Factor the expression below. Pls help!

60%

arsen [322]

Answer:

To answer the question above,

If you entered your test scores correctly, then your choices are off the wall.

The median is 87

The mode is 89

The mean is 85.833...

There is not a mode of 91 !

I hope this helps

Step-by-step explanation:

6 0

3 years ago

(CHALLENGE QUESTION FOR ALL OF YOU):

NARA [144]

Answer:

Euler's Identity

Euler's Identity is written simply as: e^(iπ) + 1 = 0, it comprises the five most important mathematical constants, and it is an equation that has been compared to a Shakespearean sonnet.

Step-by-step explanation:

5 0

3 years ago

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Solve for n. 24 = 3(n -5)

Masteriza [31]

Answer:

n=13

Step-by-step explanation:

24=3(n-5)

24=3n-15

-3n=-24-15

-3n=-39

n=13

8 0

3 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

4/7<br> of a number is 36<br> What is the number?

katrin [286]

$\text{If the number is x,}\\\\~~~~~\dfrac 47 \cdot x =36\\\\\implies 4x = 36 \times 7\\\\\implies x =\dfrac{36 \times 7}4\\ \\\implies x = 9 \times 7\\\\\implies x = 63\\\\\text{Hence, the number is 63.}$

7 0

2 years ago

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