The distance between Eagle Lake and Swallow Lake on a map is 9 cm. The scale of the map is 4 cm = 2 km.What is the actual distance between Eagle Lake and Swallow Lake?
OA. 4.5 km
OB. 5.5 km
OC. 18 km
OD. 72 km

Answers

Answer 1
Answer:

Final answer:

The actual distance between Eagle Lake and Swallow Lake is 4.5 km, calculated by using the given map scale and setting up a proportion.

Explanation:

This question is about calculating actual distances using a map scale. A map scale allows distances on a map to relate to actual distances in real life. The given map scale is 4 cm = 2 km, which means every 4 cm on the map corresponds to 2 km in real life.

Thus, to calculate the actual distance between Eagle Lake and Swallow Lake, we need to apply a simple proportion.

  • First, set up the proportion using the map scale, i.e. 4 cm/2 km = 9 cm/x km.
  • Next, solve for x, which represents the actual distance. Multiply both sides of the equation by x and then by 2 km to isolate x. The equation then becomes: 4x = 18.
  • Finally, divide both sides by 4 to solve for x, which gives us: x = 4.5 km. So, the actual distance between the two lakes is 4.5 km.

Learn more about Map Scale here:

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Answers

To do this, imagine 2/8 for 1/4

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Answers

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Answers

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First we need a conversion equations to get from pounds to ounces

1pound=16ounce

2pounds=2*16=32 ounces of cucumber

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Now that all of the vegetables are in ounces, we can solve for their total weight.


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He bought 103 ounces of vegetables in total.
If you would like to know how many ounces of vegetables did Simon buy in total, you can calculate this using the following steps:

1 pound equals to 16 ounces.
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Answers

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Answers

area of a circle=πr²
area of this cirlce=π*(12 ft)²=144π ft²≈452.39 ft².

answer 1= the area of this circle is 452.39 ft².

Perimeter of a circle=2πr.
Perimeter of this cirlce=2π(12 ft)=24π ft≈75.4 ft

answer 2= the perimeter of this cirlce is 75.4 ft.

Answer:

Area: 452.4

Perimeter: 75.4 ft

Step-by-step explanation:

Formula for area: \pi r^(2)

Formula for perimeter: \pi d or 2\pi r