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

We need to determine when $T = 80$:

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

Rearranging gives:

$e^{-kt} = \frac{190 - 80}{185} = \frac{110}{185} = \frac{22}{37}.$

Taking the natural logarithm on both sides:

$-kt = \ln(\frac{22}{37}),$

therefore:

$t = -\frac{1}{k}\ln(\frac{22}{37}).$

Considering that at $t = 1$, $T = 29$, we can determine $k$ using the equation:

$29 = 190 - 185e^{-k(1)}.$

So:

$e^{-k} = \frac{190 - 29}{185} = \frac{161}{185}.$

Taking the natural logarithm gives us $k$:

$k = -\ln(\frac{161}{185}).$

Substituting back into our equation for $t$ allows us to solve for the exact time when $T$ reaches 80°C.

If John has to go after Barbara with no one in between, we can consider them as a single unit (BA).
This unit and the six other people can be arranged in:

$7! = 5040$

ways.

Thus, the total number of arrangements is 5040.

The total arrangements for eight people are $8!$.
Since John must be after Barbara, we can split the arrangements in half:

$\text{Total ways} = \frac{8!}{2} = 20160.$

Calculate the gradient of line $OQ$ using the coordinates of points $O$ and $Q$. If $O(0,0)$ and $Q(2aq, aq^2)$ are used, the gradient can be expressed as:

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

Then, using the product of gradients property for perpendicular lines, we can verify that $pq = -2$.

Using the slopes of the tangents at points $P$ and $Q$, we can show that the product of their gradients is $-1$, which confirms that $\angle PLQ$ is a right angle.

Employ the midpoints of segments $PQ$ and the relationships of the coordinates, particularly $M(x_M, y_M)$, where $x_M = \frac{x_P + x_Q}{2}$, to establish that distances $MK$ and $ML$ are equal.