Help me pls don’t put random

Mathematics

2 answers:

Vladimir [108]2 years ago

4 0

I think it’s A
Sorry if I’m wrong

mina [271]2 years ago

3 0

The distance is 25 !!!!!

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If x= –4 and y= 3 , what is the value of x–3y2 ?

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Answer:

-31

Step-by-step explanation:

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The family of solutions to the differential equation y ′ = −4xy3 is y = 1 √ C + 4x2 . Find the solution that satisfies the initi

slega [8]

Answer:

The correct option is 4

Step-by-step explanation:

The solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}$

Now for the initial condition the value of C is calculated as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(-2)=\frac{1}{\sqrt{C+4(-2)^2}}\\4=\frac{1}{\sqrt{C+4(4)}}\\4=\frac{1}{\sqrt{C+16}}\\16=\frac{1}{C+16}\\C+16=\frac{1}{16}\\C=\frac{1}{16}-16$

So the solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}$

Simplifying the equation as

$y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1-256+64x^2}{16}}}\\y(x)=\frac{\sqrt{16}}{\sqrt{{1-256+64x^2}}}\\y(x)=\frac{4}{\sqrt{{1+64(x^2-4)}}}$