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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2274134

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Area blue total, exactly in cm² decimal is; Area trapezoid with parallel sides (80/7) cm and ((80/7)+21.4285714284) cm, and height 20 cm - Area trapezoid with parallel sides 20 cm and ((80/7)+12.2448979591) cm, and height (60/7) cm - Area triangle with height (80/7) cm and base 12.2448979591 cm. = (½*20((80/7)+((80/7)+21.4285714284))) - (½*(60/7)(20+((80/7)+12.2448979591))) - (½*(80/7)*12.2448979591) = 442.857142855 - 187.172011662 - 69.9708454806 = 185.714285712 cm² = 185.71 cm² to 2 decima...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2273967

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Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever. Communicate us to mentor/teach 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/2272792

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Side length of the square is; Let it be x. x = √((16*34)-144) x = √(400) x = 20 cm. Therefore; Area Orange exactly in decimal cm² is; Half area square with side 20 cm - Area triangle with height 7.88405797101 cm and base ⅓(32sin(2atan(3/5))) cm - Area triangle with height 8.74642784227 cm and base (12sin(atan(5/3))) cm = ½(20²) - ½(7.88405797101*⅓(32sin(2atan(3/5)))) - ½(8.74642784227*(12sin(atan(5/3)))) = 200 - 37.1014492753 - 45 = 155 - 37.1014492753 = 117.898550725 cm² = 117.8 cm² to 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2272197

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4^(2x)+2 = 3*4^(x) Let 4^(x) = p It implies; p²+2 = 3p p²-3p+2 = 0 Resolving the above quadratic equation. p²-p-2p+2 = 0 p(p-1)-2(p-1) = 0 (p-2)(p-1) = 0 Therefore; p = 1 or p = 2 And p = 4^(x) Calculating x. It implies; For p = 1 4^(x) = 1 4^(x) = 4⁰ x = 0 For p = 2 4^(x) = 2 2^(2x) = 2¹ 2x = 1 x = ½

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2271423

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Let the blue square side length be x. 2² = a²+b² b² = 4-a² b = √(4-a²) 4² = a²+(√(4-a²)+x)² 16 = a²+((4-a²)+2x√(4-a²)+x² 12 = 2x√(4-a²)+x² --- (1). 5² = (x+a)²+(4-a²) 25 = x²+2xa+a²+4-a² 21 = x²+2xa  a = (21-x²)/(2x) --- (2). Calculating x. Substituting (2) and (1) 12 = 2x√(4-((21-x²)/(2x))²)+x² 12 = 2x√(4-((441-42x²+x⁴)/(4x²))+x² (12-x²)² = 4x²((58x²-x⁴-441)/4x²) 144-24x² +x⁴ = 58x²-x⁴-441 2x⁴-82x²+585 = 0 Let x² = p. It implies; 2p²-82p+585 = 0 Resolving the above quadratic equation....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2271409

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a = ½(8+2) a = 5 cm. a is the radius of the ascribed semi circle. b = 8 2 = b. Cross Multiply. b² = 16 b = 4 cm. Therefore, area Blue is; ½(8*4) = ½(32) = 16 cm² Calculating Area Red. tanc = 4/3 c = atan(4/3)° d = (180-c) d = (180-atan(4/3))° Area Red is; Area sector with radius 5 cm and angle (180-atan(4/3))° - Area triangle with height 5 cm and base 5sin(180-atan(4/3)) cm. = ((180-atan(4/3))π*25/360) - (0.5*25sin(180-atan(4/3))) = 27.6787179449-10 = 17.6787179449 cm² It implies; A...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2267593

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Sir Mike Ambrose is the author of the question. 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 unit. (e/tan36)+(e/tan27) = d Where e is the radius of the inscribed circle. e = 0.61803398875/((1/tan36)+(1/tan27)) e = 0.18509595408 unit. f = 180-72-54 f = 54° Therefore; g = d = 0.61803398875 unit. Shaded Area Red is; Area triangle with height...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2267580

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Notice. a = 18+55 = 73° b = 28+45 = 73° It implies; The ascribed triangle is isosceles. Calculating angle x. Let the shorter length of the ascribed isosceles triangle be 1 unit. c = 180-45-55-18 c = 180-118 c = 62° (1/sin62) = (d/sin45) d = 0.800847963 units. cos73 = e/0.800847963 e = 0.2341452842 units. f = 1-2*e f = 1-2*0.2341452842 f = 0.5317094316 units. g = ½(360-73-73) g = ½(360-146) g = ½(214) g = 107° h = 180-55-73 h = 180-128 h = 52° (1/sin52) = (j/sin55) j = 1.0395188651 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2266498

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Notice. ABCD is a square. Let AB = 1 unit. cos50 = a/1 a = 0.6427876097 units. a is CE. b = ½√(1²+1²) b = ½√(2) units. b = 0.7071067812 units. b is OC. c² = 0.7071067812²+0.6427876097²-2*0.7071067812*0.6427876097cos(50+45) c = 0.9961946981 units. c is OE. (0.9961946981/sin95) = (0.6427876097/sind) d = 40° d is angle COE. e = 90-d e = 90-40 e = 50° e is angle BOF. Therefore, the required angle, x ( angle BFO) is; x = 180-45-e x = 180-45-50 x = 180-95 x = 85°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2265879

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tana = (10/5) a = atan(2)° b = ½(180-a) b = ½(180-atan(2))° cosb = c/10 cos(½(180-atan(2))) = c/10 c = 5.2573111212 cm. sinb = d/10 sin(½(180-atan(2))) = d/10 d = 8.5065080835 cm. It implies, blue area (right-angled triangle area) is; ½(cd) = 0.5*5.2573111212*8.5065080835 = 22.360679775 square units.
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