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3 years ago

Advantage of this app?

Mathematics

2 answers:

raketka [301]3 years ago

5 0

Spicy memes, my man. spicy memes

Morgarella [4.7K]3 years ago

3 0

You get an answer to your question really quick,
...like now!

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Match each system of equations to its solution represented by an augmented matrix.

SCORPION-xisa [38]

Answer:

Step-by-step explanation:

7 0

2 years ago

What is the solution for 4.2=c/8?

musickatia [10]

4.2 = c
 8

multiply both sides by 8

4.2* 8 = 8c
 8

 c = 33.6

5 0

3 years ago

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

28÷x=168 What is the value of "x"

coldgirl [10]

It is 1/6 or .16 I used Photomath

7 0

3 years ago

Multiply 2/33•1/5•11/10 in simplest form

taurus [48]

2/33 * 1/5 * 11/10 = (2 * 1 * 11) / (33 * 5 * 10) = 22/1650 reduces to 1/75

6 0

3 years ago