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4 years ago

Can someone explain to me how to solve these types of problems ? * will mark bainlyest *

Mathematics

1 answer:

Anni [7]4 years ago

6 0

Answer:

2 i believe

Step-by-step explanation:

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on the number line above, w,x,y, and z are the coordinates of the indicated points. which of the following is NOT true?

zhenek [66]

Answer:

Not true

(x/w)^2 > z

Step-by-step explanation:

(x/w)^2 will be less than 1

z is greater than 1

so (x/w)^2 < z

4 0

3 years ago

Factor this expression completely mr + ns - nr - ms

Talja [164]

(m - n)(r - s) Hope this helps and heve a nice day!

8 0

3 years ago

Read 2 more answers

???whats the answers

Aleks04 [339]

11, 3, -3, -12
Hope this helps :)

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4 years ago

How do u write the problem

mr Goodwill [35]

I don’t know ‍♀️ I am very sorry

4 0

3 years ago

Consider the differential equation: xy′(x2+7)y=cos(x)+e3xy. Put the differential equation into the form: y′+p(x)y=g(x), determin

icang [17]

Answer:

Linear and non-homogeneous.

Step-by-step explanation:

We are given that

$\frac{xy'}{(x^2+7)y}=cosx+\frac{e^{3x}}{y}$

We have to convert into y'+P(x)y=g(x) and determine P(x) and g(x).

We have also find type of differential equation.

$y'=\frac{(x^2+7)y}{x}(cosx+\frac{e^{3x}}{y}}$

$y'=\frac{(x^2+7)cosx}{x}y+\frac{(x^2+7)e^{3x}}{x}$

$y'-\frac{cosx(x^2+7)}{x}y=\frac{e^{3x}(x^2+7)}{x}$

It is linear differential equation because this equation is of the form

y'+P(x)y=g(x)

Compare it with first order first degree linear differential equation

$y'+P(x)y=g(x)$

$P(x)=-\frac{cosx (x^2+7)}{x},g(x)=\frac{e^{3x}(x^2+7)}{x}$

$\frac{dy}{dx}=\frac{(x^2+7)(ycosx+e^{3x})}{x}$

Homogeneous equation

$\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}$

Degree of f and g are same.

$f(x,y)=(x^2+7)(ycosx+e^{3x}),g(x,y)=x$

Degree of f and g are not same .

Therefore, it is non- homogeneous .

Linear and non-homogeneous.

3 0

3 years ago