Recent content by hatelove

Well I know that, set them equal to each other and it's a couple of irrational numbers (sorry I didn't come up with a cleaner problem; I just picked random numbers instead of multiplying factors), and then I just assume that it's one area and use the integral formula for finding the area of...

y = x + 10 and y = x^2 + 5, find the area between x = -7 and x = 6

I don't remember it, but I will try to find one or some up with my own, hold on

I was given a problem with two functions and two x-values for boundaries, so I found the points of intersection (there were two) and attempted to find the area between those functions, but I didn't get to finish. In any case, I would have gotten it wrong, because when graph the two functions and...

Say I have: \frac{x \cdot \frac{1}{z}}{y} How can I make it look like: \frac{x}{y \cdot \frac{1}{z}} I'm trying to figure out the algebraic rule(s) for this

\frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{2}(\frac{x - 3}{x - 2})^{-\frac{1}{2}} \cdot \frac{1}{(x - 2)^{2}} \frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{2}(\frac{x - 3}{x - 2})^{-\frac{1}{2}} \cdot \frac{1}{(x - 2)^{2}} \\ \frac{1}{(\frac{x - 3}{x -...

I drew a diagram here to try and help: I think it will have something to do with a triangle. So just to find the other side, I'll use the Pythagorean theorem and get \sqrt{3}. I already knew this because I recognized the 30-60-90 triangle, so I know the angles too, but I don't know if we...

Oh wait, I see it now...the very first negative sign at the very beginning...I completely missed that this entire time and I couldn't see it...wow

I can't see where I've missed any negative sign

After I simplify, I get this: = \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)} \\ \\ \\ = \frac{1}{x^{2}(5 - 2x)^{\frac{1}{2}}} - \frac{1}{x(5-2x)^{\frac{3}{2}}} But the solution says I'm wrong...what have I done?

f'(x) = (x\sqrt{5-2x})^{-1} \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot...

Good explanation CB, thanks. I believe I understand now. At the point (x,f(x)) the slope is f'(x) evaluated and the y-coordinate of derivative function is the slope of the line tangent at the point (x,f(x)) in the original function, and the entire equation of the line (not just the slope) is...

Here is f(x) = x^2: And the derivative of it (2x): So each point on the slope of the derivative is supposed to represent the slope of the line tangent at a certain point on the original function. Say I choose an x-value on the derivative 1, so the point on the line would be (1,2). Where...

By 5 ft./sec.? This is a constant velocity, so...this is just a slope of 0 on the y-axis since it's a horizontal line. If function 'A' is the distance traveled while going at a constant 5 ft./sec (f(A) = 5t), then the slope of a line is always going to be the same so taking the derivative of...

So here I thought it might be something to formulate a triangle, or actually two: So the problem is asking me to find the derivative of the shadow's length with respect to time: \frac{dB}{dt} I think the triangles are similar: \frac{15}{A+B} = \frac{6}{B} \\ \\ \cdots \\ B = \frac{2}{3}A...