All categories

2 years ago

What force impacts all machines and reduces its energy?

Mathematics

2 answers:

omeli [17]2 years ago

5 0

I think it’s frictional force sorry if it’s wron g

Contact [7]2 years ago

3 0

Answer:

Step-by-step explanation:

FRICTION impacts all machines and reduces energy.

You might be interested in

LAST ONE THANKSSSS

pishuonlain [190]

Answer: A

Step-by-step explanation: with my calculation, and the information you have give i have to say it is A.

6 0

2 years ago

Read 2 more answers

The answer is 288, but i don’t know how to get the answer. please give me an explanation.

andriy [413]

Sorry I don’t know the answer and also sorry about that other comment they are very rude I hope you find someone who actually knows the answer!! ❤️

3 0

2 years ago

14/15 minus 1/3 as a fraction in simplest form

eduard

First step: Create a common denominator with 1/3

1/3 x 5/5 = 5/15

Second step: Subtract

14/15

-5/15

_____

a: 9/15

4 0

3 years ago

Read 2 more answers

Tia, Veronica, Pam, and Lily are sisters. Tia is 8 years old and she is 2 years older than Pam. Pam is 5 years younger than Vero

JulijaS [17]

Answer:

Pam 6

Tia 8

Veronica 11

Lily 15

Step-by-step explanation:

Tia's age = 8

Pam's age = Tia's age - 2

8 - 2 = 6

Veronica = pam's age + 5

6 + 5 = 11

Lily's age = veronica's age + 4

11 + 4 = 15

3 0

3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago