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This is definitely not on the SAT

All six is 18𝛑(-4√3+7)

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Enjoyed. Go on.

From △OAP, we get that cos(∠OPA)=AP/OP=10/(5√5)=2/√5 But ∠OPA=x/2 i.e., cos(x/2)=2/√5 From cos(x/2), we can find the value of cos(x) using the double-angle formula, namely: cos(2θ)=2cos²(θ)-1 Thus, cos(2*x/2)=2cos²(x/2)-1=2(2/√5)²-1=8/5-1=3/5 i.e., cos(x)=3/5

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Which program do you use?

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Can it be applied to any quadrilateral?

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Height of the rectangle is (-2a+20) Multiplying by the width (a) gives (-2a+20)a The first derivative with respect to a is -4a+20. When set to 0 to find the maximum, we get a=5 We get the height by plugging a back into -2a+20 to get 10. 10*2+5*2=30 I have no clue what is going on in the explaination in the video. It hurts to look at.

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70 degree?

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Error: triangle ABD (9,10,13) is not a right triangle. It should be rectangular.

Error: triangle ABD (9,10,13) is not a right triangle. It should be rectangular.

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Hjbibj itu adalah hari yang penting

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BD^2 = 25 + 144 = 169 ; BD = 13 Triangle [BCD] Area = (5 * 12) / 2 = 30 Triangle [ABD] Area (using Heron's Formula for : 9 / 10 / 13) = 44,9 ~ 45 Quadrilateral Area = 30 + 45 = 75 Square Units (approx.)

Hello! 1) Let AB = OC = X 2) Let OA = BC = Y 3) Let OB = R = 6 4) Let 2X + 2Y = 16 ; 2*(X + Y) = 16 ; X + Y = 8 5) X + Y = 8 and X^2 + Y^2 = R^2 ; X^2 + Y^2 = 36 System of Two Equations: a) X + Y = 8 b) X^2 + Y^2 = 36 6) Positive Solutions: 7) X = 4 - sqrt(2) and Y = 4 + sqrt(2) 8) AP = 6 - (4 + sqrt(2)) Linear Units 9) CQ = 6 - (4 - sqrt(2)) Linear Units 10) AC = OB = 6 Linear Units 11) Arc PQ = 2*Pi*R / 4 ; Arc PQ = Pi*6/2 ; 3Pi Linear Units 12) Perimeter = [6 - (4 + sqrt(2)) + 6 - (4 - sqrt(2)) + 6 + (3Pi] Linear Units 13) Perimeter = 6 - 4 - sqrt(2) + 6 - 4 + sqrt(2) + 6 + (3Pi) Linear Units 14) Perimeter = (10 + 3Pi) Linear Units 15) Perimeter ~ 19,425 Linear Units.

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noice, I came up with 23 and a 1/3 as a Rough guess without doing the actual geometry...good to see I still got it after 25 years of not using it

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My way of solution is ➡ HC=x EH= y ⇒ FE= 20-y BH= 20-x GE= BH A(HCE)= 80 square units 80= HC*EH/2 160= x*y xy= 160 AC would be equal to the radius of the circle, which is also the side length of the square: a= 20 length units R= 20 according to the Pythagorean theorem: x²+y²= R² x²+y²= 20² x²+y²= 400 we know that: (x+y)²= x²+y²+2xy (x+y)²= 400+2*160 (x+y)²= 720 x+y= √720 x+y= √144*5 x+y= 12√5 the area of the green rectangle: A(GEFA)= (20-x)*(20-y) A(GEFA)= 400-20y-20x+xy = 400-20(x+y)+xy x+y= 12√5 xy= 160 ⇒ A(GEFA)= 400-20*12√5+160 A(GEFA)= 400-240√5+160 A(GEFA)= 560-240√5 square units

Прикольно ❤

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P=12√2≈16,98

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Показано отлично, но жаль канал популярным не будет. ) Математику не очень любят.

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