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Natasha_Volkova [10]
2 years ago
6

What is the mathematical sentence of "A number increasd by 7 is 15?

Mathematics
1 answer:
svet-max [94.6K]2 years ago
6 0

Answer:

B

Step-by-step explanation:

A number increased by 7 means, 7 was added to the number and the result is 15

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The wholesale price for a shirt is $5.50. A certain department store marks up the wholesale price by 40%. Find the price of the
Kazeer [188]

Answer:

$7.7

Step-by-step explanation:

Hello, I can help you with this

you can solve this by using a simple rule of three.

Step 1

define

if the wholesale price for a shirt is $5.50,it is 100 percent

$5.50⇔100%

the new price(x) is 40% more than the original, it means 140%

x  ⇔ 140%

the relation is

\frac{5.5}{100}= \frac{x}{140}\\

Step 2

solve for x(isolate x)

\frac{5.5}{100}= \frac{x}{140}\\x=\frac{5.5*140}{100}\\x=7.7

the price of the shirt in the department store is $7.7

Have a nice day.

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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.

Read more about convergence

brainly.com/question/15415793

#SPJ4

The complete question is attached below

3 0
2 years ago
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