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3 years ago

Which of the following is most likely to be a relative minimum for this graph?

Mathematics

1 answer:

olga55 [171]3 years ago

4 0

Answer: A (3, -4)

Step-by-step explanation:

A minimum is at the lowest point on the parabola. The minimum and maximum can be seen by looking at the vertex.

If you go right 3 spaces on the x-axis horizontally, then down 4 spaces vertically on the y-axis you get (3, -4).

4 is negative because you are going down below the x-axis.

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Which of the following are considered skew lines? <br> 1.cone 2. Prism 3. Cylinder 4.cube

Andrej [43]

The answer is number 4. Cube

8 0

2 years ago

Think about the function f(x) = 10 - x3. What is the input, or independent variable? f(x) x y

denpristay [2]

Answer:

The input or independent variable is x.

Step-by-step explanation:

3 0

1 year 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

−4x−2y x−2y =−2 =9

fenix001 [56]

Dy\dx=-2+y/x+1 is the answer I think

7 0

2 years ago

How many degrees has ABC been rotated counterclockwise about the<br> origin?

omeli [17]

Answer:

Step-by-step explanation:

It would take 90 degree to rotate clockwise so it will take 360-90=270 degree to rotate counterclockwise