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ivann1987 [24]
3 years ago
13

What is the standard form of the equation of the circle x2 -4x + y2 + 6y + 12 = 0?

Mathematics
1 answer:
Rashid [163]3 years ago
4 0
We take the equation
                            <span>x^2 -4x + y^2 + 6y + 12 = 0
and complete the square for x and y.
                             </span><span>x^2 -4x + 4 - 4 + y^2 + 6y + 9 - 9 + 12= 0</span>
<span>                                            (x-2)^2 - 4 + (y+3)^2 - 9 = -12</span>
<span>                                                       <span>(x-2)^2+ (y+3)^2  = 1
Therefore the answer is </span></span><span>(x-2)^2+ (y+3)^2  = 1</span>
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Answer: A

Step-by-step explanation:

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How to simplify this expression
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\frac{x^2-16}{2x-8}÷ \frac{x-4)^2}{8x-32}
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4 0
4 years ago
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kati45 [8]

120 divided by 15 =8

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4 years ago
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s2008m [1.1K]

Our goal here is to somehow "surgically remove" the repeating part of the number, so let's start by putting the original value in a variable and messing around with it a bit.

We'll let x=7.\overline{6}. We want to cut the 0.\overline{6} bit off completely, so let's create the scalpel that'll let us do that. If x=7.\overline{6}, then we can also say that 10x=76.\overline{6}. Maybe I was lying a bit: the x is our real scalpel here, and 10x is where we'll be making the cut. Mathematically, a "cut" is almost always shorthand for subtraction, so let's see what our operation (cutting x off of 10x) leaves us with:

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3 years ago
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Sloan [31]

Answer:

95%

Step-by-step explanation:

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0.95 x 100 = 95%

3 0
3 years ago
Read 2 more answers
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