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3 years ago

BRIANLIEST!!!The triangles shown are similar. Which side of triangle JGH corresponds to side PM of triangle PMO?

Mathematics

2 answers:

hjlf3 years ago

6 0

Jg because JGH is double the size of PMO. so JG would be corresponding

steposvetlana [31]3 years ago

3 0

The answer would be JG.

If we rotate triangle PMO so that the 6 cm side is on the bottom, it would look more similar to the triangle JGH, and make it clear that side PM corresponds to side JG

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Determine whether each first-order differential equation is separable, linear, both, or neither. 1. ????y????x+????xy=x2y2 2. y+

Mkey [24]

Answer:

a) Linear

b) Linear

c) Linear

d) Neither

See explanation below.

Step-by-step explanation:

a) $\frac{dy}{dx} +e^x y = x^2 y^2$

For this case the differential equation have the following general form:

$y' +p(x) y = q(x) y^n$

Where $p(x) =e^x$ and $q(x) = x^2$ and since n>1 we can see that is a linear differential equation.

b) $y + sin x = x^3 y'$

We can rewrite the following equation on this way:

$y' -\frac{1}{x^3} y= \frac{sin (x)}{x^3}$

For this case the differential equation have the following general form:

$y' +p(x) y = q(x) y^n$

Where $p(x) =-\frac{1}{x^3}$ and $q(x) = \frac{sin(x)}{x^3}$ and since n=0 we can see that is a linear differential equation.

c) $ln x -x^2 y =xy'$

For this case we can write the differential equation on this way:

$y' +xy = \frac{ln(x)}{x}$

For this case the differential equation have the following general form:

$y' +p(x) y = q(x) y^n$

Where $p(x) =x$ and $q(x) = \frac{ln(x)}{x}$ and since n=0 we can see that is a linear differential equation.

d) $\frac{dy}{dx} + cos y = tan x$

For this case we can't express the differential equation in terms:

$y' +p(x) y = q(x) y^n$

So the is not linear, and since we can separate the variables in order to integrate is not separable. So then the answer for this one is neither.

4 0

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earnstyle [38]

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kifflom [539]

Answer:

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yanalaym [24]

Answer:

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Step-by-step explanation:

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