Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2298926

6² = 4²+6²-2*6*4cosa 48cosa = 52-36 cosa = 16/48 a = acos(⅓) a = 70.5287793655° b = 180-70.5287793655 b = 109.4712206345° c = 180-2(70.5287793655) c = 38.942441269° d = ½(180-70.5287793655) d = 54.7356103172° e = 180-54.7356103172-109.4712206345 e = 15.7931690483° It implies; (4/sin15.7931690483) = (f/sin54.7356103172) f = 12 cm. g = 12+6 g = 18 cm. Therefore, the required length, x is; sin38.942441269 = x/18 x = 18sin38.942441269 x = 11.313708499 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2298875

a = 180-60 a = 120° b = 180-120-45 b = 15° (1/sin15) = (c/sin45) c = 2.7320508076 units. d² = 2.7320508076²+2²-2*2*2.7320508076cos60 d = 2.4494897428 units. It implies; Calculating angle x. (2.7320508076/sinx) = (2.4494897428/sin60) x = 75°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2297100

Equation of the curve is; y = ¼(x²)+4 Shaded area exactly in square units is; Area triangle with height 8 unit and base 4 unit - Area sector with radius 4√(5) unit and angle (2atan(0.5))° + Area triangle with height 4√(5) unit and base 4√(5)sin(2atan(0.5)) unit = (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. = 10.9081912799 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2295336

Area orange exactly in square units is; Area triangle with height ⅓(10√(2)) unit and base 5sin45 unit - Area triangle with height (8/3) unit and base 2√(2)sin45 unit - Area triangle with height ⅔√(10) unit and ⅓(7sin(atan(2)-45)) unit. = (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/2292279

Sir Mike Ambrose is the author of the question. Let the side of the square be 2 units. Therefore; a = 3√(5)/5 unit b = 2√(5)/5 unit c = √(5)/5 unit d = 2√(5)/5 unit It implies; a² = b² + c² + d² is; (3√(5)/5)² = (2√(5)/5)² + (√(5)/5)² + (2√(5)/5)² 9/5 = 4/5 + 1/5 + 4/5 9/5 = 9/5. Please, move the above question left or right one time to also review pen written solution of the question. Thank you.

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Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever. Communicate us to mentor/teach/educate your child/children Mathematics online at affordable tuition, helping them become a lover of Mathematics, and also, a Mathematician. Call/WhatsApp: +2349076614992 or +2348152827445 Email: [email protected] [email protected] [email protected] Facebook Page: Online Edumath We are super smart and amazingly, very clever!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2289190

Let the side length of the regular Pentagon be 1 unit. Therefore Area Pentagon is; (5*1)÷(4tan(180/5)) = 5/(4tan36) space unit. = 1.72047740059 space unit. Area star is; 5(area triangle with height 0.61803398875 unit and base 0.61803398875sin36 unit) + Area regular Pentagon with side 0.38196601125 unit. = 5(0.5*0.61803398875*0.61803398875sin36) + ((5*0.38196601125²)/(4tan(180/5))) = 0.56128497072 + 0.25101426986 = 0.81229924058 space unit. It implies; Area Star ÷ Area Pentagon to 3 d. p....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2289168

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 unit and 16 unit, and height 1 unit - Area triangle with height 16 unit and base (1-((3In2-1)/3In2)) = (½*1(2+16)) - (½(16)(1-((3In2-1)/3In2)) = 9 - (8(1/3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2288018

Let the side length of the be 2 unit. Therefore; BC = 2 unit. It implies; Area orange is; ½(1*2) = 1 square unit. Area green is; ½(1*2) + ½(2/√(5))(4/√(5)) = 1 + (4/5) = (9/5) Therefore; Area orange : Area green is; 1 : (9/5) =(5*1) : (5*(9/5)) = 5 : 9

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2288002

Let the side length of the square be 1 unit. Therefore; Area square is; 1² = 1 square unit. 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 unit. Therefore; Area R ÷ Area Square as a single fraction in terms of π is; (12-3√(3)-2π)/12 ÷ 1 = (12-3√(3)-2π)/12

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