Answer:
- 892 lb (right)
- 653 lb (left)
Step-by-step explanation:
The weight is in equilibrium, so the net force on it is zero. If R and L represent the tensions in the Right and Left cables, respectively ...
Rcos(45°) +Lcos(75°) = 800
Rsin(45°) -Lsin(75°) = 0
Solving these equations by Cramer's Rule, we get ...
R = 800sin(75°)/(cos(75°)sin(45°) +cos(45°)sin(75°))
= 800sin(75°)/sin(120°) ≈ 892 . . . pounds
L = 800sin(45°)/sin(120°) ≈ 653 . . . pounds
The tension in the right cable is about 892 pounds; about 653 pounds in the left cable.
_____
This suggests a really simple generic solution. For angle α on the right and β on the left and weight w, the tensions (right, left) are ...
(right, left) = w/sin(α+β)×(sin(β), sin(α))
I believe it would be 360 students.
Answer:
See below
Step-by-step explanation:
If you are squaring a number, and then taking the square root of it, you are essentially undoing the original operation:
![\displaystyle \sqrt{\biggr(\frac{4}{7}\biggr)^2}=\biggr[\biggr(\frac{4}{7}\biggr)^2\biggr]^{\frac{1}{2}}=\biggr(\frac{4}{7}\biggr)^{2*\frac{1}{2}}=\frac{4}{7}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%7B%5Cbiggr%28%5Cfrac%7B4%7D%7B7%7D%5Cbiggr%29%5E2%7D%3D%5Cbiggr%5B%5Cbiggr%28%5Cfrac%7B4%7D%7B7%7D%5Cbiggr%29%5E2%5Cbiggr%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cbiggr%28%5Cfrac%7B4%7D%7B7%7D%5Cbiggr%29%5E%7B2%2A%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cfrac%7B4%7D%7B7%7D)
Hence, we are back starting with the original number
