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hichkok12 [17]
4 years ago
11

On a road map that uses a scale of 1inch:20 miles, two cities are 7 inches apart. if you are making a map that uses a scale of 3

inches:35 miles, how far apart are the same two cities on your map?
Mathematics
2 answers:
Lady bird [3.3K]4 years ago
7 0
Well, we know that there are 2 cities. The original map is having 1 inch represent 20 miles. The cities are 7 inches apart. First off, you need to know how far apart in MILES they are.

20*7=140, so the cities are 7 inches apart, or 140 miles.

So, you want to make an identical map, but you want your measurements to be a bit more spaced. we know YOUR map will have every 3 inches representing 35 miles, but we need to know how far apart in inches will 140 miles be.
Well, we simply take 140 and divide it by 35 then!
140/35=4
Wait, your not done yet, The 4 only represents how many "35 mile" increments you need to get to 140. This means the number representing inches also has to be multiplied by this 4 to get your final answer.
3*4=12
This means on YOUR map, the 2 cities will be 12 inches apart.

~Hope this helps!
eduard4 years ago
4 0
7 X 20 is 140 so they're 140 miles apart in real life, if the scale is 3:35 then divide by 3 to make it 11.66667 then times 11.66667 by 140 I think to get the answer ;)
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Step-by-step explanation:

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3 years ago
I need help on 22 and 24
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4 years ago
Plz help me with this...there might be errors in this but I'm not sure ​
Flura [38]

Step-by-step explanation:

1.  cot(A − B)

From angle sum/difference equations, we know this equals:

cot(A − B) = (cot A cot B + 1) / (cot B − cot A)

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cot(A − B) = 1 / tan(A − B)

cot(A − B) = (1 + tan A tan B) / (tan A − tan B)

Setting the two equal to each other:

(cot A cot B + 1) / (cot B − cot A) = (1 + tan A tan B) / (tan A − tan B)

Writing tangent in terms of cotangent:

(cot A cot B + 1) / (cot B − cot A) = (1 + 1 / (cot A cot B)) / (tan A − tan B)

(cot A cot B + 1) / (cot B − cot A) = ((cot A cot B + 1) / (cot A cot B)) / (tan A − tan B)

(cot A cot B + 1) / (cot B − cot A) = (cot A cot B + 1) / ((cot A cot B) (tan A − tan B))

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Solve for cot A cot B:

cot A cot B = (cot B − cot A) / (tan A − tan B)

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cot(A − B) = (a / b + 1) / a

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2. cos A / (1 ± sin A)

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From half angle formula, this equals:

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And finally, phase shift:

tan (-45 ∓ A/2 + 90)

tan (45 ∓ A/2)

Check that you wrote the problem correctly.  Looks like you switched ∓ with ±.

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