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please help understand how to solve |x^2 - 2x| less than x  TOPIC_SOLVED

|x^2 - 2x| less than x. Please show all steps invovled in solving this problem. Thanks!
by rogermiranda
on Fri Apr 30, 2010 6:05 pm
 
Forum: Intermediate Algebra
Topic: please help understand how to solve |x^2 - 2x| less than x
Replies: 1
Views: 656

absolute value quadratic inequalities  TOPIC_SOLVED

Hi, does anyone know any websites where I can go to find examples on how to solve absolute value quadratic inequalites such as |x^2 - 1| is less than or equal to 2? Purple math has a section dealing with linear absolute value inequalites but I need to know the specific steps on how to solve quadrati...
by rogermiranda
on Sat May 01, 2010 5:24 pm
 
Forum: Intermediate Algebra
Topic: absolute value quadratic inequalities
Replies: 1
Views: 2194

Composition of functions  TOPIC_SOLVED

I am struggling with this problem. f(g(x)) = 2x-8 and g(x) = x-4 FIND f(x).

If I multiply g(x) by 2 then I get 2x-8. But the answer is 2x. So where does the x come from?
Can someone please help me understand how to solve this?
by rogermiranda
on Fri May 07, 2010 4:42 am
 
Forum: Intermediate Algebra
Topic: Composition of functions
Replies: 1
Views: 710

Need help verifying an anwer to this problem.  TOPIC_SOLVED

Complex Fraction:
a^2 / b^2 - 1 denominator 1/a + 1/b = a^2 - b^2 / b^2 denominator b+a / ab
= a^2 - b^2 / b^2 * ab / b+a = a(a-b) / b.
But my textbook has the answer a(a-b). So what happened to the denominator b?
by rogermiranda
on Sun May 23, 2010 7:35 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Need help verifying an anwer to this problem.
Replies: 5
Views: 1819

Re: Need help verifying an anwer to this problem.  TOPIC_SOLVED

Here is the Complex Fraction: Numerator= (a^2 / b^2) - 1 Denominator = (1/a + b/a) So, finding the LCD I have Numerator= (a^2 - b^2 / b^2) Denominator = (b + a / ab) Next I multiply by the reciprocal (a^2 - b^2 / b^2) * (ab / b + a) [(a + b)(a - b) / b^2)] * (ab / b + a) Simplifying, I am left with ...
by rogermiranda
on Mon May 24, 2010 3:49 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Need help verifying an anwer to this problem.
Replies: 5
Views: 1819

Re: Need help verifying an anwer to this problem.  TOPIC_SOLVED

Yes you have written the problem correctly.
Except the denomitor of the complex fraction is 1 over a + 1 over b (1/a + 1/b). So the LCD is ab.
So the denominator of the complex fraction is now b + a over ab (b + a / ab).
by rogermiranda
on Mon May 24, 2010 5:00 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Need help verifying an anwer to this problem.
Replies: 5
Views: 1819

Need help with this exponential equation.  TOPIC_SOLVED

The equation is x(raised to the one-half) - 6x(raised to the negative one-half) - 5 = 0 x^(1/2) - 6x^(-1/2) - 5 = 0 At this point what I tried to do was change it to a square root. So I get this sqr of x + sqr of 6x - 5 = 0 At this point I am confused and stuck. Please help. By the way the answer is...
by rogermiranda
on Sat Jun 12, 2010 2:31 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Need help with this exponential equation.
Replies: 1
Views: 1007

What are the steps to solving this problem?  TOPIC_SOLVED

I have 3 raised to the (x^2 - 1) / 9 raised to (x + 2) = one-nineth. So first I set up common bases. I have 3 raised to the (x^2 - 1) divided by 3 raised to the (2x + 4) = one-nineth. Now I am stuck. I have the same bases but am I suppose subtract the exponents? Can someone please help me get over t...
by rogermiranda
on Tue Jun 15, 2010 12:40 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: What are the steps to solving this problem?
Replies: 3
Views: 1271

Re: What are the steps to solving this problem?  TOPIC_SOLVED

I'm sorry, I still cannot sovle this equation. I followed your advice and converted the right side to equal bases. This is what I now have, 3 raised to (x^2 - 1) = 3^0 3 raised to (2x + 4) = 3^2 So I have x^2 - 1 = 0 and 2x + 4 = 2. So I solved both equations. x= 1 and x = -1 My textbook has the ans...
by rogermiranda
on Wed Jun 16, 2010 12:11 am
 
Forum: Advanced Algebra ("pre-calculus")
Topic: What are the steps to solving this problem?
Replies: 3
Views: 1271

How do I find the nth term of this sequence?  TOPIC_SOLVED

I have the sequence 2,8,26,80,242
So b sub n = 6,18,54,162
b sub 1 = 6
r = 3
using the geometric formula I have 6 times 3^(n-1).
Now I am stuck? What are the next steps? Please help!
by rogermiranda
on Tue Jul 06, 2010 2:13 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: How do I find the nth term of this sequence?
Replies: 2
Views: 1623
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