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:
Use inspect element then click on the question and wabam theres the answer
Step-by-step explanation:
Given an endpoint of a segment and a midpoint, the other endpoint can be obtained by manipulation of the midpoint formula. The said formula is shown below:
Let: (a,b) = coordinates of point 1 ; (c,d) = coordinates of point 2; (e,f) = coordinates of the midpoint
Midpoint = ( (a+c)/2 , (b + d)/2 )
From the formula: (a+c)/2 = e ; (b + d)/2 = f
Since we are already given an endpoint and the midpoint, we can solve for the other endpoint using the obtained equations. This is shown below:
(a+c<span>)/2 = e
</span>(3 + c)/2 = 0
c = -3
(b + d<span>)/2 = f
</span>(11 + d)/2 = 0
d = -11
Therefore, the coordinates of the other point is Q(-3,-11)
Answer:
(7-5i)( 2+3i)
=7(2)+ 7(3i)-5i(2) -(5i)(3i)
= 14 +21i -10i +15
= 29 +11i