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3 years ago

Need help ASAP. Last two problems on my study guide, can't seem to solve them?

Mathematics

1 answer:

Dimas [21]3 years ago

4 0

Answer:

9) 4

10) p¹⁵/q⁹

Step-by-step explanation:

As per the law of indices, (xᵃ)ᵇ=xᵃᵇ.

So divide 8 by 2 to get 4

10)

p(p⁻⁷q³)⁻²q⁻³

p(p¹⁴q⁻⁶)q⁻³ <em>(because (xᵃ)ᵇ=xᵃᵇ)</em>

pq⁻³(p¹⁴q⁻⁶)

p¹⁵q⁻⁹

p¹⁵/q⁹ <em>(because x⁻ᵃ =1 /xᵃ)</em>

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Find two linearly independent power series solutions about the point x0 = 0 of

aksik [14]

Assume a solution of the form

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

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting into the ODE, which appears to be

$(x^2-4)y''+3xy'+y=0,$

gives

$\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}x^{n+2}-4(n+2)(n+1)a_{n+2}x^n+3(n+1)a_{n+1}x^{n+1}+a_nx^n\bigg)=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^n-4\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n+3\sum_{n\ge1}na_nx^n+\sum_{n\g0}a_nx^n=0$

$(a_0-8a_2)+(4a_1-24a_3)x+\displaystyle\sum_{n\ge2}\bigg[(n+1)^2a_n-4(n+2)(n+1)a_{n+2}\bigg]x^n=0$

which gives the recurrence for the coefficients $a_n$ ,

$\begin{cases}a_0=a_0\\a_1=a_1\\4(n+2)a_{n+2}=(n+1)a_n&\text{for }n\ge0\end{cases}$

There's dependency between coefficients that are 2 indices apart, so we consider 2 cases.

If $n=2k$ , where $k\ge0$ is an integer, then

$k=0\implies n=0\implies a_0=a_0$

$k=1\implies n=2\implies a_2=\dfrac1{4\cdot2}a_0=\dfrac2{4\cdot2^2}a_0=\dfrac{2!}{2^4}a_0$

$k=2\implies n=4\implies a_4=\dfrac3{4\cdot4}a_2=\dfrac3{4^2\cdot4\cdot2}a_0=\dfrac{4!}{2^8(2!)^2}a_0$

$k=3\implies n=6\implies a_6=\dfrac5{4\cdot6}a_4=\dfrac{5\cdot3}{4^3\cdot6\cdot4\cdot2}a_0=\dfrac{6!}{2^{12}(3!)^2}a_0$

and so on, with the general pattern

$a_{2k}=\dfrac{(2k)!}{2^{4k}(k!)^2}a_0$

If $n=2k+1$ , then

$k=0\implies n=1\implies a_1=a_1$

$k=1\implies n=3\implies a_3=\dfrac2{4\cdot3}a_1=\dfrac{2^2}{2^2\cdot3\cdot2}a_1=\dfrac1{(3!)^2}a_1$

$k=2\implies n=5\implies a_5=\dfrac4{4\cdot5}a_3=\dfrac{4\cdot2}{4^2\cdot5\cdot3}a_1=\dfrac{(2!)^2}{5!}a_1$

$k=3\implies n=7\implies a_7=\dfrac6{4\cdot7}a_5=\dfrac{6\cdot4\cdot2}{4^3\cdot7\cdot5\cdot3}a_1=\dfrac{(3!)^2}{7!}a_1$

and so on, with

$a_{2k+1}=\dfrac{(k!)^2}{(2k+1)!}a_1$

Then the two independent solutions to the ODE are

$\boxed{y_1(x)=\displaystyle a_0\sum_{k\ge0}\frac{(2k)!}{2^{4k}(k!)^2}x^{2k}}$

and

$\boxed{y_2(x)=\displaystyle a_1\sum_{k\ge0}\frac{(k!)^2}{(2k+1)!}x^{2k+1}}$

By the ratio test, both series converge for $|x|$ , which also can be deduced from the fact that $x=\pm2$ are singular points for this ODE.

6 0

3 years ago

You can retry this question below

AysviL [449]

Answer:

305 calories, 120 from fat

Step-by-step explanation:

The ratio of the area of the larger pizza to that of the smaller pizza is the square of the ratio of the diameters. So, the larger pizza has an area that is ...

(12/6)² = 4

times that of the smaller pizza. When that area is divided into 8 parts, each part has an area that is 4/8 = 1/2 the area of the smaller pizza.

We expect a slice of the larger pizza to have 1/2 the calories of a smaller pizza, so 305 calories, 120 from fat.

610/2 = 305; 240/2 = 120.

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3 years ago

How to solve an equation for x with natural log of x

Soloha48 [4]

Question is not specific, so can only give a general answer.
As an example:
If the equation is of the form
log(x)=5
then raise each side to the power of e to give
e^(log(x))=e^5
Since e^(log(x))=x, we have
x=e^5.

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4 years ago

HELP ME AND I WILL GIVE BRAINLIST AND 15 POINTS

Yakvenalex [24]

Answer:

im not 100% sure but i think the first one is 544, the second is 549 and the last is 551

545 goes in the second one

Step-by-step explanation:

mark brainliest if you want <3

7 0

3 years ago

Read 2 more answers