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O ✪ • © 2024FA-MATH-020-104 > Assignments > 3rd In Class Extra Credit Work ... Fall 2024 Regular Home Syllabus Assignments SJCC Library K- 3rd In Class Extra Credit Work time, speed, distance A+ Due Monday by 11:59pm Points 10 Available Sep 19 at 12am - Sep 23 at 11:59pm I redrew the diagram for the two right triangle, solve the equation and the answers are whole numbers. AD-750 750 ft Letting AC-720-x CE=x -720 210 ft boardwalk AE-720 >The triangle AED is right triangle with hypotenuse AD 750, one of the leg ED 210 but to complete we need to find AE the other leg of the right triangle using Pythagoream AE 210-7502 AE² 7502-2102 (750 +210) (750-210)-960-540 AE= √√960-540 = 720, if AC-720-x, then CE-x this will simplify the equations. so the distance he walks on the sand from point C to D is the hypotenuse of the right triangle ACED whose two legs are x and 210 CD²x²+2102 thus CD-√x²+210² √x²+2102 then time walking on the sand at 2 ft per second 2 4 f plus time walking on the boardwalk at 720-x must be 4 minutes 45 seconds or 285 seconds. sec 4 D sun umbrella DE-210 Immersive Reader

O ✪ • © 2024FA-MATH-020-104 > Assignments > 3rd In Class Extra Credit Work ... Fall 2024 Regular Home Syllabus Assignments SJCC Library K- 3rd In Class Extra Credit Work time, speed, distance A+ Due Monday by 11:59pm Points 10 Available Sep 19 at 12am - Sep 23 at 11:59pm I redrew the diagram for the two right triangle, solve the equation and the answers are whole numbers. AD-750 750 ft Letting AC-720-x CE=x -720 210 ft boardwalk AE-720 >The triangle AED is right triangle with hypotenuse AD 750, one of the leg ED 210 but to complete we need to find AE the other leg of the right triangle using Pythagoream AE 210-7502 AE² 7502-2102 (750 +210) (750-210)-960-540 AE= √√960-540 = 720, if AC-720-x, then CE-x this will simplify the equations. so the distance he walks on the sand from point C to D is the hypotenuse of the right triangle ACED whose two legs are x and 210 CD²x²+2102 thus CD-√x²+210² √x²+2102 then time walking on the sand at 2 ft per second 2 4 f plus time walking on the boardwalk at 720-x must be 4 minutes 45 seconds or 285 seconds. sec 4 D sun umbrella DE-210 Immersive Reader

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 7RE
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need help solving this: A boardwalk is parallel to and 210 ft inland from a straight shoreline. A sandy beach lies between the boardwalk and the shoreline. A man is standing on the boardwalk, exactly 750 ft across the sand from his beach umbrella, which is right at the shoreline. The man walks 4 ft/s on the boardwalk and 2 ft/s on the sand. How far should he walk on the boardwalk before veering off onto the sand if he wishes to reach his umbrella in exactly 4 min 45 s?

O ✪ • © 2024FA-MATH-020-104 > Assignments > 3rd In Class Extra Credit Work ... Fall 2024 Regular Home Syllabus Assignments SJCC Library K- 3rd In Class Extra Credit Work time, speed, distance A+ Due Monday by 11:59pm Points 10 Available Sep 19 at 12am - Sep 23 at 11:59pm I redrew the diagram for the two right triangle, solve the equation and the answers are whole numbers. AD-750 750 ft Letting AC-720-x CE=x -720 210 ft boardwalk AE-720 >The triangle AED is right triangle with hypotenuse AD 750, one of the leg ED 210 but to complete we need to find AE the other leg of the right triangle using Pythagoream AE 210-7502 AE² 7502-2102 (750 +210) (750-210)-960-540 AE= √√960-540 = 720, if AC-720-x, then CE-x this will simplify the equations. so the distance he walks on the sand from point C to D is the hypotenuse of the right triangle ACED whose two legs are x and 210 CD²x²+2102 thus CD-√x²+210² √x²+2102 then time walking on the sand at 2 ft per second 2 4 f plus time walking on the boardwalk at 720-x must be 4 minutes 45 seconds or 285 seconds. sec 4 D sun umbrella DE-210 Immersive Reader
Transcribed Image Text:O ✪ • © 2024FA-MATH-020-104 > Assignments > 3rd In Class Extra Credit Work ... Fall 2024 Regular Home Syllabus Assignments SJCC Library K- 3rd In Class Extra Credit Work time, speed, distance A+ Due Monday by 11:59pm Points 10 Available Sep 19 at 12am - Sep 23 at 11:59pm I redrew the diagram for the two right triangle, solve the equation and the answers are whole numbers. AD-750 750 ft Letting AC-720-x CE=x -720 210 ft boardwalk AE-720 >The triangle AED is right triangle with hypotenuse AD 750, one of the leg ED 210 but to complete we need to find AE the other leg of the right triangle using Pythagoream AE 210-7502 AE² 7502-2102 (750 +210) (750-210)-960-540 AE= √√960-540 = 720, if AC-720-x, then CE-x this will simplify the equations. so the distance he walks on the sand from point C to D is the hypotenuse of the right triangle ACED whose two legs are x and 210 CD²x²+2102 thus CD-√x²+210² √x²+2102 then time walking on the sand at 2 ft per second 2 4 f plus time walking on the boardwalk at 720-x must be 4 minutes 45 seconds or 285 seconds. sec 4 D sun umbrella DE-210 Immersive Reader
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