Author Topic: math doubts [NEW]  (Read 16162 times)

nid404

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Re: functions
« Reply #45 on: September 27, 2009, 09:07:51 am »
whatever way u like

You can do that right??

Offline slvri

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Re: functions
« Reply #46 on: September 27, 2009, 09:10:14 am »
i hate A level...........

Offline astarmathsandphysics

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Re: functions
« Reply #47 on: September 27, 2009, 10:04:25 am »
yeah..then..how can it be less than -5 and greater than 3 at the same time??

x<-5 OR x>3


NOT x<-5 AND x>3

Offline slvri

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Re: functions
« Reply #48 on: September 27, 2009, 10:13:18 am »
x<-5 OR x>3


NOT x<-5 AND x>3
whoops.....meant to say that both of these are the possible solution sets
i hate A level...........

Offline Ghost Of Highbury

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Re: functions
« Reply #49 on: September 27, 2009, 02:03:52 pm »
ok...so the final answer is ..... x>5 and x<-3
divine intervention!

nid404

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Re: functions
« Reply #50 on: September 27, 2009, 02:07:45 pm »
ok...so the final answer is ..... x>5 and x<-3

plot it and check it if you r still not sure

Offline Ghost Of Highbury

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Re: functions
« Reply #51 on: September 27, 2009, 03:39:36 pm »
its correct...slvri PMed me abt it....
divine intervention!

Offline coolcar221

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Re: functions
« Reply #52 on: October 02, 2009, 07:48:18 pm »
Can anyone please help me to solve this problem?
find the set of values of k for which y=kx-4 intersects the curve y=x^2-2x at two distinct points

Offline astarmathsandphysics

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Offline astarmathsandphysics

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Re: functions
« Reply #54 on: October 02, 2009, 08:11:09 pm »
Can anyone please help me to solve this problem?
find the set of values of k for which y=kx-4 intersects the curve y=x^2-2x at two distinct points
kx-4=x^2-2x so x^2-x(k+2)+4=0
so b^2-4ac>0 with a=1, b=-(k+2),c=4
b^2-4ac=(-(k+2))^2-4*1*4>0
k^2+4k+4-16=k^2+4k-12=(k+6)(k-2)>0
k<-6 or k>2

Offline simba

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Re: functions
« Reply #55 on: October 13, 2009, 11:52:01 am »
guyz i need help........

Q: f(x): 3-2sinx , for 0<x<360   (its greater than or equal & less than or equal)

find the range of f

Offline slvri

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Re: functions
« Reply #56 on: October 13, 2009, 11:55:34 am »
guyz i need help........

Q: f(x): 3-2sinx , for 0<x<360   (its greater than or equal & less than or equal)

find the range of f

ok..........u know that the sine function is always between -1 and 1 (max 1 and min -1). so when sinx=1, f(x)3-2sinx=3-2(1)=3-2=1(min)
and when sinx=-1, f(x)=3-2(-1)=3+2=5(max)
so 1<=f(x)<=5
« Last Edit: October 13, 2009, 05:30:23 pm by slvri »
i hate A level...........

Offline simba

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Re: functions
« Reply #57 on: October 13, 2009, 12:16:44 pm »
Thanks

Offline simba

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Re: functions
« Reply #58 on: October 13, 2009, 05:27:17 pm »
i gt a question bt nt abt fuctions.............in june 06 .....num.3 ..how do i get the rate in decimal rather than the percentage ????

Offline slvri

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Re: functions
« Reply #59 on: October 13, 2009, 05:31:39 pm »
i gt a question bt nt abt fuctions.............in june 06 .....num.3 ..how do i get the rate in decimal rather than the percentage ????
um can u post the question here?
i hate A level...........