Following are the important terminologies for Jr 1 Math coming final exam.
Page 183 - names of plane and solid figures
Page 192 and 193- Different kinds of angles
- acute angle
- right angle
- obtuse angle
- straight angle
- reflex angle
- full turn
- complimentary angles
- supplementary angles
- conjugate angles
Page 202 and 206 - Angles relationships
- vertically opposite angles
- alternate angles
- corresponding angles
- interior angles
Page 220 - Name of polygons
Page 227 and 228 - Types of triangles
- equilateral triangle
- isosceles triangle
- scalene triangle
- acute-angled triangle
- right-angled triangle
- obtuse-angled triangle
Page 231 and 232 - Types of quadrilaterals
- trapezium
- parallelogram
- rectangle
- rhombus
- square
- kite
Finally to sum up this list, special for chapter 15 our 2 superheroes :
- Ultraman
- Superman
Hope this helps.....
Saturday, October 2, 2010
For Kai Zheng
Let the number of cows = x.
If we have 160 more cows than sheep, then the number of sheep is x - 160.
After selling off 25% of the cows, the farmer is left with 75% of the cows. So the number of cows left after the fair will be 75% multiple with x.
After selling 2 fifth of the sheeps, the farmer is left with 3 fifth of the sheeps. So the number of sheeps left after the fair will be 3/5 multiple with (x - 160).
If after the fair the farmer still has 156 more cows then sheeps, then the following equation can be built.
x times 75% = (3/5 times (x - 160)) + 156
By changing 75% into 3/4 you will get x = 400.
Therefore the farmer has 400 cows in the beginning.
Hope this is clear.
If we have 160 more cows than sheep, then the number of sheep is x - 160.
After selling off 25% of the cows, the farmer is left with 75% of the cows. So the number of cows left after the fair will be 75% multiple with x.
After selling 2 fifth of the sheeps, the farmer is left with 3 fifth of the sheeps. So the number of sheeps left after the fair will be 3/5 multiple with (x - 160).
If after the fair the farmer still has 156 more cows then sheeps, then the following equation can be built.
x times 75% = (3/5 times (x - 160)) + 156
By changing 75% into 3/4 you will get x = 400.
Therefore the farmer has 400 cows in the beginning.
Hope this is clear.
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