THERE ARE @! chickens and !$ pigs... u have to figure out the numbers.... look at the numbers at ur keyboard. look ontop of numbers... any1 here play dungeon fighters online??? A A AA AA
48 chickens 1 pig
There are 139 chickens and 30 pigs. Let c = number of chickens and p = number of pigs Each has 1 head, therefore c + p = 169 Each chicken has 2 feet and each pig has 4 feet, therefore 2c + 4p = 398 => c + 2p = 199 Subtracting the first from the second gives: c + 2p - c - p = 199 - 169 => p = 30 Substitute in the first c + 30 = 169 c = 139
8 pigs & 9 ducks
12 pigs42 divided by 7 is 6 .making 6 groups of 7 chickens .since there is 2 pigs for every 7 chickens and there are six groups of 7 chickens .6 x 2 = 12 .there are 12 pigs .
40 chickens and 30 pigsThis can be solved with 2 equations. Let p = the number of pigs and c = the number of chickens. Pigs and chickens each have 1 head, so p+c=70. Pigs have 4 legs and chickens have 2, so the total number of legs is 4p+2c=200.Solve this "system of equations" by substitution: The first equation can be written as p=70-c. Substitute the right side for p in the second equation:4(70 - c) + 2c = 200280 - 4c + 2c = 200280 - 2c = 200-2c = 200 - 280-2c = -80c = 40.Substitute this back into p=70-c: p=70-40=30.There are 40 chickens and 30 pigs.Check (in the words of the question): 40 chicken heads and 30 pig heads = 70 heads. 40x2=80 chicken legs, and 30x4=120 pig legs; 80+120=200 legs. Check!
This is not possible
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.
35 pigs, 65 chickens
Since PIgs don't fly yet, we will say they have no wings and we say chickens have 2 legs. Let C be the number of chickens and P be the number of pigs. Also, let's assume each chicken has two wings. 4P+2C=24 2C=12 so C=6 4P+12=24 4P=12 so P=3 6 chickens and 3 pigs. This answer must be changed when pigs fly!
48 chickens 1 pig
14 pigs 8 chicken's Since there are 22 heads, let the number of pigs be x and the number of chickens be 22 - x. Since pigs have 4 legs and chickens have 2 legs, and there are 72 legs in all, we have 4x + 2(22 - x) = 72 4x + 44 - 2x = 72 2x + 44 = 72 2x + 44 - 44 = 72 - 44 2x = 28 2x/2 = 28/2 x = 14 (the number of pigs) 22 - x = 22 - 14 = 8 (the number of chickens)
On this farm there would be a total of 18 cows. There would also be 18 chickens, combining to make 36 heads and 104 legs.
There are 139 chickens and 30 pigs. Let c = number of chickens and p = number of pigs Each has 1 head, therefore c + p = 169 Each chicken has 2 feet and each pig has 4 feet, therefore 2c + 4p = 398 => c + 2p = 199 Subtracting the first from the second gives: c + 2p - c - p = 199 - 169 => p = 30 Substitute in the first c + 30 = 169 c = 139
8 pigs & 9 ducks
Let c represent the number of chickens and p represent the number of pigs.Since chickens and pigs have one head each, 70 heads = c + p.Since chickens have 2 legs and pigs have 4 legs, 200 legs = 2c + 4p.Subtracting twice the first equation (140 = 2c + 2p) from the second, we get60 = 2porp = 30 pigsRewriting the first equation as c = 70 - p, we get c = 40 chickens.
14 chickens and 11 pigs.
chickens chickens