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secondary 4 | A Maths
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need help with with this qn, pls explain too as im new to this topic
You can see that in getting from O to P, we move 2 units to the right and 7 units downwards.
We represent this as OP--> = (column vector with 2 on top and -7 below).
(I write that OP with an arrow on top as OP-->)
Similarly, OQ--> = (column vector with 8 on top and 3 below)
I let you try for PQ and OM
O to P, as mentioned before, is a movement 2 units to the right and 7 units downwards.
Q to R will be the same, with 2 units to the right and 7 units downwards.
R will eventually be (10, -4).
3/(x-4)^2 - 1/4-x
= 4/(x-4)^2 + 1/x-4 and not 4/(x-4)^2 + (-1/x-4)?
3 / (x - 4)^2 - 1 / (4 - x)
= 3 / (x - 4)^2 + 1 / (x - 4)
= (make denominator (x - 4)^2)
= (numerator becomes 3 + x - 4)
= (x - 1) / (x - 4)^2
One, in changing the connector of - into + for the two fractions.
Two, in introducing the - sign in front of the second fraction when there was no - sign previously.
Third, hidden as the conversion of 4 - x into x - 4.
As such, there are three changes to the negative signs, bringing about an imbalance to the equality of the expressions between your working steps.
by the way what does the "n" in n(AnB) represent tho, is it in the syllabus? as i have never seen it before
(Hence, the idea of flipping)
The n outside the brackets represents the number of elements in the set A n B.
So let’s say A n B = {2, 5, 10}.
Then n (A n B) = 3 because there are three elements 2, 5 and 10 inside the set.
It’s well within your syllabus.
Their functions are primarily the same, but it works for specific classes of gases
Concentrated sulfuric acid should work for most gases which are non-basic or non-alkaline. It will not be able to dry ammonia gas.
Calcium chloride works for many gases, but I know that ammonia is one of them which fails.
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