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katrin2010 [14]

3 years ago

9

ABCD is a trapezium in which AB parallel DC ,DC=30cm and AB=50cm .If X and Y are respectively the mid point of AD and BC then pr

ove that area(DCYX)=7/9 area(XYBA)

1 answer:

Ratling [72]3 years ago

3 0

I have attached the answer. Hope it helps.

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A line goes through the point (2, 4) and has a slope of 3/2. Which of the following points is also on the line? CHOOSE ALL THAT

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Answer:

(4,7) and (0,1)

Step-by-step explanation:

(7-4)/(4-2) = 3/ 2, and (4-1)/(2-0) = 3/2.

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The image of a parabolic mirror is sketched on a graph. The image can be represented using the function y = x2 + 2, where x repr

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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