Answer:
gtntt4bgt4thn
Step-by-step explanation:
This question is not complete
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°
D=rt
r=(boatrate+riverrate)=(x+2)mph
d=528ft=528/5280mi=1/10mi
t=1/3hr
remember to keep the same units
so
1/10=(x+2)(1/3)
times both sides by 3
3/10=x+2
minus 2 or 20/10
-17/10=x
-1.7=x
it would be going -1.7mph (means going backwards)
Answer: 8
Step-by-step explanation:
48,498,867 divided by 9 = 5,388,763
The data for resort A shows more consistency because a larger interquartile range such as the one for resort B, shows more variation. This means that the snowfall for resort A is more likely to be close to the median.
Just did this on edg. :)
