All of the above. Sorry if this is wrong but in 99% sure it isn’t!!
Answer:
4
Step-by-step explanation:
lim (x^2 - 4) / (x - 2)
x --> 2
When we plug x =2, we get
(2^2 - 4) / (2 - 2)
= (4 - 4)/(2 - 2)
= 0 /0
Which is undefined.
Now we have to use L'hospital rule. Which says we need to differentiate the numerator and the denominator and apply the limit.
When we differentiate x^2 -4, we get 2x
When we differentiate x -2, we get 1
lim 2x/1
x --> 2
Now apply, the limit x = 2
2(2)/1
= 4/1
= 4
Therefore, limit of this function is 4, when x tends to 2.
Hope you will understand the concept.
Thank you.
Answer:
nice to meet u im get over it
Step-by-step explanation:
notice, the circle is missing 1/4, so the area of it is just 3/4 of the whole area of the circle.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=8 \end{cases}\implies A=\pi 8^2\implies A=64\pi \\\\\\ \stackrel{whole}{\cfrac{4}{4}}-\stackrel{one~quarter}{\cfrac{1}{4}}=\cfrac{3}{4}~\hfill \cfrac{3}{4}\cdot 64\pi \implies 48\pi \implies \stackrel{\pi =3.14}{150.72} \\\\\\ ~\hspace{34em}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D8%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%208%5E2%5Cimplies%20A%3D64%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bwhole%7D%7B%5Ccfrac%7B4%7D%7B4%7D%7D-%5Cstackrel%7Bone~quarter%7D%7B%5Ccfrac%7B1%7D%7B4%7D%7D%3D%5Ccfrac%7B3%7D%7B4%7D~%5Chfill%20%5Ccfrac%7B3%7D%7B4%7D%5Ccdot%2064%5Cpi%20%5Cimplies%2048%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7B150.72%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chspace%7B34em%7D)
