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4 years ago

For which academic reasons have you used the Internet? Check all that apply.

Advanced Placement (AP)

2 answers:

slava [35]4 years ago

6 0

a, b, c, d, e, f

They are all correct selections for what many people academically use the internet for.

Ede4ka [16]4 years ago

5 0

<span>A. to research topics
C. to find the definition of words
D. to search for maps or other geographic tools
</span>F. to access practice tests, quizzes, and flash cards

Hope this Helps!!

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3 years ago

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3 years ago

An architect makes a model of a new house. The model shows a tile patio in the backyard. In the model, each tile has length

STALIN [3.7K]

Answer:

$Ratio = \frac{1}{16}$ ----- Model length to Actual length

$Ratio = \frac{1}{256}$ ----- Model Area to Actual Area

Explanation:

Given

Model Measurement:

$Length = \frac{1}{2}\ in$

$Width = \frac{1}{6}\ in$

Actual Measurement:

$Length = \frac{2}{3}\ ft$

$Width = \frac{2}{9}\ ft$

Solving (a): Ratio of actual length to model length

Convert length from feet to inches

$Length = \frac{2}{3}\ ft$

$Length = \frac{2}{3}\ * 12in$

$Length = \frac{24\ in}{3}$

$Length = 8\ in$

The required ratio is calculated by dividing the model length by the actual length

$Ratio = \frac{Model}{Actual}$

$Ratio = \frac{1}{2}\ in/8\ in$

$Ratio = \frac{1}{2}/8$

$Ratio = \frac{1}{2} * \frac{1}{8}$

$Ratio = \frac{1}{16}$

Another possible way is to convert both measurement to a different unit:

Convert model length from inches to cm

$Length = \frac{1}{2}\ in$

$Model\ Length = \frac{1}{2} * 2.54cm$

$Model\ Length = \frac{1 * 2.54cm}{2}$

$Model\ Length = \frac{2.54cm}{2}$

$Model\ Length = 1.27 cm$

Convert actual length from ft to cm

$Length = \frac{2}{3}\ ft$

$Actual\ Length = \frac{2}{3} * 30.48cm$

$Actual\ Length = \frac{2 * 30.48cm}{3}$

$Actual\ Length = \frac{60.96cm}{3}$

$Actual\ Length = 20.32cm$

Then calculate the ratio as:

$Ratio = \frac{Model}{Actual}$

$Ratio = \frac{1.27cm}{20.32cm}$

$Ratio = \frac{1.27}{20.32}$

Simplify fraction

$Ratio = \frac{1.27/1.27}{20.32/1.27}$

$Ratio = \frac{1}{16}$

Solving (b): Model of Area;

Way 1:

In (a), we calculated the ratio of length to be:

$Ratio = \frac{1}{16}$

This is also the model of the width.

So, Ratio of Area is then calculated as:

$Ratio = \frac{1}{16}^2$

$Ratio = \frac{1}{256}$

Way (2): We calculate the actual area of the model and actual measurements.

Model Measurements

$Length = \frac{1}{2}\ in$

$Width = \frac{1}{6}\ in$

$Area = Length * Width$

$Area = \frac{1}{2} * \frac{1}{6} in^2$

$Area = \frac{1}{12}\ in^2$

Actual Measurements

$Length = \frac{2}{3}\ ft$

$Width = \frac{2}{9}\ ft$

$Area = Length * Width$

$Area = \frac{2}{3} * \frac{2}{9} ft^2$

$Area = \frac{4}{27} ft^2$

Convert to $in^2$

$Area = \frac{4}{27} * 144in^2$

$Area = \frac{4}{3} * 16in^2$

$Area = \frac{64}{3}in^2$

Ratio is then calculated as:

$Ratio = \frac{Model}{Actual}$

$Ratio = \frac{1}{12}/\frac{64}{3}$

$Ratio = \frac{1}{12} * \frac{3}{64}$

$Ratio = \frac{1}{4} * \frac{1}{64}$

$Ratio = \frac{1}{256}$

7 0

3 years ago