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At $v=8$ on the $v$ -axis go straight up until you reach the graph corresponding to $D=10$; then read off the answer on the $P$ -axis.

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$P=0.00013\cdot {25}^2\cdot {20}^3=650$ kW.

The heating costs when there are no sun-hours and the outside temperature is 20°C.

$k=800-60\cdot 3.5-50\cdot (-4)=790$, so € 790 per day.

When $60u+50t=800$, for example when there are $5$ sun-hours and there is an outside temperature of 30°C, or $10$ sun-hour with an outside temperature of 24°C.

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$K=340$

The costs are highest when $t=-2$ and $u=4$, i.e. € 660. The costs are lowest when $t=2$ and $u=10$, i.e. € 100. Therefore, the cost can vary between € 100 and € 660.

$d=0.5v$

The number $3.6$ arrives from the conversion of $v$ in km/h to m/s.

$t=\frac{(14.4+1.8v)}{v}$

$N=\frac{\left(60v\right)}{(14.4+1.8v)}$

The speed should be $70$ km/h.

$V=50$, so $a=65$, $b=19.25$ and $c=0.39$.
$L=1$, $S=2$ and $D=40$ gives $B=65\cdot 1+19.25\cdot 2+0.39\cdot 40=119.1$ mL.
$B_{(s\top s)}=19.25\cdot 2=38.5$ mL, so 32.2%. $B_{\left(wait\right)}=0.39\cdot 40=15.6$ mL, so 13,1%.

Car $1$ drives at $50$ km/h = $13.9$ m/s, so for $600$ m the first car takes $43.2$ s.
Car $2$ drives at $70$ km/h = $19.4$ m/s, so for $600$ m this car will take $30.9$ s.
Therefore the second car must wait for $12.3$ s.

Car 1: $V=50$, so $a=65$, $b=19.25$ and $c=0.39$. $L=0.9$, $S=0$ and $D=0$, so $B=58.5$ mL.
Car 2: $V=70$, so $a=91.6$, $b=37.73$ and $c=0.39$. $L=0.9$, $S=1$ and $D=12$, so $B=124.85$ mL.
Car $2$ needs more than twice the amount of petrol.