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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1880138

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xy = 40 ------ (1).   b = 4+x-5 b = x-1 y(x-1) = 20 ------ (2). At (1) x = 40/y ------ (3). Notice; y is the length or base of the rectangle. Substituting (3) in (2) to get y. y((40/y)-1) = 20 40-y = 20 y = 40-20 y = 20 cm.

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1879888

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Screenshot of today's teaching and learning activities with arithmetical, super smart and clever Oghenevwarhe. Oghenevwarhe is in year 8. Educator: Mr. Daniel

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1879482

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Making a circle to ascribe A, B, C and E quadrilateral with the circumference of the circle cutting across A, B, C and E. Therefore; Angle ABC will be; ½(240) = 120° Parallelogram ABCD height is; Let it be x. sin60 = x/4 ½√(3) = x/4 x= 2√(3) cm. Parallelogram ABCD Area is; 6*2√(3) = 12√(3) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1879295

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Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Where g is angle PSA. ta...

Happy Birthday Ogheneruona

https://onlineedumath.com/update/happy-birthday-ogheneruona/1878945

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The Management and Educators of Online Edumath gladly wish you, Ogheneruona a happy birthday. Thank you for allowing us to celebrate with you. Ogheneruona is a year 10 student, a Mathematician, and a proud Online Edumath learner. We encourage you to continue to work hard in your academics because working hard is never enough. We are super proud of you, we love you and we like you to know that you are smart and clever.

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1877952

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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. We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1877949

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½(360-60-60-64-36) = ½(140) = 70° Therefore; Required angle is; 180-64-70 = 180-134 = 46°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1877416

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Let the base of the quadrilateral be 1 unit. (1/sin80) = (a/sin60) a = 0.87938524157 units. (0.87938524157/sin80) = (b/sin30) b = 0.44647558819 units. c² = 0.44647558819²+1-2*0.44647558819cos110  c = 1.2266815969 units. Calculating x, the required angle. (1.2266815969/sin110) = (1/sinx) sinx = 0.76604444312 x = asin(0.76604444312 ) x = 50°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1877260

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AD = √(12²+6²) AD = 6√(5) cm. Angle AMD = atan(2√(5))° Angle EBM = (90-ata(2√(5)))° sin(90-ata(2√(5))) = EM/9 EM = 1.96396101212 cm. Angle CME = 180-atan(2√(5)) = 102.60438264838° (CE)² = 9+1.96396101212²-6*1.96396101212cos102.60438264838 CE = 3.92792202425 cm. (3.92792202425/sin102.60438264838) = (1.96396101212/sina) a = 29.20593224736° Where a is angle ECM. cos29.20593224736 = 6/CF CF = 6.87386354243 cm. EF = CF-CE  EF = 6.87386354243-3.92792202425 EF = 2.94594151818 cm. Therefore;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1876454

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Let a be the side of the inscribed square. Calculating a. tan60 = a/b b = a/(tan60) cm. Where b is BS = CR. Therefore; 2b+a = 10 2a/(tan60)+a = 10 a((2/tan60)+1) = 10 a = 10(2√(3)-3) cm. a = 4.64101615138 cm. c = ½(10-a) c = 0.5(10-4.64101615138) c = 2.67949192431 cm. Where c = b. d = c+a d = 2.67949192431+4.64101615138 d = 7.32050807569 cm. Where d is BR. e = atan(4.64101615138/7.32050807569) e = 32.37365912705° It implies; Calculating TI, let it be f. tan32.37365912705 = f/g g = f/...
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