14 beasts and 22 birds
I agree.
Let b = birds (2 legs each)
Let a = animals (4 legs each)
So 2b + 4a = 100 and b + a = 36
2b + 2a = 72
and 2a = 28 So a = 14 animals or beasts
Then b + 14 = 36 and b = 22
los pies = feet
smelly feet
feet - pieds foot - pied :)
A tape measure has a feet, inches and CM measurement.
Pies is the Spanish word for "feet".
Birds and beasts have one head each. So totally there are 43 birds & beasts. Let number of birds = x number of beasts = (43-x) Birds have 2 feet and beasts have 4 feet. Total feet = 2 * no of birds + 4 * no of beasts = 2 * x + 4 * (43 -x) = 172 - 2*x (on simplification) = 120 (given) 172 - 2*x = 120 2*x = 172 - 120 = 52 x = 52/2 = 26 number of birds = x = 26 number of beasts = 43 -x = 43 -26 = 17 (check: 26 + 17 = 43 ; 26*2 + 17*4 = 52 + 68 = 120)
There are 13 birds in the cage.
Ok, so we know altogether there are 30 animals(all animals have heads right?). Sooo... I make it, 20 land beasts and 10 birds. Sorry if its wrong! :> - that's meant to be a bird by the way.
Ships, beasts of burden, feet.
no flowing of blood in the feet of birds
Assuming the dogs have four feet and birds two, you need 15 dogs (60 feet) and 1 bird (2 feet) giving a total of 16 birds and dogs and 62 feet. None. Dogs have paws, not feet. Also, birds have claws, not feet.
Mainly webbed feet are seen on swimming birds
You can tell that birds naturally clench their feet because their feet are rolled up in a ball
Aliens.
Here's a Perl script that will answer the problem for any number of heads and feet. sub how_many { print "Number of heads: "; chomp($h = ); print "Number of feet: ";chomp($f = ); $p = $h; $c = 0; while (($p*4 + $c*2) != $f) { $p--; $c++; } } &how_many; print "There are $p pigs and $c chickens.\n"; print "They have $h heads and $f feet.\n"; Number of heads: 27 Number of feet: 78 There are 12 pigs and 15 chickens. They have 27 heads and 78 feet.
Hot Heads and Cold Feet - 1915 was released on: USA: 4 December 1915
Pigeon's feet are much the same as most other birds.