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hi evene,

That's the answer I'm getting too.

Bob

hi numbergeek

Welcome to the forum.

The total surface area of the pyramid is 260. The scale factor is .25 (.5 squared). I have 260 - 260(.25) + surface area of the smaller square. Length of the larger square is 10, so smaller square side length is 5. So I think the correct calculation is 260 - 65 + 25 which is 220.

Thanks for including your working. That makes it much easier for me to see what is going wrong.

The scale factor principle applies to similar shapes. So you can use it for the four large sloping sides and the equivalent smaller sloping sides. For the large, the total area of all four is 260 - 10x10 = 160. So the area of the smaller sloping sides is 160 x 0.25 = 40. So for the frustum the area of the sloping sides is 160 - 40 = 120. Now add in the top and bottom and you're finished.

Note: Each large sloping side is a triangle ... area = 160/4 = 40. So each equivalent triangle in the small is 40 x 0.25 = 10, so all four is 10x4 = 40. So you can see that the scale factor principle works for each triangle or all four together.

You cannot do 260 - 260 x 0.25 because it doesn't take account of the bottom square properly. It isn't scaled down. Stick to working out the sloping area first and you should be OK.

Hope that helps,

Bob

hi sassytonigirl

It would be impossible to have a normal distribution table for every mean and standard deviation, so there is just one. It has a mean of zero and a standard deviation of 1. This diagram shows the men's height curve being converted into the standard curve.

You take every z, subtract the mean, get the result and divide by the sd. The result is a new normal curve with a mean of zero and sd = 1.

To exclude the tallest 5% that green shaded area must be 0.95. So if phi is the function that converts a 'z' into a probability (of being less than z)

Because the normal curve is symmetrical, tables usually only give one half of the possible values to save space. The MIF table here:

http://www.mathsisfun.com/data/standard … table.html

starts at z=0 and just gives the probability from there onwards, so you'd use it by looking up 0.45 (because 0.5 + 0.45 = 0.95).

Before you get to the table on that page you will find an interactive graph. The default matches the table, but, on the left are three buttons for changing the way the interaction works. If you click for "Up to Z" then the shaded area can be adjusted to show the probability you want. The closest I can get by adjusting the line is 95.05 when the Z value is 1.65. I can improve the accuracy a little by looking at the table. 0.4495 gives Z = 1.64 and 0.4505 gives Z = 1.65 so I'd go halfway and say 0.45 is Z = 1.645.

So I have

and you can solve for z from that. That gives you the maximum allowable height for men.

The women's problem can be done in the same way but with a new mean and sd. Use the symmetry of the curve, so that you're working on the left hand side of zero to get the lowest height here.

Bob

hi Steve_K

**Welcome to the forum.**

One of my early posts was a long thread about the angles for making polyhedral. Also I started a geometry contents page in the Maths Teaching Resources.

Bob

hi sassytonigirl

4c ?

They've given an 'x' and you've had to work out the probability. Now they want you to work out the reverse ... given P = 0.95 what x value does that correspond to ?

Bob

hi evene,

My method did eliminate unknowns so it is a sort of elimination. I've found it's a trick that often works when you have equations like this. I call it symmetry although that's just a term I've made up myself. Each letter occurs the same number of times in similar ways. It's easy to see when it happens with a set of equations; harder to put into words what it is I'm noticing. Take two unknowns, say t and u, and swap them over. The equations are unchanged by this. That's true for any pair of swaps. Once I'd noticed that I though it would be worth adding them all up to get t+u+v+w+x+y+z. It took a few moments to see how to get a sum with any 6 of them and then it was easy.

Bob

OK, I'll have a go.

here I have put three graphs in one.

I'll start at the bottom with the blue graph.

Let's say that is part of a double differentiated graph f''

At A and B the function is zero and at C it has its lowest local value.

So the single differentiated graph must have turning points at the equivalent points to A and B and a largest negative gradient in line with C. That's the red graph.

So the function f has a certain gradient in line with A, lower values as we track right and lower still in line with B. That's the black graph.

The black graph is downwards concave.

The test is to look for where the f'' graph is fully negative in an interval. In this case, from A to B.

You can explore upwards concavity similarly by looking for positive bits of f''

Bob

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.

Bob

hi mathgogocart

I'm in the process of working out the functions for f' and f. I've got the graph for f' ..... still de-bugging the graph for f.

I'll post later when I'm more confident of what I'm saying.

Bob

I think this has been asked before (with diagram). I'll try to find it.

Bob

hi evene,

If you add up everything, you'll get t + u + v + w + x + y + z = 98.

If you add up (t+u) + (v+w) + (x+y) you get 84.

So you can get z.

Similar tricks will allow you to calculate the rest.

Bob

Hi philaretus

Using that formula of bobbym's with n = 5, the interior angle of a regular pentagon is

So angle AED = 108, and the triangle is isosceles. Use the angle sum of a triangle to work out EAD.

BAC will be the same (by symmetry) so you can work out DAC. That triangle is also isosceles so ADC = ACD and can be worked out from there.

Bob

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.

hi math9maniac

3). In triangle PQR, as A and B are midpoints, AB is parallel to PR and equal to half its length. Similarly BC, CD and DA. Also the diagonals of all rhombuses bisect each other at right angles.

So for any rhombus ABCD will always be a rectangle, with sides equal to half the diagonals of the rhombus.

So, if you are additionally told that ABCD is a square, then PR = QS and that makes the rhombus a square too.

Bob

hi math9maniac

That perimeter is correct. There are two things wrong with your angle calculation (1) To get the acute angle you need arctan(3/4). (2) Then you need to double this as the diagonal splits the angle in half.

A rhombus is a special case of a parallelogram. Opposite angles are equal but only all four if it's a square. You can think of a square as a special case of a rhombus.

An explanation could come from the unequal nature of the diagonals.

Bob

I'll call one diagonal AC.

By choosing a point, B, on the circle, you can make a new rectangle. Choose it very close to A and the area is close to zero, so a lower bound is 0.

The upper bound is when the rectangle becomes a square.

Bob

hi Hate Number Theory

Let's have a go at number 1 first.

There are two possible places for C, I've marked them C and C'.

The line BD bisects CC' at right angles.

AC + AC' = 2.AD, so you can use trig to get AD and hence answer the question.

Please have a go and post back your answer. If that is OK we can progress to the others.

Bob

hi math9maniac

Rectangles have two degrees of freedom, meaning you can choose the length, and then, independently, choose the width. So you'll never find a formula with only one variable.

These rectangles have the same diagonals, but very different areas.

Bob

HI Relentless

Let's call the bankroller A and the player B.

I agree, the expected win for B is infinite. A doesn't ever get a chance to 'win' so eventually B must win.

Conclusion. Don't be A.

Bob

hi mathgogocart

2nd: If the derivative is undefined at x=1 then it cannot have a critical point there as such points are defined in terms of the derivative.

In your last post you also asked about CPs. From a quick internet search it seems most sites use CP to mean a point where the first derivative is zero. Some also recognise a second order critical point as one where the second derivative is zero.

http://www.mathwords.com/s/second_order … _point.htm

So you need to look at how Khan is defining CPs.

Questions of definition sometimes are open to debate because there is no absolute authority on definitions. I usually check Wolfram, Wiki and Mathword.

Bob

hi reaganchoi

That's a neat suggestion. :-)

Bob

hi CoRRupt3d

Welcome to the forum.

Start by trying to solve this pair using a simultaneous technique:

multiply first by d and second by b

and

subtract

If you factorise the x out and rewrite the expressions as determinants of matrices you have the first formula. The other is obtained similarly.

Bob

hi

Been away for the New Year so I haven't had a chance to post until now.

Here's my latest analysis. (sorry bobbym, but this involves some pure maths )

On is the odd total, En the even total and Tn the grand total. fn is the nth Fibonacci number.

Using my earlier rules:

and

Therefore

As

this is beginning to look like bobbym's EM formula and could be easily proved by showing the EM version fits this version. I'll let someone with M have a go at that!

Bob