V+9/3=8
*9 *9
v+3=72
-3 -8
v=69
Answer:
Town C is approximately 58.356 kilometers from town A.
Step-by-step explanation:
According to the statement, we understand that tourist travels from town A to town B (50 kilometers) at a direction of 80º west of north and from town B to town C (40 kilometers) at a direction of 20º east of north. Vectorially speaking, the resultant from town A to town C is described by the following formula:
![\vec r = (50\,km)\cdot (-\sin80^{\circ},\cos 80^{\circ})+(40\,km)\cdot (\sin 20^{\circ}, \cos 20^{\circ})](https://tex.z-dn.net/?f=%5Cvec%20r%20%3D%20%2850%5C%2Ckm%29%5Ccdot%20%28-%5Csin80%5E%7B%5Ccirc%7D%2C%5Ccos%2080%5E%7B%5Ccirc%7D%29%2B%2840%5C%2Ckm%29%5Ccdot%20%28%5Csin%2020%5E%7B%5Ccirc%7D%2C%20%5Ccos%2020%5E%7B%5Ccirc%7D%29)
![\vec r = (-35.560,46.270)\,[km]](https://tex.z-dn.net/?f=%5Cvec%20r%20%3D%20%28-35.560%2C46.270%29%5C%2C%5Bkm%5D)
The distance from town A to town C is the magnitude of vector reported above, which is now calculated by Pythagorean Theorem:
![\|\vec r\| = \sqrt{(-35.560\,km)^{2}+(46.270\,km)^{2}}](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20r%5C%7C%20%3D%20%5Csqrt%7B%28-35.560%5C%2Ckm%29%5E%7B2%7D%2B%2846.270%5C%2Ckm%29%5E%7B2%7D%7D)
![\|\vec r\| \approx 58.356\,km](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20r%5C%7C%20%5Capprox%2058.356%5C%2Ckm)
Town C is approximately 58.356 kilometers from town A.
14.
479.76:24=19.99 each month
The slope is 2. rise over run.