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tia_tia [17]
3 years ago
12

On a city​ map, a rectangular park has a length of 2 in. If the actual length and width of the park are 400 ft and 100 ft​, ​res

pectively, how wide is the park on the​ map?

Mathematics
2 answers:
pshichka [43]3 years ago
7 0

Answer:

0.5 inches

Step-by-step explanation:

yan [13]3 years ago
5 0

Answer:

0.5 inches

Step-by-step explanation:

If 400ft = 2in

then 100ft = x

We simply cross multiply

400x = 200

x = 200/400

x = 0.5in

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Solve using long division <br> Please
madreJ [45]

1. Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}

Steps:

\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}

\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}

\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}

\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}

\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}

2. Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}

Steps:

\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}

\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}

\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

4 0
3 years ago
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