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Rasek [7]
2 years ago
6

HELP ASAP Write the equation in standard form for the circle x2 + 32 = -y² + 12x.

Mathematics
2 answers:
vagabundo [1.1K]2 years ago
7 0

Step-by-step explanation:

Move

26

to the right side of the equation because it does not contain a variable.

3x2−12x−2y2+16y=26

Olegator [25]2 years ago
7 0

Answer:

(x - 6)² + y² = 4

Step-by-step explanation:

x² + 32 = -y² + 12x

Rearrange

x² - 12x + y² = -32

Complete the square for x

(x - 6)² + y² = -32 + (-6)²

(x - 6)² + y² = -32 + 36

(x - 6)² + y² = 4

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How do u do this?!?!
soldi70 [24.7K]

Answer:

-9

Step-by-step explanation:

5-p

p=-9

-4 is greater than 5-p

8 0
2 years ago
Which equation represents the line that passes through (–6, 7) and (–3, 6)? y = –y equals negative startfraction one-third endfr
egoroff_w [7]

The equation represents the line that passes through (–6, 7) and (–3, 6) \rm y=\dfrac{-1}{3}x+5.

<h3>What is the slope of the equation?</h3>

For all lines in slope y-intercept form, it would be very simple to just find the answer by finding yourself the slope and y-intercept of the line in question.

The slope of the line is;

\rm m = \dfrac{y_2-y_1}{x_2-x_1}\\\\m =\dfrac{7-6}{-6-(-3)}\\\\m = \dfrac{1}{-3}

The equation represents the line that passes through (–6, 7) and (–3, 6) is;

\rm y=\dfrac{-1}{3}x+b\\\\6=\dfrac{-1}{3}(-3)+b\\\\6=1+b\\\\b = 6-1\\\\b=5

The required line of the equation is;

\rm  y =mx+c\\\\y=\dfrac{-1}{3}x+5

Hence, the equation represents the line that passes through (–6, 7) and (–3, 6) \rm y=\dfrac{-1}{3}x+5.

To know more about the equation of line click the link given below.

brainly.com/question/8955867

3 0
2 years ago
***Function table***
Afina-wow [57]

Answer:

y for each x value

-18

52

-13

-8

32

8 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
astra-53 [7]

Answer:

3\pi \rightarrow y=2\cos \dfrac{2x}{3}\\ \\\dfrac{2\pi }{3}\rightarrow y=6\sin 3x\\ \\\dfrac{\pi }{3}\rightarrow  y=-3\tan 3x\\ \\10\pi \rightarrow y=-\dfrac{2}{3}\sec \dfrac{x}{5}

Step-by-step explanation:

The period of the functions y=a\cos(bx+c) , y=a\sin(bx+c), y=a\sec (bx+c) or y=a\csc(bx+c) can be calculated as

T=\dfrac{2\pi}{b}

The period of the functions y=a\tan(bx+c) or y=a\cot(bx+c) can be calculated as

T=\dfrac{\pi}{b}

A. The period of the function y=-3\tan 3x is

T=\dfrac{\pi}{3}

B. The period of the function y=6\sin 3x is

T=\dfrac{2\pi}{3}

C. The period of the function y=-4\cot \dfrac{x}{4} is

T=\dfrac{\pi}{\frac{1}{4}}=4\pi

D. The period of the function y=2\cos \dfrac{2x}{3} is

T=\dfrac{2\pi}{\frac{2}{3}}=3\pi

E. The period of the function y=-\dfrac{2}{3}\sec \dfrac{x}{5} is

T=\dfrac{2\pi}{\frac{1}{5}}=10\pi

5 0
3 years ago
$132.75 discounted 40%
saw5 [17]

Answer:

$79.65

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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