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3 years ago

Which of the following criteria is excluded from national benchmark test?

Mathematics

2 answers:

Lina20 [59]3 years ago

7 0

Answer:

Quantitative literacy

Zepler [3.9K]3 years ago

4 0

The answer is all of the above so A hope this helped

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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

3 years ago

Calvina Miller makes a car payment of $214.50 every month. Her car

Akimi4 [234]

12,870 dollars. Let me know if you want me to show my work.

4 0

2 years ago

If you find this, it was an accident

Anna007 [38]

Alright, lols yes yes

6 0

3 years ago

If the numerator of a fraction is increased by 3, the fraction becomes 3/4. If the denominator is decreased by 7, the

Sauron [17]

7 over 1

Step-by-step explanation:

3 0

3 years ago

A whole number is 6 more than 2 times another number. The sum of the two numbers is less than 50. This can be written in an ineq

alukav5142 [94]

Answer:

13, 14

Step-by-step explanation:

The parameters of the numbers are;

A whole number value = 2 × Another number + 6

The sum of the two numbers is less than 50

Given that the first number is equal to more than twice the second number, we have that the first number is the larger number, while the second number is the smaller number

Where 'x' represents the second number, we get;

x + 2·x + 6 < 50

Simplifying gives;

3·x + 6 < 50

x < (50 - 6)/3 = 14. $\overline 6$

x < 14. $\overline 6$

Therefore, the numbers for which the inequality holds true are numbers less than 14. $\overline 6$ . From the given option, the numbers are 13, and 14.

3 0

3 years ago