According to the new remuneration scheme, the starting pay of a lecturer a is RM3,087 per month and the annual increment is RM195. 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?

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.

(x, y)→(x − 8, y − 7)

step-by-step explanation:

enjoy!

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)