Answer:
One
Step-by-step explanation:
Given data
The capacity of Jug= 6L
Needed capacity of Juice= 5L
Hence one Jug of Juice will be enough because one jug of juice will cover for 6L and 1 L will actually remain
Answer:
50000000
Step-by-step explanation:
Do it yourself
Answer:
<em>option</em><em> </em><em>A </em><em>is </em><em>correct</em><em> </em><em>!</em>
<em>hope </em><em>this </em><em>answer </em><em>helps </em><em>you </em><em>dear.</em><em>.</em><em>.</em><em>take </em><em>care!</em>
All you need to do is multiply it so it will be 1,791
Check the picture below, so the circle looks more or less like that one.
well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-5%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%5Ccfrac%7B%203%20-5%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%205%20%2B%209%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B-2%7D%7B2%7D~~%2C~~%5Ccfrac%7B14%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cstackrel%7Bcenter%7D%7B%28-1~~%2C~~7%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-5%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B3%20-%20%28-5%29%5D%5E2%20%2B%20%5B5%20-%209%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%283%2B5%29%5E2%2B%28-4%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B8%5E2%2B16%7D%5Cimplies%20d%3D%5Csqrt%7B80%7D%5Cimplies%20d%3D4%5Csqrt%7B5%7D~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bhalf%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%5Csqrt%7B5%7D%7D%7B2%7D%5Cimplies%20%5Cunderset%7Bradius%7D%7B2%5Csqrt%7B5%7D%7D%7D)