Answer to Question #172525 in Algebra for Vishal Manilal

Question

2. Using the formulae for geometric progression

(a) The value of a lathe originally valued at £9000 depreciates by 5% each year. Calculate its value after 6 years.

(b) The lathe will be sold when its value falls below £6000 after how many years will this happen?

3) A construction company is fined for each day of delay in the construction of a

bridge. The penalty for the delay is £3000 for the first day and then an additional

£500 for each subsequent day ( i.e. Day 1 £3000, day 2 £3500, day 3 £4000

etc.). The maximum fine the company can afford is £75000.

Use arithmetic sequences formulae to find the maximum number of days they can

afford to delay.

4) Simplify the following, using the rectangular form of complex numbers

5 + 2j / 7 - 3j  and (5+ 3j) (-8-2j)

5) Simplify, using the polar form of complex numbers

a. 5 < 55 ^degree x 8 < - 28^ degree

b. 5 < 55 ^ degree / 8 < - 28 ^ degree

Convert both answers to rectangular form

Answer 1

2. (a)

$$A_n=A_0(1-r)^n$$

$$A_6=£9000\cdot(1-0.05)^6=£6615.83$$

(b)

$$A_n=A_0(1-r)^n$$

$$9000(1-0.05)^n<6000$$

$$0.95^n<\dfrac{2}{3}$$

$$n>\dfrac{\ln(\dfrac{2}{3})}{\ln0.95}$$

$$n=8$$

$£9000(1-0.05)^7=£6285.04$

$£9000(1-0.05)^8=£5770.78$

The lathe will be sold after 8 years.

3.

$$a_1=£3000, d=£500$$

$$a_n=a_1+d(n-1)$$ $$S_n=n\cdot\dfrac{2a_1+d(n-1)}{2}$$

Given $S_n=£75000$

$$n\cdot\dfrac{2(3000)+500(n-1)}{2}\leq75000$$

$$12n+n^2-n\leq300$$

$$n^2+11n-300\leq0$$

$$n=\dfrac{-11\pm\sqrt{1321}}{2}, n>0$$

$\dfrac{-11\pm\sqrt{1321}}{12}\approx12.673$

$n=12$

The maximum number of days they can afford to delay is 12.

4)

$$\dfrac{5+2j}{7-3j}=\dfrac{(5+2j)(7+3j)}{7^2+3^2}=\dfrac{29+29j}{58}$$

$$=\dfrac{1}{2}+\dfrac{1}{2}j$$

$$(5+3j)(-8-2j)=-40-10j-24j+6$$

$$=-34-34j$$

5) Simplify, using the polar form of complex numbers

$$5e^{j55\degree}\times8e^{-j28\degree}=40e^{j27\degree}$$

$$=40\cos(27\degree)+j40\sin(27\degree)$$

b.

$$\dfrac{5e^{j55\degree}}{8e^{-j28\degree}}=\dfrac{5}{8}e^{j83\degree}$$

$$=\dfrac{5}{8}\cos(83\degree)+j\dfrac{5}{8}\sin(83\degree)$$