Advanced diploma applied electrical engineering dee

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Advanced Diploma of Applied Electrical Engineering (DEE – 52883WA)	3

For a ‘satisfactory’ result in this assessment task, all unit elements covered in this assessment task (as indicated on the cover page and in the questions section of this document) must be completed at a‘satisfactory’ standard.

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		5
	DCSBME604: Use Basic Mathematics in Engineering
Assessment type:

Calculator except for: Casio CFX-9970, Casio Algebra FX 2.0, HP-40G, HP 49G, TI-89, TI 92

Topics Covered - Use Fundamental Techniques:
Arithmetic
Measurement
Mensuration
Geometry
Algebra
Quadratics
Right angle trigonometry

Advanced Diploma of Applied Electrical Engineering (DEE – 52883WA)	6

Q2	a) A pyramid has a height of 10 cm and a square base with sides of 5 cm. Calculate the volume in litres. (Give you answer correct to 2 decimal places)


Student Answer a)height(h)=10cm base side(a)=5 cm volume of square based pyramid=(aah)/3 =(5510)/3 cm3 =83.33 cm3 b) area of rectangular plate=500300 mm2 =150000 mm2 four sectors form a circle with radius 25 mm. area of circle =π25*25mm2=1963.49 mm2 the final area of plate=(150000-1963.49)mm2=148036.50mm2

	Not Satisfactory	☐

Advanced Diploma of Applied Electrical Engineering (DEE – 52883WA)	8

A4
F4	Assessor Feedback
F4	Assessor Feedback

so, the number of houses that can be wired completely=3



		Not Satisfactory	☐

DEE_DCSBME604_WrittenAssessment_v1

	10

	a) b)


		Satisfactory	☐

	12

Q9
	Student Answer a) equilateral triangle b) octagon c) parallelogram d) obtuse-angled triangle
F9
F9		☐

	13

	b) 2+3−5−7−9+10 c) (−4)(−3)(−8)
	b) 2+3−5−7−9+10 =2+3+10-5-7-9=15-21= -6 c) (−4)(−3)(−8) = 12*(-8)= -96
			☐

DEE_DCSBME604_WrittenAssessment_v1

	15

−9 I1+4 I 2=−28........ (1)
4 I 1−5 I 2=7 …….. (2)
Using Elimination method,
multiplying equation 1 by 5 we get,
-45I1+20I2= -140……………………..(3)
multiplying equation 2 by 4 we get,
16I1-20I2=28………………………….(5)
now, adding equations 3 and 4 we get,
-45I1+20I2+16I1-20I2=28-140
or,-29I1= -112
so, I1=3.862 A
putting the value of I1=3.862 A in equation 2 we get, 4∗3.862−5I 2=7
or, 5I2 =15.448-7
or, 5I2 =8.448
so, I2 =8.448/5= 1.6896 A

DEE_DCSBME604_WrittenAssessment_v1


		Satisfactory	☐

Advanced Diploma of Applied Electrical Engineering (DEE – 52883WA)	17

DEE_DCSBME604_WrittenAssessment_v1

	18

	19

Advanced Diploma of Applied Electrical Engineering (DEE – 52883WA)	20

	a) f (x )=2 x−3 b) f (x )=−x+5

Advanced Diploma of Applied Electrical Engineering (DEE – 52883WA)	22

	23

DEE_DCSBME604_WrittenAssessment_v1

	24

	25

in graph,

DEE_DCSBME604_WrittenAssessment_v1


F19	Assessor Feedback
F19	Assessor Feedback

	27

	28

Student Answer
in tabular form,

	from graph, the approximate solutions are: x=2.3 and x=-1.3

		Not Satisfactory	☐

DEE_DCSBME604_WrittenAssessment_v1

	30

	a) Use your calculator to obtain following ratios correct to 4 decimal places: i. sin 53° ii. cos 66° 04’ 54’’ b) Calculate the value of x correct to the nearest degree given that x is an acute angle: iii. sin x = 0.2765 iv. 3tan x = 2.4568
A23

	b) iii) sin x = 0.2765 using calculator, x=sin-1(0.2765)= 16.05142°=16° iv) 3tan x = 2.4568 or, tanx=2.4568/3 or, tanx=0.818933333 or, x=tan-1(0.8189333333)=39.3151°=39°

		Not Satisfactory	☐

DEE_DCSBME604_WrittenAssessment_v1

	32

Advanced Diploma of Applied Electrical Engineering (DEE – 52883WA)	33


	Student Answer secondary output voltage(V2)=1100 V primary input voltage(V1)=25 V We know, a)constant of proportionality(K)=V2/V1 K=1100/25 K=44 b)secondary output voltage(V2)=800 V constant of proportionality(K)=44 primary input voltage(V1)=? we know, constant of proportionality(K)=V2/V1 or, 44=800/V1 or V1=800/44 so V1=18.181818=18.182 V


	Assessor Feedback	Satisfactory
	Assessor Feedback