Ifarrange increasing in order.

Ifthenincreases with increasingso increasing ofmeans increasing

Note thatand(1)

Sincewe have

Ifarrange in increasing order.

Ifthendecreases with bigger values of so a larger value ofimplies a smaller value of

Using (1) and thatand

Sincewe have

]]>{jatex options:inline}\frac{1576}{24}=65.67{/jatex} terrorist crimes per hour.

{jatex options:inline}\frac{65.67}{60}=1.1{/jatex} terrorist crimes per minute.

Ohew! There is much more terrorism in Donald Trumps imagination in the UK that implied by the four incidents involving deaths in 2017 and a score of arrests. ]]>

The line {jatex options:inline}l_2{/jatex} goes along 3-6=-3 and up 3+5=8. The gradient is {jatex options:inline}\frac{8}{-3}{/jatex}.

The angle {jatex options:inline}\alpha{/jatex} is {jatex options:inline}tan^{-1}( \frac{8}{-})- tan^{-1}(\frac{3}{10})=93.9^o{/jatex} to 1 decimal place. ]]>

Among other things, he worked on irrational numbers, the concept of infinity, an estimate of the number of grains of sand required to fill the Universe, the lever principle (supposedly he said 'Give me a place to stand on and I will move the Earth@).

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To find the area of a sector, we just find that fraction of the area of a circle.

The sector to the right is a fractionof the circle to the left so the the area of the sector is

The curved part of the sector isof the circumference of the circle but to find the perimeter of the sector we must add(the radius of the circle is) so the perimeter of the sector is

Ifandthe area of the sector isand the perimeter is

]]>If {jatex options:inline}Ab=x{/jatex} then {jatex options:inline}BC=\frac{x}{2}{/jatex} and the area of triangle ABD is {jatex options:inline}\frac{1}{2} \times x \times \frac{x}{2}=\frac{x^2}{4}{/jatex}

The areas of triangles ABC and BDC are in the ratio {jatex options:inline}(2)^2= 4{/jatex} so Area ABC is {jatex options:inline}\frac{4}{1+4} \times \frac{x^2}{4}=\frac{x^2}{5}{/jatex}.

Then the area of the green square is {jatex options:inline}x^2 - 4 \times \frac{x^2}{5}=\frac{x^2}{5}{/jatex}.

The area of the Green square is one fifth the area of the big square.]]>

{jatex options:inline}BC= \sqrt{(3-1)^2+(4-1)^2}= \sqrt{13}{/jatex}.

To find the other diagonal we use that triangles ABC and BQO are similar triangles, so {jatex options:inline}\frac{BQ}{BO}=\frac{AB}{AO}{/jatex}.

{jatex options:inline}AB=\sqrt{AO^2-BO^2}= \sqrt{13-2^2}=3{/jatex}, then {jatex options:inline}\frac{BQ}{BO}=\frac{AB}{AO} \rightarrow BQ= \frac{AB}{AO} BO=\frac{3}{\sqrt{13}} \times 2 = \frac{6}{\sqrt{13}}{/jatex}.

Then {jatex options:inline}BC=\frac{12}{\sqrt{13}}{/jatex}.

The area of the kite is {jatex options:inline}\frac{1}{2} \times \sqrt{13} \times \frac{12}{\sqrt{13}}=6{/jatex}.]]>

Cosine Rule:

Sine Rule:

For the above diagram, find

a)The distance BC

b)The bearing of A from B and the bearing of B from C.

a)Label the triangle as above, with sides labelled by littleletters opposite angles labelled by big letters.

so

b)

We draw a right angled triangle between A and B. Construct a rightangled triangle at A and a horizontal line starting from A to theright. The internal angles of the right angled triangle are 70 and 20degrees. Hence the bearing of A from B is 360-20=340 degrees.

To find the bearing of B from C, use the Sine Rule to find C.

Construct the right angled triangle as shown, with the angles at Cthen the bearing is 56+32.75=88.75 degrees.

]]>Numerator and denominator are both positive, and as {jatex options:inline}x \rightarrow \infty 0{/jatex}, the denominator tends to infinity also, so the fraction tends to zero. Hence {jatex options:inline}0 \lt \frac{2}{x^2+4x+9}{/jatex}.

Completing the square for the denominator gives {jatex options:inline}\frac{2}{x^2+4x+9}=\frac{2}{(x+2)^2+5}{/jatex}.

To maximise this fraction we must minimise the denominator, which is the sum of non negative terms. The denominator is minimised when {jatex options:inline}(x+2)^2=0 \rightarrow x=-2{/jatex}.

For this value of {jatex options:inline}x{/jatex} the fraction is equal to {jatex options:inline}0 \lt \frac{2}{(-2+2)^2+5}=\frac{2}{0+5}=\frac{2}{5}{/jatex}.

Hence {jatex options:inline}0 \lt \frac{2}{x^2+4x+9} \lt \frac{2}{5}{/jatex}.

The graph of the function is shown below.

]]>

When drawing the bars of histograms there are three simple relationships:

These equations may be used as in the below example:

Notice that the intervals or bin sizes are 9.5 -15.5, 15.5 – 18.5 and 18.5 - {} respectively since else we would have gaps: all the x's are to the nearest whole.

a)

b)

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