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VladimirAG [237]
3 years ago
14

Calculate the area of the regular pentagon below:

Mathematics
2 answers:
anyanavicka [17]3 years ago
4 0
Area of the regular pentagon is 532.40 SQUARE INCHES.

Finding the Area of a regular pentagon, use this formula:

A = 1/4 √5(5+2√5) a²

where:  a = side length.

A = 1/4 √5(5+2√5) (17.6in)²
A = 1/4 √5(5+2(2.24) 309.76 in² 
A = 1/4 √5(5+4.48) * 309.76 in²
A = 1/4 √5(9.48) *  309.76 in²
A = 1/4 √47.4 * 309.76 in²
A = 1/4 * 6.88 * 309.76 in²
A = (1*6.88*309.76in²)/4
A = 2131.15 / 4
A = 532.79 in²       
kvasek [131]3 years ago
4 0
Your pentagon can be decomposed in 10 triangles (see the attached picture). Each triangle has an area \frac{17.6*12.1}4. Hence the total area is 10*\frac{17.6*12.1}4=\boxed{532.4\text{ in}}


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Solve in attachment....​
olga2289 [7]

Answer:

A)2

Step-by-step explanation:

we would like to integrate the following definite Integral:

\displaystyle  \int_{0} ^{1} 5x \sqrt{x} dx

use constant integration rule which yields:

\displaystyle  5\int_{0} ^{1} x \sqrt{x} dx

notice that we can rewrite √x using Law of exponent therefore we obtain:

\displaystyle  5\int_{0} ^{1} x \cdot  {x}^{1/2} dx

once again use law of exponent which yields:

\displaystyle  5\int_{0} ^{1}  {x}^{ \frac{3}{2} } dx

use exponent integration rule which yields;

\displaystyle  5 \left( \frac{{x}^{ \frac{3}{2}  + 1  } }{ \frac{3}{2}  + 1} \right)  \bigg|  _{0} ^{1}

simplify which yields:

\displaystyle  2 {x}^{2}  \sqrt{x}   \bigg|  _{0} ^{1}

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our answer is A

8 0
3 years ago
Read 2 more answers
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