Split up each force into horizontal and vertical components.
• 300 N at N30°E :
(300 N) (cos(30°) i + sin(30°) j)
• 400 N at N60°E :
(400 N) (cos(60°) i + sin(60°) j)
• 500 N at N80°E :
(500 N) (cos(80°) i + sin(80°) j)
The resultant force is the sum of these forces,
∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i
… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N
∑ F ≈ (546.632 i + 988.814 j) N
so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.
Answer:
Step-by-step explanation:
Given that among 500 freshmen pursuing a business degree at a university, 315 are enrolled in an economics course, 213 are enrolled in a mathematics course, and 123 are enrolled in both an economics and a mathematics course.
From the above we find that
a) either economics of Math course is

Out of 500 students 405 have taken either Math or Economics
Hence
c) student who have taken neither = 
Exactly one course is either math or economics - both
= 
Answer:
Step-by-step explanation:
2(n+3) - 4 < 6n
2n+6-4 < 6n
2n + 2 < 6n
2 < 4n
½ < n
Hi there!
v = ±
- - - - - - - - -
K = 1/2mv²
Isolate for the variable "v". We can begin by dividing both sides by 1/2. (Multiply by the reciprocal, or 2):
2 · K = 2 · (1/2mv²)
2K = mv²
Continue isolating by dividing both sides by "m":
2K / m = v²
Take the square root of both sides. Remember that the solution can either be positive or negative since there are positive and negative roots.
v = ±