3(2x+2)+x+5=-10
6x+6+x+5=-10
7x+11=-10
7x=-21
x=-3
Hint #2 is a or b so ya, and 3 is c
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)
to find the distance between 2 points we should apply the formula
![d=\sqrt[]{(x_2-x_1)^2+(y_2-_{}y_1)^2_{}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_2-_%7B%7Dy_1%29%5E2_%7B%7D%7D)
call point q as point 1 for reference in the formula and p as point 2
replace the coordinates in the formula
![d=\sqrt[]{(3-(-1))^2+(-4-(-1))^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%283-%28-1%29%29%5E2%2B%28-4-%28-1%29%29%5E2%7D)
simplify the equation
![\begin{gathered} d=\sqrt[]{(3+1)^2+(-4+1)^2} \\ d=\sqrt[]{4^2+(-3)^2} \\ d=\sqrt[]{16+9} \\ d=\sqrt[]{25} \\ d=5 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d%3D%5Csqrt%5B%5D%7B%283%2B1%29%5E2%2B%28-4%2B1%29%5E2%7D%20%5C%5C%20d%3D%5Csqrt%5B%5D%7B4%5E2%2B%28-3%29%5E2%7D%20%5C%5C%20d%3D%5Csqrt%5B%5D%7B16%2B9%7D%20%5C%5C%20d%3D%5Csqrt%5B%5D%7B25%7D%20%5C%5C%20d%3D5%20%5Cend%7Bgathered%7D)
the distance between the 2 points is 5 units
