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3)
V = Mg√1 – r²
Square both sides
V² = m²g²(1-r²)
V²/m²g² = 1-r²
r² = 1 – v²/m²g²
r = √1-(v/mg)²
If v = 15, m = 20, and g = 10
r = √1 – (15/20*10)²
r = √1 – (0.075)²
r= √(1.075)(0.925)
r = √0.994375
r = 0.9972

10a)
Obtuse 105 + reflexReflex =255°
Now 2w = reflex2w =255°
W = 255/2 =127.5°
Also 2x = obtuse2x = 105°
X = 105/2 = 52.5°
Now EDF = y(base angles of an isosceles triangle)
BED=X=52.5°(angles in the same segment)
EFD+EDF=BED (sum of interior angles of a triangle equal exterior angle)
Y+y = 52.5°
2y = 52.5°
Y = 52.5°/2

==============================
5a)
Mode = mass with highest frequency = 35kg
Median is the 18th mass
= 40kg.
5b)
In a tabular form
Under Masses(x kg)
30,35,40,45,50,55
Under frequency(f)
5,9,7,6,4,4
Ef = 35
Under X-A
-10, -5, 0, 5, 10, 15
Under F(X-A)
-50, -45, 0, 30, 40, 60
Ef(X – A) = 35
Mean = A + (Ef(X – A)/Ef)
= 40 + 35/35
= 40 + 1

8)
x=a+by(eqi)
when y=5 and x=19
19=a+5b(eqii)
when y=10 and x=34
34=a+10b(eqiii)
solving eqii and eqiii
a+10b=34
a+5b=19
=>5b=15
b=15/5=3
putting b=3 in eqii
19=a+5(3)
19=a+15
a=19-15
a=4
8i)
Putting a=4 and b=3 in eqi
x=4+3y
This is the relationship between xand y
8ii)
When y=7
x=4+3(7)
x=4+21

x=25

1a)
Log 10(20-10)-log10(+3)=log105
(20-10/+3)=log10 =5
20-10/+3=5
5(+3)=20-10
5+15=20-10
15+10=20-5
25=15* =25/15
*=5/3=1 2/3
1b)
Discount percent =15%
Discount amount =#600
Actual amount paid on the article =?
Original amount on the article =
15%=#600
15/100 =600
15=600100
15*=60000
*=60000/15
*=#4,000
Therefore actual amount paid on the article
=#4,000-#600
=#3,400

Actual amount paid on the article =#3,400

4i)
length of Arc of the sector
Titter= 72?, r = 14cm
L= titter / 360 x 2 pie r
==> L= 72/360 x 2 x 22/7 x 14
=44352/2520
=17.6cm
4ii) perimeter of the sector
Perimeter = titter/360 x 2 pie r + 2r
=17.6 +(2×14)
=17.6+28
=45.6cm
4iii) Area of the sector
Area = Titter/360 x pie r?
=72/360 x 22/7 x (14)?
=72 x 22 x 196/2520
Area= 310464/2520
=123.2cm?

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