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3 years ago

Someone please help ;-;

Mathematics

1 answer:

egoroff_w [7]3 years ago

7 0

So for A Write all numerators above the least common denominator Add the numbers
7
-
12

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A measure of ________ for a numerical data set describes how its values vary with a single number.

cestrela7 [59]

14 because you divide

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3 years ago

What is the perimeter of a baseball diamond 90 ft on each sode

Maurinko [17]

The perimeter is 360 ft

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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

Help its for someone!!

yawa3891 [41]

Answer:

b 17/18

Step-by-step explanation:

Solve the one's in bracket then later solve everything together

5 0

3 years ago

If f(x) = 1 - x, whitch value is equivelent to f(i)

cricket20 [7]

Answer:

a). 0

Step-by-step explanation:

f(1) = 1 - 1

= 0

8 0

3 years ago

Read 2 more answers

Someone please help ;-; ​

Someone please help ;-;