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Simple trig questions

November 30, 2010 Leave a comment

Find the value of x

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Parallel lines

November 16, 2010 Leave a comment

Below is a typical diagram with parallel lines.

The parallel lines are marked with arrows and the line that crosses them is called the transversal.

Also notice that the angles are either obtuse or acute.  All the acute angles are equal and all the obtuse angles are equal.  Also any acute plus any obtuse angle equals 180^o.

Pairs of angles

Corresponding Angles

Are in the same position on each parallel line

Alternate Angles

Are on oppisite sides of the transversal touching each parallel line

Co-interior Angles

On the same side of the transversal between the parallel lines

Examples

Find the values of the unknowns

x=79^o \because \mbox{ Alternate angles are equal.}

x=37^o \because \mbox{ Corresponding angles are equal.}

x = 180^o-51^o-62^o=67^o \hspace{36 pt}\because 180^o\hspace{5 pt}\mbox{in a } \triangle

y=67^o \hspace{136 pt}\because \mbox{Corresponding angles are equal.}

z=180^o-67^o=113^o \hspace{62 pt}\because 180^o\hspace{5 pt}\mbox{on a line}

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Probability tree diagrams

November 14, 2010 Leave a comment

Let’s see how we can represent the situation of taking of taking balls out of a bag with a diagram.

The bag contains 4 red balls and 7 green balls.  We are going to take out two balls one at a time.  We will not be putting the balls back in the bag.

The diagram

This diagram shows what could happen at each stage of the trial, but so far we have not introduced any probabilities.  We know that for the first ball P(red)=\frac{4}{11}\mbox{ and }P(green)=\frac{7}{11}.  We place these values in the appropriate place.

After we have picked out a red ball we are left with a bag with 3 red balls and 7 green balls.

After we have picked out a green ball we are left with a bag with 4 red balls and 6 green balls.

We can calculate the appropriate probabilities and put them in the right place.

It is always useful to work out the values at the end.  The end of red, red means the probability of getting red AND then another red.  We learnt earlier that ‘AND’ means \times, so to work out the ends of the branches we multiply the probabilities.

Questions

P(red, red)

Here we just the end of that branch.  Therefore the answer is \frac{12}{110}

P(red and blue)

This could be P(red, blue) or P(blue, red).  ‘OR’ means add so we add the ends of those two branches.  Therefore the answer is \frac{56}{110}

P(not 2 greens)

Here we want 1-P(green,green)=\frac{68}{110}

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Questions: Solving basic equations

November 10, 2010 Leave a comment

Set A

  1. 2x+7=17
  2. 5x-4=11
  3. \frac{y}{3}+4=6
  4. \frac{t}{2}-7=-2
  5. 5x+2=3x+10
  6. 7x-3=4x+15
  7. 6(w+3)=2(2w+5)
  8. 4(3k-5)=3(2k-1)

Set B

  1. 4x+5=17
  2. \frac{f}{7}-5=9
  3. 5x+6=3x+10
  4. 7x-5=4x+7
  5. 4t+6=3t+20
  6. 3(x+4)=27
  7. 5(2x-6)=10
  8. 2(p+5)+3(2p-7)=5

Set C

  1. 8t+6=62
  2. \frac{x}{3}+5=7
  3. 5x-8=x+4
  4. 3y+6=8y-4
  5. 2y-6=5y-24
  6. 5(m+10)=55
  7. 3(3g-5)=21
  8. 3(x+4)-3(2x-5)=21

Set D

  1. \frac{g}{2}+5=11
  2. 7x-5=30
  3. 3x-9=10x-58
  4. 2w-8=4-2w
  5. 2d-5=7-d
  6. 2(x-6)=0
  7. 4(2x+1)=28
  8. 4(2d+4)=3(d+7)

Answers

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    Perpendicular Gradients

    October 27, 2010 Leave a comment

    In this small article we are going to look at the connection between two gradients which are perpendicular to each other.

    Remember that the gradient is defined as \frac{rise}{run} or \frac{\delta y}{\delta x}

    Consider the graph below

    Gradient change after 90 degree rotation

    The original gradient =\frac{\delta y}{\delta x}

    The rotated gradient can be seen to be =\frac{-\delta x}{\delta y}

    \therefore The gradient of the perpendicular = the negative reciprocal of the other line.

    Notice also that \frac{\delta y}{\delta x}\times \frac{-\delta x}{\delta y} = -1

    Some examples

    Line Perpendicular
    2 -\frac{1}{2}
    -3 \frac{1}{3}
    \frac{3}{5} -\frac{5}{3}
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    Welcome notice

    October 22, 2010 Leave a comment

    Welcome to my blog.  It is my intention to help young gain a fuller understanding of mathematics and to help them with their preparation for exams where ever they are in the world.  I would also like to produce an alternative on-line text book for mathematics.

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    Hello world!

    October 21, 2010 Leave a comment

    Welcome to WordPress.com. This is your first post. Edit or delete it and start blogging!

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