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2 years ago

13

What is the coefficient in this equation? 2x/7 + 2

2 answers:

hammer [34]2 years ago

8 0

As previously mentioned, the coefficient is indeed 2. A coefficient is any number with a variable

Svet_ta [14]2 years ago

6 0

The coefficient is 2

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Simplify: 4x + 6(3x - 2)<br> A) 10x - 12 <br> B) 18x - 12 <br> C) 22x - 12 <br> D) 30x2 - 20x

alekssr [168]

C have goood luck bye

4 0

3 years ago

Which of the following options correctly represents the complete factored form of the polynomial F(x)=x^4-3x^2-4?

Bumek [7]

Answer:

D

Step-by-step explanation:

Let's substitute a for x²:

x^4 - 3x² - 4

a² - 3a - 4

Now, this looks like something that is much more factorisable:

a² - 3a - 4 = (a - 4)(a + 1)

Plug x² back in for a:

(a - 4)(a + 1)

(x² - 4)(x² + 1)

The first one is a difference of squares, which can be factored into:

x² - 4 = (x + 2)(x - 2)

The second one can also be treated as a difference of squares:

x² + 1 = x² - (-1) = (x + √-1)(x - √-1) = (x + i)(x - i)

The answer is (x + 2)(x - 2)(x + i)(x - i), or D.

6 0

3 years ago

Read 2 more answers

What is the value of n? 3n + 16 = 7n Enter your answer in the box.

liraira [26]

Subtract 3n and then divide by 4 and you get n= 4

7 0

2 years ago

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Help help help help help

uranmaximum [27]

Answer:

This is what I got, the second one, the third one, and the fifth one.

Step-by-step explanation:

5 0

2 years ago

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Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a

Crank

I'll assume the ODE is actually

$y''+(x-2)y'+y=0$

Look for a series solution centered at $x=2$ , with

$y=\displaystyle\sum_{n\ge0}c_n(x-2)^n$

$\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n$

$\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n$

with $c_0=y(2)=2$ and $c_1=y'(2)=0$ .

Substituting the series into the ODE gives

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0$

$\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}$

If $n=2k$ for integers $k\ge0$ , then

$k=0\implies n=0\implies c_0=c_0$

$k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}$

$k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}$

$k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}$

and so on, with

$c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}$

If $n=2k+1$ , we have $c_{2k+1}=0$ for all $k\ge0$ because $c_1=0$ causes every odd-indexed coefficient to vanish.

So we have

$y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}$

Recall that

$e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}$

The solution we found can then be written as

$y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k$

$\implies\boxed{y(x)=2e^{-(x-2)^2/2}}$

6 0

3 years ago