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Brilliant_brown [7]

3 years ago

9

The function y = f(x) is graphed below. Plot a line segment connecting the points

1 answer:

Papessa [141]3 years ago

6 0

Answer:

no not really

Step-by-step explanation:

the values dont match up

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3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago

Pleaseeee pleaseee help mee with precal...One root of 2x2 + 8x + c = 0 is r1 = 1 + √5. Find r2.

igomit [66]

If a root is a+√b then another is a-√b because of quadatic formula

r2=1-√5

6 0

3 years ago

Read 2 more answers

Marisol has 486 square inches of wrapping paper

AfilCa [17]

Answer:

9 inches

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Which of the following is true about the system 4/3x−2/5y=2 and 3x−2y=−1?

lilavasa [31]

Equation 1 is $\frac{4}{3}x -\frac{2}{5}y =2$ ,

Equation 2 is [/tex] 3x-2y = -1 [/tex]

for first equation LCD= 3 *5 = 15 , So we multiply whole equation by 15

$\frac{4}{3} *15x -\frac{2}{5}*15 y =2*15$

$20x - 6y = 30$

Now multiply second equation by -3 , to make the coefficient of y equal and opposite , so that we can apply the elimination method

$-9x+6y = 3$

Add both the equations

$20x -9x -6y +6y = 30+3$

$11x= 33$

Divide both sides by 11

$x=3$

Plug in any one of the equation we get

3(3) -2y = -1

9 - 2y = -1

subtract 9 from both sides

-2y = -10

divide both sides by -2

y=5

So the solution is x= 3 , y= 5

Which means

a. The system is consistent and independent. TRUE

7 0

4 years ago

The base of a triangular prism is a right triangle with a base of 6 inches, a height of 8 inches, and a third side length of 10

Vikentia [17]

The surface area of the rectangular prism with the dimensions that are stated is: 384 in.²

<h3>What is the Surface Area of a Triangular Prism?</h3>

Surface area = perimeter of base × height of prism + 2(base area)

= (s1 + s2 + s3)L + 2(1/2bh)

Given the following:

side of base (s1) = 6 in.
side of base (s2) = 8 in.
side of base (s3) = 10 in.
Length of prism (L) = 14 in.
Triangular base length (b) = 6 in.
h = 8 in.

Surface area = (6 + 8 + 10)14 + 2(1/2 × 6 × 8)

Surface area = 384 in.²

Learn more about the surface area of a rectangular prism on:

brainly.com/question/1310421

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5 0

2 years ago