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During a rainfall, the depth of water in a rain gauge increases at a rate modeled by r(t)=0.5+tcos(πt380) r ( t ) = 0.5 + t cos ( π t 3 80 ) , where t is the time in hours since the start of the rainfall and r (t) r ( t ) is measured in centimeters per hour. how much did the depth of water in the rain gauge increase from t = 0 t = 0 to t = 3 t = 3 hours?

Respuesta :

MrRoyal

The function r(t)= 0.5 + t cos(πt³/80) is an illustration of a cosine function

The depth of water in the rain gauge increases by 1.466cm from t = 0 to t = 3

How to determine the increase in water depth?

The function is given as:

r(t)= 0.5 + t cos(πt³/80)

When t = 0, the depth of water is:

r(0)= 0.5 + 0 * cos(π *0³/80)

Evaluate

r(0) = 0.5

When t = 3, the depth of water is:

r(3)= 0.5 + 3 * cos(π *3³/80)

Evaluate

r(3)= 1.966

Calculate the difference (d) in the depths

d = r(3) - r(0)

So, we have:

d = 1.966 - 0.5

Evaluate

d = 1.466

Hence, the depth of water in the rain gauge increases by 1.466cm from t = 0 to t = 3

Read more about cosine functions at:

https://brainly.com/question/17075439

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