Prove that using: X ∩ (Y ∪ Z ) =(X∩Y) ∪ (Y ∩ Z )

Mathematics

1 answer:

sertanlavr [38]3 years ago

5 0

There should be a typo in the question: the correct formula would be

$X\cap(Y\cup Z)=(X\cap Y)\cup (X\cap Z)$

Let's compute $X\cap (Y\cup Z)$ first. We have

$(Y\cup Z)=\{1, 4, 5, 6, 7, 8, 9\}$

If we intersect this set with X, we have

$X\cap (Y\cup Z) = \{5, 8, 9\}$

So, that would be the left hand side. For the right hand side, we have

$X\cap Y = \{8\},\quad X\cap Z = \{5,9\}$

And their union is again $\{5,8,9\}$

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Hi i just need help checking if my answer is correct I have been working on a implicit diffrention problem and I need some help.

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Answer:

$\dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}$

Step-by-step explanation:

Normally, when I do this, I differentiate each term first with respect to x then with respect to y. In this solution, I differentiated the entire expression with respect to x, then with respect to y. That makes separating the dx and dy coefficients much easier, so the solution almost falls into your lap.

$-x^4 y + 2 x y^3 + x^3 y^3=0 \qquad\text{given}\\\\(-4x^3y+2y^3+3x^2y^3)dx+(-x^4+6xy^2+3x^3y^2)dy=0\\\\(-4x^3y+2y^3+3x^2y^3)dx=(x^4-6xy^2-3x^3y^2)dy\\\\\boxed{\dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}}$