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2 years ago

Does anyone know how to solve this?

Mathematics

1 answer:

KATRIN_1 [288]2 years ago

4 0

Answer:

<30,-20>

Step-by-step explanation:

4u + 2v

4< 6,-4> + 2< 3,-2>

<24,-16> + < 6,-4>

<30,-20>

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Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie

melamori03 [73]

Answer:

The complete solution is

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

$y = e^{mx}$

Differentiating with respect to x

$y'= me^{mx}$

Again differentiating with respect to x

$y''= m ^2 e^{mx}$

Putting the value of y, y' and y'' in left side of the differential equation

$3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0$

$\Rightarrow 3m^2-8m-3=0$

The auxiliary equation is

$3m^2-8m-3=0$

$\Rightarrow 3m^2 -9m+m-3m=0$

$\Rightarrow 3m(m-3)+1(m-3)=0$

$\Rightarrow (3m+1)(m-3)=0$

$\Rightarrow m = 3, -\frac13$

The complementary function is

$y= Ae^{3x}+Be^{-\frac13 x}$

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)

For particular integral

$y_p=\frac 1{(3D+1)(D-3)} .4$

$=4.\frac 1{(3D+1)(D-3)} .e^{0.x}$ [since $e^{0.x}=1$ ]

$=4\frac{1}{(3.0+1)(0-3)}$ [ replace D by 0 , since L(0)≠0]

$=-\frac43$

The complete solution is

y= C.F+P.I

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

4 0

2 years ago

If G = {(-1, 7),(-8, 2),(0, 0),(6, 6)), then the range of G is

lakkis [162]

Answer:

B) {-8,-1,0,6)

Step-by-step explanation:

range is always y

4 0

2 years ago

Multiply both sides of the each equation in order to get the coefficient to be an integer.

monitta

$45= \frac{1}{2}a-3 //+3 \\ 48= \frac{1}{2} a// : \frac{1}{2} \\ 48: \frac{1}{2}=a=\ \textgreater \ 48*2=a=\ \textgreater \ a=96$
Hope that helps.Have a nice day!! ^^

5 0

2 years ago

Read 2 more answers

N is a rational number.m is an irrational number.n2 cannot be an irrational number? why or why not

Korolek [52]

If n is rational, it means that

$n\in\mathbb{Q}\Rightarrow n=\frac{p}{q},\quad p\in\mathbb{Z},\quad q\in\mathbb{Z}^*$

Therefore when we do n² we can write it as

$n^2=\frac{p^2}{q^2},\quad p\in\mathbb{Z},\quad q\in\mathbb{Z}^*$

Remember that the product of two integer numbers is also an integer, therefore we can guarantee that

$p^2\in\mathbb{Z}\text{ and }q^2\in\mathbb{Z}^*$

Then we can confirm that n² is the quotient of two integers and the denominator is not zero, therefore, n² is always rational, it cannot be an irrational number

8 0

8 months ago

T÷5/12= -3/10<br> Thank you if you help :)

skelet666 [1.2K]

Answer:

t = -1/8.

Step-by-step explanation:

t / 5/12 = -3/10 Invert the 5/12 and multiply:

12t / 5 = -3/10

Cross multiply:

12t * 10 = 5 * -3

120t = -15

t = -15/120 = -1/8.

7 0

2 years ago