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MrMuchimi
2 years ago
5

please help! a point is chosen at random in the square, what percent of the time will the point be in the circle,round to the ne

arest tenth of a percent?

Mathematics
2 answers:
vodomira [7]2 years ago
8 0

Answer:

78.5%

Step-by-step explanation:

You take the area of the circle which is pie*r^2

and divide it by the area of the square

Lelechka [254]2 years ago
4 0
The answer to your question will be 20%
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Answer:

a_n=-3(3)^{n-1} ; {-3,-9, -27,- 81, -243, ...}

a_n=-3(-3)^{n-1} ; {-3, 9,-27, 81, -243, ...}

a_n=3(\frac{1}{2})^{n-1} ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

a_n=243(\frac{1}{3})^{n-1} ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

a_n=-3(3)^{n-1}

At n=1,

a_1=-3(3)^{1-1}=-3

At n=2,

a_2=-3(3)^{2-1}=-9

At n=3,

a_3=-3(3)^{3-1}=-27

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

a_n=-3(-3)^{n-1}

At n=1,

a_1=-3(-3)^{1-1}=-3

At n=2,

a_2=-3(-3)^{2-1}=9

At n=3,

a_3=-3(-3)^{3-1}=-27

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

a_n=3(\frac{1}{2})^{n-1}

At n=1,

a_1=3(\frac{1}{2})^{1-1}=3

At n=2,

a_2=3(\frac{1}{2})^{2-1}=1.5

At n=3,

a_3=3(\frac{1}{2})^{3-1}=0.75

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

a_n=243(\frac{1}{3})^{n-1}

At n=1,

a_1=243(\frac{1}{3})^{1-1}=243

At n=2,

a_2=243(\frac{1}{3})^{2-1}=81

At n=3,

a_3=243(\frac{1}{3})^{3-1}=27

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0
3 years ago
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