Sir Mike Ambrose is the author of the question.
x = 16
The sequence is;
16, 12, 9, (27/4), ...
a = 16
r = ¾
Calculating n.
(2187/16383) = (¾)^(n)
n = 7.
Therefore;
T...
Shaded area is;
Area trapezoid with parallel sides 58 cm and 28 cm, and height 30 cm - Area trapezoid with parallel sides 9 cm and 30 cm, and height 28 cm - Area triangle with height 12 cm and b...
Sir Mike Ambrose is the author of the question.
Radius, r of the inscribed circle is;
r = (3-√(3)) cm.
Therefore;
Area purple exactly in square cm is;
Area trapezium with parallel side...
Sir Mike Ambrose is the author of the question.
P(2, (16√(5)/5))
Gradient of the curve at the tangent is;
= 24/5√(5)
Therefore;
Gradient of the curve at the normal is;
= -5√(5)/2...
Sir Mike Ambrose is the author of the question.
AB exactly is;
= 16tan(3acos(8/9)) - 2√(17)
= 2(8tan(3acos(8/9))-√(17)) cm.
= 102.760478669 cm.
Let AB = 4 units.
BC is;
√((2√(20))²-4²)
BC = √(80-16)
BC = √(64)
BC = 8 units.
It implies;
AB ÷ BC
= 4 ÷ 8
= ½
Sir Mike Ambrose is the author of the question.
y = e, x = e²
dy/dx at the tangent is;
= 1/(y+2yIny) for y = e.
dy/dx at the tangent = 1/3e
Therefore;
dy/dx at the normal = -3e...
Sir Mike Ambrose is the author of the question.
Equation of the curve is;
y = ¼(x²)+4
Shaded area exactly in square units is;
Area triangle with height 8 units and base 4 units - Area sec...
Let the radius of the inscribed circle be 2 unit.
Therefore;
AB = 8 units.
BC = 6 units.
Area triangle ABC is;
½*6*8
= 24 square units.
It implies;
OG = ⅓(10) units.
GE =...
Sir Mike Ambrose is the author of the question.
Area orange exactly in square units is;
Area triangle with height ⅓(10√(2)) units and base 5sin45 units - Area triangle with height (8/3) units a...
