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Mathematics 23 Online
OpenStudy (anonymous):

can someone explain what s is in the following limit?

OpenStudy (anonymous):

OpenStudy (anonymous):

height of said brush at t seconds

OpenStudy (auctoratrox):

The "s(t)" stands for a position function. Does this help?

OpenStudy (anonymous):

at \(t=2,s(t)=-16\times 2^2+100=36\) and so you want \[\frac{36-(-16t^2+100)}{2-t}\]

OpenStudy (anonymous):

we get \[\frac{36+16t^2-100}{2-t}=\frac{16t^2-64}{2-t}\] \[=\frac{16(t^2-4)}{2-t}=\frac{16(t+2)(t-2)}{2-t}\] \[=-16(t+2)\] replace \(t\) by 2 to get your answer

OpenStudy (anonymous):

oh i get the s part, but if i put 2 in for t, that would give me zero...

OpenStudy (anonymous):

you get \(\frac{0}{0}\) until you factor and cancel

OpenStudy (anonymous):

okay got it! thanks! :)

OpenStudy (anonymous):

once you factor and cancel you can replace \(t\) by 2

OpenStudy (anonymous):

btw it always works this way, although in one week you will be taught a snap method and will be able to do this in your head on 3 seconds, and then get annoyed that you had to do it this way

OpenStudy (anonymous):

you will say "the derivative of \(-16t^2+100\) is \(-32t\) replace \(t\) by \(2\) get \(-64\)

OpenStudy (anonymous):

haha i think derivatives are for calc, im only in precal

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