Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2908034

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 sector with radius 4√(5) units and angle (2atan(0.5))° + Area triangle with height 4√(5) units and base 4√(5)sin(2atan(0.5)) units. = (½*4*16) - ((2atan(0.5))*π*(4√(5))²÷360) + (½*4√(5)*4√(5)sin(2atan(0.5))) = 16 - (4πatan(0.5))/9 + 32 = 48 - (4πatan(0.5))/9 = 4(108-πatan0.5)/9 square units. =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2906192

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 = ⅓(8) units. OH = ½(5) units. EH = ½(3) units. Showing that; (OG+GE)(OH+EH) = Area triangle ABC. (⅓(10)+⅓(8))(½(5)+½(3)) = 24 (⅓(18))(½(8)) = 24 6 * 4 = 24 24 = 24 proved.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2905934

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 and base 2√(2)sin45 units - Area triangle with height ⅔√(10) units and ⅓(7sin(atan(2)-45)) units. = (½*⅓(10√(2))*½(5√(2))) - (½*(8/3)*(2√(2)/√(2)) - (½*⅔√(10)*⅓(7sin(atan(2)-45))) = ⅓(25) - ⅓(8) - (7/9) = ⅓(17) - 7/9 = (51-7)/9 = 44/9 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2904569

Let the side of the regular pentagon be 1 unit. a = 360-2(108)-90 a = 54° tan72 = b/0.5 b = 1.53884176859 units. tan54 = 1.53884176859/c c = 1.11803398875 units. d = c-0.5 d = 0.61803398875 units. (e/tan36)+(e/tan27) = d Where e is the radius of the inscribed circle. e = 0.61803398875/((1/tan36)+(1/tan27)) e = 0.18509595408 units. f = 180-72-54 f = 54° Therefore; g = d = 0.61803398875 units. Shaded Area Red is; Area triangle with height 0.61803398875...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2903304

(x/√(3))+x+2x+3x+(3x/√(3)) = 1 (4√(3)x/3)+6x = 1 (4√(3)x)+18x=3 (4√(3)+18)x = 3 x = 3/(4√(3)+18) x = 3(18-4√(3))/(324-48) x = 3(15-4√(3))/276 x = (18-4√(3))/92 x = (9-2√(3))/46 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2901810

Sir Mike Ambrose is the author of the question. Area yellow in square units to 2 decimal places is; Area triangle with height (tan70+tan(450/7)) units and base (tan70+tan54)sin80 units. = ½(tan70+tan(450/7))(tan70+tan54)sin80 = 9.79563286034 square units. ≈ 9.80 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2900788

Sir Mike Ambrose is the author of the question. At t = 2, x = 3 and y = 3/4 dx/dt = 2/3 dy/dt = 13/16 Therefore; Gradient at the tangent and of the curve at t = 2 is; dy/dx = dy/dt÷dx/dt = (13/16)÷(2/3) dy/dx = 39/32 Notice; Triangle OPQ height coordinate is (3, (3/4)). Triangle OPQ base coordinate is (x1, 0). Calculating x to get X1. ((3/4)-0)/(3-x) = 39/32 x = 31/13 x1 = 3-(31/13) x1 = 18/13. It implies; Triangle OPQ is; Area triangle with...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2900060

Sir Mike Ambrose is the author of the question. At x = 0, y = 2. At y = 16, x = 1. Gradient of the curve at x = 1 or at the tangent is; dy/dx = 3In2*2^(3x+1) For x = 1 dy/dx = 48In2 Therefore coordinate Q is (x¹, 0). Calculating x¹. (0-16)/(x¹-1) = 48In2 x¹= (3In2-1)/3In2 It implies; Area Red exactly as a single fraction in its simplest form is; Area trapezoid with parallel sides 2 units and 16 units, and height 1 unit - Area triangle with height 16 units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2899870

#Sir Mike Ambrose is the author of the question. Let the side length of the regular Pentagon be 1 unit. Therefore Area Pentagon is; (5*1)÷(4tan(180/5)) = 5/(4tan36) space units. = 1.72047740059 space units. Area star is; 5(area triangle with height 0.61803398875 units and base 0.61803398875sin36 units) + Area regular Pentagon with side 0.38196601125 units. = 5(0.5*0.61803398875*0.61803398875sin36) + ((5*0.38196601125²)/(4tan(180/5))) = 0.56128497072 + 0.25101426986 = 0.812...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2898445

Sir Mike Ambrose is the author of the question. Let the side length of the square be 1 unit. Therefore; Area square is; 1² = 1 square units. It implies; Area R is; Area square - Area equilateral triangle with side 1 unit - 2(area sector with radius 1 unit and angle 30°) = 1 - ½(sin60) - 2(30π÷360) = 1 - ¼(√(3)) - (π/6) = (12-3√(3)-2π)/12 square units. Therefore; Area R ÷ Area Square as a single fraction in terms of π is; (12-3√(3)-2π)/12 ÷ 1 = (12-3√(3)-2π)/...

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