If the typos are randomly distributed over the 10 pages, find the probability of having a single typo on each page

Find the volume of this figure. round your answer to the nearest hundredth. if necessary #8

ivann1987 [24]

Answer:

30 miles²

Step-by-step explanation:

To solve the area of the prism we have to multiply the area of the base * the height of the prism. The base of the prism is a right triangle.

(b*h)÷2 = Area of a triangle

Base = 4

Height = 3

Now:

(4*3)/2 = 6 miles²

Now we have to multiply 6 miles² by 5 as our height.

We get 30 miles² as the volume of the figure.

4 0

1 year ago

What is the answer too this,it’s rational numbers - 3.3(6)= ?

Sloan [31]

Answer:

the answer is -19.8.

5 0

3 years ago

Read 2 more answers

What's the estimate for 2-9

andrew11 [14]

The estimate between 2 and 9 is 7

5 0

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

2 years ago

Given: line hg is the perpendicular bisector of segment JK. Which conclusion is valid?

Varvara68 [4.7K]

JG = GK
Angle JGH is a right angle
Angle HGK is a right angle assuming it is drawn like this (sorry i did my best to draw it):

H
l
l
l
l
J--------------l--------------K
G

6 0

3 years ago