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2 years ago

Which sequences are arithmetic? Select three options.

Mathematics

2 answers:

ArbitrLikvidat [17]2 years ago

9 0

Answer:

-6.2, -3.1, -1.55, -0.775, -0.3875...this is not an arithmetic sequence

Step-by-step explanation:

grandymaker [24]2 years ago

4 0

Answer:

it should be 2nd option, 3rd, and 5th.

Step-by-step explanation:

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6 tractors take 10days to collect the harvest. How long would it take 18 tractors to do the same amount of work?

Nutka1998 [239]

This is a question of proportionality. However, it is not direct proportionality as it is expected that as the number of tractors increases, the work is finished faster as opposed to fewer number of tractors. This is referred to as inverse proportionality.

Therefore;
6 tractors ----- 10 days
18 tractors --- x days

Then,
x = (6*10)/18 = 10/3 days = 3 days, and 8 hours.

This means that 18 tractors will take 3 days and 8 hours to collect the same harvest.

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1 year ago

Find the product of 0.094 ×0.367

Keith_Richards [23]

0.034498 is the answer

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2 years ago

Power series of y''+x^2y'-xy=0

Ray Of Light [21]

Assuming we're looking for a power series solution centered around $x=0$ , take

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Substituting into the ODE yields

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0$

The first series starts with a constant term; the second series starts at $x^2$ ; the last starts at $x^1$ . So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a $x^2$ term. We have

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}$

$\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}$

Re-index the first sum to have it start at $n=1$ (to match the the other two sums):

$\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}$

So now the ODE is

$\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0$

Consolidate into one series starting $n=1$ :

$\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0$

Suppose we're given initial conditions $y(0)=a_0$ and $y'(0)=a_1$ (which follow from setting $x=0$ in the power series representations for $y$ and $y'$ , respectively). From the above equation it follows that

$\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}$

Let's first consider what happens when $n=3k-2$ , i.e. $n\in\{1,4,7,10,\ldots\}$ . The recurrence relation tells us that

$a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0$

and so on, so that $a_{3k-2}=0$ except for when $k=1$ .

Now let's consider $n=3k-1$ , or $n\in\{2,5,8,11,\ldots\}$ . We know that $a_2=0$ , and from the recurrence it follows that $a_{3k-1}=0$ for all $k$ .

Finally, take $n=3k$ , or $n\in\{0,3,6,9,\ldots\}$ . We have a solution for $a_3$ in terms of $a_0$ , so the next few terms ( $k=2,3,4$ ) according to the recurrence would be

$a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}$
$a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}$
$a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}$

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for $n=3k$ as

$a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

So the series solution to the ODE is given by

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at $a_0=y(0)=1$ and $a_1=y'(0)=2$ overlaid with the series solution (orange) with $n=3$ and $n=6$ . (Note the rapid convergence.)

7 0

1 year ago

The International Business Club starts the school year with $250.50 in their account. They spend $35 each month on activities. T

svlad2 [7]

Answer:

250.50 - 35m = 300 - 45.25m

Step-by-step explanation:

Whenever an equation asks for when it will take x (or m in this case) amount of things or time you should know that there will be 2 different equations that have to equal each other

The equation will be 250.50 - 35m = 300 - 45.25m

The 35m is being subtracted from 250.50 because they are losing money each time they spend. This also applies to 45.25m

Also 35m and 45.25m are variables because it says they spend that much money each month. They don't specify how many months, so we have to put a variable after it.

7 0

1 year ago

You and a classmate have a bug collection for science class. You ﬁnd 5 out of every 9 bugs in the collection. You ﬁnd 4 more bug

FrozenT [24]

Answer:

The total number of bugs in the collection is 36

Step-by-step explanation:

Let

x -----> the number of bugs that you find out

y ----> the number of bugs that the classmate find out

we know that

$\frac{x}{y+x} =\frac{5}{9}$ -----> equation A

$x=y+4$ ----> equation B

Solve the system by graphing

Remember that the solution is the intersection point both graphs

The solution is the point (20,16)

The total number of bugs in the collection is

$(20+16)=36\ bugs$

5 0

1 year ago