Answer:
Enter the value of the score in the laboratory 1
L1 = digited value.
Enter the value of the score in laboratory 2
L2 = digited value.
Enter the value of the score in laboratory 3
L3 = digited value.
Enter the value of the test score 1
P1 = digited value.
Enter the value of the test score 2
P2 = digited value.
Laboratory Average = (L1 + L2 + L3) / 3
Print "Laboratory Average"
Test Average = (P1 + P2) / 2
Print "Test Average"
Course qualification = 0.55 * (Laboratory Average) + 0.45 * (Test Average)
print "Course qualification"
Step-by-step explanation:
I think i’m may be ‘B’ i’m so sorry if i’m incorrect!!
Changing B changes where it crosses on the y-axis... on the graph
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.
