<span>a.
The radius of earth is about 6400 kilometers. Find the circumference of
a great circle.
Circumference = 2π(radius) = 2π(6400 km) = 40.212,39 km
b. Write an equation for the circumference of any
latitude circle with angle theta
As stated, </span><span><span>the
length of any parallel of latitude (this is the circumference of corresponding circle) is equal to the circumference of a
great circle of Earth times the cosine of the latitude angle</span>:
=> Circumference = 2π*radius* cos(Θ) = 2 π*6400km*cos(Θ) = 40,212.39 cos(Θ)
Answer: circumference = 40,212.39 cos(Θ) km
c. Which latitude circle has a
circumference of about 3593 kilometers?
Make </span><span><span>40,212.39 cos(Θ)</span> km = 3593 km
=> cos(Θ) = 3593 / 40,212.39 = 0.08935 => Θ = arccos(0.08935) = 84.5° = 1.48 rad
Answer: 1.48
d. What is the circumference of
the Equator?
</span>
For the Equator Θ = 0°
=> circumference = 40,213.49cos(0°) km = 40,212.49 km
Answer: 40,212.49 km
Answer:
Just divide the numbers by 10 and then multiply the quotient by 2. hope this helped
Greater than 1.Because 1.25-0.25=1.23.That's it.
Answer:
Step-by-step explanation:
Represent the width by W. Then, "The length of a rectangular field is 7 m less than 4 times the width" expressed symbolically is
L = 4W - 7 (dimensions in meters)
Recall that the perimeter formula in this case is P = 2L + 2W, and recognize that the perimeter value is 136 m. After substituting 4W - 7 for L, we get:
136 m = 2(4W - 7) + 2W, or
136 = 8W - 14 + 2W, or
150 = 10W These three equations are equivalent mathematical statements.
150 = 10W reduces to W = 15 (meters).
Part A: the independent variable is W, the width of the field.
Part B: The mathematical statement is 136 m = 2(4W - 7) + 2W, which after algebraic manipulation becomes 150 = 10W.
Part C: The above equation can be solved for W: W = 15 meters. This is the value of the independent variable.
