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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 = √0.994375
r = 0.9972
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
Mode = mass with highest frequency = 35kg
Median is the 18th mass
In a tabular form
Under Masses(x kg)
Ef = 35
-10, -5, 0, 5, 10, 15
-50, -45, 0, 30, 40, 60
Ef(X – A) = 35
Mean = A + (Ef(X – A)/Ef)
= 40 + 35/35
= 40 + 1
when y=5 and x=19
when y=10 and x=34
solving eqii and eqiii
putting b=3 in eqii
Putting a=4 and b=3 in eqi
This is the relationship between xand y
Discount percent =15%
Discount amount =#600
Actual amount paid on the article =?
Original amount on the article =
Therefore actual amount paid on the article
Actual amount paid on the article =#3,400
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
4ii) perimeter of the sector
Perimeter = titter/360 x 2 pie r + 2r
4iii) Area of the sector
Area = Titter/360 x pie r?
=72/360 x 22/7 x (14)?
=72 x 22 x 196/2520
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