Answer to Question #179511 in Math for EUGINE HAWEZA

Question #179511

a. 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.


b. 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?


c. Given that 𝑢𝑖 = 5 + 2𝑖 𝑎𝑛𝑑 𝑣𝑖 = 2 − 5𝑖, find ∑5i = 2𝑢𝑖 − 𝑣𝑖 And ∑5i = 𝑢𝑖 (use the properties of the sigma notation)


d. Two competing companies, Tecno and Samsung produce mobile phones. Tecno starts production at 1000 phones per week and plans to increase output by 200 each week. Samsung start production with 500 phones per week and plans to increase output by 20% each week.


e. i). a. Calculate the weekly production in weeks 1; 5; 10; 15.


ii). Calculate the total production during the first 15 weeks for each firm 


1
Expert's answer
2021-04-15T06:31:37-0400

a. Arithmetic progression: a1=k70,d=k3,n=120a_1=k70, d=k3, n=120


S120=1202a1+d(1201)2S_{120}=120\cdot\dfrac{2a_1+d(120-1)}{2}

=1202(k70)+k3(1201)2=k29820=120\cdot\dfrac{2(k70)+k3(120-1)}{2}=k29820

The cost of boring is k29820.


b. Let t=t= the time in months, C(t)=C(t)= value of IPhone after tt months


C(t)=3250085t,t0C(t)=32500-85t, t\geq0

If the Iphone’s value falls below k12500


C(t)<12500C(t)<12500

3250085t<1250032500-85t<12500

85t>325001250085t>32500-12500

85t>2000085t>20000

t>2000085t>\dfrac{20000}{85}

t>235t>235

It will take 236 months before the Iphone’s value falls below k12500.


c.

ui=5+2i,vi=25iu_i=5+2i, v_i=2-5i

2uivi=2(5+2i)(25i)=89i2u_i-v_i=2(5+2i)-(2-5i)=8-9i

We know that

i=1n1=n\displaystyle\sum_{i=1}^n1=n

i=1ni=n(n+1)2\displaystyle\sum_{i=1}^ni=\dfrac{n(n+1)}{2}

Then

i=15ui=i=15(5+2i)\displaystyle\sum_{i=1}^5u_i=\displaystyle\sum_{i=1}^5(5+2i)=5i=151+2i=15i=5\displaystyle\sum_{i=1}^51+2\displaystyle\sum_{i=1}^5i

=5(5)+2(5(5+1)2)=55=5(5)+2(\dfrac{5(5+1)}{2})=55


i=15(2u1vi)=i=15(8+9i)\displaystyle\sum_{i=1}^5(2u_1-v_i)=\displaystyle\sum_{i=1}^5(8+9i)=8i=151+9i=15i=8\displaystyle\sum_{i=1}^51+9\displaystyle\sum_{i=1}^5i

=8(5)+9(5(5+1)2)=175=8(5)+9(\dfrac{5(5+1)}{2})=175

d. Let n=n= the week's number , An=A_n= the output of mobile phones per week.

e. Tecno


AnT=1000+200(n1)A_{nT}=1000+200(n-1)

Samsung


AnS=500(1.2)n1A_{nS}=500(1.2)^{n-1}


i)

A1T=1000+200(11)=1000A_{1T}=1000+200(1-1)=1000


A5T=1000+200(51)=1800A_{5T}=1000+200(5-1)=1800


A10T=1000+200(101)=2800A_{10T}=1000+200(10-1)=2800


A15T=1000+200(151)=3800A_{15T}=1000+200(15-1)=3800


A1S=500(1.2)11=500A_{1S}=500(1.2)^{1-1}=500


A5S=500(1.2)51=1037A_{5S}=500(1.2)^{5-1}=1037


A10S=500(1.2)101=2580A_{10S}=500(1.2)^{10-1}=2580


A15S=500(1.2)151=6420A_{15S}=500(1.2)^{15-1}=6420


ii)

Tecno


n=115(1000+200(n1))=\displaystyle\sum_{n=1}^{15}(1000+200(n-1))=

=15(1000+38002)=36000=15(\dfrac{1000+3800}{2})=36000

Samsung


n=115(500(1.2)n1)=\displaystyle\sum_{n=1}^{15}(500(1.2)^{n-1})=

=500(1(1.2)1511.2)=36018=500(\dfrac{1-(1.2)^{15}}{1-1.2})=36018



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Comments

Assignment Expert
27.04.21, 20:28

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Anthony Isaacs
26.04.21, 14:23

this is brilliant, thank you so much for helping me

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