1. **Stating the problem:** We have a triangle ABC with sides $AB=3.6$ cm and $AC=4.3$ cm. Points D and E are defined such that $\vec{AD} = 2\vec{AB}$ and $\vec{DE} = 3\vec{BC}$. We want to analyze these vectors and understand the geometric relationships. 2. **Understanding vectors and points:** - Since $\vec{AD} = 2\vec{AB}$, point D lies on the line through A and B, twice as far from A as B is. - Since $\vec{DE} = 3\vec{BC}$, vector DE is three times the vector BC, so E lies along the line through D in the direction of BC, scaled by 3. 3. **Expressing points in coordinate form:** - Let’s place point A at the origin $(0,0)$ for simplicity. - Then $B$ is at $(3.6,0)$ since AB is horizontal. - To find coordinates of C, since AC = 4.3 cm, and triangle is horizontal, assume C lies somewhere above the x-axis. Without angle info, we cannot find exact coordinates but can express vectors symbolically. 4. **Coordinates of D:** - Since $\vec{AD} = 2\vec{AB}$, and $\vec{AB} = B - A = (3.6,0)$, - Then $D = A + 2\vec{AB} = (0,0) + 2(3.6,0) = (7.2,0)$. 5. **Coordinates of E:** - Vector $\vec{BC} = C - B$. - Vector $\vec{DE} = 3\vec{BC}$, so $E = D + 3\vec{BC} = (7.2,0) + 3(C - B) = (7.2,0) + 3(C_x - 3.6, C_y - 0) = (7.2 + 3(C_x - 3.6), 3C_y)$. 6. **Summary:** - Point D is at $(7.2,0)$. - Point E depends on coordinates of C, which are not fully specified. 7. **Additional given values:** - $M=5$ (context unclear), - $A = \frac{705000}{0.000643} \approx 1,096,581,027.52$, - $B = (0.00007 + 0.0003) - 6 \times 10^{-6} = 0.000364$. These may relate to other parts of the problem but are not directly connected to the triangle vectors. **Final answer:** - $D$ is located at twice the distance along $AB$ from $A$, coordinates $(7.2,0)$. - $E$ is located starting from $D$ along vector $BC$ scaled by 3, coordinates depend on $C$. This completes the vector and coordinate analysis of the problem.