Having trouble with this- is there a faster way of doing this than counting up everything?
A 4 by 4 grid of points is shown below. Each point is 1 unit from it's nearest horizontal and vertical neighbors. What is the greatest number of segments that can be drawn using pairs of these points as endpoints, such as no two segments are the same length.
Ok. I'll post what I've got. (Bear in mind I've never heard of this concavity test.)
Step 1. What is the function for f'' ? (Assuming it is a polynomial)
It has four roots so a good trial is
There is an objection to this; it doesn't go through (0,11)
This is easily corrected by introducing a multiplier of 11/18. As we are only concerned with the shape of the graph a re-scaling such as this is irrelevant so I'll not bother.
I have tried this on the MIF function grapher and it looks ok.
Step 2. What does the graph f' look like? We have enough information from the given graph to make a good 'stab' at this.
It should have four turning points at x = -3, -2, 1 and 3. Looking closely at the behaviour close to x = -3 we can see that f' will have a positive gradient just to the left of x = -3, and a negative gradient just to the right. So f' has a local maximum at x = -3. Similarly the next three are minimum, maximum and minimum respectively.
That's enough to sketch the graph.
But I thought I would integrate f'' to get an algebraic function:
I have re-scaled to make a better image of the shape of the graph.
There could also be a constant of integration but all this does is shift the graph up or down parallel to the y axis, so it doesn't affect the shape.
Step 3. What does the graph f look like?
Depending on that constant of integration it might more than one turning point. Since the test should work for any graph that double differentiates to f'', I'll just try integrating the f' function.
f'' is negative in the interval (1,3) . To establish whether the graph is downwardly concave in this interval it is necessary to determine the exact position of the points of inflexion.
For f', the gradient is negative from 1 to 3 so that seems to confirm the concavity.
In the interval (-3,-2) the gradient is similarly negative. I think that is it although my head is aching so much I'll have to come back to this later to check what I'm saying.
Yeah but how do you do it on the go? Khan says that it can be solved by looking at the graph.
I had to look up concavity as I'd not met it before. Here's a link:
You graph may be divided into 5 sections: (1) up to -3 (2) -3 to -1 (3) -1 to +1 (4) +1 to +2 (5) above + 2
From that page it would seem that the function is concave upwards for (2) and (4) and concave downwards elsewhere. Hope that helps,
ps. The theorem only applies to open intervals as f'' is zero at the 4 points.
pps. If you attempt a sketch of the f' graph you will see that, for example, in the interval -3 to -1, the gradient is increasing meaning the f graph is concave upwards.
Ok so I was just brushing up algebra skills for the test next week and I ran in a bump with this problem.
2. The perimeter of a rectangle is 48m. The width of the rectangle is 2 more than half the length. Find length and width.
So I set up these 2 equations
(solved from w=l/2+2)
and got this
Is this wrong because all the awnsers said they were integers..?
It can but the administrator does not want a chatroom.
Two years back, I suggested that we completely archive the current forum and install a new software. He was too busy to listen.
Dang, so is there anyway we could update the forum software to be more interactive like profile pages, etc.?
bobbym if ya see this
I went there but could not get very far. What is that place all about?
bobbym if ya see this
http://www.clubpenguininsiders.com/foru … #msg442014 they have quite a few members
An astronomer knows the distances from herself to stars AA and BB, as well as the distance between them. The distances are 450450, 400400, and 9090 light years respectively.
If the astronomer's telescope is currently pointed at star AA, how many degrees must she rotate her telescope to see star BB?
Round your answer to the nearest degree.
what i did
and they got 10 degrees...