PLEASE HELP ME!!!! I will give you 45 points Because that’s all I have!!!

y = abˣ
20 = ab¹
20 = ab
b b
20/b = a

y = abˣ
4 = (20/b)b²
4 = 20b
20 20
¹/₅ = b

y = abˣ
20 = ¹/₅b¹
20 = ¹/₅b
¹/₅ ¹/₅
100 = b

y = abˣ
y = 100(0.2)ˣ

4 0

3 years ago

What is -15x(-7) ( integer operation)

MA_775_DIABLO [31]

$- 15x( - 7) = 105x$
Not much more to say about this. Just multiply the integer by the coefficient.

6 0

3 years ago

In a certain region, about 6% of a city's population moves to the surrounding suburbs each year, and about 4% of the suburban po

Sedbober [7]

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:

$\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}p_c\\p_s\end{array}\right] = \left[\begin{array}{c}p_c'\\p_s'\end{array}\right]$

- 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:

$\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_k = x_k_+_1$

- 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:

$\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o = x_1$

- 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.

$x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o$

- Where,

$x_o = \left[\begin{array}{c}10,000,000\\800,000\end{array}\right]$

- The population in 2017 x_2 would be:

$x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}10,000,000\\800,000\end{array}\right] \\\\\\x_2 = \left[\begin{array}{c}8,920,800\\1,879,200\end{array}\right]$

5 0

4 years ago