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anonymous
 one year ago
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?
anonymous
 one year ago
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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.intervisualtechnology.us/data/1/12383/texasmap.jpg is the link to the map

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.1Ok please pull out your ruler and mark it down then post it; I'll go over it

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.12.4 centimeters is what you messaged me; what's the mileage

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.1Looks like 4 miles to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats the part i'm having trouble with. The scale says 05 miles, but idk how to incorporate that with the 2.4 centimeters.

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.1Is this an essay question?

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.1I think you separate it

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3So you measured the distance between Robinson and Neale and got 2.4 cm?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3use a ruler to measure how long the "05" bar is

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3x = 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
 one year ago
Best ResponseYou've already chosen the best response.0but it says to find the distance by air and not by roads, what about that part?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3that's the actual air distance

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and it would be 5.7 for part C also?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3measure the width and height of the entire map with the ruler

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3what do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.06.8 cm length and width

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do i just multiply that now?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3well you first have to convert those distances on paper to actual distances

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3by using a proportion like shown above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean 16.19 is the length and width, and multiplying it i get 262.12

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3so you just found that 6.8 cm on paper corresponds to 16.19 miles in actual distance

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3and yeah the approximate actual area is 262.1161 square miles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea, i set it up like 5/2.1 = x/6.8 is that right?
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