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erica [24]
3 years ago
11

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:

9)

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.

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You can retry this question below
AysviL [449]

Answer:

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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 ...

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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.

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Question is not specific, so can only give a general answer.
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Yakvenalex [24]

Answer:

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Step-by-step explanation:

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