Question #179792

An oil company bores a hole 120m deep. Estimate the cost of boring if the cost is k70 for drilling the first meter with an increase in cost of k3per meter for each succeeding meter.

ii. An IPhone 12 pro who’s original Value was K32500 decreases in value by k85 per month. How long will it take before the Iphone’s value falls below k12500?


1
Expert's answer
2021-04-29T17:38:42-0400

i) use the formulai) \space use \space the \space formula

Sn=n/2[2a+(1201)3]=120/2[(2×70+(357)]=60(140+357)=29820S_n=n/2[2a+(120-1)3]=120/2[(2\times70+(357)]=60(140+357)=29820

ii) let cost in kth month=a+(n1)d=32500+(k1)(85)^{th}\space month=a+(n-1)d=32500+(k-1)(-85)

32500+(k1)(85)<1250032500+(k-1)(-85)\lt12500

(k1)(85)<20000(k-1)(-85)\lt-20000

(k1)(85)20000(k-1)(85)\geq20000

k120000/85k-1\geq20000/85

k20000/85+1=236.29k\geq20000/85+1=236.29

k=236.29 monthsk=236.29\space months


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