A turkey is taken from the refrigerator - HSC - SSCE Mathematics Extension 1 - Question 4 - 2008 - Paper 1

To find when the turkey reaches 80°C, we set:

$80 = 190 - 185e^{-kt}.$

This leads to:

$185e^{-kt} = 110 \\ e^{-kt} = \frac{110}{185} \\ -kt = \ln\left(\frac{110}{185}\right) \\ t = -\frac{1}{k}\ln\left(\frac{110}{185}\right).$

To find the time, we need the value of $k$. We use the information given for $T(1) = 29°C$:

Substituting into the equation:

$29 = 190 - 185e^{-k(1)} \\ e^{-k} = \frac{161}{185} \\ k = -\ln\left(\frac{161}{185}\right).$

Substituting $k$ back into the expression for $t$, we get:

$t = -\frac{1}{-\ln\left(\frac{161}{185}\right)}\ln\left(\frac{110}{185}\right).$

Calculating this will yield the time in hours after 9 am when it will be cooked.

We treat John and Barbara as a single entity (or block) when they pass through the doorway since John must go directly after Barbara without anyone in between. This block can have 6 other people arranged with them.

The total arrangements then will be:

$7!,$

where the "7" comes from the block of Barbara and John, plus the 6 other individuals.

Firstly, calculate the total number of arrangements for 8 people:

$8!$.

However, since John must go after Barbara, we will divide this by 2, as half of the arrangements will have Barbara before John. Hence, the total arrangements will be:

$\frac{8!}{2}.$

To find the gradient of the line segment OQ, the coordinates of O are (0,0).

The coordinates of Q are $(2aq, aq^2)$. Therefore, the gradient $m$ of the line OQ is calculated as:

$m = \frac{aq^2 - 0}{2aq - 0} = \frac{aq^2}{2aq} = \frac{q}{2}.$

Using the property of tangents, since point P lies on the parabola $x^2 = 4ay$, we find a relation that leads to:

$pq = -2$ using appropriate substitution.

To show that ∠PLQ is a right angle, we can examine the gradients of lines PO and QT. If the product of their gradients is -1, then the lines are perpendicular. The coordinates of points will assist in calculating the tangent slopes which ultimately show:

$\text{Gradient}_{PO} * \text{Gradient}_{QT} = -1.$

Let M be the midpoint of PQ, which gives $PM = MQ$.

Since this quadrilateral is symmetric, and we know the properties of midpoints and diagonals in triangles, we can conclude:

$MK = ML$ by using the bisector theorem, which guarantees equal lengths of the segments.