a = x+1+2
a = (x+3) units.
a is BC.
b = 4+2
b = 6 units.
Calculating x.
(2x)² = (x+3)²+6²
4x² = x²+6x+9+36
3x²-6x-45 = 0
x²-2x-15 = 0
x²-5x+3x-15 = 0
x(x-5)+3(x-5) = 0
(x-5)(x+3)...
½*a*b = 10
a = (20/b) cm.
a is CD.
b is OD.
b² - 30
(20/b)² - 10
Cross Multiply.
10b² = 12000/b²
10b⁴ = 12000
b⁴ = 1200
b⁴ = 1200
b² = √(1200)
b² = 20√(3)
b = √(20√(3))
b = 5.885...
a = 2+3
a = 5 units.
b²+3² = a²
b² = 25-9
b = √(16)
b = 4 units.
Let x be the radius of the ascribed semi circle.
c = (2-(x-4)) units.
c = (6-x) units.
d² = (6-x)²+3²
d² = 36-12x+...
Shaded/Area Blue is;
2(area sector with radius 4 units and angle (180-2atan(½))° - area triangle with two side 4 units respectively, and angle (180-2atan(½))°) + 2(area sector with radius 2 unit...
8^(x)+4^(x) = 36
2^(3x)+2^(2x) = 36
Let 2^(x) be p.
Therefore;
p³+p² = 36
p = 3
And p = 2^(x)
2^(x) = 3
x = log3/log2
x = 1.5849625007
Checking Accuracy.
8^(x)+4^(x) = 36...
a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the ascribed regular pentagon.
b = 360-2a-90
b = 360-306
b = 54°
c = 90-b
c = 36°
d = 180-2c
d = 108°
Notice.
The insc...
Let a be the radius of the inscribed half circle.
b = (a-1) units.
It implies, calculating a.
a² = (a-1)²+5²
a² = a²-2a+1+25
2a = 26
a = 13 units.
Again, a is the radius of the inscrib...
Sir Mike Ambrose is the author of the question.
The Equation of the curve is;
y = -¾(x²) + 6x
Therefore shaded area is;
Area under the curve at point (4, 0) and ((4+√(6)), 0) - Area recta...
Sir Mike Ambrose is the author of the question.
Area R exactly, in square cm decimal form is;
Area square with side 2√(11) cm - Area triangle with side 2√(11) cm and 4.97194902762 cm, and angle...
a² = 20²+15²
a = √(625)
a = 25 units.
25 - 20
15 - b
Cross Multiply.
25b = 15*20
b = 12 units.
b is the height of the shaded inscribed region, triangle.
Again;
25 - 20
20 - c...
