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Serhud [2]
3 years ago
13

Match the x-coordinates with their corresponding pairs of y-coordinates on the unit circle.

Mathematics
1 answer:
jeka57 [31]3 years ago
5 0
<span>y=+- square root 5 over 3

y^2 + x^2 = 1 => x^2 = 1 - y^2 = 1 - 5/9 = 4/9 => x = +/- 2/3

Answer: x = +/- 2/3

y=+- square root 7 over 3

y^2 + x^2 = 1 => x^2 = 1 - y^2  =  1 - 7/9 = 2/9 => x = +/- (√2) / 3

Answer: x = +/-(√2)/3

y=+- 3 over 3

x^2 = 1 - y^2 = 1 - 3/9 = 1 - 1/3 = 2/3 => x = +/-(√2/3)

Answer: x = +/-√(2/3)

y=+- 2 square root 2 over 2

= y = +/- 2(√2) /2 = √2  ...... these y-coordinates are out of the unit circle, then there is not a corresponding x - coordinate for them.

</span>
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Answer:

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hope that helps have a good day

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Step-by-step explanation:

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4 years ago
A surveyor wishes to lay out a square region with each side having length L. However, because of measurement error, he instead l
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Step-by-step explanation:

Area, A = Length,x × Breadth,y

A=xy

When x and y are independent, E(xy) = E(x)E(y)

As x and y have the same distribution, U[L-A,L+A], they have the same mean.

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We can also reason this from the fact that, if X ~ U[L-A, L+A], f(x) = 1/(2A) from L-A to L+A

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E(x)= \int\limits {f(x)} \, dx \\\\=\int\limits^{(L+A)}_{(L-A)} {\frac{1}{2A}x } \, dx

=\int\limits^{(L+A)}_{(L-A)} {\frac{1}{2A}x } \, dx \\\\=[\frac{1}{4A}x^2 ]\limits^{(L+A)}_{(L-A)}

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