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grandymaker [24]

2 years ago

14

Please help. If you could put your work in the answer that would Be greatly appreciated.

1 answer:

AleksandrR [38]2 years ago

6 0

Answer:

38°

because it is equal.

GJ=21

JH=21

GJ=JH

∠GFJ =∠HFJ

∠GFJ =38°

∠HFJ=38°

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Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

Can anyone give me this answere

nika2105 [10]

$\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=6\\ h= 11 \end{cases}\implies V=\cfrac{\pi (6)^2(11)}{3}\implies V=132\pi$

3 0

1 year ago

Through(-2,-3); parallel to 3x+2y=5

Vilka [71]

Step-by-step explanation:

Hey there!

<u>Firstly </u><u>find </u><u>slope </u><u>of</u><u> the</u><u> </u><u>given</u><u> equation</u><u>.</u>

Given eqaution is: 3x + 2y = 5.......(i)

Now;

$slope(m1) = \frac{ - coeff. \: of \: x}{coeff. \: of \: y}$

$or \: m1 = \frac{ - 3}{2}$

Therefore, slope (m1) = -3/2.

As per the condition of parallel lines,

Slope of the 1st eqaution (m1) = Slope of the 2nd eqaution (m2) = -3/2.

The point is; (-2,-3). From the above solution we know that the slope is (-3/2). So, the eqaution of a line which passes through the point (-2,-3) is;

(y-y1) = m2 (x-x1)

~ Keep all values.

$(y + 3) = \frac{ - 3}{2} (x + 2)$

~ Simplify it.

$2(y + 3) = - 3x - 6$

$2y + 6 = - 3x - 6$

$3x + 2y + 12 = 0$

Therefore, the eqaution of the line which passes through the point (-2,-3) and parallel to 3x + 2y= 5 is 3x + 2y +12 =0.

<em><u>Hope </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

8 0

2 years ago

Read 2 more answers

I need help please and thank you

maksim [4K]

I hope this helps you

7 0

2 years ago

Please help! Same answer choices for both of the drop down menus...

Artyom0805 [142]

the first one is left and the second one is up

5 0

3 years ago