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vitfil [10]
4 years ago
6

Two groups went hiking, leaving at the same time and arriving back at the same time. While one group moved away at the speed 6 m

ph and then moved back at the speed 4 mph, the other group moved all the way with the same speed. What was the speed of this second group, if both groups covered the same total distance, 24 miles?
Mathematics
1 answer:
loris [4]4 years ago
3 0
This uses the formula d = r * t
First group
========
This group determines the time to be used by both groups.
t = d/r
t = 24 / 6
t = 4 hours hours going there.

t = 24 / 4
t = 6 hours 
Total time taken is 6 hours + 4 hours = 10 hours.

Group two
========
t = 10 hours.
d = 48 miles (remember, they have to come back).
r = 48 miles / 10 hours
r = 4.8 miles / hour
For humans, that actually is quite quick.

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Given :

\:

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\:

\gray{ \frak{The \:  given \:  two \:   \: points  \: are(x_{1 },y_{1})=(−6 , -10)and(x_{ 2 },y_{2} )=( 2,5 )\:}}

\:

Let's solve by using midpoint formula :

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\bf \boxed{\color{red}\frak{Midpoint \:   \: Formula :  {( \: x ,\:y \:  ) = }(  \frak{ \frac{x_{1 } + y_{1}}{2} , \frac{x_{2 } + y_{2}}{2} )}}}

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\large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{ \frac{ -6 + 2}{2} , \frac{ - 10 + 5}{2} )}}

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\: \large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{ \frac{ -4}{2} , \frac{ - 5}{2} )}}

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\large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{  \cancel\frac{ -4}{2} ,  \cancel\frac{ - 5}{2} )}}

\:  \:

\underline{ \boxed{ \large \red{ \frak{option \: d }(  \frak{   - 2, - 2.5)}}}}✓

Hope Helps! :)

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