Explanation Given Information: The provided hexadecimal numbers are 2 − E . Formula used: Procedure to subtract the hexadecimal number: Find the first number in the left vertical column, find the second number in the top horizontal row, and find their sum at the intersection in the table. According to the hexadecimal form to decimal form conversion method find the place value of each digit in the given hexadecimal number and multiply by its corresponding power of 16 . The complementary pair of the hexadecimal is always consisting in positive and negative, if the digit is positive the complementary of that digit will be negative. The complementary pair of some of the digits are, Complementary of 0 is F, 1 is E, 2 is D, 3 is C, 4 is B, 5 is A, 6 is 9, 7 is 8 . Calculation: Consider the provided hexadecimal numbers 2 − E . Refer the table 16.3, and take the all values from this table, Take the compliment of the digit in the number to be subtracted, E → 1 Now, add both hexadecimal numbers, 2 1 Now, use the table for the addition of 2 + 1 which is found to be 3. Thus, 2 1 3 As, the number to be subtracted is larger so the answer will be negative, thus taking the compliment of each digit in result, 3 → C Thus, the result of the subtraction is − C

11th Edition

ISBN: 9781285199191

Author: Dale Ewen, C. Robert Nelson

Publisher: Cengage Learning

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Chapter 16.8, Problem 31E

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To calculate: The result of hexadecimal subtraction 2−E, and check the result by using decimals.

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Chapter 16.8, Problem 30E

Chapter 16.8, Problem 32E

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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 16 Solutions

Elementary Technical Mathematics

Ch. 16.1 - Change each binary number to decimal form: 11 Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E Ch. 16.1 - Prob. 5E Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E Ch. 16.1 - Prob. 8E Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E

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The result of hexadecimal subtraction 2 − E , and check the result by using decimals.

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