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3 years ago

Helpppzzzzzz

1 answer:

8 0

if the 3% comes with the car in equation you take the amount of seperate asteroids created by the moon and divide it by that 3% then add the number 25 and use that to the power of 3 and then you get the answer:

Santa is and isnt real at the same time

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Solve the following questions<br>

Simora [160]

Start with

$\dfrac{2x-1}{2x+1}-\dfrac{2x+1}{2x-1}$

Observe that this implies that the domain is

$2x+1,\ 2x-1\neq 0 \implies x \neq \pm\dfrac{1}{2}$

The least common denominator is

$\dfrac{2x-1}{2x+1}-\dfrac{2x+1}{2x-1} = \dfrac{(2x-1)^2-(2x+1)^2}{(2x+1)(2x-1)}=\dfrac{-8x}{4x^2-1}$

In the right hand side, we have

$-2\dfrac{2}{3}=-\left(2+\dfrac{2}[3}\right)=-\left(\dfrac{6}{3}+\dfrac{2}{3}\right)=-\dfrac{8}{3}$

So, the equation becomes

$-\dfrac{8x}{4x^2-1}=-\dfrac{8}{3} \iff \dfrac{x}{4x^2-1}=\dfrac{1}{3}$

Multiply both sides by 3:

$\dfrac{3x}{4x^2-1}=1 \iff 3x=4x^2-1 \iff 4x^2-3x-1=0$

This equation has solutions

$x=-\dfrac{1}{4},\quad x=1$

Which are compliant with the domain of the equation

7 0

4 years ago

What is the distance between (8,1) and (9,1)?

amid [387]

Answer:

Step-by-step explanation:

9-8=1

since it is on the same y coordinate it doesn't move much

4 0

4 years ago

Im going to cry... I need help..

Bad White [126]

Answer: 9

Step-by-step explanation:

If 12 is the whole length, and 3 is a small section of it, then 3 + j = 12

Or 12 - 3 = j

Therefore, 12 - 3 = 9

5 0

3 years ago

Read 2 more answers

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 :

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

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