CPhill

AB = AC + 14

BC  = AC + 30

And by the triangle inequality :

AB + AC > BC

[AC + 14] + AC > AC + 30

2AC + 14 > AC + 30

AC > 16

Which means that AB > 30     and BC > 46

Then, the minimum perimeter must be something greater than 16 + 30 + 46  = 92 cm

Here's my take on this one:

1. Beatriz is the oldest        2.  Ana is not the oldest        3. Celia is not the youngest

Let's assume that 1 is true.........   then 2 is false........but this is impossible because Beatriz and Ana would both be the oldest - a contradiction......so  T F F  isn't a possibiity

Let's assume 2 is true.........then Ana could be the second oldest or the youngest. Assume that she is the second oldest.......so 3 is false and Celia is the youngest.......but 1 is also false.......which means that Beatriz could be the second youngest [ violating our assumption]........or the youngest - also impossible because we know that 3 is false  and Celia is the youngest

And Ana can't be the youngest if we assume 2 is true, because 3 is false......Celia is the youngest in this scenario.......

So F T F  isn't possible, either

This leaves FFT  as the only possibility.......assume that Celia is the oldest.....then, since 2 is false, Ana would also have to be the oldest........and this is a contradiction

Assume Celia is the second oldest........then, since 2 is false, Ana is the oldest.......and since 1 is also false, Beatriz would be the youngest in this scenario

Thus

FFT   is the only possibility and the correct order  is  A - C - B

Very nice, heureka.......!!!!

Thanks, Anonymous4338......!!!!

We can use something called Heron's Formula to find this area

The area is given by :

A  = sqrt [ s (s - a) (s - b) ( s - c) ]       where a,b,c are the lengths of the triangle's sides and s is the  "semi-perimeter"   calculated as

s = [a + b + c] / 2   = [100 + 250 + 275]/2   = 312.5

So  the area is

sqrt [ 312.5 (312.5 - 100) (312.5 - 250) (312.5 - 275) ]   ≈  12,475.56 square meters

There are many solutions to this. For instance :

If   x = 1 , y = 2 and z = 0

x + y + z = 3

And

xy + xz + yz - xyz =

2 +  0   +  0  -  0   =   2  ≤  9/4

Switching x and y values and letting z remain 0  would work, as well

Let  P(F)  = the probability that the sum of the pips on the top face = 5

This is = 4/36   =  1/9

Let P(G)  = the probability that a number greater than a three is rolled.  This is the same probability that  1 - [a number 3 or less is rolled]  = 1- [ probability that a 2 is rolled + probability that a 3 is rolled] = 1 - [ 1/36 + 2/36]  = 36/36 - [ 3/36]  =  33/36  = 11/12

So

The probability that F or G occurs  =

P( F or G)   =

P(F) + P(G)  - P(F and G) =

(1/9) +  (33/36)  -  [ (1/9)(11/12)]  =

(4/36) + (33/36) - [ 11/108]  =

(12/108) + (99/108) - (11/108)  =

100/108 =

25/27

Very nice, Anonymous4338..........5 pts from me.....!!!

Very nice, heureka and Bertie......!!!!!

Vey nice, heureka......!!!!!