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Temka [501]
3 years ago
8

GEOMETRY DESPERATE HELP

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
7 0
<h3>Answer: Choice C) x^2 + (y-2)^2 = 4</h3>

==========================================================

Explanation:

The center is at (0,2). So (h,k) = (0,2) leads to h = 0 and k = 2.

The radius is 2 units, meaning r = 2.

Plug h = 0, k = 2, r = 2 into the formula below and simplify

(x-h)^2 + (y-k)^2 = r^2\\\\(x-0)^2 + (y-2)^2 = 2^2\\\\x^2 + (y-2)^2 = 4

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4 years ago
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7 0
3 years ago
PLEASE HELP!!!! WILL MARK BRAINLIEST!!!!
irga5000 [103]

Answer:

None of these.

Step-by-step explanation:

Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem.  So we know (x-6) is not a factor because the remainder wasn't zero.

Let's assume we are trying to figure out if (x^2+6) is a factor.  The quotient is (x-6) and the remainder is 13 according to the problem.  So we know (x^2+6) is not a factor because the remainder wasn't zero.

In order for 13 to be a factor of P, all the terms of P must be divisible by 13.  That just means you can reduce it to a form that is not a fraction.

If we look at the first term x^3 and we divide it by 13 we get \frac{x^3}{13} we cannot reduce it so it is not a fraction so 13 is not a factor of P

None of these is the right option.

4 0
3 years ago
Read 2 more answers
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