Solve the following math problems: 1. Rearrange the formula A = (2π+y)x² to make x the subject. 2. If n(ξ) = 20, n(P) = 10, n(Q) = 13 and n((P ∪ Q)') = 5, work out n(P ∩ Q). 3.... Solve the following math problems: 1. Rearrange the formula A = (2π+y)x² to make x the subject. 2. If n(ξ) = 20, n(P) = 10, n(Q) = 13 and n((P ∪ Q)') = 5, work out n(P ∩ Q). 3. Simplify (3+x) / (9-x²). Read more

Steps to Solve

1. Isolate the $x^2$ term

Divide both sides of the equation $A = (2\pi + y)x^2$ by $(2\pi + y)$ to isolate the $x^2$ term:

$$ \frac{A}{2\pi + y} = x^2 $$

2. Solve for x

Take the square root of both sides of the equation to solve for $x$:

$$ x = \sqrt{\frac{A}{2\pi + y}} $$

Since we are only asked to rearrange the formula, we only consider the positive square root.

3. Find $n(P \cup Q)$

We are given that the number of elements in the universal set $n(\xi) = 20$ and the number of elements outside the union of P and Q, $n((P \cup Q)') = 5$. Therefore, the number of elements inside the union of P and Q is:

$$n(P \cup Q) = n(\xi) - n((P \cup Q)') = 20 - 5 = 15$$

4. Use the inclusion-exclusion principle

We know that $n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)$. We are given $n(P) = 10$, $n(Q) = 13$, and we found that $n(P \cup Q) = 15$. Substitute these values into the equation and solve for $n(P \cap Q)$:

$$ 15 = 10 + 13 - n(P \cap Q) $$

$$ 15 = 23 - n(P \cap Q) $$

Add $n(P \cap Q)$ to both sides and subtract 15 from both sides:

$$ n(P \cap Q) = 23 - 15 = 8 $$

5. Factor the denominator

The given expression is $\frac{3+x}{9-x^2}$. The denominator $9 - x^2$ is a difference of squares, which can be factored as:

$$9 - x^2 = (3 - x)(3 + x)$$

6. Simplify the expression

Now we have:

$$ \frac{3+x}{(3-x)(3+x)} $$

Since $(3+x)$ is a common factor in the numerator and the denominator, we can cancel it out, provided that $x \ne -3$:

$$ \frac{3+x}{(3-x)(3+x)} = \frac{1}{3-x} $$