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maddison10

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ok to start
4c)

hope im right here goes....

(intergral of y^2 dx between 0- pi/3) times by pi

y^2= 4 sec^2 x
when u intergrate it u get

4tanx

between 0-pi/3
u get 4root3 x pi units ^3
 

maddison10

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5aiii)

now Sn = N/2 ( A + L) where A is 1st term and L is last term A= 30 L= 130 from part i)

50hrs = 3000minutes ( as all other parts are in minutes..)
so

3000=N/2 (30 + 160)

N= 37.5 if im not mistaken
so in the 38th lesson 50hrs will be completed...
 

maddison10

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ok ill do 7a) as it was relatively simple (i hope)

that thing means sum of ...

so 2^2 + 3^2 + 4^2

=4 +9 +16

=29
 

SpaceMonkey

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maddison10 said:
5aiii)

now Sn = N/2 ( A + L) where A is 1st term and L is last term A= 30 L= 130 from part i)

50hrs = 3000minutes ( as all other parts are in minutes..)
so

3000=N/2 (30 + 160)

N= 37.5 if im not mistaken
so in the 38th lesson 50hrs will be completed...
I tried that, then checked using:

Sn = n/2 [2a + (n-1)d]

It comes out to like 70+ hours...
 

smallcattle

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maddison10 said:
5aiii)

now Sn = N/2 ( A + L) where A is 1st term and L is last term A= 30 L= 130 from part i)

50hrs = 3000minutes ( as all other parts are in minutes..)
so

3000=N/2 (30 + 160)

N= 37.5 if im not mistaken
so in the 38th lesson 50hrs will be completed...
i think 37.5 is still in 37th lesson, you shouldnt rounded it off
 

Mystienu

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um... shouldn't it be


3000 = n / 2 ( 2 x 30 + 5 (n-1) )
6000 = 60n + 5n^2 - 5n
6000 = 55n + 5n^2



ultimately n = 30?
 

Sweet_Lemon

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oh~~ and it asked for which year....so i think i added 1991+38 or 45?? crap...think i added wrong....
 

mizz_smee

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Mystienu said:
um... shouldn't it be


3000 = n / 2 ( 2 x 30 + 5 (n-1) )
6000 = 60n + 5n^2 - 5n
6000 = 55n + 5n^2



ultimately n = 30?
i had what u say
but i couldn't get any answer for n
how did u get 30?
i spent like 30 mins all up trying to get n
but i never did and i still dunno y
 

Kazuya

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Mystienu said:
um... shouldn't it be


3000 = n / 2 ( 2 x 30 + 5 (n-1) )
6000 = 60n + 5n^2 - 5n
6000 = 55n + 5n^2



ultimately n = 30?
i think i got n = 29 point something, so i did S29 and S30, and got 48-ish hours and 51-ish hours respectively, so i took it as during the 29th lesson...dunno if that's right though..
 

Willmaker

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For q5aiii

Using the formula Sn=n/2[2a+(n-1)d]

3000=n/2[60+(n-1)5]

6000=n[60+5n-5]
6000=5n^2 + 55n

5n^2 + 55n - 6000 = 0

Therefore n^2 + 11n - 1200 = 0

Using quadratic formula n = {-11 + [root(121+ 4800)]}/2
= {-11 + [root(4921)]}/2
Using calculator n = 29.5749......
Therefore Clare's 30th lesson is when she will achieve 50hrs.
This can be checked by S30=15[60+(29)(5)]
=3075
This shows that she achieved 50hrs durign her 30th lesson.
 

Willmaker

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Mystienu said:
what did yous get for 8 a ii?
I got up to the point where i had 8sin@ - 1 = 0
I did something with difference of two cubes and I got one of the answers as @=1/2 or something like that.
 

acmilan

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SpaceMonkey said:
Yah 45 ... 1991 + 45 = 2036 AD (I wrote AD)
the answer for t = 45.4334... meaning that during the 46th year it reached 30 million meaning in the year 2037. If you sub in 45 and 46 into the equation you will see that the population for 45 is slightly under 30 million and hence it doesnt reach 30 million in 2036
 

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