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3 years ago

SOMEONE HELP ME SOLVE THis.

Mathematics

2 answers:

zhenek [66]3 years ago

4 0

Answer:

shiiiii i dont no Nlgga

Step-by-step explanation:

Sergeeva-Olga [200]3 years ago

3 0

Uhhhh , i think the answer is 41.
blue monster = 20
yellow monster = 15
green monster = 6

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Ax-bx+y=z which of the following represents the formula that could be used to find x?

Alex73 [517]

Step by step explanation:

You first subtract z on both sides of the equal sign

ax-bx=(z-y)

Since a and b both have a "x" you can subtract them

(a-b)x=(z-y)

then you divide "x" on both sides of the equal sign

$\frac{a - b}{x} = \: \frac{z - y}{x}$

4 0

4 years ago

Name two integers whose product is negative.

oksano4ka [1.4K]

You know that one integer is a negative integer, as the product is negative- and if the sum is zero, then the absolute value of the two integers are equal. Thus, the integers are +/- sqrt(36). or +/- 6.

Example: 6 and -6

Your welcome.

3 0

3 years ago

Given f(x) = 1/x+4 and <br> g(x) = 8/x-1, find the given domain of f(g(x)).

Artemon [7]

Answer:

he domain of the composition is all real x values except for x = -1

In other words: $\left \{ x \, |\, x \neq -1} \right \}$

Step-by-step explanation:

Let's find the composition $f(g(x))$ in order to answer about its domain (where on the Real number set the function is defined), give the two functions:

$f(x)= \frac{1}{x+4}$ and $g(x)=\frac{8}{x-1}$ :

$f(g(x))=\frac{1}{g(x)+4} \\f(g(x))=\frac{1}{\frac{8}{x-1} +4} \\f(g(x))=\frac{1}{\frac{8+4(x-1)}{x-1} }\\f(g(x))=\frac{x-1}{8+4x-4} \\f(g(x))=\frac{x-1}{4+4x} \\$

This rational function is defined for every real number except when the denominator adopts the value zero. Such happens when:

$4+4x=0\\4x=-4\\x=-1$

So the domain of the composition is all real x values except for x = -1

8 0

3 years ago

Can someone please help

kirill115 [55]

answer:

sen this to yo teach

7 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