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3 years ago

Write the equation in standard form for the circle with center (-5, 3) and radius 10.

Mathematics

1 answer:

Vladimir [108]3 years ago

8 0

Answer:

x² + y² +10x -6y - 69 = 0

Step-by-step explanation:

We need to write the standard equation of circle with , centre (-5,3) and radius 10 unit . The Standard equation of the circle ⭕ is ,

=> ( x - (-5))² + (y -3)² = r²

=> ( x +5)² + (y-3)² = 10²

=> x² + 25 + 10x + y² +9-6y = 100

=> x² + y² +10x -6y + 31 = 100

=> x² + y² +10x -6y - 69 = 0

<h3>• Hence the equation of the circle is x² + y² +10x -6y - 69 = 0.</h3>

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I don’t know how to solve this

Lady_Fox [76]

Answer:

$\theta =2\pi k,\ \ k\in Z\ \\\text{or}\ \\\theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

Step-by-step explanation:

Given:

$\cos \theta-\sqrt{3}\sin \theta=1$

Divide this equation by 2:

$\dfrac{1}{2}\cos \theta-\dfrac{\sqrt{3}}{2}\sin \theta=\dfrac{1}{2}$

Note that

$\cos \dfrac{\pi }{3}=\dfrac{1}{2}\\ \\\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$

So, the previous equation is

$\cos \dfrac{\pi}{3}\cdot \cos \theta-\sin \dfrac{\pi}{3}\cdot \sin \theta=\dfrac{1}{2}$

Remind that

$\cos x\cos y-\sin x\sin y=\cos (x+y),$

then

$\cos \left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{2}$

The solution of this equation is

$\dfrac{\pi}{3}+\theta=\pm \arccos \dfrac{1}{2}+2\pi k,\ \ k\in Z\\ \\\dfrac{\pi}{3}+\theta=\pm \dfrac{\pi}{3}+2\pi k,\ \ k\in Z\\ \\\theta =2\pi k,\ \ k\in Z\ \text{or}\ \theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

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3 years ago

1²+3²+4²+.......+(2n-2)²= ⅓n(2n-1)(2n+1)

sergiy2304 [10]

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26 + (4n² - 8n + 4) = ¹/₃n(4n² - 1)
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4 years ago

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Brilliant_brown [7]

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4 years ago

The volume of a cylinder is 347.6cm cube. If the area of the base 34.76cm squared, what is the height?

34kurt

Answer:

115.3333

Step-by-step explanation:

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Hope this helps. :)

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3 years ago

Find the GCF of 14a5 and 22a3

miskamm [114]

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3 years ago