Solved on Nov 26, 2023

Cooling liquid temperature model: $T(t) = 80(\frac{1}{2})^{\frac{t}{10}} + 20$ . Sketch graph, find temp. after 40 mins, time to cool to 45 degrees.

STEP 1

Assumptions1. The temperature of the liquid, $(t)$ , is given by the function $(t)=80\left(\frac{1}{}\right)^{\frac{t}{10}}+20$ . . The temperature is measured in degrees Celsius and time, $t$ , is measured in minutes.
3. The function models the cooling of the liquid over time.

STEP 2

To sketch a graph of the function, we need to understand the behavior of the function.This function is an exponential decay function, where the temperature starts at $80 +20 =100$ degrees Celsius and decreases over time, approaching a lower limit of $20$ degrees Celsius.

STEP 3

To sketch the graph, plot the function $(t)=80\left(\frac{1}{2}\right)^{\frac{t}{10}}+20$ on a graph where the x-axis represents time (in minutes) and the y-axis represents temperature (in degrees Celsius).
The graph starts at $(0,100)$ and approaches the line $y =20$ as $t$ goes to infinity.

STEP 4

To calculate the temperature after40 minutes, substitute $t =40$ into the function $(t)$ .
$(40)=80\left(\frac{1}{2}\right)^{\frac{40}{10}}+20$

STEP 5

implify the exponent in the equation.
$(40)=80\left(\frac{1}{2}\right)^{4}+20$

STEP 6

Calculate the value of the expression.
$(40)=80\left(\frac{1}{16}\right)+20$

STEP 7

implify the equation to find the temperature after40 minutes.
$(40)=5+20=25$ So, the temperature after40 minutes is25 degrees Celsius.

STEP 8

To calculate the time taken for the liquid to cool to45 degrees Celsius, set $(t) =45$ and solve for $t$ .
$45=80\left(\frac{1}{2}\right)^{\frac{t}{10}}+20$

STEP 9

Subtract20 from both sides of the equation.
$25=80\left(\frac{}{2}\right)^{\frac{t}{}}$

STEP 10

Divide both sides of the equation by80.
$\frac{25}{80}=\left(\frac{}{2}\right)^{\frac{t}{10}}$

STEP 11

Take the natural logarithm of both sides of the equation.
$\ln\left(\frac{25}{80}\right)=\frac{t}{10}\ln\left(\frac{}{}\right)$

STEP 12

olve for $t$ by multiplying both sides of the equation by $\frac{10}{\ln\left(\frac{}{2}\right)}$ .
$t=\frac{10\ln\left(\frac{25}{80}\right)}{\ln\left(\frac{}{2}\right)}$

STEP 13

Calculate the value of $t$ .
$t\approx16.609640474436812$ So, it takes approximately16.61 minutes for the liquid to cool to45 degrees Celsius.

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Cooling liquid temperature model: $T(t) = 80(\frac{1}{2})^{\frac{t}{10}} + 20$ . Sketch graph, find temp. after 40 mins, time to cool to 45 degrees.

STEP 1

Assumptions1. The temperature of the liquid, $(t)$ , is given by the function $(t)=80\left(\frac{1}{}\right)^{\frac{t}{10}}+20$ . . The temperature is measured in degrees Celsius and time, $t$ , is measured in minutes.
3. The function models the cooling of the liquid over time.

STEP 2

To sketch a graph of the function, we need to understand the behavior of the function.This function is an exponential decay function, where the temperature starts at $80 +20 =100$ degrees Celsius and decreases over time, approaching a lower limit of $20$ degrees Celsius.

STEP 3

To sketch the graph, plot the function $(t)=80\left(\frac{1}{2}\right)^{\frac{t}{10}}+20$ on a graph where the x-axis represents time (in minutes) and the y-axis represents temperature (in degrees Celsius).
The graph starts at $(0,100)$ and approaches the line $y =20$ as $t$ goes to infinity.

STEP 4

To calculate the temperature after40 minutes, substitute $t =40$ into the function $(t)$ .
$(40)=80\left(\frac{1}{2}\right)^{\frac{40}{10}}+20$

STEP 5

implify the exponent in the equation.
$(40)=80\left(\frac{1}{2}\right)^{4}+20$

STEP 6

Calculate the value of the expression.
$(40)=80\left(\frac{1}{16}\right)+20$

STEP 7

implify the equation to find the temperature after40 minutes.
$(40)=5+20=25$ So, the temperature after40 minutes is25 degrees Celsius.

STEP 8

To calculate the time taken for the liquid to cool to45 degrees Celsius, set $(t) =45$ and solve for $t$ .
$45=80\left(\frac{1}{2}\right)^{\frac{t}{10}}+20$

STEP 9

Subtract20 from both sides of the equation.
$25=80\left(\frac{}{2}\right)^{\frac{t}{}}$

STEP 10

Divide both sides of the equation by80.
$\frac{25}{80}=\left(\frac{}{2}\right)^{\frac{t}{10}}$

STEP 11

Take the natural logarithm of both sides of the equation.
$\ln\left(\frac{25}{80}\right)=\frac{t}{10}\ln\left(\frac{}{}\right)$

STEP 12

olve for $t$ by multiplying both sides of the equation by $\frac{10}{\ln\left(\frac{}{2}\right)}$ .
$t=\frac{10\ln\left(\frac{25}{80}\right)}{\ln\left(\frac{}{2}\right)}$

STEP 13

Calculate the value of $t$ .
$t\approx16.609640474436812$ So, it takes approximately16.61 minutes for the liquid to cool to45 degrees Celsius.

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Cooling liquid temperature model: T(t)=80(12)t10+20T(t) = 80(\frac{1}{2})^{\frac{t}{10}} + 20T(t)=80(21​)10t​+20. Sketch graph, find temp. after 40 mins, time to cool to 45 degrees.

STEP 1

STEP 2

To sketch a graph of the function, we need to understand the behavior of the function.This function is an exponential decay function, where the temperature starts at 80+20=10080 +20 =10080+20=100 degrees Celsius and decreases over time, approaching a lower limit of 202020 degrees Celsius.

STEP 3

STEP 4

To calculate the temperature after40 minutes, substitute t=40t =40t=40 into the function (t)(t)(t).(40)=80(12)4010+20(40)=80\left(\frac{1}{2}\right)^{\frac{40}{10}}+20(40)=80(21​)1040​+20

STEP 5

implify the exponent in the equation.(40)=80(12)4+20(40)=80\left(\frac{1}{2}\right)^{4}+20(40)=80(21​)4+20

STEP 6

Calculate the value of the expression.(40)=80(116)+20(40)=80\left(\frac{1}{16}\right)+20(40)=80(161​)+20

STEP 7

implify the equation to find the temperature after40 minutes.(40)=5+20=25(40)=5+20=25(40)=5+20=25So, the temperature after40 minutes is25 degrees Celsius.

STEP 8

To calculate the time taken for the liquid to cool to45 degrees Celsius, set (t)=45(t) =45(t)=45 and solve for ttt.45=80(12)t10+2045=80\left(\frac{1}{2}\right)^{\frac{t}{10}}+2045=80(21​)10t​+20

STEP 9

Subtract20 from both sides of the equation.25=80(2)t25=80\left(\frac{}{2}\right)^{\frac{t}{}}25=80(2​)t​

STEP 10

Divide both sides of the equation by80.2580=(2)t10\frac{25}{80}=\left(\frac{}{2}\right)^{\frac{t}{10}}8025​=(2​)10t​

STEP 11

Take the natural logarithm of both sides of the equation.ln⁡(2580)=t10ln⁡()\ln\left(\frac{25}{80}\right)=\frac{t}{10}\ln\left(\frac{}{}\right)ln(8025​)=10t​ln(​)

STEP 12

olve for ttt by multiplying both sides of the equation by 10ln⁡(2)\frac{10}{\ln\left(\frac{}{2}\right)}ln(2​)10​.t=10ln⁡(2580)ln⁡(2)t=\frac{10\ln\left(\frac{25}{80}\right)}{\ln\left(\frac{}{2}\right)}t=ln(2​)10ln(8025​)​

STEP 13

Calculate the value of ttt.t≈16.609640474436812t\approx16.609640474436812t≈16.609640474436812So, it takes approximately16.61 minutes for the liquid to cool to45 degrees Celsius.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Cooling liquid temperature model: $T(t) = 80(\frac{1}{2})^{\frac{t}{10}} + 20$ . Sketch graph, find temp. after 40 mins, time to cool to 45 degrees.

To sketch a graph of the function, we need to understand the behavior of the function.This function is an exponential decay function, where the temperature starts at $80 +20 =100$ degrees Celsius and decreases over time, approaching a lower limit of $20$ degrees Celsius.

To calculate the temperature after40 minutes, substitute $t =40$ into the function $(t)$ .
$(40)=80\left(\frac{1}{2}\right)^{\frac{40}{10}}+20$

implify the exponent in the equation.
$(40)=80\left(\frac{1}{2}\right)^{4}+20$

Calculate the value of the expression.
$(40)=80\left(\frac{1}{16}\right)+20$

implify the equation to find the temperature after40 minutes.
$(40)=5+20=25$ So, the temperature after40 minutes is25 degrees Celsius.

To calculate the time taken for the liquid to cool to45 degrees Celsius, set $(t) =45$ and solve for $t$ .
$45=80\left(\frac{1}{2}\right)^{\frac{t}{10}}+20$

Subtract20 from both sides of the equation.
$25=80\left(\frac{}{2}\right)^{\frac{t}{}}$

Divide both sides of the equation by80.
$\frac{25}{80}=\left(\frac{}{2}\right)^{\frac{t}{10}}$

Take the natural logarithm of both sides of the equation.
$\ln\left(\frac{25}{80}\right)=\frac{t}{10}\ln\left(\frac{}{}\right)$

olve for $t$ by multiplying both sides of the equation by $\frac{10}{\ln\left(\frac{}{2}\right)}$ .
$t=\frac{10\ln\left(\frac{25}{80}\right)}{\ln\left(\frac{}{2}\right)}$

Calculate the value of $t$ .
$t\approx16.609640474436812$ So, it takes approximately16.61 minutes for the liquid to cool to45 degrees Celsius.