Answer:
If your vertex is at the orgin you can't have a point on (6,0)
Step-by-step explanation:
Answer:
Randomized block design
Step-by-step explanation:
In the scenario described above, the subjects which are the different car types are grouped into fixed units of 4 called blocks with each unit then assigned randomly to a particular treatment condition, which are the 4 different types of tire. In the end, the mean mileage for each block is taken and compared to that of the other blocks. This type of research design improves the reliability of the result obtained as it also eliminates the occurrence of systematic error in the course of our experiment.
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 
Hector accidentally subtracted -2x from 7x instead of adding it. Therefore, in step 2, the correct equation should be
9x = 15
Answer:
Ok, I'm so so so sorry if I get this wrong, but I'ma try.
A.
Juan Increase 25/350 = .071 * 100 = 7 %
Rhia Increase 75/275 = .273 * 100 = 27 %
Max Decrease 40/400 = .1 * 100 = 10 %
B.
Juan sold more than $355 but his increase was less than 10% so he did not get a bonus.
Rhia had an increase of more than 10% but she did not sell more than $355 so she did not get a bonus.
Max had a decrease so he did not get a bonus.
