anonymous
  • anonymous
During the first part of a trip, a conoeist travels 34 miles at a certain speed. The conoeist travels 6 miles on the second part of the trip at a speed of 5mph slower. The total time for the trip is 3 hrs. What was the speed on each part of the trip?
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!
anonymous
  • anonymous
yeh, its very similar to all questions asked one use pronumerals to represent unknowns
anonymous
  • anonymous
let the first speed be x
anonymous
  • anonymous
then distance = rate x time

Looking for something else?

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

More answers

anonymous
  • anonymous
time = distance/rate
anonymous
  • anonymous
time for first section of trip is T1 = 34/x now, the speed they travel for second part of trip is (x-5) { 5mph slower} so time for second part = T2 = 6/ (x-5) and we are told that the total trip = 3hrs so the sum of the times = 3hrs
anonymous
  • anonymous
\[\frac{34}{x} + \frac{6}{x-5} = 3\]
anonymous
  • anonymous
solve for x,
anonymous
  • anonymous
this gives 2 solutions x= 3.94 and x= 14.4
anonymous
  • anonymous
now, the answer cannot be x=3.94 , because if we try to determine the speed of the second part of the trip then we would get a negative value ( remember the speed for second trip was (x-5)mph )
anonymous
  • anonymous
therefore x=14.4mph is the speed for first section , and 9.4mph is the speed for the second section
anonymous
  • anonymous
\[\sqrt{9x+67}=x+5\]
anonymous
  • anonymous
what ?
anonymous
  • anonymous
I need to solve....well see what the heck I did wrong anyway..
anonymous
  • anonymous
then I have one more..if that is ok...then I will leave everyone alone...I hate that I get so confused
anonymous
  • anonymous
square both sides 9x + 67 = x^2 +10x +25 x^2 +x +25-67 =0 x^2 + x -42=0 (x+7)(x-6) =0
anonymous
  • anonymous
x=6, x=-7 but check your solns
anonymous
  • anonymous
x=-7 doesnt work, so x=6 is only soln
anonymous
  • anonymous
ok so one more if you have time?
anonymous
  • anonymous
might as well, shouldnt take long
anonymous
  • anonymous
Write a quadratic equation in the variable x having the given numbers as solutions. The equation is standard form, ax^2+bx+c=0 \[-\sqrt{3}, 6\sqrt{3}\]
anonymous
  • anonymous
\[(x+\sqrt{3}) ( x-6\sqrt{3}) =0\]
anonymous
  • anonymous
expand
anonymous
  • anonymous
x^2 -6sqrt(3)x +sqrt(3)x -6sqrt(3)sqrt(3) =0 \[x^2 - 6\sqrt{3}x - 18 =0 \]
anonymous
  • anonymous
wait no
anonymous
  • anonymous
\[x^2 -5\sqrt{3}x -18 =0 \] thats it
anonymous
  • anonymous
if you have a quadratic eqn with roots a and b , then the formula for that quadratic is (x-a)(x-b) =0
anonymous
  • anonymous
ok..thanks:)

Looking for something else?

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