

For Ashley:

• She drove from the gas station to her sister’s house from noon to 2:30 p.m., which is a total of 2.5 hours.
• The distance Ashley traveled can be expressed as: \[ d_{Ashley} = r \times 2.5 \]

Distance Relationship: According to the problem, the total distance from the motel to Ashley's sister's house is 341 miles. Therefore, we have: \[ d_{Doug} + d_{Ashley} = 341 \]

Substituting the expressions for distance gives us: \[ (r \times 3) + (r \times 2.5) = 341 \]

Now, combine the terms: \[ 3r + 2.5r = 341 \] \[ 5.5r = 341 \]

Solving for \( r \): Now, divide both sides by 5.5: \[ r = \frac{341}{5.5} \] \[ r = 61.027272727272726 \text{ miles per hour} \]

To simplify the answer: \[ r \approx 61.03 \text{ miles per hour} \]

Thus, both Doug and Ashley drove at approximately \( 61.03 \) miles per hour.

Answer 4

That's wrong

Answer 5

Let's go through the calculations step-by-step once more to ensure correctness.

Given:

• Doug drove for 3 hours (from noon to 3 p.m.).
• Ashley drove for 2.5 hours (from noon to 2:30 p.m.).
• The total distance from the motel (where Doug went) to Ashley's sister’s house is 341 miles.

Let \( r \) be the speed in miles per hour.

1. Distance traveled by Doug: \[ \text{Distance}_{Doug} = r \times 3 \]

2. Distance traveled by Ashley: \[ \text{Distance}_{Ashley} = r \times 2.5 \]

3. Total distance relationship: Since the distances traveled by both add up to the total distance: \[ \text{Distance}{Doug} + \text{Distance}{Ashley} = 341 \] Substituting the expressions derived above: \[ (r \times 3) + (r \times 2.5) = 341 \]

4. Combining terms: \[ 3r + 2.5r = 341 \] \[ 5.5r = 341 \]

5. Solving for \( r \): Divide both sides by 5.5: \[ r = \frac{341}{5.5} \]

Now let's compute this division accurately: \[ r = 341 \div 5.5 \approx 61.02727272727273 \]

To perform it with more precision:

• Convert 5.5 into a fraction: \[ r = \frac{3410}{55} = 61.0 \]

So, multiplying both numerator and denominator by 10 gives: \[ r = \frac{3410}{55} = 61.0 \]

Thus: The driving speed of both Doug and Ashley is \( 61 \) miles per hour (exact value).

If there is a specific aspect you would like me to clarify further or check again, please let me know!

Answer 6

You're wrong, it's 62

Answer 7

Apologies for any confusion in the calculations. Let’s recheck everything step-by-step.

Given Information:

• Doug drove west to the motel from noon to 3 p.m. (3 hours).
• Ashley drove east to her sister's house from noon to 2:30 p.m. (2.5 hours).
• The distance from the motel to Ashley's sister's house is 341 miles.
• Both drove at the same speed \( r \).

Equations Setup:

1. Distance Doug traveled: \[ \text{Distance}_{Doug} = r \times 3 \]

2. Distance Ashley traveled: \[ \text{Distance}_{Ashley} = r \times 2.5 \]

Total Distance Relationship:

The combined distances from Doug and Ashley to the motel and sister's house is: \[ \text{Distance}{Doug} + \text{Distance}{Ashley} = 341 \] Substituting the expressions for the distances: \[ (r \times 3) + (r \times 2.5) = 341 \] \[ 3r + 2.5r = 341 \] Combine terms: \[ 5.5r = 341 \]

Solving for \( r \):

Now, divide both sides by 5.5: \[ r = \frac{341}{5.5} \]

Calculating this more accurately: \[ r = \frac{341 \times 10}{55} = \frac{3410}{55} \] Now, performing the division: \[ r = 62 \]

Conclusion:

You are correct! The speed at which both Doug and Ashley drove is indeed 62 miles per hour. Thank you for your patience, and I apologize for the previous error.