Presentation DBM-20023.KUMP 5

PRESENTATION DBM-20023
KUMPULAN 5

Member of group’s: 06DET21F1035
1.ADAM IRSYAD BIN RAZALI 06DET21F1011
2.AHMAD SYAHMI BIN AHMAD IZHAM 06DET21F1044
3.MUHAMMAD ZAHIRUL HAQIEM BIN

HAIRUL AZHAR

CASE 1

1560(1)2+23400
1 = (12 + 45)2

 = 104 u + 1560 + 23400
= (42)2
= 104 1 + 5
 2+45 2460
2 = 2116

 = 104 + 504 = 520 2 +( 2+45)(520) = 11.80
2+45 ( 2+45)2
1560(2)2+23400
 = 520 = 1040 2+520 2+23400 2 = (22 + 45)2
( 2+45)2
26520
= (49)2

 = 520 = 1560 2+23400 26520
( 2+45)2 = 2401

= 11.05

 = 2 + 45 Assume that, 3 = 15660(3)2 + 23400
= 1, = 2, = 3
(32 + 45)2

 = 2 14040 + 23400
= 2916

37440
= 2916

= 12.84

CASE 2

p=pollution of virus (2.5 , 829) is max point
t=time in hour GRAPH FOR CASE 2:

P(t)=-100 ² + 500 +M
M=204

=-200 ² +500+204

200t=500

t=


t=2.5

P=-100(2.5)²+500(2.5)+204

=-625+1250+204

= 829

X,y=2.5,829
²
² =-200t + 500

=-200

=-200>0

SUMMARY/CONCLUSION

 Considering all of facts, in case 1, we able to identify the initial population of
algae,N. Then, using the formula: = ( + + ) , =population of algae and
=time in hour, we able to determine the groeth of algae in 8 to 10 hours. In
case 2, using the formula: = − + + , =population of virus
while =time in hour, we can state either it is minimum or maximum, and draw
the graph to demonstrate the growth of algae