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Over [174]
2 years ago
12

PLS HELP ASAP

Mathematics
1 answer:
Contact [7]2 years ago
3 0

Answers:

Equation is (x+1)^2 + (y+2)^2 = 25

Center is (-1, -2)

Radius = 5

===========================================

Work Shown:

x^2+2x+y^2+4y=20\\\\(x^2+2x)+(y^2+4y)=20\\\\(x^2+2x+1)+(y^2+4y)=20+1\\\\(x^2+2x+1)+(y^2+4y+4)=20+1+4\\\\(x+1)^2 + (y+2)^2 = 25\\\\

center = (h,k) = (-1,-2)

radius = 5

-----------------------------

Explanation:

I grouped up the x and y terms separately. Then I added 1 to both sides to complete the square for the x terms. I cut the 2 from 2x in half, then squared it to get 1. In the next step, I cut the 4 from 4y in half to get 2, which squares to 4. So that's why I added 4 to both sides to complete the square for the y terms.

Each piece is factored using the perfect squares factoring rule which is a^2+2ab+b^2 = (a+b)^2

The last equation is in the form (x-h)^2 + (y-k)^2 = r^2

We can think of x+1 as x - (-1) to show that h = -1

Similarly, y+2 = y-(-2) = y-k to show that k = -2

The center is (h,k) = (-1,-2)

The radius is r = 5 because r^2 = 5^2 = 25 is on the right hand side in the last equation above.

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Simplify. x + y - 0 x 2 - y 2
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-y^(2)+x+y

Step-by-step explanation:

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3 years ago
Which expression is a fourth root of -1+isqrt3?
aleksklad [387]

Answer:

Step-by-step explanation:

\sf n^{th} roots of a complex number is given by DeMoivre's formula.

   \sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}

Here, k lies between 0 and (n -1) ; n is the exponent.

\sf -1 + i\sqrt{3}

a = -1 and b = √3

\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}

\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}

                   \sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})

                   \sf = \dfrac{-\pi }{3}

n = 4

For k = 0,

          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

For k =1,

         \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]

For k =2,

       z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]

For k = 3,

      \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]

For k = 4,

      \sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]

4 0
1 year ago
On Monday, there was no snow on the ground in Buffalo, New York. On Tuesday, three inches of snow fell.On Wensday a half an inch
mario62 [17]
Hi there!

Monday- 0 in.

Tuesday- 3 in.

Wedsnday- 2 1/2 in.

Thursday- 5 in.

Friday- 3 1/2 in.

Therefore, the answer is - 3 1/2 inches of snow was left on Friday.

Hope this helps you!

~DL
7 0
3 years ago
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