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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°
Multiplying 20×1/2 is the same as dividing 20 by 2. Which means the answer is 10.
Answer:
To quickly solve this problem, we can use a graphing tool or a calculator to plot the equation.
Please see the attached image below, to find more information about the graph
The equation is:
y = tan(2x - π)
y = tan (2*(x-π/2))
We can compare it to its parent function
g(x) = tan(x)
The answer is
Option a.
Horizontal shrink by 1/2 and horizontal shift of π/2 to the right
Answer: MUV OR VUM
Step-by-step explanation:
It is a line so anyway you name it is fine as lomg as its in it's respected order.
Each qt has 2 pts in it.
13 qts add up to 26 pts.
That's more than 25 pts.
