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CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.
An alternative form of the law of sines is
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Chapter 7 Solutions
Trigonometry, Books a la Carte Edition plus MyLab Math with Pearson eText -- Access Card Package (11th Edition)
- T2.3: Prove that there exists a connected graph with degrees d₁ ≥ d₂ >> dn if and only if d1, d2,..., dn is graphic, d ≥ 1 and di≥2n2. That is, some graph having degree sequence with these conditions is connected. Hint - Do not attempt to directly prove this using Erdos-Gallai conditions. Instead work with a realization and show that 2-switches can be used to make a connected graph with the same degree sequence. Facts that can be useful: a component (i.e., connected) with n₁ vertices and at least n₁ edges has a cycle. Note also that a 2-switch using edges from different components of a forest will not necessarily reduce the number of components. Make sure that you justify that your proof has a 2-switch that does decrease the number of components.arrow_forwardT2.2 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_forward
- 4 sin 15° cos 15° √2 cos 405°arrow_forward2 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_forward
- Draw 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_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_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_forward
- X 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_forward2 d) Draw the following graph on the interval k 5π Next Questionarrow_forwardDraw the following graph on the interval 5л Next Questionarrow_forward
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