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WITCHER [35]
3 years ago
11

Please answer #1 and #2

Mathematics
1 answer:
DIA [1.3K]3 years ago
6 0
1)
x^2 + (3y/2z) = 7
2x^2z + 3y = 14z
3y = 14z - 2x^2z
3y = 2z(7 - x^2)
  y = 2/3(z)(7 - x^2)

2)
(3zx^4) /(5+z) = 2y
3zx^4 = 2y(5+z)
3zx^4 = 10y + 2yz

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Hello I need help Finding the probability
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Each petal of the region R is the intersection of two circles, both of diameter 10. Each petal in turn is twice the area of a circular segment bounded by a chord of length 5\sqrt2, which implies the segment is subtended by an angle of \dfrac\pi2. This means the area of the segment is

\text{area}_{\text{segment}}=\text{area}_{\text{sector}}-\text{area}_{\text{triangle}}
\text{area}_{\text{segment}}=\dfrac{25\pi}4-\dfrac{25}2

This means the area of one petal is \dfrac{25\pi}2-25, and the area of R is four times this, or 50\pi-100.

Meanwhile, the area of G is simply the area of the square minus the area of R, or 10^2-(50\pi-100)=200-50\pi.

So

\mathbb P(X=R)=\dfrac{50\pi-100}{100}=\dfrac\pi2-1
\mathbb P(X=G)=\dfrac{200-50\pi}{100}=2-\dfrac\pi2
\mathbb P((X=R)\land(X=G))=0 (provided these regions are indeed disjoint; it's hard to tell from the picture)
\mathbb P((X=R)\lor(X=G))=\mathbb P(X=R)+\mathbb P(X=G)=1

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Answer:

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Step-by-step explanation:

hey! The formula for the equation of a circle is r^2= (x-h)^2 + (y-k)^2

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(h,k)= is the center coordinates, so your center coordinate would be (4, -9). You now have your h (h=4) and your k (k=-9) so you can plug it in the formula! :)

a nice rating would help out

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