By OnlineEdumath   |  5th January, 2025
Sir Mike Ambrose is the author of the question. AB exactly is; = 16tan(3acos(8/9)) - 2√(17) = 2(8tan(3acos(8/9))-√(17)) cm. = 102.760478669 cm.
By OnlineEdumath   |  4th January, 2025
Let AB = 4 units. BC is; √((2√(20))²-4²) BC = √(80-16) BC = √(64) BC = 8 units. It implies; AB ÷ BC = 4 ÷ 8 = ½
By OnlineEdumath   |  4th January, 2025
Sir Mike Ambrose is the author of the question.  y = e, x = e² dy/dx at the tangent is; = 1/(y+2yIny) for y = e. dy/dx at the tangent = 1/3e Therefore; dy/dx at the normal = -3e...
By OnlineEdumath   |  3rd January, 2025
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 sec...
By OnlineEdumath   |  2nd January, 2025
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 =...
By OnlineEdumath   |  2nd January, 2025
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 a...
By OnlineEdumath   |  1st January, 2025
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 = 0.618...
By OnlineEdumath   |  31st December, 2024
(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))...
By OnlineEdumath   |  30th December, 2024
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(...
By OnlineEdumath   |  29th December, 2024
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÷...
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