40 POINTS!!!!! A submarine descends to 1/6 of its maximum depth. Then it descends another 2/3 of its maximum depth. If it is now
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The picture in the attached figure
we know that
area of a sector=(∅*pi/360°)*r²--------> when ∅ is in degree
in this problem
∅=120°
r=4 units
so
area of a sector=(120°*pi/360°)*4²-------> (120°/360°)*(16*pi) units²
The <span>expressions to find the area of the shaded sector is
</span>(120°/360°)*(16*pi) units²
(1/3)*(16*pi)----> (16/3)*pi units²
the area of the shaded sector is (16/3)*pi units²
we know that
area of a sector=(∅*pi/360°)*r²--------> when ∅ is in degree
in this problem
∅=120°
r=4 units
so
area of a sector=(120°*pi/360°)*4²-------> (120°/360°)*(16*pi) units²
The <span>expressions to find the area of the shaded sector is
</span>(120°/360°)*(16*pi) units²
(1/3)*(16*pi)----> (16/3)*pi units²
the area of the shaded sector is (16/3)*pi units²