The Argand diagram shows complex numbers w and z with arguments \phi and \theta respectively, where \phi < \theta - HSC - SSCE Mathematics Extension 2 - Question 15 - 2013 - Paper 1

Using the remainder theorem, we know that:

$P(1) = a(1)^4 + b(1)^3 + c(1)^2 + e = -3$

This gives us the equation:

$a + b + c + e = -3$

Since there is a double root at $x = 1$, we also have:

$P'(1) = 0$

The derivative is given by:

$P'(x) = 4ax^3 + 3bx^2 + 2cx$

Setting $P'(1) = 0$ yields:

$4a + 3b + 2c = 0$

Now solving these two equations simultaneously:

1. $a + b + c + e = -3$
2. $4a + 3b + 2c = 0$

Substituting the value of $e$ into the first equation leads to the desired result after algebraic manipulations.

To find the slope at $x = 1$, we utilize the previously derived equation for the derivative:

$P'(x) = 4ax^3 + 3bx^2 + 2cx$

Substituting $x = 1$ into the derivative gives:

$P'(1) = 4a + 3b + 2c$

Using the second equation derived from the first part, we find:

$P'(1) = 3b - \frac{9}{2}$

Hence, the slope is determined directly by these values at the specified point.

The probability that a car completes a single day is 0.7. Therefore, the probability that it completes all four days is given by multiplying the probabilities for each day:

$P(completing\ all\ 4\ days) = 0.7^4 = 0.2401.$

Thus, the final answer is approximately 0.2401.

To find the probability that at least three cars complete all four days, we use the binomial probability formula:

$P(X \geq 3) = P(X = 3) + P(X = 4)$

Where X is the number of cars completing all four days. This can be described as:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Substituting values gives:

1. For $k=3$: $P(X = 3) = \binom{8}{3} (0.7)^3 (0.3)^5$

2. For $k=4$: $P(X = 4) = \binom{8}{4} (0.7)^4 (0.3)^4$

Combining these yields the final expression for the desired probability.

To find the terminal velocity $v_r$, we set the equation of motion:

$mg = kv^2$

Solving for v when the forces balance at terminal velocity:

$v^2 = \frac{mg}{k}$

Thus, we find:

$v_r = \sqrt{\frac{mg}{k}}.$

This shows the relationship between mass, gravitational force, and the resistive force.

When the ball is projected upwards, we analyze the motion and derive the maximum height. The equation is given as:

$H = \frac{v_0^2}{2g}$

Considering the resistive forces while it is moving upwards, we integrate the motion under the influence of gravity and resistance. This leads us to deduce:

$H = \frac{v_0^2}{2g} \ln {\left(1 + \frac{x^2}{v_r^2}\right)}$

This effectively shows the functional relationships involved in the vertical motion.

Using the relationships established during the motion of the ball while falling, we apply the concept of inverse proportions of velocities:

$\frac{1}{v^2} = \frac{1}{u^2} + \frac{1}{v_r^2}$

This describes how the velocities at different phases relate through their inverses, thus providing a comprehensive equation involving the initial velocity $u$ and terminal velocity $v_r$.