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Yakvenalex [24]
3 years ago
11

The base of a building is shaped like a parallelogram. The first floor has an area of 1575 square feet. If the base of this para

llelogram is 75 feet, can its height be 25 feet? Explain
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
8 0

Answer:

No, the height cannot be 25 feet because the resulting area is inconsistent with the information given.

Step-by-step explanation:

Given that the base of the first floor is 75 feet, we can answer this question in two different ways:

1) Solve for Area using the given height and compare the results to that of the question.

Using the formula for Area:

<em> A = b x h,</em>

<em>A = (75) x (25) = </em><em>1875 </em>

Because the given height results in an area of 1875 feet, and not 1575 feet, the base of this parallelogram cannot be 25 feet.

2) Solve for Height using the given base and Area

Using the formula for Height:

<em>h = A / b,</em>

<em>h = (1575) / (75) = </em><em>21</em><em> ft</em>

The base of this parallelogram could not be 25 feet because when solving for height using the given base and area, the resulting measurement is 21 feet.

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Step-by-step explanation:

Given expression is:

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Putting y = 1/4 in expression

<u></u>16y^2 = 16 *(\frac{1}{4})^2 = 16 * \frac{1}{16} = 1<u></u>

<u></u>

<u>If y=3/8, the values of 16y^2 is 4/9</u>

Putting y = 3/8 in expression

<u></u>16y^2 = 16 *(\frac{3}{8})^2 = 16 * \frac{9}{64} = \frac{9}{4}<u></u>

<u></u>

<u>If y=1/8, the values of 16y^2 is 1/4</u>

Putting y = 1/8 in expression

<u></u>16y^2 = 16 *(\frac{1}{8})^2 = 16 * \frac{1}{64} = \frac{1}{4}<u></u>

<u></u>

<u></u>

<u>If y=2/3, the values of 16y^2 is 3/4</u>

Putting y = 2/3 in expression

<u></u>16y^2 = 16 *(\frac{2}{3})^2 = 16 * \frac{4}{9} = \frac{64}{9}<u></u>

<u></u>

<u>Hence,</u>

The correct statement is:

If y=1/8, the values of 16y^2 is 1/4

Keywords: Expressions, variables

Learn more about expressions at:

  • brainly.com/question/10772025
  • brainly.com/question/10879401

#LearnwithBrainly

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Given:

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In triangle SPT and triangle QPR,

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