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?

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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
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yeh, its very similar to all questions asked one use pronumerals to represent unknowns
let the first speed be x
then distance = rate x time

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Other answers:

time = distance/rate
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
\[\frac{34}{x} + \frac{6}{x-5} = 3\]
solve for x,
this gives 2 solutions x= 3.94 and x= 14.4
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 )
therefore x=14.4mph is the speed for first section , and 9.4mph is the speed for the second section
\[\sqrt{9x+67}=x+5\]
what ?
I need to solve....well see what the heck I did wrong anyway..
then I have one more..if that is ok...then I will leave everyone alone...I hate that I get so confused
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
x=6, x=-7 but check your solns
x=-7 doesnt work, so x=6 is only soln
ok so one more if you have time?
might as well, shouldnt take long
Write a quadratic equation in the variable x having the given numbers as solutions. The equation is standard form, ax^2+bx+c=0 \[-\sqrt{3}, 6\sqrt{3}\]
\[(x+\sqrt{3}) ( x-6\sqrt{3}) =0\]
expand
x^2 -6sqrt(3)x +sqrt(3)x -6sqrt(3)sqrt(3) =0 \[x^2 - 6\sqrt{3}x - 18 =0 \]
wait no
\[x^2 -5\sqrt{3}x -18 =0 \] thats it
if you have a quadratic eqn with roots a and b , then the formula for that quadratic is (x-a)(x-b) =0
ok..thanks:)

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