## Welcome to H3 Maths

Blog Support for Growing Mathematicians

## Archive for October, 2009

### Thanks Mr Juang

October30

Mr Juang – thank you so much for doing your Prac with our Year 9 Maths class. We really enjoyed your teaching and help for each student. God bless you in your teaching and we hope to have you contribute ono this blog site.

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### Interact!

October30

It was great to see some comments coming through. We had a test on Straight Lines today and a few struggled. Don’t give up! A test is designed to help you find out what needs more revision, and to confirm what you have learned in class. Of course, it helped that the test today was […]

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### Finding Gradient when given Two Points

October27

A cool trick is to find the gradient when we are just given two points. e.g. Point A (3,-5) and Point B (0,4) The gradient = rise/run or (in maths language) difference in y values divided by difference in x values Therefore, starting at the same point (A), rise/run = -5-4/3-0 = -9/3, or, gradient […]

### Tuesday’s Amazing Lesson

October27

Today you were challenged to find the equation of a straight line when you were given the following information – the gradient and a point the line goes through. For example, Find the equation of a straight line that has a gradient of -2 and that goes through the point (5,7) Solution: If the gradient […]

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### Straight Line Geometry

October26

We have covered most of the basic rules regarding straight lines. To check out your knowledge of using the straight line equation (y=mx+c) check out this website – in particular, Straight Line Graphs 1 and Straight Line Graphs 2. You can discover what happens when you change the “m” value, and “c” value to see how […]

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#### Post Support

The graph on the left (Coronavirus) is for a time period of 30 days, while the one on the right (SARS) is for 8 months! Very poor graphical comparison and hardly relevant, unless it is attempting to downplay the seriousness of the coronavirus?

10 x 9 x 8 + (7 + 6) x 5 x 4 x (3 + 2) x 1 = 2020

NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]