Trigonometry (MindTap Course List)
8th Edition
ISBN: 9781305652224
Author: Charles P. McKeague, Mark D. Turner
Publisher: Cengage Learning
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Chapter 1.5, Problem 1PS
To determine
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Question 9
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- The second solution I got is incorrect. What is the correct solution? The other thrree with checkmarks are correct Question 19 Score on last try: 0.75 of 1 pts. See Details for more. Get a similar question You can retry this question below Solve 3 sin 2 for the four smallest positive solutions 0.75/1 pt 81 99 Details T= 1.393,24.666,13.393,16.606 Give your answers accurate to at least two decimal places, as a list separated by commas Question Help: Message instructor Post to forum Submit Questionarrow_forwardd₁ ≥ ≥ dn ≥ 0 with di even. di≤k(k − 1) + + min{k, di} vi=k+1 T2.5: Let d1, d2,...,d be integers such that n - 1 Prove the equivalence of the Erdos-Gallai conditions: for each k = 1, 2, ………, n and the Edge-Count Criterion: Σier di + Σjeл(n − 1 − d;) ≥ |I||J| for all I, JC [n] with In J = 0.arrow_forwardT2.4: Let d₁arrow_forwardT2.3: 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_forward4 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_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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