Excursions in Modern Mathematics (9th Edition)
Excursions in Modern Mathematics (9th Edition)
9th Edition
ISBN: 9780134468372
Author: Peter Tannenbaum
Publisher: PEARSON
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Chapter 13, Problem 28E
To determine

(a)

To find:

The solutions of the given equation.

To determine

(b)

To find:

The sum of the two solutions.

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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.
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.

Chapter 13 Solutions

Excursions in Modern Mathematics (9th Edition)

Ch. 13 - Prob. 11ECh. 13 - Using a good calculator an online calculator if...Ch. 13 - Consider the following sequence of equations...Ch. 13 - Consider the following sequence of equations...Ch. 13 - Fact: If we make a list of any four consecutive...Ch. 13 - Fact: If we make a list of any 10 consecutive...Ch. 13 - Express each of the following as a single...Ch. 13 - Prob. 18ECh. 13 - Prob. 19ECh. 13 - Prob. 20ECh. 13 - Prob. 21ECh. 13 - Prob. 22ECh. 13 - Prob. 23ECh. 13 - Prob. 24ECh. 13 - Consider the quadratic equation x2=x+1. a. Use the...Ch. 13 - Prob. 26ECh. 13 - Consider the quadratic equation 3x2=8x+5. a. Use...Ch. 13 - Prob. 28ECh. 13 - Prob. 29ECh. 13 - Prob. 30ECh. 13 - Consider the quadratic equation 21x2=34x+55. a....Ch. 13 - Prob. 32ECh. 13 - Prob. 33ECh. 13 - Consider the quadratic equation (FN2)x2=(FN1)x+FN,...Ch. 13 - The reciprocal of =1+52 is the rational number...Ch. 13 - The square of the golden ratio is the irrational...Ch. 13 - Given that F4998.61710103, a. find an approximate...Ch. 13 - Prob. 38ECh. 13 - Prob. 39ECh. 13 - Prob. 40ECh. 13 - Prob. 41ECh. 13 - Prob. 42ECh. 13 - Triangles T and T shown in Fig. 13-23 are similar...Ch. 13 - Polygons P and P shown in Fig. 13-24 are similar...Ch. 13 - Find the value of x so that the shaded rectangle...Ch. 13 - Find the value of x so that the shaded figure in...Ch. 13 - Prob. 47ECh. 13 - Prob. 48ECh. 13 - Prob. 49ECh. 13 - Prob. 50ECh. 13 - In Fig. 13-31 triangles BCA is a 36-36-108...Ch. 13 - Prob. 52ECh. 13 - Find the value of x of y so that in Fig. 13-33 the...Ch. 13 - Prob. 54ECh. 13 - Prob. 55ECh. 13 - Consider the sequence of ratios FN2FN. a. Using a...Ch. 13 - Prob. 57ECh. 13 - Prob. 58ECh. 13 - Prob. 59ECh. 13 - a.Explain what happens to the values of (152)N as...Ch. 13 - Prob. 61ECh. 13 - Prob. 62ECh. 13 - Prob. 63ECh. 13 - Prob. 64ECh. 13 - Prob. 65ECh. 13 - Find the value of x of y so that in Fig. 13-37 the...Ch. 13 - Prob. 67ECh. 13 - In Fig. 13-39 triangle BCD is a 727236 triangle...Ch. 13 - Prob. 69ECh. 13 - Prob. 70ECh. 13 - Prob. 71ECh. 13 - Prob. 72ECh. 13 - Prob. 73ECh. 13 - Prob. 74ECh. 13 - Prob. 75ECh. 13 - Prob. 76ECh. 13 - During the time of the Greeks the star pentagram...
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