28 pounds of recyle trash is what percent of the barrel of 57 pounds of trash

KatRina [158]

The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for $y$

$\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0$

which indeed gives the recurrence you found,

$a_{n+3}=-\dfrac{n-3}{n+3}a_n$

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that $a_2=0$ , and substituting this into the recurrence, you find that $a_2=a_5=a_8=\cdots=a_{3k-1}=0$ for all $k\ge1$ .

Next, the linear term tells you that $6a_0+6a_3=0$ , or $a_3=a_0$ .

Now, if $a_0$ is the first term in the sequence, then by the recurrence you have

$a_3=a_0$
$a_6=-\dfrac{3-3}{3+3}a_3=0$
$a_9=-\dfrac{6-3}{6+3}a_6=0$

and so on, such that $a_{3k}=0$ for all $k\ge2$ .

Finally, the quadratic term gives $6a_1-12a_4=0$ , or $a_4=\dfrac12a_1$ . Then by the recurrence,

$a_4=\dfrac12a_1$
$a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1$
$a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1$
$a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1$

and so on, such that

$a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}$

for all $k\ge2$ .

Now, the solution was proposed to be

$y=\displaystyle\sum_{n\ge0}a_nx^n$

so the general solution would be

$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots$
$y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)$
$y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)$

4 0

3 years ago

What is the value of x? sin 55' = COS 3 Fnter your answer in the box

vivado [14]

Bro be careful with the link

6 0

3 years ago

Select all ratios equivalent to 4:3.

valentina_108 [34]

Answer:

8:6 and 16:12

Step-by-step explanation:

5 0

3 years ago

Nivyana and Ana are selling their apparel to earn money for a cruise. Knitted scarves are $50 each in Mittens are $25 per pair.

Katarina [22]

Answer:

The number of knitted scarves is 10

The number of knitted mitten is 20

Step-by-step explanation:

Given as :

The cost of each knitted scarves = $50

The cost of pairs of mittens = $25

Total number of scarves and mitten bought = 30

The total money invested to pay for items = $1000

Let The number of knitted scarves = k

Let The number of mitten = m

Now, According to question

The total money invested to pay for items = cost of each knitted scarves × number of knitted scarves + cost of pairs of mittens × number of mitten

i.e $1000 = $50 × k + $25 × m

Or, 40 = 2 k + m

i.e 2 k + m = 40 ........A

And

Total number of scarves and mitten bought = number of knitted scarves + number of mitten

i.e k + m = 30 ...........B

Now, solving eq A and eq B

So, (2 k + m ) - ( k + m ) = 40 - 30

Or, (2 k - k) + ( m - m) = 10

Or, k + 0 = 10

∴ k = 10

So, The number of knitted scarves = k = 10

Put the value of k into eq B

∵ k + m = 30

So, m = 30 - k

or, m = 30 - 10

∴ m = 20

The number of knitted mitten = m = 20

Hence, The number of knitted scarves is 10 and The number of knitted mitten is 20 . Answer

7 0

3 years ago

Write a unit test for addinventory(). call redsweater.addinventory() with parameter sweatershipment. print the shown error if th

Katena32 [7]

#include <iostream> using namespace std; class InventoryTag { public: InventoryTag(); int getQuantityRemaining() const; void addInventory(int numItems); private: int quantityRemaining; }; InventoryTag::InventoryTag() { quantityRemaining = 0; } int InventoryTag::getQuantityRemaining() const { return quantityRemaining; } void InventoryTag::addInventory(int numItems) { if (numItems > 10) { quantityRemaining = quantityRemaining + numItems; } } int main() { InventoryTag redSweater; int sweaterShipment = 0; int sweaterInventoryBefore = 0; sweaterInventoryBefore = redSweater.getQuantityRemaining(); sweaterShipment = 25; cout << "Beginning tests." << endl; // FIXME add unit test for addInventory /* Your solution goes here */ cout << "Tests complete." << endl; return 0; }

5 0

3 years ago