MaxWong

Note that when x = -2 or x = 2, the denominator will become 0.

Things inside any square root must be nonnegative and the denominator is nonzero.

\(x^2 - 4 \geq 0\text{ and } \sqrt{x^2 - 4} \neq 0 \\ x^2 - 4 \geq 0 \text{ and }x^2 - 4 \neq 0\\ x^2 - 4 > 0\)

Continue to solve the inequality, and you will get the domain of the function. 

\(\operatorname{Domain}(f) = \{x \in \mathbb R:-2 < x < 2\}\)

-3(x - 3) > 5 - 5x

3(x - 3) + 5 - 5x < 0

-2x - 4 < 0

-2x < 4

Now please finish it. Be careful that when you divide by a negative number on both sides of an inequality, you need to invert the inequality sign.

Suppose that \(y = r + x\), where \(0 \leq x < 1\) and r is an integer.

Then \(\lfloor y \rfloor = r\).

\(r(r +x) = 42\\ x = \dfrac{42}r - r\)

Then x has to satisfy the inequality \(0 \leq x < 1\).

We find the range of values of r such that \(0 \leq \dfrac{42}r - r < 1\). 

Since y > 0, r >= 0. We multiply each term by r to get \(0 \leq 42 - r^2 < r\).

Solving \(0 \leq 42 - r^2\) gives \(-\sqrt{42} \leq r \leq \sqrt{42}\).

Solving \(42 - r^2 < r\) gives \(r > 6\text{ or }r < -7\).

Combining the solutions of the two inequalities gives \(6 < r \leq \sqrt{42}\). But this inequality has no integer solutions.

Therefore, there are no such value of y such that y * floor(y) = 42.

Suppose x <= -7. When f(x) = 0, -x - 3 = 0. This means x = -3, which is not <= -7. This root is rejected.

Suppose x > -7. When f(x) = 0, 2x + 1 = 0. This means \(x = -\dfrac{\boxed{\phantom{\text{aaaaa}}}}{\boxed{\phantom{\text{aaaaa}}}}\).

Therefore, the sum of all values of x such that f(x) = 0 is \(-\dfrac{\boxed{\phantom{\text{aaaaa}}}}{\boxed{\phantom{\text{aaaaa}}}}\), as it is the only solution to f(x) = 0.

Please fill in the blanks.

Evaluate each of the terms separately. Note the formulas \(\sqrt{\dfrac ab} = \dfrac{\sqrt a}{\sqrt b}\) and \(\left(\dfrac ab\right)^2 = \dfrac{a^2}{b^2}\). Please try to take it from here.

Since b is positive, the common ratio \(\dfrac b{12}\) is also positive. Then all the numbers in the sequence are positive.

Using geometric mean, \(a = \sqrt{16 \cdot 12} = 8\sqrt3\).

We can find the common ratio by calculating the second term divided by the first term, so the common ratio is \(\dfrac{8\sqrt 3}{16} = \dfrac{\sqrt 3}2\).

Therefore, \(\dfrac b{12} = \dfrac{\sqrt 3}2\). Multiply both sides by 12 to get \(b = 6\sqrt 3\).

\(m + n \equiv 4 + 10 = 14 \equiv 0 \pmod {14} \)

The remainder is 0.

Dividing by 2 on both sides gives x^2 + y^2 + 2x + 6y + 3 = 0.

Complete the square to get (x + 1)^2 + (y + 3)^2 = 7. This is now in center-radius form. Can you proceed from here?

If you reflect (0, -5) across x = 5, you get another point on the graph. Can you proceed from here?