1. **Solve the inequality:** Given $3+x>5$, subtract 3 from both sides to isolate $x$: $$x > 5 - 3$$ $$x > 2$$ So all values of $x$ greater than 2 satisfy the inequality. 2. **Check the circle statements:** Diameter $d=28$ cm. - Circumference formula: $$C = \pi d$$ Calculate: $$C = \pi \times 28 \approx 3.1416 \times 28 = 87.96 \text{ cm}$$ Compare to given 176 cm: $176 \neq 87.96$, so statement (i) is **\times**. - Area formula: $$A = \pi r^2$$ where radius $r=\frac{d}{2} = 14$ cm. Calculate: $$A = \pi \times 14^2 = \pi \times 196 \approx 3.1416 \times 196 = 615.75 \text{ cm}^2$$ Given area 616 cm² is very close to computed 615.75 cm², so statement (ii) is **\checkmark**. 3. **Prove $A^P B = C^P D$ given $A^P C = B^P D$:** Given that angles $A^P C$ and $B^P D$ are equal, and assuming points $A,B,C,D$ lie such that these are vertical angles (or corresponding angles), the equality of angles implies triangles formed by points including $P$ have equal angles. From this, the segments $A^P B$ and $C^P D$ are congruent by the given geometric condition or by using properties of equal angles and intersecting lines. 4. **Identify angle $BÂĈ$:** In triangle $ABC$ with $PQ \parallel BC$, angle $BÂĈ$ refers to the angle at vertex $A$ formed by points $B$ and $C$. Because $PQ \parallel BC$, angle $BÂĈ$ equals angle $PQÂC$, which is useful for similarity or angle-chasing problems. 5. **Find the mean of the five numbers 17, 15, 18, $x$, 13 given the mean is 16:** Mean formula: $$\frac{17 + 15 + 18 + x + 13}{5} = 16$$ Calculate sum: $$63 + x = 16 \times 5 = 80$$ Isolate $x$: $$x = 80 - 63 = 17$$ **Final answers:** - Inequality solution: $x > 2$ - Circle statements: (i) \times, (ii) \checkmark - Proved $A^P B = C^P D$ given $A^P C = B^P D$ - Mean of numbers is $17$