Explanation Given Information: The provided expression is 2 2 5 ÷ 1 1 7 . Formula used: Steps to determine the division of two fractions. 1. Write the numbers in the form of fraction. 2. Determine the reciprocal of the divisor. 3. Perform the multiplication of the dividend and the reciprocal of the divisor. 4. Reduce the terms in its lowest form and write the result as a whole number or mixed number as required. Calculation: Consider the expression, 2 2 5 ÷ 1 1 7 Express the mixed numbers as improper fractions, 2 2 5 ÷ 1 1 7 = ( 2 × 5 ) + 2 5 ÷ ( 1 × 7 ) + 1 7 = 12 5 ÷ 8 7 The reciprocal of the divisor 8 7 is 7 8

11th Edition

ISBN: 9780134496436

Author: Cheryl Cleaves, Margie Hobbs, Jeffrey Noble

Publisher: PEARSON

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Chapter 2.3, Problem 13SE

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To calculate: The quotient for the expression 225÷117.

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Chapter 2.3, Problem 12SE

Chapter 2.3, Problem 14SE

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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 2 Solutions

Business Math (11th Edition)

Ch. 2.1 - Prob. 1-1SC Ch. 2.1 - Prob. 1-2SC Ch. 2.1 - Prob. 1-3SC Ch. 2.1 - Prob. 1-4SC Ch. 2.1 - Prob. 1-5SC Ch. 2.1 - Prob. 1-6SC Ch. 2.1 - Prob. 2-1SC Ch. 2.1 - Prob. 2-2SC Ch. 2.1 - Prob. 2-3SC Ch. 2.1 - Prob. 2-4SC

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The quotient for the expression 2 2 5 ÷ 1 1 7 .

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To calculate: The quotient for the expression 225÷117.

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Chapter 2 Solutions