Sunday, August 31, 2014
Extraneous solutions to radical equations
Extraneous solutions - a solution of a simplified version of an equation that does not satisfy the original equation. Referred from http://www.mathwords.com/e/extraneous_solution.htm
Extraneous solutions to radical equations: Extraneous Solutions to Radical Equations
Friday, August 29, 2014
Dependent and independent variables
For p=5q
p is dependent on what happens to q so p is the dependent variable.
Dependent and independent variables exercise: the basics: Here we have a problem that asks us to identify which variables are dependent and independent. Hint: independent variables are not influenced and remain unchanged by the other variable.
p is dependent on what happens to q so p is the dependent variable.
Dependent and independent variables exercise: the basics: Here we have a problem that asks us to identify which variables are dependent and independent. Hint: independent variables are not influenced and remain unchanged by the other variable.
Wednesday, August 27, 2014
Surface Area
The problems for this topic may be quite hard to get through.
You may need to draw the object or the net of it.
Write down the dimensions one by one if you are having difficulty, calculating the area first, then the number of sides.
Read the question carefully!
Here is one of the odd objects you may encounter
-tetrahedron

You may need to draw the object or the net of it.
Write down the dimensions one by one if you are having difficulty, calculating the area first, then the number of sides.
Read the question carefully!
Here is one of the odd objects you may encounter
-tetrahedron

Triangle inequality theorem
Tuesday, August 26, 2014
Monday, August 25, 2014
Limits
From Khan Academy
Evaluatelimx→11−x√1−x using algebraic methods.
Evaluate
If we try substitution, we obtain the indeterminate form
.
.
The first solution requires rationalizing the numerator
The rest is not too difficult. Sometimes, have to factorise expressions like 25-x.
This becomes (5-x^0.5)(5+x^0.5). The above case can be done with this method.
1-x becomes (1-x^0.5)(1+x^0.5)
This becomes (5-x^0.5)(5+x^0.5). The above case can be done with this method.
1-x becomes (1-x^0.5)(1+x^0.5)
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