## Search found 8 matches

Tue Jul 19, 2011 9:59 pm
Topic: Help with solving equations: (x+3)(x-3)>0, x^2-2x-15</= 0,
Replies: 3
Views: 2122

### Re: Help with solving equations: (x+3)(x-3)>0, x^2-2x-15</=

Thank you it all helped. I am stuck however on the trig equation. I am not sure what to do after I get 2sin(x)cos(x)=sin(x).
Tue Jul 19, 2011 2:53 am
Forum: Calculus
Topic: help with evaluation
Replies: 2
Views: 2007

### Re: help with evaluation

thankyou
Mon Jul 18, 2011 7:31 am
Forum: Calculus
Topic: help with evaluation
Replies: 2
Views: 2007

### help with evaluation

Mon Jul 18, 2011 3:04 am
Topic: Help with solving equations: (x+3)(x-3)>0, x^2-2x-15</= 0,
Replies: 3
Views: 2122

### Help with solving equations: (x+3)(x-3)>0, x^2-2x-15</= 0,

less than or equal to = </=

(x+3)(x-3)>0

x^2-2x-15</= 0

sin2x=sinx, 0</=x</=2pi

logx+log(x-3)=1
Mon Jul 18, 2011 1:55 am
Forum: Intermediate Algebra
Topic: help with operations with functions
Replies: 6
Views: 6137

### Re: help with operations with functions

so would it be 1/5,1/4,1/7. and by the way thankyou
Sun Jul 17, 2011 11:57 pm
Forum: Intermediate Algebra
Topic: help with operations with functions
Replies: 6
Views: 6137

### Re: help with operations with functions

ok so that helped alot but i am having trouble understanding how to do the function operation with 1/f(x) due to the fact that f(x) has a list of points, f(x)={(3,5)(2,4)(1,7)}.
Sun Jul 17, 2011 1:34 am
Forum: Intermediate Algebra
Topic: help with operations with functions
Replies: 6
Views: 6137

### Re: help with operations with functions

Thank you very much. I have two more questions if someone is willing to help. Using the same f(x), how would I go about solving 1/f(x). And if k(x)=x^2+5 and g(x)=sqrt(x-3) how would I figure out (kg)(x).
Sun Jul 17, 2011 12:08 am
Forum: Intermediate Algebra
Topic: help with operations with functions
Replies: 6
Views: 6137

### help with operations with functions

Ok so f(x)={(3,5)(2,4)(1,7)}, and g(x)=sqrt(x-3). I have to figure out (f+g)(1). Any ideas? Thanks in advance.