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2 years ago

-1/3x+1=-7 negative one third x plus one equals negative seven

2 answers:

4 0

Multiply both sides by -3 to get rid of the negative and the fraction.

X - 3 = 21

Move the constants over to isolate the variable

X = 24

Vera_Pavlovna [14]2 years ago

4 0

Answer:
x=24

Solving:

-1/3x + 1 = -7
- 1 -1
-1/3x = -8
3(-1/3x) = 3(-8)
-1x = -24
x = 24

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How many 7-digit numbers can be

padilas [110]

Answer: 5040 7-digit numbers

Step-by-step explanation:

Permutations:

1: 1 ==> 1 permutation ==> 1 ==> 1!

12: 12, 21 ==> 2 permutations ==> 1*2 ==> 2!

123: 123, 132, 213, 231, 312, 321 ==> 6 permutations ==> 1*2*3 ==> 3!

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2 years ago

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

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Check attachment.

Step-by-step explanation:

I graphed it.

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3 years ago

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