If she arrives by boat;
The distance she needs to row = Lake diameter
Distance = Lake diameter = 2 x Radius
Boat distance = 2 x 2 miles = 4 miles
So;
When moving on foot
The distance that needs to be moved D = Half of the circle circle
∴ D = π × Radius
The distance that needs to be moved D = π × 2 miles = 2π Miles
Arc length = Radius × θ
String length = radius x 2 x sin (θ / 2)
0 ≤ θ ≤ π, 0 ≤ sin (θ / 2) ≤ 1
Learn more here shore of a circular at
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Answer:
The following are the solution to the given points:
Step-by-step explanation:
for point A:
The set A is not part of the subspace 
for point B:
The set B is part of the subspace
for point C:
In this, the scalar multiplication can't behold
∉ C
this inequality is not hold
The set C is not a part of the subspace
for point D:
The scalar multiplication s is not to hold
∉ D
this is an inequality, which is not hold
The set D is not part of the subspace
For point E:
The 
is the arbitrary, in which 
is arbitrary
The set E is the part of the subspace
For point F:
The arbitrary so, they have as the arbitrary 
The set F is the subspace of
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Answer:
d(32) = 28.80
the price Marcus pays on an item with an original price of 32
Step-by-step explanation:
d(32) = 32 -0.1(32)
d(32) = 28.8
The problem statement tells you d(32) is the price Marcus pays when the original price is 32.
Answer
∴ The true statement is Answer:
The true statement is BD ≅ CE ⇒ 3rd answer
Step-by-step explanation:
- There is a line contained points B , C , D , E
- All points are equal distance from each other
- That means the distance of BC equal the distance of CD and equal
the distance of DE
∴ BC = CD = DE
- That means the line id divided into 3 equal parts, each part is one
third the line
∴ BC = 1/3 BE
∴ CD = 1/3 BE
∴ DE = 1/3 BE
∵ BC = CD
∴ C is the mid-point of BD
∴ BC = 1/2 BD
∵ CD = DE
∴ D is the mid-point of DE
∴ CD = 1/2 CE
- Lets check the answers
* BD = one half BC is not true because BC = one half BD
* BC = one half BE is not true because BC = one third BE
* BD ≅ CE is true because
BD = BC + CD
CE = CD + DE
BC ≅ DE and CD is common
then BD ≅ CE
* BC ≅ BD is not true because BC is one half BD
∴ The true statement is BD ≅ CE
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