Happy Birthday, Oghenevwarhe.

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The Management and Educators of Online Edumath kindly, joyfully wish Oghenevwarhe, one of our super smart and amazingly, very clever learners a happy birthday.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2308770

a² = (2+1)²+2² a² = 9+4 a = √(13) units. Where a is blue length + red length. b² = 1+2² b = √(5) units. tanc = 2/3 c = atan(2/3)° tand = 2/1 d = atan(2)° e = 90-d e = 90-atan(2) e = atan(½)° f = 180-e-c f = 180-atan(½)-atan(2/3) f = 119.7448812969° (2/sin119.7448812969) = (g/sin(atan(2/3))) g = 1.27775313 units. h² = 1.27775313²+√(5)² h = 2.5753937682 units. Where h is blue length. Therefore; Length red is; √(13)-2.5753937682 = 1.0301575073 units. It implies; Length blue : Length re...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2308096

Notice. Radius of the semi circle is 5 cm. Let the width of the blue rectangle be a. Therefore; Blue rectangle length is 2a. b² = a²+a² b = √(2a²) b = √(2)a c = ½(b) c = ½(√(2)a) d²+c² = 5² d² = 25-(½(√(2)a))² d² = 25-½(a²) d = √(25-½(a²)) e = d-b e = √(25-½(a²))-½(√(2)a) r² = b²+e² Where r is the radius of the semi circle (5 cm). 5² = (√(2)a)²+(√(25-½(a²))-½(√(2)a))² 25 = 2a²+25-½(a²)-a√(50-a²)+½(a)² 0 = 2a²-a√(50-a²) a√(50-a²) = 2a² √(50-a²) = 2a 50-a² = 4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2307392

Calculating the length and width of the rectangle. Let the length be a. Let the width be b. Notice; ½(a) = b Or 2b = a It implies; b²+b² = 4² 2b² = 16 b² = 8 b = 2√(2) cm. Therefore; a = 2b a = 4√(2) cm. Let the semi circle radius be c. Calculating c. d²+2² = c² d = √(c²-4) e = d+2 e = √(c²-4)+2 It implies; c² = (4√(2))²+(√(c²-4)+2)²-2*4√(2)*(√(c²-4)+2)cos45 c² = 32+c²-4+4√(c²-4)+4-8(√(c²-4)+2) 0 = 32+4√(c²-4)-8√(c²-4)-16 0 = 16-4√(c²-4) 16 = 4√(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2304845

Let the base of the triangle be 1 unit. a = 20+100 = 120° b = 10+10 = 20° c = 190-a-b c = 180-120-20 c = 40° It implies; (1/sin120) = (d/sin40) d = 0.742227199 units. e = 180-10-20 e = 180-30 e = 150° It implies; (0.742227199/sin150) = (f/sin20) f = 0.5077133059 units. g² = 0.5077133059²+1²-2*0.5077133059cos10 g = 0.507713306 units. Therefore, the required angle, angle x is; (0.507713306/sin10) = (0.507713306/sinx) sin10 = sinx It implies; x = 10°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2303187

Sir Mike Ambrose is the author of the question. Radius, r of the inscribed circle is; r = (3-√(3)) cm Therefore; Area purple exactly in square cm is; Area trapezium with parallel sides (3-√(3)) cm and 4 cm, and height 3+√(3) cm - 2(area triangle with height (3-√(3)) cm and base (3√(3)-3) cm). = ½*(3+√(3))(4+3-√(3)) - 2(½*(3-√(3))*(3√(3)-3)) = (9+2√(3)) - (12√(3)-18) = (27-10√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2303172

Shaded area is; Area trapezoid with parallel sides 58 cm and 28 cm, and height 30 cm - Area trapezoid with parallel sides 9 cm and 30 cm, and height 28 cm - Area triangle with height 12 cm and base 12 cm. = ½*30(28+58) - ½*28(9+30) - ½*12*12 = (15*86) - (14*39) - 72 = 1290 - 546 - 72 = 1290 - 618 = 672 cm²

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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/2301614

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

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2301600

Sir Mike Ambrose is the author of the question. P(2, (16√(5)/5)) Gradient of the curve at the tangent is; = 24/5√(5) Therefore; Gradient of the curve at the normal is; = -5√(5)/24 It implies; Coordinate Q is; Q(4, (167√(5)/60)) Therefore area R as a single fraction in square units is; (Area under the curve at x = 2 and x = 4) - (Area trapezium with parallel sides 167√(5)/60 unit and 16√(5)/5 unit, and height 2 unit). = (24-⅓(8√(5))) - ½*2((167√(5)/60 )+(16√(5)/5)) =...

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Happy Birthday, Oghenevwarhe.

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

We are smart and clever!

Mathematics Question and Solution

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