What is the standard form of this equation of a circle?

2 answers:

5 0

<span>x² + y² + 14x − 4y − 28 = 0

x² +14x +y² - 4y =28
x²+2*7x +7² -7² + y² - 2*2y +2² - 2² = 28
(x+7)² + (y-2)² -7²-2² =28
</span>(x+7)² + (y-2)²=28+49+4

(x+7)² + (y-2)² =81 is the answer.

Marina CMI [18]3 years ago

5 0

(x2 + 14x + 49) + (y2 − 4y + 4) − 28 = 49 + 4

(x + 7)2 + (y − 2)2 − 28 = 49 + 4

(x + 7)2 + (y − 2)2 = 49 + 4 + 28 = 81.

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If -xy – 5+ y^2+ x^2= 0 and it is known that dy/dx= y-2x/-x+2y, find all

sleet_krkn [62]

Answer:

There is no point of the form (-1, y) on the curve where the tangent is horizontal

Step-by-step explanation:

Notice that when x = - 1. then dy/dx becomes:

dy/dx= (y+2) / (2y+1)

therefore, to request that the tangent is horizontal we ask for the y values that make dy/dx equal to ZERO:

0 = ( y + 2) / (2 y + 1)

And we obtain y = -2 as the answer.

But if we try the point (-1, -2) in the original equation, we find that it DOESN'T belong to the curve because it doesn't satisfy the equation as shown below:

(-1)^2 + (-2)^2 - (-1)*(-2) - 5 = 1 + 4 + 2 - 5 = 2 (instead of zero)

Then, we conclude that there is no horizontal tangent to the curve for x = -1.

3 0

3 years ago

Arnold is planning on painting a mural. Before painting the mural, Arnold painted the picture on a paper that is 5 inches wide a

Finger [1]

Answer: Mural height: 22ft and the Mural width is 10ft

Step-by-step explanation: I took the test and got it wrong, but they told me the correct answers after I took the test. (I got to K-12) Proof: