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HACTEHA [7]
3 years ago
5

Sibal started with $500 in a bank account that does not earn interest. In the middle of every month, she withdraws 15 of the acc

ount balance. Which recursive function rule models Sibal’s balance at the end of each month?
an=4/5⋅an−1 and a1=500 ​
an=500⋅an−1 ​ and a1=1/5
an=1/5⋅an−1 ​ and a1=500
an=500⋅an−1 and a1=4/5
Mathematics
1 answer:
Otrada [13]3 years ago
4 0
After every month's withdrawals, 4/5 of the original amount at the start of the month will remain. The amount at the start of every month will change. Thus:
an = 4/5 (an-1) ; where a1 = 500.
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Answer:

5^20

Step-by-step explanation:

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u>

\displaystyle \large{ \frac{ {a}^{m} }{ {a}^{n} }  =  {a}^{m - n} }

Therefore:

\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{8 - 3})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{5})^{4}  }

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u><u>I</u>

\displaystyle \large{( {a}^{m} ) ^{n} =  {a}^{m \times n}  }

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\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{8 - 3})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{5})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  = {5}^{5 \times 4}   } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =   {5}^{20} }

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