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3 years ago

5

Distance: (-2,7) and (2,-5)

1 answer:

scZoUnD [109]3 years ago

6 0

Does it say anything about slope?

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One of the factor of x² +3x+2 is x+1 then the other factor is …..

azamat

Hi there!

$\large\boxed{(x + 2)}$

x² + 3x + 2

We know that x + 1 is a factor, so:

We must find another number that adds up to 3 when added to 1 and multiplies into 2 with 1. We get:

x + 2

(x + 1)(x + 2)

6 0

3 years ago

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

Write (8,3)m=-7 in point slope and slope intercept form

Lerok [7]

Answer:

The slope intercept form is y=-7x+59

Step-by-step explanation:

y-y=m(x-x1)

y-3=-7(x-8)

y-3=-7x+56

+3 +3

y=-7x+59

8 0

3 years ago

Given <img src="https://tex.z-dn.net/?f=f%28x%29%20%3Dx%5E2%20-6x%2B8%20and%20g%28x%29%20%3Dx-2" id="TexFormula1" title="f(x) =x

Eva8 [605]

For this case we have the following functions:

$f (x) = x ^ 2-6x + 8\\g (x) = x-2$

We must subtract the functions:

$f (x) -g (x) = x ^ 2-6x + 8- (x-2)\\f (x) -g (x) = x ^ 2-6x + 8-x + 2\\f (x) -g (x) = x ^ 2-7x + 10$

We build a table of values for $x = 0,1,2,3.$

$x = 0, f (x) -g (x) = 0 ^ 2-7 (0) + 10 = 10\\x = 1, f (x) -g (x) = 1 ^ 2-7 (1) + 10 = 1-7 + 10 = 4\\x = 2, f (x) -g (x) = 2 ^ 2-7 (2) + 10 = 4-14 + 10 = 0\\x = 3, f (x) -g (x) = 3 ^ 2-7 (3) + 10 = 9-21 + 10 = -2$

Answer:

$f (x) -g (x) = x ^ 2-7x + 10$

8 0

3 years ago

Solve the following equation for the

Akimi4 [234]

Divide -3
-3/-3a = 15/-3
a = -5

5 0

3 years ago

Read 2 more answers