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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:
-8/9 yw
Step-by-step explanation:
just divide the numbers by each other:D
Answer:
It's 1/2 I found this out by subtracting 3/4 - 1/3= 5/12 and if you put that into a decimal it would be .41. So that 1/4= .25 and 1/2= .50. You could figure out that it’s closed to 1/2.
<u>1/3; 2/7; 2/9</u>
Lets find least common multiple
The least common multiple is 63
1/3 --> 21/63
2/7 --> 18/63
2/9 --> 14/63
As you can see I am correct, but I found the LCM by multiplying 1/3 by 21 to get to 63 and the numerator as well
