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stich3 [128]
2 years ago
15

The equation 2x² + 2y2 – 8x - 12y + 16 = 0 defines a circle.

Mathematics
1 answer:
mixas84 [53]2 years ago
3 0

Answer:

  C, E

Step-by-step explanation:

Here, we want to change the equation of a circle from general form to standard form. This is done by making the leading coefficients 1, completing the squares, and then rewriting the equation in standard form.

The leading coefficients of the given equation are 2, so we first want to divide by 2. This gives ...

  x² +y² -4x -6y +8 = 0

Subtracting 8 puts us in better position to complete the squares.

  x² +y² -4x -6y = -8

Now, we can add the squares of half the coefficients of the linear terms.

  (x² -4x +4) +(y² -6x +9) = -8 +13 . . . . . . matches C

And we can simplify this to the standard form equation:

  (x -2)² +(y -3)² = 5 . . . . . matches E

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Step-by-step explanation:

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2 years ago
24=3x-16-8x thank you, sorry not to good at math
antiseptic1488 [7]
First, simplify the equation 3X-8X is -5X. So 24=16+-5X. Then subtract 16 from both sides. So you get 8=-5X. Then divide by -5.
So X= -8/5
7 0
3 years ago
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The sum of two numbers is 35. The greater number is 4 less than twice the lesser number. What is the greater number?
zubka84 [21]

Answer: x=13

Step-by-step explanation: ?+?=35

(2x-4)+x=35

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5 0
3 years ago
M is the midpoint of AN, A has coordinates
maksim [4K]

Step-by-step explanation:

According to Question , M is the midpoint of AN, A has coordinates (-7,4), and M has coordinates (-4, -1).

<u>Figure</u><u> </u><u>:</u><u>-</u>

\setlength{\unitlength}{1 cm}\begin{picture}(20,12) \put(4,0.2){\line(0,-1){0.4}}\put(1,0){\line(1,0){6}} \put(3.8,-0.6){$\bf M(-4,-1) $} \put(1,-0.6){$\bf A (-7,4)$}   \put(6.8,-0.6){$\bf N(x,y)$} \end{picture}

Let the coordinates of N be ( x , y )

Now , according to Midpoint Formula , the midpoint of points say A(x , y) and B (x' , y') is given by ,

\boxed{\blue{ \bf Midpoint =\bigg( \dfrac{x+x'}{2} , \dfrac{y+y'}{2} \bigg) }}

<u>On</u><u> </u><u>using</u><u> </u><u>this</u><u> </u><u>formula</u><u> </u><u>,</u><u> </u>

=> Midpoint = \bigg( \dfrac{x+x'}{2} , \dfrac{y+y'}{2} \bigg)  \\ \\ => (-4, 1) = \bigg( \dfrac{x -7 }{2} , \dfrac{y + 4 }{2}\bigg) \\ \\ => -4 = \dfrac{x-7}{2} \\\\ => x - 7 = -8 \\\\ => x = 7 -8 \\\\ \boxed{\red{\sf=> x = -1 }} \\\\ => \dfrac{y+4}{2}=-1 \\\\=> y +4 = -2 \\ \\ y = -4-2 \\\\\boxed{\red{\sf=> y = -6 }}

8 0
2 years ago
Read 2 more answers
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