easy algebra question help plz
If itss JUST SOOOO EASY....... jkjk
Whats the q?
the question?
\[\frac{ 9 }{ h^3 }\]
simplify tht
ever heard of mathway?
Alright
thts the question
alright
ok so do you how to evaluate th eproblem or no?
not at all lol
I have like 6 questions just like this one
any answer choices
yea hold on
9 - h^3
for those who cant see it clearly
here is an example problem that i just did? idk if it will help or not but in my opinion it should So does this help? There are at least two ways of doing this: Either you can use de l'Hôpital's rule, and as I pointed out in the comments the third example on Wikipedia gives the details. I think a better way of doing this (and Jonas seems to agree, as I saw after posting) is to write f(h)=3h=elog3⋅h and write the limit as limh→0f(h)−f(0)h and recall the definition of a derivative. What comes out is f′(0)=log3.
@chrisdbest i know ur obssesed with her but im going to ask you to stay on topic
@Alexandra675 do you know what the l'Hôpital's rule, is?
How
@Muzzack How
@chrisdbest please discuss that somewhere else
\[\frac{ h^3 }{ 9}\] \[\frac{ 9 }{ h^3 }\] \[9h^3\] This expression is already simplified.
@Alexandra675 you didnt answer the question do you know what l'Hôpital's rule, is or means?
no
sorry I was busy writing the euations for answer choices
here is the definiton of it In calculus, l'Hôpital's rule uses derivatives to help evaluate limits involving indeterminate forms. Application of the rule often converts an indeterminate form to a determinate form, allowing easy evaluation of the limit. That should clear up some of the confusion?
ummmm
im not in calculus
Why are we diccussing calculus on an algebra q?
): sorry ill be going now
@gamer56 & ur telling me....
lol
@chrisdbest kk stalker. uhum
Whats with people defending themselves? Its kindof funny
The given expression is already simplified. I don't see what you can do to it to simplify it more.
its \[\frac{ 9 }{h^3 }\] @mathstudent55
Yes, that's what I meant. \(\dfrac{9}{h^3} \) is a simplified expression. I don't see what can be done to simplify it further.
the starting equation was \[\frac{ 9 }{ h^{-3}}\]
@mathstudent55
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