Answer:
The number of different computer systems possible is 1440.
Step-by-step explanation:
For each computer, there are 10 options of monitor.
For each monitor, there are 8 printers.
For each printer, there are 2 scanners.
There are 9 computers.
Determine the number of different computer systems possible.
9*10*8*2 = 1440
The number of different computer systems possible is 1440.
Answer:
10a + 6
Step-by-step explanation:
3a + 4 + 2a - 1
5a + 3
2( 5a + 3)
10a + 6
<h3>
Answer: 3 m/s^2</h3>
=======================================================
According to Newton's Second Law, we know that
F = m*a
where F is the force applied, m is the mass and 'a' is the acceleration.
We see that this is a direct variation equation for F and a, such that m is the constant of variation. It's similar to how y = kx is also a direct variation equation.
Plug in F = 35 and a = 5 to find m
F = ma
35 = m*5
35/5 = m
7 = m
m = 7
The object has a mass of 7 kg
Our equation F = ma updates to F = 7a
Now plug in the force F = 21 to find 'a'
F = 7a
21 = 7a
21/7 = a
3 = a
a = 3
The acceleration will be 3 m/s^2
Notice how a smaller force applied means that the acceleration has also gone down as well.
Step-by-step explanation:
Below is an attachment containing the solution.
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 
