A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
A piece of land is shaped like a right triangle. Two people start at the right angle vertex of the triangle at the same time and walk the same speed along different legs of the triangle. If the area formed by the positions of the two ppl and their starting point (the right angle) is changing at 4m²/sec, then how fast are the ppl moving when they are 5m from the right angle?
anonymous
 5 years ago
A piece of land is shaped like a right triangle. Two people start at the right angle vertex of the triangle at the same time and walk the same speed along different legs of the triangle. If the area formed by the positions of the two ppl and their starting point (the right angle) is changing at 4m²/sec, then how fast are the ppl moving when they are 5m from the right angle?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Think of this as a difference in areas problem: subtract one right triangle having two legs of 5m from another having two legs each 5m+dx, where dx is the rate at which the two individuals are travelling. This difference is equal to 4m^2/s: ((5+dx)^2)/2(5^2)/2=4m^2/s (25+10*dx+dx^225)/2=4 (dx^2+10*dx)/2=4 dx^2+10*dx8=0 Use the quadratic equation to solve for dx. Choose the positive answer.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.