anonymous
  • anonymous
a. Use a metric ruler to measure the distances between Robinson and Neale on the map. b. Using the scale of the map, find the approximate actual distance by air (not by roads), between Robinson and Neale. c. Approximately how many square miles are shown on this map?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
http://www.intervisualtechnology.us/data/1/12383/texasmap.jpg is the link to the map
misssunshinexxoxo
  • misssunshinexxoxo
Ok please pull out your ruler and mark it down then post it; I'll go over it
misssunshinexxoxo
  • misssunshinexxoxo
2.4 centimeters is what you messaged me; what's the mileage

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misssunshinexxoxo
  • misssunshinexxoxo
Looks like 4 miles to me
anonymous
  • anonymous
Thats the part i'm having trouble with. The scale says 0-5 miles, but idk how to incorporate that with the 2.4 centimeters.
misssunshinexxoxo
  • misssunshinexxoxo
Is this an essay question?
anonymous
  • anonymous
no
misssunshinexxoxo
  • misssunshinexxoxo
I think you separate it
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
So you measured the distance between Robinson and Neale and got 2.4 cm?
anonymous
  • anonymous
yea
jim_thompson5910
  • jim_thompson5910
use a ruler to measure how long the "0-5" bar is
anonymous
  • anonymous
2.1 cm
jim_thompson5910
  • jim_thompson5910
x = actual distance from Robinson to Neale solve the proportion for x \[\Large \frac{\text{Actual Distance}}{\text{Distance on paper}} = \frac{\text{Actual Distance}}{\text{Distance on paper}}\] \[\Large \frac{x}{2.4} = \frac{5}{2.1}\] \[\Large x = ???\]
anonymous
  • anonymous
5.7
anonymous
  • anonymous
but it says to find the distance by air and not by roads, what about that part?
jim_thompson5910
  • jim_thompson5910
that's the actual air distance
anonymous
  • anonymous
and it would be 5.7 for part C also?
jim_thompson5910
  • jim_thompson5910
measure the width and height of the entire map with the ruler
jim_thompson5910
  • jim_thompson5910
what do you get?
anonymous
  • anonymous
6.8 cm length and width
anonymous
  • anonymous
so do i just multiply that now?
jim_thompson5910
  • jim_thompson5910
well you first have to convert those distances on paper to actual distances
jim_thompson5910
  • jim_thompson5910
by using a proportion like shown above
anonymous
  • anonymous
so then 16.19
anonymous
  • anonymous
i mean 16.19 is the length and width, and multiplying it i get 262.12
jim_thompson5910
  • jim_thompson5910
so you just found that 6.8 cm on paper corresponds to 16.19 miles in actual distance
jim_thompson5910
  • jim_thompson5910
and yeah the approximate actual area is 262.1161 square miles
anonymous
  • anonymous
yea, i set it up like 5/2.1 = x/6.8 is that right?
jim_thompson5910
  • jim_thompson5910
yes it is
anonymous
  • anonymous
cool, thank you!
jim_thompson5910
  • jim_thompson5910
no problem

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