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Longitude Longitude is the angular distance (expressed in degrees) East or West of the prime meridian, which goes from the North Pole to the South Pole through Greenwich, England. Arcs of 1° longitude are 110 km apart at the equator, and therefore 15° arcs subtend 15 (110) km, or 1650 km, at the equator.
Because Earth rotates 15° per hr, longitude is found by taking the difference between time zones multiplied by 15°. For example, if it is 12 noon where we are (in the United States) and 5 p.m. in Greenwich, we are located at longitude 5(15°), or 75° W.
(a) What is the longitude at Greenwich, England?
(b) Use time zones to determine the longitude where you live.
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Chapter 3 Solutions
Trigonometry plus MyLab Math with Pearson eText -- Access Card Package (11th Edition)
- T2.2 Prove that a sequence s d₁, d₂,..., dn with n ≥ 3 of integers with 1≤d; ≤ n − 1 is the degree sequence of a connected unicyclic graph (i.e., with exactly one cycle) of order n if and only if at most n-3 terms of s are 1 and Σ di = 2n. (i) Prove it by induction along the lines of the inductive proof for trees. There will be a special case to handle when no d₂ = 1. (ii) Prove it by making use of the caterpillar construction. You may use the fact that adding an edge between 2 non-adjacent vertices of a tree creates a unicylic graph.arrow_forward= == T2.1: Prove that the necessary conditions for a degree sequence of a tree are sufficient by showing that if di 2n-2 there is a caterpillar with these degrees. Start the construction as follows: if d1, d2,...,d2 and d++1 = d = 1 construct a path v1, v2, ..., vt and add d; - 2 pendent edges to v, for j = 2,3,..., t₁, d₁ - 1 to v₁ and d₁ - 1 to v₁. Show that this construction results vj in a caterpillar with degrees d1, d2, ..., dnarrow_forward4 sin 15° cos 15° √2 cos 405°arrow_forward
- 2 18-17-16-15-14-13-12-11-10 -9 -8 -6 -5 -4-3-2-1 $ 6 8 9 10 -2+ The curve above is the graph of a sinusoidal function. It goes through the points (-10, -1) and (4, -1). Find a sinusoidal function that matches the given graph. If needed, you can enter π-3.1416... as 'pi' in your answer, otherwise use at least 3 decimal digits. f(x) = > Next Questionarrow_forwardketch a graph of the function f(x) = 3 cos (표) 6. x +1 5 4 3 3 80 9 2+ 1 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 -1 -2 -3+ -4 5 -6+ Clear All Draw: пи > Next Questionarrow_forwardDraw the following graph on the interval πT 5π < x < 2 2 y = 2 sin (2(x+7)) 6. 5. 4 3 3 2 1 +3 /2 -π/3 -π/6 π/6 π/3 π/2 2π/3 5π/6 π 7π/6 4π/3 3π/2 5π/311π/6 2π 13π/67π/3 5π Clear All Draw:arrow_forward
- ketch a graph of the function f(x) = 3 cos (표) 6. x +1 5 4 3 3 80 9 2+ 1 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 -1 -2 -3+ -4 5 -6+ Clear All Draw: пи > Next Questionarrow_forward3 2 20-10-18-17-16-15-14-13-12-11-10-9 -8 -7 -6 -$4-3-2-1 -1 -2 -3 4- -5+ The curve above is the graph of a sinusoidal function. It goes through the points (-8, -4) and (6,-4). Find a sinusoidal function that matches the given graph. If needed, you can enter π=3.1416... as 'pi' in your answer, otherwise use at least 3 decimal digits. f(x) = > Next Question Barrow_forwardX Grades for X Assignmen X A-Z Datab XE Biocultural X EBSCO-Ful X Review es/119676/assignments/3681238 Review Quiz 8.1-p2 points possible Answered: 3/5 ● Question 1 4+ 3. 2 1 13 /12-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 -1 -2 -3 -4- 5 2 6 The curve above is the graph of a sinusoidal function. It goes through the points (-7,0) and (3,0). Find a sinusoidal function that matches the given graph. If needed, you can enter π=3.1416... as 'pi' in your answer, otherwise use at least 3 decimal digits. f(x) = > Next Question 申 J % F 刀 Q Search S € t ח Y 7 I * 00 J ப I Darrow_forward
- 2 d) Draw the following graph on the interval k 5π Next Questionarrow_forwardDraw the following graph on the interval 5л Next Questionarrow_forwardDraw the following graph on the interval πT 5π < x < x≤ 2 2 y = 2 cos(3(x-77)) +3 6+ 5 4- 3 2 1 /2 -π/3 -π/6 Clear All Draw: /6 π/3 π/2 2/3 5/6 x 7/6 4/3 3/2 5/311/6 2 13/67/3 5 Question Help: Video Submit Question Jump to Answerarrow_forward
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