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vitfil [10]
2 years ago
6

How do you put 2hundreds 9tens 18ones in standard form

Mathematics
2 answers:
alexira [117]2 years ago
5 0

three hundred seven

or

307

Step-by-step explanation:

I can't remember which is standard form. But hope it helps.

nadya68 [22]2 years ago
4 0

Answer:

308

Step-by-step explanation:

200+90+18=200+108=308

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E(x) = A(x) - S(x) = 2400x + 40x^2 - (250x + 500) = 2400x + 40x^2 - 250x - 500 = 40x^2 + 2150x - 500

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22. An employee joined a company in 2017 with a starting salary of $50,000. Every year this employee receives a raise of $1000 p
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Answer:

(a) The recurrence relation for the salary is

S_{n+1}=1.05*S_n+1000\\\\S_0=50000

(b) The salary 25 years after 2017 will be $217044.85.

(c) S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n

Step-by-step explanation:

We can define the next year salary S_{n+1} as

S_{n+1}=S_n+1000+0.05*S_n=1.05*S_n+1000

wit S0=$50000

If we extend this to 2 years from 2017 (n+2), we have

S_{n+2}=1.05*S_{n+1}+1000=1.05*(1.05*S_n+1000)+1000\\S_{n+2} =1.05^2*S_n+1.05*1000+1000\\S_{n+2}=1.05^2*S_n+1000*(1.05^1+1)

Extending to 3 years (n+3)

S_{n+3}=1.05*S_{n+2}+1000=1.05(1.05^2*S_n+1000*(1.05^1+1))+1000\\\\S_{n+3}=1.05^3S_n+1.05*1000*(1.05^1+1)+1000\\\\S_{n+3}=1.05^3*S_n+1000*(1.05^2+1.05^1+1)

Extending to 4 years (n+4)

S_{n+4}=1.05*S_{n+3}+1000=1.05*(1.05^3*S_n+1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1.05*1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1000*(1.05^3+1.05^2+1.05^1+1.05^0)

We can now express a general equation for S_n (salary at n years from 2017)

S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n

The salary at 25 years from 2017 (n=25) will be

S_{25}=1.05^{25}S_0+1000*\sum_{0}^{24}1.05^i\\\\S_{25}=3.386*50000+1000*47.72=217044.85

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4 years ago
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There’s no question here!
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