All categories

3 years ago

What is the area of a pentagon with apothem of about 10.3

1 answer:

7 0

knowing the pentagon has 5 sides, we can draw 5 equilateral triangles within the pentagon, find the area of one triangle, and multiply that area by5

the length of a side of the pentagon is 2(tan(108)). we multiply 2(tan(108)) with the apothem 10.3 then divide by 2 to get the area of 10.3tan(108). multiply this area by 5 to get the answer of 51.5tan(108) = 127.11, rounded to the nearest hundreth

You might be interested in

HELP!!!!!!!!!!!!!!!!!!!!!!!ME PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

horrorfan [7]

I believe this is the answer
A) 24
B) -3

7 0

3 years ago

12. dress: $70<br> discount: 30%<br> tax: 7%

anyanavicka [17]

Answer:

All you have to do is multiply 45 by 0.70. You will get 31.5, then you subtract 31.5 and 45 to get the answer: $13.50

Step-by-step explanation:

5 0

4 years ago

Find the percent of decrease from 270 to 200. Round to the nearest 10th of a percent if necessary.

choli [55]

270-200 = 70

70/200 = 0.35 = 3.5%

3 0

3 years ago

Verify that each equation is an identity (1 - sin^(2)((x)/(2)))/(1+sin^(2)((x)/(2)))= (1+cosx)/(3-cosX)

Allisa [31]

Answer:

Given that we have;

$sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }$

By the application of the law of indices and algebraic process of adding a and subtracting a fraction from a whole number, we have;

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) } =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

Step-by-step explanation:

An identity is a valid or true equation for all variable values

The given equation is presented as follows;

$\dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

From trigonometric identities, we have;

$sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }$

$\therefore sin^2 \left (\dfrac{x}{2} \right ) = \dfrac{1 - cos (x)}{2}$

$1 - sin^2 \left (\dfrac{x}{2} \right ) = 1 - \dfrac{1 - cos (x)}{2} = \dfrac{2 - (1 - cos (x))}{2} = \dfrac{1 + cos (x))}{2}$

$1 + sin^2 \left (\dfrac{x}{2} \right ) = 1 + \dfrac{1 - cos (x)}{2} = \dfrac{2 + 1 - cos (x))}{2} = \dfrac{3 - cos (x))}{2}$

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

3 0

3 years ago

Find the least common denominator (LCD) for the fractions. 1/12 and 3/4 A. 12 B. 4

kirill [66]

'12' and '48' (choices 'A' and 'D') will both work as common denominators.

The least one is '12' .

Since 4 is a factor of 12, any multiple of 12 is also a multiple of 4 , and can
serve as a common denominator. The smallest multiple of 12 is 12 itself.

4 0

4 years ago

Read 2 more answers