Question

A lake 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.

Answer 1

y = 2 - 0.25x

Explanation:

A lake near the Arctic Circle is covered by a 2-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.

S(t) denote the ice sheet's thickness S ( measured in meters) as a function of time (measured in weeks).

Therefore the equation formed will be linear.

The equation will be of the form y = mx + b

Here S(t) = mt + b

Here m is the slope which is the rate at which ice is melting.

Putting t = 0

S(t) = 2

Putting t = 3,

S(t) = 1.25

Therefore, m*0 + b = 2 or, b = 2

and 3m + b = 1.25

or, 3m = 1.25 - 2 = -0.75

or, t = -0.25

Hence, function's formula = S(t) = -0.25*t + 2

i.e. S(t) = 2 - 0.25*t

Answer 2

S(t) = 2 - 0.25*t

Explanation:

A lake near the Arctic Circle is covered by a 2-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.

S(t) denote the ice sheet's thickness S ( measured in meters) as a function of time (measured in weeks).

Therefore the equation formed will be linear.

The equation will be of the form y = mx + b

Here S(t) = mt + b

Here m is the slope which is the rate at which ice is melting.

Putting t = 0

S(t) = 2

Putting t = 3,

S(t) = 1.25

Therefore, m*0 + b = 2 or, b = 2

and 3m + b = 1.25

or, 3m = 1.25 - 2 = -0.75

or, t = -0.25

Hence, function's formula = S(t) = -0.25*t + 2

i.e. S(t) = 2 - 0.25*t