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2 years ago

Can anyone help, please and thank you

Mathematics

1 answer:

Akimi4 [234]2 years ago

4 0

A is the center of the dilation because it is being delarged(is that even a word) by a factor of 2 downwards, and the only place possible to be the center of this "delargation" is A

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Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

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

Then the ODE can be written as

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

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

3 years ago

Select all the correct answers. Which two inequalities can be used to find the solution to this absolute value inequality? 3|x +

ziro4ka [17]

The absolute value inequality can be decomposed into two simpler ones.

x < 0

x > -8

<h3></h3><h3>Which two inequalities can be used?</h3>

Here we start with the inequality:

3|x + 4| - 5 < 7

First we need to isolate the absolute value part:

3|x + 4| < 7 + 5

|x + 4| < (7 + 5)/3

|x + 4| < 12/3

|x + 4| < 4

The absolute value inequality can now be decomposed into two simpler ones:

x + 4 < 4

x + 4 > - 4

Solving both of these we get:

x < 4 - 4

x > -4 - 4

x < 0

x > -8

These are the two inequalities.

Learn more about inequalities:

brainly.com/question/24372553

#SPJ1

6 0

1 year ago

the length of a rectangle is 5 times as long as its breadth and its area is 1620 m² find area of square whose perimeter is equal

Rus_ich [418]

Answer:

Area of the square is 2916 $m^{2}$ .

Step-by-step explanation:

Let the breadth of the rectangle be represented by w. So that;

length of rectangle = 5w

Area of rectangle = length x breadth

= 5w x w

1620 = 5 $w^{2}$

divide through by 5 to have,

$w^{2}$ = 324

w = $\sqrt{324}$

= 18

Thus, the breadth of the rectangle is 18 m.

length = 5 w = 5 x 18

= 90 m

The length of the rectangle is 90 m.

Perimeter of the rectangle = 2(l + w)

= 2( 90 + 18)

= 216 m

Perimeter of a square = 4l

where l is the length of its side.

Given that the perimeters of the rectangle and square are equal, then;

4l = 216

l = $\frac{216}{4}$

= 54

length of the side of square is 54 m.

Therefore,

Area of square = $l^{2}$

= $54^{2}$

= 2916

Area of the square whose perimeter is equal to that of the rectangle is 2916 $m^{2}$ .

6 0

3 years ago

Binomial expression

LiRa [457]

Answer

2.(2,1)

3. (2,2)

Step-by-step explanation:

Given,

2.2x+y=4

x+3y=_3

sol

2x+y=4..... equation i

x+3y=_3 ....eq ii

eq ii of 2 multiple the eq i

2x+y=4

2x+6y=_6

_ _ +

__________

o+ 5y = 10

y =10÷5

y = 2 ans

y valu put the eq 1

2x+y=4

2x+2=4

2x=4_2

2x=2

x=2÷2

x=1

sol

3x+5y=4..... eq 1

x+3y=4 ..... eq 2

eq 2 multiple eq 1. 3

3x+5y=4

3x+9y=12

_ _ _

___________

0x+4y=8

4y=8

y=8÷4

y=2 ans

the y value put the eq 1

3x+5y=4

3x+5×2=4

3x+10=4

3x=10_4

3x=6

x=6÷3

×=2 ans

5 0

3 years ago

Read 2 more answers

Enzo and Beatriz are playing games at their local arcade. Incredibly, Enzo wins 5 tickets from every game, and Beatriz wins 11 t

Hoochie [10]

Given :Enzo wins 5 tickets from every game, and Beatriz wins 11 tickets from every game.We need to find the minimum number of games that Enzo could have played to win the same number of tickets.

The minimum number of games that Enzo could have played to win the same number of tickets Will be the least common multiple of 11 and 5.

The factors of 11 and 5 are

11=11x1

5= 5x1

Least common multiply = 11x5=5.

The minimum number of games that Enzo could have played to win the same number of tickets is 55.

3 0

3 years ago

Read 2 more answers