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Concept explainers
Sundials The Moorish sundial is designed so that the shadow of the gnomon (the vertical triangular piece) is consistent at each hour from one day to the next. This allows the hours to be marked on the sundial using a single scale that is not affected by the changes in the Sun’s path during the year. If the sundial is positioned so that the gnomon is aligned along a longitudinal line from north to south, then at exactly noon the gnomon will cast a shadow due north.
Figure 7
As the sun moves, the shadow will sweep out an
where
The latitude of Goose Pimple Junction,w Virginia, is 36.597°. Find the shadow angle for a sundial in this town at 5:00 p.m. Round to the nearest tenth of a degree.
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Chapter 2 Solutions
Trigonometry
- Solve for theta 3 and 4arrow_forwardC III https://www-awu.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNslkr7j8P3jH-li-WkWxK Zm85LW27IRVU66k591 O Trigonometric Functions Sketching the graph of y = a sin(x) or y = a cos(x) Graph the trigonometric function. 3 =sin.x 2 Plot all points corresponding to x-intercepts, minima, and maxima within one cycle. Then cli Explanation Check Esc F1 Search F2 #3 72 F3 4 F4 DII F5 % 5 A G F6 لarrow_forwardIf 0 = 0 = 10元 3 10元 then find exact values for the following. If the trigonometric function is undefined fo enter DNE. > 3 sec(0) equals csc(0) equals tan(0) equals cot (0) equals من Question Help: Video B من B Submit Question Jump to Answerarrow_forward
- Question 9 1 5 4 3 2 1 -8 -7 -05 -4 -3 -2 1 1 2 3 4 5 6 7 8 -1 7 -2 -3 -4 -5+ 1-6+ For the graph above, find the function of the form -tan(bx) + c f(x) =arrow_forwardQuestion 8 5 4 3 2 1 -8 -7 -6 -5/-4 -3 -2 -1, 1 2 3 4 5 6 7/8 -1 -2 -3 -4 -5 0/1 pt 3 98 C -6 For the graph above, find the function of the form f(x)=a tan(bx) where a=-1 or +1 only f(x) = = Question Help: Video Submit Question Jump to Answerarrow_forward6+ 5 -8-7-0-5/-4 -3 -2 -1, 4 3+ 2- 1 1 2 3/4 5 6 7.18 -1 -2 -3 -4 -5 -6+ For the graph above, find the function of the form f(x)=a tan(bx) where a=-1 or +1 only f(x) =arrow_forward
- Question 10 6 5 4 3 2 -π/4 π/4 π/2 -1 -2 -3- -4 -5- -6+ For the graph above, find the function of the form f(x)=a tan(bx)+c where a=-1 or +1 only f(x) = Question Help: Videoarrow_forwardThe 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_forward
- T2.4: 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_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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