HW2 - Pickering, Tracey Homework 2 Due: Jan 25 2006, noon...

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Pickering, Tracey – Homework 2 – Due: Jan 25 2006, noon – Inst: Drummond 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points A high Fountain oF water is located at the center oF a circular pool as in the fgure. A student walks around the pool and estimates its circumFerence to be 197 m. Next, the student stands at the edge oF the pool and uses a protractor to gauge the angle oF elevation oF the top oF the Fountain to be 51 . 5 . How high is the Fountain? Correct answer: 39 . 4168 m. Explanation: R h θ The circumFerence oF the pool is C = 2 πR so its radius is R = C 2 π = 197 m 2 π = 31 . 3535 m The radius and the height oF the Fountain Form the adjacent and opposite sides oF the angle oF elevation, respectively. Thus the tangent oF the angle can be used, and tan θ = h R so that h = R tan θ = 39 . 4168 m 002 (part 1 oF 1) 10 points A graph oF a straight line going through two points is shown below. y - 5 - 3 - 1 0 1 2 3 4 5 x - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 What is the equation y = f ( x ) oF this line? 1. y = + 7 4 x - 2 2. y = - 4 7 x - 8 7 3. y = + 4 7 x - 8 7 correct 4. y = + 4 7 x + 8 7 5. y = + 4 7 x + 2 6. y = - 4 7 x + 2 7. y = + 7 4 x - 8 7 8. y = - 7 4 x + 8 7 9. y = + 4 7 x - 2 10. y = - 7 4 x - 8 7 Explanation: Solution: Let ( x 1 ,y 1 ) = ( - 5 , - 4) ( x 2 ,y 2 ) = (2 , 0) . The equation y = f ( x ) oF a line going through these two points is y - y 1 x - x 1 = y 2 - y 1 x 2 - x 1 = s, (1)
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Pickering, Tracey – Homework 2 – Due: Jan 25 2006, noon – Inst: Drummond 2 where s is the slope of the line. s = y 2 - y 1 x 2 - x 1 = (0) - ( - 4) (2) - ( - 5) = + 4 7 . (2) Rewriting Eq. 1 using the slope, we have y = s ( x - x 1 ) + y 1 = µ + 4 7 ( x + 5) - 4 = + 4 7 x - 8 7 , (3) where the slope of the line is + 4 7 and the y -intercept is - 8 7 (as can be seen from the graph). Note: This problem has a diFerent line for each student. 003 (part 1 of 1) 10 points On a drive from your ranch to Austin, you wish to average 48 mph. The distance from your ranch to Austin is 100 miles. However, at 50 miles (half way), you ±nd you have averaged only 36 mph. What average speed must you maintain in the remaining distance in order to have an overall average speed of 48 mph? Correct answer: 72 mph. Explanation: Let t denote the total time, d the total distance, and v the average velocity over the total distance. Let t 1 denote the time over the ±rst half, d 1 the ±rst half distance, and v 1 the average velocity of the ±rst half. Let t 2 denote the time over the second half, d 2 the second
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HW2 - Pickering, Tracey Homework 2 Due: Jan 25 2006, noon...

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