Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2482618

a² = 4²+3² a = √(25) a = 5 units. a is the radius of the semi circle. tanb = (3/4) b = atan(3/4)° c = 90-b c = (atan(4/3))° tan(atan(3/4)) = d/5 (3/4) = d/5 d = 3.75 units. d = ¼(15) units. tan(atan(4/3)) = e/5 (4/3) = e/5 e = ⅓(20) units. Area 1 is; Area triangle with height 5 units and base ¼(15) units - Area sector with radius 5 units and angle atan(3/4)° = ½*5*¼(15)-(atan(3/4)π*25/360) = ⅛(75)-(25atan(3/4)π/360) = 25(135-atan(3/4)π)/360 square units. Area 2 is; Area triangle with...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2481368

Calculating x (radius of the small inscribed circle) 28² + x² = (56-x)² 784 + x² = 3136 - 112x + x² 112x = 2351 x = 21 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2479474

a = 108-90 a = 18° cos18 = b/1 b = 0.9510565163 units. c = 1+0.9510565163 c = 1.9510565163 units. c is the side length of the regular ascribed pentagon. sin18 = d/1 d = 0.3090169944 units. e² = 2*1.9510565163²-2*1.9510565163²cos108 e = 3.1568757573 units. Therefore, green area is; Area regular pentagon with side length 1.9510565163 cm - 5(area square with side length 1 unit) - 5(area triangle with height 0.9510565163 units and base 0.3090169944 units. = 2(0.5*1.9510565163*1.9510565163si...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2479159

a = 120π*2*30÷360 a = ⅓(60)π a = 20π cm. a = 62.8318530718 cm. a is the length of arc of the sector and also, the base circumference of the cone. 20π = 2πb b = 10 cm. b is the radius of the base of the circular cone. 30² = 10²+c² c² = 900-100 c = √(800) c = 20√(2) cm. c is the height of the cone. Volume of the cone is; ⅓*10*10π*20√(2) = ⅓(2000√(2))π cm³ = 2961.9219587722 cm³ Calculating the volume outside the sphere. 2*30d+20d = 20*20√(2) 80d = 400√(2) d = 5√(2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2479061

Let the square side be 2a cm. It implies; a is the radius of the inscribed semi circle. Let b equal the radius of the big inscribed quarter circle. Therefore; (a+b)² = a²+(2a)² a²+2ab+b² = 5a² 2ab+b² = 4a² --- (1). (2a-2)² = 2²+b² 4a²-8a+4 = 4+b² b² = 4a²-8a --- (2) From (2). b = √(4a²-8a) --- (3). Substituting (2) and (3) in (1) to get a. 2ab+b² = 4a² 2a√(4a²-8a)+4a²-8a = 4a² 2a√(4a²-8a) = 8a 4a²(4a²-8a) = 64a² 4a²-8a = 16 a²-2a-4 = 0 Resolving the abo...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2478964

Let the two equal lengths of the isosceles triangle be 1 unit each. a = 180-2(12+36) a = 180-96 a = 84° b² = 2-2cos84 b = 1.3382612127 units. c = 180-12-30 c = 138° (d/sin30) = (1.3382612127/sin138) d = 1 unit. It implies; The required angle x, is; x = ½(180-36) x = ½(144) x = 72°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2478768

Let the side length of the regular pentagon be 1 unit. a² = 2-2cos108 108° is the single interior angle of the regular pentagon. a = 1.6180339887 units. a is BE. b = 108-60 b = 48° c = ½(180-48) c = 66° d = 180-60-c d = 54° e = 180-d-b e = 78° f = 180-e f = 102° g = 180-f-72 g = 6° g is angle BEG = angle BGE. h = 180-2g h = 168° h is angle EBG. j = h-72 j = 96° j is angle CBG. k = a-1 k = 1.6180339887-1 k = 0.6180339887 units.. k is BF l² = 0....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2478557

a² = 3²+1² a = √(10) units. a is AB. It implies; AC = ½√(10) units. tanb = 1/3 b = atan(1/3)° tanc = ½√(10)/√(10) c = atan(½)° d² = (½√(10))²+√(10)² d² = ½(5)+10 d = √(25/2) d = ½(5√(2)) units. Observing similar right-angled triangle side length ratios. ½√(10) - ½(5√(2)) e - √(10) Cross Multiply. 10 = 5√(2)e e = √(2) units. sinf = √(2)/(0.5√(10)) f = 63.4349488229° tanh = 3/1 h = atan(3)° h = 71.5650511771° j = 63.4349488229+71.5650511771 j...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2478126

tan30 = a/12 a = 12tan30 a = 4√(3) cm. a is the side length of the square. b = ½(12)+a b = (6+4√(3)) cm. c = ½(a) c = 2√(3) cm. Observing similar right-angled triangle side length ratios. 2√(3) - d (6+4√(3)) - 4√(3) Cross Multiply. 24 = d(6+4√(3)) d = 12/(3+2√(3)) d = 12(3-2√(3))/(-3) d = 4(2√(3)-3) cm. Shaded Area exactly is; Area square with side 4√(3) cm - Area triangle with height 4√(3) cm and base 4(2√(3)-3) cm. = (4√(3))²-½*4(2√(3)-3)*4√(3) = 48-48+24√(3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2477392

x+50+180-a = 180 a = x+50 --- (1). x+30+180-b = 180 b = x+30 --- (2). 3x+180-2a+180-2b = 180 3x+180-2(a+b) = 0 2(a+b) = 3x+180 --- (3). Therefore, substituting (1) and (2) in (3) to get x. 2(x+50+x+30) = 3x+180 4x+160 = 3x+180 x = 20° Therefore; 3x is; = 3*20 = 60°

You Searched for "delta state office ekrejeta house 2 zenith bank abraka delta state nigeria"

2620 result found.

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

OnlineEdumath

Socials

Facebook Page

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 "delta state office ekrejeta house 2 zenith bank abraka delta state nigeria"

2620 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?