The position of an object at time t is given by s(t) = -2 - 6t. Find the instantaneous velocity at t = 2 by finding the derivative.
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OpenStudy (michele_laino):
please, you have to calculate the first derivative of s(t), so
what is:
\[\frac{ ds }{ dt }=...\]
OpenStudy (anonymous):
is the answer -6?
OpenStudy (anonymous):
@SolomonZelman
OpenStudy (michele_laino):
yes!
OpenStudy (anonymous):
ok. just changed the -6t to x and did d/dx(-6x) and got -6 and the -2 is 0
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OpenStudy (anonymous):
can i ask another question? @Michele_Laino
Use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 7 squared as x approaches 7.
OpenStudy (michele_laino):
ok!
OpenStudy (anonymous):
or 1/(x-7)^2
OpenStudy (anonymous):
finding the limit and and vertical asymptotes. i got 1 as the limit and 7 as any vertical asymptote
OpenStudy (anonymous):
is that correct?
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OpenStudy (michele_laino):
vertical asymptote x=7 is right!
limit when x is going to?
OpenStudy (anonymous):
x->7
OpenStudy (anonymous):
\[x\]
OpenStudy (anonymous):
\[x \rightarrow 7\]
OpenStudy (michele_laino):
we can only calculate this limit:
\[\lim _{x \rightarrow 7+}\frac{ 1 }{ \sqrt{x-7} }=+ \infty\]
since domain of your function is the set of all real x such that: x>7
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OpenStudy (michele_laino):
namely limit when x goes to 7 from right!
OpenStudy (anonymous):
ok. thanks for the help. i'll give you a medal and fan you if you want.
OpenStudy (michele_laino):
thanks!
OpenStudy (anonymous):
that sounds wrong. *i will become a fan
OpenStudy (michele_laino):
ok!
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