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image_path = "/mnt/data/1732473406007yh056734.jpg" image = Image.open(image_path) image.show()successFor question 10, let’s solve step by step:

1. Rectangle Information:

   • Perimeter $P = 46 \, \text{cm}$
   • Area $A = 120 \, \text{cm}^2$
   • Let the length be $l$ and the width be $w$.

2. Forming equations:

   • From the perimeter: $2l + 2w = 46$, which simplifies to: $l + w = 23 \quad \text{(1)}$
   • From the area: $l \cdot w = 120 \quad \text{(2)}$

3. Substitute $l$ from equation (1) into equation (2): From equation (1): $l = 23 - w$. Substitute into equation (2): $(23 - w) \cdot w = 120$ Simplify: $23w - w^2 = 120$ Rearrange into standard quadratic form: $w^2 - 23w + 120 = 0$

4. Solve the quadratic equation: Using the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -23$, and $c = 120$. Substitute: $w = \frac{-(-23) \pm \sqrt{(-23)^2 - 4(1)(120)}}{2(1)}$ $w = \frac{23 \pm \sqrt{529 - 480}}{2}$ $w = \frac{23 \pm \sqrt{49}}{2}$ $w = \frac{23 \pm 7}{2}$

   Two solutions: $w = \frac{23 + 7}{2} = 15 \quad \text{or} \quad w = \frac{23 - 7}{2} = 8$

5. Find $l$: If $w = 15$, then $l = 23 - 15 = 8$.
   If $w = 8$, then $l = 23 - 8 = 15$.

6. Final dimensions: The rectangle's dimensions are $15 \, \text{cm} \times 8 \, \text{cm}$.

Let me know if you want further clarifications or a graph of this quadratic!

Questions for further exploration:

1. How can the quadratic formula be derived?
2. What other methods can solve quadratic equations besides factoring or the quadratic formula?
3. Can the relationship between area and perimeter be generalized for any rectangle?
4. How do the roots of the quadratic equation relate to the dimensions of the rectangle?
5. What would change in this problem if the perimeter or area were given in a different unit?

Tip:

Always check both roots of a quadratic equation to ensure they make sense in the context of the problem.