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Neporo4naja [7]
3 years ago
15

According to the new remuneration scheme, the starting pay of a lecturer a is RM3,087 per month and the annual increment is RM19

5. Ahmad, who is 25 years old, becomes a lecturer in a university. (a) What will his monthly salary be when he is 45 years old? (b) What will his age be when he gets a monthly payment of RM8,937?​
Mathematics
1 answer:
vichka [17]3 years ago
7 0

Step-by-step explanation:

arithmetic sequence : every new item of the sequence is created by adding a constant term to the previous item - in this case 195.

a1 = 3087 (that's our starting value)

a2 = a1 + 195

a3 = a2 + 195 = a1 + 2×195

an = a1 + (n-1)×195

a)

when he is 45 years old, that is 20 years (and 20 annual increments) plus to our starting value.

so, n = 1+20 = 21

a21 = a1 + 20×195 = 3087 + 20×195 = 6987

so, when he is 45 years old, his monthly salary will be RM6,987

b)

how many years (= how big is n) until he gets 8937 ?

8937 = a1 + (n-1)×195 = 3087 + (n-1)×195 =

= 3087 + n×195 - 195 = 2892 + n×195

6045 = n×195

n = 6045/195 = 31

so, a31 = 8937

and that means, he has to work 30 additional years (31 minus the starting level 1) to earn monthly RM8,937.

that means he will be 25+30 = 55 years old.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

<u><em>Explanation</em></u>

Given  

       (\frac{8^4}{10^4} )^{-\frac{1}{4} }

By using algebra formulas

  (\frac{a}{b} )^{m} = \frac{a^m}{b^n}

(\frac{8^4}{10^4} )^{-\frac{1}{4} } = \frac{(8^{4})^{\frac{-1}{4} }  }{(10^{4} )^{\frac{-1}{4} } }

(a^{m} )^{n} = a^{mn}

\frac{(8^{4})^{\frac{-1}{4} }  }{(10^{4} )^{\frac{-1}{4} } } = \frac{8^{4X\frac{-1}{4} } }{10^{4 X\frac{-1}{4} } }

          = \frac{8^{-1} }{10^{-1} }

           = \frac{10}{8} \\= \frac{5}{4}

<u><em>Final answer:-</em></u>

 (\frac{8^4}{10^4} )^{-\frac{1}{4} }  = \frac{5}{4}

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