All categories

2 years ago

Help Me, Will give 100 points.

Mathematics

1 answer:

Bad White [126]2 years ago

6 0

Answer:

baka

Step-by-step explanation:

You might be interested in

Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

4 0

2 years ago

32 3/4 take away 12 1/2

Otrada [13]

Answer:

The answer is 20 and 1/4

7 0

3 years ago

Read 2 more answers

Find the product of (2x-3) and (x+3)

miss Akunina [59]

Answer:

(2x - 3) * (x + 3)

=> (2x * x) + (-3 * 3)

=> 2x - 9

=> 2x -9

If my answer helped, kindly mark me as the brainliest!!

Thank You!!

3 0

3 years ago

In a certain neighborhood there are 25 dogs and 10 cats. In a neighborhood across the way, the ratio of cats to dogs is the same

Akimi4 [234]

28 because 2.5 times 8 equals 20 plus that 8 equals 28.

Step-by-step explanation:

6 0

2 years ago

Last question and then I’ll take care of the rest on my own ^^^^

Bond [772]

Answer:

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

a^-b = 1/a^b

Using these rules, we can simplify the expression as follows:

(5^-3)(2^2)(5^6)/((2^3)(5^2)) = (5^(-3+6-2))(2^(2-3)) = (5^1)(2^-1)

= 5/2

6 0

2 years ago