Is a relation always a function ? is a function always a relation ? explain

1 answer:

3 0

It just so happens that it's always the same y for each x, but it is only that one y. So this is a function; it's just an extremely boring function! ... So this is a relation, but it is not a function.

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Which statement about y= 3x^2 + 11x + 6 is true?

Eva8 [605]

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2 years ago

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

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