AsKevin Selma madelyn Louis

elena-s [515]

Answer: Velveeta

Step-by-step explanation:

Unit rate is basically the price for a single slice. This being said you would divide 2.50 by 10 and you would get .25, then you would do 4.00 by 20 which would be .20. So the better deal would be Velveeta.

6 0

3 years ago

Identify the independent and dependent variables in the following situation Julie notices that her plant grows one inch for ever

LUCKY_DIMON [66]

Answer:

Independent variable: amount of water

Dependent variable: growth of the plant

Step-by-step explanation:

In the statement "Julie notices that her plant grows one inch for every liter of water that it receives" it is implied that the growth of the plant is related proportionally to the amount of water it receives.

We know the amount of growth in function of the amount of water

The dependant variable, the result, is the growth of the plant.

Then, the independent variable is the amount of water, as it is the input to calculate the amount of growth.

4 0

3 years ago

What is the slope of the line given by the following equation? Give an exact number. -13x 5y=-7

Black_prince [1.1K]

The slope should be 13/5x

8 0

3 years ago

What are three ratios that are equivalent to 6/7

Tasya [4]

14 and 21 and 18

is three of them

5 0

2 years ago

Read 2 more answers

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