Complete the table below by filling in the missing lengths - Leaving Cert Mathematics - Question 7 - 2021

Using the formula for the arc length: $T_{25} = 45(0.9)^{24}$ Calculating: $T_{25} ≈ 3.6 ext{ cm}$

The distance can be found by summing up the lengths for the first 40 swings: $S_{40} = 45 + 40.5 + 36.45 + ... + \left( 45(0.9)^{39} \right)$ Using the series sum formula: $S_n = \frac{a(1-r^n)}{1-r}$ where $a = 45$ and $r = 0.9$, we find: $S_{40} ≈ 443 ext{ cm}$

Set the equation $45(0.9)^{p-1} < 2$. Solving for $p$ gives: $p = \lceil \log_{0.9}(\frac{2}{45}) + 1 \rceil$ Calculating gives approximately: $p ≈ 31$

Using the formula for the arc length: $L = r \theta$ Here, $r = 100$ cm and $L = 45$ cm: $45 = 100\theta \Rightarrow \theta = \frac{45}{100} = 0.45 ext{ radians}$ Converting to degrees: $\theta = 0.45 \times \frac{180}{\pi} \approx 26°$

The total angle for swings 1 to 40 is: $\sum = 26 + 26(0.9) + 26(0.9^2) + ... + 26(0.9^{39})$ This series sums to: $\text{Total} = 26 \times \frac{1 - (0.9)^{40}}{1 - 0.9} ≈ 260°$

Half of the total accumulated angle is: $\frac{260}{2} = 130°$ Distance travelled is: $S = 2\pi(1) \times \frac{130}{360} ≈ 225 ext{ cm}$