Answer to Question #179511 in Math for EUGINE HAWEZA

Question

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 ∑⁵_i = 2𝑢𝑖 − 𝑣𝑖 And ∑⁵_i = 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 

Answer 1

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

$$S_{120}=120\cdot\dfrac{2a_1+d(120-1)}{2}$$

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

The cost of boring is k29820.

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

$$C(t)=32500-85t, t\geq0$$

If the Iphone’s value falls below k12500

$$C(t)<12500$$

$$32500-85t<12500$$

$$85t>32500-12500$$

$$85t>20000$$

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

$$t>235$$

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

c.

$$u_i=5+2i, v_i=2-5i$$

$$2u_i-v_i=2(5+2i)-(2-5i)=8-9i$$

We know that

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

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

Then

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

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

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

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

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

e. Tecno

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

Samsung

$$A_{nS}=500(1.2)^{n-1}$$

i)

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

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

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

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

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

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

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

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

ii)

Tecno

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

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

Samsung

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

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