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3 years ago

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in triangle abc , the length of ab is 3 cm and the length of bc is 7 cm. triangle abc will be dilated with the center at the ori

gin to create a’ b’ c’. if image of point a (-3,5) after the dilation is a’ (-12,20) what is the length of b’c’ ?

1 answer:

lys-0071 [83]3 years ago

3 0

Answer:

What are the options?

Step-by-step explanation:

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7 pounds of rice is 4$ how much is one pound of rice?

blagie [28]

Answer:

$0.57

Step-by-step explanation:

7 pound = 4 $

1 pound =4/7 $

=0.57$

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1 year ago

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Please hurry. it's geometry.

Ainat [17]

That would be option A as the angles and sides ( the AS) have already been stated.

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2 years ago

javon is helping his dad build a tree house. he has a piece of trim that is 13 ft long. how many pieces can Javon cut that are 1

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3 years ago

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Find total surface area

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2 years ago

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