Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2372583

It implies; 2^(3x)-2^(2x) = 0 Let 2^(x) = p Therefore; p³-p² = 0 p³ = p² p = 1 And p = 2^(x) It implies; 2^(x) = 1 2^(x) = 2⁰ x = 0

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2372555

Area blue is; Area square with side 10 cm - Area quarter circle with radius 10 cm - Area square with side 5 cm + Area quarter circle with radius 5 cm. = (10*10) - (¼*10*10*π) - (5*5) + (¼*5*5*π) = 100 - 25π - 25 + ¼(25π) = 75 - 25π + ¼(25π) = ¼(300 - 100π + 25π) = ¼(300-75π) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2372103

Let the two equal lengths be 1 unit. a = 180-20-80 a = 80° (1/sin20) = (b/sin80) b = 2.8793852416 units. c = 2.8793852416-1 c = 1.8793852416 units. d² = 1.8793852416²+1²-2*1.8793852416cos80 d = 1.969615506 units. It implies, required angle x is; (1.969615506/sin80) = (1/sinx) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2371610

Calculating angle BCD. Let AC = 2 units. Therefore; AB = BC = 1 unit. a = 60-45 a = 15° a is angle ADB. It implies; (1/sin15) = (b/sin30) b = 1.9318516526 units. b is BD. 1.9318516526² = 2c² c = 1.3660254038 units. d = 1.3660254038-1 d = 0.3660254038 units. tane = (0.3660254038/1.3660254038) e = 15° f = 45-15 f = 30° f is angle BDC. It implies, the required angle, angle BCD is; 180-45-30 = 180-75 = 105°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2371578

Calculating area shaded. tana = 2/4 a = atan(½)° b = 180-2a b = (180-2atan(½))° c = 180-b c = 180-(180-2atan(½)) c = 2atan(½)° Notice. Radius of the inscribed half circle r, is; r = ½(4) r = 2 units. Area Shaded is; Area triangle with height and base 4 units and 2 units respectively - Area triangle with height 2 units and base 2sin(180-2atan(½)) units - Area sector with radius 2 units and angle 2atan(½)° = ½(2*4) - ½*2*2sin(180-2atan(½)) - (2atan(½)π*2*2/360) = 4-1.6-1.854590436 = 0.54...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2371253

Sir Mike Ambrose is the author of the question. Yellow area exactly in decimal cm² is; Area triangle with height 7.49903313849 cm and base 8sin(32.2356103172) cm + Area triangle with height 7.49903313849 cm and base 8sin(12.764389683) cm. = (½*7.49903313849*8sin(32.2356103172)) + (½*7.49903313849*8sin(12.764389683)) = 16 + 6.62741699809 = 22.6274169981 cm² ≈ 22.63 cm² to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2370714

1. The proof that triangle EGH is equilateral. Let the side length of the regular hexagon be 1 unit. a = ⅙(180(6-2)) a = 30*4 a = 120° a is single interior angle of the regular hexagon. b² = 1²+0.5²-2*0.5cos120 b = 1.3228756555 units. b is EG. (1.3228756555/sin120) = (0.5/sinc) c = 19.1066053509° c is angle DEG. d² = 2-2cos120 d = 1.7320508076 units. d is BF. e = ½(d) e = ½(1.7320508076) e = 0.8660254038 units. e is FH. g² = 1²+0.8660254038² g = 1.3228756555 units. g is EH. Notice! EH...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2370512

a² = (2r)²+r² a² = 5r² a = √(5)r units. a is BD. Where r is the radius of the circle. b = ½(a) b = ½(√(5)r) units because BD, points B and D are symmetry with respect to length AC. tanc = 2r/r c = atan(2)° c is angle DBF. d = 90-c d = (90-atan(2))° tane = 2/1 e = atan(2)° e is angle BAC. tanf = 1/2 f = atan(½)° f is angle ACB. Notice! c (angle DBF) + f (angle ACB) = 90° Or atan(2) + atan(½) = 90° It implies; (½(BD))BC is a right-angled triangle. Notice! ½(BD) = ½(√(5)r) units. Calcula...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2370478

Let the side length of the regular octagon be 1 unit. a = ⅛(180(8-2)) a = ⅛(180*6) a = 3*45 a = 135° a is the single interior angle of the regular octagon. b = ½(360-135-135) b = ½(360-270) b = ½(90) b = 45° c = ½(180-135) c = ½(45) c = 22.5° d = 135-45-22.5 d = 90-22.5 d = 67.5° e = (180(5-2)) e = (180*3) e = 540° e is the sum of interior angles of a regular pentagon. f = ½(540-3(135)) f = ½(540-405) f = ½(135) f = 67.5° g = f-c g = 67.5-22.5 g = 45° h = 180-d-g h = 180-67.5-45 h = 13...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2368954

Sir Mike Ambrose is the author of the question. Area ABC (yellow area) exactly is; Area triangle with height ⅓√(265) cm and base ((5√(26)/6)sin84.3306169587) cm. = ½*⅓√(265)*((5√(26)/6)sin84.3306169587) = 11.4722222222 cm² = 413/36 cm²

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