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Mathematics 15 Online
OpenStudy (anonymous):

Suppose a runner takes 45 min to run a route a 8 mi/h at the beginning of training season. By the end of training season, she can run the same route in 38 min. What is her speed at the end of training season?

OpenStudy (anonymous):

@amistre64 @mertsj @hartnn @Hero @ParthKohli @ivettef365

hartnn (hartnn):

equate the distances, use Distance = Speed / Time

Parth (parthkohli):

Her route's length is ...?

OpenStudy (anonymous):

Well, 45/8=

OpenStudy (anonymous):

5.6

hartnn (hartnn):

so, speed1/ time1 = speed2/time2 time 1 = 45, speed1 = 8 time2 =38 find speed2, you will get that in mi/h

OpenStudy (anonymous):

mi.

Parth (parthkohli):

\[\dfrac{8 \rm ~miles}{\rm hour} \div 45 ~ \rm minutes\]

Parth (parthkohli):

is the distance.

OpenStudy (anonymous):

So multiply those 2? to reverse the equation?

hartnn (hartnn):

no need to find distance...to get exact distance we have to do conversions...if we do the proportionality, we directly get speed in mi/hr

Parth (parthkohli):

When you get the distance, you know that the runner covers the same distance.

hartnn (hartnn):

45/8 = 35/x just find x

hartnn (hartnn):

**38/x

OpenStudy (anonymous):

x=6.2repeating

OpenStudy (anonymous):

ooh 38..

OpenStudy (anonymous):

45/38=1.18421526 8/1.18421526=6.755528551

OpenStudy (anonymous):

So 6.8

hartnn (hartnn):

yeah, thats her speed. in mi/h

OpenStudy (anonymous):

45/38=1.2 8/1.2=6.6repeating

OpenStudy (anonymous):

ok thnks all done...gee i havent gone to bed since like 7:00 am yesterday morning

OpenStudy (anonymous):

:z

OpenStudy (anonymous):

I need to catch some Zzzzzzzzzz's

OpenStudy (anonymous):

-_- *snoring*

hartnn (hartnn):

sleep well :) sweet dreams :)

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