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Pachacha [2.7K]
3 years ago
6

The Identity function i behaves just like the number 1. That is (F o i)=f and (i o g)= g

Mathematics
1 answer:
andrezito [222]3 years ago
3 0

I'll denote the identity function by \mathrm{Id}. Then for any functions f with inverse f^{-1},

\begin{cases}f\circ\mathrm{Id}=f\\\mathrm{Id}\circ f=f\\f\circ f^{-1}=\mathrm{Id}\end{cases}

One important fact is that composition is associative, meaning for functions f,g,h, we have

(f\circ g)\circ h=f\circ(g\circ h)

So given

h=f\circ g

we can compose the functions on either side with g^{-1}:

h\circ g^{-1}=(f\circ g)\circ g^{-1}=f\circ(g\circ g^{-1})

then apply the rules listed above:

h\circ g^{-1}=f\circ\mathrm{Id}=f

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Find d for the arithmetic series with S17=-170 and a1=2
Irina18 [472]
So, we know the sum of the first 17 terms is -170, thus S₁₇ = -170, and we also know the first term is 2, well

\bf \textit{ sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n(a_1+a_n)}{2}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
----------\\
n=17\\
S_{17}=-170\\
a_1=2
\end{cases}
\\\\\\
-170=\cfrac{17(2+a_{17})}{2}\implies \cfrac{-170}{17}=\cfrac{(2+a_{17})}{2}
\\\\\\
-10=\cfrac{(2+a_{17})}{2}\implies -20=2+a_{17}\implies -22=a_{17}

well, since the 17th term is that much, let's check what "d" is then anyway,

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=17\\
a_{17}=-22\\
a_1=2
\end{cases}
\\\\\\
-22=2+(17-1)d\implies -22=2+16d\implies -24=16d
\\\\\\
\cfrac{-24}{16}=d\implies -\cfrac{3}{2}=d
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Please see attached photo.

Step-by-step explanation:

Step 1:

Analyze what was given

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Please attached photo for explanation

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