rectangle has length $$ 2 x \mathrm{~cm} $$, perimeter 18 cm and area $$ 10 \mathrm{~cm}^{2} $$. i) Show that $$ 2 x^{2}-9 x+5=0 $$ $$ x $$

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To show that $$ 2x^2 - 9x + 5 = 0 $$, we will use the known conditions of the rectangle: its length, perimeter, and area.

1. **Known Conditions:**
   - Length $$ L = 2x $$ cm
   - Perimeter $$ P = 18 $$ cm
   - Area $$ A = 10 $$ cm²

2. **Perimeter Equation:**
   The formula for the perimeter of a rectangle is given by:
   $$
   P = 2(L + W)
   $$
   where $$ W $$ is the width. Substituting the known values:
   $$
   18 = 2(2x + W)
   $$
   Simplifying this equation:
   $$
   18 = 4x + 2W
   $$
   Dividing the entire equation by 2:
   $$
   9 = 2x + W
   $$
   Rearranging gives us:
   $$
   W = 9 - 2x
   $$

3. **Area Equation:**
   The area of a rectangle is given by:
   $$
   A = L \times W
   $$
   Substituting the known values:
   $$
   10 = (2x)(9 - 2x)
   $$
   Expanding this equation:
   $$
   10 = 18x - 4x^2
   $$
   Rearranging gives us:
   $$
   4x^2 - 18x + 10 = 0
   $$
   Dividing the entire equation by 2:
   $$
   2x^2 - 9x + 5 = 0
   $$

Thus, we have shown that:
$$
2x^2 - 9x + 5 = 0
$$
$$ 2x^2 - 9x + 5 = 0 $$

rectangle has length , perimeter 18 cm and area .
i) Show that

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rectangle has length , perimeter 18 cm and area . i) Show that