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marishachu [46]
4 years ago
6

Show work and simplify the equation

Mathematics
1 answer:
disa [49]4 years ago
5 0
Hello,

if \ a \neq -3, a \neq 1, a \neq -5\\

 \dfrac{2a^2+5a-3}{a^2+8a+15}  *  \dfrac{a^2+4a-5}{3a^2-a-2} \\


= \dfrac{(a+3)(2a-1)}{(a+5)(a+3)} *  \dfrac{(a+5)(a-1)}{(a-1)(3a+2)} \\

= \dfrac{2a-1}{3a+2}

explains:

1)
a^2+4a-5=a^2+5a-a-5\\
=a(a+5)-(a+5)\\
=(a+5)(a-1)




2)
a^2+8a+15=a^2+5a+3a+15\\
=a(a+5)+3(a+5)\\
=(a+5)(a+3)


3)
a^2+4a-5=a^2+5a-a-5\\
=a(a+5)-(a+5)\\
=(a+5)(a-1)


4)
3a^2-a-2=3a^2-3a+2a-2\\
=3a(a-1)+2(a-1)\\
=(a-1)(3a+2)



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Answer:

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age  \leq 3}\end{array}\right

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The current age distribution vector is as follows:

x = \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right] \left[\begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 0 \  \leq  age   \leq  2 }\\{0 \  \leq  age   \leq 3}\end{array}\right]

Also , the age transition matrix is as follows:

L = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]

After 1 year ; the age distribution vector will be :

x_2 =Lx_1 = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]  \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right]

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq 1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age   \leq  3}\end{array}\right

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3 years ago
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Answer:

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Add them up to get $22.34 if I’m right.

Hope this helped.

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3 years ago
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