A cup of tea has a temperature of 100 degrees Celsius. it is cooling in a room that has a temperature of 20 degrees Celsius. it cools in accordance with the following model..
T = temp in degrees Celsius
t = time in minutes
simplify the equation; `T(t)=100-70(1-(1/2)^(t/3))`
, then using the simplified equation from state the transformations based on ``
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We are given the cooling model of `T(t)=100-70(1-(1/2)^(t/3))`
We can rewrite this:
So the simplified model is `T(t)=70(1/2)^(t/3)+30`
The domain is `t>=0` as the physical model only makes sense for positive time.
The range is `30<T(t)<=100` . The function starts at 100, and approaches 30 as t increases without bound. Note that `70(1/2)^(t/3)>0` for all t.
** As stated, your problem probably has a typo. If the room is `20^@` C then the liquid will approach 20 degrees, not 30 degrees as t increases. **
The T intercept is where t=0 so it is 100.
There is no t-intercept as the function is always greater than 30.
There is a horizontal asymptote at T=30.
Using `(1/2)^t` as the base function the transformations are as follows:
A vertical stretch of factor 70.
A horizontal stretch of factor 1/3.
A vertical translation (shift) up 30 units.
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