Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3110074

Length x is; 15² = 9*x x = ⅑(15*15) x = 25 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3109784

Sir Mike Ambrose is the author of the question. Equation of the curve is; y = ⅔(8x-x²) Calculating Area E. (Area under the curve at x = (3/2) and x = 4) - Area trapezium with parallel sides (13/2) units and (10⅔) units, and height (5/3) units. =(835/36) - ½*(5/2)*((13/2)+(32/3)) = (835/36) - (515/24) = (125/72) square units. Calculating Area F. ½(Area under the curve at x = 0 and x = 8) - Area triangle with height (32/3) units and base 4 units - Area triangle with hei...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3109459

Notice; Blue area = yellow area. Let the single side length of blue area be 2 units. Let the single side length of red area be y. Therefore; Tanx=y/(4+y) ------ (1) Tanx=2/(2+y) ------ (2) Equating (1) and (2) y/(4+y)=2/(2+y) 8+2y=2y+y² y²=8 y=2√(2). Therefore; Considering equation (2), and y=2√(2). Angle a will be; 180-2(angle x) And angle x is; Tan-¹(2/(2√(2)+2)) It implies; 180 - 2(tan–¹(2/(2√(2)+2))) = 180 - 2tan-¹(1/(√(2)+1))...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3109445

a = ½(6) a = 3 units. a is the radius of the circle. b = (5-c) units. 6² = d²+(5-c)² d² = 36-(25-10c+c²) d = √(11+10c-c²) units. e = ½(d) e = ½√(11+10c-c²) units. f = (3-c) units. Calculating c. Notice. f = ½(b) 3-c = ½(5-c) 6-2c = 5-c It implies; c = 1 unit. Therefore; For c = 1 unit. e = ½√(11+10c-c²) e = ½√(11+10(1)-1²) e = ½√(20) e = √(5) units. Therefore, length x is; x² = 5²+e² x² = 5²+√(5)² x² = 25+5 x = √(30) units. x = 5.4772255...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3109067

Sir Mike Ambrose is the author of the question. Let the side length of the square be 1 unit. Area square is; 1² = 1 square unit. Calculating Area Orange. a = 45+60 a = 105° b = ½(180-a) b = ½(75) b = 37.5° c = 60-b c = 22.5° d = ½(90-45) d = 22.5° e = 60+d e = 82.5° f = 180-c-e f = 180-22.5-82.5 f = 180-105 f = 75° It implies; (1/sin75) = (g/sin82.5) g = 1.02641924917 units. Therefore, area orange is; Area triangle with height 1.02641924917 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3108812

Calculating angle x. Let the base of the ascribed quadrilateral be 1 unit. a = 180-20-60-50 a = 50° (1/sin50) = (b/sin80) b = 1.28557521937 units. (1/sin40) = (c/sin60) c = 1.34729635533 units. d = 180-60-40-50 d = 30° e² = 1.28557521937²+1.34729635533²-2*1.28557521937*1.34729635533cos30 e = 0.68404028665 units. Therefore, the required angle x is; (1/sinx) = (0.68404028665/sin20) sinx = sin20/0.68404028665 x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3108530

Let the single side length of the inscribed blue square area be x. Therefore; x+(x÷(tan(tan–¹(39/52))))+(x÷(tan(90-tan–¹(39/52))))=65 x=65÷(1+(1÷(tan(tan–¹(39/52))))+(1÷(tan(90-tan–¹(39/52))))) x²=(65÷(1+(1÷(tan(tan–¹(39/52))))+(1÷(tan(90-tan–¹(39/52))))))² Therefore inscribed blue area is; 444.411979548 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3108517

Sir Mike Ambrose is the author of the question. a = ⅛*180(8-2) a = 135° a is the single interior angle of the regular octagon. 2² = 2b² b = √(2) units. c = 2+b c = (2+√(2)) units. tand = (2+√(2))/2 d = atan(1+½√(2))° tand = 2/e 1+½√(2) = 2/e e = 4/(2+√(2)) units. f = c-e f = (2+√(2))-4/(2+√(2)) f = ((2+√(2))²-4)/(2+√(2)) units. 2/(2+√(2)) = g/((2+√(2))²-4)/(2+√(2)) 2/(2+√(2)) = g(2+√(2))/((2+√(2))²-4) g(2+√(2))² = 2(2+√(2))²-8 g = 2-(8/(2+√(2))²) units. Shade...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3108507

Sir Mike Ambrose is the author of the question. 2² = 2(3²)-2(3²)cosa 18cosa = 18-4 a = acos(7/9)° a is angle ABC. Let b be the radius of the inscribed circle. Calculating b. c²+1² = 3² c = √(8) c = 2√(2) units. 2b+3b+3b = 2*2√(2) 8b = 4√(2) b = ½√(2) units. Again b is the radius of the inscribed circle. d = 2√(2)-½√(2) d = ½(3√(2)) units. e²+(0.5√(2))² = (0.5*3√(2))² e² = ½(9)-½ e = √(4) e = 2 units. tanf = 2/(0.5√(2)) f = atan(2√(2))° Therefore, area T e...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3107062

Sir Mike Ambrose is the author of the question. Let CD = 1 units. It implies; DE = 1 units. AD = 2 units. AE = 2 units. 1² = 2²+2²-2*2*2cosa 1 = 8-8cosa cosa = 7/8 a = acos(⅞)° b = ½(180-acos(⅞)) b = (90-0.5acos(⅞))° sinb = c/2 c = 2sin(90-0.5acos(⅞)) c = 1.9364916731 units. c = CF. cosb = d/2 d = 2cos(90-0.5acos(⅞)) d = ½ units. d = EF. AF = AE-EF AF = 2-½ AF = ½(3) units. tan(acos(⅞)) = e/AF e = 1.5tan(acos(⅞)) e = 0.82992500276 units. e = BF The...

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