# Search results

1. ### Operations of Functions

Yeah, but see it didn't say If, it said at. But I did look on the graph for f(4). :)
2. ### Operations of Functions

OK, thanks. On the 2nd part, it has nothing to do with graphing. There is just a graph with f and g drawn on it and you just have to find the value based on that graph. All the other questions were normal but I don't understand the whole at x = whatever business. That why I asked if it was...
3. ### Operations of Functions

Little help please and some work checking. Given f(x)=3x+2 and g(x)=\frac{x-4}{2x} It asks for g(\frac{1}{x}) So substitute in: \frac{\frac{1}{x} - 4}{2(\frac{1}{x})} Simplify: \frac{\frac{1}{x} - 4}{\frac{2}{x}} I multiply top and bottom by x right? That would give: \frac{1 -...
4. ### Confusion on a subtraction problem

Whoops, I see where I went wrong. Messed up when I subtracted the numerators. :( Thanks for the clarification.
5. ### Confusion on a subtraction problem

\frac{x}{x^2-9} - \frac{1}{2x-6} When I first worked this problem I found the answer to be: \frac{1}{2x-6} However, in my English class we had this Vietnamese substitute who took my worksheet and did the problems on it in his head and pointed at the answer I wrote for this problem and...
6. ### Conics: The Ellipse - Practice

You have to get it in standard form first. Divide both sides by 144. That gives: \frac{x^2}{16} + \frac{y^2}{9} = 1 a = length of the major axis b= length of the minor axis
7. ### Solve This

? Justin answered your question...taught me a thing or two as well...
8. ### Semi Latus Rectum Proof - b^2/a

Ok, I'll just write out my whole proof here because I like the equations better than my handwriting. e = \frac{c}{a} so c = ae The line of the Semi Latus Rectum is x = \pm ae. So you plug this in for x in the equation: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \frac{(ae)^2}{a^2} +...
9. ### Semi Latus Rectum Proof - b^2/a

So I just do: \frac{(ae)^2}{a^2} + \frac{y^2}{b^2} = 1 But what I don't get is what happens to y if you said to solve for y, but the answer is in terms of b? :confused:
10. ### Semi Latus Rectum Proof - b^2/a

:confused: Why has no one responded? Doesn't anyone know how to do this?
11. ### Semi Latus Rectum Proof - b^2/a

I need to prove that the length of the semi latus rectum = b^2/a I am assuming that you make some sort of a triangle within the ellipse to prove this, can someone point out which one it is?
12. ### Extra Cedit Algebra II Problems (2 of them)

Actually I didn't figure it out the first time I read you two's posts, so I just now figured it out. So I did get the Ah-Ha. :p x^2 - 30 = \sqrt{30 + \sqrt{30 + \sqrt{30 \ldots } } } x^2 - 30 = x x^2-x-30=0 (x-6)(x+5)=0 So: x=6 and x=-5 And the answer is x=6 since the square root of a...
13. ### Extra Cedit Algebra II Problems (2 of them)

How could I forget the original equation??? If I knew what to do when it looks the same I would have done it already. How do I get a solvabale equation out of that mess?
14. ### Extra Cedit Algebra II Problems (2 of them)

Ya it does :p But when I end up with: x^2 - 30 = \sqrt{30 + \sqrt{30 + \sqrt{30 \ldots } } } then what? Do I take the square root of the left side?
15. ### Extra Cedit Algebra II Problems (2 of them)

I still don't get all this 4 numbers business and I certainly don't know where you got that polar form equation because I've never learned anything about it. Looks more like pre-cal to me. But forget that problem right now. What do I do with the other one after I square both sides?
16. ### Extra Cedit Algebra II Problems (2 of them)

Yes, that's what I mean. I'll edit the first post in a minute. Well I guess you could square both sides and that would give you: x^2 = 30 + \sqrt{30 + \sqrt{30 + \sqrt{30 \ldots } } } For the 2nd one I still don't understandd how there are 4 answers. If you simplify the cube root of...
17. ### Extra Cedit Algebra II Problems (2 of them)

My bad for the first one it's x = sqrt30 + sqrt 30... starting under one sqare root. For the 2nd I don't know what your talking about but my teacher basically said just prove that the sqrt of i is equal to sqrt2/2 + (sqrt2/2)i.
18. ### Extra Cedit Algebra II Problems (2 of them)

Little help please, I'm failing and need the extra credit and have no idea how to work these problems. 1. x = \sqrt{30 + \sqrt{30 + \sqrt{30 + \ldots } } } 2. Find the 4th root of -1. I know that this is the same thing as the square root of i. I also know that the answer is...