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3 years ago

PLS HELP WITH THIS!!!!!

Mathematics

1 answer:

Butoxors [25]3 years ago

4 0

Answer:

Step-by-step explanation:

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Please solve and show work! Thanks!

notsponge [240]

Answer:

1) x < 1, x > 9

2) 2 < x ≤ 6, -4 ≤ x < 0

Step-by-step explanation:

1) lx - 5l > 4

x - 5 = 4

x = 9

-(x - 5) = 4

x = -4 + 5 = 1

x < 1, x > 9

3) 1 < lx-1l ≤ 5

1 < x-1 ≤ 5

2 < x ≤ 6

1 < -(x-1) ≤ 5

-5 ≤ x-1 < -1

-4 ≤ x < 0

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3 years ago

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1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

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3 years ago

How is everybody doing in there life?

denis23 [38]

Answer:

I'm doing okay i miss school and my friends

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2 years ago

HELPP! BRAINLIEST AND 10 POINTS !!

Scrat [10]

Answer:

I think it's 3 units so you got that right

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3 years ago

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Grade 8 Math

Roman55 [17]

725 minus the 125 registration fee is 600. 600 divided by 12(months) equals 50 dollars a month for your answer.

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3 years ago