Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2074713

Let the small square side be 2 units. Area Blue is; 0.5*2*1 = 1 square unit. a² = 2²+1 a = √(5) units. a is the side of the big square. b = (90-atan(½))° c = 180-(90-atan(½))-60 c = (30+atan(½))° d = 180-90-(30+atan(½)) d = (60-atan(½))° Let the side length of the inscribed regular hexagon be e. (e/sin(90-atan(½))) = (f/ sin60) f = 0.96824583655e units. g = (√(5)-0.96824583655e) units. cos(60-atan(½)) = (√(5)-0.96824583655e)/h h = (2.67949192431-1.16025403784e) units. j = 2(e/2)+e j =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2074369

Calculating p. p^(⅓)+1 = 6p^(⅙) Let p^(⅓) = x x + 1 = 6x^(½) (x+1)² = 36x x²+2x+1 = 36x x²-34x+1 = 0 (x-17)² = -1+(-17)² x = 17±√(288) x = 17-12√(2) Or x = 17+12√(2) It implies; p^(⅓) = x, for x = 17-12√(2) p = (17-12√(2))³ p = 0.0000255089 p^(⅓) = x, for x = 17+12√(2) p = (17+12√(2))³ p = 39202 Where 39202 is the maximum value of p. Therefore, maximum value of; p^(¼)-(1/p^(¼)) is; 39202^(¼)-(1/39202^(¼)) = 14 Therefore, the minimum value of p^(¼)-(1/p^(¼)) will be; For x = 0.00002550...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2073877

Sir Mike Ambrose is the author of the question. P(2, (8/3)) Q(4, (32/15)) Gradient of the curve is; dy/dx = -4x/(x²-1)² For x = 2. dy/dx = -8/9 It implies; R(5, 0) Area S exactly in square units is; Area under the curve at x = 2 and x = 4 - Area trapezoid with parallel sides (8/3) units and (8/9) unit, and height 2 units. = 4 + ln(9/5) - ½*2((8/3)+(8/9)) = 4 + ln(9/5) - (32/9) = (4/9) + ln(9/5) = (4+9(ln(9/5)))/9 square units.

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Our certified, experienced, resilient, productive and committed Math Educators will be glad to be at your service. We make teaching and learning Mathematics fun for learners, we expose the simplicity and beauty of Mathematics to our learners, and mentor them to appreciate and love Mathematics. Communicate us to mentor/educate you Mathematics better online at affordable tuition. We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2072034

Let the square side be 2 units. Area Square is; 2² = 4 square units. tana = ½ a = atan(½)° b = 2a b = 2atan(½)° c = (180-2atan(½))° d = atan(2)° e = (180-2atan(½))-atan(2) e = (90-atan(½))° tan(0.5atan(2)) = f/1 f = 0.61803398875 units. tan(0.5(90-atan(½))) = g/1 g = 0.61803398875 units. Therefore; Area Lilac is; 2*0.5(2*1)-4*0.5(1*0.61803398875 ) = 2-2*0.6180339887 = 0.7639320226 square units. Area Lilac ÷ Area Square to 2 decimal places is; 0.7639320226÷4 = 0.19098300565 ≈ 0.1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2071369

Sir Mike Ambrose is the author of the question. a = atan(2)° b = (45+atan(½))° (a) Calculating Alpha. Let alpha be c. c = 180-b c = 180-45-atan(½) c = (135-atan(½))° Or c = 180-(135-atan2) c = (45+atan2)° Let the green square side be d. Calculating d. 2d+½(d) = 30 d = 12 cm. e = atan(⅓)° f² = 18²+6² f = 6√(10) cm. g² = 18²+56² g = 2√(865) cm. h² = 42²+54² h = 6√(130) cm. i = 90-atan(9/28)-atan(7/9) (b) Calculating Length CD exactly in cm. Let it be j. j² = (2√(865))²+(6√(130))²-2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2069771

Let the side length of the regular hexagon be 1 unit. Calculating the side length of the regular pentagon. Let it be b. (2/sin72) = (a/sin60) a = 1.82118599462 units. (1.82118599462/sin108) = (b/sin36) b = 1.12555484451 units. Notice; b is the side length of the regular pentagon. d = 180-60-72 d = 48° (2/sin72) = (e/sin48) e = 1.56277742226 units. f² = 1.56277742226²+1.12555484451²-2*1.56277742226*1.12555484451cos144 f = 2.56032328694 units. (2.56032328694/sin144) = (1.56277742226/sing)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2069018

Sir Mike Ambrose is the author of the question. a² = 8²+12²-2*8*12cos60 a = 4√(7) cm. b = 2a b = 8√(7) cm. (4√(7)/sin60) = (8/sinc) c = 40.89339464928° (16/sin40.89339464928) = (8√(7)/Sind) d = 60° e² = 12²+16²-2*12*16cos60 e = 4√(13) cm. (a) Let beta be x. Calculating tanx. (12/sinx) = (4√(13)/sin60) x = asin(3√(39)/26) It implies; tanx is; tan(asin(3√(39)/26)) = ⅕(3√(3)) (b) Calculating Orange hi Area. f² = 6²+8²-2*6*8cos60 f = 2√(13) cm. g² = 8²+24²-2*8*24cos60 g = 8√(7) cm. (8/...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2067532

20² = 12²+a² a² = 400-144 a = √(256) a = 16 cm. b = 16-9 b = 7 cm. c = atan(3/4)° Let Alpha be d. d = ½(c) d = 0.5atan(¾)° tan(0.5atan(¾)) = e/16 e = ⅓(16) cm. tan(0.5atan(¾)) = f/7 f = ⅓(7) cm. g = 12-⅓(16) g = ⅓(36-16) g = ⅓(20) cm. cos(0.5atan(¾)) = 7/h h = 7.37864787373 cm. cos(0.5atan(¾)) = 16/j j = 16.86548085423 cm. k = 16.86548085423-7.37864787373 k = 9.4868329805 cm. l = ½(k) l = 4.74341649025 units. ¾ = m/7 m = ¼(21) cm. n = ¼(21)-⅓(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2066258

Sir Mike Ambrose is the author of the question. a² = 32-32cos120 a = 4√(3) units. b = 2a b = 8√(3) units. c² =32-32cos150 c = 7.72740661031 units. sin30 = d/4 d = 2 units. sin60 = e/4 e = 2√(3) units. tan15 = 2√(3)/f f = (6+4√(3)) units. g =f-(8√(3)-2) g = (8-4√(3)) units. It implies; Area yellow rectangle is; 4*(8-4√(3)) = (32-16√(3)) square units. Area yellow pentagon is; 2(area triangle with height 4 units and base 4sin120 units) + Area triangle with...

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