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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:
1120479037157389478073281965092374871290432
Step-by-step explanation:
have a great day
Answer:
308.5 lol
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given Equation:
y=? at x= 1
Putting the given value of 'x' in the equation:
At 
, the value of y is
Step-by-step explanation:
Tn=ar^(n-1)
r=-10/-5=2
a=-5
Tn=-5×(2)^(n-1)
Tn=-5•(2)^(n-1)
an=5•(2)^(n-1)
