Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2724171

Sir Mike Ambrose is the author of the question. Area R exactly, in square cm decimal form is; Area square with side 2√(11) cm - Area triangle with side 2√(11) cm and 4.97194902762 cm, and angle 15° - Area trapezoid with parallel sides 1.97064750999 cm and 2√(11) cm, and height 2√(11) cm. = (2√(11))² - ½(2√(11)*4.97194902762)sin15 - ½(1.97064750999+2√(11))*2√(11) = 44 - 4.26794919252 - 28.5358983847 = 11.1961524228 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2663292

a² = 11²+8² a = √(121+64) a = √(185) units. a² = 4²+b² √(185)² = 4²+b² b² = 185-16 b = √(169) b = 13 units. b is the length of the 2 congruent rectangles. c = b-11 c = 2 units. Therefore, shaded area is; Area trapezoid with parallel sides 8 units and 4 units, and height 13 units + Area trapezoid with parallel sides 13 units and 2 units, and height 8 units - Area triangle with height 4 units and base 13 units - Area triangle with height 8 units and base 11 units. ½(4+8)*...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2660870

Cross multiply. 5x-3(4-x) = 20x Expand. 5x-12+3x = 20x Collect like terms and simplify. -12 = 12x Divide both sides by 12, the coefficient of x. x = -1 Checking accuracy. (5(-1)-3(4-(-1)))/(2(-1)) = ? (-5-3(5))/(-2) = ? (-5-15)/(-2) = ? -20/-2 = 10

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2640156

Sir Mike Ambrose is the author of the question. Calculating k. 3k² = 108 k = 6 units. The equation of the curve is; y = 4x-⅓(x²) Area R is; Area under the curve at x=0 and x=(6-3√(2)). = 2(6-3√(2))²-(6-3√(2))²/9 = 6(8-5√(2)) square units. Area S is; Area rectangle with length 12 units and width 6 units - Area trapezoid with parallel sides 12 units and 9 units, and height 3 units - Area under the curve at x=3 and x=6. = (12*6) - ½*3(12+9) - 33 = 72 - ½(63) - 33 =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2636430

Let Alpha be x. Calculating x; ½(180-atan2)-atan(0.5) x = ½(90-atan(0.5)) It implies; tan x exactly is; = tan(½(90-atan(0.5))) Calculating Area Orange. Notice, Area Orange is an isosceles triangle with one of the two congruent sides 1.01857603 units. It implies; Area Orange exactly in decimal is; Area triangle with height 1.01857603 units and base 1.01857603sin(atan2) units. = ½*1.01857603*1.01857603sin(atan2) = 0.46398282133 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2619978

Sir Mike Ambrose is the author of the question. Area D exactly in cm² is; Area trapezoid with parallel sides 4 cm and 8 cm, and height 4 cm - Area triangle with height 4 cm and base 3 cm - Area sector with radius 4 cm and angle atan(4/3). = (½*4(4+8)) - (½*4*3) - (atan(4/3)*π*4²÷360) = 24 - 6 - (2πatan(4/3))/45 = 18 - (2πatan(4/3))/45 = (810 - (2πatan(4/3)))/45 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585239

a = ⅓(90) a = 30° sin30 = b/4 b = 2 cm. sin60 = c/4 c = 2√(3) cm. d = √(4²-2²)-√(4²-(2√(3))²) d = √(12)-√(4) d = (2√(3)-2) cm. d = 1.4641016151 units. Area Blue is; Area trapezoid with parallel sides 2 cm and 2√(3) cm respectively and height (2√(3)-2) cm + Area sector with radius 4 cm and angle 30° - Area triangle with height 4 cm and base 4sin30 cm. = (½*(2+2√(3))*(2√(3)-2))+(30π*4*4/360)+(½*4*4sin30) = (½*8)+⅓(4π)-4 = 4+⅓(4π)-4 = ⅓(4π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2562085

Equation of the curve is; y = ½(x^(3/2)) It implies; PR = √(4²+6²) PR = 2√(13) units QR = (2√(13))/(cos(90-atan(⅔))) = 13 units. Therefore; Shaded Area is; (Area under the curve at x = 4 and x = 10) - Area trapezium with parallel sides 4 units and 13 units, and height 6 units. = 20√(10) - (32/5) - ½*6(4+13) = 20√(10) - (32/5) - 51 = (100√(10)-287)/5 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2561633

Let CE be 3 units. Therefore: Area Orange is; Area triangle with height √(19) units and base 6sin(60-acos(4/√(19))) units - Area triangle with height (12√(19)/19) units and base (36√(57)/133) units. = ½*√(19)*6sin(60-acos(4/√(19))) - ½*(12√(19)/19)*(36√(57)/133) = ½(9√(3)) - (216√(3))/133 = (765√(3)/266) square units. Area ACEB is; Area trapezium with parallel sides 3 units and 6 units, and height ½(3√(3)) units. = ½*½(3√(3))*(3+6) = ¼*9*3√(3) = ¼(27√(3)) square unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2558651

Shaded area will be; Area quarter circle of radius 2 unit + 2(area quarter circle of radius √(2) unit - area triangle of height and base √(2) unit respectively). = ¼(4π) + 2(¼(2π) - ½(2)) = π + π - 2 = 2(π-1) square units. Or a² = 2(2)² a = √(8) a = 2√(2) units. b = ½(a) b = √(2) units Shaded Area is; Area semi circle with radius √(2) units - Area isosceles right-angled triangle with adjacent sides 2 units each + Area quarter circle with radius 2 units. (½*π*√(2)²)-...

You Searched for "fin des temps"

170 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 "fin des temps"

170 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?