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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2342889

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a = ½(4√(6)-2√(6)) a = √(6) cm. tan60 = b/√(6) b = 3√(2). b is the height of the trapezoid. c² = (√(6))²+(3√(2))² c² = 6+18 c = √(24) c = 2√(6) cm. Or cos60 = √(6)/c c = √(6)/(½) c = 2√(6) cm. c is the forward and backward equal slant lengths of the trapezoid. d = 180-45-60 d = 75° e = 75+60 e = 135° tanf = 3√(2)/(3√(6)) f = 30° g = 180-135-30 g = 15° (4√(6)/sin15) = (h/sin30) h = 18.9282032303 cm. (4√(6)/sin15) = (j/sin135) j = 26.7685217196 cm. k = 60-30 k = 30° It implies; Area Bl...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2337411

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Above is screenshot of teaching and learning activities with Adesuwa. She is one of Online Edumath learners, hardworking, resilient and an amazing year 6 Mathematician. The Management and Educators of Online Edumath are very proud of you, Adesuwa. Adesuwa is super smart and amazingly, very clever!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2327464

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Sir Mike Ambrose is the author of the question. y = 2^(x) P coordinate is; P(1, 2) dy/dx = 2^(x)In2 For x = 1 dy/dx = 2In2 It implies; Q coordinate is; Q(0, (2-2In2)) Therefore; Shaded area exactly in square units is; (Area under the curve at x = 0 and x = 1) - (Area trapezoid with parallel sides (2-2In2) unit and 2 unit, and height 1 unit). = (1/(In2)) - (½*1*(2+2-2In2)) = (1/(In2)) - (2-In2) = (In2-1)²/In2 square units.

Mathematics Question and Solution

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

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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

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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

Mathematics Question and Solution

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

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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)) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2299144

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Area Orange exactly in square cm is; Area trapezium with parallel sides 10 cm and 5 cm, and height 5 cm - Area triangle with height ½(5) cm and base ⅓(10) cm. = ½*10(10+5) - ½*½(5)*⅓(10) = 5(15) - (25/6) = 75 - (25/6) = 425/6 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2289168

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At x = 0, y = 2. At y = 16, x = 1. Gradient of the curve at x = 1 or at the tangent is; dy/dx = 3In2*2^(3x+1) For x = 1 dy/dx = 48In2 Therefore coordinate Q is (x¹, 0). Calculating x¹. (0-16)/(x¹-1) = 48In2 x¹= (3In2-1)/3In2 It implies; Area Red exactly as a single fraction in its simplest form is; Area trapezoid with parallel sides 2 unit and 16 unit, and height 1 unit - Area triangle with height 16 unit and base (1-((3In2-1)/3In2)) = (½*1(2+16)) - (½(16)(1-((3In2-1)/3In2)) = 9 - (8(1/3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2276031

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Area green exactly in cm² is; Area ABCD - Area trapezoid with parallel sides (9-2√(3)) cm and (24√(3)-39) cm, and height 6 cm. = (54-6√(3)) - ½*6((9-2√(3))+(24√(3)-39)) = (54-6√(3)) - 3(22√(3)-30) = 54-6√(3)-66√(3)+90 = 144-72√(3) = 72(2-√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2274134

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Area blue total, exactly in cm² decimal is; Area trapezoid with parallel sides (80/7) cm and ((80/7)+21.4285714284) cm, and height 20 cm - Area trapezoid with parallel sides 20 cm and ((80/7)+12.2448979591) cm, and height (60/7) cm - Area triangle with height (80/7) cm and base 12.2448979591 cm. = (½*20((80/7)+((80/7)+21.4285714284))) - (½*(60/7)(20+((80/7)+12.2448979591))) - (½*(80/7)*12.2448979591) = 442.857142855 - 187.172011662 - 69.9708454806 = 185.714285712 cm² = 185.71 cm² to 2 decima...
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