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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:
2x
Step-by-step explanation:
when the base is the same, we subtract when dividing, so
3-2=1, 2x^1=2x
Answer:
14.25 yd^2
Step-by-step explanation:
The triangle area is given by the formula ...
A = 1/2bh
Here, the base is 10 yd, and the height is 5 yd, so ...
triangle area = (1/2)(10 yd)(5 yd) = (5 yd)^2 = 25 yd^2
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The semicircle is half the area of a circle with radius 5 yd, so its area is ...
semicircle area = (1/2)π(5 yd)^2 = 12.5π yd^2 ≈ 39.25 yd^2
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The shaded region is the difference between these areas:
shaded area = 39.25 yd^2 -25 yd^2 = 14.25 yd^2
<span> allows you to </span>do<span> repeated addition quickly and efficiently</span>
Answer:
313 shapes
Step-by-step explanation:
The nth term is 3n+1
so 3x104=312+1= 313 shapes
