Question

The rectangle is
  
enter your response here units long and
  
enter your response here units wide.
​(Round to the nearest tenth as​ needed.)

Answer 1

I don’t have the numbers or full problem statement — please paste the rest of the problem (area, perimeter, diagonal, a ratio, or any other information given).

Meanwhile, here are the common methods so you know what I’ll do once you give the missing data.

1) If you are given area A and perimeter P:
- Let width = w and length = L. Then
wL = A and 2(w + L) = P.
- Substitute L = A/w into the perimeter equation and solve the quadratic:
2(w + A/w) = P → 2w^2 − Pw + 2A = 0.
- Solve for w: w = [P ± sqrt(P^2 − 16A)] / 4, then L = A/w. (Round to nearest tenth.)

2) If you are given area A and diagonal d:
- wL = A and w^2 + L^2 = d^2.
- Let S = w + L = sqrt(d^2 + 2A) (because S^2 = d^2 + 2A).
- Then w and L are the roots of t^2 − S t + A = 0: t = [S ± sqrt(S^2 − 4A)]/2.

3) If you are given a ratio (e.g., length = k × width) and one other quantity:
- Substitute L = k w into the given equation (area, perimeter, or diagonal) and solve for w, then L = k w.

4) If you are given perimeter P and diagonal d:
- Use w + L = P/2 and w^2 + L^2 = d^2. From (w + L)^2 = w^2 + L^2 + 2wL, you can find wL then solve the quadratic t^2 − (P/2) t + wL = 0.

If you paste the exact numbers from your problem, I’ll compute the length and width and round to the nearest tenth.