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2 years ago

Sam bought 6 pens for an unknown price plus a

1 answer:

5 0

So sam bought SIX pens for a TOTAL of $28.50. we have to figure out how much money each of the six pens cost.

steps:

1. divide the total by the amount of items purchased:

28.50 / 6 = 4.75

and there’s your answer! each pen cost $4.75

steps to check your work:

1. multiply the single cost by the amount of items purchased:

4.75 ( 6 ) = 28.50

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Translate the sentence into an inequality.<br> The sum of 3 and c is less than -29.

Solnce55 [7]

Answer:

3+c < -29 or c+3 < -29 because of communitive property

Step-by-step explanation:

First, the sum means addition, you could write 3+c or c+3 because addition is communitive property.

Then the sum is less than -29 so we need to use the less than symbol which is this symbol: <

Now write the inequality! 3+c < -29

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3 years ago

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What effect does the elliptical orbit of the moon around Earth have?

Levart [38]

It causes very low gravity but it still has a bit of gravity. Hope it helped! ;-) <3

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3 years ago

I’ll give points + brainalist for the correct answer (:

Harlamova29_29 [7]

Answer:

3 examples of mechanical digestion:

Mastication

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Peristalsis

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3 years ago

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Y < 3 includes three as a possible solution

irakobra [83]

False...it does not include 3....it includes only numbers less then 3.

For it to include 3, u would need an equal sign in there..
Y < = 3...(thats less then or equal)...this would include 3

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3 years ago

Find the taylor series centered at the given value of a and find the associated radius of convergence. (1) f(x) = 1 x , a = 1 (2

Kitty [74]

a) The radius of convergence is calculated as

R=1.

b) Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.

<h3>What is the associated radius of convergence.?</h3>

(a)

Take into consideration the function f with respect to the number a,

$f(x)=\frac{1}{x}, \quad a=1$

In case you forgot, the Taylor series for the function $f$ at the number a looks like this:

$\begin{aligned}f(x) &=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n !}(x-a)^{n} \\&=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{*}(a)}{2 !}(x-a)^{2}+\ldots\end{aligned}$

Determine the function f as well as any derivatives of the function $f by setting a=1 and working backward from there.

$\begin{aligned}f(x) &=\frac{1}{x} & f(1)=\frac{1}{1}=1 \\\\f^{\prime}(x) &=-\frac{1}{x^{2}} & f^{\prime}(1)=-\frac{1}{(1)^{2}}=-1 \\\\f^{\prime \prime}(x) &=\frac{2}{x^{3}} & f^{\prime \prime}(1)=\frac{2}{(1)^{3}}=2 \\\\f^{\prime \prime}(x) &=-\frac{2 \cdot 3}{x^{4}} & f^{\prime \prime}(1)=-\frac{2 \cdot 3}{(1)^{4}}=-2 \cdot 3 \\\\f^{(*)}(x) &=\frac{2 \cdot 3 \cdot 4}{x^{5}} & f^{(n)}(1)=\frac{2 \cdot 3 \cdot 4}{(1)^{5}}=2 \cdot 3 \cdot 4\end{aligned}$

At the point when a = 1, the Taylor series for the function f looks like this:

$f(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}+\cdots \\\\&=f(1)+\frac{f^{\prime}(1)}{1 !}(x-1)+\frac{f^{\prime}(1)}{2 !}(x-1)^{2}+\frac{f^{\prime \prime}(1)}{3 !}(x-1)^{3}+\cdots \\$

$&=1+\frac{-1}{1 !}(x-1)+\frac{2}{2 !}(x-1)^{2}+\frac{-2 \cdot 3}{3 !}(x-1)^{3}+\frac{2 \cdot 3 \cdot 4}{4 !}(x-1)^{4}+\cdots \\\\&=1-(x-1)+(x-1)^{2}-(x-1)^{3}+(x-1)^{4}+\cdots \\\\&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}$

In conclusion,

$&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}$

Find the radius of convergence by using the Ratio Test in the following manner:

$\begin{aligned}L &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \\&=\lim _{n \rightarrow \infty} \frac{(-1)^{n+1}(x-1)^{n+1}}{(-1)^{n}(x-1)^{n}} \mid \\&=\lim _{n \rightarrow \infty}|x-1| \\&=|x-1|\end{aligned}$

The convergence of the series when L<1, that is, |x-1|<1.

The radius of convergence is calculated as

R=1.

For B

Take into consideration the function f with respect to the number a,

$a_{n}=(-1)^{n}(x-1)^{n}$

$f(x)=\left(x^{2}+2 x\right) e^{x}, a=0$ The Taylor series for f(x)=e^{x} at a=0 is,

$e^{2}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots$

$f(x) &=\left(x^{2}+2 x\right) e^{x} \\&=\left(x^{2}+2 x\right)\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+2 x\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right) \\&=x^{2}\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+\left(\frac{x^{4}}{2 !}+\frac{x^{5}}{3 !}+\frac{x^{6}}{4 !}+\ldots\right)+\left(2 x+2 x^{2}+\frac{2 x^{3}}{2 !}+\frac{2 x^{4}}{3 !}+\frac{2 x^{5}}{4 !}+\ldots\right) \\$

$&=\left(x^{2}+x^{3}+\frac{x^{4}}{2 !}\right) \\&=2 x+3 x^{2}+\left(1+\frac{2}{2 !}\right) x^{3}+\left(\frac{1}{2 !}+\frac{2}{3 !}\right) x^{4}+\left(\frac{1}{4 !}\right) x^{5}+\ldots \\&=2 x+3 x^{2}+2 x^{3}+\frac{5}{6} x^{4}+\frac{1}{4} x^{5}+\ldots$

Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.