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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°
This is a fraction, so treat it like one. Divide 43 and 25 and you'll get 1.72. Multiply that decimal value by 100 and that is your percent, 172%
Answer:
10 big and 27 small
Step-by-step explanation:
First do a random amount and move closer to the answer by changing the amount of bottles
110/5=22
3 x 7 = 21
Bottles too little so less big water bottles
50/5=10
81/3=27
37 bottles so Liz sold 10 big bottles and 27 small bottles.
He should have worked out 7h(2 - h) first before adding the 3h. (first step)
2nd step he multiplied the 2 by -10h but he also should have multiplied the -h by the -10h ( Distributive law)
