Recent content by ://Justice

Which means that the line is going down. So if you go to the left, you're going up. So the father left, the higher up you are (assuming that there is not another change in slope). Umm... I'm afraid I'm a bit lost by attempting to visualize the answer. I think I would understand this more easily...

The top of the hill, of course :) And I understand the there is surely a "top" to the hill, however, how can I prove that it is on the interval [-5,6], and not somewhere farther to the left? How can I tell that x=-5 is not near the bottom of the hill?

Down, of course And I can see that x=-5 is a "high" point, but I don't see how to prove that it is the highest

Ummm... attempting to visualize it, but having some trouble. I do believe that the y value for x=-5 has to be greater than at x=-3 EDIT: Soo.. the slope is negative at -5, and still negative until 1. It is only positive on the interval (1,3). Logically, it seems that the y value could not go...

Umm... The slope is negative, but is increasing, correct? Not sure if that was what you were looking for

That the slope is negative at that point on f(x)

It is negative

But it is an endpoint! Endpoints are also possible candidates for maximum values on a closed interval. I do not know how to determine that it is the maximum value, however. EDIT: To clarify, -5 is an endpoint on the closed interval [-5,6]. The question asks for the maximum value on the interval...

Homework Statement The specific problem can be found here: http://www.cbsd.org/sites/teachers/hs/cmcglone/Student%20Documents/Chapter%204%20(Application%20of%20Derivatives)/Section%204.3%20-%20Olsen%20Curve%20Sketching%20Answers.pdf" The above link also gives the answer. I am not sure on how...

I know it's no excuse, but I swear I have a horrible physics teacher... (by the way, this isn't just some new homework, but a take home test (so we are supposed to already know all of this)). I think I probably have a basic knowledge of it, I just do not know it... basically, we have probably...

AH, I get it now... Sorry I'm so slow... Okay, thanks. Then T1 is easy enough. Thanks. Haha... now (c)... I have absolutely no idea where to begin...

But would 3.01 not equal 2T2? but I kind of get it...

Which, I thought, was essentially what I did... T1 = T2*Cos(37) / Cos(53), then plug that in... (T2*Cos(37) / Cos(53))*Sin(53)+T2*Sin(37)=5N So, then set it equal to T2, as I did above. And then you have your answer... which I got as 1.213N. Where did I go wrong?

AH, now it all flashes back to me! haha. Thanks. Good old systems of equations and such. Really not sure if I did this right. Basically, after setting the first equation to equal T2, I got this: 2(T2)= (5N*Cos(theta1)) / (Cos(theta2)*Sin(theta1))+Sin(theta2)) wow, long... And so I got T2 =...

Homework Statement A ball of weight 5 N is suspended by two strings as shown [linked below]. (a) Draw and label all forces that act on the ball. (b)Determine the magnitude of each of the forces indicated in part (a). Suppose that the ball swings as a pendulum perpendicular to the plane of the...