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please post the link of the paper
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you can thank me for the formulaabcde said:AoA!sspigcse said:i have a doubt in the transformation matrices and my exam is on monday please help me with the topic.
Also ae there any formula for quadratic sequence.
For help on transformation matrices, please check my previous post here.
The formula to find the nth term of a quadratic sequence is: a + (n - 1)d1 + 1/2 (n - 1)(n - 2)d2 ,
where: a is the 1st term of the sequence
d1 is the difference b/w the first two terms of the sequence
d2 is the constant difference b/w any two terms of the derived sequence.
How come? Elucidate!sspigcse said:you can thank me for the formulaabcde said:AoA!sspigcse said:i have a doubt in the transformation matrices and my exam is on monday please help me with the topic.
Also ae there any formula for quadratic sequence.
For help on transformation matrices, please check my previous post here.
The formula to find the nth term of a quadratic sequence is: a + (n - 1)d1 + 1/2 (n - 1)(n - 2)d2 ,
where: a is the 1st term of the sequence
d1 is the difference b/w the first two terms of the sequence
d2 is the constant difference b/w any two terms of the derived sequence.
I missed that. Why ask if you knew the formula already?sspigcse said:i posted the same thing before ya @abcde
please post a particular question so I may be able to help u.princex said:i d0nt understand bearing plz help..!!!
princex said:i d0nt understand bearing plz help..!!!
This one is easy!Martee100 said:Could somebody help me in this?
http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s02_qp_2.pdf
Q14 part B especially! Thanks 8)
SalmanPakRocks said:please post a particular question so I may be able to help u.princex said:i d0nt understand bearing plz help..!!!
P.S Bearings is VERY easy.
SalmanPakRocks said:This one is easy!Martee100 said:Could somebody help me in this?
http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s02_qp_2.pdf
Q14 part B especially! Thanks 8)
ok so in part b we are given that y < -1
thus therefore we know that all the values of y are less then -1.
therefore to arrange it in increasing order
first number will be y^3 as its the smallest. just choose 1 number and keep on using that number throughout the pattern. If we have choosed y=-1 then y^3 = -3.
then for the next y^-1 = -1^-1 and that = -1
for the next y^0 = -1^0 = 1
so now the arrangement will be the following
y^3, y^-1, y^0, y^2.
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