how do you have a reflection in the x axis, if the shape is already partially in the third and second quadrent of the graph
1 answer:
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The answer is: " 17.35 " .
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→ The circumference of the circle is: " 17.35 units " .
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The area, "A", of a circle:
A = * r² ;
Note: .
The area of the circumference, "C", of a circle:
C = 2 r .
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Given:
A = 24 ;
Use the formula for the area, "A" of a circle, to find the radius, "r" .
Then, plug the value obtained for "r" into the formula for the circumference, "C" , of a circle:
C = 2 r l
To solve for the circumference, "C", of the circle.
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A = ;
24 = 3.14 * r² ;
↔ 3.14 * r² = 24 ;
Divide each side of the equation by "(3.14)" ;
[3.14 * r²] / 3.14 = (24) / (3.14) ;
to get:
→ r² = 7.64331210191 ;
Take the positive square root of each side of the equation;
to isolate "r" on one side of the equation; & to solve for the radius, "r" ;
→ ⁺√(r²) = ⁺√(7.64331210191) ;
to get:
→ r = 2.76281034802 ;
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Now, plug this "obtained value" for the radius, "r" into the equation for the formula for the circumference, "C" of a circle:
→ C = 2 * r ;
to solve for the circumference, "C", of the circle:
→ C = 2 * (3.14) * (2.76281034802) ;
= (6.28) * (2.76281034802) ;
= 17.3504489856 ;
→ round to: " 17.35 units" .
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The answer is: " 17.35 " .
__________________________________
→ The circumference of the circle is: " 17.35 units " .
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