#help plz

116 messages · Page 1 of 1 (latest)

coarse needleBOT
edgy trench
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!status

coarse needleBOT
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
obsidian bobcat
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3

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my answer was 94

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but turns out the answer is wrong

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so i was hoping you could please help me out

edgy trench
coarse needleBOT
obsidian bobcat
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i did the average temperature formula using 1/b-a ∫^b and a below T(t) dt

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∫e^-kt/2 dt= -2/k e^-kt/2

edgy trench
obsidian bobcat
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and i then substituted k= 1/32

obsidian bobcat
edgy trench
obsidian bobcat
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52

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yea sorry

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amd then i simplified after getting the answer e^-29/64 = to 0.641

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and added the inrgrals

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which got be 2575

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and then i divided the interval 2575/29 which got me 94

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but its wrong

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so

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so what did i do wrong?

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@edgy trench

edgy trench
obsidian bobcat
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i did that because it was already part of the functions exponent

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the rule for exponentials adds a factor

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did i do that wrong?

edgy trench
edgy trench
obsidian bobcat
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the exponent

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i used that

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i did it wrong

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nvm

edgy trench
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But where is the division by 2 coming from?

obsidian bobcat
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shooy

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i was told to divided the exponent by 2 by someone

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they said something like the division comes from the exponent -kt/2

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guess they were wrong

edgy trench
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Yeah that makes no sense. Sorry

obsidian bobcat
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its okay

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so what am i supposed to do then?

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should i ask somebody else?

edgy trench
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Just get rid of that division by 2

obsidian bobcat
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∫e^-kt/2 dt= -2/k e^-kt/2

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so this

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but just take out the divided by 2

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∫e^-kt dt= -2/k e^-kt

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like thid?

edgy trench
obsidian bobcat
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i wrote it wrong then

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because i think i got confused with the -kt/2

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so i might have flipped it

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i messed up

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i thought i was doing it right with the -kt/2

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and the -2

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can you please guide me to the right formula by any chance

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so i can plug in the numbers and try to answer ir correctly

edgy trench
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$\overline{T}=\frac 1{t_2-t_1}\int_{t_1}^{t_2} T(t)dt$

tropic spearBOT
edgy trench
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$t_1=0$\
$t_2=29$\
$T(t)=27+78e^{-kt}$

tropic spearBOT
edgy trench
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@obsidian bobcat ☝️

obsidian bobcat
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ok ok

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let me try it out

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okay so i think i got the answer

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1/29-0 ∫29 above 0 (27+78e^-kt) dt

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which = 1/29 ∫29 above 0 (27+78e^-t/52) dt

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and then i split the intergal

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which got me 783

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and the second term which got me 78 times 52 (1-e^-29/52)

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and after the substituing

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i got 1720.7

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and i added both terms up

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783 + 1720.7

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which gives me 2503.7

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and i did 2503.7/29

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which got me 86.34

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is that correct

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or still wrong?

edgy trench
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,calc 29*27

tropic spearBOT
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Result:

783
edgy trench
obsidian bobcat
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are you sure

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that looks right does it not?

edgy trench
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No it does not

edgy trench
obsidian bobcat
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ooh okay

edgy trench
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@obsidian bobcat mevermind you double canceled the negative

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I did not notice

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You did that correctly

obsidian bobcat
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omg

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okay good

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thanks

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lol

edgy trench
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My mistake

obsidian bobcat
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no worries

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i made lots of mistakes too lol

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so is my answer correct?

edgy trench
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,calc 7852(1-e^(-29/52))

tropic spearBOT
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Result:

1733.8233463855
edgy trench
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But maybe not enough to worry

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,calc (1733.8+783)/29

tropic spearBOT
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Result:

86.786206896552
edgy trench
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Your answer is accurate to the correct number of significant figures

obsidian bobcat
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so 86.34 if correct?

edgy trench
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It is an acceptable answer. It should be at least

obsidian bobcat
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okay perfect thank you so much

edgy trench
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Happy to help

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Sorry for the delays, as i am presently occupied elsewhere

obsidian bobcat
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its alright