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ki77a [65]
2 years ago
12

What’s the radius of center (-2,1) and contains the point (2,-1)

Mathematics
1 answer:
solniwko [45]2 years ago
8 0
<h2>Radius of a Circle Given Center and Point</h2>

To find the radius of a circle when we're given the coordinates of its center and a point it contains, we can use the following formula for distance:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

  • (x_1,y_1) is one point and (x_2,y_2) is another

<h2>Solving the Question</h2>

We're given:

  • Center: (-2,1)
  • Point: (2,-1)

Plug these given points into the formula for distance:

d=\sqrt{(2-(-2))^2+(-1-1)^2}\\d=\sqrt{(2+2)^2+(-1-1)^2}\\d=\sqrt{(4)^2+(-2)^2}\\d=\sqrt{16+4}\\d=\sqrt{20}\\d\approx4.47

<h2>Answer</h2>

Therefore, the radius of the circle is \sqrt{20} units, or approximately 4.47 units.

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The pair which are inverse of each other are:

Option: C

f(x)=11x-4\ and\ g(x)=\dfrac{x+4}{11}

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A)

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Now we calculate fog(x):

fog(x)=f(g(x))\\\\fog(x)=f(12x-15)\\\\fog(x)=\dfrac{12x-15}{12}+15\\\\fog(x)=x-\dfrac{15}{12}+15\\\\fog(x)=x-\dfrac{55}{4}\neq x

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B)

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Now we calculate fog(x):

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Hence, option: B is incorrect.

D)

f(x)=9+\sqrt[3]{x}\ and\ g(x)=9-x^3

Now we calculate fog(x):

fog(x)=f(g(x))\\\\fog(x)=f(9-x^3)

fog(x)=9+\sqrt[3]{9-x^3}\neq x

Hence, option: D is incorrect.

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Now we calculate fog(x):

fog(x)=f(g(x))\\\\fog(x)=f(\dfrac{x+4}{11})

fog(x)=11\times (\dfrac{x+4}{11})-4\\\\fog(x)=x+4-4\\\\fog(x)=x

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