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3 years ago

Please helpppp! im struggling so bad rn

Mathematics

1 answer:

Mrrafil [7]3 years ago

8 0

X=50 January school is a school

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I was doing this task in math and I came across this:

Galina-37 [17]

Answer:

mx+b+a= x-12x=0.88x

Step-by-step explanation:

4 0

3 years ago

How to factor out expressions

Brut [27]

To factor an expression:

1: find a common factor

2: divide a common factor by a term

3: divide any other terms after putting in parenthesis

4: simplify

Sorry if I just confused you. >///< Ty for reading.

6 0

3 years ago

Find the solution to the system of equations. 4x + 3y = –1 3x – 9y = 33

alexandr1967 [171]

Answer:

Step-by-step explanation:

4x + 3y = -1 --------------(i)

3x - 9y = 33 -------------(ii)

multiply equation (i) by 3

(i)* 3 12x + 9y = -3

(ii) <u> 3x - 9y = 33</u>

add, 15x = 30

x = 30/15

x = 2

Put x =2 in Equation (i)

4*2 + 3y = -1

8 +3y = -1

3y = -1-8

3y = -9

y = -9/3

y= -3

5 0

4 years ago

Help me on this math question

Blababa [14]

Answer:

$\huge \boxed{X=30}$

B. 30

Step-by-step explanation:

First, you multiply by 5 from both sides of equation.

$\displaystyle \frac{x}{5}=6\times5$

Simplify, to find the answer.

$\displaystyle 6\times5=30$

$\huge \boxed{X=30}$ , which is our answer.

6 0

3 years ago

Read 2 more answers

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

7 0

3 years ago