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Further Mathematics: Number Systems Essay Example
 Category:Engineering and Construction
 Document type:Math Problem
 Level:Undergraduate
 Page:2
 Words:786
Assignment title 
Assignment 1 

Task (LO 1.1) The number of moles () of an ideal gas are obtained using the equation is the pressure measured in Pascals (Pa), is the volume measured in cubic metres (m^{3}), is the universal gas constant (8.31447JK^{1}mol^{1}), is the temperature measured in Kelvin (K) In a repeated experiment, the temperature of the gas was measured by a digital meter to the nearest Celsius. The following repeated measurements were obtained : 32C, 33C, 32C, 34C, 32C, 33C, 32C, 32C, 32C, 31C
Working = {(32+33+32+34+32+33+32++32+32+31)/(10)} Max = 32.35 Min = 32.295 Error = [(MaxMin)/working]*100 [(32.3532.295)/32.3]*100 Error= 0.17027863
Working = {(32+33+32+34+32+33+32++32+32+31)/(10)} The temperature of the gas must be in Kelvin before it is used in the equation. This is obtained by adding 273 to the Celsius temperature. The measured values of the other quantities are: Pa cm^{3} JK^{1}mol^{1}
N= [][750] [][305.3] 1237.5*10^5 2538.407691 =48751.03413 

Task (LO 1.2)
1101= .64 67521.64
1101111010101101 Now assign each group its corresponding hex value 1101 = F, 1110 = 5, 1010 = 4, 1101 = 14 When put together, we get F54E_{16}


Task (LO 1.3) Two complex numbers are such that and
a + bj = a(b + j) + b(a + j) = b + j + a + j = b – aj = (a – j)(b + j) = c – j + aj – dj^{2} = c + bj – a(– 1) = c + aj on remembering that j^{2} = – 1
The
If we substitute z = j into this we get e^{j}^{}= 1 + j + + + … + + … = 1 + j – – j+ + … on using j^{2} = – 1 repeatedly. Gathering real and imaginary terms together then gives e^{j}^{}= 1 – + + … + j (remember is in radians) cos = 1 – + + … and sin = – + … to obtain the required relation: e^{j}^{} = cos + j sin
Using this we have
=
For
= . = cos For z = 1 we get = .= = z (Check: z1 = z2z = – z) For z = 2 we get = .=
= cos 

Task (LO 1.4) Find the m complex roots of the equation [For example, for n=9 then n modulo 5 = 9 modulo 5 = 4. Therefore 3 + n modulo 5 = 3+4 = 7. And the question would be to find the 7 roots of ()^{5} = [3(cos = 2^{5}(cos z + i sin = 32(cos = 32(3/2 + 1/2 i) = 16√3 + 16 i. 
Task (LO 1.5) The complex impedance of a circuit containing a resistor of resistance R, inductor of inductance L and a capacitor of capacitance C is given by
L=240(r+c) = = 500 600 = 30000 = = = ^{2} = 500^{2} = 2500
Arg (z) = Arg (z_{1}) + Arg (z_{2}) = +
Arg = Arg _{1 } = – Arg z_{1} = – 