My question is: Prove that 1/9 <= sqrt(66)-8 <= 1/8, without solving for sqrt(66). The only hint I have is to use the mean value theorem. Please don't cheat, as I have to show my work, and this question is due in 24 hours.
My question is: Prove that 1/9 <= sqrt(66)-8 <= 1/8, without solving for sqrt(66). The only hint I have is to use the mean value theorem. Please don't cheat, as I have to show my work, and this question is due in 24 hours.
I'm not sure how to do this using the 'mean value theorem', I'm pretty sure the mean value theorem goes something like: (roughly) when considering an arc between two endpoints of a function, at least one point exists where tangent to the arc is parallel to the line connecting the endpoints.
However, this is how I would approach your problem:
With the expression in this form it's easier to argue the following:
As well as:
So:
It must be true that:
Thanks a lot! I imagine that you're still on the site, so here's another:
The area between two varying concentric circles is at all times 9pi inches^2. The rate of change of the area of the larger circle is 10pi inches^2/sec. How fast is the circumference of the smaller circle changing when it has area 16pi inches^2?
Im using large letters for larger circle and small letters for smaller circle
A-a=9pi dA/dt-da/dt=0 dA/dt=10 So da/dt=10 Whena=16pi pi(r^2)=16pi So r=4
We want dc/dt
dc/dt=da/dt*dc/da c=2pir and a=pir^2 so get dc/da from dc/dr*dr/da and finish off