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Exercise 1

Solve the equations below algebraically.

$4 t + 50 = 200$

$4 t^{2} + 50 = 150$

$6 p - 20 = 12 + 4 p$

$x^{2} + 4 = 20$

${(x - 5)}^{2} = 4$

$4 (a - 2) - 20 = 0$

$\frac{12}{V} = 400$

$2 x^{2} - 2 = 6 x^{2} + 14$

Exercise 2

Solve the equations below your graphic calculator. Make sure you find all the solutions.

$\sqrt{x} = 6 - x$

$x^{4} = 2 + x$

Exercise 3

For being connected to the water supply and the possibility to use water you pay an amount of money per year. This is called a standing charge. Over and above you pay a fixed amount for every m³ of water you use. A waterworks company has set the charges for this as follows:

standing charge: € 38
price per `m^3` : € 1.75

The total costs per year $T K$ for water usage therefore depends on the total volume of water in `m^3` . This volume is denoted by `a` .

Explain why $T K = 38 + 1.75 a$ describes the total cost for this year.

The waterworks company estimates that a family uses between $140$ and $160$ m³ of water this year. Find out between which two values the total cost for this family will be.

The family wishes to reduce the costs for the use of water to below € 250. What is the maximum volume of water they can use?

Exercise 4

For the formulas below calculate each variable in turn, given that the other variable is $0$ .

$2 p - 3 q = 600$

$W = - 0.25 q (0.5 q - 100)$

$k^{2} + {(l + 2)}^{2} = 100$

$a = \frac{1200}{(600 + 0.2 d)}$

Exercise 5

Some candles are nearly completely cylindrical. Suppose you wish to make such a candle with a length of $20$ cm. You would start with a wick with a diameter of $3$ mm which you dip into a container with molten candle wax. With every dip the diameter of the candle grows by $1$ mm. The total volume of candle wax $V$ in the candle depends on the number of dips $a$ .

Give a formula for $V$ , depending on $a$ .

Plot the corresponding graph in your graphic calculator, showing all characteristics.

After how many dips is the volume of candle wax approximately $106$ `cm^3` ? Find the answer by means of the graph(s), then calculate this answer by solving the corresponding equation algebraically.

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