Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2318812

Notice! Length 3 units and length 6 units are both tangent to the green inscribed circle. It implies; Radius r, of the ascribed semi circle is; r = ½(3+6) r = ½(9) r = 4.5 units. Where 9 units is the diameter of the ascribed semi circle. 9² = 5²+6²-2*5*6cosa 81 = 61-60cosa 60cosa = 61-81 cosa = (-20/60) a = acos(-1/3) a = 109.4712206345° a is angle ADB. b = 180-a b = 180-109.4712206345 b = 70.5287793655° b is angle BDC. Let length CD, the required length be x. It implies; cos70.5287793...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2315557

Let the side length of the regular pentagon be a. b = ⅕(180(5-2)) b = 108° b is the single interior angle of the regular pentagon. c = 108-90 c = 18° d = (180-2(18)) d = (180-36) d = 144° e = ½(f) e = ½(144) e = 72° notice! Radius of the circle is 1 unit. It implies; sin72 = 0.5a/1 0.5a = sin72 a = 2sin72 units. a is the side length of the regular pentagon which is also the side length of the inscribed square. It implies; Area square exactly is; a² = (2sin72)² = 4sin²(72) square units...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2315187

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/2315105

Let AE = CE = BE = b. Let OE = a. 4² = a²+b²-2abcos45 16 = a²+b²-√(2)ab --- (1). 6² = a²+b²-2abcos135 36 = a²+b²+√(2)ab --- (2). Solving (1) and (2) simultaneously. Subtracting (1) from (2). (36 = a²+b²+√(2)ab) -(16 = a²+b²-√(2)ab) 20 = 2√(2)ab --- (3). From (3). a = 20/(2√(2)b) a = 10/(√(2)b) a = (5√(2)/b) --- (4). Therefore, Substituting (4) in (1) to get b. 16 = ((5√(2)/b))²+b²-√(2)((5√(2)/b))b 16 = (50/b²)+b²-10 26-b² = (50/b²) 26b²-b⁴ = 50 b⁴-26b²-50...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2313943

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/2312970

Area of coloured region is; Area triangle with height 53.3479501083 m and base 60sin(102.042575143) m + Area triangle with height 100.027406452 m and base √(5924)sin(12.5245961777) m = (½*53.3479501083*60sin(102.042575143)) + (½*100.027406452*√(5924)sin(12.5245961777)) = 1565.21739131 + 834.782608691 = 2400 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2312798

Observing Pythagoras Rule. (n³)² = (n²)²+n² n⁶ = n⁴+n² n⁶-n⁴-n² = 0 n²(n⁴-n²-1) = 0 Let n² = p. It implies; p(p²-p-1) = 0 p²-p-1 = 0 Resolving the above quadratic equation, applying completing the square approach. (p-½)² = 1+(½)² (p-½)² = 1+¼ p-½ = ±√(5/4) p = ½±½√(5) p = ½(1±√(5)) Therefore; p = ½(1-√(5)) Or p = ½(1+√(5)) Notice. p = n² Therefore; For p = ½(1-√(5)) n ≠ √(½(1-√(5))) For p = ½(1+√(5)) n = √(½(1+√(5))) n = 1.2720196495 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2312738

Sir Mike Ambrose is the author of the question. Let beta be x. x = (½*atan(5/4)) Let alpha be y. y = (½*atan(4/5))° Therefore; Area blue exactly in cm² decimal is; Area triangle with height (10/cos(0.5atan(4/5))) cm and base (8/cos(0.5atan(5/4)))sin45 = ½* (10/cos(0.5atan(4/5)))*(8/cos(0.5atan(5/4)))sin45 = √(2)(5/cos(0.5atan(4/5)))(4/cos(0.5atan(5/4))) = 33.2562362737 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2312121

Let the two equal red lengths of the triangle, ABC be 1 unit. Therefore; AB = CE = 1 unit. a = 180-100-20 a = 60° (b/sin100) = (1/sin20) b = 2.8793852416 units. b is BC. c = 2.8793852416-1 c = 1.8793852416 units. c is BE. Notice; BE = CD = the blue equal lengths. It implies; d² = 1+1.8793852416²-2*1*1.8793852416cos20 d = 1 unit. d is DE. Notice; CE = DE = 1 unit. Therefore; e = 180-2(20) e = 180-40 e = 140° e is angle CED. It implies; The required angle, x is; x+140 = 180 x = 180-14...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2311442

Sir Mike Ambrose is the author of the question. Calculating a. ½((0, 6√(3)) (-3, 3√(3)) (a, (16a/3)) (0, 6√(3))) = ¼(9+20√(3)) Therefore; a = x = ½√(3) unit y = 16a/3 = (16/3)(√(3)/2) y = ⅓(8√(3)) unit. Therefore; Area green is; Area triangle with height (√(697-72√(3))/2√(3)) unit and base 6sin(120-atan(16√(3)/(36-3√(3))) unit = ½*(√(697-72√(3))/2√(3))*6sin(120-atan(16√(3)/(36-3√(3))) = ½(√(2091-216√(3)))sin(120-atan(16√(3)/(36-3√(3))) = ½√(2091-216√(3)sin(120-atan((64√(3)+16)/141) squa...

You Searched for "post"

2242 result found.

Mathematics Question and Solution

Mathematics Question and Solution

We are smart and clever!

Mathematics Question and Solution

We are smart and clever!

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

OnlineEdumath

Socials

Useful Tags

Quick Links

Reach Us

Safety and Abuse Reporting

Thanks for being awesome!

Are you sure you want to report abuse against this website?

Search

You Searched for "post"

2242 result found.

Subscribe to our mailing list for latest updates.

Safety and Abuse Reporting

Thanks for being awesome!

Are you sure you want to report abuse against this website?