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Sidana [21]
3 years ago
14

Devy likes to learn! Could someone please tell me how to answer this question?

Mathematics
1 answer:
Ronch [10]3 years ago
4 0

Answer:

B

Step-by-step explanation:

Recall that if two functions, <em>f</em> and <em>g</em>, are inverses, then by definition:

\displaystyle f(g(x)) = g(f(x)) = x

Hence, the graph of f(g(x)) should be simply <em>y</em> = <em>x</em>.

Therefore, our answer is B, as both coordinates are equivalent for all three points.

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In 2010, a city's population was 1,405,233 and it was decreasing at a rate of 1.1%. At this rate when will the city's population fall below 1,200,000?
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A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.
olga55 [171]
We know that:

(x-a)^2+(y-b)^2=r^2

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}

and all we have to do is solve it for a, b and r.

There will be:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\&#10;\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\&#10;\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\&#10;

From equations (II) and (III) we have:

\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}

and from (I) and (II):

\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}

Now we can easly calculate a and b:

\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}

Finally we calculate r^2:

a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}

And the equation of the circle is:

(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}
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LenaWriter [7]

Answer: $6197.77 would be the right answer

hope it helps :]

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