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$\noindent We square both sides of $PA = PB$ because both $PA$ and $PB$ are square roots as found through the distance formula. By squaring, we immediately undo the root.

$\noindent The number of digits in $N \in \mathbb{N}$ can be found using $\log_{10}{N}$ (or, rather ($\log_{10}{N} + 1))$. The number of digits in $2^{2005}$ is $\log_{10}{2^{2005}} = 2005\log_{10}{2} = 603.565 \ldots$ meaning the answer is (C)

$\noindent The train travels 32.5 kilometres in 0.25 hours. This is 130 km/h.

One method is to apply the distance formula and discover two equal sides.

$\noindent $f(-1) = 3(-1) = -3$ and so $g(f(-1)) = g(-3) = (-3) + 2 = -1$.

$\noindent The coordinates of the given points must satisfy $y = ax^3 + bx^2 + cx + d$. \begin{align*}-a + b - c + d &= -6 \quad (1)\\ d &= -6 \quad (2) \\ a + b + c +d &= -2 \quad (3) \\ 8a + 4b + 2c + d &=2 \hspace{3mm} \quad (4) \end{align*} \\ Use $(2)$ to reduce (1), (3) and (4) to a set of...

$\noindent Yes, $-8$ is a solution because the equation can be written as $x^2 = 64$. If $b \leq 0$ or $b=1$ there would be few values of $x$ that satisfies $b^y = x$ (ie. $\log_bx = y)$ and so logarithms with base $b\leq0$ or $b=1$ are not as useful as those with bases $b>0$ $(b\neq1).

$\noindent Integer powers are exponents that belong to the set $\mathbb{Z} = \{\ldots -3, -2 , -1, 0, 1, 2, 3 \ldots \}$. But polynomials only have \textbf{non-negative} integer exponents, $\{0, 1, 2, 3, \ldots \}

$\noindent $\mathbb{R}^+ \cup \{0\}$ denotes all non-negative real numbers.

$\noindent $\tan{30} = \frac{h}{18}$ so $h = \frac{18}{\sqrt{3}} = 6\sqrt{3}$. Since the person is 2 metres tall then the height of the building is $(2 + 6\sqrt{3})$ metres.

$\noindent The endpoint is $(-1, -4)$ and the domain is $x \geq -1$ and the range is $y \geq -4$

$\noindent I believe a rational root is a number $x \in \mathbb{Q}$ that satisfies $P(x) = 0$

$\noindent Well $x^2 = y^{\frac{2}{3}}$ and so a rotation about the $y$-axis means, \\ \begin{align*} \quad V &= \pi \int_1^8 y^{\frac{2}{3}}\mathrm{d}y - \pi (1)^2(7) \\ &= \pi \left[\frac{3}{5}y^{\frac{5}{3}}\right]^8_1 - 7\pi \\ &= \pi \left(\frac{3}{5}(8)^{\frac{5}{3}} -...

$\noindent \textbf{Q10.} When the sector of a circle is curved around to form a cone, the sector's arc length becomes the cone's base circumference. The circumference, $c$ is $c = r\theta = \frac{5\pi}{3}$. We know that $r = \frac{c}{2\pi}$, and so the radius of the circular base $r =...

That isn't an equation.

What is the 83/125?

$\noindent The gradient of a line of the general form $ax + by + c = 0$ is $m = -\frac{a}{b}$. If two straight lines are perpendicular then the product of their gradients is $-1$. Thus, $\frac{-8}{q} \times \frac{3}{4} = -1$. That is, $\frac{8}{q} = \frac{4}{3} \implies \frac{q}{8} =...

$\noindent $v = \frac{1}{15x^2} \Rightarrow \frac{dx}{dt} = \frac{1}{15x^2} \Rightarrow \frac{dt}{dx} = 15x^2.$ Integrating, $t = 5x^3 + C$ but for $t = 0, x = 0$ as the particle is initially at the origin (ie. zero displacement) and thus $C = 0$. For $t = 4, x^3 = \frac{4}{5}$ and so $x =...

Re: maths questions help $\noindent \textbf{Q3.} $O(0,0)$ and $P(3,4)$ and so $m_{OP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4}{3}$ $\noindent Since the point $P$ lies on the circle, $OP$ is a radius and the tangent to the circle at $P$ is perpendicular to $OP$. So $m_{OP} \cdot m_t = -1...

$\noindent \textbf{Q1.} $a = n+ 1$ is correct. \\ \textbf{Q2.} $a = 8\sqrt{x^{11}}$ is also correct.