Answer:
(x, y) = (4, 9)
Step-by-step explanation:
You can use the second equation to write an expression for y.
y = 10 - 1/4x . . . . subtract 1/4x from the given 2nd equation
Now, substitute that for y in the first equation:
1/2x + 1/3(10 -1/4x) = 5
1/2x + 10/3 -1/12x = 5 . . . . eliminate parentheses
You can work this a couple of ways from here. One is to multiply by 12 to eliminate fractions. Another is to work with the numbers as they are. We'll do that, because we don't get enough practice doing arithmetic with fractions.
5/12x = 5/3 . . . . . subtract 10/3
x = (5/3)/(5/12) = 12/3 = 4
From above, we can use our expression for y:
y = 10 - 1/4·4 = 9
The solution is (x, y) = (4, 9).
Answer:
230
Step-by-step explanation:
54+4×22×2
54+88×2
54+176
230
Step-by-step explanation:
Correct option is
D
a
=0,b
=0
The condition for ax+by+c=0 to be a linear eqaution in two variables is
a
=0,b
=0
Answer:
City @ 2017 = 8,920,800
Suburbs @ 2017 = 1, 897, 200
Step-by-step explanation:
Solution:
- Let p_c be the population in the city ( in a given year ) and p_s is the population in the suburbs ( in a given year ) . The first sentence tell us that populations p_c' and p_s' for next year would be:
0.94*p_c + 0.04*p_s = p_c'
0.06*p_c + 0.96*p_s = p_s'
- Assuming 6% moved while remaining 94% remained settled at the time of migrations.
- The matrix representation is as follows:
- In the sequence for where x_k denotes population of kth year and x_k+1 denotes population of x_k+1 year. We have:
![\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_k = x_k_+_1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20x_k%20%3D%20x_k_%2B_1)
- Let x_o be the populations defined given as 10,000,000 and 800,000 respectively for city and suburbs. We will have a population x_1 as a vector for year 2016 as follows:
![\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o = x_1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20x_o%20%3D%20x_1)
- To get the population in year 2017 we will multiply the migration matrix to the population vector x_1 in 2016 to obtain x_2.
![x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20x_o)
- Where,
![x_o = \left[\begin{array}{c}10,000,000\\800,000\end{array}\right]](https://tex.z-dn.net/?f=x_o%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D10%2C000%2C000%5C%5C800%2C000%5Cend%7Barray%7D%5Cright%5D)
- The population in 2017 x_2 would be:
![x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}10,000,000\\800,000\end{array}\right] \\\\\\x_2 = \left[\begin{array}{c}8,920,800\\1,879,200\end{array}\right]](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D10%2C000%2C000%5C%5C800%2C000%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cx_2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%2C920%2C800%5C%5C1%2C879%2C200%5Cend%7Barray%7D%5Cright%5D)