Use the distance formula to find 2 sides. O through A and O through B.
distance formula is in the image.
OA
root[(x2-x1)^2+(y2-y1)^2]
root[(4-0)^2+(0-0)^2]
root[16]
4
OB
root[(x2-x1)^2+(y2-y1)^2]
root[(0-6)^2+(6-0)^2]
root[36+36]
root[72]
6root[2]
multiply the two numbers and divide by 2.
4 * 6root[2]
24root[2]
24root[2]/2
12root[2]
I think the answer is £53, sorry if I am wrong.
Write each vector as a linear combination of the vectors in s. s = {(2, 0, 7), (2, 4, 5), (2, −12, 13)} (a) u = (−1, 7, −7)
Paladinen [302]
The vectors in s are linearly dependent, but you can make u from
.. (2, 4, 5)/16 -9*(2, -12, 13)/16 = (-1, 7, -7)
<h2>
Option C is the correct answer,</h2>
Step-by-step explanation:
The distance of planet Jupiter from the Sun is approximately 7.8 x 10⁸ kilometers.
Distance from Jupiter to Sun = 7.8 x 10⁸ kilometers
The distance of planet Saturn from the Sun is 1.5 x 10⁹ kilometers.
Distance from Saturn to Sun = 1.5 x 10⁹ kilometers
Difference in distances = 1.5 x 10⁹ - 7.8 x 10⁸ = 15 x 10⁸ - 7.8 x 10⁸
Difference in distances = 7.2 x 10⁸ kilometers
Option C is the correct answer,
Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20J%28%5Cstackrel%7Bx_1%7D%7B-3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5Cqquad%20K%28%5Cstackrel%7Bx_2%7D%7B1%7D~%2C~%5Cstackrel%7By_2%7D%7B3%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%201%20-3%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%203%20%2B1%7D%7B2%7D%20%5Cright%29%20%5Cimplies%20%5Cleft%28%5Ccfrac%7B%20-2%20%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%204%20%7D%7B2%7D%20%5Cright%29%5Cimplies%20JK%3D%28-1~~%2C~~2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20L%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-1%7D%29%5Cqquad%20M%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%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%20-1%20%2B5%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-3%20-1%7D%7B2%7D%20%5Cright%29%20%5Cimplies%20%5Cleft%28%5Ccfrac%7B%204%20%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-4%20%7D%7B2%7D%20%5Cright%29%5Cimplies%20LM%3D%282~~%2C~~-2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
now, let's check the other path, JM and KL
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20J%28%5Cstackrel%7Bx_1%7D%7B-3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5Cqquad%20M%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%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%20-1%20-3%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-3%20%2B1%7D%7B2%7D%20%5Cright%29%20%5Cimplies%20%5Cleft%28%5Ccfrac%7B%20-4%20%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-2%20%7D%7B2%7D%20%5Cright%29%5Cimplies%20JM%3D%28-2~~%2C~~-1%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20K%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20L%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%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%205%20%2B1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-1%20%2B3%7D%7B2%7D%20%5Cright%29%20%5Cimplies%20%5Cleft%28%5Ccfrac%7B%206%20%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%202%20%7D%7B2%7D%20%5Cright%29%5Cimplies%20KL%3D%283~~%2C~~1%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
so the red path will be 
