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Complete Question
A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point
Answer:
a)8.717km
b) 54.146°
Step-by-step explanation:
(a)how far is the boat from its starting point.
We solve this question using resultant vectors
= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)
Where
Rcos θ = x
Rsinθ = y
= (4cos38,4sin38) + (5cos67,5sin67)
= (3.152, 2.4626) + (1.9536, 4.6025)
= (5.1056, 7.065)
x = 5.1056
y = 7.065
Distance = √x² + y²
= √(5.1056²+ 7.065²)
= √75.98137636
= √8.7167296826
Approximately = 8.717 km
Therefore, the boat is 8.717km its starting point.
(b)calculate the bearing of the boat from its starting point.
The bearing of the boat is calculated using
tan θ = y/x
tan θ = 7.065/5.1056
θ = arc tan (7.065/5.1056)
= 54.145828196°
θ ≈ 54.146°
Answer:
7.24pm
Step-by-step explanation:
196+50=3.9h
0.9h=0.9x60mnt=54mnt
3.9h=3h 54mnt
3:30pm+3h 54mnt
3:30pm+3h 54mnt=7.2pm
Answer:
Ella incorrectly combined the r coefficients. The correctly simplified expression is b Plus 3r.
Step-by-step explanation:
(3b+4r)+(-2b-r)
Open the bracket
=3b+4r-2b-r
Combine the like coefficient
=3b-2b+4r-r
=b+3r
Ella incorrectly combined the r coefficients. The correctly simplified expression is b Plus 3r.
