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sashaice [31]
3 years ago
8

The fifth term of a non-linear sequence is 9.

Mathematics
1 answer:
kozerog [31]3 years ago
5 0

Answer:

a₂ = 243

Step-by-step explanation:

The term to term rule from left to right in the sequence is divide by 3

The term to term rule from right to left is the opposite, multiply by 3

Given

a₅ = 9, then

a₄ = 9 × 3 = 27

a₃ = 27 × 3 = 81

a₂ = 81 × 3 = 243 ← required term

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Solve it by substitution method:- 5x - 6y=-9 and 3x + 4y= 25
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\left\{\begin{array}{ccc}-5x-6y=-9&/\cdot2\\3x+4y=25&/\cdot3\end{array}\right\\+\left\{\begin{array}{ccc}-10x-12y=-18\\9x+12y=75\end{array}\right\\------------\\.\ \ \ \ \ \ \ -x=57\\.\ \ \ \ \ \ \ \ \ x=-57\\\\3\cdot(-57)+4y=25\\-171+4y=25\\4y=25+171\\4y=196\ \ \ \ /:4\\y=49\\\\Solution:x=-57;\ y=49.
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FURTHER MATHEMATICS Use determinants to solve the systems of equation:
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\left\{\begin{array}{ccc}2x+1y+2z&=&13\\1x+1y-2z&=&8\\1x+2y+1z&=&11\\\end{array}\right.\\\\\\\Delta=\left| \begin{array}{ccc}2&1&2\\1&1&-2\\1&2&1\end{array}\right| =2*\left| \begin{array}{ccc}1&\frac{1}{2}&1\\1&1&-2\\1&2&1\end{array}\right| =2*\left| \begin{array}{ccc}1&\frac{1}{2}&1\\0&\frac{1}{2}&-3\\0&1&3\end{array}\right| =2*(\frac{3}{2}+3)=9\\\\

\Delta_1=\left| \begin{array}{ccc}13&1&2\\8&1&-2\\11&2&1\end{array}\right| =2*\left| \begin{array}{ccc}13&1&2\\8&1&-2\\\frac{11}{2}& 1&\frac{1}{2}\end{array}\right| =2*\left| \begin{array}{ccc}13&1&2\\-5&0&-4\\\frac{-5}{2}& 0&\frac{5}{2}\end{array}\right| \\\\=2*(-1)*(\frac{-25}{2}-\frac{20}{2}) =45\\

\Delta_2=\left| \begin{array}{ccc}2&13&2\\1&8&-2\\1&11&1\end{array}\right| \\\\\\=\left| \begin{array}{ccc}3&21&0\\3&30&0\\1&11&1\end{array}\right| \\\\\\=1*(90-63) =27\\

\Delta_3=\left| \begin{array}{ccc}2&1&13\\1&1&8\\1&2&11\end{array}\right| \\\\\\=\left| \begin{array}{ccc}0&-1&-3\\0&-1&-3\\1&2&11\end{array}\right| \\\\\\=0\\

\left\{\begin{array}{ccc}x=\dfrac{\Delta_1}{\Delta}=\dfrac{45}{9}=5\\\\y=\dfrac{\Delta_2}{\Delta}=\dfrac{27}{9}=3\\\\z=\dfrac{\Delta_3}{\Delta}=\dfrac{0}{9}=0\\\end{array}\right.

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