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10 Pages

### hw7

Course: M 408K, Spring 2007
School: University of Texas
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Word Count: 2086

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Yvette Granillo, Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the derivative of f when 2 f (x) = 2x sin 5x + cos 5x. 5 1. f (x) = 10x cos 5x correct 2. f (x) = 10x cos 5x + 4 sin 5x 3. f (x) = 10x...

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Yvette Granillo, Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the derivative of f when 2 f (x) = 2x sin 5x + cos 5x. 5 1. f (x) = 10x cos 5x correct 2. f (x) = 10x cos 5x + 4 sin 5x 3. f (x) = 10x cos 5x - 4 sin 5x 4. f (x) = 10 cos 5x 5. f (x) = -10x cos 5x Explanation: Since d sin x = cos x, dx it follows that f (x) = 2 sin 5x + 10x cos 5x - 2 sin 5x. Consequently, f (x) = 10x cos 5x . d cos x = - sin x, dx keywords: Stewart5e, 003 (part 1 of 1) 10 points Determine f (x) when f (x) = 1. f (x) = - 2. f (x) = - 3. f (x) = 1+x . 1 - 2x2 4(x - 2) (x + 1)3 1 2. f (x) = - 3. f (x) = 4. f (x) = 6(x - 2) (x + 1)3 4(x - 1) correct (x + 1)3 6(x + 2) (x - 1)3 5. f (x) = - 6. f (x) = 6(x + 1) (x - 1)3 Explanation: By the Chain and Quotient Rules, f (x) = 2 Consequently, f (x) = 4(x - 1) . (x + 1)3 x - 1 (x + 1) - (x - 1) . x+1 (x + 1)2 1 - 2x (1 - 2x2 )1/2 1 + 2x (1 - 2x2 )1/2 keywords: Stewart5e, 002 (part 1 of 1) 10 points Find f (x) when f (x) = x-1 x+1 2 1 + 2x (1 - 2x2 )1/2 4. f (x) = - . 5. f (x) = 6. f (x) = 1. f (x) = - 4(x + 1) (x - 1)3 1 + 2x correct (1 - 2x2 )3/2 1 - 2x (1 - 2x2 )3/2 1 + 2x (1 - 2x2 )3/2 Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell Explanation: By the Product and Chain Rules, f (x) = 1 4x(1 + x) + 2 )1/2 (1 - 2x 2(1 - 2x2 )3/2 = Consequently, f (x) = 1 + 2x . (1 - 2x2 )3/2 (1 - 2x2 ) + 2x(1 + x) . (1 - 2x2 )3/2 keywords: Stewart5e, 005 (part 1 of 1) 10 points Find the value of f (1) when f (x) = (x2 + 8)1/2 + 1 3 1 . x 2 1. f (1) = - 2. f (1) = 2 3 (Note: the Quotient Rule could have been used, but it's simpler to use the Product Rule.) keywords: Stewart5e, 004 (part 1 of 1) 10 points Determine f (x) when f (x) = 2 cos2 x + 3 sin2 x . 1. f (x) = -4 sin x - 6 cos x 2. f (x) = 2 sin x cos x correct 3. f (x) = -10 sin x cos x 4. f (x) = 4 cos x + 6 sin x 3. f (1) = 0 2 4. f (1) = - correct 3 5. f (1) = 1 3 Explanation: Using the Chain Rule and the fact that d r x = r xr-1 dx holds for all values of r, we see that f (x) = x 1 - 2. 1/2 x + 8) 5. f (x) = -2 sin x cos x 6. f (x) = -4 sin x + 6 cos x 7. f (x) = 10 sin x cos x Explanation: Since d d sin x = cos x , cos x = - sin x , dx dx the Chain Rule ensures that f (x) = -4 cos x sin x + 6 sin x cos x . Consequently, f (x) = 2 sin x cos x . (x2 At x = 1, therefore, f (1) = - 2 . 3 keywords: Stewart5e, 006 (part 1 of 1) 10 points Find the derivative of f when f (x) = 1 (3 - 4x) 3 + 8x . 4 Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell 1. f (x) = - 2. f (x) = - 3. f (x) = - 4. f (x) = 5. f (x) = 4x 3 + 8x 12x correct 3 + 8x 12x 3 + 4x 4 f (x + 8)2 x+4 x+8 3 6. F (x) = - Explanation: By the Chain Rule, F (x) = f x+4 x+8 d dx x+4 x+8 . 12x 3 + 8x 4x 3 + 8x But by the Quotient Rule, dx dx x+4 x+8 = (x + 8) - (x + 4) . (8 + x)2 Explanation: By the Product and Chain Rules, 3 - 4x f (x) = - 3 + 8x + . 3 + 8x Bringing the right hand side to a common denominator, we thus see that: (3 - 4x) - (3 + 8x) 12x f (x) = = - . 3 + 8x 3 + 8x Consequently, F (x) = 4 f (x + 8)2 x+4 x+8 . keywords: Stewart5e, 008 (part 1 of 1) 10 points Find the derivative of f when f (x) = sin 5+x 2-x . keywords: Stewart5e, 007 (part 1 of 1) 10 points Find the derivative of F when F (x) = f and f is differentiable. 1. F (x) = f 2. F (x) = 3. F (x) = 4 (x + 8)2 x+4 x+8 x+4 x+8 correct x+4 x+8 1. f (x) = 7 sin (2 - x)2 7 cos (2 - x)2 7 (2 - x)2 7 (2 - x)2 5+x 2-x 5+x 2-x 2. f (x) = - 3. f (x) = sin 4. f (x) = cos 5. f (x) = - 6. f (x) = x+4 x+8 4 f (x + 8)2 4 f (x + 8)2 4 (x + 8)2 4 f (x + 8)2 7 sin (2 - x)2 5+x 2-x 5+x 2-x correct 4. F (x) = f 5. F (x) = - 7 cos (2 - x)2 Explanation: Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell By the Chain and Quotient rules, f (x) = cos 5+x 2-x 5+x 2-x d dx 5+x 2-x . keywords: Stewart5e, 010 (part 1 of 1) 10 points . Find the value of F (2) when F (x) = f (g(x)) keywords: Stewart5e, 009 (part 1 of 1) 10 points Find an equation for the tangent line to the graph of f (x) = 8 sin(sin(x)) at the point P (, f ()). 1. y + 8x = 8 correct 2. y + = 8x 3. y + 8x + = 0 4. y = 8x + 5. y = 8x - 8 Explanation: Since f () = sin(sin()) = sin 0 = 0 , we have to find the equation of the tangent line to the graph of f (x) = 8 sin(sin(x)) at P (, 0). Now f (x) = (8 cos x) cos(sin x)) , so at P (, 0), the tangent line has slope = 8(cos ) cos 0 = -8 . Thus and g(2) = 6, f (2) = 5, 1. F (2) = 10 2. F (2) = 14 3. F (2) = 12 correct 4. F (2) = 11 5. F (2) = 13 Explanation: By the Chain Rule, F (x) = f (g(x))g (x) . F (2) = f (g(2))g (2) = f (6)g (2) . Consequently, when g(2) = 6, f (2) = 5, we see that F (2) = 12 . g (2) = 4 , f (6) = 3 , g (2) = 4 , f (6) = 3 . 4 By the point-slope formula, therefore, the tangent line is y = -8(x - ), i.e., y + 8x = 8 . = cos Thus (2 - x) + (5 + x) (2 - x)2 5+x 2-x 7 f (x) = cos (2 - x)2 Notice that the value of f (2) was not needed. Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell keywords: Stewart5e, 011 (part 1 of 1) 10 points 3. Find dy/dx when 3x2 + 2y 2 = 1 . dy = -3xy dx 3x dy = - correct dx 2y 3x dy = dx 2y 3x dy = - dx y x dy = dx 2y dy = 2xy dx 4. 5. 2. dy = cot x cot y dx dy = tan x tan y correct dx dy = tan xy dx dy = cot x tan y dx 5 1. 2. 3. 4. 5. 6. Explanation: Differentiating implicitly with respect to x we see that 2 cos x cos y Thus dy cos x cos y = sin x sin y . dx Consequently, dy sin x sin y = = tan x tan y . dx cos x cos y dy - sin y sin x dx = 0. Explanation: Diferentiating 3x2 + 2y 2 = 1 implicitly with respect to x we see 6x + 4y Consequently, 6x 3x dy = - = - . dx 4y 2y dy = 0. dx keywords: Stewart5e, 013 (part 1 of 1) 10 points dy when dx tan(x y) - = x . dy x2 1. = correct dx 1 + x2 2. 3. 4. dy x2 = dx 1 + y2 dy x2 = - dx 1 + y2 dy x2 = - dx 1 + x2 Find keywords: Stewart5e, 012 (part 1 of 1) 10 points Determine dy/dx when 2 cos x sin y = 5 . dy = tan x 1. dx Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell dy y2 = dx 1 + x2 dy y2 = - dx 1 + x2 1. 2. 3. 4. 5. = 1. dy dx dy dx dy dx dy dx dy dx = = 7 11 1 7 6 5. 6. P P Explanation: Differentiating tan(x - y) = x implicitly with respect to x we see that sec2 (x - y) 1 - Thus - sec2 (x - y) in which case dy = dx sec2 (x - y) - 1 sec2 (x - y) 2 P = 7 = -7 correct = - 1 7 P dy dx P Explanation: Differentiating implicitly with respect to x we see that 15x2 + 9y 2 Consequently, dy dy = 9y + 9x . dx dx dy = 1 - sec2 (x - y) , dx . But at 2 But sec2 = 1 + tan2 , so sec (x - y) = 1 + tan (x - y) = 1 + x , while sec2 (x - y) - 1 = tan2 (x - y) = x2 . Consequently, dy x2 = . dx 1 + x2 keywords: Stewart5e, 014 (part 1 of 1) 10 points Find the slope of the tangent line to the graph of 5x3 + 3y 3 = 9xy at the point P = 9 9 , . 8 8 2 dy 3y - 5x2 = . dx 3y 2 - 3x 9 9 , , 8 8 P = simple algebra shows that 3y - 5x2 = while 3y 2 - 3x = Hence at P , dy dx = -7. 243 27 27 - = . 64 8 64 189 27 405 - = - , 8 64 64 P keywords: Stewart5e, 015 (part 1 of 1) 10 points Determine F (x) when F (x) = -2xf (x) + 3f (x) Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell and f is a twice-differentiable function. 1. F (x) = -5 f (x) + 2 x f (x) 2. F (x) = f (x) - 3 x f (x) 3. F (x) = f (x) - 2 x f (x) correct 4. F (x) = f (x) + 3 x f (x) 5. F (x) = -5 f (x) - 2 x f (x) Explanation: After differentiation using the Product Rule we see that F (x) = -2f (x) - 2xf (x) + 3f (x) . Consequently, F (x) = f (x) - 2xf (x) . keywords: Stewart5e, 016 (part 1 of 1) 10 points The figure below shows the graphs of three functions of time t: 6. F (x) = -5 f (x) - 3 x f (x) 4. s : 5. s : 6. s : v: v: v: a: a: a: correct 1. s : 2. s : 3. s : v: v: v: a: a: a: 7 Explanation: Experience tells us that the car is (i) moving forwards when its velocity is positive, (ii) moving backwards when its velocity is negative, (iii) speeding up when its velocity is positive and increasing, i.e., when both velocity and acceleration are positive, (iv) slowing down when its velocity is positive but decreasing, i.e., when its velocity is positive but its acceleration is negative. Now ds dv v = a = , dt dt so we need to look at the slope of the tangent line to three graphs to determine which graph is that of position, that of velocity and that of acceleration. Inspection of the graphs thus shows that s: v: a: . t keywords: Stewart5e, One is the graph of the position function s of a car, one is its velocity v, and one is its acceleration a. Identify which graph goes with which function. 017 (part 1 of 1) 10 points Determine d2 y/dx2 when 4x2 + 2y 2 = 2. Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell d2 y 1 1. = - 3 2 dx y 2. 3. 4. 5. 2 d2 y = 3 2 dx y 2 d2 y = - 2 2 dx y d2 y 2 = - 3 correct 2 dx y d2 y 2 = 2 2 dx y 8 Find the value of R (0) by first finding R (0). 1. R (0) = 2. R (0) = 3. R (0) = 4. R (0) = 5. R (0) = 3 14 1 4 3 28 1 7 5 correct 28 Explanation: Differentiating implicitly with respect to x we see that 8x + 4y so x dy = -2 . dx y But then d2 y d x = - 2 2 dx dx y = - Thus 1 d2 y = - 3 4x2 + 2y 2 2 dx y keywords: Stewart5e, 018 (part 1 of 1) 10 points A special function known as the Rivaldo function satisfies both the equation 3xR (x) + 4R (x) + R(x) = 0 and the equation R(0) = 5 . 2 = - 3. y = - 8y - 8x 4y 2 dy dx dy = 0, dx Explanation: Setting x = 0 in the equation 3xR (x) + 4R (x) + R(x) = 0 , we see that 4R (0) + R(0) = 0 , so 5 R (0) = - . 4 () On the other hand, by differentiating () implicitly with respect to x we see also that 3R (x) + 3xR (x) + 4R (x) + R (x) = 0 . At x = 0, therefore, 3R (0) + 4R (0) + R (0) = 0 . Consequently, R (0) = 5 . . 28 4x2 1 2y + . y2 y keywords: Stewart5e, 019 (part 1 of 1) 10 points Determine the third derivative, f (x), of f when f (x) = 4x + 1 . Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell 1. f (x) = -24(4x + 1) -5/2 9 2. f (x) = -24(4x + 1)-3/2 3. f (x) = 24(4x + 1)-5/2 correct 4. f (x) = -4(4x + 1)-3/2 5. f (x) = 4(4x + 1)-3/2 6. f (x) = 4(4x + 1)-5/2 Explanation: To use the Chain Rule successively it's more convenient to write f (x) = 4x + 1 = (4x + 1)1/2 . For then f (x) = 1 4 (4x + 1)-1/2 2 = 2(4x + 1) while 1 f (x) = - 4 2(4x + 1)-3/2 2 = -4(4x + 1)-3/2 , and f (x) = Consequently, f (x) = 24(4x + 1)-5/2 . 3 4 4(4x + 1)-5/2 . 2 -1/2 where A is the amplitude of its oscillations and is a constant. Find the velocity and acceleration as functions of time. 1. v(t) = - A cos t, a(t) = -A 2 cos t 2. v(t) = A sin t, a(t) = -A cos t 3. v(t) = - A sin t, a(t) = -A 2 cos t correct 4. v(t) = A sin t, a(t) = -A 2 cos t 5. v(t) = - A sin t, a(t) = -A cos t Explanation: y(t) = A cos t v(t) = y (t) = - A sin t , a(t) = v (t) = -A 2 cos t keywords: Stewart5e, 021 (part 1 of 1) 10 points If P is a second degree polynomial such that P (2) = 8, P (2) = 5, P (2) = 5, find the value of P (1). 1. P (1) = 6 2. P (1) = 7 3. P (1) = 11 correct 2 13 2 keywords: Stewart5e, 020 (part 1 of 1) 10 points A mass attached to a vertical spring has position function given by y(t) = A cos t, 4. P (1) = 5. P (1) = 5 Explanation: Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell The most general second degree polynomial has the form P (x) = ax2 + bx + c . But then P (x) = 2ax + b, Thus P (2) = 5 in which case P (x) = 5x + b . The condition on P (2) now ensures that P (2) = 10 + b = 5 , i.e., b = -5. Thus P (x) = in which case P (2) = 8 Consequently, P (x) = At x = 1, therefore, P (1) = 11 . 2 5 2 x - 5x + 8 . 2 = c = 8. 5 2 x - 5x + c , 2 = a = 5 , 2 P (x) = 2a . 10 keywords: Stewart5e,
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University of Texas - M - 408K
Granillo, Yvette Homework 8 Due: Oct 20 2005, 3:00 am Inst: Edward Odell This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 0
Pepperdine - HUM - 111
March 29, 2006Etruscans Three named divinities Last king overthrown in 509 BC, hence the founding of RomeRoman RepublicOVERVIEW Beginning in 509 BC, and subsequently the founding of Rome took Place The republic is a form of democracy. It was
University of Texas - M - 408K
Granillo, Yvette Homework 9 Due: Oct 27 2005, 3:00 am Inst: Edward Odell This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 0
University of Texas - M - 408K
Granillo, Yvette Homework 10 Due: Nov 3 2005, 3:00 am Inst: Edward Odell This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 0
University of Texas - M - 408K
Granillo, Yvette Homework 11 Due: Nov 11 2005, 3:00 am Inst: Edward Odell This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time.
University of Texas - M - 408K
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time.
University of Texas - M - 408K
Granillo, Yvette Homework 13 Due: Nov 24 2005, 3:00 am Inst: Edward Odell This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time.
University of Texas - M - 408K
Granillo, Yvette Homework 14 Due: Dec 1 2005, 3:00 am Inst: Edward Odell This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 0
University of Texas - M - 408K
Granillo, Yvette Homework 15 Due: Dec 9 2005, 3:00 am Inst: Edward Odell This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 0
University of Texas - M - 408K
Granillo, Yvette Review 1 Due: Dec 9 2005, 6:00 pm Inst: Edward Odell This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Revi
University of Texas - M - 408K
Granillo, Yvette Review 2 Due: Dec 9 2005, 6:00 pm Inst: Edward Odell This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Revi
University of Texas - M - 408K
Granillo, Yvette Review 3 Due: Dec 9 2005, 6:00 pm Inst: Edward Odell This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001
University of Texas - M - 408K
Cantu, Corina Final 1 Due: May 10 2006, 10:00 pm Inst: Castravet This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - M - 408K
Granillo, Yvette Final 1 Due: May 10 2006, 10:00 pm Inst: E Schultz This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (p
University of Texas - M - 408K
Granillo, Yvette Final 1 Due: May 10 2006, 10:00 pm Inst: E Schultz This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (p
University of Texas - M - 408K
Granillo, Yvette Final 1 Due: May 10 2006, 10:00 pm Inst: E Schultz This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (p
Pepperdine - HUM - 111
The Age of Augustus and the Roman Empire Age of AugustusAGUSTUS' ACCOMPLISHMENTS Says that he found Rome a city of brick, and left it a city of marble How the Romans used public monuments to show what Rome and Romans should be; a walk through impe
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
Charisma I. How do we define it? A. Perceived People dont have charisma.we perceive it from them. We can shape peoples perception of us. B. Hard to define Humor is part of charisma. Even though there are broad characteristics of charisma but there
University of Texas - CMS - 315M
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University of Texas - CMS - 315M
Charisma, part 3 (review)III. Dimensions of Charisma A. Credibility B. Similarity C. AttractionD. Perceived Sensitivity 1. Listening In a relationship, the single most important factor. -Listen to tell someone else what was said If you want to
University of Texas - CMS - 315M
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Pepperdine - HUM - 111
The Roman EmpireRomanization and Urbanization Benefits of Rome extended to the provinces Gave them cities in order to keep track of them, extended government, roman law courts, roman citizenship, built new cities and gave them benefits of living i
University of Texas - CMS - 315M
Enduring RelationshipsI. Characteristics A. Variability Seeing the person in many different states/situations. When you see them in lots of situations you know them better. B. Duration How long you see someone on average. Long distance relationsh
University of Texas - CMS - 315M
We will rely on the nonverbal communication messages more. Old Adage Its not WHAT you say that matters, its how you say it. Sometimes. Its WHAT you say that matters and HOW you say it. How: Organize Construct Tailor Words have Power Impact thoughts a
University of Texas - CMS - 315M
Pick-ups (Relationships) I. Stages A. Assess qualifiers Look around the room and see people that ,qualify. Is this someone Im interested in? 1. Extraordinary Qualifiers- Extraordinary qualities that very few people have and everyone knows what they
University of Texas - CMS - 315M
Relational ControlI. Principles of Control relationships involve the battle of control. A. Least Interest The more you care, the stupider you are. B. Alternatives When you have more alternative you have more power. The more choices you have the m
University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
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University of Texas - CMS - 315M
Fair Fighting I. Basic skills A. Editing watching what you are saying. 1. take a pause every so often. If you do the it can reduce the intensity. 2. remember, &quot;words hurt deeply&quot; Words do hurt! Bruising in the inside not on the outside. &quot;Do I real
University of Texas - CMS - 315M
Jealousy I. Social Comparison Jealousy (Envy) A. Definition Want something that someone has and you want it. B. Why? 1. Negative it says something negative about you. 2. Relevant it has to be relevant to you. 3. Similar if they are similar it mat
University of Texas - CMS - 315M
Break Ups I. Stages of Breaking Up A. Denial &amp; Disillusionment When you claim that you are happy when you aren't and when you are alone you are cynical. Denying anything is wrong initially. B. Erosion &amp; Anger as a relationship ends it starts to wea
University of Texas - CMS - 315M
Death &amp; Dying I. Fear of Death A. Suffering Death isnt pleasant. B. Humiliation The process of death takes away ,esteem. C. Extinction If you are very religious and good on earth then youll go to a better place after. Or like just being gone, no m
University of Texas - CMS - 315M
Charisma, not a physical feature but a characteristic that we perceive from people that we admire, is an important trait to possess. Charisma is indefinable; each person can have their interpretation of what charisma is. To some charisma is the abili
University of Texas - CMS - 315M
CMS 315M Paper TopicsSpring 2006Papers are due AT 2PM, Thursday, April 6th, 2006Choose ONE of the topics listed below for your paper. Each of the topics requires you to use information presented in class or in your textbook. You do NOT need to do
Pepperdine - HUM - 111
The Rise of Christianity and the Decline of RomePeriod of PersecutionJesus of Nazareth Romans were polytheistic; believed spirits were in everything. In your household there was a spirit of the heart, spirit of the doorwayJesus of Nazareth
Pepperdine - HUM - 111
The Early Middle Ages: Instability in RomeTerms &amp; Review Byzantine Empire : later historian's word for the Eastern Roman Empire Late Roman Empire Problems Overexpansion, disunity, migration/invitations, loss of tax base Diocletian Divided Empire:
Pepperdine - HUM - 111
The Middle Ages and the Search for OrderEarly Middle Ages &quot;Dark Ages&quot; vs. &quot;Middle Ages&quot; or the &quot;Medieval Era&quot; Over Expansion of Roman Empire, weakening of strong government Diocletian divided Empire in East and West Changes in climate in Europe, no