Green = 6
Yellow = 2
6 + 2 = 8
8/20 can be simplified to 2/5
Answer:
C. The ratio of the area to the circumference is equal to half the radius.
Step-by-step explanation:
The area of a circle can be written as;
Area A = πr^2
The circumference of a circle is;
Circumference C = 2πr
Using the formula, w can derive the relationship between the two variables.
A = kC
k = A/C
Substituting the two formulas;
k = (πr^2)/(2πr) = r/2
So,
A = (r/2)C
A/C = r/2
The ratio of the area to the circumference is equal to half the radius.
Given;
Area = 200.96
Circumference = 50.24
Radius = 8
To confirm;
k = r/2 = 8/2 = 4
Also,
A/C = 200.96/50.24
A/C = 4
Answer:
perimeter :
4+10+4+ (pi * radius) .........."radius =5"
area:
(10*4)+(0.5 * pi * r square)
Answer:
![\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E6y%5E4%7D%7B81x%5E2y%5E8%7D%7D%5Crightarrow%5Cfrac%7B2x%7D%7B3y%7D%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B81x%5E2y%5E%7B10%7D%7D%7B81x%5E6y%5E6%7D%7D%20%5Crightarrow%5Cfrac%7B3y%7D%7B2x%7D%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7B64x%5E8y%5E7%7D%7B125x%5E2y%5E%7B10%7D%7D%7D%5Crightarrow%5Cfrac%7B4x%5E2%7D%7B5y%7D%5C%5C%5Csqrt%5B5%5D%7B%5Cfrac%7B243x%5E%7B17%7Dy%5E%7B16%7D%7D%7B32x%5E7y%5E%7B21%7D%7D%7D%5Crightarrow%5Cfrac%7B3x%5E2%7D%7B2y%7D%5C%5C%5Csqrt%5B5%5D%7B%5Cfrac%7B32x%5E%7B12%7Dy%5E%7B15%7D%7D%7B243x%5E7y%5E%7B10%7D%7D%7D%20%5Crightarrow%5Cfrac%7B2xy%7D%7B3%7D%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B10%7Dy%5E%7B9%7D%7D%7B256x%5E2y%5E%7B17%7D%7D%7D%5Crightarrow%5Cfrac%7Bx%7D%7B2y%7D)
Step-by-step explanation:
![\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E6y%5E4%7D%7B81x%5E2y%5E8%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B%282%5E4%29%28x%5E%7B6-2%7D%29%28y%5E%7B4-8%7D%29%7D%7B%283%5E4%29%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B2%5E4x%5E4y%5E%7B-4%7D%7D%7B3%5E4%7D%7D%20%3D%5Cfrac%7B2xy%5E%7B-1%7D%7D%7B3%7D%3D%5Cfrac%7B2x%7D%7B3y%7D)
![\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B81x%5E2y%5E%7B10%7D%7D%7B81x%5E6y%5E6%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B%283%5E4%29%28x%5E%7B2-6%7D%29%28y%5E%7B10-6%7D%29%7D%7B%282%5E4%29%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B3%5E4x%5E%7B-4%7Dy%5E%7B4%7D%7D%7B2%5E4%7D%7D%20%3D%5Cfrac%7B3x%5E%7B-1%7Dy%5E1%7D%7B3%7D%3D%5Cfrac%7B3y%7D%7B2x%7D)
![\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B64x%5E8y%5E7%7D%7B125x%5E2y%5E%7B10%7D%7D%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%284%5E3%29%28x%5E%7B8-2%7D%29%28y%5E%7B7-10%7D%29%7D%7B%285%5E3%29%7D%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4%5E3x%5E6y%5E%7B-3%7D%7D%7B5%5E3%7D%7D%20%3D%5Cfrac%7B4x%5E2y%5E%7B-1%7D%7D%7B5%7D%3D%5Cfrac%7B4x%5E2%7D%7B5y%7D)
![\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%5Cfrac%7B243x%5E%7B17%7Dy%5E%7B16%7D%7D%7B32x%5E7y%5E%7B21%7D%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B%283%5E5%29%28x%5E%7B17-7%7D%29%28y%5E%7B16-21%7D%29%7D%7B%282%5E5%29%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B3%5E5x%5E%7B10%7Dy%5E%7B-5%7D%7D%7B2%5E5%7D%7D%20%3D%5Cfrac%7B3x%5E2y%5E%7B-1%7D%7D%7B2%7D%3D%5Cfrac%7B3x%5E2%7D%7B2y%7D)
![\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%5Cfrac%7B32x%5E%7B12%7Dy%5E%7B15%7D%7D%7B243x%5E7y%5E%7B10%7D%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B%282%5E5%29%28x%5E%7B12-7%7D%29%28y%5E%7B15-10%7D%29%7D%7B%283%5E5%29%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B2%5E5x%5E%7B5%7Dy%5E%7B5%7D%7D%7B3%5E5%7D%7D%20%3D%5Cfrac%7B2x%5E1y%5E%7B1%7D%7D%7B3%7D%3D%5Cfrac%7B2xy%7D%7B3%7D)
![\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B10%7Dy%5E%7B9%7D%7D%7B256x%5E2y%5E%7B17%7D%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B%282%5E4%29%28x%5E%7B10-2%7D%29%28y%5E%7B9-17%7D%29%7D%7B%284%5E4%29%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B2%5E4x%5E%7B8%7Dy%5E%7B-8%7D%7D%7B4%5E4%7D%7D%20%3D%5Cfrac%7B2x%5E%7B1%7Dy%5E%7B-1%7D%7D%7B4%7D%3D%5Cfrac%7Bx%7D%7B2y%7D)
Thus,
The derivative of f(x) at x=3 is 2x=6 approaching from the left side (apply power rule to y=x^2). The derivative of f(x) at x=3 is m approaching from the right side. In order for the function to be differentiable, the limit of derivative at x=3 must be the same approaching from both sides, so m=6. Then, x^2=mx+b at x=3, plug in m=6, 9=18+b, so b=-9.
