I'm stuck on this one as well The function f(t) = 65 sin (pi over 5t) + 35 models the temperature of a periodic chemical reaction where t represents time in hours. What are the maximum and minimum temperatures of the reaction, and how long does the entire cycle take? maximum: 65°; minimum: 35°; period: pi over 5 hours maximum: 100°; minimum: −30°; period: pi over 5 hours maximum: 65°; minimum: 35°; period: 10 hours maximum: 100°; minimum: −30°; period: 10 hours
I dont really want to download anything
maximum of sine is when the input is \(\frac{\pi}{2}\) (you get 1 in that case, which is the largest sine can be)
set \[\frac{\pi}{5}t=\frac{\pi}{2}\] and solve for \(t\)
where would I start?
ok now i have a question what moron wrote this problem? the question is posed with the measure in radians, but the answer is in degrees!!
another copy and paste for my bad math folder hold on, we can answer in a second
Okay
oh wait, i am a moron, not the person who wrote the problem degrees is the output, it is temperature!!
\[ f(t) = 65 \sin (\frac{\pi}{5}t) + 35 \] we need very little to find the maximum value
I just saw that too
the maximum value of sine is 1 the maximum value of 65 times sine is 65 the maximum value of 65 times sine plus 35 is 100
i will let you find the minimum on your own, starting with the minimum value of sine is -1
Where would I plug it in on the equation you gave me?
as for the period, the period of \(\sin(bx)\) is \(\frac{2\pi}{b}\) put \(b=\frac{\pi}{5}\) so get your answer
not sure what you are asking did you understand these lines: the maximum value of sine is 1 the maximum value of 65 times sine is 65 t he maximum value of 65 times sine plus 35 is 100
I am so lost lol,sorry..math is my weak point
ok lets go real slow
the largest \(\sin(x)\) can ever be is \(1\) , it can never be bigger
okay
if the largest \(\sin(x)\) can be is \(1\), what is the largest \(65\sin(x)\) can be?
the largest would be 1 as well?
let me rewrite it like this suppose \(\sin(x)=1\) then what is \[65\times \sin(x)\]?
it would be 35?
let me try again what is \[65\times 1\]?
65
k good
then finally, what is \[65\times 1+35\]?
100
got it
so 100 is the maximum
now lets see what we know the biggest sine can be is \(1\) so the biggest \[f(t) = 65 \sin (\frac{\pi}{5}t) + 35\]can be is \[65\times1+35=100\]
right
now lets find the minimum, knowing that the minimum value of sine is \(-1\) what do you think the value of \[f(t) = 65 \sin (\frac{\pi}{5}t) + 35\] is if you replace sine by \(-1\) ?
t could be -30?
now we got it!
So how would I apply that to the answer options I have?
i forget what they were, but i hope one had "maximum is 100, minimum is -30"
Yes I had two of those
maximum: 100°; minimum: −30°; period: pi over 5 hours maximum: 100°; minimum: −30°; period: 10 hours
ok so it is one of those now you need the period
I think it would be the first one
would you like to know how to find it?
Yes please
the period of \[\sin(bx)\] is \[\frac{2\pi}{b}\] it is NOT \(b\)
therefore the period of \[\sin(\frac{\pi}{5}t)\] is \[2\pi\div \frac{\pi}{5}=2\pi\times \frac{5}{\pi}=10\]
maximum: 65°; minimum: 35°; period: 10 hours
hmm no
we already found the max and min
ssorry not that one
maximum: 100°; minimum: −30°; period: 10 hours
got it
THANK YOU...you help me more than my online teacher
more importantly i hope the steps were more or less clear on line teacher?
It was more clear..Thank you again!!
yw
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