Check the picture below.
now, let's notice the larger "yellow" semicircle, it has a gap, the gap on the right is of a semicircle with a diameter of 10, BUT it also has a descender on the left, a part that's hanging out, that part is also a semicircle.
so if we use the descending semicircle to fill up the gap on the right, we'll end up with a filled up larger semicircle, whose diameter is 20, and whose radius is 10 cm.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=10 \end{cases}\implies A=\pi 10^2\implies A=100\pi \\\\\\ \stackrel{\textit{half of that for a semicircle}}{A=\cfrac{100\pi }{2}}\implies A=50\pi \implies \stackrel{\pi =3.142}{A=157.1}](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%3D10%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%2010%5E2%5Cimplies%20A%3D100%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%20for%20a%20semicircle%7D%7D%7BA%3D%5Ccfrac%7B100%5Cpi%20%7D%7B2%7D%7D%5Cimplies%20A%3D50%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Cpi%20%3D3.142%7D%7BA%3D157.1%7D)
Answer:
3.8
Step-by-step explanation:
Because its right ur welcome
Answer:
There is a 3/5 chance of the first ball being white, and a 3/10 chance the second one is black.
Step-by-step explanation:
There are 5 balls, of which 3 are white, so you have a 3/5 chance of the first one being white. Then you have 2 white and 2 black balls. There is a 2/4 chance of picking a black ball. Multiply 3/5 and 2/4 to get 6/20, or 3/10 for choosing a white ball then a black ball.
Answer is C
Y-y1=m(x-x1)
y1=-1/2 x=4/5, m=3/4
