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MAVERICK [17]
2 years ago
8

on a map with a scale 1:100,000 the distance between two cities is 12cm. what would be the distance between these two cities on

a different map with a scale 1:300,000?
Mathematics
2 answers:
navik [9.2K]2 years ago
8 0
First scale is 1 -> 100,000 each centimeter in map represent 100,000 in the real world.  If you have 12 cm, in reality the distance is (12 cm)(100,000)
12 cm -> 1 200 000 cm,  each m = 100 cm
Distance: 1 200 000 cm / 100 cm/m = 1 200 m
Distance: 1 200 m / 1000 m/km = 1,2 km

Second scale is 1 -> 300,000 each centimeter in map represent 300,000 in the real world.  If you have 12 cm, in reality the distance is (12 cm)(300,000)
12 cm -> 3 600 000 cm,  each m = 100 cm and each km = 1000 m
Distance: 3 600 000 cm / 100 cm/m = 3 600 m
Distance: 3 600 m / 1000 m/km = 3,6 km

300 000 is three times 100 000

The real distance is three times. first case 1,2 second case 3,6

Your anwer is 3,6 in kilometers, 3 600 in meters or 3 600 000 in cm
klio [65]2 years ago
5 0

Part 1)

we have

the scale is 1:100,000

the distance on a map is 12\ cm

we know that

The scale is equal to the distance on a map divided by the real distance

Let

x------> distance on a map

y-------> real distance

S-------> scale

S=\frac{x}{y}

In this part we have

S=\frac{1}{100,000}

x=12\ cm  

Find the value of y

y=\frac{x}{S}

Substitute the values

y=\frac{12}{(1/100,000)}

y=1,200,000\ cm

Convert to kilometers

y=12\ Km

Part 2)

In this part we have

S=\frac{1}{300,000}

y=12\ km  

Find the value of x

x=S*y

Substitute the values

x=(1/300,000)*12

x=0.00004 Km

Convert to centimeters

x=4\ cm

therefore

the answer is

The distance is 4\ cm

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1.) y=3
2.) y=13
     2y + y = 3y , 22+17= 39
     3y = 39 , divide 3 on each side (which cancels out the 3 with the y which leaves you with y)
3.) y=10
     3y - y = 2y , 35-15= 20
     2y=20 , do the same as # 2.
4.) y=4
     9y - 5y = 4y , 48-32= 16
     4y=16, do the same as #2 & #3
5.) (4,5)
     x + 3y = 19
   2x - 3y = -7
    solve for x first ; x + 2x= 3x , 19-7 = 12; 3x = 12 ; divide 3 ; x= 4. You know have to make the x's cancel out. so I will multiply -2 on the first equation. -2(x+3y=19)
     -2x - 6y = -38
      2x - 3y = -7
    Now solve for y ; -6y + -3y = -9y , -38 + -7 = -45; -9y=-45 ; divide by -9 (since they are both negative your answer will be positive) ; y= 5
6.) (6,8)
   x + 4y = 38
  -x - 3y = -30
solve for y first ; 4y - 3y = y , 38-30 = 8; y=8 (that simple!) Now you need to find a common multiple for 4 & 3 which is 12. So you will have to multiply each equation by either 4 or 3.
  (x + 4y = 38)*3      =     3x + 12y = 114         
 (-x - 3y = -30)*4     =     -4x - 12y = -120
solve for x ; 3x - 4x = -x , 114-120= -6 ; -x=-6. Since the x has a negative that means there's still a 1 there so -1x=-6 ; you will need to divide this which makes the 6 a positive; x=6


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