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?
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...
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" [Broken]
The above link also gives the answer. I am not sure...
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...
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...
Homework Statement
So, we have a physics project, in which we must build a water balloon launcher and launch the balloon at a stationary target about 50 yds away (normally we use meter, however we are firing on a football field). I have created my launcher (it is basically a slingshot), but...
Homework Statement
haha, so hopefully my last question for the night!
Find the initial velocity, when the object is shot at 70 degrees and took 6.5s to land. (hint: look at the y direction first this time)
(For more information regarding my question, visit my earlier question from tonight...
Which is why I excluded 'a' when simplifying the equation, because a=0. Let me try that equation. Hmm... that seemed to yield a much more reasonable answer, 3.356s. That seems to be correct. Yeah, I get it now. It makes sense. Funny how things can just click and then you get it. Thank's for the...
Homework Statement
So, this is pretty easy and simple, I am just missing something obvious here I am pretty sure. This stems from a question I asked earlier tonight.
Calculate the time it took from launch to land, given a velocity of 30.197m/s, and a distance traveled of 85m. (other...
Hmm... do you think I could have help with that last question that was supposed to be easy enough, Calculate the time it took from launch to land? Sorry, but I am a bit stuck.
So, I would use the formula X=Xo+Vot+.5at^2, correct? But then, I get a bit confused...
-.5t^2=Vot-X
divide by t...
Ah, oops. Technicalities always seem to mess me up, haha. Thanks. This forum seems really cool. My physics teacher doesn't seem to know how to teach, so I'll probably be here a lot in the future, haha. Thanks again, you really saved me.
V2=Vo2+2a(x-xo) ??
If V is 0, Vois 27.586, a is g, X is our unknown, and Xo is 0.
So, 0=27.5862+(2)(9.8)(X)
So, 27.5862 / (2)(9.8)= X
Correct?
Then, 38.826 would be the answer. I think I got it. Thanks :D
Really, so 27.586 m/s would be correct, then, assuming I solved correctly for Vo? Great.
Yes, so the object is at it's maximum height when V=0. V is decelerating at 9.8 m/s due to gravity. Now, I know that the angle, 33 degrees, is somehow involved here, no? But then what would my equation be...
Homework Statement
An object is shot (from a cannon) at an angle of 33 degrees and landed 85 m away. Calculate the magnitude of the initial velocity (Hint:Look at the x direction and solve for Vox)
Homework Equations
Other questions I must answer. If you have time, help with these would...