anonymous
  • anonymous
Newport & Vernonville are 200 miles apart. A car leaves Newport traveling towards Vernonville, and another car leaves Vernonville the same time, traveling towards newport. The car leaving Newport averages 10mph more than other, and they meet after 1 hour 36. What are the average speeds of the car? The car leaving Newport averages _______MPH? The car leaving Vernonville averages _____MPH?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
radar
  • radar
The distance that each car travels sums up to 200 miles. distance = rate times time. I am thinkin you might could use that info.
radar
  • radar
Let x = distance of the Newport bound car Let y =distance of the veronville bound car x+y=200 We are given time t as 1 hour 36 min, which we need to convert to hour units 1 + 36/60 =1.6 hours. Do I need to go on?
radar
  • radar
Mph of Newport bound car = x/1.6 mph of Vernonville bound car = y/1.6

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Yes, so my 1st equation is 1.6x+1.6y=200 my second equation is x-y-10
radar
  • radar
I don't think so, lets see what robtobey has to say.
anonymous
  • anonymous
r t = d Let r be speed of the slower car. Solve the following for r:\[(r+10)\frac{8}{5}+r\frac{8}{5} =200\]\[r=\frac{115}{2} \]\[r+10=\frac{135}{2} \]
radar
  • radar
I may have got off to a wrong startt.
radar
  • radar
Thanks robtobey.
anonymous
  • anonymous
\[\{67.5 \text{ mph},57.5\text{ mph}\} \]
anonymous
  • anonymous
Your welcome.
radar
  • radar
LynDor please note, 8/5 is the time 1.6 hr
radar
  • radar
As you can see I should of Let x =rate of slower car rather than distant, I also should of kept it in terms of one unknown (x+10) rate of faster car.
anonymous
  • anonymous
Thanks. I would never have answered it. So 67.5 is the car leaving newport?
radar
  • radar
Yes, the faster car is leaving Newport.
anonymous
  • anonymous
OK, thanks. Is there a limit of questions per day?
anonymous
  • anonymous
How do I give a medal?
radar
  • radar
I don't think there is a limit to the number of questions. To award a medal click on the "Good Answer" button. Good luck with your studies.

Looking for something else?

Not the answer you are looking for? Search for more explanations.