document.write( "Question 1199938: Cliff left point A at 8:00 A.M walking east at 3 kph. Renz left point A at 9:00 A.M walking north at 4 kph. The distance between the two boys since 8:00 A.M may be expressed as d(t)= √(〖at〗^2-2βt+β) where d and t denote the respective distance and time. Find a+β. \n" ); document.write( "
Algebra.Com's Answer #848203 by textot(100)![]() ![]() ![]() You can put this solution on YOUR website! **1. Determine the Distances Traveled**\r \n" ); document.write( "\n" ); document.write( "* **Cliff:** Since Cliff started 1 hour earlier and walks at 3 kph, he has traveled 3 km east by 9:00 AM. \n" ); document.write( "* **Renz:** Let 't' be the time in hours since 9:00 AM. Renz travels 4t kilometers north.\r \n" ); document.write( "\n" ); document.write( "**2. Use Pythagorean Theorem**\r \n" ); document.write( "\n" ); document.write( "* The distance between Cliff and Renz at time 't' can be found using the Pythagorean theorem:\r \n" ); document.write( "\n" ); document.write( " d(t) = √[(Cliff's distance)² + (Renz's distance)²] \n" ); document.write( " d(t) = √[(3 + 3t)² + (4t)²] \n" ); document.write( " d(t) = √(9 + 18t + 9t² + 16t²) \n" ); document.write( " d(t) = √(25t² + 18t + 9)\r \n" ); document.write( "\n" ); document.write( "**3. Compare with the Given Expression**\r \n" ); document.write( "\n" ); document.write( "* Compare the derived expression with the given expression: \n" ); document.write( " * d(t) = √(at² + 2βt + β) \n" ); document.write( " * d(t) = √(25t² + 18t + 9)\r \n" ); document.write( "\n" ); document.write( "* We can see that: \n" ); document.write( " * a = 25 \n" ); document.write( " * 2β = 18 \n" ); document.write( " * β = 9\r \n" ); document.write( "\n" ); document.write( "**4. Calculate a - β**\r \n" ); document.write( "\n" ); document.write( "* a - β = 25 - 9 = 16\r \n" ); document.write( "\n" ); document.write( "**Therefore, a - β = 16.** \n" ); document.write( " \n" ); document.write( " |