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scoundrel [369]

3 years ago

5

2. José is using 12 brown tiles and

1 answer:

Karolina [17]3 years ago

3 0

Answer:

12:20

Step-by-step explanation:

your trying to find the ratio to brown tiles to one section and theirs 12 brown tiles per section so you would start off with 12. and to find how many tiles would be in each section you would have to add 8 and 12 and you would end up with 20 which the number of tiles in each section

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Y=2× Y=×+1<br> A: (2/3, 1/3)<br> B: (1,2)<br> C: (2,1)<br> D: (1/3, 2/3)

FrozenT [24]

Answer:

The answer is B: (1,2)

Step-by-step explanation:

(solved using elimination method)

-2x+y=0

y=x+1

-x+y=1

-2x+y=0

x+y=-1

-x=-1

x=1

-1+y=1

y=2

(x,y)=(1,2)

5 0

2 years ago

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Svetradugi [14.3K]

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7 0

2 years ago

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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]

If

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

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

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

7 0

3 years ago

Twice the difference of x and 5

Novosadov [1.4K]

Answer:

2*(x-5) = -33, so x-5 = -16.5, so x = -11.5

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7 0

3 years ago