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by buddy
Fri Feb 13, 2015 1:48 pm
Forum: Geometry
Topic: Scale position of points in a circle so it looks like a regular scaling?
Replies: 1
Views: 142

Re: Scale position of points in a circle so it looks like a regular scaling?

It doesn't look like P' is half the height of P. Are you maybe scaling the ellipse's height to be half that of the circle, and you're wanting to connect P' on the ellipse to P on the circle?
by buddy
Fri Feb 06, 2015 12:04 am
Forum: Intermediate Algebra
Topic: Selection the solution and graph the inequality
Replies: 4
Views: 158

Re: Selection the solution and graph the inequality

3/8x - 1/4x < 4/5 (5/32x - 1) I need help in finding the solution. I tried finding LCD and it wasn't helping at all. Please write back showing what you tried so we can see if its wrong or if you were nearly there. When you do, please say if you mean this for your inequality: \dfrac{3}{8x}\, -\, \df...
by buddy
Mon Feb 02, 2015 1:44 pm
Forum: Intermediate Algebra
Topic: Formulas and Functions problem
Replies: 3
Views: 183

Re: Formulas and Functions problem

Sleepy Ash wrote:I did mean the second one and how did you get the 11 when finding the lowest common denominator?

What did YOU get when you did the factorization to find the LCM of 7, 8 = 2*2*2, 13, 22 = 2*11, and 25 = 5*5?
by buddy
Sat Jan 17, 2015 3:36 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Help needed to find domain
Replies: 4
Views: 295

Re: Help needed to find domain

f(x) = square root of (2x-6) (the whole expression is under a root, so I had to find (fof)(x)= square root of [2*(square root of (2x-6)) - 6]. So it is root over another root. Please, forgive me if the way I wrote is not clear... Now I have to find Domain of the whole expression Do the regular thin...
by buddy
Sat Jan 17, 2015 3:30 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Graph evaluations
Replies: 2
Views: 406

Re: Graph evaluations

I don't understand how evaluate these graphs with the information they gave. The red graph is f(x) and blue is g(x) . http://i84.photobucket.com/albums/k5/Kohleria1/DSC08637_zps6ee38766.jpg http://i84.photobucket.com/albums/k5/Kohleria1/DSC08636_zpsc497dd68.jpg 1. f(g(2)) 2. limit (x => 2 - ) f(x)*...
by buddy
Fri Jan 16, 2015 3:21 am
Forum: Beginning Algebra
Topic: Domain and Range of a relation
Replies: 5
Views: 540

Re: Domain and Range of a relation

x = y squared I know that D = x > or = 0. I just don't understand why. I also know that R = all real numbers. That one makes sense. Are you working with y as a function of x, or with x and a function of y? Does "D" stand for "domain"? Does "R" stand for "range&quo...
by buddy
Thu Dec 18, 2014 12:55 pm
Forum: Intermediate Algebra
Topic: Algebraic LCDs
Replies: 6
Views: 659

Re: Algebraic LCDs

Is "-5/2n+1" (the five being negative only) the same as "-(5/2n+1) [the whole thing with a negative sign in front] Yes. Do the "minus" separately if you're not sure. \dfrac{-5}{2n\, +\, 1}\, =\, \dfrac{(-1)(5)}{2n\, +\, 1}\, =\, (-1)\dfrac{5}{2n\, +\...
by buddy
Wed Dec 17, 2014 3:50 pm
Forum: Intermediate Algebra
Topic: Algebraic LCDs
Replies: 6
Views: 659

Re: Algebraic LCDs

You look at what's different. (a+b)/(a^2b) + (a-b)/(ab^2): denoms a^2b and ab^2, common denom a^2b^2, so what's different is a b for the 1st fraction & an a for the 2nd (x+1)/(x-8) - (x)/(8 - x): denoms x - 8 and 8 - x = -1(x - 8), common denom x - 8, so what's different is a -1 for the 2nd If w...
by buddy
Wed Dec 17, 2014 1:41 pm
Forum: Intermediate Algebra
Topic: Algebraic LCDs
Replies: 6
Views: 659

Re: Algebraic LCDs

Why do some algebraic equations regarding LCDs require you to multiply each side by a fraction like "a/a" and others, just by a single number? If you multiply a fraction by "just a single number" then you change it. So you can't do that. Examples: a+b/a2b + a-b/ab2 x+1/x-8 - x/8...
by buddy
Wed Dec 17, 2014 1:38 pm
Forum: Geometry
Topic: Find the area of polygon
Replies: 3
Views: 553

Re: Find the area of polygon

Smrithi wrote:Can u see the example diagram in the easycalculation site.

Not without knowing where you're looking. Go to the page and copy the URL from the location bar in your browser. Then paste it here. Thnx

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