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3 years ago

What is the standard form of the equation of the circle x^2 - 4x +y^2 +6y + 4 = 0 ?

Mathematics

1 answer:

dybincka [34]3 years ago

5 0

Answer: (x-2)^2+(y+3)^2 = 9

Side notes
1) This circle has a center of (2,-3)
2) The radius of this circle is 3
3) The graph is shown in the attached image

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Work Shown:

x^2-4x+y^2+6y+4=0
x^2-4x+y^2+6y+4-4=0-4
x^2-4x+y^2+6y = -4
x^2-4x+4+y^2+6y = -4+4 ... see note 1 below
(x^2-4x+4)+y^2+6y = 0
(x-2)^2+y^2+6y = 0
(x-2)^2+y^2+6y+9 = 0+9 ... see note 2 below
(x-2)^2+(y^2+6y+9) = 9
(x-2)^2+(y+3)^2 = 9

note 1: I'm adding 4 to both sides to complete the square for the x terms. You do this by first taking half of the x (not x^2) coefficient which in this case is -4. So take half of -4 to get -2. Then square this result to get 4

note 2: Like with note 1, I'm completing the square. What's different this time is that this is for the y terms now. The y coefficient is 6. Half of this is 3. Square 3 to get 9. So this is why we add 9 to both sides.

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So the equation in standard form is (x-2)^2+(y+3)^2 = 9

Note how
(x-2)^2+(y+3)^2 = 9
is equivalent to
(x-2)^2+(y-(-3))^2 = 3^2

So that second equation listed above is in the form (x-h)^2+(y-k)^2 = r^2
where
h = 2
k = -3
r = 3

making the center to be (h,k) = (2,-3) and the radius to be r = 3

The graph is attached.

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Step-by-step explanation:

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Cross multiply:

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Divide both sides by 0.68 to solve for XY

$\frac{XY*0.68}{0.68} = \frac{24*0.92}{0.68}$

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Step 2: find the area using the formula, ½*XY*WY*sin(Y).

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Area = ½*32.5*24*0.94

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3 years ago

Jacob wants to know the favorite sport among 6th grade students. Which types of survey below would give Jacob valid results?

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2 years ago

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miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

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2 years ago

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Step-by-step explanation:

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3 years ago

Need help please, does any one know how to do this

ziro4ka [17]

Answer:

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Step-by-step explanation:

There are a few ways you can write the equivalent of this.

1) Distribute the minus sign. The starting numerator is -(u-6). After you distribute the minus sign, you get -u+6. You can leave it like that, so that your equivalent form is ...

(-u+6)/(u+2)

Or, you can rearrange the terms so the leading coefficient is positive:

(6 -u)/(u +2)

2) You can perform the division and express the result as a quotient and a remainder. Once again, you can choose to make the leading coefficient positive or not.

-(u -6)/(u +2) = (-(u +2)-8)/(u +2) = -(u+2)/(u+2) +8/(u+2) = -1 + 8/(u+2)

8/(u+2) -1

Of course, anywhere along the chain of equal signs the expressions are equivalent.

3) You can separate the numerator terms, expressing each over the denominator:

(-u +6)/(u+2) = -u/(u+2) +6/(u+2)

4) You can also multiply numerator and denominator by some constant, say 3:

-(3u -18)/(3u +6)

You could do the same thing with a variable, as long as you restrict the variable to be non-zero. Or, you could use a non-zero expression, such as 1+x^2:

(1+x^2)(6 -u)/((1+x^2)(u+2))

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3 years ago