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3 years ago

69 predicted the pandemic????

Mathematics

2 answers:

prohojiy [21]3 years ago

7 0

Yes it’s just wowwowowow

LenaWriter [7]3 years ago

7 0

Answer:

I guess lol.....

You might be interested in

SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).

Ann [662]

Answer:

Required series is:

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

Step-by-step explanation:

Given that

$f'(x) = -\frac{1}{1 + x^{2}}$ ---(1)

We know that:

$\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}}$ ---(2)

Comparing (1) and (2)

$f'(x)=-(tan^{-1}x)$ ---- (3)

Using power series expansion for $tan^{-1}x$

$f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx$

$= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]$

$=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}$

$=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

$tan^{-1}(0)=0 \implies C=0$

Hence,

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

7 0

4 years ago

To find amplitude, measure (blank). A. the horizontal distance between consecutive crest positions B. the horizontal distance be

Oksanka [162]

The vertical distance from a crest to the rest position ! I know I'm a little late but I just took this.

5 0

4 years ago

Read 2 more answers

Write an equation for this $13 more than twice the cost of a video game is $67. Find the cost of the video game.

Nezavi [6.7K]

Answer:

13+2x=67 soooooo x = 27 so the video game is $27 and thats the equation

Step-by-step explanation:

6 0

3 years ago

Find the average value of f(x)=e2x over the interval [2, 4].

shusha [124]

<h2>Option B is the correct answer.</h2>

Step-by-step explanation:

We need to find average value of $e^{2x}$ in [2,4]

Area of $e^{2x}$ in [2,4] is given by

$\int_{2}^{4}e^{2x}dx=\frac{1}{2}\times \left [ e^{2x}\right ]^4_2\\\\\int_{2}^{4}e^{2x}dx=\frac{1}{2}\times(e^8-e^4)=1463.18$

Area of $e^{2x}$ in [2,4] = 1463.18

Difference = 4 - 2 = 2

Average value = Area of $e^{2x}$ in [2,4] ÷ Difference

Average value = 1463.18 ÷ 2

Average value = 731.59

Option B is the correct answer.

5 0

3 years ago

HELP IM IN DESPERATE NEED!!!<br> What is the solution to this system of equations?

Paladinen [302]

[Given]

{ x + y = 6

{ x = y + 4

[Plug-in our x value & solve]

[Given] x + y = 6

[ Plug-in] (y + 4) + y = 6

[Distribute] y + 4 + y = 6

[Combine like terms] 2y + 4 = 6

[Subtract 4 from both sides] 2y = 2

[Divide both sides by 2] y = 1

[Answer]

Third option - (5, 1)

-> You do not need to solve for x since this is the only option that has y = 1, but to solve for x we would do y + 4 = 1 + 4 = 5, so this answer fully checks correctly

Have a nice day!

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- Heather

6 0

3 years ago