a rectangular picture frame has a perimeter of 46 inches. the length is 7 inches more than the width, what are the dimensions of the table?

Let's define the dimensions of the rectangular picture frame:

- Let the width of the frame be \( w \) inches.
- The length of the frame is 7 inches more than the width, so the length is \( w + 7 \) inches.

The formula for the perimeter \( P \) of a rectangle is:
\[
P = 2 \times (\text{length} + \text{width})
\]

Given that the perimeter is 46 inches, we can set up the equation:
\[
46 = 2 \times ((w + 7) + w)
\]

Simplifying the equation:
\[
46 = 2 \times (2w + 7)
\]
\[
46 = 4w + 14
\]

Subtract 14 from both sides:
\[
32 = 4w
\]

Now divide by 4:
\[
w = 8
\]

So the width is 8 inches. Since the length is 7 inches more than the width:
\[
\text{Length} = 8 + 7 = 15 \, \text{inches}
\]

Thus, the dimensions of the frame are:
- Width: 8 inches
- Length: 15 inches

Would you like further details or have any questions?

### Related questions:
1. What would the dimensions be if the perimeter were 60 inches?
2. How would the solution change if the length were 5 inches more than the width?
3. Can you calculate the area of this rectangular picture frame?
4. How would the dimensions change if the width was doubled but the perimeter stayed the same?
5. How do you calculate the diagonal of a rectangle given the width and length?

### Tip:
Always check the dimensions by plugging them back into the perimeter formula to ensure they satisfy the given conditions.

A rectangular picture frame has a perimeter of 46 inches. The length is 7 inches more than the width. What are the dimensions of the frame?

Learn how to calculate the dimensions of a rectangular picture frame with a perimeter of 46 inches, where the length is 7 inches more than the width. This problem involves basic algebra and geometry concepts suitable for students in Grades 6-8.

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