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2 years ago

Hey can some one please help me !!!

Mathematics

2 answers:

natulia [17]2 years ago

6 0

1 is C- Algebraic expression: it consist of numbers and variables

2 is D- the number of times Chris washes the car

trasher [3.6K]2 years ago

5 0

1)c because it has a variable so it is an algebraic expression

2)d because 10 represents how much money he is making per wash so X must be the number of times he washes the car

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Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

2 years ago

Is the collection of US museums well-defined and therefore is not a set

taurus [48]

Answer:

be “well-defined” the collection description would have to settle all such questions. ... All these questions indicate the statement is ambiguous, i.e., it is not clear which students are members of this collection, hence, the collection is not well-defined.

4 0

2 years ago

Use the given graph of f to state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)

alex41 [277]

Answer:

Hence the answer is given as follows,

Step-by-step explanation:

Graph of y = f(x) given,

(a) $\lim_{x\rightarrow 2^{-}}f(x)=3$

(b) $\lim_{x\rightarrow 2^{+}}f(x)=1$

(c) $\lim_{x\rightarrow 2}f(x)= DNE \left \{ \therefore \lim_{x\rightarrow 2^{-}} f(x)\neq \lim_{x\rightarrow 2^{+}}f(x) \right.$

(d) $f(2)=3$

(e) $\lim_{x\rightarrow 4}f(x) = 4$

(f) $f(4)= DNE.$ { Hole in graph}

Hence solved.

3 0

2 years ago

........\........................

Vikki [24]

Part a: 6:3 > 2:1
part b: 9:3> 3:1

4 0

2 years ago

Can you explain please I don’t understand?

Angelina_Jolie [31]

don't be alarmed the two lines just mean absolute value, the problem is asking what is m. the problem is asking what number multipyed by 2, subtracted by 1 will give you 3?

2x2=4 right?

so 4-1=3 there!

m=4

so the problem is 2(4)-1=3

i hope i explained well enough!

5 0

3 years ago

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