Answer:
a. y equals one third times x plus 10
= y = 1/3(x) + 10
Step-by-step explanation:
Let us represent:
Let the original final plan = x
Let the current flight plan = y
The initial time of departure = 4.00pm
Her flight was then delayed for 10 minutes
We are told in the question that:
The current flight plan allows her arrive at her destination three times faster.
This means y= (1/3)x
y = x/3
Hence the equation generated =
y = x/3 +10
y = 1/3(x) + 10
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°
Answer:
m∠ABC = 125° and AB ≅ DB, ΔACD is isosceles with base AD, CD = 52 cm
Step-by-step explanation:
i just took the quiz
Answer:
Change = -(1/2) * 30 = -15 inches
Step-by-step explanation:
If the water level decreases by 1/2 inch each day, the change in the water level in one day is -1/2 inch.
So, after t days, the total change in the water level can be calculated as:
Total change = -(1/2) * t inches
So, after 30 summer days, the total change is:
Change = -(1/2) * 30 = -15 inches
So the water level of the pool after 30 summer days has decreased by 15 inches.
