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2 years ago

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Select the correct answer. A circle is described by the equation x2 + y2 + 14x + 2y + 14 = 0. What are the coordinates for the c

enter of the circle and the length of the radius? A. (-7, -1), 36 units B. (7, 1), 36 units C. (7, 1), 6 units D. (-7, -1), 6 units

1 answer:

son4ous [18]2 years ago

4 0

Answer:

The correct answer is D.

Step-by-step explanation:

Given:

General equation of second degree

x² + y² + 14 x + 2 y + 14 = 0

We must transform given equation to the canonical form from which we will read requested data.

The canonical form of the circle equation is:

(x - p)² + (y - q)² = r²

Where p and q are the coordinates of the center of the circle and r are radius. (p,q) = (x,y)

x² + 2 · x ·7 + 7² - 7² + y² + 2 · y · 1 + 1 - 1 + 14 = (x+7)² + (y+1) - 49 - 1 + 14 = 0

(x + 7)² + (y + 1)² = 36

We see that p = - 7 , q = - 1 and r = 6

God with you!!!

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2x+3y=1 and y=-2 - 9 solve using linear combination method. Please explain steps.

kaheart [24]

Answer:

$x=17$

Step-by-step explanation:

$2x+3y=1$

$y=-2-9$

In order to combine these two equations, an idea you need to keep in mind is finding a way of setting these equations as equal to each other. I saw that each equation shared a common value, $y$ . In this case, we need to isolate $y$ in the first equation so that both equations $=y$ .

$2x+3y=1$

$3y=-2x+1$

$\frac{3y}{3}=\frac{-2x+1}{3}$

$y=-\frac{2}{3}x+\frac{1}{3}$

With this, we now know that both $-2-9$ and $-\frac{2}{3}x+\frac{1}{3}$ are equal to $y$ , so we can set them equal to each other.

$y=-\frac{2}{3}x+\frac{1}{3}$

$y=-11$

$-\frac{2}{3}x+\frac{1}{3}=-11$

Reply to this if anything I'm saying or doing is confusing in any way, or if you find a mistake. :) Solve for $x$ .

$-\frac{2}{3}x+\frac{1}{3}=-11$

$-\frac{2}{3}x=-11-\frac{1}{3}$

$-\frac{2}{3}x =-\frac{33}{3}-\frac{1}{3}$

$-\frac{2}{3}x =-\frac{34}{3}$

$-\frac{2}{3}x(-\frac{3}{2})=-\frac{34}{3}(-\frac{3}{2})$

$x=\frac{102}{6}=17$

$x=17$

Hopefully this answer is correct AND makes sense in terms of how I achieved it. Again, reply to this with any questions or mistakes I made and I'll do my best to answer or fix them.

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2 years ago

Arthur hs 20/8 cups of dish wahing detergent, he uses 1/4 cup for each load, how many loads can he do

scoray [572]

Answer:

10

Step-by-step explanation:

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2 years ago

Triangle PQR is transformed to similar triangle P′Q′R′ below:

morpeh [17]

1 over 3 should be the answer I think

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2 years ago

Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega

AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

To find the simplified form of the given expression :

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

$=(2^6)(3^{-12})\times (2^{-6})(3^4)$ ( using the property $(a^m)^n=a^{mn}$

$=(2^6)(2^{-6})(3^{-12})(3^4)$ ( combining the like powers )

$=2^{6-6}3^{-12+4}$ ( using the property $a^m.a^n=a^{m+n}$ )

$=2^03^{-8}$

$=\frac{1}{3^8}$ ( using the property $a^{-m}=\frac{1}{a^m}$ )

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

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3 years ago

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maria [59]

Answer:

Addition property of equality

Step-by-step explanation:

3/4 was added to each side

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3 years ago