12 y − 18 6 = 4 y + 3 12 y - 18 6 = 4 y + 3

Mathematics

1 answer:

Burka [1]3 years ago

3 0

Step-by-step explanation:

I didn't understood thr question and I'm pretty sure that u made a mistake while writing it

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What's the approximate area of a segment of a circle with a height 6 m and the length of the chord is 20 m? Round your answer to

yuradex [85]

Hello!

The Correct Answer to this would be 100%:

Option "85.4".

(Work Below)

Given:
height = 6m
chord = 20 m

We need to find the radius of the circle.

20 m = 2 √ [ 6m( 2 x radius - 6 m ) ]
20 m / 2 = 2 √[ 6m( 2 x radius - 6 m ) ] / 2
10 m = √ [ 6m( 2 x radius - 6 m ) ]
(10 m)² = √[ 6m( 2 x radius - 6 m ) ] ²
100 m² = 6 m( 2 x radius - 6 m )
100 m² = 12 m x radius - 36 sq m
100 m² + 36 m² = 12 m x radius - 36 m² + 36 m²
136 m² = 12 m x radius
136 m² / 12 m = 12 m x radius / 12 m
11.333 m = radius

the area beneath an arc:

Area = r² x arc cosine [ ( r - h ) / r ] - ( r - h ) x √( 2 x r x h - h² ).

r² = (11.333 m)² = 128.444 m²
r - h= 11.333 m - 6 m = 5.333 m
r * h = 11.333 m x 6 m = 68 m²

Area = 128.444 m² x arc cosine [ 5.333 m / 11.333 m ] - 5.333 m x √[ 2 x 68 m² - 36 m² ]

Area = 128.444 m² x arc cosine [ 0.4706 ] - 5.333 m x √ [ 100m² ]

Area = 128.444 m² x 1.0808 radians - 5.333 m x 10 m

Area = 138.828 m² - 53.333 m²

Area = 85.4 m²


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3 years ago

Read 2 more answers

For circle O, mCD=125° and m

Burka [1]

CA is a diameter of the circle, so $m\widehat{AC}=180^\circ$ , which means $m\angle AOB=m\angle AOD=m\widehat{AD}=180^\circ-m\widehat{CD}=55^\circ$ . Then $m\boxed{\angle ABO}=90^\circ-55^\circ=35^\circ$ .

This means $m\angle CBO=55^\circ-35^\circ=20^\circ$ . Also, if $m\angle AOB=55^\circ$ , then $m\angle BOC=180^\circ-55^\circ=125^\circ$ , which in turns tells us that $m\boxed{\angle BCO}=180^\circ-20^\circ-125^\circ=35^\circ$ .