Mathematics teaching and learning

https://onlineedumath.com/update/mathematics-teaching-and-learning/2971120

Area ABED is; Area trapezium with parallel sides (4√(2)sin15) cm and (4+4√(2)sin15) cm, and height 4√(3) cm. = ½*4√(3)((4√(2)sin15)+(4+4√(2)sin15)) = 2√(3)((4√(2)sin15)+(4+4√(2)sin15)) = 24 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2953008

Sir Mike Ambrose is the author of the question. Inscribed square single side length is; 7.80540954246 cm Area green in cm² to 1 decimal place is; Area trapezoid with parallel sides 12 cm and 7.25948291225 cm, and height 12 cm - Area square with side 7.80540954246 cm - Area triangle with height 7.25948291225 cm and base 2.86780856615 cm. = ½*12(12+7.25948291225) - 7.80540954246² - ½*7.25948291225*2.86780856615 = 115.556897474 - 60.9244181255 - 10.4094036408 = 44.2230757077 c...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2951172

Sir Mike Ambrose is the author of the question. Let the side let of the square be 2 units. Therefore; Area square is; 2² = 4 square units. It implies; Area R is; Area trapezoid with parallel sides 1 unit and 1.44948974279 unit, and height 2 unit - Area square with side 1 unit - Area triangle with base and height 1 unit respectively - Area sector with radius √(2) unit and angle 32.3335306935° - Area triangle with height √(2) unit and base 0.3178372452 unit = (½*2(1+1.4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2937623

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) units and 2 unit, and height 1 units). = (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/2913587

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/2913582

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/2911913

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 units and 16√(5)/5 units, and height 2 units). = (24-⅓(8√(5))) - ½*2((167√(5)/60 )+...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2900060

Sir Mike Ambrose is the author of the question. 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 units and 16 units, and height 1 unit - Area triangle with height 16 units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2890287

Sir Mike Ambrose is the author of the question. 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/2887086

Sir Mike Ambrose is the author of the question. 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 -...

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Mathematics teaching and learning

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Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

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Mathematics Question and Solution

Mathematics Question and Solution

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