Melody

If 2x − 2 ≤ f(x) ≤ x^2 − 2x + 2 for x ≥ 0,

find  lim x→2 f(x).

sub in x=2

\(4-2\le f(2)\le4-4+2\\ 2\le f(2)\le2\\ f(2)=2\)

I missed the word 'complex' and soved for reals.

U got that the value coudl be 6.5 or 20.5 so the smallest is 6.5.

I have not checked this answer and I  know it is not what you asked for anyway.  

We do like people to show appreciation, that is true.

And I thank you for yours.

HOWEVER

We also want to help people learn. When people ask a lot of questions in a short period of time as you have done it is fairly clear to me that you are primarily using this site as a cheat site. You are getting our well meaning answerers to do your homework for you.

This assessment is backed up by your not asking ANY of the answerers any questions at all about their solutions. You have not interacted with them in any way.

It depresses me when this site is used/abused like this.

I do not think that you have posted the whole question.

Thanks Max, I am really pleased that you have answered this :))

Это будет зависеть от того, какие месяцы.
Но 60 месяцев - это 5 лет

5*365 = 1825

У февраля есть дополнительный день, поэтому вы должны добавить 1 или 2 к этому числу.

1826     или же      1827

Thanks Chris and Alan. 

Asad,

Rom was not trying to bully you. 

He was responding in anger to what he saw as rudeness from you.

I thought what you wrote was rude too but I do not think you intended to be rude.

However, no matter what degree of intent you had, your post was rude.

It was understandable for Rom to respond the way he did.

Now accept Rom's response as a lesson and learn from it !

Hi Lightning,

A right square pyramid with base edges of length units each and slant edges of length 10 units each is cut by a plane that is parallel to its base and 3 units above its base. What is the volume, in cubic units, of the new pyramid that is cut off by this plane?

You have the edge slant height 10, and the base length  8sqrt2 

Using pythagoras you can get the slant height in the middle of each triangle (not that the sides are isoceles triangles)

this will be 

\(\sqrt{10^2-(4\sqrt2)^2}=\sqrt{100-32}=\sqrt{68}\;units\)

Now you can use pythagoras again to get the perpendicular height of the entire pyramid.

\(=\sqrt{68-(4\sqrt2)^2}=\sqrt{68-32}=\sqrt{36}=6 units.\)

sionce you are halving the height when you cut the top off, you can halve all the other dimensions for the little pyramid too.

\(volume\;\;of \;\;little\;\;pyramid\\ =\frac{1}{3}*(4\sqrt2)^2*3\\ =\frac{1}{3}*32*3\\ =32\;\;units^3\)

See if you can draw the pics that should go with tihs explanation :)