1a) 8m + 2³m = 1/4
2²m + 2²m = 2-²
4³m = 2-²
Equate the base
6m = - 2
m = -2/4 = -1/3
4) Given y= 5/x² + 3
Y = 5(x² +3)-1
dy/dx = anxn-1
y = 5(x² + 3)-1
dy = 5(-1) (x2 +3)-1^1
dy/dx = 10x (x² + 3)-²
dy/dx = - 10/(x² +3)²
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1a) 8m + 2³m = 1/4
2²m + 2²m = 2-²
4³m = 2-²
Equate the base
6m = - 2
m = -2/4 = -1/3
1b) log15 base 4 = x
x = log15 base/log4 base 10 = 1.1761/0.6020
= 1.9536
therefore x approximately 1.954
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4) Given y= 5/x² + 3
Y = 5(x² +3)-1
dy/dx = anxn-1
y = 5(x² + 3)-1
dy = 5(-1) (x2 +3)-1^1
dy/dx = 10x (x² + 3)-²
dy/dx = - 10/(x² +3)²
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7a)Given
h=15.4t-4.9t
Velocity V=dh/dt =15.4-9.8t
At maximum height V=0
Therefore 15.4-9.8t=0
9.8t=15.4
t=15.4/9.8
t=1.6secs
Time to reach maximum height is 1.6secs
14a) m=0.5kg
F-r=ma
0.5*10-R=0.5*2
5-r=1
R=5-1
R=4n
bi)Ra+Rb=80+M+100
Ra+Rb=180M
Rb=180+M-90
Rb=90+m -------(i)
Taking moment about B
A.C.M=80*1.7*2*M
=136+2m
C.W.M=100*1.2+90+2*4
=120+216
=336
A.C.W=C.W.M
136+2M=336
2M=336-136
2M=200
M=200/2
=100N
-: M= 100N
ii)
Rb=90+M
=90+100
=190N
-: the reaction at B is 190N
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