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Amiraneli [1.4K]
3 years ago
7

Find an equation equivalent to r = 10 sin theta rectangular coordinates.

Mathematics
2 answers:
Aleksandr [31]3 years ago
7 0

Answer:

x^2+y^2=10y

Step-by-step explanation:

d on edg

bazaltina [42]3 years ago
4 0

We have been an equation in polar coordinates r=10\text{sin}(\theta). We are asked to write our equation in rectangular coordinates.  

We know that the equation r=2b\text{sin}(\theta) is equation of a circle with a radius |b| and center at (0,b).

Let us find the value of b.

2b\text{sin}(\theta)=10\text{sin}(\theta)

\frac{2b\text{sin}(\theta)}{2\text{sin}(\theta)}=\frac{10\text{sin}(\theta)}{2\text{sin}(\theta)}

b=5

We know that equation of a circle in rectangular coordinates is (x-h)^2+(y-k)^2=r^2

Since b=5, so radius is 5 and center is at point (0,5).

(x-0)^2+(y-5)^2=5^2

x^2+(y-5)^2=25

Therefore, our required equation would be x^2+(y-5)^2=25.

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Write an equation of the line that passes through (2,−5) and is parallel to the line 2y=3x+10
LenaWriter [7]

Step-by-step explanation:

Divide two on both sides to get rid of it and the make the equation in the form y = mx + c

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y = 3/2x + 5

since both lines are parallel they must have the same gradient which is 3/2

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All you have to do now is to replace x and y with (2, -5) to find c

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Hello there! :)

\sqrt{81}

It's rational

it's an integer

Hope it helps!

~Just an emotional teen listening to her favorite song "Try Everything"

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You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.


-x^6 + 3x^4 + 4x^2

-x^2(x^4 - 3x^2 - 4)


Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.


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-x^2(x^2 - 4)(x^2 + 1)


Now we can use the factors of two perfect squares rule to factor the middle parenthesis.


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We would also want to split the term in the front.


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Now we would set each portion equal to 0 and solve.


First root

x = 0 ---> no work needed


Second root

-x = 0 ---> divide by -1

x = 0


Third root

x - 2 = 0

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Forth root

x + 2 = 0

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Fifth and Sixth roots

x^2 + 1 = 0

x^2 = -1

x = +/- \sqrt{-1}

x = +/- i

7 0
3 years ago
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