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...
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 -...
\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...
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
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} +...
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:
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?
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...
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?
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?
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...
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.
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...