Ask question

Ask question

All categories

3 years ago

8

Add.

1 answer:

Phoenix [80]3 years ago

4 0

The answers is 11x+2 I’m pretty sure. I can’t tell if your x3 was an exponent or multiplying.

You might be interested in

What is the value of 10 to the 3rd power

Novosadov [1.4K]

10 x 10 x 10 = 1000 your welcome

3 0

3 years ago

Read 2 more answers

Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

$y'' - 3y' + 2y = e^x$

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

6 0

1 year ago

To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equati

murzikaleks [220]

Answer:

$x-4y+4=0$

$f(x)=\sqrt x$ and x=4

Step-by-step explanation:

We are given that a curve

$y=\sqrt x$

We have to find the equation of tangent at point (4,2) on the given curve.

Let y=f(x)

Differentiate w.r.t x

$f'(x)=\frac{dy}{dx}=\frac{1}{2\sqrt x}$

By using the formula $\frac{d(\sqrt x)}{dx}=\frac{1}{2\sqrt x}$

Substitute x=4

Slope of tangent

$m=f'(x)=\frac{1}{2\sqrt 4}=\frac{1}{2\times 2}=\frac{1}{4}$

In given question

$m=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$

$\frac{1}{4}=\lim_{x\rightarrow 4}\frac{f(x)-f(4)}{x-4}$

By comparing we get a=4

Point-slope form

$y-y_1=m(x-x_1)$

Using the formula

The equation of tangent at point (4,2)

$y-2=\frac{1}{4}(x-4)$

$4y-8=x-4$

$x-4y-4+8=0$

$x-4y+4=0$

8 0

3 years ago

I am simplifying radicals and I've come across one that has me a bit confused.

zhenek [66]

The sqrt of (34 m^4) is sqrt(34) times sqrt(m^4).

The sqrt of m^4 is easy ... that's just m^2 .

The other part is more messy. There's not much you can do with it.

sqrt(34) = sqrt(17) times sqrt(2).

That doesn't help make it any simpler.
You might as well just leave it as sqrt(34).

Then the final, simplified form of the original expression is

m^2 sqrt(34)

5 0

3 years ago

Value of x for 12(x-2)+3x=1/2(x+6)+2

Damm [24]

Answer:

The value of x = 2

Step-by-step explanation:

Given the equation

$12\left(x-2\right)+3x=\frac{1}{2}\left(x+6\right)+2$

Expand 12(x-2) = 12x - 24

Expand 1/2(x+6) = 1/2x + 3

so the equation becomes

$12x-24+3x=\frac{1}{2}x+3+2$

simplifying

$15x-24=\frac{1}{2}x+5$

$15x=\frac{1}{2}x+29$

multiplying both sides by 2

$30x = x + 58$

subtracting x from both sides

$30x - x = 58 - x$

$29x = 58$

divide both sides by 29

$\frac{29x}{29}=\frac{58}{29}$

simplifying

$x = 2$

Therefore, the value of x = 2

4 0

2 years ago