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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 at affordable tuition. We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1995689

Let the small inscribed circle's radius be x. x = 6 cm. Let the big inscribed circle's radius be y. y = 8 cm. Let the ascribed semi circle's radius be z. = 18 cm. a² = 10²-8² a = 6 cm. b² = 12²-6² b = 6√(3) cm c = atan(z/b)° c = atan(18/6√(3))° c = atan√(3)° c = 60° d = atan(z/a)° d = atan(18/6) d = atan(3)° (a) Angle theta exactly is; Let it be e. e = 180-c-d e = 180-60-atan3 e = (120-atan3)° e = (atan(⅓)+30)° (b) Calculating Area Shaded to 1 decimal place cm². f = atan(1/√(3)) = 3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1994045

Let r be the radius of the circle. 3x+3 = 2r --- (1). r² = (r-1)²+(r-2x)² r² = r²-2r+1+r²-4xr+4x² 4xr+2r = r²+4x²+1 --- (2). From (1). 3x+3 = 2r x = ⅓(2r-3) --- (3). Calculating r. Therefore, Substituting (3) in (2). 4( ⅓(2r-3))r+2r = r²+4(⅓(2r-3))²+1 ⅓(8r²-6r) = r²+4(⅑(4r²-12r+9))+1 ⅓(8r²-6r) = ⅑(9r²+16r²-48r+36+9) ⅓(8r²-6r) = ⅑(25r²-48r+45) 24r²-18r = 25r²-48r+45 r²-30r+45 = 0 Solving the quadratic equation. r ≠ (15+6√(5)) units. r = (15-6√(5)) units. r = 1.583592135 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1993663

Notice; Hexagon Area = 24√(3) square units. Calculating hexagon side length. Let it be a. (6a²)/(4tan30) = 24√(3) ½(3√(3))a² = 24√(3) 3a² = 48 a = 4 units. Equation of the curve is; y = ½(x²) ------ (1). sin60 = b/4 b = 2√(3) units. cos60 = c/4 c = 2 units. Gradient = ((4√(3)+2)-(2√(3)+2))/(2-4) Gradient = -√(3) (y-(4√(3)+2))/(x-2) = -√(3) y = -√(3)x+6√(3)+2 ---- (2). Equation (1) and (2). ½(x)² = -√(3)x+6√(3)+2 x = 3.53906 y = ½(3.53906)² y = 6.2624728418 d = 2(3.53906) d = 7....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1993189

Let b = 1 unit. (1/sin72) = (a/sin48) a = 0.78138871113 units. BD = 1-0.78138871113 BD = 0.21861128887 units. (AD)² = 1+0.21861128887²-2*0.21861128887cos60 AD = 0.91059299731 units. Calculating the required angle x. (0.91059299731/sin60) = (0.21861128887/sinx) x = 12°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1992768

a = atan(3/2)° b = atan(2/3)° b = Angle BEG. tan(atan(2/3)) = c/2 ⅔ = c/2 c = (4/3) cm. c = BG. OG = 2-(4/3) OG = ⅔ cm. Notice; Radius of the circle is 2 cm. Angle OGH = 180-atan(3/2) = 123.69006752598° (2/sin123.69006752598) = ((⅔)/sind) d = 16.10211375199° d = Angle OHG. Angle GOH = 180-16.1021137519-123.69006752598 = 40.20781872212° It implies; Length GH is; (GH/sin40.20781872212) = ((2/3)/sin16.10211375199) GH = 1.55173771485 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1992764

Let DE be 1 unit. AC = 4 units. Area ABC is; 8sin60 = 4√(3) square units. a² = 1+2²-4cos45 a = 1.47362575821 units. (1.47362575821/sin45) = (1/sinb) b = 28.6750500631° c = 60-b c = 31.3249499369° d = 360--60-150 d = 150° e = 240-d e = 90° f² = 1²+12 f = √(2) units. g = 180-15-45-60 g = 60° h² = 4²+1.47362575821²-8*1.47362575821cos31.3249499369 h = 2.84622917461 units. (2.84622917461/sin31.3249499369) = (1.47362575821/sini) i = 15.61499741821° j = 60-i j = 44.38500258179° k = 120...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1991137

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 at affordable tuition. We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1990242

a² = √(8) a = 2√(2) units. a is the diameter of the orange half circle. b = ½(a) b = √(2) units. b is the radius of the orange half circle. Calculating c, radius of the inscribed purple circle. 2c+2-√(2) = √(2) 2c+2 = 2√(2) c = (√(2)-1) units. Therefore, Area Orange is; Area half circle with radius √(2) units - Area quarter circle with radius 2 units + Area isosceles right-angled triangle with base and height 2 units respectively - Area circle with radius (√(2)-1) units. = ½*π√(2)² - ¼*2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1990128

Let the equal lengths be 1 unit. (1/sin54) = (a/sin84) a = 1.22929666779 units. b² = 1.22929666779+1-2*1.22929666779cos126 b = 1.98904379074 units. Calculating the required angle x. (1.98904379074/sin126) = (1.22929666779/sinx) x = 30°

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