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2 years ago

Evaluate the function y= -3 (1/4) for x =-3 on top of the 1/4 is where the x goes. so it is 1/4 x

1 answer:

7 0

$\text{Consider the function}\\
\\
y=-3\left ( \frac{1}{4} \right )^x\\
\\
\text{now to evaluate the value of the function at }x=-3, \text{ we substitute}\\
x=-3 \text{ in the given function and solve further for y.}\\
\\
\text{so we get}\\
\\
y=-3\left ( \frac{1}{4} \right )^{-3}\\
\\
\text{now using the exponent property, }\left ( \frac{1}{a} \right )^{-m}=(a)^m, \text{ we get}\\
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y=-3(4)^{3}\\
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\Rightarrow y=-3(64)\\
\\
\Rightarrow y=-192$

so the value of the function at x=-3 is -192

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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 :)

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3 years ago

How many solutions does this<br> graph have?

OlgaM077 [116]

Answer:10

Step-by-step explanation:

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2 years ago

A car travels at a speed of s miles per hour. It covers 126 miles in 3 hours.

makvit [3.9K]

Answer:

s=42

Step-by-step explanation: 126 divided by 3 = 42

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1 year ago

S see Use the graph y=e^x to evaluate these expression e^0

nika2105 [10]

Answer:

$y= e^x$

And we want to evaluate e^0 using the graph

And as we can see in the plot the y intercept is the blue point with y=0 and that correspond with:

$y(0)= e^0 =1$

Step-by-step explanation:

For this problem we know the following function:

$y= e^x$

And we want to evaluate e^0 using the graph

And as we can see in the plot the y intercept is the blue point with y=0 and that correspond with:

$y(0)= e^0 =1$

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2 years ago

Write an expression for the calculation the difference of the products of 5 and 2 and 5 and 1

kifflom [539]

(5x2) - (5-1)
10 - 4
6

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2 years ago