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jasenka [17]
3 years ago
9

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

es into the city. A 2019 census measures 10,000,000 residents in the city and 800,000 in the suburbs. (a) Set up a difference equation (of the form ~xn+1 = A~xn) that describes this situation, where ~x0 is the initial population in 2019. (b) Estimate the populations in the city and suburbs two years later, in 2021.
Mathematics
1 answer:
Charra [1.4K]3 years ago
3 0

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Following are the differential equation:

\to P(C,S)t+1 = P(C,S)t+(- 0.06Ct +0.04St, 0.06Ct - 0.04St ) \ \ \ \\\\or \ \ \ \ \ \ \ \\\\ \to (0.94Ct + 0.04St, 0.96St + 0.06Ct)

In equation:

\to t = 0, \ as \ 2019\\\\\to Ct = 10,000,000  \\\\\to St = 800,000,000\\

\to P(C, S)1 = (0.94Ct + 0.04St, 0.96St + 0.06Ct) \\\\

                   = (0.94(10000000) + 0.04(800000), 0.96(800000) + 0.06(10000000))\\\\ = (9,400,000 + 32,000), (768,000 +6000000))\\\\ = (9432000, 1368000)

\to P(C, S)2 = (0.94(9432000) + 0.04(1368000), 0.96(1368000) + 0.06(9432000))

                  =(8,866,080 + 54,720), (1,313,280 + 565,920)\\\\=(8,920,800, 1,879,200)

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Find all solutions of the equation: 2cos^2x-cosx=1
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/3166243

——————————

Solve the trigonometric equation:

     \mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}


Make a substitution:

     \mathsf{cos\,x=t\qquad (-1\le t\le 1)}

and the equation becomes

     \mathsf{2t^2-t-1=0}


Rewrite conveniently  – t  as  + t – 2t,  and then factor the left-hand side by grouping:

      \mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}


Factor out  2t + 1:

     \mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}


Substitute back for  t = cos x:

     \begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}


Therefore,

     \begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}

where  k  is an integer.


Solution set:   

\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}


I hope this helps. =)

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