even i have bio p6 tomorrow. I hope its easy!
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- Thread starter notnek01
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same here!!
best of luck!!
best of luck!!
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I need help with 0580/42/M/J/13 Question 11, page 18, the weird triangle question, how do you do it?? It's really twisting :/
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Use the Pythogoras' theorem
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when will u post the solution of P22 / 2014?
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first solve the equation: just like any other equation.
move 3x to the right
make the denominator same on the left side.
cross multipication
and rearrange.
Just like this in the uploaded file. It's really messy I did it in Paint
but i hope you get the idea.
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Need help in oct/nov 2013 question 10 part ( v )
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http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w11_qp_43.pdf
Question 3c!
Please help meee!
Question 3c!
Please help meee!
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look , the main info is that O is the center. therefore, by using the property angle at center is twice at circumference, which in turn also means that angle at the circumference is half that of the angle at center.
now we got 25 degrees which is OAB, due to alternate angles are equal.
again angle at center twice, using at ACO its given 25.
so twice of 25 is 50
which AOC = 50 ,, why? BECAUSE center is twice at circumference..
now, use the add up 180 method 50 + 25 u get 105
now look carefully on the straight line AC so u minus 180 from 105 = 75
now oac is again half of 50 (center)
so we got AOC as 80
finishing it now 80 + 50 is 130
again use the same property opp of the 130 is our angle that we have to find
130/2 is 65
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A student is chosen at random. The probability that the student estimates that the mass is greater than M grams is 0.3. Find the value of M.?? HELP PLZZ
There are 200 students in total so the number of students who estimate more than M grams must be 0.3 x 200 = 60.
You can check this: 60/200 = 0.3
So 60 people estimate more than M.
Now work backwards : how do you use a cumulative freqency graph to find the number of people who estimate more than some number?
Have a go at this and post your ideas/working if you get stuck.
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please solve this nth term question
find the nth term of the following sequence:
5,11,21,35,53
find the nth term of the following sequence:
5,11,21,35,53
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2n^2 + 3
Was this an exam question? Was there a question part that came before it?
For these questions there's usually another part that will help you.
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No...it was a stand alone question
If this was stand-alone exam question then the syllabus must have changed - it would not have been a question a few years ago. I teach GCSE now so I can't say.
For this question, first look at the differences between terms:
6, 10, 14, 18, ...
And the difference of these is 4 so '4' is the second difference and so this must be a quadratic sequence.
Quadratic sequences have nth terms : an^2 + bn + c
The coefficient 'a' is always the second difference divided by 2. So here, a=4/2 = 2.
So the nth term is 2n^2 + bn + c
From here I would write out the sequence with nth term 2n^2 :
2, 8, 18, 32, 50, ...
And notice that each term is 3 away from the original sequence so the nth term of the original sequence must be:
2n^2 + 3
Your teacher should have taught you how to do these.