Ask question

Ask question

All categories

3 years ago

10

Please help as soon as possible

2 answers:

Rzqust [24]3 years ago

4 0

Answer:

The last one doesn't satisfy! for example if y=2 and x=1 then 2<0 false.

Step-by-step explanation:

Kobotan [32]3 years ago

3 0

The bottom right one.

Using

2 < 1 - 1

2 < 0

That's false, of course

You might be interested in

What is the rule of this question

NARA [144]

It adds 13
11 + 13 =24

6 0

4 years ago

Read 2 more answers

Find a closed-form solution to the integral equation y(x) = 3 + Z x e dt ty(t) , x > 0. In other words, express y(x) as a fun

MrMuchimi

Answer:

$y{x} = \sqrt{7+2Inx}$

Step-by-step explanation:

$y(x)= 3 + \int\limits^x_e {dx}/ \, ty(t) , x>0}$

Let say; By y(x)= y(e)

we have;

$y(e)= 3 + \int\limits^e_e {dt}/ \, ty= 3+0$

Using Fundamental Theorem of Calculus and differentiating by Lebiniz Rule:

$y^{1} (x) = 0 + 1/ xy$

$y^{1} = 1/xy$

dy/dx = 1/xy

$\int\limits {y} \, dxy = \int\limits \, dx/x$

$y^{2}/2 Inx + C$

RECALL: y(e) = 3

$(3)^{2} / 2 = In (e) + C$

$\frac{9}{2} =In(e)+C$

$\frac{9}{2} - 1 = C$

$\frac{7}{2} = C$

$y^{2} / 2 = In x +C$

$y^{2} / 2 = In x +7/2$

MULTIPLYING BOTH SIDE BY 2 , TO ELIMINATE THE DENOMINATOR, WE HAVE;

$y^{2} = {7+2Inx}$

$y{x} = \sqrt{7+2Inx}$

8 0

3 years ago

100.00=137.76/(1+r)^5<br> solve for r<br><br> thanks much

Feliz [49]

(1+r)^5=137.76/100
1+r=(137.76/100)^(1/5)
R=(137.76/100)^(1/5))-1

5 0

3 years ago

Melanie has a $60 so far to buy a lawn mower. This is 20% of the price of the lawn mower. What is the full price of the lawn mow

andreev551 [17]

--Percent means out of 100, so 20% = 20/100

The prob in other words is asking
-- $60 is 20% of what number?

'$60'   -- ---> 60
'is'    -- ---> =
20%    -- ---> 20/100
'of'    -- ---> * (multiply)
'what number' ---> x (variable)

4 0

3 years ago

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]

If

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

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