To ease your problem, consider "L" as you x-axis
Then the coordinate become:
A(- 4 , 3) and B(1 , 2) [you notice that just the y's changed]
This is a reflection problem.
Reflect point B across the river line "L" to get B', symmetric of B about L.
The coordinates of B'(1 , -1) [remember L is our new x-axis]
JOIN A to B' . AB' intersect L, say in H
We have to find the shortest way such that AH + HB = shortest.
But HB = HB' (symmetry about L) , then I can write instead of
AH + HB →→ AH + HB'. This is the shortest since the shortest distance between 2 points is the straight line and H is the point requiered
10×sine of 60 degrees=8.66 which is the same as B. you can check by finding the square root of 3 and then multiplying that by 5.
Answer:
vgkf çµr nm 5hr 6tqy7gu8hibyg7tf6r5d4es3
Step-by-step explanation:
Q=M-2P/3
because in order to solve for a variable you have to get alone. Make sense?
assuming that the bases are placed correctly, the bases connected draw a square, each side being 90 ft long. We multiply 90 by the amount of sides(4) to find the total distance/perimeter = 360.
