these are just two semi-circles, one with a diameter of 22 and the other with a diameter of 30 - 22, or 8, and therefore their radius are 11 and 4 respectively.

$\bf \textit{area of a semi-circle}\\\\ A=\cfrac{\pi r^2}{2}~~,~~ \stackrel{r=11}{A=\cfrac{\pi 11^2}{2}}\implies A=\cfrac{121\pi }{2}~~,~~ \stackrel{r=4}{A=\cfrac{\pi 4^2}{2}}\implies A=8\pi \\\\ -------------------------------\\\\ \stackrel{\textit{total area of the figure}}{\cfrac{121\pi }{2}+8\pi }\implies \cfrac{121\pi +16\pi }{2}\implies \cfrac{137\pi }{2}$

4 0

2 years ago

How to determine if a function is one to one algebraically?

Mkey [24]

All you have to do is multiply the least commen donomenator by 2 and that will be your answer

4 0

3 years ago

56+3<br><img src="https://tex.z-dn.net/?f=56%20%2B%203%20%3D%20" id="TexFormula1" title="56 + 3 = " alt="56 + 3 = " align="absmi

tangare [24]

sorry if I'm wrong. I got confused because you put 56+3 twice.×= 63
56+3=59=x-4 (this is with subtraction sign)

I didnt know if that a subtracting or negative so here both

56+3=x-4 (this is with the negative sign)
x= -14.75

6 0

3 years ago