Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3104876

Calculating x. Notice. The composite plane shape is drawn to scale. 2(60+x)+x = 180 120+2x+x = 180 3x = 60 x = 20° Or (60-x)+2(60+10) = 180 60-x+140 = 18 200-x = 180 x = 200-180 x = 20°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3104249

It will be; Let the single side length of the regular heptagon be 5 units. Therefore; (1/x) + (1/y) will be; (1÷(√(50-50cos129))) + (1÷(5+10cos51)) = 0.19934168501≈ 0.2 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3103921

Sir Mike Ambrose is the author of the question. Let the side length of the introduced equilateral triangle inscribing the circle be 2 units. Calculating r, radius of the circle. cos30 = 1/r r = 1/cos30 r = 2/√(3) r = ⅓(2√(3)) units. Area circle is; πr² = π(⅓(2√(3)))² = ⅓(4π) square units. = 4.18879020479 square units. tan30 = a/1 a = ⅓√(3) units. b = r-a b = ⅓(2√(3))-⅓√(3) b = ⅓√(3) units. tan60 = ⅓√(3)/c c = ⅓ units. cos60 = (⅓)/d d = ⅔ units. or si...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3103006

Sir Mike Ambrose is the author of the question. Let the ascribed circle's radius be 1 unit. Area ascribed circle is; π(1)² = π square units. Calculating shaded area. Notice. The side length of the inscribed regular heptagon is equal the radius of the circle, 1 unit. Therefore, shaded area is; 6(area equilateral triangle with side length 1 unit)-Area sector with radius 1 unit and angle 120°-2(area sector with radius 0.5 units and angle 120°)-2(area equilateral triangle with...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3102550

Sir Mike Ambrose is the author of the question. Let AB be 2 units. Area ABCD is; 2² = 4 square units. Let the side length of the ascribed square be x. Calculating x. a = (x-2) units. b = (x-1) units. It implies; x² = a²+b² x² = (x-2)²+(x-1)² x² = x²-4x+4+x²-2x+1 x²-6x+5 = 0 Therefore; x = 5 units. Again, x is the side length of the ascribed square. c = x-2 c = 3 units. d = x-1 d = 4 units. tane = 3/4 e = atan(3/4)° f = 90-2e f = 90-2atan(3/4) f...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3102466

Let the two equal smaller inscribed circle radius be r. Let the radius of the smallest inscribed circle be q. The radius is the biggest inscribed circle is 0.5 units. Therefore; x+2r=0.5 r=½(0.5-x) ----- (1) (0.5+r)²=(0.5+x)²+(x+r)² 0.25+r+r²=0.25+x+x²+x²+2xr+r² r=x+2x²+2xr r(1-2x)=x+2x² r=(x+2x²)/(1-2x) ------- (2) Equating (1) and (2). (x+2x²)/(1-2x)=½(0.5-x) 4x²+2x=(0.5-x)(1-2x) 4x²+2x=0.5-x-x+2x² 2x²+4x-0.5=0 x²+2x-0.25=0 x²+x=0.25+1² (x...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3100770

Length EF will be; √(49+50-70√(2)cos(tan-¹(¾)+45)) EF = √(85) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3099214

Notice. r, radius of the ascribed half circle is 5 units. Calculating the area of the inscribed red circle. Let it's radius be x. Calculating x. a = (5-x) units. tan15 = x/b b = (x/tan15) units. c = b-5 c = ((x/tan15)-5) units. It implies; a² = c²+x² (5-x)² = ((x/tan15)-5)²+x² 25-10x+x² = (x²/(tan15)²)-(10x/tan15)+25+x² -10x = (x²/(tan15)²)-(10x/tan15) (10x/tan15)-10x = (x²/(tan15)²) (10/tan15)-10 = (x/(tan15)²) x = 10tan15-10(tan15)² x = 1.96...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3099190

Area Black will be; Area triangle BCD - area sector of radius 4m and angle 45° + area sector of radius 4m and angle 45° - quarter circle of radius 2m - 2(right-angled triangle of height and base 2m each) + quarter circle of radius 2m. = ½(16) - (16π)/8 + (16π)/8 - ¼(4π) - 2(½(4)) + ¼(4π) = 8 - 2π + 2π - π - 4 + π = 8 - 4 = 4 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3098663

Let MN=NP=PD=MD=y. Let AR=x. Therefore; AD=AB=BC=CD=2+x. RD=2. RM=2-y. And; RC=2+2x. There deriving an equation. Considering similar triangle RCD and RNM. It implies; 2-y=2 y =2+x, cross multiply. 2y=(2-y)(2+x) 2y=4+2x-2y-xy 4y+xy=4+2x y(4+x)=4+2x Therefore; y=(4+2x)/(4+x)--- (1) Calculating x, considering triangle RCD, applying Pythagoras rule. Therefore; (2+2x)²=2²+(2+x)² 4+8x+4x²=4+4+4x+x² 3x²+4x-4=0 3x²+6x-2x-4=0 3x(x+2)-2(x+2)=0 (3x-...

You Searched for "update"

182 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 "update"

182 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?