Alake near the arctic circle is covered by a 222-meter-thick sheet of ice during the cold winter months. when spring arrives, the warm air gradually melts the ice, causing its thickness to decrease at a constant rate. after 333 weeks, the sheet is only 1.251.251, point, 25 meters thick. let s(t)s(t)s, left parenthesis, t, right parenthesis denote the ice sheet's thickness sss (measured in meters) as a function of time ttt (measured in weeks).

S(t) = 3.82 - 0.2t

Step-by-step explanation:

After 7 weeks, the ice is 2.42 meters thick. The ice loses 0.2 meters of thickness per week; this means on the 7th week, it has lost 7(0.2) = 1.4 meters of thickness.

This means the ice started at 2.42+1.4 = 3.82 meters thick.

Our function will start at the original thickness of the ice, 3.82 meters.

Since the ice is losing thickness, we will subtract; it loses at a rate of 0.2 meters per week (t), which gives us 0.2t. This is subtracted from the original, 3.82 meters, giving us

S(t) = 3.82 - 0.2t

S(t)=−0.25t+2

Step-by-step explanation:

if you're on khan academy this is correct!

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐