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3 years ago

What family ruled russia for 300 years

Mathematics

2 answers:

Aleksandr [31]3 years ago

6 0

The Romanov family ruled for 300 years.

solong [7]3 years ago

4 0

The Romanov family was the family that ruled Russia for 300 years. Their last ruling was in the year 1918 by Nicholas the second.

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0.2.0.4, 0.6, 0.8, 1....<br> Arithmetic or geometric

Basile [38]

Arithmetic sequences have a common difference between consecutive terms.

Geometric sequences have a common ratio between consecutive terms.

Let's compute the differences and ratios between consecutive terms:

Differences:

$0.4-0.2 = 0.2,\quad 0.6-0.4=0.2,\quad 0.8-0.6=0.2,\quad 1-0.8=0.2$

Ratios:

$\dfrac{0.4}{0.2}=2,\quad \dfrac{0.6}{0.4} = 1.5,\quad \dfrac{0.8}{0.6} = 1.33\ldots, \quad\dfrac{1}{0.8}=1.25$

So, as you can see, the differences between consecutive terms are constant, whereas ratios vary.

So, this is an arithmetic sequence.

3 0

3 years ago

It has been estimated that a student's final grade in a course decreases by 5 points for every 2 days absent from class. If a st

OverLord2011 [107]

Answer:

the final grade will decrease by 10 points

3 0

3 years ago

If the domain of f (x)=x^2 + 1 to {0,1,2,3}, what is the maximum value of the range?

Naddik [55]

I hope this helps you

x=0 f (0)=0^2+1=1

x=1 f (1)=1^2+1=2

x=2 f (2)=2^2+1=5

x=3 f (3)=3^2+1=10

3 0

4 years ago

Read 2 more answers

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago

20 to 8 in simpliest form

Mashcka [7]

Answer:

in it's lowest form its 5/2

8 0

3 years ago