# Search results

1. ### Related Rates Problem (Check work)

OK. Thanks :)
2. ### Related Rates Problem (Check work)

A square is inscribed in a circle. As the square expands, the circle expands to maintain the four points of intersection. The perimeter of the square is expanding at the rate of 8 inches per second. Find the rate at which the circumference of the circle is increasing. Perimeter = p...
3. ### Ellipse Word Problem

Oh, I figured it out. Turns out that 141.7 = a . So then using the eccentricity I solved for c and then for b using the relationship between the 3. Then the max is a + c and the min is b - c. :)
4. ### Ellipse Word Problem

Oh, I took the square root to get that. Anyways, what am I suppsed to do then?
5. ### Ellipse Word Problem

OK, so .093 = 1 - \frac{b}{a} and .093^2 = 1 - \frac{b^2}{a^2} But if I rewrite the first I get a = .093a+b That can't be right because I still have 2 variables.
6. ### Ellipse Word Problem

Yeah, I still don't know why that person said that >_>
7. ### Ellipse Word Problem

So what you're saying is that: .093 = \sqrt{1 - \frac{b^2}{a^2}} and .093^2 = 1 - \frac{b^2}{a^2} Right?
8. ### Ellipse Word Problem

Um, a^2 = b^2 + c^2 is the same thing as the formula I posted... But, either way, yout idea is not right. 141.7 million is the mean distance not the distance to any point on the orbit. I think the sun is supposed to be at the focus.
9. ### Ellipse Word Problem

Yeah c^2 = a^2 - b^2
10. ### Ellipse Word Problem

But there is 3 variables...right?
11. ### Ellipse Word Problem

Having trouble with this problem. "The mean distance from the sun to Mars is 141.7 million miles. If the eccentricity of the orbit of Mars is .093, determine the maximum distance that Mars orbits from the sun." So basically what it is asking for is half the length of the major axis right...
12. ### Rectangular Equation to Polar

Yes it does. So what do I do next?
13. ### Rectangular Equation to Polar

I don't want coordinates for when the value equals 0. All I want, is to convert the rectangular equation into polar form. So instead of having x and y I need r and \Theta. Currently, I have r^2-3cos\Theta+4sin\Theta=0 but I don't know how to convert the sin and cos into polar form.
14. ### Rectangular Equation to Polar

Well, actually, I guess they are the same thing, but either way, I don't know how to do it :( As for your question, I don't see a purpose, but my teacher sure does.
15. ### Rectangular Equation to Polar

Who said anything about coordinates. The polar form of the equation is what I need to achieve.
16. ### Rectangular Equation to Polar

I need to convert x^2+y^2-3cos\Theta+4sin\Theta=0 to polar. Obviously the x^2+y^2 part would = r^2, but how can I get the cos and sin part to simplify?
17. ### Imaginary multiplication with answer to be in polar (variables)

That would give -(ac) -(adi) +(bci) -(bd)? But how do I find a modulus and argument with that?
18. ### Imaginary multiplication with answer to be in polar (variables)

I have to find (a+bi)(c+di) in polar form given that b,c,d>0 and a<0. So I convert each one to polar first. ( (a+b)cis(\arctan(-b/a) + \pi) ) ( (c+d)cis(\arctan(d/c)) ) That's as far as I got. Little help please?
19. ### Trigonometric Equations

Yeah, I figured out the tan(x) thing. These were due today. I forget how I did the 2nd one because I didn't do it like you said. It worked some other way. What is the purpose of learning precal when the calculator already told me the answer :p
20. ### Trigonometric Equations

Um HallsofIvy how was the 3 seemingly unaffected by you multiplying through by sin and cos? And when I type in sin^-1(+/- root 5 / 2) I get domain error.... ? What do you mean by let y=sin(x)?
21. ### Trigonometric Equations

Oh, my teacher told me sin and cos usually mess things up, but I guess that is the only option. Edit: I ended up with 2sin(x)^2 - 2cos(x)^2 + 3cox(x)sin(x) = 0 for the first one. That can't be right can it? And using the identity on the 2nd one I get: -sin(x)^2 + sin(x) +1 = 0 Then what...
22. ### Trigonometric Equations

I'm having trouble with two problems: 2tan(x) - 2cot(x) = -3 and cos(x)^2 + sin(x) = 0 On the 2nd one, I can substitute 1-sin(x)^2 for cos(x)^2 right? I tried that, but it didn't work. And I have no clue what to do on the first one. Little help please?
23. ### Operations of Functions

Yeah, but see it didn't say If, it said at. But I did look on the graph for f(4). :)
24. ### 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...
25. ### 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 -...
26. ### Confusion on a subtraction problem

Whoops, I see where I went wrong. Messed up when I subtracted the numerators. :( Thanks for the clarification.
27. ### 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...
28. ### Yellowstone docu-drama.

I havn't seen the one you're talking about with the pyroclastic car chase. (o_0 ?) I saw one like last month on the Discovery Channel but I found that it was actually made by the BBC back in around 2001 or something, so I guess they have been making other ones.
29. ### Controlling the Weather

OK, thanks for the help guys, I'll look into these sites. I had been to the disasterrelief one already, found that on google :)
30. ### 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