What is the best approximation for the perimeter of a semicircle with a diameter of 64 meters?

2 answers:

OverLord2011 [107]3 years ago

7 0

By definition, the perimeter of a semicircle is given by:
 $P = ( \pi ) * (r) + d$
 Where,
d: diameter of the semicircle
r: radius of the semicircle.
The radius of the semicircle is given by:
 $r = d / 2 
$
 Where,
d: diameter of the semicircle
Substituting values:
 $r = 64/2 

r = 32 m$ 
Then, the perimeter is:
 $P = (3.14) * (32) + 64 

P = 
 164.48 m$ 
Answer:
The perimeter of the semicurcle is given by:
P = 164.48 m

Natasha_Volkova [10]3 years ago

6 0

Answer: 164.48

Step-by-step explanation:

and don't forget we all have different answers :D

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Solve the following equation; domain 0< x < 360; 3 cot x + sqrt 3 = 0

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ANSWER

$x =120 \degree \: or \: 300 \degree$

EXPLANATION

We want to solve the trigonometric equation

$3 \cot(x) + \sqrt{3} = 0$

We group like terms to obtain,

$3\cot(x) = - \sqrt{3}$

Divide both sides of the equation by 3 to get,

$\cot(x) = - \frac{ \sqrt{3} }{3}$

We can reciprocate both sides to get,

$\tan(x) = - \frac{3}{ \sqrt{3} }$

We simplify the right hand side to get,

$\tan(x) = - \sqrt{3}$

Since the tangent ratio is negative, it implies that it is either in the second quadrant or fourth quadrant.

In the second quadrant,

$x = 180 \degree - arctan( \sqrt{3})$

$x = 180 \degree - 60 \degree = 120 \degree$

In the fourth quadrant,

$x = 360 \degree - arctan( \sqrt{3})$

$x = 360 \degree - 60 \degree$

$x = 300 \degree$