Explanation Given: The number of people P ( t ) (in hundreds) infected t days after an epidemic begins is approximated by: P ( t ) = 10 ln ( .19 t + 1 ) .19 t + 1 Formula used: The power, and the quotient formula of differentiation: d d x x n = n x n − 1 d d x ln x = 1 x d d x ( f ( x ) g ( x ) ) = g ( x ) ⋅ d d x f ( x ) − f ( x ) ⋅ d d x g ( x ) ( g ( x ) ) 2 And According to the first derivative test, c is the only critical number of the given function for an interval [ a , b ] such that a < c < b . Consider, f is differential for every value of x in the interval except at x = c . 1. For f ′ ( a ) > 0 and f ′ ( b ) < 0 , there will be a local maximum at c . 2. For f ′ ( a ) < 0 and f ′ ( b ) > 0 , there will be a local minimum at c . 3. There will be no local extrema when f ′ ( a ) and f ′ ( b ) are either both positive or both negative Calculation: Determine the critical numbers by finding the derivative of the provided function: P ( t ) = 10 ln ( .19 t + 1 ) .19 t + 1 The derivative of the function P ( t ) : P ′ ( t ) = d d t [ 10 ln ( .19 t + 1 ) .19 t + 1 ] = ( .19 t + 1 ) d d t [ 10 ln ( .19 t + 1 ) ] − [ 10 ln ( .19 t + 1 ) ] d d t ( .19 t + 1 ) ( .19 t + 1 ) 2 = ( .19 t + 1 ) [ 10 ⋅ 1 (

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ISBN: 9780135904282

Author: Lial, HUNGERFO

Publisher: PEARSON

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Chapter 12.1, Problem 68E

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To calculate: The time at which number of people infected starts to decline if the number of people P(t) (in hundreds) infected t days after an epidemic begins is approximated by:

P(t)=10ln(.19t+1).19t+1

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Chapter 12.1, Problem 66E

Chapter 12.1, Problem 67E

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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 12 Solutions

MYMATHLAB ACCESS CODE - MATHEMATICS WIT

Ch. 12.1 - Checkpoint 1 For what values of x is the function...Ch. 12.1 - Checkpoint 2 Find all intervals on which is...Ch. 12.1 - Checkpoint 3 Identity the x-values of all points...Ch. 12.1 - Checkpoint 4 Find the critical numbers for each of...Ch. 12.1 - Prob. 5CP Ch. 12.1 - Prob. 6CP Ch. 12.1 - Checkpoint 7 Find the locations of the local...Ch. 12.1 - Prob. 8CP Ch. 12.1 - Checkpoint 9 If a sales function is given by...Ch. 12.1 - Prob. 1E

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The time at which number of people infected starts to decline if the number of people P ( t ) (in hundreds) infected t days after an epidemic begins is approximated by: P ( t ) = 10 ln ( .19 t + 1 ) .19 t + 1

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To calculate: The time at which number of people infected starts to decline if the number of people P(t) (in hundreds) infected t days after an epidemic begins is approximated by: P(t)=10ln(.19t+1).19t+1

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To calculate: The time at which number of people infected starts to decline if the number of people P(t) (in hundreds) infected t days after an epidemic begins is approximated by:

P(t)=10ln(.19t+1).19t+1