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lana [24]
2 years ago
7

the length of the rectangle is 10 centimetre more than its breadth if the perimeter of a rectangle is 30 cm find the length and

breadth of the rectangle the answer is Length-12.5 Bredth-2.5​
Mathematics
1 answer:
Artemon [7]2 years ago
3 0

Step-by-step explanation:

if you have the answer, why do you ask ?

anyway,

l = length

b = breadth

perimeter = 2×l + 2×b = 30 cm

l = b + 10

now we use this identity in the perimeter equation :

2×(b+10) + 2b = 30

2b + 20 + 2b = 30

4b = 10

b = 2.5 cm

l = b + 10 = 2.5 + 10 = 12.5 cm

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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]

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City @ 2017 = 8,920,800

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- 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'

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                         \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]          

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                         \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

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                         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]

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