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Mazyrski [523]
3 years ago
10

PLEASE HELP

Mathematics
1 answer:
GuDViN [60]3 years ago
7 0
The equation of a circle with center (h,k) and radius r is
(x-h)²+(y-k)²=r²
so
given center is (3,-4)
(x-3)²+(y-(-4))²=r²
(x-3)²+(y+4)²=r²
find r² by subsitutiong the point taht it passes through which is (6,5)

so
(6-3)²+(5+4)²=r²
3²+9²=r²
9+81=r²
90=r²
so

(x-3)²+(y+4)²=90 is da equation

4th equation
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Step-by-step explanation:

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For each function below, determine whether or not the function is injective and whether or not the function is surjective. (You
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(a) f:R----> + R given by f(x) = x2 (i guess you written x^{2})

The function is not injective, because f(-2) = f(2), and is surjective, because the codomain is +R, and x^{2} is always a positive real number.

(b) f:N----> + N given by f(n) = n2  (i guess you written n^{2})

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(c) f: Zx Z → Z given by f(n, k) = n +k:

here the domain is of the form (n,k) and the function is n +k, so the numbers (z,w) and (w,z) return the same value when evaluated in f, then f is not injective. And is easy to see that f is surjective, because in the sum you can reach al the integers on the codomain.

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3 years ago
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