Given
Height (length) of the anchor dropped from the boat = 75 feet
Distance of the treasure from the anchor = 40 feet
Solution
The dropped at the bottom makes an angle of 90 degree . Therefore making an right triangle with the chest and boat
By Pythagoras theorem
AC^2 = AB^2 + BC^2
(AC = distance of the treasure from the boat
AB = height of the anchor dropped from the boat
BC = the distance of treasure from the anchor)
AC^2 = 75^2 + 40^2
AC^2 = 5325 +1600
AC^2 = 6925
AC = 25 * (under root 11)
Therefore they have cover a distance of 25 * (under root 11)
You can take the first point as the origin O(0,0) and the second one (-2,0.5) for easier calculations.
Find the slope, m = (y2 - y1)/(x2 - x1)
= -2-0 / -0.5-0 = 2/0.5 = 4
now take (x,y) = (0,0) and m= 4
use the point slope form
(y - y1) = m(x - x1)
(y - 0) = (4)(x - 0) ......taking (x1,y1) = (0,0)
y = 4x + 0 ...... in the form of y = mx + b
I hope I was helpful:)
Answer:
this can be solved by simple unitary method
What diagram are you talking about because I don’t see one
Answer:
96 is your answer
Step-by-step explanation:
mark brainliest
