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Rina8888 [55]
4 years ago
7

A circle centered at (-2,3) and contains point (5,9) which represents the equation

Mathematics
1 answer:
SIZIF [17.4K]4 years ago
8 0
The standard form of the equation of a circle is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle, (x,y) is a point of the circle, and r is the length of the radius of the circle. When the equation of a circle is written, h,k, and r are numbers, while x and y are still variables. (x-2)^2 + (y-k)^2 = 16 is an example of a circle. The problem gives us two of the three things that a circle has, a point (5,9) and the center (-2,3). We need to find the radius in order to write the equation. We substitute -2 for h, 3 for k, 5 for x, and 9 for y to get (5 - (-2))^2 + (9 - 3)^2 = r^2 We simplify: 49 + 36 = r^2, r^2 = 85. We only need to know r^2 because the equation of a circle has r^2. We now have all the information to write the equation of a circle. (x + 2)^2 + (y - 3)^2 = 85.
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Which function has an inverse function?
daser333 [38]

Answer: Option B

Step-by-step explanation:

By definition, only those functions that are one to one have an inverse function.

A function is one by one if there are not two different input values, x_1 and x_2, that have the same output value y

Note that the function f(x)= \frac{|x+3|}{5}  is not a one-to-one function

When x=2  f(x)= \frac{|2+3|}{5}=1\ ,\ \ y=1

When x=8  f(x)= \frac{|-8+3|}{5}=1\ \ ,\ y=1

Note that the function f(x)= \frac{x^4}{7}+ 27  is not a one-to-one function

When x=1 f(x)= \frac{(1)^4}{7}+27\ ,\ \ y=\frac{190}{7}

When x=-1  f(x)= \frac{(-1)^4}{7}+27\ ,\ \ y=\frac{190}{7}

Note that the function f(x)= \frac{1}{x^2}  is not a one-to-one function

When x=1 f(x)= \frac{1}{(1)^2}\ ,\ \ y=1

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Then the answer is the option B.

You can verify that The function f (x) = x ^ 5-3 is a one-to-one function and therefore its inverse is a function

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3 years ago
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