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July 15, 2025, 07:42:55 pm

Author Topic: onur369's Methods Question Thread :)  (Read 30839 times)  Share 

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onur369

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Re: onur369's Methods Question Thread :)
« Reply #15 on: February 06, 2011, 04:57:46 pm »
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f(-1)=0

(-1)^4 +a(-1)^3 - (-1)^2 + b(-1) -12=0
1-a-1-b-12=0
-a-b-12=0
a=-b-12

g(-1)=0

(-1)^4 (-b-12)(-1)^3 -9(-1) +b(-1)^2 +29(-1)+3- =0
1+b+12+9+b-29+30=0
14+2b=0
2b=-14
b=-7

sub in b into a

a= -(-7)-12
a= 7-12
a=-5.

Thats what I did.
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Re: onur369's Methods Question Thread :)
« Reply #16 on: February 06, 2011, 05:08:13 pm »
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f(-1)= 1 + (-a) -1+(-b)-12=0 thus -a-b=12 {1}
g(-1)= 1+-(a-9) + (b+9) -29=30=0 => -a+b=-20 {2}
f: a=-b-12 rearragne {1} sub into {2}
-(-b-12)+b=-20 => 2b+12=-20 => 2b=-32 => b=-16 sub into {1}
a=-(-16)-12=4
a=4 and b=-16
« Last Edit: February 06, 2011, 06:07:03 pm by swarley »

luken93

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Re: onur369's Methods Question Thread :)
« Reply #17 on: February 06, 2011, 05:15:44 pm »
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If the functions and both cross the x-axis at -1, determine the values for a and b.
















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onur369

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Re: onur369's Methods Question Thread :)
« Reply #18 on: February 06, 2011, 05:43:20 pm »
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Cheers.

Also swarley might want to double check you calculations, especially towards the end of your working out. Seems wrong to me. Since when does -b +b = -2b :/
-(b+12)+b=20 => -2b=32 => b=-16 sub into {1}
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Re: onur369's Methods Question Thread :)
« Reply #19 on: February 06, 2011, 06:07:33 pm »
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Cheers.

Also swarley might want to double check you calculations, especially towards the end of your working out. Seems wrong to me. Since when does -b +b = -2b :/
-(b+12)+b=20 => -2b=32 => b=-16 sub into {1}
haha yeah bad working there :x fixed.

Andiio

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Re: onur369's Methods Question Thread :)
« Reply #20 on: February 06, 2011, 07:17:26 pm »
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Ah it's cool.. It's good to help everyone out. :)

My only requirement is that you now get 50 :) (seriously, ask andiio)

LOL OMG m@tty you still remember that? :P
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onur369

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Re: onur369's Methods Question Thread :)
« Reply #21 on: February 06, 2011, 07:19:35 pm »
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Please explain :p
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Halil

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Re: onur369's Methods Question Thread :)
« Reply #22 on: February 06, 2011, 08:06:20 pm »
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Onur, study together, find solutions together! -_-.
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onur369

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Re: onur369's Methods Question Thread :)
« Reply #23 on: February 06, 2011, 09:43:02 pm »
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Of course bro. We need 94+
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MrIraq

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Re: onur369's Methods Question Thread :)
« Reply #24 on: February 06, 2011, 09:51:09 pm »
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Onur, study together, find solutions together! -_-.
Of course bro. We need 94+

nerds..
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onur369

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Re: onur369's Methods Question Thread :)
« Reply #25 on: February 06, 2011, 09:59:30 pm »
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Lol
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m@tty

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Re: onur369's Methods Question Thread :)
« Reply #26 on: February 06, 2011, 10:41:07 pm »
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Ah it's cool.. It's good to help everyone out. :)

My only requirement is that you now get 50 :) (seriously, ask andiio)

LOL OMG m@tty you still remember that? :P

You didn't think you were going to get out of it that easily did you.

Btw, your aims equate to a 99.90 :P haha (aggregate of 207.2)

But with the 50 for Spesh you break the 99.95 (209.2)
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Andiio

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Re: onur369's Methods Question Thread :)
« Reply #27 on: February 06, 2011, 10:50:03 pm »
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Ah it's cool.. It's good to help everyone out. :)

My only requirement is that you now get 50 :) (seriously, ask andiio)

LOL OMG m@tty you still remember that? :P

You didn't think you were going to get out of it that easily did you.

Btw, your aims equate to a 99.90 :P haha (aggregate of 207.2)

But with the 50 for Spesh you break the 99.95 (209.2)

You atar-calc'ed me? LOL

I only aspire to get into my course though, nothing else :)
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onur369

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Re: onur369's Methods Question Thread :)
« Reply #28 on: February 07, 2011, 08:54:22 pm »
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One of the factors in order to be successful is to be quick with answering your questions right?

Ok here is one ridiculously easy question, I know how to do, but I dont want to spend more than 3-4minutes doing.

Expand: (2y-3x)^5

Any quicker ways, rather than doing it the long way and wasting valuable test time.
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xZero

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Re: onur369's Methods Question Thread :)
« Reply #29 on: February 07, 2011, 09:01:26 pm »
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if this is exam 2, which i assume it is since i dont think you are expected to expand that by hand, use the calculator!
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