Answer:
CORRECTED QUESTION:
Two cities have nearly the same north-south line of 110 degrees Upper W. The latitude of the first city is 23 degrees Upper N, and the latitude of the second city is 36 degrees N. Approximate the distance between the cities if the average radius of Earth is 6400 km.
ANSWER: 1452.11 km
Step-by-step explanation:
Since the two cities both lies on the Northern latitude of the sphere along the same longitude, we are going to subtract the angles the latitude that each city subtend at the equator.
36 - 23 = 13 degrees i.e the angles between the with two cities on a cross section the large circle formed by the longitude and its center.
Applying the formula for the length of an arc on a sector on the large circle
(∅/ 360) x 2πR
where, ∅ = is the angle between the two cities
R = radius of the Earth.
13/360 x 2 x π x 6400 = 1452.11 km
If you meant y = 27x - 3, OK. Then, if x = 4, y = 27(4) - 3, or 108-3 = 105.
Answer:
0.64
Step-by-step explanation:
Answer:
21 ft by 66 ft
Step-by-step explanation:
From the question,
P = 2(L+W)............... Equation 1
Where P = Perimeter of the playing Field, L = Length of the playing Field, W = width of the playing Field.
If the Length of the Field is 45 ft longer than the width,
L = 45+W............ Equation 2
Substitute Equation 2 into equation 1
P = 2(45+W+W)
P = 90+4W............. Equation 3
Given: P = 174 ft.
Substitute into equation 3
174 = 90+4W
4W = 174-90
4W = 84
W = 84/4
W = 21 ft.
Substituting the value of W into equation 2
L = 45+21
L = 66 ft.
Hence the dimensions of the playing field is 21 ft by 66 ft
