All categories

2 years ago

Zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

Mathematics

1 answer:

Kisachek [45]2 years ago

3 0

Mmmmmmmmmmmmmmmmkkkkkkkkkkkkkkk

You might be interested in

Which number is equivalent to 0.3

Strike441 [17]

3/10 that's equivalent

8 0

3 years ago

Read 2 more answers

Samuel walks forward 19 steps. He represents this movement with a positive 19. How would he represent the opposite of this numbe

spayn [35]

Answer: -38

Step-by-step explanation:

First he would have to walk back 19 steps to get back to his orginal position, then he would have to further go back another 19 steps.

7 0

3 years ago

Mike pays $285 in rent every week.<br> How much will he pay in rent for the entire year?

sergejj [24]

Answer:

14,820

Step-by-step explanation:

A year has 52 weeks. 285 x 52 = 14820

6 0

1 year ago

Read 2 more answers

5 friends split 1/2 of a chocolate bar. How<br> much does each friend eat?<br> Draw a model.

Vinvika [58]

Answer:

1/10

Step-by-step explanation:

the friends eat 1/10 of the whole bar or 1/5 of half a bar

3 0

2 years ago

An=an-1+6an-2 <br> n>2 a0=3 a1=6

Svetach [21]

$\begin{cases}a_0=3\\a_1=6\\a_n=a_{n-1}+6a_{n-2}&\text{for }n>2\end{cases}$

Let the generating function for $a_n$ be

$A(x)=\displaystyle\sum_{n\ge0}a_nx^n$

Multiplying both sides by $x^{n-2}$

$a_nx^{n-2}=a_{n-1}x^{n-2}+6a_{n-2}x^{n-2}$

Summing both sides over the non-negative integers greater than or equal to 2 gives

$\displaystyle\sum_{n\ge2}a_nx^{n-2}=\sum_{n\ge2}a_{n-1}x^{n-2}+6\sum_{n\ge2}a_{n-2}x^{n-2}$
$\displaystyle\frac1{x^2}\sum_{n\ge2}a_nx^n=\frac1x\sum_{n\ge1}a_nx^n+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(\sum_{n\ge0}a_nx^n-a_0-a_1x\right)=\frac1x\left(\sum_{n\ge0}a_nx^n-a_0\right)+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(A(x)-3-6x\right)=\frac1x\left(A(x)-3\right)+6A(x)$
$A(x)=\dfrac{3x+3}{1-x-6x^2}$
$A(x)=\dfrac3{5(1+2x)}+\dfrac{12}{5(1-3x)}$

For $|x|$ , the two series converge to

$\displaystyle A(x)=\frac35\sum_{n\ge0}(-2x)^n+\frac{12}5\sum_{n\ge0}(3x)^n$
$\implies A(x)=\displaystyle\sum_{n\ge0}\dfrac{3(-2)^n+12(3)^n}5x^n$
$a_n=\dfrac{3(-2)^n+12(3)^n}5=\dfrac{3(-2)^n+4(3)^{n+1}}5$

5 0

3 years ago