1. The problem involves evaluating two summations and expressing one summation in terms of another. 2. First, consider the summation $$\sum_{r=1}^3 (3r + 4)$$. This can be separated by the linearity of summation as: $$\sum_{r=1}^3 (3r + 4) = \sum_{r=1}^3 3r + \sum_{r=1}^3 4$$ 3. Compute each summation individually. - $$\sum_{r=1}^3 3r = 3(1) + 3(2) + 3(3) = 3 + 6 + 9 = 18$$ - $$\sum_{r=1}^3 4 = 4 + 4 + 4 = 12$$ 4. Adding these gives: $$18 + 12 = 30$$ 5. Next, consider the summation $$\sum_{r=1}^4 (4r)$$ which can be factored as: $$\sum_{r=1}^4 (4r) = 4 \sum_{r=1}^4 r$$ 6. The summation $$\sum_{r=1}^4 r$$ is an arithmetic series equal to: $$1 + 2 + 3 + 4 = 10$$ 7. Therefore, $$4 \times 10 = 40$$ 8. Finally, the graph description of a right triangle with legs 3 cm and 4 cm suggests a hypotenuse length by the Pythagorean theorem: $$c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm}$$ Answer summary: - $$\sum_{r=1}^3 (3r + 4) = 30$$ - $$\sum_{r=1}^4 (4r) = 40$$ - Hypotenuse of the right triangle = $$5 \text{ cm}$$