Wouldn’t you just for part a, insert different x values and solve to get your y value. and do that over and over again with different numbers to create a data table. and then for part b you’ll just make a graph and put down the points. so for the first slot, do 0 for x. y will equal -1 after solving, so your first plot for part b can be (0,-1) and continue on from there

5 0

3 years ago

What is the constant of proportionality?

beks73 [17]

Answer:

A. 12

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Compute the sum:

Nady [450]

You could use perturbation method to calculate this sum. Let's start from:

$S_n=\sum\limits_{k=0}^nk!\\\\\\\(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}$

On the other hand, we have:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!=\\\\\\=1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!\\\\\\(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}$

So from (1) and (2) we have:

$\begin{cases}S_{n+1}=S_n+(n+1)!\\\\S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases}\\\\\\
S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\\\\\\
(\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}$

Now, let's try to calculate sum $\sum\limits_{k=0}^{n}k\cdot k!$ , but this time we use perturbation method.

$S_n=\sum\limits_{k=0}^nk\cdot k!\\\\\\
\boxed{S_{n+1}=S_n+(n+1)(n+1)!}\\\\\\
$

but:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\=
\sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\=
\sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\
\boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}$

When we join both equation there will be:

$\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\
S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\
\sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\=
(n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\=
n(n+1)!+1$

So the answer is:

$\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}$

Sorry for my bad english, but i hope it won't be a big problem :)

8 0

4 years ago

Write Thirty-four thousand.six hundred fifty-two as a base ten number​

Write Thirty-four thousand.six hundred fifty-two as a base ten number