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iren2701 [21]
3 years ago
8

Which is the standard equation for a circle centered at the origin with radius r?

Mathematics
2 answers:
scoundrel [369]3 years ago
6 0
D. This is the base equation. 
Nookie1986 [14]3 years ago
6 0

The standard equation of a circle is given by :

(x-h)^{2} + (y-k)^{2} =r^{2}

where (h,k) is the center and r is the radius of circle.

Now in the question we are given that the center is at origin.

So coordinate of center becomes (h,k) =(0,0)

radius is r.

So plugging these values in the standard equation, we get,

(x-0)^{2} + (y-0)^{2} =r^{2}

Simplifying,

(x)^{2} + (y)^{2} =r^{2}

Answer is option first:

x² + y² =r²

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3 years ago
R = sec(θ) − 2cos(θ), where -π/2 < θ < π/2
Alex

Answer:

  y = (x/(1-x))√(1-x²)

Step-by-step explanation:

The equation can be translated to rectangular coordinates by using the relationships between polar and rectangular coordinates:

  x = r·cos(θ)

  y = r·sin(θ)

  x² +y² = r²

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  r = sec(θ) -2cos(θ)

  r·cos(θ) = 1 -2cos(θ)² . . . . . . . . multiply by cos(θ)

  r²·r·cos(θ) = r² -2r²·cos(θ)² . . . multiply by r²

  (x² +y²)x = x² +y² -2x² . . . . . . . substitute rectangular relations

  x²(x +1) = y²(1 -x) . . . . . . . . . . . subtract xy²-x², factor

  y² = x²(1 +x)/(1 -x) = x²(1 -x²)/(1 -x)² . . . . multiply by (1-x)/(1-x)

  \boxed{y=\dfrac{x\sqrt{1-x^2}}{1-x}}

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The attached graph shows the equivalence of the polar and rectangular forms.

4 0
3 years ago
Is the relation a function? {(-1,7), (9,4), (3,-2), (5,3), (9,1)}
guapka [62]

Answer:

Yes. it is a function

Step-by-step explanation:

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7 0
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Wjat are the x intercepts if y=x squared minus 100​
Effectus [21]

Answer:

x = ± 10

Step-by-step explanation:

Given

y = x² - 100 ← a difference of squares , that is

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To find the x- intercepts let y = 0 , that is

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Equate each factor to zero and solve for x

x - 10 = 0 ⇒ x = 10

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Which of the following sets of ordered pairs is NOT a function?
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Step-by-step explanation:

7 0
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