Excursions in Modern Mathematics (9th Edition)
9th Edition
ISBN: 9780134468372
Author: Peter Tannenbaum
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Videos
Textbook Question
Chapter 7, Problem 11E
In Exercises 11 through 24 you are given information about a network. Choose one of the following three options:
The network has
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
For each real-valued nonprincipal character x mod k, let
A(n) = x(d) and F(x) = Σ
:
dn
* Prove that
F(x) = L(1,x) log x + O(1).
n
By considering appropriate series expansions,
e². e²²/2. e²³/3.
....
=
= 1 + x + x² + ·
...
when |x| < 1.
By expanding each individual exponential term on the left-hand side
the coefficient of x- 19 has the form
and multiplying out,
1/19!1/19+r/s,
where 19 does not divide s. Deduce that
18! 1 (mod 19).
Proof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.
Chapter 7 Solutions
Excursions in Modern Mathematics (9th Edition)
Ch. 7 - A computer lab has seven computers labeled A...Ch. 7 - The following is a list of the electrical power...Ch. 7 - Consider the network shown in Fig.720_. a. How...Ch. 7 - Consider the network shown in Fig.721_. a. How...Ch. 7 - Consider once again the network shown in. Fig720_....Ch. 7 - Consider once again the network shown in. Fig721_....Ch. 7 - Consider the network shown in. Fig722. This is the...Ch. 7 - Consider the network shown in. Fig723_. This is...Ch. 7 - Consider the tree shown in. Fig724_. a. How many...Ch. 7 - Consider the tree shown in. Fig725. a. How many...
Ch. 7 - In Exercises 11 through 24 you are given...Ch. 7 - Prob. 12ECh. 7 - Prob. 13ECh. 7 - Prob. 14ECh. 7 - In Exercises 11 through 24 you are given...Ch. 7 - Prob. 16ECh. 7 - Prob. 17ECh. 7 - Prob. 18ECh. 7 - Prob. 19ECh. 7 - In Exercises 11 through 24 you are given...Ch. 7 - Prob. 21ECh. 7 - Prob. 22ECh. 7 - Prob. 23ECh. 7 - Prob. 24ECh. 7 - Prob. 25ECh. 7 - Consider the network shown in Fig.727_. a. Find a...Ch. 7 - Prob. 27ECh. 7 - Consider the network shown in Fig.729_. a. Find a...Ch. 7 - Prob. 29ECh. 7 - Prob. 30ECh. 7 - Prob. 31ECh. 7 - Prob. 32ECh. 7 - Prob. 33ECh. 7 - Prob. 34ECh. 7 - Prob. 35ECh. 7 - The 4 by 5 grid shown in Fig. 7-37 represents a...Ch. 7 - Prob. 37ECh. 7 - Find the MST of the network shown in Fig. 7-39...Ch. 7 - Find the MST of the network shown in Fig. 7-40...Ch. 7 - Find the MST of the network shown in Fig. 7-41...Ch. 7 - Prob. 41ECh. 7 - Find the MaxST of the network shown in Fig. 7-39...Ch. 7 - Find the MaxST of the network shown in Fig. 7-40...Ch. 7 - Prob. 44ECh. 7 - The mileage chart in Fig. 742 shows the distances...Ch. 7 - Figure 7-43a shows a network of roads connecting...Ch. 7 - Prob. 47ECh. 7 - Prob. 48ECh. 7 - Prob. 49ECh. 7 - This exercise refers to weighted networks where...Ch. 7 - Prob. 51ECh. 7 - Prob. 52ECh. 7 - Prob. 53ECh. 7 - Prob. 54ECh. 7 - Prob. 55ECh. 7 - Prob. 56ECh. 7 - A bipartite graph is a graph with the property...Ch. 7 - Prob. 58ECh. 7 - Prob. 59E
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.Similar questions
- By considering appropriate series expansions, ex · ex²/2 . ¸²³/³ . . .. = = 1 + x + x² +…… when |x| < 1. By expanding each individual exponential term on the left-hand side and multiplying out, show that the coefficient of x 19 has the form 1/19!+1/19+r/s, where 19 does not divide s.arrow_forwardLet 1 1 r 1+ + + 2 3 + = 823 823s Without calculating the left-hand side, prove that r = s (mod 823³).arrow_forwardFor each real-valued nonprincipal character X mod 16, verify that L(1,x) 0.arrow_forward
- *Construct a table of values for all the nonprincipal Dirichlet characters mod 16. Verify from your table that Σ x(3)=0 and Χ mod 16 Σ χ(11) = 0. x mod 16arrow_forwardFor each real-valued nonprincipal character x mod 16, verify that A(225) > 1. (Recall that A(n) = Σx(d).) d\narrow_forward24. Prove the following multiplicative property of the gcd: a k b h (ah, bk) = (a, b)(h, k)| \(a, b)' (h, k) \(a, b)' (h, k) In particular this shows that (ah, bk) = (a, k)(b, h) whenever (a, b) = (h, k) = 1.arrow_forward
- 20. Let d = (826, 1890). Use the Euclidean algorithm to compute d, then express d as a linear combination of 826 and 1890.arrow_forwardLet 1 1+ + + + 2 3 1 r 823 823s Without calculating the left-hand side, Find one solution of the polynomial congruence 3x²+2x+100 = 0 (mod 343). Ts (mod 8233).arrow_forwardBy considering appropriate series expansions, prove that ez · e²²/2 . e²³/3 . ... = 1 + x + x² + · ·. when <1.arrow_forward
- Prove that Σ prime p≤x p=3 (mod 10) 1 Р = for some constant A. log log x + A+O 1 log x ,arrow_forwardLet Σ 1 and g(x) = Σ logp. f(x) = prime p≤x p=3 (mod 10) prime p≤x p=3 (mod 10) g(x) = f(x) logx - Ր _☑ t¯¹ƒ(t) dt. Assuming that f(x) ~ 1½π(x), prove that g(x) ~ 1x. 米 (You may assume the Prime Number Theorem: 7(x) ~ x/log x.) *arrow_forwardLet Σ logp. f(x) = Σ 1 and g(x) = Σ prime p≤x p=3 (mod 10) (i) Find ƒ(40) and g(40). prime p≤x p=3 (mod 10) (ii) Prove that g(x) = f(x) logx – [*t^¹ƒ(t) dt. 2arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Elementary Geometry For College Students, 7eGeometryISBN:9781337614085Author:Alexander, Daniel C.; Koeberlein, Geralyn M.Publisher:Cengage,
Elementary Geometry For College Students, 7e
Geometry
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Cengage,
Minimum cuts and maximum flow rate; Author: Juddy Productions;https://www.youtube.com/watch?v=ylxhl1ipWss;License: Standard YouTube License, CC-BY