The capacity of a bottle is measured as 1.35 litres correct to the nearest 10 millilitres.
What is the percentage error for this measurement, correct to two significant figures? (2 marks)
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The capacity of a bottle is measured as 1.35 litres correct to the nearest 10 millilitres.
What is the percentage error for this measurement, correct to two significant figures? (2 marks)
`text(0.37%)`
`text{1.35 litres = 1350 milliliters (mL)}`
`text(A) text(bsolute error) = 1/2 xx\ text{precision} = 1/2xx10=5\ text(mL)`
`:.\ text(% error)` | `= 5/1350 xx 100` |
`=0.3703… %` | |
`=0.37%\ text{(to 2 sig.fig.)}` |
A dinosaur fossil is measured to be 1.3 metres in length.
What is the percentage error in this measurement, giving your answer correct to two decimal places? (2 marks)
`3.85%`
`text{Absolute error} = 1/2 xx \text{precision} = 1/2 xx 0.1 = 0.05\ text{m}`
`% text(error)` | `= frac(0.05)(1.3) xx 100` |
`= 3.84615%` | |
`=3.85%\ text{(to 2 d.p.)}` |
A puppy's weight is measured at 5.2 kilograms, to the nearest 100 grams.
Calculate the percentage error in this measurement, correct to one significant figure? (3 marks)
`1%`
`text{Using 1 kilogram = 1000 grams}`
`=>\ text{5.2 kilograms = 5200 grams}`
`text{Absolute error}\ =1/2 xx text{precision}\ = 1/2 xx 100 = 50\ text{g}`
`text{% error}` | `=\ frac{text{absolute error}}{text{measurement}} xx 100%` | |
`=50/5200 xx 100%` | ||
`=0.961… %` | ||
`=1%\ text{(to 1 sig. fig.)}` |
The height of palm tree is measured at 8 metres, to the nearest metre.
What is the percentage error in this measurement? (2 marks)
`6.25%`
`text{Absolute error}\ =1/2 xx text{precision}\ = 1/2 xx 1 = 0.5\ text{m}`
`text{% error}` | `=\ frac{text{absolute error}}{text{measurement}} xx 100%` | |
`=0.5/8 xx 100%` | ||
`=6.25%` |
The width of a hockey field is measured to be 45 metres, correct to the nearest metre.
What is the upper limit for the width of the hockey field? (2 marks)
`45.5\ text(m)`
`text{Absolute error}\ =1/2 xx text{precision}\ = 1/2 xx 1 = 0.5\ text{m}`
`=>\ text{True length could lie 0.5 metres either side of 45 m measurement.}`
`:.\ text(Upper limit)` | `= 45+0.5` |
`= 45.5\ text(m)` |
The width of a soccer field is measured to be 50.60 metres, correct to the nearest centimetre.
What is the lower limit for the length of the netball court?
`C`
`text{1 cm = 0.01 metre}`
`text{Absolute error}\ =1/2 xx text{precision}\ = 1/2 xx 0.01 = 0.005\ text{m}`
`:.\ text(Lower limit)` | `= 50.60-0.005` |
`= 50.595\ text(m)` |
`=>C`
A cockroach is measured in a school science experiment and its length is recorded as 5.2 cm.
What is the upper limit of accuracy of this measurement? (2 marks)
`5.25\ text(cm)`
`text{Absolute error} = 1/2 xx \text{precision} = 1/2 xx 0.1 = 0.05\ text{cm}`
`=>\ text{True length could lie 0.05 cm either side of 5.2 cm measurement.}`
`text(Upper limit)` | `= 5.2 + 0.05` |
`= 5.25\ text(cm)` |
Find the centre and radius of the circle with the equation
`x^2+6x+y^2-y+3=0` (2 marks)
`text{Centre}\ (-3,1/2),\ text{Radius}\ = 5/2`
`x^2+6x+y^2-y+3` | `=0` | |
`x^2+6x+9+y^2-y+1/4-25/4` | `=0` | |
`(x+3)^2+(y-1/2)^2` | `=25/4` | |
`(x+3)^2+(y-1/2)^2` | `=(5/2)^2` |
`text{Centre}\ (-3,1/2),\ text{Radius}\ = 5/2`
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Find the centre and radius of the circle with the equation
`x^2+ y^2+8y= 0` (2 marks)
`text(Centre)\ (0,-4)`
`text(Radius = 4)`
`x^2+ y^2+8y` | `= 0` |
`x^2+ y^2+8y+16-16` | `= 0` |
`x^2+(y+4)^2` | `= 4^2` |
`:.\ text(Centre)\ (0,-4)`
`:.\ text(Radius = 4)`
Find the centre and radius of the circle with the equation
`x^2+10x + y^2-6y+33 = 0` (2 marks)
`text(Centre)\ (-5,3)`
`text(Radius = 1)`
`x^2+10x + y^2-6y+33` | `= 0` |
`x^2+10x + 25 + y^2-6y+9-1` | `= 0` |
`(x+5)^2 + (y-3)^2` | `= 1` |
`:.\ text(Centre)\ (-5,3)`
`:.\ text(Radius = 1)`
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a. `(x+3)^2 + (y+2)^2 = 3^2`
`:.\ text(Centre:)\ (-3,-2)\text(, Radius:)\ 3`
b.
a. | `x^2+6x+y^2+4y+4` | `=0` |
`x^2+6x+9+y^2+4y+4-9` | `=0` | |
`(x+3)^2+(y+2)^2` | `=3^2` |
`:.\ text(Centre:)\ (-3,-2)\text(, Radius:)\ 3`
b.
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a. `(x-1)^2 + (y+2)^2 = 4`
b.
a. `text{Circle with centre}\ (1,-2),\ r = 2:`
`(x-1)^2 + (y+2)^2 = 4`
b.
Write down the equation of the circle with centre `(0, -3)` and radius 4. (1 mark)
`x^2 + (y+3)^2 = 16`
`text{Circle with centre}\ (0, -3),\ r = 4:`
`x^2 + (y+3)^2 = 16`
Find the centre and radius of the circle with the equation
`x^2-12x + y^2 + 2y-12 = 0` (2 marks)
`text(Centre)\ (6, −1)`
`text(Radius = 7)`
`x^2-12x + y^2 + 2y-12` | `= 0` |
`(x-6)^2 + (y + 1)^2-36-1-12` | `= 0` |
`(x-6)^2 + (y + 1)^2` | `= 49` |
`:.\ text(Centre)\ (6, −1)`
`:.\ text(Radius = 7)`
A circle with centre `(a,-2)` and radius 5 units has equation
`x^2-6x + y^2 + 4y = b` where `a` and `b` are real constants.
The values of `a` and `b` are respectively
`B`
`x^2-6x + y^2 + 4y=b`
`text(Completing the squares:)`
`x^2-6x + 3^2-9 + y^2 + 4y + 2^2-4` | `= b` |
`(x-3)^2 + (y + 2)^2-13` | `= b` |
`(x-3)^2 + (y + 2)^2` | `= b + 13` |
`:. a=3`
`:. b+13=25\ \ =>\ \ b=12`
`=> B`
The diagram shows the curve with equation `y = x^2-7x + 10`. The curve intersects the `x`-axis at points `A and B`. The point `C` on the curve has the same `y`-coordinate as the `y`-intercept of the curve.
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i. `y` | `= x^2-7x + 10` |
`= (x-2) (x-5)` |
`:.x = 2 or 5`
`:.\ \ x text(-coordinate of)\ \ A = 2`
`x text(-coordinate of)\ \ B = 5`
ii. `y\ text(intercept occurs when)\ \ x = 0`
`=>y text(-intercept) = 10`
`C\ text(occurs at intercept:)`
`y` | `= x^2-7x + 10` | `\ \ \ \ \ text{… (1)}` |
`y` | `= 10` | `\ \ \ \ \ text{… (2)}` |
`(1) = (2)`
`x^2-7x + 10` | `= 10` |
`x^2-7x` | `= 10` |
`x (x-7)` | `= 10` |
`x = 0 or 7`
`:.\ C\ \ text(is)\ \ (7, 10)`
The parabola `y = −2x^2 + 8x` and the line `y = 2x` intersect at the origin and at the point `A`.
Find the `x`-coordinate of the point `A`. (2 marks)
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`x=3`
Sketch the graph of `y=4/(x-3)`. (3 marks)
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`text{Vertical asymptote at}\ \ x=3`
`text{As}\ \ x->oo, \ y->0`
`text{Horizontal asymptote at}\ \ y=0`
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1 & \ \ 0\ \ & \ \ 2\ \ & \ \ 4\ \ & \ \ 5\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -1 & -\frac{4}{3} & -4 & 4 & 2\\
\hline
\end{array}
Sketch the graph of `y=2/(3-x)`. (3 marks)
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`text{Vertical asymptote at}\ \ x=3`
`text{As}\ \ x->oo, \ y->0`
`text{Horizontal asymptote at}\ \ y=0`
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1 & \ \ 0\ \ & \ \ 2\ \ & \ \ 4\ \ & \ \ 5\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & \frac{1}{2} & \frac{2}{3} & 2 & -2 & -1\\
\hline
\end{array}
Sketch the graph of `y=3/(x+1)`. (2 marks)
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`text{Vertical asymptote at}\ \ x=-1`
`text{As}\ \ x->oo, \ y->0`
`text{Horizontal asymptote at}\ \ y=0`
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -3 & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -\frac{3}{2} & -3 & ∞ & 3 & \frac{3}{2} \\
\hline
\end{array}
Sketch the graph of `y=1/(x-2)`. (2 marks)
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`text{Vertical asymptote at}\ \ x=2`
`text{As}\ \ x->oo, \ y->0`
`text{Horizontal asymptote at}\ \ y=0`
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3\ \ & \ \ 4\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -\frac{1}{2} & -1 & ∞ & 1 & \frac{1}{2} \\
\hline
\end{array}
Sketch the graph of `f(x) = (2x+1)/(x-1)`. Label the axis intercepts with their coordinates and label any asymptotes with the appropriate equation. (4 marks)
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`(2x+1)/(x-1)` | `=(2x-2+3)/(x-1)` | |
`=(2(x-1)+3)/(x-1)` | ||
`=2 + 3/(x-1)` |
`text(Asymptotes:)\ \ x = 1,\ \ y = 2`
`text(As)\ \ x->oo,\ \ y->2(+)`
`text(As)\ \ x->-oo,\ \ y->2(-)`
`text(As)\ \ x->-1 (-),\ \ y->-oo`
`text(As)\ \ x->-1 (+),\ \ y->oo`
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i. `y=2-1/x`
`text{Vertical asymptote at}\ \ x=0`
`text{As}\ x->oo, \ 1/x -> 0\ \ => 2-1/x -> 2`
`text{Horizontal asymptote at}\ \ y=2`
ii.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & \frac{5}{2} & 3 & ∞ & 1 & \frac{3}{2} \\
\hline
\end{array}
Sketch the graph of `y=2/x+2`.
Clearly mark all asymptotes. (3 marks)
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\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & 1 & 0 & ∞ & 4 & 3 \\
\hline
\end{array}
Sketch the graph of `y=-2/x`. (2 marks)
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\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -\frac{3}{2} & -3 & ∞ & 3 & \frac{3}{2} \\
\hline
\end{array}
Factorise the parabola described by the equation `y=-x^2-x+12` and find its vertex. (3 marks)
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`y=(3-x)(x+4)`
`text{Vertex}\ = (-1/2,12 1/4)`
`y` | `=-x^2-x+12` | |
`=-(x^2+x-12)` | ||
`=-(x+4)(x-3)` | ||
`=(3-x)(x+4)` |
`text{Solutions at}\ \ x=3, -4`
`text{Line of symmetry at mid-point of solutions.}`
`x=(3+(-4))/2=-1/2`
`text{Substitute}\ \ x=-1/2\ \ text{into}\ \ y=-x^2-x+12`
`y=-1/4+1/2+12=12 1/4`
`:.\ text{Vertex}\ = (-1/2,12 1/4)`
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i. `y` | `=x^2-8x+15` | |
`=(x-5)(x-3)` |
ii. `text{Solutions at}\ \ x=3,5.`
`text{Line of symmetry at mid-point of solutions.}`
`x=(3+5)/2=4`
`text{Substitute}\ \ x=4\ \ text{into}\ \ y=x^2-8x+15`
`y=4^2-8xx4+15=-1`
`:.\ text{Vertex}\ = (4,-1)`
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i. `y` | `=2x^2+5x-3` | |
`=(2x-1)(x+3)` |
ii. `text{Solutions at}\ \ x=1/2,-3.`
`text{Line of symmetry at mid-point of solutions.}`
`x=(1/2+(-3))/2=-5/4`
`text{Substitute}\ \ x=-5/4\ \ text{into}\ \ y=2x^2+5x-3`
`y=2xx(-5/4)^2-5xx5/4-3=-49/8`
`:.\ text{Vertex}\ = (-5/4,-49/8)`
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i. `y` | `=6-x-x^2` | |
`=-(x^2+x-6)` | ||
`=-(x-2)(x+2)` | ||
`=(2-x)(x+3)` |
ii. `text{Solutions at}\ \ x=2,-3.`
`text{Line of symmetry at mid-point of solutions.}`
`x=(2+(-3))/2=-1/2`
`text{Substitute}\ \ x=-1/2\ \ text{into}\ \ y=6-x-x^2`
`y=6+1/2-1/4=6 1/4`
`:.\ text{Vertex}\ = (-1/2,6 1/4)`
By completing the square, find the coordinates of the vertex of the parabola with equation
`y=x^2-3x+1` (3 marks)
`(3/2,-5/4)`
`y` | `=x^2-3x+1` | |
`=x^2-3x+9/4-5/4` | ||
`=(x-3/2)^2-5/4` |
`:.\ text{Vertex}\ = (3/2,-5/4)`
By completing the square, find the coordinates of the vertex of the parabola with equation
`y=x^2+8x+9` (3 marks)
`(-4,-7)`
`y` | `=x^2+8x+9` | |
`=x^2+8x+16-7` | ||
`=(x+4)^2-7` |
`:.\ text{Vertex}\ = (-4,-7)`
By completing the square, find the coordinates of the vertex of the parabola with equation
`y=x^2-6x-4` (3 marks)
`(3,-13)`
`y` | `=x^2-6x-4` | |
`=x^2-6x+9-13` | ||
`=(x-3)^2-13` |
`:.\ text{Vertex}\ = (3,-13)`
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & 2 & & 0 \\
\hline
\end{array}
i.
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & \frac{3}{2} & 2 & \frac{3}{2} & 0 \\
\hline
\end{array}
ii.
iii. `text{The parabola is concave down for all values of}\ x.`
`=>\ text{There are no values of}\ x\ text{where the graph is concave up.}`
i.
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & \frac{3}{2} & 2 & \frac{3}{2} & 0 \\
\hline
\end{array}
ii.
iii. `text{The parabola is concave down for all values of}\ x.`
`=>\ text{There are no values of}\ x\ text{where the graph is concave up.}`
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & 6 & & & & -10 \\
\hline
\end{array}
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i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -10 & 0 & 6 & 8 & 6 & 0 & -10 \\
\hline
\end{array}
ii.
iii. `text{The parabola is concave down for all values of}\ x.`
i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -10 & 0 & 6 & 8 & 6 & 0 & -10 \\
\hline
\end{array}
ii.
iii. `text{The parabola is concave down for all values of}\ x.`
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & & 3 & \\
\hline
\end{array}
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By completing the table of values, sketch the graph of `y=2x^2-3`. (3 marks)
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & & -1 & \\
\hline
\end{array}
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By completing the table of values, sketch the graph of `y=x^2+3`. (3 marks)
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & & 4 & \\
\hline
\end{array}
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By completing the table of values, sketch the graph of `y=x^2-2`. (3 marks)
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & -1 & & & \\
\hline
\end{array}
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Sketch the graph of `y=3^x+2`, clearly labelling any asymptotes. (3 marks)
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`y=3^x+2`
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 2 \frac{1}{9} & 2 \frac{1}{3} & 3 & 5 & 11\\
\hline
\end{array}
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 5 & 2 & \frac{1}{2} & -\frac{1}{4} & -\frac{5}{8}\\
\hline
\end{array}
By completing the table of values, sketch the graph of `y=2^(-x)` (3 marks)
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ x\ \ \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} & & & 1 & & \\
\hline
\end{array}
--- 0 WORK AREA LINES (style=lined) ---
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 4 & 2 & 1 & \frac{1}{2} & \frac{1}{4}\\
\hline
\end{array}
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & -\frac{1}{2} & & & 3\\
\hline
\end{array}
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In 2004 there were 13.5 million registered motor vehicles in Australia. The number of registered motor vehicles is increasing at a rate of 2.3% per year.
Find an expression that represents the number (in millions) of registered motor vehicles `(V)`, if `y` represents the number of years after 2004? (2 marks)
`V = 13.5 xx (1.023)^y`
`text(In 2004, 13.5 million)`
`Vtext{(1 year later)} = 13.5 xx (1.023)`
`Vtext{(2 years later)}= 13.5 xx (1.023) xx (1.023)= 13.5 xx (1.023)^2`
`:. V(y\ text{years later}) = 13.5 xx (1.023)^y`
Points `P` and `Q`, shown on the Cartesian plane diagram, are rotated 180° about the origin and become points `P^(′)` and `Q^(′)`.
Plot the points `P^(′)` and `Q^(′)` on the diagram. (3 marks)
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`P^(′)(2,1)`
`Q^(′)(-4,2)`
Point `Q(3,1)` on the Cartesian plane is rotated 180° about the origin in a clockwise direction to become point `Q^(′)`.
What are the coordinates of `Q^(′)`. (2 marks)
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`Q^(′)(-3,-1)`
Gabby put 5 points on a grid and labelled them `A` to `E`, as shown on the diagram below.
Point `A` is 35 millimetres from point `D.`
Gabby adds a sixth point, `F` so that the arrangement of points has one line of symmetry.
How far is point `F` from point `B?` (3 marks)
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`42\ text{mm}`
The trapezium `ABCD` is moved to the new position shown by trapezium `SRQP.`
Which of these transformations resulted in the new position?
`C`
`text(Reflection in the)\ xtext(-axis:)`
`ABCD -> A^{prime}B^{prime}C^{prime}D^{prime}`
`text(Translate 8 units left:)`
`A^{prime}B^{prime}C^{prime}D^{prime} -> SRQP`
`=>C`
Rochelle drew a pattern which is pictured below.
Rochelle rotates the pattern.
How much does Rochelle to turn the pattern until it looks exactly the same?
`B`
`text(Outer pattern looks the same every)\ \ 1/8\ \ text(turn).`
`text(Inner cross pattern looks the same every)\ \ 1/4\ \ text(turn).`
`:.\ text(Whole pattern looks the same every)\ \ 1/4\ \ text(turn).`
`=>B`