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Final Exam Review Quiz Part 1
Question 17 of 18 (1 point) | Question Attempt: 1 of 1
Garrett

-=8
9
10

-=11

-=12

-=13
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-=15

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Español
18
Henry places a bottle of water inside a cooler. As the water cools, its temperature 
C(t) in degrees Celsius is given by the following function, where 
t is the number of minutes since the bottle was placed in the cooler.

C(t)=3+19e^(-0.045 t)
Henry wants to drink the water when it reaches a temperature of 16 degrees Celsius. How many minutes should he leave it in the cooler?
Round your answer to the nearest tenth, and do not round any intermediate computations.
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Chrome\newlineFile\newlineEdit\newlineView\newlineHistory\newlineBookmarks\newlineProfiles\newlineTab\newlineWindow\newlineHelp\newlineMon Apr 2222 33:1717 AM\newlineDashboard\newlineWilliams - MAT 117117: Colleg:\newline(s) Go To ALEKS Homepage\newlineALEKS - Garrett Slaton - Fi\newlineByteLearn\newlineScreenshot 2024202404-0422-22 a\newlinewww-awu.aleks.com/alekscgi/x/IsI.exe/11o_U-IgNslkr77j88P33jH-lis11WHQv77rHh00j22icTeSIC44Q00KrKw00C-WYA33I11XBIkqCRq88Ed101101jSq7171rFf99EB_M...\newlineYouTube home page\newlineChrome Web Store\newlineAll Bookmarks\newlineFinal Exam Review Quiz Part 11\newlineQuestion 1717 of 1818 (11 point) | Question Attempt: 11 of 11\newlineGarrett\newline8 \equiv 8 \newline99\newline1010\newline11 \equiv 11 \newline12 \equiv 12 \newline13 \equiv 13 \newline1414\newline15 \equiv 15 \newline16 \equiv 16 \newline1717\newlineEspañol\newline1818\newlineHenry places a bottle of water inside a cooler. As the water cools, its temperature C(t) C(t) in degrees Celsius is given by the following function, where t t is the number of minutes since the bottle was placed in the cooler.\newlineC(t)=3+19e0.045t C(t)=3+19 e^{-0.045 t} \newlineHenry wants to drink the water when it reaches a temperature of 1616 degrees Celsius. How many minutes should he leave it in the cooler?\newlineRound your answer to the nearest tenth, and do not round any intermediate computations.\newlineminutes\newlineContinue\newlineSave For Later\newlineSubmit Assignment\newline[i. 20242024 McGraw Hill LLC. All Rights Reserved. Terms of Use I Privacy Center I Accessibility

Full solution

Q. Chrome\newlineFile\newlineEdit\newlineView\newlineHistory\newlineBookmarks\newlineProfiles\newlineTab\newlineWindow\newlineHelp\newlineMon Apr 2222 33:1717 AM\newlineDashboard\newlineWilliams - MAT 117117: Colleg:\newline(s) Go To ALEKS Homepage\newlineALEKS - Garrett Slaton - Fi\newlineByteLearn\newlineScreenshot 2024202404-0422-22 a\newlinewww-awu.aleks.com/alekscgi/x/IsI.exe/11o_U-IgNslkr77j88P33jH-lis11WHQv77rHh00j22icTeSIC44Q00KrKw00C-WYA33I11XBIkqCRq88Ed101101jSq7171rFf99EB_M...\newlineYouTube home page\newlineChrome Web Store\newlineAll Bookmarks\newlineFinal Exam Review Quiz Part 11\newlineQuestion 1717 of 1818 (11 point) | Question Attempt: 11 of 11\newlineGarrett\newline8 \equiv 8 \newline99\newline1010\newline11 \equiv 11 \newline12 \equiv 12 \newline13 \equiv 13 \newline1414\newline15 \equiv 15 \newline16 \equiv 16 \newline1717\newlineEspañol\newline1818\newlineHenry places a bottle of water inside a cooler. As the water cools, its temperature C(t) C(t) in degrees Celsius is given by the following function, where t t is the number of minutes since the bottle was placed in the cooler.\newlineC(t)=3+19e0.045t C(t)=3+19 e^{-0.045 t} \newlineHenry wants to drink the water when it reaches a temperature of 1616 degrees Celsius. How many minutes should he leave it in the cooler?\newlineRound your answer to the nearest tenth, and do not round any intermediate computations.\newlineminutes\newlineContinue\newlineSave For Later\newlineSubmit Assignment\newline[i. 20242024 McGraw Hill LLC. All Rights Reserved. Terms of Use I Privacy Center I Accessibility
  1. Set Function Equal: Set the function C(t)C(t) equal to 1616 degrees Celsius to solve for tt.C(t)=3+19e(0.045t)=16C(t) = 3 + 19e^{(-0.045t)} = 16
  2. Subtract to Isolate: Subtract 33 from both sides to isolate the exponential term.\newline19e(0.045t)=1319e^{(-0.045t)} = 13
  3. Divide to Solve: Divide both sides by 1919 to solve for the exponential part.\newlinee(0.045t)=1319e^{(-0.045t)} = \frac{13}{19}
  4. Calculate Decimal Value: Calculate 13/1913 / 19 to get a decimal value.\newlinee(0.045t)0.6842e^{(-0.045t)} \approx 0.6842
  5. Take Natural Logarithm: Take the natural logarithm of both sides to solve for 0.045t-0.045t.\newlineln(e0.045t)=ln(0.6842)\ln(e^{-0.045t}) = \ln(0.6842)
  6. Simplify Left Side: Simplify the left side using the property ln(ex)=x\ln(e^x) = x.0.045t=ln(0.6842)-0.045t = \ln(0.6842)
  7. Calculate Decimal Value: Calculate ln(0.6842)\ln(0.6842) to get a decimal value.\newline0.045t0.3802-0.045t \approx -0.3802
  8. Divide to Solve: Divide both sides by 0.045-0.045 to solve for tt.t0.38020.045t \approx \frac{-0.3802}{-0.045}
  9. Calculate Value of t: Calculate 0.3802/0.045-0.3802 / -0.045 to get the value of tt.\newlinet8.4489t \approx 8.4489

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