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Agnishom wrote:

You want to count the number of elements in the set { d : d = x^2 + y^2 ; 3 >= x,y >= 0}

This was in my math HW.

**mathgogocart**- Replies: 12

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.

Nehushtan wrote:

mathgogocart wrote:y=x+32

y+4=5(x+4)

Shouldn't it be 36 because 32+4 years is 36?

Sorry if I seem kinda brushed up on this.

**mathgogocart**- Replies: 6

Help, I keep getting negative results

A father is 32 years older than his son. In 4 years, the father will be 5 years older than his son. How old is each now?

I have made these

y=x+32

y+4=5(x+36)

in 4 years

y= fathers age

x=sons age?

Can someone help me?

bob bundy wrote:

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:

http://i.imgur.com/6uembH5.gif

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.

http://i.imgur.com/d03BPqo.gif

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.

Bob

Yeah but how do you do it on the go? Khan says that it can be solved by looking at the graph.

**mathgogocart**- Replies: 4

http://prntscr.com/9w849a

http://prntscr.com/9w846x Would I be correct on this? Also how do you tell wether a (-infinity,-3) is concave up or down

bob bundy wrote:

hi mathgogocart

I had to look up concavity as I'd not met it before. Here's a link:

https://www.math.hmc.edu/calculus/tutor … condderiv/

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,

Bob

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.

Thank you!

**mathgogocart**- Replies: 2

-3 to -1 seems to be downwards(concave) so I chose that

1 to 2 is also downward(concave)

and 2 to ∞ is downwards

Am I right?

**mathgogocart**- Replies: 3

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

2l+2w=48

and

2w-4=l

(solved from w=l/2+2)

and got this

2(2w-4)+2w=48

4w-8+2w=48

6w-8=48

6w=56

w=28/3

Is this wrong because all the awnsers said they were integers..?

**mathgogocart**- Replies: 1

Flossing skills, is this right? http://prntscr.com/9v3kw8

**mathgogocart**- Replies: 3

Is this the right awnser?

at -5 the result is 0 so I chose that it was a critical point.

and number 2 question is TRUE, edited because undefined./

Nehushtan wrote:

There should be five critical points altogether: two local maxima, one local minimum, and two non-differentiable points of inflection.

I understand, already got it.

**mathgogocart**- Replies: 2

I was doing some Khan to refresh my calc and this happened

I see there are ciritical points at -1 and 3 however there is no -1.

Agnishom wrote:

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.?

Agnishom wrote:

We need o improve on the forum software

I agree. Could a chatroom at least be implemented?

**mathgogocart**- Replies: 1

Yes, so?

We need more interactive profiles

bobbym wrote:

Hi mathgogocart;

Something wrong with your profile?

what do you mean?

bobbym wrote:

bobbym if ya see this

I went there but could not get very far. What is that place all about?

A game

also..http://www.mathisfunforum.com/profile.p … &id=188456

I'm just saying the forum looks more light and interactive, tbh.

bobbym if ya see this

http://www.clubpenguininsiders.com/foru … #msg442014 they have quite a few members

As

Relentless wrote:

I believe he means that the forum is not very active or popular, and wants to discuss ways to improve this

Same

**mathgogocart**- Replies: 56

I believe we should go to a new style?

The forum seems to be dead..?

I mean 450.... mistake

Another problem

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?

^\circ

∘

Round your answer to the nearest degree.

Well

what i did

I got

76 degrees...

and they got 10 degrees...