1.how high is a flagpole that caists a shadow of 15 meters of a nearby post 6 meters high casts a shadow of 9m? 2.the width and lenght of a 6cm by 8cm picture frame are enlarged propotionally. if the length of the enlargement is 24cm,what is its width? 3.devone jesse drove for three consecutive days.His cars odometer read is 84.6 km in the first day,93.85 km in the second day,and 17 km in the third day.How for did he travel in those three days? please show the answer and solution..

1
by abegonia

2014-07-27T11:08:42+05:30

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1.  A shadow of 9m cast by a pole of 6m,
a shadow of 1m cast by a pole of    pole height / shadow length = ( 6 / 9 ) m

Shadow of 15 m is cast by a pole of   (6 / 9 ) * 15 = 10m

2.  Width = 6cm       Length = 8 cm        Enlarged length = 24 cm
Proportion (or ratio or magnification) of enlargement =  24 cm / 8 cm  = 3
So width is also enlarged by a factor of 3.
Enlarged width  =  6 cm * 3 = 18 cm

First day when he started odometer read  0 km.
he continued for 3 days.  Last day odometer read 17km. This is because
odometer incremented reading upto 99.99 km and later showed 0
onwards. Because it shows only between 0 and 99.99.  So 101 km is
So last day reading is actually  100 + 17 =  117 km.
this is the total distance traveled in all the 3 days.

Detailed solution:  day by day traveling.

1st day,  when he started odometer read 0.  WHen he finished, its 84.6 km
He traveled  84.6 - 0 = 84.6 km    (final reading - initial reading of odometer)

second day he traveled:  initial reading = 84.6 km & final reading = 93.85 km
2nd day he traveled = 93.85 km - 84.6 km =   9.25 km

3rd day : initial reading 93.85 km & final reading 17 km.  Odometer shows
reading upto 99.99 km & afterwards it shows a reading of 0 km. Because, it
has a range from 0.00 to 100.00 only. So 100 km is shown as 0km and
101 km is shown as 1km.
So final reading is actually = 100 + 17 = 117 km
3rd day he traveled = 117 - 93.85 =  23.15 km

So, all the days he traveled =  84.6 km + 9.25 km + 23.15 km  =  117 km.